Solucionario Johnstoon and beers 2

PROBLEM 6.1
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
AB = 32 + 1.252 = 3.25 m
BC = 32 + 42 = 5 m
Reactions:
ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0
C = 48 kN
ΣFx = 0: Ax − C = 0
A x = 48 kN
ΣFy = 0: Ay = 84 kN = 0
A y = 84 kN
Joint A:
ΣFx = 0: 48 kN −
12
FAB = 0
13
FAB = +52 kN
ΣFy = 0: 84 kN −
FAB = 52 kN T
5
(52 kN) − FAC = 0
13
FAC = +64.0 kN
FAC = 64.0 kN T
!
Joint C:
FBC 48 kN
=
5
3
FBC = 80.0 kN C
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737
PROBLEM 6.2
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss
ΣFx = 0: Bx = 0
ΣM B = 0: C (15.75 ft) − (945 lb)(12 ft) = 0
C = 720 lb
ΣFy = 0: By + 720 lb − 945 lb = 0
B y = 225 lb
Free body: Joint B:
FAB FBC 225 lb
=
=
5
4
3
FAB = 375 lb C
FBC = 300 lb T
Free body: Joint C:
FAC FBC 720 lb
=
=
9.75 3.75
9
FBC = 300 lb T
FAC = 780 lb C
(Checks)
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738
!
PROBLEM 6.3
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss
ΣFx = 0: C x = 0 C x = 0
ΣM B = 0: (1.92 kN)(3 m) + C y (4.5 m) = 0
C y = −1.28 kN C y = 1.28 kN
ΣFy = 0: B − 1.92 kN − 1.28 kN = 0
B = 3.20 kN
Free body: Joint B:
FAB FBC 3.20 kN
=
=
5
3
4
FAB = 4.00 kN C
FBC = 2.40 kN C
Free body: Joint C:
ΣFx = 0: −
7.5
FAC + 2.40 kN = 0
8.5
FAC = +2.72 kN
!
! ΣFy =
FAC = 2.72 kN T
!
4
(2.72 kN) − 1.28 kN = 0 (Checks)
8.5
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739
PROBLEM 6.4
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is
in tension or compression.
SOLUTION
Reactions:
ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0
Fy = 4.2 kips
ΣFx = 0: Fx = 0
ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0
D = 4.2 kips
Joint A:
ΣFx = 0: FAB = 0
FAB = 0
ΣFy = 0 : −1 − FAD = 0
FAD = −1 kip
Joint D:
ΣFy = 0: − 1 + 4.2 +
ΣFx = 0:
8
FBD = 0
17
FBD = −6.8 kips
15
(−6.8) + FDE = 0
17
FDE = +6 kips
FAD = 1.000 kip C
FBD = 6.80 kips C
FDE = 6.00 kips T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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740
PROBLEM 6.4 (Continued)
Joint E:
ΣFy = 0 : FBE − 2.4 = 0
FBE = +2.4 kips
FBE = 2.40 kips T
Truss and loading symmetrical about cL
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741
PROBLEM 6.5
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss
ΣFx = 0: A x = 0
ΣM A = 0: D (22.5) − (10.8 kips)(22.5) − (10.8 kips)(57.5) = 0
D = 38.4 kips
ΣFy = 0: A y = 16.8 kips
Free body: Joint A:
FAB
F
16.8 kips
= AD =
22.5 25.5
12
FAB = 31.5 kips T
FAD = 35.7 kips C
Free body: Joint B:
ΣFx = 0:
FBC = 31.5 kips T
ΣFy = 0:
FBD = 10.80 kips C
Free body: Joint C:
FCD FBC 10.8 kips
=
=
37
35
12
FBC = 31.5 kips T
FCD = 33.3 kips C
(Checks)
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742
!
PROBLEM 6.6
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Free Body: Truss
ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0
F = 2025 N
ΣFx = 0: Ex + 900 N + 900 N = 0
Ex = −1800 N E x = 1800 N
ΣFy = 0: E y + 2025 N = 0
E y = −2025 N E y = 2025 N
FAB = FBD = 0
We note that AB and BD are zero-force members:
Free body: Joint A:
FAC FAD 900 N
=
=
2.25 3.75
3
FAC = 675 N T
FAD = 1125 N C
Free body: Joint D:
FCD FDE 1125 N
=
=
3
2.23
3.75
FCD = 900 N T
FDF = 675 N C
Free body: Joint E:
ΣFx = 0: FEF − 1800 N = 0
FEF = 1800 N T
ΣFy = 0: FCE − 2025 N = 0
FCE = 2025 N T
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743
PROBLEM 6.6 (Continued)
Free body: Joint F:
ΣFy = 0:
2.25
FCF + 2025 N − 675 N = 0
3.75
FCF = −2250 N
ΣFx = −
FCF = 2250 N C
3
( −2250 N) − 1800 N = 0 (Checks)
3.75
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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744
PROBLEM 6.7
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss
ΣFy = 0: B y = 0
ΣM B = 0: D(4.5 m) + (8.4 kN)(4.5 m) = 0
D = −8.4 kN
D = 8.4 kN
ΣFx = 0: Bx − 8.4 kN − 8.4 kN − 8.4 kN = 0
Bx = +25.2 kN B x = 25.2 kN
Free body: Joint A:
FAB FAC 8.4 kN
=
=
5.3
4.5
2.8
FAB = 15.90 kN C
FAC = 13.50 kN T
Free body: Joint C:
ΣFy = 0: 13.50 kN −
4.5
FCD = 0
5.3
FCD = +15.90 kN
ΣFx = 0: − FBC − 8.4 kN −
FCD = 15.90 kN T
2.8
(15.90 kN) = 0
5.3
FBC = −16.80 kN
FBC = 16.80 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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745
PROBLEM 6.7 (Continued)
Free body: Joint D:
FBD 8.4 kN
=
4.5
2.8
FBD = 13.50 kN C
We can also write the proportion
FBD 15.90 kN
=
4.5
5.3
FBD = 13.50 kN C (Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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746
PROBLEM 6.8
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
AD = 52 + 122 = 13 ft
BCD = 122 + 162 = 20 ft
Reactions:
ΣFx = 0: Dx = 0
ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0
ΣFy = 0: 165 lb − 693 lb + E = 0
Joint D:
D y = 165 lb
E = 528 lb
ΣFx = 0:
5
4
FAD + FDC = 0
13
5
(1)
ΣFy = 0:
12
3
FAD + FDC + 165 lb = 0
13
5
(2)
Solving (1) and (2), simultaneously:
FAD = −260 lb
FAD = 260 lb C
FDC = +125 lb
FDC = 125 lb T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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747
PROBLEM 6.8 (Continued)
Joint E:
ΣFx = 0:
5
4
FBE + FCE = 0
13
5
(3)
ΣFy = 0:
12
3
FBE + FCE + 528 lb = 0
13
5
(4)
Solving (3) and (4), simultaneously:
FBE = −832 lb
FBE = 832 lb C
FCE = +400 lb
FCE = 400 lb T
Joint C:
Force polygon is a parallelogram (see Fig. 6.11 p. 209)
FAC = 400 lb T
FBC = 125 lb T
Joint A:
ΣFx = 0:
5
4
(260 lb) + (400 lb) + FAB = 0
13
5
FAB = −420 lb
ΣFy = 0:
FAB = 420 lb C
12
3
(260 lb) − (400 lb) = 0
13
5
0 = 0 (Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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748
!
PROBLEM 6.9
Determine the force in each member of the Pratt roof
truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss
ΣFx = 0: Ax = 0
Due to symmetry of truss and load
Ay = H =
1
total load = 21 kN
2
Free body: Joint A:
FAB FAC 15.3 kN
=
=
37
35
12
FAB = 47.175 kN FAC = 44.625 kN
FAB = 47.2 kN C
FAC = 44.6 kN T
!
Free body: Joint B:
From force polygon:
FBD = 47.175 kN, FBC = 10.5 kN
FBC = 10.50 kN C
FBD = 47.2 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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749
PROBLEM 6.9 (Continued)
Free body: Joint C:
ΣFy = 0:
3
FCD − 10.5 = 0
5
ΣFx = 0: FCE +
FCD = 17.50 kN T
4
(17.50) − 44.625 = 0
5
FCE = 30.625 kN
Free body: Joint E:
DE is a zero-force member
FCE = 30.6 kN T
!
FDE = 0
!
Truss and loading symmetrical about cL
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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750
PROBLEM 6.10
Determine the force in each member of the fan roof
truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss
ΣFx = 0 : A x = 0
From symmetry of truss and loading:
Ay = I =
1
2
Total load
A y = I = 6 kN
Free body: Joint A:
F
FAB
5 kN
= AC =
9.849
9
4
FAB = 12.31 kN C
FAC = 11.25 kN T
Free body: Joint B:
ΣFx =
9
(12.31 kN + FBD + FDC ) = 0
9.849
FBD + FBC = −12.31 kN
or
ΣFy =
or
(1)
4
(12.31 kN + FBD − FBC ) − 2 kN = 0
9.849
FBD − FBC = −7.386 kN
(2)
Add (1) and (2):
2 FBD = −19.70 kN
FBD = 9.85 kN C
Subtract (2) from (1):
2 FBC = −4.924 kN
FBC = 2.46 kN C
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751
PROBLEM 6.10 (Continued)
Free body: Joint D:
FCD = 2.00 kN C
From force polygon:
FDE = 9.85 kN C
Free body: Joint C:
ΣFy =
4
4
FCE −
(2.46 kN) − 2 kN = 0
5
9.849
FCE = 3.75 kN T
3
9
ΣFx = 0 : FCG + (3.75 kN) +
(2.46 kN) − 11.25 kN = 0
5
9.849
FCG = +6.75 kN
FCG = 6.75 kN T
From the symmetry of the truss and loading:
FEF = FDE
FEF = 9.85 kN C
FEG = FCE
FEG = 3.75 kN T
FFG = FCD
FFG = 2.00 kN C
FFH = FBD
FFH = 9.85 kN C
FGH = FBC
FGH = 2.46 kN C
FGI = FAC
FGI = 11.25 kN T
FHI = FAB
FHI = 12.31 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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752
PROBLEM 6.11
Determine the force in each member of the Howe roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss
ΣFx = 0: H x = 0
Because of the symmetry of the truss and loading:
A = Hy =
1
2
Total load
A = H y = 1200 lb
Free body: Joint A:
FAB FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C
FAC = 1200 lb T
Free body: Joint C:
BC is a zero-force member
FCE = 1200 lb T
FBC = 0
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753
!
PROBLEM 6.11 (Continued)
Free body: Joint B:
ΣFx = 0:
or
FBD + FBE = −1500 lb
ΣFy = 0:
or
Add Eqs. (1) and (2):
Subtract (2) from (1):
4
4
4
FBD + FBC + (1500 lb) = 0
5
5
5
(1)
3
3
3
FBD − FBE + (1500 lb) − 600 lb = 0
5
5
5
FBD − FBE = −500 lb
2 FBD = −2000 lb
2 FBE = −1000 lb
(2)
FBD = 1000 lb C
FBE = 500 lb C
Free Body: Joint D:
4
4
(1000 lb) + FDF = 0
5
5
ΣFx = 0:
FDF = −1000 lb
FDF = 1000 lb C
3
3
(1000 lb) − ( −1000 lb) − 600 lb − FDE = 0
5
5
ΣFy = 0:
FDE = +600 lb
FDE = 600 lb T
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 500 lb C
FEG = FCE
FEG = 1200 lb T
FFG = FBC
FFG = 0
FFH = FAB
FFH = 1500 lb C
FGH = FAC
FGH = 1200 lb T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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754
PROBLEM 6.12
Determine the force in each member of the Gambrel roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss
ΣFx = 0: H x = 0
Because of the symmetry of the truss and loading
A = Hy =
1
2
Total load
A = H y = 1200 lb
Free body: Joint A:
FAB FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C
FAC = 1200 lb T
Free body: Joint C:
BC is a zero-force member
FBC = 0
FCE = 1200 lb T
Free body: Joint B:
ΣFx = 0:
24
4
4
FBD + FBE + (1500 lb) = 0
25
5
5
24 FBD + 20 FBE = −30, 000 lb
or
ΣFy = 0:
(1)
7
3
3
FBD − FBE + (1500) − 600 = 0
25
5
5
7 FBD − 15FBE = −7,500 lb
or
(2)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
755
PROBLEM 6.12 (Continued)
Multiply (1) by (3), (2) by 4, and add:
100 FBD = −120, 000 lb
FBD = 1200 lb C
500 FBE = −30,000 lb
FBE = 60.0 lb C
Multiply (1) by 7, (2) by –24, and add:
Free body: Joint D:
24
24
FDF = 0
(1200 lb) +
25
25
ΣFx = 0:
FDF = −1200 lb
ΣFy = 0:
FDF = 1200 lb C
7
7
(1200 lb) − (−1200 lb) − 600 lb − FDE = 0
25
25
FDE = 72.0 lb
FDE = 72.0 lb T
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 60.0 lb C
FEG = FCE
FEG = 1200 lb T
FFG = FBC
FFG = 0
FFH = FAB
FFH = 1500 lb C
FGH = FAC
FGH = 1200 lb T
Note: Compare results with those of Problem 6.9.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
756
PROBLEM 6.13
Determine the force in each member of the truss shown.
SOLUTION
12.5 kN FCD FDG
=
=
2.5
6
6.5
Joint D:
FCD = 30 kN T
FDG = 32.5 kN C
ΣF = 0: FCG = 0
Joint G:
ΣF = 0: FFG = 32.5 kN C
Joint C:
BF =
BF
2
(2.5 m) = 1.6667 m β = ∠BCF = tan −1
= 39.81°
3
2
ΣFy = 0: − 12.5 kN − FCF sin β = 0
−12.5 kN − FCF sin 39.81° = 0
FCF = −19.526 kN
FCF = 19.53 kN C
ΣFx = 0: 30 kN − FBC − FCF cos β = 0
30 kN − FBC − (−19.526 kN) cos 39.81° = 0
FBC = +45.0 kN
Joint F:
ΣFx = 0: −
FBC = 45.0 kN T
6
6
FEF −
(32.5 kN) − FCF cos β = 0
6.5
6.5
6.5 !
FEF = −32.5 kN − "
# (19.526 kN) cos 39.81°
$ 6 %
FEF = −48.75 kN
ΣFy = 0: FBF −
FEF = 48.8 kN C
2.5
2.5
(32.5 kN) − (19.526 kN)sin 39.81° = 0
FEF −
6.5
6.5
FBF −
2.5
( −48.75 kN) − 12.5 kN − 12.5 kN = 0
6.5
FBF = +6.25 kN
FBF = 6.25 kN T
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you are using it without permission.
757
PROBLEM 6.13 (Continued)
Joint B:
tan α =
2.5 m
; γ = 51.34°
2m
ΣFy = 0: − 12.5 kN − 6.25 kN − FBE sin 51.34° = 0
FBE = −24.0 kN
FBE = 24.0 kN C
ΣFx = 0: 45.0 kN − FAB + (24.0 kN) cos 51.34° = 0
FAB = +60 kN
FAB = 60.0 kN T
Joint E:
γ = 51.34°
ΣFy = 0: FAE − (24 kN) sin 51.34° − (48.75 kN)
FAE = +37.5 kN
2.5
=0
6.5
FAE = 37.5 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
758
PROBLEM 6.14
Determine the force in each member of the roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss
ΣFx = 0: A x = 0
From symmetry of loading:
Ay = E =
1
Total load
2
A y = E = 3.6 kN
We note that DF is a zero-force member and that EF is aligned with the load. Thus
FDF = 0
FEF = 1.2 kN C
Free body: Joint A:
FAB FAC 2.4 kN
=
=
13
12
5
FAB = 6.24 kN C
FAC = 2.76 kN T
Free body: Joint B:
ΣFx = 0:
3
12
12
FBC + FBD + (6.24 kN) = 0
3.905
13
13
ΣFy = 0:
−
2.5
5
5
FBC + FBD + (6.24 kN) − 2.4 kN = 0
3.905
13
13
(1)
(2)
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759
PROBLEM 6.14 (Continued)
Multiply (1) by 2.5, (2) by 3, and add:
45
45
FBD + (6.24 kN) − 7.2 kN = 0, FBD = −4.16 kN,
13
13
FBD = 4.16 kN C
Multiply (1) by 5, (2) by –12, and add:
45
FBC + 28.8 kN = 0, FBC = −2.50 kN,
3.905
FBC = 2.50 kN C
Free body: Joint C:
ΣFy = 0:
5
2.5
(2.50 kN) = 0
FCD −
5.831
3.905
FCD = 1.867 kN T
ΣFx = 0: FCE − 5.76 kN +
3
3
(2.50 kN) +
(1.867 kN) = 0
3.905
5.831
FCE = 2.88 kN T
Free body: Joint E:
ΣFy = 0:
5
FDE + 3.6 kN − 1.2 kN = 0
7.81
FDE = −3.75 kN
ΣFx = 0: − FCE −
FDE = 3.75 kN C
6
( −3.75 kN) = 0
7.81
FCE = +2.88 kN FCE = 2.88 kN T
(Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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760
PROBLEM 6.15
Determine the force in each member of the Warren bridge
truss shown. State whether each member is in tension or
compression.
SOLUTION
ΣFx = 0: Ax = 0
Free body: Truss
Due to symmetry of truss and loading
Ay = G =
Free body: Joint A:
1
Total load = 6 kips
2
FAB FAC 6 kips
=
=
5
3
4
FAB = 7.50 kips C
FAC = 4.50 kips T
!
Free body: Joint B:
FBC FBD 7.5 kips
=
=
5
6
5
FBC = 7.50 kips T
FBD = 9.00 kips C
Free body: Joint C:
ΣFy = 0:
4
4
(7.5) + FCD − 6 = 0
5
5
FCD = 0
3
ΣFx = 0: FCE − 4.5 − (7.5) = 0
5
FCE = +9 kips
FCE = 9.00 kips T
Truss and loading symmetrical about cL
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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761
!
PROBLEM 6.16
Solve Problem 6.15 assuming that the load applied at E has
been removed.
PROBLEM 6.15 Determine the force in each member of the
Warren bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss
ΣFx = 0: Ax = 0
ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips
ΣFy = 0: 4 − 6 + G = 0 G = 2 kips
Free body: Joint A:
FAB FAC 4 kips
=
=
5
3
4
FAB = 5.00 kips C
FAC = 3.00 kips T
Free body Joint B:
FBC FBD 5 kips
=
=
5
6
5
FBC = 5.00 kips T
FBD = 6.00 kips C
Free body Joint C:
ΣM y = 0:
4
4
(5) + FCD − 6 = 0
5
5
3
3
ΣFx = 0: FCE + (2.5) − (5) − 3 = 0
5
5
FCD = 2.50 kips T
FCE = 4.50 kips T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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762
!
PROBLEM 6.16 (Continued)
Free body: Joint D:
4
4
ΣFy = 0: − (2.5) − FDE = 0
5
5
FDE = −2.5 kips
FDE = 2.50 kips C
3
3
ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0
5
5
Free body: Joint F:
FDF = −3 kips
FDF = 3.00 kips C
FEF FFG 3 kips
=
=
5
5
6
FEF = 2.50 kips T
FFG = 2.50 kips C
!
Free body: Joint G:
FEG 2 kips
=
3
4
Also:
FFG 2 kips
=
5
4
FEG = 1.500 kips T
FFG = 2.50 kips C (Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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763
!
PROBLEM 6.17
Determine the force in member DE and in each of the
members located to the left of DE for the inverted Howe
roof truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
ΣM H = 0: (400 lb)(4d ) + (800 lb)(3d ) + (800 lb)(2d ) + (800 lb)d − Ay (4d ) = 0
Ay = 1600 lb
Angles:
6.72
α = 32.52°
10.54
6.72
tan β =
β = 16.26°
23.04
tan α =
Free body: Joint A:
FAC
FAB
1200 lb
=
=
sin 57.48° sin106.26° sin16.26°
FAB = 3613.8 lb C
FAC = 4114.3 lb T
FAB = 3610 lb C , FAC = 4110 lb T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
764
PROBLEM 6.17 (Continued)
Free body: Joint B:
ΣFy = 0: − FBC − (800 lb) cos16.26° = 0
FBC = −768.0 lb
FBC = 768 lb C
ΣFx = 0: FBD + 3613.8 lb + (800 lb) sin16.26° = 0
FBD = −3837.8 lb
FBD = 3840 lb C
Free body: Joint C:
ΣFy = 0: − FCE sin 32.52° + (4114.3 lb) sin 32.52° − (768 lb) cos16.26° = 0
FCE = 2742.9 lb
FCE = 2740 lb T
ΣFx = 0: FCD − (4114.3 lb) cos 32.52° + (2742.9 lb) cos 32.52° − (768 lb) sin16.26° = 0
FCD = 1371.4 lb
FCD = 1371 lb T
Free body: Joint E:
FDE
2742.9 lb
=
sin 32.52° sin73.74°
FDE = 1536 lb C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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765
PROBLEM 6.18
Determine the force in each of the members located to the
right of DE for the inverted Howe roof truss shown. State
whether each member is in tension or compression.
SOLUTION
Free body: Truss
ΣM A = 0: H (4d ) − (800 lb)d − (800 lb)(2d ) − (800 lb)(3d ) − (400 lb)(4d ) = 0
H = 1600 lb
Angles:
6.72
α = 32.52°
10.54
6.72
tan β =
β = 16.26°
23.04
tan α =
Free body: Joint H:
FGH = (1200 lb) cot16.26°
FGH = 4114.3 lb T
FFH =
1200 lb
= 4285.8 lb
sin16.26°
FGH = 4110 lb T
FFH = 4290 lb C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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766
PROBLEM 6.18 (Continued)
Free body Joint F:
ΣFy = 0: − FFG − (800 lb) cos16.26° = 0
FFG = −768.0 lb
FFG = 768 lb C
ΣFx = 0: − FDF − 4285.8 lb + (800 lb) sin 16.26° = 0
FDF = −4061.8 lb
FDF = 4060 lb C
Free body: Joint G:
ΣFy = 0: FDG sin 32.52° − (768.0 lb) cos16.26° = 0
FDG = +1371.4 lb
FDG = 1371 lb T
!
ΣFx = 0: − FEG + 4114.3 lb − (768.0 lb)sin16.26° − (1371.4 lb) cos 32.52° = 0
FEG = +2742.9 lb
FEG = 2740 lb T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
767
!
PROBLEM 6.19
Determine the force in each of the members located to the
left of FG for the scissors roof truss shown. State whether
each member is in tension or compression.
SOLUTION
Free Body: Truss
ΣFx = 0: A x = 0
ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0
A y = 4.50 kN
FBC = 0
We note that BC is a zero-force member:
Also:
FCE = FAC
Free body: Joint A:
ΣFx = 0:
ΣFy = 0:
(1)
1
2
1
2
FAB +
FAB +
2
5
1
5
FAC = 0
(2)
FAC + 3.50 kN = 0
(3)
Multiply (3) by –2 and add (2):
−
1
2
FAB − 7 kN = 0
FAB = 9.90 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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768
PROBLEM 6.19 (Continued)
Subtract (3) from (2):
1
5
FAC − 3.50 kN = 0 FAC = 7.826 kN
FCE = FAC = 7.826 kN
From (1):
FAC = 7.83 kN T
FCE = 7.83 kN T
Free body: Joint B:
ΣFy = 0:
1
2
FBD +
1
2
(9.90 kN) − 2 kN = 0
FBD = −7.071 kN
1
ΣFx = 0: FBE +
2
(9.90 − 7.071)kN = 0
FBE = −2.000 kN
Free body: Joint E:
ΣFx = 0:
2
5
FBD = 7.07 kN C
FBE = 2.00 kN C
( FEG − 7.826 kN) + 2.00 kN = 0
FEG = 5.590 kN
ΣFy = 0: FDE −
1
5
FEG = 5.59 kN T
(7.826 − 5.590)kN = 0
FDE = 1.000 kN
FDE = 1.000 kN T
Free body: Joint D:
ΣFx = 0:
2
5
1
2
(7.071 kN)
FDF + FDG = −5.590 kN
or
ΣFy = 0:
or
( FDF + FDG ) +
1
5
( FDF − FDG ) +
1
2
(4)
(7.071 kN) = 2 kN − 1 kN = 0
FDE − FDG = −4.472
(5)
Add (4) and (5):
2 FDF = −10.062 kN
FDF = 5.03 kN C
Subtract (5) from (4):
2 FDG = −1.1180 kN
FDG = 0.559 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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769
PROBLEM 6.20
Determine the force in member FG and in each of the
members located to the right of FG for the scissors roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss
ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0
L = 1.500 kN
Angles:
tan α = 1
α = 45°
1
tan β =
2
β = 26.57°
Zero-force members:
Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI.
FHI = FIJ = FJK = 0
Thus:
We also note that
and
FGI = FIK = FKL
(1)
FHJ = FJL
(2)
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770
PROBLEM 6.20 (Continued)
Free body: Joint L:
FJL
F
1.500 kN
= KL =
sin116.57° sin 45° sin18.43°
FJL = 4.2436 kN
FJL = 4.24 kN C
FKL = 3.35 kN T
From Eq. (1):
FGI = FIK = FKL
FGI = FIK = 3.35 kN T
From Eq. (2):
FHJ = FJL = 4.2436 kN
FHJ = 4.24 kN T
FGH
FFH
4.2436
=
=
sin108.43° sin18.43° sin 53.14°
FFH = 5.03 kN C
Free body: Joint H:
FGH = 1.677 kN T
Free body: Joint F:
ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0
FDF = −5.03 kN
ΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57° − ( −5.03 kN)sin 26.57° = 0
FFG = 3.500 kN
FFG = 3.50 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
771
PROBLEM 6.21
Determine the force in each of the members located to the left
of line FGH for the studio roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss
ΣFx = 0: A x = 0
Because of symmetry of loading:
Ay = L =
1
Total load
2
A y = L = 1200 lb
Zero-Force Members. Examining joints C and H, we conclude that
BC, EH, and GH are zero-force members. Thus
FBC = FEH = 0
Also,
FCE = FAC
Free body: Joint A
FAB
(1)
FAC 1000 lb
=
2
1
5
FAB = 2236 lb C
=
FAB = 2240 lb C
FAC = 2000 lb T
FCE = 2000 lb T
From Eq. (1):
Free body: Joint B
ΣFx = 0:
2
5
FBD +
2
5
FBE +
2
5
(2236 lb) = 0
FBD + FBE = −2236 lb
or
ΣFy = 0:
1
5
FBD −
1
5
FBE +
1
FBD − FBE = −1342 lb
or
(2)
5
(2236 lb) − 400 lb = 0
(3)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
772
PROBLEM 6.21 (Continued)
Add (2) and (3):
2 FBD = −3578 lb
FBD = 1789 lb C
Subtract (3) from (1):
2 FBE = − 894 lb
FBE = 447 lb C
Free body: Joint E
2
ΣFx = 0:
5
FEG +
2
5
(447 lb) − 2000 lb = 0
FEG = 1789 lb T
ΣFy = 0: FDE +
1
5
1
(1789 lb) −
5
FDE = − 600 lb
(447 lb) = 0
FDE = 600 lb C
Free body: Joint D
2
ΣFx = 0:
5
FDF +
2
5
2
FDG +
5
(1789 lb) = 0
FDF + FDG = −1789 lb
or
ΣFy = 0:
1
FDF −
1
FDG +
1
5
5
5
+ 600 lb − 400 lb = 0
(4)
(1789 lb)
FDF − FDG = −2236 lb
or
Add (4) and (5):
2 FDF = − 4025 lb
Subtract (5) from (4):
2 FDG = 447 lb
(5)
FDF = 2010 lb C
FDG = 224 lb T
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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773
PROBLEM 6.22
Determine the force in member FG and in each of the members
located to the right of FG for the studio roof truss shown. State
whether each member is in tension or compression.
SOLUTION
Reaction at L: Because of the symmetry of the loading,
L=
1
Total load,
2
L = 1200 lb
(See F, B, diagram to the left for more details)
Free body: Joint L
9
= 26.57°
18
3
β = tan −1 = 9.46°
18
FJL
FKL
1000 lb
=
=
sin 63.43° sin 99.46° sin17.11°
α = tan −1
FKL = 3352.7 lb C
FJL = 3040 lb T
FKL = 3350 lb C
Free body: Joint K
ΣFx = 0: −
2
5
FIK −
2
5
FJK −
2
5
(3352.7 lb) = 0
FIK + FJK = − 3352.7 lb
or:
ΣFy = 0:
or:
Add (1) and (2):
1
5
FIK −
1
5
FJK +
(1)
1
5
FIK − FJK = − 2458.3 lb
(2)
2 FIK = − 5811.0
FJK = − 2905.5 lb
Subtract (2) from (1):
(3352.7) − 400 = 0
FIK = 2910 lb C
2 FJK = − 894.4
FJK = − 447.2 lb
FJK = 447 lb C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
774
PROBLEM 6.22 (Continued)
Free body: Joint J
2
ΣFx = 0: −
13
3
ΣFy = 0:
FIJ +
13
6
FIJ −
37
1
FGJ −
37
6
FGJ +
37
1
5
1
(3040 lb) −
37
2
(3040 lb) −
5
(447.2) = 0
(447.2) = 0
(3)
(4)
Multiply (4) by 6 and add to (3):
16
13
FIJ −
8
5
(447.2) = 0
FIJ = 360.54 lb
FIJ = 361 lb T
Multiply (3) by 3, (4) by 2, and add:
−
16
37
( FGJ − 3040) −
8
5
(447.2) = 0
FGJ = 2431.7 lb
FGJ = 2430 lb T
Free body: Joint I
2
ΣFx = 0: −
5
2
FFI −
5
2
FGI −
5
(2905.5) +
2
13
(360.54) = 0
FFI + FGI = − 2681.9 lb
or
ΣFy = 0:
1
5
FFI −
1
5
FGI +
1
5
(2905.5) −
(5)
3
13
(360.54) − 400 = 0
FFI − FGI = −1340.3 lb
or
(6)
2 FFI = −4022.2
Add (5) and (6):
Subtract (6) from (5):
FFI = −2011.1 lb
FFI = 2010 lb C
2 FGI = −1341.6 lb
FGI = 671 lb C
Free body: Joint F
From
ΣFx = 0: FDF = FFI = 2011.1 lb C
!
1
2011.1 lb # = 0
ΣFy = 0: FFG − 400 lb + 2 "
$ 5
%
FFG = + 1400 lb
FFG = 1400 lb T
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
775
PROBLEM 6.23
Determine the force in each of the members connecting joints A
through F of the vaulted roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss
ΣFx = 0: A x = 0
ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a
− Ay (6a) = 0 A y = 5.40 kN
Free body: Joint A
FAB
FAC 4.20 kN
=
2
1
5
FAB = 9.3915 kN
=
FAB = 9.39 kN C
FAC = 8.40 kN T
Free body: Joint B
2
ΣFx = 0:
5
1
ΣFy = 0:
5
FBD +
FBD −
1
2
1
2
FBC +
FBC +
2
5
1
5
(9.3915) = 0
(1)
(9.3915) − 2.4 = 0
(2)
Add (1) and (2):
3
5
FBD +
3
5
(9.3915 kN) − 2.4 kN = 0
FBD = − 7.6026 kN
FBD = 7.60 kN C
Multiply (2) by –2 and add (1):
3
2
FB + 4.8 kN = 0
FBC = −2.2627 kN
FBC = 2.26 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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776
PROBLEM 6.23 (Continued)
Free body: Joint C
1
ΣFx = 0:
5
2
ΣFy = 0:
5
4
FCD +
17
1
FCD +
17
FCE +
FCE −
1
2
1
2
(2.2627) − 8.40 = 0
(3)
(2.2627) = 0
(4)
Multiply (4) by −4 and add (1):
−
7
5
FCD +
5
(2.2627) − 8.40 = 0
2
FCD = − 0.1278 kN
FCD = 0.128 kN C
Multiply (1) by 2 and subtract (2):
7
17
FCE +
3
2
(2.2627) − 2(8.40) = 0
FCE = 7.068 kN
FCE = 7.07 kN T
Free body: Joint D
ΣFx = 0:
2
5
1
+
ΣFy = 0:
5
1
5
+
FDF +
(0.1278) = 0
FDF −
2
5
1
2
FDE +
(7.6026)
1.524
5
(5)
1.15
1
FDE +
(7.6026)
1.524
5
(0.1278) − 2.4 = 0
(6)
Multiply (5) by 1.15 and add (6):
3.30
5
FDF +
3.30
5
(7.6026) +
3.15
5
(0.1278) − 2.4 = 0
FDF = − 6.098 kN
FDF = 6.10 kN C
Multiply (6) by –2 and add (5):
3.30
3
FDE −
(0.1278) + 4.8 = 0
1.524
5
FDE = − 2.138 kN
FDE = 2.14 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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777
PROBLEM 6.23 (Continued)
Free body: Joint E
ΣFx = 0:
0.6
4
1
FEF +
( FEH − FCE ) +
(2.138) = 0
2.04
1.524
17
(7)
ΣFy = 0:
1.95
1
1.15
FEF +
( FEH − FCE ) −
(2.138) = 0
2.04
1.524
17
(8)
Multiply (8) by 4 and subtract (7):
7.2
FEF − 7.856 kN = 0
2.04
FEF = 2.23 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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778
!
PROBLEM 6.24
The portion of truss shown represents the upper part of a power
transmission line tower. For the given loading, determine the
force in each of the members located above HJ. State whether
each member is in tension or compression.
SOLUTION
Free body: Joint A
F
FAB
1.2 kN
= AC =
2.29 2.29
1.2
FAB = 2.29 kN T
FAC = 2.29 kN C
Free body: Joint F
FDF
F
1.2 kN
= EF =
2.29 2.29
2.1
FDF = 2.29 kN T
FEF = 2.29 kN C
Free body: Joint D
FBD FDE 2.29 kN
=
=
2.21 0.6
2.29
FBD = 2.21 kN T
FDE = 0.600 kN C
Free body: Joint B
ΣFx = 0:
4
2.21
(2.29 kN) = 0
FBE + 2.21 kN −
5
2.29
FBE = 0
3
0.6
(2.29 kN) = 0
ΣFy = 0: − FBC − (0) −
5
2.29
FBC = − 0.600 kN
FBC = 0.600 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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779
PROBLEM 6.24 (Continued)
Free body: Joint C
ΣFx = 0: FCE +
2.21
(2.29 kN) = 0
2.29
FCE = − 2.21kN
ΣFy = 0: − FCH − 0.600 kN −
FCE = 2.21 kN C
0.6
(2.29 kN) = 0
2.29
FCH = −1.200 kN
FCH = 1.200 kN C
Free body: Joint E
ΣFx = 0: 2.21 kN −
2.21
4
(2.29 kN) − FEH = 0
2.29
5
FEH = 0
ΣFy = 0: − FEJ − 0.600 kN −
FEJ = −1.200 kN
0.6
(2.29 kN) − 0 = 0
2.29
FEJ = 1.200 kN C
!
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780
PROBLEM 6.25
For the tower and loading of Problem 6.24 and knowing that
FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in
member HJ and in each of the members located between HJ and
NO. State whether each member is in tension or compression.
PROBLEM 6.24 The portion of truss shown represents the
upper part of a power transmission line tower. For the given
loading, determine the force in each of the members located
above HJ. State whether each member is in tension or
compression.
SOLUTION
Free body: Joint G
FGH
F
1.2 kN
= GI =
3.03 3.03
1.2
FGH = 3.03 kN T
FGI = 3.03 kN C
Free body: Joint L
FJL
F
1.2 kN
= KL =
3.03 3.03
1.2
FJL = 3.03 kN T
FKL = 3.03 kN C
Free body: Joint J
ΣFx = 0: − FHJ +
2.97
(3.03 kN) = 0
3.03
FHJ = 2.97 kN T
0.6
(3.03 kN) = 0
3.03
= −1.800 kN
FJK = 1.800 kN C
Fy = 0: − FJK − 1.2 kN −
FJK
Free body: Joint H
ΣFx = 0:
4
2.97
(3.03 kN) = 0
FHK + 2.97 kN −
5
3.03
FHK = 0
0.6
3
(3.03) kN − (0) = 0
3.03
5
= −1.800 kN
FHI = 1.800 kN C
ΣFy = 0: − FHI − 1.2 kN −
FHI
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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781
PROBLEM 6.25 (Continued)
Free body: Joint I
ΣFx = 0: FIK +
2.97
(3.03 kN) = 0
3.03
FIK = −2.97 kN
ΣFy = 0: − FIN − 1.800 kN −
FIN = −2.40 kN
FIK = 2.97 kN C
0.6
(3.03 kN) = 0
3.03
FIN = 2.40 kN C
Free body: Joint K
ΣFx = 0: −
4
2.97
(3.03 kN) = 0
FKN + 2.97 kN −
5
3.03
FKN = 0
ΣFy = 0: − FKD −
0.6
3
(3.03 kN) − 1.800 kN − (0) = 0
3.03
5
FKD = −2.40 kN
FKD = 2.40 kN C
!
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782
PROBLEM 6.26
Solve Problem 6.24 assuming that the cables hanging from the
right side of the tower have fallen to the ground.
PROBLEM 6.24 The portion of truss shown represents the
upper part of a power transmission line tower. For the given
loading, determine the force in each of the members located
above HJ. State whether each member is in tension or
compression.
SOLUTION
Zero-Force Members.
Considering joint F, we note that DF and EF are zero-force members:
FDF = FEF = 0
Considering next joint D, we note that BD and DE are zero-force
members:
FBD = FDE = 0
Free body: Joint A
F
FAB
1.2 kN
= AC =
2.29 2.29
1.2
FAB = 2.29 kN T
FAC = 2.29 kN C
Free body: Joint B
ΣFx = 0:
4
2.21
(2.29 kN) = 0
FBE −
5
2.29
FBE = 2.7625 kN
ΣFy = 0: − FBC −
FBE = 2.76 kN T
0.6
3
(2.29 kN) − (2.7625 kN) = 0
2.29
5
FBC = −2.2575 kN
FBC = 2.26 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
783
PROBLEM 6.26 (Continued)
Free body: Joint C
ΣFx = 0: FCE +
2.21
(2.29 kN) = 0
2.29
FCE = 2.21 kN C
ΣFy = 0: − FCH − 2.2575 kN −
FCH = −2.8575 kN
0.6
(2.29 kN) = 0
2.29
FCH = 2.86 kN C
Free body: joint E
ΣFx = 0: −
4
4
FEH − (2.7625 kN) + 2.21 kN = 0
5
5
FEH = 0
3
3
ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0
5
5
FEJ = +1.6575 kN
FEJ = 1.658 kN T
!
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784
PROBLEM 6.27
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Free body: Truss
ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0
G = 42 kips
ΣFx = 0: Fx + 15 kips = 0
Fx = 15 kips
ΣFy = 0: Fy − 40 kips + 42 kips = 0
Fy = 2 kips
Free body: Joint F
ΣFx = 0:
1
5
FDF − 15 kips = 0
FDF = 33.54 kips
ΣFy = 0: FBF − 2 kips +
2
5
FDF = 33.5 kips T
(33.54 kips) = 0
FBF = − 28.00 kips
FBF = 28.0 kips C
Free body: Joint B
ΣFx = 0:
ΣFy = 0:
5
29
2
29
FAB +
FAB −
5
61
6
61
FBD + 15 kips = 0
(1)
FBD + 28 kips = 0
(2)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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785
PROBLEM 6.27 (Continued)
Multiply (1) by 6, (2) by 5, and add:
40
29
FAB + 230 kips = 0
FAB = −30.96 kips
FAB = 31.0 kips C
Multiply (1) by 2, (2) by –5, and add:
40
FBD − 110 kips = 0
61
FBD = 21.48 kips
FBD = 21.5 kips T
Free body: joint D
2
ΣFy = 0:
5
2
FAD −
5
(33.54) +
6
61
(21.48) = 0
FAD = 15.09 kips T
1
ΣFx = 0: FDE +
5
(15.09 − 33.54) −
5
(21.48) = 0
61
FDE = 22.0 kips T
Free body: joint A
5
ΣFx = 0:
29
ΣFy = 0: −
1
FAC +
2
29
5
FAC −
FAE +
2
5
5
FAE +
29
(30.36) −
2
29
1
5
(15.09) = 0
2
(30.96) −
5
(15.09) = 0
(3)
(4)
Multiply (3) by 2 and add (4):
8
29
FAC +
12
29
(30.96) −
4
5
FAC = −28.27 kips,
(15.09) = 0
FAC = 28.3 kips C
Multiply (3) by 2 (4) by 5 and add:
−
8
5
FAE +
20
29
(30.96) −
12
5
(15.09) = 0
FAE = 9.50 kips T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
786
PROBLEM 6.27 (Continued)
Free body: Joint C
From force triangle
FCE
61
=
FCG 28.27 kips
=
8
29
FCE = 41.0 kips T
FCG = 42.0 kips C
Free body: Joint G
ΣFx = 0:
ΣFy = 0: 42 kips − 42 kips = 0
FEG = 0
(Checks) !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
787
PROBLEM 6.28
Determine the force in each member of the truss shown. State
whether each member is in tension or compression.
SOLUTION
Free body: Truss
ΣFx = 0: H x = 0
ΣM H = 0: 48(16) − G (4) = 0 G = 192 kN
ΣFy = 0: 192 − 48 + H y = 0
H y = −144 kN H y = 144 kN
Zero-Force Members:
Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE,
DE, DG, FG
Thus,
FBC = FBE = FDE = FDG = FFG = 0
Also note:
FAB = FBD = FDF = FFH
(1)
FAC = FCE = FEG
(2)
Free body: Joint A:
FAB FAC 48 kN
=
=
8
3
73
FAB = 128 kN
FAB = 128.0 kN T
FAC = 136.704 kN
FAC = 136.7 kN C
From Eq. (1):
FBD = FDF = FFH = 128.0 kN T
From Eq. (2):
FCE = FEG = 136.7 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
788
!
PROBLEM 6.28 (Continued)
Free body: Joint H
Also
FGH
145
=
144 kN
9
FGH = 192.7 kN C
FFH 144 kN
=
8
9
FFH = 128.0 kN T
(Checks) !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
789
PROBLEM 6.29
Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a
are simple trusses.
SOLUTION
Truss of Problem 6.31a
Starting with triangle ABC and adding two members at a time, we obtain joints D,
E, G, F, and H, but cannot go further thus, this truss
is not a simple truss
Truss of Problem 6.32a
Starting with triangle HDI and adding two members at a time, we obtain
successively joints A, E, J, and B, but cannot go further. Thus, this truss
is not a simple truss
Truss of Problem 6.33a
Starting with triangle EHK and adding two members at a time, we obtain
successively joints D, J, C, G, I, B, F, and A, thus completing the truss.
Therefore, this truss
is a simple truss
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
790
!
PROBLEM 6.30
Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b
are simple trusses.
SOLUTION
Truss of Problem 6.31b.
Starting with triangle ABC and adding two members at a time, we obtain
successively joints E, D, F, G, and H, but cannot go further. Thus, this truss
is not a simple truss
Truss of Problem 6.32b.
Starting with triangle CGH and adding two members at a time, we obtain
successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing
the truss.
Therefore, this truss is a simple truss
Truss of Problem 6.33b.
Starting with triangle GLM and adding two members at a time, we obtain joints K
and I but cannot continue, starting instead with triangle BCH, we obtain joint D
but cannot continue, thus, this truss
is not a simple truss
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
791
PROBLEM 6.31
For the given loading, determine the zero-force members in each
of the two trusses shown.
SOLUTION
Truss (a)
FB: Joint B: FBC = 0
FB: Joint C: FCD = 0
FB: Joint J : FIJ = 0
FB: Joint I : FIL = 0
FB: Joint N : FMN = 0
(a)
FB: Joint M : FLM = 0
The zero-force members, therefore, are
Truss (b)
BC , CD, IJ , IL, LM , MN
FB: Joint C: FBC = 0
FB: Joint B: FBE = 0
FB: Joint G: FFG = 0
FB: Joint F : FEF = 0
FB: Joint E: FDE = 0
FB: Joint I : FIJ = 0
(b)
FB: Joint M : FMN = 0
FB: Joint N : FKN = 0
The zero-force members, therefore, are
BC , BE , DE , EF , FG , IJ , KN , MN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
792
PROBLEM 6.32
For the given loading, determine the zero-force members in each
of the two trusses shown.
SOLUTION
Truss (a)
FB: Joint B: FBJ = 0
FB: Joint D: FDI = 0
FB: Joint E: FEI = 0
FB: Joint I : FAI = 0
(a)
FB: Joint F : FFK = 0
FB: Joint G: FGK = 0
FB: Joint K : FCK = 0
The zero-force members, therefore, are
Truss (b)
AI , BJ , CK , DI , EI , FK , GK
FB: Joint K : FFK = 0
FB: Joint O: FIO = 0
(b)
The zero-force members, therefore, are
FK and IO
All other members are either in tension or compression.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
793
PROBLEM 6.33
For the given loading, determine the zero-force members in each of the two trusses shown.
SOLUTION
Truss (a)
FB: Joint F : FBF = 0
FB: Joint B: FBG = 0
FB: Joint G: FGJ = 0
FB: Joint D: FDH = 0
(a)
FB: Joint J : FHJ = 0
FB: Joint H : FEH = 0
The zero-force members, therefore, are
Truss (b)
BF , BG , DH , EH , GJ , HJ
FB: Joint A: FAB = FAF = 0
FB: Joint C: FCH = 0
FB: Joint E: FDE = FEJ = 0
FB: Joint L: FGL = 0
FB: Joint N : FIN = 0
The zero-force members, therefore, are
(b)
AB, AF , CH , DE , EJ , GL, IN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
794
PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.
SOLUTION
(a)
Truss of Problem 6.23
FB : Joint I : FIJ = 0
FB : Joint J : FGJ = 0
FB : Joint G: FGH = 0
The zero-force members, therefore, are
(b)
GH , GJ , IJ
Truss of Problem 6.28
FB : Joint C: FBC = 0
FB : Joint B: FBE = 0
FB : Joint E: FDE = 0
FB : Joint D: FDG = 0
FB : Joint F : FFG = 0
The zero-force members, therefore, are
BC , BE , DE , DG, FG
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
795
PROBLEM 6.35*
The truss shown consists of six members and is supported by a
short link at A, two short links at B, and a ball and socket at D.
Determine the force in each of the members for the given loading.
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx
and
Dy = B y
ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0
A = −960 lb
ΣFx = 0: Bx + Dx + A = 0
2 Bx − 960 lb = 0, Bx = 480 lb
ΣFy = 0: B y + Dy − 400 lb = 0
2 By = 400 lb
By = +200 lb
B = (480 lb)i + (200 lb) j "
Thus
Free body: C
FCA = FAC
FCB = FBC
FCD = FCD
!
CA FAC
=
( −24i + 10 j)
CA 26
!
CB FBC
=
(−24i + 7k )
CB
25
!
CD FCD
=
(−24i − 7k )
CD
25
ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
796
PROBLEM 6.35* (Continued)
Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k :
24
24
FAC − ( FBC + FCD ) = 0
26
25
i:
−
j:
10
FAC − 400 lb = 0
26
k:
7
( FBC − FCD ) = 0 FCD = FBC
25
(1)
FAC = 1040 lb T
Substitute for FAC and FCD in Eq. (1):
−
24
24
(10.40 lb) − (2 FBC ) = 0 FBC = −500 lb
26
25
FBC = FCD = 500 lb C
Free body: B
FBC
FBA
FBD
!
CB
= (500 lb)
= −(480 lb)i + (140 lb)k
CB
!
BA FAB
= FAB
=
(10 j − 7k )
BA 12.21
= − FBD k
ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0
Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k:
j:
10
FAB + 200 lb = 0 FAB = −244.2 lb
12.21
k: −
7
FAB − FBD + 140 lb = 0
12.21
FBD = −
From symmetry:
7
(−244.2 lb) + 140 lb = +280 lb
12.21
FAD = FAB
FAB = 244 lb C
FBD = 280 lb T
FAD = 244 lb C
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797
PROBLEM 6.36*
The truss shown consists of six members and is supported by a
ball and socket at B, a short link at C, and two short links at D.
Determine the force in each of the members for P = (−2184 N)j
and Q = 0.
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx and Dy = By
ΣFx = 0: 2 Bx = 0
Bx = Dx = 0
ΣFz = 0: Bz = 0
ΣM cz = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0
By = 780 N
B = (780 N) j "
Thus
Free body: A
!
AB FAB
FAB = FAB
=
( −0.8i − 4.8 j + 2.1k )
AB 5.30
!
AC FAC
=
FAC = FAC
(2i − 4.8 j)
AC 5.20
!
AD FAD
=
FAD = FAD
(+0.8i − 4.8 j − 2.1k )
AD 5.30
ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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798
PROBLEM 6.36* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k :
i:
−
0.8
2
( FAB + FAD ) +
FAC = 0
5.30
5.20
(1)
j:
−
4.8
4.8
( FAB + FAD ) −
FAC − 2184 N = 0
5.30
5.20
(2)
2.1
( FAB − FAD ) = 0
5.30
k:
FAD = FAB
Multiply (1) by –6 and add (2):
16.8 !
−"
# FAC − 2184 N = 0, FAC = −676 N
$ 5.20 %
FAC = 676 N C
Substitute for FAC and FAD in (1):
0.8 !
2 !
−"
# 2 FAB + "
# (−676 N) = 0, FAB = −861.25 N
$ 5.30 %
$ 5.20 %
FAB = FAD = 861 N C
Free body: B
!
AB
= −(130 N)i − (780 N) j + (341.25 N)k
FAB = (861.25 N)
AB
2.8i − 2.1k !
FBC = FBC "
# = FBC (0.8i − 0.6k )
3.5
$
%
FBD = − FBD k
ΣF = 0: FAB + FBC + FBD + (780 N) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k :
i:
k:
−130 N + 0.8 FBC = 0 FBC = +162.5 N
341.25 N − 0.6 FBC − FBD = 0
FBD = 341.25 − 0.6(162.5) = +243.75 N
From symmetry:
FBC = 162.5 N T
FCD = FBC
FBD = 244 N T
FCD = 162.5 N T
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799
PROBLEM 6.37*
The truss shown consists of six members and is supported
by a ball and socket at B, a short link at C, and two short
links at D. Determine the force in each of the members for
P = 0 and Q = (2968 N)i.
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx and Dy = By
ΣFx = 0: 2 Bx + 2968 N = 0
Bx = Dx = −1484 N
ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0
By = −2544 N
B = −(1484 N)i − (2544 N) j "
Thus
Free body: A
FAB = FAB
!
AB
AB
FAB
(−0.8i − 4.8 j + 2.1k )
5.30
!
AC FAC
= FAC
=
(2i − 4.8 j)
AC 5.20
!
AD
= FAD
AD
FAD
=
(−0.8i − 4.8 j − 2.1k )
5.30
=
FAC
FAD
ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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800
PROBLEM 6.37* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k :
0.8
2
FAC + 2968 N = 0
( FAB + FAD ) +
5.30
5.20
(1)
4.8
4.8
FAC = 0
( FAB + FAD ) −
5.30
5.20
(2)
i:
−
j:
−
k:
2.1
( FAB − FAD ) = 0
5.30
FAD = FAB
Multiply (1) by –6 and add (2):
16.8 !
−"
# FAC − 6(2968 N) = 0, FAC = −5512 N
$ 5.20 %
FAC = 5510 N C
Substitute for FAC and FAD in (2):
4.8 !
4.8 !
−"
# 2 FAB − "
# (−5512 N) = 0, FAB = +2809 N
$ 5.30 %
$ 5.20 %
FAB = FAD = 2810 N T
Free body: B
FAB
FBC
FBD
!
BA
= (2809 N)
= (424 N)i + (2544 N) j − (1113 N)k
BA
2.8 i − 2.1k !
= FBC "
# = FBC (0.8i − 0.6k )
3.5
$
%
= − FBD k
ΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k:
i:
k:
From symmetry:
+24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N
FBC = 1325 N T
−1113 N − 0.6 FBC − FBD = 0
FBD = −1113 N − 0.6(1325 N) = −1908 N,
FBD = 1908 N C
FCD = FBC
FCD = 1325 N T
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you are using it without permission.
801
PROBLEM 6.38*
The truss shown consists of nine members and is
supported by a ball and socket at A, two short links at B,
and a short link at C. Determine the force in each of the
members for the given loading.
SOLUTION
Free body: Truss
From symmetry:
Az = Bz = 0
ΣFx = 0: Ax = 0
ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0
A = −(2000 lb)j "
Ay = −2000 lb
By = C
From symmetry:
ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0
B = (1800 lb) j "
By = 1800 lb
ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0
Free body: A
FAB
i+k
+ FAC
2
i−k
2
+ FAD (0.6 i + 0.8 j) − (2000 lb) j = 0
Factoring i, j, k and equating their coefficient to zero:
1
2
FAB +
1
2
FAC + 0.6 FAD = 0
(1)
0.8FAD − 2000 lb = 0
1
2
FAB −
1
2
FAC = 0
FAD = 2500 lb T
FAC = FAB
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802
PROBLEM 6.38* (Continued)
Substitute for FAD and FAC into (1):
2
2
Free body: B
FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb,
FBA
FBC
FAB = FAC = 1061 lb C
!
BA
i+k
= FAB
= + (1060.7 lb)
= (750 lb)(i + k )
BA
2
= − FBC k
FBD = FBD (0.8 j − 0.6k )
!
BE FBE
FBE = FBE
=
(7.5i + 8 j − 6k )
BE 12.5
ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0
Substituting for FBA , FBC , FBD + FBE and equate to zero the coefficients of i, j, k :
i:
7.5 !
750 lb + "
# FBE = 0, FBE = −1250 lb
$ 12.5 %
FBE = 1250 lb C
j:
8 !
0.8 FBD + "
# (−1250 lb) + 1800 lb = 0
$ 12.5 %
FBD = 1250 lb C
k:
750 lb − FBC − 0.6( −1250 lb) −
6
( −1250 lb) = 0
12.5
FBC = 2100 lb T
FBD = FCD = 1250 lb C
From symmetry:
Free body: D
ΣF = 0: FDA + FDB + FDC + FDE i = 0
We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its
coefficient is
−0.6 FAD = −0.6(2500 lb) = −1500 lb
i:
−1500 lb + FDE = 0
FDE = 1500 lb T
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803
PROBLEM 6.39*
The truss shown consists of nine members and is supported by a ball
and socket at B, a short link at C, and two short links at D. (a) Check
that this truss is a simple truss, that it is completely constrained,
and that the reactions at its supports are statically determinate.
(b) Determine the force in each member for P = (−1200 N)j and Q = 0.
SOLUTION
Free body: Truss
ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dk )
+ (0.6i − 0.75k ) × (−1200 j) = 0
−1.8 Ck + 1.8 D y k − 1.8 Dz j
+ 3D y i − 720k − 900i = 0
Equate to zero the coefficients of i, j, k :
i:
3Dy − 900 = 0, D y = 300 N
j:
Dz = 0,
D = (300 N) j "
k:
1.8C + 1.8(300) − 720 = 0
C = (100 N) j "
ΣF = 0: B + 300 j + 100 j − 1200 j = 0
B = (800 N) j "
Free body: B
ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with
!
BA FAB
FBA = FAB
(0.6i + 3j − 0.75k )
=
BA 3.15
FBE = − FBE k
FBC = FBC i
Substitute and equate to zero the coefficient of j, i, k :
j:
3 !
"
# FAB + 800 N = 0, FAB = −840 N,
$ 3.315 %
i:
0.6 !
"
# (−840 N) + FBC = 0
$ 3.15 %
FBC = 160.0 N T
k:
0.75 !
"−
# (−840 N) − FBE = 0
$ 3.15 %
FBE = 200 N T
FAB = 840 N C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
804
PROBLEM 6.39* (Continued)
Free body C:
ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with
!
F
CA
FCA = FAC
= AC (−1.2i + 3j − 0.75 k )
CA 3.317
FCB = −(160 N) i
FCD = − FCD k FCE = FCE
!
F
CE
= CE (−1.8i − 3k )
CE 3, 499
Substitute and equate to zero the coefficient of j, i, k :
3 !
"
# FAC + 100 N = 0, FAC = −110.57 N
$ 3.317 %
j:
−
i:
k:
−
FAC = 110.6 N C
1.2
1.8
(−110.57) − 160 −
FCE = 0, FCE = −233.3
3.317
3.499
FCE = 233 N C
0.75
3
(−110.57) − FCD −
(−233.3) = 0
3.317
3.499
FCD = 225 N T
Free body: D
ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with
!
F
DA
FDA = FAD
= AD (−1.2i + 3j + 2.25k )
DA 3.937
FDC = FCD k = (225 N)k
FDE = − FDE i
Substitute and equate to zero the coefficient of j, i, k :
j:
i:
k:
3 !
"
# FAD + 300 N = 0,
$ 3.937 %
FAD = −393.7 N,
1.2 !
"−
# ( −393.7 N) − FDE = 0
$ 3.937 %
FAD = 394 N C
FDE = 1200 N T
2.25 !
"
# (−393.7 N) + 225 N = 0 (Checks)
$ 3.937 %
Free body: E
Member AE is the only member at E which does not lie in the xz plane.
Therefore, it is a zero-force member.
FAE = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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805
PROBLEM 6.40*
Solve Problem 6.39 for P = 0 and Q = (−900 N)k.
PROBLEM 6.39* The truss shown consists of nine members
and is supported by a ball and socket at B, a short link at C,
and two short links at D. (a) Check that this truss is a simple
truss, that it is completely constrained, and that the reactions
at its supports are statically determinate. (b) Determine the
force in each member for P = (−1200 N)j and Q = 0.
SOLUTION
Free body: Truss
ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k )
+ (0.6i + 3j − 0.75k ) × (−900N)k = 0
1.8 Ck + 1.8D y k − 1.8Dz j
+ 3D y i + 540 j − 2700i = 0
Equate to zero the coefficient of i, j, k :
3Dy − 2700 = 0 Dy = 900 N
−1.8Dz + 540 = 0 Dz = 300 N
1.8C + 1.8D y = 0 C = − Dy = −900 N
Thus:
C = −(900 N) j D = (900 N) j + (300 N)k
"
ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0
B = (600 N)k "
Free body: B
Since B is aligned with member BE:
FAB = FBC = 0, FBE = 600 N T
Free body: C
!
ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, !with
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806
PROBLEM 6.40* (Continued)
!
FCA
FCD
!
F
CA
= FAC
= AC ( −1.2i + 3j − 0.75k ) !
CA 3.317
!
F
CE
= − FCD k FCE = FCE
= CE (−1.8i − 3k )
CE 3.499
Substitute and equate to zero the coefficient of j, i, k:
j:
3 !
"
# FAC − 900 N = 0, FAC = 995.1 N
$ 3.317 %
FAC = 995 N T
i:
−
1.2
1.8
(995.1) −
FCE = 0, FCE = −699.8 N
3.317
3.499
FCE = 700 N C
k:
−
0.75
3
(995.1) − FCD −
( −699.8) = 0
3.317
3.499
FCD = 375 N T
Free body: D
ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0
!
F
DA
FDA = FAD
= AD (−1.2i + 3j + 2.25k )
DA 3.937
with
FDE = − FDE i
and
Substitute and equate to zero the coefficient j, i, k:
j:
3 !
"
# FAD + 900 N = 0, FAD = −1181.1 N
$ 3.937 %
FAD = 1181N C
i:
1.2 !
−"
# (−1181.1 N) − FDE = 0
$ 3.937 %
FDE = 360 N T
k:
2.25 !
"
# (−1181.1 N + 375 N + 300 N = 0)
$ 3.937 %
(Checks)
Free body: E
Member AE is the only member at E which
does not lie in the XZ plane. Therefore, it is
a zero-force member.
FAE = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
807
PROBLEM 6.41*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in
each of the six members joined at E.
SOLUTION
(a)
Check simple truss.
(1) start with tetrahedron BEFG
(2) Add members BD, ED, GD joining at D.
(3) Add members BA, DA, EA joining at A.
(4) Add members DH, EH, GH joining at H.
(5) Add members BC, DC, GC joining at C.
Truss has been completed: It is a simple truss
Free body: Truss
Check constraints and reactions:
Six unknown reactions-ok-however supports at A and B
constrain truss to rotate about AB and support at G
prevents such a rotation. Thus,
Truss is completely constrained and reactions are statically determinate
Determination of Reactions:
ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j
+ (10.08 j − 9.6k ) × (275i + 240k ) = 0
11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k
+ (10.08)(240)i − (9.6)(275) j = 0
Equate to zero the coefficient of i, j, k:
i : 9.6G + (10.08)(240) = 0 G = −252 lb
G = (−252 lb) j "!
j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb
k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb
B = (504 lb) j − (240 lb)k "
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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808
PROBLEM 6.41* (Continued)
ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j
+ (275 lb)i + (240 lb)k = 0
A = −(275 lb)i − (252 lb) j "
Zero-force members
The determination of these members will facilitate our solution.
FB: C. Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
Yields FBC = FCD = FCG = 0 "!
FB: F. Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
Yields FBF = FEF = FFG = 0 "
FB: A: Since
FB: H: Writing
Az = 0, writing ΣFz = 0
ΣFy = 0
Yields FAD = 0 "
Yields FDH = 0 "
FB: D: Since FAD = FCD = FDH = 0, we need
consider only members DB, DE, and DG.
Since FDE is the only force not contained in plane BDG, it must be
zero. Simple reasonings show that the other two forces are also zero.
FBD = FDE = FDG = 0 "
The results obtained for the reactions at the supports and for the zero-force members are shown on the
figure below. Zero-force members are indicated by a zero (“0”).
(b)
Force in each of the members joined at E
FDE = FEF = 0
We already found that
Free body: A
ΣFy = 0
Yields FAE = 252 lb T
Free body: H
ΣFz = 0
Yields FEH = 240 lb C
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809
PROBLEM 6.41* (Continued)
ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0
Free body: E
F
FBE
(11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0
14.92
14.6
Equate to zero the coefficient of y and k:
10.08 !
j: − "
# FBE − 252 = 0
14.92
$
%
FBE = 373 lb C
9.6 !
−"
# FEG + 240 = 0
14.6
$
%
FEG = 365 lb T
k:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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810
PROBLEM 6.42*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at G.
SOLUTION
See solution to Problem 6.41 for Part (a) and for reactions and zero-force members.
(b)
Force in each of the members joined at G.
We already know that
FCG = FDG = FFG = 0
Free body: H
ΣFx = 0
Free body: G
ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0
!
Yields FGH = 275 lb C
FBG
F
(−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0
13.92
14.6
Equate to zero the coefficient of i, j, k:
11 !
i: − "
# FEG + 275 = 0
14.6
$
%
FEG = 365 lb T
10.08 !
j: − "
# FBG − 252 = 0
$ 13.92 %
FBG = 348 lb C
9.6 !
9.6 !
k: "
# (−348) + "
# (365) = 0
$ 13.92 %
$ 14.6 %
(Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
811
!
PROBLEM 6.43
A Warren bridge truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Truss
ΣFx = 0: k x = 0
ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft)
− (6000 lb)(25 ft) = 0
k = k y = 3600 lb "!
ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0
A = 8400 lb "
We pass a section through members CE, DE, and DF and use the
free body shown.
ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft)
+ (6000 lb)(6.25 ft) = 0
FCE = +8000 lb
ΣFy = 0: 8400 lb − 6000 lb −
FCE = 8000 lb T
15
FDE = 0
16.25
FDE = +2600 lb
FDE = 2600 lb T
ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft)
− FDF (15 ft) = 0
FDF = −9000 lb
FDF = 9000 lb C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
812
!
PROBLEM 6.44
A Warren bridge truss is loaded as shown. Determine the force in
members EG, FG, and FH.
SOLUTION
See solution of Problem 6.43 for free-body diagram of truss and determination of reactions:
A = 8400 lb
k = 3600 lb "
We pass a section through members EG, FG, and FH, and use the free body shown.
ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0
FEG = +7500 lb
+ΣFy = 0:
FEG = 7500 lb T
!
15
FFG + 3600 lb = 0
16.25
FFG = −3900 lb
FFG = 3900 lb C
ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0
FFH = −6000 lb
FFH = 6000 lb C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
813
PROBLEM 6.45
Determine the force in members BD and DE of the truss shown.
SOLUTION
Member BD:
ΣM E = 0: FBD (4.5 m) − (135 kN)(4.8 m) − (135 kN)(2.4 m) = 0
FBD = +216 kN
FBD = 216 kN T
Member DE:
ΣFx = 0: 135 kN + 135 kN − FDE = 0
FDE = +270 kN
FDE = 270 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
814
PROBLEM 6.46
Determine the force in members DG and EG of the truss shown.
SOLUTION
Member DG:
ΣFx = 0: 135 kN + 135 kN + 135 kN +
FDG = −459 kN
15
FDG = 0
17
FDG = 459 kN C
Member EG:
ΣM D = 0: (135 kN)(4.8 m) + (135 kN)(2.4 m)
+ FEG (4.5 m) = 0
FEG = −216 kN
FEG = 216 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
815
PROBLEM 6.47
A floor truss is loaded as shown. Determine the force in
members CF, EF, and EG.
SOLUTION
Free body: Truss
ΣFx = 0: k x = 0
ΣM A = 0: k y (4.8 m) − (4 kN)(0.8 m)
− (4 kN)(1.6 m) − (3 kN)(2.4 m)
− (2 kN)(3.2 m) − (2 kN)(4 m) − (1 kN)(4.8 m) = 0
k y = 7.5 kN
k = 7.5 kN "
Thus:
ΣFy = 0: A + 7.5 kN − 18 kN = 0
A = 10.5 kN
A = 10.5 kN "
We pass a section through members CF, EF, and EG and use the free body shown.
ΣM E = 0: FCF (0.4 m) − (10.5 kN)(1.6 m) + (2 kN)(1.6 m)
+ (4 kN)(0.8 m) = 0
FCF = +26.0 kN
ΣFy = 0: 10.5 kN − 2 kN − 4 kN − 4 kN −
FCF = 26.0 kN T
1
5
FEF = +1.118 kN
FEF = 0
FEF = 1.118 kN T
ΣM F = 0: (2 kN)(2.4 m) + (4 kN)(1.6 m) + (4 kN)(0.8 m)
− (10.5 kN)(2.4 m) − FEG (0.4 m) = 0
FEG = −27.0 kN
FEG = 27.0 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
816
PROBLEM 6.48
A floor truss is loaded as shown. Determine the force in
members FI, HI, and HJ.
SOLUTION
See solution of Problem 6.47 for free-body diagram of truss and determination of reactions:
A = 10.5 kN , k = 7.5 kN "
We pass a section through members FI, HI, and HJ, and use the free body shown.
ΣM H = 0: (7.5 kN)(1.6 m) − (2 kN)(0.8 m) − (1 kN)(1.6 m)
− FFI (0.4 m) = 0
FFI = +22.0 kN
ΣFy = 0:
1
5
FFI = 22.0 kN T
FHI − 2 kN − 1 kN + 7.5 kN = 0
FHI = −10.06 kN
FHI = 10.06 kN C
ΣM I = 0: FHJ (0.4 m) + (7.5 kN)(0.8 m) − (1 kN)(0.8 m) = 0
FHJ = −13.00 kN
FHJ = 13.00 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
817
PROBLEM 6.49
A pitched flat roof truss is loaded as shown. Determine the force
in members CE, DE, and DF.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
Ax = 0
Ay = I =
1
1
(Total load) = (8 kN)
2
2
A = I = 4 kN "
We pass a section through members CD, DE, and DF, and use the free body shown.
(We moved FDE to E and FDF to F)
Slope
BJ =
Slope
DE =
2.16 m 9
=
9.6 m 40
−1 m −5
=
2.4 m 12
0.46 m
0.46 m
a=
=
= 2.0444 m
9
Slope BJ
40
ΣM D = 0: FCE (1 m) + (1 kN)(2.4 m) − (4 kN)(2.4 m) = 0
FCE = 7.20 kN T
FCE = +7.20 kN
ΣM K = 0: (4 kN)(2.0444 m) − (1 kN)(2.0444 m)
5
!
− (2 kN)(4.4444 m) − " FDE # (6.8444 m) = 0
$ 13
%
FDE = −1.047 kN
FDE = 1.047 kN C
ΣM E = 0: (1 kN)(4.8 m) + (2kN)(2.4 m) − (4 kN)(4.8 m)
40
!
− " FDF # (1.54 m) = 0
41
$
%
FDF = −6.39 kN
FDF = 6.39 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
818
PROBLEM 6.50
A pitched flat roof truss is loaded as shown. Determine the force
in members EG, GH, and HJ.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
Ax = 0
Ay = I =
1
1
(Total load) = (8 kN)
2
2
A = I = 4 kN "
We pass a section through members EG, GH, and HJ, and use the free body shown.
ΣM H = 0: (4 kN)(2.4 m) − (1 kN)(2.4 m) − FEG (2.08 m) = 0
FEG = +3.4615 kN
FEG = 3.46 kN T
ΣM J = 0: − FGH (2.4 m) − FEG (2.62 m) = 0
FGH = −
2.62
(3.4615 kN)
2.4
FGH = −3.7788 kN
ΣFx = 0: − FEG −
FGH = 3.78 kN C
2.4
FHJ = 0
2.46
FHJ = −
2.46
2.46
(3.4615 kN)
FEG = −
2.4
2.4
FHJ = −3.548 kN
FHJ = 3.55 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
819
PROBLEM 6.51
A Howe scissors roof truss is loaded as shown. Determine
the force in members DF, DG, and EG.
SOLUTION
Reactions at supports.
Because of symmetry of loading.
Ax = 0,
Ay = L =
1
1
(Total load) = (9.60 kips) = 4.80 kips
2
2
A = L = 4.80 kips
We pass a section through members DF, DG, and EG, and use the free body shown.
We slide FDF to apply it at F:
ΣM G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft)
8FDF
− (4.8 kips)(24 ft) −
82 + 3.52
(6 ft) = 0
FDF = −10.48 kips, FDF = 10.48 kips C
ΣM A = 0: − (1.6 kips)(8 ft) − (1.6 kips)(16 ft)
−
2.5 FDG
2
8 + 2.5
2
(16 ft) −
8 FDG
82 + 2.52
(7 ft) = 0
FDG = −3.35 kips, FDG = 3.35 kips C
ΣM D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) − (4.8 kips)(16 ft) −
8 FEG
82 + 1.52
(4 ft) = 0
FEG = +13.02 kips, FEG = 13.02 kips T
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
820
!
PROBLEM 6.52
A Howe scissors roof truss is loaded as shown. Determine
the force in members GI, HI, and HJ.
SOLUTION
Reactions at supports. Because of symmetry of loading:
1
(Total load)
2
1
= (9.60 kips)
2
Ay = L =
Ax = 0,
= 4.80 kips
A = L = 4.80 kips
We pass a section through members GI, HI, and HJ, and use the free body shown.
ΣM H = 0: −
16 FGI
162 + 32
(4 ft) + (4.8 kips)(16 ft) − (0.8 kip)(16 ft) − (1.6 kips)(8 ft) = 0
FGI = +13.02 kips
FGI = 13.02 kips T
ΣM L = 0: (1.6 kips)(8 ft) − FHI (16 ft) = 0
FHI = +0.800 kips
FHI = 0.800 kips T
We slide FHG to apply it at H.
ΣM I = 0:
8FHJ
82 + 3.52
(4 ft) + (4.8 kips)(16 ft) − (1.6 kips)(8 ft) − (0.8 kip)(16 ft) = 0
FHJ = −13.97 kips
FHJ = 13.97 kips C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
821
!
PROBLEM 6.53
A Pratt roof truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Entire truss
ΣFx = 0: Ax = 0
Total load = 5(3 kN) + 2(1.5 kN)
= 18 kN
By symmetry:
Ay = L =
1
(18 kN)
2
A = L = 9 kN
Free body: Portion ACD
Note:
Slope of ABDF is
6.75 3
=
9.00 4
Force in CE:
2
!
ΣM D = 0: FCE " × 6.75 m # − (9 kN)(6 m) + (1.5 kN)(6 m) + (3 kN)(3 m) = 0
$3
%
FCE (4.5 m) − 36 kN ⋅ m = 0
FCE = +8 kN
FCE = 8 kN T
Force in DE:
ΣM A = 0: FDE (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0
FDE = −4.5 kN
FDE = 4.5 kN C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
822
PROBLEM 6.53 (Continued)
Force in DF:
Sum moments about E where FCE and FDE intersect.
ΣM E = 0: (1.5 kN)(6 m) − (9 kN)(6 m) + (3 kN)(3 m) +
4
2
!
FCE " × 6.75 m # = 0
5
$3
%
4
FCE (4.5 m) − 36 kN ⋅ m
5
FCE = 10.00 kN C
FCE = −10.00 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
823
!
PROBLEM 6.54
A Pratt roof truss is loaded as shown. Determine the force in
members FH, FI, and GI.
SOLUTION
Free body: Entire truss
ΣFx = 0: Ax = 0
Total load = 5(3 kN) + 2(1.5 kN) = 18 kN
By symmetry:
Ay = L =
1
(18) = 9 kN
2
Free body: Portion HIL
Slope of FHJL
6.75 3
=
9.00 4
tan α =
FG 6.75 m
=
α = 66.04°
GI
3m
Force in FH:
ΣM I = 0:
4
2
!
FFH " × 6.75 m # + (9 kN)(6 m) − (1.5 kN)(6 m) − (3 kN)(3 m) = 0
5
$3
%
4
FFH (4.5 m) + 36 kN ⋅ m
5
FFH = −10.00 kN
FFH = 10.00 kN C
Force in FI:
ΣM L = 0: FFI sin α (6 m) − (3 kN)(6 m) − (3 kN)(3 m) = 0
FFI sin 66.04°(6 m) = 27 kN ⋅ m
FFI = +4.92 kN
FFI = 4.92 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
824
!
PROBLEM 6.54 (Continued)
Force in GI:
ΣM H = 0: FGI (6.75 m) + (3 kN)(3 m) + (3 kN)(6 m) + (1.5 kN)(9 m) − (9 kN)(9 m) = 0
FGI (6.75 m) = +40.5 kN ⋅ m
FGI = 6.00 kN T
FGI = +6.00 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
825
!
PROBLEM 6.55
Determine the force in members AD, CD, and CE of the
truss shown.
SOLUTION
Reactions:
ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0
ΣFx = 0: −36 + K x = 0
B = 26.4 kN
K x = 36 kN
ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN
ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0
FAD = −13.5 kN
8
!
ΣM A = 0: " FCD # (4.5) = 0
17
$
%
FAD = 13.5 kN C
FCD = 0
15
!
ΣM D = 0: " FCE # (2.4) − 26.4(4.5) = 0
17
$
%
FCE = +56.1 kN
FCE = 56.1 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
826
!
PROBLEM 6.56
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
See the solution to Problem 6.55 for free-body diagram and analysis to determine the reactions at the supports
B and K.
B = 26.4 kN ;
K x = 36.0 kN
;
K y = 13.60 kN
ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0
FDG = −75 kN
FDG = 75.0 kN C
8
!
ΣM D = 0: − 26.4(4.5) + " FFG # (4.5) = 0
$ 17
%
FFG = +56.1 kN
!
15
ΣM G = 0: 20(4.5) − 26.4(9) + " FFH
17
$
!
# (2.4) = 0 !
%
FFH = +69.7 kN !
!
FFG = 56.1 kN T
FFH = 69.7 kN T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
827
!
PROBLEM 6.57
A stadium roof truss is loaded as shown. Determine the force
in members AB, AG, and FG.
SOLUTION
We pass a section through members AB, AG, and FG, and use the free body shown.
40
!
ΣM G = 0: " FAB # (6.3 ft) − (1.8 kips)(14 ft) − (0.9 kips)(28 ft) = 0
$ 41
%
FAB = +8.20 kips
FAB = 8.20 kips T
3
!
ΣM D = 0: − " FAG # (28 ft) + (1.8 kips)(28 ft) + (1.8 kips)(14 ft) = 0
$5
%
FAG = +4.50 kips
FAG = 4.50 kips T
M A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
FFG = −11.60 kips
FFG = 11.60 kips C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
828
!
PROBLEM 6.58
A stadium roof truss is loaded as shown. Determine the force
in members AE, EF, and FJ.
SOLUTION
We pass a section through members AE, EF, and FJ, and use the free body shown.
!
8
ΣM F = 0: "
FAE # (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
" 82 + 9 2
#
$
%
FAE = +17.46 kips
FAE = 17.46 kips T
ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
FEF = −11.60 kips
FEF = 11.60 kips C
ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0
FFJ = −18.45 kips
FFJ = 18.45 kips C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
829
!
PROBLEM 6.59
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members DF, EF,
and EG.
SOLUTION
Free body: Truss
ΣFx = 0: N x = 0
ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a)+(350 lb)(4a ) + (300 lb)(3a + 2a + a) − A(8a ) = 0
A = 1500 lb "
ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0
N y = 1300 lb N = 1300 lb "
We pass a section through DF, EF, and EG, and use the free body shown.
(We apply FDF at F)
ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft)
!
18
F # (4.5 ft) = 0
−"
" 182 + 4.52 DF #
$
%
FDF = −3711 lb FDF = 3710 lb C
ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0
FEF = +400 lb
FEF = 400 lb T
ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0
FEG = +3600 lb FEG = 3600 lb T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
830
!
PROBLEM 6.60
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members HI, GI,
and GJ.
SOLUTION
See solution of Problem 6.59 for reactions:
A = 1500 lb ,
N = 1300 lb "
We pass a section through HI, GI, and GJ, and use the free body shown.
(We apply FHI at H.)
!
6
ΣM G = 0: "
FHI # (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft)
" 2
#
2
$ 6 +4
%
− (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0
FHI = −2375.4 lb FHI = 2375 lb C
ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft)
− (150 lb)(18 ft) − FGJ (4.5 ft) = 0
FGJ = +3400 lb FGJ = 3400 lb T
ΣFx = 0: −
4
6
(−2375.4 lb) − 3400 lb = 0
FGI −
2
5
6 + 42
FGI = −1779.4 lb
FGI = 1779 lb C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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831
!
PROBLEM 6.61
Determine the force in members AF and EJ of the truss shown
when P = Q = 1.2 kN. (Hint: Use section aa.)
SOLUTION
Free body: Entire truss
ΣM A = 0: K (8 m) − (1.2 kN)(6 m) − (1.2 kN)(12 m) = 0
K = +2.70 kN
K = 2.70 kN
"
Free body: Lower portion
ΣM F = 0: FEJ (12 m) + (2.70 kN)(4 m) − (1.2 kN)(6 m) − (1.2 kN)(12 m) = 0
FEJ = +0.900 kN
FEJ = 0.900 kN T
ΣFy = 0: FAF + 0.9 kN − 1.2 kN − 1.2 kN = 0
FAF = +1.500 kN
FAF = 1.500 kN T
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832
PROBLEM 6.62
Determine the force in members AF and EJ of the truss shown
when P = 1.2 kN and Q = 0. (Hint: Use section aa.)
SOLUTION
Free body: Entire truss
ΣM A = 0: K (8 m) − (1.2 kN)(6 m) = 0
K = +0.900 kN
K = 0.900 kN
"
Free body: Lower portion
ΣM F = 0: FEJ (12 m) + (0.900 kN)(4 m) − (1.2 kN)(6 m) = 0
FEJ = +0.300 kN
FEJ = 0.300 kN T
ΣFy = 0: FAF + 0.300 kN − 1.2 kN = 0
FAF = +0.900 kN
FAF = 1.900 kN T
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833
PROBLEM 6.63
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
SOLUTION
Reactions:
ΣFx = 0: Ax = 0
ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0
A y = 12 kips
ΣFy = 0: 12 − 12 − 12 − 12 + P = 0
P = 24 kips
ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0
FEH = −22.5 kips
ΣFx = 0: FGI − 22.5 kips = 0
FEH = 22.5 kips C
FGI = 22.5 kips T
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834
PROBLEM 6.64
Determine the force in members HJ and IL of the truss
shown. (Hint: Use section bb.)
SOLUTION
See the solution to Problem 6.63 for free body diagram and analysis to determine the reactions at supports A
and P.
A x = 0; A y = 12.00 kips ;
P = 24.0 kips
ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0
FHJ = −33.75 kips
FHJ = 33.8 kips C
ΣFx = 0 : 33.75 kips − FIL = 0
FIL = +33.75 kips
FIL = 33.8 kips T
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835
PROBLEM 6.65
The diagonal members in the center panels of
the power transmission line tower shown are
very slender and can act only in tension; such
members are known as counters. For the given
loading, determine (a) which of the two
counters listed below is acting, (b) the force
in that counter.
Counters CJ and HE.
SOLUTION
Free body: Portion ABDFEC of tower
We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ.
ΣFx = 0:
4
FCJ − 2(1.2 kN) sin 20° = 0 FCJ = +1.026 kN
5
Since CJ is found to be in tension, our assumption was correct. Thus, the answers are
(a) CJ
(b) 1.026 kN T
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836
PROBLEM 6.66
The diagonal members in the center panels of
the power transmission line tower shown are
very slender and can act only in tension; such
members are known as counters. For the given
loading, determine (a) which of the two
counters listed below is acting, (b) the force
in that counter.
Counters IO and KN.
SOLUTION
Free body: Portion of tower shown
We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO.
ΣFx = 0 :
4
FIO − 4(1.2 kN)sin 20° = 0 FIO = +2.05 kN
5
Since IO is found to be in tension, our assumption was correct. Thus, the answers are
(a) IO
(b) 2.05 kN T
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837
PROBLEM 6.67
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
are acting under the given loading.
SOLUTION
Free body: Truss
ΣFx = 0: Fx = 0
ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0
Fy = +13.20 kips
F = 13.20 kips "
ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0
H = +6.00 kips
H = 6.00 kips "
Free body: ABF
We assume that counter BG is acting.
ΣFy = 0: −
9.6
FBG + 13.20 − 2(4.8) = 0
14.6
FBG = +5.475
FBG = 5.48 kips T
Since BG is in tension, our assumption was correct
Free body: DEH
We assume that counter DG is acting.
+ΣFy = 0: −
9.6
FDG + 6.00 − 2(2.4) = 0
14.6
FDG = 1.825 kips T
FDG = +1.825
Since DG is in tension, O.K.
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838
PROBLEM 6.68
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
are acting under the given loading.
SOLUTION
Free body: Truss
ΣFx = 0: Fx = 0
ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a ) = 0
Fy = 7.20
F = 7.20 kips "
Free body: ABF
We assume that counter CF is acting.
ΣFy = 0 :
9.6
FCF + 7.20 − 2(4.8) = 0
14.6
FCF = 3.65 kips T
FCF = +3.65
Since CF is in tension, O.K.
Free body: DEH
We assume that counter CH is acting.
+ΣFy = 0:
9.6
FCH − 2(2.4 kips) = 0
14.6
FCH = +7.30
FCH = 7.30 kips T
Since CH is in tension, O.K.
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839
PROBLEM 6.69
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m = 16
Number of joints:
n = 10
Reaction components:
r=4
(a)
m + r = 20, 2n = 20
m + r = 2n "
Thus:
To determine whether the structure is actually completely constrained and determinate, we must try to find the
reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of
each truss.
This is an improperly supported simple
truss. (Reaction at C passes through B. Thus,
Eq. ΣM B = 0 cannot be satisfied.)
This is a properly supported simple truss – O.K.
Structure is improperly constrained
!
Structure (b)
m = 16
n = 10
r=4
m + r = 20, 2n = 20
(b)
m + r = 2n "
Thus:
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840
PROBLEM 6.69 (Continued)
We must again try to find the reactions at the supports dividing the structure as shown.
Both portions are simply supported simple trusses.
Structure is completely constrained and determinate
Structure (c)
m = 17
n = 10
r=4
m + r = 21, 2n = 20
(c)
m + r > 2n "
Thus:
This is a simple truss with an extra support which causes reactions (and forces in members) to be
indeterminate.
Structure is completely constrained and indeterminate
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841
PROBLEM 6.70
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Non-simple truss with r = 4, m = 16, n = 10
so m + r = 20 = 2n, but must examine further.
FBD Sections:
FBD
I:
ΣM A = 0 &
T1
II:
ΣFx = 0 &
T2
I:
ΣFx = 0
&
Ax
I:
ΣFy = 0
&
Ay
II:
ΣM E = 0 &
Cy
II:
ΣFy = 0 &
Ey
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate.
Non-simple truss with r = 3, m = 16, n = 10
Structure (b):
so
m + r = 19 < 2n = 20
structure is partially constrained
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842
PROBLEM 6.70 (Continued)
Structure (c):
Simple truss with r = 3, m = 17, n = 10 m + r = 20 = 2n, but the horizontal reaction forces Ax and E x
are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained
and indeterminate
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843
PROBLEM 6.71
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Non-simple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n,
check for determinacy:
One can solve joint F for forces in EF, FG and then solve joint
E for E y and force in DE.
This leaves a simple truss ABCDGH with
r = 3, m = 9, n = 6 so r + m = 12 = 2n
Structure is completely constrained and determinate
Structure (b):
Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8
so
r + m = 17 > 2n = 16
Constrained but indeterminate
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844
PROBLEM 6.71 (Continued)
Structure (c):
Non-simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in
Part (a) above to get truss at left.
Since F1 ≠ 0 (from solution of joint F),
ΣM A = aF1
≠ 0 and there is no equilibrium.
Structure is improperly constrained
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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845
PROBLEM 6.72
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m = 12
Number of joints:
n=8
Reaction components:
r =3
m + r = 15,
Thus:
2n = 16
m + r , 2n
"
Structure is partially constrained
Structure (b)
m = 13, n = 8
r =3
m + r = 16, 2n = 16
Thus:
m + r = 2n
"
To verify that the structure is actually completely constrained and determinate, we observe that it is a simple
truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus:
Structure is completely constrained and determinate
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846
PROBLEM 6.72 (Continued)
Structure (c)
m = 13,
r=4
n=8
m + r = 17, 2n = 16
Thus:
m + r . 2n
"
Structure is completely constrained and indeterminate
This result can be verified by observing that the structure is a simple truss (follow lettering to check thus),
therefore rigid, and that its supports involve 4 unknowns.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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847
PROBLEM 6.73
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Rigid truss with r = 3, m = 14, n = 8
so
r + m = 17 . 2n = 16
so completely constrained but indeterminate
Structure (b):
Simple truss (start with ABC and add joints alphabetically), with
r = 3, m = 13, n = 8 so r + m = 16 = 2n
so completely constrained and determinate
Structure (c):
Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear
so cannot be resolved by any equilibrium equation.
structure is improperly constrained
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848
PROBLEM 6.74
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
No. of members
m = 12
No. of joints
n = 8 m + r = 16 = 2n
No. of react. comps.
r = 4 unks = eqns
FBD of EH:
ΣM H = 0
FDE ; ΣFx = 0
FGH ; ΣFy = 0
Hy
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate.
Structure (b):
No. of members
m = 12
No. of joints
n = 8 m + r = 15 , 2n = 16
No. of react. comps.
r = 3 unks , eqns
partially constrained
Note: Quadrilateral DEHG can collapse with joint D moving downward: in (a) the roller at F prevents this
action.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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849
PROBLEM 6.74 (Continued)
Structure (c):
No. of members
m = 13
No. of joints
n = 8 m + r = 17 . 2n = 16
No. of react. comps.
r = 4 unks . eqns
completely constrained but indeterminate
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850
PROBLEM 6.75
For the frame and loading shown, determine the force
acting on member ABC (a) at B, (b) at C.
SOLUTION
FBD ABC:
Note: BD is two-force member
(a)
(b)
3
!
ΣM C = 0: (0.09 m)(200 N) − (2.4 m) " FBD # = 0
5
$
%
4
ΣFx = 0: 200 N − (125 N) − C x = 0
5
ΣFy = 0:
3
FBD − C y = 0
5
FBD = 125.0 N
36.9°
C = 125.0 N
36.9°
C x = 100 N
3
C y = (125 N) = 75 N
5
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851
PROBLEM 6.76
Determine the force in member BD and the components of the
reaction at C.
SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along the BD.
Free body: ABC
Attaching FBD at D and resolving it into components, we write
ΣM C = 0:
450
!
FBD # (240 mm) = 0
(400 N)(135 mm) + "
510
$
%
FBD = −255 N
ΣFx = 0: C x +
FBD = 255 N C
240
(−255 N) = 0
510
C x = +120.0 N
ΣFy = 0: C y − 400 N +
C x = 120.0 N
450
(−255 N) = 0
510
C y = +625 N
C y = 625 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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852
PROBLEM 6.77
Rod CD is fitted with a collar at D that can be moved along
rod AB, which is bent in the shape of an arc of circle. For the
position when θ = 30°, determine (a) the force in rod CD,
(b) the reaction at B.
SOLUTION
FBD:
(a)
ΣM C = 0: (15 in.)(20 lb − By ) = 0
ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0
(b)
B y = 20 lb
FCD = 80.0 lb T
ΣFx = 0: (80 lb) cos 30° − Bx = 0
B x = 69.282 lb
so B = 72.1 lb
16.10°
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853
PROBLEM 6.78
Solve Problem 6.77 when θ = 150°.
PROBLEM 6.77 Rod CD is fitted with a collar at D that can be moved
along rod AB, which is bent in the shape of an arc of circle. For the
position when θ = 30°, determine (a) the force in rod CD, (b) the
reaction at B.
SOLUTION
Note that CD is a two-force member, FCD must be directed along DC.
(a)
ΣM B = 0: (20 lb)(2 R ) − ( FCD sin 30°) R = 0
FCD = 80 lb
(b)
FCD = 80.0 lb T
ΣM C = 0: (20 lb) R + ( By ) R = 0
By = −20 lb
B y = 20.0 lb
ΣFx = 0: − FCD cos 30° + Bx = 0
−(20 lb) cos 30° + Bx = 0
Bx = 69.28 lb
B x = 69.28 lb
B = 72.1 lb
16.10°
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854
PROBLEM 6.79
Determine the components of all forces acting on member
ABCD when θ = 0.
SOLUTION
Free body: Entire assembly
ΣM B = 0: A(8 in.) − (60 lb)(20 in.) = 0
A = 150 lb
A = 150.0 lb
ΣFx = 0: Bx + 150.0 lb = 0 Bx = −150 lb
B x = 150.0 lb
ΣFy = 0: By − 60.0 lb = 0
Free body: Member ABCD
By = +60.0 lb
B y = 60.0 lb
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (12) − (60 lb)(4) + (150 lb)(8) = 0
D = −80 lb
ΣFx = 0: C x + 150.0 − 150.0 = 0
Cx = 0
ΣFy = 0: C y + 60.0 − 80.0 = 0
C y = +20.0 lb
D = 80.0 lb
C = 20.0 lb
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855
PROBLEM 6.80
Determine the components of all forces acting on member
ABCD when θ = 90°.
SOLUTION
Free body: Entire assembly
ΣM B = 0: A(8 in.) − (60 lb)(8 in.) = 0
A = +60.0 lb
A = 60.0 lb
ΣFx = 0: Bx + 60 lb − 60 lb = 0 Bx = 0
ΣFy = 0: By = 0
Free body: Member ABCD
B=0
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D(12 in.) + (60 lb)(8 in.) = 0
D = −40.0 lb
D = 40.0 lb
ΣFx = 0: C x + 60 lb = 0
C x = −60 lb
C x = 60.0 lb
ΣFy = 0: C y − 40 lb = 0
C y = +40 lb
C y = 40.0 lb
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856
PROBLEM 6.81
For the frame and loading shown, determine the components of all forces acting
on member ABC.
SOLUTION
Free body: Entire frame
ΣFx = 0: Ax + 18 kN = 0
Ax = −18 kN
A x = 18.00 kN
ΣM E = 0: −(18 kN)(4 m) − Ay (3.6 m) = 0
Ay = −20 kN
A y = 20.0 kN
ΣFy = 0: − 20 kN + F = 0
F = +20 kN F = 20 kN
Free body: Member ABC
Note: BE is a two-force member. Thus B is directed along line BE.
ΣM C = 0: B(4 m) − (18 kN)(6 m) + (20 kN)(3.6 m) = 0
B = 9 kN
B = 9.00 kN
ΣFx = 0: C x − 18 kN + 9 kN = 0
C x = 9 kN
C x = 9.00 kN
ΣFy = 0: C y − 20 kN = 0
C y = 20 kN
C y = 20.0 kN
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857
PROBLEM 6.82
Solve Problem 6.81 assuming that the 18-kN load is replaced by a
clockwise couple of magnitude 72 kN · m applied to member CDEF at
Point D.
PROBLEM 6.81 For the frame and loading shown, determine the
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame
ΣFx = 0: Ax = 0
ΣM F = 0: − 72 kN ⋅ m − Ay (3.6 m) = 0
Ay = −20 kN A y = 20 kN
A = 20.0 kN
Free body: Member ABC
Note: BE is a two-force member. Thus B is directed along line BE.
ΣM C = 0: B(4 m) + (20 kN)(3.6 m) = 0
B = −18 kN
B = 18.00 kN
Fx = 0: − 18 kN + C x = 0
C x = 18 kN
C x = 18.00 kN
ΣFy = 0: C y − 20 kN = 0
C y = 20 kN
C y = 20.0 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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858
PROBLEM 6.83
Determine the components of the reactions at A and E if a 750-N
force directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free-body: Entire Frame
The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial.
ΣM E = 0: −(750 N)(240 mm) − Ax (400 mm) = 0
Ax = −450 N A x = 450 N
ΣFx = 0: Ex − 450 N = 0 Ex = 450 N E x = 450 N
ΣFy = 0: Ay + E y − 750 N = 0
(a)
(1)
Load applied at B.
Free body: Member CE
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
=
450 N
From Eq. (1):
240 mm
; E y = 225 N
480 mm
Ay + 225 − 750 = 0; A y = 525 lb
Thus, reactions are:
(b)
A x = 450 N
,
A y = 525 lb
E x = 450N
,
E y = 225 lb
A x = 450 N
,
A y = 150.0 N
E x = 450 N
,
E y = 600 N
Load applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
450 N
From Eq. (1):
=
160 mm
480 mm
A y = 150.0 N
Ay + E y − 750 N = 0
150 N + E y − 750 N = 0
E y = 600 N E y = 600 N
Thus, reactions are:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
859
!
PROBLEM 6.84
Determine the components of the reactions at A and E if a 750-N
force directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame
The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial.
ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0
Ax = − 300 N A x = 300 N
ΣFx = 0: Ex − 300 N = 0 Ex = 300 N E x = 300 N
ΣFy = 0: Ay + E y − 750 N = 0
(a)
(1)
Load applied at B.
Free body: Member CE
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
=
300 N
From Eq. (1):
75 mm
250 mm
E y = 90 N
Ay + 90 N − 750 N = 0
Ay = 660 N
Thus reactions are:
(b)
A x = 300 N
,
A y = 660 N
E x = 300 N
,
E y = 90.0 N
Load applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
300 N
=
125 mm
250 mm
Ay = 150 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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860
PROBLEM 6.84 (Continued)
Ay + E y − 750 N = 0
From Eq. (1):
150 N + E y − 750 N = 0
E y = 600 N E y = 600 N
Thus, reactions are:
A x = 300 N
,
A y = 150.0 N
E x = 300 N
,
E y = 600 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
861
!
PROBLEM 6.85
Determine the components of the reactions at A and E if the frame is
loaded by a clockwise couple of magnitude 36 N · m applied (a) at B,
(b) at D.
SOLUTION
Free body: Entire frame
The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial.
ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0
Ax = − 90 N A x = 90.0 N
ΣFx = 0: −90 +Ex = 0
Ex = 90 N E x = 90.0 N
ΣFy = 0: Ay + E y = 0
(a)
(1)
Couple applied at B.
Free body: Member CE
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
=
90 N
From Eq. (1):
0.24 m
; E y = 45.0 N
0.48 m
Ay + 45 N = 0
Ay = − 45 N A y = 45.0 N
Thus, reactions are
(b)
A x = 90.0 N
,
A y = 45.0 N
E x = 90.0 N
,
E y = 45.0 N
Couple applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
90 N
=
0.16 m
; A y = 30 N
0.48 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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862
PROBLEM 6.85 (Continued)
From Eq. (1):
Ay + E y = 0
30 N + E y = 0
E y = − 30 N E y = 30 N
A x = 90.0 N
Thus, reactions are:
E x = − 90.0 N
A y = 30.0 N
,
,
E y = 30.0 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
863
!
PROBLEM 6.86
Determine the components of the reactions at A and E if the frame is
loaded by a clockwise couple of magnitude 36 N · m applied (a) at B,
(b) at D.
SOLUTION
Free body: Entire frame
The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial.
ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0
Ax = −180 N A x = 180 N
ΣFx = 0: −180 N + Ex = 0
Ex = 180 N E x = 180 N
ΣFy = 0: Ay + E y = 0
(a)
(1)
Couple applied at B.
Free body: Member CE
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
=
180 N
From Eq. (1):
0.075 m
0.25 m
E y = 54 N
Ay + 54 N = 0
Ay = − 54 N A y = 54.0 N
Thus reactions are
(b)
A x = 180.0 N
,
A y = 54.0 N
E x = 180.0 N
,
E y = 54.0 N
Couple applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be
directed along EC.
Ay
180 N
=
0.125 m
0.25 m
Ay = 90 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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864
PROBLEM 6.86 (Continued)
From Eq. (1):
Ay + E y = 0
90 N + E y = 0
E y = − 90 N E y = 90 N
Thus, reactions are
A x = 180.0 N
E x = −180.0 N
A y = 90.0 N
,
,
E y = 90.0 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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865
!
PROBLEM 6.87
Determine the components of the reactions at A and B, (a) if the
500-N load is applied as shown, (b) if the 500-N load is moved
along its line of action and is applied at Point F.
SOLUTION
Free body: Entire frame
Analysis is valid for either (a) or (b), since position of 100-lb load on its line of action is immaterial.
ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb
ΣFy = 0: Ay + 60 − 100 = 0
Ay = + 40 lb
ΣFx = 0: Ax + Bx = 0
(a)
(1)
Load applied at E.
Free body: Member AC
Since AC is a two-force member, the reaction at A must be directed along CA. We have
Ax
40 lb
=
10 in. 5 in.
From Eq. (1):
A x = 80.0 lb
,
A y = 40.0 lb
B x = 80.0 lb
,
B y = 60.0 lb
− 80 + Bx = 0 Bx = + 80 lb
Thus,
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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866
PROBLEM 6.87 (Continued)
(b)
Load applied at F.
Free body: Member BCD
Since BCD is a two-force member (with forces applied at B and C only),
the reaction at B must be directed along CB. We have therefore
Bx = 0
The reaction at B is
From Eq. (1):
The reaction at A is
Bx = 0
Ax + 0 = 0
B y = 60.0 lb
Ax = 0
A y = 40.0 lb
Ax = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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867
!
PROBLEM 6.88
The 48-lb load can be moved along the line of action shown and
applied at A, D, or E. Determine the components of the reactions at B
and F if the 48-lb load is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b) and (c) since the position of the load
along its line of action is immaterial.
ΣM F = 0: (48lb)(8 in.) − Bx (12 in.) = 0
Bx = 32 lb B x = 32 lb
ΣFx = 0: 32 lb + Fx = 0
Fx = − 32 lb Fx = 32 lb
ΣFy = 0: By + Fy − 48 lb = 0
(a)
(1)
Load applied at A.
Free body: Member CDB
CDB is a two-force member. Thus, the reaction at B must be directed along BC.
By
32 lb
From Eq. (1):
=
5 in.
16 in.
B y = 10 lb
10 lb + Fy − 48 lb = 0
Fy = 38 lb Fy = 38 lb
Thus reactions are:
B x = 32.0 lb
,
B y = 10.00 lb
Fx = 32.0 lb
,
Fy = 38.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
868
!
PROBLEM 6.88 (Continued)
(b)
Load applied at D.
Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be
directed along CF.
Fy
32 lb
From Eq. (1):
=
7 in.
16 in.
Fy = 14 lb
By + 14 lb − 48 lb = 0
By = 34 lb By = 34 lb
Thus, reactions are:
(c)
B x = 32.0 lb
,
B y = 34.0 lb
Fx = 32.0 lb
,
Fy = 14.00 lb
Load applied at E.
Free body: Member CDB
This is the same free body as in Part (a).
Reactions are same as (a)
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869
!
PROBLEM 6.89
The 48-lb load is removed and a 288-lb · in. clockwise couple is applied
successively at A, D, and E. Determine the components of the reactions at B
and F if the couple is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b), and (c), since the point of
application of the couple is immaterial.
ΣM F = 0: − 288 lb ⋅ in. − Bx (12 in.) = 0
Bx = − 24 lb B x = 24 lb
ΣFx = 0: − 24 lb + Fx = 0
Fx = 24 lb Fx = 24 lb
ΣFy = 0: By + Fy = 0
(a)
(1)
Couple applied at A.
Free body: Member CDB
CDB is a two-force member. Thus, reaction at B must be directed along BC.
By
24 lb
From Eq. (1):
=
5 in.
16 in.
B y = 7.5 lb
− 7.5 lb + Fy = 0
Fy = 7.5 lb Fy = 7.5 lb
Thus, reactions are:
B x = 24.0 lb
,
B y = 7.50 lb
Fx = 24.0 lb
,
Fy = 7.50 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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870
!
PROBLEM 6.89 (Continued)
(b)
Couple applied at D.
Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be directed along CF.
Fy
24 lb
From Eq. (1):
=
7 in.
16 in.
Fy = 10.5 lb
By − 10.5 lb:
By = + 10.5 lb B y = 10.5 lb
Thus, reactions are:
(c)
B x = 24.0 lb
,
B y = 10.50 lb
Fx = 24.0 lb
,
Fy = 10.50 lb
Couple applied at E.
Free body: Member CDB
This is the same free body as in Part (a).
Reactions are same as in (a)
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871
!
PROBLEM 6.90
(a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its
component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to
the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at
a point B, a force of magnitude equal to the tension in the cable should also be applied at B.
SOLUTION
First note that, when a cable or cord passes over a frictionless, motionless pulley,
the tension is unchanged.
ΣM C = 0: rT1 − rT2 = 0 T1 = T2
(a)
Replace each force with an equivalent force-couple.
(b)
Cut cable and replace forces on pulley with equivalent pair of forces at A as above.
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
872
PROBLEM 6.91
Knowing that the pulley has a radius of 50 mm, determine the
components of the reactions at B and E.
SOLUTION
Free body: Entire assembly
ΣM E = 0 : − (300 N)(350 mm) − Bx (150 mm) = 0
Bx = −700 N
B x = 700 N
ΣFx = 0: − 700 N + Ex = 0
Ex = 700 N
E x = 700 N
ΣFy = 0: By + E y − 300 N = 0
(1)
Free body: Member ACE
ΣM C = 0: (700 N)(150 mm) − (300 N)(50 mm) − E y (180 mm) = 0
E y = 500 N
From Eq. (1):
E y = 500 N
By + 500 N − 300 N = 0
By = −200 N
B y = 200 N
Thus, reactions are:
B x = 700 N
,
B y = 200 N
E x = 700 N
,
E y = 500 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
873
PROBLEM 6.92
Knowing that each pulley has a radius of 250 mm,
determine the components of the reactions at D and E.
SOLUTION
Free body: Entire assembly
ΣM E = 0: (4.8 kN)(4.25 m) − Dx (1.5 m) = 0
Dx = +13.60 kN
D x = 13.60 kN
ΣFx = 0: Ex + 13.60 kN = 0
Ex = −13.60 kN
E x = 13.60 kN
ΣFy = 0: D y + E y − 4.8 kN = 0
(1)
Free body: Member ACE
ΣM A = 0 : (4.8 kN)(2.25 m) + E y (4 m) = 0
E y = −2.70 kN
From Eq. (1):
E y = 2.70 kN
Dy − 2.70 kN − 4.80 kN = 0
Dy = +7.50 kN
D y = 7.50 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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874
PROBLEM 6.93
Two 9-in.-diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft
by a small frame like that shown. Knowing that the combined weight of
each pipe and its contents is 30 lb/ft and assuming frictionless surfaces,
determine the components of the reactions at A and G.
SOLUTION
Free-body: Pipe 2
W = (30 lb/ft)(7.5 ft) = 225 lb
F D 225 lb
=
=
8 17
15
F = 120 lb
D = 255 lb
r = 4.5 in.
Geometry of pipe 2
CF = CD
By symmetry:
(1)
Equate horizontal distance:
r+
8
15 !
r = CD " #
17
$ 17 %
25
15 !
r = CD " #
17
17
$ %
25
5
CD =
r= r
15
3
From Eq. (1):
5
5
CF = r = (4.5 in.)
3
3
CF = 7.5 in.
Free-body: Member CFG
ΣM C = 0: (120 lb)(7.5 in.) − Gx (16 in.) = 0
Gx = 56.25 lb
G x = 56.3 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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875
PROBLEM 6.93 (Continued)
Free body: Frame and pipes
Note: Pipe 2 is similar to pipe 1.
AE = CF = 7.5 in.
E = F = 120 lb
ΣM A = 0: G y (15 in.) − (56.25 lb)(24 in.) − (225 lb)(4.5 in.)
−(225 lb)(19.5 in.) − (120 lb)(7.5 in.) = 0
G y = 510 lb
G y = 510 lb
!
ΣFx = 0: Ax + 120 lb + 56.25 lb = 0
Ax = 176.25 lb
A x = 176.3 lb
ΣFy = 0: Ay + 510 lb − 225 lb − 225 lb = 0
Ay = −60 lb
A y = 60.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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876
PROBLEM 6.94
Solve Problem 6.93 assuming that pipe 1 is removed and that only
pipe 2 is supported by the frames.
PROBLEM 6.93 Two 9-in.-diameter pipes (pipe 1 and pipe 2) are
supported every 7.5 ft by a small frame like that shown. Knowing
that the combined weight of each pipe and its contents is 30 lb/ft and
assuming frictionless surfaces, determine the components of the
reactions at A and G.
SOLUTION
Free-body: Pipe 2
W = (30 lb/ft)(7.5 ft) = 225 lb
F D 225 lb
=
=
8 17
15
F = 120 lb
D = 255 lb
r = 4.5 in.
Geometry of pipe 2
CF = CD
By symmetry:
(1)
Equate horizontal distance:
r+
8
15 !
r = CD " #
17
$ 17 %
25
15 !
r = CD " #
17
17
$ %
25
5
CD =
r= r
15
3
From Eq. (1):
5
5
CF = r = (4.5 in.)
3
3
CF = 7.5 in.
Free body: Member CFG
ΣM C = 0: (120 lb)(7.5 in.) − Gx (16 in.) = 0
Gx = 56.25 lb
G x = 56.3 lb
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877
PROBLEM 6.94 (Continued)
Free body: Frame and pipe 2
G x = 56.3 lb
ΣM A = 0: G y (15 in.) − (56.25 lb)(24 in.) − (225 lb)(19.5 in.) = 0
G y = 382.5 lb
G y = 383 lb
!
ΣFx = 0: Ax + 56.25 lb
Ax = −56.25 lb
A x = 56.3 lb
ΣFy = 0: Ay + 382.5 lb − 225 lb = 0
Ay = −157.5 lb
A y = 157.5
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878
PROBLEM 6.95
A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D.
Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional
load on each of the truck wheels due to the trailer.
SOLUTION
(a)
Free body: Trailer
(We shall denote by A, B, C the reaction at one wheel)
ΣM A = 0: − (2400 lb)(2 ft) + D(11 ft) = 0
D = 436.36 lb
ΣFy = 0: 2 A − 2400 lb + 436.36 lb = 0
A = 981.82 lb
A = 982 lb
Free body: Truck
ΣM B = 0: (436.36 lb)(3 ft) − (2900 lb)(5 ft) + 2C (9 ft) = 0
C = 732.83 lb
C = 733 lb
ΣFy = 0: 2 B − 436.36 lb − 2900 lb + 2(732.83 lb) = 0
B = 935.35 lb
(b)
B = 935 lb
Additional load on truck wheels
Use free body diagram of truck without 2900 lb.
ΣM B = 0: (436.36 lb)(3 ft) + 2C (9 ft) = 0
C = −72.73 lb
∆C = −72.7 lb
ΣFy = 0: 2 B − 436.36 lb − 2(72.73 lb) = 0
B = 290.9 lb
∆B = +291 lb
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879
PROBLEM 6.96
In order to obtain a better weight distribution over the
four wheels of the pickup truck of Problem 6.95, a
compensating hitch of the type shown is used to attach
the trailer to the truck. The hitch consists of two bar
springs (only one is shown in the figure) that fit into
bearings inside a support rigidly attached to the truck.
The springs are also connected by chains to the trailer
frame, and specially designed hooks make it possible to
place both chains in tension. (a) Determine the tension T
required in each of the two chains if the additional load
due to the trailer is to be evenly distributed over the four
wheels of the truck. (b) What are the resulting reactions at
each of the six wheels of the trailer-truck combination?
PROBLEM 6.95 A trailer weighing 2400 lb is attached
to a 2900-lb pickup truck by a ball-and-socket truck hitch
at D. Determine (a) the reactions at each of the six wheels
when the truck and trailer are at rest, (b) the additional
load on each of the truck wheels due to the trailer.
SOLUTION
(a)
We small first find the additional reaction ∆ at each wheel due the trailer.
Free body diagram (Same ∆ at each truck wheel)
ΣM A = 0: − (2400 lb)(2 ft) + 2∆(14 ft) + 2∆(23 ft) = 0
∆ = 64.86 lb
ΣFY = 0: 2 A − 2400 lb + 4(64.86 lb) = 0;
A = 1070 lb;
A = 1070 lb
Free body: Truck
(Trailer loading only)
ΣM D = 0: 2∆(12 ft) + 2∆(3 ft) − 2T (1.7 ft) = 0
T = 8.824∆
= 8.824(64.86 lb)
T = 572.3 lb
T = 572 lb
Free body: Truck
(Truck weight only)
ΣM B = 0: − (2900 lb)(5 ft) + 2C ′(9 ft) = 0
C ′ = 805.6 lb
C′ = 805.6 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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880
PROBLEM 6.96 (Continued)
ΣFy = 0: 2 B′ − 2900 lb + 2(805.6 lb) = 0
B′ = 644.4 lb
B′ = 644.4 lb
Actual reactions
B = B′ + ∆ = 644.4 lb + 64.86 = 709.2 lb
B = 709 lb
!
C = C ′ + ∆ = 805.6 lb + 64.86 = 870.46 lb
C = 870 lb
!
A = 1070 lb
(From Part a):
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881
PROBLEM 6.97
The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the
cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm,
while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl.
Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four
wheels, (b) the forces exerted on the motor unit at C and D.
SOLUTION
(a)
Free body: Entire machine
A = Reaction at each front wheel
B = Reaction at each rear wheel
ΣM A = 0: 75(3.2 m) − 100(1.2 m) + 2 B(4.8 m) − 300(5.6 m) = 0
2 B = 325 kN
B = 162.5 kN
!
A = 75.0 kN
!
ΣFy = 0: 2 A + 325 − 75 − 100 − 300 = 0
2 A = 150 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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882
PROBLEM 6.97 (Continued)
(b)
Free body: Motor unit
ΣM D = 0: C (1 m) + 2 B(2.8 m) − 300(3.6 m) = 0
C = 1080 − 5.6 B
Recalling
B = 162.5 kN, C = 1080 − 5.6(162.5) = 170 kN
ΣFx = 0: Dx − 170 = 0
!
!
(1)
! ΣFy = 0: 2(162.5) − Dy − 300 = 0
C = 170.0 kN
!
D x = 170.0 kN
!
D y = 25.0 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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883
PROBLEM 6.98
Solve Problem 6.97 assuming that the 75-kN load has been removed.
PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin
located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN
motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located,
respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the
reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.
SOLUTION
(a)
Free body: Entire machine
A = Reaction at each front wheel
B = Reaction at each rear wheel
ΣM A = 0: 2 B(4.8 m) − 100(1.2 m) − 300(5.6 m) = 0
2 B = 375 kN
B = 187.5 kN
!
A = 12.50 kN
!
ΣFy = 0: 2 A + 375 − 100 − 300 = 0
2 A = 25 kN
(b)
Free body: Motor unit
See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B = 187.5 kN,
we have
C = 1080 − 5.6(187.5) = 30 kN
C = 30.0 kN
ΣFx = 0: Dx − 30 = 0
D x = 30.0 kN
ΣFy = 0: 2(187.5) − Dy − 300 = 0
D y = 75.0 kN
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884
PROBLEM 6.99
For the frame and loading shown, determine the components of the
forces acting on member CFE at C and F.
SOLUTION
Free body: Entire frame
ΣM D = 0: (40 lb)(13 in.) + Ax (10 in.) = 0
Ax = −52 lb,
A x = 52 lb
!
Free body: Member ABF
ΣM B = 0: − (52 lb)(6 in.) + Fx (4 in.) = 0
Fx = +78 lb
Free body: Member CFE
Fx = 78.0 lb
From above:
!
ΣM C = 0: (40 lb)(9 in.) − (78 lb)(4 in.) − Fy (4 in.) = 0
Fy = +12 lb
Fy = 12.00 lb
ΣFx = 0: Cx − 78 lb = 0
Cx = +78 lb
ΣFy = 0: − 40 lb + 12 lb + C y = 0; C y = +28 lb
C x = 78.0 lb
!
C y = 28.0 lb
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885
PROBLEM 6.100
For the frame and loading shown, determine the components of the
forces acting on member CDE at C and D.
SOLUTION
Free body: Entire frame
ΣM y = 0: Ay − 25 lb = 0
Ay = 25 lb
A y = 25 lb !
ΣM F = 0: Ax (6.928 + 2 × 3.464) − (25 lb)(12 in.) = 0
Ax = 21.651 lb
A y = 21.65 lb
ΣFx = 0: F − 21.651 lb = 0
F = 21.651 lb
F = 21.65 lb
Free body: Member CDE
ΣM C = 0: Dy (4 in.) − (25 lb)(10 in.) = 0
Dy = +62.5 lb
D y = 62.5 lb
!
ΣFy = 0: −C y + 62.5 lb − 25 lb = 0
C y = +37.5 lb
C y = 37.5 lb
Free body: Member ABD
ΣM B = 0: Dx (3.464 in.) + (21.65 lb)(6.928 in.)
−(25 lb)(4 in.) − (62.5 lb)(2 in.)
Dx = +21.65 lb
Return to free body: Member CDE
From above
Dx = +21.65 lb
D x = 21.7 lb
ΣFx = 0: Cx − 21.65 lb
Cx = +21.65 lb
C x = 21.7 lb
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886
PROBLEM 6.101
For the frame and loading shown, determine the components of all forces
acting on member ABE.
SOLUTION
FBD Frame:
ΣM E = 0: (1.8 m) Fy − (2.1 m)(12 kN) = 0
Fy = 14.00 kN
ΣFy = 0: − E y + 14.00 kN − 12 kN = 0
E y = 2 kN
E y = 2.00 kN
FBD member BCD:
ΣM B = 0: (1.2 m)C y − (12 kN)(1.8 m) = 0 C y = 18.00 kN
But C is ⊥ ACF, so C x = 2C y ; C x = 36.0 kN
ΣFx = 0: − Bx + C x = 0 Bx = C x = 36.0 kN
Bx = 36.0 kN
on BCD
ΣFy = 0: − B y + 18.00 kN − 12 kN = 0 By = 6.00 kN on BCD
B x = 36.0 kN
On ABE:
B y = 6.00 kN
FBD member ABE:
ΣM A = 0: (1.2 m)(36.0 kN) − (0.6 m)(6.00 kN)
+ (0.9 m)(2.00 kN) − (1.8 m)( E x ) = 0
E x = 23.0 kN
ΣFx = 0: − 23.0 kN + 36.0 kN − Ax = 0
ΣFy = 0: − 2.00 kN + 6.00 kN − Ay = 0
A x = 13.00 kN
A y = 4.00 kN
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887
PROBLEM 6.102
For the frame and loading shown, determine the components of all
forces acting on member ABE.
PROBLEM 6.101 For the frame and loading shown, determine the
components of all forces acting on member ABE.
SOLUTION
FBD Frame:
ΣM F = 0: (1.2 m)(2400 N) − (4.8 m) E y = 0
E y = 600 N
FBD member BC:
Cy =
4.8
8
Cx = Cx
5.4
9
ΣM C = 0: (2.4 m) By − (1.2 m)(2400 N) = 0 B y = 1200 N
B y = 1200 N
on ABE:
ΣFy = 0: −1200 N + C y − 2400 N = 0 C y = 3600 N
so
9
Cx = C y
8
ΣFx = 0: − Bx + C x = 0 Bx = 4050 N
C x = 4050 N
on BC
B x = 4050 N
on ABE:
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888
PROBLEM 6.102 (Continued)
FBD member AB0E:
ΣM A = 0: a(4050 N) − 2aEx = 0
Ex = 2025 N
E x = 2025 N
ΣFx = 0 : − Ax + (4050 − 2025) N = 0
A x = 2025 N
ΣFy = 0: 600 N + 1200 N − Ay = 0
A y = 1800 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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889
!
PROBLEM 6.103
Knowing that P = 15 lb and Q = 65 lb, determine the components of
the forces exerted (a) on member BCDF at C and D, (b) on member
ACEG at E.
SOLUTION
Free body: Entire frame
ΣM E = 0: ( P + Q)(25 in.) − ( P + Q )(15 in.) − Dx (8 in.) = 0
10
8
10
D x = ( P + Q)
8
Dx = ( P + Q)
ΣFx = 0: − Ex + ( P + Q )
10
=0
8
Ex = ( P + Q)
10
10
; E x = ( P + Q)
8
8
Free body: Member BCDF
From above:
ΣFx = 0: −C x + Dx = 0
ΣM D = 0: − ( P + Q )
D x = ( P + Q)
10
8
v
C x = ( P + Q)
10
8
v
10
(4 in.) − P(15 in.) + Q (25 in.) + C y (10 in.)
8
Cy =
1
(− 20 P + 20 Q )
10
C y = 2Q − 2 P
C y = 2Q − 2 P v
ΣFy = 0: D y + (2Q − 2 P ) = P − Q = 0
D y = − Q + 3P
D y = − Q + 3P v
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890
PROBLEM 6.103 (Continued)
Free body: Member ACEG
From above
E x = ( P + Q)
10
8
v
Cy = 2Q − P
and
ΣFy = 0: E y − C y − P − Q = 0
E y − (2Q − 2 P ) − P − Q = 0
E y = 3Q − P
E y = 3Q − P v
P = 15 lb and Q = 65 lb
C x = ( P + Q)
10
10
= (15 + 65) = +100 lb
8
8
C y = 2 Q − 2 P = 2(65) − 2(15) = +100 lb
Dx = ( P + Q )
10
10
= (15 + 65) = +100 lb
8
8
Dy = − Q + 3P = − 65 + 3(15) = − 20 lb
E x = ( P + Q)
10
10
= (15 + 65) = +100 lb
8
8
E y = 3Q − P = 3(65) − 15 = + 180 lb
C x = 100.0 lb
C y = 100.0 lb
D x = 100.0 lb
D y = 20.0 lb
E x = 100.0 lb
E y = 180.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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891
!
PROBLEM 6.104
Knowing that P = 25 lb and Q = 55 lb, determine the components of
the forces exerted (a) on member BCDF at C and D, (b) on member
ACEG at E.
SOLUTION
Free body: Entire frame
ΣM E = 0: ( P + Q)(25 in.) − ( P + Q )(15 in.) − Dx (8 in.) = 0
10
8
10
D x = ( P + Q)
8
Dx = ( P + Q)
ΣFx = 0: − Ex + ( P + Q )
10
=0
8
Ex = ( P + Q)
10
10
; E x = ( P + Q)
8
8
Free body: Member BCDF
From above:
ΣFx = 0: − C x + Dx = 0
ΣM D = 0: − ( P + Q )
D x = ( P + Q)
10
8
v
C x = ( P + Q)
10
8
v
10
(4 in.) − P(15 in.) + Q (25 in.) + C y (10 in.)
8
Cy =
1
(− 20 P + 20 Q )
10
C y = 2Q − 2 P
C y = 2Q − 2 P v
ΣFy = 0: D y + (2Q − 2 P ) − P − Q = 0
D y = − Q + 3P
D y = − Q + 3P v
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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892
PROBLEM 6.104 (Continued)
Free body: Member ACEG
From above:
E x = ( P + Q)
10
8
v
Cy = 2Q − P
and
ΣFy = 0: E y − C y − P − Q = 0
E y − (2Q − 2 P ) − P − Q = 0
E y = 3Q − P
E y = 3Q − P v
P = 25 lb and Q = 55 lb
C x = ( P + Q)
10
10
= (25 + 55) = +100 lb
8
8
C y = 2 Q − 2 P = 2(55) − 2(25) = +60 lb
Dx = ( P + Q )
10
10
= (25 + 55) = +100 lb
8
8
Dy = − Q + 3P = − 55 + 3(25) = +20 lb
E x = ( P + Q)
10
10
= (25 + 55) = +100 lb
8
8
E y = 3Q − P = 3(55) − 25 = +140 lb
C x = 100.0
C y = 60.0 lb
D x = 100.0 lb
D y = 20.0 lb
E x = 100.0 lb
E y = 140.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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893
!
PROBLEM 6.105
For the frame and loading shown, determine the components of the
forces acting on member DABC at B and D.
SOLUTION
Free body: Entire frame
ΣM G = 0: H (0.6 m) − (12 kN)(1 m) − (6 kN)(0.5 m) = 0
H = 25 kN H = 25 kN
Free body: Member BEH
ΣM F = 0: Bx (0.5 m) − (25 kN)(0.2 m) = 0
Bx = + 10 kN
Free body: Member DABC
B x = 10.00 kN
From above:
ΣM D = 0: − By (0.8 m) + (10 kN + 12 kN)(0.5 m) = 0
By = + 13.75 kN
B y = 13.75 kN
ΣFx = 0: − Dx + 10 kN + 12 kN = 0
Dx = + 22 kN
D x = 22.0 kN
ΣFy = 0: − D y + 13.75 kN = 0
Dy = + 13.75 kN
D y = 13.75 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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894
!
PROBLEM 6.106
Solve Problem 6.105 assuming that the 6-kN load has been removed.
PROBLEM 6.105 For the frame and loading shown, determine the
components of the forces acting on member DABC at B and D.
SOLUTION
Free body: Entire frame
ΣM G = 0: H (0.6 m) − (12 kN)(1 m) = 0
H = 20 kN H = 20 kN
Free body: Member BEH
ΣM E = 0: Bx (0.5 m) − (20 kN)(0.2 m) = 0
Bx = + 8 kN
Free body: Member DABC
B x = −8.00 kN
From above:
ΣM D = 0: − By (0.8 m) + (8 kN + 12 kN)(0.5 m) = 0
By = + 12.5 kN
B y = 12.50 kN
ΣFx = 0: − Dx + 8 kN + 12 kN = 0
Dx = + 20 kN
ΣFy = 0: − D y + 12.5 kN = 0; Dy = + 12.5 kN
D x = 20.0 kN
D y = 12.50 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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895
!
PROBLEM 6.107
The axis of the three-hinge arch ABC is a parabola
with vertex at B. Knowing that P = 112 kN and
Q = 140 kN, determine (a) the components of the
reaction at A, (b) the components of the force
exerted at B on segment AB.
SOLUTION
Free body: Segment AB:
ΣM A = 0: Bx (3.2 m) − By (8 m) − P(5 m) = 0
(1)
Bx (2.4 m) − By (6 m) − P(3.75 m) = 0
(2)
0.75 (Eq. 1)
Free body: Segment BC:
ΣM C = 0: Bx (1.8 m) + By (6 m) − Q(3 m) = 0
Add (2) and (3):
(3)
4.2 Bx − 3.75P − 3Q = 0
Bx = (3.75 P + 3Q)/4.2
Eq. (1):
(3.75P + 3Q)
(4)
3.2
− 8B y − 5 P = 0
4.2
By = (− 9P + 9.6Q )/33.6
(5)
Given that P = 112 kN and Q = 140 kN
(a)
Reaction at A:
Considering again AB as a free body
ΣFx = 0: Ax − Bx = 0;
Ax = Bx = 200 kN
A x = 200 kN
ΣFy = 0: Ay − P − By = 0
Ay − 112 kN − 10 kN = 0
Ay = + 122 kN
A y = 122.0 kN
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896
PROBLEM 6.107 (Continued)
(b)
Force exerted at B on AB
Eq. (4):
Bx = (3.75 × 112 + 3 × 140)/4.2 = 200 kN
B x = 200 kN
Eq. (5):
By = (− 9 × 112 + 9.6 × 140)/33.6 = + 10 kN
B y = 10.00 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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897
!
PROBLEM 6.108
The axis of the three-hinge arch ABC is a parabola
with vertex at B. Knowing that P = 140 kN and
Q = 112 kN, determine (a) the components of the
reaction at A, (b) the components of the force exerted
at B on segment AB.
SOLUTION
Free body: Segment AB:
ΣM A = 0: Bx (3.2 m) − By (8 m) − P(5 m) = 0
(1)
Bx (2.4 m) − By (6 m) − P(3.75 m) = 0
(2)
0.75 (Eq. 1)
Free body: Segment BC:
ΣM C = 0: Bx (1.8 m) + By (6 m) − Q(3 m) = 0
(3)
4.2 Bx − 3.75P − 3Q = 0
Add (2) and (3):
Bx = (3.75 P + 3Q)/4.2
(3.75P + 3Q)
Eq. (1):
(4)
3.2
− 8B y − 5 P = 0
4.2
By = (− 9P + 9.6Q )/33.6
(5)
Given that P = 140 kN and Q = 112 kN
(a)
Reaction at A:
ΣFx = 0: Ax − Bx = 0;
Ax = Bx = 205 kN
A x = 205 kN
ΣFy = 0: Ay − P − By = 0
Ay − 140 kN − (− 5.5 kN) = 0
Ay = 134.5 kN
A y = 134.5 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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898
PROBLEM 6.108 (Continued)
(b)
Force exerted at B on AB
Eq. (4):
Bx = (3.75 × 140 + 3 × 112)/4.2 = 205 kN
B x = 205 kN
Eq. (5):
By = (− 9 × 140 + 9.6 × 112)/33.6 = − 5.5 kN
B y = 5.5 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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899
!
PROBLEM 6.109
Knowing that the surfaces at A and D are frictionless,
determine the forces exerted at B and C on member BCE.
SOLUTION
Free body of Member ACD
ΣM H = 0: C x (2 in.) − C y (18 in.) = 0 C x = 9C y
(1)
Free body of Member BCE
ΣM B = 0: C x (6 in.) + C y (6 in.) − (50 lb)(12 in.) = 0
9C y (6) + C y (6) − 600 = 0
Substitute from (1):
C y = + 10 lb; C x = 9C y = 9(10) = + 90 lb
C = 90.6 lb
ΣFx = 0: Bx − 90 lb = 0
Bx = 90 lb
ΣFy = 0: By + 10 lb − 50 lb = 0
By = 40 lb
B = 98.5 lb
6.34°
24.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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900
!
PROBLEM 6.110
For the frame and loading shown, determine (a) the reaction at C,
(b) the force in member AD.
SOLUTION
Free body: Member ABC
ΣM C = 0: + (100 lb)(45 in.) +
4
4
FAD (45 in.) + FBE (30 in.) = 0
5
5
3FAD + 2 FBF = − 375 lb
(1)
Free Body: Member DEF
ΣM F = 0:
4
4
FAD (30 in.) + FBF (15 in.) = 0
5
5
FBE = −2 FAD
(a)
(2)
Substitute from (2) into (1)
3FAD + 2( − 2 FAD ) = − 375 lb
FAD = + 375 lb
(2)
FAD = 375 lb ten.
FBE = − 2 FAD = − 2(375 lb)
FBE = −750 lb FBE = 750 lb comp.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
901
PROBLEM 6.110 (Continued)
(b)
Return to free body of member ABC
ΣFx = 0: C x + 100 lb +
C x + 100 +
4
4
FAD + FBE = 0
5
5
4
4
(375) + (−750) = 0
5
5
C x = + 200 lb C x = 200 lb
3
3
ΣFy = 0: C y − FAD − FBF = 0
5
5
3
3
C y − (375) − (− 750) = 0
5
5
C y = − 225 lb C y = 225 lb
α = 48.37°
C = 301.0 lb
C = 301 lb
48.4°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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902
!
PROBLEM 6.111
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each link.
SOLUTION
Member FBDs:
I
II
From FBD I:
ΣM J = 0:
a
3a
a
C x + C y − P = 0 C x + 3C y = P
2
2
2
FBD II:
ΣM K = 0:
a
3a
C x − C y = 0 C x − 3C y = 0
2
2
Solving:
Cx =
FBD I:
P
P
; C y = as drawn
2
6
ΣM B = 0: aC y − a
ΣFx = 0: −
1
2
FAG = 0 FAG = 2C y =
P
6
2
FAG =
1
1
2
2
FAG +
FBF − C x = 0 FBF = FAG + C x 2 =
P+
P
6
2
2
2
FBF =
FBD II:
ΣM D = 0: a
2
P C
6
1
2
FEH + aC y = 0 FEH = − 2C y = −
P
6
2
ΣFx = 0: C x −
2 2
P C
3
FEH =
2
P T
6
1
1
2
2
FDI +
FEH = 0 FDI = FEH + C x 2 = −
P+
P
6
2
2
2
FDI =
2
P C
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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903
PROBLEM 6.112
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each link.
SOLUTION
Member FBDs:
I
FBD I:
FBD II:
FBDs combined:
ΣM B = 0:
II
1
aC y − a
1
M D = 0: aC y − a
ΣM G = 0:
aP − a
FAF = 0 FAF = 2C y
2
2
1
2
FEH = 0 FEH = 2C y
1
FAF − a
2
FEH = 0 P =
Cy =
FBD I:
FBD II:
ΣFy = 0:
1
2
FAF +
ΣFy = 0: − C y +
1
2
1
2
P
2
FBG − P + C y = 0
FDG −
1
2
FEH = 0 −
1
2
2C y +
1
2C y
2
so FAF =
2
P C
2
FEH =
2
P T
2
!
FBG = 0
!
P
1
P
+
FBG − P + = 0
2
2
2
P
1
P
+
FDG − = 0
2
2
2
FDG = 2 P C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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904
PROBLEM 6.113
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each link.
SOLUTION
Member FBDs:
I
II
FBD I:
ΣM I = 0: 2aC y + aC x − aP = 0 2C y + C x = P
FBD II:
ΣM J = 0: 2aC y − aC x = 0 2C y − C x = 0
Solving:
FBD I:
Cx =
P
P
; C y = as shown
2
4
ΣFx = 0: −
1
2
FBG + C x = 0 FBG = C x 2
ΣFy = 0: FAF − P +
FBD II:
ΣFx = 0: − Cx +
ΣFy = 0: − C y +
1
2
2 ! P
P# + = 0
""
2 $ 2 #% 4
1
FDG = 0 FDG = Cx 2
2 !
P P
P
P # + FEH = 0 FEH = − = −
""
#
4 2
4
2$ 2 %
1
FBG =
P
2
P
4
FAF =
FDG =
P
FEH =
2
C
C
C
P
T
4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
905
PROBLEM 6.114
Members ABC and CDE are pin-connected at C and supported by
the four links AF, BG, DG, and EH. For the loading shown,
determine the force in each link.
SOLUTION
Free body: Member ABC
ΣM H = 0: Cx (a ) − C y (2a) = 0
Cx = 2C y
Note: This checks that for 3-force member the forces are concurrent.
Free body: Member CDE
ΣM F = 0: Cx (2a ) − C y (a) + P( a) = 0
2Cx − C y + P = 0
1
Cy = − P v
3
2(2C y ) − C y + P = 0
2
Cx = − P v
3
Cx = 2C y :
ΣF = 0: Cx −
E
2
= 0; −
E=−
E
2
P−
=0
3
2
2 2
P
3
FEH =
2 2
P comp.
3
ΣM E = 0: D (2a ) + C y (3a) − P(3a) = 0
D (2a) −
P
(3a) − P(3a) = 0
3
D = + 2P
FDG = 2 P T
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
906
PROBLEM 6.114 (Continued)
Return to free body of ABC
ΣFy = 0:
A
+ Cy = 0
2
A
2
−
P
=0
3
A=+
2
P
3
FAF =
2
P T
3
ΣM A = 0: B(2a) − Cx (3a) = 0
B(2a ) +
2
P(3a) = 0
3
B = −P
FBG = P C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
907
PROBLEM 6.115
Solve Problem 6.113 assuming that the force P is replaced by a
clockwise couple of moment M0 applied to member CDE at D.
PROBLEM 6.113 Members ABC and CDE are pin-connected at C
and supported by four links. For the loading shown, determine the
force in each link.
SOLUTION
Free body: Member ABC
ΣM J = 0: C y (2a) + Cx ( a) = 0
Cx = − 2C y
Free body: Member CDE
ΣM K = 0: C y (2a ) − Cx (a ) − M 0 = 0
C y (2a) − (− 2C y )(a) − M 0 = 0
Cx = − 2C y :
D
ΣFx = 0:
2
D
+ Cx = 0;
D=
2
−
M0
M0
v
4a
M
Cx = − 0 v
2a
Cy =
M0
=0
2a
FDG =
2a
M0
2a
T
ΣM D = 0: E (a) − C y ( a) + M 0 = 0
M !
E (a) − " 0 # ( a) + M 0 = 0
$ 4a %
E=−
3 M0
4 a
FEH =
3 M0
4 a
C
Return to Free Body of ABC
ΣFx = 0:
B
2
B
+ C x = 0;
B=
2
−
M0
=0
2a
M0
FBG =
2a
ΣM B = 0: A( a) + C y (a);
A(a) +
M0
2a
T
M0
(a ) = 0
4a
A=−
M0
4a
FAF =
M0
4a
C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
908
PROBLEM 6.116
Solve Problem 6.114 assuming that the force P is replaced by a
clockwise couple of moment M0 applied to member CDE at D.
PROBLEM 6.114 Members ABC and CDE are pin-connected at C
and supported by the four links AF, BG, DG, and EH. For the
loading shown, determine the force in each link.
SOLUTION
Free body: Member ABC
ΣM H = 0: C x (a ) − C y (2a) = 0
C x = 2C y
Free body: Member CDE
ΣM F = 0: C x (2a ) − C y (a) − M 0 = 0
(2C y )(2a ) − C y (a) − M 0 = 0
C x = 2C y :
ΣFx = 0: C x −
E
2
= 0;
E=
ΣFy = 0: D +
D+
E
2
Cy =
Cx =
M0
v
3a
2M 0
v
3a
2M 0
E
−
=0
3a
2
2 2 M0
3 a
FEH =
2 2 M0
3 a
+ Cy = 0
M
2 2 M0 1
+ 0 =0
3 a 2 3a
D=−
M0
a
FDG =
M0
a
C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
909
PROBLEM 6.116 (Continued)
Return to free body of ABC
ΣFy = 0:
A
2
+ C y = 0;
A=−
A
2
+
M0
=0
3a
2 M0
3 a
FAF =
2 M0
3 a
C
M0
a
T
ΣM A = 0: B(2a) − C x (3a) = 0
2 M0 !
B(2a ) − "
# (3a) = 0
$3 a %
B=+
M0
a
FBG =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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910
PROBLEM 6.117
Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a
support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are
exerted at the connections, determine the vertical reactions at E, F, G, and H.
SOLUTION
We shall draw the free body of each member. Force P will be applied to member AFB. Starting with member
AED, we shall express all forces in terms of reaction E.
Member AFB:
ΣM D = 0: A(3a) + E (a ) = 0
E
3
ΣM A = 0: − D (3a ) − E (2a ) = 0
A=−
D=−
2E
3
Member DHC:
2E !
ΣM C = 0: " −
# (3a) − H (a) = 0
$ 3 %
H = − 2E
(1)
2E !
ΣM H = 0: " −
# (2a) + C (a ) = 0
$ 3 %
C=+
4E
3
Member CGB:
4E !
ΣM B = 0: + "
# (3a ) − G (a ) = 0
$ 3 %
G = + 4E
(2)
4E !
ΣM G = 0: + "
# (2a) + B (a ) = 0
$ 3 %
B=−
8E
3
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you are using it without permission.
911
PROBLEM 6.117 (Continued)
Member AFB:
ΣFy = 0: F − A − B − P = 0
8E !
E!
F −"− # −"−
#−P =0
$ 3% $ 3 %
F = P − 3E
(3)
ΣM A = 0: F ( a) − B(3a) = 0
8E !
( P − 3E )(a) − " −
# (3a ) = 0
$ 3 %
P − 3E + 8 E = 0; E = −
P
5
E=
P
5
Substitute E = − P5 into Eqs. (1), (2), and (3).
P!
H = −2 E = − 2 " − #
$ 5%
H =+
2P
5
H=
2P
5
P!
G = + 4E = 4 " − #
$ 5%
G=−
4P
5
G=
4P
5
P!
F = P − 3E = P − 3 " − #
$ 5%
F =+
8P
5
F=
8P
5
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
912
!
PROBLEM 6.118
Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown.
Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E,
and H.
SOLUTION
Note that, if we assume P is applied to EG, each individual member FBD looks like
so
2 Fleft = 2 Fright = Fmiddle
Labeling each interaction force with the letter corresponding to the joint of its application, we see that
B = 2 A = 2F
C = 2B = 2D
G = 2C = 2 H
P + F = 2G ( = 4C = 8 B = 16 F ) = 2 E
From
P + F = 16 F , F =
P
15
!
so A =
P
15
D=
2P
15
H=
4P
15
E=
8P
15
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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913
!
!
PROBLEM 6.119
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame,
determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the
third force, at B, must be parallel to the link forces.
(a)
FBD whole:
ΣM A = 0: − 2aP −
ΣFx = 0: Ax −
ΣFy = 0:
a 4
1
B + 5a
B = 0 B = 2.06 P
4 17
17
4
17
Ay − P +
B=0
1
17
B = 2.06P
14.04°
A = 2.06P
14.04°
A x = 2P
B = 0 Ay =
P
2
rigid
(b)
FBD whole:
Since B passes through A,
ΣM A = 2aP = 0 only if P = 0
no equilibrium if P ≠ 0
not rigid
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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914
PROBLEM 6.119 (Continued)
(c)
FBD whole:
ΣM A = 0: 5a
1
17
ΣFx = 0: Ax +
B+
4
17
ΣFy = 0: Ay − P +
3a 4
17
B − 2aP = 0 B =
P
4 17
4
B=0
1
17
B = 1.031P
14.04°
A = 1.250P
36.9°
Ax = − P
B=0
Ay = P −
P 3P
=
4
4
System is rigid
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
915
PROBLEM 6.120
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame,
determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
(a)
Member FBDs:
I
II
FBD I:
ΣM A = 0: aF1 − 2aP = 0 F1 = 2 P; ΣFy = 0:
FBD II:
ΣM B = 0: − aF2 = 0 F2 = 0
Ay − P = 0 A y = P
ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2 P B x = 2 P
ΣFy = 0: B y = 0
FBD I:
ΣFx = 0: − Ax − F1 + F2 = 0
so B = 2P
Ax = F2 − F1 = 0 − 2 P A x = 2 P
so A = 2.24P
26.6°
frame is rigid
(b)
FBD left:
FBD whole:
I
FBD I:
ΣM E = 0:
FBD II:
ΣM B = 0:
II
a
a
5a
P + Ax −
Ay = 0 Ax − 5 Ay = − P
2
2
2
3aP + aAx − 5aAy = 0 Ax − 5 Ay = −3P
This is impossible unless P = 0
not rigid
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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916
!
PROBLEM 6.120 (Continued)
(c)
Member FBDs:
I
FBD I:
II
ΣFy = 0: A − P = 0
A=P
!
C=P
!
ΣM D = 0: aF1 − 2aA = 0 F1 = 2 P
ΣFx = 0: F2 − F1 = 0 F2 = 2 P !
FBD II:
ΣM B = 0: 2aC − aF1 = 0 C =
F1
=P
2
ΣFx = 0: F1 − F2 + Bx = 0 Bx = P − P = 0 !
ΣFx = 0: By + C = 0 B y = − C = − P
B=P
Frame is rigid
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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917
PROBLEM 6.121
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame,
determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
(a)
Member FBDs:
I
FBD II:
FBD I:
ΣFy = 0: B y = 0
II
ΣM B = 0: aF2 = 0
F2 = 0
ΣM A = 0: aF2 − 2aP = 0 but F2 = 0
so P = 0
(b)
not rigid for P ≠ 0
Member FBDs:
Note: 7 Unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ) but only 6 independent equations.
!
System is statically indeterminate
!
System is, however, rigid
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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918
PROBLEM 6.121 (Continued)
(c)
FBD whole:
FBD right:
I
FBD I:
II
ΣM A = 0: 5aBy − 2aP = 0
ΣFy = 0: Ay − P +
2
P=0
5
a
5a
Bx −
By = 0 Bx = 5B y
2
2
FBD II:
ΣM c = 0:
FBD I:
ΣFx = 0: Ax + Bx = 0
Ax = − Bx
By =
2
P !
5
Ay =
3
P
5
B x = 2P
A x = 2P
A = 2.09P
16.70°
B = 2.04P
11.31°
System is rigid
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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919
PROBLEM 6.122
An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°,
determine (a) the vertical force exerted on the block at D, (b) the force
exerted on member ABC at B.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC
BD = (7) 2 + (24)2 = 25 in.
We have
( FBD ) x =
7
24
FBD , ( FBD ) y =
FBD
25
25
ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0
7
24
!
!
" FBD # (24) + " FBD # (7) = 84(16)
25
25
$
%
$
%
336
FBD = 1344
25
FBD = 100 lb
tan α =
FBD = 100.0 lb
(b)
Force exerted at B:
(a)
Vertical force exerted on block
( FBD ) y =
24
α = 73.7°
7
73.7°
24
24
FBD =
(100 lb) = 96 lb
25
25
(FBD ) y = 96.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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920
PROBLEM 6.123
Solve Problem 6.122 when θ = 0.
PROBLEM 6.122 An 84-lb force is applied to the toggle vise at C.
Knowing that θ = 90°, determine (a) the vertical force exerted on
the block at D, (b) the force exerted on member ABC at B.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC
BD = (7) 2 + (24)2 = 25 in.
We have
( FBD ) x =
7
24
FBD , ( FBD ) y =
FBD
25
25
ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(40) = 0
7
24
!
!
" FBD # (24) + " FBD # (7) = 84(40)
25
25
$
%
$
%
336
FBD = 3360
25
FBD = 250 lb
tan α =
FBD = 250.0 lb
(b)
Force exerted at B:
(a)
Vertical force exerted on block
( FBD ) y =
24
α = 73.7°
7
73.7°
24
24
FBD =
(250 lb) = 240 lb
25
25
(FBD ) y = 240 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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921
PROBLEM 6.124
The control rod CE passes through a horizontal hole in the body
of the toggle system shown. Knowing that link BD is 250 mm
long, determine the force Q required to hold the system in
equilibrium when β = 20°.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC
Dimensions in mm
Since
BD = 250, θ = sin −1
33.404
; θ = 7.679°
250
ΣM C = 0: ( FBD sin θ )187.94 − ( FBD cos θ )68.404 + (100 N)328.89 = 0
FBD [187.94sin 7.679° − 68.404 cos 7.679°] = 32889
FBD = 770.6 N
ΣFx = 0: (770.6 N) cos 7.679° = C x = 0
C x = + 763.7 N
Member CQ:
ΣFx = 0: Q = C x = 763.7 N
Q = 764 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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922
PROBLEM 6.125
Solve Problem 6.124 when (a) β = 0, (b) β = 6°.
PROBLEM 6.124 The control rod CE passes
through a horizontal hole in the body of the
toggle system shown. Knowing that link BD is
250 mm long, determine the force Q required to
hold the system in equilibrium when β = 20°.
SOLUTION
We note that BD is a two-force member.
(a)
β = 0:
Free body: Member ABC
BD = 250 mm, sinθ =
Since
35 mm
; θ = 8.048°
250 mm
ΣM C = 0: (100 N)(350 mm) − FBD sin θ (200 mm) = 0
FBD = 1250 N
ΣFx = 0: FBD cos θ − Cx = 0
(1250 N)(cos 8.048°) − C x = 0
C x = 1237.7 N
Member CE:
ΣFx = 0: (1237.7 N) − Q = 0
Q = 1237.7 N
(b)
β = 6°
Q = 1238 N
Free body: Member ABC
Dimensions in mm
Since
14.094 mm
250 mm
θ = 3.232°
BD = 250 mm, θ = sin −1
ΣM C = 0: ( FBD sin θ )198.90 + ( FBD cos θ )20.906 − (100 N)348.08 = 0
FBD [198.90sin 3.232° + 20.906 cos 3.232°] = 34808
FBD = 1084.8 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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923
PROBLEM 6.125 (Continued)
ΣFx = 0: FBD cos θ − Cx = 0
(1084.8 N) cos 3.232° − C x = 0
C x = + 1083.1 N
Member DE:
ΣFx = 0: Q = C x
Q = 1083.1 N
Q = 1083 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
924
!
PROBLEM 6.126
The press shown is used to emboss a small seal at E. Knowing
that P = 250 N, determine (a) the vertical component of the force
exerted on the seal, (b) the reaction at A.
SOLUTION
FBD Stamp D:
ΣFy = 0: E − FBD cos 20° = 0, E = FBD cos 20°
FBD ABC:
ΣM A = 0: (0.2 m)(sin 30°)( FBD cos 20°) + (0.2 m)(cos 30°)( FBD sin 20°)
− [(0.2 m)sin 30° + (0.4 m) cos15°](250 N) = 0
FBD = 793.64 N C
and, from above,
E = (793.64 N) cos 20°
(a)
E = 746 N
ΣFx = 0: Ax − (793.64 N)sin 20° = 0
A x = 271.44 N
ΣFy = 0: Ay + (793.64 N) cos 20° − 250 N = 0
A y = 495.78 N
so
(b)
A = 565 N
61.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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925
PROBLEM 6.127
The press shown is used to emboss a small seal at E. Knowing
that the vertical component of the force exerted on the seal must
be 900 N, determine (a) the required vertical force P, (b) the
corresponding reaction at A.
SOLUTION
FBD Stamp D:
ΣFy = 0: 900 N − FBD cos 20° = 0, FBD = 957.76 N C
(a)
FBD ABC:
ΣM A = 0: [(0.2 m)(sin 30°)](957.76 N) cos 20° + [(0.2 m)(cos 30°)](957.76 N) sin 20°
− [(0.2 m)sin 30° + (0.4 m) cos15°]P = 0
P = 301.70 N,
(b)
P = 302 N
ΣFx = 0: Ax − (957.76 N)sin 20° = 0
A x = 327.57 N
ΣFy = 0: −Ay + (957.76 N) cos 20° − 301.70 N = 0
A y = 598.30 N
so
A = 682 N
61.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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926
PROBLEM 6.128
Water pressure in the supply system exerts a downward force of
135 N on the vertical plug at A. Determine the tension in the fusible
link DE and the force exerted on member BCE at B.
SOLUTION
Free body: Entire linkage
+ ΣFy = 0: C − 135 = 0
C = + 135 N
Free body: Member BCE
ΣFx = 0: Bx = 0
ΣM B = 0: (135 N)(6 mm) − TDE (10 mm) = 0
TDE = 81.0 N
ΣFy = 0: 135 + 81 − By = 0
By = + 216 N
B = 216 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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927
!
PROBLEM 6.129
A couple M of magnitude 1.5 kN ⋅ m is applied to the crank of the engine system shown. For each of the two
positions shown, determine the force P required to hold the system in equilibrium.
SOLUTION
(a)
FBDs:
50 mm
175 mm
2
=
7
Dimensions in mm
Note:
tan θ =
FBD whole:
ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kN
FBD piston:
ΣFy = 0: C y − FBC sin θ = 0 FBC =
Cy
sin θ
=
6.00 kN
sinθ
ΣFx = 0: FBC cos θ − P = 0
P = FBC cos θ =
6.00 kN
= 7 kips
tan θ
P = 21.0 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
928
PROBLEM 6.129 (Continued)
(b)
FBDs:
Dimensions in mm
2
as above
7
Note:
tan θ =
FBD whole:
ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN
ΣFy = 0: C y − FBC sin θ = 0 FBC =
ΣFx = 0:
Cy
sin θ
FBC cos θ − P = 0
P = FBC cos θ =
Cy
tan θ
=
15 kN
2/7
P = 52.5 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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929
PROBLEM 6.130
A force P of magnitude 16 kN is applied to the piston of the engine system shown. For each of the two
positions shown, determine the couple M required to hold the system in equilibrium.
SOLUTION
(a)
FBDs:
Note:
FBD piston:
50 mm
tan θ =
175 mm
2
=
7
Dimensions in mm
ΣFx = 0: FBC cos θ − P = 0 FBC =
P
cos θ
ΣFy = 0: C y − FBC sin θ = 0 C y = FBC sin θ = P tan θ =
FBD whole:
2
P
7
ΣM A = 0: (0.250 m)C y − M = 0
2!
M = (0.250 m) " # (16 kN)
7
$ %
= 1.14286 kN ⋅ m
M = 1143 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
930
PROBLEM 6.130 (Continued)
(b)
FBDs:
Dimensions in mm
tan θ =
Note:
FBD piston: as above
FBD whole:
2
as above
7
C y = P tan θ =
2
P
7
2
ΣM A = 0: (0.100 m)C y − M = 0 M = (0.100 m) (16 kN)
7
M = 0.45714 kN ⋅ m
M = 457 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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931
PROBLEM 6.131
The pin at B is attached to member ABC and can slide freely along
the slot cut in the fixed plate. Neglecting the effect of friction,
determine the couple M required to hold the system in equilibrium
when θ = 30°.
SOLUTION
Free body: Member ABC
ΣM C = 0: (25 lb)(13.856 in.) − B(3 in.) = 0
B = +115.47 lb
ΣFy = 0: − 25 lb + C y = 0
C y = +25 lb
ΣFx = 0: 115.47 lb − Cx = 0
Cx = +115.47 lb
Free body: Member CD
β = sin −1
5.196
; β = 40.505°
8
CD cos β = (8 in.) cos 40.505° = 6.083 in.
ΣM D = 0: M − (25 lb)(5.196 in.) − (115.47 lb)(6.083 in.) = 0
M = +832.3 lb ⋅ in.
M = 832 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
932
PROBLEM 6.132
The pin at B is attached to member ABC and can slide freely along
the slot cut in the fixed plate. Neglecting the effect of friction,
determine the couple M required to hold the system in equilibrium
when θ = 60°.
SOLUTION
Free body: Member ABC
ΣM C = 0: (25 lb)(8 in.) − B(5.196 in.) = 0
B = +38.49 lb
ΣFx = 0: 38.49 lb − C x = 0
Cx = +38.49 lb
ΣFy = 0: − 25 lb + C y = 0
C y = +25 lb
Free body: Member CD
3
8
β = sin −1 ; β = 22.024°
CD cos β = (8 in.) cos 22.024° = 7.416 in.
ΣM D = 0: M − (25 lb)(3 in.) − (38.49 lb)(7.416 in.) = 0
M = +360.4 lb ⋅ in.
M = 360 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
933
PROBLEM 6.133
Arm ABC is connected by pins to a collar at B and to crank CD at C.
Neglecting the effect of friction, determine the couple M required to
hold the system in equilibrium when θ = 0.
SOLUTION
Free body: Member ABC
ΣFx = 0: Cx − 240 N = 0
Cx = +240 N
ΣM C = 0: (240 N)(500 mm) − B(160 mm) = 0
B = +750 N
ΣFy = 0: C y − 750 N = 0
C y = +750 N
Free body: Member CD
ΣM D = 0: M + (750 N)(300 mm) − (240 N)(125 mm) = 0
M = −195 × 103 N ⋅ mm
M = 195.0 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
934
PROBLEM 6.134
Arm ABC is connected by pins to a collar at B and to crank CD at C.
Neglecting the effect of friction, determine the couple M required to
hold the system in equilibrium when θ = 90°.
SOLUTION
Free body: Member ABC
ΣFx = 0: Cx = 0
ΣM B = 0: C y (160 mm) − (240 N)(90 mm) = 0
C y = +135 N
Free body: Member CD
ΣM D = 0: M − (135 N)(300 mm) = 0
M = +40.5 × 103 N ⋅ mm
M = 40.5 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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935
PROBLEM 6.135
Two rods are connected by a slider block as shown. Neglecting the
effect of friction, determine the couple MA required to hold the
system in equilibrium.
SOLUTION
D ⊥ AC
Note:
Member FBDs:
ΣM B = 0:
lD sin115° − 250 lb ⋅ in. = 0
D=
250 lb ⋅ in.
2(15 in.) cos 25° sin115°
= 10.1454 lb
ΣM A = 0:
M A − (15 in.)(10.1454 lb) = 0
M A = 152.2 lb ⋅ in.
!
!
!
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
936
PROBLEM 6.136
Two rods are connected by a slider block as shown. Neglecting the
effect of friction, determine the couple MA required to hold the
system in equilibrium.
SOLUTION
C ⊥ BD
Note:
Member FBD’s:
ΣM B = 0: lC − 250 lb ⋅ in. = 0
C=
250 lb ⋅ in.
2(15 in.) cos 25°
C = 9.1948 lb
ΣM A = 0: M A − (15 in.)C sin 65° = 0
M A = (15 in.)(9.1948 lb) sin 65°
M A = 125.0 lb ⋅ in.
!
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
937
PROBLEM 6.137
Rod CD is attached to the collar D and passes through a
collar welded to end B of lever AB. Neglecting the effect of
friction, determine the couple M required to hold the system
in equilibrium when θ = 30°.
SOLUTION
B ⊥ CD
Note:
FBD DC:
ΣFx′ = 0: D y sin 30° − (300 N) cos 30° = 0
Dy =
300 N
= 519.62 N
tan 30°
FBD machine:
ΣM A = 0:
0.200 m
519.62 N − M = 0
sin 30°
M = 207.85 N ⋅ m
M = 208 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
938
PROBLEM 6.138
Rod CD is attached to the collar D and passes through a collar welded
to end B of lever AB. Neglecting the effect of friction, determine the
couple M required to hold the system in equilibrium when θ = 30°.
SOLUTION
FBD DC:
ΣFx′ = 0:
D y sin 30° − (150 N) cos 30° = 0
Dy = (150 N) ctn 30° = 259.81 N
FBD machine:
ΣM A = 0: (0.100 m)(150 N) + d (259.81 N) − M = 0
d = b − 0.040 m
b=
so
0.030718 m
tan 30
b = 0.053210 m
d = 0.0132100 m
M = 18.4321 N ⋅ m
M = 18.43 N ⋅ m.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
939
PROBLEM 6.139
Two hydraulic cylinders control the position
of the robotic arm ABC. Knowing that in the
position shown the cylinders are parallel,
determine the force exerted by each cylinder
when P = 160 N and Q = 80 N.
SOLUTION
Free body: Member ABC
ΣM B = 0:
4
FAE (150 mm) − (160 N)(600 mm) = 0
5
FAE = +800 N
FAE = 800 N T
3
ΣFx = 0: − (800 N) + Bx − 80 N = 0
5
Bx = +560 N
4
ΣFy = 0: − (800 N) + By − 160 N = 0
5
By = +800 N
Free body: Member BDF
ΣM F = 0: (560 N)(400 mm) − (800 N)(300 mm) −
FDG = −100 N
4
FDG (200 mm) = 0
5
FDG = 100.0 N C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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940
PROBLEM 6.140
Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are
parallel and both are in tension. Knowing the FAE = 600 N and FDG = 50 N, determine the forces P and Q applied
at C to arm ABC.
SOLUTION
Free body: Member ABC
ΣM B = 0:
4
(600 N)(150 mm) − P(600 mm) = 0
5
P = +120 N
ΣM C = 0:
P = 120.0 N
4
(600)(750 mm) − By (600 mm) = 0
5
By = +600 N
Free body: Member BDF
ΣM F = 0: Bx (400 mm) − (600 N)(300 mm)
4
− (50 N)(200 mm) = 0
5
Bx = +470 N
Return to free body: Member ABC
3
ΣFx = 0: − (600 N) + 470 N − Q = 0
5
Q = +110 N
Q = 110.0 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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941
!
PROBLEM 6.141
A log weighing 800 lb is lifted by a pair of tongs as shown. Determine
the forces exerted at E and F on tong DEF.
SOLUTION
FBD AB:
Ay = By = 400 lb
By symmetry:
Ax = Bx
and
6
(400 lb)
5
= 480 lb
=
D = −B
Note:
Dx = 480 lb
so
FBD DEF:
Dy = 400 lb
ΣM F = (10.5 in.)(400 lb) + (15.5 in.)(480 lb) − (12 in.) Ex = 0
Ex = 970 lb
E = 970 lb
ΣFx = 0: − 480 lb + 970 lb − Fx = 0
Fx = 490 lb
ΣFy = 0: 400 lb − Fy = 0
Fy = 400 lb
F = 633 lb
39.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
942
PROBLEM 6.142
A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs
shown. Determine the forces exerted at D and F on tong BDF.
SOLUTION
Free body: Rail
W = (39 ft)(44 lb/ft) = 1716 lb
Free body: Upper link
By symmetry
1
E y = Fy = W = 858 lb
2
By symmetry,
1
( FAB ) y = ( FAC ) y = W = 858 lb
2
Since AB is a two-force member,
( FAB ) x ( FAB ) y
=
9.6
6
Free Body: Tong BDF
( FAB ) x =
9.6
(858) = 1372.8 lb
6
ΣM D = 0: (Attach FAB at A)
Fx (8) − ( FAB ) x (18) − Fy (0.8) = 0
Fx (8) − (1372.8 lb)(18) − (858 lb)(0.8) = 0
Fx = +3174.6 lb
F = 3290 lb
15.12°
ΣFx = 0: − Dx + ( FAB ) x + Fx = 0
Dx = ( FAB ) x + Fx = 1372.8 + 3174.6 = 4547.4 lb
ΣFy = 0: D y + ( FAB ) y − Fy = 0
Dy = 0
D = 4550 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
943
!
PROBLEM 6.143
The tongs shown are used to apply a total upward force of 45 kN on a pipe cap.
Determine the forces exerted at D and F on tong ADF.
SOLUTION
FBD whole:
By symmetry
FBD ADF:
A = B = 22.5 kN
ΣM F = 0: (75 mm)CD − (100 mm)(22.5 kN) = 0
CD = 30.0 kN
ΣFx = 0: Fx − CD = 0
Fx = CD = 30 kN
ΣFy = 0: 22.5 kN − Fy = 0
Fy = 22.5 kN
F = 37.5 kN
so
36.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
944
!
PROBLEM 6.144
If the toggle shown is added to the tongs of Problem 6.143 and a single
vertical force is applied at G, determine the forces exerted at D and F
on tong ADF.
SOLUTION
Free body: Toggle
By symmetry
Ay =
1
(45 kN) = 22.5 kN
2
AG is a two-force member
Ax
22.5 kN
=
22 mm 55 mm
Ax = 56.25 kN
Free body: Tong ADF
ΣFy = 0: 22.5 kN − Fy = 0
Fy = +22.5 kN
ΣM F = 0 : D (75 mm) − (22.5 kN)(100 mm) − (56.25 kN)(160 mm) = 0
D = +150 kN
D = 150.0 kN
!
ΣFx = 0: 56.25 kN − 150 kN + Fx = 0
Fx = 93.75 kN
F = 96.4 kN
13.50°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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945
PROBLEM 6.145
The pliers shown are used to grip a 0.3-in.-diameter rod.
Knowing that two 60-lb forces are applied to the handles,
determine (a) the magnitude of the forces exerted on the rod,
(b) the force exerted by the pin at A on portion AB of the
pliers.
SOLUTION
Free body: Portion AB
(a)
ΣM A = 0: Q (1.2 in.) − (60 lb)(9.5 in.) = 0
Q = 475 lb
(b)
!
ΣFx = 0: Q(sin 30°) + Ax = 0
(475 lb)(sin 30°) + Ax = 0
Ax = −237.5 lb
A y = 237.5 lb
ΣFy = 0: − Q (cos 30°) + Ay − 60 lb = 0
−(475 lb)(cos 30°) + Ay − 60 lb = 0
Ay = +471.4 lb
A y = 471.4 lb
A = 528 lb
63.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
946
PROBLEM 6.146
In using the bolt cutter shown, a worker applies two 300-N
forces to the handles. Determine the magnitude of the forces
exerted by the cutter on the bolt.
SOLUTION
FBD cutter AB:
I
FBD I:
FBD handle BC:
II
Dimensions in mm
ΣFx = 0: Bx = 0
FBD II:
ΣM C = 0: (12 mm)By − (448 mm)300 N = 0
By = 11, 200.0 N
Then
FBD I:
ΣM A = 0: (96 mm) By − (24 mm) F = 0
F = 4 By
F = 44,800 N = 44.8 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
947
PROBLEM 6.147
Determine the magnitude of the gripping forces exerted along
line aa on the nut when two 50-lb forces are applied to the
handles as shown. Assume that pins A and D slide freely in
slots cut in the jaws.
SOLUTION
FBD jaw AB:
ΣFx = 0: Bx = 0
ΣM B = 0: (0.5 in.)Q − (1.5 in.) A = 0
A=
Q
3
ΣFy = 0: A + Q − B y = 0
By = A + Q =
FBD handle ACE:
4Q
3
By symmetry and FBD jaw DE:
Q
3
E x = Bx = 0
D= A=
E y = By =
4Q
3
ΣM C = 0: (5.25 in.)(50 lb) + (0.75 in.)
Q
4Q
− (0.75 in.)
=0
3
3
Q = 350 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
948
PROBLEM 6.148
Determine the magnitude of the gripping forces produced
when two 300-N forces are applied as shown.
SOLUTION
We note that AC is a two-force member
FBD handle CD:
ΣM D = 0: − (126 mm)(300 N) − (6 mm)
2.8
8.84
A
!
1
+ (30 mm) "
A# = 0
$ 8.84 %
A = 2863.6 8.84 N
Dimensions in mm
FBD handle AB:
ΣM B = 0: (132 mm)(300 N) − (120 mm)
1
8.84
(2863.6 8.84 N)
+ (36 mm) F = 0
F = 8.45 kN
!
Dimensions in mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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949
PROBLEM 6.149
Knowing that the frame shown has a sag at B of a = 1 in., determine
the force P required to maintain equilibrium in the position shown.
SOLUTION
We note that AB and BC are two-force members
Free body: Toggle
Cy =
By symmetry:
P
2
Cy
Cx
=
10 in. a
10
10 P 5 P
Cx = C y = ⋅ =
a
a 2
a
Free body: Member CDE
ΣM E = 0: C x (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0
P
5P
(b) − (20) = 500
a
2
30
!
P " − 10 # = 500
$ a
%
For
(1)
a = 1.0 in.
30
!
P " − 10 # = 500
$ 1
%
20 P = 500
P = 25.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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950
PROBLEM 6.150
Knowing that the frame shown has a sag at B of a = 0.5 in.,
determine the force P required to maintain equilibrium in the
position shown.
SOLUTION
We note that AB and BC are two-force members
Free body: Toggle
Cy =
By symmetry:
P
2
Cy
Cx
=
10 in. a
10
10 P 5 P
Cx = C y = ⋅ =
a
a 2
a
Free body: Member CDE
ΣM E = 0: C x (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0
5P
P
(6) − (20) = 500
a
2
30
!
P " − 10 # = 500
$ a
%
For
(1)
a = 0.5 in.
30
!
− 10 # = 500
P"
$ 0.5
%
50 P = 500
P = 10.00 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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951
PROBLEM 6.151
The garden shears shown consist of two blades and two handles. The two handles are connected by pin C and
the two blades are connected by pin D. The left blade and the right handle are connected by pin A; the right
blade and the left handle are connected by pin B. Determine the magnitude of the forces exerted on the small
branch at E when two 80-N forces are applied to the handles as shown.
SOLUTION
By symmetry vertical components Cy, Dy, Ey are 0. Then by considering ΣFy = 0 on the blades or handles, we
find that Ay and By are 0.
Thus forces at A, B, C, D, and E are horizontal.
Free body: Right handle
ΣM C = 0: A(30 mm) − (80 N)(270 mm) = 0
A = +720 N
ΣFx = 0: C − 720 N − 80 N = 0
C = +800 N
Free body: Left blade
ΣM D = 0: E (180 mm) − (720 N)(60 mm) = 0
E = 240 N !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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952
PROBLEM 6.152
The telescoping arm ABC is used to provide an elevated platform for
construction workers. The workers and the platform together have a
mass of 200 kg and have a combined center of gravity located
directly above C. For the position when θ = 20°, determine (a) the
force exerted at B by the single hydraulic cylinder BD, (b) the force
exerted on the supporting carriage at A.
SOLUTION
a = (5 m) cos 20° = 4.6985 m
Geometry:
b = (2.4 m) cos 20° = 2.2553 m
c = (2.4 m)sin 20° = 0.8208 m
d = b − 0.5 = 1.7553 m
e = c + 0.9 = 1.7208 m
tan β =
e 1.7208
; β = 44.43°
=
d 1.7553
Free body: Arm ABC
We note that BD is a two-force member
W = (200 kg)(9.81 m/s 2 ) = 1.962 kN
(a)
ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 44.43°(2.2553 m) + FBD cos 44.43(0.8208 m) = 0
9.2185 − FBD (0.9927) = 0: FBD = 9.2867 kN
FBD = 9.29 kN
(b)
44.4° !
ΣFx = 0: Ax − FBD cos β = 0
Ax = (9.2867 kN) cos 44.43° = 6.632 kN
A x = 6.632 kN
ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0
Ay = 1.962 kN − (9.2867 kN)sin 44.43° = −4.539 kN
A y = 4.539 kN
A = 8.04 kN
34.4° !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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953
PROBLEM 6.153
The telescoping arm ABC can be lowered until end C is close
to the ground, so that workers can easily board the platform.
For the position when θ = −20°, determine (a) the force exerted
at B by the single hydraulic cylinder BD, (b) the force exerted
on the supporting carriage at A.
SOLUTION
Geometry:
a = (5 m) cos 20° = 4.6985 m
b = (2.4 m) cos 20° = 2.2552 m
c = (2.4 m)sin 20° = 0.8208 m
d = b − 0.5 = 1.7553 m
e = 0.9 − c = 0.0792 m
tan β =
e 0.0792
; β = 2.584°
=
d 1.7552
Free body: Arm ABC
We note that BD is a two-force member
W = (200 kg)(9.81 m/s 2 )
W = 1962 N = 1.962 kN
(a)
ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 2.584°(2.2553 m) − FBD cos 2.584°(0.8208 m) = 0
9.2185 − FBD (0.9216) = 0 FBD = 10.003 kN
FBD = 10.00 kN
(b)
2.58° !
ΣFx = 0: Ax − FBD cos β = 0
Ax = (10.003 kN) cos 2.583° = 9.993 kN
A x = 9.993 kN
ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0
Ay = 1.962 kN − (10.003 kN)sin 2.583° = −1.5112 kN
A y = 1.5112 kN
A = 10.11 kN
8.60° !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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954
PROBLEM 6.154
The position of member ABC is controlled by the hydraulic
cylinder CD. Knowing that θ = 30°, determine for the loading
shown (a) the force exerted by the hydraulic cylinder on pin C,
(b) the reaction at B.
SOLUTION
Geometry: In DBCD
(CD ) 2 = (0.5)2 + (1.5) 2 − 2(0.5)(1.5) cos 60°
CD = 1.3229 m
Law of cosines
sin β
sin 60°
=
0.5 m 1.3229 m
sin β = 0.3273
β = 19.107°
Law of sines
Free body: Entire system
Move force FCD along its line of action so it acts at D.
(a)
ΣM B = 0: (10 kN)(0.69282 m) − FCD sin β (1.5 m) = 0
6.9282 kN ⋅ m − FCD sin 19.107°(1.5 m) = 0
FCD = 14.111 kN
FCD = 14.11 kN
(b)
19.11° !
ΣFx = 0: Bx + FCD cos β = 0
Bx + (14.111 kN) cos 19.107° = 0
B x = 13.333 kN
Bx = −13.333 kN
ΣFy = 0: By − 10 kN − FCD sin 19.107° = 0
By − 10 kN − (14.111 kN)sin 19.107° = 0
By = +14.619 kN
B y = 14.619 kN
B = 19.79 kN
47.6° !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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955
PROBLEM 6.155
The motion of the bucket of the front-end loader shown is
controlled by two arms and a linkage that are pin-connected
at D. The arms are located symmetrically with respect to
the central, vertical, and longitudinal plane of the loader;
one arm AFJ and its control cylinder EF are shown. The
single linkage GHDB and its control cylinder BC are located
in the plane of symmetry. For the position and loading
shown, determine the force exerted (a) by cylinder BC,
(b) by cylinder EF.
SOLUTION
Free body: Bucket
ΣM J = 0: (4500 lb)(20 in.) − FGH (22 in.) = 0
FGH = 4091 lb
Free body: Arm BDH
ΣM D = 0: − (4091 lb)(24 in.) − FBC (20 in.) = 0
FBC = −4909 lb
FBC = 4.91 kips C !
Free body: Entire mechanism
(Two arms and cylinders AFJE)
Note: Two arms thus 2 FEF
18 in.
65 in.
β = 15.48°
tan β =
ΣM A = 0: (4500 lb)(123 in.) + FBC (12 in.) + 2 FEF cos β (24 in.) = 0
(4500 lb)(123 in.) − (4909 lb)(12 in.) + 2 FEF cos 15.48°(24 in.) = 0
FEF = −10.690 lb
FEF = 10.69 kips C !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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956
PROBLEM 6.156
The bucket of the front-end loader shown carries a 3200-lb
load. The motion of the bucket is controlled by two
identical mechanisms, only one of which is shown.
Knowing that the mechanism shown supports one-half
of the 3200-lb load, determine the force exerted (a) by
cylinder CD, (b) by cylinder FH.
SOLUTION
Free body: Bucket (One mechanism)
ΣM D = 0: (1600 lb)(15 in.) − FAB (16 in.) = 0
FAB = 1500 lb
Note: There are 2 identical support mechanisms.
Free body: One arm BCE
8
20
β = 21.8°
tan β =
ΣM E = 0: (1500 lb)(23 in.) + FCD cos 21.8°(15 in.) − FCD sin 21.8°(5 in.) = 0
FCD = −2858 lb
FCD = 2.86 kips C !
Free body: Arm DFG
ΣM G = 0: (1600 lb)(75 in.) + FFH sin 45°(24 in.) − FFH cos 45°(6 in.) = 0
FFH = −9.428 kips
FFH = 9.43 kips C !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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957
PROBLEM 6.157
The motion of the backhoe bucket shown is
controlled by the hydraulic cylinders AD, CG, and
EF. As a result of an attempt to dislodge a portion
of a slab, a 2-kip force P is exerted on the bucket
teeth at J. Knowing that θ = 45°, determine the
force exerted by each cylinder.
SOLUTION
Free body: Bucket
ΣM H = 0
(Dimensions in inches)
4
3
FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0
5
5
FCG = −
P
(16 cos θ + 8 sin θ )
14
(1)
Free body: Arm ABH and bucket
(Dimensions in inches)
ΣM B = 0:
4
3
FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0
5
5
FAD = −
P
(86 cos θ − 42 sin θ )
15.6
(2)
Free body: Bucket and arms IEB + ABH
Geometry of cylinder EF
16 in.
40 in.
β = 21.801°
tan β =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
958
PROBLEM 6.157 (Continued)
ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0
FEF =
=
P(120 sin θ − 28 cos θ )
cos 21.8°(18)
P
(120 sin θ − 28 cos θ )
16.7126
(3)
For P = 2 kips, θ = 45°
Eq. (1):
FCG = −
2
(16 cos 45° + 8 sin 45°) = −2.42 kips
14
FCG = 2.42 kips C !
Eq. (2):
FAD = −
2
(86 cos 45° − 42 sin 45°) = −3.99 kips
15.6
FAD = 3.99 kips C !
Eq. (3):
FEF =
2
(120 sin 45° − 28 cos 45°) = +7.79 kips
16.7126
FEF = 7.79 kips T !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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959
PROBLEM 6.158
Solve Problem 6.157 assuming that the 2-kip force P
acts horizontally to the right (θ = 0).
PROBLEM 6.157 The motion of the backhoe bucket
shown is controlled by the hydraulic cylinders AD,
CG, and EF. As a result of an attempt to dislodge a
portion of a slab, a 2-kip force P is exerted on the
bucket teeth at J. Knowing that θ = 45°, determine the
force exerted by each cylinder.
SOLUTION
Free body: Bucket
ΣM H = 0
(Dimensions in inches)
4
3
FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0
5
5
FCG = −
P
(16 cos θ + 8 sin θ )
14
(1)
Free body: Arm ABH and bucket
(Dimensions in inches)
ΣM B = 0:
4
3
FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0
5
5
FAD = −
P
(86 cos θ − 42 sin θ )
15.6
(2)
Free body: Bucket and arms IEB + ABH
Geometry of cylinder EF
16 in.
40 in.
β = 21.801°
tan β =
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960
PROBLEM 6.158 (Continued)
ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0
FEF =
=
P(120 sin θ − 28 cos θ )
cos 21.8°(18)
P
(120 sin θ − 28 cos θ )
16.7126
(3)
For P = 2 kips, θ = 0
Eq. (1):
FCG = −
2
(16 cos 0 + 8 sin 0) = −2.29 kips
14
Eq. (2):
FAD = −
2
(86 cos 0 − 42 sin 0) = −11.03 kips
15.6
FAD = 11.03 kips C !
Eq. (3):
FEF =
2
(120 sin 0 − 28 cos 0) = −3.35 kips
16.7126
FEF = 3.35 kips C !
FCG = 2.29 kips C !
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961
PROBLEM 6.159
The gears D and G are rigidly attached to shafts that
are held by frictionless bearings. If rD = 90 mm and
rG = 30 mm, determine (a) the couple M0 that must
be applied for equilibrium, (b) the reactions at A and B.
SOLUTION
(a)
Projections on yz plane
Free body: Gear G
ΣM G = 0: 30 N ⋅ m − J (0.03 m) = 0; J = 1000 N
Free body: Gear D
ΣM D = 0: M 0 − (1000 N)(0.09 m) = 0
M 0 = 90 N ⋅ m
(b)
M 0 = (90.0 N ⋅ m)i !
Gear G and axle FH
ΣM F = 0: H (0.3 m) − (1000 N)(0.18 m) = 0
H = 600 N
ΣFy = 0: F + 600 − 1000 = 0
F = 400 N
Gear D and axle CE
ΣM C = 0: (1000 N)(0.18 m) − E (0.3 m) = 0
E = 600 N
ΣFy = 0: 1000 − C − 600 = 0
C = 400 N
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962
PROBLEM 6.159 (Continued)
Free body: Bracket AE
ΣFy = 0: A − 400 + 400 = 0
A=0 !
ΣM A = 0: M A + (400 N)(0.32 m) − (400 N)(0.2 m) = 0
M A = −48 N ⋅ m
M A = −(48.0 N ⋅ m)i !
Free body: Bracket BH
ΣFy = 0: B − 600 + 600 = 0
B=0 !
ΣM B = 0: M B + (600 N)(0.32 m) − (600 N)(0.2 m) = 0
M B = −72 N ⋅ m
M B = −(72.0 N ⋅ m)i !
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963
PROBLEM 6.160
In the planetary gear system shown, the radius of the central gear A is a = 18 mm,
the radius of each planetary gear is b, and the radius of the outer gear E is (a + 2b).
A clockwise couple of magnitude MA = 10 N ⋅ m is applied to the central gear
A and a counterclockwise couple of magnitude MS = 50 N ⋅ m is applied to the
spider BCD. If the system is to be in equilibrium, determine (a) the required
radius b of the planetary gears, (b) the magnitude ME of the couple that must
be applied to the outer gear E.
SOLUTION
FBD Central Gear:
By symmetry:
F1 = F2 = F3 = F
ΣM A = 0: 3(rA F ) − 10 N ⋅ m = 0,
ΣM C = 0: rB ( F − F4 ) = 0,
FBD Gear C:
F=
10
N⋅m
3rA
F4 = F
ΣFx′ = 0: Cx′ = 0
ΣFy ′ = 0: C y′ − 2 F = 0,
C y′ = 2 F
Gears B and D are analogous, each having a central force of 2F
ΣM A = 0: 50 N ⋅ m − 3(rA + rB )2 F = 0
FBD Spider:
50 N ⋅ m − 3(rA + rB )
20
N⋅m = 0
rA
rA + rB
r
= 2.5 = 1 + B ,
rA
rA
rB = 1.5rA
Since rA = 18 mm,
FBD Outer Gear:
rB = 27.0 mm !
(a)
ΣM A = 0: 3(rA + 2rB ) F − M E = 0
3(18 mm + 54 mm)
10 N ⋅ m
− ME = 0
54 mm
M E = 40.0 N ⋅ m
(b)
!
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964
PROBLEM 6.161*
Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings
at B and D do not exert any axial force. A couple of magnitude 500 lb ⋅ in. (clockwise when viewed from the
positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is
horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain
equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be
zero.)
SOLUTION
We recall from Figure 4.10, that a universal joint exerts on members it connects a force of unknown direction
and a couple about an axis perpendicular to the crosspiece.
Free body: Shaft DF
ΣM x = 0: M C cos 30° − 500 lb ⋅ in. = 0
M C = 577.35 lb ⋅ in.
Free body: Shaft BC
We use here x′, y ′, z with x′ along BC
ΣM C = 0: − M R i ′ − (577.35 lb ⋅ in)i ′ + (−5 in)i ′ × ( By j ′ + Bz k ) = 0
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965
PROBLEM 6.161* (Continued)
Equate coefficients of unit vectors to zero:
i:
M A − 577.35 lb ⋅ in = 0
M A = 577.35 lb ⋅ in.
j:
Bz = 0
k:
By = 0
M A = 577 lb ⋅ in. !
B=0
ΣF = 0:
B + C = 0,
B=0
since B = 0,
C=0
Return to free body of shaft DF
ΣM D = 0
(Note that C = 0 and M C = 577.35 lb ⋅ in. )
(577.35 lb ⋅ in.)(cos 30°i + sin 30° j) − (500 lb ⋅ in.)i
+ (6 in.)i × ( Ex i + E y j + Ez k ) = 0
(500 lb ⋅ in.)i + (288.68 lb ⋅ in.) j − (500 lb ⋅ in.)i
+ (6 in.) E y k − (6 in.) Ez j = 0
Equate coefficients of unit vectors to zero:
j:
288.68 lb ⋅ in. − (6 in.)Ez = 0 Ez = 48.1 lb
k:
Ey = 0
ΣF = 0: C + D + E = 0
0 + Dy j + Dz k + Ex i + (48.1 lb)k = 0
i:
Ex = 0
j:
Dy = 0
k:
Dz + 48.1 lb = 0
Dz = −48.1 lb
B=0 !
Reactions are:
D = −(48.1 lb)k !
E = (48.1 lb)k !
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966
PROBLEM 6.162*
Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical.
PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal
joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb · in. (clockwise
when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece
attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC
at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the
crosspiece must be zero.)
SOLUTION
Free body: Shaft DF
ΣM x = 0: M C − 500 lb ⋅ in. = 0
M C = 500 lb ⋅ in.
Free body: Shaft BC
We resolve −(520 lb ⋅ in.)i into components along x′ and y ′ axes:
−M C = −(500 lb ⋅ in.)(cos 30°i ′ + sin 30° j′)
ΣM C = 0: M Ai ′ − (500 lb ⋅ in.)(cos 30°i ′ + sin 30° j′) + (5 in.)i ′ × ( By ′ j′ + Bz k ) = 0
M A i ′ − (433 lb ⋅ in.)i ′ − (250 lb ⋅ in.) j + (5 in.) By′k − (5 in.) Bz j′ = 0
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967
PROBLEM 6.162* (Continued)
Equate to zero coefficients of unit vectors:
i ′: M A − 433 lb ⋅ in. = 0
M A = 433 lb ⋅ in. !
j′: − 250 lb ⋅ in. − (5 in.) Bz = 0
Bz = −50 lb
k : B y′ = 0
B = −(50 lb)k
Reactions at B:
ΣF = 0: B − C = 0
−(50 lb)k − C = 0
C = −(50 lb)k
Return to free body of shaft DF:
ΣM D = 0: (6 in.)i × ( Ex i + E y j + Ez k ) − (4 in.)i × (−50 lb)k
− (500 lb ⋅ in.)i + (500 lb ⋅ in.)i = 0
(6 in.) E y k − (6 in.) Ez j − (200 lb ⋅ in.) j = 0
k : Ey = 0
j: −(6 in.) Ez − 200 lb ⋅ in. = 0
Ez = −33.3 lb
ΣF = 0: C + D + E = 0
−(50 lb)k + D y j + Dz k + Ex i − (33.3 lb)k = 0
i : Ex = 0
k : −50 lb − 33.3 lb + Dz = 0
Dz = 83.3 lb
B = −(50 lb)k !
Reactions are:
D = (83.3 lb)k !
E = −(33.3 lb)k !
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968
PROBLEM 6.163*
The large mechanical tongs shown are used to grab and lift a thick 7500-kg
steel slab HJ. Knowing that slipping does not occur between the tong grips
and the slab at H and J, determine the components of all forces acting on
member EFH. (Hint: Consider the symmetry of the tongs to establish
relationships between the components of the force acting at E on EFH and
the components of the force acting at D on CDF.)
SOLUTION
Free body: Pin A
T = W = mg = (7500 kg)(9.81 m/s 2 ) = 73.575 kN
ΣFx = 0: ( FAB ) x = ( FAC ) x
1
ΣFy = 0: ( FAB ) y = ( FAC ) y = W
2
Also:
( FAC ) x = 2( FAC ) y = W
Free body: Member CDF
1
ΣM D = 0: W (0.3) + W (2.3) − Fx (1.8) − Fy (0.5 m) = 0
2
or
1.8 Fx + 0.5Fy = 1.45W
(1)
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969
PROBLEM 6.163* (Continued)
ΣFx = 0: Dx − Fx − W = 0
Ex − Fx = W
or
(2)
1
ΣFy = 0: Fy − Dy + W = 0
2
1
E y − Fy = W
2
or
(3)
Free body: Member EFH
1
ΣM E = 0: Fx (1.8) + Fy (1.5) − H x (2.3) + W (1.8 m) = 0
2
1.8 Fx + 1.5Fy = 2.3H x − 0.9W
or:
(4)
ΣFx = 0: Ex + Fx − H x = 0
Ex + Fx = H x
or
2 Fx = H x − W
Subtract (2) from (5):
Subtract (4) from 3 × (1):
3.6 Fx = 5.25W − 2.3H x
Add (7) to 2.3 × (6):
8.2 Fx = 2.95W
Fx = 0.35976W
(5)
(6)
(7)
(8)
Substitute from (8) into (1):
(1.8)(0.35976W ) + 0.5Fy = 1.45W
0.5Fy = 1.45W − 0.64756W = 0.80244W
Fy = 1.6049W
(9)
Substitute from (8) into (2):
Ex − 0.35976W = W ; Ex = 1.35976W
Substitute from (9) into (3):
1
E y − 1.6049W = W
2
From (5):
H x = Ex + Fx = 1.35976W + 0.35976W = 1.71952W
Recall that:
1
Hy = W
2
E y = 2.1049W
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970
PROBLEM 6.163* (Continued)
Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams.
Substitute
W = 73.575 kN
E x = 100.0 kN
E y = 154.9 kN !
Fx = 26.5 kN
Fy = 118.1 kN !
H x = 126.5 kN
H y = 36.8 kN !
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971
PROBLEM 6.164
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss.
C = D = 600 lb
From the symmetry of the truss and loading, we find
Free body: Joint B
FAB
5
=
FBC 300 lb
=
2
1
FAB = 671 lb T
FBC = 600 lb C !
Free body: Joint C
ΣFy = 0:
3
FAC + 600 lb = 0
5
FAC = −1000 lb
ΣFx = 0:
4
( −1000 lb) + 600 lb + FCD = 0
5
FAC = 1000 lb C !
FCD = 200 lb T !
From symmetry:
FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C !
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972
PROBLEM 6.165
Using the method of joints, determine the force in each
member of the double-pitch roof truss shown. State whether
each member is in tension or compression.
SOLUTION
Free body: Truss
ΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m)
− (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0
H = 4.50 kN
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + H − 9 = 0
Ay = 9 − 4.50,
Ay = 4.50 kN
Free body: Joint A
FAB
5
FAB
FAC 3.50 kN
=
2
1
= 7.8262 kN C
=
FAB = 7.83 kN C !
FAC = 7.00 kN T !
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973
PROBLEM 6.165 (Continued)
Free body: Joint B
ΣFx = 0:
2
5
FBD +
2
5
(7.8262 kN) +
1
2
FBC = 0
FBD + 0.79057 FBC = −7.8262 kN
or
ΣFy = 0:
1
5
FBD +
1
5
(7.8262 kN) −
1
2
(1)
FBC − 2 kN = 0
FBD − 1.58114 FBC = −3.3541
or
(2)
Multiply (1) by 2 and add (2):
3FBD = −19.0065
FBD = 6.34 kN C !
FBD = −6.3355 kN
Subtract (2) from (1):
2.37111FBC = −4.4721
FBC = −1.8861 kN
FBC = 1.886 kN C !
Free body: Joint C
2
ΣFy = 0:
FCD −
5
1
2
(1.8861 kN) = 0
FCD = +1.4911 kN
ΣFx = 0: FCE − 7.00 kN +
1
2
FCD = 1.491 kN T !
(1.8861 kN) +
1
5
(1.4911 kN) = 0
FCE = 5.000 kN
FCE = 5.00 kN T !
Free body: Joint D
ΣFx = 0:
2
5
FDF +
FDF
or
ΣFy = 0:
or
Add (1) and (2):
1
5
1
FDE +
2
(6.3355 kN) −
2
5
+ 0.79057 FDE = −5.5900 kN
FDF −
1
2
FDE +
1
5
(6.3355 kN) −
5
(1.4911 kN) = 0
(1)
2
5
(1.4911 kN) − 2 kN = 0
FDF − 0.79057 FDE = −1.1188 kN (2)
2 FDF = −6.7088 kN
FDF = −3.3544 kN
Subtract (2) from (1):
1
FDF = 3.35 kN C !
1.58114 FDE = −4.4712 kN
FDE = −2.8278 kN
FDE = 2.83 kN C !
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974
PROBLEM 6.165 (Continued)
Free body: Joint F
ΣFx = 0:
1
FFG +
2
2
(3.3544 kN) = 0
5
FFG = −4.243 kN
ΣFy = 0: −FEF − 1.75 kN +
1
5
FFG = 4.24 kN C !
(3.3544 kN) −
1
2
FEF = 2.750 kN
(−4.243 kN) = 0
FEF = 2.75 kN T !
Free body: Joint G
ΣFx = 0:
1
FGH −
2
ΣFy = 0: −
1
2
1
2
(4.243 kN) = 0
FGH −
1
2
FEG −
1
2
(1)
(4.243 kN) − 1.5 kN = 0
FGH + FEG = −6.364 kN
or:
(2)
2 FGH = −10.607
FGH = −5.303
Subtract (1) from (2):
2
FEG +
FGH − FEG = −4.243 kN
or:
Add (1) and (2):
1
FGH = 5.30 kN C !
2 FEG = −2.121 kN
FEG = −1.0605 kN
FEG = 1.061 kN C !
Free body: Joint H
FEH 3.75 kN
=
1
1
We can also write:
FGH
2
=
FEH = 3.75 kN T !
3.75 kN
1
FGH = 5.30 kN C (Checks)
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975
PROBLEM 6.166
The truss shown was designed to support the roof of
a food market. For the given loading, determine the
force in members FG, EG, and EH.
SOLUTION
Reactions at supports. Because of the symmetry of the loading
Ax = 0,
Ay = O =
1
(Total load)
2
A = O = 4.48 kN !
We pass a section through members FG, EG, and EH, and use the Free body shown.
Slope FG = Slope FI =
Slope EG =
1.75 m
6m
5.50 m
2.4 m
ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m)
− (4.48 kN)(7.44 m)
6
!
FFG # (4.80 m) = 0
−"
6.25
$
%
FFG = −5.231 kN
FFG = 5.23 kN C !
ΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m)
+ (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m)
−(4.48 kN)(9.84 m) = 0
ΣFy = 0:
FEH = 5.08 kN T !
5.50
1.75
FEG +
(−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0
6.001
6.25
FEG = −0.1476 kN
FEG = 0.1476 kN C !
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976
PROBLEM 6.167
The truss shown was designed to support the roof of
a food market. For the given loading, determine the
force in members KM, LM, and LN.
SOLUTION
Because of symmetry of loading,
O=
1
(Load)
2
O = 4.48 kN
We pass a section through KM, LM, LN, and use free body shown
3.84
!
ΣM M = 0: "
FLN # (3.68 m)
4
$
%
+ (4.48 kN − 0.6 kN)(3.6 m) = 0
FLN = −3.954 kN
FLN = 3.95 kN C !
ΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m)
+ (4.48 kN − 0.6 kN)(7.44 m) = 0
FKM = +5.022 kN
+ΣFy = 0:
FKM = 5.02 kN T !
4.80
1.12
(−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0
FLM +
6.147
4
FLM = −1.963 kN
FLM = 1.963 kN C !"
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
977
PROBLEM 6.168
For the frame and loading shown, determine the components of all
forces acting on member ABC.
SOLUTION
Free body: Entire frame
ΣM E = 0: − Ax (4) − (20 kips)(5) = 0
Ax = −25 kips,
A x = 25.0 kips
!
ΣFy = 0: Ay − 20 kips = 0
Ay = 20 kips
A y = 20.0 kips !
Free body: Member ABC
Note: BE is a two-force member, thus B is directed along line BE and By =
2
Bx
5
ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0
−100 kip ⋅ ft + Bx (2 ft) +
2
Bx (5 ft) = 0
5
Bx = 25 kips
By =
B x = 25.0 kips
2
2
( Bx ) = (25) = 10 kips
5
5
!
B y = 10.00 kips !
ΣFx = 0: C x − 25 kips − 25 kips = 0
Cx = 50 kips
C x = 50.0 kips
!
ΣFy = 0: C y + 20 kips − 10 kips = 0
C y = −10 kips
C y = 10.00 kips !"
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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978
PROBLEM 6.169
Solve Problem 6.168 assuming that the 20-kip load is replaced by a
clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC
at Point D.
PROBLEM 6.168 For the frame and loading shown, determine the
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame
ΣFy = 0: Ay = 0
ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0
Ax = −25 kips
A x = 25.0 kips
A = 25.0 kips
!
B x = 25.0 kips
!
Free Body: Member ABC
Note: BE is a two-force member, thus B is directed along line BE and By =
2
Bx
5
ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0
100 kip ⋅ ft + Bx (2 ft) +
2
Bx (5 ft) = 0
5
Bx = −25 kips
By =
2
2
Bx = (−25) = −10 kips;
5
5
ΣFx = 0: − 25 kips + 25 kips + Cx = 0
B y = 10.00 kips !
Cx = 0
ΣFy = 0: +10 kips + C y = 0
C y = −10 kips
C y = 10 kips
C = 10.00 kips !"
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
979
PROBLEM 6.170
Knowing that the pulley has a radius of 0.5 m, determine the
components of the reactions at A and E.
SOLUTION
FBD Frame:
ΣM A = 0: (7 m) E y − (4.5 m)(700 N) = 0
ΣFy = 0: Ay − 700 N + 450 N = 0
ΣFx = 0: Ax − Ex = 0
E y = 450 N !
A y = 250 N !
Ax = Ex
Dimensions in m
FBD Member ABC:
ΣM C = 0: (1 m)(700 N) − (1 m)(250 N) − (3 m) Ax = 0
A x = 150.0 N
so E x = 150.0 N
!
!"
"
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
980
PROBLEM 6.171
For the frame and loading shown, determine the reactions at A, B,
D, and E. Assume that the surface at each support is frictionless.
SOLUTION
Free body: Entire frame
ΣFx = 0: A − B + (1000 lb) sin 30° = 0
A − B + 500 = 0
(1)
ΣFy = 0: D + E − (1000 lb) cos 30° = 0
D + E − 866.03 = 0
(2)
Free body: Member ACE
ΣM C = 0: − A(6 in.) + E (8 in.) = 0
E=
3
A
4
(3)
Free body: Member BCD
ΣM C = 0: − D(8 in.) + B(6 in.) = 0
D=
3
B
4
(4)
Substitute E and D from (3) and (4) into (2):
−
3
3
A + B − 866.06 = 0
4
4
A + B − 1154.71 = 0
(5)
(1)
A − B + 500 = 0
(5) + (6)
2 A − 654.71 = 0
A = 327.4 lb
A = 327 lb
!
(5) − (6)
2 B − 1654.71 = 0
B = 827.4 lb
B = 827 lb
!
(4)
D=
3
(827.4)
4
D = 620.5 lb
D = 621 lb !
(3)
E=
3
(327.4)
4
E = 245.5 lb
E = 246 lb !
(6)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
981
PROBLEM 6.172
For the system and loading shown, determine (a) the force P required
for equilibrium, (b) the corresponding force in member BD, (c) the
corresponding reaction at C.
SOLUTION
Member FBDs:
FBD I:
I:
ΣM C = 0: R( FBD sin 30°) − [ R(1 − cos 30°)](100 N) − R(50 N) = 0
FBD = 126.795 N
ΣFx = 0: −Cx + (126.795 N) cos 30° = 0
(b) FBD = 126.8 N T !"
C x = 109.808 N
ΣFy = 0: C y + (126.795 N) sin 30° − 100 N − 50 N = 0
C y = 86.603 N
II:
(c) so C = 139.8 N
38.3° !"
FBD II:
ΣM A = 0: aP − a[(126.795 N) cos 30°] = 0
(a ) P = 109.8 N
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
982
PROBLEM 6.173
A small barrel weighing 60 lb is lifted by a pair of tongs as
shown. Knowing that a = 5 in., determine the forces exerted at B
and D on tong ABD.
SOLUTION
We note that BC is a two-force member.
Free body: Tong ABD
Bx By
=
15
5
Bx = 3By
ΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0
By = 30 lb
Bx = 3By : Bx = 90 lb
ΣFx = 0: −90 lb + Dx = 0
ΣFy = 0: 60 lb − 30 lb − D y = 0
D x = 90 lb
D y = 30 lb
B = 94.9 lb
18.43° !"
D = 94.9 lb
18.43° !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
983
PROBLEM 6.174
A 20-kg shelf is held horizontally by a self-locking brace that consists
of two Parts EDC and CDB hinged at C and bearing against each other
at D. Determine the force P required to release the brace.
SOLUTION
Free body: Shelf
W = (20 kg)(9.81 m/s 2 ) = 196.2 N
ΣM A = 0: By (200 mm) − (196.2 N)(125 mm) = 0
By = 122.63 N
Free body: Portion ACB
ΣM A = 0: − Bx (150 mm) − P(150 mm) − (122.63 N)(200 mm) = 0
Bx = −163.5 − P
(1)
ΣM C = 0: + (122.63 N)(92 mm) + Bx (94 mm) = 0
+ (122.63 N)(92 mm) + (−163.5 − P)(44 mm) = 0
P = 92.9 N
P = 92.9 N
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
984
PROBLEM 6.175
The specialized plumbing wrench shown is used in confined areas (e.g., under a
basin or sink). It consists essentially of a jaw BC pinned at B to a long rod.
Knowing that the forces exerted on the nut are equivalent to a clockwise (when
viewed from above) couple of magnitude 135 lb ⋅ in., determine (a) the magnitude
of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the
wrench.
SOLUTION
Free body: Jaw BC
This is a two-force member
Cy
1.5 in.
Free body: Nut
=
Cx
C y = 2.4 C x
in.
5
8
ΣFx = 0: Bx = C x
(1)
ΣFy = 0: By = C y = 2.4 Cx
(2)
ΣFx = 0: Cx = Dx
ΣM = 135 lb ⋅ in.
Cx (1.125 in.) = 135 lb ⋅ in.
Cx = 120 lb
(a)
Eq. (1):
Bx = Cx = 120 lb
Eq. (2):
By = C y = 2.4(120 lb) = 288 lb
(
B = Bx2 + By2
(b)
1/ 2
)
= (1202 + 2882 )1/ 2
B = 312 lb !"
Free body: Rod
ΣM D = 0: − M 0 + By (0.625 in.) − Bx (0.375 in.) = 0
− M 0 + (288)(0.625) − (120)(0.375) = 0
M 0 = 135.0 lb ⋅ in.
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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985
CHAPTER 7
PROBLEM 7.1
Determine the internal forces (axial force, shearing force,
and bending moment) at Point J of the structure indicated.
Frame and loading of Problem 6.79.
SOLUTION
From Problem 6.79: On JD
D = 80 lb
FBD of JD:
ΣFy = 0: V − 80 lb = 0
ΣFx = 0: F = 0
V = 80.0 lb
F =0
ΣM J = 0: M − (80 lb)(6 in.) = 0
M = 480 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
989
PROBLEM 7.2
Determine the internal forces (axial force, shearing force, and
bending moment) at Point J of the structure indicated.
Frame and loading of Problem 6.80.
SOLUTION
From Problem 6.80: On JD
D = 40 lb
FBD of JD:
ΣFy = 0: V − 40 lb = 0
ΣFx = 0: F = 0
V = 40.0 lb
F =0
ΣM J = 0: M − (40 lb)(6 in.) = 0
M = 240 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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990
PROBLEM 7.3
Determine the internal forces at Point J when α = 90°.
SOLUTION
Free body: Entire bracket
ΣM D = 0: (1.4 kN)(0.6 m) − A(0.175 m) = 0
A = + 4.8 kN A = 4.8 kN
ΣFx = 0: Dx − 4.8 = 0
D x = 4.8 kN
ΣFy = 0: D y − 1.4 = 0
D y = 1.4 kN
Free body: JCD
ΣFx = 0: 4.8 kN − F = 0
F = 4.80 kN
ΣFy = 0: 1.4 kN − V = 0
V = 1.400 kN
ΣM J = 0: (4.8 kN)(0.2 m) + (1.4 kN)(0.3 m) − M = 0
M = + 1.38 kN ⋅ m
M = 1.380 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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991
PROBLEM 7.4
Determine the internal forces at Point J when α = 0.
SOLUTION
Free body: Entire bracket
ΣFy = 0: D y = 0
Dy = 0
ΣM A = 0: (1.4 kN)(0.375 m) − Dx (0.175 m) = 0
Dx = + 3 kN
D x = 3 kN
Free body: JCD
ΣFx = 0: 3 kN − F = 0
F = 3.00 kN
ΣFy = 0: − V = 0
V =0
ΣM J = 0: (3 kN)(0.2 m) − M = 0
M = + 0.6 kN ⋅ m
M = 0.600 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
992
PROBLEM 7.5
Determine the internal forces at Point J of the structure shown.
SOLUTION
FBD ABC:
ΣM D = 0: (0.375 m)(400 N) − (0.24 m)C y = 0
C y = 625 N
ΣM B = 0: − (0.45 m)C x + (0.135 m)(400 N) = 0
C x = 120 N
FBD CJ:
ΣFy = 0: 625 N − F = 0
ΣFx = 0: 120 N − V = 0
F = 625 N
V = 120.0 N
!
!
ΣM J = 0: M − (0.225 m)(120 N) = 0
M = 27.0 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
993
PROBLEM 7.6
Determine the internal forces at Point K of the structure shown.
SOLUTION
FBD AK:
ΣFx = 0: V = 0
V=0
ΣFy = 0: F − 400 N = 0
F = 400 N
ΣM K = 0: (0.135 m)(400 N) − M = 0
M = 54.0 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
994
PROBLEM 7.7
A semicircular rod is loaded as shown. Determine the internal forces at Point J.
SOLUTION
FBD Rod:
ΣM B = 0: Ax (2r ) = 0
Ax = 0
ΣFx′ = 0: V − (120 N) cos 60° = 0
V = 60.0 N
!
FBD AJ:
ΣFy′ = 0: F + (120 N)sin 60° = 0
F = −103.923 N
F = 103.9 N
ΣM J = 0: M − [(0.180 m)sin 60°](120 N) = 0
M = 18.7061
M = 18.71
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
995
!
PROBLEM 7.8
A semicircular rod is loaded as shown. Determine the internal forces at
Point K.
SOLUTION
FBD Rod:
ΣFy = 0: B y − 120 N = 0 B y = 120 N
ΣM A = 0: 2rBx = 0 B x = 0
ΣFx′ = 0: V − (120 N) cos 30° = 0
V = 103.923 N
V = 103.9 N
FBD BK:
ΣFy′ = 0: F + (120 N)sin 30° = 0
F = − 60 N
F = 60.0 N
ΣM K = 0: M − [(0.180 m)sin 30°](120 N) = 0
M = 10.80 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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996
PROBLEM 7.9
An archer aiming at a target is pulling with a 45-lb force on the bowstring.
Assuming that the shape of the bow can be approximated by a parabola,
determine the internal forces at Point J.
SOLUTION
FBD Point A:
By symmetry
T1 = T2
3 !
ΣFx = 0: 2 " T1 # − 45 lb = 0 T1 = T2 = 37.5 lb
$5 %
Curve CJB is parabolic: x = ay 2
FBD BJ:
At B : x = 8 in.
y = 32 in.
8 in.
1
a=
=
(32 in.) 2 128 in.
x=
y2
128
Slope of parabola = tan θ =
At J:
So
dx 2 y
y
=
=
dy 128 64
& 16 '
) = 14.036°
* 64 +
θ J = tan −1 (
α = tan −1
4
− 14.036° = 39.094°
3
ΣFx′ = 0: V − (37.5 lb) cos(39.094°) = 0
V = 29.1 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
997
!
PROBLEM 7.9 (Continued)
ΣFy ′ = 0: F + (37.5 lb) sin (39.094°) = 0
F = − 23.647
F = 23.6 lb
&3
'
&4
'
ΣMJ = 0: M + (16 in.) ( (37.5 lb) ) + [(8 − 2) in.] ( (37.5 lb) ) = 0
5
5
*
+
*
+
M = 540 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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998
!
PROBLEM 7.10
For the bow of Problem 7.9, determine the magnitude and location
of the maximum (a) axial force, (b) shearing force, (c) bending
moment.
PROBLEM 7.9 An archer aiming at a target is pulling with a 45-lb
force on the bowstring. Assuming that the shape of the bow can be
approximated by a parabola, determine the internal forces at Point J.
SOLUTION
Free body: Point A
3 !
ΣFx = 0: 2 " T # − 45 lb = 0
$5 %
T = 37.5 lb v
Free body: Portion of bow BC
ΣFy = 0: FC − 30 lb = 0
FC = 30 lb v
ΣFx = 0: VC − 22.5 lb = 0
VC = 22.5 lb
v
M C = 960 lb ⋅ in.
v
ΣM C = 0: (22.5 lb)(32 in.) + (30 lb)(8 in.) − M C = 0
Equation of parabola
x = ky 2
At B:
8 = k (32) 2
Therefore, equation is
x=
k=
1
128
y2
128
(1)
The slope at J is obtained by differentiating (1):
dx =
2 y dy
dx
y
=
, tan θ =
128
dy 64
(2)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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999
PROBLEM 7.10 (Continued)
(a)
Maximum axial force
ΣFV = 0: − F + (30 lb) cos θ − (22.5 lb)sin θ = 0
Free body: Portion bow CK
F = 30 cos θ − 22.5sin θ
F is largest at C (θ = 0)
Fm = 30.0 lb at C
(b)
Maximum shearing force
ΣFV = 0: −V + (30 lb)sin θ + (22.5 lb) cos θ = 0
V = 30sin θ + 22.5cos θ
V is largest at B (and D)
Where
1
θ = θ max = tan −1 " !# = 26.56°
$2%
Vm = 30sin 26.56° + 22.5cos 26.56°
(c)
Vm = 33.5 lb at B and D
Maximum bending moment
ΣM K = 0: M − 960 lb ⋅ in. + (30 lb) x + (22.5 lb) y = 0
M = 960 − 30 x − 22.5 y
M is largest at C, where x = y = 0.
M m = 960 lb ⋅ in. at C
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1000
PROBLEM 7.11
Two members, each consisting of a straight and a
quarter-circular portion of rod, are connected as
shown and support a 75-lb load at A. Determine the
internal forces at Point J.
SOLUTION
Free body: Entire frame
ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0
F = 100 lb v
ΣFx = 0: C x = 0
ΣFy = 0: C y − 75 lb − 100 lb = 0
C y = + 175 lb
C = 175 lb v
Free body: Member BEDF
ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0
D = 125 lb v
ΣFx = 0: Bx = 0
ΣFy = 0: By + 125 lb − 100 lb = 0
By = − 25 lb
B = 25 lb v
Free body: BJ
ΣFx = 0: F − (25 lb)sin 30° = 0
F = 12.50 lb
30.0°
V = 21.7 lb
60.0°
ΣFy = 0: V − (25 lb) cos 30° = 0
ΣM J = 0: − M + (25 lb)(3 in.) = 0
M = 75.0 lb ⋅ in.
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1001
PROBLEM 7.12
Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and
support a 75-lb load at A. Determine the internal forces at
Point K.
SOLUTION
Free body: Entire frame
ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0
F = 100 lb v
ΣFx = 0: C x = 0
ΣFy = 0: C y − 75 lb − 100 lb = 0
C y = + 175 lb
C = 175 lb v
Free body: Member BEDF
ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0
D = 125 lb v
ΣFx = 0: Bx = 0
ΣFy = 0: By + 125 lb − 100 lb = 0
By = − 25 lb
B = 25 lb v
Free body: DK
We found in Problem 7.11 that
D = 125 lb on BEDF.
D = 125 lb on DK. v
Thus
ΣFx = 0: F − (125 lb) cos 30° = 0
F = 108.3 lb
60.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1002
PROBLEM 7.12 (Continued)
ΣFy = 0: V − (125 lb)sin 30° = 0
V = 62.5 lb
30.0°
ΣM K = 0: M − (125 lb)d = 0
M = (125 lb)d = (125 lb)(0.8038 in.)
= 100.5 lb ⋅ in.
M = 100.5 lb ⋅ in.
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1003
PROBLEM 7.13
A semicircular rod is loaded as shown. Determine the internal
forces at Point J knowing that θ = 30°.
SOLUTION
FBD AB:
4 !
3 !
ΣM A = 0: r " C # + r " C # − 2r (280 N) = 0
$5 %
$5 %
C = 400 N
ΣFx = 0: − Ax +
4
(400 N) = 0
5
A x = 320 N
3
ΣFy = 0: Ay + (400 N) − 280 N = 0
5
A y = 40.0 N
FBD AJ:
ΣFx′ = 0: F − (320 N) sin 30° − (40.0 N) cos 30° = 0
F = 194.641 N
F = 194.6 N
60.0°
!
30.0°
!
ΣFy ′ = 0: V − (320 N) cos 30° + (40 N) sin 30° = 0
V = 257.13 N
V = 257 N
ΣM 0 = 0: (0.160 m)(194.641 N) − (0.160 m)(40.0 N) − M = 0
M = 24.743
M = 24.7 N ⋅ m
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1004
PROBLEM 7.14
A semicircular rod is loaded as shown. Determine the magnitude
and location of the maximum bending moment in the rod.
SOLUTION
Free body: Rod ACB
4
3
!
!
ΣM A = 0: " FCD # (0.16 m) + " FCD # (0.16 m)
$5
%
$5
%
− (280 N)(0.32 m) = 0
ΣFx = 0: Ax +
FCD = 400 N
v
A x = 320 N
v
4
(400 N) = 0
5
Ax = − 320 N
3
+ ΣFy = 0: Ay + (400 N) − 280 N = 0
5
Ay = + 40.0 N
A y = 40.0 N v
Free body: AJ (For θ , 90°)
ΣM J = 0: (320 N)(0.16 m) sin θ − (40.0 N)(0.16 m)(1 − cos θ ) − M = 0
M = 51.2sin θ + 6.4 cos θ − 6.4
(1)
For maximum value between A and C:
dM
= 0: 51.2 cos θ − 6.4sin θ = 0
dθ
tan θ =
51.2
=8
6.4
θ = 82.87° v
Carrying into (1):
M = 51.2sin 82.87° + 6.4cos82.87° − 6.4 = + 45.20 N ⋅ m
v
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1005
PROBLEM 7.14 (Continued)
Free body: BJ (For θ . 90°)
ΣM J = 0: M − (280 N)(0.16 m)(1 − cos φ ) = 0
M = (44.8 N ⋅ m)(1 − cos φ )
Largest value occurs for φ = 90°, that is, at C, and is
M C = 44.8 N ⋅ m v
We conclude that
M max = 45.2 N ⋅ m
for
θ = 82.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1006
PROBLEM 7.15
Knowing that the radius of each pulley is 150 mm,
that α = 20°, and neglecting friction, determine the
internal forces at (a) Point J, (b) Point K.
SOLUTION
Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter
computations: (cf. Problem 6.90)
(a)
Free body: AJ
ΣFx = 0: 500 N − F = 0
F = 500 N
F = 500 N
ΣFy = 0: V − 500 N = 0
V = 500 N
V = 500 N
ΣM J = 0: (500 N)(0.6 m) = 0
M = 300 N ⋅ m
(b)
M = 300 N ⋅ m
Free body: ABK
ΣFx = 0: 500 N − 500 N + (500 N)sin 20° − V = 0
V = 171.01 N
V = 171.0 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1007
PROBLEM 7.15 (Continued)
ΣFy = 0: − 500 N − (500 N) cos 20° + F = 0
F = 969.8 N
F = 970 N
ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 20°(0.9 m) − M = 0
M = 446.1 N ⋅ m
M = 446 N ⋅ m
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1008
PROBLEM 7.16
Knowing that the radius of each pulley is 150 mm,
that α = 30°, and neglecting friction, determine the
internal forces at (a) Point J, (b) Point K.
SOLUTION
Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter
computations: (cf. Problem 6.90)
(a)
Free body: AJ:
ΣFx = 0: 500 N − F = 0
F = 500 N
F = 500 N
ΣFy = 0: V − 500 N = 0
V = 500 N
V = 500 N
ΣM J = 0: (500 N)(0.6 m) = 0
M = 300 N ⋅ m
(b)
M = 300 N ⋅ m
FBD: Portion ABK
ΣFx = 0: 500 N − 500 N + (500 N)sin 30° − V
ΣFy = 0: − 500 N − (500 N) cos 30° + F = 0
ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 30°(0.9 m) − M = 0
V = 250 N
F = 933 N
M = 375 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1009
PROBLEM 7.17
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point J
of the frame shown.
SOLUTION
Free body: Frame and pulleys
ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0
Bx = − 520 N
B x = 520 N
v
A x = 520 N
v
ΣFx = 0: Ax − 520 N = 0
Ax = + 520 N
ΣFy = 0: Ay + By − 360 N = 0
Ay + By = 360 N
(1)
Free body: Member AE
ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0
Ay = − 240 N
From (1):
A y = 240 N v
By = 360 N + 240 N
By = + 600 N
B y = 600 N v
Free body: BJ
We recall that the forces applied to a pulley may be applied directly to its axle.
ΣFy = 0:
3
4
(600 N) + (520 N)
5
5
3
− 360 N − (360 N) − F = 0
5
F = + 200 N
F = 200 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1010
PROBLEM 7.17 (Continued)
ΣFx = 0:
4
3
4
(600 N) − (520 N) − (360 N) + V = 0
5
5
5
V = + 120.0 N
V = 120.0 N
ΣM J = 0: (520 N)(1.2 m) − (600 N)(1.6 m) + (360 N)(0.6 m) + M = 0
M = + 120.0 N ⋅ m
M = 120.0 N ⋅ m
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1011
PROBLEM 7.18
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point K
of the frame shown.
SOLUTION
Free body: Frame and pulleys
ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0
Bx = − 520 N
B x = 520 N
v
A x = 520 N
v
ΣFx = 0: Ax − 520 N = 0
Ax = + 520 N
ΣFy = 0: Ay + By − 360 N = 0
Ay + By = 360 N
(1)
Free body: Member AE
ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0
Ay = − 240 N
From (1):
A y = 240 N v
By = 360 N + 240 N
By = + 600 N
B y = 600 N v
Free body: AK
ΣFx = 0: 520 N − F = 0
F = + 520 N
F = 520 N
ΣFy = 0: 360 N − 240 N − V = 0
V = + 120.0 N
V = 120.0 N
ΣMK = 0: (240 N)(1.6 m) − (360 N)(0.8 m) − M = 0
M = + 96.0 N ⋅ m
M = 96.0 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1012
PROBLEM 7.19
A 5-in.-diameter pipe is supported every 9 ft by a small frame
consisting of two members as shown. Knowing that the combined
weight of the pipe and its contents is 10 lb/ft and neglecting the
effect of friction, determine the magnitude and location of the
maximum bending moment in member AC.
SOLUTION
Free body: 10-ft section of pipe
4
ΣFx = 0: D − (90 lb) = 0
5
D = 72 lb
v
3
ΣFy = 0: E − (90 lb) = 0
5
E = 54 lb
v
Free body: Frame
ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.)
+ (54 lb)(8.75 in.) = 0
Ay = + 34.8 lb
A y = 34.8 lb v
4
3
ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0
5
5
By = + 55.2 lb
B y = 55.2 lb v
3
4
ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0
5
5
Ax + Bx = 0
(1)
Free body: Member AC
ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0
Ax = − 26.4 lb
A x = 26.4 lb
v
B x = 26.4 lb
v
Bx = − Ax = + 26.4 lb
From (1):
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1013
PROBLEM 7.19 (Continued)
Free body: Portion AJ
For x # 12.5 in. ( AJ # AD ) :
3
4
ΣM J = 0: (26.4 lb) x − (34.8 lb) x + M = 0
5
5
M = 12 x
M max = 150 lb ⋅ in. for x = 12.5 in.
M max = 150.0 lb ⋅ in. at D v
For x . 12.5 in.( AJ . AD):
3
4
ΣM J = 0: (26.4 lb) x − (34.8 lb) x + (72 lb)( x − 12.5) + M = 0
5
5
M = 900 − 60 x
M max = 150 lb ⋅ in. for x = 12.5 in.
M max = 150.0 lb ⋅ in. at D
Thus:
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1014
PROBLEM 7.20
For the frame of Problem 7.19, determine the magnitude and
location of the maximum bending moment in member BC.
PROBLEM 7.19 A 5-in.-diameter pipe is supported every 9 ft
by a small frame consisting of two members as shown. Knowing
that the combined weight of the pipe and its contents is 10 lb/ft
and neglecting the effect of friction, determine the magnitude
and location of the maximum bending moment in member AC.
SOLUTION
Free body: 10-ft section of pipe
4
ΣFx = 0: D − (90 lb) = 0
5
D = 72 lb
v
3
ΣFy = 0: E − (90 lb) = 0
5
E = 54 lb
v
Free body: Frame
ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.)
+ (54 lb)(8.75 in.) = 0
Ay = + 34.8 lb
A y = 34.8 lb v
4
3
ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0
5
5
By = + 55.2 lb
B y = 55.2 lb v
3
4
ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0
5
5
Ax + Bx = 0
(1)
Free body: Member AC
ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.)
− Ax (9 in.) = 0
Ax = − 26.4 lb
From (1):
A x = 26.4 lb
v
B x = 26.4 lb
v
Bx = − Ax = + 26.4 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1015
PROBLEM 7.20 (Continued)
Free body: Portion BK
For x # 8.75 in.( BK # BE ):
3
4
ΣM K = 0: (55.2 lb) x − (26.4 lb) x − M = 0
5
5
M = 12 x
M max = 105.0 lb ⋅ in. for
x = 8.75 in.
M max = 105.0 lb ⋅ in. at E v
For x . 8.75 in.( BK . BE ):
3
4
ΣM K = 0: (55.2 lb) x − (26.4 lb) x − (54 lb)( x − 8.75 in.) − M = 0
5
5
M = 472.5 − 42 x
M max = 105.0 lb ⋅ in. for
x = 8.75 in.
M max = 105.0 lb ⋅ in. at E
Thus
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you are using it without permission.
1016
PROBLEM 7.21
A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.
SOLUTION
(a)
FBD Rod:
ΣFx = 0:
Ax = 0
ΣM D = 0: aP − 2aAy = 0
Ay =
P
2
ΣFx = 0: V = 0
FBD AJ:
ΣFy = 0:
P
−F =0
2
F=
P
2
ΣM J = 0: M = 0
(b)
FBD Rod:
4 !
3 !
ΣM A = 0: 2a " D # + 2a " D # − aP = 0
5
5
$
%
$
%
ΣFx = 0: Ax −
4 5
P=0
5 14
ΣFy = 0: Ay − P +
3 5
P=0
5 14
D=
5P
14
Ax =
2P
7
Ay =
11P
14
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1017
PROBLEM 7.21 (Continued)
FBD AJ:
!
!
!
(c)
ΣFx = 0:
2
P −V = 0
7
ΣFy = 0:
11P
−F =0
14
ΣM J = 0: a
2P
−M =0
7
ΣM A = 0:
a 4D !
− aP = 0
2 "$ 5 #%
V=
2P
7
F=
M=
11P
14
2
aP
7
FBD Rod:
D=
5P
2
Ax −
4 5P
=0
5 2
Ax = 2 P
ΣFy = 0: Ay − P −
3 5P
=0
5 2
Ay =
ΣFx = 0:
5P
2
FBD AJ:
ΣFx = 0:
2P − V = 0
ΣFy = 0:
5P
−F =0
2
ΣM J = 0: a(2 P) − M = 0
V = 2P
F=
5P
2
M = 2aP
!
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you are using it without permission.
1018
!
PROBLEM 7.22
A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.
SOLUTION
(a)
FBD Rod:
ΣM D = 0: aP − 2aA = 0
A=
ΣFx = 0: V −
FBD AJ:
P
=0
2
V=
ΣFy = 0:
P
2
F=0
ΣM J = 0: M − a
(b)
P
2
P
=0
2
M=
aP
2
FBD Rod:
ΣM D = 0: aP −
a 4 !
A =0
2 "$ 5 #%
A=
5P
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1019
PROBLEM 7.22 (Continued)
FBD AJ:
(c)
ΣFx = 0:
3 5P
−V = 0
5 2
ΣFy = 0:
4 5P
−F =0
5 2
V=
3P
2
F = 2P
M=
3
aP
2
V=
3P
14
!
FBD Rod:
3 !
4 !
ΣM D = 0: aP − 2a " A # − 2a " A # = 0
$5 %
$5 %
A=
3 5P !
ΣFx = 0: V − "
#=0
$ 5 14 %
ΣFy = 0:
4 5P
−F =0
5 14
3 5P !
ΣM J = 0: M − a "
#=0
5
$ 14 %
5P
14
F=
M=
2P
7
3
aP
14
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you are using it without permission.
1020
!
PROBLEM 7.23
A semicircular rod of weight W and uniform cross section is supported as
shown. Determine the bending moment at Point J when θ = 60°.
SOLUTION
FBD Rod:
ΣM A = 0:
2r
π
W − 2rB = 0
B=
ΣFy′ = 0: F +
W
π
W
W
sin 60° − cos 60° = 0
3
π
F = − 0.12952W
FBD BJ:
W ! 3r W !
ΣM 0 = 0: r " F − # +
+M =0
π % 2π "$ 3 #%
$
1
1 !
= 0.28868Wr
M = Wr " 0.12952 + −
2
π
π #%
$
On BJ
M J = 0.289Wr
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1021
!
PROBLEM 7.24
A semicircular rod of weight W and uniform cross section is supported as
shown. Determine the bending moment at Point J when θ = 150°.
SOLUTION
FBD Rod:
ΣFy = 0: Ay − W = 0
ΣM B = 0:
2r
π
Ay = W
W − 2rAx = 0
Ax =
W
π
FBD AJ:
ΣFx′ = 0:
W
π
cos 30° +
5W
sin 30° − F = 0 F = 0.69233W
6
W!
W!
ΣM 0 = 0: 0.25587 r " # + r " F − # − M = 0
π %
6
$ % $
1'
& 0.25587
M = Wr (
+ 0.69233 − )
6
π+
*
M = Wr (0.4166)
On AJ
M = 0.417Wr
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1022
!
PROBLEM 7.25
A quarter-circular rod of weight W and uniform cross section is supported
as shown. Determine the bending moment at Point J when θ = 30°.
SOLUTION
FBD Rod:
ΣFx = 0: A x = 0
2r
ΣM B = 0:
FBD AJ:
π
W − rAy = 0
α = 15°, weight of segment = W
r=
r
α
ΣFy ′ = 0:
sin α =
r
π
Ay =
2W
π
30° W
=
90° 3
sin15° = 0.9886r
12
2W
π
cos 30° −
W
cos 30° − F = 0
3
F=
W 3 2 1!
" − #
2 $π 3%
W
2W !
ΣM 0 = M + r " F −
# + r cos15° 3 = 0
π
$
%
M = 0.0557 Wr
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1023
PROBLEM 7.26
A quarter-circular rod of weight W and uniform cross section is supported as
shown. Determine the bending moment at Point J when θ = 30°.
SOLUTION
FBD Rod:
ΣM A = 0: rB −
2r
π
W =0
B=
α = 15° =
FBD BJ:
r=
r
π
2W
π
π
12
sin15° = 0.98862r
12
Weight of segment = W
30° W
=
90° 3
ΣFy′ = 0: F −
W
2W
cos 30° −
sin 30° = 0
3
π
3 1!
W
+
F="
" 6 π ##
$
%
ΣM 0 = 0: rF − ( r cos15°)
W
−M =0
3
3 1!
cos15° !
M = rW "
+
− 0.98862
#Wr
" 6 π ## "$
3 %
$
%
M = 0.289Wr
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1024
PROBLEM 7.27
For the rod of Problem 7.25, determine the magnitude and location of the
maximum bending moment.
PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross
section is supported as shown. Determine the bending moment at Point J
when θ = 30°.
SOLUTION
ΣFx = 0: Ax = 0
FBD Rod:
ΣM B = 0:
2r
π
W − rAy = 0
α=
θ
2
Ay =
r=
,
Weight of segment = W
2α
π
=
F=
FBD AJ:
2W
ΣM 0 = 0: M + " F −
π
$
M=
But,
so
sin α cos α =
M=
2W
π
4α
π
W cos 2α +
2W
π
π
r
α
4α
2
ΣFx′ = 0: − F −
2W
π
2W
π
sin α
W
cos 2α = 0
(1 − 2α ) cos 2α =
2W
π
(1 − θ ) cos θ
4α
!
# r + (r cos α ) π W = 0
%
(1 + θ cos θ − cos θ ) r −
4αW r
π
α
sin α cos α
1
1
sin 2α = sin θ
2
2
2r
π
W (1 − cos θ + θ cos θ − sin θ )
dM 2rW
=
(sin θ − θ sin θ + cos θ − cos θ ) = 0
π
dθ
for
(1 − θ )sin θ = 0
dM
= 0 for θ = 0, 1, nπ ( n = 1, 2, )
dθ
Only 0 and 1 in valid range
At
θ = 0 M = 0, at θ = 1 rad
at
θ = 57.3°
M = M max = 0.1009Wr
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1025
PROBLEM 7.28
For the rod of Problem 7.26, determine the magnitude and location of the maximum
bending moment.
PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at Point J when θ = 30°.
SOLUTION
FBD Bar:
ΣM A = 0: rB −
2r
π
W =0
α=
B=
θ
r=
α
π
0#α #
so
2
r
2W
π
4
sin α
Weight of segment = W
2α
π
2
=
ΣFx′ = 0: F −
F=
=
4α
π
π
W
W cos 2α −
2W
π
2W
π
4α
2W
π
sin 2α = 0
(sin 2α + 2α cos 2α )
(sin θ + θ cos θ )
FBD BJ:
ΣM 0 = 0: rF − (r cos α )
M=
But,
sin α cos α =
2
π
4α
π
W −M =0
r
! 4α
Wr (sin θ + θ cos θ ) − " sin α cos α #
W
α
$
% π
1
1
sin 2α = sin θ
2
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1026
PROBLEM 7.28 (Continued)
so
M=
or
M=
2Wr
π
2
π
(sin θ + θ cos θ − sin θ )
Wrθ cos θ
dM 2
= Wr (cos θ − θ sin θ ) = 0 at θ tan θ = 1
dθ π
Solving numerically
θ = 0.8603 rad
M = 0.357Wr
and
at θ = 49.3°
(Since M = 0 at both limits, this is the maximum)!
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1027
!
PROBLEM 7.29
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
(a)
From A to B:
ΣFy = 0:
V = −P
ΣM1 = 0: M = − Px
From B to C:
ΣFy = 0: − P − P − V = 0
V = −2 P
ΣM 2 = 0: Px + P( x − a) + M = 0
(b)
|V | max = 2 P;
M = −2 Px + Pa
| M | max = 3Pa
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1028
!
PROBLEM 7.30
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
Reactions:
A=
2P
3
C=
P
3
From A to B:
ΣFy = 0: V = +
2P
3
ΣM1 = 0: M = +
2P
x
3
From B to C:
ΣFy = 0: V = −
P
3
ΣM 2 = 0: M = +
(b)
|V | max =
2P
;
3
P
( L − x)
3
|M | max =
2 PL
9
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1029
PROBLEM 7.31
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
FBD beam:
(a)
Ay = D =
By symmetry:
1
L
( w)
2
2
Ay = D =
wL
4
Along AB:
ΣFy = 0:
wL
−V = 0
4
ΣM J = 0: M − x
V=
wL
4
wL
=0
4
wL
M=
x (straight)
4
Along BC:
ΣFy = 0:
wL
− wx1 − V = 0
4
wL
V=
− wx1
4
Straight with
ΣM k = 0: M +
Parabola with
V = 0 at
x1 =
L
4
x1
L
! wL
wx1 −
+ x1 !
=0
2
"4
# 4
M=
%
w $ L2 L
+ x1 − x12 !!
2" 8 2
#
M=
3
L
wL2 at x1 =
32
4
Section CD by symmetry
(b)
|V |max =
From diagrams:
wL
on AB and CD
4
| M |max =
3wL2
at center
32
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1030
PROBLEM 7.32
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
Along AB:
ΣFy = 0: − wx − V = 0
Straight with
V =−
wL
2
ΣM J = 0: M +
Parabola with
x=
at
V = − wx
L
2
x
1
wx = 0 M = − wx 2
2
2
M =−
wL2
8
at
x=
L
2
Along BC:
ΣFy = 0: − w
L
−V = 0
2
1
V = − wL
2
L% L
$
ΣM k = 0: M + x1 + ! w = 0
4# 2
"
M =−
Straight with
(b)
wL $ L
%
+ x1 !
2 "4
#
3
M = − wL2
8
From diagrams:
at
x1 =
L
2
|V |max =
wL
on BC
2
|M |max =
3wL2
at C
8
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you are using it without permission.
1031
PROBLEM 7.33
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Portion AJ
ΣFy = 0: − P − V = 0
ΣM J = 0: M + Px − PL = 0
(a)
V = −P !
M = P( L − x) !
The V and M diagrams are obtained by plotting the functions V and M.
|V |max = P
(b)
| M |max = PL
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1032
PROBLEM 7.34
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
(a)
FBD Beam:
ΣM C = 0: LAy − M 0 = 0
Ay =
M0
L
ΣFy = 0: − Ay + C = 0
C=
M0
L
Along AB:
ΣFy = 0: −
M0
−V = 0
L
V =−
ΣM J = 0: x
M0
L
M0
+M =0
L
Straight with
M =−
M0
x
L
M =−
M0
at B
2
Along BC:
ΣFy = 0: −
M0
−V = 0
L
ΣM K = 0: M + x
Straight with
(b)
M=
V =−
M0
− M0 = 0
L
M0
at B
2
M0
L
x%
$
M = M0 1− !
L#
"
M = 0 at C
|V |max = P everywhere
From diagrams:
| M |max =
#
M0
at B
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1033
PROBLEM 7.35
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
Just to the right of A:
ΣFy = 0 V1 = +15 kN M1 = 0
Just to the left of C:
V2 = +15 kN M 2 = +15 kN ⋅ m
Just to the right of C:
V3 = +15 kN M 3 = +5 kN ⋅ m
Just to the right of D:
V4 = −15 kN M 4 = +12.5 kN ⋅ m
Just to the right of E:
V5 = −35 kN M 5 = +5 kN ⋅ m
At B:
(b)
M B = −12.5 kN ⋅ m
|V |max = 35.0 kN
| M |max = 12.50 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1034
PROBLEM 7.36
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(3.2 m) − (40 kN)(0.6 m) − (32 kN)(1.5 m) − (16 kN)(3 m) = 0
B = +37.5 kN
B = 37.5 kN !#
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 37.5 kN − 40 kN − 32 kN − 16 kN = 0
Ay = +50.5 kN
(a)
A = 50.5 kN !
Shear and bending moment.
Just to the right of A:
V1 = 50.5 kN
M1 = 0 !
Just to the right of C:
ΣFy = 0: 50.5 kN − 40 kN − V2 = 0
ΣM 2 = 0: M 2 − (50.5 kN)(0.6 m) = 0
V2 = +10.5 kN !
M 2 = +30.3 kN ⋅ m !#
Just to the right of D:
ΣFy = 0: 50.5 − 40 − 32 − V3 = 0
ΣM 3 = 0: M 3 − (50.5)(1.5) + (40)(0.9) = 0
V3 = −21.5 kN !
M 3 = +39.8 kN ⋅ m !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1035
PROBLEM 7.36 (Continued)
Just to the right of E:
ΣFy = 0: V4 + 37.5 = 0
V4 = −37.5 kN !
ΣM 4 = 0: − M 4 + (37.5)(0.2) = 0
M 4 = +7.50 kN ⋅ m !
VB = M B = 0
At B:
!
|V |max = 50.5 kN
(b)
|M |max = 39.8 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1036
PROBLEM 7.37
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: E (6 ft) − (6 kips)(2 ft) − (12 kips)(4 ft) − (4.5 kips)(8 ft) = 0
E = +16 kips
E = 16 kips !#
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 16 kips − 6 kips − 12 kips − 4.5 kips = 0
Ay = +6.50 kips
(a)
A = 6.50 kips !
Shear and bending moment
Just to the right of A:
V1 = +6.50 kips M1 = 0
!
Just to the right of C:
ΣFy = 0: 6.50 kips − 6 kips − V2 = 0
ΣM 2 = 0: M 2 − (6.50 kips)(2 ft) = 0
V2 = +0.50 kips !
M 2 = +13 kip ⋅ ft !#
Just to the right of D:
ΣFy = 0: 6.50 − 6 − 12 − V3 = 0
ΣM 3 = 0: M 3 − (6.50)(4) − (6)(2) = 0
V3 = +11.5 kips !
M 3 = +14 kip ⋅ ft !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1037
PROBLEM 7.37 (Continued)
Just to the right of E:
ΣFy = 0: V4 − 4.5 = 0
ΣM 4 = 0: − M 4 − (4.5)2 = 0
V4 = +4.5 kips !
M 4 = −9 kip ⋅ ft !
VB = M B = 0
At B:
!
|V |max = 11.50 kips
(b)
| M |max = 14.00 kip ⋅ ft
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1038
PROBLEM 7.38
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM C = 0: (120 lb)(10 in.) − (300 lb)(25 in.) + E (45 in.) − (120 lb)(60 in.) = 0
E = +300 lb
E = 300 lb !#
ΣFx = 0: Cx = 0
ΣFy = 0: C y + 300 lb − 120 lb − 300 lb − 120 lb = 0
C y = +240 lb
(a)
C = 240 lb !
Shear and bending moment
Just to the right of A:
ΣFy = 0: − 120 lb − V1 = 0
V1 = −120 lb, M 1 = 0 !
Just to the right of C:
ΣFy = 0: 240 lb − 120 lb − V2 = 0
ΣM C = 0: M 2 + (120 lb)(10 in.) = 0
V2 = +120 lb !
M 2 = −1200 lb ⋅ in. !#
Just to the right of D:
ΣFy = 0: 240 − 120 − 300 − V3 = 0
ΣM 3 = 0: M 3 + (120)(35) − (240)(25) = 0,
V3 = −180 lb !
M 3 = +1800 lb ⋅ in. !
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1039
PROBLEM 7.38 (Continued)
Just to the right of E:
+ΣFy = 0: V4 − 120 lb = 0
ΣM 4 = 0: − M 4 − (120 lb)(15 in.) = 0
V4 = +120 lb !
M 4 = −1800 lb ⋅ in. !
VB = M B = 0 !
At B:
|V |max = 180.0 lb
(b)
| M |max = 1800 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1040
PROBLEM 7.39
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(5 m) − (60 kN)(2 m) − (50 kN)(4 m) = 0
B = +64.0 kN
B = 64.0 kN !#
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 64.0 kN − 6.0 kN − 50 kN = 0
Ay = +46.0 kN
(a)
A = 46.0 kN !
Shear and bending-moment diagrams.
From A to C:
ΣFy = 0: 46 − V = 0
ΣM y = 0: M − 46 x = 0
V = +46 kN !
M = (46 x)kN ⋅ m !
From C to D:
ΣFy = 0: 46 − 60 − V = 0
V = −14 kN !
ΣM j = 0: M − 46 x + 60( x − 2) = 0
M = (120 − 14 x)kN ⋅ m
For
x = 2 m: M C = +92.0 kN ⋅ m
!#
For
x = 3 m: M D = +78.0 kN ⋅ m
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1041
PROBLEM 7.39 (Continued)
From D to B:
ΣFy = 0: V + 64 − 25µ = 0
V = (25µ − 64)kN
ΣM j = 0: 64 µ − (25µ )
$µ%
!−M =0
"2#
M = (64µ − 12.5µ 2 )kN ⋅ m
For
µ = 0: VB = −64 kN
MB = 0 !
|V |max = 64.0 kN
(b)
| M |max = 92.0 kN ⋅ m
#
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1042
PROBLEM 7.40
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM B = 0: (60 kN)(6 m) − C (5 m) + (80 kN)(2 m) = 0
C = +104 kN
From A to C:
C = 104 kN
ΣFy = 0: 104 − 60 − 80 + B = 0
B = 36 kN
ΣFy = 0: − 30 x − V = 0
V = −30 x
ΣM1 = 0: (30 x)
$x%
!+M =0
"2#
M = −15 x 2
From C to D:
ΣFy = 0: 104 − 60 − V = 0
ΣM 2 = 0: (60)( x − 1) − (104)( x − 2) + M = 0
V = +44 kN
M = 44 x − 148
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1043
PROBLEM 7.40 (Continued)
From D to B:
ΣFy = 0: V = −36 kN
ΣM 3 = 0: (36)(7 − x) − M = 0
M = −36 x + 252
(b)
|V |max = 60.0 kN
| M |max = 72.0 kN ⋅ m.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1044
PROBLEM 7.41
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
By symmetry:
Ay = B = 8 kips +
1
(4 kips)(5 ft) A y = B = 18 kips
2
Along AC:
ΣFy = 0 : 18 kips − V = 0 V = 18 kips
ΣM J = 0: M − x(18 kips) M = (18 kips) x
M = 36 kip ⋅ ft at C ( x = 2 ft)
Along CD:
ΣFy = 0: 18 kips − 8 kips − (4 kips/ft) x1 − V = 0
V = 10 kips − (4 kips/ft) x1
V = 0 at x1 = 2.5 ft (at center)
ΣM K = 0: M +
x1
(4 kips/ft) x1 + (8 kips) x1 − (2 ft + x1 )(18 kips) = 0
2
M = 36 kip ⋅ ft + (10 kips/ft) x1 − (2 kips/ft) x12
M = 48.5 kip ⋅ ft at x1 = 2.5 ft
Complete diagram by symmetry
(b)
|V |max = 18.00 kips on AC and DB
From diagrams:
|M |max = 48.5 kip ⋅ ft at center
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1045
PROBLEM 7.42
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B (10 ft) − (15 kips)(3 ft) − (12 kips)(6 ft) = 0
B = +11.70 kips
B = 11.70 kips !
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 15 − 12 + 11.70 = 0
Ay = +15.30 kips
(a)
A = 15.30 kips !
Shear and bending-moment diagrams
From A to C:
ΣFy = 0: 15.30 − 2.5 x − V = 0
V = (15.30 − 2.5 x) kips
x!
ΣM J = 0: M + (2.5 x) " # − 15.30 x = 0
2
$ %
M = 15.30 x − 1.25 x 2
For x = 0:
VA = +15.30 kips
For x = 6 ft:
VC = +0.300 kip
MA = 0 !
M C = +46.8 kip ⋅ ft !
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1046
PROBLEM 7.42 (Continued)
From C to B:
ΣFy = 0: V + 11.70 = 0
V = −11.70 kips !
ΣM J = 0: 11.70µ − M = 0
M = (11.70µ ) kip ⋅ ft
M C = +46.8 kip ⋅ ft !
For µ = 4 ft:
For µ = 0:
MB = 0 !
(b)
|V |max = 15.30 kips
"
"
"
"
"
"
"
|M |max = 46.8 kip ⋅ ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1047
PROBLEM 7.43
Assuming the upward reaction of the ground on beam AB to be uniformly
distributed and knowing that a = 0.3 m, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear
and bending moment.
SOLUTION
(a)
FBD Beam:
ΣFy = 0: w(1.5 m) − 2(3.0 kN) = 0
w = 4.0 kN/m
Along AC:
ΣFy = 0: (4.0 kN/m) x − V = 0
V = (4.0 kN/m) x
x
(4.0 kN/m) x = 0
2
M = (2.0 kN/m) x 2
ΣM J = 0: M −
Along CD:
ΣFy = 0: (4.0 kN/m) x − 3.0 kN − V = 0
V = (4.0 kN/m) x − 3.0 kN
ΣM K = 0: M + ( x − 0.3 m)(3.0 kN) −
x
(4.0 kN/m) x = 0
2
M = 0.9 kN ⋅ m − (3.0 kN) x + (2.0 kN/m) x 2
Note: V = 0 at x = 0.75 m, where M = −0.225 kN ⋅ m
Complete diagrams using symmetry.
|V |max = 1.800 kN at C and D
(b)
|M |max = 0.225 kN ⋅ m at center
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1048
PROBLEM 7.44
Solve Problem 7.43 knowing that a = 0.5 m.
PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB
to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute values
of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (1.5 m) − 3 kN − 3 kN = 0
(a)
wg = 4 kN/m !
Shear and bending moment
From A to C:
ΣFy = 0: 4 x − V = 0 V = (4 x) kN
ΣM J = 0: M − (4 x)
x
= 0, M = (2 x 2 ) kN ⋅ m
2
For x = 0:
For x = 0.5 m:
From C to D:
VA = M A = 0 !
VC = 2 kN,
M C = 0.500 kN ⋅ m !"
ΣFy = 0: 4 x − 3 kN − V = 0
V = (4 x − 3) kN
ΣM J = 0: M + (3 kN)( x − 0.5) − (4 x)
x
=0
2
M = (2 x 2 − 3 x + 1.5) kN ⋅ m
For x = 0.5 m:
VC = −1.00 kN,
For x = 0.75 m:
VC = 0,
For x = 1.0 m:
VC = 1.00 kN,
M C = 0.500 kN ⋅ m !
M C = 0.375 kN ⋅ m !"
M C = 0.500 kN ⋅ m !
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1049
PROBLEM 7.44 (Continued)
From D to B:
ΣFy = 0: V + 4µ = 0 V = −(4µ ) kN
ΣM J = 0: (4 µ )
µ
2
− M = 0, M = 2µ 2
For µ = 0:
For µ = 0.5 m:
VB = M B = 0 !
VD = −2 kN,
M D = 0.500 kN ⋅ m !
(b)
|V |max = 2.00 kN
"
"
|M |max = 0.500 kN ⋅ m
"
"
"
"
"
"
"
"
"
"
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1050
PROBLEM 7.45
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear
and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0
(a)
wg = 1.5 kips/ft !
Shear and bending-moment diagrams.
From A to C:
ΣFy = 0: 1.5 x − 3x − V = 0
V = (−1.5 x) kips
x
x
− (1.5 x) = 0
2
2
2
M = ( −0.75 x ) kip ⋅ ft
ΣM J = 0: M + (3x)
For x = 0:
VA = M A = 0 !
For x = 3 ft:
VC = −4.5 kips M C = −6.75 kip ⋅ ft !"
From C to D:
ΣFy = 0: 1.5 x = 9 − V = 0, V = (1.5 x − 9) kips
ΣM J = 0: M + 9( x − 1.5) − (1.5 x)
x
=0
2
M = 0.75 x 2 − 9 x + 13.5
For x = 3 ft:
VC = −4.5 kips,
For x = 6 ft:
VC = 0,
M C = −6.75 kip ⋅ ft !
M C = −13.50 kip ⋅ ft !
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1051
PROBLEM 7.45 (Continued)
For x = 9 ft :
VD = 4.5 kips, M D = −6.75 kip ⋅ ft !
VB = M B = 0 !
At B:
(b)
|V |max = 4.50 kips
"
|M |max = 13.50 kip ⋅ ft
"
"
"
"
Bending moment diagram consists of three distinct arcs of parabola.
Since entire diagram is below the x axis:
M # 0 everywhere
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1052
PROBLEM 7.46
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed, (a) draw the shear and bending-moment diagrams,
(b) determine the maximum absolute values of the shear and bending
moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0
wg = 1.5 kips/ft !
(a)
Shear and bending-moment diagrams from A to C:
ΣFy = 0: 1.5 x − V = 0
ΣM J = 0: M − (1.5 x)
x
2
V = (1.5 x)kips
M = (0.75 x 2 )kip ⋅ ft
For x = 0:
VA = M A = 0 !
For x = 3 ft: VC = 4.5 kips,
M C = 6.75 kip ⋅ ft !
From C to D:
ΣFy = 0: 1.5 x − 3( x − 3) − V = 0
V = (9 − 1.5 x)kips
ΣM J = 0: M + 3( x − 3)
x−3
x
− (1.5 x) = 0
2
2
M = [0.75 x 2 − 1.5( x − 3)2 ]kip ⋅ ft
For x = 3 ft: VC = 4.5 kips,
M C = 6.75 kip ⋅ ft !
For x = 6 ft: VcL = 0,
McL= 13.50 kip ⋅ ft !
For x = 9 ft: VD = −4.5 kips,
M D = 6.75 kip ⋅ ft !
VB = M B = 0 !
At B:
|V |max = 4.50 kips
(b)
| M | max = 13.50 kip ⋅ ft
Bending-moment diagram consists or three distinct arcs of parabola,
all located above the x axis.
Thus: M $ 0 everywhere "
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1053
PROBLEM 7.47
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed and knowing that P = wa, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (4a) − 2wa − wa = 0
wg =
(a)
3
w !
4
Shear and bending-moment diagrams
From A to C:
ΣFy = 0:
3
wx − wx − V = 0
4
1
V = − wx
4
ΣM J = 0 : M + ( wx)
x
3 !x
− " wx # = 0
2 $4 %2
1
M = − wx 2
8
For x = 0:
VA = M A = 0 !
1
VC = − wa
4
For x = a :
1
M C = − wa 2 !
8
From C to D:
ΣFy = 0:
3
wx − wa − V = 0
4
3
!
V = " x − a#w
4
$
%
a! 3
x!
ΣM J = 0: M + wa " x − # − wx " # = 0
2
4
2
$
%
$ %
3
a!
M = wx 2 − wa " x − #
8
2%
$
(1)
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1054
PROBLEM 7.47 (Continued)
For x = a :
1
VC = − wa
4
1
M C = − wa 2 !
8
For x = 2a :
1
VD = + wa
2
MD = 0 !
Because of the symmetry of the loading, we can deduce the values of V and
M for the right-hand half of the beam from the values obtained for its
left-hand half.
|V |max =
(b)
To find | M |max , we differentiate Eq. (1) and set
dM
dx
1
wa
2
= 0:
dM 3
4
= wx − wa = 0, x = a
dx 4
3
2
wa 2
3
4 !
4 1!
M = w " a # − wa 2 " − # = −
8 $3 %
6
$3 2%
|M |max =
1 2
wa
6
Bending-moment diagram consists of four distinct arcs of parabola."
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1055
PROBLEM 7.48
Solve Problem 7.47 knowing that P = 3wa.
PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB
to be uniformly distributed and knowing that P = wa, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (4a ) − 2wa − 3wa = 0
wg =
(a)
5
w !
4
Shear and bending-moment diagrams
From A to C:
ΣFy = 0:
5
wx − wx − V = 0
4
1
V = + wx
4
ΣM J = 0 : M + ( wx)
x
5 !x
− " wx # = 0
2 $4 %2
1
M = + wx 2
8
For x = 0:
VA = M A = 0 !
1
VC = + wa
4
For x = a :
1
M C = + wa 2 !
8
From C to D:
ΣFy = 0:
5
wx − wa − V = 0
4
5
!
V = " x − a#w
$4
%
a! 5
x!
ΣM J = 0: M + wa " x − # − wx " # = 0
2% 4 $2%
$
M=
5 2
a!
wx − wa " x − #
8
2%
$
(1)
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1056
PROBLEM 7.48 (Continued)
1
1
VC = + wa, M C = + wa 2 !
8
4
For x = a :
For x = 2a :
3
VD = + wa, M D = + wa 2 !
2
Because of the symmetry of the loading, we can deduce the values
of V and M for the right-hand half of the beam from the values
obtained for its left-hand half.
|V |max =
(b)
To find | M |max , we differentiate Eq. (1) and set
dM
dx
3
wa
2
= 0:
dM 5
= wx − wa = 0
dx 4
4
x = a , a (outside portion CD )
5
The maximum value of | M | occurs at D:
| M |max = wa 2
Bending-moment diagram consists of four distinct arcs of parabola."
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1057
PROBLEM 7.49
Draw the shear and bending-moment diagrams for the beam AB, and
determine the shear and bending moment (a) just to the left of C, (b) just
to the right of C.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(0.6 m) − (600 N)(0.2 m) = 0
B = +200 N
B = 200 N !
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 600 N + 200 N = 0
Ay = +400 N
A = 400 N !
We replace the 600-N load by an equivalent force-couple system at C
Just to the right of A:
V1 = +400 N , M1 = 0 !
(a)
Just to the left of C:
V2 = +400 N
M 2 = (400 N)(0.4 m)
(b)
Just to the right of C:
M 3 = (200 N)(0.2 m)
Just to the left of B:
M 2 = +160.0 N ⋅ m
V3 = −200 N
M 3 = + 40.0 N ⋅ m
V4 = −200 N , M 4 = 0 !
"
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1058
PROBLEM 7.50
Two small channel sections DF and EH have been
welded to the uniform beam AB of weight W = 3 kN to
form the rigid structural member shown. This member is
being lifted by two cables attached at D and E. Knowing
that θ = 30° and neglecting the weight of the channel
sections, (a) draw the shear and bending-moment
diagrams for beam AB, (b) determine the maximum
absolute values of the shear and bending moment in the
beam.
SOLUTION
FBD Beam + channels:
(a)
T1 = T2 = T
By symmetry:
ΣFy = 0: 2T sin 60° − 3 kN = 0
T=
T1x =
3
3
3
kN
2 3
3
T1 y = kN
2
M = (0.5 m)
FBD Beam:
3
2 3
= 0.433 kN ⋅ m
kN
With cable force replaced by equivalent force-couple system at F and G
Shear Diagram:
V is piecewise linear
dV
!
" dx = −0.6 kN/m # with 1.5 kN
$
%
discontinuities at F and H.
VF − = −(0.6 kN/m)(1.5 m) = 0.9 kN
+
V increases by 1.5 kN to +0.6 kN at F
VG = 0.6 kN − (0.6 kN/m)(1 m) = 0
Finish by invoking symmetry
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you are using it without permission.
1059
PROBLEM 7.50 (Continued)
Moment diagram: M is piecewise parabolic
dM
!
" dx decreasing with V #
$
%
with discontinuities of .433 kN at F and H.
1
M F − = − (0.9 kN)(1.5 m)
2
= −0.675 kN ⋅ m
M increases by 0.433 kN m to –0.242 kN ⋅ m at F+
M G = −0.242 kN ⋅ m +
1
(0.6 kN)(1 m)
2
= 0.058 kN ⋅ m
Finish by invoking symmetry
|V |max = 900 N
(b)
+
−
at F and G
M
max
= 675 N ⋅ m
at F and G
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1060
PROBLEM 7.51
Solve Problem 7.50 when θ = 60°.
PROBLEM 7.50 Two small channel sections DF and EH
have been welded to the uniform beam AB of weight
W = 3 kN to form the rigid structural member shown. This
member is being lifted by two cables attached at D and E.
Knowing that θ = 30° and neglecting the weight of the
channel sections, (a) draw the shear and bending-moment
diagrams for beam AB, (b) determine the maximum absolute
values of the shear and bending moment in the beam.
SOLUTION
Free body: Beam and channels
From symmetry:
E y = Dy
E x = Dx = Dy tan θ
Thus:
(1)
ΣFy = 0: D y + E y − 3 kN = 0
D y = E y = 1.5 kN !
D x = (1.5 kN) tan θ
From (1):
E = (1.5 kN) tan θ
!
We replace the forces at D and E by equivalent force-couple systems at F and H, where
M 0 = (1.5 kN tan θ )(0.5 m) = (750 N ⋅ m) tan θ
(2)
We note that the weight of the beam per unit length is
w=
(a)
W 3 kN
=
= 0.6 kN/m = 600 N/m
L
5m
Shear and bending moment diagrams
From A to F:
ΣFy = 0: − V − 600 x = 0 V = (−600 x)N
ΣM J = 0: M + (600 x)
x
= 0, M = ( −300 x 2 )N ⋅ m
2
VA = M A = 0 !
For x = 0:
For x = 1.5 m:
VF = −900 N, M F = −675 N ⋅ m !
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1061
PROBLEM 7.51 (Continued)
From F to H:
ΣFy = 0: 1500 − 600 x − V = 0 "
"
V = (1500 − 600 x)N "
"
ΣM J = 0: M + (600 x)
"
x
− 1500( x − 1.5) − M 0 = 0
2
M = M 0 − 300 x 2 + 1500( x − 1.5)N ⋅ m "
For x = 1.5 m:
VF = +600 N, M F = ( M 0 − 675) N ⋅ m !
For x = 2.5 m:
VG = 0, M G = ( M 0 − 375) N ⋅ m !
From G To B, The V and M diagrams will be obtained by symmetry,
(b)
Making θ = 60° in Eq. (2):
|V |max = 900 N
M 0 = 750 tan 60° = 1299 N ⋅ m
M = 1299 − 675 = 624 N ⋅ m !
Thus, just to the right of F:
M G = 1299 − 375 = 924 N ⋅ m !
and
(b)
|V |max = 900 N
"
| M |max = 924 N ⋅ m
"
"
"
"
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you are using it without permission.
1062
PROBLEM 7.52
Draw the shear and bending-moment diagrams for the
beam AB, and determine the maximum absolute values
of the shear and bending moment.
SOLUTION
FBD whole:
ΣM D = 0: (1.2 ft)(1.5 kips) − (1.2 ft)(6 kips)
− (3.6 ft)(1.5 kips) + (1.6 ft)G = 0
G = 6.75 kips
(Dimensions in ft., loads in kips, moments in kips ⋅ ft)
ΣFx = 0: − Dx + G = 0
D x = 6.75 kips
ΣFy = 0: D y − 1.5 kips − 6 kips − 1.5 kips = 0
D y = 9 kips
Beam AB, with forces D and G replaced by equivalent force/couples at C and F
Along AD:
ΣFy = 0: − 1.5 kips − V = 0 V = −1.5 kips
ΣM J = 0: x(1.5 kips) + M = 0 M = −(1.5 kips) x
M = −1.8 kips at x = 1.2 ft
Along DE:
ΣFy = 0: −1.5 kips + 9 kips − V = 0
V = 7.5 kips
ΣM K = 0: M + 5.4 kip ⋅ ft − x1 (9 kips)
+ (1.2 ft + x1 )(1.5 kips) = 0
M = 7.2 kip ⋅ ft + (7.5 kips) x1
M = 1.8 kip ⋅ ft at x1 = 1.2 ft
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1063
PROBLEM 7.52 (Continued)
Along EF:
ΣFy = 0: V − 1.5 kips = 0 V = 1.5 kips
ΣM N = 0: − M + 5.4 kip ⋅ ft − ( x4 + 1.2 ft)(1.5 kips)
M = 3.6 kip ⋅ ft − (1.5 kips) x4
M = 1.8 kip ⋅ ft at x4 = 1.2 ft
M = 3.6 kip ⋅ ft at x4 = 0
Along FB:
ΣFy = 0: V − 1.5 kips = 0
V = 1.5 kips
ΣM L = 0: − M − x3 (1.5 kips) = 0
M = (−1.5 kips) x3
M = −1.8 kip ⋅ ft at x3 = 1.2 ft
|V |max = 7.50 kips on DE
From diagrams:
| M |max = 7.20 kip ⋅ ft at D +
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1064
PROBLEM 7.53
Draw the shear and bending-moment diagrams for the
beam AB, and determine the maximum absolute values
of the shear and bending moment.
SOLUTION
ΣFG = 0: T (9 in.) − (45 lb)(9 in.) − (120 lb)(21 in.) = 0
T = 325 lb
ΣFx = 0: − 325 lb + Gx = 0
G x = 325 lb
ΣFy = 0: G y − 45 lb − 120 lb = 0
G y = 165 lb
Equivalent loading on straight part of beam AB
From A to E:
ΣFy = 0: V = +165 lb
ΣM1 = 0: + 1625 lb ⋅ in. − (165 lb) x + M = 0
M = −1625 + 165 x
From E to F:
ΣFy = 0: 165 − 45 − V = 0
V = +120 lb
ΣM 2 = 0 : + 1625 lb ⋅ in. − (165 lb) x + (45 lb)( x − 9) + M = 0
M = −1220 + 120 x
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1065
PROBLEM 7.53 (Continued)
From F to B:
ΣFy = 0 V = +120 lb
ΣM 3 = 0: − (120)(21 − x) − M = 0
M = −2520 + 120 x
|V |max = 165.0 lb
| M |max = 1625lb ⋅ in.
"
"
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1066
PROBLEM 7.54
Draw the shear and bending-moment diagrams for the beam AB, and
determine the maximum absolute values of the shear and bending
moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(0.9 m) − (400 N)(0.3 m) − (400 N)(0.6 m)
− (400 N)(0.9 m) = 0
B = +800 N
B = 800 N !
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 800 N − 3(400 N) = 0
Ay = +400 N
A = 400 N !
We replace the loads by equivalent force-couple systems at C, D, and E.
We consider successively the following F-B diagrams.
V5 = −400 N
V1 = +400 N
M1 = 0
M 5 = +180 N ⋅ m
V2 = +400 N
V6 = −400 N
M 2 = +60 N ⋅ m
M 6 = +60 N ⋅ m
V3 = 0
V7 = −800 N
M 3 = +120 N ⋅ m
M 7 = +120 N ⋅ m
V8 = −800 N
V4 = 0
M 4 = +120 N ⋅ m
M8 = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1067
PROBLEM 7.54 (Continued)
(b)
|V |max = 800 N
|M |max = 180.0 N ⋅ m
"
"
"
"
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1068
PROBLEM 7.55
For the structural member of Problem 7.50, determine (a) the angle
θ for which the maximum absolute value of the bending moment in
beam AB is as small as possible, (b) the corresponding value of
| M |max. (Hint: Draw the bending-moment diagram and then equate
the absolute values of the largest positive and negative bending
moments obtained.)
PROBLEM 7.50 Two small channel sections DF and EH have
been welded to the uniform beam AB of weight W = 3 kN to form
the rigid structural member shown. This member is being lifted by
two cables attached at D and E. Knowing that θ = 30° and neglecting
the weight of the channel sections, (a) draw the shear and bendingmoment diagrams for beam AB, (b) determine the maximum absolute
values of the shear and bending moment in the beam.
SOLUTION
See solution of Problem 7.50 for reduction of loading or beam AB to the following:
M 0 = (750 N ⋅ m) tan θ !
where
[Equation (2)]
The largest negative bending moment occurs Just to the left of F:
ΣM1 = 0: M 1 + (900 N) "
$
1.5 m !
=0
2 #%
M1 = −675 N ⋅ m !
The largest positive bending moment occurs
At G:
ΣM 2 = 0: M 2 − M 0 + (1500 N)(1.25 m − 1 m) = 0
M 2 = M 0 − 375 N ⋅ m !
Equating M 2 and − M 1 :
M 0 − 375 = +675
M 0 = 1050 N ⋅ m
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1069
PROBLEM 7.55 (Continued)
(a)
From Equation (2):
tan θ =
1050
= 1.400
750
θ = 54.5°
|M |max = 675 N ⋅ m
(b)
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1070
PROBLEM 7.56
For the beam of Problem 7.43, determine (a) the distance a for which
the maximum absolute value of the bending moment in the beam is as
small as possible, (b) the corresponding value of | M |max. (See hint for
Problem 7.55.)
PROBLEM 7.43 Assuming the upward reaction of the ground on
beam AB to be uniformly distributed and knowing that a = 0.3 m, (a)
draw the shear and bending-moment diagrams, (b) determine the
maximum absolute values of the shear and bending moment.
SOLUTION
Force per unit length exerted by ground:
wg =
6 kN
= 4 kN/m
1.5 m
The largest positive bending moment occurs Just to the left of C:
ΣM1 = 0: M 1 = (4a)
a
2
M 1 = 2a 2 !
The largest negative bending moment occurs
At the center line:
ΣM 2 = 0: M 2 + 3(0.75 − a ) − 3(0.375) = 0
Equating M1 and − M 2 :
M 2 = −(1.125 − 3a) !
2a 2 = 1.125 − 3a
a 2 + 1.5a − 0.5625 = 0
(a)
Solving the quadratic equation:
(b)
Substituting:
a = 0.31066,
| M |max = M1 = 2(0.31066) 2
a = 0.311 m
| M |max = 193.0 N ⋅ m
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1071
"
PROBLEM 7.57
For the beam of Problem 7.47, determine (a) the ratio k = P/wa for
which the maximum absolute value of the bending moment in the beam
is as small as possible, (b) the corresponding value of Mmax. (See
hint for Problem 7.55.)
PROBLEM 7.47 Assuming the upward reaction of the ground on
beam AB to be uniformly distributed and knowing that P = wa, (a)
draw the shear and bending-moment diagrams, (b) determine the
maximum absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (4a) − 2wa − kwa = 0
wg =
wg
Setting
w
w
(2 + k )
4
=α
(1)
k = 4α − 2
We have
(2)
Minimum value of B.M. For M to have negative values, we must have wg , w. We verify that M will then be
negative and keep decreasing in the portion AC of the beam. Therefore, M min will occur between C and D.
From C to D:
a!
x!
ΣM J = 0: M + wa " x − # − α wx " # = 0
2%
$
$2%
M=
We differentiate and set
dM
= 0:
dx
1
w(α x 2 − 2ax + a 2 )
2
αx − a = 0
xmin =
a
α
(3)
(4)
Substituting in (3):
1 2 1 2
!
wa " − + 1#
2
$α α
%
1−α
= − wa 2
2α
M min =
M min
(5)
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1072
PROBLEM 7.57 (Continued)
Maximum value of bending moment occurs at D
3a !
ΣM D = 0: M D + wa " # − (2α wa)a = 0
$ 2 %
3!
M max = M D = wa 2 " 2α − #
2%
$
(6)
Equating − M min and M max :
wa 2
1−α
3!
= wa 2 " 2α − #
2α
2%
$
4α 2 − 2α − 1 = 0
α=
(a)
Substitute in (2):
(b)
Substitute for α in (5):
2 + 20
8
α=
1+ 5
= 0.809
4
k = 4(0.809) − 2
|M |max = − M min = − wa 2
Substitute for α in (4):
k = 1.236
1 − 0.809
2(0.809)
|M |max = 0.1180wa 2
xmin =
a
1.236a !
0.809
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1073
PROBLEM 7.58
A uniform beam is to be picked up by crane cables attached at A and B.
Determine the distance a from the ends of the beam to the points where
the cables should be attached if the maximum absolute value of the
bending moment in the beam is to be as small as possible. (Hint: Draw
the bending-moment diagram in terms of a, L, and the weight w per
unit length, and then equate the absolute values of the largest positive
and negative bending moments obtained.)
SOLUTION
w = weight per unit length
To the left of A:
x!
ΣM1 = 0: M + wx " # = 0
$2%
1
M = − wx 2
2
1 2
M A = − wa
2
Between A and B:
1
1 !
!
ΣM 2 = 0: M − " wL # ( x − a) + ( wx) " x # = 0
2
$
%
$2 %
1
1
1
M = − wx 2 + wLx − wLa
2
2
2
At center C:
x=
L
2
2
L!
L! 1
1
1
M C = − w " # + wL " # − wLa
2 $2%
2
$2% 2
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1074
PROBLEM 7.58 (Continued)
We set
| M A| = | M C | :
1
1
1
1
1
1
− wa 2 = wL2 − wLa + wa 2 = wL2 − wLa
2
8
2
2
8
2
a 2 + La − 0.25L2 = 0
1
1
( L ± L2 + L2 ) = ( 2 − 1) L
2
2
1
= w(0.207 L)2 = 0.0214 wL2
2
a=
M max
a = 0.207 L
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1075
PROBLEM 7.59
For the beam shown, determine (a) the magnitude P of the two
upward forces for which the maximum absolute value of the bending
moment in the beam is as small as possible, (b) the corresponding
value of |M |max . (See hint for Problem 7.55.)
SOLUTION
Ay = B = 60 kips − P
By symmetry:
Along AC:
ΣM J = 0: M − x(60 kips − P) = 0
M = (60 kips − P) x
M = 120 kips ⋅ ft − (2 ft) P at
x = 2 ft
Along CD:
ΣM K = 0: M + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0
M = 120 kip ⋅ ft − Px
M = 120 kip ⋅ ft − (4 ft) P at
x = 4 ft
Along DE:
ΣM L = 0: M − ( x − 4 ft) P + ( x − 2 ft)(60 kips)
− x(60 kips − P) = 0
M = 120 kip ⋅ ft − (4 ft) P (const)
Complete diagram by symmetry
For minimum |M |max , set M max = − M min
120 kip ⋅ ft − (2 ft) P = − [120 kip ⋅ ft − (4 ft) P]
M min = 120 kip ⋅ ft − (4 ft) P
(a)
P = 40.0 kips
(b)
| M |max = 40.0 kip ⋅ ft
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1076
PROBLEM 7.60
Knowing that P = Q = 150 lb, determine (a) the distance a for which
the maximum absolute value of the bending moment in beam AB is as
small as possible, (b) the corresponding value of |M |max . (See hint for
Problem 7.55.)
SOLUTION
Free body: Entire beam
ΣM A = 0: Da − (150)(30) − (150)(60) = 0
D=
13,500
a
!
Free body: CB
ΣM C = 0: − M C − (150)(30) +
13,500
(a − 30) = 0
a
45 !
M C = 9000 "1 − #
a %
$
!
Free body: DB
ΣM D = 0: − M D − (150)(60 − a) = 0
M D = −150(60 − a)
(a)
!"
We set
45 !
M max = | M min | or M C = − M D : 9000 "1 − # = 150(60 − a)
a %
$
60 −
2700
= 60 − a
a
a 2 = 2700 a = 51.96 in.
(b)
a = 52.0 in.
| M |max = − M D = 150(60 − 51.96)
| M | max = 1206 lb ⋅ in.
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1077
PROBLEM 7.61
Solve Problem 7.60 assuming that P = 300 lb and Q = 150 lb.
PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the
distance a for which the maximum absolute value of the bending
moment in beam AB is as small as possible, (b) the corresponding value
of |M |max . (See hint for Problem 7.55.)
SOLUTION
Free body: Entire beam
ΣM A = 0: Da − (300)(30) − (150)(60) = 0
D=
18,000
!
a
Free body: CB
ΣM C = 0: − M C − (150)(30) +
18,000
(a − 30) = 0
a
40 !
M C = 13,500 "1 − # !
a %
$
Free body: DB
ΣM D = 0: − M D − (150)(60 − a) = 0
M D = −150(60 − a) !
(a)
We set
40 !
M max = | M min | or M C = − M D : 13,500 "1 − # = 150(60 − a )
a %
$
90 −
3600
= 60 − a
a
a 2 + 30a − 3600 = 0
a=
−30 + 15.300
= 46.847
2
a = 46.8 in.
(b)
| M |max = − M D = 150(60 − 46.847)
|M | max = 1973 lb ⋅ in.
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1078
PROBLEM 7.62*
In order to reduce the bending moment in the cantilever beam
AB, a cable and counterweight are permanently attached at end
B. Determine the magnitude of the counterweight for which the
maximum absolute value of the bending moment in the beam is
as small as possible and the corresponding value of Mmax.
Consider (a) the case when the distributed load is permanently
applied to the beam, (b) the more general case when the
distributed load may either be applied or removed.
SOLUTION
M due to distributed load:
ΣM J = 0: − M −
x
wx = 0
2
1
M = − wx 2
2
M due to counter weight:
ΣM J = 0: − M + xw = 0
M = wx
(a)
Both applied:
M = Wx −
w 2
x
2
dM
W
= W − wx = 0 at x =
dx
w
And here M =
W2
> 0 so Mmax; Mmin must be at x = L
2w
So M min = WL −
so
1 2
wL . For minimum | M |max set M max = − M min ,
2
W2
1
= − WL + wL2
2w
2
or
W 2 + 2 wLW − w2 L2 = 0
W = − wL ± 2w2 L2 (need + )
W = ( 2 − 1) wL = 0.414 wL
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1079
PROBLEM 7.62* (Continued)
(b)
w may be removed
M max =
Without w,
With w (see Part a)
W 2 ( 2 − 1) 2 2
=
wL
2w
2
M max = 0.858 wL2
M = Wx
M max = WL at A
M = Wx −
w 2
x
2
W2
W
at x =
2w
w
1 2
= WL − wL at x = L
2
M max =
M min
For minimum M max , set M max (no w) = − M min (with w)
WL = − WL +
1 2
1
wL → W = wL →
2
4
With
M max =
1 2
wL
4
W=
1
wL
4
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1080
PROBLEM 7.63
Using the method of Section 7.6, solve Problem 7.29.
PROBLEM 7.29 For the beam and loading shown, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: C − P − P = 0
C = 2P
ΣM C = 0: P(2a) + P(a) − M C = 0
M C = 3Pa
Shear diagram
VA = − P
At A:
|V | max = 2 P
Bending-moment diagram
MA = 0
At A:
| M | max = 3Pa
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1081
PROBLEM 7.64
Using the method of Section 7.6, solve Problem 7.30.
PROBLEM 7.30 For the beam and loading shown, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
2L !
ΣM C = 0: P "
# − A( L) = 0
$ 3 %
A=
2
P
3
2
P−P+C =0
3
ΣFY = 0:
C=
1
P
3
Shear diagram
VA =
At A:
2
P
3
|V | max =
2
P
3
Bending-moment diagram
MA = 0
At A:
|M | max =
2
PL
9
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1082
PROBLEM 7.65
Using the method of Section 7.6, solve Problem 7.31.
PROBLEM 7.31 For the beam and loading shown, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
Reactions at A and D
Because of the symmetry of the supports and loading.
A=D=
1
L! 1
w # = wL
"
2$ 2 % 4
A=D=
1
wL v
4
|V | max =
1
wL
4
Shear diagram
1
VA = + wL
4
At A:
From B to C:
Oblique straight line
Bending-moment diagram
MA = 0
At A:
From B to C: ARC of parabola
|M | max =
3
wL2
32
Since V has no discontinuity at B nor C, the slope of the parabola at these
points is the same as the slope of the adjoining straight line segment.
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1083
PROBLEM 7.66
Using the method of Section 7.6, solve Problem 7.32.
PROBLEM 7.32 For the beam and loading shown, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: C − w
L
=0
2
C=
1
wL
2
1
! 3L !
ΣM C = 0: " wL #" # = M C = 0
$2
%$ 4 %
3
M C = wL2
8
Shear diagram
VA = 0
At A:
|V | max =
1
wL
2
Bending-moment diagram
M =0
At A:
dM
=V = 0
dx
3
|M | max = wL2
8
From A to B: ARC of parabola
Since V has no discontinuity at B, the slope of the parabola at B is equal to
the slope of the straightline segment.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1084
PROBLEM 7.67
Using the method of Section 7.6, solve Problem 7.33.
PROBLEM 7.33 For the beam and loading shown, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: B − P = 0
B=P
Σ M B = 0: M B − M 0 + PL = 0
MB = 0
Shear diagram
VA = − P
At A:
|V | max = P
Bending-moment diagram
M A = M 0 = PL
At A:
|M | max = PL
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1085
PROBLEM 7.68
Using the method of Section 7.6, solve Problem 7.34.
PROBLEM 7.34 For the beam and loading shown, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: A = C
ΣM C = 0: Al − M 0 = 0
A=C =
M0
L
Shear diagram
VA = −
At A:
M0
L
|V | max =
M0
L
Bending-moment diagram
MA = 0
At A:
At B, M increases by M0 on account of applied couple.
|M | max = M 0 /2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1086
PROBLEM 7.69
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
ΣM A = 0: (8)(2) + (4)(3.2) − 4C = 0
C = 7.2 kN
ΣFy = 0: A = 4.8 kN
(a)
Shear diagram
Similar Triangles:
x
3.2 − x 3.2
=
=
; x = 2.4 m
4.8
1.6
6.4
↑
Add num. & den.
Bending-moment diagram
(b)
|V | max = 7.20 kN
!
|M | max = 5.76 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1087
PROBLEM 7.70
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM D = 0: (500 N)(0.8 m) + (500 N)(0.4 m)
− (2400 N/m)(0.3 m)(0.15 m) − A(1.2 m) = 0
A = 410 N
ΣFy = 0: 410 − 2(500) − 2400(0.3) + D = 0
D = 1310 N
Shear diagram
At A:
VA = + 410 N
|V | max = 720 N
Bending-moment diagram
At A:
MA = 0
| M | max = 164.0 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1088
PROBLEM 7.71
Using the method of Section 7.6, solve Problem 7.39.
PROBLEM 7.39 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM B = 0: (60 kN)(3 m) + (50 kN)(1 m) − Ay (5 m) = 0
Ay = + 46.0 kN v
ΣFy = 0: B + 46.0 kN − 60 kN − 50 kN = 0
B = + 64.0 kN v
Shear diagram
At A:
VA = Ay = + 46.0 kN
|V |max = 64.0 kN
Bending-moment diagram
At A:
MA = 0
| M |max = 92.0 kN ⋅ m
Parabola from D to B. Its slope at D is same as that of straight-line segment
CD since V has no discontinuity at D.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1089
PROBLEM 7.72
Using the method of Section 7.6, solve Problem 7.40.
PROBLEM 7.40 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute
values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM B = 0: (30 kN/m)(2 m)(6 m) − C (5 m) + (80 kN)(2 m) = 0
C = 104 kN
ΣFy = 0: 104 − 30(2) − 80 + B = 0
B = 36 kN
Shear diagram
VA = 0
At A:
|V |max = 60.0 kN
Bending-moment diagram
MA = 0
At A:
| M |max = 72.0 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1090
PROBLEM 7.73
Using the method of Section 7.6, solve Problem 7.41.
PROBLEM 7.41 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Reactions at supports.
Because of the symmetry:
A= B=
1
(8 + 8 + 4 × 5) kips
2
A = B = 18 kips v
Shear diagram
At A:
VA = + 18 kips
|V |max = 18.00 kips
Bending-moment diagram
At A:
MA = 0
| M |max = 48.5 kip ⋅ ft
Discontinuities in slope at C and D, due to the discontinuities of V.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1091
PROBLEM 7.74
Using the method of Section 7.6, solve Problem 7.42.
PROBLEM 7.42 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM B = 0: (12 kips)(4ft) + (15 kips)(7 ft) − Ay (10 ft) = 0
Ay = + 15.3 kips v
ΣFy = 0: B + 15.3 − 15 − 12 = 0
B = + 11.7 kips v
Shear diagram
At A:
VA = Ay = 15.3 kips
|V |max = 15.30 kips
Bending-moment diagram
At A:
MA = 0
| M |max = 46.8 kip ⋅ ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1092
PROBLEM 7.75
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Beam
ΣM B = 0: 5 kip ⋅ ft + (1 kips)(10 ft) + (4 kips)(5 ft) − Ay (15 ft) = 0
Ay = + 3.00 kips v
ΣFx = 0: Ax = 0
Shear diagram
At A:
VA = Ay = + 3.00 kips
|V | max = 3.00 kips
Bending-moment diagram
At A:
M A = − 5 kip ⋅ ft
| M |max = 15.00 kip ⋅ ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1093
PROBLEM 7.76
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Beam
ΣM B = 0: 6 kip ⋅ ft + 12 kip ⋅ ft + (2 kips)(18 ft)
+ (3 kips)(12 ft) + (4 kips)(6 ft) − Ay (24 ft) = 0
Ay = + 4.75 kips v
ΣFx = 0: Ax = 0
Shear diagram
At A:
VA = Ay = + 4.75 kips
|V |max = 4.75 kips
Bending-moment diagram
At A:
M A = − 6 kip ⋅ ft
| M |max = 39.0 kip ⋅ ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1094
PROBLEM 7.77
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM B = 0: (19.2 kN)(1.2 m) − Ay (3.2 m) = 0
Ay = + 7.20 kN v
Shear diagram
VA = VC = Ay = + 7.20 kN
To determine Point D where V = 0, we write
VD − VC = wu
0 − 7.20 kN = − (8 kN/m)u
u = 0.9 m v
We next compute all areas
Bending-moment diagram
At A:
MA = 0
Largest value occurs at D with
AD = 0.8 + 0.9 = 1.700 m
| M |max = 9.00 kN ⋅ m
!
!
1.700 m from A
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1095
PROBLEM 7.78
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum
bending moment.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM B = 0: (40 kN)(1.5 m) − Ay (3.75 m) = 0
Ay = + 16.00 kN v
Shear diagram
VA = VC = Ay = + 16.00 kN
To determine Point E where V = 0, we write
VE − VC = − wu
0 − 16 kN = − (20 kN/m)u
u = 0.800 m v
We next compute all areas
Bending-moment diagram
At A:
MA = 0
Largest value occurs at E with
AE = 1.25 + 0.8 = 2.05 m
| M |max = 26.4 kN ⋅ m
2.05 m from A
From A to C and D to B: Straight line segments. From C to D: Parabola
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1096
PROBLEM 7.79
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.
SOLUTION
Free body: Entire beam
ΣM C = 0: − (27 kip ⋅ ft) + A(22.5 ft) − (1 kip/ft)(18 ft)(13.5 ft) = 0
A = 12 kips
ΣFy = 0: 12 − 1(18) + C = 0
C = 6 kips
Shear diagram
VA = + 12 kips
At A:
To locate Point D (where V = 0)
VD − VA = − wu
0 − 12 kips = − (1 kip/ft)u
u = 12 ft
Bending-moment diagram
MA = 0
At A:
| M |max = 45.0 kip ⋅ ft
12.00 ft from A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1097
PROBLEM 7.80
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum absolute
value of the bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: (6 kips)(10 ft) − (9 kips)(7 ft) + 8(4 ft) = 0
B = + 0.75 kips
B = 0.75 kips
ΣFy = 0: A + 0.75 kips − 9 kips + 6 kips = 0
A = + 2.25 kips
A = 2.25 kips
M max = 12.00 kip ⋅ ft
6.00 ft from A
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1098
PROBLEM 7.81
(a) Draw the shear and bending-moment diagrams for beam AB, (b) determine
the magnitude and location of the maximum absolute value of the bending
moment.
SOLUTION
Reactions at supports
Because of symmetry of load
A= B=
1
(300 × 8 + 300)
2
A = B = 1350 lb v
Load diagram for AB
The 300-lb force at D is replaced by an equivalent force-couple system at C.
Shear diagram
VA = A = 1350 lb
At A:
To determine Point E where V = 0:
VE − VA = − wx
0 − 1350 lb = − (300 lb/ft) x
x = 4.50 ft v
We compute all areas
Bending-moment diagram
At A:
MA = 0
Note 600 − lb ⋅ ft drop at C due to couple
| M |max = 3040 lb ⋅ ft
!
!
4.50 ft from A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1099
PROBLEM 7.82
Solve Problem 7.81 assuming that the 300-lb force applied at D is directed
upward.
PROBLEM 7.81 (a) Draw the shear and bending-moment diagrams for
beam AB, (b) determine the magnitude and location of the maximum absolute
value of the bending moment.
SOLUTION
Reactions at supports
Because of symmetry of load:
A= B=
1
(300 × 8 − 300)
2
A = B = 1050 lb v
Load diagram
The 300-lb force at D is replaced by an equivalent force-couple
system at C.
Shear diagram
VA = A = 1050 lb
At A:
To determine Point E where V = 0:
VE − VA = − wx
0 − 1050 lb = − (300 lb/ft) x
x = 3.50 ft v
We compute all areas
Bending-moment diagram
MA = 0
At A:
Note 600 − lb ⋅ ft increase at C due to couple
| M |max = 1838 lb ⋅ ft
!
!
3.50 ft from A
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1100
PROBLEM 7.83
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.
SOLUTION
Free body: Beam
ΣM A = 0: B (4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0
B = + 55 kN v
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 55 − 100 = 0
Ay = + 45 kN v
Shear diagram
VA = Ay = + 45 kN
At A:
To determine Point C where V = 0:
VC − VA = − wx
0 − 45 kN = − (25 kN ⋅ m) x
x = 1.8 m v
We compute all areas bending-moment
Bending-moment diagram
At A:
MA = 0
At B:
M B = − 20 kN ⋅ m
| M |max = 40.5 kN ⋅ m
!
!
1.800 m from A
!
Single arc of parabola
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1101
PROBLEM 7.84
Solve Problem 7.83 assuming that the 20-kN ⋅ m couple applied at B is
counterclockwise.
PROBLEM 7.83 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the magnitude and location
of the maximum absolute value of the bending moment.
SOLUTION
Free body: Beam
ΣM A = 0: B (4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0
B = + 45 kN v
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 45 − 100 = 0
Ay = + 55 kN v
Shear diagram
VA = Ay = + 55 kN
At A:
To determine Point C where V = 0 :
VC − VA = − wx
0 − 55 kN = − (25 kN/m) x
x = 2.20 m v
We compute all areas bending-moment
Bending-moment diagram
At A:
MA = 0
At B:
M B = + 20 kN ⋅ m
| M |max = 60.5 kN ⋅ m
!
!
2.20 m from A
!
Single arc of parabola
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1102
PROBLEM 7.85
For the beam and loading shown, (a) write the equations of the
shear and bending-moment curves, (b) determine the magnitude and
location of the maximum bending moment.
SOLUTION
Distributed load
1
x!
!
w = w0 "1 − # " Total = w0 L #
2
L% $
$
%
L 1
!
w0 L # − LB = 0
"
3 $2
%
ΣM A = 0:
ΣFy = 0: Ay −
B=
w0 L
6
wL
wL
1
w0 L + 0 = 0 A y = 0
2
6
3
w0 L
3
Shear:
VA = Ay =
Then
dV
= −w → V
dx
x!
w0 "1 − # dx
L%
$
w0 L !
1 w0 2
V ="
x
− w0 x +
#
2 L
$ 3 %
= VA −
,
x
0
& 1 x 1 x !2 '
= w0 L ( − + " # )
(* 3 L 2 $ L % )+
Note: At
x=L
V =−
w0 L
6
2
x!
x! 2
V = 0 at " # − 2 " # +
$L%
$L% 3
=0→
x
1
=1−
L
3
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1103
PROBLEM 7.85 (Continued)
Moment:
Then
MA = 0
x
x/ L
dM !
x!
x!
"
# = V → M = 0 Vdx = L 0 V " # d " #
$ dx %
$ L% $ L%
2
x/ L & 1
x 1 x! '
x!
M = w0 L2
( − + " # )d" #
0 (3
L
L
L
2
$
%
$
%
*
+)
,
,
,
& 1 x ! 1 x ! 2 1 x !3 '
M = w0 L ( " # − " # + " # )
6 $ L % )+
(* 3 $ L % 2 $ L %
1!
x
2
M max " at = 1 −
## = 0.06415w0 L
" L
3
$
%
2
& 1 x 1 x !2 '
V = w0 L ( − + " # )
(* 3 L 2 $ L % )+
!
& 1 x ! 1 x ! 2 1 x !3 '
M = w0 L ( " # − " # + " # )
6 $ L % )+
(* 3 $ L % 2 $ L %
!
M max = 0.0642 w0 L2
!
(a)
2
(c)
at x = 0.423L
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1104
PROBLEM 7.86
For the beam and loading shown, (a) write the equations of the
shear and bending-moment curves, (b) determine the magnitude and
location of the maximum bending moment.
SOLUTION
(a)
We note that at
Load:
B( x = L): VB = 0, M B = 0
(1)
4x !
x! 1
x!
w( x) = w0 "1 − # − w0 " # = w0 "1 −
#
L% 3 $L%
$
$ 3L %
Shear: We use Eq. (7.2) between C ( x = x) and B( x = L):
VB − VC = −
,
V ( x) = w0
L
x
,
w( x) dx 0 − V ( x) = −
L
x
,
L
x
w( x)dx
4x !
"1 − 3L # dx
$
%
L
&
2x2 '
2L
2x2 !
= w0 ( x −
−x+
#
) = w0 "" L −
3L + x
3
3L %#
*
$
V ( x) =
w0
(2 x 2 − 3Lx + L2 )
3L
(2)
Bending moment: We use to Eq. (7.4) between C ( x = x) and B( x = L) :
M B − MC =
=
,
L
x
w0
3L
V ( x)dx 0 − M ( x)
,
L
x
(2 x 2 − 3Lx + L2 )dx
L
w0 & 2 3 3 2
'
x − Lx + L2 x )
(
3L * 3
2
+
w
= − 0 [4 x3 − 9 Lx 2 + 6 L2 x]Lx
18L
w
= − 0 [(4 L3 − 9 L3 + 6 L3 ) − (4 x3 − 9 Lx 2 + 6 L2 x)]
18L
M ( x) = −
M ( x) =
w0
(4 x3 − 9 Lx 2 + 6 L2 x − L3 )
18L
(3)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1105
PROBLEM 7.86 (Continued)
(b)
Maximum bending moment
dM
=V = 0
dx
Eq. (2):
2 x 2 − 3Lx + L2 = 0
x=
Carrying into (3):
Note:
M max =
| M |max =
3− 9−8
L
L=
4
2
w0 L2
, At
72
w0 L2
18
At
x=
L
2
x =0!
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1106
PROBLEM 7.87
For the beam and loading shown, (a) write the equations of the shear
and bending-moment curves, (b) determine the magnitude and location
of the maximum bending moment.
SOLUTION
π x
dv
= − w = w0 cos
2 L
dx
2L ! π x
V = − wdx = − w0 "
# sin 2 L + C1
$ π %
,
(1)
2L ! π x
dM
= V = − w0 "
# sin 2 L + C1
dx
$ π %
2
2L !
πx
cos
M = Vdx = + w0 "
+ C1 x + C2
#
2L
$ π %
,
(2)
Boundary conditions
At x = 0:
V = C1 = 0 C1 = 0
At x = 0:
2L !
M = + w0 "
# cos(0) + C2 = 0
$ π %
2
2L !
C2 = − w0 "
#
$ π %
2
2L ! π x
V = − w0 "
# sin 2 L
$ π %
Eq. (1)
2
πx!
2L !
−1 + cos
M = w0 "
#
"
2 L #%
$ π % $
2
2L !
4
| M max | = w0 "
| −1 + 0|= 2 w0 L2
#
π
$ π %
M max at x = L :
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1107
!
PROBLEM 7.88
The beam AB, which lies on the ground, supports the parabolic
load shown. Assuming the upward reaction of the ground to be
uniformly distributed, (a) write the equations of the shear and
bending-moment curves, (b) determine the maximum bending
moment.
SOLUTION
ΣFy = 0: wg L −
(a)
,
L
0
4 w0
L2
( Lx − x 2 ) dx = 0
4 w0 1 2 1 3 ! 2
" LL − 3 L # = 3 w0 L
L2 $ 2
%
2 w0
wg =
3
wg L =
2
&x
x! ' 2
net load w = 4 w0 ( − " # ) − w0
(* L $ L % )+ 3
Define
x
dx
ξ = so d ξ =
L
L
or
1
!
w = 4 w0 " − + ξ − ξ 2 #
6
$
%
1
!
4w0 L " − + ξ − ξ 2 # d ξ
6
$
%
1
1 2 1 3!
2
V = w0 L (ξ − 3ξ 2 + 2ξ 3 )
= 0 + 4w0 L " ξ + ξ − ξ #
2
3 %
3
$6
x
ξ
2
M = M 0 + Vdx = 0 + w0 L2 (ξ − 3ξ 2 + 2ξ 3 )d ξ
0
0
3
2
1
1 ! 1
= w0 L2 " ξ 2 − ξ 3 + ξ 4 # = w0 L2 (ξ 2 − 2ξ 3 + ξ 4 )
3
2 % 3
$2
V = V (0) −
,
ξ
0
,
(b)
Max M occurs where
V =0
,
1 − 3ξ + 2ξ 2 = 0
ξ=
1
2
1! 1
1 2 1 ! w L2
M " ξ = # = w0 L2 " − + # = 0
2% 3
48
$
$ 4 8 16 %
M max =
w0 L2
at center of beam
48
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1108
!
PROBLEM 7.89
The beam AB is subjected to the uniformly distributed load shown and to two
unknown forces P and Q. Knowing that it has been experimentally
determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E,
(a) determine P and Q, (b) draw the shear and bending-moment diagrams for
the beam.
SOLUTION
(a)
Free body: Portion AD
ΣFx = 0: Cx = 0
ΣM D = 0: −C y (0.3 m) + 0.800 kN ⋅ m + (6 kN)(0.45 m) = 0
C y = +11.667 kN
C = 11.667 kN "
Free body: Portion EB
ΣM E = 0: B(0.3 m) − 1.300 kN ⋅ m = 0
B = 4.333 kN "
Free body: Entire beam
ΣM D = 0: (6 kN)(0.45 m) − (11.667 kN)(0.3 m)
− Q (0.3 m) + (4.333 kN)(0.6 m) = 0
Q = 6.00 kN
ΣM y = 0: 11.667 kN + 4.333 kN
−6 kN − P − 6 kN = 0
P = 4.00 kN
Load diagram
(b)
Shear diagram
At A: VA = 0
|V |max = 6 kN "!
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1109
PROBLEM 7.89 (Continued)
Bending-moment diagram
MA = 0
At A:
| M |max = 1300 N ⋅ m "!
We check that
M D = +800 N ⋅ m and M E = +1300 N ⋅ m
As given:
M C = −900 N ⋅ m
At C:
!
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1110
PROBLEM 7.90
Solve Problem 7.89 assuming that the bending moment was found to
be +650 N ⋅ m at D and +1450 N ⋅ m at E.
PROBLEM 7.89 The beam AB is subjected to the uniformly
distributed load shown and to two unknown forces P and Q. Knowing
that it has been experimentally determined that the bending moment
is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q,
(b) draw the shear and bending-moment diagrams for the beam.
SOLUTION
!
(a)
Free body: Portion AD
ΣFx = 0: Cx = 0
ΣM D = 0: −C (0.3 m) + 0.650 kN ⋅ m + (6 kN)(0.45 m) = 0
C y = +11.167 kN
C = 11.167 kN "
Free body: Portion EB
ΣM E = 0: B(0.3 m) − 1.450 kN ⋅ m = 0
!
!
B = 4.833 kN "
!
Free body: Entire beam
ΣM D = 0: (6 kN)(0.45 m) − (11.167 kN)(0.3 m)
−Q (0.3 m) + (4.833 kN)(0.6 m) = 0
Q = 7.50 kN
!
!
ΣM y = 0: 11.167 kN + 4.833 kN
− 6 kN − P − 7.50 kN = 0
P = 2.50 kN
Load diagram
!
!
(b)
Shear diagram
VA = 0
At A:
|V |max = 6 kN "!
!
!
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1111
PROBLEM 7.90 (Continued)
!
!
Bending-moment diagram
MA = 0
At A:
| M |max = 1450 N ⋅ m "!
We check that
M D = +650 N ⋅ m and M E = +1450 N ⋅ m
As given:
!
M C = −900 N ⋅ m
At C:
!
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1112
PROBLEM 7.91*
The beam AB is subjected to the uniformly distributed load shown and
to two unknown forces P and Q. Knowing that it has been
experimentally determined that the bending moment is +6.10 kip ⋅ ft at
D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and
bending-moment diagrams for the beam.
SOLUTION
(a)
Free body: Portion DE
ΣM E = 0: 5.50 kip ⋅ ft − 6.10 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0
VD = +0.350 kip
ΣFy = 0: 0.350 kip − 1kip − VE = 0
VE = −0.650 kip
Free body: Portion AD
ΣM A = 0: 6.10 kip ⋅ ft − P(2ft) − (1 kip)(2 ft) − (0.350 kip)(4 ft) = 0
P = 1.350 kips
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 1 kip − 1.350 kip − 0.350 kip = 0
Ay = +2.70 kips
A = 2.70 kips
Free body: Portion EB
ΣM B = 0: (0.650 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 5.50 kip ⋅ ft = 0
Q = 0.450 kip
ΣFy = 0: B − 0.450 − 1 − 0.650 = 0
B = 2.10 kips
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1113
PROBLEM 7.91* (Continued)
(b)
Load diagram
Shear diagram
VA = A = +2.70 kips
At A:
To determine Point G where V = 0, we write
VG − VC = − wµ
0 − 0.85 kips = −(0.25 kip/ft)µ
µ = 3.40 ft
|V |max = 2.70 kips at A
We next compute all areas
Bending-moment diagram
At A: M A = 0
Largest value occurs at G with
AG = 2 + 3.40 = 5.40 ft
| M |max = 6.345 kip ⋅ ft
5.40 ft from A
Bending-moment diagram consists of 3 distinct arcs of parabolas.!
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1114
PROBLEM 7.92*
Solve Problem 7.91 assuming that the bending moment was found to
be +5.96 kip ⋅ ft at D and +6.84 kip ⋅ ft at E.
PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load
shown and to two unknown forces P and Q. Knowing that it has been
experimentally determined that the bending moment is +6.10 kip ⋅ ft at D
and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.
SOLUTION
(a)
Free body: Portion DE
ΣM E = 0: 6.84 kip ⋅ ft − 5.96 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0
VD = +0.720 kip
ΣFy = 0: 0.720 kip − 1kip − VE = 0
VE = −0.280 kip
Free body: Portion AD
ΣM A = 0: 5.96 kip ⋅ ft − P (2 ft) − (1 kip)(2 ft) − (0.720 kip)(4 ft) = 0
P = 0.540 kip
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 1 kip − 0.540 kip − 0.720 kip = 0
Ay = +2.26 kips
A = 2.26 kips "
Free body: Portion EB
ΣM B = 0: (0.280 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 6.84 kip ⋅ ft = 0
Q = 1.860 kips
ΣFy = 0: B − 1.860 − 1 − 0.280 = 0
B = 3.14 kips
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1115
PROBLEM 7.92* (Continued)
(b)
Load diagram
Shear diagram
VA = A = +2.26 kips
At A:
To determine Point G where V = 0, we write
VG − VC = − wµ
0 − (1.22 kips) = −(0.25 kip/ft)µ
µ = 4.88 ft "
|V |max = 3.14 kips at B
We next compute all areas
Bending-moment diagram
At A: M A = 0
Largest value occurs at G with
AG = 2 + 4.88 = 6.88 ft
| M |max = 6.997 kip ⋅ ft
6.88 ft from A
Bending-moment diagram consists of 3 distinct arcs of parabolas.!
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1116
PROBLEM 7.93
Two loads are suspended as shown from the cable ABCD.
Knowing that hB = 1.8 m, determine (a) the distance hC , (b) the
components of the reaction at D, (c) the maximum tension in the
cable.
SOLUTION
ΣFx = 0: − Ax + Dx = 0
FBD Cable:
Ax = Dx
ΣM A = 0: (10 m)Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0
D y = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
FDB AB:
ΣM B = 0: (1.8 m) Ax − (3 m)(8.2 kN) = 0
Ax =
From above
Dx = Ax =
41
kN
3
41
kN
3
FBD CD:
41
!
ΣM C = 0: (4 m)(7.8 kN) − hC "
kN # = 0
$ 3
%
hC = 2.283 m
hC = 2.28 m
(a)
D x = 13.67 kN
(b)
D y = 7.80 kN
Since Ax = Bx and Ay . B y , max T is TAB
2
TAB =
Ax2
+
Ay2
41
!
kN # + (8.2 kN)2
= "
$ 3
%
Tmax = 15.94 kN
(c)
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1117
!
PROBLEM 7.94
Knowing that the maximum tension in cable ABCD is 15 kN,
determine (a) the distance hB , (b) the distance hC .
SOLUTION
ΣFx = 0: − Ax + Dx = 0
FBD Cable:
Ax = Dx
ΣM A = 0: (10 m) Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0
D y = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
Since
Ax = Dx
and
Ay > Dy , Tmax = TAB
FBD Pt A:
ΣFy = 0: 8.2 kN − (15 kN)sin θ A = 0
θ A = sin −1
8.2 kN
= 33.139°
15 kN
ΣFx = 0: − Ax + (15 kN) cos θ A = 0
Ax = (15 kN) cos(33.139°) = 12.56 kN
FBD CD:
From FBD cable:
hB = (3 m) tan θ A
= (3 m) tan(33.139°)
hB = 1.959 m
(a)
ΣM C = 0: (4 m)(7.8 kN) − hC (12.56 kN) = 0
hC = 2.48 m
(b)
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1118
!
PROBLEM 7.95
If dC = 8 ft, determine (a) the reaction at A, (b) the reaction at E.
SOLUTION
Free body: Portion ABC
ΣM C = 0
2 Ax − 16 Ay + 300(8) = 0
Ax = 8 Ay − 1200
(1)
Free body: Entire cable
ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0
3 Ax + 16 Ay − 6400 = 0
Substitute from Eq. (1):
3(8 Ay − 1200) + 16 Ay − 6400 = 0
Eq. (1)
Ax = 8(250) − 1200
A y = 250 lb
A x = 800 lb
ΣFx = 0: − Ax + Ex = 0 − 800 lb + Ex = 0
E x = 800 lb
ΣFy = 0: 250 + E y − 300 − 200 − 300 = 0
E y = 550 lb
A = 838 lb
E = 971 lb
17.4°
34.5°
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1119
!
PROBLEM 7.96
If dC = 4.5 ft, determine (a) the reaction at A, (b) the reaction at E.
SOLUTION
Free body: Portion ABC
ΣM C = 0: − 1.5 Ax − 16 Ay + 300 × 8 = 0
Ax =
Free body: Entire cable
(2400 − 16 Ay )
(1)
1.5
ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0
3 Ax + 16 Ay − 6400 = 0
Substitute from Eq. (1):
3(2400 − 16 Ay )
1.5
+ 16 Ay − 6400 = 0
Ay = −100 lb
A y = 100 lb
Thus Ay acts downward
Eq. (1)
(2400 − 16( −100))
= 2667 lb
1.5
A x = 2667 lb
ΣFx = 0: − Ax + Ex = 0 − 2667 + Ex = 0
E x = 2667 lb
Ax =
ΣFy = 0: Ay + E y − 300 − 200 − 300 = 0
−100 lb + E y − 800 lb = 0
E y = 900 lb
A = 2670 lb
2.10°
E = 2810 lb
18.6°
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1120
!
PROBLEM 7.97
Knowing that dC = 3 m, determine (a) the distances dB and dD (b) the
reaction at E.
SOLUTION
Free body: Portion ABC
ΣM C = 0: 3 Ax − 4 Ay + (5 kN)(2 m) = 0
Ax =
4
10
Ay −
3
3
(1)
Free body: Entire cable
ΣM E = 0: 4 Ax − 10 Ay + (5 kN)(8 m) + (5 kN)(6 m) + (10 kN)(3 m) = 0
4 Ax − 10 Ay + 100 = 0
Substitute from Eq. (1):
4
10 !
4 " Ay − # − 10 Ay + 100 = 0
3
3%
$
Ay = +18.571 kN
Eq. (1)
A y = 18.571 kN
4
10
(18.511) −
= +21.429 kN
3
3
A x = 21.429 kN
ΣFx = 0: − Ax + Ex = 0 − 21.429 + Ex = 0
E x = 21.429 kN
ΣFy = 0: 18.571 kN + E y + 5 kN + 5 kN + 10 kN = 0
E y = 1.429 kN
Ax =
E = 21.5 kN
3.81°
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1121
PROBLEM 7.97 (Continued)
Portion AB
ΣM B = 0: (18.571 kN)(2 m) − (21.429 kN)d B = 0
d B = 1.733 m
Portion DE
Geometry
h = (3 m) tan 3.8°
= 0.199 m
d D = 4 m + 0.199 m
d D = 4.20 m
!
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1122
!
PROBLEM 7.98
Determine (a) distance dC for which portion DE of the cable is
horizontal, (b) the corresponding reactions at A and E.
SOLUTION
Free body: Entire cable
ΣFy = 0: Ay − 5 kN − 5 kN − 10 kN = 0
A y = 20 kN
ΣM A = 0: E (4 m) − (5 kN)(2 m) − (5 kN)(4 m) − (10 kN)(7m) = 0
E = +25 kN
E = 25.0 kN
ΣFx = 0: − Ax + 25 kN = 0
A x = 25 kN
A = 32.0 kN
38.7°
!
Free body: Portion ABC
ΣM C = 0: (25 kN)dC − (20 kN)(4 m) + (5 kN)(2 m) = 0
25dC − 70 = 0
dC = 2.80 m
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1123
PROBLEM 7.99
If dC = 15 ft, determine (a) the distances dB and dD,
(b) the maximum tension in the cable.
SOLUTION
Free body: Entire cable
ΣM A = 0: E y (30 ft) − Ex (7.5 ft) − (2 kips)(6 ft) − (2 kips)(15 ft) − (2 kips)(21 ft) = 0
7.5Ex − 30 E y + 84 = 0
(1)
Free body: Portion CDE
ΣM C = 0: E y (15 ft) − Ex (15 ft) − (2 kips)(6 ft) = 0
15Ex − 15E y + 12 = 0
(2)
1
Eq. (1) × : 3.75 Ex − 15E y + 42 = 0
2
(2) – (3):
(3)
11.25Ex − 30 = 0
Ex = 2.6667 kips
Eq. (1):
7.5(2.6667) − 30 E y + 84 = 0 E y = 3.4667 kips
Tm = Ex2 + E y2 = (2.6667) 2 + (3.4667)2
Tm = 4.37 kips
!
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1124
PROBLEM 7.99 (Continued)
Free body: Portion DE
ΣM D = 0: (3.4667 kips)(9 ft) − (2.6667 kips) d D = 0
Return to free body of entire cable
d D = 11.70 ft
(with Ex = 2.6667 kips, E y = 3.4667 kips)
ΣFy = 0: Ay − 3(2 kips) + 3.4667 kips = 0
Ay = 2.5333 kips
ΣFx = 0: 2.6667 kips − Ax = 0
Ax = 2.6667 kips
Free body: Portion AB
ΣM B = 0: Ax ( d B − 7.5) − Ay (6) = 0
(2.6667)(d B − 7.5) − (2.5333)(6) = 0
d B = 13.20 ft
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1125
PROBLEM 7.100
Determine (a) the distance dC for which portion BC of the
cable is horizontal, (b) the corresponding components of
the reaction at E.
SOLUTION
Free body: Portion CDE
ΣFy = 0: E y − 2(2 kips) = 0 E y = 4 kips
ΣM C = 0: (4 kips)(15 ft) − Ex dC − (2 kips)(6 ft) = 0
Ex dC = 48 kip ⋅ ft
(1)
Free body: Entire cable
ΣM A = 0: (4 kips)(30 ft) − Ex (7.5 ft) − (2 kips)(6 ft) − (2 kips)(15 ft) − (2 kips)(21 ft) = 0
Ex = 4.8 kips
From Eq. (1):
dC =
48 48
=
Ex 4.8
dC = 10.00 ft
E x = 4.80 kips
Components of reaction at E:
;
E y = 4.00 kips
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1126
PROBLEM 7.101
Cable ABC supports two loads as shown. Knowing that b = 4 ft, determine
(a) the required magnitude of the horizontal force P, (b) the corresponding
distance a.
SOLUTION
FBD ABC:
ΣFy = 0: − 40 lb − 80 lb + C y = 0
C y = 120 lb
FBD BC:
ΣM B = 0: (4 ft)(120 lb) − (10 ft)Cx = 0
C x = 48 lb
From ABC:
ΣFx = 0: − P + Cx = 0
P = Cx = 48 lb
(a)
P = 48.0 lb
(b)
a = 10.00 ft
!
ΣM C = 0: (4 ft)(80 lb) + a(40 lb) − (15 ft)(48 lb) = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1127
PROBLEM 7.102
Cable ABC supports two loads as shown. Determine the distances a and b
when a horizontal force P of magnitude 60 lb is applied at A.
SOLUTION
FBD ABC:
ΣFx = 0: Cx − P = 0
C x = 60 lb
ΣFy = 0: C y − 40 lb − 80 lb = 0
C y = 120 lb
FBD BC:
ΣM B = 0: b(120 lb) − (10 ft)(60 lb) = 0
b = 5.00 ft
FBD AB:
ΣM B = 0: (a − b)(40 lb) − (5 ft)60 lb = 0
a − b = 7.5 ft
a = b + 7.5 ft
= 5 ft + 7.5 ft
a = 12.50 ft
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1128
PROBLEM 7.103
Knowing that mB = 70 kg and mC = 25 kg, determine the
magnitude of the force P required to maintain equilibrium.
SOLUTION
Free body: Portion CD
ΣM C = 0: D y (4 m) − Dx (3 m) = 0
Dy =
3
Dx
4
Free body: Entire cable
ΣM A = 0:
3
Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0
4
ΣM B = 0:
3
Dx (10 m) − Dx (5 m) − WC (6 m) = 0
4
(1)
Free body: Portion BCD
Dx = 2.4WC
For
mB = 70 kg mC = 25 kg
g = 9.81 m/s 2 :
WB = 70 g
Eq. (2):
Dx = 2.4WC = 2.4(25g ) = 60 g
Eq. (1):
(2)
WC = 25 g
3
60 g (14) − 70 g (4) − 25 g (10) − 5P = 0
4
100 g − 5 P = 0: P = 20 g
P = 20(9.81) = 196.2 N
P = 196.2 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1129
PROBLEM 7.104
Knowing that mB = 18 kg and mC = 10 kg, determine the magnitude
of the force P required to maintain equilibrium.
SOLUTION
Free body: Portion CD
ΣM C = 0: D y (4 m) − Dx (3 m) = 0
Dy =
3
Dx
4
Free body: Entire cable
ΣM A = 0:
3
Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0
4
ΣM B = 0:
3
Dx (10 m) − Dx (5 m) − WC (6 m) = 0
4
(1)
Free body: Portion BCD
Dx = 2.4WC
For
mB = 18 kg mC = 10 kg
g = 9.81 m/s 2 :
WB = 18 g
Eq. (2):
Dx = 2.4WC = 2.4(10 g ) = 24 g
Eq. (1):
3
24 g (14) − (18 g )(4) − (10 g )(10) − 5P = 0
4
(2)
WC = 10 g
80 g − 5P : P = 16 g
P = 16(9.81) = 156.96 N
P = 157.0 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1130
PROBLEM 7.105
If a = 3 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.
SOLUTION
Free body: Portion DE
ΣM D = 0: E y (4 m) − Ex (5 m) = 0
Ey =
5
Ex
4
Free body: Portion CDE
ΣM C = 0:
5
Ex (8 m) − Ex (7 m) − P(2 m) = 0
4
Ex =
2
P
3
(1)
Free body: Portion BCDE
ΣM B = 0:
5
Ex (12 m) − Ex (5 m) − (120 kN)(4 m) = 0
4
10 Ex − 480 = 0; Ex = 48 kN
Eq. (1):
48 kN =
2
P
3
P = 72.0 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1131
PROBLEM 7.105 (Continued)
Free body: Entire cable
ΣM A = 0:
5
Ex (16 m) − Ex (2 m) + P(3 m) − Q (4 m) − (120 kN)(8 m) = 0
4
(48 kN)(20 m − 2 m) + (72 kN)(3 m) − Q(4 m) − 960 kN ⋅ m = 0
4Q = 120
Q = 30.0 kN
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1132
PROBLEM 7.106
If a = 4 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.
SOLUTION
Free body: Portion DE
ΣM D = 0: E y (4 m) − Ex (6 m) = 0
Ey =
3
Ex
2
Free body: Portion CDE
ΣM C = 0:
3
Ex (8 m) − Ex (8 m) − P(2 m) = 0
2
Ex =
1
P
2
(1)
Free body: Portion BCDE
ΣM B = 0:
3
Ex (12 m) − Ex (6 m) + (120 kN)(4 m) = 0
2
12 Ex = 480
Eq (1):
Ex =
1
P;
2
Ex = 40 kN
40 kN =
1
P
2
P = 80.0 kN
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1133
PROBLEM 7.106 (Continued)
Free body: Entire cable
ΣM A = 0:
3
Ex (16 m) − Ex (2 m) + P(4 m) − Q(4 m) − (120 kN)(8 m) = 0
2
(40 kN)(24 m − 2 m) + (80 kN)(4 m) − Q (4 m) − 960 kN ⋅ m = 0
4Q = 240
Q = 60.0 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1134
PROBLEM 7.107
A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation that
are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum tension in
the wire.
SOLUTION
Eq. 7.16:
Solve by trial and error:
Eq. 7.15:
xB
c
60
30m + c = c cosh
c
yB = c cosh
c = 64.459 m
sB = c sin h
xB
c
sB = (64.456 m) sinh
60 m
64.459 m
sB = 69.0478 m
Length = 2 sB = 2(69.0478 m) = 138.0956 m
Eq. 7.18:
L = 138.1 m
Tm = wyB = w(h + c)
= (0.65 kg/m)(9.81 m/s 2 )(30 m + 64.459 m)
Tm = 602.32 N
Tm = 602 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1135
PROBLEM 7.108
Two cables of the same gauge are attached to a transmission tower
at B. Since the tower is slender, the horizontal component of the
resultant of the forces exerted by the cables at B is to be zero.
Knowing that the mass per unit length of the cables is 0.4 kg/m,
determine (a) the required sag h, (b) the maximum tension in each
cable.
SOLUTION
W = wxB
ΣM B = 0: T0 yB − ( wxB )
Horiz. comp. = T0 =
(a)
Cable AB
w(45 m) 2
2h
xB = 30 m,
T0 =
Equate T0 = T0
wxB2
2 yB
xB = 45 m
T0 =
Cable BC
yB
=0
2
yB = 3 m
w(30 m) 2
2(3 m)
w (45 m) 2 w (30 m) 2
=
2h
2(3 m)
h = 6.75 m
Tm2 = T02 + W 2
(b)
Cable AB:
w = (0.4 kg/m)(9.81 m/s) = 3.924 N/m
xB = 45 m,
T0 =
yB = h = 6.75 m
wxB2 (3.924 N/m)(45 m)2
=
= 588.6 N
2 yB
2(6.75 m)
W = wxB = (3.924 N/m)(45 m) = 176.58 N
Tm2 = (588.6 N) 2 + (176.58 N)2
Tm = 615 N
For AB:
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1136
PROBLEM 7.108 (Continued)
Cable BC
xB = 30 m,
T0 =
yB = 3 m
wxB2 (3.924 N/m)(30 m) 2
=
= 588.6 N (Checks)
2 yB
2(3 m)
W = wxB = (3.924 N/m)(30 m) = 117.72 N
Tm2 = (588.6 N) 2 + (117.72 N) 2
Tm = 600 N
For BC
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1137
PROBLEM 7.109
Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft along the horizontal. Knowing that the
span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length
of each cable.
SOLUTION
Eq. (7.8) Page 386:
At B:
(a)
yB =
wxB2
2T0
T0 =
wxB2 (11.1 kip/ft)(2075 ft)2
=
2 yB
2(464 ft)
T0 = 51.500 kips
W = wxB = (11.1 kips/ft)(2075 ft) = 23.033 kips
Tm = T02 + W 2 = (51.500 kips)2 + (23.033 kips) 2
Tm = 56, 400 kips
(b)
&
sB = xB (1 +
(
*
2
4
2 yB !
2 yB !
"
# − "
# +
3 $ xB %
5 $ yB %
'
)
)
+
yB
464 ft
=
= 0.22361
xB 2075 ft
2
& 2
sB = (2075 ft) (1 + (0.22361) 2 − (0.22361) 4 +
3
5
*
'
) = 2142.1 ft
+
Length = 4280 ft
Length = 2sB = 2(2142.1 ft)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1138
PROBLEM 7.110
The center span of the George Washington Bridge, as originally constructed, consisted of a uniform roadway
suspended from four cables. The uniform load supported by each cable was w = 9.75 kips/ft along the
horizontal. Knowing that the span L is 3500 ft and that the sag h is 316 ft, determine for the original
configuration (a) the maximum tension in each cable, (b) the length of each cable.
SOLUTION
W = wxB = (9.75 kips/ft) (1750 ft)
W = 17,063 kips
ΣM B = 0: T0 (316 ft) − (17063 kips) (875 ft) = c
T0 = 47, 247 kips
(a)
Tm = T02 + W 2
= (47, 247 kips) 2 + (17, 063 kips) 2
Tm = 50, 200 kips
(b)
& 2 y !2 2 y !4
sB = xB (1 + " B # − " B # +
5 $ xB %
( 3 $ xB %
*
yB
316 ft
=
= 0.18057
xB 1750 ft
!
'
)
)
+
2
& 2
= (1750 ft) (1+ (0.18057)2 − (0.18057) 4 +
3
5
*
sB = 1787.3 ft; Length = 2 sB = 3579.6 ft
'
)
+
Length = 3580 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1139
PROBLEM 7.111
The total mass of cable AC is 25 kg. Assuming that the mass of the
cable is distributed uniformly along the horizontal, determine the
sag h and the slope of the cable at A and C.
SOLUTION
Cable:
m = 25 kg
W = 25 (9.81)
= 245.25 N
Block:
m = 450 kg
W = 4414.5 N
ΣM B = 0: (245.25) (2.5) + (4414.5)(3) − C x (2.5) = 0
C x = 5543 N
ΣFx = 0: Ax = C x = 5543 N
ΣM A = 0: C y (5) − (5543) (2.5) − (245.25) (2.5) = 0
C y = 2894 N
+ ΣFy = 0: C y − Ay − 245.25N = 0
2894 N − Ay − 245.25 N = 0
Point A:
tan φ A =
Point C:
tan φC =
Free body: Half cable
Ay
Ax
Cy
Cx
A y = 2649 N
=
2649
= 0.4779;
5543
φ A = 25.5°
=
2894
= 0.5221;
5543
φC = 27.6°
!
W = (12.5 kg) g = 122.6
ΣM 0 = 0: (122.6 N) (1.25 M) + (2649 N) (2.5 m) − (5543 N) yd = 0
yd = 1.2224 m; sag = h = 1.25 m − 1.2224 m
h = 0.0276 m = 27.6 mm
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1140
PROBLEM 7.112
A 50.5-m length of wire having a mass per unit length of 0.75 kg/m is used to span a horizontal distance of
50 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use only
the first two terms of Eq. (7.10).]
SOLUTION
First two terms of Eq. 7.10
1
(50.5 m) = 25.25 m,
2
1
xB = (50 m) = 25 m
2
yB = h
sB =
(a)
&
sB = xB (1 +
(
*
2
2 yB ! '
"
# )
3 $ xB % )
+
& 2 y !2 '
25.25 m = 25 m (1 − " B # )
( 3 $ xB % )
*
+
2
2
yB !
3!
"
# = 0.01" # = 0.015
$2%
$ xB %
yB
= 0.12247
xB
h
= 0.12247
25m
h = 3.0619 m
(b)
h = 3.06 m "
Free body: Portion CB
w = (0.75 kg/m) (9.81m) = 7.3575 N/m
W = sB w = (25.25 m) (7.3575 N/m)
W = 185.78 N
ΣM 0 = 0: T0 (3.0619 m) − (185.78 N) (12.5 m) = 0
T0 = 758.4 N
Bx = T0 = 758.4 N
+ ΣFy = 0: By − 185.78 N = 0 By = 185.78 N
Tm = Bx2 + By2 = (758.4 N) 2 + (185.78 N) 2
Tm = 781 N "
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1141
PROBLEM 7.113
A cable of length L + ∆ is suspended between two points that are at the same elevation and a distance L apart.
(a) Assuming that ∆ is small compared to L and that the cable is parabolic, determine the approximate sag in
terms of L and ∆. (b) If L = 100 ft and ∆ = 4 ft, determine the approximate sag. [Hint: Use only the first two
terms of Eq. (7.10).
SOLUTION
Eq. 7.10
(First two terms)
(a)
&
sB = xB (1 +
(
*
2
2 yB ! '
"
# )
3 $ xB % )
+
xB = L/2
1
( L + ∆)
2
yB = h
sB =
2'
&
1
L( 2 h! )
( L + ∆) = 1 + " L #
2
2 ( 3 "$ 2 #% )
*
+
∆ 4 h2
3
=
; h 2 = L∆;
2 3 L
8
(b)
L = 100 ft, h = 4 ft.
h=
h=
3
(100)(4);
8
3
L∆
8
h = 12.25 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1142
PROBLEM 7.114
The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four
cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of the
center span to vary from hw = 386 ft in winter to hs = 394 ft in summer. Knowing that the span is L = 4260 ft,
determine the change in length of the cables due to extreme temperature changes.
SOLUTION
Eq. 7.10.
&
sB = xB (1 +
(
*
Winter:
yB = h = 386 ft, xB =
2
4
2 yB !
2 yB !
"
# − "
# +
3 $ xB %
5 $ xB %
'
)
)
+
1
L = 2130 ft
2
& 2 386 !2 2 386 !4
sB = (2130) (1 + "
# − 5 " 2130 # +
$
%
(* 3 $ 2130 %
Summer:
yB = h = 394 ft, xB =
&
sB = (2130) (1 +
(*
'
) = 2175.715 ft
)+
1
L = 2130 ft
2
2
4
2 394 !
2 394 !
− "
+
3 "$ 2130 #%
5 $ 2130 #%
'
) = 2177.59 ft
)+
∆ = 2( ∆ sB ) = 2(2177.59 ft − 2175.715 ft) = 2(1.875 ft)
Change in length = 3.75 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1143
PROBLEM 7.115
Each cable of the side spans of the Golden Gate Bridge
supports a load w = 10.2 kips/ft along the horizontal.
Knowing that for the side spans the maximum vertical
distance h from each cable to the chord AB is 30 ft and
occurs at midspan, determine (a) the maximum tension
in each cable, (b) the slope at B.
SOLUTION
FBD AB:
ΣM A = 0: (1100 ft)TBy − (496 ft)TBx − (550 ft)W = 0 !
11TBy − 4.96TBx = 5.5W
(1)!
FBD CB:
ΣM C = 0: (550 ft)TBy − (278 ft)TBx − (275 ft)
W
=0
2
11TBy − 5.56TBx = 2.75W
Solving (1) and (2)
TBy = 28, 798 kips
Solving (1) and (2
TBx = 51, 425 kips !
Tmax = TB = TB2x + TB2y
So that
tan θ B =
TBy
(2)
!
TBx
(a)
Tmax = 58,940 kips
(b)
θ B = 29.2°
!
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1144
PROBLEM 7.116
A steam pipe weighting 45 lb/ft that passes between two
buildings 40 ft apart is supported by a system of cables as
shown. Assuming that the weight of the cable system is
equivalent to a uniformly distributed loading of 5 lb/ft,
determine (a) the location of the lowest Point C of the
cable, (b) the maximum tension in the cable.
SOLUTION
Note:
xB − x A = 40 ft
or
x A = xB − 40 ft
(a)
Use Eq. 7.8
Point A:
yA =
wx A2
w ( xB − 40) 2
; 9=
2T0
2T0
(1)
Point B:
yB =
wxB2
wx 2
; 4= B
2T0
2T0
(2)
Dividing (1) by (2):
(b)
9 ( xB − 40) 2
=
; xB = 16 ft
4
xB2
Point C is 16 ft to left of B
Maximum slope and thus Tmax is at A
x A = xB − 40 = 16 − 40 = −24 ft
yA =
wx A2
(50 lb/ft)(−24 ft) 2
; 9 ft =
; T0 = 1600 lb
2T0
2T0
WAC = (50 lb/ft)(24 ft) = 1200 lb
Tmax = 2000 lb
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1145
PROBLEM 7.117
Cable AB supports a load uniformly distributed along the
horizontal as shown. Knowing that at B the cable forms
an angle θB = 35° with the horizontal, determine (a) the
maximum tension in the cable, (b) the vertical distance a
from A to the lowest point of the cable.
SOLUTION
Free body: Entire cable
By = Bx tan 35°
W = (45 kg/m)(12 m)(9.81 m/s 2 )
W = 5297.4 N
ΣM A = 0 : W (6 m) + Bx (1.8 m) − By (12 m) = 0
(5297.4)(6) + 1.8Bx − Bx tan 35°(12) = 0
Bx = 4814 N By = (4814 N) tan 35° = 3370.8 N
Free body: Portion CB
ΣFy = 0: By − WBC = 0
WBC = By = 3370.8 N
WBC = (45 kg/m)(9.81 m/s 2 )b
3370.8 N = (441.45 N/m)b
b = 7.6357 m
1 !
ΣM B = 0: T0 dC − WBC " b # = 0
$2 %
1
(4814 N)dC − (3370.8 N) (7.6357 m) = 0
2
dC = 2.6733 m
(a)
dC = 1.8 m + a; 2.6733 m = 1.8 m + a;
(b)
Tm = B = Bx2 + By2 = (4814 N) 2 + (3370.8 N)
Tm = 5877 N
a = 0.873 m
Tm = 5880 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1146
PROBLEM 7.118
Cable AB supports a load uniformly distributed along
the horizontal as shown. Knowing that the lowest point
of the cable is located at a distance a = 0.6 m below A,
determine (a) the maximum tension in the cable, (b) the
angle θB that the cable forms with the horizontal at B.
SOLUTION
Note:
xB − x A = 12 m
or
x A = xB − 12 m
Point A:
yA =
wx A2
w( xB − 12) 2
; 0.6 =
2T0
2T0
(1)
Point B:
yB =
wxB2
wx 2
; 2.4 = B
2T0
2T0
(2)
Dividing (1) by (2):
0.6 ( xB − 12) 2
; xB = 8 m
=
2.4
xB2
(a)
2.4 =
Eq. (2):
w(8) 2
; T0 = 13.333w
2T0
Free body: Portion CB
ΣFy = 0 By = wxB
B y = 8w
Tm2 = Bx2 + By2 ; Tm2 = (13.333w) 2 + (8w) 2
Tm = 15.549w = 15.549(45)(9.81)
θ B = tan −1 By /Bx = tan −1 8w/13.333w
Tm = 6860 N
θ B = 31.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1147
PROBLEM 7.119*
A cable AB of span L and a simple beam A′B′ of the same span are
subjected to identical vertical loadings as shown. Show that the
magnitude of the bending moment at a point C′ in the beam is equal to
the product T0h, where T0 is the magnitude of the horizontal component
of the tension force in the cable and h is the vertical distance between
Point C and the chord joining the points of support A and B.
SOLUTION
ΣM B = 0: LACy + aT0 − ΣM B loads = 0
(1)
FBD Cable:
(Where ΣM B loads includes all applied loads)
x!
ΣM C = 0: xACy − " h − a # T0 − ΣM C left = 0
L%
$
(2)
FBD AC:
(Where ΣM C left includes all loads left of C)
x
x
(1) − (2): hT0 − ΣM B loads + ΣM C left = 0
L
L
(3)
ΣM B = 0: LABy − ΣM B loads = 0
(4)
ΣM C = 0: xABy − ΣM C left − M C = 0
(5)
FBD Beam:
FBD AC:
Comparing (3) and (6)
x
x
(4) − (5): − ΣM B loads + ΣM C left + M C = 0
L
L
M C = hT0
(6)
Q.E.D.
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1148
PROBLEM 7.120
Making use of the property established in Problem 7.119, solve
the problem indicated by first solving the corresponding beam
problem.
PROBLEM 7.94 (a) Knowing that the maximum tension in cable
ABCD is 15 kN, determine the distance hB .
SOLUTION
ΣM B = 0: A(10 m) − (6 kN)(7 m) − (10 kN)(4 m) = 0
A = 8.2 kN
ΣFy = 0: 8.2 kN − 6 kN − 10 kN + B = 0
B = 7.8 kN
At A:
Tm2 = T02 + A2
152 = T02 + 8.2
T0 = 12.56 kN
At B:
M B = T0 hB ; 24.6 kN ⋅ m = (12.56 kN)hB ;
hB = 1.959 m
At C:
M C = T0 hC ; 31.2 kN ⋅ m = (12.56 kN)hC ;
hC = 2.48 m
!
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1149
PROBLEM 7.121
Making use of the property established in Problem 7.119, solve the problem
indicated by first solving the corresponding beam problem.
PROBLEM 7.97 (a) Knowing that dC = 3 m, determine the distances dB
and dD .
SOLUTION
ΣM B = 0: A(10 m) − (5 kN)(8 m) − (5 kN)(6 m) − (10 kN)(3 m) = 0
A = 10 kN
Geometry:
dC = 1.6 m + hC
3 m = 1.6 m + hC
hC = 1.4 m
Since M = T0h, h is proportional to M, thus
h
hB
h
= C = D ;
M B MC M D
hB
hD
1.4 m
=
=
20 kN ⋅ m 30 kN ⋅ m 30 kN ⋅ m
20 !
hB = 1.4 " # = 0.9333 m
30
$ %
30 !
hD = 1.4 " # = 1.4 m
$ 30 %
d B = 0.8 m + 0.9333 m
d D = 2.8 m + 1.4 m
d B = 1.733 m
!
d D = 4.20 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1150
PROBLEM 7.122
Making use of the property established in Problem 7.119,
solve the problem indicated by first solving the corresponding
beam problem.
PROBLEM 7.99 (a) If dC = 15 ft, determine the distances dB
and dD.
SOLUTION
Free body: Beam AE
ΣM E = 0: − A(30) + 2(24) + 2(15) + 2(9) = 0
A = 3.2 kips
+ΣFy = 0: 3.2 − 3(2) + B = 0
B = 2.8 kips
Geometry:
Given:
dC = 15 ft
Then,
hC = dC − 3.75 ft = 11.25 ft
Since M = T0 h, h is proportional to M. Thus,
h
hB
h
= C = D
M B MC M D
Then,
or,
hB
h
11.25 ft
=
= D
19.2
30
25.2
or,
hB = 7.2 ft hD = 9.45 ft
d B = 6 + hB = 6 + 7.2
d B = 13.20 ft
d D = 2.25 + hD = 2.25 + 9.45
d D = 11.70 ft
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1151
PROBLEM 7.123
Making use of the property established in Problem 7.119,
solve the problem indicated by first solving the corresponding
beam problem.
PROBLEM 7.100 (a) Determine the distance dC for which
portion BC of the cable is horizontal.
SOLUTION
Free body: Beam AE
ΣM E = 0: − A(30) + 2(24) + 2(15) + 2(9) = 0
A = 3.2 kips
ΣFy = 0: 3.2 − 3(2) + B = 0
B = 2.8 kips
Geometry:
Given:
Then,
dC = d B
3.75 + hC = 6 + hB
hC = 2.25 + hB
(1)
Since M = T0h, h is proportional to M. Thus,
h
hB
= C
M B MC
or,
h
hB
= C
19.2 30
hB = 0.64hC
(2)
Substituting (2) into (1):
hC = 2.25 + 0.64hC
Then,
hC = 6.25 ft
dC = 3.75 + hC = 3.75 + 6.25
dC = 10.00 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1152
PROBLEM 7.124*
Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential
equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.
SOLUTION
FBD Elemental segment:
ΣFy = 0: Ty ( x + ∆ x) − Ty ( x) − w( x)∆ x = 0
So
Ty ( x + ∆ x )
T0
−
Ty ( x )
Ty
But
So
In lim :
∆ x →0
T0
T0
dy
dx
−
x +∆x
∆x
dy
dx
x
=
w( x)
∆x
T0
=
dy
dx
=
w( x)
T0
d 2 y w( x)
=
T0
dx 2
Q.E.D.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1153
PROBLEM 7.125*
Using the property indicated in Problem 7.124, determine the curve assumed by a cable of span L and sag h
carrying a distributed load w = w0 cos (πx/L), where x is measured from mid-span. Also determine the
maximum and minimum values of the tension in the cable.
PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by
the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.
SOLUTION
w( x) = w0 cos
πx
L
From Problem 7.124
d 2 y w( x) w0
πx
cos
=
=
2
T0
T0
L
dx
So
!
dy
= 0#
" using
dx 0
$
%
πx
dy W0 L
=
sin
dx T0π
L
y=
But
And
w0 L2
πx!
"1 − cos L # [using y (0) = 0]
T0π $
%
2
!
w L2
w L2
π!
L!
y " # = h = 0 2 "1 − cos # so T0 = 02
2%
π h
T0π $
$2%
T0 = Tmin
Tmax = TA = TB :
TBy =
Tmin =
so
TBy
T0
=
w0 L
dy
dx
=
x = L/2
π 2h
w0 L
T0π
2
TB = TBy
+ T02 =
π
w0 L2
w0 L
π
L !
1+ "
#
π
$ h%
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1154
!
PROBLEM 7.126
If the weight per unit length of the cable AB is w0/cos2 θ, prove that the curve
formed by the cable is a circular arc. (Hint: Use the property indicated in
Problem 7.124.)
PROBLEM 7.124 Show that the curve assumed by a cable that carries a
distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0,
where T0 is the tension at the lowest point.
SOLUTION
Elemental Segment:
Load on segment*
w ( x)dx =
w0
cos 2 θ
ds
dx = cos θ ds, so w( x) =
But
w0
cos3 θ
From Problem 7.119
w0
d 2 y w( x)
=
=
T0
dx 2
T0 cos3 θ
In general
d 2 y d dy ! d
dθ
= " # = (tan θ ) = sec2 θ
2
dx $ dx % dx
dx
dx
So
or
Giving r =
w0
w0
dθ
=
=
dx T0 cos3 θ sec 2 θ T0 cos θ
T0
cos θ dθ = dx = rdθ cos θ
w0
T0
= constant. So curve is circular arc
w0
Q.E.D.!
*For large sag, it is not appropriate to approximate ds by dx.
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1155
PROBLEM 7.127
A 30-m cable is strung as shown between two buildings. The
maximum tension is found to be 500 N, and the lowest point of
the cable is observed to be 4 m above the ground. Determine
(a) the horizontal distance between the buildings, (b) the total
mass of the cable.
SOLUTION
sB = 15 m
Tm = 500 N
Eq. 7.17:
yB2 − sB2 = c 2 ; (6 + c) 2 − 152 = c 2
36 + 12c + c 2 − 225 = c 2
12c = 189 c = 15.75 m
Eq. 7.15:
xB = 0.8473(15.75) = 13.345 m; L = 2 xB
(a)
(b)
xB
x
; 15 = (15.75) sinh B
c
c
xB
xB
sinh
= 0.95238
= 0.8473
c
c
sB = c sinh
Eq. 7.18:
L = 26.7 m
Tm = wyB ; 500 N = w(6 + 15.75)
w = 22.99 N/m
W = 2sB w = (30 m)(22.99 N/m)
= 689.7 N
689.7 N
W
=
m=
g 9.81 m/s 2
Total mass = 70.3 kg
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1156
PROBLEM 7.128
A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched between two points at the same elevation and
pulled until the tension at each end is 16 lb, determine the horizontal distance between the ends of the tape.
Neglect the elongation of the tape due to the tension.
SOLUTION
sB = 100 ft
4 lb !
w="
= 0.02 lb/ft Tm = 16 m
200
ft #%
$
Eq. 7.18:
Eq. 7.17:
Eq. 7.15:
Tm = wyB ; 16 lb = (0.02 lb/ft) yB ;
yB = 800 ft
yB2 − sB2 = c 2 ; (800)2 − (100) 2 = c 2 ; c = 793.73 ft
sB = c sinh
xB
x
; 100 = 793.73 sinh B
c
c
xB
= 0.12566; xB = 99.737 ft
c
L = 2 xB = 2(99.737 ft)
L = 199.5 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1157
PROBLEM 7.129
A 200-m-long aerial tramway cable having a mass per unit length of 3.5 kg/m is suspended between two
points at the same elevation. Knowing that the sag is 50 m, find (a) the horizontal distance between the
supports, (b) the maximum tension in the cable.
SOLUTION
Given:
Length = 200 m
Unit mass = 3.5 kg/m
h = 50 m
w = (3.5 kg/m)(9.81 m/s 2 ) = 34.335 N/m
Then,
sB = 100 m
yB = h + c = 50 m + c
yB2
− sB2 = c 2 ; (50 + c)2 − (100) 2 = c 2
502 + 100c + c 2 − 1002 = c 2
c = 75 m
sB = c sinh
xB
x
; 100 = 75sinh B
c
75
xB = 82.396 m
span = L = 2 xB = 2(82.396 m)
Tm = wyB = (34.335 N/m)(50 m + 75 m)
L = 164.8 m
Tm = 4290 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1158
PROBLEM 7.130
An electric transmission cable of length 400 ft weighing 2.5 lb/ft is suspended between two points at the same
elevation. Knowing that the sag is 100 ft, determine the horizontal distance between the supports and the
maximum tension.
SOLUTION
sB = 200 ft
Eq. 7.17:
yB2 − sB2 = c 2 ; (100 + c) 2 − 2002 = c 2
10000 + 200c + c 2 − 40000 = c 2 ; c = 150 ft
sB = c sinh
Eq. 7.15:
sinh
xB
x
; 200 = 150sinh B
c
c
xB 4
xB
= ;
= 1.0986
3
c
c
xB = (150)(1.0986) = 164.79 ft
L = 2 xB = 2(164.79 ft) = 329.58 ft
Eq. 7.18:
Tm = wyB = (2.5 lb/ft)(100 ft + 150 ft)
L = 330 ft
Tm = 625 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1159
PROBLEM 7.131
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the force P for which h = 8 m, (b) the
corresponding span L.
SOLUTION
FBD Cable:
sT = 20 m " so sB =
$
20 m
!
= 10 m #
2
%
w = (0.2 kg/m)(9.81 m/s 2 )
= 1.96200 N/m
hB = 8 m
yB2 = (c + hB )2 = c 2 + sB2
So
c=
sB2 − hB2
2hB
(10 m)2 − (8 m)2
2(8 m)
= 2.250 m
c=
Now
xB
s
→ xB = c sinh −1 B
c
c
10 m !
= (2.250 m)sinh −1 "
#
2.250
m%
$
sB = c sinh
xB = 4.9438 m
P = T0 = wc = (1.96200 N/m)(2.250 m)
L = 2 xB = 2(4.9438 m)
(a)
(b)
P = 4.41 N
L = 9.89 m
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1160
!
PROBLEM 7.132
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Knowing that the
magnitude of the horizontal force applied to the collar is P = 20 N,
determine (a) the sag h, (b) the span L.
SOLUTION
FBD Cable:
sT = 20 m, w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m
P = T0 = wc c =
c=
P
w
20 N
= 10.1937 m
1.9620 N/m
yB2 = (hB + c) 2 = c 2 + sB2
h 2 + 2ch − sB2 = 0 sB =
20 m
= 10 m
2
h 2 + 2(10.1937 m)h − 100 m 2 = 0
h = 4.0861 m
sB = c sinh
(a)
h = 4.09 m
xA
s
10 m !
→ xB = c sinh −1 B = (10.1937 m)sinh −1 "
#
c
c
$ 10.1937 m %
= 8.8468 m
L = 2 xB = 2(8.8468 m)
(b)
L = 17.69 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1161
PROBLEM 7.133
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the sag h for which L = 15 m, (b) the
corresponding force P.
SOLUTION
FBD Cable:
20 m
= 10 m
2
w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m
L = 15 m
sT = 20 m → sB =
L
xB
= c sinh 2
c
c
7.5 m
10 m = c sinh
c
sB = c sinh
Solving numerically:
c = 5.5504 m
x
yB = c cosh " B
$ c
yB = 11.4371 m
7.5 !
!
# = (5.5504) cosh " 5.5504 #
$
%
%
hB = yB − c = 11.4371 m − 5.5504 m
(a)
P = wc = (1.96200 N/m)(5.5504 m)
(b)
hB = 5.89 m
P = 10.89 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1162
PROBLEM 7.134
Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.
SOLUTION
30 ft
= 15 ft
2
L
xB = = 10 ft
2
x
sB = c sinh B
c
10 ft
15 ft = c sinh
c
sB =
Solving numerically:
L = 20 ft
c = 6.1647 ft
yB = c cosh
xB
c
= (6.1647 ft)cosh
10 ft
6.1647 ft
= 16.2174 ft
hB = yB − c = 16.2174 ft − 6.1647 ft
hB = 10.05 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1163
PROBLEM 7.135
A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the
maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.
SOLUTION
sB = 45 m
Eq. 7.17:
Eq. 7.16:
xB
c
30
45 = c sinh ; c = 18.494 m
c
sB = c sinh
yB = c cosh
xB
c
yB = (18.494) cosh
30
18.494
yB = 48.652 m
yB = h + c
48.652 = h + 18.494
h = 30.158 m
Eq. 7.18:
h = 30.2 m
Tm = wyB
300 N = w(48.652 m)
w = 6.166 N/m
Total weight of cable
W = w(Length)
= (6.166 N/m)(90 m)
= 554.96 N
Total mass of cable
m=
W 554.96 N
=
= 56.57 kg
9.81 m/s
g
m = 56.6 kg
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1164
PROBLEM 7.136
A counterweight D is attached to a cable that passes over a small pulley at A
and is attached to a support at B. Knowing that L = 45 ft and h = 15 ft,
determine (a) the length of the cable from A to B, (b) the weight per unit
length of the cable. Neglect the weight of the cable from A to D.
SOLUTION
Given:
L = 45 ft
h = 15 ft
TA = 80 lb
xB = 22.5 ft
By symmetry:
TB = TA = Tm = 80 lb
We have
yB = c cosh
and
yB = h + c = 15 + c
Then
or
Solve by trial for c:
(a)
xB
22.5
= c cosh
c
c
c cosh
22.5
= 15 + c
c
cosh
22.5 15
= +1
c
c
c = 18.9525 ft
sB = c sinh
xB
c
= (18.9525 ft) sinh
22.5
18.9525
= 28.170 ft
Length = 2sB = 2(28.170 ft) = 56.3 ft
(b)
Tm = wyB = w(h + c)
80 lb = w(15 ft + 18.9525 ft)
w = 2.36 lb/ft
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1165
PROBLEM 7.137
A uniform cord 50 in. long passes over a pulley at B and is attached to a
pin support at A. Knowing that L = 20 in. and neglecting the effect of
friction, determine the smaller of the two values of h for which the cord
is in equilibrium.
SOLUTION
b = 50 in. − 2sB
Length of overhang:
Weight of overhang equals max. tension
Tm = TB = wb = w(50 in. − 2sB )
Eq. 7.15:
sB = c sinh
xB
c
Eq. 7.16:
yB = c cosh
xB
c
Eq. 7.18:
Tm = wyB
w(50 in. − 2sB ) = wyB
x !
x
w " 50 in. − 2c sinh B # = wc cosh B
c
c
$
%
xB = 10: 50 − 2c sinh
Solve by trial and error:
For c = 5.549 in.
c = 5.549 in.
and
10
10
= c cosh
c
c
c = 27.742 in.
10 in.
= 17.277 in.
5.549 in.
yB = h + c; 17.277 in. = h + 5.549 in.
yB = (5.549 in.) cosh
h = 11.728 in.
For c = 27.742 in.
h = 11.73 in.
10 in.
= 29.564 in.
27.742 in.
yB = h + c; 29.564 in. = h + 27.742 in.
yB = (27.742 in.) cosh
h = 1.8219 in.
h = 1.822 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1166
PROBLEM 7.138
A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart.
Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.
SOLUTION
Eq. 7.18:
Eq. 7.16:
Tm = wyB ; 400 lb = (2 lb/ft) yB ;
yB = 200 ft
xB
c
80 ft
200 ft = c cosh
c
yB = c cosh
Solve for c: c = 182.148 ft and c = 31.592 ft
yB = h + c; h = yB − c
For
c = 182.148 ft;
h = 200 − 182.147 = 17.852 ft "
For
c = 31.592 ft;
h = 200 − 31.592 = 168.408 ft "
For
smallest h = 17.85 ft
Tm # 400 lb:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1167
PROBLEM 7.139
A motor M is used to slowly reel in the cable shown. Knowing that the
mass per unit length of the cable is 0.4 kg/m, determine the maximum
tension in the cable when h = 5 m.
SOLUTION
w = 0.4 kg/m L = 10 m hB = 5 m
xB
c
L
hB + c = c cosh
2c
5m !
5 m = c " cosh
− 1#
c
$
%
yB = c cosh
Solving numerically:
c = 3.0938 m
yB = hB + c = 5 m + 3.0938 m
Tmax
= 8.0938 m
= TB = wyB
= (0.4 kg/m)(9.81 m/s 2 )(8.0938 m)
Tmax = 31.8 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1168
PROBLEM 7.140
A motor M is used to slowly reel in the cable shown. Knowing that the
mass per unit length of the cable is 0.4 kg/m, determine the maximum
tension in the cable when h = 3 m.
SOLUTION
w = 0.4 kg/m, L = 10 m, hB = 3 m
xB
L
= c cosh
c
2c
5m !
− 1#
3 m = c " c cosh
c
$
%
yB = hB + c = c cosh
Solving numerically:
c = 4.5945 m
yB = hB + c = 3 m + 4.5945 m
Tmax
= 7.5945 m
= TB = wyB
= (0.4 kg/m)(9.81 m/s 2 )(7.5945 m)
Tmax = 29.8 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1169
PROBLEM 7.141
A uniform cable weighing 3 lb/ft is held in the position shown by a
horizontal force P applied at B. Knowing that P = 180 lb and θA = 60°,
determine (a) the location of Point B, (b) the length of the cable.
SOLUTION
T0 = P = cw
Eq. 7.18:
c=
c = 60 ft
P
cos 60°
cw
=
= 2cw
0.5
Tm =
At A:
(a)
P 180 lb
=
w 3 lb/ft
Tm = w(h + c)
Eq. 7.18:
2cw = w(h + c)
2c = h + c h = b = c
b = 60.0 ft
xA
c
x
h + c = c cosh A
c
y A = c cosh
Eq. 7.16:
(60 ft + 60 ft) = (60 ft) cosh
cosh
xA
=2
60 m
xA
60
xA
= 1.3170
60 m
x A = 79.02 ft
(b)
Eq. 7.15:
a = 79.0 ft
xB
79.02 ft
= (60 ft)sinh
c
60 ft
s A = 103.92 ft
s A = c sinh
length = s A
s A = 103.9 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1170
PROBLEM 7.142
A uniform cable weighing 3 lb/ft is held in the position shown by a
horizontal force P applied at B. Knowing that P = 150 lb and θA = 60°,
determine (a) the location of Point B, (b) the length of the cable.
SOLUTION
Eq. 7.18:
T0 = P = cw
c=
P
cos 60°
cw
=
= 2 cw
0.5
Tm =
At A:
(a)
P 150 lb
=
= 50 ft
w 3 lb/ft
Tm = w(h + c)
Eq. 7.18:
2cw = w(h + c)
2c = h + c h = c = b
Eq. 7.16:
xA
c
xA
h + c = c cosh
c
y A = c cosh
(50 ft + 50 ft) = (50 ft) cosh
cosh
xA
=2
c
xA
c
xA
= 1.3170
c
x A = 1.3170(50 ft) = 65.85 ft
(b)
Eq. 7.15:
b = 50.0 ft
s A = c sinh
a = 65.8 ft
xA
65.85 ft
= (50 ft)sinh
c
50 ft
s A = 86.6 ft
length = s A = 86.6 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1171
PROBLEM 7.143
To the left of Point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the mass per unit length of
the cable is 2 kg/m, determine the force F when a = 3.6 m.
SOLUTION
xD = a = 3.6 m h = 4 m
xD
c
a
h + c = c cosh
c
3.6 m !
4 m = c " cosh
− 1#
c
$
%
yD = c cosh
Solving numerically
Then
c = 2.0712 m
yB = h + c = 4 m + 2.0712 m = 6.0712 m
F = Tmax = wyB = (2 kg/m)(9.81 m/s 2 )(6.0712 m)
F = 119.1 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1172
PROBLEM 7.144
To the left of Point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the mass per unit length of
the cable is 2 kg/m, determine the force F when a = 6 m.
SOLUTION
xD = a = 6 m h = 4 m
xD
c
a
h + c = c cosh
c
6m !
4 m = c " cosh
− 1#
c
$
%
y D = c cosh
Solving numerically
c = 5.054 m
yB = h + c = 4 m + 5.054 m = 9.054 m
F = TD = wyD = (2 kg/m)(9.81 m/s 2 )(9.054 m)
F = 177.6 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1173
PROBLEM 7.145
The cable ACB has a mass per unit length of 0.45 kg/m. Knowing
that the lowest point of the cable is located at a distance a = 0.6 m
below the support A, determine (a) the location of the lowest
Point C, (b) the maximum tension in the cable.
SOLUTION
xB − x A = 12 m
Note:
− x A = 12 m − xB
or,
Point A:
y A = c cosh
12 − xB
− xA
; c + 0.6 = c cosh
c
c
(1)
Point B:
yB = c cosh
xB
x
; c + 2.4 = c cosh B
c
c
(2)
From (1):
c + 0.6 !
12 xB
−
= cosh −1 "
#
c
c
$ c %
(3)
From (2):
xB
c + 2.4 !
= cosh −1 "
#
c
$ c %
(4)
Add (3) + (4):
12
c + 0.6 !
c + 2.4 !
= cosh −1 "
+ cosh −1 "
#
#
c
$ c %
$ c %
c = 13.6214 m
Solve by trial and error:
Eq. (2)
13.6214 + 2.4 = 13.6214 cosh
cosh
xB
= 1.1762;
c
xB
c
xB
= 0.58523
c
xB = 0.58523(13.6214 m) = 7.9717 m
Point C is 7.97 m to left of B
yB = c + 2.4 = 13.6214 + 2.4 = 16.0214 m
Eq. 7.18:
Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(16.0214 m)
Tm = 70.726 N
Tm = 70.7 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1174
PROBLEM 7.146
The cable ACB has a mass per unit length of 0.45 kg/m.
Knowing that the lowest point of the cable is located at a
distance a = 2 m below the support A, determine (a) the
location of the lowest Point C, (b) the maximum tension in
the cable.
SOLUTION
Note:
xB − x A = 12 m
or
− x A = 12 m − xB
Point A:
y A = c cosh
12 − xB
− xA
; c + 2 = c cosh
c
c
(1)
Point B:
yB = c cosh
xB
x
; c + 3.8 = c cosh B
c
c
(2)
c+2!
12 xB
−
= cosh −1 "
#
c
c
$ c %
From (1):
(3)
From (2):
xB
c + 3.8 !
= cosh −1 "
#
c
$ c %
Add (3) + (4):
12
c+2!
c + 3.8 !
= cosh −1 "
+ cosh −1 "
#
#
c
$ c %
$ c %
c = 6.8154 m
Solve by trial and error:
Eq. (2):
(4)
6.8154 m + 3.8 m = (6.8154 m) cosh
cosh
xB
c
xB
xB
= 1.5576
= 1.0122
c
c
xB = 1.0122(6.8154 m) = 6.899 m
Point C is 6.90 m to left of B
yB = c + 3.8 = 6.8154 + 3.8 = 10.6154 m
Eq. (7.18):
Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(10.6154 m)
Tm = 46.86 N
Tm = 46.9 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1175
PROBLEM 7.147*
The 10-ft cable AB is attached to two collars as shown. The collar at
A can slide freely along the rod; a stop attached to the rod prevents
the collar at B from moving on the rod. Neglecting the effect of
friction and the weight of the collars, determine the distance a.
SOLUTION
Collar at A: Since µ = 0, cable ' rod
Point A:
x dy
x
= sinh
y = c cosh ;
c
dx
c
xA
dy
tan θ =
= sinh
c
dx A
xA
= sinh(tan (90° − θ ))
c
x A = c sinh(tan (90° − θ ))
Length of cable = 10 ft
(1)
10 ft = AC + CB
xA
x
+ c sinh B
c
c
xB 10
xA
=
− sinh
sinh
c
c
c
10 = c sinh
x '
&10
xB = c sinh −1 ( − sinh A )
c
c +
*
x
x
y A = c cosh A yB = c cosh B
c
c
In ∆ ABD:
tan θ =
yB − y A
xB + x A
(2)
(3)
(4)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1176
PROBLEM 7.147* (Continued)
Method of solution:
For given value of θ, choose trial value of c and calculate:
From Eq. (1): xA
Using value of xA and c, calculate:
From Eq. (2): xB
From Eq. (3): yA and yB
Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ
Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ.
For θ = 30°
Result of trial and error procedure:
c = 1.803 ft
x A = 2.3745 ft
xB = 3.6937 ft
y A = 3.606 ft
yB = 7.109 ft
a = yB − y A
= 7.109 ft − 3.606 ft
= 3.503 ft
a = 3.50 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1177
PROBLEM 7.148*
Solve Problem 7.147 assuming that the angle θ formed by the rod and
the horizontal is 45°.
PROBLEM 7.147 The 10-ft cable AB is attached to two collars as
shown. The collar at A can slide freely along the rod; a stop attached
to the rod prevents the collar at B from moving on the rod. Neglecting
the effect of friction and the weight of the collars, determine the
distance a.
SOLUTION
Collar at A: Since µ = 0, cable ' rod
Point A:
x dy
x
= sinh
y = c cosh ;
c
dx
c
x
dy
tan θ =
= sinh A
c
dx A
xA
= sinh(tan (90° − θ ))
c
x A = c sinh(tan (90° − θ ))
Length of cable = 10 ft
(1)
10 ft = AC + CB
x
x
10 = c sinh A + c sinh B
c
c
xB 10
xA
sinh
=
− sinh
c
c
c
x '
&10
xB = c sinh −1 ( − sinh A )
c +
*c
y A = c cosh
xA
c
yB = c cosh
(2)
xB
c
(3)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1178
PROBLEM 7.148* (Continued)
In ∆ ABD:
tan θ =
yB − y A
xB + x A
(4)
Method of solution:
For given value of θ, choose trial value of c and calculate:
From Eq. (1): xA
Using value of xA and c, calculate:
From Eq. (2): xB
From Eq. (3): yA and yB
Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ
Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ.
For θ = 45°
Result of trial and error procedure:
c = 1.8652 ft
x A = 1.644 ft
xB = 4.064 ft
y A = 2.638 ft
yB = 8.346 ft
a = yB − y A
= 8.346 ft − 2.638 ft
= 5.708 ft
a = 5.71 ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1179
PROBLEM 7.149
Denoting by θ the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan θ,
(b) y = c sec θ.
SOLUTION
dy
x
= sinh
dx
c
x
s = c sinh = c tan θ
c
tan θ =
(a)
(b)
Q.E.D.
Also
y 2 = s 2 + c 2 (cosh 2 x = sinh 2 x + 1)
so
y 2 = c 2 (tan 2 θ + 1)c 2 sec2 θ
and
y = c secθ
Q.E.D.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1180
PROBLEM 7.150*
(a) Determine the maximum allowable horizontal span for a uniform cable of weight per unit length w if the
tension in the cable is not to exceed a given value Tm. (b) Using the result of part a, determine the maximum
span of a steel wire for which w = 0.25 lb/ft and Tm = 8000 lb.
SOLUTION
Tm = wyB
(a)
xB
c
!
x
1 #
# cosh B
xB #
c
#
c %
= wc cosh
"
= wxB "
""
$
We shall find ratio
( ) for when T
xB
c
m
is minimum
2
&
'
!
(
"
#
d Tm
x
xB )
1
1
= wxB (
sinh B − "
# cosh ) = 0
( xB
x !
c " xB #
c )
d" B #
"
#
(
)
$ c %
$ c %
* c
+
x
sinh B
c = 1
xB
xB
cosh
c
c
xB
c
=
tanh
c
xB
Solve by trial and error for:
sB = c sinh
Eq. 7.17:
xB
= c sinh(1.200):
c
xB
= 1.200
c
(1)
sB
= 1.509
c
yB2 − sB2 = c 2
2
&
s ! '
yB2 = c 2 (1 + " B # ) = c 2 (1 + 1.5092 )
(* $ c % )+
yB = 1.810c
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1181
PROBLEM 7.150* (Continued)
Eq. 7.18:
Tm = wyB
= 1.810 wc
Tm
c=
1.810 w
Eq. (1):
Span:
(b)
xB = 1.200c = 1.200
Tm
T
= 0.6630 m
w
1.810 w
L = 2 xB = 2(0.6630)
Tm
w
L = 1.326
Tm
w
For w = 0.25lb/ft and Tm = 8000 lb
8000 lb
0.25lb/ft
= 42, 432ft
L = 1.326
L = 8.04 miles
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1182
PROBLEM 7.151*
A cable has a mass per unit length of 3 kg/m and is supported
as shown. Knowing that the span L is 6 m, determine the two
values of the sag h for which the maximum tension is 350 N.
SOLUTION
L
=h+c
2c
w = (3 kg/m)(9.81 m/s 2 ) = 29.43 N/m
ymax = c cosh
Tmax = wymax
Tmax
w
350 N
=
= 11.893 m
29.43 N/m
ymax =
ymax
c cosh
Solving numerically
3m
= 11.893 m
c
c1 = 0.9241 m
c2 = 11.499 m
h = ymax − c
h1 = 11.893 m − 0.9241 m
h1 = 10.97 m
h2 = 11.893 m − 11.499 m
h2 = 0.394 m
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1183
PROBLEM 7.152*
Determine the sag-to-span ratio for which the maximum
tension in the cable is equal to the total weight of the entire
cable AB.
SOLUTION
Tmax = wyB = 2wsB
y B = 2 sB
L
2c
L
tanh
2c
L
2c
hB
c
c cosh
= 2c sinh
=
L
2c
1
2
1
= 0.549306
2
y −c
L
= B
= cosh
− 1 = 0.154701
2c
c
= tanh −1
h
B
hB
0.5(0.154701)
= cL =
= 0.14081
0.549306
L 2 ( 2c )
hB
= 0.1408
L
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1184
PROBLEM 7.153*
A cable of weight w per unit length is suspended between
two points at the same elevation that are a distance L apart.
Determine (a) the sag-to-span ratio for which the maximum
tension is as small as possible, (b) the corresponding values
of θB and Tm.
SOLUTION
L
2c
L
L
L!
= w " cosh
− sinh #
2c 2c
2c %
$
Tmax = wyB = wc cosh
(a)
dTmax
dc
For
min Tmax ,
dTmax
=0
dc
tanh
L 2c
=
2c L
L
= 1.1997
2c
yB
L
= cosh
= 1.8102
2c
c
h yB
=
− 1 = 0.8102
c
c
0.8102
h & 1 h 2c ! '
=(
= 0.3375
)=
L * 2 c "$ L #% + 2(1.1997)
T0 = wc Tmax = wc cosh
(b)
But
T0 = Tmax cos θ B
So
θ B = sec −1 "
L
2c
!
θ B = 56.5°
!
Tmax
L yB
= cosh
=
2c
T0
c
Tmax
= sec θ B
T0
yB !
−1
# = sec (1.8102) = 56.46°
$ c %
Tmax = wyB = w
h
= 0.338
L
y B 2c ! L !
L
= w(1.8102)
"
#"
#
c $ L %$ 2 %
2(1.1997)
Tmax = 0.755wL
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1185
PROBLEM 7.154
It has been experimentally determined that the bending moment at Point K of
the frame shown is 300 N ⋅ m. Determine (a) the tension in rods AE and FD,
(b) the corresponding internal forces at Point J.
SOLUTION
Free body: ABK
(a)
ΣM K = 0: 300 N ⋅ m −
8
15
T (0.2 m) − T (0.12 m) = 0
17
17
T = 1500 N
Free body: AJ
8
8
T = (1500 N)
17
17
= 705.88 N
Tx =
15
15
T = (1500 N)
17
17
= 1323.53 N
Ty =
Internal forces on ABJ
(b)
ΣFx = 0: 705.88 N − V = 0
V = + 705.88 N
V = 706 N
ΣFy = 0: F − 1323.53 N = 0
F = + 1323.53 N
F = 1324 N
ΣM J = 0: M − (705.88 N)(0.1m) − (1323.53 N)(0.12 m) = 0
M = + 229.4 N ⋅ m
M = 229 N ⋅ m
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1186
PROBLEM 7.155
Knowing that the radius of each pulley is 200 mm and neglecting
friction, determine the internal forces at Point J of the frame shown.
SOLUTION
FBD Frame with pulley and cord:
ΣM A = 0: (1.8 m) Bx − (2.6 m)(360 N)
− (0.2 m)(360 N) = 0
B x = 560 N
FBD BE:
Note: Cord forces have been moved to pulley hub as per Problem 6.91.
ΣM E = 0: (1.4 m)(360 N) + (1.8 m)(560 N)
− (2.4 m) By = 0
B y = 630 N
FBD BJ:
3
ΣFx′ = 0: F + 360 N − (630 N − 360 N)
5
4
− (560 N) = 0
5
F = 250 N
ΣFy′ = 0: V +
4
3
(630 N − 360 N) − (560 N) = 0
5
5
V = 120.0 N
ΣM J = 0: M + (0.6 m)(360 N) + (1.2 m)(560 N)
− (1.6 m)(630 N) = 0
M = 120.0 N ⋅ m
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1187
PROBLEM 7.156
A steel channel of weight per unit length w = 20 lb/ft forms one side of a flight of stairs. Determine the
internal forces at the center C of the channel due to its own weight for each of the support conditions shown.
SOLUTION
(a)
Free body: AB
AB = 92 + 122 = 15 ft
W = (20 lb/ft)(15 ft) = 300 lb
ΣM B = 0: A(12 ft) − (300 lb)(6 ft) = 0
A = +150 lb
A = 150 lb
Free body: AC
(150-lb Forces form a couple)
ΣF = 0
F=0
ΣF
V=0
ΣM C = 0: M − (150 lb)(3 ft) = 0
M = +450 lb ⋅ ft
M = 450 lb ⋅ ft
(b)
Free body: AB
ΣM A = 0: B(9 ft) − (300 lb)(6 ft) = 0
B = +200 lb
B = 200 lb
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1188
PROBLEM 7.156 (Continued)
Free body: CB
ΣM C = 0: (200 lb)(4.5 ft) − (150 lb)(3 ft) − M = 0
M = +450 lb ⋅ ft
M = 450 lb ⋅ ft
tan θ =
9 3
3
4
= ; sin θ = ; cos θ =
12 4
5
5
3
4
ΣF = 0: F − (150 lb) − (200 lb) = 0
5
5
F = +250 lb F = 250 lb
4
3
ΣF = 0: V − (150 lb) + (200 lb) = 0
5
5
V =0 V=0
M = 450 lb ⋅ ft , F = 250 lb
On portion AC internal forces are
, V=0
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1189
!
PROBLEM 7.157
For the beam shown, determine (a) the magnitude P of the two concentrated
loads for which the maximum absolute value of the bending moment is as
small as possible, (b) the corresponding value of | M |max .
SOLUTION
Free body: Entire beam
By symmetry
1
(16 lb/in.)(30 in.) + P
2
D = 240 lb + P
D=E=
Free body: Portion AD
ΣM D = 0: M D = −10 P
Free body: Portion ADC
ΣM C = 0: P(25 in.) − (240 lb + P)(15 in.)
+ (240 lb)(7.5 in.) + M C = 0
M C = 1800 lb ⋅ in. − (10 in.)P
(a)
We equate:
|M D | = | MC |
| − 10 P | = |1800 − 10 P |
10 P = 1800 − 10 P
(b)
For P = 90 lb:
M D = −10 lb (90 lb)
P = 90.0 lb
!
| M |max = 900 lb ⋅ in.
!
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1190
PROBLEM 7.158
Knowing that the magnitude of the concentrated loads P is 75 lb, (a) draw
the shear and bending-moment diagrams for beam AB, (b) determine the
maximum absolute values of the shear and bending moment.
SOLUTION
By symmetry:
D=E
1
(16 lb/in.)(30 in.) + 75 lb
2
= 315 lb
=
!
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1191
PROBLEM 7.159
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM B = 0: (24 kN)(3 m)
+ (24 kN)(2.4 m) + (21.6 kN)(0.9 m)
− Ay (3.6 m) = 0
Ay = +91.4 kN
ΣFx = 0: Ax = 0
Shear diagram
At A:
VA = Ay = +41.4 kN
|V |max = 41.4 kN
Bending-moment diagram
At A:
| M |max = 35.3 kN ⋅ m
MA = 0
The slope of the parabola at E is the same as that of the segment DE!
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1192
PROBLEM 7.160
For the beam shown, draw the shear and bending-moment diagrams, and
determine the magnitude and location of the maximum absolute value of
the bending moment, knowing that (a) P = 6 kips, (b) P = 3 kips.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM A = 0: C (6 ft) − (12 kips)(3 ft) − P(8 ft) = 0
C = 6 kips +
4
P
3
(1) "
4 !
ΣFy = 0: Ay + " 6 + P # − 12 − P = 0
3 %
$
1
Ay = 6 kips − P
3
(a)
(2) "
P = 6 kips.
Load diagram
Substituting for P in Eqs. (2) and (1):
1
Ay = 6 − (6) = 4 kips
3
4
C = 6 + (6) = 14 kips
3
Shear diagram
VA = Ay = +4 kips
To determine Point D where V = 0:
VD − VA = − wx
0 − 4 kips = (2 kips/ft)x
x = 2 ft "
We compute all areas
Bending-moment diagram
At A:
MA = 0
| M |max = 12.00 kip ⋅ ft, at C
Parabola from A to C
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1193
PROBLEM 7.160 (Continued)
(b)
P = 3 kips
Load diagram
Substituting for P in Eqs. (2) and (1):
1
A = 6 − (3) = 5 kips
3
4
C = 6 + (3) = 10 kips
3
Shear diagram
VA = Ay = +5 kips
To determine D where V = 0:
VD − VA = − wx
0 − (5 kips) = −(2 kips/ft) x
x = 2.5 ft "
We compute all areas
Bending-moment diagram
At A:
MA = 0
| M |max = 6.25 kip ⋅ ft
2.50 ft from A
!
Parabola from A to C.
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1194
PROBLEM 7.161
For the beam and loading shown, (a) write the equations of the
shear and bending-moment curves, (b) determine the maximum
bending moment.
SOLUTION
(a)
Reactions at supports:
1
W
A = B = W , where
= Total load
2
L
W=
-
L
wdx = w0
0
-
L
0
πx!
"1 − sin
# dx
L %
$
L
π x'
L
&
= w0 ( x + cos )
x
L +0
*
2!
= w0 L "1 − #
$ π%
1
1
2!
VA = A = W = w0 L "1 − #
2
2
$ π%
Thus
MA = 0
(1)
πx!
w( x) = w0 "1 − sin
L #%
$
Load:
Shear: From Eq. (7.2):
V ( x ) − VA = −
-
x
w( x) dx
0
= − w0
-
x
0
πx!
"1 − sin L # dx
$
%
Integrating and recalling first of Eqs. (1),
x
L
1
2!
π x'
&
V ( x) − w0 L "1 − # = − w0 ( x + cos )
L +0
2
π
$ π%
*
V ( x) =
1
2!
L
L
πx!
w0 L "1 − # − w0 " 2 + cos
# + w0
2
L
π
π
π
$
%
$
%
L
L
πx!
V ( x) = w0 " − x − cos
L #%
π
$2
(2)
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you are using it without permission.
1195
PROBLEM 7.161 (Continued)
Bending moment: From Eq. (7.4) and recalling that M A = 0.
M ( x) − M A =
-
x
V ( x) dx
0
x
2
&L
L!
π x'
1
= w0 ( x − x 2 − " # sin
)
L )+
2
$π %
(* 2
0
M ( x) =
(b)
πx!
1
2 L2
w0 "" Lx − x 2 − 2 sin
#
L #%
2 $
π
(3)
Maximum bending moment
dM
= V = 0.
dx
This occurs at x =
L
2
as we may check from (2):
L!
L L L
π!
V " # = w0 " − − cos # = 0
2%
$2%
$2 2 π
From (3):
π!
L! 1
L2 L2 2 L2
M " # = w0 "" −
− 2 sin ##
4 π
2%
$2% 2 $ 2
1
8 !
= w0 L2 "1 − 2 #
8
$ π %
= 0.0237 w0 L2
M max = 0.0237 w0 L2 , at
x=
L
2
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1196
!
PROBLEM 7.162
An oil pipeline is supported at 6-ft intervals by vertical hangers
attached to the cable shown. Due to the combined weight of the
pipe and its contents the tension in each hanger is 400 lb.
Knowing that dC = 12 ft, determine (a) the maximum tension in
the cable, (b) the distance dD.
SOLUTION
FBD Cable: Hanger forces at A and F act on the supports, so A y and Fy act on the cable.
ΣM F = 0: (6 ft + 12 ft + 18 ft + 24 ft)(400 lb)
− (30 ft)Ay − (5 ft) Ax = 0
Ax + 6 Ay = 4800 lb
FBD ABC:
(1)
ΣM C = 0: (7 ft)Ax − (12 ft)Ay + (6 ft)(400 lb) = 0
(2)
Solving (1) and (2)
A x = 800 lb
Ay =
From FBD Cable:
2000
lb
3
ΣFx = 0: − 800 lb + Fx = 0
Fx = 800 lb
FBD DEF:
ΣFy = 0:
200
lb − 4 (400 lb) + Fy = 0
3
Fy =
Since Ax = Fx and Fy . Ay ,
2800 !
Tmax = TEF = (800 lb) 2 + "
lb #
$ 3
%
2800
lb
3
2
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1197
PROBLEM 7.162 (Continued)
(a)
Tmax = 1229.27 lb,
Tmax = 1229 lb
2800 !
lb # − d D (800 lb) − (6 ft)(400 lb) = 0
ΣM D = 0: (12 ft) "
$ 3
%
d D = 11.00 ft
(b)
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1198
!
PROBLEM 7.163
Solve Problem 7.162 assuming that dC = 9 ft.
PROBLEM 7.162 An oil pipeline is supported at 6-ft intervals by
vertical hangers attached to the cable shown. Due to the combined
weight of the pipe and its contents the tension in each hanger is
400 lb. Knowing that dC = 12 ft, determine (a) the maximum
tension in the cable, (b) the distance dD.
SOLUTION
FBD CDEF:
ΣM C = 0: (18 ft)Fy − (9 ft)Fy − (6 ft + 12 ft)(400 lb) = 0
Fx − 2 Fy = −800 lb
(1)
EM A = 0: (30 ft)Fy − (5 ft)Fx
FBD Cable:
− (6 ft)(1 + 2 + 3 + 4)(400 lb) = 0
Fx − 6 Fy = −4800 lb
(2)
Solving (1) and (2),
Fx = 1200 lb
, Fy = 1000 lb
ΣFx = 0: − Ax + 1200 lb = 0, A x = 1200 lb
Point F:
ΣFy = 0: Ay + 1000 lb − 4(400 lb) = 0,
A y = 600 lb
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1199
PROBLEM 7.163 (Continued)
Since
Ax = Ay
and Fy . Ay , Tmax = TEF
Tmax = (1 kip)2 + (1.2 kips) 2
Tmax = 1.562 kips
(a)
FBD DEF:
ΣM D = 0: (12 ft)(1000 lb) − d D (1200 lb)
− (6 ft)(400 lb) = 0
d D = 8.00 ft
(b)
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1200
!
PROBLEM 7.164
A transmission cable having a mass per unit length of 0.8 kg/m is strung between two insulators at the same
elevation that are 75 m apart. Knowing that the sag of the cable is 2 m, determine (a) the maximum tension in
the cable, (b) the length of the cable.
SOLUTION
w = (0.8 kg/m)(9.81 m/s 2 )
= 7.848 N/m
W = (7.848 N/m)(37.5 m)
W = 294.3 N
(a)
1
!
ΣM B = 0: T0 (2 m) − W " 37.5 m # = 0
$2
%
1
T0 (2 m) − (294.3 N) (37.5 m) = 0
2
T0 = 2759 N
Tm2 = (294.3 N) 2 + (2759 N)2
(b)
Tm = 2770 N
& 2 y !2
'
sB = xB (1 + " B # + !)
( 3 $ xB %
)
*
+
& 2 2 m !2
'
= 37.5 m (1+ "
+ !)
#
(* 3 $ 37.5 m %
)+
= 37.57 m
Length = 2sB = 2(37.57 m)
Length = 75.14 m
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1201
!
PROBLEM 7.165
Cable ACB supports a load uniformly distributed along the
horizontal as shown. The lowest Point C is located 9 m to the right
of A. Determine (a) the vertical distance a, (b) the length of the
cable, (c) the components of the reaction at A.
SOLUTION
Free body: Portion AC
ΣFy = 0: Ay − 9 w = 0
A y = 9w
ΣM A = 0: T0 a − (9 w)(4.5 m) = 0
(1)
Free body: Portion CB
ΣFy = 0: By − 6w = 0
B y = 6w
Free body: Entire cable
ΣM A = 0: 15w (7.5 m) − 6 w (15 m) − T0 (2.25 m) = 0
T0 = 10 w
(a)
Eq. (1):
T0 a − (9w)(4.5 m) = 0
10wa = (9w)(4.5) = 0
a = 4.05 m
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1202
PROBLEM 7.165 (Continued)
(b)
Length = AC + CB
Portion AC:
x A = 9 m,
s AC
s AC
Portion CB:
y A = a = 4.05 m;
y A 4.05
=
= 0.45
9
xA
& 2 y !2 2 y !4
'
= xB (1 + " A # − " B # + !)
5 $ xA %
( 3 $ xA %
)
*
+
2
2
!
= 9 m "1+ 0.452 − 0.454 + ! # = 10.067 m
5
$ 3
%
xB = 6 m,
yB = 4.05 − 2.25 = 1.8 m;
yB
= 0.3
xB
2
2
!
sCB = 6 m "1 + 0.32 − 0.34 + ! # = 6.341 m
3
5
$
%
Total length = 10.067 m + 6.341 m
(c)
Total length = 16.41 m
Components of reaction at A.
Ay = 9 w = 9(60 kg/m)(9.81 m/s 2 )
= 5297.4 N
Ax = T0 = 10 w = 10(60 kg/m)(9.81 m/s 2 )
= 5886 N
A x = 5890 N
A y = 5300 N
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1203
!
CHAPTER 8
PROBLEM 8.1
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when θ = 25° and P = 150 lb.
SOLUTION
Assume equilibrium:
ΣFx = 0: F + (240 lb) sin 25° − (150 lb) cos 25° = 0
F = +34.518 lb
F = 34.518 lb
ΣFy = 0: N − (240 lb) cos 25° − (150 lb)sin 25° = 0
N = +280.91 lb
Maximum friction force:
N = 280.91 lb
Fm = µ s N = 0.35(280.91 lb) = 98.319 lb
Block is in equilibrium
F = 34.5 lb
Since F , Fm ,
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1207
PROBLEM 8.2
Determine whether the block shown is in equilibrium and find the magnitude and
direction of the friction force when θ = 30° and P = 30 lb.
SOLUTION
Assume equilibrium:
ΣFx = 0: F + (240 lb) sin 30° − (30 lb) cos 30° = 0
F = −94.019 lb
F = 94.019 lb
ΣFy = 0: N − (240 lb) cos 30° − (30 lb) sin 30° = 0
N = +222.85 lb
Maximum friction force:
N = 222.85 lb
Fm = µ s N
= 0.35(222.85 lb)
= 77.998 lb
Since F is
and F . Fm ,
Actual friction force:
Block moves down
F = Fk = µ k N = 0.25(222.85 lb)
F = 55.7 lb
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1208
!
PROBLEM 8.3
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when θ = 40° and
P = 400 N.
SOLUTION
Assume equilibrium:
ΣFy = 0: N − (800 N) cos 25° + (400 N) sin 15° = 0
N = +621.5 N
N = 621.5 N
ΣFx = 0: − F + (800 N) sin 25° − (400 N) cos 15° = 0
F = +48.28 N
F = 48.28 N
Maximum friction force:
Fm = µ s N
= 0.20(621.5 N)
= 124.3 N
Block is in equilibrium
F = 48.3 N
Since F , Fm , !
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1209
!
PROBLEM 8.4
Determine whether the block shown is in equilibrium and find the
magnitude and direction of the friction force when θ = 35° and
P = 200 N.
SOLUTION
Assume equilibrium:
ΣFy = 0: N − (800 N) cos 25° + (200 N) sin 10° = 0
N = 690.3 N
N = 690.3 N
ΣFx = 0: − F + (800 N) sin 25° − (200 N) cos 10° = 0
F = 141.13 N
F = 141.13 N
Maximum friction force:
Fm = µs N
= (0.20)(690.3 N)
= 138.06 N
Since F . Fm ,
Block moves down
Friction force:
F = µk N
!
= (0.15)(690.3 N)
= 103.547 N !
F = 103.5 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1210
!
PROBLEM 8.5
Knowing that θ = 45°, determine the range of values of P for which
equilibrium is maintained.
SOLUTION
To start block up the incline:
µs = 0.20
φs = tan −1 0.20 = 11.31°
Force triangle:
P
800 N
=
sin 36.31° sin 98.69°
P = 479.2 N "!
To prevent block from moving down:
Force triangle:
P
800 N
=
sin 13.69° sin 121.31°
P = 221.61 N "
222 N # P # 479 N
Equilibrium is maintained for!
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1211
PROBLEM 8.6
Determine the range of values of P for which equilibrium of the block
shown is maintained.
SOLUTION
FBD block:
(Impending motion down):
φs = tan −1 µs = tan −1 0.25
P = (500 lb) tan (30° − tan −1 0.25)
= 143.03 lb
(Impending motion up):
P = (500 lb) tan (30° + tan −1 0.25)
= 483.46 lb
143.0 lb # P # 483 lb
Equilibrium is maintained for
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1212
PROBLEM 8.7
Knowing that the coefficient of friction between the 15-kg block and the
incline is µ s = 0.25, determine (a) the smallest value of P required to
maintain the block in equilibrium, (b) the corresponding value of β.
SOLUTION
FBD block (Impending motion downward):
W = (15 kg)(9.81 m/s 2 )
= 147.150 N
φs = tan −1 µ s
= tan −1 (0.25)
= 14.036°
(a)
Note: For minimum P,
P
So
β =α
R
= 90° − (30° + 14.036°)
= 45.964°
and
P = (147.150 N)sin α
= (147.150 N)sin (45.964°)
P = 108.8 N
β = 46.0°
(b)
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1213
PROBLEM 8.8
Considering only values of θ less than 90°, determine the smallest value
of θ required to start the block moving to the right when (a) W = 75 lb,
(b) W = 100 lb.
SOLUTION
FBD block (Motion impending):
φs = tan −1 µs = 14.036°
W
30 lb
=
sin φs sin(θ − φs )
sin(θ − φs ) =
or
sin(θ − 14.036°) =
W sin 14.036°
30 lb
W
123.695 lb
(a)
W = 75 lb:
θ = 14.036° + sin −1
75 lb
123.695 lb
θ = 51.4°
!
(b)
W = 100 lb: θ = 14.036° + sin −1
100 lb
123.695 lb
θ = 68.0°
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1214
PROBLEM 8.9
The coefficients of friction between the block and the rail are µ s = 0.30
and µk = 0.25. Knowing that θ = 65°, determine the smallest value of
P required (a) to start the block moving up the rail, (b) to keep it from
moving down.
SOLUTION
(a)
To start block up the rail:
µs = 0.30
φs = tan −1 0.30 = 16.70°
Force triangle:
P
500 N
=
sin 51.70° sin(180° − 25° − 51.70°)
(b)
P = 403 N
To prevent block from moving down:
Force triangle:
P
500 N
=
sin 18.30° sin(180° − 25° − 18.30°)
P = 229 N
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1215
!
PROBLEM 8.10
The 80-lb block is attached to link AB and rests on a moving belt.
Knowing that µ s = 0.25 and µk = 0.20, determine the magnitude of
the horizontal force P that should be applied to the belt to maintain its
motion (a) to the right, (b) to the left.
SOLUTION
We note that link AB is a two-force member, since there is motion between belt and block µk = 0.20
and φk = tan −1 0.20 = 11.31°
(a)
Belt moves to right
Free body: Block
Force triangle:
R
80 lb
=
sin 120° sin 48.69°
R = 92.23 lb
Free body: Belt
ΣFx = 0: P − (92.23 lb) sin 11.31°
P = 18.089 lb
P = 18.09 lb
(b)
Belt moves to left
Free body: Block
Force triangle:
R
80 lb
=
sin 60° sin 108.69°
R = 73.139 lb
Free body: Belt
ΣFx = 0: (73.139 lb)sin 11.31° − P = 0
P = 14.344 lb
P = 14.34 lb
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1216
PROBLEM 8.11
The coefficients of friction are µ s = 0.40 and µk = 0.30 between all
surfaces of contact. Determine the smallest force P required to start the
30-kg block moving if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a)
Free body: 20-kg block
W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N
F1 = µ s N1 = 0.4(196.2 N) = 78.48 N
ΣF = 0: T − F1 = 0 T = F1 = 78.48 N
Free body: 30-kg block
W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N
N 2 = 196.2 N + 294.3 N = 490.5 N
F2 = µ s N 2 = 0.4(490.5 N) = 196.2 N
ΣF = 0: P − F1 − F2 − T = 0
P = 78.48 N + 196.2 N + 78.48 N = 353.2 N
P = 353 N
(b)
Free body: Both blocks
Blocks move together
W = (50 kg)(9.81 m/s 2 ) = 490.5 N
ΣF = 0: P − F = 0
P = µ s N = 0.4(490.5 N) = 196.2 N
P = 196.2 N
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1217
PROBLEM 8.12
The coefficients of friction are µ s = 0.40 and µk = 0.30 between all
surfaces of contact. Determine the smallest force P required to start the
30-kg block moving if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a)
Free body: 20-kg block
W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N
F1 = µ s N1 = 0.4(196.2 N) = 78.48 N
ΣF = 0: T − F1 = 0 T = F1 = 78.48 N
Free body: 30-kg block
W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N
N 2 = 196.2 N + 294.3 N = 490.5 N
F2 = µ s N 2 = 0.4(490.5 N) = 196.2 N
ΣF = 0: P − F1 − F2 = 0
P = 78.48 N + 196.2 N = 274.7 N
P = 275 N
(b)
Free body: Both blocks
Blocks move together
W = (50 kg)(9.81 m/s 2 )
= 490.5 N
ΣF = 0: P − F = 0
P = µ s N = 0.4(490.5 N) = 196.2 N
P = 196.2 N
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1218
PROBLEM 8.13
Three 4-kg packages A, B, and C are placed on a conveyor belt that is at
rest. Between the belt and both packages A and C the coefficients of
friction are µ s = 0.30 and µk = 0.20; between package B and the belt
the coefficients are µ s = 0.10 and µk = 0.08. The packages are placed
on the belt so that they are in contact with each other and at rest.
Determine which, if any, of the packages will move and the friction
force acting on each package.
SOLUTION
Consider C by itself: Assume equilibrium
ΣFy = 0: N C − W cos 15° = 0
N C = W cos 15° = 0.966W
ΣFx = 0: FC − W sin 15° = 0
FC = W sin 15° = 0.259W
But
Fm = µ s NC
= 0.30(0.966W )
= 0.290W
Package C does not move
Thus, FC , Fm
FC = 0.259W
= 0.259(4 kg)(9.81 m/s 2 )
= 10.16 N
FC = 10.16 N
Consider B by itself: Assume equilibrium. We find,
FB = 0.259W
N B = 0.966W
But
Fm = µ s N B
= 0.10(0.966W )
= 0.0966W
Package B would move if alone
Thus, FB . Fm .
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1219
PROBLEM 8.13 (Continued)
Consider A and B together: Assume equilibrium
FA = FB = 0.259W
N A = N B = 0.966W
FA + FB = 2(0.259W ) = 0.518W
( FA ) m + ( FB ) m = 0.3 N A + 0.1N B = 0.386W
Thus,
FA + FB . ( FA ) m + ( FB )m
!
FA = µk N A = 0.2(0.966)(4)(9.81) !
!
!
A and B move
FA = 7.58 N
FB = µk N B = 0.08(0.966)(4)(9.81)
FB = 3.03 N
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1220
!
PROBLEM 8.14
Solve Problem 8.13 assuming that package B is placed to the right of
both packages A and C.
PROBLEM 8.13 Three 4-kg packages A, B, and C are placed on a
conveyor belt that is at rest. Between the belt and both packages A and
C the coefficients of friction are µ s = 0.30 and µk = 0.20; between
package B and the belt the coefficients are µ s = 0.10 and µk = 0.08.
The packages are placed on the belt so that they are in contact with
each other and at rest. Determine which, if any, of the packages will
move and the friction force acting on each package.
SOLUTION
Consider package B by itself: Assume equilibrium
ΣFy = 0: N B − W cos 15° = 0
N B = W cos 15° = 0.966W
ΣFx = 0: FB − W sin 15° = 0
FB = W sin 15° = 0.259W
But
Fm = µ s N B
= 0.10(0.966W )
= 0.0966W
Thus, FB . Fm . Package B would move if alone.
Consider all packages together: Assume equilibrium. In a manner similar to above, we find
N A = N B = N C = 0.966W
FA = FB = FC = 0.259W
FA + FB + FC = 3(0.259W )
= 0.777 W
But
( FA ) m = ( FC ) m = µ s N
= 0.30(0.966W )
= 0.290W
and
( FB ) m = 0.10(0.966W )
= 0.0966W
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you are using it without permission.
1221
PROBLEM 8.14 (Continued)
Thus,
( FA ) m + ( FC ) m + ( FB ) m = 2(0.290W ) + 0.0966W
= 0.677 W
and we note that
FA + FB + FC . ( FA ) m + ( FC ) m + ( FB ) m
All packages move
FA = FC = µk N
= 0.20(0.966)(4 kg)(9.81 m/s 2 )
= 7.58 N
FB = µk N
= 0.08(0.966)(4 kg)(9.81 m/s 2 )
= 3.03 N
FA = FC = 7.58 N
;
FB = 3.03 N
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you are using it without permission.
1222
PROBLEM 8.15
A 120-lb cabinet is mounted on casters that can be locked to prevent their
rotation. The coefficient of static friction between the floor and each caster is
0.30. If h = 32 in., determine the magnitude of the force P required to move
the cabinet to the right (a) if all casters are locked, (b) if the casters at B are
locked and the casters at A are free to rotate, (c) if the casters at A are locked
and the casters at B are free to rotate.
SOLUTION
FBD cabinet: Note: for tipping,
N A = FA = 0
ΣM B = 0: (12 in.)W − (32 in.)Ptip = 0
Ptip = 2.66667
(a)
All casters locked. Impending slip:
FA = µ s N A
FB = µ s N B
ΣFy = 0 : N A + N B − W = 0
W = 120 lb
N A + NB = W
So
FA + FB = µ sW
µs = 0.3
ΣFx = 0: P − FA − FB = 0
P = FA + FB = µ sW
P = 0.3(120 lb) or P = 36.0 lb
( P = 0.3W , Ptip
(b)
Casters at A free, so
FA = 0
Impending slip:
FB = µ s N B
OK)
ΣFx = 0: P − FB = 0
P = FB = µ s N B
NB =
P
µs
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you are using it without permission.
1223
PROBLEM 8.15 (Continued)
ΣM A = 0: (32 in.)P + (12 in.)W − (24 in.)N B = 0
8 P + 3W − 6
P
= 0 P = 0.25W
0.3
( P = 0.25W , Ptip
OK)
P = 0.25(120 lb)
(c)
Casters at B free, so
FB = 0
Impending slip:
FA = µ s N A
or P = 30.0 lb
ΣFx = 0: P − FA = 0 P = FA = µ s N A
NA =
P
µs
=
P
0.3
ΣM B = 0: (12 in.)W − (32 in.)P − (24 in.)N A = 0
3W − 8P − 6
( P , Ptip
P
=0
0.3
P = 0.107143W = 12.8572
OK)
P = 12.86 lb
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1224
PROBLEM 8.16
A 120-lb cabinet is mounted on casters that can be locked to prevent their
rotation. The coefficient of static friction between the floor and each caster is
0.30. Assuming that the casters at both A and B are locked, determine (a) the
force P required to move the cabinet to the right, (b) the largest allowable value
of h if the cabinet is not to tip over.
SOLUTION
FBD cabinet:
(a)
ΣFy = 0: N A + N B − W = 0
N A + NB = W
Impending slip:
FA = µ s N A
FB = µ s N B
So
FA + FB = µ sW
W = 120 lb
µs = 0.3
ΣFx = 0: P − FA − FB = 0
P = FA + FB = µ sW
P = 0.3(120 lb) = 141.26 N
P = 36.0 lb
(b)
For tipping,
N A = FA = 0
ΣM B = 0: hP − (12 in.)W = 0
hmax = (12 in.)
W
1 12 in.
= (12 in.)
=
P
0.3
µs
hmax = 40.0 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1225
!
PROBLEM 8.17
The cylinder shown is of weight W and radius r, and the coefficient of static
friction µ s is the same at A and B. Determine the magnitude of the largest
couple M that can be applied to the cylinder if it is not to rotate.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
FA = µ s N A
FB = µ s N B
ΣFx = 0: N A − FB = 0
N A = FB = µ s N B
FA = µ s N A = µ s2 N B
ΣFy = 0: N B + FA − W = 0
N B + µ s2 N B = W
or
NB =
and
FB =
FA =
W
1 + µ s2
µsW
1 + µs2
µs2W
1+ µ2
ΣM C = 0: M − r ( FA + FB ) = 0
(
M = r µs + µs2
) 1 +Wµ
2
s
M max = Wr µ s
1 + µs
1 + µ s2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1226
!
PROBLEM 8.18
The cylinder shown is of weight W and radius r. Express in terms W and r the
magnitude of the largest couple M that can be applied to the cylinder if it is not to
rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B,
(b) 0.25 at A and 0.30 at B.
SOLUTION
FBD cylinder:
For maximum M, motion impends at both A and B
FA = µ A N A
FB = µ B N B
ΣFx = 0: N A − FB = 0
N A = FB = µ B N B
FA = µ A N A = µ A µ B N B
ΣFy = 0: N B + FA − W = 0
N B (1 + µ A µ B ) = W
or
and
NB =
1
1 + µ A µB
W
µB
W
1 + µ AµB
µ A µB
FA = µ A µ B N B =
W
1 + µ A µB
FB = µ B N B =
ΣM C = 0: M − r ( FA + FB ) = 0
M = Wr µ B
(a)
1 + µA
1 + µ A µB
For µ A = 0 and µ B = 0.30:
M = 0.300Wr
(b)
For µ A = 0.25 and µ B = 0.30:
M = 0.349Wr
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1227
!
PROBLEM 8.19
The hydraulic cylinder shown exerts a force of 3 kN directed
to the right on Point B and to the left on Point E. Determine
the magnitude of the couple M required to rotate the drum
clockwise at a constant speed.
SOLUTION
Free body: Drum
ΣM C = 0: M − (0.25 m)( FL + FR ) = 0
M = (0.25 m)( FL + FR )
(1)
Since drum is rotating
FL = µk N L = 0.3N L
FR = µk N R = 0.3 N R
Free body: Left arm ABL
ΣM A = 0: (3 kN)(0.15 m) + FL (0.15 m) − N L (0.45 m) = 0
0.45 kN ⋅ m + (0.3N L )(0.15 m) − N L (0.45 m) = 0
0.405 N L = 0.45
N L = 1.111 kN
FL = 0.3N L = 0.3(1.111 kN)
= 0.3333 kN
Free body: Right arm DER
(2)
ΣM D = 0: (3 kN)(0.15 m) − FR (0.15 m) − N R (0.45 m) = 0
0.45 kN ⋅ m − (0.3N R )(0.15 m) − N R (0.45 m) = 0
0.495 N R = 0.45
N R = 0.9091 kN
FR = µk N R = 0.3(0.9091 kN)
= 0.2727 kN
Substitute for FL and FR into (1):
(3)
M = (0.25 m)(0.333 kN + 0.2727 kN)
M = 0.1515 kN ⋅ m
M = 151.5 N ⋅ m
!
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1228
PROBLEM 8.20
A couple M of magnitude 100 N ⋅ m is applied to the drum as
shown. Determine the smallest force that must be exerted by the
hydraulic cylinder on joints B and E if the drum is not to rotate.
SOLUTION
Free body: Drum
ΣM C = 0: 100 N ⋅ m − (0.25 m)( FL + FR ) = 0
FL + FR = 400 N
(1)
Since motion impends
FL = µs N L = 0.4 N L
FR = µs N R = 0.4 N R
Free body: Left arm ABL
ΣM A = 0: T (0.15 m) + FL (0.15 m) − N L (0.45 m) = 0
0.15T + (0.4 N L )(0.15 m) − N L (0.45 m) = 0
0.39 N L = 0.15T ; N L = 0.38462T
FL = 0.4 N L = 0.4(0.38462T )
FL = 0.15385T
(2)
Free body: Right arm DER
ΣM D = 0: T (0.15 m) − FR (0.15 m) − N R (0.45 m) = 0
0.15T − (0.4 N R )(0.15 m) − N R (0.45 m) = 0
0.51N R = 0.15T ; N R = 0.29412T
FR = 0.4 N R = 0.4(0.29412T )
FR = 0.11765T
(3)
Substitute for FL and FR into Eq. (1):
0.15385T + 0.11765T = 400
T = 1473.3 N
T = 1.473 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1229
!
PROBLEM 8.21
A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction µs is zero at B, determine the smallest value of µs at A for
which equilibrium is maintained.
SOLUTION
Free body: Ladder
Three-force body.
Line of action of A must pass through D, where W and B intersect.
At A:
µs = tan φs =
1.25 m
= 0.2083
6m
µs = 0.208
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1230
!
PROBLEM 8.22
A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of
static friction µs is the same at A and B, determine the smallest value of µs for
which equilibrium is maintained.
SOLUTION
Free body: Ladder
Motion impending:
FA = µs N A
FB = µ s N B
ΣM A = 0: W (1.25 m) − N B (6 m) − µ s N B (2.5 m) = 0
NB =
1.25W
6 + 2.5µs
(1)
ΣFy = 0: N A + µs N B − W = 0
N A = W − µs N B
NA = W −
1.25µ sW
6 + 2.5µ s
(2)
ΣFx = 0: µ s N A − N B = 0
Substitute for NA and NB from Eqs. (1) and (2):
µsW −
1.25µ s2W
1.25W
=
6 + 2.5µ s 6 + 2.5µ s
6µ s + 2.5µ s2 − 1.25µ s2 = 1.25
1.25µ s2 + 6µ s − 1.25 = 0
µ s = 0.2
and
µs = −5 (Discard)
µs = 0.200
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1231
!
PROBLEM 8.23
End A of a slender, uniform rod of length L and weight W bears on a
surface as shown, while end B is supported by a cord BC. Knowing that
the coefficients of friction are µs = 0.40 and µk = 0.30, determine (a) the
largest value of θ for which motion is impending, (b) the corresponding
value of the tension in the cord.
SOLUTION
Free-body diagram
Three-force body. Line of action of R must pass through D, where T and R intersect.
Motion impends:
tan φs = 0.4
φs = 21.80°
(a)
Since BG = GA, it follows that BD = DC and AD bisects ∠BAC
θ
2
θ
2
+ φs = 90°
+ 21.8° = 90°
θ = 136.4°
(b)
Force triangle (right triangle):
T = W cos 21.8°
T = 0.928W
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1232
!
PROBLEM 8.24
End A of a slender, uniform rod of length L and weight W bears on a surface as
shown, while end B is supported by a cord BC. Knowing that the coefficients of
friction are µs = 0.40 and µk = 0.30, determine (a) the largest value of θ for
which motion is impending, (b) the corresponding value of the tension in the cord.
SOLUTION
Free-body diagram
Rod AB is a three-force body. Thus, line of action of R must pass through D,
where W and T intersect.
Since AG = GB, CD = DB and the median AD of the isosceles
triangle ABC bisects the angle θ.
(a)
Thus,
1
2
φs = θ
Since motion impends,
φs = tan −1 0.40 = 21.80°
θ = 2φs = 2(21.8°)
θ = 43.6°
(b)
Force triangle:
This is a right triangle.
T = W sin φs
= W sin 21.8°
T = 0.371W
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1233
!
PROBLEM 8.25
A window sash weighing 10 lb is normally supported by two 5-lb sash
weights. Knowing that the window remains open after one sash cord
has broken, determine the smallest possible value of the coefficient of
static friction. (Assume that the sash is slightly smaller than the frame
and will bind only at Points A and D.)
SOLUTION
FBD window:
T = 5 lb
ΣFx = 0: N A − N D = 0
N A = ND
Impending motion:
FA = µs N A
FD = µs N D
ΣM D = 0: (18 in.)W − (27 in.) NA − (36 in.) FA = 0
W = 10 lb
3
N A + 2µs N A
2
2W
NA =
3 + 4µs
W=
ΣFy = 0: FA − W + T + FD = 0
FA + FD = W − T =
Now
W
2
FA + FD = µs ( N A + N D )
= 2µs N A
Then
W
2W
= 2µs
2
3 + 4µs
µs = 0.750
or
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1234
!
PROBLEM 8.26
A 500-N concrete block is to be lifted by the pair of tongs shown.
Determine the smallest allowable value of the coefficient of static
friction between the block and the tongs at F and G.
SOLUTION
Free body: Members CA, AB, BD
C y = Dy =
By symmetry:
1
(500) = 250 N
2
Since CA is a two-force member,
Cy
Cx
250 N
=
=
90 mm 75 mm 75 mm
Cx = 300 N
ΣFx = 0: Dx = C x
Dx = 300 N
Free body: Tong DEF
ΣM E = 0: (300 N)(105 mm) + (250 N)(135 mm)
+ (250 N)(157.5 mm) − Fx (360 mm) = 0
Fx = +290.625 N
Minimum value of µs :
µs =
Fy
Fx
=
250 N
290.625 N
µs = 0.860
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1235
!
PROBLEM 8.27
The press shown is used to emboss a small seal at E. Knowing
that the coefficient of static friction between the vertical guide
and the embossing die D is 0.30, determine the force exerted by
the die on the seal.
SOLUTION
Free body: Member ABC
ΣM A = 0: FBD cos 20°(4) + FBD sin 20°(6.9282)
− (50 lb)(4 + 15.4548) = 0
FBD = 158.728 lb
Free body: Die D
φs = tan −1 µs
= tan −1 0.3
= 16.6992°
Force triangle:
158.728 lb
D
=
sin 53.301° sin 106.6992°
D = 132.869 lb
D = 132.9 lb
On seal:
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1236
PROBLEM 8.28
The 100-mm-radius cam shown is used to control the motion of the
plate CD. Knowing that the coefficient of static friction between the
cam and the plate is 0.45 and neglecting friction at the roller
supports, determine (a) the force P required to maintain the motion
of the plate, knowing that the plate is 20 mm thick, (b) the largest
thickness of the plate for which the mechanism is self locking (i.e.,
for which the plate cannot be moved however large the force P
may be).
SOLUTION
Free body: Cam
Impending motion:
F = µs N
ΣM A = 0: QR − NR sin θ + ( µs N ) R cos θ = 0
N=
Free body: Plate
ΣFx = 0 P = µs N
Q
sin θ − µs cos θ
(1)
(2)
Geometry in ∆ABD with R = 100 mm and d = 20 mm
R−d
R
80 mm
=
100 mm
= 0.8
cos θ =
sin θ = 1 − cos 2 θ = 0.6
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1237
PROBLEM 8.28 (Continued)
(a)
Eq. (1) using
Q = 60 N and µs = 0.45
60 N
0.6 − (0.45)(0.8)
60
=
= 250 N
0.24
N=
Eq. (2)
P = µs N = (0.45)(250 N)
P = 112.5 N
(b)
For P = ∞, N = ∞. Denominator is zero in Eq. (1).
sin θ − µ s cos θ = 0
tan θ = µ s = 0.45
θ = 24.23°
R−d
R
100 − d
cos 24.23 =
100
cos θ =
d = 8.81 mm
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1238
!
PROBLEM 8.29
A slender rod of length L is lodged between peg C and the vertical wall and
supports a load P at end A. Knowing that the coefficient of static friction is 0.20 at
both B and C, find the range of values of the ratio L/a for which equilibrium is
maintained.
SOLUTION
We shall first assume that the motion of end B is impending upward. The friction forces at B and C will have
the values and directions indicated in the FB diagram.
a !
ΣM B = 0: PL sin θ − N C "
#=0
$ sin θ %
NC =
PL 2
sin θ
a
(1)
ΣFx = 0: N C cos θ + µ N C sin θ − N B = 0
(2)
ΣFy = 0: N C sin θ − µ N C cos θ − µ N B − P = 0
(3)
Multiply Eq. (2) by µ and subtract from Eq. (3):
N C (sin θ − µ cos θ ) − µ N C (cos θ + µ sin θ ) − P = 0
P = N C [sin θ (1 − µ 2 ) − 2 µ cos θ ]
Substitute for N C from Eq. (1) and solve for a/L:
a
= sin 2 θ [(1 − µ 2 )sin θ − 2µ cos θ ]
L
(4)
Making θ = 35° and µ = 0.20 in Eq. (4):
a
= sin 2 35°[(1 − 0.04)sin 35° − 2(0.20) cos 35°]
L
= 0.07336
L
= 13.63 #
a
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1239
PROBLEM 8.29 (Continued)
Assuming now that the motion at B is impending downward, we should reverse the direction of FB and FC in.
the FB diagram. The same result may be obtained by making θ = 35° and µ = −0.20 in Eq. (4):
a
= sin 2 35°[(1 − 0.04)sin 35° − 2(−0.20) cos 35°]
L
= 0.2889
L
= 3.461 #
a
Thus, the range of values of L/a for which equilibrium is maintained is
3.46 #
L
# 13.63
a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1240
!
PROBLEM 8.30
The 50-lb plate ABCD is attached at A and D to collars that can slide on
the vertical rod. Knowing that the coefficient of static friction is 0.40
between both collars and the rod, determine whether the plate is in
equilibrium in the position shown when the magnitude of the vertical
force applied at E is (a) P = 0, (b) P = 20 lb.
SOLUTION
(a)
P=0
ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) = 0
N A = 75 lb
ΣFx = 0: N D = N A = 75 lb
ΣFy = 0: FA + FD − 50 lb = 0
FA + FD = 50 lb
But:
( FA ) m = µs N A = 0.40(75 lb) = 30 lb
(FD ) m = µ s N D = 0.40(75 lb) = 30 lb
(b)
Thus:
( FA ) m + ( FD )m = 60 lb
and
( FA ) m + ( FD )m . FA + FD
P = 20 lb
Plate is in equilibrium
ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + (20 lb)(5 ft) = 0
N A = 25 lb
ΣFx = 0: N D = N A = 25 lb
ΣFy = 0: FA + FD − 50 lb + 20 lb = 0
FA + FD = 30 lb
But:
( FA ) m = µs N A = 0.4(25 lb) = 10 lb
(FD ) m = µ s N D = 0.4(25 lb) = 10 lb
Thus:
and
( FA ) m + ( FD )m = 20 lb
FA + FD . ( FA ) m + ( FD ) m
Plate moves downward
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1241
!
PROBLEM 8.31
In Problem 8.30, determine the range of values of the magnitude P of the
vertical force applied at E for which the plate will move downward.
PROBLEM 8.30 The 50-lb plate ABCD is attached at A and D to collars
that can slide on the vertical rod. Knowing that the coefficient of static
friction is 0.40 between both collars and the rod, determine whether the
plate is in equilibrium in the position shown when the magnitude of the
vertical force applied at E is (a) P = 0, (b) P = 20 lb.
SOLUTION
We shall consider the following two cases:
(1)
0 , P , 30 lb
ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0
N A = 75 lb − 2.5P
(Note: N A $ 0 and directed
for P # 30 lb as assumed here)
ΣFx = 0: N A = N D
ΣFy = 0: FA + FD + P − 50 = 0
FA + FD = 50 − P
But:
( FA ) m = ( F0 ) m = µ s N A
= 0.40(75 − 2.5P)
= 30 − P
Plate moves if:
or
(2)
FA + FD . ( FA ) m + ( FD ) m
50 − P . (30 − P) + (30 − P)
P . 10 lb
30 lb , P , 50 lb
ΣM D = 0: − N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0
N A = 2.5P − 75
(Note: NA . and directed
for P . 30 lb as assumed)
ΣFx = 0: N A = N D
ΣFy = 0: FA + FD + P − 50 = 0
FA + FD = 50 − P
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1242
PROBLEM 8.31 (Continued)
But:
( FA ) m = ( FD ) m = µ s N A
= 0.40(2.5 P − 75)
= P − 30 lb
Plate moves if:
FA + FD . ( FA ) m + ( FD ) m
50 − P . ( P − 30) + ( P − 30)
P,
110
= 36.7 lb
3
10.00 lb , P , 36.7 lb !
Thus, plate moves downward for:
(Note: For P . 50 lb, plate is in equilibrium)
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1243
PROBLEM 8.32
A pipe of diameter 60 mm is gripped by the stillson wrench shown. Portions AB
and DE of the wrench are rigidly attached to each other, and portion CF is
connected by a pin at D. If the wrench is to grip the pipe and be self-locking,
determine the required minimum coefficients of friction at A and C.
SOLUTION
FBD ABD:
ΣM D = 0: (15 mm)N A − (110 mm)FA = 0
FA = µ A N A
Impending motion:
15 − 110µ A = 0
Then
µ A = 0.13636
or
µ A = 0.1364 !
ΣFx = 0: FA − Dx = 0 Dx = FA
FBD pipe:
ΣFy = 0: N C − N A = 0
NC = N A
FBD DF:
ΣM F = 0: (550 mm)FC − (15 mm)N C − (500 mm)Dx = 0
Impending motion:
Then
But
So
FC = µC N C
550 µC − 15 = 500
NC = N A
FA
NC
and
FA
= µ A = 0.13636
NA
550 µC = 15 + 500(0.13636)
µC = 0.1512 ! "
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you are using it without permission.
1244
PROBLEM 8.33
Solve Problem 8.32 assuming that the diameter of the pipe is 30 mm.
PROBLEM 8.32 A pipe of diameter 60 mm is gripped by the stillson wrench
shown. Portions AB and DE of the wrench are rigidly attached to each other,
and portion CF is connected by a pin at D. If the wrench is to grip the pipe and
be self-locking, determine the required minimum coefficients of friction at
A and C.
SOLUTION
FBD ABD:
ΣM D = 0: (15 mm)N A − (80 mm)FA = 0
FA = µ A N A
Impending motion:
Then
15 mm − (80 mm)µ A = 0
µ A = 0.1875 !
ΣFx = 0: FA − Dx = 0
Dx = FA = 0.1875 N A
So that
FBD pipe:
ΣFy = 0: N C − N A = 0
NC = N A
FBD DF:
ΣM F = 0: (550 mm)FC − (15 mm)N C − (500 mm)Dx = 0
Impending motion:
FC = µC N C
550 µC − 15 = 500(0.1875)
But N A = N C (from pipe FBD) so
NA
NC
NA
=1
NC
µC = 0.1977 !"
and
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1245
PROBLEM 8.34
A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the
platform. A horizontal force P is applied to the dolly, which is
mounted on frictionless wheels. The coefficients of friction between
all surfaces are µs = 0.30 and µk = 0.25, and initially x = 2 ft.
Knowing that the top surface of the dolly is slightly higher than the
platform, determine the force P required to start moving the beam.
(Hint: The beam is supported at A and D.)
SOLUTION
FBD beam:
ΣM A = 0: N D (8ft) − (1200 lb)(5ft) = 0
N D = 750 lb
ΣFy = 0: N A − 1200 + 750 = 0
N A = 450 lb
( FA ) m = µ s N A = 0.3(450) = 135.0 lb
( FD ) m = µ s N D = 0.3(750) = 225 lb
Since ( FA ) m , ( FD )m , sliding first impends at A with
FA = ( FA ) m = 135 lb
ΣFx = 0: FA − FD = 0
FD = FA = 135.0 lb
FBD dolly:
From FBD of dolly:
ΣFx = 0: FD − P = 0
P = FD = 135.0 lb
P = 135.0 lb !"
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1246
PROBLEM 8.35
(a) Show that the beam of Problem 8.34 cannot be moved if the top
surface of the dolly is slightly lower than the platform. (b) Show
that the beam can be moved if two 175-lb workers stand on the
beam at B and determine how far to the left the beam can be moved.
PROBLEM 8.34 A 10-ft beam, weighing 1200 lb, is to be moved
to the left onto the platform. A horizontal force P is applied to the
dolly, which is mounted on frictionless wheels. The coefficients of
friction between all surfaces are µs = 0.30 and µk = 0.25, and
initially x = 2 ft. Knowing that the top surface of the dolly is
slightly higher than the platform, determine the force P required to
start moving the beam. (Hint: The beam is supported at A and D.)
SOLUTION
(a)
Beam alone
ΣM C = 0: N B (8 ft) − (1200 lb)(3 ft) = 0
N B = 450 lb
ΣFy = 0: N C + 450 − 1200 = 0
N C = 750 lb
( FC ) m = µs N C = 0.3(750) = 225 lb
(FB ) m = µs N B = 0.3(450) = 135 lb
Beam cannot be moved !
Since ( FB ) m , ( FC ) m , sliding first impends at B, and
(b)
Beam with workers standing at B
ΣM C = 0: N B (10 − x) − (1200)(5 − x) − 350(10 − x) = 0
NB =
9500 − 1550 x
10 − x
ΣM B = 0: (1200)(5) − N C (10 − x) = 0
NC =
6000
10 − x
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1247
PROBLEM 8.35 (Continued)
Check that beam starts moving for x = 2 ft:
For x = 2 ft:
9500 − 1550(2)
= 800 lb
10 − 2
6000
NC =
= 750 lb
10 − 2
( FC ) m = µs NC = 0.3(750) = 225 lb
NB =
( FB ) m = µs N B = 0.3(800) = 240 lb
Since ( FC ) m , ( FB ) m , sliding first impends at C,
Beam moves !
How far does beam move?
Beam will stop moving when
FC = ( FB ) m
But
FC = µk NC = 0.25
and
( FB ) m = µ s N B = 0.30
Setting FC = ( FB ) m :
6000 1500
=
10 − x 10 − x
9500 − 1550 x 2850 − 465 x
=
10 − x
10 − x
1500 = 2850 − 465 x
x = 2.90ft !"
(Note: We have assumed that, once started, motion is continuous and uniform (no acceleration).)
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1248
PROBLEM 8.36
Knowing that the coefficient of static friction between the collar and the rod
is 0.35, determine the range of values of P for which equilibrium is maintained
when θ = 50° and M = 20 N ⋅ m.
SOLUTION
Free body member AB:
BC is a two-force member.
ΣM A = 0: 20 N ⋅ m − FBC cos 50°(0.1 m) = 0
FBC = 311.145 N
Motion of C impending upward:
ΣFx = 0: (311.145 N) cos 50° − N = 0
N = 200 N
ΣFy = 0: (311.145 N) sin 50° − P − (0.35)(200 N) = 0
P = 168.351 N
Motion of C impending downward:
ΣFx = 0: (311.145 N) cos 50° − N = 0
N = 200 N
ΣFy = 0: (311.145 N) sin 50° − P + (0.35)(200 N) = 0
P = 308.35 N
168.4 N # P # 308 N !
Range of P:
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1249
PROBLEM 8.37
Knowing that the coefficient of static friction between the collar and the
rod is 0.40, determine the range of values of M for which equilibrium is
maintained when θ = 60° and P = 200 N.
SOLUTION
Free body member AB:
BC is a two-force member.
ΣM A = 0: M − FBC cos 60°(0.1 m) = 0
M = 0.05FBC
(1)
Motion of C impending upward:
ΣFx = 0: FBC cos 60° − N = 0
N = 0.5FBC
ΣFy = 0: FBC sin 60° − 200 N − (0.40)(0.5FBC ) = 0
FBC = 300.29 N
Eq. (1):
M = 0.05(300.29)
M = 15.014 N ⋅ m
Motion of C impending downward:
ΣFx = 0: FBC cos 60° − N = 0
N = 0.5FBC
ΣFy = 0: FBC sin 60° − 200 N + (0.40)(0.5 FBC ) = 0
FBC = 187.613 N
Eq. (1):
M = 0.05(187.613)
M = 9.381 N ⋅ m
9.38 N ⋅ m # M # 15.01 N ⋅ m !
Range of M:
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1250
PROBLEM 8.38
The slender rod AB of length l = 600 mm is attached to a collar at B and
rests on a small wheel located at a horizontal distance a = 80 mm from
the vertical rod on which the collar slides. Knowing that the coefficient
of static friction between the collar and the vertical rod is 0.25 and
neglecting the radius of the wheel, determine the range of values of P for
which equilibrium is maintained when Q = 100 N and θ = 30°.
SOLUTION
For motion of collar at B impending upward:
F = µs N
ΣM B = 0: Ql sin θ −
Ca
=0
sin θ
l!
C = Q " # sin 2 θ
a
$ %
l!
ΣFx = 0: N = C cos θ = Q " # sin 2 θ cos θ
$a%
ΣFy = 0: P + Q − C sin θ − µ s N = 0
l!
l!
P + Q − Q " # sin 3 θ − µ s Q " # sin 2 θ cos θ = 0
a
a
$ %
$ %
l
&
'
P = Q ( sin 2 θ (sin θ − µ s cosθ ) − 1)
*a
+
Substitute data:
(1)
& 600 mm 2
'
P = (100 N) (
sin 30°(sin 30° − 0.25cos 30°) − 1)
* 80 mm
+
P = −46.84 N (P is directed )
P = −46.8 N
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1251
PROBLEM 8.38 (Continued)
For motion of collar, impending downward:
F = µs N
In Eq. (1) we substitute − µ s for µ s .
&l
'
P = Q ( sin 2 θ (sin θ + µ s cos θ ) − 1)
*a
+
& 600 mm 2
'
P = (100 N) (
sin 30°(sin 30° + 0.25cos θ ) − 1)
80
mm
*
+
P = + 34.34 N
−46.8 N # P # 34.3 N !"
For equilibrium:
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1252
PROBLEM 8.39
Two 10-lb blocks A and B are connected by a slender rod of negligible
weight. The coefficient of static friction is 0.30 between all surfaces of
contact, and the rod forms an angle θ = 30°. with the vertical. (a) Show that
the system is in equilibrium when P = 0. (b) Determine the largest value of
P for which equilibrium is maintained.
SOLUTION
FBD block B:
(a)
Since P = 2.69 lb to initiate motion,
equilibrium exists with P = 0 !
(b)
For Pmax , motion impends at both surfaces:
ΣFy = 0: N B − 10 lb − FAB cos 30° = 0
Block B:
N B = 10 lb +
Impending motion:
3
FAB
2
(1)
FB = µs N B = 0.3N B
ΣFx = 0: FB − FAB sin 30° = 0
FAB = 2 FB = 0.6 N B
Solving Eqs. (1) and (2):
N B = 10 lb +
(2)
3
(0.6 N B ) = 20.8166 lb
2
FBD block A:
FAB = 0.6 N B = 12.4900 lb
Then
Block A:
ΣFx = 0: FAB sin 30° − N A = 0
NA =
Impending motion:
1
1
FAB = (12.4900 lb) = 6.2450 lb
2
2
FA = µs N A = 0.3(6.2450 lb) = 1.8735 lb
ΣFy = 0: FA + FAB cos 30° − P − 10 lb = 0
3
FAB − 10 lb
2
3
(12.4900 lb) − 10 lb
= 1.8735 lb +
2
= 2.69 lb
P = FA +
P = 2.69 lb !"
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1253
PROBLEM 8.40
Two identical uniform boards, each of weight 40 lb, are temporarily leaned
against each other as shown. Knowing that the coefficient of static friction
between all surfaces is 0.40, determine (a) the largest magnitude of the
force P for which equilibrium will be maintained, (b) the surface at which
motion will impend.
SOLUTION
Board FBDs:
Assume impending motion at C, so
FC = µs N C = 0.4 N C
FBD II:
ΣM B = 0: (6 ft)N C − (8 ft)FC − (3 ft)(40 lb) = 0
[6 ft − 0.4(8 ft)]N C = (3 ft)(40 lb)
or
N C = 42.857 lb
and
FC = 0.4 N C = 17.143 lb
ΣFx = 0: N B − FC = 0
N B = FC = 17.143 lb
ΣM y = 0: − FB − 40 lb + N C = 0
FB = N C − 40 lb = 2.857 lb
Check for motion at B:
FBD I:
FB
2.857 lb
=
= 0.167 , µs , OK, no motion.
N B 17.143 lb
ΣM A = 0: (8 ft)N B + (6 ft)FB − (3 ft)(P + 40 lb) = 0
(8 ft)(17.143 lb) + (6 ft)(2.857 lb)
− 40 lb
3 ft
= 11.429 lb
P=
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1254
PROBLEM 8.40 (Continued)
Check for slip at A (unlikely because of P):
ΣFx = 0: FA − N B = 0 or
FA = N B = 17.143 lb
ΣFy = 0: N A − P − 40 lb + FB = 0
or
NA = 11.429 lb + 40 lb − 2.857 lb
= 48.572 lb
Then
FA 17.143 lb
=
= 0.353 , µ s
N A 48.572 lb
OK, no slip , assumption is correct.
Therefore
(a)
Pmax = 11.43 lb !
(b)
Motion impends at C !
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1255
PROBLEM 8.41
Two identical 5-ft-long rods connected by a pin at B are placed
between two walls and a horizontal surface as shown. Denoting
by µs the coefficient of static friction at A, B, and C, determine
the smallest value of µs for which equilibrium is maintained.
SOLUTION
Sense of impending motion:
ΣM B = 0: 2W − 3N A − 4 µs N A = 0
ΣM B = 0: 1.5W − 4 NC + 3µ s N C = 0
2W
(3 + 4µs )
(1)
ΣFy : N AB = W − µs N A
(3)
NA =
ΣFx = 0: Bx + µs N BA − N A = 0
Bx = N A − µs N BA
Equate (5) and (6):
Substitute from Eqs. (3) and (4):
1.5W
(4 − 3µs )
(2)
ΣFy : N BC = W + µs NC
(4)
NC =
ΣFx = 0: Bx − NC − µs N BC = 0
(5)
Bx = NC + µs N BC
(6)
N A − µs N BA = NC + µs N BC
N A − µ s (W − µ s N A ) = NC + µ s (W + µ s NC )
N A (1 + µ s2 ) − µ sW = NC (1 + µ s2 ) + µ sW
Substitute from Eqs. (1) and (2):
2W
1.5W
(1 + µ s2 ) − µ sW =
(1 + µ s2 ) + µsW
3 + 4µs
4 − 3µ s
2µs
2
1.5
−
=
3 + 4µ s 4 − 3µ s 1 + µ s2
Solve for µ s :
µs = 0.0949 !
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1256
PROBLEM 8.42
Two 8-kg blocks A and B resting on shelves are connected by a rod of
negligible mass. Knowing that the magnitude of a horizontal force P
applied at C is slowly increased from zero, determine the value of P for
which motion occurs, and what that motion is, when the coefficient of
static friction between all surfaces is (a) µ s = 0.40, (b) µ s = 0.50.
SOLUTION
(a)
µs = 0.40:
Assume blocks slide to right.
W = mg = (8 kg)(9.81 m/s 2 ) = 78.48 N
FA = µ s N A
FB = µ s N B
ΣFy = 0: N A + N B − 2W = 0
N A + N B = 2W
ΣFx = 0: P − FA − FB = 0
P = FA + FB = µ s ( N A + N B ) = µ s (2W )
(1)
P = 0.40(2)(78.48 N) = 62.78 N
ΣM B = 0: P (0.1 m) − ( N A − W )(0.09326 m) + FA (0.2 m) = 0
(62.78)(0.1) − ( N A − 78.48)(0.09326) + (0.4)( N B )(0.2) = 0
0.17326 N A = 1.041
N A = 6.01 N . 0 OK
System slides: P = 62.8 N !
(b)
µs = 0.50: See part a.
Eq. (1):
P = 0.5(2)(78.48 N) = 78.48 N
ΣM B = 0: P(0.1 m) + ( N A − W )(0.09326 m) + FA (0.2 m) = 0
(78.48)(0.1) + ( N A − 78.48)(0.09326) + (0.5) N A (0.2) = 0
0.19326 N A = −0.529
N A = −2.73 N , 0 uplift, rotation about B
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1257
PROBLEM 8.42 (Continued)
For N A = 0:
ΣM B = 0: P(0.1 m) − W (0.09326 m) = 0
P = (78.48 N)(0.09326 m)/(0.1) = 73.19
System rotates about B: P = 73.2 N !
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1258
PROBLEM 8.43
A slender steel rod of length 225 mm is placed inside a pipe as shown.
Knowing that the coefficient of static friction between the rod and the
pipe is 0.20, determine the largest value of θ for which the rod will not
fall into the pipe.
SOLUTION
Motion of rod impends down at A and to left at B.
FA = µ s N A FB = µ s N B
ΣFx = 0: N A − N B sin θ + FB cos θ = 0
N A − N B sin θ + µ s N B cos θ = 0
N A = N B (sin θ − µ s cos θ )
(1)
ΣFy = 0: FA + N B cos θ + FB sin θ − W = 0
µs N A + N B cos θ + µs N B sin θ − W = 0
(2)
Substitute for NA from Eq. (1) into Eq. (2):
µs N B (sin θ − µ s cos θ ) + N B cos θ + µs N B sin θ − W = 0
NB =
W
(1 −
µs2 ) cos θ
+ 2µ s sin θ
=
W
(1 − 0.2 ) cos θ + 2(0.2) sin θ
2
(3)
75 !
ΣM A = 0: N B "
# − W (112.5cos θ ) = 0
$ cos θ %
Substitute for N B from Eq. (3), cancel W, and simplify to find
9.6 cos3 θ + 4sin θ cos 2 θ − 6.6667 = 0
cos3 θ (2.4 + tan θ ) = 1.6667
θ = 35.8° !
Solve by trial & error:
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1259
PROBLEM 8.44
In Problem 8.43, determine the smallest value of θ for which the rod will
not fall out the pipe.
PROBLEM 8.43 A slender steel rod of length 225 mm is placed inside a
pipe as shown. Knowing that the coefficient of static friction between the
rod and the pipe is 0.20, determine the largest value of θ for which the rod
will not fall into the pipe.
SOLUTION
Motion of rod impends up at A and right at B.
FA = µs N A
FB = µ s N B
ΣFx = 0: N A − N B sin θ − FB cos θ = 0
N A − N B sin θ − µ s N B cos θ = 0
N A = N B (sin θ + µ s cos θ )
(1)
ΣFy = 0: −FA + N B cos θ − FB sin θ − W = 0
− µ s N A + N B cos θ − µs N B sin θ − W = 0
(2)
Substitute for N A from Eq. (1) into Eq. (2):
− µ s N B (sin θ + µ s cos θ ) + N B cos θ − µ s N B sin θ − W = 0
NB =
W
(1 −
µs2 ) cos θ
− 2µs sin θ
=
W
(1 − 0.2 ) cos θ − 2(0.2)sin θ
(3)
2
75 !
ΣM A = 0: N B "
− W (112.5cos θ ) = 0
cos
θ #%
$
Substitute for NB from Eq. (3), cancel W, and simplify to find
9.6 cos3 θ − 4sin θ cos 2 θ − 6.6667 = 0
cos3 θ (2.4 − tan θ ) = 1.6667
θ = 20.5°
Solve by trial + error:
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1260
PROBLEM 8.45
Two slender rods of negligible weight are pin-connected at C and
attached to blocks A and B, each of weight W. Knowing that θ = 80°
and that the coefficient of static friction between the blocks and the
horizontal surface is 0.30, determine the largest value of P for which
equilibrium is maintained.
SOLUTION
FBD pin C:
FAC = P sin 20° = 0.34202 P
FBC = P cos 20° = 0.93969 P
ΣFy = 0: N A − W − FAC sin 30° = 0
N A = W + 0.34202 P sin 30° = W + 0.171010 P
or
FBD block A:
ΣFx = 0: FA − FAC cos 30° = 0
or
FA = 0.34202 P cos 30° = 0.29620 P
For impending motion at A:
FA = µs N A
Then
NA =
FA
µs
: W + 0.171010 P =
0.29620
P
0.3
P = 1.22500W
or
ΣFy = 0: N B − W − FBC cos 30° = 0
N B = W + 0.93969 P cos 30° = W + 0.81380 P
ΣFx = 0: FBC sin 30° − FB = 0
FB = 0.93969 P sin 30° = 0.46985P
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you are using it without permission.
1261
PROBLEM 8.45 (Continued)
FBD block B:
For impending motion at B:
FB = µs N B
Then
NB =
FB
µs
: W + 0.81380 P =
0.46985P
0.3
P = 1.32914W
or
Pmax = 1.225W
Thus, maximum P for equilibrium
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1262
PROBLEM 8.46
The machine part ABC is supported by a frictionless hinge at B and a
10° wedge at C. Knowing that the coefficient of static friction is 0.20
at both surfaces of the wedge, determine (a) the force P required to
move the wedge to the left, (b) the components of the corresponding
reaction at B.
SOLUTION
µs = 0.20
φs = tan −1 µs = tan −1 0.20 = 11.3099°
Free body: ABC
10° + 11.3099° = 21.3099°
ΣM B = 0: ( RC cos 21.3099°)(10) − (120 lb)(8) = 0
RC = 103.045 lb
Free body: Wedge
Force triangle:
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1263
PROBLEM 8.46 (Continued)
Law of sines:
( R = 103.045 lb)
P
= C
sin 32.6198°
sin 78.690°
P = 56.6 lb
(a)
(b)
Returning to free body of ABC:
ΣFx = 0: Bx + 120 − (103.045) sin 21.3099° = 0
Bx = −82.552 lb
B x = 82.6 lb
ΣFy = 0: By + (103.045) cos 21.3099° = 0
By = −96.000 lb
B y = 96.0 lb
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1264
!
PROBLEM 8.47
Solve Problem 8.46 assuming that the wedge is to be moved to the
right.
PROBLEM 8.46 The machine part ABC is supported by a
frictionless hinge at B and a 10° wedge at C. Knowing that the
coefficient of static friction is 0.20 at both surfaces of the wedge,
determine (a) the force P required to move the wedge to the left,
(b) the components of the corresponding reaction at B.
SOLUTION
µs = 0.20
φs = tan −1 µ s = tan −1 0.20 = 11.30993°
Free body: ABC
11.30993° − 10° = 1.30993°
ΣM B = 0: ( RC cos 1.30993°)(10) − (120 lb)(8) = 0
RC = 96.025 lb
Free body: Wedge
Force triangle:
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1265
PROBLEM 8.47 (Continued)
Law of sines:
( R = 96.025 lb)
P
= C
sin 12.6198°
sin 78.690°
P = 21.4 lb
(a)
(b)
Returning to free body of ABC:
ΣFx = 0: Bx + 120 + (96.025)sin 1.30993° = 0
Bx = −122.195 lb
B x = 122.2 lb
ΣFy = 0: By + (96.025) cos 1.30993° = 0
By = −96.000 lb
B y = 96.0 lb
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you are using it without permission.
1266
!
PROBLEM 8.48
Two 8° wedges of negligible weight are used to move and position the 800-kg
block. Knowing that the coefficient of static friction is 0.30 at all surfaces of
contact, determine the smallest force P that should be applied as shown to one
of the wedges.
SOLUTION
µs = 0.30 φs = tan −1 0.30 = 16.70°
Free body: 800-kg block and right-hand wedge
W = (800 kg)(9.81 m/s2 ) = 7848 N
Force triangle:
α = 90° − 16.70° − 24.70°
= 48.60°
Law of sines:
R1
7848 N
=
sin 16.70° sin 48.60°
R1 = 3006.5 N
Free body: Left-hand wedge
Force triangle:
β = 16.70° + 24.70°
= 41.40°
Law of sines:
R1
P
=
sin 41.40° sin(90° − 16.70°)
P
3006.5 N
=
sin 41.40° cos 16.70°
P = 2080 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1267
!
PROBLEM 8.49
Two 8° wedges of negligible weight are used to move and position the
800-kg block. Knowing that the coefficient of static friction is 0.30 at all
surfaces of contact, determine the smallest force P that should be applied
as shown to one of the wedges.
SOLUTION
µs = 0.30 φs = tan −1 0.30 = 16.70°
Free body: 800-kg block
Force triangle:
W = (800 kg)(9.81 m/s 2 ) = 7848 N
α = 90° − 2φs = 90° − 2(16.70°)
= 56.60°
Law of sines:
R1
7848 N
=
sin 16.70° sin 56.60°
R1 = 2701 N
Free body: Right-hand wedge
Force triangle:
β = 16.70° + 24.70°
= 41.40°
Law of sines:
R1
P
=
sin 41.40° sin(90° − 24.70°)
P
2701 N
=
sin 41.40° cos 24.70°
P = 1966 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1268
!
PROBLEM 8.50
The elevation of the end of the steel beam supported by a
concrete floor is adjusted by means of the steel wedges E and F.
The base plate CD has been welded to the lower flange of the
beam, and the end reaction of the beam is known to be 100 kN.
The coefficient of static friction is 0.30 between two steel
surfaces and 0.60 between steel and concrete. If the horizontal
motion of the beam is prevented by the force Q, determine
(a) the force P required to raise the beam, (b) the corresponding
force Q.
SOLUTION
Free body: Beam and plate CD
(100 kN)
cos16.7°
R1 = 104.4 kN
R1 =
Free body: Wedge E
P
104.4 kN
=
sin 43.4° sin 63.3°
(a)
P = 80.3 kN
!
φs = tan −1 0.3 = 16.7°
(b)
Q = (100 kN) tan16.7°
Q = 30 kN
Free body: Wedge F
(To check that it does not move.)
Since wedge F is a two-force body, R2 and R3 are colinear
Thus
But
θ = 26.7°
φconcrete = tan −1 0.6 = 31.0° . θ
OK!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1269
PROBLEM 8.51
The elevation of the end of the steel beam supported by a
concrete floor is adjusted by means of the steel wedges E and F.
The base plate CD has been welded to the lower flange of
the beam, and the end reaction of the beam is known to be
100 kN. The coefficient of static friction is 0.30 between two
steel surfaces and 0.60 between steel and concrete. If the
horizontal motion of the beam is prevented by the force Q,
determine (a) the force P required to raise the beam, (b) the
corresponding force Q.
SOLUTION
Free body: Wedge F
φs = tan −1 0.30 = 16.7°
(a)
P = (100 kN) tan 26.7° + (100 kN) tan φs
P = 50.29 kN + 30 kN
P = 80.29 kN
R1 =
P = 80.3 kN
(100 kN)
= 111.94 kN
cos 26.7°
Free body: Beam, plate, and wedge E
(b)
Q = W tan 26.7° = (100 kN) tan 26.7°
Q = 50.29 kN
Q = 50.3 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1270
!
PROBLEM 8.52
A wedge A of negligible weight is to be driven between two 100-lb
plates B and C. The coefficient of static friction between all surfaces
of contact is 0.35. Determine the magnitude of the force P required
to start moving the wedge (a) if the plates are equally free to move,
(b) if plate C is securely bolted to the surface.
SOLUTION
(a)
With plates equally free to move
Free body: Plate B
φs = tan −1 µs = tan −1 0.35 = 19.2900°
Force triangle:
α = 180° − 124.29° − 19.29° = 36.42°
Law of sines:
R1
100 lb
=
sin 19.29° sin 36.42°
R1 = 55.643 lb
Free body: Wedge A
Force triangle:
By symmetry,
R3 = R1 = 55.643 lb
β = 19.29° + 15° = 34.29°
(b)
Then
P = 2 R1 sin β
or
P = 2(55.643) sin 34.29°
P = 62.7 lb
With plate C bolted
The free body diagrams of plate B and wedge A (the only members to move) are same as above.
Answer is thus the same.
P = 62.7 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1271
!
PROBLEM 8.53
Block A supports a pipe column and rests as shown on wedge B.
Knowing that the coefficient of static friction at all surfaces of contact
is 0.25 and that θ = 45°, determine the smallest force P required to raise
block A.
SOLUTION
φs = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
3 kN
=
sin 104.036° sin 16.928°
R2 = 10.0000 kN
FBD wedge B:
P
10.0000 kN
=
sin 73.072° sin 75.964°
P = 9.8611 kN
P = 9.86 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1272
!
PROBLEM 8.54
Block A supports a pipe column and rests as shown on wedge B. Knowing
that the coefficient of static friction at all surfaces of contact is 0.25 and
that θ = 45°, determine the smallest force P for which equilibrium is
maintained.
SOLUTION
φs = tan −1 µ s = tan −1 0.25 = 14.036°
FBD block A:
R2
3 kN
=
sin(75.964°) sin(73.072°)
R2 = 3.0420 kN
FBD wedge B:
P
3.0420 kN
=
sin 16.928° sin 104.036°
P = 0.91300 kN
P = 913 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1273
!
PROBLEM 8.55
Block A supports a pipe column and rests as shown on wedge B.
The coefficient of static friction at all surfaces of contact is 0.25.
If P = 0, determine (a) the angle θ for which sliding is impending,
(b) the corresponding force exerted on the block by the vertical wall.
SOLUTION
Free body: Wedge B
(a)
φs = tan −1 0.25 = 14.04°
Since wedge is a two-force body, R 2 and R 3 must be equal and opposite.
Therefore, they form equal angles with vertical
β = φs
and
θ − φs = φs
θ = 2φs = 2(14.04°)
θ = 28.1°
Free body: Block A
R1 = (3 kN)sin 14.04° = 0.7278 kN
(b)
R1 = 728 N
Force exerted by wall:
14.04°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1274
!
PROBLEM 8.56
A 12° wedge is used to spread a split ring. The coefficient of static
friction between the wedge and the ring is 0.30. Knowing that a force P
of magnitude 25 lb was required to insert the wedge, determine the
magnitude of the forces exerted on the ring by the wedge after insertion.
SOLUTION
Free body: Wedge
φs = tan −1 µ s = tan −1 0.30 = 16.6992°
Force triangle:
α = 6° + φ s
= 6° + 16.6992°
= 22.6992°
Q = Horizontal component of R
Q=
1
2
(25 lb)
tan 22.6992°
= 29.9 lb
Free body: After wedge has been inserted
Wedge is now a two-force body with forces shown.
Q = 29.9 lb
(Note: Since angles between force Q and normal to wedge is 6° , φs , wedge stays in place.)!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1275
PROBLEM 8.57
A 10° wedge is to be forced under end B of the 5-kg rod AB. Knowing
that the coefficient of static friction is 0.40 between the wedge and the
rod and 0.20 between the wedge and the floor, determine the smallest
force P required to raise end B of the rod.
SOLUTION
FBD AB:
W = mg
W = (5 kg)(9.81 m/s 2 )
W = 49.050 N
φs1 = tan −1 ( µ s )1 = tan −1 0.40 = 21.801°
ΣM A = 0: rR1 cos(10° + 21.801°) − rR1 sin(10° + 21.801°)
−
2r
π
(49.050 N) = 0
R1 = 96.678 N
FBD wedge:
φs 2 = tan −1 ( µ s ) 2 = tan −1 0.20 = 11.3099
P
96.678 N
=
sin(43.111°) sin 78.690°
P = 67.4 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1276
!
PROBLEM 8.58
A 10° wedge is used to split a section of a log. The coefficient of static friction
between the wedge and the log is 0.35. Knowing that a force P of magnitude
600 lb was required to insert the wedge, determine the magnitude of the forces
exerted on the wood by the wedge after insertion.
SOLUTION
FBD wedge (impending motion ):
φs = tan −1 µs
= tan −1 0.35
= 19.29°
By symmetry:
R1 = R2
ΣFy = 0: 2 R1 sin(5° + φs ) − 600 lb = 0
or
R1 = R2 =
300 lb
= 729.30 lb
sin (5°+19.29°)
When P is removed, the vertical components of R1 and R2 vanish, leaving the horizontal components
R1x = R2 x
= R1 cos(5° + φs )
= (729.30 lb) cos(5° + 19.29°)
R1x = R2 x = 665 lb
(Note that φs . 5°, so wedge is self-locking.)!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1277
PROBLEM 8.59
A conical wedge is placed between two horizontal plates that are then
slowly moved toward each other. Indicate what will happen to the wedge
(a) if µs = 0.20, (b) if µs = 0.30.
SOLUTION
As the plates are moved, the angle θ will decrease.
(a)
φs = tan −1 µs = tan −1 0.2 = 11.31°.
As θ decrease, the minimum angle at the contact approaches
12.5° . φs = 11.31°, so the wedge will slide up and out from the slot.
(b)
φs = tan −1 µs = tan −1 0.3 = 16.70°.
As θ decreases, the angle at one contact reaches 16.7°. (At this
time the angle at the other contact is 25° − 16.7° = 8.3° , φs ). The wedge binds in the slot.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1278
!
PROBLEM 8.60
A 15° wedge is forced under a 50-kg pipe as shown. The coefficient of static
friction at all surfaces is 0.20. (a) Show that slipping will occur between the pipe
and the vertical wall. (b) Determine the force P required to move the wedge.
SOLUTION
Free body: Pipe
ΣM B = 0: Wr sin θ + FA r (1 + sin θ ) − N A r cos θ = 0
Assume slipping at A:
FA = µs NA
NA cos θ − µs NA (1 + sin θ ) = W sin θ
NA =
W sin θ
cos θ − µs (1 + sin θ )
W sin 15°
cos 15° − (0.20)(1 + sin 15°)
= 0.36241W
NA =
ΣFx = 0: − FB − W sin θ − FA sin θ + N A cos θ = 0
FB = N A cos θ − µs N A sin θ − W sin θ
FB = 0.36241W cos 15° − 0.20(0.36241W )sin 15° − W sin 15°
FB = 0.072482W
ΣFy = 0: N B − W cos θ − FA cos θ − NA sin θ = 0
N B = N A sin θ + µ s N A cos θ + W cos θ
N B = (0.36241W )sin 15° + 0.20(0.36241W ) cos 15° + W cos 15°
N B = 1.12974W
Maximum available:
(a)
FB = µs N B = 0.22595W
No slip at B
We note that FB , Fmax
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1279
PROBLEM 8.60 (Continued)
(b)
Free body: Wedge
ΣFy = 0: N 2 − N B cos θ + FB sin θ = 0
N 2 = N B cos θ − FB sin θ
N 2 = (1.12974W ) cos15° − (0.07248W ) sin 15°
N 2 = 1.07249W
ΣFx = 0: FB cos θ + N B sin θ + µs N 2 − P = 0
P = FB cos θ + N B sin θ + µ s N 2
P = (0.07248W ) cos 15° + (1.12974W ) sin 15° + 0.2(1.07249W )
P = 0.5769W
W = mg : P = 0.5769(50 kg)(9.81 m/s 2 )
P = 283 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1280
!
PROBLEM 8.61
A 15° wedge is forced under a 50-kg pipe as shown. Knowing that the coefficient
of static friction at both surfaces of the wedge is 0.20, determine the largest
coefficient of static friction between the pipe and the vertical wall for which
slipping will occur at A.
SOLUTION
Free body: Pipe
ΣM A = 0: N B r cos θ − µ B N B r − ( µ B N B sin θ )r − Wr = 0
NB =
W
cos θ − µ B (1 + sin θ )
W
cos15° − 0.2(1 + sin15°)
N B = 1.4002W
NB =
ΣFx = 0: N A − N B sin θ − µ B N B cos θ = 0
N A = N B (sin θ + µ B cos θ )
= (1.4002W )(sin15° + 0.2 × cos15°)
N A = 0.63293W
ΣFy = 0: − FA − W + N B cos θ − µ B N B sin θ = 0
FA = N B (cos θ − µ B sin θ ) − W
FA = (1.4002W )(cos15° − 0.2 × sin θ ) − W
FA = 0.28001W
For slipping at A:
FA = µ A N A
µA =
FA 0.28001W
=
N A 0.63293W
µ A = 0.442
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1281
PROBLEM 8.62
An 8° wedge is to be forced under a machine base at B.
Knowing that the coefficient of static friction at all surfaces of
contact is 0.15, (a) determine the force P required to move the
wedge, (b) indicate whether the machine base will slide on the
floor.
SOLUTION
Free body: Machine base
ΣM B = 0: (200 lb)(3 ft) + (400 lb)(1.5 ft) − Ay (6 ft) = 0
Ay = 200 lb
ΣFy = 0: Ay + By − 200 lb − 400 lb = 0
200 lb + By − 200 lb − 400 lb = 0
By = 400 lb
Free body: Wedge
(Assume machine base will not move)
µs = 0.15, φs = tan −1 0.15 = 8.53°
We know that
Force triangle:
By = 400 lb
8° + φs = 8° + 8.53° = 16.53°
P = (400 lb) tan16.53° + (400 lb) tan 8.53°
P = 178.7 lb
Total maximum friction force at A and B:
Fm = µ sW = 0.15(200 lb + 400 lb) = 90 lb
If machine moves with wedge:
P = Fm = 90 lb
Using smaller P, we have
(a)
P = 90.0 lb
(b)
Machine base moves
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1282
PROBLEM 8.63
Solve Problem 8.62 assuming that the wedge is to be forced under
the machine base at A instead of B.
PROBLEM 8.62 An 8° wedge is to be forced under a machine
base at B. Knowing that the coefficient of static friction at all
surfaces of contact is 0.15, (a) determine the force P required to
move the wedge, (b) indicate whether the machine base will slide
on the floor.
SOLUTION
FBD: Machine base
ΣM B = 0: (200 lb)(3 ft) + (400 lb)(1.5 ft) − Ay (6 ft) = 0
A y = 200 lb
ΣFy = 0: Ay + B y − 200 lb − 400 lb = 0
B y = 400lb
φs = tan −1 0.15 = 8.53°
We known that
Ay = 200 lb
FBD: Wedge
Force triangle:
P = (200 lb) tan 8.53° + (200 lb) tan 16.53°
(a)
(b)
P = 89.4 lb
Total maximum friction force at A and B:
Fm = µs (W ) = 0.15(200 lb + 400 lb) = 90 lb
Since P , Fm ,
machine base will not move
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1283
PROBLEM 8.64*
A 200-N block rests as shown on a wedge of negligible weight. The coefficient
of static friction µs is the same at both surfaces of the wedge, and friction
between the block and the vertical wall may be neglected. For P = 100 N,
determine the value of µs for which motion is impending. (Hint: Solve the
equation obtained by trial and error.)
SOLUTION
Free body: Wedge
Force triangle:
Law of sines:
R2
P
=
sin(90° − φs ) sin(15° + 2φs )
R2 = P
sin(90° − φs )
sin(15° + 2φs )
(1)
Free body: Block
Fy = 0
Vertical component of R2 is 200 N
Return to force triangle of wedge. Note P = 100 N
100 N = (200 N) tan φ + (200 N) tan(15° + φs )
0.5 = tan φ + tan(15° + φs )
Solve by trial and error
φs = 6.292
µs = tan φs = tan 6.292°
µs = 0.1103
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1284
PROBLEM 8.65*
Solve Problem 8.64 assuming that the rollers are removed and that µs is the
coefficient of friction at all surfaces of contact.
PROBLEM 8.64* A 200-N block rests as shown on a wedge of negligible
weight. The coefficient of static friction µs is the same at both surfaces of the
wedge, and friction between the block and the vertical wall may be neglected.
For P = 100 N, determine the value of µs for which motion is impending.
(Hint: Solve the equation obtained by trial and error.)
SOLUTION
Free body: Wedge
Force triangle:
Law of sines:
R2
P
=
sin (90° − φs ) sin (15° + 2φs )
R2 = P
sin (90° − φs )
sin (15° + 2φs )
(1)
Free body: Block (Rollers removed)
Force triangle:
Law of sines:
R2
W
=
sin (90° + φs ) sin (75° − 2φs )
R2 = W
sin (90° + φs )
sin (75° − 2θ s )
(2)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1285
PROBLEM 8.65* (Continued)
Equate R2 from Eq. (1) and Eq. (2):
P
sin (90° − φs )
sin (90° + φs )
=W
sin (15° + 2φs )
sin (75° − 2φs )
P = 100 lb
W = 200 N
sin (90° + φs )sin (15° + 2φs )
0.5 =
sin (75° − 2φs )sin (90° − φs )
Solve by trial and error:
φs = 5.784°
µs = tan φs = tan 5.784°
µs = 0.1013
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1286
PROBLEM 8.66
Derive the following formulas relating the load W and the force P exerted on the handle of the jack discussed
in Section 8.6. (a) P = (Wr/a) tan (θ + φs), to raise the load; (b) P = (Wr/a) tan(φs − θ ), to lower the load if the
screw is self-locking; (c) P = (Wr/a) tan (θ − φs), to hold the load if the screw is not self-locking.
SOLUTION
FBD jack handle:
See Section 8.6.
ΣM C = 0: aP − rQ = 0 or
P=
r
Q
a
FBD block on incline:
(a)
Raising load
Q = W tan (θ + φs )
(b)
Lowering load if screw is self-locking (i.e.,: if φs > θ )
Q = W tan (φs − θ )
(c)
r
P = W tan (θ + φs )
a
r
P = W tan (φs − θ )
a
Holding load is screw is not self-locking (i.e., if φs , θ )
Q = W tan (θ − φs )
r
P = W tan (θ − φs )
a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1287
PROBLEM 8.67
The square-threaded worm gear shown has a mean radius of 1.5 in. and a
lead of 0.375 in. The large gear is subjected to a constant clockwise
couple of 7.2 kip ⋅ in. Knowing that the coefficient of static friction between
the two gears is 0.12, determine the couple that must be applied to shaft AB
in order to rotate the large gear counterclockwise. Neglect friction in the
bearings at A, B, and C.
SOLUTION
FBD large gear:
ΣM C = 0: (12 in.)W − 7.2 kip ⋅ in. = 0
W = 0.600 kips
= 600 lb
Block on incline:
θ = tan −1
0.375 in.
= 2.2785°
2π (1.5 in.)
φs = tan −1 µs = tan −1 0.12
= 6.8428°
Q = W tan (θ + φs )
= (600 lb) tan 9.1213°
= 96.333lb
FBD worm gear:
r = 1.5 in.
ΣM B = 0: (1.5in.)(96.333lb) − M = 0
M = 144.5 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1288
PROBLEM 8.68
In Problem 8.67, determine the couple that must be applied to shaft AB
in order to rotate the large gear clockwise.
PROBLEM 8.67 The square-threaded worm gear shown has a mean
radius of 1.5 in. and a lead of 0.375 in. The large gear is subjected to a
constant clockwise couple of 7.2 kip ⋅ in. Knowing that the coefficient of
static friction between the two gears is 0.12, determine the couple that must
be applied to shaft AB in order to rotate the large gear counterclockwise.
Neglect friction in the bearings at A, B, and C.
SOLUTION
FBD large gear:
ΣM C = 0: (12 in.)W − 7.2 kip ⋅ in. = 0
W = 0.600 kips = 600 lb
Block on incline:
θ = tan −1
0.375 in.
2π (1.5 in.)
= 2.2785°
φs = tan −1 µs = tan −1 0.12
= 6.8428°
Q = W tan (φs − θ )
= (600 lb) tan 4.5643°
= 47.898 lb
FBD worm gear:
r = 1.5 in.
ΣM B = 0: M − (1.5in.)(47.898 lb) = 0
M = 71.8lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1289
PROBLEM 8.69
High-strength bolts are used in the construction of many steel structures. For a 24-mmnominal-diameter bolt the required minimum bolt tension is 210 kN. Assuming the
coefficient of friction to be 0.40, determine the required couple that should be applied to
the bolt and nut. The mean diameter of the thread is 22.6 mm, and the lead is 3 mm.
Neglect friction between the nut and washer, and assume the bolt to be square-threaded.
SOLUTION
FBD block on incline:
3 mm
(22.6 mm)π
= 2.4195°
θ = tan −1
φs = tan −1 µ s = tan −1 0.40
φs = 21.801°
Q = (210 kN) tan (21.801° + 2.4195°)
Q = 94.468 kN
d
Torque = Q
2
22.6 mm
=
(94.468 kN)
2
= 1067.49 N ⋅ m
Torque = 1068 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1290
PROBLEM 8.70
The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius
6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of
static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that
must be applied to the sleeve in order to draw the rods closer together.
SOLUTION
To draw rods together:
Screw at A
tan θ =
2 mm
2π (6 mm)
θ = 3.037°
φs = tan −1 0.12
= 6.843°
Q = (2 kN) tan 9.88°
= 348.3 N
Torque at A = Qr
= (348.3 N)(6 mm)
= 2.09 N ⋅ m
Total torque = 4.18 N ⋅ m
Same torque required at B
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1291
PROBLEM 8.71
Assuming that in Problem 8.70 a right-handed thread is used on both rods A and B, determine the magnitude
of the couple that must be applied to the sleeve in order to rotate it.
PROBLEM 8.70 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of
mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The
coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the
couple that must be applied to the sleeve in order to draw the rods closer together.
SOLUTION
From the solution to Problem 8.70,
Torque at A = 2.09 N ⋅ m
Screw at B: Loosening
θ = 3.037°
φs = 6.843°
Q = (2 kN) tan 3.806°
= 133.1 N
Torque at B = Qr
= (133.1 N)(6 mm)
= 0.798 N ⋅ m
Total torque = 2.09 N ⋅ m + 0.798 N ⋅ m
Total torque = 2.89 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1292
PROBLEM 8.72
In the machinist’s vise shown, the movable jaw D is rigidly attached
to the tongue AB that fits loosely into the fixed body of the vise.
The screw is single-threaded into the fixed base and has a mean
diameter of 0.75 in. and a pitch of 0.25 in. The coefficient of static
friction is 0.25 between the threads and also between the tongue and
the body. Neglecting bearing friction between the screw and the
movable head, determine the couple that must be applied to the
handle in order to produce a clamping force of 1 kip.
SOLUTION
Free body: Jaw D and tongue AB
P is due to elastic forces in clamped object.
W is force exerted by screw.
ΣFy = 0: N H − N J = 0 N J = N H = N
For final tightening,
FH = FJ = µs N = 0.25 N
ΣFx = 0: W − P − 2(0.25 N) = 0
N = 2(W − P)
(1)
ΣM H = 0: P(3.75) − W (2) − N (3) + (0.25 N)(1.25) = 0
3.75P − 2W − 2.6875 N = 0
(2)
3.75 P − 2W − 2.6875[2(W − P )] = 0
Substitute Eq. (1) into Eq. (2):
7.375W = 9.125P = 9.125(1 kip)
W = 1.23729 kips
Block-and-incline analysis of screw:
tan φs = µ s = 0.25
φs = 14.0362°
tan θ =
0.25 in.
π (0.75 in.)
θ = 6.0566°
θ + φs = 20.093°
Q = (1.23729 kips) tan 20.093°
= 0.45261 kip
! 0.75 in. "
T = Qr = (452.61 lb) #
$
% 2 &
T = 169.7 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1293
PROBLEM 8.73
In Problem 8.72, a clamping force of 1 kip was obtained by
tightening the vise. Determine the couple that must be applied
to the screw to loosen the vise.
PROBLEM 8.72 In the machinist’s vise shown, the movable
jaw D is rigidly attached to the tongue AB that fits loosely into
the fixed body of the vise. The screw is single-threaded into the
fixed base and has a mean diameter of 0.75 in. and a pitch of
0.25 in. The coefficient of static friction is 0.25 between the
threads and also between the tongue and the body. Neglecting
bearing friction between the screw and the movable head,
determine the couple that must be applied to the handle in order
to produce a clamping force of 1 kip.
SOLUTION
Free body: Jaw D and tongue AB
P is due to elastic forces in clamped object.
W is force exerted by screw.
ΣFy = 0: N H − N J = 0 N J = N H = N
As vise is just about to loosen,
FH = FJ = µs N = 0.25 N
ΣFx = 0: W − P + 2(0.25 N) = 0
N = 2( P − W )
(1)
ΣM H = 0: P(3.75) − W (2) − N (3) − (0.25 N)(1.25) = 0
3.75P − 2W − 3.3125 N = 0
Substitute Eq. (1) into Eq. (2):
(2)
3.75 P − 2W − 3.3125[2( P − W )] = 0
4.625W = 2.875P = 2.875(1 kip)
W = 0.62162 kip
Block-and-incline analysis of screw:
tan φs = µs = 0.25
0.25in.
π (0.75in.)
φs − θ = 7.9796°
tan θ =
φs = 14.0362°
θ = 6.0566°
Q = (621.62 lb) tan 7.9796° = 87.137 lb
! 0.75 in. "
M = Qr = (87.137 lb) #
$
% 2 &
M = 32.7 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1294
PROBLEM 8.74
In the gear-pulling assembly shown the square-threaded screw AB has a mean
radius of 15 mm and a lead of 4 mm. Knowing that the coefficient of static
friction is 0.10, determine the couple that must be applied to the screw in order
to produce a force of 3 kN on the gear. Neglect friction at end A of the screw.
SOLUTION
Block/Incline:
0.25 in.
1.875π in.
= 2.4302°
θ = tan −1
φs = tan −1 µs
= tan −1 (0.10)
= 5.7106°
Q = (3000 N) tan (8.1408°)
= 429.14 N
Couple = rQ
= (0.015 m)(429.14 N)
= 6.4371 N ⋅ m
M = 6.44 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1295
PROBLEM 8.75
A 6-in.-radius pulley of weight 5 lb is attached to a 1.5-in.-radius shaft that fits
loosely in a fixed bearing. It is observed that the pulley will just start rotating if a
0.5-lb weight is added to block A. Determine the coefficient of static friction
between the shaft and the bearing.
SOLUTION
ΣM D = 0: (10.5 lb)(6 in. − rf ) − (10 lb)(6 in. + rf ) = 0
(0.5 lb)(6 in.) = (20.5 lb)r f
rf = 0.146341 in.
r f = r sin φs
0.146341 in.
1.5 in.
= 0.097561
sin φs =
φs = 5.5987°
µs = tan φs
µs = 0.0980
= tan 5.5987°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1296
PROBLEM 8.76
The double pulley shown is attached to a 10-mm-radius shaft that
fits loosely in a fixed bearing. Knowing that the coefficient of
static friction between the shaft and the poorly lubricated bearing is
0.40, determine the magnitude of the force P required to start
raising the load.
SOLUTION
ΣM D = 0: P(45 − r f ) − W (90 + rf ) = 0
P =W
90 + rf
45 − rf
= (196.2 N)
P = 449.82 N
90 mm + 4 mm
45 mm − 4 mm
P = 450 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1297
!
PROBLEM 8.77
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the
coefficient of static friction between the shaft and the poorly
lubricated bearing is 0.40, determine the magnitude of the
force P required to start raising the load.
SOLUTION
Find P required to start raising load
ΣM D = 0: P(45 − r f ) − W (90 − rf ) = 0
P =W
90 − r f
45 − rf
= (196.2 N)
P = 411.54 N
90 mm − 4 mm
45 mm − 4 mm
P = 412 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1298
!
PROBLEM 8.78
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the coefficient
of static friction between the shaft and the poorly lubricated
bearing is 0.40, determine the magnitude of the smallest force P
required to maintain equilibrium.
SOLUTION
Find smallest P to maintain equilibrium
ΣM D = 0: P(45 + rf ) − W (90 − rf ) = 0
P =W
90 − rf
45 + rf
= (196.2 N)
P = 344.35 N
90 mm − 4 mm
45 mm + 4 mm
P = 344 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1299
!
PROBLEM 8.79
The double pulley shown is attached to a 10-mm-radius shaft
that fits loosely in a fixed bearing. Knowing that the coefficient
of static friction between the shaft and the poorly lubricated
bearing is 0.40, determine the magnitude of the smallest force P
required to maintain equilibrium.
SOLUTION
Find smallest P to maintain equilibrium
ΣM D = 0: P(45 + rf ) − W (90 + r f ) = 0
P =W
90 + rf
45 + rf
= (196.2 N)
P = 376.38 N
90 mm + 4 mm
45 mm + 4 mm
P = 376 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1300
!
PROBLEM 8.80
A lever of negligible weight is loosely fitted onto a 75-mmdiameter fixed shaft. It is observed that the lever will just start
rotating if a 3-kg mass is added at C. Determine the coefficient
of static friction between the shaft and the lever.
SOLUTION
ΣM O = 0: WC (150) − WD (100) − Rr f = 0
WC = (23 kg)(9.81 m/s 2 )
But
WD = (30 kg)(9.81 m/s 2 )
R = WC + WD = (53 kg)(9.81)
Thus, after dividing by 9.81,
23(150) − 30(100) − 53 rf = 0
rf = 8.49 mm
But
µs ≈
rf
r
=
8.49 mm
37.5 mm
µs ≈ 0.226
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1301
!
PROBLEM 8.81
The block and tackle shown are used to raise a 150-lb load. Each of the
3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that
the coefficient of static friction is 0.20, determine the tension in each
portion of the rope as the load is slowly raised.
SOLUTION
For each pulley:
Axle diameter = 0.5 in.
! 0.5 in. "
r f = r sin φs ≈ µs r = 0.20 #
$ = 0.05 in.
% 2 &
Pulley BC:
ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. + rf ) = 0
1
TCD = (150 lb)(1.5 in. + 0.05 in.)
3
ΣFy = 0: TAB + 77.5 lb − 150 lb = 0
Pulley DE:
TCD = 77.5 lb
TAB = 72.5 lb
ΣM B = 0: TCD (1.5 + rf ) − TEF (1.5 − rf ) = 0
TEF = TCD
1.5 + rf
1.5 − rf
= (77.5 lb)
1.5 in. + 0.05 in.
1.5 in. − 0.05 in.
TEF = 82.8 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1302
!
PROBLEM 8.82
The block and tackle shown are used to lower a 150-lb load. Each of the
3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that
the coefficient of static friction is 0.20, determine the tension in each
portion of the rope as the load is slowly lowered.
SOLUTION
For each pulley:
Pulley BC:
! 0.5 in. "
rf = r µs = #
$ 0.2 = 0.05 in.
% 2 &
ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. − rf ) = 0
TCD =
(150 lb)(1.5 in. − 0.05 in.)
3 in.
ΣFy = 0: TAB + 72.5 lb − 150 lb = 0
Pulley DE:
TCD = 72.5 lb
TAB = 77.5 lb
TCD (1.5 in. − rf ) − TEF (1.5 in. + rf ) = 0
TEF = TCD
1.5 in. − rf
1.5 in. + r f
= (72.5 lb)
1.5 in. − 0.05 in.
1.5 in. + 0.05 in.
TEF = 67.8 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1303
!
PROBLEM 8.83
A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mmdiameter axles. Knowing that the coefficients of friction are µs = 0.020 and µk = 0.015, determine the
horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect
rolling resistance between the wheels and the track.
SOLUTION
rf = µ r ; R = 400 mm
sin θ = tan θ =
rf
=
µr
R
µr
P = W tan θ = W
R
62.5 mm
P =Wµ
400 mm
= 0.15625Wµ
For one wheel:
For eight wheels of rail road car:
(a)
To start motion:
R
1
W = (30 mg)(9.81 m/s 2 )
8
1
= (294.3 kN)
8
1
ΣP = 8(0.15625) (294.3 kN) µ
8
= (45.984µ ) kN
µs = 0.020
ΣP = (45.984)(0.020)
= 0.9197 kN
(b)
To maintain motion:
ΣP = 920 N
µk = 0.015
ΣP = (45.984)(0.015)
= 0.6897 kN
ΣP = 690 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1304
!
PROBLEM 8.84
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter
fixed shaft. Knowing that the coefficient of static friction between the
fixed shaft and the lever is 0.15, determine the force P required to start
the lever rotating counterclockwise.
SOLUTION
rf = µs r
= 0.15(1.25 in.)
= 0.1875 in.
ΣM D = 0: P(5 in. + rf ) − (50 lb) rf = 0
50(0.1875)
5.1875
= 1.807 lb
P=
P = 1.807 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1305
!
PROBLEM 8.85
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter
fixed shaft. Knowing that the coefficient of static friction between the
fixed shaft and the lever is 0.15, determine the force P required to start
the lever rotating counterclockwise.
SOLUTION
rf = µ s r
= 0.15(1.25 in.)
rf = 0.1875 in.
2 in.
5 in .
γ = 21.801°
tan γ =
In ∆EOD:
OD = (2 in.) 2 + (5 in.) 2
= 5.3852 in.
rf
OE
sin θ =
=
OD OD
0.1875 in.
=
5.3852 in.
θ = 1.99531°
Force triangle:
P = (50 lb) tan (γ + θ )
= (50 lb) tan 23.796°
= 22.049 lb
P = 22.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1306
!
PROBLEM 8.86
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter
fixed shaft. Knowing that the coefficient of static friction between the
fixed shaft and the lever is 0.15, determine the force P required to start
the lever rotating clockwise.
SOLUTION
rf = µs r
= 0.15(1.25 in.)
r f = 0.1875 in.
ΣM D = 0: P (5 in. − rf ) − (50 lb)r f = 0
50(0.1875)
5 − 0.1875
= 1.948 lb
P=
P = 1.948 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1307
!
PROBLEM 8.87
A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter
fixed shaft. Knowing that the coefficient of static friction between the
fixed shaft and the lever is 0.15, determine the force P required to start
the lever rotating clockwise.
SOLUTION
rf = µ s r
= 0.15(1.25 in.)
= 0.1875 in.
5 in.
tan β =
2 in.
β = 68.198°
In ∆EOD:
OD = (2 in.) 2 + (5 in.) 2
OD = 5.3852 in.
OE 0.1875 in.
sin θ =
=
OD 5.3852 in.
θ = 1.99531°
Force triangle:
P=
50
50 lb
=
tan ( β + θ ) tan 70.193°
P = 18.01 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1308
!
PROBLEM 8.88
The link arrangement shown is frequently used in highway bridge construction
to allow for expansion due to changes in temperature. At each of the 60-mmdiameter pins A and B the coefficient of static friction is 0.20. Knowing that the
vertical component of the force exerted by BC on the link is 200 kN, determine
(a) the horizontal force that should be exerted on beam BC to just move the link,
(b) the angle that the resulting force exerted by beam BC on the link will form
with the vertical.
SOLUTION
Bearing:
r = 30 mm
rf = µs r
= 0.20(30 mm)
= 6 mm
Resultant forces R must be tangent to friction circles at Points C and D.
(a)
Ry = Vertical component = 200 kN
Rx = Ry tan θ
= (200 kN) tan 1.375°
= 4.80 kN
Horizontal force = 4.80 kN
(b)
6 mm
250 mm
sin θ = 0.024
sin θ =
θ = 1.375°
!
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1309
PROBLEM 8.89
A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of
kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter
of the wheels. Neglect the rolling resistance between the wheels and the ground.
SOLUTION
tan θ =
2
= 0.02
100
Since a scooter rolls at constant speed, each wheel is in equilibrium. Thus, W and R must have a common line
of action tangent to the friction circle.
rf = µ k r = (0.10)(12.5 mm)
= 1.25 mm
rf
OB
1.25 mm
=
=
tan θ tan θ
0.02
= 62.5 mm
OA =
Diameter of wheel = 2(OA) = 125.0 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1310
!
PROBLEM 8.90
A 50-lb electric floor polisher is operated on a surface for which
the coefficient of kinetic friction is 0.25. Assuming that the
normal force per unit area between the disk and the floor is
uniformly distributed, determine the magnitude Q of the horizontal
forces required to prevent motion of the machine.
SOLUTION
See Figure 8.12 and Eq. (8.9).
Using:
R = 9 in.
P = 50 lb
and
µk = 0.25
2
2
µk PR = (0.25)(50 lb)(9 in.)
3
3
= 75 lb ⋅ in.
M =
ΣM y = 0 yields:
M = Q(20 in.)
75 lb ⋅ in. = Q(20 in.)
Q = 3.75 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1311
!
PROBLEM 8.91
Knowing that a couple of magnitude 30 N ⋅ m is required to start the vertical
shaft rotating, determine the coefficient of static friction between the annular
surfaces of contact.
SOLUTION
For annular contact regions, use Equation 8.8 with impending slipping:
M=
So,
30 N ⋅ m =
R3 − R13
2
µ s N 22
3
R2 − R12
2
(0.06 m)3 − (0.025 m)3
µ s (4000 N)
3
(0.06 m) 2 − (0.025 m) 2
µs = 0.1670
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1312
PROBLEM 8.92*
The frictional resistance of a thrust bearing decreases as the shaft and bearing surfaces wear out. It is generally
assumed that the wear is directly proportional to the distance traveled by any given point of the shaft and thus
to the distance r from the point to the axis of the shaft. Assuming, then, that the normal force per unit area is
inversely proportional to r, show that the magnitude M of the couple required to overcome the frictional
resistance of a worn-out end bearing (with contact over the full circular area) is equal to 75 percent of the
value given by Eq. (8.9) for a new bearing.
SOLUTION
Using Figure 8.12, we assume
∆N =
k
∆A: ∆A = r ∆θ ∆ r
r
∆N =
P = Σ∆N
We write
P=
∆N =
2π
0
or
R
0
k
r ∆θ ∆r = k ∆θ ∆r
r
P = dN
k ∆θ ∆r = 2π Rk ; k =
P
2π R
P∆θ ∆r
2π R
P∆θ ∆r
2π R
R µ P
2πµk P R 2 1
k
rdrdθ =
⋅
= µk PR
0 2π R
2π R
2
2
∆M = r ∆F = r µ k ∆N = r µk
M=
2π
0
From Eq. (8.9) for a new bearing,
Thus,
M New =
2
µk PR
3
M
=
M New
1
2
2
3
=
3
4
M = 0.75M New
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1313
!
PROBLEM 8.93*
Assuming that bearings wear out as indicated in Problem 8.92, show that the magnitude M of the couple
required to overcome the frictional resistance of a worn-out collar bearing is
M=
1
µk P( R1 + R2 )
2
where P = magnitude of the total axial force
R1, R2 = inner and outer radii of collar
SOLUTION
Let normal force on ∆A be ∆N , and
∆N k
= .
∆A r
As in the text,
∆F = µ∆N
∆M = r ∆F
The total normal force P is
P = lim Σ∆N =
∆A→ 0
P = 2π
Total couple:
R2
R1
!
#
%
k
"
rdr $ dθ
r
&
R2
R1
kdr = 2π k ( R2 − R1 ) or k =
M worn = lim Σ∆M =
∆A→ 0
M worn = 2πµ k
2π
0
R2
R1
2π
0
!
#
%
R2
R1
(rdr ) = πµ k
(
rµ
R22
P
2π ( R2 − R1 )
k
"
rdr $ dθ
r
&
−
R12
)=
(
πµ P R22 − R12
)
2π ( R2 − R1 )
M worn =
1
µ P( R2 + R1 )
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1314
!
PROBLEM 8.94*
Assuming that the pressure between the surfaces of contact is uniform, show that
the magnitude M of the couple required to overcome frictional resistance for the
conical bearing shown is
2 µ k P R23 − R13
3 sin θ R22 − R12
M=
SOLUTION
Let normal force on ∆A be ∆N and
So
∆N = k ∆A
∆N
= k.
∆A
∆A = r ∆s ∆φ
∆s =
∆r
sin θ
where φ is the azimuthal angle around the symmetry axis of rotation.
∆Fy = ∆N sin θ = kr ∆r ∆φ
Total vertical force:
P = lim Σ∆Fy
∆A→0
P=
2π
0
!
#
%
R2
R1
"
krdr $ dφ = 2π k
&
(
P = π k R22 − R12
Friction force:
Moment:
Total couple:
)
or k =
R2
R1
rdr
P
π
(
R22
− R12
)
∆F = µ∆N = µ k ∆A
∆M = r ∆F = r µ kr
M = lim Σ∆M =
∆A→0
M = 2π
µk
sin θ
R2
R1
∆r
∆φ
sin θ
2π !
R2
0
R1
#
%
r 2 dr =
µk 2 "
r dr $ dφ
sin θ
&
2 πµ
P
3 sin θ π R22 − R32
(
)
(R
3
2
− R33
)
M=
2 µ P R23 − R13
3 sin θ R22 − R12
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1315
!
PROBLEM 8.95
Solve Problem 8.90 assuming that the normal force per unit
area between the disk and the floor varies linearly from a
maximum at the center to zero at the circumference of
the disk.
PROBLEM 8.90 A 50-lb electric floor polisher is operated
on a surface for which the coefficient of kinetic friction is
0.25. Assuming that the normal force per unit area between
the disk and the floor is uniformly distributed, determine
the magnitude Q of the horizontal forces required to prevent
motion of the machine.
SOLUTION
r"
∆N
!
= k #1 − $ .
R
∆A
%
&
Let normal force on ∆A be ∆N and
r"
r"
!
!
∆F = µ∆N = µ k #1 − $ ∆A = µ k #1 − $ r ∆r ∆θ
R
R
%
&
%
&
P = lim Σ∆N =
∆A→0
P = 2π k
0
'
)
+
R
0
(
r"
!
k #1 − $ rdr * dθ
R&
%
,
! R 2 R3 "
r"
rdr
k
π
1
2
−
=
#
$
## 2 − 3R $$
R&
%
%
&
R!
0
1
P = π kR 2
3
or k =
3P
π R2
(
r"
!
r µ k #1 − $ rdr * dθ
R
%
&
,
3
R!
r "
= 2πµ k ## r 2 − $$ dr
0
R&
%
M = lim Σr ∆F =
∆A→0
2π
2π
0
'
)
+
R
0
! R3 R 4 " 1
3
= 2πµ k ##
−
$$ = πµ kR
R
3
4
6
%
&
πµ 3P 3 1
R = µ PR
=
6 π R2
2
where
µ = µk = 0.25 R = 9 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1316
PROBLEM 8.95 (Continued)
P = W = 50 lb
1
(0.25)(50 lb)(9 in.)
2
= 56.250 lb ⋅ in.
Then
M=
Finally
Q=
M 56.250 lb ⋅ in.
=
d
20 in.
Q = 2.81 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1317
PROBLEM 8.96
A 900-kg machine base is rolled along a concrete floor using a series
of steel pipes with outside diameters of 100 mm. Knowing that the
coefficient of rolling resistance is 0.5 mm between the pipes and
the base and 1.25 mm between the pipes and the concrete floor,
determine the magnitude of the force P required to slowly move the
base along the floor.
SOLUTION
FBD pipe:
W = mg
= (900 kg)(9.81 m/s 2 )
= 8829.0 N
.5 mm + 1.25 mm
θ = sin −1
100 mm
= 1.00273°
P = W tan θ for each pipe, so also for total
P = (8829.0 N) tan (1.00273°)
P = 154.4 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1318
PROBLEM 8.97
Knowing that a 6-in.-diameter disk rolls at a constant velocity down a 2 percent incline, determine the
coefficient of rolling resistance between the disk and the incline.
SOLUTION
FBD disk:
tan θ = slope = 0.02
b = r tan θ = (3 in.)(0.02)
b = 0.0600 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1319
PROBLEM 8.98
Determine the horizontal force required to move a 2500-lb automobile with 23-in.-diameter tires along a
horizontal road at a constant speed. Neglect all forms of friction except rolling resistance, and assume the
coefficient of rolling resistance to be 0.05 in.
SOLUTION
FBD wheel:
r = 11.5 in.
b = 0.05 in.
b
θ = sin −1
r
b"
!
P = W tan θ = W tan # sin −1 $ for each wheel, so for total
r&
%
0.05 "
!
P = 2500 lb tan # sin −1
11.5 $&
%
P = 10.87 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1320
!
PROBLEM 8.99
Solve Problem 8.83 including the effect of a coefficient of rolling resistance of 0.5 mm.
PROBLEM 8.83 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels
with 125-mm-diameter axles. Knowing that the coefficients of friction are µs = 0.020 and µk = 0.015, determine
the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed.
Neglect rolling resistance between the wheels and the track.
SOLUTION
For one wheel:
rf = µ r
tan θ ≈ sin θ ≈
tan θ =
rf + b
a
µr + b
a
W
W µr + b
Q = tan θ =
a
8
8
For eight wheels of car:
P =W
µr + b
a
W = mg = (30 Mg)(9.81 m/s 2 ) = 294.3 kN
a = 400 mm, r = 62.5 mm, b = 0.5 mm
(a)
To start motion:
µ = µs = 0.02
P = (294.3 kN)
(b)
To maintain constant speed
(0.020)(62.5 mm) + 0.5 mm
400 mm
P = 1.288 kN
!
P = 1.058 kN
!
µ = µk = 0.015
P = (294.3 kN)
(0.015)(62.5 mm) + 0.5 mm
400 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1321
PROBLEM 8.100
Solve Problem 8.89 including the effect of a coefficient of rolling resistance of 1.75 mm.
PROBLEM 8.89 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming
that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine
the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.
SOLUTION
Since the scooter rolls at a constant speed, each wheel is in equilibrium. Thus, W and R must have a common
line of action tangent to the friction circle.
a = Radius of wheel
tan θ =
2
= 0.02
100
Since b and r f are small compared to a,
tan θ ≈
Data:
rf + b
a
=
µk r + b
a
= 0.02
µk = 0.10, b = 1.75 mm, r = 12.5 mm
(0.10)(12.5 mm) + 1.75 mm
= 0.02
a
a = 150 mm
Diameter = 2a = 300 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1322
PROBLEM 8.101
A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser,
a dockworker can resist a force of 5000 lb on the other end of the hawser. Determine (a) the coefficient of
static friction between the hawser and the bollard, (b) the number of times the hawser should be wrapped
around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force.
SOLUTION
β = 2 turns = 2(2π ) = 4π
(a)
T1 = 80 lb,
ln
T2 = 5000 lb
T2
= µs β
T1
µs =
µs =
1
β
ln
T2
1
5000 lb
ln
=
T1 4π
80 lb
1
4.1351
ln 62.5 =
4π
4π
µs = 0.329
T1 = 80 lb, T2 = 20,000 lb, µs = 0.329
(b)
ln
T2
= µs β
T1
β=
β=
1
µ
ln
T2
1
20, 000 lb
=
ln
T1 0.329
80 lb
1
5.5215
= 16.783
ln(250) =
0.329
0.329
Number of turns =
16.783
2π
Number of turns = 2.67
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1323
PROBLEM 8.102
A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of
static friction is 0.25, determine (a) the smallest value of the mass m for which
equilibrium is possible, (b) the corresponding tension in portion BC of the rope.
SOLUTION
We apply Eq. (8.14) to pipe B and pipe C.
T2
= eµs β
T1
Pipe B:
(8.14)
T2 = WA , T1 = TBC
µs = 0.25, β =
2π
3
WA
= e0.25(2π /3) = eπ /6
TBC
Pipe C:
(1)
T2 = TBC , T1 = WD , µ s = 0.25, β =
π
3
TBC
= e0.025(π /3) = eπ /12
WD
(a)
(2)
Multiplying Eq. (1) by Eq. (2):
WA
= eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ / 4 = 2.193
WD
WA
WD =
WA
W
mA
50 kg
g
m= D =
=
=
g
2.193
2.193 2.193 2.193
m = 22.8 kg
(b)
TBC =
From Eq. (1):
WA
eπ /6
=
(50 kg)(9.81 m/s 2 )
= 291 N
1.688
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1324
PROBLEM 8.103
A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of
static friction is 0.25, determine (a) the largest value of the mass m for which
equilibrium is possible, (b) the corresponding tension in portion BC of the rope.
SOLUTION
See FB diagrams of Problem 8.102. We apply Eq. (8.14) to pipes B and C.
Pipe B:
T1 = WA , T2 = TBC , µ s = 0.25, β =
2π
3
T
T2
= e µs β : BC = e0.25(2π /3) = eπ /6
T1
WA
Pipe C:
T1 = TBC , T2 = WD , µ s = 0.25, β =
(1)
π
3
T2
WD
= e µs β :
= e0.25(π /3) = eπ /12
T1
TBC
(a)
(2)
Multiply Eq. (1) by Eq. (2):
WD
= eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ /4 = 2.193
WA
W0 = 2.193WA
(b)
From Eq. (1):
m = 2.193mA = 2.193(50 kg)
m = 109.7 kg
TBC = WAeπ /6 = (50 kg)(9.81 m/s 2 )(1.688) = 828 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1325
PROBLEM 8.104
A 300-lb block is supported by a rope that is wrapped 1 12 times around a
horizontal rod. Knowing that the coefficient of static friction between the rope
and the rod is 0.15, determine the range of values of P for which equilibrium
is maintained.
SOLUTION
β = 1.5 turns = 3π rad
For impending motion of W up,
P = We µs β = (300 lb)e(0.15)3π
= 1233.36 lb
For impending motion of W down,
P = We− µs β = (300 lb)e− (0.15)3π
= 72.971 lb
73.0 lb # P # 1233 lb
For equilibrium,
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1326
PROBLEM 8.105
The coefficient of static friction between block B and the horizontal
surface and between the rope and support C is 0.40. Knowing that
mA = 12 kg, determine the smallest mass of block B for which equilibrium
is maintained.
SOLUTION
Support at C:
FBD block B:
WA = mg = (12 kg) g
ΣFy = 0: N B − WB = 0 or
N B = WB
FB = µs N B = 0.4 N B = 0.4WB
Impending motion:
ΣFx = 0: FB − TB = 0 or TB = FB = 0.4WB
At support, for impending motion of WA down:
WA = TB e µs β
so
TB = WAe − µs β = (12 kg) g − (0.4)π /2 = (6.4019 kg) g
Now
WB =
TB ! 6.4019 kg "
=
$ g = 16.0048 g
0.4 #%
0.4
&
so that
mB =
WB 16.0048 g
=
g
g
mB = 16.00 kg
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1327
PROBLEM 8.106
The coefficient of static friction µs is the same between block B
and the horizontal surface and between the rope and support C.
Knowing that mA = mB, determine the smallest value of µs for
which equilibrium is maintained.
SOLUTION
Support at C:
FBD B:
ΣFy = 0: N B − W = 0 or
FB = µs N B = µsW
Impending motion:
ΣFx = 0: FB − TB = 0 or
Impending motion of rope on support:
or
or
NB = W
TB = FB = µ sW
W = TB e µs β = µsWe µs β
1 = µ s e µs β
µs eπ /2 µs = 1
µs = 0.475
Solving numerically:
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1328
PROBLEM 8.107
A flat belt is used to transmit a couple from drum B to drum A. Knowing
that the coefficient of static friction is 0.40 and that the allowable belt
tension is 450 N, determine the largest couple that can be exerted on drum A.
SOLUTION
FBD’s drums:
7π
6
π 5π
β B = 180° − 30° = π − =
6
6
β A = 180° + 30° = π +
π
6
=
Since β B , β A , slipping will impend first on B (friction coefficients being equal)
So
T2 = Tmax = T1e µs β B
450 N = T1e(0.4)5π /6
or T1 = 157.914 N
ΣM A = 0: M A + (0.12 m)(T1 − T2 ) = 0
M A = (0.12 m)(450 N − 157.914 N) = 35.05 N ⋅ m
M A = 35.1 N ⋅ m
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1329
!
PROBLEM 8.108
A flat belt is used to transmit a couple from pulley A to
pulley B. The radius of each pulley is 60 mm, and a force
of magnitude P = 900 N is applied as shown to the axle
of pulley A. Knowing that the coefficient of static friction
is 0.35, determine (a) the largest couple that can be
transmitted, (b) the corresponding maximum value of the
tension in the belt.
SOLUTION
Drum A:
T2
= e µsπ = e(0.35)π
T1
T2 = 3.0028 T1
β = 180° = π radians
(a)
Torque:
ΣM A = 0: M − (675.15 N)(0.06 m) + (224.84 N)(0.06 m)
M = 27.0 N ⋅ m
(b)
ΣFx = 0: T1 + T2 − 900 N = 0
T1 + 3.0028 T1 − 900 N = 0
4.00282 T1 = 900
T1 = 224.841 N
T2 = 3.0028(224.841 N) = 675.15 N
Tmax = 675 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1330
!
PROBLEM 8.109
Solve Problem 8.108 assuming that the belt is looped around
the pulleys in a figure eight.
PROBLEM 8.108 A flat belt is used to transmit a couple from
pulley A to pulley B. The radius of each pulley is 60 mm, and a
force of magnitude P = 900 N is applied as shown to the axle
of pulley A. Knowing that the coefficient of static friction is
0.35, determine (a) the largest couple that can be transmitted,
(b) the corresponding maximum value of the tension in the belt.
SOLUTION
Drum A:
β = 240° = 240°
60 1
=
120 2
θ = 30°
π
sin θ =
4
= π
180° 3
T2
= e µs β = e0.35(4/3π )
T1
T2 = 4.3322 T1
(a)
Torque:
ΣM B = 0: M − (844.3 N)(0.06 m) + (194.9 N)(0.06 m) = 0
M = 39.0 N ⋅ m
(b)
ΣFx = 0: (T1 + T2 ) cos30° − 900 N
(T1 + 4.3322 T1 ) cos 30° = 900
T1 = 194.90 N
T2 = 4.3322(194.90 N) = 844.3 N
Tmax = 844 N
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1331
PROBLEM 8.110
In the pivoted motor mount shown the weight W of the
175-lb motor is used to maintain tension in the drive belt.
Knowing that the coefficient of static friction between the
flat belt and drums A and B is 0.40, and neglecting the
weight of platform CD, determine the largest couple that
can be transmitted to drum B when the drive drum A is
rotating clockwise.
SOLUTION
FBD motor and mount:
Impending belt slip: cw rotation
T2 = T1e µs β = T1e0.40π = 3.5136T1
ΣM D = 0: (12 in.)(175 lb) − (7 in.)T2 − (13 in.)T1 = 0
2100 lb = [(7 in.)(3.5136) + 13 in.]T1
T1 = 55.858 lb, T2 = 3.5136 T1 = 196.263 lb
FBD drum at B:
ΣM B = 0: M B − (3 in.)(196.263 lb − 55.858 lb) = 0
r = 3 in.
M B = 421 lb ⋅ in.
(Compare to 857 lb ⋅ in. using V-belt, Problem 8.130)
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1332
PROBLEM 8.111
Solve Problem 8.110 assuming that the drive drum A is
rotating counterclockwise.
PROBLEM 8.110 In the pivoted motor mount shown the
weight W of the 175-lb motor is used to maintain tension in
the drive belt. Knowing that the coefficient of static friction
between the flat belt and drums A and B is 0.40, and neglecting
the weight of platform CD, determine the largest couple that
can be transmitted to drum B when the drive drum A is rotating
clockwise.
SOLUTION
FBD motor and mount:
Impending belt slip: ccw rotation
T1 = T2 e µs β = T2 e0.40π = 3.5136T2
ΣM D = 0: (12 in.)(175 lb) − (13 in.)T1 − (7 in.)T2 = 0
2100 lb = [(13 in.)(3.5136) + 7 in.]T2 = 0
T2 = 39.866 lb, T1 = 3.5136 T2 = 140.072 lb
FBD drum at B:
ΣM B = 0: (3 in.)(140.072 lb − 39.866 lb) − M B = 0
M B = 301 lb ⋅ in.
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1333
PROBLEM 8.112
A band brake is used to control the speed of a flywheel as shown. The
coefficients of friction are µs = 0.30 and µk = 0.25. Determine the magnitude
of the couple being applied to the flywheel, knowing that P = 45 N and that
the flywheel is rotating counterclockwise at a constant speed.
SOLUTION
Free body: Cylinder
Since slipping of band relative to cylinder is clockwise, T1 and T2 are located as shown.
From free body: Lever ABC
ΣM C = 0: (45 N)(0.48 m) − T2 (0.12 m) = 0
T2 = 180 N
Free body: Lever ABC
From free body: Cylinder
Using Eq. (8.14) with µk = 0.25 and β = 270° =
3π
rad:
2
T2
= e µs β = e(0.25)(3π / 2) = e3π /8
T1
T1 =
T2
e
3π /8
=
180 N
= 55.415 N
3.2482
ΣM D = 0: (55.415 N)(0.36 m) − (180 N)(0.36 m) + M = 0
M = 44.9 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1334
PROBLEM 8.113
The speed of the brake drum shown is controlled by a belt attached to the
control bar AD. A force P of magnitude 25 lb is applied to the control bar
at A. Determine the magnitude of the couple being applied to the drum,
knowing that the coefficient of kinetic friction between the belt and the
drum is 0.25, that a = 4 in., and that the drum is rotating at a constant
speed (a) counterclockwise, (b) clockwise.
SOLUTION
(a)
Counterclockwise rotation
Free body: Drum
r = 8 in. β = 180° = π radians
T2
= e µk β = e0.25π = 2.1933
T1
T2 = 2.1933T1
Free body: Control bar
ΣM C = 0: T1 (12 in.) − T2 (4 in.) − (25 lb)(28 in.) = 0
T1 (12) − 2.1933T1 (4) − 700 = 0
T1 = 216.93 lb
T2 = 2.1933(216.93 lb) = 475.80 lb
Return to free body of drum
ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0
M + (216.96 lb)(8 in.) − (475.80 lb)(8 in.) = 0
M = 2070.9 lb ⋅ in.
(b)
M = 2070 lb ⋅ in.
Clockwise rotation
r = 8 in. β = π rad
T2
= e µk β = e0.25π = 2.1933
T1
T2 = 2.1933T1
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1335
PROBLEM 8.113 (Continued)
Free body: Control rod
ΣM C = 0: T2 (12 in.) − T1 (4 in.) − (25 lb)(28 in.) = 0
2.1933T1 (12) − T1 (4) − 700 = 0
T1 = 31.363 lb
T2 = 2.1933(31.363 lb)
T2 = 68.788 lb
Return to free body of drum
ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0
M + (31.363 lb)(8 in.) − (68.788 lb)(8 in.) = 0
M = 299.4 lb ⋅ in.
M = 299 lb ⋅ in.
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1336
PROBLEM 8.114
Knowing that a = 4 in., determine the maximum value of the coefficient
of static friction for which the brake is not self-locking when the drum
rotates counterclockwise.
SOLUTION
r = 8 in., β = 180° = π radians
T2
= e µ s β = e µsπ
T1
T2 = e µsπ T1
Free body: Control rod
ΣM C = 0: P(28 in.) − T1 (12 in.) + T2 (4 in.) = 0
28 P − 12T1 + e µπ T1 (4) = 0
P=0
For self-locking brake:
12 T1 = 4 T1e µsπ
e µs π = 3
µ sπ = ln 3 = 1.0986
µs =
1.0986
π
= 0.3497
µs = 0.350
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1337
PROBLEM 8.115
Knowing that the coefficient of static friction is 0.30 and that the brake
drum is rotating counterclockwise, determine the minimum value of a for
which the brake is not self-locking.
SOLUTION
r = 8 in., β = π radians
T2
= e µs β = e0.30π = 2.5663
T1
T2 = 2.5663T1
Free body: Control rod
b = 16 in. − a
ΣM C = 0: P(16 in. + b) − T1b + T2 a = 0
For brake to be self-locking,
P=0
T2 a = T1b ; 2.5663T1a = T1 (16 − a)
2.5663a = 16 − a
3.5663a = 16
a = 4.49 in.
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1338
PROBLEM 8.116
Bucket A and block C are connected by a cable that passes over
drum B. Knowing that drum B rotates slowly counterclockwise and
that the coefficients of friction at all surfaces are µs = 0.35 and
µk = 0.25, determine the smallest combined mass m of the bucket
and its contents for which block C will (a) remain at rest, (b) start
moving up the incline, (c) continue moving up the incline at a
constant speed.
SOLUTION
Free body: Drum
2
µ π
T2
=e 3
mg
T2 = mge 2 µπ /3
(a)
(1)
Smallest m for block C to remain at rest
Cable slips on drum.
Eq. (1) with µk = 0.25; T2 = mge2(0.25)π /3 = 1.6881mg
Block C: At rest, motion impending
ΣF = 0: N − mC g cos 30°
N = mC g cos 30°
F = µ s N = 0.35 mC g cos 30°
mC = 100 kg
ΣF = 0: T2 + F − mC g sin 30° = C
1.6881 mg + 0.35mC g cos 30° − mC g sin 30° = 0
1.6881m = 0.19689mC
m = 0.11663mC = 0.11663(100 kg);
(b)
m = 11.66 kg
Smallest m to start block moving up
No slipping at both drum and block:
µs = 0.35
Eq. (1):
T2 = mge 2(0.35)π /3 = 2.0814 mg
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1339
PROBLEM 8.116 (Continued)
Block C:
Motion impending
mC = 100 kg
ΣF = 0: N − mg cos 30°
N = mC g cos30°
F = µs N = 0.35mC g cos 30°
ΣF = 0: T2 − F − mC g sin 30° = 0
2.0814mg − 0.35mC g cos 30° − mC g sin 30° = 0
2.0814m = 0.80311mC
m = 0.38585mC = 0.38585(100 kg)
m = 38.6 kg
(c)
Smallest m to keep block moving up drum: No slipping: µs = 0.35
Eq. (1) with µs = 0.35
T2 = mg 2 µsπ /3 = mge2(0.35)π /3
T2 = 2.0814mg
Block C: Moving up plane, thus µk = 0.25
Motion up
ΣF = 0: N − mC g cos 30° = 0
N = mC g cos 30°
F = µk N = 0.25 mC g cos 30°
ΣF = 0: T2 − F − mC g sin 30° = 0
2.0814mg − 0.25mC g cos 30° − mC g sin 30° = 0
2.0814m = 0.71651mC
m = 0.34424mC = 0.34424 (100 kg)
m = 34.4 kg
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1340
PROBLEM 8.117
Solve Problem 8.116 assuming that drum B is frozen and cannot rotate.
PROBLEM 8.116 Bucket A and block C are connected by a cable that passes
over drum B. Knowing that drum B rotates slowly counterclockwise and that
the coefficients of friction at all surfaces are µs = 0.35 and µk = 0.25,
determine the smallest combined mass m of the bucket and its contents for
which block C will (a) remain at rest, (b) start moving up the incline,
(c) continue moving up the incline at a constant speed.
SOLUTION
(a)
Block C remains at rest: Motion impends
T2
= e µk β = e0.35(2π /3)
mg
T2 = 2.0814mg
Drum:
Block C:
Motion impends
ΣF = 0: N − mC g cos 30° = 0
N = mC g cos30°
F = µs N = 0.35mC g cos 30°
ΣF = 0: T2 + F − mC g sin 30° = 0
2.0814mg + 0.35mC g cos 30° − mC g sin 30° = 0
2.0814mg = 0.19689mC
m = 0.09459mC = 0.09459(100 kg)
(b)
Block C: Starts moving up
m = 9.46 kg
µs = 0.35
Drum: Impending motion of cable
T2
= e µs β
T1
mg
= e0.35(2/3π )
T1
mg
2.0814
= 0.48045mg
T1 =
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1341
PROBLEM 8.117 (Continued)
Block C: Motion impends
ΣF = 0: N − mC g cos 30°
N = mC g cos30°
F = µs N = 0.35mC g cos 30°
ΣF = 0: T1 − F − mC g sin 30° = 0
0.48045mg − 0.35mC g cos 30° − 0.5mC g = 0
0.48045m = 0.80311mC
m = 1.67158mC = 1.67158(100 kg)
(c)
m = 167.2 kg
Smallest m to keep block moving
Drum: Motion of cable
µk = 0.25
T2
= e µk β = e0.25(2/3π )
T1
mg
= 1.6881
T1
T1 =
mg
= 0.59238mg
1.6881
Block C: Block moves
ΣF = 0: N − mC g cos 30° = 0
N = mC g cos30°
F = µk N = 0.25mC g cos 30°
ΣF = 0: T1 − F − mC g sin 30° = 0
0.59238mg − 0.25mC g cos 30° − 0.5mC g = 0
0.59238m = 0.71651mC
m = 1.20954mC = 1.20954(100 kg)
m = 121.0 kg
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1342
!
PROBLEM 8.118
A cable is placed around three parallel pipes. Knowing that the
coefficients of friction are µs = 0.25 and µk = 0.20, determine
(a) the smallest weight W for which equilibrium is maintained,
(b) the largest weight W that can be raised if pipe B is slowly
rotated counterclockwise while pipes A and C remain fixed.
SOLUTION
(a)
Smallest W for equilibrium
B = π , µ = µs
TAC
= e0.25π
50 lb
TBC
= e0.25π
TAC
W
= e0.25π
TBC
TAC TBC W
⋅
⋅
= eπ /4 ⋅ eπ / 4 ⋅ eπ / 4 = e3π / 4 = 10.551
50 lb TAC TBC
W
= 10.551;
50 lb
(b)
W = 4.739 lb
W = 4.74 lb
Largest W which can be raised by pipe B rotated
β = π , µ = µk
β = π , µ = µk
β = π , µ = µs
50 lb
= e0.2π
TAC
TAC
= e0.2π
TBC
W
= e0.25π
TBC
50 lb TAC TBC
⋅
⋅
= eπ /5 ⋅ eπ /5 ⋅ e−π /4 = eπ (1/5+1/5−1/4)
TAC TBC W
= e3π / 20 = 1.602
50 lb
= 1.602;
W
W=
50 lb
= 31.21 lb
1.602
W = 31.2 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1343
!
PROBLEM 8.119
A cable is placed around three parallel pipes. Two of the pipes are
fixed and do not rotate; the third pipe is slowly rotated. Knowing
that the coefficients of friction are µs = 0.25 and µk = 0.20,
determine the largest weight W that can be raised (a) if only pipe A
is rotated counterclockwise, (b) if only pipe C is rotated clockwise.
SOLUTION
(a)
Pipe a rotates
β = π , µ = µs
β = π , µ = µk
TAC
= e0.25π
50 lb
TAC
= e0.2π
TBC
β = π , µ = µk
TBC
= e0.2π
W
TAC TBC W
⋅
⋅
= eπ /4 ⋅ e−π /5 ⋅ e−π /5
50 lb TAC TBC
= eπ (1/ 4−1/5−1/5) = e−3π / 20 = 0.62423
W
= 0.62423; W = 31.21 lb
50 lb
(b)
Pipe C rotates
β = π , µ = µk
50 lb
= e0.2π
TAC
β = π , µ = µs
TBC
= e0.25π
TAC
W = 31.2 lb
β = π , µ = µk
TBC
= e0.2π
W
50 lb TAC TBC
⋅
⋅
= eπ /5 ⋅ e −π / 4 ⋅ eπ /5 = eπ (1/5−1/4 +1/5) = e3π / 20
TAC TBC W
50 lb
= e3π /20 = 1.602
W
50 lb
W=
= 31.21 lb
1.602
W = 31.2 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1344
!
PROBLEM 8.120
A cable is placed around three parallel pipes. Knowing that the
coefficients of friction are µs = 0.25 and µk = 0.20, determine
(a) the smallest weight W for which equilibrium is maintained,
(b) the largest weight W that can be raised if pipe B is slowly
rotated counterclockwise while pipes A and C remain fixed.
SOLUTION
(a)
µ = µ s = 0.25 at all pipes.
TAC
= e0.25π
TBC
50 lb
= e0.25π /2
TAB
TBC
= e0.25π / 2
W
50 lb TAB TBC
⋅
⋅
= eπ /8 ⋅ eπ /4 ⋅ eπ /8 = eπ /8+π / 4+π /8 = eπ / 2 = 4.8105
TAB TBC W
50 lb
= 4.8015; W = 10.394 lb
W
(b)
Pipe B rotated
β=
π
2
; µ = µk
50 lb
= e0.2π /2
TAB
β = π ; µ = µs
TBC
= e0.25π
TAB
W = 10.39 lb
β=
π
2
; µ = µk
TBC
= e0.2π / 2
W
50 lb TAB TBC
⋅
⋅
= eπ /10 ⋅ e −π /4 ⋅ eπ /10
TAB TBC W
= eπ /10 −π /4 +π /10 = e −π / 20 = 0.85464
50 lb
= 0.85464
W
50 lb
W=
= 58.504 lb
0.85464
W = 58.5 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1345
!
PROBLEM 8.121
A cable is placed around three parallel pipes. Two of the pipes are
fixed and do not rotate; the third pipe is slowly rotated. Knowing
that the coefficients of friction are µs = 0.25 and µk = 0.20,
determine the largest weight W that can be raised (a) if only pipe A
is rotated counterclockwise, (b) if only pipe C is rotated clockwise.
SOLUTION
(a)
Pipe a rotates
β=
π
2
; µ = µs
β = π , µ = µk
TAB
= e0.25π /2
50 lb
β=
π
2
; µ = µk
TBC
= e0.2π /2
W
TAB
= e0.2π
TBC
TAB TBC W
⋅
⋅
= eπ /8 ⋅ e−π /5 ⋅ e−π /10
50 lb TAB TBC
= eπ (1/8−1/5−1/10) = e −7π / 40 = 0.57708
W
= 0.57708; W = 28.854 lb
50 lb
(b)
Pipe C rotates
β=
π
; µ = µk
β = π ; µ = µk
50 lb
= e0.2π / 2
TAB
TAB
= e0.2π
TBC
2
β=
π
2
W = 28.9 lb
, µ = µs
W
= e0.25π /2
TBC
50 lb TAB TBC
⋅
⋅
= eπ /10 ⋅ eπ /5 ⋅ e −π /8 = e7π /40 = 0.57708
TAB TBC W
50 lb
= 0.57708
W
W = 28.854 lb
W = 28.9 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1346
!
PROBLEM 8.122
A recording tape passes over the 20-mm-radius drive drum B and
under the idler drum C. Knowing that the coefficients of friction
between the tape and the drums are µs = 0.40 and µk = 0.30 and
that drum C is free to rotate, determine the smallest allowable value
of P if slipping of the tape on drum B is not to occur.
SOLUTION
FBD drive drum:
ΣM B = 0: r (TA − T ) − M = 0
TA − T =
Impending slipping:
M 300 N ⋅ mm
=
= 15.0000 N
r
20 mm
TA = Te µs β = Te0.4π
So
T (e0.4π − 1) = 15.0000 N
or
T = 5.9676 N
If C is free to rotate, P = T
P = 5.97 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1347
!
PROBLEM 8.123
Solve Problem 8.122 assuming that the idler drum C is frozen and
cannot rotate.
PROBLEM 8.122 A recording tape passes over the 20-mm-radius
drive drum B and under the idler drum C. Knowing that the
coefficients of friction between the tape and the drums are µs = 0.40
and µk = 0.30 and that drum C is free to rotate, determine the smallest
allowable value of P if slipping of the tape on drum B is not to occur.
SOLUTION
FBD drive drum:
ΣM B = 0: r (TA − T ) − M = 0
TA − T =
Impending slipping:
M
= 300 N ⋅ mm = 15.0000 N
r
TA = Te µs β = Te0.4π
So
(e0.4π − 1)T = 15.000 N
or
T = 5.9676 N
If C is fixed, the tape must slip
So
P = Te µk βC = (5.9676 N)e0.3π / 2 = 9.5600 N
P = 9.56 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1348
!
PROBLEM 8.124
The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius
drum. Vertical motion of end E of the bar is prevented by the two stops
shown. Knowing that µs = 0.30 between the cable and the drum, determine
(a) the largest counterclockwise couple M 0 that can be applied to the drum
if slipping is not to occur, (b) the corresponding force exerted on end E of
the bar.
SOLUTION
Drum: Slipping impends
µs = 0.30
T2
T
= e µβ : D = e0.30π = 2.5663
T1
TB
TD = 2.5663TB
(a)
Free-body: Drum and bar
ΣM C = 0: M 0 − E (8 in.) = 0
M 0 = (3.78649 lb)(8 in.)
= 30.27 lb ⋅ in.
(b)
Bar AE:
M 0 = 30.3 lb ⋅ in.
ΣFy = 0: TB + TD − E − 10 lb = 0
TB + 2.5663TB − E − 10 lb = 0
3.5663TB − E − 10 lb = 0
E = 3.5663TB − 10 lb
(1)
ΣM D = 0: E (3 in.) − (10 lb)(5 in.) + TB (10 in.) = 0
(3.5663TB − 10 lb)(3 in.) − 50 lb ⋅ in. + TB (10 in.) = 0
20.699TB = 80 TB = 3.8649 lb
Eq. (1):
E = 3.5663(3.8649 lb) − 10 lb
E = +3.78347 lb
E = 3.78 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1349
!
PROBLEM 8.125
Solve Problem 8.124 assuming that a clockwise couple M 0 is applied to the
drum.
PROBLEM 8.124 The 10-lb bar AE is suspended by a cable that passes over a
5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two
stops shown. Knowing that µs = 0.30 between the cable and the drum,
determine (a) the largest counterclockwise couple M 0 that can be applied to
the drum if slipping is not to occur, (b) the corresponding force exerted on end
E of the bar.
SOLUTION
Drum: Slipping impends
µs = 0.30
T2
= e µβ
T1
TB
= e0.30π = 2.5663
TD
TB = 2.5663TD
(a)
Free body: Drum and bar
ΣM C = 0: M 0 − E (8 in.) = 0
M 0 = (2.1538 lb)(8 in.)
M 0 = 17.23 lb ⋅ in.
(b)
Bar AE:
ΣFy = 0: TB + TD + E − 10 lb = 0
= 2.5663TD + TD + E − 10 lb
E = −3.5663TD + 10 lb
(1)
ΣM B = 0: TD (10 in.) − (10 lb)(5 in.) + E (13 in.) = 0
TD (10 in.) − 50 lb ⋅ in. + ( − 3.5663TD + 10 lb)(13 in.) = 0
−36.362TD + 80 lb ⋅ in. = 0; TD = 2.200 lb
E = −3.5633(2.200 lb) + 10 lb
Eq. (1):
E = + 2.1538 lb
E = 2.15 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1350
!
PROBLEM 8.126
The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe.
Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest
value of µs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.
SOLUTION
For wrench to be self-locking ( P = 0), the value of µs must prevent slipping of strap which is in contact with
the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase
from zero to the minimum tension required to develop “belt friction” between strap and pipe.
Free body: Wrench handle
Geometry
In ∆CDH:
On wrench handle
a
tan θ
a
CD =
sin θ
DE = BH = CH − BC
a
−r
DE =
tan θ
a
−r
AD = CD − CA =
sin θ
CH =
ΣM D = 0: TB ( DE ) − F ( AD ) = 0
a
−r
TB AD sin θ
=
=
a
F DE
−r
tan θ
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1351
PROBLEM 8.126 (Continued)
Free body: Strap at Point A
ΣF = 0: T1 − 2 F = 0
T1 = 2 F
(2)
Pipe and strap
β = (2π − θ ) radians
µs β = ln
Eq. (8.13):
µs =
1
β
T2
T1
ln
TB
2F
(3)
Return to free body of wrench handle
ΣFx = 0: N sin θ + F cos θ − TB = 0
T
N
sin θ = B − cos θ
F
F
Since F = µ s N , we have
or
1
µs
sin θ =
µs =
TB
− cos θ
F
sin θ
(4)
TB
− cos θ
F
(Note: For a given set of data, we seek the larger of the values of µs from Eqs. (3) and (4).)
For
Eq. (1):
a = 200 mm, r = 30 mm, θ = 65°
200 mm
− 30 mm
TB
= sin 65°
200 mm
F
− 30 mm
tan 65°
190.676 mm
=
= 3.0141
63.262 mm
β = 2π − θ = 2π − 65°
π
180°
= 5.1487 radians
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1352
PROBLEM 8.126 (Continued)
Eq. (3):
1
3.0141
ln
5.1487 rad
2
0.41015
=
5.1487
µs =
= 0.0797
Eq. (4):
#
sin 65°
3.0141 − cos 65°
0.90631
=
2.1595
µs =
= 0.3497
#
µs = 0.350
We choose the larger value:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1353
!
PROBLEM 8.127
Solve Problem 8.126 assuming that θ = 75°.
PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external
surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact,
determine the smallest value of µs for which the wrench will be self-locking when a = 200 mm, r = 30 mm,
and θ = 65°.
SOLUTION
For wrench to be self-locking ( P = 0), the value of µs must prevent slipping of strap which is in contact with
the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase
from zero to the minimum tension required to develop “belt friction” between strap and pipe.
Free body: Wrench handle
Geometry
In ∆CDH:
On wrench handle
a
tan θ
a
CD =
sin θ
DE = BH = CH − BC
a
DE =
−r
tan θ
a
AD = CD − CA =
−r
sin θ
CH =
ΣM D = 0: TB ( DE ) − F ( AD ) = 0
a
−r
TB AD sin θ
=
=
a
F DE
−r
tan θ
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1354
PROBLEM 8.127 (Continued)
Free body: Strap at Point A
ΣF = 0: T1 − 2 F = 0
T1 = 2 F
(2)
Pipe and strap
β = (2π − θ ) radians
µs β = ln
Eq. (8.13):
µs =
1
β
T2
T1
ln
TB
2F
(3)
Return to free body of wrench handle
ΣFx = 0: N sin θ + F cos θ − TB = 0
T
N
sin θ = B − cos θ
F
F
Since F = µ s N , we have
or
1
µs
sin θ =
µs =
TB
− cos θ
F
sin θ
TB
− cos θ
F
(4)
(Note: For a given set of data, we seek the larger of the values of µs from Eqs. (3) and (4).)
For
Eq. (1):
a = 200 mm, r = 30 mm, θ = 75°
200 mm
− 30 mm
TB
= sin 75°
200 mm
F
− 30 mm
tan 75°
177.055 mm
=
= 7.5056
23.590 mm
β = 2π − θ = 2π − 75°
π
180°
= 4.9742
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1355
PROBLEM 8.127 (Continued)
Eq. (3):
1
7.5056
ln
4.9742 rad
2
1.3225
=
4.9742
µs =
= 0.2659
Eq. (4):
#
sin 75°
7.5056 − cos 75°
0.96953
=
7.2468
µs =
= 0.1333
#
µs = 0.266
We choose the larger value:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1356
!
PROBLEM 8.128
Prove that Eqs. (8.13) and (8.14) are valid for any shape of surface provided
that the coefficient of friction is the same at all points of contact.
SOLUTION
ΣFn = 0: ∆N − [T + (T + ∆T )]sin
∆N = (2T + ∆T ) sin
or
ΣFt = 0: [(T + ∆T ) − T ]cos
∆F = ∆T cos
or
∆T cos
So
So
∆θ
− ∆F = 0
2
∆θ
2
∆θ
∆θ
∆θ
sin ∆θ
= µ s 2T sin
+ µs ∆T
2
2
2
0: dT = µsTdθ
T2
T1
and
∆θ
2
∆F = µ s ∆N
Impending slipping:
In limit as
∆θ
=0
2
ln
dT
=
T
β
0
or
dT
= µ s dθ
T
µ s dθ
T2
= µs β
T1
or T2 = T1e µs β
(Note: Nothing above depends on the shape of the surface, except it is assumed to be a smooth curve.)!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1357
PROBLEM 8.129
Complete the derivation of Eq. (8.15), which relates the tension in both parts of a V belt.
SOLUTION
Small belt section:
Side view:
End view:
ΣFy = 0: 2
∆N
α
∆θ
sin − [T + (T + ∆T )]sin
=0
2
2
2
ΣFx = 0: [(T + ∆T ) − T ]cos
∆F = µs ∆N . ∆T cos
Impending slipping:
In limit as
∆θ
0: dT =
µ sTdθ
α
sin
so
or
T2
T1
2
µs
dT
=
α
T
sin
2
ln
or
β
0
∆θ
− ∆F = 0
2
∆θ
2T + ∆T
∆θ
sin
= µs
α
2
2
sin
2
µs
dT
dθ
=
α
T
sin
2
dθ
µβ
T2
= s
T1 sin α
2
T2 = T1e
or
µ s β / sin α2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1358
!
PROBLEM 8.130
Solve Problem 8.107 assuming that the flat belt and drums are replaced by a
V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.)
PROBLEM 8.107 A flat belt is used to transmit a couple from drum B to
drum A. Knowing that the coefficient of static friction is 0.40 and that the
allowable belt tension is 450 N, determine the largest couple that can be
exerted on drum A.
SOLUTION
Since β is smaller for pulley B. The belt will slip first at B.
! π rad " 5
$ = π rad
% 180° & 6
β = 150° #
T2
µ β / sin α2
=e s
T1
450 N
(0.4) 5 π / sin18°
=e 6
= e3.389
T1
450 N
= 29.63; T1 = 15.187 N
T1
Torque on pulley A:
ΣM B = 0: M − (Tmax − T1 )(0.12 m) = 0
M − (450 N − 15.187 N)(0.12 m) = 0
M = 52.18 N ⋅ m
M = 52.2 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1359
!
PROBLEM 8.131
Solve Problem 8.108 assuming that the flat belt and pulleys are
replaced by a V belt and V pulleys with α = 36°. (The angle α is
as shown in Figure 8.15a.)
PROBLEM 8.108 A flat belt is used to transmit a couple from
pulley A to pulley B. The radius of each pulley is 60 mm, and a
force of magnitude P = 900 N is applied as shown to the axle of
pulley A. Knowing that the coefficient of static friction is 0.35,
determine (a) the largest couple that can be transmitted, (b) the
corresponding maximum value of the tension in the belt.
SOLUTION
Pulley A:
β = π rad
T2
µ β / sin α2
=e s
T1
T2
= e0.35π / sin18°
T1
T2
= e3.558 = 35.1
T1
T2 = 35.1T1
ΣFx = 0: T1 + T2 − 900 N = 0
T1 + 35.1T1 − 900 N = 0
T1 = 24.93 N
T2 = 35.1(24.93 N) = 875.03 N
ΣM A = 0: M − T2 (0.06 m) + T1 (0.06 m) = 0
M − (875.03 N)(0.06 m) + (24.93 N)(0.06 m) = 0
M = 51.0 N ⋅ m
Tmax = T2
Tmax = 875 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1360
!
PROBLEM 8.132
Knowing that the coefficient of friction between the 25-kg block and the
incline is µs = 0.25, determine (a) the smallest value of P required to start the
block moving up the incline, (b) the corresponding value of β.
SOLUTION
FBD block (Impending motion up)
W = mg
= (25 kg)(9.81 m/s 2 )
= 245.25 N
φs = tan −1 µs
= tan −1 (0.25)
= 14.04°
(a)
(Note: For minimum P, P
R so β = φs .)
Then
P = W sin (30° + φs )
= (245.25 N)sin 44.04°
Pmin = 170.5 N
(b)
We have β = φs
β = 14.04°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1361
PROBLEM 8.133
The 20-lb block A and the 30-lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of static
friction is 0.15 between all surfaces of contact, determine the value of θ
for which motion is impending.
SOLUTION
Since motion impends, F = µs N at all surfaces.
Free body: Block A
Impending motion:
ΣFy = 0: N1 = 20cos θ
ΣFx = 0: T − 20sin θ − µ s N1 = 0
T = 20sin θ + 0.15(20 cos θ )
T = 20sin θ + 3cos θ
(1)
Free body: Block B
Impending motion:
ΣFy = 0: N 2 − 30 cos θ − N1 = 0
N 2 = 30 cos θ + 20cos θ = 50 cos θ
F2 = µ s N 2 = 0.15(50 cos θ ) = 6cos θ
ΣFx = 0: T − 30sin θ + µ s N1 + µs N 2 = 0
T = 30sin θ − 0.15(20 cos θ ) − 0.15(50cos θ )
T = 30sin θ − 3cos θ − 7.5cos θ
Eq. (1) subtracted by Eq. (2):
(2)
20sin θ + 3cos θ − 30sin θ + 3cos θ + 7.5cos θ = 0
13.5cos θ = 10sin θ , tan θ =
13.5°
10
θ = 53.5°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1362
!
PROBLEM 8.134
A worker slowly moves a 50-kg crate to the left along a loading dock
by applying a force P at corner B as shown. Knowing that the crate
starts to tip about the edge E of the loading dock when a = 200 mm,
determine (a) the coefficient of kinetic friction between the crate and
the loading dock, (b) the corresponding magnitude P of the force.
SOLUTION
Free body: Crate Three-force body.
Reaction E must pass through K where P and W intersect.
Geometry:
HK = (0.6 m) tan15° = 0.16077 m
(a)
JK = 0.9 m + HK = 1.06077 m
0.4 m
= 0.37708
tan φs =
1.06077 m
µs = tan φs = 0.377
φs = 20.66°
Force triangle:
W = (50 kg)(9.81 m/s)
= 490.5 N
(b)
Law of sines:
P
490.5 N
=
sin 20.66° sin 84.34°
P = 173.91 N
P = 173.9 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1363
!
PROBLEM 8.135
A slender rod of length L is lodged between peg C and the vertical wall and
supports a load P at end A. Knowing that the coefficient of static friction between
the peg and the rod is 0.15 and neglecting friction at the roller, determine the range
of values of the ratio L/a for which equilibrium is maintained.
SOLUTION
FBD rod:
Free-body diagram: For motion of B impending upward:
! a
ΣM B = 0: PL sin θ − N C #
% sin θ
"
$=0
&
NC =
PL 2
sin θ
a
(1)
ΣFy = 0: NC sin θ − µ s N C cos θ − P = 0
NC (sin θ − µ cos θ ) = P
Substitute for NC from Eq. (1), and solve for a/L.
a
= sin 2 θ (sin θ − µ s cos θ )
L
For θ = 30° and µ s = 0.15:
(2)
a
= sin 2 30°(sin 30° − 0.15cos 30°)
L
a
L
= 0.092524
= 10.808
L
a
For motion of B impending downward, reverse sense of friction force FC . To do this we make
µs = −0.15 in. Eq. (2).
Eq. (2):
a
= sin 2 30°(sin 30° − (−0.15) cos 30°)
L
a
L
= 0.15748
= 6.350
L
a
Range of values of L/a for equilibrium:
6.35 #
L
# 10.81
a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1364
!
PROBLEM 8.136
A safety device used by workers climbing ladders fixed to high structures
consists of a rail attached to the ladder and a sleeve that can slide on the
flange of the rail. A chain connects the worker’s belt to the end of an
eccentric cam that can be rotated about an axle attached to the sleeve at C.
Determine the smallest allowable common value of the coefficient of
static friction between the flange of the rail, the pins at A and B, and the
eccentric cam if the sleeve is not to slide down when the chain is pulled
vertically downward.
SOLUTION
Free body: Cam
ΣM C = 0: N D (0.8 in.) − µ s N D (3in.) − P (6 in.) = 0
ND =
6P
0.8 − 3µ s
(1)
Free body: Sleeve and cam
ΣFx = 0: N D − N A − N B = 0
N A + NB = ND
(2)
ΣFy = 0: FA + FB + FD − P = 0
µs ( N A + N B + N D ) = P
or
µs (2 N D ) = P
Substitute from Eq. (2) into Eq. (3):
ND =
(3)
P
2µ s
(4)
Equate expressions for N D from Eq. (1) and Eq. (4):
P
6P
; 0.8 − 3µ s = 12 µ s
=
2µ s 0.8 − 3µ s
µs =
0.8
15
µs = 0.0533
(Note: To verify that contact at pins A and B takes places as assumed, we shall check that N A . 0 and N B = 0.)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1365
PROBLEM 8.136 (Continued)
From Eq. (4):
ND =
P
P
=
= 9.375P
2µ s 2(0.0533)
From free body of cam and sleeve:
ΣM B = 0: N A (8 in.) − N D (4 in.) − P(9 in.) = 0
8 N A = (9.375 P)(4) + 9 P
N A = 5.8125P . 0 OK
From Eq. (2):
N A + NB = ND
5.8125P + N B = 9.375P
N B = 3.5625P . 0 OK
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1366
PROBLEM 8.137
To be of practical use, the safety sleeve described in the preceding
problem must be free to slide along the rail when pulled upward.
Determine the largest allowable value of the coefficient of static
friction between the flange of the rail and the pins at A and B if the
sleeve is to be free to slide when pulled as shown in the figure,
assuming (a) θ = 60°, (b) θ = 50°, (c) θ = 40°.
SOLUTION
Note the cam is a two-force member.
Free body: Sleeve
We assume contact between rail and pins as shown.
ΣM C = 0: FA (3 in.) + FB (3 in.) − N A (4 in.) − N B (4 in.) = 0
But
FA = µ s N A
FB = µ s N B
We find
3µ s ( N A + N B ) − 4( N A + N B ) = 0
µs =
4
= 1.33333
3
We now verify that our assumption was correct.
ΣFx = 0: N A − N B + P cos θ = 0
N B − N A = P cos θ
(1)
ΣFy = 0: − FA − FB + P sin θ = 0
µ s N A + µ s N B = P sin θ
N A + NB =
Add Eqs. (1) and (2):
P sin θ
µs
(2)
!
sin θ "
2 N B = P # cos θ +
$ . 0 OK
µs &
%
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1367
PROBLEM 8.137 (Continued)
Subtract Eq. (1) from Eq. (2):
NA . 0 only if
! sin θ
"
2N A = P #
− cos θ $
% µs
&
sin θ
− cos θ . 0
µs
tan θ . µ s = 1.33333
θ = 53.130°
(a)
For case (a): Condition is satisfied, contact takes place as shown. Answer is correct.
µ s = 1.333
But for (b) and (c): θ , 53.130° and our assumption is wrong, N A is directed to left.
ΣFx = 0: − NA − N B + P cos θ = 0
N A + N B = P cos θ
(3)
ΣFy = 0: − FA − FA + P sin θ = 0
µ s ( N A + N B ) = P sin θ
(4)
Divide Eq. (4) by Eq. (3):
µ s = tan θ
(b)
(c)
(5)
We make θ = 50° in Eq. (5):
µ s = tan 50°
µ s = 1.192
µ s = tan 40°
µ s = 0.839
We make θ = 40° in Eq. (5):
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1368
PROBLEM 8.138
Bar AB is attached to collars that can slide on the inclined rods shown.
A force P is applied at Point D located at a distance a from end A.
Knowing that the coefficient of static friction µs between each collar
and the rod upon which it slides is 0.30 and neglecting the weights of
the bar and of the collars, determine the smallest value of the ratio a/L
for which equilibrium is maintained.
SOLUTION
FBD bar and collars:
Impending motion:
φs = tan −1 µs
= tan −1 0.3
= 16.6992°
Neglect weights: 3-force FBD and " ACB = 90°
so
AC =
a
cos (45° + φs )
= l sin (45° − φs )
a
= sin (45° − 16.6992°) cos (45° + 16.6992°)
l
a
= 0.225
l
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1369
PROBLEM 8.139
The machine part ABC is supported by a frictionless hinge at B
and a 10° wedge at C. Knowing that the coefficient of static
friction at both surfaces of the wedge is 0.20, determine (a) the
force P required to move the wedge, (b) the components of the
corresponding reaction at B.
SOLUTION
φs = tan −1 0.20 = 11.31°
Free body: Part ABC
ΣM B = 0 (1800 N)(0.35 m) − R cos 21.31°(0.6 m) = 0
RC = 1127.1 N
Force triangle:
Free body: Wedge
(a)
Law of sines:
P
1127.1 N
=
sin(11.31° + 21.31°) sin 78.69°
P = 619.6 N
(b)
P = 620 N
Return to part ABC:
ΣFx = 0: Bx + 1800 N − RC sin 21.31° = 0
Bx + 1800 N − (1127.1 N) sin 21.31°
Bx = −1390.4 N
B x = 1390 N
ΣFy = 0: By + RC cos 21.31° = 0
By + (1127.1 N) cos 21.31° = 0
By = −1050 N
B y = 1050 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1370
!
PROBLEM 8.140
A wedge A of negligible weight is to be driven between two 100-lb
blocks B and C resting on a horizontal surface. Knowing that the
coefficient of static friction at all surfaces of contact is 0.35,
determine the smallest force P required to start moving the wedge
(a) if the blocks are equally free to move, (b) if block C is securely
bolted to the horizontal surface.
SOLUTION
Wedge angle θ :
(a)
0.75 in.
4 in.
θ = 10.62°
θ = tan −1
Free body: Block B
φs = tan −1 0.35 = 19.29°
R1
100 lb
=
sin19.29° sin 40.80°
R1 = 50.56 lb
Free body: Wedge
By symmetry:
R3 = R1
P = 2 R1 sin (θ + φs ) = 2(50.56)sin 29.91°
P = 50.42 lb
(b)
P = 50.4 lb
P = 50.4 lb
Free bodies unchanged: Same result.
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1371
!
PROBLEM 8.141
The position of the automobile jack shown is controlled by a
screw ABC that is single-threaded at each end (right-handed
thread at A, left-handed thread at C). Each thread has a pitch of
0.1 in. and a mean diameter of 0.375 in. If the coefficient of static
friction is 0.15, determine the magnitude of the couple M that
must be applied to raise the automobile.
SOLUTION
Free body: Parts A, D, C, E
Two-force members
Joint D:
FAD = FCD
Symmetry:
ΣFy = 0: 2 FCD sin 25° − 800 lb = 0
FCD = 946.5 lb
Joint C:
FCE = FCD
Symmetry:
ΣFx = 0: 2 FCD cos 25° − FAC = 0
FAC = 2(946.5 lb) cos 25°
FAC = 1715.6 lb
Block-and-incline analysis of one screw:
tan θ =
0.1 m
π (0.375 in.)
θ = 4.852°
φs = tan −1 0.15
= 8.531°
Q = (1715.6 lb) tan13.383°
Q = 408.2 lb
But, we have two screws:
! 0.375 in. "
Torque = 2Qr = 2(408.2 lb) #
$
2
%
&
Torque = 153.1 lb ⋅ in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1372
PROBLEM 8.142
A lever of negligible weight is loosely fitted onto a 30-mmradius fixed shaft as shown. Knowing that a force P of
magnitude 275 N will just start the lever rotating clockwise,
determine (a) the coefficient of static friction between the
shaft and the lever, (b) the smallest force P for which the
lever does not start rotating counterclockwise.
SOLUTION
(a)
Impending motion
W = (40 kg)(9.81 m/s 2 ) = 392.4 N
ΣM D = 0: P(160 − rf ) − W (100 + rf ) = 0
160 P − 100W
P +W
(160 mm)(275 N) − (100 mm)(392.4 N)
rf =
275 N + 392.4 N
r f = 7.132 mm
rf =
r f = r sin φs = r µ s
µs =
(b)
Impending motion
rf
r
=
7.132 mm
= 0.2377
30 mm
µs = 0.238
r f = r sin φs = r µ s
= (30 mm)(0.2377)
r f = 7.132 mm
ΣM D = 0: P(160 + rf ) − W (100 − rf ) = 0
P =W
100 − r f
160 + rf
P = (392.4 N)
100 mm − 7.132 mm
160 mm + 7.132 mm
P = 218 N
P = 218.04 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1373
PROBLEM 8.143
A couple MB is applied to the drive drum B to maintain
a constant speed in the polishing belt shown. Knowing
that µk = 0.45 between the belt and the 15-kg block
being polished and µ s = 0.30 between the belt and the
drive drum B, determine (a) the couple MB, (b) the
minimum tension in the lower portion of the belt if no
slipping is to occur between the belt and the drive drum.
SOLUTION
Block:
Portion of belt located under block:
ΣFx = 0: T2 − T1 − 66.217 N = 0
(1)
Drum B:
T2
= e µsπ = e0.3π = 2.5663
T1
T2 = 2.5663T1
Eq. (1):
(2)
2.5663T1 − T1 − 66.217 N = 0
1.5663T1 = 66.217 N
T1 = 42.276 N
Eq. (2):
Tmin = 42.3 N
T2 = 2.5663(42.276 N) = 108.493 N
ΣM B = 0: M B − (108.493 N)(0.075 m) + (42.276 N)(0.075 m) = 0
M B = 4.966 N ⋅ m
M B = 4.97 N ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1374
CHAPTER 9
PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
By observation
Now
y=
h
x
b
dI y = x 2
dA = x 2 [(h − y ) dx]
x!
= hx 2 "1 − # dx
$ b%
Then
&
I y = dI y =
&
x!
hx 2 "1 − # dx
0
$ b%
b
'1
b
x4 (
=h x − !
4b # 0
"3
3
or
Iy =
1 3
bh
12
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1377
!
PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
a=
k
a
Then
y=
a2
x
Now
dIy = x 2
At x = a,
y = a:
or k = a 2
dA = x 2 ( y dx)
$ a2 %
= x 2 && dx '' = a 2 x dx
( x
)
Then
,
Iy = dI y =
,
2a
a
2a
a 2 x dx = a 2
a2
*1 2+
x ! = [(2a) 2 − (a) 2 ]
2
"2 #a
or
Iy =
3 4
a
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1378
!
PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
y = kx 2
For x = a :
Thus:
b = ka 2
k=
b
a2
y=
b 2
x
a2
dA = (b − y )dx
b
$
%
dIy = x 2 dA = x 2 (b − y )dx = x 2 & b − 2 x 2 ' dx
a
(
)
a$
1
1
b
%
Iy = dI y = & bx 2 − 2 x 4 ' dx = a 3b − a 3b
0 (
3
5
a
)
,
,
Iy =
2 3
ab
15
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1379
!
PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded
area with respect to the y axis.
SOLUTION
$ x x2 %
y = 4h && − 2 ''
(a a )
dA = y dx
$ x x2
dI y = x 2 dA = 4hx 2 && − 2
(a a
a $ x3
x4 %
I y = 4h && − 2 '' dx
0
( a a )
%
'' dx
)
,
a
* x4
$ a3 a3 %
x5 +
− 2 ! = 4h && − ''
I y = 4h
5 )
" 4 a 5a # 0
( 4
1
I y = ha3
5
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1380
!
PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
By observation
y=
h
x
b
or
x=
b
y
h
Now
dI x = y 2 dA = y 2 ( x dy )
$b
%
= y 2 & y dy '
(h
)
b
= y 3 dy
h
Then
,
I x = dI x =
,
h
0
b 3
y dy
h
h
=
b *1 4 +
y
h " 4 !# 0
or
Ix =
1 3
bh
4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1381
!
PROBLEM 9.6
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
At x = a,
Then
Now
y = a:
a=
k
a
y=
a2
x
1
1 $ a2
dI x = y 3 dx = &&
3
3( x
=
Then
or k = a 2
3
%
'' dx
)
1 a6
dx
3 x3
,
I x = dI x =
,
a
a
2a
1 a6
1 * 1 1+
dx = a 6 −
3 x3
3 " 2 x 2 !# a
1 * 1
1 +
= − a6
−
!
2
6 " (2a)
(a) 2 #
or
1
I x = a4
8
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1382
!
PROBLEM 9.7
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
See figure of solution of Problem 9.3.
y=
b
x
a2
1
1
1
1 b3 6
dI x = b3 dx − y 3 dx = b3 dx −
x dx
3
3
3
3 a6
,
I x = dI x =
,
a$ 1
0
1 b3 6 %
3
x ' dx
&& b −
3 a 6 ')
(3
1
1 b3 a 7 $ 1 1 % 3
= b3 a −
= & − ' ab
3
3 a 6 7 ( 3 21 )
1 %
6
$ 7
= & − ' ab3 = ab3
21
( 21 21 )
Ix =
2 3
ab
7
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1383
!
PROBLEM 9.8
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
See figure of solution of Problem 9.4.
y = h1 + ( h2 − h1 )
dI x =
x
a
1 3
y dx
3
,
I x = dI y =
1
3
,
3
a*
0
x+
h1 + ( h2 − h1 ) ! dx
a#
"
4
a
x+ $ a %
1 *
h1 + ( h2 − h1 ) ! &
=
'
a # ( h2 − h1 )
12 "
0
=
2
2
a
a ( h + h1 ) (h2 + h1 )(h2 − h1 )
h24 − h14 = ⋅ 2
12(h2 − h1 )
12
h2 − h1
(
)
Ix =
a 2
h1 + h22 (h1 + h2 )
12
(
)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1384
!
PROBLEM 9.9
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
At x = 0, y = b :
b = c(1 − 0)
or
c=b
At x = a, y = 0:
0 = b(1 − ka1/2 )
or k =
Then
$
x1/2 %
y = b &&1 − 1/ 2 ''
( a )
1
a
1/2
3
Now
1
1* $
x1/ 2
dI x = y 3 dx = b &&1 − 1/ 2
3
3" ( a
or
1 $
x1/ 2
x x3/ 2
dI x = b3 &&1 − 3 1/2 + 3 − 3/2
3 (
a a
a
Then
,
I x = dI x = 2
,
%+
'' ! dx
) #!
%
'' dx
)
$
x1/2
x x3/2
b3 &&1 − 3 1/ 2 + 3 − 3/2
a a
3 (
a
a1
0
%
'' dx
)
a
2 *
x3/2 3 x 2 2 x5/2 +
= b3 x − 2 1/ 2 +
−
!
3 "
2 a 5 a3/ 2 # 0
a
or
Ix =
1 3
ab
15
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1385
!
PROBLEM 9.10
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
At x = a, y = b :
b = ke a/a
Then
y=
Now
dI x =
Then
or k =
b
e
b x/a
e = be x/a −1
e
1 3
1
y dx = (be x/a −1 )3 dx
3
3
1 3 3( x/a −1)
= be
dx
3
,
I x = dI x =
,
a
0
a
1 3 3( x/a −1)
b3 * a 3( x/a −1) +
be
dx =
e
!
3
3 "3
#0
1
= ab3 (1 − e−3 )
9
or
I x = 0.1056ab3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1386
!
PROBLEM 9.11
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
At x = 3a,
y = b:
b = k (3a − a)3
or
k=
b
8a 3
Then
y=
b
( x − a )3
8a 3
Now
dI x =
1 3
y dx
3
3
Then
=
1* b
+
( x − a)3 ! dx
3 " 8a 3
#
=
b3
( x − a )9 dx
1536a9
,
I x = dI x =
=
,
3a
a
3a
b3
b3 * 1
+
9
−
=
(
x
a
)
dx
( x − a )10 !
9
9
1536a
1536a "10
#a
b3
[(3a − a )10 − 0]
15,360a9
or
Ix =
1 3
ab
15
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1387
!
PROBLEM 9.12
Determine by direct integration the moment of inertia of the shaded
area with respect to the y axis.
SOLUTION
At x = 0,
y = b:
b = c(1 − 0)
or c = b
x = a,
y = 0:
0 = c(1 − ka1/ 2 )
or k =
1
a1/2
$
x1/2
y = b &&1 − 1/ 2
( a
Then
%
''
)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1388
!
PROBLEM 9.13
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
At x = a,
b = kea/a
y = b:
or
k=
b
e
Then
y=
b x/a
e = be x/a −1
e
dI y = x 2 dA = x 2 ( y dx)
Now
= x 2 (be x/a −1dx)
,
I y = dI y =
Then
,
a
bx 2 e x/a −1dx
0
Now use integration by parts with
dv = e x/a −1dx
u = x2
v = ae x/a −1
du = 2 x dx
Then
,
a
0
a
x 2 e x/a −1dx = *" x 2 ae x/a −1 +# −
0
= a 3 − 2a
,
a
0
,
a
0
(ae x/a −1 )2 x dx
xe x/a −1dx
Using integration by parts with
Then
u=x
dv = e x/a −1dx
du = dx
v = ae x/a −1
*
I y = b / a3 − 2a ( xae x/a −1 ) |0a −
"
1
{
= b {a
,
a
0
+.
( ae x/a −1 ) dx ! 0
#2
}
) +#}
= b a3 − 2a *" a 2 − ( a 2 e x/a −1 ) |0a +#
3
− 2a *" a 2 − ( a 2 − a 2 e−1
or
I y = 0.264a3b
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1389
!
PROBLEM 9.14
Determine by direct integration the moment of inertia of the
shaded area with respect to the y axis.
SOLUTION
At x = 3a,
y = b:
b = k (3a − a)3
or
k=
b
8a 3
Then
y=
b
( x − a )3
8a 3
Now
dI y = x 2 dA = x 2 ( y dx) = x 2
=
Then
* b
+
( x − a)3 dx !
3
" 8a
#
b 2 3
x ( x − 3x 2 a + 3xa 2 − a 3 )dx
8a 3
,
I y = dI y =
,
3a
a
b 5
( x − 3x 4 a + 3x3 a 2 − a3 x 2 )dx
3
8a
3a
b *1 6 3 5 3 2 4 1 3 3 +
= 3
x − ax + a x − a x !
5
4
3
8a " 6
#a
=
3
3
1
b -* 1
+
(3a)6 − a(3a)5 + a 2 (3a) 4 − a3 (3a )3 !
3 /
5
4
3
8a 1 " 6
#
3
3
1
*1
+.
− ( a )6 − a ( a )5 + a 2 ( a ) 4 − a 3 ( a )3 ! 0
5
4
3
"6
#2
or I y = 3.43a3b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1390
!
PROBLEM 9.15
Determine the moment of inertia and the radius of gyration of
the shaded area shown with respect to the x axis.
SOLUTION
At x = a,
y1 = y2 = b :
b
a
b
or k = 3
a
y1 : b = ma or m =
y2 : b = ka3
Then
y1 =
b
x
a
or
x1 =
a
y
b
and
y2 =
b 3
x
a3
or
x2 =
a 1/3
y
b
Now
dA = ( x2 − x1 ) dy
1/3
a %
$ a
= & 1/3 y1/3 − y ' dy
b )
(b
Then
,
A = dA = 2
,
b
0
$ y1/3 1 %
a && 1/3 − y '' dy
b )
(b
b
= 2a
Now
1
* 3 1 4/3 1 2 +
y −
y ! = ab
1/3
2b # 0 2
"4 b
*$ a
a % +
dI x = y 2 dA = y 2 & 1/3 y1/3 − y ' dy !
b ) #
"( b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1391
PROBLEM 9.15 (Continued)
Then
Ix = 2
,
b
0
1 %
$ 1
a & 1/3 y 7/3 − y 3 ' dy
b
b
(
)
b
= 2a
and
k x2 =
* 3 1 10/3 1 4 +
y
y
−
1/3
4b !# 0
"10 b
Ix
=
A
1
ab3
10
1 ab
2
1
= b2
5
or
or
Ix =
1 3
ab
10
kx =
b
5
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1392
PROBLEM 9.16
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
y1 = k1 x 2
For
y2 = k2 x1/ 2
x = 0 and
y1 = y2 = b
b = k1a 2
b = k2 a1/ 2
k1 =
b
a2
y1 =
b 2
x
a2
k2 =
b
a
1/ 2
Thus,
y2 =
b 1/ 2
x
a
1/2
dA = ( y2 − y1 )dx
A=
,
a*
0
b 1/2 b 2 +
x − 2 x ! dx
1/ 2
a
a
"
#
2 ba 3/2 ba 3
−
3 a1/2 3a 2
1
A = ab
3
A=
1 3
1
y2 dx − y13 dx
3
3
3
1 b
1 b3 6
3/ 2
x
dx
x dx
=
−
3 a3/ 2
3 a6
dI x =
,
b3 a 3/ 2
b3 a 6
−
x
dx
x dx
3a 3/2 0
3a 6 0
1 %
b3 a 5/2 b3 a 7 $ 2
= 3/ 2 5 − 6
= & − ' ab3
7
15
21
3a
3
a
(
)
(2)
,
I x = dI x =
k x2 =
Ix
=
A
(
3
35
ab3
ab
b
)
,
Ix =
3 3
ab
35
kx = b
9
35
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1393
PROBLEM 9.17
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
At x = a,
Then
Now
Then
y1 = y2 = b :
b
a3
b
y2 : b = ma or m =
a
y1 : b = ka3
or k =
b 3
x
a3
b
y2 = x
a
y1 =
b
$b
%
dA = ( y2 − y1 )dx = & x − 3 x3 ' dx
a
a
(
)
,
A = dA = 2
,
a
0
1 3%
b$
& x − 2 x ' dx
a(
a
)
a
=2
Now
Then
1
1
b *1 2
+
x − 2 x 4 ! = ab
a "2
4a
#0 2
*$ b
b
% +
dI y = x 2 dA = x 2 & x − 3 x3 ' dx !
a
) #
"( a
,
I y = dI y = 2
,
a
0
b 2$
1
%
x x − 2 x3 ' dx
a &(
a
)
a
=2
and
k y2 =
Iy
A
=
b *1 4 1 1 6+
x −
x
a "4
6 a 2 !# 0
1 3
ab
6
1 ab
2
1
= a2
3
or
or
Iy =
1 3
ab
6
ky =
a
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1394
PROBLEM 9.18
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 9.16.
1
A = ab
3
a
dI y = x 2 dA = x 2 ( y2 − y1 )dx
b
b
$ b
%
x 2 & 1/2 x1/ 2 − 2 x 2 ' dx = 1/2
a
a
(a
)
Iy =
,
Iy =
b b7/2 b a5 $ 2 1 % 3
⋅
−
⋅
= & − 'a b
a1/ 2 ( 72 ) a 2 5 ( 7 5 )
k y2
=
0
Iy
A
(
=
3
35
a 3b
)
,
a
0
x5/ 2 dx −
b
a2
,
a
0
x 4 dx
Iy =
3 3
ab
35
ky = a
ab
3
9
35
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1395
PROBLEM 9.19
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
y1 :
At x = 2a,
y = 0:
0 = c sin k (2a)
2ak = π
y2 :
At x = a,
y = h:
At x = a,
y = 2h :
At x = 2a,
y = 0:
or k =
h = c sin
π
2a
π
2a
( a) or
2h = ma − b
0 = m(2a) + b
2h
, b = 4h
a
Solving yields
m=−
Then
y1 = h sin
Now
dA = ( y2 − y1 )dx =
Then
c=h
,
π
2a
A = dA =
,
2h
x + 4h
a
2h
( − x + 2a )
=
a
y2 = −
x
2a
a
h
π +
* 2h
(− x + 2a) − h sin
x dx
2a !#
"a
π +
*2
(− x + 2a) − sin
x dx
2a !#
"a
2a
2a
π +
* 1
cos
x
= h − ( − x + 2a ) 2 +
2a !# a
π
" a
*$ 2 a % 1
+
2%
$
= h & − ' + (− a + 2a ) 2 ! = ah &1 − '
( π)
"( π ) a
#
= 0.36338ah
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1396
PROBLEM 9.19 (Continued)
Find:
We have
I x and k x
3
3
1 %
1 -3 * 2h
π % .3
$1
+ $
dI x = & y2 − y1 ' dx = /
(− x + 2a) ! − & h sin
x 0 dx
3 )
3 13 " a
2a )' 23
(3
# (
=
Then
Now
Then
h3 * 8
π +
( − x + 2a )3 − sin 3
x
3 " a3
2a !#
,
I x = dI x =
,
2a
a
h3 * 8
π +
(− x + 2a)3 − sin 3
x dx
3
3 "a
2a !#
sin 3 θ = sin θ (1 − cos 2 θ ) = sin θ − sin θ cos 2 θ
Ix =
h3
3
,
2a *
a
8
π
π
π %+
$
( − x + 2a )3 − & sin
x − sin
x cos 2
x ! dx
3
2a
2a
2a ') #
(
"a
2a
=
2a
2a
π
π +
h3 * 2
cos
cos3
− ( − x + 2a ) 4 +
x−
x
3 " a3
2a
3π
2a !# a
π
=
h3
3
=
2 3$
2 %
ah &1 −
'
3
( 3π )
*$ 2a 2a % 2
4+
& − π + 3π ' + 3 (− a + 2a) !
) a
"(
#
I x = 0.52520ah3
and
k x2 =
I x 0.52520ah3
=
0.36338ah
A
or
I x = 0.525ah3
or
k x = 1.202h
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1397
!
PROBLEM 9.20
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
y1 :
At x = 2a,
y = 0:
0 = c sin k (2a)
2ak = π
y2 :
At x = a,
y = h:
At x = a,
y = 2h :
At x = 2a,
y = 0:
or k =
h = c sin
π
2a
π
2a
( a) or
2h = ma − b
0 = m(2a) + b
2h
, b = 4h
a
Solving yields
m=−
Then
y1 = h sin
Now
dA = ( y2 − y1 )dx =
Then
c=h
,
π
2a
A = dA =
,
2h
x + 4h
a
2h
( − x + 2a )
=
a
y2 = −
x
2a
a
h
π +
* 2h
(− x + 2a) − h sin
x dx
2
a
a !#
"
π +
*2
x dx
(− x + 2a) − sin
a
a !#
2
"
2a
2a
π +
* 1
cos
x
= h − ( − x + 2a ) 2 +
2a !# a
π
" a
*$ 2 a % 1
+
2%
$
= h & − ' + (− a + 2a ) 2 ! = ah &1 − '
π
π
a
(
)
(
)
"
#
= 0.36338ah
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1398
PROBLEM 9.20 (Continued)
Find: I y and k y
- * 2h
π + .
dI y = x 2 dA = x 2 /
(− x + 2a) − h sin
x dx 0
2a !# 2
1" a
π +
*2
= h ( − x3 + 2ax 2 ) − x 2 sin
x dx
2a !#
"a
We have
,
I y = dI y =
Then
,
2a
h
a
π +
*2
x dx
( − x3 + 2ax 2 ) − x 2 sin
a
a !#
2
"
Now using integration by parts with
u = x2
dv = sin
du = 2 x dx
,x
Then
2
sin
π
2a
,x
2
sin
π
2a
2a
π
x dx
cos
π
2a
x
,
dv = cos
du = dx
Then
2a
π % $ 2a
π %
$ 2a
x dx = x 2 & − cos
x ' − & − cos
x (2 x dx)
a
a ')
π
π
2
2
(
)
(
u=x
Now let
v=−
π
x dx = −
Iy = h
2a
π
x 2 cos
v=
π
2a
2a
π
x+
π
2a
sin
x dx
π
2a
x
4a * $ 2a
π % $ 2a
π % +
x & sin
x ' − & sin
x dx !
2a )
2a ') #
π " (π
(π
,
*2$ 1 4 2 3%
& − x + 3 ax '
)
"a ( 4
2a
$ 2a 2
π
π
π %+
8a 2
16a 3
− && −
x cos
x + 2 x sin
x + 3 cos
x '!
2a
2a
2a ') !#
π
π
( π
a
-3 2 * 1
2
16a3 .3
+ 2a
(2a )2 + 3 0
= h / − (2a )4 + a (2a)3 ! −
3
π 23
# π
31 a " 4
2
2
3- 2 * 1
+ 8a
3.
− h / − ( a ) 4 + a ( a )3 ! − 2 ( a ) 0
3
# π
31 a " 4
32
= 0.61345a3 h
and
k y2 =
Iy
A
=
0.61345a 3 h
0.36338ah
or
I y = 0.613a3 h
or
k y = 1.299a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1399
!
PROBLEM 9.21
Determine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point P.
SOLUTION
We have
Then
dI x = y 2 dA = y 2 [( x2 − x1 )dy ]
2a
* a 2
+
y (2a − a) dy +
y 2 (2a − 0)dy !
a
2
"
#
2a .
-3 * 1 3 + a
*1 + 3
= 2 / a y ! + 2a y 3 ! 0
" 3 # a 23
13 " 3 # 0
,
Ix = 2
,
- *1
.
+ 2
= 2 / a ( a)3 ! + a *"(2a )3 − ( a)3 +# 0
3
3
#
1 "
2
= 10a 4
Also
Then
dI y = x 2 dA = x 2 [( y2 − y1 )dx]
Iy = 2
*
"
,
a
0
x 2 (2a − a)dx +
,
2a
a
+
x 2 (2a − 0)dx !
#
= 10a 4
Now
JP = I x + I y = 10a 4 + 10a 4
and
k P2 =
=
or
J P = 20a 4
or
k P = 1.826a
JP
20a 4
=
A (4a)(2a ) − (2a)(a)
10 2
a
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1400
!
PROBLEM 9.22
Determine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point P.
SOLUTION
x=
a a y 1
+
= (a + y )
2 2a 2
dA = x dy =
1
A=
2
,
a
0
dA =
,
1
I x = y 2 dA =
2
,
a
0
,
1
1
y2
(a + y )dy = ay +
2
2
2
a
0
1 y3 y4
= a
+
2 3
4
1
Iy =
2
,
a
0
1
1
Iy =
2
24
,
0
1
1
(a + y )dy =
2
2
y2
a
=
0
1 3
1
x dy =
3
3
a
1
(a + y )dy
2
,
a
0
=
0
1 $ 2 a2 % 3 2
& a + '' = a
2 &(
2 ) 4
A=
Ix =
7 4
a "
12
Iy =
5 4
a "
16
3
1
%
& 2 (a + y ) ' dy
(
)
a
(a + y )3 dy =
1 1
1 *
15 4
4
4+
(a + y ) 4 =
"(2a) − a # = 96 a
24 4
96
0
1
5 4
Iy =
a
2
32
From Eq. (9.4):
3 2
a "
2
(ay 2 + y 3 )dy
1$1 1% 4 1 7 4
a =
a
+
2 &( 3 4 ')
2 12
a$
0
,
a
JO = I x + I y =
J O = kO2 A
7 4 5 4 $ 28 + 15 % 4
a + a =&
'a
12
16
( 48 )
kO2 =
JO
=
A
43 4
a
48
3 2
a
2
=
43 2
a
72
JO =
kO = a
43
72
43 4
a
48
kO = 0.773a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1401
!
PROBLEM 9.23
(a) Determine by direct integration the polar moment of inertia of the annular
area shown with respect to Point O. (b) Using the result of Part a, determine
the moment of inertia of the given area with respect to the x axis.
SOLUTION
dA = 2π u du
(a)
dJ O = u 2 dA = u 2 (2π u du )
= 2π u 3 du
,
J O = dJ O = 2π
= 2π
(b)
From Eq. (9.4):
(Note by symmetry.)
,
R2
R1
u 3 du
1 4
u
4
R2
JO =
π* 4
R2 − R14 +
Ix =
π* 4
R2 − R14 +
R1
2"
#
Ix = Iy
JO = I x + I y = 2I x
Ix =
1
π
J O = *" R24 − R14 +#
2
4
4"
#
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1402
PROBLEM 9.24
(a) Show that the polar radius of gyration kO of the annular area shown is
approximately equal to the mean radius Rm = ( R1 + R2 ) 2 for small values
of the thickness t = R2 − R1. (b) Determine the percentage error introduced
by using Rm in place of kO for the following values of t/Rm: 1, 12 , and 101 .
SOLUTION
(a)
From Problem 9.23:
JO =
kO2
π* 4
R2 − R14 +
2"
#
4
4
π
J O 2 *" R2 − R1 +#
=
=
A π R22 − R12
(
)
kO2 =
1 2
* R2 + R12 +
#
2"
Mean radius:
Rm =
1
( R1 + R2 )
2
Thus,
1
1
R1 = Rm − t and R2 = Rm + t
2
2
Thickness: t = R2 − R1
kO2 =
2
2
1 *$
1 % $
1 % +
1
R
+
t
+
R
−
t
! = Rm2 + t 2
m
m
&
'
&
'
2 "(
2 ) (
2 ) #!
4
For t small compared to Rm : kO2 ≈ Rm2
(b)
Percentage error =
kO ≈ Rm
(exact value) − (approximation value)
100
(exact value)
P.E. = −
Rm2 + 14 t 2 − Rm
Rm2 + 14 t 2
(100) = −
2
( ) − 1100
1+ ( )
1+
t
Rm
1
4
1
4
t
Rm
2
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1403
PROBLEM 9.24 (Continued)
For
t
= 1:
Rm
t
1
For
= :
Rm 2
1
t
For
= :
Rm 10
P.E. =
1 + 14 − 1
1+
P.E. = −
P.E. =
1
4
(100) = −10.56%
2
( 12 ) − 1
(100) = −2.99%
1 1 2
1+ 4 ( 2 )
1+
1
4
2
( ) −1
(100) = −0.125%
2
1 + 14 ( 101 )
1+
1 1
4 10
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1404
PROBLEM 9.25
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown with respect to Point P.
SOLUTION
y1 :
At x = 2a,
y = 2a :
2a = k1 (2a) 2
y2 :
At x = 0,
At x = 2a,
y = a:
or k1 =
a=c
y = 2a :
2a = a + k2 (2a )2
Then
y1 =
1 2
x
2a
Then
Now
k2 =
or
y2 = a +
=
Now
1
2a
1
4a
1 2
x
4a
1
(4a 2 + x 2 )
4a
1 2+
*1
(4a 2 + x 2 ) −
x dx
2a !#
" 4a
1
(4a 2 − x 2 ) dx
=
4a
dA = ( y2 − y1 )dx =
,
A = dA = 2
,
2a
0
2a
1
1 * 2
1 +
8
(4a 2 − x 2 )dx =
4a x − x 3 ! = a 2
4a
2a "
3 #0
3
3
3
1 %
1 -3 * 1
$1
+ * 1 2 + .3
(4a 2 + x 2 ) ! −
dI x = & y23 − y13 ' dx = /
x ! 0 dx
3 )
3 13 " 4a
(3
# " 2a # 23
1* 1
1
+
(64a 6 + 48a 4 x 2 + 12a 2 x 4 + x6 ) − 3 x6 ! dx
3
3 " 64a
8a
#
1
(64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx
=
3
192a
=
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1405
PROBLEM 9.25 (Continued)
Then
,
I x = dI x = 2
,
2a
0
1
(64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx
192a3
2a
=
1 *
12
+
64a 6 x + 16a 4 x3 + a 2 x5 − x 7 !
3
5
96a "
#0
1 *
12
+
64a 6 (2a) + 16a 4 (2a )3 + a 2 (2a)5 − (2a )7 !
5
96a3 "
#
1 4$
12
% 32 4
a &128 + 128 + × 32 − 128 ' =
a
=
96 (
5
) 15
=
Also
Then
*1
+
(4a 2 − x 2 )dx !
" 4a
#
dI y = x 2 dA = x 2
,
I y = dI y = 2
=
,
2a
0
2a
1 2
1 *4 2 3 1 5+
x (4a 2 − x 2 )dx =
a x − x !
4a
2a " 3
5 #0
1 *4 2
1
+ 32 4 $ 1 1 % 32 4
a (2a )3 − (2a )5 ! =
a & − '=
a
2a " 3
5
# 2
( 3 5 ) 15
Now
JP = I x + I y =
and
k P2 =
JP
=
A
32 4 32 4
a + a
15
15
64 4
a
15
8 2
a
3
=
8 2
a
5
or
JP =
64 4
a
15
or k P = 1.265a
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1406
PROBLEM 9.26
Determine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point P.
SOLUTION
The equation of the circle is
x2 + y 2 = r 2
So that
Now
x = r 2 − y2
dA = x dy = r 2 − y 2 dy
,
,
r
r 2 − y 2 dy
Then
A = dA = 2
Let
y = r sin θ ; dy = r cos θ dθ
Then
A=2
=2
π /2
,π
− /6
π /2
,π
− /6
= 2r 2
− r/2
r 2 − (r sin θ ) 2 r cos dθ
π /2
r 2 cos 2 θ dθ = 2r 2
* π2
$ − π sin − π3
− && 6 +
4
"2 ( 2
* θ sin 2θ +
+
4 !# −π /6
"2
$
%+
3%
2 π
'
'' ! = 2r && +
8 )'
) #!
(3
= 2.5274r 2
Now
dI x = y 2 dA = y 2
Then
I x = dI x = 2
Let
Then
,
r
− r/2
r 2 − y 2 dy
)
y 2 r 2 − y 2 dy
y = r sin θ ; dy = r cos θ dθ
Ix = 2
=2
Now
,
(
π /2
,π
− /6
π /2
,π
− /6
(r sin θ )2 r 2 − (r sin θ ) 2 r cos θ dθ
r 2 sin 2 θ (r cos θ )r cos θ dθ
sin 2θ = 2sin θ cos θ 4 sin 2 θ cos 2 θ =
1
sin 2θ
4
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1407
PROBLEM 9.26 (Continued)
Then
Ix = 2
,
π /2
r 4 * θ sin 4θ +
$1
%
r 4 & sin 2 2θ ' dθ =
−
−π /6
2 "2
8 #! −π /6
(4
)
π /2
r 4 * π2 $ π6 sin − 23π
−& −
2 " 2 &( 2
8
3%
r4 $ π
= & −
'
&
2 ( 3 16 ')
=
Also
dI y =
1 3
1
x dy =
3
3
(
r 2 − y2
,
r
1
,
(r 2 − y 2 )3/2 dy
I y = dI y = 2
Let
y = r sin θ ; dy = r cos θ dθ
Then
Iy =
2
3
,π
Iy =
2
3
,π
Now
Then
π /2
− /6
π /2
− /6
3
) dy
Then
− r/ 2 3
%+
'' !
) !#
[r 2 − (r sin θ ) 2 ]3/ 2 r cos θ dθ
( r 3 cos3 θ )r cos θ dθ
1
cos 4 θ = cos 2 θ (1 − sin 2 θ ) = cos 2 θ − sin 2 2θ
4
Iy =
=
2
3
π /2
,π
− /6
1
$
%
r 4 & cos 2 θ − sin 2 2θ ' dθ
4
(
)
2 4 *$ θ sin 2θ
r & +
3 "( 2
4
% 1 $ θ sin 4θ
'− & −
8
) 4( 2
π /2
%+
'!
) # −π / 6
2 4 -3 * π2 1 $ π2 % + * − π6 sin − π3 1 $ − π6 sin − 23π
+
− &
−
r / − & '! −
3 3 " 2 4 &( 2 ') #! " 2
4
4 (& 2
8
1
2 *π π π 1 $ 3 % π
1 $ 3 %+
= r4
− + + &
+ &
'' −
'!
&
3 " 4 16 12 4 ( 2 ) 48 32 &( 2 ') #!
2 $π 9 3 %
= r4 & +
'
3 &( 4
64 ')
=
% + .3
'' ! 0
) !# 32
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1408
PROBLEM 9.26 (Continued)
Now
JP = Ix + Iy =
r4
2
$π
3 % 2 4$π 9 3 %
&& −
'' + r && +
'
64 ')
( 3 16 ) 3 ( 4
$π
3%
= r4 & +
= 1.15545r 4
& 3 16 ''
(
)
and
k P2 =
J P 1.15545r 4
=
A
2.5274r 2
or
J P = 1.155r 4
or k P = 0.676r
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1409
!
PROBLEM 9.27
Determine the polar moment of inertia and the polar radius of gyration
of the shaded area shown with respect to Point O.
SOLUTION
At θ = π , R = 2a :
2a = a + k (π )
a
or
k=
Then
R=a+
π
a
π
$ θ%
θ = a &1 + '
π
(
)
dA = (dr )(r dθ )
= rdr dθ
Now
,
A = dA =
Then
π
a (1+θ /π )
0
0
, ,
rdr dθ =
,
π
0
a (1+θ /π )
*1 2 +
r !
" 2 #0
dθ
π
A=
,
π
0
2
3
1 2$ θ %
1 *π $ θ % +
a & 1 + ' dθ = a 2
1
+
!
2 ( π)
2 " 3 &( π ') !#
0
*$ π % 3
+ 7
1
= π a 2 &1 + ' − (1)3 ! = π a 2
6
"( π )
#! 6
Now
Then
dJ O = r 2 dA = r 2 (rdr dθ )
,
J O = dJ O =
=
,
π
0
π
a (1+θ /π )
0
0
, ,
r 3 dr dθ
a (1+θ /π )
*1 4 +
r !
" 4 #0
dθ =
,
π
0
4
1 4$ θ %
a & 1 + ' dθ
4 ( π)
π
5+
=
and
kO2 =
5
*
+
π%
1 4 *π $ θ %
1
4 $
a
a
π
1
1
+
=
+
− (1)5 !
!
&
'
&
'
4 " 5 ( π ) #!
20
"( π )
#!
0
JO
=
A
31
π a4
20
7
π a2
6
=
93 2
a
70
31 4
πa
20
or
JO =
or
kO = 1.153a
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1410
!
PROBLEM 9.28
Determine the polar moment of inertia and the polar radius of gyration of
the isosceles triangle shown with respect to Point O.
SOLUTION
By observation:
y=
or
x=
h
b
2
x
b
y
2h
Now
$ b
dA = xdy = &
( 2h
and
dI x = y 2 dA =
Then
,
%
y ' dy
)
b 3
y dy
2h
I x = dI x = 2
,
h
0
b 3
y dy
2h
b y4
=
h 4
From above:
Now
y=
=
0
1 3
bh
4
2h
x
b
2h %
$
dA = (h − y )dx = & h −
x dx
b ')
(
=
and
h
dI y = x 2 dA = x 2
h
(b − 2 x)dx
b
h
(b − 2 x)dx
b
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1411
PROBLEM 9.28 (Continued)
Then
,
I y = dI y = 2
,
b/2
0
h 2
x (b − 2 x)dx
b
b/2
=2
h *1 3 1 4 +
bx − x !
b "3
2 #0
h *b $ b % 1 $ b %
=2
−
b " 3 &( 2 ') 2 &( 2 ')
3
Now
JO = I x + I y =
and
kO2 =
JO
=
A
bh
48
4+
1
! = b3 h
48
!#
1 3 1 3
bh + b h
4
48
(12h 2 + b 2 )
1
2
bh
=
1
(12h 2 + b 2 )
24
or
or
JO =
bh
(12h 2 + b 2 )
48
kO =
12h 2 + b 2
24
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1412
PROBLEM 9.29*
Using the polar moment of inertia of the isosceles triangle of
Problem 9.28, show that the centroidal polar moment of inertia
of a circular area of radius r is π r 4 /2. (Hint: As a circular area
is divided into an increasing number of equal circular sectors,
what is the approximate shape of each circular sector?)
PROBLEM 9.28 Determine the polar moment of inertia and
the polar radius of gyration of the isosceles triangle shown with
respect to Point O.
SOLUTION
First the circular area is divided into an increasing number of identical circular sectors. The sectors can be
approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of the
isosceles triangles are as shown.
For an isosceles triangle (see Problem 9.28):
JO =
b = r ∆θ
Then with
=
dJ O sector
dθ
and h = r
1
(r ∆θ )(r ) *"12r 2 + (r ∆θ ) 2 +#
48
(∆ J O )sector
Now
bh
(12h 2 + b 2 )
48
1 4
r ∆θ *"(12 + ∆θ 2 ) +#
48
$ ∆ J O sector
= lim &
∆θ →0
( ∆θ
%
-1 4
2 .
' = lim / r *"12 + ( ∆θ ) +# 0
∆θ →0 1 48
2
)
=
Then
,
( J O )circle = dJ O sector =
,
2π
0
1 4
r
4
1 4
1
2π
r dθ = r 4 [θ ]0
4
4
or
( J O )circle =
π
2
r4
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1413
!
PROBLEM 9.30*
Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 /2π . (Hint: Compare
the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the
same centroid.)
SOLUTION
From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is
( J C )cir =
π
2
r4
The area of the circle is
Acir = π r 2
So that
[ J C ( A)]cir =
A2
2π
Two methods of solution will be presented. However, both methods depend upon the observation that as a
given element of area dA is moved closer to some Point C, The value of J C will be decreased ( J C = , r 2 dA;
as r decreases, so must J C ).
Solution 1
Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its
area is equal to A. To minimize the value of ( J C ) A, the area would have to be distributed as closely as
possible about C. This is accomplished by winding the strip into a tightly wound roll with C as its center; any
voids in the roll would place the corresponding area farther from C than is necessary, thus increasing the value
of ( J C ) A. (The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll
is circular, with the centroid of its area at C, it follows that
( JC ) A ≥
A2
Q.E.D.
2π
!
!
!
where the equality applies when the original area is circular.
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1414
PROBLEM 9.30* (Continued)
Solution 2
Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid) at
Point C. Without loss of generality, assume that
A1 = A2
A3 = A4
It then follows that
( J C ) A = ( J C )cir + [ J C ( A1 ) − J C ( A2 ) + J C ( A3 ) − J C ( A4 )]
Now observe that
J C ( A1 ) − J C ( A2 ) $ 0
J C ( A3 ) − J C ( A4 ) $ 0
since as a given area is moved farther away from C its polar moment of inertia with respect to C must
increase.
( J C ) A $ ( J C )cir
or ( J C ) A $
A2
Q.E.D.
2π
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you are using it without permission.
1415
!
PROBLEM 9.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 + A2 + A3
= [(24)(6) + (8)(48) + (48)(6)] mm 2
= (144 + 384 + 288) mm 2
= 816 mm 2
Now
I x = ( I x )1 + ( I x )2 + ( I x )3
where
1
(24 mm)(6 mm)3 + (144 mm 2 )(27 mm)2
12
= (432 + 104,976) mm 4
( I x )1 =
= 105, 408 mm 4
1
(8 mm)(48 mm)3 = 73,728 mm 4
12
1
( I x )3 = (48 mm)(6 mm)3 + (288 mm 2 )(27 mm)2
12
= (864 + 209,952) mm 4 = 210,816 mm 4
( I x )2 =
Then
I x = (105, 408 + 73, 728 + 210,816) mm 4
= 389,952 mm 4
and
k x2 =
I x 389,952 mm 4
=
A
816 mm 2
or
I x = 390 × 103 mm 4
or
k x = 21.9 mm
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1416
!
PROBLEM 9.32
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 − A2 − A3
= [(5)(6) − (4)(2) − (4)(1)] in.2
= (30 − 8 − 4) in.2
= 18 in.2
Now
where
I x = ( I x )1 − ( I x ) 2 − ( I x )3
( I x )1 =
1
(5 in.)(6 in.)3 = 90 in.4
12
1
(4 in.)(2 in.)3 + (8 in.2 )(2 in.) 2
12
2
= 34 in.4
3
( I x )2 =
1
$3
%
(4 in.)(1 in.)3 + (4 in.2 ) & in. '
12
2
(
)
1 4
= 9 in.
3
2
( I x )3 =
Then
2
1%
$
I x = & 90 − 34 − 9 ' in.4
3
3)
(
and
k x2 =
I x 46.0 in.4
=
A
18 in.4
or
I x = 46.0 in.4
or k x = 1.599 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1417
PROBLEM 9.33
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 + A2 + A3
= [(24 × 6) + (8)(48) + (48)(6)] mm 2
= (144 + 384 + 288) mm 2
= 816 mm 2
Now
I y = ( I y )1 + ( I y ) 2 + ( I y )3
where
1
(6 mm)(24 mm)3 = 6912 mm 4
12
1
( I y ) 2 = (48 mm)(8 mm)3 = 2048 mm 4
12
1
( I y )3 = (6 mm)(48 mm)3 = 55, 296 mm 4
12
( I y )1 =
Then
and
I y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4
k y2 =
Iy
A
=
64, 256 mm 4
816 mm 2
or
I y = 64.3 × 103 mm 4
or
k y = 8.87 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1418
PROBLEM 9.34
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 − A2 − A3
= [(5)(6) − (4)(2) − (4)(1)] in.2
= (30 − 8 − 4) in.2
= 18 in.2
Now
I y = ( I y )1 − ( I y ) 2 − ( I y )3
where
( I y )1 =
1
(6 in.)(5 in.)3 = 62.5 in.4
12
( I y )2 =
1
2
(2 in.)(4 in.)3 = 10 in.4
12
3
( I y )3 =
1
1
(1 in.)(4 in.)3 = 5 in.4
12
3
Then
2
1%
$
I y = & 62.5 − 10 − 5 ' in.4
3
3)
(
and
k y2 =
Iy
A
=
46.5 in.4
18 in.2
or
I y = 46.5 in.4
or k y = 1.607 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1419
PROBLEM 9.35
Determine the moments of inertia of the shaded area shown with respect to the x
and y axes when a = 20 mm.
SOLUTION
We have
where
I x = ( I x )1 + 2( I x ) 2
1
(40 mm)(40 mm)3
12
= 213.33 × 103 mm 4
( I x )1 =
( I x )2 =
*π
"8
(20 mm)4 −
2
$ 4 × 20
% +
(20 mm) 2 &
mm ' !
2
( 3π
) !#
π
*$ 4 × 20
+
%
+ (20 mm) &
+ 20 ' mm !
2
)
"( 3π
#
π
2
2
= 527.49 × 103 mm 4
Then
I x = [213.33 + 2(527.49)] × 103 mm 4
or
Also
where
I y = ( I y )1 + 2( I y )2
( I y )1 =
( I y )2 =
Then
I x = 1.268 × 106 mm 4
1
(40 mm)(40 mm)3 = 213.33 × 103 mm 4
12
π
8
(20 mm)4 = 62.83 × 103 mm 4
I y = [213.33 + 2(62.83)] × 103 mm 4
or
I y = 339 × 103 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1420
!
PROBLEM 9.36
Determine the moments of inertia of the shaded area shown with
respect to the x and y axes when a = 20 mm.
SOLUTION
Area = Square − 2(Semicircles)
Given:
For a = 20 mm, we have
Ix = Iy =
Square:
Semicircle
1
(60) 4 = 1080 × 103 mm 4
12
:
Ix =
π
8
(20) 4 = 62.83 × 103 mm 4
I AA′ = I y′ + Ad 2 ;
π
8
$π %
(20)4 = I y′ + & ' (20) 2 (8.488) 2
(2)
I y′ = 62.83 × 103 − 45.27 × 103
I y′ = 17.56 × 103 mm 4
I y = I y′ + A(30 − 8.488)2 = 17.56 × 103 +
π
2
(20) 2 (21.512)2
I y = 308.3 × 103 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1421
PROBLEM 9.36 (Continued)
Semicircle !: Same as semicircle
.
Entire Area:
I x = 1080 × 103 − 2(62.83 × 103 )
= 954.3 × 103 mm 4
I x = 954 × 103 mm 4
I y = 1080 × 103 − 2(308.3 × 103 )
= 463.3 × 103 mm 4
I y = 463 × 103 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1422
!
PROBLEM 9.37
For the 4000-mm 2 shaded area shown, determine the distance d2
and the moment of inertia with respect to the centroidal axis
parallel to AA′ knowing that the moments of inertia with respect to
AA′ and BB′ are 12 × 106 mm 4 and 23.9 × 106 mm4, respectively, and
that d1 = 25 mm.
SOLUTION
We have
I AA′ = I + Ad12
and
I BB′ = I + Ad 22
Subtracting
or
(
I AA′ − I BB′ = A d12 − d 22
d 22 = d12 −
(1)
)
I AA′ − I BB′
A
= (25 mm)2 −
(12 − 23.9)106 mm 4
4000 mm 2
= 3600 mm 2
Using Eq. (1):
or d 2 = 60.0 mm
I = 12 × 106 mm 4 − (4000 mm 2 )(25 mm) 2
or
I = 9.50 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1423
!
PROBLEM 9.38
Determine for the shaded region the area and the moment of inertia
with respect to the centroidal axis parallel to BB′, knowing that
d1 = 25 mm and d 2 = 15 mm and that the moments of inertia with
respect to AA′ and BB′ are 7.84 × 106 mm 4 and 5.20 × 106 mm4,
respectively.
SOLUTION
We have
I AA′ = I + Ad12
and
I BB′ = I + Ad 22
Subtracting
or
(
I AA′ − I BB′ = A d12 − d 22
A=
I AA′ − I BB′
d12 − d 22
(1)
)
=
(7.84 − 5.20)106 mm 4
(25 mm) 2 − (15 mm)2
or
Using Eq. (1):
A = 6600 mm 2
I = 7.84 × 106 mm 4 − (6600 mm 2 )(25 mm)2
= 3.715 × 106 mm 4
or I = 3.72 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1424
!
PROBLEM 9.39
The shaded area is equal to 50 in.2 . Determine its centroidal moments
of inertia I x and I y , knowing that I y = 2 I x and that the polar moment
of inertia of the area about Point A is JA = 2250 in.4 .
SOLUTION
Given:
A = 50 in.2
I y = 2 I x , J A = 2250 in.4
J A = J C + A(6 in.) 2
2250 in.4 = J C + (50 in.2 )(6 in.) 2
J C = 450 in.4
JC = I x + I y
with
450 in.4 = I x + 2 I x
I y = 2I x
I x = 150.0 in.4
I y = 2 I x = 300 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1425
!
PROBLEM 9.40
The polar moments of inertia of the shaded area with respect to
Points A, B, and D are, respectively, JA = 2880 in.4, JB = 6720 in.4,
and JD = 4560 in.4. Determine the shaded area, its centroidal moment
of inertia JC , and the distance d from C to D.
SOLUTION
See figure at solution of Problem 9.39.
Given:
J A = 2880 in.4 , J B = 6720 in.4 , J D = 4560 in.4
J B = J C + A(CB ) 2 ; 6720 in.4 = J C + A(62 + d 2 )
(1)
J D = J C + A(CD) 2 ; 4560 in.4 = J C + Ad 2
(2)
Eq. (1) subtracted by Eq. (2): J B − J D = 2160 in.4 = A(6) 2
J A = J C + A( AC )2 ; 2880 in.4 = J C + (60 in.2 )(6 in.) 2
Eq. (2):
4560 in.4 = 720 in.4 + (60 in.2 )d 2
A = 60.0 in.2
J C = 720 in.4
d = 8.00 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1426
!
PROBLEM 9.41
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
Dimensions in mm
First locate centroid C of the area.
Symmetry implies Y = 30 mm.
1
Then
A, mm 2
x , mm
108 × 60 = 6480
54
349,920
46
–59,616
2
1
− × 72 × 36 = −1296
2
Σ
5184
290,304
X ΣA = Σ xA : X (5184 mm 2 ) = 290,304 mm3
or
X = 56.0 mm
Now
I x = ( I x )1 − ( I x )2
where
xA, mm3
1
(108 mm)(60 mm)3 = 1.944 × 106 mm 4
12
*1
+
$1
%
( I x )2 = 2
(72 mm)(18 mm)3 + & × 72 mm × 18 mm ' (6 mm) 2 !
(2
)
" 36
#
( I x )1 =
= 2(11, 664 + 23,328) mm 4 = 69.984 × 103 mm 4
[( I x ) 2 is obtained by dividing A2 into
Then
]
I x = (1.944 − 0.069984) × 106 mm 4
or
I x = 1.874 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1427
PROBLEM 9.41 (Continued)
Also
where
I y = ( I y )1 − ( I y ) 2
1
(60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2
12
= (6, 298,560 + 25,920) mm 4 = 6.324 × 106 mm 4
( I y )1 =
1
(36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2
36
= (373, 248 + 129, 600) mm 4 = 0.502 × 106 mm 4
( I y )2 =
Then
I y = (6.324 − 0.502)106 mm 4
or I y = 5.82 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1428
!
PROBLEM 9.42
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
First locate C of the area:
Symmetry implies X = 12 mm.
Then
A, mm 2
y , mm
yA, mm3
1
12 × 22 = 264
11
2904
2
1
(24)(18) = 216
2
28
6048
Σ
480
8952
Y ΣA = Σ yA: Y (480 mm 2 ) = 8952 mm3
Y = 18.65 mm
Now
where
I x = ( I x )1 + ( I x ) 2
1
(12 mm)(22 mm)3 + (264 mm 2 )[(18.65 − 11) mm]2
12
= 26, 098 mm 4
( I x )1 =
1
(24 mm)(18 mm)3 + (216 mm 2 )[(28 − 18.65) mm]2
36
= 22, 771 mm 4
( I x )2 =
Then
I x = (26.098 + 22.771) × 103 mm 4
or
I x = 48.9 × 103 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1429
PROBLEM 9.42 (Continued)
Also
where
I y = ( I y )1 + ( I y ) 2
1
(22 mm)(12 mm)3 = 3168 mm 4
12
*1
+
$1
%
( I y )2 = 2
(18 mm)(12 mm)3 + & × 18 mm × 12 mm ' (4 mm)2 !
36
2
(
)
"
#
( I y )1 =
= 5184 mm 4
[( I y )2 is obtained by dividing A2 into
Then
]
I y = (3168 + 5184) mm 4
or I y = 8.35 × 103 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1430
!
PROBLEM 9.43
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid C of the area.
Then
A, in.2
x , in.
1
5 × 8 = 40
2.5
2
−2 × 5 = −10
1.9
Σ
30
where
4
100
160
4.3
–19
–43
81
117
X = 2.70 in.
Y ΣA = Σ yA: Y (30 in.2 ) = 117 in.3
Y = 3.90 in.
or
Now
yA, in.3
X ΣA = Σ xA: X (30 in.2 ) = 81 in.3
or
and
xA, in.3
y , in.
I x = ( I x )1 − ( I x )2
1
(5 in.)(8 in.)3 + (40 in.2 )[(4 − 3.9) in.]2
12
= (213.33 + 0.4) in.4 = 213.73 in.4
( I x )1 =
1
(2 in.)(5 in.)3 + (10 in.2 )[(4.3 − 3.9) in.]2
12
= (20.83 + 1.60) = 22.43 in.4
( I x )2 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1431
PROBLEM 9.43 (Continued)
Then
I x = (213.73 − 22.43) in.4
Also
I y = ( I y )1 − ( I y ) 2
where
or I x = 191.3 in.4
1
(8 in.)(5 in.)3 + (40 in.2 )[(2.7 − 2.5) in.]2
12
= (83.333 + 1.6)in.4 = 84.933 in.4
( I y )1 =
1
(5 in.)(2 in.)3 + (10 in.2 )[(2.7 − 1.9) in.]2
12
= (3.333 + 6.4)in.4 = 9.733 in.4
( I y )2 =
Then
I y = (84.933 − 9.733)in.4
or
I y = 75.2 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1432
!
PROBLEM 9.44
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
First locate centroid C of the area.
A, in.2
Then
or
and
or
x , in.
y , in.
xA, in.3
yA, in.3
1
3.6 × 0.5 = 1.8
1.8
0.25
3.24
0.45
2
0.5 × 3.8 = 1.9
0.25
2.4
0.475
4.56
3
1.3 × 1 = 1.3
0.65
4.8
0.845
6.24
Σ
5.0
4.560
11.25
X ΣA = Σ xA: X (5 in.2 ) = 4.560 in.3
X = 0.912 in.
Y ΣA = Σ yA: Y (5 in.2 ) = 11.25 in.3
Y = 2.25 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1433
PROBLEM 9.44 (Continued)
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
1
(3.6 in.)(0.5 in.)3 + (1.8 in.2 )[(2.25 − 0.25)in.]2
12
= (0.0375 + 7.20)in.4 = 7.2375 in.4
( I x )1 =
1
(0.5 in.)(3.8 in.)3 + (1.9 in.2 )[(2.4 − 2.25)in.]2
12
= (2.2863 + 0.0428)in.4 = 2.3291 in.4
( I x )2 =
1
(1.3 in.)(1 in.)3 + (1.3 in.2 )[(4.8 − 2.25 in.)]2
12
= (0.1083) + 8.4533) in.4 = 8.5616 in.4
( I x )3 =
Then
I x = (7.2375 + 2.3291 + 8.5616) in.4 = 18.1282 in.4
or
Also
where
I x = 18.13 in.4
I y = ( I y )1 + ( I y ) 2 + ( I y )3
1
(0.5 in.)(3.6 in.)3 + (1.8 in.2 )[(1.8 − 0.912)in.]2
12
= (1.9440 + 1.4194)in.4 = 3.3634 in.4
( I y )1 =
1
(3.8 in.)(0.5 in.)3 + (1.9 in.2 )[(0.912 − 0.25) in.]2
12
= (0.0396 + 0.8327) in.4 = 0.8723 in.4
( I y )2 =
1
(1 in.)(1.3 in.)3 + (1.3 in.2 )[(0.912 − 0.65) in.]2
12
= (0.1831 + 0.0892) in.4 = 0.2723 in.4
( I y )3 =
Then
I y = (3.3634 + 0.8723 + 0.2723) in.4 = 4.5080 in.4
or
I y = 4.51 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1434
!
PROBLEM 9.45
Determine the polar moment of inertia of the area shown with
respect to (a) Point O, (b) the centroid of the area.
SOLUTION
Symmetry: X = Y
Dimensions in mm
Determination of centroid, C, of entire section.
Area, mm 2
Section
π
1
4
(100) 2 = 7.854 × 103
y , mm
yA, mm3
42.44
333.3 × 103
2
(50)(100) = 5 × 103
50
250 × 103
3
(100)(50) = 5 × 103
–25
−125 × 103
Σ
17.854 × 103
458.3 × 103
Y ΣA = Σ yA: Y (17.854 × 103 mm 2 ) = 458.3 × 103 mm3
Y = 25.67 mm
Distance O to C:
(a)
X = Y = 25.67 mm
OC = 2Y = 2(25.67) = 36.30 mm
π
(100) 4 = 39.27 × 106 mm 4
Section 1:
JO =
Section 2:
JO = J + A(OD) 2 =
8
*$ 50 %2 $ 100 %2 +
1
(50)(100)[502 + 1002 ] + (50)(100) & ' + &
' !
12
"( 2 ) ( 2 ) #!
JO = 5.208 × 106 + 15.625 × 106 = 20.83 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1435
PROBLEM 9.45 (Continued)
Section 3: Same as Section 2;
J O = 20.83 × 106 mm 4
Entire section:
J O = 39.27 × 106 + 2(20.83 × 106 )
JO = 80.9 × 106 mm 4
= 80.94 × 106
(b)
Recall that,
OC = 36.30 mm and
A = 17.854 × 103 mm 2
J O = J C + A(OC ) 2
80.94 × 106 mm 4 = J C + (17.854 × 103 mm 2 )(36.30 mm) 2
J C = 57.41 × 106 mm 4
J C = 57.4 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1436
!
PROBLEM 9.46
Determine the polar moment of inertia of the area shown with
respect to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
Note that symmetry implies Y = 0.
Dimensions in mm
A, mm 2
π
1
2
(84)(42) = 5541.77
π
2
2
−
3
π
2
−
4
X , mm
−
2
π
56
(42)2 = 2770.88
π
(54)(27) = −2290.22
π
112
−
72
π
36
(27) 2 = −1145.11
π
= −35.6507
−197,568
= 17.8254
49,392
= −22.9183
52,488
= 11.4592
4877.32
Σ
XA, mm3
–13,122
–108,810
X ΣA = Σ xA: X (4877.32 mm 2 ) = −108,810 mm3
Then
X = −22.3094
or
J O = ( J O )1 + ( J O ) 2 − ( J O )3 − ( J O )4
(a)
where
π
(84 mm)(42 mm)[(84 mm) 2 + (42 mm)2 ]
8
= 12.21960 × 106 mm 4
( J O )1 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1437
PROBLEM 9.46 (Continued)
π
(42 mm) 4
4
= 2.44392 × 106 mm 4
( J O )2 =
π
(54 mm)(27 mm)[(54 mm)2 + (27 mm)2 ]
8
= 2.08696 × 106 mm 4
( J O )3 =
π
(27 mm) 4
4
= 0.41739 × 106 mm 4
( J O )4 =
Then
J O = (12.21960 + 2.44392 − 2.08696 − 0.41739) × 106 mm 4
= 12.15917 × 106 mm 4
or
J O = 12.16 × 106 mm 4
J O = J C + AX 2
(b)
or
J C = 12.15917 × 106 mm 4 − (4877.32 mm 2 )(−22.3094 mm) 2
or
J C = 9.73 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1438
!
PROBLEM 9.47
Determine the polar moment of inertia of the area shown with respect to (a) Point O,
(b) the centroid of the area.
SOLUTION
Area, in.2
Section
π
1
2
(4.5) 2 = 15.904
4
−
π
4
xA, in.3
1.9099
30.375
1.2732
–9.00
8.835
Σ
Then
(3) 2 = −7.069
x , in.
21.375
XA = Σ xA: X (8.835 in.2 ) = 21.375 in.3
X = 2.419 in.
Then
JO =
π
8
(4.5 in.)4 −
π
8
(3 in.)4 = 129.22 in.4
J O = 129.2 in.4
OC = 2 X = 2(2.419 in.) = 3.421 in.
J O = J C + A(OC ) 2 :
129.22 in.4 = J C + (8.835 in.)(3.421 in.)2
J C = 25.8 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1439
!
PROBLEM 9.48
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
A, in.2
1
π
2
y , in.
8
(6)2 = 56.5487
2
1
− (12)(4.5) = −27
2
Σ
29.5487
π
= 2.5465
1.5
yA, in.3
144
–40.5
103.5
Y ΣA = Σ yA: Y (29.5487 in.2 ) = 103.5 in.3
Then
Y = 3.5027 in.
or
J O = ( J O )1 − ( J O ) 2
(a)
π
where
(6 in.) 4 = 107.876 in.4
4
( J O ) 2 = ( I x′ ) 2 + ( I y ′ ) 2
Now
( I x′ ) 2 =
and
( I y′ ) 2 = 2
[Note: ( I y ′ ) 2 is obtained using
( J O )1 =
1
(12 in.)(4.5 in.)3 = 91.125 in.4
12
*1
+
(4.5 in.)(6 in.)3 ! = 162.0 in.4
12
"
#
]
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1440
PROBLEM 9.48 (Continued)
Then
Finally
( J O ) 2 = (91.125 + 162.0) in.4
= 253.125 in.4
J O = (1017.876 − 253.125) in.4 = 764.751 in.4
or
J O = 765 in.4
J O = J C + AY 2
(b)
or
J C = 764.751 in.4 − (29.5487 in.2 )(3.5027 in.) 2
or
J C = 402 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1441
!
PROBLEM 9.49
Two 20-mm steel plates are welded to a rolled S section as shown.
Determine the moments of inertia and the radii of gyration of the
combined section with respect to the centroidal x and y axes.
SOLUTION
S section:
A = 6010 mm 2
I x = 90.3 × 106 mm 4
I y = 3.88 × 106 mm 4
Note:
Atotal = AS + 2 Aplate
= 6010 mm 2 + 2(160 mm)(20 mm)
= 12, 410 mm 2
Now
where
I x = ( I x )S + 2( I x ) plate
( I x )plate = I xplate + Ad 2
1
(160 mm)(20 mm)3 + (3200 mm 2 )[(152.5 + 10) mm]2
12
= 84.6067 × 106 mm 4
=
Then
I x = (90.3 + 2 × 84.6067) × 106 mm 4
= 259.5134 × 106 mm 4
and
k x2 =
Ix
259.5134 × 106 mm 4
=
Atotal
12410 mm 2
or I x = 260 × 106 mm 4
or
k x = 144.6 mm
!
!
!
!
!
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1442
!
PROBLEM 9.49 (Continued)
Also
I y = ( I y )S + 2( I y ) plate
= 3.88 × 106 mm 4 + 2
*1
+
(20 mm)(160 mm)3 !
12
"
#
or I y = 17.53 × 106 mm 4
= 17.5333 × 106 mm 4
and
k y2 =
Iy
Atotal
=
17.5333 × 106 mm 4
12, 410 mm 2
or
k y = 37.6 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1443
!
PROBLEM 9.50
Two channels are welded to a rolled W section as shown.
Determine the moments of inertia and the radii of gyration of the
combined section with respect to the centroidal x and y axes.
SOLUTION
W section:
A = 9.12 in.2
I x = 110 in.4
I y = 37.1 in.4
Channel:
A = 3.37 in.2
I x = 1.31 in.4
I y = 32.5 in.4
Atotal = AW + 2Achan
= 9.12 + 2(3.37) = 15.86 in.2
Now
where
I x = ( I x ) W + 2( I x )chan
( I x )chan = I xchan + Ad 2
= 1.31 in.4 + (3.37 in.2 )(4.572 in.)2 = 71.754 in.4
Then
I x = (110 + 2 × 71.754) in.4 = 253.51 in.4
and
k x2 =
Also
I y = ( I y ) W + 2( I y )chan
Ix
253.51 in.4
=
Atotal 15.86 in.2
= (37.1 + 2 × 32.5) in.4 = 102.1 in.4
and
k y2 =
Iy
Atotal
=
102.1 in.4
15.86 in.2
I x = 254 in.4
k x = 4.00 in.
I y = 102.1 in.4
k y = 2.54 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1444
!
PROBLEM 9.51
To form a reinforced box section, two rolled W sections and two plates
are welded together. Determine the moments of inertia and the radii of
gyration of the combined section with respect to the centroidal axes
shown.
SOLUTION
W section:
A = 5880 mm 2
I x = 15.4 × 106 mm 4
I y = 45.8 × 106 mm 4
Note:
Atotal = 2 AW + 2 Aplate
= 2(5880 mm 2 ) + 2(203 mm)(6 mm)
= 14,196 mm 2
Now
where
I x = 2( I x ) W + 2( I x )plate
(I x )W
203
!
mm !
= I x + Ad = 15.4 × 10 mm + (5880 mm )
" 2
#
2
6
4
2
2
= 75.9772 × 106 mm 4
( I x )plate = I xplate + Ad 2
1
(203 mm)(6 mm)3 + (1218 mm 2 )[(203 + 3) mm]2
12
= 51.6907 × 106 mm 4
=
Then
I x = [2(75.9772) + 2(51.6907)] × 106 mm 4
= 255.336 × 106 mm 4
and
k x2 =
or
I x = 255 × 106 mm 4
or
k x = 134.1 mm
Ix
255.336 × 106 mm 4
=
Atotal
14,196 mm 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1445
PROBLEM 9.51 (Continued)
also
where
I y = 2( I y ) W + 2( I y ) plate
(I y )W = I y
( I y ) plate =
Then
1
(6 mm)(203 mm)3 = 4.1827 × 106 mm 4
12
I y = [2(45.8) + 2(4.1827)] × 106 mm 4
= 99.9654 × 106 mm 4
or I y = 100.0 × 106 mm 4
and
k y2 =
Iy
Atotal
=
99.9654 × 106 mm 4
14,196 mm 2
or
k y = 83.9 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1446
PROBLEM 9.52
Two channels are welded to a d × 12-in. steel plate as shown.
Determine the width d for which the ratio I x /I y of the centroidal
moments of inertia of the section is 16.
SOLUTION
Channel:
A = 4.48 in.2
I x = 67.3 in.4
I y = 2.27 in.4
Now
I x = 2( I x )C + ( I x )plate
1
(d in.)(12 in.)3
12
= (134.6 + 144d )in.4
= 2(67.3 in.4 ) +
and
where
I y = 2( I y )C + ( I y ) plate
( I y )C = I y + Ad 2
$& d
' %
+ 0.634 ! in.)
( I y )C = 2.27 in. + (4.48 in. ) (
# +
*" 2
4
( I y ) plate
Then
2
2
= (1.1200d 2 + 2.84032d + 4.07076) in.4
1
= (12 in.)(d in.)3 = d 3 in.4
12
I y = [2(1.1200d 2 + 2.84032d + 4.07076) + d 3 ] in.4
= ( d 3 + 2.240d 2 + 5.68064d + 8.14152) in.4
Now
or
Ix
= 16: (134.6 + 144d ) = 16( d 3 + 2.240d 2 + 5.68064d + 8.14152)
Iy
d 3 + 2.240d 2 − 3.31936d − 0.27098 = 0
Solving yields d = 1.07669 in. (the other roots are negative).
d = 1.077 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1447
PROBLEM 9.53
Two L76 × 76 × 6.4-mm angles are welded to a C250 × 22.8
channel. Determine the moments of inertia of the combined
section with respect to centroidal axes respectively parallel and
perpendicular to the web of the channel.
SOLUTION
A = 929 mm 2
Angle:
I x = I y = 0.512 × 106 mm 4
A = 2890 mm 2
Channel:
I x = 0.945 × 106 mm 4
I y = 28.0 × 106 mm 4
First locate centroid C of the section
Then
A, mm 2
y , mm
yA, mm3
Angle
2(929) = 1858
21.2
39,389.6
Channel
2890
−16.1
−46,529
Σ
4748
−7139.4
Y Σ A = Σ y A: Y (4748 mm 2 ) = −7139.4 mm3
Y = −1.50366 mm
or
Now
where
I x = 2( I x ) L + ( I x )C
( I x ) L = I x + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(21.2 + 1.50366) mm]2
= 0.990859 × 106 mm 4
( I x )C = I x + Ad 2 = 0.949 × 106 mm 4 + (2890 mm 2 )[(16.1 − 1.50366) mm]2
= 1.56472 × 106 mm 4
Then
I x = [2(0.990859) + 1.56472 × 106 mm 4
or
I x = 3.55 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1448
PROBLEM 9.53 (Continued)
Also
where
I y = 2( I y ) L + ( I y )C
( I y ) L = I y + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(127 − 21.2) mm]2
= 10.9109 × 106 mm 4
( I y )C = I y
Then
I y = [2(10.9109) + 28.0] × 106 mm 4
or
I y = 49.8 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1449
PROBLEM 9.54
Two L4 × 4 × 12 -in. angles are welded to a steel plate as shown.
Determine the moments of inertia of the combined section with
respect to centroidal axes respectively parallel and perpendicular
to the plate.
SOLUTION
1
For 4 × 4 × -in. angle:
2
A = 3.75 in.2 , I x = I y = 5.52 in.4
YA = Σ y A
Y (12.5 in.2 ) = 33.85 in.3
Y = 2.708 in.
Area, in.2
y in.
yA, in.3
Plate
(0.5)(10) = 5
5
25
Two angles
2(3.75) = 7.5
1.18
8.85
Section
Σ
12.5
33.85
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1450
PROBLEM 9.54 (Continued)
Entire section:
$1
%
I x = Σ( I x′ + Ad 2 ) = ( (0.5)(10)3 + (0.5)(10)(2.292) 2 ) + 2[5.52 + (3.75)(1.528) 2 ]
*12
+
= 41.667 + 26.266 + 1604 + 17.511 = 96.48 in.4
Iy =
I x = 96.5 in.4
1
(10)(0.5)3 + 2[5.52 + (3.75)(1.43) 2 ]
12
= 0.104 + 11.04 + 15.367 = 26.51 in.4
I y = 26.5 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1451
PROBLEM 9.55
The strength of the rolled W section shown is increased by welding a channel
to its upper flange. Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.
SOLUTION
A = 14, 400 mm 2
W Section:
I x = 554 × 106 mm 4
I y = 63.3 × 106 mm 4
A = 2890 mm 2
Channel:
I x = 0.945 × 106 mm 4
I y = 28.0 × 106 mm 4
First locate centroid C of the section.
W Section
Channel
Σ
Then
A, mm2
y , mm
yA, mm3
14400
−231
−3326400
2890
49.9
144211
17290
−3182189
Y Σ A = Σ y A: Y (17, 290 mm 2 ) = −3,182,189 mm3
Y = −184.047 mm
or
Now
where
I x = ( I x ) W + ( I x )C
( I x ) W = I x + Ad 2
= 554 × 106 mm 4 + (14, 400 mm 2 )(231 − 184.047) 2 mm 2
= 585.75 × 106 mm 4
( I x )C = I x − Ad 2
= 0.945 × 106 mm 4 + (2890 mm 2 )(49.9 + 184.047) 2 mm 2
= 159.12 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1452
PROBLEM 9.55 (Continued)
Then
also
I x = (585.75 + 159.12) × 106 mm 4
or
I x = 745 × 106 mm 4
or
I y = 91.3 × 106 mm 4
I y = ( I y ) W + ( I y )C
= (63.3 + 28.0) × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1453
PROBLEM 9.56
Two L5 × 3 × 12 -in. angles are welded to a 12 -in. steel plate.
Determine the distance b and the centroidal moments of inertia I x
and I y of the combined section, knowing that I y = 4 I x .
SOLUTION
A = 3.75in.2
Angle:
I x = 9.43 in.4
I y = 2.55 in.4
First locate centroid C of the section.
Area, in.2
y , in.
yA, in.3
Angle
2(3.75) = 7.50
1.74
13.05
Plate
(10)(0.5) = 5
−0.25
−1.25
Σ
Then
12.50
11.80
Y Σ A = Σ yA: Y (12.50 in.2 ) = 11.80 in.3
Y = 0.944 in.
or
Now
where
I x = 2( I x )angle + ( I x ) plate
( I x )angle = I x + Ad 2 = 9.43 in.4 + (3.75 in.2 )[(1.74 − 0.944) in.]2
( I x ) plate
Then
= 11.8061 in.4
1
= I x + Ad 2 = (10 in.)(0.5 in.)3 + (5 in.2 )[(0.25 + 0.944) in.]2
12
= 7.2323 in.4
I x = [2(11.8061) + 7.2323] in.4 = 30.8445 in.4
or
I x = 30.8 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1454
PROBLEM 9.56 (Continued)
We have
I y = 4 I x = 4(30.8445 in.4 ) = 123.378 in.4
or
Now
where
I y = 2( I y )angle + ( I y )plate
( I y )angle = I y + Ad 2 = 2.55 in.4 + (3.75 in.2 )[(b − 0.746) in.]2
( I y ) plate =
Then
I y = 123.4 in.4
1
(0.5 in.)(10 in.)3 = 41.6667 in.4
12
123.378 in.4 = 2[2.55 + 3.75(b − 0.746)2 ] in.4 + 41.6667 in.4
or
b = 3.94 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1455
PROBLEM 9.57
The panel shown forms the end of a trough that is filled with water
to the line AA′. Referring to Section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).
SOLUTION
From Section 9.2:
,
,
R = γ y dA, M AA′ = γ y 2 dA
Let yP = distance of center of pressure from AA′. We must have
,
,
2
M AA′ γ y dA I AA′
RyP = M AA′ : yP =
=
=
R
γ y dA y A
(1)
For semicircular panel:
I AA′ =
π
8
r4
y=
4r
3π
A=
π
2
r2
π
r4
I AA′
8
yP =
=
yA & 4r ' π 2
! r
" 3π # 2
yP =
3π
r
16
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1456
PROBLEM 9.58
The panel shown forms the end of a trough that is filled with water
to the line AA′. Referring to Section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).
SOLUTION
See solution of Problem 9.57 for derivation of Eq. (1):
yP =
I AA′
yA
(1)
For quarter ellipse panel:
I AA′ =
π
16
ab3
y=
4b
3π
A=
π
4
ab
π
ab3
I AA′
16
yP =
=
yA & 4b '& π '
ab !
!
" 3π #" 4 #
yP =
3π
b
16
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1457
PROBLEM 9.59
The panel shown forms the end of a trough that is filled with water
to the line AA′. Referring to Section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text:
yP =
Now
I AA′
yA
YA = Σ yA
4 &1
h
'
(2b × h) + h × 2b × h !
2
3 "2
#
7 2
= bh
3
=
and
I AA′ = ( I AA′ )1 + ( I AA′ ) 2
where
1
2
( I AA′ )1 = (2b)(h)3 = bh3
3
3
1
&1
'& 4 '
(2b)( h)3 +
× 2b × h ! h !
36
"2
#" 3 #
11
= bh3
6
2
( I AA′ ) 2 = I x + Ad 2 =
Then
Finally,
I AA′ =
2 3 11 3 5 3
bh + bh = bh
3
6
2
5 3
bh
yP = 2
7 2
bh
3
or
yP =
15
h
14
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1458
PROBLEM 9.60*
The panel shown forms the end of a trough that is filled with water
to the line AA′. Referring to Section 9.2, determine the depth of the
point of application of the resultant of the hydrostatic forces acting
on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text:
For a parabola:
Now
By observation
yP =
I AA′
yA
y=
2
h
5
A=
4
ah
3
1
dI AA′ = (h − y )3 dx
3
y=
h 2
x
a2
3
So that
Then
dI AA′ =
1&
1 h3 2
h
'
(a − x 2 )3 dx
h − 2 x 2 ! dx =
6
3"
3
a
a
#
1 h 6
(a − 3a 4 x 2 + 3a 2 x 4 − x 6 )dx
=
6
3a
I AA′ = 2
,
a
0
1 h3 6
(a − 3a 4 x 2 + 3a 2 x 4 − x 6 )dx
6
3a
a
2 h3 $ 6
3
1 %
a x − a 4 x3 + a 2 x5 − x 7 )
3 a 6 (*
5
7 +0
32 3
=
ah
105
=
Finally,
32 3
ah
105
yp =
2
4
h × ah
5
3
or
yP =
4
h
7
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1459
PROBLEM 9.61
The cover for a 0.5-m-diameter access hole in a water storage
tank is attached to the tank with four equally spaced bolts as
shown. Determine the additional force on each bolt due to the
water pressure when the center of the cover is located 1.4 m
below the water surface.
SOLUTION
From Section 9.2:
R = γ yA
yP =
I AA′
yA
where R is the resultant of the hydrostatic forces acting on the cover
and yP is the depth to the point of application of R.
Recalling that γ = p ⋅ y , we have
R = (103 kg/m3 × 9.81 m/s 2 )(1.4 m)[π (0.25 m) 2 ]
= 2696.67 N
I AA′ = I x + Ay 2 =
Also
π
(0.25 m) 4 + [π (0.25 m)2 ](1.4 m) 2
4
= 0.387913 m 4
yP =
Then
0.387913 m 4
= 1.41116 m
(1.4 m)[π (0.25 m) 2 ]
Now note that symmetry implies
FA = FB
FC = FD
Next consider the free-body of the cover.
We have
ΣM CD = 0: [2(0.32 m) sin 45°](2 FA )
− [0.32 sin 45° − (1.41116 − 1.4)] m
× (2696.67 N) = 0
or
Then
or
FA = 640.92 N
ΣFz = 0: 2(640.92 N) + 2 FC − 2696.67 N = 0
FC = 707.42 N
FA = FB = 641 N
FC = FD = 707 N
and
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1460
PROBLEM 9.62
A vertical trapezoidal gate that is used as an automatic valve is held
shut by two springs attached to hinges located along edge AB.
Knowing that each spring exerts a couple of magnitude 1470 N · m,
determine the depth d of water for which the gate will open.
SOLUTION
From Section 9.2:
R = γ yA
yP =
I SS ′
yA
where R is the resultant of the hydrostatic forces acting on the gate
and yP is the depth to the point of application of R. Now
&1
'
&1
'
× 1.2 m × 0.51 m ! + [( h + 0.34) m] × 0.84 × 0.51 m !
"2
#
"2
#
3
= (0.5202h + 0.124848) m
yA = Σ yA = [(h + 0.17) m]
Recalling that γ = py, we have
R = (103 kg/m3 × 9.81 m/s 2 )(0.5202h + 0.124848) m3
= 5103.162(h + 0.24) N
Also
where
I SS ′ = ( I SS ′ )1 + ( I SS ′ ) 2
( I SS ′ )1 = I x + Ad 2
=
1
&1
'
(1.2 m)(0.51 m)3 +
× 1.2 m × 0.51 m ! [(h + 0.17) m]2
36
2
"
#
= [0.0044217 + 0.306(h + 0.17)2 ] m 4
= (0.306h2 + 0.10404h + 0.0132651) m 4
( I SS ′ ) 2 = I X 2 + Ad 2
=
1
&1
'
(0.84 m)(0.51 m)3 +
× 0.84 m × 0.51 m ! [(h + 0.34) m]2
36
"2
#
= [0.0030952 + 0.2142( h + 0.34) 2 ] m 4
= (0.2142h 2 + 0.145656h + 0.0278567) m 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1461
PROBLEM 9.62 (Continued)
I SS ′ = ( I SS ′ )1 + ( I SS ′ ) 2
Then
= (0.5202h 2 + 0.249696h + 0.0411218) m 4
yP =
and
=
(0.5202h2 + 0.244696h + 0.0411218) m 4
(0.5202h + 0.124848) m3
h 2 + 0.48h + 0.07905
m
h + 0.24
For the gate to open, require that
ΣM AB : M open = ( yP − h) R
Substituting
or
2940 N ⋅ m =
& h 2 + 0.48h + 0.07905
'
− h !! m × 5103.162( h + 0.24) N
h + 0.24
"
#
5103.162(0.24h + 0.07905) = 2940
or
h = 2.0711 m
Then
d = (2.0711 + 0.79) m
or
d = 2.86 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1462
PROBLEM 9.63*
Determine the x coordinate of the centroid of the volume shown.
(Hint: The height y of the volume is proportional to the x coordinate;
consider an analogy between this height and the water pressure on a
submerged surface.)
SOLUTION
y=
First note that
Now
where
Then
,
h
x
a
,
x dV = xEL dV
xEL = x dV = y dA =
&h '
x ! dA
"a #
, x ( x dA) = , x dA = ( I )
x=
, x dA , x dA ( xA)
2
h
a
z A
h
a
A
where ( I z ) A and ( x A ) A pertain to area.
OABC: ( I z ) A is the moment of inertia of the area with respect to the z axis, x A is the x coordinate
of the centroid of the area, and A is the area of OABC. Then
( I z ) A = ( I z ) A1 + ( I z ) A2
1
1
= (b)(a)3 + (b)( a)3
3
12
5 3
= ab
12
and
( xA) A = Σ xA
$& a '
% $& a '& 1
'%
= ( ! ( a × b) ) + ( ! × a × b ! )
#+
*" 2 #
+ *" 3 #" 2
2
= a 2b
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1463
PROBLEM 9.63* (Continued)
Finally,
x=
5 3
ab
12
2 a 2b
3
or
5
x= a
8
Analogy with hydrostatic pressure on a submerged plate:
Recalling that P = γ y, it follows that the following analogies can be established.
Height y ~ P
dV = y dA ~ p dA = dF
xdV = x( y dA) ~ y dF = dM
Recalling that
It can then be concluded that
yP =
&
dM '
Mx
!
=
R
dF !!
"
#
,
,
x ~ yP
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1464
!
PROBLEM 9.64*
Determine the x coordinate of the centroid of the volume shown;
this volume was obtained by intersecting an elliptic cylinder with
an oblique plane. (Hint: The height y of the volume is
proportional to the x coordinate; consider an analogy between this
height and the water pressure on a submerged surface.)
SOLUTION
Following the “Hint,” it can be shown that (see solution to Problem 9.63)
x=
(I z ) A
x
( xA) A
where ( I z ) A and ( xA) A are the moment of inertia and the first moment of the area, respectively, of the
elliptical area of the base of the volume with respect to the z axis. Then
( I z ) A = I z + Ad 2
π
(39 mm)(64 mm)3 + [π (64 mm)(39 mm)](64 mm) 2
4
= 12.779520π × 106 mm 4
=
( xA) A = (64 mm)[π (64 mm)(39 mm)]
= 0.159744π × 106 mm3
Finally
x=
12.779520π × 106 mm 4
0.159744π × 106 mm 4
or
x = 80.0 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1465
!
PROBLEM 9.65*
Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a force P at
the centroid C of the area and two couples. The force P is perpendicular to the area and is of magnitude
P = γ Ay sin θ , where γ is the specific weight of the liquid, and the couples are M x′ = (γ I x′ sin θ )i and
M y′ = (γ I x′y ′ sin θ ) j, where I x′y′ = , x′y ′ dA (see Section 9.8). Note that the couples are independent of the
depth at which the area is submerged.
SOLUTION
The pressure p at an arbitrary depth ( y sin θ ) is
p = γ ( y sin θ )
so that the hydrostatic force dF exerted on an infinitesimal area dA is
dF = (γ y sin θ )dA
Equivalence of the force P and the system of infinitesimal forces dF requires
,
,
,
ΣF : P = dF = γ y sin θ dA = γ sin θ y dA
or
P = γ Ay sin θ
Equivalence of the force and couple (P, M x′ + M y′ ) and the system of infinitesimal hydrostatic
forces requires
,
ΣM x : − yP − M x′ = (− y dF )
Now
,
,
,
− y dF = − y (γ y sin θ ) dA = −γ sin θ y 2 dA
= −(γ sin θ ) I x
Then
or
− yP − M x′ = −(γ sin θ ) I x
M x′ = (γ sin θ ) I x − y (γ Ay sin θ )
= γ sin θ ( I x − Ay 2 )
or
M x′ = γ I x′ sin θ
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1466
PROBLEM 9.65* (Continued)
,
ΣM y : xP + M y′ = x dF
Now
, x dF = , x(γ y sin θ )dA = γ sin θ , xy dA
= (γ sin θ ) I xy
Then
or
(Equation 9.12)
xP + M y′ = (γ sin θ ) I xy
M y′ = (γ sin θ ) I xy − x (γ Ay sin θ )
= γ sin θ ( I xy − Ax y )
or
M y′ = γ I x′y′ sin θ
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1467
!
PROBLEM 9.66*
Show that the resultant of the hydrostatic forces acting on a
submerged plane area A is a force P perpendicular to the area and
of magnitude P = γ Ay sin θ = pA, where γ is the specific weight of
the liquid and p is the pressure at the centroid C of the area. Show
that P is applied at a Point CP, called the center of pressure, whose
coordinates are xP = I xy /Ay and yP = I x /Ay , where I xy = , xy dA
(see Section 9.8). Show also that the difference of ordinates yP − y
is equal to k x2′ / y and thus depends upon the depth at which the area
is submerged.
SOLUTION
The pressure P at an arbitrary depth ( y sin θ ) is
P = γ ( y sin θ )
so that the hydrostatic force dP exerted on an infinitesimal area dA is
dP = (γ y sin θ )dA
The magnitude P of the resultant force acting on the plane area is then
,
,
,
P = dP = γ y sin θ dA = γ sin θ y dA
= γ sin θ ( yA)
Now
p = γ y sin θ
P = pA
Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then
requires
,
ΣM x : − yP P = − y dP
Now
, y dP = , y(γ y sin θ )dA = γ sin θ , y dA
2
= (γ sin θ ) I x
Then
or
yP P = (γ sin θ ) I x
yP =
(γ sin θ ) I x
γ sin θ ( yA)
or
yP =
Ix
Ay
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1468
PROBLEM 9.66* (Continued)
,
ΣM y : xP P = x dP
Now
, x dP = , x(γ y sin θ )dA = γ sin θ , xy dA
= (γ sin θ ) I xy
Then
or
(Equation 9.12)
xP P = (γ sin θ ) I xy
xP =
(γ sin θ ) I xy
γ sin θ ( yA)
or
Now
I x = I x′ + A y 2
From above
I x = ( A y ) yP
By definition
I x′ = k x′ A
Substituting
( Ay ) yP = k x2′ A + Ay 2
xP =
I xy
!
Ay
2
yP − y =
Rearranging yields
k x2′
y
Although k x′ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on
the depth.
( yP − y ) = f (depth) !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1469
!
PROBLEM 9.67
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.
SOLUTION
y = a 1−
First note
=
x2
4a 2
1
4a 2 − x 2
2
We have
dI xy = dI x′y′ + xEL yEL dA
where
dI x′y′ = 0 (symmetry) xEL = x
yEL =
1
1
4a 2 − x 2
y=
2
4
dA = ydx =
Then
1
4a 2 − x 2 dx
2
,
2a
I xy = dI xy =
,
=
1
8
=
a4
8
0
,
x
&1
'& 1
4a 2 − x 2 !
4a 2 − x 2
4
2
"
#"
2a
0
'
! dx
#
2a
1$
1 %
(4a 2 x − x3 )dx = ( 2a 2 x 2 − x 4 )
8*
4 +0
1 4%
$
2
( 2(2) − 4 (2) )
*
+
or
I xy =
1 4
a
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1470
PROBLEM 9.68
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.
SOLUTION
At
x = a, y = b: a = kb 2
or
k=
a
b2
Then
x=
a 2
y
b2
We have
dI xy = dI x′y′ + xEL yEL dA
where
dI x′y′ = 0 (symmetry) xEL =
yEL = y dA = x dy =
Then
,
I xy = dI xy =
=
,
1
a
x = 2 y2
2
2b
a 2
y dy
b2
b&
0
a 2' & a 2 '
y ! ( y ) 2 y dy !
2
" 2b
# "b
#
a2
2b 4
,
b
0
y 5 dy =
a2
2b4
b
$1 6 %
(6 y )
*
+0
or
I xy =
1 2 2
a b
12
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1471
PROBLEM 9.69
Determine by direct integration the product of inertia of the given area with respect
to the x and y axes.
SOLUTION
First note that
h
y=− x
b
Now
dI xy = dI x′y′ + xEL yEL dA
where
dI x′y′ = 0 (symmetry)
xEL = x
yEL =
1
1h
y=−
x
2
2b
h
dA = y dx = − x dx
b
Then
,
I xy = dI xy =
=
,
& 1 h '& h
'
x −
x ! − x dx !
−b " 2 b #" b
#
0
1 h2
2 b2
,
0
−b
0
x3 dx =
1 h2 $ 1 4 %
x
2 b 2 (* 4 )+ −b
or
1
I xy = − b 2 h 2
8
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1472
PROBLEM 9.70
Determine by direct integration the product of inertia of the given area
with respect to the x and y axes.
SOLUTION
y = h1 + ( h2 − h1 )
x
a
dI xy = dI x′y′ + xEL yEL dA
dI x′y′ = 0 (by symmetry)
xEL = x
yEL =
,
I xy = dI xy −
=
1
2
=
1
2
,
,
a
0
,
a
0
x
1
y
2
dA = y dx
1
&1 '
y ! y dx =
2
"2 #
,
a
0
xy 2 dx
2
x%
$
x ( h1 + (h2 − h1 ) ) dx
a+
*
a$
0
3
x2
2
2 x %
( h1 x + 2h1 (h2 − h1 ) + (h2 − h1 ) 2 ) dx
a
a +
*
1 $ 2 a2
a3
a4 %
+ 2h1 ( h2 − h1 ) + (h2 − h1 ) 2 2 )
( h1
2*
2
3a
4a +
%
1 $ a2 2
2
1
1
1
= ( h12
+ h1h2 a 2 − h12 a 2 + h22 a 2 − h2 h1a + h12 a 2 )
2*
2 3
3
4
2
4
+
2
a $ 2&1 2 1'
& 2 1 ' 2 & 1 '%
=
− + ! + h1h2
− ! + h2
( h1
!)
2 * "2 3 4#
"3 2#
" 4 #+
=
=
a2
2
$ 2& 1 '
&1'
& 1 '%
+ h1h2 ! + h22
( h1
!
!)
"6#
" 4 #+
* " 12 #
I xy =
a2 2
h1 + 2h1h2 + 3h22
24
(
)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1473
PROBLEM 9.70 (Continued)
The following table is provided for the convenience of the instructor, as many problems in this and the next
lesson are related.
Type of
Problem
Compute
I x and I y
Figure 9.12
Compute
I xy
9.67
9.72
9.73
9.74
9.75
9.78
9.79
9.80
9.81
9.83
9.82
9.84
9.85
9.86
9.87
9.89
9.88
9.90
9.91
9.92
9.93
9.95
9.94
9.96
9.97
9.98
9.100
9.101
9.103
9.106
I x′, I y′, I x′y′
by equations
Principal axes
by equations
Figure 9.13B
Figure 9.13A
I x′, I y′, I x′y′
by
Mohr's circle
Principal axes
by Mohr’s
circle
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1474
PROBLEM 9.71
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.
SOLUTION
Dimensions in mm
I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3
We have
Now symmetry implies
( I xy ) 2 = 0
and for the other rectangles
I xy = I x′y′ + x yA
I x′y′ = 0 (symmetry)
where
I xy = ( x yA)1 + ( x yA)3
Thus
A, mm 2
x , mm
y , mm
x yA, mm 4
1
10 × 80 = 800
−55
20
−880, 000
3
10 × 80 = 800
55
−20
−880, 000
−1, 760,000
Σ
I xy = −1.760 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1475
!
PROBLEM 9.72
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
Dimensions in mm
I xy = ( I xy )1 − ( I xy ) 2 − ( I xy )3
We have
Now symmetry implies
and for each triangle
( I xy )1 = 0
I xy = I x′y′ + x yA
1 2 2
b h for both triangles. Note that the sign of I x′y′
where, using the results of Sample Problem 9.6, I x′y′ = − 72
is unchanged because the angles of rotation are 0° and 180° for triangles 2 and 3, respectively. Now
A, mm 2
x , mm
x yA, mm 4
2
1
(120)(60) = 3600
2
−80
−40
11.520 × 106
3
1
(120)(60) = 3600
2
80
40
11.520 × 106
23.040 × 106
Σ
Then
y , mm
- $ 1
.
%
I xy = − /2 ( − (120 mm) 2 (60 mm) 2 ) + 23.040 × 106 mm 4 0
72
+
1 *
2
or
I xy = −21.6 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1476
!
PROBLEM 9.73
Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2
For each semicircle
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
Thus
I xy = Σ x yA
A, in.2
1
2
π
2
π
2
x , in.
(6) 2 = 18π
−3
(6) 2 = 18π
3
Σ
y , in.
−
8
π
8
π
xyA
432
432
864
I xy = 864 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1477
!
PROBLEM 9.74
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy ) 2
We have
For each rectangle
I xy = I x′y′ + x yA
I x′y′ = 0 (symmetry)
and
I xy = Σ x yA
Thus
A, in.2
x , in.
y , in.
x yA, in.4
1
3 × 0.25 = 0.75
−0.520
0.362
−0.141180
2
0.25 × 1.75 = 0.4375
0.855
−0.638
−0.238652
−0.379832
Σ
I xy = −0.380 in.4
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1478
!
PROBLEM 9.75
Using the parallel-axis theorem, determine the product of inertia
of the area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3
We have
( I xy )1 = 0
Now symmetry implies
and for the other rectangles
I xy = I x′y′ + x yA
where
I x′y′ = 0 (symmetry)
Thus
I xy = ( x yA) 2 + ( x yA)3
A, mm 2
X , mm
y , mm
x yA, mm 4
2
8 × 32 = 256
–46
–20
235,520
3
8 × 32 = 256
46
20
235,520
Σ
471,040
I xy = 471 × 103 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1479
!
PROBLEM 9.76
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3
We have
Now, symmetry implies
( I xy )1 = 0
I xy = I x′y′ + x yA
and for each triangle
1 2 2
where, using the results of Sample Problem 9.6, I x′y′ = − 72
b h for both triangles. Note that the sign of I x′y′
is unchanged because the angles of rotation are 0° and 180° for triangles 2 and 3, respectively.
Now
A, in.2
x , in.
y , in.
x yA, in.4
2
1
(9)(15) = 67.5
2
–9
7
–4252.5
3
1
(9)(15) = 67.5
2
9
–7
–4252.5
Σ
Then
–8505
$ 1
%
I xy = 2 ( − (9 in.) 2 (15 in.) 2 ) − 8505 in.4
72
*
+
= −9011.25 in.4
or
I xy = −9010 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1480
!
PROBLEM 9.77
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3
For each rectangle
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
I xy = Σ x yA
Thus
A, in.2
x , in.
y , in.
1
3.6 × 0.5 = 1.8
0.888
–2.00
2
0.5 × 3.8 = 1.9
–0.662
0.15
–0.18867
3
1.3 × 1.0 = 1.3
–0.262
2.55
–0.86853
Σ
x yA, in.4
–3.196
–4.25320
I xy = −4.25 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1481
!
PROBLEM 9.78
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2
For each rectangle
I xy = I x′y′ + x yA
and
I x′y′ = 0 (symmetry)
Thus
I xy = Σ x yA
x yA, mm 4
A, mm 2
x , mm
y , mm
1
76 × 12.7 = 965.2
–19.1
–37.85
697,777
2
12.7 × (127 − 12.7) = 1451.61
12.55
25.65
467,284
Σ
1,165,061
I xy = 1.165 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1482
!
PROBLEM 9.79
Determine for the quarter ellipse of Problem 9.67 the moments of inertia
and the product of inertia with respect to new axes obtained by rotating
the x and y axes about O (a) through 45° counterclockwise, (b) through
30° clockwise.
SOLUTION
From Figure 9.12:
Ix =
=
Iy =
=
From Problem 9.67:
First note
π
16
π
8
a4
π
16
π
2
I xy =
(2a )( a)3
(2a )3 (a)
a4
1 4
a
2
1
1&π 4 π 4 ' 5
(I x + I y ) =
a + a ! = π a4
2
2" 8
2 # 16
1
1&π 4 π 4 '
3
(I x − I y ) =
a − a ! = − π a4
2
2" 8
2 #
16
Now use Equations (9.18), (9.19), and (9.20).
Equation (9.18):
I x′ =
=
Equation (9.19):
I y′ =
=
Equation (9.20):
I x′y′ =
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
5
3
1
π a 4 − π a 4 cos 2θ − a 4 sin 2θ
16
16
2
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
5
3
1
π a 4 + π a 4 cos 2θ + a 4 sin 2θ
16
16
2
1
( I x − I y ) sin 2θ + I xy cos 2θ
2
=−
3
1
π a 4 sin 2θ + a 4 cos 2θ
16
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1483
PROBLEM 9.79 (Continued)
(a)
θ = +45°:
I x′ =
I y′ =
5
3
1
π a 4 − π a 4 cos 90° − a 4 sin 90°
16
16
2
or
I x′ = 0.482a 4
or
I y′ = 1.482a 4
5
3
1
π + π a 4 cos 90° + a 4
16
16
2
I x′y′ = −
3
1
π a 4 sin 90° + a 4 cos 90°
16
2
or I x′y′ = −0.589a 4
(b)
θ = −30° :
I x′ =
I y′ =
5
3
1
π a 4 − π a 4 cos(−60°) − a 4 sin( −60°)
16
16
2
or
I x′ = 1.120a 4
or
I y′ = 0.843a 4
or
I x′y′ = 0.760a 4
5
3
1
π a 4 + π a 4 cos(−60°) + a 4 sin( −60°)
16
16
2
I x′y′ = −
3
1
π a 4 sin(−60°) + a 4 cos(−60°)
16
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1484
!
PROBLEM 9.80
Determine the moments of inertia and the product of inertia of the
area of Problem 9.72 with respect to new centroidal axes obtained
by rotating the x and y axes 30° counterclockwise.
SOLUTION
From Problem 9.72:
I xy = −21.6 × 106 mm 4
Now
I x = ( I x )1 − ( I x ) 2 − ( I x )3
where
1
(240 mm)(160 mm)3
12
= 81.920 × 106 mm 4
( I x )1 =
( I x ) 2 = ( I x )3 =
1
(120 mm)(60 mm)3
36
$1
%
+ ( (120 mm)(60 mm) ) (40 mm) 2
*2
+
= 6.480 × 106 mm 4
Then
I x = [81.920 − 2(6.480)] × 106 mm 4 = 68.96 × 106 mm 4
Also
I y = ( I y )1 − ( I y ) 2 − ( I y )3
where
1
(160 mm)(240 mm)3 = 184.320 × 106 mm 4
12
1
( I y ) 2 = ( I y )3 = (60 mm)(120 mm)3
36
$1
%
+ ( (120 mm)(60 mm) ) (80 mm) 2
2
*
+
( I y )1 =
= 25.920 × 106 mm 4
Then
I y = [184.320 − 2(25.920)] × 106 mm 4 = 132.48 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1485
PROBLEM 9.80 (Continued)
Now
1
(I x + I y ) =
2
1
(I x − I y ) =
2
1
(68.96 + 132.48) × 106 = 100.72 × 106 mm 4
2
1
(68.96 − 132.48) × 106 = −31.76 × 106 mm 4
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [100.72 + (−31.76) cos 60° − (−21.6)sin 60°] × 106 mm 4
Ix =
or
Eq. (9.19):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
= [100.72 − ( −31.76) cos 60° + (−21.6) sin 60°] × 106 mm 4
I y′ =
or
Eq. (9.20):
I x′ = 103.5 × 106 mm 4
I y′ = 97.9 × 106 mm 4
1
( I x − I y ) sin 2θ + I xy sin 2θ
2
= [−31.76sin 60° + (−21.6) cos 60°] × 106 mm 4
I x′y′ =
or
I x′y′ = −38.3 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1486
!
PROBLEM 9.81
Determine the moments of inertia and the product of inertia of the
area of Problem 9.73 with respect to new centroidal axes obtained
by rotating the x and y axes 60° counterclockwise.
SOLUTION
From Problem 9.73:
I xy = 864 in.4
I x = ( I x )1 + ( I x ) 2
Now
π
(6 in.) 4
8
= 162π in.4
( I x )1 = ( I x ) 2 =
where
Then
I x = 2(162π in.4 ) = 324π in.4
Also
I y = ( I y )1 + ( I y ) 2
( I y )1 = ( I y ) 2 =
where
8
$π
%
(6 in.) 4 + ( (6 in.) 2 ) (3 in.) 2 = 324π in.4
*2
+
I y = 2(324π in.4 ) = 648π in.4
Then
Now
π
1
(I x + I y ) =
2
1
(I x − I y ) =
2
1
(324π + 648π ) = 486π in.4
2
1
(324π − 648π ) = −162π in.4
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [486π + (−162π ) cos120° − 864sin120°] in.4
I x′ =
or
I x′ = 1033 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1487
PROBLEM 9.81 (Continued)
Eq. (9.19):
Eq. (9.20):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
= [486π − (−162π ) cos120° + 864sin120°] in.4
I y′ =
or
I y′ = 2020 in.4
or
I x′y′ = −873 in.4
1
( I x − I y )sin 2θ + I xy cos 2θ
2
= [(−162π ) sin120° + 864cos120°] in.4
I x′y′ =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1488
!
PROBLEM 9.82
Determine the moments of inertia and the product of inertia
of the area of Problem 9.75 with respect to new centroidal
axes obtained by rotating the x and y axes 45° clockwise.
SOLUTION
From Problem 9.75:
I xy = 471, 040 mm 4
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
1
(100 mm)(8 mm)3
12
= 4266.67 mm 4
( I x )1 =
1
(8 mm)(32 mm)3 + [(8 mm)(32 mm)](20 mm) 2
12
= 124, 245.33
( I x ) 2 = ( I x )3 =
Then
I x = [4266.67 + 2(124, 245.33)] mm 4 = 252, 757 mm 4
Also
I y = ( I y )1 + ( I y ) 2 + ( I y )3
where
Then
Now
1
(8 mm)(100 mm)3 = 666, 666.7 mm 4
12
1
( I y ) 2 = ( I y )3 = (32 mm)(8 mm)3 + [(8 mm)(32 mm)](46 mm) 2
12
= 543,061.3 mm 4
( I y )1 =
I y = [666, 666.7 + 2(543,061.3)] mm 4 = 1, 752, 789 mm 4
1
(I x + I y ) =
2
1
(I x − I y ) =
2
1
(252, 757 + 1, 752, 789) mm 4 = 1, 002,773 mm 4
2
1
(252, 757 − 1, 752, 789) mm 4 = −750, 016 mm 4
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [1, 002,773 + (−750,016) cos(−90°) − 471, 040sin( −90°)]
I x′ =
or
I x′ = 1.474 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1489
!
PROBLEM 9.82 (Continued)
Eq. (9.19):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
= [1,002, 773 − (−750, 016) cos(−90°) + 471,040sin(−90°)]
I y′ =
or
Eq. (9.20):
I y′ = 0.532 × 106 mm 4
1
( I x − I y ) sin 2θ + I xy cos 2θ
2
= [( −750, 016) sin(−90°) + 471,040cos( −90°)]
I x′y′ =
or
I x′y′ = 0.750 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1490
!
PROBLEM 9.83
Determine the moments of inertia and the product of inertia of
the L3 × 2 × 14 -in. angle cross section of Problem 9.74 with
respect to new centroidal axes obtained by rotating the x and
y axes 30° clockwise.
SOLUTION
From Figure 9.13:
I x = 0.390 in.4
I y = 1.09 in.4
From Problem 9.74:
I xy = −0.37983 in.4
Now
1
(I x + I y ) =
2
1
(I x − I y ) =
2
1
(0.390 + 1.09) in.4 = 0.740 in.4
2
1
(0.390 − 1.09) in.4 = −0.350 in.4
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [0.740 + (−0.350) cos(−60°) − (−0.37983) sin( −60°)]
I x′ =
or
Eq. (9.19):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
= [0.740 − (−0.350) cos(−60°) + ( −0.37983)sin(−60°)]
I y′ =
or
Eq. (9.20):
I x′ = 0.236 in.4
I y′ = 1.244 in.4
1
( I x − I y )sin 2θ + I xy cos 2θ
2
= [(−0.350) sin(−60°) + (−0.37983) cos(−60°)]
I x′y′ =
or I x′y′ = 0.1132 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1491
!
PROBLEM 9.84
Determine the moments of inertia and the product of inertia of the
L127 ×76 × 12.7-mm angle cross section of Problem 9.78 with
respect to new centroidal axes obtained by rotating the x and y axes
45° counterclockwise.
SOLUTION
From Figure 9.13:
I x = 3.93 × 106 mm 4
I y = 1.06 × 106 mm 4
From Problem 9.78:
I xy = 1.165061 × 106 mm 4
Now
1
1
( I x + I y ) = (3.93 + 1.06) × 106 mm 4
2
2
= 2.495 × 106 mm 4
1
1
( I x − I y ) = (3.93 − 1.06) × 106 mm 4 = 1.435 × 106 mm 4
2
2
Using Eqs. (9.18), (9.19), and (9.20):
Eq. (9.18):
1
1
( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ
2
2
= [2.495 + 1.435cos 90° − 1.165061sin 90°] × 106 mm 4
I x′ =
or
Eq. (9.19):
1
1
( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ
2
2
= [2.495 − 1.435cos 90° + 1.165061sin 90°] × 106 mm 4
I y′ =
or
Eq. (9.20):
I x′ = 1.330 × 106 mm 4
I y′ = 3.66 × 106 mm 4
1
( I x − I y ) sin 2θ + I xy cos 2θ
2
= [(1.435sin 90° + 1.165061cos 90°] × 106 mm 4
I x′y′ =
or
I x′y′ = 1.435 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1492
!
PROBLEM 9.85
For the quarter ellipse of Problem 9.67, determine the orientation of the
principal axes at the origin and the corresponding values of the moments
of inertia.
SOLUTION
From Problem 9.79:
Ix =
Problem 9.67:
I xy =
Now Eq. (9.25):
π
8
a4
Iy =
2
a4
1 4
a
2
tan2θ m = −
2 I xy
Ix − I y
=−
2
π
8
=
Then
π
(
4
1
2
a4
)
a − 2 a4
π
8
= 0.84883
3π
2θ m = 40.326° and 220.326°
or
Also Eq. (9.27):
I max, min =
=
Ix + I y
2
θ m = 20.2° and 110.2°
2
±
& Ix − Iy '
2
! + I xy
2
"
#
1&π 4 π 4 '
a + a !
2" 8
2 #
2
$1 & π 4 π 4 % & 1 4 '
a − a ) +
a !
± (
2 + "2 #
*2" 8
2
= (0.981, 748 ± 0.772,644)a 4
or
I max = 1.754a 4
and
I min = 0.209a 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1493
!
PROBLEM 9.86
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments of
inertia.
Area of Problem 9.72.
SOLUTION
From Problem 9.80:
I x = 68.96 × 106 mm 4
I y = 132.48 × 106 mm 4
Problem 9.72:
Now Eq. (9.25):
I xy = −21.6 × 106 mm 4
tan2θ m = −
2 I xy
Ix − I y
=−
2( −21.6 × 106 )
(68.96 − 132.48) × 106
= −0.68010
Then
2θ m = −34.220° and 145.780°
or
Ix + Iy
θ m = −17.11° and 72.9°
2
& Ix − I y '
2
±
! + I xy
!
2
"
#
Also Eq. (9.27):
I max, min =
Then
2
$ 68.96 + 132.48
%
& 68.96 − 132.48 '
2)
(
21.6)
I max, min = (
±
+
−
× 106
!
(
)
2
2
"
#
*
+
2
= (100.72 ± 38.409) × 106 mm 4
or
and
I max = 139.1 × 106 mm 4
I min = 62.3 × 106 mm 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1494
!
PROBLEM 9.87
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments of
inertia.
Area of Problem 9.73.
SOLUTION
From Problem 9.81:
I x = 324π in.4
Problem 9.73:
I xy = 864 in.4
Now Eq. (9.25):
tan2θ m = −
2 I xy
Ix − I y
I y = 648π in.4
=−
2(864)
324π − 648π
= 1.69765
Then
2θ m = 59.500° and 239.500°
θ m = 29.7° and 119.7°
or
Also Eq. (9.27):
I max, min =
Then
I max, min =
Ix + Iy
2
2
& Ix − I y '
±
! + I xy2
!
2
"
#
2
324π + 648π
& 324π − 648π '
2
±
! + 864
2
2
"
#
= (1526.81 ± 1002.75) in.4
or
I max = 2530 in.4
and
I min = 524 in.4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1495
PROBLEM 9.88
For the area indicated, determine the orientation of the principal
axes at the origin and the corresponding values of the moments
of inertia.
Area of Problem 9.75.
SOLUTION
From Problem 9.82:
I x = 252, 757 mm 4
I y = 1, 752,789 mm 4
Problem 9.75:
Now Eq. (9.25):
I xy = 471, 040 mm 4
tan 2θ m = −
2 I xy
Ix − Iy
2(471, 040)
252, 757 − 1, 752, 789
= 0.62804
=−
Then
2θ m = 32.130° and 212.130°
or
Also Eq. (9.27):
I max, min =
Then
I max, min =
Ix − I y
2
θ m = 16.07° and 106.1°
2
& Ix − Iy '
±
! + I xy2
!
2
"
#
2
252, 757 + 1,752, 789
& 252, 757 − 1,752, 789 '
2
±
! + 471040
2
2
"
#
= (1,002, 773 ± 885, 665) mm 4
or
I max = 1.888 × 106 mm 4
and
I min = 0.1171 × 106 mm 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1496
PROBLEM 9.89
For the angle cross section indicated, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
The L3 × 2 × 14 -in. angle cross section of Problem 9.74.
SOLUTION
From Problem 9.83:
I x = 0.390 in.4
Problem 9.74:
I xy = −0.37983 in.4
Now Eq. (9.25):
tan 2θ m = −
2 I xy
Ix − I y
I y = 1.09 in.4
=−
2(−0.37983)
0.390 − 1.09
= −1.08523
Then
2θ m = −47.341° and 132.659°
or
Also Eq. (9.27):
I max, min =
Then
I max, min =
Ix + I y
2
θ m = −23.7° and 66.3°
2
& Ix − I y '
±
! + I xy2
!
2
"
#
2
0.390 + 1.09
& 0.390 − 1.09 '
2
±
! + (−0.37983)
2
2
"
#
= (0.740 ± 0.51650)2 in.4
or
I max = 1.257 in.4
and
I min = 0.224 in.4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1497
!
PROBLEM 9.90
For the angle cross section indicated, determine the orientation of
the principal axes at the origin and the corresponding values of the
moments of inertia.
The L127 × 76 × 12.7-mm angle cross section of Problem 9.78.
SOLUTION
From Problem 9.84:
I x = 3.93 × 106 mm 4
I y = 1.06 × 106 mm 4
Problem 9.78:
Now Eq. (9.25):
I xy = 1.165061 × 106 mm 4
tan 2θ m = −
2 I xy
Ix − Iy
=−
2(1.165061 × 106 )
(3.93 − 1.06) × 106
= −0.81189
Then
2θ m = −39.073° and 140.927°
or θ m = −19.54° and 70.5°
Also Eq. (9.27):
Then
I max, min =
I max, min
Ix + I y
2
2
& Ix − I y '
±
! + I xy2
!
2
"
#
2
$ 3.93 + 1.06
%
& 3.93 − 1.06 '
(
=
±
+ 1.1650612 ) × 106 mm 4
!
(
)
2
2
"
#
*
+
= (2.495 ± 1.84840) × 106 mm 4
or
I max = 4.34 × 106 mm 4
and
I min = 0.647 × 106 mm 4
By inspection, the a axis corresponds to I max and the b axis corresponds to I min .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1498
!
PROBLEM 9.91
Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67
the moments of inertia and the product of inertia with respect to new
axes obtained by rotating the x and y axes about O (a) through 45°
counterclockwise, (b) through 30° clockwise.
SOLUTION
Ix =
From Problem 9.79:
π
8
π
Iy =
2
a4
1 4
a
2
I xy =
Problem 9.67:
a4
The Mohr’s circle is defined by the diameter XY, where
X
Now
&π 4 1 4 '
&π 4 1 4'
a , a ! and Y
a ,− a !
2 #
2 #
"8
"2
I ave =
1
1&π 4 π 4 ' 5
(I x + I y ) =
a + a ! = π a 4 = 0.98175a 4
2
2" 8
2 # 16
R=
2
& Ix − Iy '
$ 1 & π 4 π 4 '% & 1 4 '
2
+
=
−
+
I
a
a
a
!
xy
(
)
2 !# + " 2 !#
*2 " 8
" 2 #
2
and
2
= 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
=−
2 I xy
Ix − I y
2
π
8
(
1
2
a4
4
)
a − 2 a4
π
= 0.84883
or
2θ m = 40.326°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1499
PROBLEM 9.91 (Continued)
α = 90° − 40.326°
Then
= 49.674°
β = 180° − (40.326° + 60°)
= 79.674°
(a)
θ = +45°:
I x′ = I ave − R cos α = 0.98175a 4 − 0.77264a 4 cos 49.674°
or
I x′ = 0.482a 4
I y′ = I ave + R cos α = 0.98175a 4 + 0.77264a 4 cos 49.674°
or
I y′ = 1.482a 4
or
I x′y′ = −0.589a 4
I x′y′ = − R sin α = −0.77264a 4 sin 49.674°
(b)
θ = −30°:
!
I x′ = I ave + R cos β = 0.98175a 4 + 0.77264a 4 cos 79.674°
or
I x′ = 1.120a 4
I y′ = I ave − R cos β = 0.98175a 4 − 0.77264a 4 cos 79.674°
or
I y′ = 0.843a 4
or
I x′y′ = 0.760a 4
I x′y′ = R sin β = 0.77264a 4 sin 79.674°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1500
!
PROBLEM 9.92
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the area of Problem 9.72 with respect to new centroidal
axes obtained by rotating the x and y axes 30° counterclockwise.
SOLUTION
I x = 68.96 × 106 mm 4
From Problem 9.80:
I y = 132.48 × 106 mm 4
I xy = −21.6 × 106 mm 4
Problem 9.72:
The Mohr’s circle is defined by the diameter XY, where X (68.96 × 106 , −21.6 × 106 )
and Y (132.48 × 106 , 21.6 × 106 ).
Now
and
I ave =
1
1
( I x − I y ) = (68.96 + 132.48) × 106 = 100.72 × 106 mm 4
2
2
2
2
- 1
.
4 $
4
$1
%
%
2
R = ( ( I x − I y ) ) + I xy = / ( (68.96 − 132.48) ) + (−21.6) 2 0 × 106 mm 4
*2
+
+
41 * 2
42
= 38.409 × 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−21.6 × 106 )
(68.96 − 132.48) × 106
= −0.68010
=−
or
Then
2θ m = −34.220°
α = 180° − (34.220° + 60°)
= 85.780°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1501
PROBLEM 9.92 (Continued)
Then
I x′ = I ave + R cos α = (100.72 + 38.409 cos85.780°) × 106
or
I x′ = 103.5 × 106 mm 4
I y′ = I ave − R cos α = (100.72 − 38.409 cos85.780°) × 106
or
I y′ = 97.9 × 106 mm 4
or
I x′y′ = −38.3 × 106 mm 4
I x′y′ = − R sin α = −(38.409 × 106 ) sin 85.780°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1502
PROBLEM 9.93
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the area of Problem 9.73 with respect to new centroidal
axes obtained by rotating the x and y axes 60° counterclockwise.
SOLUTION
I x = 324π in.4
From Problem 9.81:
I y = 648π in.4
I xy = 864 in.4
Problem 9.73:
The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864).
Now
I ave =
1
1
( I x + I y ) = (324π + 648π ) = 1526.81 in.4
2
2
2
and
$1
$1
%
R = ( ( I x − I y ) 2 + I xy2 = ( (324π + 648π ) ) + 8642
2
2
*
*
+
= 1002.75 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(864)
324π − 648π
= 1.69765
=−
or
Then
2θ m = 59.500°
α = 120° − 59.500°
= 60.500°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1503
PROBLEM 9.93 (Continued)
Then
!
I x′ = I ave − R cos α = 1526.81 − 1002.75cos 60.500°
or
I x′ = 1033 in.4
or
I y′ = 2020 in.4
or
I x′y′ = −873 in.4
I y = I ave + R cos α = 1526.81 + 1002.75cos 60.500° !
I x′y′ = − R sin α = −1002.75sin 60.500°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1504
!
PROBLEM 9.94
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the area of Problem 9.75 with respect to
new centroidal axes obtained by rotating the x and y axes
45° clockwise.
SOLUTION
From Problem 9.82:
I x = 252, 757 mm 4
I y = 1, 752,789 mm 4
Problem 9.75:
I xy = 471, 040 mm 4
The Mohr’s circle is defined by the diameter XY, where X (252,757; 471,040) and Y (1,752,789; −471, 040).
Now
1
1
( I x + I y ) = (252, 757 + 1, 752, 789)
2
2
= 1, 002, 773 mm 4
I ave =
2
and
$1
%
R = ( ( I x − I y ) ) + I xy2
*2
+
$1
%
= ( (252, 757 − 1,752, 789) 2 ) + 471, 0402
2
*
+
= 885,665 mm 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(471, 040)
252, 757 − 1, 752, 789
= 0.62804
=−
or
Then
2θ m = 32.130°
α = 180° − (32.130 + 90°)
= 57.870°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1505
PROBLEM 9.94 (Continued)
Then
I x′ = I ave + R cos α = 1,002, 773 + 885, 665cos 57.870°
I x′ = 1.474 × 106 mm 4
or
I y′ = I ave − R cos α = 1, 002,773 − 885, 665cos 57.870°
or
I y′ = 0.532 × 106 mm 4
or
I x′y′ = 0.750 × 106 mm 4
I x′y′ = R sin α = 885, 665sin 57.870°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1506
PROBLEM 9.95
Using Mohr’s circle, determine the moments of inertia and the product
of inertia of the L3 × 2 × 14 -in. angle cross section of Problem 9.74 with
respect to new centroidal axes obtained by rotating the x and y axes 30°
clockwise.
SOLUTION
From Problem 9.83:
I x = 0.390 in.4
I y = 1.09 in.4
Problem 9.74:
I xy = −0.37983 in.4
The Mohr’s circle is defined by the diameter XY, where X (0.390, − 0.37983) and Y (1.09, 0.37983).
Now
1
(I x + I y )
2
1
= (0.390 + 1.09)
2
= 0.740 in.4
I ave =
2
$1
%
R = ( ( I x − I y ) ) + I xy2
*2
+
and
2
$1
%
= ( (0.390 − 1.09) ) + (−0.37983) 2
*2
+
= 0.51650 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2( −0.37983)
0.390 − 1.09
= −1.08523
=−
or
Then
2θ m = −47.341°
α = 60° − 47.341° = 12.659°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1507
PROBLEM 9.95 (Continued)
Then
I x′ = I ave − R cos α = 0.740 − 0.51650 cos 12.659°
or
I x′ = 0.236 in.4
I y′ = I ave + R cos α = 0.740 + 0.51650cos 12.659°
or
I y = 1.244 in.4
or
I x′y′ = 0.1132 in.4
I x′y′ = R sin α = 0.51650 sin 12.659°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1508
!
PROBLEM 9.96
Using Mohr’s circle, determine the moments of inertia and the
product of inertia of the L127 × 76 × 12.7-mm angle cross section
of Problem 9.78 with respect to new centroidal axes obtained by
rotating the x and y axes 45° counterclockwise.
SOLUTION
From Problem 9.84:
I x = 3.93 × 106 mm 4
I y = 1.06 × 106 mm 4
Problem 9.78:
I xy = 1.165061 × 106 mm 4
The Mohr’s circle is defined by the diameter XY, where X (3.93 × 106 , 1.165061 × 106 ),
Y (1.06 × 106 , −1.165061 × 106 ).
Now
I ave =
1
1
( I x + I y ) = (3.93 + 1.06) × 106 = 2.495 × 106 mm 4
2
2
2
and
$1
%
R = ( ( I x − I y ) ) + I xy2
*2
+
2
- 1
.
4 $
%
4
= / ( (3.93 − 1.06) ) + 1.1650612 0 × 106 mm 4
+
41 * 2
42
= 1.84840 × 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(1.165061 × 106 )
(3.93 − 1.06) × 106
= −0.81189
=−
or
2θ m = −39.073°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1509
PROBLEM 9.96 (Continued)
Then
α = 180° − (39.073° + 90°)
= 50.927°
Then
I x′ = I ave − R cos α = (2.495 − 1.84840 cos 50.927°) × 106
or
I x′ = 1.330 × 106 mm 4
I y′ = I ave + R cos α = (2.495 + 1.84840 cos 50.927°) × 106
or
I y′ = 3.66 × 106 mm 4
or
I x′y′ = 1.435 × 106 mm 4
I x′y′ = R sin α = (1.84840 × 106 ) sin 50.927°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1510
!
PROBLEM 9.97
For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine
the orientation the principal axes at the origin and the corresponding
values of the moments of inertia.
SOLUTION
From Problem 9.79:
Ix =
Problem 9.67:
I xy =
π
8
a4
Iy =
π
2
a4
1 4
a
2
The Mohr’s circle is defined by the diameter XY, where
X
Now
1 '
&π 4 1 4 '
&π 4
a , a ! and Y
a , − a4 !
8
2
2
2 #
"
#
"
I ave =
1
1&π 4 π 4 '
a + a ! = 0.98175a 4
(I x + I y ) =
2
2" 8
2 #
and
2
$1
%
2
R = ( ( I x − I y ) ) + I xy
*2
+
2
$ 1 & π 4 π 4 '% & 1 4 '
= (
a − a !) +
a !
2 #+ " 2 #
*2 " 8
2
= 0.77264a 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
=−
2 I xy
Ix − I y
2
π
8
(
1
2
a4
4
)
a − 2 a4
π
= 0.84883
or
and
2θ m = 40.326°
θ m = 20.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1511
PROBLEM 9.97 (Continued)
The Principal axes are obtained by rotating the xy axes through
20.2° counterclockwise
!
!
about O.
!
Now
I max, min = I ave ± R = 0.98175a 4 ± 0.77264a 4
or
I max = 1.754 a 4
!
and
I min = 0.209 a 4
!
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1512
PROBLEM 9.98
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.72.
SOLUTION
I x = 68.96 × 106 mm 4
From Problem 9.80:
I y = 132.48 × 106 mm 4
I xy = −21.6 × 106 mm 4
Problem 9.72:
The Mohr’s circle is defined by the diameter XY, where X (68.96 × 106 , − 21.6 × 106 )
and Y (132.48 × 106 , 21.6 × 106 )
Now
1
(I x + I y )
2
1
= (68.96 + 132.48) × 106
2
= 100.72 × 106 mm 4
I ave =
2
and
$1
%
R = ( ( I x − I y ) ) + I xy2
*2
+
2
- 1
.
4 $
%
4
= / ( (68.96 − 132.48) ) + (−21.6) 2 0 × 106
+
41 * 2
42
= 38.409 × 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−21.6 × 106 )
(68.96 − 132.48) × 106
= −0.68010
=−
or
and
2θ m = −34.220°
θ m = −17.11°
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1513
PROBLEM 9.98 (Continued)
The principal axes are obtained by rotating the xy axes through
17.11° clockwise
about C.
Now
I max, min = I ave ± R = (100.72 + 38.409) × 106
or
I max = 139.1 × 106 mm 4
and
I min = 62.3 × 106 mm 4
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1514
!
PROBLEM 9.99
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.76.
SOLUTION
From Problem 9.76:
I xy = −9011.25 in.4
Now
where
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )1 =
1
(24 in.)(4 in.)3 = 128 in.4
12
( I x ) 2 = ( I x )3 =
1
$1
%
(9 in.)(15 in.)3 + ( (9 in.)(15 in.) ) (7 in.)2
36
*2
+
= 4151.25 in.4
Then
I x = [128 + 2(4151.25)] in.4
= 8430.5 in.4
Also
where
I y = ( I y )1 + ( I y ) 2 + ( I y )3
( I y )1 =
1
(4 in.)(24 in.) = 4608 in.4
12
( I y ) 2 = ( I y )3 =
1
$1
%
(15 in.)(9 in.)3 + ( (9 in.)(15 in.) ) (9 in.) 2
36
2
*
+
= 5771.25 in.4
Then
I y = [4608 + 2(5771.25)] in.4 = 16150.5 in.4
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1515
PROBLEM 9.99 (Continued)
The Mohr’s circle is defined by the diameter XY, where X (8430.5, − 9011.25) and Y (16150.5, 9011.25).
Now
I ave =
1
1
( I x + I y ) = (8430.5 + 16150.5) = 12290.5 in.4
2
2
2
and
$1
%
R = ( ( I x − I y ) ) + I xy2
*2
+
2
$1
%
= ( (8430.5 − 16150.5) ) + (−9011.25) 2
*2
+
= 9803.17 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − I y
2( −9011.25)
8430.5 − 16150.5
= −2.33452
=−
or
and
2θ m = −66.812°
θ m = −33.4°
The principal axes are obtained by rotating the xy axes through
33.4° clockwise
!
Now
about C.
I max, min = I ave ± R = 12290.5 ± 9803.17
or
I max = 22.1 × 103 in.4
and
I min = 2490 in.4
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1516
!
PROBLEM 9.100
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.73
SOLUTION
From Problem 9.81:
I x = 324π in.4
Problem 9.73:
I xy = 864 in.4
I y = 648π in.4
The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864).
Now
and
I ave =
1
1
( I x + I y ) = (324π + 648π ) = 1526.81 in.4
2
2
$1
% 2
R = ( ( I x − I y ) 2 ) + I xy
*2
+
2
$1
%
= ( (324π − 648π ) ) + 8642
*2
+
= 1002.75 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(864)
324π − 648π
= 1.69765
=−
or
2θ m = 59.4998°
and
θ m = 29.7°
The principal axes are obtained by rotating the xy axes through
29.7° counterclockwise
about C.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1517
!
PROBLEM 9.100 (Continued)
Now
I max, min = I ave ± R = 1526.81 ± 1002.75
or
I max = 2530 in.4
and
I min = 524 in.4
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1518
PROBLEM 9.101
Using Mohr’s circle, determine for the area indicated the orientation
of the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.74.
SOLUTION
I x = 0.390 in.4
From Problem 9.83:
I y = 1.09 in.4
I xy = −0.37983 in.4
Problem 9.74:
The Mohr’s circle is defined by the diameter XY, where X (0.390, − 0.37983) and Y (1.09, 0.37983).
Now
I ave =
1
1
( I x + I y ) = (0.390 + 1.09) = 0.740 in.4
2
2
2
and
1
!
R = " ( I x − I y ) # + I xy2
2
$
%
2
1
!
= " (0.390 − 1.09) # + (−0.37983) 2
2
$
%
= 0.51650 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2( −0.37983)
0.390 − 1.09
= −1.08523
=−
Then
and
2θ m = −47.341°
θ m = −23.7°
The principal axes are obtained by rotating the xy axes through
23.7° clockwise
about C.
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1519
!
PROBLEM 9.101 (Continued)
Now
I max, min = I ave ± R = 0.740 ± 0.51650
or
I max = 1.257 in.4
and
I min = 0.224 in.4
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1520
PROBLEM 9.102
Using Mohr’s circle, determine for the area indicated the orientation of
the principal centroidal axes and the corresponding values of the
moments of inertia.
Area of Problem 9.77
(The moments of inertia I x and I y of the area of Problem 9.102
were determined in Problem 9.44).
SOLUTION
From Problem 9.44:
I x = 18.1282 in.4
I y = 4.5080 in.4
Problem 9.77:
I xy = −4.25320 in.4
The Mohr’s circle is defined by the diameter XY, where X (18.1282, − 4.25320) and Y (4.5080, 4.25320).
Now
I ave =
1
1
( I x + I y ) = (18.1282 + 4.5080) = 11.3181 in.4
2
2
2
1
!
R = " ( I x − I y ) # + I xy2
$2
%
and
2
1
!
= " (18.1282 − 4.5080) # + ( −4.25320)2
$2
%
= 8.02915 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−4.25320)
18.1282 − 4.5080
= 0.62454
=−
or
and
2θ m = 31.986°
θ m = 15.99°
The principal axes are obtained by rotating the xy axes through
15.99° counterclockwise !
about C.
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1521
PROBLEM 9.102 (Continued)
Now
I max, min = I ave ± R = 11.3181 ± 8.02915
I max = 19.35 in.4
or
I min = 3.29 in.4
and
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1522
PROBLEM 9.103
The moments and product of inertia of an L4 × 3 × 14 -in. angle cross section with respect to two rectangular
axes x and y through C are, respectively, I x = 1.33 in.4, I y = 2.75 in.4, and I xy , 0, with the minimum value
of the moment of inertia of the area with respect to any axis through C being I min = 0.692 in.4. Using Mohr’s
circle, determine (a) the product of inertia I xy of the area, (b) the orientation of the principal axes, (c) the value
of I max .
SOLUTION
(Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when
the 4-in. leg of the angle is parallel to the x axis. Further, for I xy , 0, the angle must be oriented as shown.)
1
1
( I x + I y ) = (1.33 + 2.75) = 2.040 in.4
2
2
Now
I ave =
and
I min = I ave − R or
R = 2.040 − 0.692
= 1.348 in.4
Using I ave and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, X (1.33, I xy ) and
Y (2.75, |I xy |).
2
(a)
We have
1
!
R 2 = " ( I x − I y ) # + I xy2
$2
%
or
I xy2
1
!
= 1.348 − " (1.33 − 2.75) #
$2
%
2
2
Solving for I xy and taking the negative root (since I xy , 0) yields I xy = −1.14586 in.4 .
I xy = −1.146 in.4
(b)
We have
tan 2θ m = −
2 I xy
Ix − Iy
=−
2(−1.14586)
1.33 − 2.75
= −1.61389
or
2θ m = −58.217°
θ m = −29.1°
The principal axes are obtained by rotating the xy axes through
29.1° clockwise
about C.
(c)
We have
I max = I ave + R = 2.040 + 1.348
or
I max = 3.39 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1523
!
PROBLEM 9.104
Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes
and the corresponding values of the moments of inertia. (Properties
of the cross sections are given in Figure 9.13.)
SOLUTION
I x = 0.162 × 106 mm 4
From Figure 9.13B:
I y = 0.454 × 106 mm 4
We have
I xy = ( I xy )1 + ( I xy ) 2
For each rectangle
I xy = I x′y′ + x yA
and
I x′y′ = 0
I xy = Σ x yA
(symmetry)
A, mm 2
x , mm
y , mm
x yA, mm 4
1
76 × 6.4 = 486.4
−13.1
9.2
−58620.93
2
6.4 × (51 − 6.4) = 285.44
21.7
–16.3
–100962.98
Σ
–159583.91
I xy = −159584 mm 4
The Mohr’s circle is defined by the diameter XY where X (0.162 × 106 , − 0.159584 × 106 )
and Y (0.454 × 106 , 0.159584 × 106 )
Now
1
1
( I x + I y ) = (0.162 + 0.454) × 106
2
2
= 0.3080 × 106 mm 4
I ave =
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1524
PROBLEM 9.104 (Continued)
and
2
2
& 1
'
1
(
(
!
!
2
= " (0.162 − 0.454) # + ( −0.159584) 2 ! × 106
R = " ( I x − I y ) # + I xy
$2
%
%
&' $ 2
&(
= 0.21629 × 106 mm 4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2(−0.159584 × 106 )
(0.162 − 0.454) × 106
= −1.09304
=−
or
and
2θ m = −47.545
θ m = −23.8°
The principal axes are obtained by rotating the xy axes through
23.8° clockwise
About C.
Now
I max, min = I ave ± R = (0.3080 ± 0.21629) × 106
I max = 0.524 × 106 mm 4
or
I min = 0.0917 × 106 mm 4
and
From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1525
PROBLEM 9.105
Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes and
the corresponding values of the moments of inertia. (Properties of
the cross sections are given in Figure 9.13.)
SOLUTION
I x = 3.93 × 106 mm 4
From Figure 9.13B:
I y = 1.06 × 106 mm 4
I xy = −1.165061 × 106 mm 4
Problem 9.7B:
(Note that the figure of Problem 9.105 is obtained by replacing x with –x in the figure of Problem 9.78;
thus the change in sign of I xy .)
6
6
The Mohr’s circle is defined by the diameter XY, where X (3.93 × 10 , − 1.165061 × 10 )
6
6
and Y (1.06 × 10 , 1.165061 × 10 ).
Now
1
(I x + I y )
2
1
= (3.93 + 1.06) × 106
2
= 2.495 × 106 mm 4
I ave =
2
and
)1
*
R = " ( I x − I y ) # + I xy2
2
$
%
=
2
+ 1
,
& )
*
2&
−
+
−
× 106
(3.93
1.06)
(
1.165061)
!
"2
#
$
%
&'
(&
= 1.84840 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1526
PROBLEM 9.105 (Continued)
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2( −1.165061 × 106 )
(3.93 − 1.06) × 106
= 0.81189
=−
or
and
2θ m = 39.073°
θ m = 19.54°
The principal axes are obtained by rotating the xy axes through
19.54° counterclockwise
about C.
Now
I max, min = I ave ± R = (2.495 ± 1.84840) × 106
or
I max = 4.34 × 106 mm 4
and
I min = 0.647 × 106 mm 4
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1527
PROBLEM 9.106*
For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are
I x = 1200 in.4 and I y = 300 in.4, respectively. Knowing that after rotating the x and y axes about the centroid
30° counterclockwise, the moment of inertia relative to the rotated x axis is 1450 in.4, use Mohr’s circle to
determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.
SOLUTION
We have
I ave =
1
1
( I x + I y ) = (1200 + 300) = 750 in.4
2
2
Now observe that I x . I ave , I x′ . I x , and 2θ = +60°. This is possible only if I xy < 0. Therefore, assume
I xy , 0 and (for convenience) I x′y′ . 0. Mohr’s circle is then drawn as shown.
We have
Now using DABD:
2θ m + α = 60°
R=
I x − I ave 1200 − 750
=
cos 2θ m
cos 2θ m
=
Using DAEF:
R=
450
cos 2θ m
(in.4 )
I x′ − I ave 1450 − 750
=
cos α
cos α
700
(in.4 )
=
cos α
Then
450
700
=
cos 2θ m cos α
or
9cos (60° − 2θ m ) = 14 cos 2θ m
Expanding:
9(cos 60° cos 2θ m + sin 60° sin 2θ m ) = 14 cos 2θ m
tan 2θ m =
or
α = 60° − 2θ m
14 − 9 cos 60°
= 1.21885
9 sin 60°
2θ m = 50.633° and θ m = 25.3°
or
(Note: 2θ m , 60° implies assumption I x′y′ . 0 is correct.)
Finally,
(a)
R=
450
= 709.46 in.4
cos 50.633°
From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given centroidal x
and y axes through θ m about the centroid C or
25.3° counterclockwise
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1528
!
PROBLEM 9.106* (Continued)
(b)
We have
I max, min = I ave ± R = 750 ± 709.46
or
I max = 1459 in.4
and
I min = 40.5 in.4
From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1529
PROBLEM 9.107
It is known that for a given area I y = 48 × 106 mm4 and I xy = –20 × 106 mm4, where the x and y axes are
rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating
the x axis 67.5° counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia I x of the
area, (b) the principal centroidal moments of inertia.
SOLUTION
First assume I x . I y and then draw the Mohr’s circle as shown. (Note: Assuming I x , I y is not
consistent with the requirement that the axis corresponding to ( I xy ) max is obtained after rotating
the x axis through 67.5° CCW.)
From the Mohr’s circle we have
2θ m = 2(67.5°) − 90° = 45°
(a)
From the Mohr’s circle we have
Ix = Iy + 2
(b)
We have
and
Now
| I xy |
tan 2θ m
= 48 × 106 + 2
20 × 106
tan 45°
or
I x = 88.0 × 106 mm 4
or
I max = 96.3 × 106 mm 4
and
I min = 39.7 × 106 mm 4
1
1
( I x + I y ) = (88.0 + 48) × 106
2
2
6
= 68.0 × 10 mm 4
I ave =
R=
| I xy |
sin 2θ m
=
20 × 106
= 28.284 × 106 mm 4
sin 45°
I max, min = I ave ± R = (68.0 ± 28.284) × 106
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1530
PROBLEM 9.108
Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with
respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of
rectangular axes through the centroid is zero.
SOLUTION
Consider the regular pentagon shown, with centroidal axes x and y.
Because the y axis is an axis of symmetry, it follows that I xy = 0. Since I xy = 0, the x and y axes must be
principal axes. Assuming I x = I max and I y = I min , the Mohr’s circle is then drawn as shown.
Now rotate the coordinate axes through an angle α as shown; the resulting moments of inertia, I x′ and I y′ ,
and product of inertia, I x′y′ , are indicated on the Mohr’s circle. However, the x′ axis is an axis of symmetry,
which implies I x′y′ = 0. For this to be possible on the Mohr’s circle, the radius R must be equal to zero (thus,
the circle degenerates into a point). With R = 0, it immediately follows that
(a)
I x = I y = I x′ = I y ′ = I ave (for all moments of inertia with respect to an axis through C)
(b)
I xy = I x′y′ = 0 (for all products of inertia with respect to all pairs of rectangular axes with origin at C)
!
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1531
PROBLEM 9.109
Using Mohr’s circle, prove that the expression I x′ I y′ − I x2′y′ is independent of the orientation of the x′ and y′
axes, where Ix′, Iy′, and Ix′y′ represent the moments and product of inertia, respectively, of a given area with
respect to a pair of rectangular axes x′ and y′ through a given Point O. Also show that the given expression is
equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr’s
circle.
SOLUTION
First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R
are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the moments of
inertia, I x′ and I y′ , and the product of inertia, I x′y′ , having been computed for an arbitrary orientation of the
x′y ′ axes.
From the Mohr’s circle
I x′ = I ave + R cos 2θ
I y′ = I ave − R cos 2θ
I x′y′ = R sin 2θ
Then, forming the expression
I x′ I y′ − I x2′y′
I x′ I y′ − I x2′y′ = ( I ave + R cos 2θ )( I ave − R cos 2θ ) − ( R sin 2θ ) 2
(
)
2
= I ave
− R 2 cos 2 2θ − ( R 2 sin 2 2θ )
2
= I ave
− R2
which is a constant
I x′ I y′ − I x2′y′ is independent of the orientation of the coordinate axes Q.E.D.
Shown is a Mohr’s circle, with line OA, of length L, the required tangent.
Noting that
OAC is a right angle, it follows that
2
L2 = I ave
− R2
or
L2 = I x′ I y′ − I x2′y′ Q.E.D.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1532
!
PROBLEM 9.110
Using the invariance property established in the preceding problem,
express the product of inertia Ixy of an area A with respect to a pair of
rectangular axes through O in terms of the moments of inertia Ix and Iy
of A and the principal moments of inertia Imin and Imax of A about O.
Use the formula obtained to calculate the product of inertia Ixy of the
L3 × 2 × 14 -in. angle cross section shown in Figure 9.13A, knowing
that its maximum moment of inertia is 1.257 in4.
SOLUTION
Consider the following two sets of moments and products of inertia, which correspond to two different
orientations of the coordinate axes whole origin is at Point O.
Case 1:
I x′ = I x , I y ′ = I y , I x′y′ = I xy
Case 2:
I x′ = I max , I y′ = I min , I x′y′ = 0
The invariance property then requires
2
I x I y − I xy
= I max I min
From Figure 9.13A:
or I xy = ± I x I y − I max I min
I x = 1.09 in.4
I y = 0.390 in.4
Using Eq. (9.21):
Substituting
or
Then
I x + I y = I max + I min
1.09 + 0.390 = 1.257 + I min
I min = 0.223 in.4
I xy = (1.09)(0.390) − (1.257)(0.223)
= ±0.381 in.4
The two roots correspond to the following two orientations of the cross section.
I xy = −0.381 in.4
For
I xy = 0.381 in.4
and for
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1533
!
PROBLEM 9.111
Determine the mass moment of inertia of a ring of mass m, cut from a
thin uniform plate, with respect to (a) the axis AA′, (b) the centroidal
axis CC′ that is perpendicular to the plane of the ring.
SOLUTION
Mass = m = ρV = ρ tA
I mass = ρ t I area =
m
I area
A
We first determine
(
A = π r22 − π r12 = π r22 − r12
I AA′,area =
I AA′, mass =
(a)
(b)
π
4
r24 −
π
4
r14 =
π
4
(r
4
2
)
− r14
)
m
m
π 4 4
1
I AA′, area =
r2 − r1 = m r22 + r12
2
2
4
A
π r2 − r1 4
(
)
(
By symmetry:
I BB′ = IAA′
Eq. (9.38):
I CC ′ = IAA′ + I BB′ = 2 IAA′
)
(
)
I AA′ =
1
m r12 + r22
4
)
I CC ′ =
1
m r12 + r22
2
)
(
(
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1534
PROBLEM 9.112
A thin semicircular plate has a radius a and a mass m. Determine the
mass moment of inertia of the plate with respect to (a) the centroidal
axis BB′, (b) the centroidal axis CC′ that is perpendicular to the plate.
SOLUTION
mass = m = ρ tA
I mass = ρ t I area =
m
I area
A
1
A = π a2
2
Area:
I AA′,area = I DD′,area =
1-π 4. 1 4
/ a 0 = πa
21 4 2 8
I AA′, mass = I DD′,mass =
m
m -1 4. 1 2
I AA′,area =
/ π a 0 = ma
1
A
π a2 1 8
2 4
2
I BB′ = I DD′ − m( AC ) 2 =
(a)
1 2
- 4a .
ma − m /
0
4
1 3π 2
= (0.25 − 0.1801) ma 2
(b)
Eq. (9.38):
I CC ′ = I AA′ + I BB′ =
2
I BB′ = 0.0699ma 2
1
ma 2 + 0.0699ma 2
4
I CC ′ = 0.320ma 2
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1535
PROBLEM 9.113
The quarter ring shown has a mass m and was cut from a thin, uniform
plate. Knowing that r1 = 34 r2, determine the mass moment of inertia of the
quarter ring with respect to (a) the axis AA′, (b) the centroidal axis CC′ that
is perpendicular to the plane of the quarter ring.
SOLUTION
First note
mass = m = ρV = ρ tA
= ρt
(r
2
2
4
− r12
=
m
π
4
Using Figure 9.12,
I AA′,area =
Then
I AA′, mass =
(
π
r22 − r12
(r
16
4
2
− r14
(
r22
−
r12
(
=
)
I area
)
m
π
4
m 2
r2 + r12
=
4
(b)
)
I mass = ρ t I area
Also
(a)
π
)
×
π
16
(r
4
2
− r14
)
)
2
m) 2 -3 . *
" r2 + / r2 0 #
4 $"
1 4 2 %#
or
I AA′ =
25 2
mr2
64
Symmetry implies
I BB′, mass = I AA′,mass
Then
I DD′ = I AA′ + I BB′
- 25
.
= 2 / mr22 0
64
1
2
25 2
mr2
=
32
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1536
PROBLEM 9.113 (Continued)
Now locate centroid C.
X Σ A = Σx A
or
π . 4r - π . 4r - π .
-π
X / r22 − r12 0 = 2 / r22 0 − 1 / r12 0
4
4 2 3π 1 4 2 3π 1 4 2
1
4 r23 − r13
3π r22 − r12
or
X=
Now
r=X 2
3
-3 .
r23 − / r2 0
4 2
14 2
=
3π 2 - 3 .2
r2 − / r2 0
14 2
=
Finally,
or
37 2
r2
21π
I DD′ = I CC ′ + mr 2
- 37 2 .
25 2
mr2 = I CC ′ + m /
r
/ 21π 2 00
32
1
2
2
or
I CC ′ = 0.1522mr22
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1537
PROBLEM 9.114
The parabolic spandrel shown was cut from a thin, uniform plate.
Denoting the mass of the spandrel by m, determine its mass moment
of inertia with respect to (a) the axis BB′, (b) the axis DD′ that is
perpendicular to the spandrel. (Hint: See Sample Problem 9.3.)
SOLUTION
First note
mass = m = ρV = ρ tA
-π .
= ρ t / ab 0
12 2
I mass = ρ t I area
Also
=
(a)
2m
I
π ab area
We have
I x,area = I BB′,area + Ay 2
or
(a)
I BB′,area =
- π .- 4b .
ab3 − / ab 0/
0
8
1 2 21 3π 2
π
2
From Sample Problem 9.3:
1 3
ab
21
3m 1 3
=
× ab
ab 21
I BB′,area =
Then
I BB′, mass
or
(b)
I BB′ =
1 2
mb
7
From Sample Problem 9.3:
I AA′,area =
Now
1 3
ab
5
-3 .
I AA′,area = I11′,area + A / a 0
14 2
2
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1538
PROBLEM 9.114 (Continued)
and
-1 .
I 22′,area = I11′,area + A / a 0
14 2
2
)- 1 . - 3 .
I 22′,area = I AA′,area + A "/ a 0 − / a 0
"$1 4 2 1 4 2
1
1 - 1
9
.
= a3b + ab / a 2 − a 2 0
5
3 1 16
16 2
1 3
=
ab
30
2
Then
and
Finally,
2*
#
#%
3m 1 3
× ab
ab 30
1
= ma 2
10
I 22′, mass =
I DD′,mass = I BB′,mass + I 22′, mass
=
1 2 1
mb + ma 2
7
10
or
I DD′ =
1
m(7a 2 + 10b 2 )
70
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1539
PROBLEM 9.115
A thin plate of mass m was cut in the shape of a parallelogram as
shown. Determine the mass moment of inertia of the plate with respect
to (a) the x axis, (b) the axis BB′, which is perpendicular to the plate.
SOLUTION
mass = m = ρ t A
I mass = ρ t I area =
(a)
m
I area
A
Consider parallelogram as made of horizontal strips and slide strips to form a square since distance
from each strip to x axis is unchanged.
1
I x,area = a 4
3
I x, mass =
(b)
m
m
I x,area = 2
A
a
-1 4.
/3a 0
1
2
1
I x = ma 2
3
For centroidal axis y′:
)1 * 1
I y′,area = 2 " a 4 # = a 4
$12 % 6
m
m -1 . 1
I y′,mass = I y′,area = 2 / a 4 0 = ma 2
A
a 16 2 6
1
7
I y′′ = I y′ + ma 2 = ma 2 + ma 2 = ma 2
6
6
For axis BB′ ⊥ to plate, Eq. (9.38):
1
7
I BB′ = I x + I y′′ = ma 2 + ma 2
3
6
I BB′ =
3 2
ma
2
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1540
PROBLEM 9.116
A thin plate of mass m was cut in the shape of a parallelogram as
shown. Determine the mass moment of inertia of the plate with
respect to (a) the y axis, (b) the axis AA′, which is perpendicular to
the plate.
SOLUTION
See sketch of solution of Problem 9.115.
(a)
From Part b of solution of Problem 9.115:
I y′ =
1 2
ma
6
I y = I y′ + ma 2 =
(b)
1 2
ma + ma 2
6
Iy =
7
ma 2
6
I AA′ =
1
ma 2
2
From solution of Problem 9.115:
I y′ =
Eq. (9.38):
1 2
ma
6
1
and I x = ma 2
3
I AA′ = I y′ + I x
=
1 2 1 2
ma + ma
6
3
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1541
PROBLEM 9.117
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the x axis,
(b) the y axis.
SOLUTION
First note
mass = m = ρV = ρ tA
1
)
*
= ρ t "(2a )( a) + (2a)(a) # = 3ρ ta 2
2
$
%
I mass = ρ tI area =
Also
(a)
Now
m
I area
3a 2
I x,area = ( I x )1,area + ( I x ) 2,area
1
1
= (2a)(a)3 + (2a )(a )3
3
12
5 4
= a
6
Then
(b)
I x, mass =
m 5 4
× a
3a 2 6
or
I x , mass =
5
ma 2
18
We have
I z ,area = ( I z )1,area + ( I z )2, area
2
)1
1
1
)1
. *
3*
3
= " (a )(2a) # + " ( a)(2a) + (2a)(a ) / 2a + × 2a 0 # = 10a 4
2
3
$3
% "$ 36
1
2 #%
m
10
× 10a 4 = ma 2
2
3
3a
Then
I x, mass =
Finally,
I y ,mass = I x ,mass + I z ,mass
=
5
10
ma 2 + ma 2
18
3
=
65 2
ma
18
or
I y , mass = 3.61ma 2
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1542
PROBLEM 9.118
A thin plate of mass m has the trapezoidal shape shown. Determine
the mass moment of inertia of the plate with respect to (a) the
centroidal axis CC′ that is perpendicular to the plate, (b) the axis AA′
that is parallel to the x axis and is located at a distance 1.5a from the
plate.
SOLUTION
First locate the centroid C.
1
.
X ΣA = Σ xA: X (2a 2 + a 2 ) = a(2a 2 ) + / 2a + × 2a 0 ( a 2 )
3
1
2
X=
or
14
a
9
-1 .
-1 .
Z ΣA = Σ zA: Z (2a 2 + a 2 ) = / a 0 (2a 2 ) + / a 0 (a 2 )
2
1
2
13 2
Z=
or
(a)
We have
4
a
9
I y , mass = I CC ′, mass + m( X 2 + Z 2 )
From the solution to Problem 9.117:
I y ,mass =
Then
I cc′,mass =
65 2
ma
18
)- 14 .2 - 4 . 2 *
65 2
ma − m "/ a 0 + / a 0 #
18
"$1 9 2 1 9 2 #%
or
(b)
We have
and
Then
I cc′ = 0.994ma 2
I x, mass = I BB′,mass + m( Z ) 2
I AA′, mass = I BB′, mass + m(1.5a) 2
I AA′, mass = I x, mass
2
)
-4 . *
2
+ m "(1.5a) − / a 0 #
1 9 2 #%
"$
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1543
!
PROBLEM 9.118 (Continued)
From the solution to Problem 9.117:
I x, mass =
Then
I AA′, mass =
5
ma 2
18
2
)
5
-4 . *
ma 2 + m "(1.5a) 2 − / a 0 #
18
1 9 2 #%
"$
or
I AA′ = 2.33ma 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1544
PROBLEM 9.119
The area shown is revolved about the x axis to form a homogeneous
solid of revolution of mass m. Using direct integration, express the
mass moment of inertia of the solid with respect to the x axis in
terms of m and h.
SOLUTION
We have
y=
2h − h
x+h
a
so that
r=
h
( x + a)
a
For the element shown:
dm = ρπ r 2 dx dI x =
1 2
r dm
2
2
)h
*
= ρπ " ( x + a) # dx
$a
%
h2
h2
1
( x + a ) 2 dx = ρπ 2 [( x + a)3 ]0a
2
0
3
a
a
2
h
1
7
= ρπ 2 (8a 3 − a3 ) = ρπ ah 2
3
3
a
3
Then
m = dm =
Now
I x = dI x =
3
3
a
3
1 2
1
r ( ρπ r 2 dx) = ρπ
2
2
ρπ
3
a)h
0
4
*
" a ( x + a ) # dx
$
%
h4
1
1 h4
1
[( x + a)5 ]0a = ρπ 4 (32a5 − a 5 )
ρπ ×
4
2
5a
10
a
31
4
= ρπ ah
10
=
From above:
Then
3
7
ρπ ah 2 = m
Ix =
31 - 3 . 2 93 2
m h =
mh
10 /1 7 02
70
or
I x = 1.329 mh 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1545
PROBLEM 9.120
Determine by direct integration the mass moment of inertia with
respect to the y axis of the right circular cylinder shown, assuming
that it has a uniform density and a mass m.
SOLUTION
For element shown:
dm = ρ dV = ρπ a 2 dx
dI y = dI y + x 2 dm
1 2
a dm + x 2 dm
4
-1
.
= / a 2 + x 2 0 ρπ a 2 dx
14
2
+ L/2
-1
.
ρπ a 2 / a 2 + x 2 0 dx
I y = dI y =
− L/2
14
2
=
3
= ρπ a 2
3
1 2
x3
a x+
4
3
+ L/2
− L/2
) 1 L 1 - L .3 *
= 2 ρπ a 2 " a 2 + / 0 #
"$ 4 2 3 1 2 2 #%
Iy =
But for whole cylinder:
1
ρπ a 2 L(3a 2 + L2 ) "
12
m = ρV = ρπ a 2 L
Iy =
Thus,
1
m(3a 2 + L2 )
12
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you are using it without permission.
1546
PROBLEM 9.121
The area shown is revolved about the x axis to form a
homogeneous solid of revolution of mass m. Determine by
direct integration the mass moment of inertia of the solid
with respect to (a) the x axis, (b) the y axis. Express your
answers in terms of m and the dimensions of the solid.
SOLUTION
We have at
(a, h): h =
k
a
k = ah
or
For the element shown:
r=4
dm = ρπ r 2 dx
2
- ah .
= ρπ / 0 dx
1 x 2
3
m = dm =
Then
3
3a
a
2
- ah .
ρπ / 0 dx
1 x 2
3a
) 1*
= ρπ a 2 h2 " − #
$ x %a
) 1 - 1 .* 2
= ρπ a 2 h2 " − − / − 0 # = ρπ ah 2
$ 3a 1 a 2 % 3
(a)
For the element:
Then
dI x =
1 2
1
r dm = ρπ r 4 dx
2
2
3
I x = dI x =
3
3a
a
3a
4
1
1
- ah .
) 1 1*
ρπ / 0 dx = ρπ a 4 h 4 " − 3 #
2
2
1 x 2
$ 3 x %a
)- 1 .3 - 1 .3 * 1 26
1
= − ρπ a 4 h 4 "/ 0 − / 0 # = ×
ρπ ah4
6
3
6
27
a
a
$"1 2 1 2 %#
=
1 2
13
13 2
mh
× ρπ ah 2 × h 2 =
6 3
9
54
or
I x = 0.241 mh 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1547
PROBLEM 9.121 (Continued)
(b)
For the element:
dI y = dI y + x 2 dm
=
Then
1 2
r dm + x 2 dm
4
3
I y = dI y =
3
3a ) 1 a
2
2
*
ah .
- ah .
" / 0 + x 2 # ρπ / 0 dx
1 x 2
"$ 4 1 x 2
#%
2 2
= ρπ a h
3
3a
3a a
.
*
1 a 2 h2
1 a 2 h2
2 2)
+
=
−
+ x#
ρπ
dx
a
h
1
//
0
"
4
3
0
14 x
2
$ 12 x
%a
2 2
* ) 1 a 2 h2
* ,&
- 3 . +& ) 1 a h
= / m 0 a "−
+
+ a# !
a
3
# − "−
3
3
1 2 2 '& $ 12 (3a)
% $ 12 (a )
% &(
2
- 1 26 h
.
3
- 13 2
.
h + 3a 2 0
= ma // ×
+ 2a 00 = m /
2
1 108
2
1 12 27 a
2
or
I y = m(3a 2 + 0.1204h 2 )
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1548
PROBLEM 9.122
Determine by direct integration the mass moment of inertia with
respect to the x axis of the tetrahedron shown, assuming that it has a
uniform density and a mass m.
SOLUTION
We have
x=
a
y.
y + a = a /1 + 0
h
h2
1
and
z=
b
y.
y + b = b /1 + 0
h
h2
1
2
For the element shown:
Then
y.
-1
. 1
dm = ρ / xz dy 0 = ρ ab /1 + 0 dy
h
2
2
1
2
1
2
3
m = dm =
3
2
y.
1
ρ ab /1 + 0 dy
−h 2
h2
1
0
0
3
h )y. *
1
= ρ ab × "/1 + 0 #
h 2 %#
2
3 $"1
−h
1
ρ abh )$(1)3 − (1 − 1)3 *%
6
1
= ρ abh
6
=
Now, for the element:
Then
I AA′,area =
y.
1 3 1 3xz =
ab /1 + 0
36
36
h2
1
4
4
)1
y. *
dI AA′,mass = ρ tI AA′,area = ρ ( dy ) " ab3 /1 + 0 #
h 2 #%
1
"$ 3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1549
PROBLEM 9.122 (Continued)
Now
2
)
-1 . *
dI x = dI AA′, mass + " y 2 + / z 0 # dm
1 3 2 %#
$"
4
=
1
y.
ρ ab3 /1 + 0 dy
36
h
1
2
2
2
*
)1 y . * &, ) 1
y.
&+
+ y 2 + " b /1 + 0 # ! " ρ ab /1 + 0 dy #
h 2 % & $" 2
h2
1
$3 1
%#
'&
(
=
Now
Then
and
m=
4
1
1
y.
y3 y4
ρ ab3 /1 + 0 dy + ρ ab // y 2 + 2 + 2
12
2
h2
h h
1
1
.
00 dy
2
1
ρ abh
6
) 1 b2
dI x = " m
$" 2 h
3
I x = dI x =
4
y.
3m - 2
y3 y 4 .*
+
+
+
+
1
y
2
/
0 # dy
/
h 02
h 1/
h h2 20 %#
1
3
4
m ) 2y.
y3 y 4 .*
+
"b /1 + 0 + 6 // y 2 + 2
0 #dy
− h 2h "
h2
h h 2 20 #%
1
$ 1
0
5
-1
m ) 2 hy.
y5
1 y4
=
+ 2
"b × /1 + 0 + 6 // y 3 +
h2
2h $"
51
2 h 5h
13
=
0
.*
00 #
2 %# − h
m +1 2
1
1
)1
*,
b h(1)5 − 6 " (− h)3 +
( − h ) 4 + 2 ( − h )5 # !
2h ' 5
2h
5h
$3
%(
or
Ix =
1
m(b 2 + h 2 )
10
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1550
PROBLEM 9.123
Determine by direct integration the mass moment of inertia with
respect to the y axis of the tetrahedron shown, assuming that it has a
uniform density and a mass m.
SOLUTION
We have
x=
a
y.
y + a = a /1 + 0
h
h2
1
and
z=
b
y.
y + b = b /1 + 0
h
h2
1
2
For the element shown:
Then
y.
-1
. 1
dm = ρ / xz dy 0 = ρ ab /1 + 0 dy
2
2
h
1
2
1
2
3
m = dm =
3
2
1
y.
ρ ab /1 + 0 dy
−h 2
h2
1
0
0
3
1
h )y. *
= ρ ab × "/1 + 0 #
2
3 $"1
h 2 %#
−h
1
ρ abh )$(1)3 − (1 − 1)3 *%
6
1
= ρ abh
6
=
Also
Then, using
we have
I BB′,area =
1 3
xz
12
I DD′,area =
1 3
zx
12
I mass = ρ tI area
- 1
.
dI BB′,mass = ρ (dy ) / xz 3 0
1 12
2
- 1
.
dI DD′,mass = ρ (dy ) / zx3 0
1 12
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1551
PROBLEM 9.123 (Continued)
Now
dI y = dI BB′, mass + dI DD′, mass
1
ρ xz ( x 2 + z 2 )dy
12
2
2
y. )
y. *
1
= ρ ab /1 + 0 "(a 2 + b2 ) /1 + 0 # dy
h 2 "$
h 2 #%
12
1
1
=
4
We have
m=
1
m
y.
ρ abh 5 dI y = (a 2 + b 2 ) /1 + 0 dy
6
2h
h
1
2
Then
3
I y = dI y =
3
4
m 2
y.
(a + b 2 ) /1 + 0 dy
− h 2h
h
1
2
0
0
5
m 2
h )y. *
a + b 2 × "/ 1 + 0 #
=
h 2 #%
2h
5 "$1
−h
m 2
a + b 2 )$(1)5 − (1 − 1)5 *%
=
10
(
)
(
)
or
Iy =
1
m(a 2 + b 2 )
10
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1552
!
PROBLEM 9.124*
Determine by direct integration the mass moment of inertia with
respect to the z axis of the semiellipsoid shown, assuming that it
has a uniform density and a mass m.
SOLUTION
First note that when
For the element shown:
Then
1/2
x2
z = 0: y = b //1 − 2
1 a
.
00
2
x2
y = 0: z = c //1 − 2
1 a
.
00
2
1/2
x2
dm = ρ (π yzdx) = πρ bc //1 − 2
1 a
3
m = dm =
3
a
0
-
πρ bc //1 −
1
x2
a2
.
00 dx
2
.
00 dx
2
a
1
2
)
*
= πρ bc " x − 2 x3 # = πρ abc
3a
$
%0 3
For the element:
Then
Now
I AA′,area =
π
4
zy 3
-π
.
dI AA′,mass = ρ tI AA′,area = ρ (dx) / zy 3 0
14
2
dI z = dI AA′, mass + x 2 dm
x2
= ρ b c //1 − 2
4
1 a
π
=
3
3m ) b 2
"
2a "$ 4
2
)
.
x2
2
00 dx + x "πρ bc //1 − 2
"$
2
1 a
. *
00 dx #
2 #%
x2 x4 . - 2 x4 . *
//1 − 2 2 + 4 00 + // x − 2 00 # dx
a
a 2 1
a 2 #%
1
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1553
PROBLEM 9.124* (Continued)
Finally,
3
I z = dI z
a ) b2
x2 x4 . - 2 x4 .*
" //1 − 2 2 + 4 00 + // x − 2 00 # dx
a
a 2 1
a 2 #%
"$ 4 1
=
3m
2a
=
3m ) b 2
"
2a "$ 4
2 x3 1 x5
+
// x −
3 a2 5 a4
1
=
3 ) b2
m"
2 $4
1 .*
- 2 1.
2-1
/1 − + 0 + a / − 0 #
1 3 52
1 3 5 2%
3
0
. - 1 3 1 x5
00 + // x −
5 a2
2 13
a
.*
00 #
2 #% 0
or
Iz =
1
m a 2 + b2
5
(
)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1554
!
PROBLEM 9.125*
A thin steel wire is bent into the shape shown. Denoting the mass
per unit length of the wire by m′, determine by direct integration the
mass moment of inertia of the wire with respect to each of the
coordinate axes.
SOLUTION
dy
= − x −1/3 (a 2/3 − x 2/3 )1/2
dx
First note
2
Then
- dy .
1 + / 0 = 1 + x −2/3 (a 2/3 − x 2/3 )
1 dx 2
-a.
=/ 0
1 x2
2/3
2
- dy .
dm = m′dL = m′ 1 + / 0 dx
1 dx 2
For the element shown:
1/3
-a.
= m′ / 0
1 x2
3
3
a
Then
m = dm =
Now
I x = y 2 dm =
3
0
m′
dx
a
a1/3
3
3
dx = m′a1/3 $) x 2/3 %* = m′a
1/3
0
2
2
x
- a1/3 .
(a 2/3 − x 2/3 )3 // m′ 1/3 dx 00
0
1 x
2
2
a- a
.
= m′a1/3 // 1/3 − 3a 4/3 x1/3 + 3a 2/3 x − x5/3 00 dx
0 x
1
2
3
a
3
a
9
3
3
)3
*
= m′a1/3 " a 2 x 2/3 − a 4/3 x 4/3 + a 2/3 x 2 − x8/3 #
2
4
2
8
$
%0
3
9
3
3
3
.
= m′a3 / − + − 0 = m′a3
1 2 4 2 82 8
or
Symmetry implies
Ix =
1 2
ma
4
Iy =
1 2
ma
4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1555
!
PROBLEM 9.125* (Continued)
Alternative solution:
- a1/3 .
x 2 // m′ 1/3 dx 00 = m′a1/3
0
1 x
2
a
3
3
= m′a1/3 × )$ x8/3 *% = m′a3
0
8
8
1
= ma 2
4
3
Also
I y = x 2 dm =
3
3(x
)
Iz =
2
a
3
a
x5/3 dx
0
+ y 2 dm = I y + I x
or
Iz =
1 2
ma
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1556
!
PROBLEM 9.126
A thin triangular plate of mass m is welded along its base AB to a block
as shown. Knowing that the plate forms an angle θ with the y axis,
determine by direct integration the mass moment of inertia of the plate
with respect to (a) the x axis, (b) the y axis, (c) the z axis.
SOLUTION
For line BC:
ζ =−
=
h
x+h
a
2
h
(a − 2 x)
a
-1 .
m = ρV = ρ t / ah 0
12 2
1
= ρ tah
2
Also
2
(a)
We have
where
Then
1
-ζ .
dI x = ζ 2 dm′ + / 0 dm′
12
122
1 2
= ζ dm′
3
dm′ = ρ tζ dx
3
I x = dI x = 2
3
a/2 1
3
0
ζ 2 ( ρ tζ dx)
a/2 ) h
3
*
" a (a − 2 x) # dx
$
%
=
2
ρt
3
=
a/2
2 h3 1 - 1 .
ρ t 3 × / − 0 )$(a − 2 x) 4 *%
0
3 a 41 22
3
0
1
h3
ρ t 3 )$(a − a ) 4 − (a )4 *%
12 a
1
= ρ tah3
12
=−
or
Ix =
1 2
mh
6
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1557
PROBLEM 9.126 (Continued)
3
Now
Iζ = x 2 dm
and
Iζ = x 2 dm′ = 2
3
3
a/2
0
x 2 ( ρ tζ dx)
a/2
)h
*
x 2 " (a − 2 x) # dx
$a
%
= 2ρt
3
= 2ρt
h )a 3 1 4*
x − x #
a "$ 3
4 %0
0
a/2
h )a - a . 1 - a .
= 2ρt " / 0 − / 0
a "$ 3 1 2 2 4 1 2 2
3
=
(b)
We have
3
4*
#
#%
1
1
ρ ta3 h = ma 2
48
24
3
I y = ry2 dm = )$ x 2 + (ζ sin θ )2 *% dm
3
3
= x 2 dm + sin 2 θ ζ 2 dm
Now
3
I x = ζ 2 dm 5 I y = Iζ + I x sin 2 θ
=
(c)
We have
3
1
1
ma 2 + mh 2 sin 2 θ
24
6
or
Iy =
m 2
(a + 4h2 sin 2 θ )
24
or
Iz =
m 2
( a + 4h 2 cos 2 θ )
24
3
I z = rz2 dm = ( x 2 + y 2 )dm
3
= 3 x dm + cos θ 3 ζ
= )$ x 2 + (ζ cos θ ) 2 *% dm
2
2
2
dm
= Iζ + I x cos 2 θ
=
1
1
ma 2 + mh2 cos 2 θ
24
6
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1558
!
PROBLEM 9.127
Shown is the cross section of a molded flat-belt pulley.
Determine its mass moment of inertia and its radius of
gyration with respect to the axis AA′. (The density of
brass is 8650 kg/m3 and the density of the fiber-reinforced
polycarbonate used is 1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that
m = ρV = ρ t ×
π
4
(d
2
2
− d12
)
and, using Figure 9.28, that
2
2
1 - d2 .
1 -d .
m2
− m1 / 1 0
2 /1 2 02
2 1 22
π .
π . *
1 )= "/ ρ t × d 22 0 d 22 − / ρ t × d12 0 d12 #
8 $1
4 2
4 2 %
1
1-π .
= / ρ t 0 d 24 − d14
81 4 2
1-π .
= / ρ t 0 d 22 − d12 d 22 + d12
81 4 2
1
= m d12 + d 22
8
I AA′ =
(
)
(
)(
(
)
)
Now treat the pulley as four concentric rings and, working from the brass outward, we have
m=
π
{
8650 kg/m3 × (0.0175 m) × (0.0112 − 0.0052 ) m 2
4
+ 1250 kg/m3 )$(0.0175 m) × (0.017 2 − 0.0112 ) m 2
+ (0.002 m) × (0.0222 − 0.017 2 ) m 2
}
+ (0.0095 m) × (0.0282 − 0.0222 ) m 2 *%
= (11.4134 + 2.8863 + 0.38288 + 2.7980) × 10−3 kg
= 17.4806 × 10−3 kg
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1559
PROBLEM 9.127 (Continued)
and
I AA′ =
1
)(11.4134)(0.0052 + 0.0112 ) + (2.8863)(0.0112 + 0.017 2 )
8$
+ (0.38288)(0.017 2 + 0.0222 )
+(2.7980)(0.0222 + 0.0282 ) *% × 10−3 kg × m 2
= (208.29 + 147.92 + 37.00 + 443.48) × 10−9 kg ⋅ m 2
= 836.69 × 10−9 kg ⋅ m 2
or
Now
2
k AA
′ =
I AA′ = 837 × 10−9 kg ⋅ m 2
I AA′ 836.69 × 10−9 kg ⋅ m 2
=
= 47.864 × 10−6 m 2
−3
m
17.4806 × 10 kg
or
k AA′ = 6.92 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1560
!
PROBLEM 9.128
Shown is the cross section of an idler roller. Determine its mass
moment of inertia and its radius of gyration with respect to the
axis AA′. (The specific weight of bronze is 0.310 lb/in3; of
aluminum, 0.100 lb/in3; and of neoprene, 0.0452 lb/in3.)
SOLUTION
First note for the cylindrical ring shown that
m = ρV = ρ t ×
π
(d
4
2
2
)
− d12 =
π
4
(
ρ t d 22 − d12
)
and, using Figure 9.28, that
2
2
1 - d2 .
1 -d .
m2
− m1 / 1 0
2 /1 2 02
2 1 22
1 )π .
π . *
= "/ ρ t × d 22 0 d 22 − / ρ t × d12 0 d12 #
8 $1
4 2
4 2 %
1
I AA′ =
1-π .
= / ρ t 0 d 24 − d14
81 4 2
1-π .
= / ρ t 0 d 22 − d12
81 4 2
1
= m d12 + d 22
8
(
)
(
)( d
(
2
2
+ d12
)
)
Now treat the roller as three concentric rings and, working from the bronze outward, we have
m=
2
2
+&
1
- 13 . )- 3 . - 1 . *
ft/s 2 (0.310 lb/in.3 ) / in. 0 "/ 0 − / 0 # in.2
4 32.2
1 16 2 $"1 8 2 1 4 2 %#
&'
2
2
- 11 . )- 1 . - 3 . *
+ 0.100 lb/in.3 / in. 0 "/ 0 − / 0 # in.2
1 16 2 "$1 2 2 1 8 2 #%
2
2
,&
- 11 . )- 1 . - 1 . *
+ 0.0452 lb/in.3 / in. 0 "/1 0 − / 0 # in.2 !
1 16 2 "$1 8 2 1 2 2 #%
&(
π
×
(
(
)
)
= ( 479.96 + 183.41 + 769.80 ) × 10 −6 lb ⋅ s 2 /ft
= 1.4332 × 10−3 lb ⋅ s 2 /ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1561
PROBLEM 9.128 (Continued)
and
)- 1 . 2 - 3 . 2 *
)- 3 . 2 - 1 . 2 *
1 +&
I AA′ = (479.96) "/ 0 + / 0 # + (183.41) "/ 0 + / 0 #
8&
"$1 4 2 1 8 2 #%
"$1 8 2 1 2 2 #%
'
)- 1 .2 - 1 . 2 * ,&
1 ft 2
+ (769.80) "/1 0 + / 0 # ! × 10−6 lb ⋅ s 2 /ft × in.2 ×
144 in.2
"$1 8 2 1 2 2 #% &(
= (84.628 + 62.191 + 1012.78) × 10−9 lb ⋅ ft ⋅ s 2
= 1.15960 × 10−6 lb ⋅ ft ⋅ s 2
or
I AA′ = 1.160 × 10−6 lb ⋅ ft ⋅ s 2
I AA′ 1.15960 × 10 −6 lb ⋅ ft ⋅ s 2
=
= 809.09 × 10−6 ft 2
m
1.4332 × 10 −3 lb ⋅ s 2 /ft
Now
2
k AA
′ =
Then
k AA′ = 28.445 × 10−3 ft
or
k AA′ = 0.341 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1562
!
PROBLEM 9.129
Knowing that the thin cylindrical shell shown is of mass m, thickness t, and
height h, determine the mass moment of inertia of the shell with respect to
the x axis. (Hint: Consider the shell as formed by removing a cylinder of
radius a and height h from a cylinder of radius a + t and height h; then
neglect terms containing t2 and t3 and keep those terms containing t.)
SOLUTION
From Figure 9.28 for a solid cylinder (change orientation of axes):
I x′ =
1
m(3a 2 + h 2 )
12
-h.
I x = I x′ + m / 0
122
2
1
h2
m(3a 2 + h 2 ) + m
12
4
- 3 2 h2 .
= m // a + 00
3 2
1 12
=
Shell = Cylinder
Ix =
1
m(3a 2 + 4h2 ) "
12
– Cylinder !
1
1
m1[3(a + t ) 2 + 4h 2 ] − m2 (3a 2 + 4h 2 )
12
12
1
1
2
= [ ρπ (a + t ) h][3(a + t ) 2 + 4h 2 ] − ( ρπ a 2 h)(3a 2 + 4h 2 )
12
12
Ix =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1563
PROBLEM 9.129 (Continued)
Expand (a + t ) factors and neglecting terms in t 2, t 3, and t 4 :
ρπ h )
(
)(
)
a 2 + 2at + t 2 3a 2 + 6at + 3t 2 + 4h 2 * − 3a 4 − 4a 2 h2
%
12 $
ρπ h
3a 4 + 6a3t + 3a 2 t 2 + 4a 2 h2 + 6a 3t + 8ath 2 − 3a 4 − 4a 2 h 2
=
12
ρπ h
12a 3t + 8ath 2
=
12
ρπ hat
12a 2 + 8h 2
=
12
Ix =
(
(
)
(
Mass of shell:
)
)
m = ρ (2π ath) = 2 ρπ hat
2 ρπ hat
12a 2 + 8h 2
24
4m
3a 2 + 2h 2
=
24
(
Ix =
(
)
)
or
Ix =
m
3a 2 + 2h2
6
(
)
Checks:
1
,
a = 0; I x = mh 2 − slender rod &
&
3
! OK !
1
2
h = 0; I x = ma − thin ring &
&(
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1564
!
PROBLEM 9.130
The machine part shown is formed by machining a conical
surface into a circular cylinder. For b = 12 h, determine the mass
moment of inertia and the radius of gyration of the machine part
with respect to the y axis.
SOLUTION
Mass:
mcyl = ρπ a 2 h
1
mcyl a 2
2
1
= ρπ a 4 h
2
I y : I cyl =
mcone =
h 1
1
ρπ a 2 = ρπ a 2 h
3
2 6
3
mcone a 2
10
1
=
ρπ a 4 h
20
I cone =
For entire machine part:
1
5
ρπ a 2 h = ρπ a 2 h
6
6
1
1
9
ρπ a 4 h = ρπ a 4 h
= ρπ a 4 h −
2
20
20
m = mcyl − mcone = ρπ a 2 h −
I y = I cyl − I cone
or
-5
.- 6 .- 9 .
I y = / ρπ a 2 h 0/ 0/ 0 a 2
16
21 5 21 20 2
Then
k y2 =
I 27 2
a
=
m 50
Iy =
27 2
ma
50
k y = 0.735a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1565
!
PROBLEM 9.131
After a period of use, one of the blades of a shredder has been worn
to the shape shown and is of mass 0.18 kg. Knowing that the mass
moments of inertia of the blade with respect to the AA′ and BB′ axes
are 0.320 g ⋅ m 2 and 0.680 g ⋅ m 2 , respectively, determine (a) the
location of the centroidal axis GG′, (b) the radius of gyration with
respect to axis GG′.
SOLUTION
(a)
d B = (0.08 − d A ) m
We have
and, using the parallel axis
I AA′ = I GG′ + md A2
I BB′ = I GG′ + md B2
Then
(
I BB′ − I AA′ = m d B2 − d A2
)
= m "(0.08 − d A ) 2 − d A2 !#
= m(0.0064 − 0.16d A )
Substituting:
(0.68 − 0.32) × 10−3 kg ⋅ m 2 = 0.18 kg(0.0064 − 0.16d A ) m 2
or
(b)
We have
I AA′ = I GG′ + md A2
or
I GG ′ = 0.32 × 10−3 kg ⋅ m 2 − 0.18 kg ⋅ (0.0275 m)2
d A = 27.5 mm
= 0.183875 × 10−3 kg ⋅ m 2
Then
2
kGG
′ =
I GG′ 0.183875 × 10−3 kg ⋅ m 2
=
= 1.02153 × 10−3 m 2
m
0.18 kg
or
kGG ′ = 32.0 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1566
!
PROBLEM 9.132
Determine the mass moment of inertia of the 0.9-lb machine component shown
with respect to the axis AA′.
SOLUTION
First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger
cone as indicated.
h
h + 2.4
=
0.4
1.2
Now
or h = 1.2 in.
The weight of the body is given by
W = mg = g (m1 + m2 − m3 ) = ρ g (V1 + V2 − V3 )
or
0.9 lb = ρ × 32.2 ft/s 2
2
(π (0.8) (2.4) +
"
π
3
(0.2) 2 (1.2) −
!
$ 1 ft %
(0.6)2 (3.6) ) in 3 × &
'
3
#
* 12 in. +
π
3
= ρ × 32.2 ft/s 2 (2.79253 + 0.02909 − 0.78540) × 10−3 ft 3
or
Then
ρ = 13.7266 lb ⋅ s 2 /ft 4
m1 = (13.7266)(2.79253 × 10 −3 ) = 0.038332 lb ⋅ s 2 /ft
m2 = (13.7266)(0.02909 × 10 −3 ) = 0.000399 lb ⋅ s 2 /ft
m3 = (13.7266)(0.78540 × 10−3 ) = 0.010781 lb ⋅ s 2 /ft
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1567
PROBLEM 9.132 (Continued)
Finally, using Figure 9.28, we have
I AA′ = ( I AA′ )1 + ( I AA′ ) 2 − ( I AA′ )3
=
1
3
3
m1a12 + m2 a22 − m3 a32
2
10
10
1
3
3
!
$ 1 ft %
= ( (0.038332)(0.8)2 + (0.000399)(0.2)2 − (0.010781)(0.6)2 ) (lb ⋅ s2 /ft) × in 2 × &
'
10
10
"2
#
* 12 in. +
2
= (85.1822 + 0.0333 − 8.0858) × 10−6 lb ⋅ ft ⋅ s 2
or
I AA′ = 77.1 × 10−6 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1568
!
PROBLEM 9.133
A square hole is centered in and extends through the aluminum
machine component shown. Determine (a) the value of a for which
the mass moment of inertia of the component with respect to the axis
AA′, which bisects the top surface of the hole, is maximum, (b) the
corresponding values of the mass moment of inertia and the radius of
gyration with respect to the axis AA′. (The specific weight of
aluminum is 0.100 lb/in.3.)
SOLUTION
First note
m1 = ρV1 = ρ b 2 L
and
m2 = ρV2 = ρ a 2 L
(a)
Using Figure 9.28 and the parallel-axis theorem, we have
I AA′ = ( I AA′ )1 − ( I AA′ ) 2
1
$a%
= ( m1 (b 2 + b 2 ) + m1 & '
12
*2+
("
2!
)
)#
2
1
$a% !
− ( m2 (a 2 + a 2 ) + m2 & ' )
* 2 + )#
("12
1 %
$1
$ 5
%
= ( ρ b 2 L) & b 2 + a 2 ' − ( ρ a 2 L) & a 2 '
4 +
*6
* 12 +
ρL
(2b 4 + 3b 2 a 2 − 5a 4 )
=
12
Then
dI AA′ ρ L
(6b 2 a − 20a 3 ) = 0
=
da
12
or
a=0
d 2 I AA′
Also
da 2
and for a = b
=
ρL
12
d 2 I AA′
Now for a = 0,
da 2
3
,
10
a=b
and
d 2 I AA′
da 2
3
10
(6b 2 − 60a 2 ) =
1
ρ L(b2 − 10a 2 )
2
.0
,0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1569
PROBLEM 9.133 (Continued)
( I AA′ )max occurs when
Then
a=b
3
10
a = (4.2 in.)
3
10
or
(b)
a = 2.30 in.
From Part a:
( I AA′ ) max =
=
2
4!
$
$
3 %
3 % ) 49
− 5&b
=
2b 4 + 3b 2 & b
ρ Lb 4
'
'
& 10 '
& 10 ' ) 240
12 (
*
+
*
+ #
"
ρL (
$ 1 ft %
49 γ AL 4 49 0.100 lb/in.3
×
× (15 in.)(4.2 in.)4 &
Lb =
'
2
240 g
240
32.2 ft/s
* 12 in. +
or
Now
where
2
k AA
′ =
2
(I AA′ )max = 20.6 × 10−3 lb ⋅ ft ⋅ s 2
( I AA′ ) max
m
m = m1 − m2 = ρ L(b 2 − a 2 )
2
$
3 % !)
2
(
= ρL b − &b
& 10 '' )
(
*
+ #
"
7
= ρ Lb2
10
Then
2
k AA
′ =
49
ρ Lb 4
240
7
ρ Lb 2
10
=
7 2 7
b =
(4.2 in.)2
24
24
or
k AA′ = 2.27 in.
!
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1570
PROBLEM 9.134
The cups and the arms of an anemometer are fabricated from a
material of density ρ. Knowing that the mass moment of inertia
of a thin, hemispherical shell of mass m and thickness t with
respect to its centroidal axis GG′ is 5ma 2 /12, determine (a) the
mass moment of inertia of the anemometer with respect to the
axis AA′, (b) the ratio of a to l for which the centroidal moment
of inertia of the cups is equal to 1 percent of the moment of
inertia of the cups with respect to the axis AA′.
SOLUTION
(a)
First note
and
marm = ρVarm = ρ ×
π
4
d 2l
dmcup = ρ dVcup
= ρ[(2π a cos θ )(t )(adθ )]
Then
,
mcup = dmcup =
,
π /2
0
2πρ a 2 t cos θ dθ
= 2πρ a 2 t[sin θ ]π0 /2
= 2πρ a 2 t
Now
( I AA′ )anem = ( I AA′ )cups + ( I AA′ )arms
Using the parallel-axis theorem and assuming the arms are slender rods, we have
2 !
!
( I AA′ )anem = 3 "( I GG ′ )cup + mcup d AG
# + 3 " I arm + marm d AGarm #
2
2
1
/- 5
$ a % ! /.
$l% !
2
2
2
= 3 0 mcup a + mcup ((l + a) + & ' ) 1 + 3 ( marm l + marm & ' )
* 2 + )# /3
* 2 + )#
("
(" 2
/212
$5
%
= 3mcup & a 2 + 2la + l 2 ' + marm l 2
3
*
+
%
$5
% $π
= 3(2πρ a 2 t ) & a 2 + 2la + l 2 ' + & ρ d 2 l ' (l 2 )
*3
+ *4
+
or
$ 5 a2
a % d 2l !
( I AA′ )anem = πρ l 2 (6a 2 t && 2 + 2 + 1'' +
)
l
*3 l
+ 4 )#
"(
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you are using it without permission.
1571
!
PROBLEM 9.134 (Continued)
(b)
( I GG ′ )cup
We have
( I AA′ )cup
5
$5
%
mcup a 2 = 0.01mcup & a 2 + 2la + l 2 ' (from Part a )
12
*3
+
or
Now let ζ =
= 0.01
a
.
l
Then
$5
%
5ζ 2 = 0.12 & ζ 2 + 2ζ + 1'
3
*
+
or
40ζ 2 − 2ζ − 1 = 0
2 ± (−2) 2 − 4(40)(−1)
2(40)
Then
ζ =
or
ζ = 0.1851 and ζ = − 0.1351
a
= 0.1851
l
To the instructor:
The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at
this time for use in the solutions of Problems 9.135 through 9.140.
Thin rectangular plate
( I x )m = ( I x′ ) m + md 2
2
=
1
$b% $h%
m(b 2 + h 2 ) + m (& ' + & '
12
("* 2 + * 2 +
2!
)
)#
1
= m(b2 + h 2 )
3
( I y ) m = ( I y′ ) m + md 2
2
=
1
1
$b%
mb 2 + m & ' = mb 2
12
2
3
* +
I z = ( I z′ ) m + md 2
2
=
1
1
$h%
mh 2 + m & ' = mh 2
12
2
3
* +
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1572
PROBLEM 9.134 (Continued)
Thin triangular plate
We have
$1
%
m = ρV = ρ & bht '
*2
+
1 3
bh
36
and
I z ,area =
Then
I z ,mass = ρ tI z ,area
= ρt ×
1 3
bh
36
=
1
mh 2
18
Similarly,
I y ,mass =
1
mb2
18
Now
I x, mass = I y ,mass + I z ,mass =
1
m(b2 + h 2 )
18
Thin semicircular plate
We have
$π
%
m = ρV = ρ & a 2 t '
*2
+
π
a4
and
I y ,area = I z ,area =
Then
I y ,mass = I z ,mass = ρ tI y ,area
8
= ρt ×
=
π
8
a4
1 2
ma
4
1
ma 2
2
Now
I x, mass = I y ,mass + I z ,mass =
Also
I x, mass = I x′,mass + my 2
or
$ 1 16
I x′,mass = m & − 2
* 2 9π
% 2
'a
+
and
I z ,mass = I z′,mass + my 2
or
$ 1 16
I z ′,mass = m & − 2
* 4 9π
% 2
'a
+
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1573
PROBLEM 9.134 (Continued)
y=z =
4a
3π
Thin Quarter-Circular Plate
We have
$π
%
m = ρV = ρ & a 2 t '
*4
+
π
a4
and
I y ,area = I z ,area =
Then
I y ,mass = I z ,mass = ρ tI y ,area
16
= ρt ×
=
π
16
a4
1
ma 2
4
1
ma 2
2
Now
I x, mass = I y ,mass + I z ,mass =
Also
I x, mass = I x′,mass + m( y 2 + z 2 )
or
$ 1 32
I x′,mass = m & − 2
* 2 9π
and
I y ,mass = I y′,mass + mz 2
or
$ 1 16 %
I y′,mass = m & − 2 ' a 2
* 4 9π +
% 2
'a
+
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1574
PROBLEM 9.135
A 2-mm thick piece of sheet steel is cut and bent into the
machine component shown. Knowing that the density of steel
is 7850 kg/m3, determine the mass moment of inertia of the
component with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m = ρSTV = ρST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.3 m) 2
= 1.413 kg
m2 = m3 = (7850 kg/m3 )(0.002 m) × (0.15 × 0.12) m 2
= 0.2826 kg
Using Figure 9.28 and the parallel-axis theorem, we have
I x = ( I x )1 + 2( I x )2
1
!
= ( (1.413 kg)(0.3 m) 2 )
"12
#
1
!
+ 2 ( (0.2826 kg)(0.12 m) 2 + (0.2828 kg)(0.152 + 0.06 2 )m 2 )
12
"
#
= [(0.0105975) + 2(0.0003391 + 0.0073759)] kg ⋅ m 2
= [(0.0105975) + 2(0.0077150)] kg ⋅ m 2
or
I x = 26.0 × 10−3 kg ⋅ m 2
I y = ( I y )1 + 2( I y )2
1
!
= ( (1.413 kg)(0.32 + 0.32 ) m 2 )
12
"
#
1
!
2
+ 2 ( (0.2826 kg)(0.15 m) + (0.2826 kg)(0.0752 + 0.152 ) m 2 )
"12
#
= [(0.0211950) + 2(0.0005299 + 0.0079481)] kg ⋅ m 2
= [(0.0211950) + 2(0.0084780)] kg ⋅ m 2
or
I y = 38.2 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1575
PROBLEM 9.135 (Continued)
I z = ( I z )1 + 2( I z ) 2
1
!
= ( (1.413 kg)(0.3 m) 2 )
12
"
#
1
!
+ 2 ( (0.2826 kg)(0.152 + 0.122 ) m 2 + (0.2826 kg)(0.0752 + 0.062 ) m 2 )
"12
#
= [(0.0105975) + 2(0.0008690 + 0.0026070)] kg ⋅ m 2
= [(0.0105975) + 2(0.0034760)] kg ⋅ m 2
or
I z = 17.55 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1576
!
PROBLEM 9.136
A 2-mm thick piece of sheet steel is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass moment of inertia of the component with respect
to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m = ρSTV = ρST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.35 × 0.39) m 2
= 2.14305 kg
$π
%
m2 = (7850 kg/m3 )(0.002 m) & × 0.1952 ' m 2 = 0.93775 kg
2
*
+
$1
%
m3 = (7850 kg/m3 )(0.002 m) & × 0.39 × 0.15 ' m 2 = 0.45923 kg
2
*
+
Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have
I x = ( I x )1 + ( I x )2 + ( I x )3
2
1
$ 0.39 % !
2
m' )
= ( (2.14305 kg)(0.39 m) + (2.14305 kg) &
* 2
+ )#
("12
2
-/$ 1 16 %
! ./
$ 4 × 0.195 %
+ 0& − 2 ' (0.93775 kg)(0.195 m) 2 + (0.93775 kg) (&
+ (0.195) 2 ) m 2 1
'
+
/2* 2 9π +
"(* 3π
#) /3
2
2
-/ 1
$ 0.39 % $ 0.15 % ! 2 ./
+ 0 (0.45923 kg)[(0.39) 2 + (0.15)2 ] m 2 + (0.45923 kg) (&
+
' &
' )m 1
("* 3 + * 3 + #) /3
/218
= [(0.027163 + 0.081489) + (0.011406 + 0.042081) + (0.004455 + 0.008909)] kg ⋅ m 2
= (0.108652 + 0.053487 + 0.013364) kg ⋅ m 2
= 0.175503 kg ⋅ m 2
or
I x = 175.5 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1577
PROBLEM 9.136 (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3
2
2
-/ 1
$ 0.35 % $ 0.39 % ! 2 ./
= 0 (2.14305 kg)[(0.35) 2 + (0.39) 2 ] m 2 + (2.14305 kg) (&
+
' &
' )m 1
("* 2 + * 2 + )# 3/
/212
1
!
+ ( (0.93775 kg)(0.195 m)2 + (0.93775 kg)(0.195 m) 2 )
"4
#
2
-/ 1
$ 0.39 % ! 2 ./
2
2
+ 0 (0.45923 kg)(0.39 m) + (0.45923 kg) ((0.35) + &
' )m 1
* 3 + )# /3
("
/218
= [(0.049040 + 0.147120) + (0.008914 + 0.035658)
+ (0.003880 + 0.064017)] kg ⋅ m 2
= (0.196160 + 0.044572 + 0.067897) kg ⋅ m 2
= 0.308629 kg ⋅ m 2
or
I y = 309 × 10−3 kg ⋅ m 2
I z = ( I z )1 + ( I z ) 2 + ( I z )3
2
1
$ 0.35 % !
m' )
= ( (2.14305 kg)(0.35 m) 2 + (2.14305 kg) &
* 2
+ )#
("12
1
!
+ ( (0.93775 kg)(0.195 m)2 )
"4
#
2
/- 1
$ 0.15 % ! 2 /.
+ 0 (0.45923 kg)(0.15 m) 2 + (0.45923 kg) ((0.35) 2 + &
' )m 1
* 3 + #) 3/
("
/218
= [(0.021877 + 0.065631) + 0.008914) + (0.000574 + 0.057404)] kg ⋅ m 2
= (0.087508 + 0.008914 + 0.057978) kg ⋅ m 2
= 0.154400 kg ⋅ m 2
or
I z = 154.4 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1578
!
PROBLEM 9.137
The cover for an electronic device is formed from sheet aluminum
that is 0.05 in. thick. Determine the mass moment of inertia of the
cover with respect to each of the coordinate axes. (The specific
weight of aluminum is 0.100 lb/in.3 .)
SOLUTION
First compute the mass of each component. We have
m = ρV =
Then
γ
g
tA
m1 =
0.100 lb/in.3
× 0.05 in. × (3 × 2.4) in.2 = 1.11801 × 10 −3 lb ⋅ s 2 /ft
2
32.2 ft/s
m2 =
0.100 lb/in.3
× 0.05 in. × (3 × 6.2) in.2 = 2.88820 × 10−3 lb ⋅ s 2 /ft
32.2 ft/s 2
m3 = m4 =
0.100 lb/in.3
× 0.05 in. × (2.4 × 6.2) in.2 = 2.31056 × 10−3 lb ⋅ s 2 /ft
2
32.2 ft/s
Using Figure 9.28 and the parallel-axis theorem, we have
( I x )3 = ( I x ) 4
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4
1
= ( (1.11801 × 10 −3 lb ⋅ s 2 /ft)(2.4 in.) 2
"12
2
2
$ 2.4
% ! $ 1 ft %
in. ' ) &
+ (1.11801 × 10−3 lb ⋅s 2 /ft) &
'
* 2
+ )# * 12 in. +
1
+ ( (2.88820 × 10−3 lb ⋅ s 2 /ft)(6.2 in.) 2
12
"
2
2
$ 6.2
% ! $ 1 ft %
in. ' ) &
+ (2.88820 × 10 lb ⋅ s /ft) &
'
* 2
+ )# * 12 in. +
-1
+ 2 0 (2.31056 × 10−3 lb ⋅ s 2 /ft) "(2.4)2 + (6.2) 2 #! in.2
212
−3
2
2
2
2
$ 2.4 % $ 6.2 % ! 2 /. $ 1 ft %
+ (2.31056 × 10−3 lb ⋅ s 2 /ft) (&
+
in.
1&
'
' &
' )
/3 * 12 in. +
"(* 2 + * 2 + #)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1579
PROBLEM 9.137 (Continued)
= [(3.7267 + 11.1801) + (64.2491 + 192.7472)
+ 2(59.1011 + 177.3034)] × 10 −6 lb ⋅ ft ⋅ s 2
= [14.9068 + 256.9963 + 2(236.4045)] × 10 −6 lb ⋅ ft ⋅ s 2
or
I x = 745 × 10−6 lb ⋅ ft ⋅ s 2
I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4
1
= ( (1.11801 × 10−3 lb ⋅ s 2 /ft)(3 in.) 2
"12
2
2
$3
% ! $ 1 ft %
2
−3
+ (1.11801 × 10 lb ⋅ s /ft) & in. ' ) &
'
*2
+ )# * 12 in. +
-1
+ 0 (2.88820 × 10−3 lb ⋅ s 2 /ft)[(3)2 + (6.2) 2 ] in.2
212
2
2
2
$ 3 % $ 6.2 % ! 2 ./ $ 1 ft %
+ (2.88820 × 10−3 lb ⋅ s 2 /ft) (& ' + &
in.
)
1
'
&
'
/3 * 12 in. +
"(* 2 + * 2 + #)
1
+ ( (2.31056 × 10−3 lb ⋅ s 2 /ft)(6.2 in.)2
"12
2
2
$ 6.2
% ! $ 1 ft %
−3
2
+ (2.31056 × 10 lb ⋅ s /ft) &
in. ' ) &
'
* 2
+ )# * 12 in. +
-1
+ 0 (2.31056 × 10−3 lb ⋅ s 2 /ft)(6.2 in.)2
212
2
2
$ 6.2 % ! 2 ./ $ 1 ft %
+ (2.31056 × 10−3 lb ⋅ s 2 /ft) ((3)2 + &
in.
)
1
'
&
'
* 2 + #)
/3 * 12 in. +
"(
= [(5.8230 + 17.4689) + (79.2918 + 237.8754) + (51.3993 + 154.1978)
+ (51.3993 + 298.6078)] × 10−6 lb ⋅ ft ⋅ s 2
= (23.2919 + 317.1672 + 205.5971 + 350.0071) × 10−6 lb ⋅ ft ⋅ s 2
or
I y = 896 × 10−6 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1580
PROBLEM 9.137 (Continued)
I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4
-1
= 0 (1.11801 × 10 −3 lb ⋅ s 2 /ft)[(3) 2 + (2.4)2 ] in.2
212
2
$ 3 % $ 2.4 %
+ (1.11801 × 10−3 lb ⋅ s 2 /ft) (& ' + &
'
("* 2 + * 2 +
1
+ ( (2.88820 × 10 −3 lb ⋅ s 2 /ft)(3 in.) 2
12
"
./ $ 1 ft % 2
) in.2 1 &
'
)#
/3 * 12 in. +
2!
2
2
$3
% ! $ 1 ft %
+ (2.88820 × 10−3 lb ⋅ s 2 /ft) & in. ' ) &
'
*2
+ )# * 12 in. +
1
+ ( (2.31056 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2
"12
2
2
$ 2.4 % ! $ 1 ft %
+ (2.31056 × 10−3 lb ⋅ s 2 / ft) &
in. ' ) &
'
* 2
+ )# * 12 in. +
-1
+ 0 (2.31056 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2
212
$ 2.4 %
+ (2.31056 × 10−3 lb ⋅ s 2 /ft) ( (3) 2 + &
'
* 2 +
("
2
! 2 ./ $ 1 ft % 2
) in. 1 &
'
12 in. +
)#
3/ *
= [(9.5497 + 28.6490) + (15.0427 + 45.1281)
+ (7.7019 + 23.1056) + (7.7019 + 167.5156)] × 10−6 lb ⋅ ft ⋅ s 2
= (38.1987 + 60.1708 + 30.8075 + 175.2175) × 10−6 lb ⋅ ft ⋅ s 2
or
I z = 304 × 10−6 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1581
!
PROBLEM 9.138
A framing anchor is formed of 0.05-in.-thick galvanized steel. Determine the
mass moment of inertia of the anchor with respect to each of the coordinate
axes. (The specific weight of galvanized steel is 470 lb/ft 3 .)
SOLUTION
First compute the mass of each component. We have
m = ρV =
γ G.S.
g
tA
Then
470 lb/ft 3
$ 1 ft %
× 0.05 in. × (2.25 × 3.5) in.2 × &
'
2
32.2 ft/s
* 12 in. +
= 3325.97 × 10−6 lb ⋅ s 2 /ft
3
m1 =
470 lb/ft 3
$ 1 ft %
m2 =
× 0.05 in. × (2.25 × 1) in.2 × &
'
2
32.2 ft/s
* 12 in. +
3
= 950.28 × 10−6 lb ⋅ s 2 /ft
470 lb/ft 3
$1
%
$ 1 ft %
× 0.05 in. × & × 2 × 4.75 ' in.2 × &
'
2
32.2 ft/s
*2
+
* 12 in. +
= 2006.14 × 10−6 lb ⋅ s 2 /ft
3
m3 =
Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
I x = ( I x )1 + ( I x ) 2 + ( I x )3
1
= ( (3325.97 × 10−6 lb ⋅ s 2 /ft)(3.5 in.) 2
"12
$ 3.5 2 % ! $ 1 ft %
in. ' ) &
+ (3325.97 × 10−6 lb ⋅ s 2 /ft) &
'
* 2
+ # * 12 in. +
-1
+ 0 (950.28 × 10−6 lb ⋅ s 2 /ft)(1 in.) 2
212
2
2
./ $ 1 ft %2
$1% !
+ (950.28 × 10−6 lb ⋅ s 2 /ft) ((3.5) 2 + & ' ) in.2 1 &
'
* 2 + )#
/3 * 12 in. +
"(
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1582
PROBLEM 9.138 (Continued)
-1
+ 0 (2006.14 × 10−6 lb ⋅ s 2 /ft)[(4.75) 2 + (2)2 ] in.2
218
2
$2
% $1
%
+ (2006.14 × 10−6 lb ⋅ s 2 /ft) (& × 4.75 ' + & × 2 '
+ *3 +
("* 3
./ $ 1 ft % 2
) in.2 1 &
'
)#
/3 * 12 in. +
2!
= [(23.578 + 70.735) + (0.550 + 82.490)
+ (20.559 + 145.894)] × 10−6 lb ⋅ ft ⋅ s 2
= (94.313 + 83.040 + 166.453) × 10−6 lb ⋅ ft ⋅ s 2
I x = 344 × 10−6 lb ⋅ ft ⋅ s 2
or
I y = ( I y )1 + ( I y ) 2 + ( I y )3
1
= ( (3325.97 × 10−6 lb ⋅ s 2 /ft)(2.25 in.)2
"12
2
2
$ 2.25
% ! $ 1 ft %
in. ' ) &
+ (3325.97 × 10 lb ⋅ s /ft) &
'
* 2
+ )# * 12 in. +
−6
2
-1
+ 0 (950.28 × 10 −6 lb ⋅ s 2 /ft)[(2.25) 2 + (1) 2 ] in.2
212
2
$ 2.25 % $ 1 %
+ (950.28 × 10−6 lb ⋅ s 2 /ft) (&
' +& '
("* 2 + * 2 +
./ $ 1 ft % 2
) in.2 1 &
'
)#
/3 * 12 in. +
2!
-1
+ 0 (2006.14 × 10−6 lb ⋅ s 2 /ft)(2 in.)2
218
$1
%
+ (2006.14 × 10−6 lb ⋅ s 2 /ft) ((2.25) 2 + & × 2 '
*3
+
("
./ $ 1 ft %2
) in.2 1 &
'
)#
/3 * 12 in. +
2!
= [(9.744 + 29.232) + (3.334 + 10.002)
+ (3.096 + 76.720)] × 10−6 lb ⋅ ft ⋅ s 2
= (38.976 + 13.336 + 79.816) × 10−6 lb ⋅ ft ⋅ s 2
or
I y = 132.1 × 10−6 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1583
PROBLEM 9.138 (Continued)
I z = ( I z )1 + ( I z ) 2 + ( I z )3
-1
= 0 (3325.97 × 10 −6 lb ⋅ s 2 /ft)[(2.25) 2 + (3.5)2 ] in.2
212
2
$ 2.25 % $ 3.5 %
+ (3325.97 × 10−6 lb ⋅ s 2 /ft) (&
' +&
'
("* 2 + * 2 +
./ $ 1 ft % 2
) in.2 1 &
'
)#
/3 * 12 in. +
2!
-1
+ 0 (950.28 × 10−6 lb ⋅ s 2 /ft)(2.25 in.) 2
212
2
! 2 ./ $ 1 ft % 2
$ 2.25 %
2
+ (950.28 × 10 lb ⋅ s /ft) (&
' + (3.5) ) in. 1 &
'
("* 2 +
)#
/3 * 12 in. +
-1
+ 0 (2006.14 × 10−6 lb ⋅ s 2 /ft)(4.75 in.) 2
218
−6
2
2
./ $ 1 ft %2
$2
% !
+ (2006.14 × 10−6 lb ⋅ s 2 /ft) ((2.25) 2 + & × 4.75 ' ) in.2 1 &
'
*3
+ )#
/3 * 12 in. +
"(
= [(33.322 + 99.967) + (2.784 + 89.192)
+ (17.463 + 210.231)] × 10−6 lb ⋅ ft ⋅ s 2
= (133.289 + 91.976 + 227.694) × 10−6 lb ⋅ ft ⋅ s 2
or
I z = 453 × 10−6 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1584
PROBLEM 9.139
A subassembly for a model airplane is fabricated from three pieces of 1.5-mm
plywood. Neglecting the mass of the adhesive used to assemble the three
pieces, determine the mass moment of inertia of the subassembly with respect
to each of the coordinate axes. (The density of the plywood is 780 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = ρV = ρ tA
Then
$1
%
m1 = m2 = (780 kg/m3 )(0.0015 m) & × 0.3 × 0.12 ' m 2
*2
+
= 21.0600 × 10−3 kg
$π
%
m3 = (780 kg/m3 )(0.0015 m) & × 0.122 ' m 2 = 13.2324 × 10−3 kg
*4
+
Using the equations derived above and the parallel-axis theorem, we have
( I x )1 = ( I x ) 2
I x = ( I x )1 + ( I x ) 2 + ( I x )3
2
1
$ 0.12 % !
m' )
= 2 ( (21.0600 × 10−3 kg)(0.12 m)2 + (21.0600 × 10−3 kg) &
* 3
+ )#
("18
1
!
+ ( (13.2324 × 10 −3 kg)(0.12 m) 2 )
"2
#
= [2(16.8480 + 33.6960) + (95.2733)] × 10−6 kg ⋅ m 2
= [2(50.5440) + (95.2733)] × 10−6 kg ⋅ m 2
or
I x = 196.4 × 10−6 kg ⋅ m 2
!
!
!
!
!
!
!
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1585
!
PROBLEM 9.139 (Continued)
!
!
I y = ( I y )1 + ( I y )2 + ( I y )3
!
2!
1
$ 0.3 %
= ( (21.0600 × 10−3 kg)(0.3 m) 2 + (21.0600 kg) &
m' )
* 3 + )#
("18
!
-1
+ 0 (21.0600 × 10−3 kg)[(0.3)2 + (0.12) 2 ] m 2 !
218
2
2
$ 0.3 % $ 0.12 % ! 2 ./
+ (21.0600 × 10−3 kg) (&
+
' &
' )m 1
/3
"(* 3 + * 3 + #)
!
!
1
!
+ ( (13.2324 × 10−3 kg)(0.12 m) 2 ) !
"4
#
= [(105.300 + 210.600) + (122.148 + 244.296) !
+ (47.637)] × 10−6 kg ⋅ m 2
!
= (315.900 + 366.444 + 47.637) × 10−6 kg ⋅ m 2 !
or
I y = 730 × 10−6 kg ⋅ m 2
Symmetry implies I y = I z
I z = 730 × 10−6 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1586
PROBLEM 9.140*
A farmer constructs a trough by welding a rectangular piece of
2-mm-thick sheet steel to half of a steel drum. Knowing that
the density of steel is 7850 kg/m3 and that the thickness of the
walls of the drum is 1.8 mm, determine the mass moment of
inertia of the trough with respect to each of the coordinate
axes. Neglect the mass of the welds.
SOLUTION
First compute the mass of each component. We have
m = ρSTV = ρST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.84 × 0.21) m 2
= 2.76948 kg
m2 = (7850 kg/m3 )(0.0018 m)(π × 0.285 × 0.84) m 2
= 10.62713 kg
$π
%
m3 = m4 = (7850 kg/m3 )(0.0018 m) & × 0.2852 ' m 2
2
*
+
= 1.80282 kg
Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have
( I x )3 = ( I x ) 4
I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4
2
1
0.21 % 2 !
$
2
= ( (2.76948 kg)(0.21 m) + (2.76948 kg) & 0.285 −
m )
2 '+
*
("12
)#
1
!
+ [(10.62713 kg)(0.285 m)2 ] + 2 ( (1.80282 kg)(0.285 m) 2 )
"2
#
= [(0.01018 + 0.08973) + (0.86319) + 2(0.07322)] kg ⋅ m 2
= [(0.09991 + 0.86319 + 2(0.07322)] kg ⋅ m 2
or
I x = 1.110 kg ⋅ m 2
!
!
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1587
!
PROBLEM 9.140* (Continued)
I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4
-1
= 0 (2.76948 kg)[(0.84) 2 + (0.21) 2 ] m 2
212
2
0.21 %
$ 0.84 % $
+ (2.76948 kg) (&
+ & 0.285 −
'
2 '+
("* 2 + *
./
) m2 1
)#
/3
2!
2
/- 1
$ 0.84 % /.
+ 0 (10.62713 kg)[(0.84) 2 + 6(0.285) 2 ] m 2 + (10.62713 kg) &
m' 1
* 2
+ /3
2/12
1
!
+ ( (1.80282 kg)(0.285 m) 2 )
"4
#
1
!
+ ( (1.80282 kg)(0.285 m) 2 + (1.80282 kg)(0.84 m) 2 )
"4
#
= [(0.17302 + 0.57827) + (1.05647 + 1.87463)
+ (0.03661) + (0.03661 + 1.27207)] kg ⋅ m 2
= (0.75129 + 2.93110 + 0.03661 + 1.30868) kg ⋅ m 2
or
I y = 5.03 kg ⋅ m 2
I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4
2
1
$ 0.84 % !
m' )
= ( (2.76948 kg)(0.84 m) 2 + (2.76948 kg) &
* 2
+ )#
("12
2
-/ 1
$ 0.84 % ./
m' 1
+ 0 (10.62713 kg)[(0.84)2 + 6(0.285) 2 ] m 2 + (10.62713 kg) &
* 2
+ /3
2/12
1
!
+ ( (1.80282 kg)(0.285 m) 2 )
"4
#
1
!
+ ( (1.80282 kg)(0.285 m)2 + (1.80282 kg)(0.84 m)2 )
4
"
#
= [(0.16285 + 0.48854) + (1.05647 + 1.87463)
+ (0.03661) + (0.03661 + 1.27207)] kg ⋅ m 2
= (0.65139 + 2.93110 + 0.03661 + 1.30868) kg ⋅ m 2
or
I z = 4.93 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1588
PROBLEM 9.141
The machine element shown is fabricated from steel.
Determine the mass moment of inertia of the assembly
with respect to (a) the x axis, (b) the y axis, (c) the z axis.
(The density of steel is 7850 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
Then
m1 = (7850 kg/m3 )(π (0.08 m) 2 (0.04 m)]
= 6.31334 kg
m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg
m3 = (7850 kg/m3 )[π (0.02 m)2 (0.04 m)] = 0.39458 kg
Using Figure 9.28 and the parallel-axis theorem, we have
(a)
I x = ( I x )1 + ( I x )2 − ( I x )3
-1
.
= 0 (6.31334 kg)[3(0.08)2 + (0.04)2 ] m 2 + (6.31334 kg)(0.02 m) 2 1
212
3
-1
.
+ 0 (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)(0.03 m) 2 1
212
3
-1
.
− 0 (0.39458 kg)[3(0.02)2 + (0.04) 2 ] m 2 + (0.39458 kg)(0.02 m) 2 1
212
3
= [(10.94312 + 2.52534) + (0.23675 + 0.53269)
− (0.09207 + 0.15783)] × 10−3 kg ⋅ m 2
= (13.46846 + 0.76944 − 0.24990) × 10−3 kg ⋅ m 2
= 13.98800 × 10−3 kg ⋅ m 2
or I x = 13.99 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1589
!
PROBLEM 9.141 (Continued)
(b)
I y = ( I y )1 + ( I y )2 − ( I y )3
1
!
= ( (6.31334 kg)(0.08 m) 2 )
"2
#
1
!
+ ( (0.59188 kg)(0.02 m) 2 + (0.59188 kg)(0.04 m) 2 )
2
"
#
1
!
− ( (0.39458 kg)(0.02 m 2 ) + (0.39458 kg)(0.04 m) 2 )
"2
#
= [(20.20269) + (0.11838 + 0.94701)
− (0.07892 + 0.63133)] × 10−3 kg ⋅ m 2
= (20.20269 + 1.06539 − 0.71025) × 10−3 kg ⋅ m 2
= 20.55783 × 10−3 kg ⋅ m 2
or
(c)
I y = 20.6 × 10−3 kg ⋅ m 2
I z = ( I z )1 + ( I z ) 2 − ( I z )3
-1
.
= 0 (6.31334 kg)[3(0.08)2 + (0.04)2 ] m 2 + (6.31334 kg)(0.02 m) 2 1
212
3
-1
.
+ 0 (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)[(0.04) 2 + (0.03) 2 ] m 2 1
212
3
-1
.
− 0 (0.39458 kg)[3(0.02) 2 + (0.04) 2 ] m 2 + (0.03958 kg)[(0.04)2 + (0.02) 2 ] m 2 1
12
2
3
= [(10.94312 + 2.52534) + (0.23675 + 1.47970)
− (0.09207 + 0.78916)] × 10−3 kg ⋅ m 2
= (13.46846 + 1.71645 − 0.88123) × 10−3 kg ⋅ m 2
= 14.30368 × 10−3 kg ⋅ m 2
or
I z = 14.30 × 10−3 kg ⋅ m 2
To the Instructor:
The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the
solutions of Problems 9.142 through 9.145.
From Figure 9.28:
Cylinder
( I x ) cyl =
1
mcyl a 2
2
( I y )cyl = ( I z )cyl =
1
mcyl (3a 2 + L2 )
12
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you are using it without permission.
1590
PROBLEM 9.141 (Continued)
Symmetry and the definition of the mass moment of inertia ( I = , r 2 dm) imply
( I )semicylinder =
( I x )sc =
and
However,
1
( I )cylinder
2
1$1
%
mcyl a 2 '
2 &* 2
+
( I y )sc = ( I z )sc =
msc =
1 1
!
mcyl (3a 2 + L2 ) )
2 ("12
#
1
mcyl
2
1
msc a 2
2
Thus,
( I x )sc =
and
( I y )sc = ( I z )sc =
1
msc (3a 2 + L2 )
12
Also, using the parallel axis theorem find
$ 1 16 %
I x′ = msc & − 2 ' a 2
* 2 9π +
$ 1 16
I z′ = msc (& − 2
"* 4 9π
% 2 1 2!
' a + 12 L )
+
#
where x′ and z ′ are centroidal axes.!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1591
PROBLEM 9.142
Determine the mass moment of inertia of the steel
machine element shown with respect to the y axis.
(The specific weight of steel is 490 lb/ft 3 .)
SOLUTION
First compute the mass of each component. We have
m = ρSTV =
Then
m1 =
γ ST
g
V
490 lb/ft 3
$ 1 ft %
× (2.7 × 3.7 × 9) in.3 × &
'
32.2 ft/s 2
* 12 in. +
3
= 791.780 × 10−3 lb ⋅ s 2 /ft
490 lb/ft 3 $ π
%
$ 1 ft %
m2 =
× & × 1.352 × 1.4 ' in.3 × &
'
2
32.2 ft/s * 2
+
* 12 in. +
3
= 35.295 × 10−3 lb ⋅ s 2 /ft
m3 =
490 lb/ft 3
$ 1 ft %
× (π × 0.62 × 1.2) in.3 × &
'
2
32.2 ft/s
* 12 in. +
3
= 11.952 × 10−3 lb ⋅ s 2 /ft
3
m4 =
490 lb/ft 3
$ 1 ft %
2
−3
× (0.9 × 0.6 × 9) in.3 × &
' = 42.799 × 10 lb ⋅ s /ft
2
32.2 ft/s
* 12 in. +
The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations
derived above (component 2).
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1592
PROBLEM 9.142 (Continued)
Find: I y
We have
I y = ( I y )1 + ( I y ) 2 + ( I y )3 − ( I y ) 4
-1
= 0 (791.780 × 10−3 lb ⋅ s 2 /ft)[(2.7) 2 + (9) 2 ] in.2
212
2
2
2
$ 2.7 % $ 9 % ! 2 ./ $ 1 ft %
+ (791.780 × 10−3 lb ⋅ s 2 /ft) (&
+
×
in.
)
1 &
' & '
'
("* 2 + * 2 + )#
/3 * 12 in. +
-1
+ 0 (35.295 × 10 −3 lb ⋅ s 2 /ft)[3(1.35) 2 + (1.4) 2 ] in.2
212
1.4 %
$
+ (35.295 × 10−3 lb ⋅ s 2 /ft) ( (1.35) 2 + & 9 −
'
2 +
*
("
./ $ 1 ft %2
) in.2 1 × &
'
)#
/3 * 12 in. +
2!
-1
+ 0 (11.952 × 10 −3 lb ⋅ s 2 /ft)[3(0.6) 2 + (1.2) 2 ] in.2
212
2
2
.
1.2 %
$
2 / $ 1 ft %
in.
+ (11.952 × 10 −3 lb ⋅ s 2 /ft)[(1.35) 2 + & 9 +
×
1 &
'
2 '+
*
/3 * 12 in. +
-1
− 0 (42.799 × 10−3 lb ⋅ s 2 /ft)[(0.9) 2 + (9) 2 ] in.2
212
2
2
$ 9 % ! 2 ./ $ 1 ft %
+ (42.799 × 10 lb ⋅ s /ft) ((1.35) + & ' ) in. 1 × &
'
* 2 + )#
("
/3 * 12 in. +
= [(40.4550 + 121.3650) + (0.1517 + 17.3319)
−3
2
2
+ (0.0174 + 7.8005) − (2.0263 + 6.5603)] × 10−3 lb ⋅ ft ⋅ s 2
= (161.8200 + 17.4836 + 7.8179 − 8.5866) × 10−3 lb ⋅ ft ⋅ s 2
or
I y = 0.1785 lb ⋅ ft ⋅ s 2
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1593
!
PROBLEM 9.143
Determine the mass moment of inertia of the steel
machine element shown with respect to the z axis.
(The specific weight of steel is 490 lb/ft 3 .)
SOLUTION
First compute the mass of each component. We have
m = ρSTV =
γ ST
g
V
Then
m1 =
490 lb/ft 3
$ 1 ft %
× (2.7 × 3.7 × 9) in.3 × &
'
2
32.2 ft/s
* 12 in. +
3
= 791.780 × 10−3 lb ⋅ s 2 /ft
m2 =
490 lb/ft 3 $ π
%
$ 1 ft %
× & × 1.352 × 1.4 ' in.3 × &
'
2
32.2 ft/s * 2
+
* 12 in. +
3
= 35.295 × 10−3 lb ⋅ s 2 /ft
m3 =
490 lb/ft 3
$ 1 ft %
× (π × 0.62 × 1.2) in.3 × &
'
2
32.2 ft/s
* 12 in. +
3
= 11.952 × 10−3 lb ⋅ s 2 /ft
3
m4 =
490 lb/ft 3
$ 1 ft %
2
−3
× (0.9 × 0.6 × 9) in.3 × &
' = 42.799 × 10 lb ⋅ s /ft
2
32.2 ft/s
* 12 in. +
The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations
derived above (component 2).
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1594
PROBLEM 9.143 (Continued)
Find: I z
We have
I z = ( I z )1 + ( I z ) 2 + ( I z )3 − ( I z ) 4
-/ 1
= 0 (791.780 × 10 −3 lb ⋅ s 2 /ft)[(2.7)2 + (3.7) 2 ] in.2
/2 12
2
$ 2.7 % $ 3.7 %
+ (791.780 lb ⋅ s /ft) (&
' +&
'
("* 2 + * 2 +
2
-/
$ 1 16
+ 0 (35.295 × 10−3 lb ⋅ s 2 /ft) & − 2
* 2 9π
/2
. $ 1 ft % 2
) in. 1 × &
'
)#
/3 * 12 in. +
2!
2/
%
2
' (1.35 in.)
+
4 × 1.35 %
$
+ (35.295 × 10−3 lb ⋅ s 2 /ft) ((1.35) 2 + & 3.7 +
'
3π +
*
("
2
.
2 / $ 1 ft %
×
in.
)
1 &
'
)#
/3 * 12 in. +
2!
-1
+ 0 (11.952 × 10−3 lb ⋅ s 2 /ft)(0.6 in.) 2
212
./ $ 1 ft %2
+ (11.952 × 10−3 lb ⋅ s 2 /ft)[(1.35)2 + (3.7)2 ] in.2 1 × &
'
/3 * 12 in. +
-/ 1
− 0 (42.799 × 10−3 lb ⋅ s 2 /ft)[(0.9) 2 + (0.6) 2 ] in.2
/2 12
$ 0.6 %
+ (42.799 × 10−3 lb ⋅ s 2 /ft) ((1.35) 2 + &
'
* 2 +
"(
I z = [(9.6132 + 28.8395) + (0.1429 + 4.9219)
2
./ $ 1 ft %2
!
) in.2 1 × &
'
/3 * 12 in. +
#)
+ (0.0149 + 1.2875) − (0.0290 + 0.5684)] × 10−3 lb ⋅ ft ⋅ s 2
= (38.4527 + 5.0648 + 1.3024 − 0.5974) × 10−3 lb ⋅ ft ⋅ s 2
or
I z = 0.0442 lb ⋅ ft ⋅ s 2
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you are using it without permission.
1595
!
PROBLEM 9.144
Determine the mass moment of inertia and the radius of
gyration of the steel machine element shown with respect to
the x axis. (The density of steel is 7850 kg/m3 .)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
Then
m1 = (7850 kg/m3 )(0.09 × 0.03 × 0.15) m3
= 3.17925 kg
π
!
m2 = (7850 kg/m3 ) ( (0.045) 2 × 0.04 ) m3
"2
#
= 0.998791 kg
m3 = (7850 kg/m3 )[π (0.038) 2 × 0.03] m3 = 1.06834 kg
Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to
Problem 9.142) for a semicylinder, we have
I x = ( I x )1 + ( I x )2 − ( I x )3
1
!
= ( (3.17925 kg)(0.032 + 0.152 ) m 2 + (3.17925 kg)(0.075 m)2 )
"12
#
-/
$ 1 16
+ 0(0.998791 kg) (& − 2
"* 4 9π
2/
1
%
2
2!
' (0.045 m) + 12 (0.04 m) )
+
#
$ 4 × 0.045
%
+ (0.998791 kg) ((0.13) + &
+ 0.015 '
* 3π
+
("
2
./
) m2 1
)#
/3
2!
-1
.
− 0 (1.06834 kg)[3(0.038 m) 2 + (0.03 m)2 ] + (1.06834 kg)(0.06 m)2 1
212
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1596
PROBLEM 9.144 (Continued)
= [(0.0061995 + 0.0178833) + (0.0002745 + 0.0180409)
− (0.0004658 + 0.0038460)] kg ⋅ m 2
= (0.0240828 + 0.0183154 − 0.0043118) kg ⋅ m 2
= 0.0380864 kg ⋅ m 2
or
Now
and
I x = 38.1 × 10−3 kg ⋅ m 2
m = m1 + m2 − m3 = (3.17925 + 0.998791 − 1.06834) kg
= 3.10970 kg
k x2 =
I x 0.0380864 kg ⋅ m 2
=
m
3.10970 kg
or
k x = 110.7 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1597
!
PROBLEM 9.145
Determine the mass moment of inertia of the steel fixture shown
with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The
density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
Then
m1 = 7850 kg/m3 × (0.08 × 0.05 × 0.160) m3
= 5.02400 kg
m2 = 7850 kg/m3 × (0.08 × 0.038 × 0.07) m3 = 1.67048 kg
m3 = 7850 kg/m3 ×
$π
%
× 0.0242 × 0.04 ! m3 = 0.28410 kg
"2
#
Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have
(a)
I x = ( I x )1 − ( I x ) 2 − ( I x )3
&) 0.05 *2 ) 0.16 *2 ' 2 %(
($ 1
= - (5.02400 kg)[(0.05)2 + (0.16) 2 ] m 2 + (5.02400 kg) +
! +
! ,m .
(112
/+" 2 # " 2 # 0,
2(
2
2
&)
0.038 * )
0.07 * ' 2 (%
($ 1
− - (1.67048 kg)[(0.038)2 + (0.07) 2 ] m 2 + (1.67048 kg) + 0.05 −
+
+
0.05
,m .
2 !# "
2 !# ,0
+/"
(112
(2
&) 1 16 *
'
1
($
− -(0.28410 kg) + − 2 ! (0.024) 2 + (0.04) 2 , m 2
12
/" 4 9π #
0
1(
2
2
&)
4 × 0.024 * )
0.04 * ' 2 %(
0.16
+ (0.28410 kg) + 0.05 −
+
−
,m .
!
3π
2 !# ,0
# "
+/"
2(
= [(11.7645 + 35.2936) − (0.8831 + 13.6745) − (0.0493 + 6.0187)] × 10−3 kg ⋅ m 2
= (47.0581 − 14.5576 − 6.0680) × 10−3 kg ⋅ m 2
= 26.4325 × 10−3 kg ⋅ m 2
or
I x = 26.4 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1598
PROBLEM 9.145 (Continued)
(b)
I y = ( I y )1 − ( I y ) 2 − ( I y )3
&) 0.08 *2 ) 0.16 * 2 ' 2 %(
($ 1
2
2
2
= - (5.02400 kg)[(0.08) + (0.16) ] m + (5.02400 kg) +
! +
! ,m .
12
(2
/+" 2 # " 2 # 0,
1(
2
&) 0.08 *2 )
0.07 * ' 2 (%
($ 1
2
2
2
− - (1.67048 kg)[(0.08) + (0.07) ] m + (1.67048 kg) +
! + 0.05 + 2 ! , m .
12
# ,0
+/" 2 # "
1(
2(
2
$( 1
&) 0.08 *2 )
0.04 * ' 2 %(
2
2
2
− - (0.28410 kg)[3(0.024) + (0.04) ] m + (0.28410 kg) +
! + 0.16 − 2 ! , m .
# ,0
+/" 2 # "
(112
(2
= [(13.3973 + 40.1920) − (1.5730 + 14.7420) − (0.0788 + 6.0229)] × 10−3 kg ⋅ m 2
= (53.5893 − 16.3150 − 6.1017) × 10−3 kg ⋅ m 2
= 31.1726 × 10−3 kg ⋅ m 2
(c)
or
I y = 31.2 × 10−3 kg ⋅ m 2
I z = ( I z )1 − ( I z ) 2 − ( I z )3
$( 1
&) 0.08 *2 ) 0.05 *2 ' 2 %(
= - (5.02400 kg)[(0.08) 2 + (0.05) 2 ] m 2 + (5.02400 kg) +
! +
! ,m .
+/" 2 # " 2 # 0,
(112
2(
2
&) 0.08 *2 )
0.038 * ' 2 (%
($ 1
0.05
− - (1.67048 kg)[(0.08) 2 + (0.038) 2 ] m 2 + (1.67048 kg) +
+
−
,m .
!
2 !# 0,
+/" 2 # "
(112
2(
$(
) 1 16
− -(0.28410 kg) − 2
" 2 9π
1(
2
&) 0.08 *2 )
4 × 0.024 * ' 2 %(
*
2
(0.024
m)
(0.28410
kg)
0.05
+
+
−
+
!
!
! ,m .
3π
#
# ,0
+/" 2 # "
2(
= [(3.7261 + 11.1784) − (1.0919 + 4.2781) − (0.0523 + 0.9049)] × 10−3 kg ⋅ m 2
= (14.9045 − 5.3700 − 0.9572) × 10−3 kg ⋅ m 2
= 8.5773 × 10−3 kg ⋅ m 2
or
I z = 8.58 × 10−3 kg ⋅ m 2
To the instructor:
The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use
in the solutions of Problems 9.146 through 9.148.
Slender Rod
Ix = 0
I y′ = I z′ =
1
mL2 (Figure 9.28)
12
1
I y = I z = mL2 (Sample Problem 9.9)
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1599
PROBLEM 9.145 (Continued)
Circle
3
We have
I y = r 2 dm = ma 2
Now
Iy = Ix + Iz
And symmetry implies
Ix = Iz
Ix = Iz =
1 2
ma
2
Semicircle
Following the above arguments for a circle, We have
Ix = Iz =
1 2
ma
2
I y = ma 2
Using the parallel-axis theorem
I z = I z ′ + mx 2
or
I z′ = m
)1 4
− 2
"2 π
x=
2a
π
* 2
!a !
#
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1600
PROBLEM 9.146
Aluminum wire with a weight per unit length of 0.033 lb/ft is used
to form the circle and the straight members of the figure shown.
Determine the mass moment of inertia of the assembly with respect
to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m=
Then
W 1 )W *
L
=
g g " L !#AL
1
1 ft
× 0.033 lb/ft × (2π × 16 in.) ×
12 in.
32.2 ft/s 2
= 8.5857 × 10−3 lb ⋅ s 2 /ft
1
1 ft
m2 = m3 = m4 = m5 =
× 0.033 lb/ft × 8 in. ×
2
12 in.
32.2 ft/s
2
= 0.6832 lb ⋅ s /ft
m1 =
Using the equations given above and the parallel-axis theorem, we have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 + ( I x )5
&1
' ) 1 ft *
= + (8.5857 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 , ×
!
2
/
0 " 12 in. #
&1
' ) 1 ft *
+ + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 , ×
!
/3
0 " 12 in. #
+ [0 + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 ] ×
2
2
) 1 ft *
!
" 12 in. #
2
&1
' ) 1 ft *
+ + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(42 + 162 ) in.2 , ×
!
12
/
0 " 12 in. #
&1
' ) 1 ft *
+ + (0.6832 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(82 + 122 ) in.2 , ×
!
12
/
0 " 12 in. #
2
2
= [(7.6315) + (0.1012) + (0.3036) + (0.0253 + 1.2905) + (0.0253 + 0.9868)] × 10−3 lb ⋅ ft ⋅ s 2
= (7.6315 + 0.1012 + 0.3036) + 1.3158 + 1.0121) × 10−3 lb ⋅ ft ⋅ s 2
= 10.3642 × 10−3 lb ⋅ ft ⋅ s 2
or
I x = 10.36 × 10−3 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1601
PROBLEM 9.146 (Continued)
( I y )2 = ( I y )4 ,
( I y )3 = ( I y )5
I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5
= /&(8.5857 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 0' ×
) 1 ft *
!
" 12 in. #
+2[0 + (0.6832 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 ] ×
2
) 1 ft *
!
" 12 in. #
2
&1
' ) 1 ft *
+2 + (0.6832 × 10−3 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(12 in.) 2 , ×
!
12
/
0 " 12 in. #
2
= [(15.2635) + 2(1.2146) + 2(0.0253 + 0.6832)] × 10−3 lb ⋅ ft ⋅ s 2
= [15.2635 + 2(1.2146) + 2(0.7085)] × 10−3 lb ⋅ ft ⋅ s 2
= 19.1097 × 10−3 lb ⋅ ft ⋅ s 2
Symmetry implies
or
I y = 19.11 × 10−3 lb ⋅ ft ⋅ s 2
I z = 10.36 × 10−3 lb ⋅ ft ⋅ s 2
Ix = Iz
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1602
!
PROBLEM 9.147
The figure shown is formed of 18 -in.-diameter steel wire. Knowing that the
specific weight of the steel is 490 lb/ft3, determine the mass moment of inertia
of the wire with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. We have
m = ρSTV =
γ ST
g
AL
Then
2
3
490 lb/ft 3 & π ) 1 * '
) 1 ft *
in. ! , × (π × 18 in.) ×
×+
m1 = m2 =
!
32.2 ft/s 2 +/ 4 " 8 # ,0
" 12 in. #
= 6.1112 × 10−3 lb ⋅ s 2 /ft
m3 = m4 =
2
3
490 lb/ft 3 & π ) 1 * '
) 1 ft *
in.
18
in.
=
×
×
+
,
!
!
32.2 ft/s 2 /+ 4 " 8 # 0,
" 12 in. #
= 1.9453 × 10−3 lb ⋅ s 2 /ft
Using the equations given above and the parallel-axis theorem, we have
( I x )3 = ( I x ) 4
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4
&1
' ) 1 ft *
= + (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 , ×
!
/2
0 " 12 in. #
2
&1
' ) 1 ft *
+ + (6.1112 × 10 −3 lb ⋅ s 2 /ft)(18 in.) 2 + (6.1112 × 10 −3 lb ⋅ s 2 /ft)(18 in.) 2 , ×
!
/2
0 " 12 in. #
2
&1
' ) 1 ft *
+ 2 + (1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 + (6.1112 × 10 −3 lb ⋅ s 2 /ft)(92 + 182 ) in.2 , ×
!
/12
0 " 12 in. #
2
= [(6.8751) + (6.8751 + 13.1502) + 2(0.3647 + 5.4712)] × 10 −3 lb ⋅ ft ⋅ s 2
= [6.8751 + 20.6252 + 2(5.8359)] × 10−3 lb ⋅ ft ⋅ s 2
= 39.1721 × 10−3 lb ⋅ ft ⋅ s 2
or
I x = 0.0392 lb ⋅ ft ⋅ s 2
!
!
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1603
!
PROBLEM 9.147 (Continued)
( I y )1 = ( I y ) 2 , ( I y )3 = ( I y ) 4
I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4
) 1 ft *
= 2 &/(6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.)2 '0 ×
!
" 12 in. #
+ 2[(0 + 1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 ] ×
2
) 1 ft *
!
" 12 in. #
2
= [2(13.7502) + 2(4.3769)] × 10−3 lb ⋅ ft ⋅ s 2
= 36.2542 × 10−3 lb ⋅ ft ⋅ s 2
or
I y = 0.0363 lb ⋅ ft ⋅ s 2
or
I z = 0.0304 lb ⋅ ft ⋅ s 2
( I z )3 = ( I z ) 4
I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4
&1
' ) 1 ft *
= + (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.)2 , ×
!
/2
0 " 12 in. #
$
)1 4
+ -(6.1112 × 10−3 lb ⋅ s 2 /ft) − 2
"2 π
1
2
*
2
! (18 in.)
#
&) 2 × 18 *2
' 2 (% ) 1 ft *2
2
+ (6.1112 × 10−3 lb ⋅ s 2 /ft) +
+
(18)
in. . ×
,
!
12 in. !#
,0
/+" π #
2( "
&1
' ) 1 ft *
+ 2 + (1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 , ×
!
3
/
0 " 12 in. #
2
= [(6.8751) + (1.3024 + 19.3229) + 2(1.4590)] × 10−3 lb ⋅ ft ⋅ s 2
= [6.8751 + 20.6253 + 2(1.4590)] × 10−3 lb ⋅ ft ⋅ s2
= 30.4184 × 10−3 lb ⋅ ft ⋅ s 2
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1604
!
PROBLEM 9.148
A homogeneous wire with a mass per unit length of 0.056 kg/m is
used to form the figure shown. Determine the mass moment of inertia
of the wire with respect to each of the coordinate axes.
SOLUTION
First compute the mass m of each component. We have
m = (m/L) L
= 0.056 kg/m × 1.2 m
= 0.0672 kg
Using the equations given above and the parallel-axis theorem, we have
I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4 + ( I x )5 + ( I x )6
Now
Then
( I x )1 = ( I x )3 = ( I x ) 4 = ( I x )6
and ( I x )2 = ( I x )5
&1
'
I x = 4 + (0.0672 kg)(1.2 m) 2 , + 2[0 + (0.0672 kg)(1.2 m) 2 ]
/3
0
= [4(0.03226) + 2(0.09677)] kg ⋅ m 2
= 0.32258 kg ⋅ m 2
or
I x = 0.323 kg ⋅ m 2
I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5 + ( I y )6
Now
Then
( I y )1 = 0, ( I y ) 2 = ( I y )6 , and ( I y )4 = ( I y )5
&1
'
I y = 2 + (0.0672 kg)(1.2 m) 2 , + [0 + (0.0672 kg)(1.2 m) 2 ]
/3
0
&1
'
+ 2 + (0.0672 kg)(1.2 m) 2 + (0.0672 kg)(1.22 + 0.62 ) m 2 ,
12
/
0
= [2(0.03226) + (0.09677) + 2(0.00806 + 0.12096)] kg ⋅ m 2
= [2(0.03226) + (0.09677) + 2(0.12902)] kg ⋅ m 2
= 0.41933 kg ⋅ m 2
Symmetry implies
or
I y = 0.419 kg ⋅ m 2
I z = 0.419 kg ⋅ m 2
I y = Iz
!
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1605
PROBLEM 9.149
Determine the mass products of inertia Ixy, Iyz, and Izx of the steel
fixture shown. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = ρV
m1 = 7850 kg/m3 × (0.08 × 0.05 × 0.16) m3 = 5.02400 kg
Then
m2 = 7850 kg/m3 × (0.08 × 0.038 × 0.07) m3 = 1.67048 kg
m3 = 7850 kg/m3 ×
)π
*
× 0.0242 × 0.04 ! m3 = 0.28410 kg
2
"
#
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of
symmetry. Now
0.038 *
)
m = 0.031 m
y2 = 0.05 −
2 !#
"
4 × 0.024 *
)
y3 = 0.05 −
! m = 0.039814 m
3π
"
#
and then
mx y , kg ⋅ m 2
my z , kg ⋅ m 2
mz x , kg ⋅ m 2
0.08
5.0240 × 10−3
10.0480 × 10−3
16.0768 × 10−3
0.031
0.085
2.0714 × 10−3
4.4017 × 10−3
5.6796 × 10−3
0.039814
0.14
0.4524 × 10−3
1.5836 × 10−3
1.5910 × 10−3
m, kg
x, m
y, m
1
5.02400
0.04
0.025
2
1.67048
0.04
3
0.28410
0.04
z, m
Finally,
I xy = ( I xy )1 − ( I xy ) 2 − ( I xy )3 = [(0 + 5.0240) − (0 + 2.0714) − (0 + 0.4524)] × 10−3
= 2.5002 × 10−3 kg ⋅ m 2
or
I xy = 2.50 × 10−3 kg ⋅ m 2
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1606
PROBLEM 9.149 (Continued)
I yz = ( I yz )1 − ( I yz ) 2 − ( I yz )3 = [(0 + 10.0480) − (0 + 4.4017) − (0 + 1.5836)] × 10−3
= 4.0627 × 10−3 kg ⋅ m 2
or
I yz = 4.06 × 10−3 kg ⋅ m 2
I zx = ( I zx )1 − ( I zx ) 2 − ( I zx )3 = [(0 + 16.0768) − (0 + 5.6796) − (0 + 1.5910)] × 10−3
= 8.8062 × 10−3 kg ⋅ m 2
or
I zx = 8.81 × 10−3 kg ⋅ m 2
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1607
PROBLEM 9.150
Determine the mass products of inertia Ixy, Iyz, and Izx of
the steel machine element shown. (The density of steel
is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. We have
m = ρSTV
m1 = 7850 kg/m3 × (0.1 × 0.018 × 0.09) m3 = 1.27170 kg
Then
m2 = 7850 kg/m3 × (0.016 × 0.06 × 0.05) m3 = 0.37680 kg
m3 = 7850 kg/m3 × (π × 0.0122 × 0.01) m3 = 0.03551 kg
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of
symmetry. Now
m, kg
x, m
y, m
z, m
mx y, kg ⋅ m 2
my z , kg ⋅ m 2
mz x , kg ⋅ m 2
1
1.27170
0.05
0.009
0.045
0.57227 × 10−3
0.51504 × 10−3
2.86133 × 10−3
2
0.37680
0.092
0.048
0.045
1.66395 × 10−3
0.81389 × 10−3
1.55995 × 10−3
3
0.03551
0.105
0.054
0.045
0.20134 × 10−3
0.08629 × 10−3
0.16778 × 10−3
2.43756 × 10−3
1.41522 × 10−3
4.58906 × 10−3
Σ
Then
0
+ mx y )
or
I xy = 2.44 × 10−3 kg ⋅ m 2
0
= Σ( I y′z ′ + my z )
or
I yz = 1.415 × 10−3 kg ⋅ m 2
or
I zx = 4.59 × 10−3 kg ⋅ m 2
I xy = Σ( I x′y′
I yz
0
I zx = Σ( I z′x′ + mz x )
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1608
PROBLEM 9.151
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast
aluminum machine component shown. (The specific weight of
aluminum is 0.100 lb/in.3)
SOLUTION
First compute the mass of each component. We have
m = ρ ALV =
Then
m1 =
γ AL
g
V
0.100 lb/in.3
× (7.8 × 0.8 × 1.6)in.3
32.2 ft/s 2
= 31.0062 × 10−3 lb ⋅ s 2 /ft
m2 =
0.100 lb/in.3
× (2.4 × 0.8 × 2) in.3
2
32.2 ft/s
= 11.9255 × 10−3 lb ⋅ s 2 /ft
m3 =
0.100 lb/in.3
π
!
× " (0.8)2 × 0.8# in.3 = 2.4977 lb ⋅ s 2 /ft
2
32.2 ft/s 2
$
%
m4 =
0.100 lb/in.3
× [π (0.55)2 × 0.6]in.3 = 1.7708 lb ⋅ s 2 /ft
2
32.2 ft/s
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry. Now
4 × 0.8 '
&
x3 = − ( 7.8 +
in. = −8.13953 in.
3π )+
*
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1609
PROBLEM 9.151 (Continued)
and then
m, lb ⋅ s 2 /ft
x , ft
y , ft
z , ft
mx y , lb ⋅ ft ⋅ s 2
my z , lb ⋅ ft ⋅ s 2
mz x , lb ⋅ ft ⋅ s 2
1
31.0062 × 10−3
−
3.9
12
0.4
12
−
0.8
12
− 335.901 × 10−6
− 68.903 × 10−6
671.801 × 10−6
2
11.9255 × 10−3
−
1.2
12
0.4
12
−
2.6
12
− 39.752 × 10−6
−86.129 × 10−6
258.386 × 10−6
3
2.4977 × 10−3
8.13953
12
0.4
12
−
0.8
12
− 56.473 × 10−6
− 5.550 × 10−6
112.945 × 10−6
4
1.7708 × 10−3
7.8
12
1.1
12
−
0.8
12
−105.511 × 10−6
−10.822 × 10−6
76.735 × 10−6
− 537.637 × 10−6
−171.404 × 10−6
1119.867 × 10−6
Σ
Then
−
−
0
I xy = Σ( I x′y′ + mx y )
0
I yz = Σ( I y ′z ′ + m y z )
or
I xy = −538 × 10−6 lb ⋅ ft ⋅ s 2
or
I yz = −171.4 × 10−6 lb ⋅ ft ⋅ s 2
0
I zx = Σ( I z ′x′ + mz x )
or
I zx = 1120 × 10−6 lb ⋅ ft ⋅ s 2
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1610
PROBLEM 9.152
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast
aluminum machine component shown. (The specific weight of
aluminum is 0.100 lb/in.3)
SOLUTION
First compute the mass of each component. We have
m = ρ ALV =
Then
m1 =
γ AL
g
V
0.100 lb/in.3
× (5.9 × 1.8 × 2.2) in.3
32.2 ft/s 2
= 72.5590 × 10−3 lb ⋅ s 2 /ft
m2 =
0.100 lb/in.3
π
!
× " (1.1)2 × 1.8# in.3
2
32.2 ft/s
$2
%
= 10.6248 × 10−3 lb ⋅ s 2 /ft
m3 =
0.100 lb/in.3
× (1.4 × 1.1 × 1.2) in.3
32.2 ft/s 2
= 5.7391 × 10−3 lb ⋅ s 2 /ft
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry. Now
4 × 1.1 '
&
in.
x2 = − ( 5.9 +
3π )+
*
= − 6.36685 in.
1.1 '
&
in.
y3 = (1.8 −
2 )+
*
= 1.25 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1611
PROBLEM 9.152 (Continued)
and then
m, lb ⋅ s 2 /ft
1
72.5590 × 10−3
2
10.6248 × 10−3
3
5.7391 × 10−3
Finally,
x , ft
y , ft
z , ft
mx y , lb ⋅ ft ⋅ s 2
my z , lb ⋅ ft ⋅ s 2
mz x , lb ⋅ ft ⋅ s 2
2.95
12
0.9
12
1.1
12
−1.33781 × 10−3
0.49884 × 10−3
−1.63510 × 10−3
6.36685
12
0.9
12
1.1
12
− 0.42279 × 10−3
0.07305 × 10−3
− 0.51674 × 10−3
0.7
12
1.25
12
1.3
12
− 0.03487 × 10−3
0.06476 × 10−3
− 0.03627 × 10−3
−
−
−
I xy = ( I xy )1 + ( I xy ) 2 − ( I xy )3
= [(0 − 1.33781) + (0 − 0.42279) − (0 − 0.03487)] × 10−3
or
I xy = −1.726 × 10−3 lb ⋅ ft ⋅ s 2
I yz = ( I yz )1 + ( I yz ) 2 − ( I yz )3
= [(0 + 0.49884) + (0 + 0.07305) − (0 + 0.06476)] × 10−3
or
I yz = 0.507 × 10−3 lb ⋅ ft ⋅ s 2
I zx = ( I zx )1 + ( I zx )2 − ( I zx )3
= [(0 − 1.63510) + (0 − 0.51674) − (0 − 0.3627)] × 10 −3
or
I zx = −2.12 × 10 −3 lb ⋅ ft ⋅ s 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1612
PROBLEM 9.153
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of
steel is 7850 kg/m3, determine the mass products of inertia
Ixy, Iyz, and Izx of the component.
SOLUTION
m1 = ρV = 7850 kg/m3 (0.2 m)(0.3 m)(0.002 m) = 0.942 kg
m2 = ρV = 7850 kg/m3 (0.4 m)(0.3 m)(0.002 m) = 1.884 kg
For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes.
!
Σ
m, kg
x, m
y, m
z, m
mx y
my z
mz x
0.942
0.15
0.1
0.4
+ 14.13 × 10−3
+ 37.68 × 10−3
+ 56.52 × 10−3
1.884
0.15
0
0.2
0
0
+ 56.52 × 10−3
+ 14.13 × 10−3
+ 37.68 × 10−3
+ 113.02 × 10−3
I xy = + 14.13 × 10−3 kg ⋅ m 2
I xy = + 14.13 g ⋅ m 2
I yz = + 37.68 × 10−3 kg ⋅ m 2
I yz = + 37.7 g ⋅ m 2
I zx = + 113.02 × 10−3 kg ⋅ m 2
I zx = + 113.0 g ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1613
PROBLEM 9.154
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of steel
is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz,
and Izx of the component.
SOLUTION
First compute the mass of each component.
We have
Then
m = ρSTV = ρST tA
m1 = (7850 kg/m3 )[(0.002)(0.4)(0.45)]m3
= 2.8260 kg
&1
'!
m2 = 7850 kg/m3 "(0.002) ( × 0.45 × 0.18 ) # m3
2
*
+%
$
= 0.63585 kg
Now observe that
( I x′y′ )1 = ( I y′z ′ )1 = ( I z′x′ )1 = 0
( I y′z′ ) 2 = ( I z′x′ ) 2 = 0
From Sample Problem 9.6:
( I x′y′ )2,area = −
1 2 2
b2 h2
72
Then
1
& 1
'
( I x′y′ ) 2 = ρST t ( I x′y′ ) 2,area = ρST t ( − b22 h22 ) = − m2 b2 h2
72
36
*
+
Also
x1 = y1 = z2 = 0
0.45 '
&
x2 = ( − 0.225 +
m = − 0.075 m
3 )+
*
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1614
PROBLEM 9.154 (Continued)
Finally,
1
I xy = Σ( I xy + mx y ) = (0 + 0) + " − (0.63585 kg)(0.45 m)(0.18 m)
$ 36
& 0.18 m ' !
+ (0.63585 kg)(− 0.075 m) (
)#
* 3 +%
= (−1.43066 × 10−3 − 2.8613 × 10−3 ) kg ⋅ m 2
and
or
I xy = − 4.29 × 10−3 kg ⋅ m 2
or
I yz = 0
or
I zx = 0
I yz = Σ( I y′z′ + m y z ) = (0 + 0) + (0 + 0) = 0
I zx = Σ( I z ′x′ + m z x ) = (0 + 0) + (0 + 0) = 0
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1615
PROBLEM 9.155
A section of sheet steel 2 mm thick is cut and bent into the
machine component shown. Knowing that the density of steel is
7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and
Izx of the component.
SOLUTION
First compute the mass of each component. We have
m = ρSTV = ρST tA
Then
m1 = (7850 kg/m3 )(0.002 m)(0.35 × 0.39) m 2
= 2.14305 kg
&π
'
m2 = (7850 kg/m3 )(0.002 m) ( × 0.1952 ) m 2
*2
+
= 0.93775 kg
&1
'
m3 = (7850 kg/m3 )(0.002 m) ( × 0.39 × 0.15 ) m 2
*2
+
= 0.45923 kg
Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero and
( I x′y′ )3 = ( I z′x′ )3 = 0
Also
( I y′z ′ )3,mass = ρST t ( I y′z′ )3,area
Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to
a 90° rotation, we have
( I y′z ′ )3,area =
Then
Also
1 2 2
b3 h3
72
& 1
' 1
m3b3 h3
( I y′z ′ )3 = ρST t ( b32 h32 ) =
72
*
+ 36
y1 = x2 = 0
y2 =
4 × 0.195
m = 0.082761 m
3π
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1616
PROBLEM 9.155 (Continued)
Finally,
I xy = Σ( I x′y′ + mx y )
& 0.15 ' !
= (0 + 0) + (0 + 0) + "0 + (0.45923 kg)(0.35 m) ( −
m )#
3
*
+%
$
= − 8.0365 × 10−3 kg ⋅ m 2
or I xy = − 8.04 × 10−3 kg ⋅ m 2
I yz = Σ( I y′z′ + my z )
= (0 + 0) + [0 + (0.93775 kg)(0.082761 m)(0.195 m)]
1
& 0.15 '& 0.39 ' !
m )(
m )#
+ " (0.45923 kg)(0.39 m)(0.15 m) + (0.45923 kg) ( −
3
*
+* 3
+%
$ 36
= [(15.1338) + (0.7462 − 2.9850)] × 10−3 kg ⋅ m 2
= (15.1338 − 2.2388) × 10−3 kg ⋅ m 2
= 12.8950 × 10−3 kg ⋅ m 2
or I yz = 12.90 × 10−3 kg ⋅ m 2
I zx = Σ( I z ′x′ + mz x )
= [0 + (2.14305 kg)(0.175 m)(0.195 m)] + (0 + 0)
!
& 0.39 '
m ) (0.35 m) #
+ "0 + (0.45923 kg) (
3
*
+
$
%
= (73.1316 + 20.8950) × 10−3 kg ⋅ m 2
= 94.0266 × 10−3 kg ⋅ m 2
or I zx = 94.0 × 10−3 kg ⋅ m 2
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1617
!
PROBLEM 9.156
A section of sheet steel 2 mm thick is cut and bent into the machine
component shown. Knowing that the density of steel is 7850 kg/m3,
determine the mass products of inertia Ixy, Iyz, and Izx of the component.
SOLUTION
First compute the mass of each component. We have
m = ρSTV = ρST tA
Then
&1
'
m1 = m3 = (7850 kg/m3 )(0.002 m) ( × 0.225 × 0.135 ) m 2 = 0.23844 kg
*2
+
&π
'
m2 = (7850 kg/m3 )(0.002 m) ( × 0.2252 ) m 2 = 0.62424 kg
*4
+
π
&
'
m4 = (7850 kg/m3 )(0.002 m) ( × 0.1352 ) m 2 = 0.22473 kg
*4
+
Now observe that the following centroidal products of inertia are zero because of symmetry.
Also
Now
( I x′y′ )1 = ( I y′z′ )1 = 0
( I x′y′ ) 2 = ( I y′z′ )2 = 0
( I x′y′ )3 = ( I z′x′ )3 = 0
( I x′y′ ) 4 = ( I z′x′ ) = 0
y1 = y2 = 0
x3 = x4 = 0
I xy = Σ( I x′y′ + mx y )
I xy = 0
so that
Using the results of Sample Problem 9.6, we have
I uv ,area =
Now
1 2 2
b h
24
I uv ,mass = ρST t I uv ,area
& 1
'
= ρST t ( b2 h 2 )
24
*
+
1
= mbh
12
Thus,
( I zx )1 =
1
m1b1h1
12
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1618
PROBLEM 9.156 (Continued)
( I yz )3 = −
While
1
m3b3 h3
12
because of a 90° rotation of the coordinate axes.
To determine I uv for a quarter circle, we have
0
dI uv = d I u ′v′ + uEL vEL dm
1
1 2
v=
a − u2
2
2
uEL = u vEL =
Where
dm = ρST t dA = ρST tv du = ρST t a 2 − u 2 du
,
I uv = dI uv =
Then
=
1
ρST t
2
,
,
a
0
a
0
&1 2
'
(u ) (
a − u 2 ) ρST t a 2 − u 2 du
2
*
+
(
)
u (a 2 − u 2 ) du
a
1
1
1 !
1
1
= ρST t " a 2 u 2 − u 4 # = ρST t a 4 =
ma 4
2
4 %0 8
2π
$2
( I zx ) 2 = −
Thus
1
m2 a22
2π
because of a 90° rotation of the coordinate axes. Also
( I yz ) 4 =
Finally,
I yz
1
m4 a42
2π
0
0
= Σ( I yz ) = [( I y′z′ ) + m1 y1 z1 ] + [( I y′z′ ) + m2 y2 z2 ] + ( I yz )3 + ( I yz ) 4
1
1
!
!
= " − (0.23844 kg)(0.225 m)(0.135 m) # + " (0.22473 kg)(0.135 m) 2 #
$ 12
% $ 2π
%
= (− 0.60355 + 0.65185) × 10−3 kg ⋅ m 2
or
I yz = 48.3 × 10−6 kg ⋅ m 2
0
0
I zx = Σ( I zx ) = ( I zx )1 + ( I zx ) 2 + [( I z ′x′ )3 + m3 z3 x3 ] + [( I z′x′ ) 4 + m4 z4 x4 ]
1
1
!
!
= " (0.23844 kg)(0.225 m)(0.135 m) # + " −
(0.62424 kg)(0.225 m)2 #
$12
% $ 2π
%
= (0.60355 − 5.02964) × 10−3 kg ⋅ m 2
or
I zx = − 4.43 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1619
PROBLEM 9.157
Brass wire with a weight per unit length w is used to form the figure shown.
Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m=
W 1
= wL
g g
W
w
(2π × a ) = 2π a
g
g
w
w
m2 = (a) = a
g
g
w
w
m3 = (2a) = 2 a
g
g
3 '
w&
w
m4 = ( 2π × a ) = 3π a
2 +
g*
g
m1 =
Then
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry.
m
1
2
2π
w
a
g
3
2
4
3π
Σ
w
a
g
w
a
g
w
a
g
x
y
z
2a
a
−a
2a
1
a
2
0
w 3
a
g
0
2a
0
a
0
0
2a
3
− a
2
2a
mx y
w 3
a
g
4π
− 9π
w 3
a
g
w
(1 − 5π )a 3
g
my z
mz x
w 3
a
g
−2π
− 4π
w 3
a
g
0
4
− 9π
w 3
a
g
−11π
w 3
a
g
w 3
a
g
12π
4
w 3
a
g
w
(1 + 2π )a3
g
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1620
PROBLEM 9.157 (Continued)
Then
0
I xy = Σ( I x′y′ + mx y )
or
0
= Σ( I y′z′ + my z )
or
I yz
I xy =
I yz = −11π
0
I zx = Σ( I z′x′ + mz x )
or
w 3
a (1 − 5π )
g
I zx = 4
w 3
a
g
w 3
a (1 + 2π )
g
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1621
PROBLEM 9.158
Brass wire with a weight per unit length w is used to form the figure shown.
Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
m=
W 1
= wL
g g
W&
3 ' 3 w
(π × a ) = π a
2 + 2 g
g*
w
w
m2 = (3a ) = 3 a
g
g
w
w
m3 = (π × a ) = π a
g
g
m1 =
Then
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of
symmetry.
m
1
3 w
π a
2 g
2
3
w
a
g
3
π
w
a
g
x
−
y
z
2a
1
a
2
0
1
a
2
2a
(a )
−a
a
2&3 '
a
π (* 2 )+
2
π
mx y
w 3
a
g
−9
0
−2
w 3
a
g
−11
Σ
w 3
a
g
my z
3 w 3
π a
2 g
3
mz x
−
9w 3
a
4g
w 3
a
g
−π
w 3
a
g
w&π
'
+ 3 ) a3
(
g*2
+
0
2
−
w 3
a
g
1w 3
a
4g
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1622
PROBLEM 9.158 (Continued)
Then
0
I xy = Σ( I x′y′ + mx y )
I xy = −11
or
0
I yz = Σ( I y′z′ + my z )
or
0
+ mz x )
or
I zx = Σ( I z′x′
I yz =
w 3
a
g
1w 3
a (π + b)
2g
I zx = −
1w 3
a
4g
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1623
PROBLEM 9.159
The figure shown is formed of 1.5-mm-diameter aluminum
wire. Knowing that the density of aluminum is 2800 kg/m3,
determine the mass products of inertia Ixy, Iyz, and Izx of the
wire figure.
SOLUTION
First compute the mass of each component. We have
m = ρ ALV = ρ AL AL
Then
π
!
m1 = m4 = (2800 kg/m3 ) " (0.0015 m) 2 # (0.25 m)
$4
%
= 1.23700 × 10−3 kg
π
!
m2 = m5 = (2800 kg/m3 ) " (0.0015 m)2 # (0.18 m)
$4
%
= 0.89064 × 10−3 kg
π
!
m3 = m6 = (2800 kg/m3 ) " (0.0015 m) 2 # (0.3 m)
$4
%
= 1.48440 × 10−3 kg
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1624
PROBLEM 9.159 (Continued)
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry.
m, kg
x, m
y, m
z, m
mx y , kg ⋅ m 2
my z , kg ⋅ m 2
mz x , kg ⋅ m 2
1
1.23700 × 10−3
0.18
0.125
0
27.8325 × 10−6
0
0
2
0.89064 × 10−3
0.09
0.25
0
20.0394 × 10−6
0
0
3
1.48440 × 10−3
0
0.25
0.15
0
55.6650 × 10−6
0
4
1.23700 × 10−3
0
0.125
0.3
0
46.3875 × 10−6
0
5
0.89064 × 10−3
0.09
0
0.3
0
0
24.0473 × 10−6
6
1.48440 × 10−3
0.18
0
0.15
0
0
40.0788 × 10−6
47.8719 × 10−6
102.0525 × 10−6
64.1261 × 10−6
Σ
Then
0
I xy = Σ( I x′y′ + mx y )
0
I yz = Σ( I y′z′ + my z )
0
I zx = Σ( I z′x′ + mz x )
or
I xy = 47.9 × 10−6 kg ⋅ m 2
or
I yz = 102.1 × 10−6 kg ⋅ m 2
or
I zx = 64.1 × 10−6 kg ⋅ m 2
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1625
PROBLEM 9.160
Thin aluminum wire of uniform diameter is used to form the figure shown.
Denoting by m′ the mass per unit length of the wire, determine the mass
products of inertia Ixy, Iyz, and Izx of the wire figure.
SOLUTION
First compute the mass of each component. We have
&m'
m = ( ) L = m′L
*L+
Then
&π ' π
m1 = m5 = m′ ( R1 ) = m′R1
*2 + 2
m2 = m4 = m′( R2 − R1 )
&π
' π
m3 = m′ ( R2 ) = m′R2
*2 + 2
Now observe that because of symmetry the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of components
2 and 4 are zero and
( I x′y′ )1 = ( I z′x′ )1 = 0
( I x′y′ )3 = ( I y′z′ )3 = 0
( I y′z ′ )5 = ( I z′x′ )5 = 0
Also
x1 = x2 = 0
y2 = y3 = y4 = 0
z4 = z5 = 0
Using the parallel-axis theorem [Equations (9.47)], it follows that I xy = I yz = I zx for components 2 and 4.
To determine I uv for one quarter of a circular arc, we have dI uv = uvdm
where
and
u = a cos θ
v = a sin θ
dm = ρ dV = ρ [ A( adθ )]
where A is the cross-sectional area of the wire. Now
&π '
&π '
m = m′ ( a ) = ρ A ( a )
2
*
+
*2 +
so that
and
dm = m′adθ
dI uv = (a cos θ )(a sin θ )(m′adθ )
= m′a3 sin θ cos θ dθ
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1626
PROBLEM 9.160 (Continued)
,
I uv = dI uv =
Then
,
π /2
0
m′a3 sin θ cos θ dθ
π /2
1
!
= m′a3 " sin 2 θ #
$2
%0
=
1
m′a3
2
1
m′R13
2
Thus,
( I yz )1 =
and
1
( I zx )3 = − m′R23
2
1
( I xy )5 = − m′R13
2
because of 90° rotations of the coordinate axes. Finally,
I xy
0
0
= Σ( I xy ) = [( I x′y ′ )1 + m1 x1 y1 ] + [( I x′y′ )3 + m3 x3 y3 ] + ( I xy )5
or
0
0
1
I xy = − m′R13
2
I yz = Σ( I yz ) = ( I yz )1 + [( I y′z′ )3 + m3 y3 z3 ] + [( I y′z′ )5 + m5 y5 z5 ]
or
0
0
I zx = Σ( I zx ) = [( I z′x′ )1 + m1 z1 x1 ] + ( I zx )3 + [( I z ′x′ )5 + m5 z5 x5 ]
or
I yz =
1
m′R13
2
1
I zx = − m′R23
2
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1627
PROBLEM 9.161
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of inertia.
SOLUTION
We have
and
Consider
,
,
,
I xy = xydm I yz = yzdm I zx = zxdm
x = x′ + x
y = y′ + y
(9.45)
z = z′ + z
(9.31)
,
I xy = xydm
Substituting for x and for y
,
= , x′y ′dm + y , x′dm + x , y ′dm + x y , dm
I xy = ( x′ + x )( y ′ + y ) dm
By definition
and
,
I x′y′ = x′y ′dm
, x′dm = mx ′
, y′dm = my ′
However, the origin of the primed coordinate system coincides with the mass center G, so that
x′ = y′ = 0
I xy = I x′y′ + mx y Q.E.D.
The expressions for I yz and I zx are obtained in a similar manner.
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1628
PROBLEM 9.162
For the homogeneous tetrahedron of mass m shown, (a) determine
by direct integration the mass product of inertia Izx, (b) deduce Iyz
and Ixy from the result obtained in Part a.
SOLUTION
(a)
First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown.
a
z'
&
z + a = a (1 − )
c
c
*
+
Now
x=−
and
b
z'
&
y = − z + b = b (1 − )
c
* c+
The mass dm of the slab is
2
z'
&1
' 1
&
dm = ρ dV = ρ ( xydz ) = ρ ab (1 − ) dz
2
2
c
*
+
*
+
Then
,
m = dm =
,
c
0
2
z'
1
&
ρ ab (1 − ) dz
c
2
*
+
z'
1
& c '&
= ρ ab "( − )(1 − )
2
3
c
+*
+
"$*
Now
where
3 !c
1
# = ρ abc
#% 0 6
dI zx = dI z′x′ + zEL xEL dm
dI z′x′ = 0 (symmetry)
and
zEL = z
Then
I zx = dI zx =
,
xEL =
1
1 &
z'
x = a (1 − )
3
3 * c+
2
!
z '! 1
z'
1 &
&
z " a (1 − ) # " ρ ab (1 − ) dz #
0
* c+
#%
$ 3 * c + % $" 2
2
3
4
c&
1
z
z
z '
= ρ a 2 b (( z − 3 + 3 2 − 3 )) dz
0
6
c
c
c +
*
,
c
,
c
m 1
z3 3 z 4 1 z5 !
= a " z2 − +
−
#
c $2
c 4 c 2 5 c3 % 0
or
I zx =
1
mac
20
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1629
PROBLEM 9.162 (Continued)
(b)
Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular
permutation of ( x, y, z ) and (a, b, c) .
Thus,
!
I xy =
1
mab
20
!
I yz =
1
mbc
20
!
Alternative solution for Part a:
First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown.
Now
x=−
a
y'
&
y + a = a (1 − )
b
b+
*
and
z=−
c
y'
&
y + c = c (1 − )
b
* b+
The mass dm of the slab is
2
y'
&1
' 1
&
dm = ρ dV = ρ ( xzdy ) = ρ ac (1 − ) dy
*2
+ 2
* b+
Now
where
and dI zx,area =
Then
dI zx = ρ tdI zx ,area
t = dy
1 2 2
x z from the results of Sample Problem 9.6.
24
2
2
-/ 1
y' ! &
y ' ! ./
&
dIzx = ρ (dy ) 0 " a (1 − ) # "c (1 − ) # 1
b + % $ * b +% /
/2 24 $ *
3
4
=
Finally,
4
1
1m &
y'
y'
&
ac (1 − ) dy
ρ a 2 c 2 (1 − ) dy =
24
4b *
b+
b+
*
,
I zx = dI zx =
,
b
0
4
1m &
y'
ac (1 − ) dy
4 b * b+
b
5
y' !
1m
& b '&
ac "( − )(1 − ) #
=
b + %0
4 b $* 5 +*
or
I zx =
1
mac
20
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1630
PROBLEM 9.162 (Continued)
Alternative solution for Part a:
The equation of the included face of the tetrahedron is
x y z
+ + =1
a b c
x z'
&
y = b (1 − − )
a
c+
*
so that
For an infinitesimal element of sides dx, dy, and dz :
dm = ρ dV = ρ dydxdz
z'
&
x = a (1 − )
* c+
From Part a
Now
,
I zx = zxdm =
c
a (1− z/c )
0
0
,,
=ρ
= ρb
= ρb
=
b (1− x/a − z/c )
0
c
a (1− z/c )
0
0
,,
= ρb
,
,
,
,
c
0
c
0
c
0
zx( ρ dydxdz )
zx $b (1 − ax −
z
c
)!% dxdz
a (1− z/c )
1
1 x3 1 z 2 !
z " x2 −
x #
−
3 a 2 c %0
$2
2
dz
3
1 &
1
z'
z' 1 z 2&
z'
&
z " a 2 (1 − ) − a3 (1 − ) −
a (1 − )
2
3
2
c
a
c
c
c
*
+
*
+
*
+
"$
2!
# dz
#%
3
1 2 &
z'
a z (1 − ) dz
6
* c+
1 2
ρa b
6
,
c&
z2
z3 z4 '
(( z − 3 + 3 2 − 3 )) dz
0
c
c
c +
*
c
m 1
z3 3 z4 1 z5 !
= a " z2 − +
−
#
c $2
c 4 c 2 5 c3 % 0
or I zx =
1
mac
20
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1631
PROBLEM 9.163
The homogeneous circular cylinder shown has a mass m. Determine the mass
moment of inertia of the cylinder with respect to the line joining the origin O
and Point A that is located on the perimeter of the top surface of the cylinder.
SOLUTION
Iy =
From Figure 9.28:
1
ma 2
2
and using the parallel-axis theorem
2
Ix = Iz =
Symmetry implies
1
1
&h'
m(3a 2 + h 2 ) + m ( ) = m(3a 2 + 4h 2 )
12
2
12
* +
I xy = I yz = I zx = 0
For convenience, let Point A lie in the yz plane. Then
λOA =
1
2
h + a2
( hj + ak )
With the mass products of inertia equal to zero, Equation (9.46) reduces to
0
I OA = I x λx2 + I y λ y2 + I z λz2
&
1
h
= ma 2 (
( h2 + a 2
2
*
2
'
&
1
a
) + m(3a 2 + 4h2 ) (
) 12
( h2 + a 2
+
*
or
'
)
)
+
2
I OA =
1
10h 2 + 3a 2
ma 2
12
h2 + a 2
Note: For Point A located at an arbitrary point on the perimeter of the top surface, λOA is given by
λOA =
1
2
h + a2
( a cos φ i + hj + a sin φ k )
which results in the same expression for I OA . !
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1632
PROBLEM 9.164
The homogeneous circular cone shown has a mass m. Determine
the mass moment of inertia of the cone with respect to the line
joining the origin O and Point A.
SOLUTION
2
9
&3 '
= ( a ) + (−3a) 2 + (3a) 2 = a
2
*2 +
First note that
dOA
Then
λOA =
1 &3
' 1
ai − 3aj + 3ak ) = (i − 2 j + 2k )
(
a*2
+ 3
9
2
For a rectangular coordinate system with origin at Point A and axes aligned with the given x, y , z axes,
we have (using Figure 9.28)
3
1
!
I x = I z = m " a 2 + (3a) 2 #
5 $4
%
=
Also, symmetry implies
Iy =
3
ma 2
10
111 2
ma
20
I xy = I yz = I zx = 0
With the mass products of inertia equal to zero, Equation (9.46) reduces to
I OA = I x λx2 + I y λ y2 + I z λz2
2
2
111 2 & 1 '
3
& 2 ' 111 2 & 2 '
=
ma ( ) + ma 2 ( − ) +
ma ( )
20
20
* 3 + 10
* 3+
*3+
=
193 2
ma
60
or
2
I OA = 3.22ma 2
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1633
!
PROBLEM 9.165
Shown is the machine element of Problem 9.141. Determine its
mass moment of inertia with respect to the line joining the origin O
and Point A.
SOLUTION
First compute the mass of each component.
We have
m = ρSTV =
0.284 lb/in.3
V = (0.008819 lb ⋅ s 2 /ft ⋅ in.3 )V
2
32.2 ft/s
Then
m1 = (7850 kg/m3 )[π (0.08 m)2 (0.04 m)] = 6.31334 kg
m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg
m3 = (7850 kg/m3 )[π (0.02 m) 2 (0.04 m)] = 0.39458 kg
Symmetry implies
and
Now
I yz = I zx = 0
( I xy )1 = 0
( I x′y′ ) 2 = ( I x′y′ )3 = 0
I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 − m3 x3 y3
= [0.59188 kg (0.04 m)(0.03 m)] − [0.39458 kg (− 0.04 m)(− 0.02 m)]
= (0.71026 − 0.31566) × 10−3 kg ⋅ m 2
= 0.39460 × 10−3 kg ⋅ m 2
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1634
PROBLEM 9.165 (Continued)
From the solution to Problem 9.141, we have
I x = 13.98800 × 10−3 kg ⋅ m 2
I y = 20.55783 × 10−3 kg ⋅ m 2
I z = 14.30368 × 10−3 kg ⋅ m 2
By observation
OA
=
1
13
(2i + 3j)
Substituting into Eq. (9.46)
I OA =
I x λx2
+
I y λ y2
+
0
0
0
− 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
I z λz2
2
2
& 2 '
& 3 '
= "(13.98800) (
) + (20.55783) (
)
13
"$
*
+
* 13 +
& 2 '& 2 ' !
2
−3
− 2(0.39460) (
)(
) # × 10 kg ⋅ m
* 13 +* 13 + %
= (4.30400 + 14.23234 − 0.36425) × 10−3 kg ⋅ m 2
or
I OA = 18.17 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1635
!
PROBLEM 9.166
Determine the mass moment of inertia of the steel fixture of
Problems 9.145 and 9.149 with respect to the axis through the
origin that forms equal angles with the x, y, and z axes.
SOLUTION
From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
I x = 26.4325 × 10−3 kg ⋅ m 2
I y = 31.1726 × 10−3 kg ⋅ m 2
I z = 8.5773 × 10−3 kg ⋅ m 2
Problem 9.149:
I xy = 2.5002 × 10−3 kg ⋅ m 2
I yz = 4.0627 × 10−3 kg ⋅ m 2
I zx = 8.8062 × 10−3 kg ⋅ m 2
From the problem statement it follows that
λx = λ y = λz
Now
or
λx2 + λ y2 + λz2 = 1 4 3λx2 = 1
λx = λ y = λz =
1
3
Substituting into Eq. (9.46)
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
Noting that
We have
λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =
1
3
1
I OL = [26.4325 + 31.1726 + 8.5773
3
− 2(2.5002 + 4.0627 + 8.8062)] × 10−3 kg ⋅ m 2
or
I OL = 11.81 × 10−3 kg ⋅ m 2
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1636
!
PROBLEM 9.167
The thin bent plate shown is of uniform density and weight W. Determine
its mass moment of inertia with respect to the line joining the origin O and
Point A.
SOLUTION
First note that
and that
m1 = m2 =
λOA =
1
3
1W
2 g
(i + j + k )
Using Figure 9.28 and the parallel-axis theorem, we have
I x = ( I x )1 + ( I x )2
1 &1W
=" (
12
$" * 2 g
' 2 1W
)a +
2 g
+
-/ 1 & 1 W
+0 (
/212 * 2 g
2
&a' !
( ) #
* 2 + %#
2
' 2
1W
2
) (a + a ) +
2 g
+
&a' &a'
"( ) + ( )
"$* 2 + * 2 +
1 W & 1 1 ' 2 & 1 1 ' 2! 1 W 2
a
=
"( + ) a + ( + ) a # =
2 g $* 12 4 +
*6 2+ % 2 g
2 !.
/
#1
#% /3
I y = ( I y )1 + ( I y )2
-/ 1 & 1 W '
1W
2
2
=0 (
) (a + a ) +
12 2 g +
2 g
2/ *
2
&a' &a'
"( ) + ( )
"$* 2 + * 2 +
2
-/ 1 & 1 W ' 2 1 W
& a ' ! /.
2
+0 (
+
+
(
)
a
a
"
)
( ) #1
2 g "$
* 2 + #% /3
/212 * 2 g +
1 W & 1 1 ' 2 & 1 5' 2! W 2
=
a + ( + )a # = a
" +
2 g $(* 6 2 )+
* 12 4 + % g
2 !.
/
#1
#% 3/
I z = ( I z )1 + ( I z ) 2
1 &1W
=" (
12
$" * 2 g
' 2 1W
)a +
2 g
+
-/ 1 & 1 W
+0 (
2/12 * 2 g
2
&a' !
( ) #
* 2 + %#
2 !.
' 2 1W
&a'
" (a ) 2 + ( )
)a +
2
g
*2+
+
$"
1 W & 1 1 ' 2 & 1 5' 2!
=
+
a + ( + )a # =
"
2 g $(* 12 4 )+
* 12 4 + %
/
#1
%# /3
5W 2
a
6 g
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1637
PROBLEM 9.167 (Continued)
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because
of symmetry. Also, y1 = 0
0
1W
&a' 1W 2
(a) ( ) =
I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 =
a
2 g
*2+ 4 g
Then
0
1 W & a '& a ' 1 W 2
I yz = Σ( I y′z′ + my z ) = m2 y2 z2 =
a
)=
2 g (* 2 )(
+* 2 + 8 g
0
I zx = Σ( I z ′x′ + mz x ) = m1z1 x1 + m2 z2 x2
=
1 W & a '& a ' 1 W & a '
3W 2
a
+
(a) =
2 g (* 2 )+ (* 2 )+ 2 g (* 2 )+
8 g
Substituting into Equation (9.46)
I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
Noting that
λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =
1
3
We have
& 1 W 2 1 W 2 3 W 2 '!
1 1W 2 W 2 5W 2
I OA = "
a + a +
a − 2(
a +
a +
a )#
3 $2 g
6 g
8 g
8 g
g
*4 g
+%
1 14
& 3 '! W
= " − 2 ( )# a 2
3$ 6
* 4 +% g
or
I OA =
5 W 2
a
18 g
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you are using it without permission.
1638
!
PROBLEM 9.168
A piece of sheet steel of thickness t and specific weight γ is cut and
bent into the machine component shown. Determine the mass
moment of inertia of the component with respect to the joining the
origin O and Point A.
SOLUTION
First note that
OA
1
=
6
(i + 2 j + k )
Next compute the mass of each component. We have
m = ρV =
Then
m1 =
m2 =
γ
g
γ
g
(tA)
t (2a × 2a) = 4
γt
g
a2
γ &π
' π γt 2
× a2 ) =
t
a
g (* 2
+ 2 g
Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem 9.134)
for a semicircular plate for component 2, we have
I x = ( I x )1 + ( I x )2
-/ 1 & γ t '
./
γt
= 0 ( 4 a 2 ) [(2a) 2 + (2a )2 ] + 4 a 2 (a 2 + a 2 ) 1
g
/212 * g
/3
+
/- 1 & π γ t 2 ' 2 π γ t 2
/.
a )a +
a [(2a) 2 + (a) 2 ]1
+0 (
2 g
/2 4 * 2 g +
/3
π γt 2 & 1
&2
'
'
a2 ( + 2 ) a2 +
a
+ 5 ) a2
g
2 g (* 4
*3
+
+
γt
= 18.91335 a 4
g
=4
γt
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you are using it without permission.
1639
PROBLEM 9.168 (Continued)
I y = ( I y )1 + ( I y ) 2
!
1 & γt '
γt
= " ( 4 a 2 ) (2a )2 + 4 a 2 ( a 2 ) #
g
$12 * g +
%
2
-/ π γ t & 1 16 '
! ./
π γ t 2 & 4a '
a2 ( − 2 ) a2 +
a "(
+0
+ (a) 2 # 1
)
2 g
"$* 3π +
#% /3
/2 2 g * 2 9π +
16
γt & 1 '
π γ t 2 & 1 16
'
a
= 4 a 2 ( + 1) a 2 +
−
+
+ 1) a 2
g *3 +
2 g (* 2 9π 2 9π 2
+
γt
= 7.68953 a 4
g
I z = ( I z )1 + ( I z ) 2
!
γt
1 & γt '
= " ( 4 a 2 ) (2a) 2 + 4 a 2 ( a 2 ) #
g
$12 * g +
%
2
-/ π γ t & 1 16 '
! ./
π γ t 2 & 4a '
+0
+ (2a) 2 # 1
a2 ( − 2 ) a2 +
a "(
)
2 g
* 4 9π +
"$* 3π +
#% /3
/2 2 g
γt &1 '
π γ t 2 & 1 16
16
'
= 4 a 2 ( + 1) a 2 +
a ( − 2 + 2 + 4 ) a2
2 g * 4 9π
g
9π
*3 +
+
γt
= 12.00922 a 4
g
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because
of symmetry. Also x1 = 0.
0
π γ t 2 & 4a '
I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 =
a (
Then
) (2a)
2 g * 3π +
γt
= 1.33333 a 4
g
I yz = Σ( I y′z′
=4
γt
g
0
+ my z ) = m1 y1 z1 + m2 y2 z2
a 2 ( a)(a) +
= 7.14159
γt
g
π γt
2 g
a 2 (2a)(a)
a4
0
π γ t 2 & 4a '
I zx = Σ( I z ′x′ + mz x ) = m2 z2 x2 =
a (a) (
)
2 g
* 3π +
γt
= 0.66667 a 4
g
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1640
PROBLEM 9.168 (Continued)
Substituting into Eq. (9.46)
I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
= 18.91335
2
& 1 '
γt 4 & 2 '
a4 (
) + 7.68953 a (
)
g
g * 6+
* 6+
γt
2
2
&
& 1 '
γ t 4 ' & 1 '& 2 '
a4 (
) − 2 (1.33333 a ) (
)(
)
g * 6+
g
*
+ * 6 +* 6 +
&
γ t ' & 2 '& 2 ' &
γ t 4 ' & 1 '& 1 '
− 2 ( 7.14159 a 4 ) (
)(
) − 2 ( 0.66667 a ) (
)(
)
g
g + * 6 +* 6 +
*
+ * 6 +* 6 +
*
γt
= (3.15223 + 5.12635 + 2.00154 − 0.88889 − 4.76106 − 0.22222) a 4
g
+ 12.00922
γt
or
I OA = 4.41
γt
g
a4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1641
!
PROBLEM 9.169
Determine the mass moment of inertia of the machine component
of Problems 9.136 and 9.155 with respect to the axis through the
origin characterized by the unit vector λ = (− 4i + 8j + k )/9.
SOLUTION
From the solutions to Problems 9.136 and 9.155. We have
Problem 9.136:
I x = 175.503 × 10−3 kg ⋅ m 2
I y = 308.629 × 10−3 kg ⋅ m 2
I z = 154.400 × 10−3 kg ⋅ m 2
Problem 9.155:
I xy = −8.0365 × 10−3 kg ⋅ m 2
I yz = 12.8950 × 10−3 kg ⋅ m 2
I zx = 94.0266 × 10−3 kg ⋅ m 2
Substituting into Eq. (9.46)
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
2
2
& 4'
&8'
&1'
= "175.503 ( − ) + 308.629 ( ) + 154.400 ( )
* 9+
*9+
*9+
"$
& 4 '& 8 '
& 8 '& 1 '
− 2(−8.0365) ( − )( ) − 2(12.8950) ( )( )
9
9
*
+* +
* 9 +* 9 +
& 1 '& 4 ' !
− 2(94.0266) ( )( − ) # × 10−3 kg ⋅ m 2
* 9 +* 9 + %
2
= (34.6673 + 243.855 + 1.906 − 6.350
− 2.547 + 9.287) × 10−3 kg ⋅ m 2
or
I OL = 281 × 10−3 kg ⋅ m 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1642
!
PROBLEM 9.170
For the wire figure of Problem 9.148, determine the mass moment
of inertia of the figure with respect to the axis through the origin
characterized by the unit vector λ = (−3i − 6j + 2k )/7.
SOLUTION
First compute the mass of each component. We have
&m'
m = ( ) L = 0.056 kg/m × 1.2 m
*L+
= 0.0672 kg
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each
component are zero because of symmetry.
Also
x1 = x6 = 0
Then
y4 = y5 = y6 = 0
z1 = z2 = z3 = 0
0
I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3
= (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m)
= 0.096768 kg ⋅ m 2
0
I yz = Σ( I y′z′ + my z ) = 0
0
I zx = Σ( I z ′x′ + mz x ) = m4 z4 x4 + m5 z5 x5
= (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m)
= 0.096768 kg ⋅ m 2
From the solution to Problem 9.148, we have
I x = 0.32258 kg ⋅ m 2
I y = I z = 0.41933 kg ⋅ m 2
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1643
PROBLEM 9.170 (Continued)
Substituting into Eq. (9.46)
0
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
2
2
& 3'
& 6'
&2'
= "0.32258 ( − ) + 0.41933 ( − ) + 0.41933 ( )
7
7
*
+
*
+
*7+
"$
2
& 3 '& 6 '
& 2 '& 3 ' !
− 2(0.096768) ( − )( − ) − 2(0.096768) ( )( − ) # kg ⋅ m 2
* 7 +* 7 +
* 7 +* 7 + %
I OL = (0.059249 + 0.30808 + 0.034231 − 0.071095 + 0.023698) kg ⋅ m 2
or
I OL = 0.354 kg ⋅ m 2
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1644
!
PROBLEM 9.171
For the wire figure of Problem 9.147, determine the mass moment of inertia of
the figure with respect to the axis through the origin characterized by the unit
vector λ = (−3i − 6j + 2k )/7.
SOLUTION
First compute the mass of each component. We have
m = ρSTV =
γ ST
g
AL
Then
m1 = m2 =
3
2
& 1 ft '
π &1 ' !
490 lb/ft 3
×
×
×
×
π
in.
(
18
in.)
"
#
(
)
(
)
32.2 ft/s 2 $" 4 * 8 + %#
* 12 in. +
= 6.1112 × 10−3 lb ⋅ s 2 /ft
π &1 '
490 lb/ft 3
× " ( in. )
m3 = m4 =
2
32.2 ft/s
$" 4 * 8 +
2!
& 1 ft '
# × 18 in. × (
)
* 12 in. +
%#
3
= 1.9453 lb ⋅ s 2 /ft
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each component are zero because
of symmetry.
Also
x3 = x4 = 0
y1 = 0
z1 = z2 = 0
Then
0
I xy = Σ( I x′y′ + mx y ) = m2 x2 y2
& 1 ft '
& 2 × 18 '
in. ) (18 in.) (
= (6.1112 × 10−3 lb ⋅ s 2 /ft) ( −
)
π
*
+
* 12 in. +
2
= −8.75480 × 10−3 lb ⋅ ft ⋅ s 2
0
I yz = Σ( I y ′z′ + my z ) = m3 y3 z3 + m4 y4 z4
Now
m3 = m4 ,
y3 = y4 ,
z 4 = − z3
0
I zx = Σ( I z′x′ + mz x ) or
I yz = 0
I zx = 0
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1645
PROBLEM 9.171 (Continued)
From the solution to Problem 9.147, we have
I x = 39.1721 × 10 −3 lb ⋅ ft ⋅ s 2
I y = 36.2542 × 10−3 lb ⋅ ft ⋅ s 2
I z = 30.4184 × 10−3 lb ⋅ ft ⋅ s 2
Substituting into Eq. (9.46)
0
0
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
2
2
& 3'
& 6'
&2'
= "39.1721( − ) + 36.2542 ( − ) + 30.4184 ( )
* 7+
* 7+
*7+
"$
& 3 '& 6 ' !
− 2(−8.75480) ( − )( − ) # × 10−3 lb ⋅ ft ⋅ s 2
* 7 +* 7 + %
2
= (7.19488 + 26.6357 + 2.48313 + 6.43210) × 10−3 lb ⋅ ft ⋅ s 2
or
I OL = 0.0427 lb ⋅ ft ⋅ s 2
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1646
!
PROBLEM 9.172
For the wire figure of Problem 9.46, determine the mass moment of
inertia of the figure with respect to the axis through the origin
characterized by the unit vector λ = (−3i − 6j + 2k )/7.
SOLUTION
First compute the mass of each component. We have
m=
W 1
= (W/L) AL L
g g
m1 =
1
1 ft
(0.033 lb/ft)(2π × 16 in.) ×
2
12 in.
32.2 ft/s
Then
= 8.5857 × 10 −3 lb ⋅ s 2 /ft
m2 = m3 = m4 = m5 =
1
1 ft
(0.033 lb/ft)(8 in.) ×
12 in.
32.2 ft/s 2
= 0.6832 × 10−3 lb ⋅ ft/s 2
Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because
of symmetry. Also
Then
x1 = x4 = x5 = 0
y1 = 0
z1 = z2 = z3 = 0
0
I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3
& 16 '& 4 '
= 0.6832 × 10 −3 lb ⋅ s 2 /ft ( ft )( ft )
* 12 +* 12 +
& 12 '& 8 '
+ 0.6832 × 10−3 lb ⋅ s 2 /ft ( ft )( ft )
* 12 +* 12 +
= (0.30364 + 0.45547) × 10−3 lb ⋅ ft ⋅ s 2
= 0.75911 × 10−3 lb ⋅ ft ⋅ s 2
Symmetry implies
I yz = I xy
I yz = 0.75911 × 10−3 lb ⋅ ft ⋅ s 2
0
I zx = Σ( I z ′x′ + mz x ) = 0
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1647
PROBLEM 9.172 (Continued)
From the solution to Problem 9.146, we have
I x = I z = 10.3642 × 10−3 lb ⋅ ft ⋅ s 2
I y = 19.1097 × 10−3 lb ⋅ ft ⋅ s 2
Substituting into Eq. (9.46)
0
I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx
2
2
& 3'
& 6'
&2'
= "10.3642 ( − ) + 19.1097 ( − ) + 10.3642 ( )
7
7
*
+
*
+
*7+
"$
2
& 3 '& 6 '
& 6 '& 2 ' !
− 2(0.75911) ( − )( − ) − 2(0.75911) ( − )( ) # × 10−3 lb ⋅ ft ⋅ s 2
* 7 +* 7 +
* 7 +* 7 + %
= (1.90663 + 14.03978 + 0.84606 − 0.55771 + 0.37181) × 10−3 lb ⋅ ft ⋅ s 2
or
I OL = 16.61 × 10−3 lb ⋅ ft ⋅ s 2
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1648
!
PROBLEM 9.173
For the rectangular prism shown, determine the values of the
ratios b/a and c/a so that the ellipsoid of inertia of the prism is
a sphere when computed (a) at Point A, (b) at Point B.
SOLUTION
(a)
Using Figure 9.28 and the parallel-axis theorem, we have
at Point A
1
I x′ = m(b 2 + c 2 )
12
1
&a'
I y ′ = m( a 2 + c 2 ) + m ( )
12
* 2+
1
= m(4a 2 + c 2 )
12
2
2
I z′ =
1
1
&a'
m(a 2 + b 2 ) + m ( ) = m(4a 2 + b 2 )
12
* 2 + 12
Now observe that symmetry implies
I x′y′ = I y′z′ = I z′x′ = 0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x′ x 2 + I y′ y 2 + I z′ z 2 = 1:
1
1
1
m(b 2 + c 2 ) x 2 + m(4a 2 + c 2 ) y 2 + m(4a 2 + b 2 ) z 2 = 1
12
12
12
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore
1
1
m(b 2 + c 2 ) = m(4a 2 + c 2 )
12
12
1
= m(4a 2 + b 2 )
12
Then
b 2 + c 2 = 4a 2 + c 2
or
b
=2
a
and
b 2 + c 2 = 4a 2 + b 2
or
c
=2
a
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1649
PROBLEM 9.173 (Continued)
(b)
Using Figure 9.28 and the parallel-axis theorem, we have at Point B
2
I x′′ =
1
1
&c'
m(b 2 + c 2 ) + m ( ) = m(b 2 + 4c 2 )
12
12
*2+
2
1
1
&c'
m ( a 2 + c 2 ) + m ( ) = m ( a 2 + 4c 2 )
12
* 2 + 12
1
I z′′ = m(a 2 + b 2 )
12
I y′′ =
Now observe that symmetry implies
I x′′y′′ = I y′′z′′ = I z′′x′′ = 0
From Part a it then immediately follows that
1
1
1
m(b 2 + 4c 2 ) = m(a 2 + 4c 2 ) = m(a 2 + b 2 )
12
12
12
Then
b 2 + 4c 2 = a 2 + 4c 2
or
b
=1
a
and
b 2 + 4c 2 = a 2 + b 2
or
c 1
=
a 2
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1650
!
PROBLEM 9.174
For the right circular cone of Sample Problem 9.11, determine the value
of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere
when computed (a) at the apex of the cone, (b) at the center of the base of
the cone.
SOLUTION
(a)
From Sample Problem 9.11, we have at the apex A
3
ma 2
10
3 &1
'
I y = I z = m ( a 2 + h2 )
5 *4
+
Ix =
I xy = I yz = I zx = 0
Now observe that symmetry implies
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x x 2 + I y y 2 + I z z 2 = 1:
3
3 &1
3 &1
'
'
ma 2 x 2 + m ( a 2 + h 2 ) y 2 + m ( a 2 + h 2 ) z 2 = 1
10
5 *4
5
4
+
*
+
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore,
3
3 &1
'
ma 2 = m ( a 2 + h 2 )
10
5 *4
+
(b)
or
a
=2
h
From Sample Problem 9.11, we have
and at the centroid C
I x′ =
3
ma 2
10
I y′′ =
3 & 2 1 2'
m a + h )
20 (*
4 +
2
Then
I y′ = I z′ =
3 & 2 1 2'
1
&h'
m( a + h ) + m( ) =
m(3a 2 + 2h 2 )
20 *
4 +
20
*4+
Now observe that symmetry implies
I x′y′ = I y′z′ = I z′x′ = 0
From Part a it then immediately follows that
3
1
ma 2 =
m(3a 2 + 2h 2 )
10
20
or
a
=
h
2
3
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1651
!
PROBLEM 9.175
For the homogeneous circular cylinder shown, of radius a and length L,
determine the value of the ratio a/L for which the ellipsoid of inertia of
the cylinder is a sphere when computed (a) at the centroid of the
cylinder, (b) at Point A.
SOLUTION
(a)
From Figure 9.28:
1 2
ma
2
1
I y = I z = m(3a 2 + L2 )
12
Ix =
Now observe that symmetry implies
I xy = I yz = I zx = 0
Using Eq. (9.48), the equation of the ellipsoid of inertia is then
I x x 2 + I y y 2 + I z z 2 = 1:
1 2 2 1
1
ma x + m(3a 2 + L2 ) y 2 + m(3a 2 + L2 ) = 1
2
12
12
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore
1 2 1
ma = m(3a 2 + L2 )
2
12
(b)
or
a
1
=
L
3
Using Figure 9.28 and the parallel-axis theorem, we have
I x′ =
1 2
ma
2
2
1
7 2'
&L'
&1
I y′ = I z′ = m(3a 2 + L2 ) + m ( ) = m ( a 2 +
L )
12
48 +
*4+
*4
Now observe that symmetry implies
I x′y′ = I y′z′ = I z′x′ = 0
From Part a it then immediately follows that
1 2
7 2'
&1
ma = m ( a 2 +
L )
2
4
48
*
+
or
a
7
=
L
12
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1652
!
PROBLEM 9.176
Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of the
body with respect to any one of the three axes cannot be larger than the sum of the mass moments of inertia of
the body with respect to the other two axes. That is, prove that the inequality I x # I y + I z and the two similar
inequalities are satisfied. Further, prove that I y $ 12 I x if the body is a homogeneous solid of revolution, where
x is the axis of revolution and y is a transverse axis.
SOLUTION
(i)
Iy + Iz $ Ix
To prove
,
= , (x
I y = ( z 2 + x 2 )dm
By definition
Iz
2
+ y 2 ) dm
,
,
I y + I z = ( z 2 + x 2 ) dm + ( x 2 + y 2 ) dm
Then
,
,
= ( y 2 + z 2 )dm + 2 x 2 dm
Now
,(y
2
+ z 2 )dm = I x
, x dm $ 0
2
and
Iy + Iz $ Ix
Q.E.D.
The proofs of the other two inequalities follow similar steps.
(ii)
If the x axis is the axis of revolution, then
I y = Iz
and from Part (i)
or
Iy + Iz $ Ix
2I y $ I x
Iy $
or
1
Ix
2
Q.E.D. !
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1653
PROBLEM 9.177
Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a
sphere, and use this property to determine the moment of inertia of the cube with respect to one of its
diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution,
and determine the principal moments of inertia of the cube at that point.
SOLUTION
(a)
At the center of the cube have (using Figure 9.28)
Ix = I y = Iz =
1
1
m(a 2 + a 2 ) = ma 2
12
6
Now observe that symmetry implies
I xy = I yz = I zx = 0
Using Equation (9.48), the equation of the ellipsoid of inertia is
&1 2' 2 &1 2' 2 &1 2' 2
( 6 ma ) x + ( 6 ma ) y + ( 6 ma ) z = 1
*
+
*
+
*
+
or
x2 + y2 + z 2 =
6
(= R 2 )
ma 2
which is the equation of a sphere.
Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through the
center O of the cube must always
&
1 '
1
).
be the same ( R =
I OL = ma 2
(
6
I OL )+
*
(b)
The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes
through corner A. For a rectangular coordinate system at A and with one of the coordinate axes aligned
with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then rotated 120°
about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid will also remain
unchanged after it is rotated. As noted at the end of Section 9.17, this is possible only if the ellipsoid is
an ellipsoid of revolution, where the diagonal is both the axis of revolution and a principal axis.!
1
It then follows that
I x′ = I OL = ma 2 !
6
In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal and
any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallel-axis
theorem between the center of the cube and corner A for any perpendicular axis
I y′ = I z′ =
& 3 '
1 2
ma + m (
a
( 2 ))
6
*
+
2
or
I y′ = I z′ =
11 2
ma
12
(Note: Part b can also be solved using the method of Section 9.18.)
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1654
PROBLEM 9.177 (Continued)
First note that at corner A
2 2
ma
3
1
= ma 2
4
Ix = I y = Iz =
I xy = I yz = I zx
Substituting into Equation (9.56) yields
k 3 − 2ma 2 k 2 +
55 2 6
121 3 9
m a k−
m a =0
48
864
For which the roots are
1 2
ma
6
!
11
k2 = k3 = ma 2
12
k1 =
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1655
PROBLEM 9.178
Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin
at O, prove that the sum Ix + Iy + Iz of the mass moments of inertia of the body cannot be smaller than the
similar sum computed for a sphere of the same mass and the same material centered at O. Further, using the
result of Problem 9.176, prove that if the body is a solid of revolution, where x is the axis of revolution, its
mass moment of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10, where a is the radius of
the sphere of the same mass and the same material.
SOLUTION
(i)
Using Equation (9.30), we have
,
,
,
I x + I y + I z = ( y 2 + z 2 )dm + ( z 2 + x 2 ) dm + ( x 2 + y 2 )dm
,
= 2 ( x 2 + y 2 + z 2 )dm
,
= 2 r 2 dm
where r is the distance from the origin O to the element of mass dm. Now assume that the given body
can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m
and radius a which is centered at O; it then follows that
m1 = m2 (mbody = msphere = m)
Then
( I x + I y + I z ) body = ( I x + I y + I z )sphere + ( I x + I y + I z )V1
− ( I x + I y + I z )V2
or
( I x + I y + I z ) body = ( I x + I y + I z )sphere + 2
,
m1
r 2 dm − 2
,
m2
r 2 dm
Now, m1 = m2 and r1 $ r2 for all elements of mass dm in volumes 1 and 2.
,
m1
r 2 dm −
,
m2
r 2 dm $ 0
( I x + I y + I z ) body $ ( I x + I y + I z )sphere
so that
Q.E.D.
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you are using it without permission.
1656
PROBLEM 9.178 (Continued)
(ii)
First note from Figure 9.28 that for a sphere
Ix = Iy = Iz =
Thus
( I x + I y + I z )sphere =
2 2
ma
5
6 2
ma
5
For a solid of revolution, where the x axis is the axis of revolution, we have
I y = Iz
Then, using the results of Part (i)
6
( I x + 2 I y ) body $ ma 2
5
From Problem 9.178 we have
or
Iy $
1
Ix
2
(2 I y − I x ) body $ 0
Adding the last two inequalities yields
6
(4 I y ) body $ ma 2
5
or
( I y ) body $
3
ma 2
10
Q.E.D.
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1657
PROBLEM 9.179*
The homogeneous circular cylinder shown has a mass m, and
the diameter OB of its top surface forms 45° angles with the
x and z axes. (a) Determine the principal mass moments of
inertia of the cylinder at the origin O. (b) Compute the angles
that the principal axes of inertia at O form with the coordinate
axes. (c) Sketch the cylinder, and show the orientation of the
principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
First compute the moments of inertia using Figure 9.28 and the
parallel-axis theorem.
2
2
" a # " a # ! 13 2
1
+
m(3a 2 + a 2 ) + m $&
' & ' % = ma
12
$*( 2 ) ( 2 ) %+ 12
1
3
I y = ma 2 + m(a) 2 = ma 2
2
2
Ix = Iz =
Next observe that the centroidal products of inertia are zero because of symmetry. Then
0
" a #" a #
1
I xy = I x′y′ + mx y = m &
ma 2
'& − ' = −
2
2
2
2
(
)
(
)
0
1
" a #" a #
I yz = I y′z ′ + my z = m & − ' &
ma 2
'=−
2 2
( 2 )( 2 )
0
" a #" a # 1 2
I zx = I z′x′ + mz x = m &
'&
' = ma
( 2 )( 2 ) 2
Substituting into Equation (9.56)
" 13 3 13 #
K 3 − & + + ' ma 2 K 2
( 12 2 12 )
2
2
1 # "
1 # "1#
" 13 3 # " 3 13 # " 13 13 # "
+ $& × ' + & × ' + & × ' − & −
' −&−
' −& '
*( 12 2 ) ( 2 12 ) ( 12 12 ) ( 2 2 ) ( 2 2 ) ( 2 )
2
1 # " 3 #" 1 #
" 13 3 13 # " 13 # "
− $& × × '− & ' & −
' − & '& '
*( 12 2 12 ) ( 12 ) ( 2 2 ) ( 2 )( 2 )
2!
% (ma 2 ) 2 K
%+
2
2
"
1 #
1 #"
1 # " 1 #!
" 13 # "
2 3
− & '& −
' − 2& −
'& −
' & ' % ( ma ) = 0
12
2
( )( 2 2 )
( 2 2 )( 2 2 ) ( ) +
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1658
PROBLEM 9.179* (Continued)
Simplifying and letting K = ma 2ζ yields
ζ3−
11 2 565
95
ζ +
ζ−
=0
3
144
96
Solving yields
ζ 1 = 0.363383
ζ2 =
19
12
ζ 3 = 1.71995
The principal moments of inertia are then
K1 = 0.363ma 2
K 2 = 1.583ma 2
K 3 = 1.720ma 2
(b)
To determine the direction cosines λx , λ y , λz of each principal axis, we use two of the equations of
Equations (9.54) and (9.57).
Thus,
( I x − K )λx − I xy λ y − I zx λz = 0
(9.54a)
− I zx λx − I yz λ y + ( I z − K )λz = 0
(9.54c)
λx2 + λ y2 + λz2 = 1
(9.57)
(Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of λ y
during the solution process.)
Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)
"
#
1
" 13 2
#
"1
#
ma 2 ' λ y − & ma 2 ' λz = 0
& 12 ma − K ' λx − & −
(
)
(2
)
( 2 2
)
"
#
1
"1
#
" 13
#
ma 2 ' λ y + & ma 2 − K ' λz = 0
− & ma 2 ' λx − & −
12
2
2
(2
)
(
)
(
)
or
and
" 13
& 12 − ζ
(
1
1
#
λ y − λz = 0
' λx +
2
2 2
)
1
1
" 13
λy + & − ζ
− λx +
2
2 2
( 12
#
' λz = 0
)
(i)
(ii)
Observe that these equations will be identical, so that one will need to be replaced, if
13
1
−ζ = −
12
2
ζ =
or
19
12
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1659
PROBLEM 9.179* (Continued)
Thus, a third independent equation will be needed when the direction cosines associated with K 2 are
determined. Then for K1 and K 3
Eq. (i) through Eq. (ii):
13
1 " 13
" 1 #!
#!
$ − ζ − & − ' % λx + $ − − & − ζ ' % λz = 0
( 2 )+
)+
*12
* 2 ( 12
λz = λx
or
Substituting into Eq. (i):
" 13
& 12 − ζ
(
1
1
#
λ y − λx = 0
' λx +
2
2 2
)
"
7 #
λ y = 2 2 & ζ − ' λx
12 )
(
or
Substituting into Equation (9.57):
"
!
7 #
2
λx2 + $ 2 2 & ζ − ' λx % + (λx ) 2 = 1
12 ) +
(
*
or
2
7# !
"
$ 2 + 8 & ζ − ' % λx2 = 1
12 ) %+
(
*$
(iii)
K1 : Substituting the value of ζ 1 into Eq. (iii):
2
7 # !
2
"
$ 2 + 8 & 0.363383 − ' % ( λx )1 = 1
12
(
)
%+
*$
or
and then
(λx )1 = (λz )1 = 0.647249
7 #
"
(λ y )1 = 2 2 & 0.363383 − ' (0.647249)
12
(
)
= −0.402662
(θ x )1 = (θ z )1 = 49.7°
(θ y )1 = 113.7°
K 3 : Substituting the value of ζ 3 into Eq. (iii):
2
7 # !
"
$ 2 + 8 &1.71995 − ' % (λx )32 = 1
12 ) %+
(
*$
or
and then
(λx )3 = (λz )3 = 0.284726
7 #
"
(λ y )3 = 2 2 &1.71995 − ' (0.284726)
12 )
(
= 0.915348
(θ x )3 = (θ z )3 = 73.5°
(θ y )3 = 23.7°
!
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1660
!
PROBLEM 9.179* (Continued)
!
K2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57).
Now
− I xy λx + ( I y − K )λ y − I yz λz = 0
(9.54b)
Substituting for the moments and products of inertia.
"
#
"
#
1
1
"3
#
ma 2 ' λx + & ma 2 − K ' λ y − & −
ma 2 ' λz = 0
−& −
(2
)
( 2 2
)
( 2 2
)
"3
1
#
1
λx + & − ξ ' λ y +
λz = 0
2 2
2 2
(2
)
or
(iv)
Substituting the value of ξ 2 into Eqs. (i) and (iv):
1
1
" 13 19 #
(λ y ) 2 − (λz ) 2 = 0
& 12 − 12 ' (λx ) 2 +
2
2
2
(
)
1
1
" 3 19 #
(λx ) 2 + & − ' (λ y ) 2 +
(λ z ) 2 = 0
2 2
2 2
( 2 12 )
or
− (λx ) 2 +
and
1
2
(λx ) 2 −
(λ y ) 2 − (λ z ) 2 = 0
2
(λ y ) 2 + (λz ) 2 = 0
6
Adding yields
(λ y ) 2 = 0
and then
(λ y ) 2 = −(λx ) 2
Substituting into Equation (9.57)
or
0
(λx ) 22 + (λ y )22 + (−λx ) 22 = 1
(λx ) 2 =
1
2
and (λz ) 2 = −
1
2
(θ x ) 2 = 45.0° (θ y )2 = 90.0° (θ z ) 2 = 135.0°
(c)
Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B. Principal
axis 2 lies in the xz plane.
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1661
PROBLEM 9.180
For the component described in Problem 9.165, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes
of inertia at the origin. Sketch the body and show the orientation of
the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.141 and 9.165 we have
I x = 13.98800 × 10−3 kg ⋅ m 2
Problem 9.141:
I y = 20.55783 × 10−3 kg ⋅ m 2
I z = 14.30368 × 10−3 kg ⋅ m 2
I yz = I zx = 0
Problem 9.165:
I xy = 0.39460 × 10−3 kg ⋅ m 2
Eq. (9.55) then becomes
Thus
Substituting:
Ix − K
− I xy
0
− I xy
Iy − K
0
0
0
Iz − K
2
= 0 or ( I x − K )( I y − K )( I z − K ) − ( I z − K ) I xy
=0
Iz − K = 0
2
I x I y − ( I x + I y ) K + K 2 − I xy
=0
K1 = 14.30368 × 10−3 kg ⋅ m 2
or
K1 = 14.30 × 10−3 kg ⋅ m 2
−3
−3
−3
2
−3 2
and (13.98800 × 10 )(20.55783 × 10 ) − (13.98800 + 20.55783)(10 ) K + K − (0.39460 × 10 ) = 0
K 2 − (34.54583 × 10−3 ) K + 287.4072 × 10−6 = 0
or
Solving yields
and
K 2 = 13.96438 × 10−3 kg ⋅ m 2
K 3 = 20.58145 × 10−3 kg ⋅ m 2
or
K 2 = 13.96 × 10−3 kg ⋅ m 2
or
K 3 = 20.6 × 10−3 kg ⋅ m 2
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1662
PROBLEM 9.180 (Continued)
(b)
To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K1:
Begin with Eqs. (9.54a) and (9.54b) with I yz = I zx = 0.
( I x − K1 )(λx )1 − I xy (λ y )1 = 0
− I xy (λx )1 + ( I y − K1 )(λ y )1 = 0
Substituting:
[(13.98800 − 14.30368) × 10 −3 ](λx )1 − (0.39460 × 10 −3 )(λ y )1 = 0
−(0.39460 × 10−3 )(λx )1 + [(20.55783 − 14.30368) × 10−3 ](λ y )1 = 0
(λx )1 = (λ y )1 = 0
Adding yields
Then using Eq. (9.57)
0
0
(λx )12 + (λ y )12 + (λz )12 = 1
(λz )1 = 1
or
(θ x )1 = 90.0° (θ y )1 = 90.0° (θ z )1 = 0°
K2: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0.
− I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 = 0
(i)
( I z − K 2 )(λz ) 2 = 0
I z ≠ K 2 , (λz ) 2 = 0
Now
Substituting into Eq. (i):
−(0.39460 × 10−3 )(λx ) z + [(20.55783 − 13.96438) × 10−3 ](λ y ) 2 = 0
or
(λ y ) 2 = 0.059847(λx ) 2
Using Eq. (9.57):
(λx ) 22
or
(λx ) 2 = 0.998214
and
(λ y ) 2 = 0.059740
2
+ [0.05984](λx ) 2 ] +
(λz ) 22
(θ x ) 2 = 3.4°
0
=1
(θ y ) 2 = 86.6°
(θ z ) 2 = 90.0°
K3: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0.
− I xy (λx )3 + ( I y − K 3 )(λ y )3 = 0
(ii)
( I z − K 3 )(λz )3 = 0
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1663
PROBLEM 9.180 (Continued)
I z ≠ K 3 , (λz )3 = 0
Now
Substituting into Eq. (ii):
−(0.39460 × 10−3 )(λx )3 + [(20.55783 − 20.58145) × 10−3 ](λ y )3 = 0
(λ y )3 = −16.70618(λx )3
or
Using Eq. (9.57):
(λx )32
2
+ [−16.70618(λx )3 ]
+ (λz )32
0
=1
or
(λx )3 = −0.059751 (axes right-handed set , "_")
and
(λ y )3 = 0.998211
(θ x )3 = 93.4°
(θ y )3 = 3.43°
(θ z )3 = 90.0°
Note: Since the principal axes are orthogonal and (θ z ) 2 = (θ z )3 = 90°, it follows that
|(λx ) 2 | = |(λ y )3 |
|(λ y ) 2 | = | (λz )3 |
The differences in the above values are due to round-off errors.
(c)
Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.
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1664
PROBLEM 9.181*
For the component described in Problems 9.145 and 9.149,
determine (a) the principal mass moments of inertia at the
origin, (b) the principal axes of inertia at the origin.
Sketch the body and show the orientation of the principal
axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.145 and 9.149, we have
Problem 9.145:
I x = 26.4325 × 10−3 kg ⋅ m 2
Problem 9.149:
I xy = 2.5002 × 10−3 kg ⋅ m 2
I y = 31.1726 × 10−3 kg ⋅ m 2
I yz = 4.0627 × 10−3 kg ⋅ m 2
I z = 8.5773 × 10−3 kg ⋅ m 2
I zx = 8.8062 × 10−3 kg ⋅ m 2
Substituting into Eq. (9.56):
K 3 − [(26.4325 + 31.1726 + 8.5773)(10−3 )]K 2
+ [(26.4325)(31.1726) + (31.1726)(8.5773) + (8.5773)(26.4325)
− (2.5002) 2 − (4.0627)2 − (8.8062)2 ](10−6 ) K
− [(26.4325)(31.1726)(8.5773) − (26.4325)(4.0627)2
− (31.1726)(8.8062)2 − (8.5773)(2.5002) 2
− 2(2.5002)(4.0627)(8.8062)](10−9 ) = 0
or
K 3 − (66.1824 × 10−3 ) K 2 + (1217.76 × 10−6 ) K − (3981.23 × 10−9 ) = 0
Solving yields
!
K1 = 4.1443 × 10−3 kg ⋅ m 2
or
K1 = 4.14 × 10−3 kg ⋅ m 2
K 2 = 29.7840 × 10−3 kg ⋅ m 2
or
K 2 = 29.8 × 10−3 kg ⋅ m 2
K3 = 32.2541 × 10−3 kg ⋅ m 2 !
or
K3 = 32.3 × 10−3 kg ⋅ m 2
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1665
!
PROBLEM 9.181* (Continued)
(b)
To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K 1: Begin with Eqs. (9.54a) and (9.54b).
( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0
I xy (λx )1 + ( I y − K1 )(λ y )1 − I yz (λz )1 = 0
Substituting:
[(26.4325 − 4.1443)(10−3 )](λx )1 − (2.5002 × 10−3 )(λ y )1 − (8.8062 × 10−3 )(λz )1 = 0
−(2.5002 × 10−3 )(λx )1 + [(31.1726 − 4.1443)(10 −3 )](λ y )1 − (4.0627 × 10−3 )(λz )1 = 0
Simplifying
8.9146(λx )1 − (λ y )1 − 3.5222(λz )1 = 0
−0.0925(λx )1 + (λ y )1 − 0.1503(λz )1 = 0
Adding and solving for (λz )1:
(λz )1 = 2.4022(λx )1
(λ y )1 = [8.9146 − 3.5222(2.4022)](λx )1
and then
= 0.45357(λx )1
Now substitute into Eq. (9.57):
(λx )12 + [0.45357(λx )1 ]2 + [2.4022(λx )1 ]2 = 1
(λx )1 = 0.37861
or
and
(λ y )1 = 0.17173
(λz )1 = 0.90950
(θ x )1 = 67.8° (θ y )1 = 80.1° (θ z )1 = 24.6°
K 2 : Begin with Eqs. (9.54a) and (9.54b).
( I x − K 2 )(λx )2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0
− I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0
Substituting:
[(26.4325 − 29.7840)(10−3 )](λx )2 − (2.5002 × 10−3 )(λ y ) 2 − (8.8062 × 10−3 )(λz )2 = 0
−(2.5002 × 10 −3 )(λx ) 2 + [(31.1726 − 29.7840)(10−3 )](λ y ) 2 − (4.0627 × 10−3 )( λz ) 2 = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1666
PROBLEM 9.181* (Continued)
Simplifying
−1.3405(λx )2 − (λ y ) 2 − 3.5222(λz ) 2 = 0
−1.8005(λx )2 + (λ y ) 2 − 2.9258(λz ) 2 = 0
Adding and solving for (λz ) 2 :
(λz )2 = −0.48713(λx ) 2
(λ y ) 2 = [−1.3405 − 3.5222(−0.48713)](λx )2
and then
= 0.37527(λx ) 2
Now substitute into Eq. (9.57):
(λx ) 22 + [0.37527(λx ) 2 ]2 + [−0.48713(λx ) 2 ]2 = 1
(λx ) 2 = 0.85184
or
(λ y ) 2 = 0.31967
and
(λz )2 = −0.41496
(θ x )1 = 31.6° (θ y ) 2 = 71.4° (θ z ) 2 = 114.5°
K 3 : Begin with Eqs. (9.54a) and (9.54b).
( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0
− I xy (λx )3 + ( I y − K3 )(λ y )3 − I yz (λz )3 = 0
Substituting:
[(26.4325 − 32.2541)(10−3 )](λx )3 − (2.5002 × 10−3 )(λ y )3 − (8.8062 × 10−3 )(λz )3 = 0
−(2.5002 × 10−3 )(λx )3 + [(31.1726 − 32.2541)(10−3 )](λ y )3 − (4.0627 × 10−3 )(λz )3 = 0
Simplifying
−2.3285(λx )3 − (λ y )3 − 3.5222(λz )3 = 0
2.3118(λx )3 + (λ y )3 + 3.7565(λz )3 = 0
Adding and solving for (λz )3 :
(λz )3 = 0.071276(λx )3
and then
(λ y )3 = [ −2.3285 − 3.5222(0.071276)](λx )3
= −2.5795(λx )3
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1667
PROBLEM 9.181* (Continued)
Now substitute into Eq. (9.57):
(λx )32 + [−2.5795(λx )3 ]2 + [0.071276(λx )3 ]2 = 1
(λx )3 = 0.36134
or
and
(i)
(λ y )3 = 0.93208
(λz )3 = 0.025755
(θ x )3 = 68.8° (θ y )3 = 158.8° (θ z )3 = 88.5°
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,
(λx )3 = −0.36134.
(θ x )3 = 111.2° (θ y )3 = 21.2° (θ z )3 = 91.5°
Then
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1668
PROBLEM 9.182*
For the component described in Problem 9.167, determine (a) the principal
mass moments of inertia at the origin, (b) the principal axes of inertia at the
origin. Sketch the body and show the orientation of the principal axes of
inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solution of Problem 9.167, we have
1W 2
a
2 g
W
I y = a2
g
5W 2
Iz =
a
6 g
Ix =
1W 2
a
4 g
1W 2
I yz =
a
8 g
3W 2
I zx =
a
8 g
I xy =
Substituting into Eq. (9.56):
5 # " W #!
"1
K 3 − $& + 1 + ' & a 2 ' % K 2
6 )( g
)+
*( 2
2
2
"1#
" 5 # " 5 #" 1 # " 1 # " 1 # " 3 #
+ $& ' (1) + (1) & ' + & '& ' − & ' − & ' − & '
( 6 ) ( 6 )( 2 ) ( 4 ) ( 8 ) ( 8 )
*$( 2 )
2!
2
"W #
% & a2 ' K
)
+% ( g
3
2
2
2
" 1 # " 5 # " 1 #" 1 #
" 3 # " 5 #" 1 #
" 1 #" 1 #" 3 # ! " W 2 #
− $& ' (1) & ' − & '& ' − (1) & ' − & '& ' − 2 & '& '& ' % & a ' = 0
( 8 ) ( 6 )( 4 )
( 4 )( 8 )( 8 ) %+ ( g
$*( 2 ) ( 6 ) ( 2 )( 8 )
)
Simplifying and letting K =
W 2
a K yields
g
K 3 − 2.33333K 2 + 1.53125K − 0.192708 = 0
Solving yields
K1 = 0.163917 K 2 = 1.05402 K3 = 1.11539
The principal moments of inertia are then
K1 = 0.1639
W 2
a
g
K 2 = 1.054
W 2
a
g
K 3 = 1.115
W 2
a
g
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1669
!
PROBLEM 9.182* (Continued)
(b)
To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K 1:
Begin with Eqs. (9.54a) and (9.54b).
( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0
− I xy (λx )1 + ( I y − K 2 )(λ y )1 − I yz (λz )1 = 0
Substituting
"1W 2#
"3W 2#
"1
# " W 2 #!
a ' (λ y )1 − &
a ' (λz )1 = 0
$& − 0.163917 ' & a ' % (λx )1 − &
)( g
)+
(4 g
)
(8 g
)
*( 2
"1W 2#
" W #!
"1W 2#
−&
a ' (λx )1 + $(1 − 0.163917) & a 2 ' % (λ y )1 − &
a ' (λz )1 = 0
(4 g
)
( g
)+
(8 g
)
*
Simplifying
1.34433(λx )1 − (λ y )1 − 1.5(λz )1 = 0
−0.299013(λx )1 + (λ y )1 − 0.149507(λz )1 = 0
Adding and solving for (λz )1:
(λz )1 = 0.633715(λx )1
(λ y )1 = [1.34433 − 1.5(0.633715)](λx )1
and then
= 0.393758(λx )1
Now substitute into Eq. (9.57):
(λx )12 + [0.393758(λx )1 ]2 + (0.633715(λx )1 ]2 = 1
(λx )1 = 0.801504 !
or
and
(λ y )1 = 0.315599
(λz )1 = 0.507925
(θ x )1 = 36.7° (θ y )1 = 71.6° (θ z )1 = 59.5°
K 2:
Begin with Eqs. (9.54a) and (9.54b):
( I x − k2 )(λx ) 2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0
− I xy (λx ) 2 + ( I y − k2 )(λ y )2 − I yz (λz ) 2 = 0
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1670
PROBLEM 9.182* (Continued)
Substituting
"1W 2#
"3W 2#
"1
# " W 2 #!
a ' (λ y ) 2 − &
a ' (λz ) 2 = 0
$& − 1.05402 ' & a ' % (λx ) 2 − &
)( g
)+
(4 g
)
(8 g
)
*( 2
"1W 2#
" W #!
"1W 2#
−&
a ' (λx )2 + $(1 − 1.05402) & a 2 ' % (λ y ) 2 − &
a ' (λz ) 2 = 0
(4 g
)
( g
)+
(8 g
)
*
Simplifying
−2.21608(λx ) 2 − (λ y ) 2 − 1.5(λz ) 2 = 0
4.62792(λx ) 2 + (λ y ) 2 + 2.31396(λz )2 = 0
Adding and solving for (λz ) 2
(λz )2 = −2.96309(λx ) 2
(λ y ) 2 = [−2.21608 − 1.5(−2.96309)](λx ) 2
and then
= 2.22856(λx ) 2
Now substitute into Eq. (9.57):
(λx ) 22 + [2.22856(λx ) 2 ]2 + [−2.96309(λx ) 2 ]2 = 1
(λx ) 2 = 0.260410
or
and
(λ y ) 2 = 0.580339
(λz ) 2 = −0.771618
(θ x ) 2 = 74.9° (θ y ) 2 = 54.5° (θ z )2 = 140.5°
!
K 3 : Begin with Eqs. (9.54a) and (9.54b):
( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0
− I xy (λx )3 + ( I y − K3 )(λ y )3 − I yz (λz )3 = 0
Substituting
"1W 2#
"3W 2#
"1
# " W 2 #!
a ' (λ y )3 − &
a ' (λz )3 = 0
$& − 1.11539 ' & a ' % (λx )3 − &
)( g
)+
(4 g
)
(8 g
)
*( 2
"1W 2#
" W #!
"1W 2#
−&
a ' (λx )3 + $(1 − 1.11539) & a 2 ' % (λ y )3 − &
a ' (λ z )3 = 0
(4 g
)
( g
)+
(8 g
)
*
Simplifying
−2.46156(λx )3 − (λ y )3 − 1.5(λz )3 = 0
2.16657(λx )3 + (λ y )3 + 1.08328(λz )3 = 0
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1671
PROBLEM 9.182* (Continued)
Adding and solving for (λz )3
(λz )3 = −0.707885(λx )3
and then
(λ y )3 = [ −2.46156 − 1.5(−0.707885)](λx )3
= −1.39973(λx )3
Now substitute into Eq. (9.57):
(λx )32 + [ −1.39973(λx )3 ]2 + [−0.707885(λx )3 ]2 = 1
or
(λx )3 = 0.537577
and
(λ y )3 = −0.752463
(i)
(λz )3 = −0.380543
(θ x )3 = 57.5° (θ y )3 = 138.8° (θ z )3 = 112.4°
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,
(λx )3 = −0.537577
(θ x )3 = 122.5° (θ y )3 = 41.2° (θ z )3 = 67.6°
Then
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1672
!
PROBLEM 9.183*
For the component described in Problem 9.168, determine (a) the
principal mass moments of inertia at the origin, (b) the principal axes
of inertia at the origin. Sketch the body and show the orientation of
the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solution to Problem 9.168, we have
I x = 18.91335
γt
a4
g
γt
I y = 7.68953 a 4
g
γt
I z = 12.00922 a 4
g
I yz
I zx
γt
a4
g
γt
= 7.14159 a 4
g
γt
= 0.66667 a 4
g
I xy = 1.33333
Substituting into Eq. (9.56):
" γ t #!
K 3 − $(18.91335 + 7.68953 + 12.00922) & a 4 ' % K 2
(g
)+
*
+ [(18.91335)(7.68953) + (7.68953)(12.00922) + (12.00922)(18.91335)
2
" γt #
− (1.33333) − (7.14159) + (0.66667) ] & a 4 ' K
(g )
2
2
2
− [(18.91335)(7.68953)(12.00922) − (18.91335)(7.14159) 2
− (7.68953)(0.66667)2 − (12.00922)(1.33333) 2
3
"γt #
− 2(1.33333)(7.14159)(0.66667)] & a 4 ' = 0
(g
)
Simplifying and letting
K=
γt
g
a4 K
yields
K 3 − 38.61210 K 2 + 411.69009 K − 744.47027 = 0
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1673
PROBLEM 9.183* (Continued)
Solving yields
K1 = 2.25890 K 2 = 17.27274 K3 = 19.08046
K1 = 2.26
The principal moments of inertia are then
K 2 = 17.27
K 3 = 19.08
(b)
γt
g
γt
g
γt
g
a4
a4
a4
To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of
Eqs. (9.54) and Eq. (9.57). Then
K 1 : Begin with Eqs. (9.54a) and (9.54b):
( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0
− I xy (λx )1 + ( I y − K 2 )(λ y )1 − I yz (λz )1 = 0
Substituting
" γ t 4 #!
"
γt 4 #
γt 4 #
"
$(18.91335 − 2.25890) & a ' % (λx )1 − &1.33333 a ' (λ y )1 − & 0.66667 a ' (λz )1 = 0
a )
g
(
( g )+
(
)
*
"
" γ t #!
"
γt #
γt #
− &1.33333 a 4 ' (λx )1 + $(7.68953 − 2.25890) & a 4 ' % (λ y )1 − & 7.14159 a 4 ' (λz )1 = 0
g )
g )
(
( g )+
(
*
Simplifying
12.49087(λx )1 − (λ y )1 − 0.5(λz )1 = 0
−0.24552(λx )1 + (λ y )1 − 1.31506(λz )1 = 0
Adding and solving for (λz )1
(λz )1 = 6.74653(λx )1
and then
(λ y )1 = [12.49087 − (0.5)(6.74653)](λx )1
= 9.11761(λx )1
Now substitute into Eq. (9.57):
(λx )12 + [9.11761(λx )1 ]2 + [6.74653(λx )1 ]2 = 1
or
(λx )1 = 0.087825
and
(λ y )1 = 0.80075
(λz )1 = 0.59251
(θ x )1 = 85.0° (θ y )1 = 36.8° (θ z )1 = 53.7°
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1674
PROBLEM 9.183* (Continued)
K 2 : Begin with Eqs. (9.54a) and (9.54b):
( I x − K 2 )(λx )2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0
− I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0
Substituting
"γt #
"
"
γt #
γt #
[(18.91335 − 17.27274) & a 4 ' (λx )2 − & 1.33333 a 4 ' (λ y ) 2 − & 0.66667 a 4 ' (λz ) 2 = 0
g
g )
(g )
(
)
(
"
" γ t #!
"
γt #
γt #
− &1.33333 a 4 ' (λx ) 2 + $(7.68953 − 17.27274) & a 4 ' % (λ y )2 − & 7.14159 a 4 ' (λz ) 2 = 0
g )
g )
(
(g
)+
(
*
Simplifying
1.23046(λx )2 − (λ y )2 − 0.5(λz ) 2 = 0
0.13913(λx )2 + (λ y ) 2 + 0.74522(λz ) 2 = 0
Adding and solving for (λz ) 2
(λz )2 = −5.58515(λx )2
and then
(λ y ) 2 = [1.23046 − (0.5)(−5.58515)](λx )2
= 4.02304 (λx ) 2
Now substitute into Eq. (9.57):
(λx ) 22 + [4.02304(λx ) 2 ]2 + [−5.58515(λx ) 2 ]2 = 1
or
(λx ) 2 = 0.14377
and
(λ y ) 2 = 0.57839
(λz )2 = −0.80298
(θ x ) 2 = 81.7° (θ y ) 2 = 54.7° (θ z ) 2 = 143.4°
K3 :
Begin with Eqs. (9.54a) and (9.54b):
( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0
− I xy (λx )3 + ( I y − K 3 )(λ y )3 − I yz (λ y )3 = 0
Substituting
"γt #
"
"
γt #
γt #
[(18.91335 − 19.08046) & a 4 ' (λx )3 − & 1.33333 a 4 ' (λ y )3 − & 0.66667 a 4 ' (λz )3 = 0
g )
g )
(g
)
(
(
"
" γ t #!
"
γt #
γt #
− &1.33333 a 4 ' (λx )3 + $(7.68953 − 19.08046) & a 4 ' % (λ y )3 − & 7.14159 a 4 ' (λz )3 = 0
g
g
g )
(
)
(
)+
(
*
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1675
PROBLEM 9.183* (Continued)
Simplifying
−0.12533(λx )3 − (λ y )3 − 0.5(λz )3 = 0
0.11705(λx )3 + (λ y )3 + 0.62695(λz )3 = 0
Adding and solving for (λz )3
(λz )3 = 0.06522(λx )3
and then
(λ y )3 = [ −0.12533 − (0.5)(0.06522)](λx )3
= −0.15794(λx )3
Now substitute into Eq. (9.57):
(λx )32 + [−0.15794(λx )3 ]2 + [0.06522(λx ]3 ]2 = 1
or
(λx )3 = 0.98571
and
(λ y )3 = −0.15568
(i)
(λz )3 = 0.06429
(θ x )3 = 9.7° (θ y )3 = 99.0° (θ z )3 = 86.3°
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is,
(λx )3 = −0.98571.
(θ x )3 = 170.3° (θ y )3 = 81.0° (θ z )3 = 93.7°
Then
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1676
PROBLEM 9.184*
For the component described in Problems 9.148 and 9.170,
determine (a) the principal mass moments of inertia at the
origin, (b) the principal axes of inertia at the origin. Sketch
the body and show the orientation of the principal axes of
inertia relative to the x, y, and z axes.
SOLUTION
(a)
From the solutions to Problems 9.148 and 9.170. We have
I x = 0.32258 kg ⋅ m 2
I y = I z = 0.41933 kg ⋅ m 2
I xy = I zx = 0.096768 kg ⋅ m 2
I yz = 0
Substituting into Eq. (9.56) and using
Iy = Iz
I xy = I zx
I yz = 0
K 3 − [0.32258 + 2(0.41933)]K 2
+ [2(0.32258)(0.41933) + (0.41933)2 − 2(0.096768)2 ]K
− [(0.32258)(0.41933) 2 − 2(0.41933)(0.096768)2 ] = 0
Simplifying
K 3 − 1.16124 K 2 + 0.42764 K − 0.048869 = 0
Solving yields
(b)
K1 = 0.22583 kg ⋅ m 2
or
K1 = 0.226 kg ⋅ m 2
K 2 = 0.41920 kg ⋅ m 2
or
K 2 = 0.419 kg ⋅ m 2
K3 = 0.51621 kg ⋅ m 2
or
K3 = 0.516 kg ⋅ m 2
To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of
Eqs. (9.54) and (9.57). Then
K1 : Begin with Eqs. (9.54b) and (9.54c):
0
− I xy (λx )1 + ( I y − K1 )(λ y )1 − I yz (λz )3 = 0
0
− I zx (λx )1 − I yz (λ y )1 + ( I z − K1 )(λz )3 = 0
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you are using it without permission.
1677
PROBLEM 9.184* (Continued)
Substituting
−(0.096768)(λx )1 + (0.41933 − 0.22583)(λ y )1 = 0
−(0.096768)(λx )1 + (0.41933 − 0.22583)(λz )1 = 0
Simplifying yields
(λ y )1 = (λz )1 = 0.50009(λx )1
Now substitute into Eq. (9.54):
(λx ) 2 + 2[0.50009(λx )1 ]2 = 1
or
(λx )1 = 0.81645
and
(λ y )1 = (λz )1 = 0.40830
(θ x )1 = 35.3° (θ y )1 = (θ z )1 = 65.9°
K 2 : Begin with Eqs. (9.54a) and (9.54b)
( I x − K 2 )(λx ) 2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0
0
− I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0
Substituting
(0.32258 − 0.41920)(λx ) 2 − (0.096768)(λ y ) 2 − (0.096768)(λz ) 2 = 0
(i)
−(0.96768)(λx ) 2 + (0.41933 − 0.41920)(λ y ) 2 = 0
(ii)
Eq. (ii) , (λx ) 2 = 0
and then Eq. (i) , (λz )2 = −(λ y )2
Now substitute into Eq. (9.57):
0
(λx ) 22 + (λ y ) 22 + [−(λ y ) 2 ]2 = 1
1
or
(λ y ) 2 =
and
(λz ) 2 = −
2
1
2
(θ x ) 2 = 90° (θ y ) 2 = 45° (θ z )2 = 135°
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1678
PROBLEM 9.184* (Continued)
K3 :
Begin with Eqs. (9.54b) and (9.54c):
0
I xy (λx )3 + ( I y − K3 )(λ y )3 + I yz (λz )3 = 0
0
− I zx (λx )3 − I yz (λ y )3 + ( I z − K3 )(λz )3 = 0
Substituting
−(0.096768)(λx )3 + (0.41933 − 0.51621)(λ y )3 = 0
−(0.096768)(λx )3 + (0.41933 − 0.51621)(λz )3 = 0
Simplifying yields
(λ y )3 = (λz )3 = − (λx )3
Now substitute into Eq. (9.57):
(λx )32 + 2[−(λx )3 ]2 = 1
(i)
1
or
(λx )3 =
and
(λ y )3 = (λz ) = −
3
1
3
(θ x )3 = 54.7° (θ y )3 = (θ z )3 = 125.3°
!
(c)
Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the
direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen;
That is,
(λx )3 = −
1
3
(θ x )3 = 125.3°
Then
(θ y )3 = (θ z )3 = 54.7°
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you are using it without permission.
1679
!
PROBLEM 9.185
Determine by direct integration the moments of inertia of the shaded area
with respect to the x and y axes.
SOLUTION
x1 = a,
At
Then
Now
Then
y1 = y2 = b
y1 :
b = ka 2
y2 :
b = ma or m =
y1 =
b 2
x
a2
y2 =
b
x
a
or k =
b
a2
b
a
1 3
y2 − y13 dx
3
b3 #
1 " b3
= && 3 x3 − 6 x6 '' dx
3( a
a
)
(
dI x =
-
I x = dI x =
)
-
a
0
b3 " 1 3 1 6 #
x − 6 x ' dx
3 &( a3
a
)
a
3
=
Also
Then
1 4
1
b
!
x − 6 x7 %
3 $* 4a 3
7a
+0
or
Ix =
1 3
ab
28
or
Iy =
1 3
ab
20
b
"b
# !
dI y = x 2 dA = x 2 [( y2 − y1 )dx = x 2 $& x − 2 x 2 ' dx %
a
a
) +
*(
-
I y = dI y =
-
a
0
1
"1
#
b & x3 − 2 x 4 ' dx
a
(a
)
a
1
1
!
= b $ x 4 − 2 x5 %
5a
* 4a
+0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1680
PROBLEM 9.186
Determine the moments of inertia and the radii of gyration of the shaded
area shown with respect to the x and y axes.
SOLUTION
1/2 !
We have
Then
" x#
dA = ydx = b $1 − & '
$* ( a )
-
A = dA =
-
a
0
% dx
%+
1/ 2 !
" x#
b $1 − & '
$* ( a )
% dx
%+
a
2 3/ 2 !
1
= b $x −
x % = ab
3 a
*
+0 3
3
Now
1/2
1
1 .0
" x # ! /0
dI x = y 3 dx = 1b $1 − & ' % 2 dx
3
3 0 $* ( a ) %+ 0
3
4
1/ 2
b3
" x#
= $1 − 3 & '
3 $*
(a)
Then
-
I x = dI x =
-
a
0
"x# " x#
+ 3& ' − & '
(a) (a)
1/ 2
b3
" x#
$1 − 3 & '
3 $*
(a)
3/2 !
% dx
%+
" x# " x#
+ 3& ' − & '
(a) (a)
3/ 2 !
% dx
%+
a
!
2 3/ 2 3 2
2
b3
= $x −
x +
x + 3/ 2 x5/2 %
3 *
2a
5a
a
+0
or
and
k x2 =
Ix
=
A
1 3
ab
30
1
ab3
30
1 ab
3
or
Also
Ix =
kx =
b
10
1/2
.0
" x # ! /0
dI y = x 2 dA = x 2 ( ydx) = x 2 1b $1 − & ' % dx 2
03 *$ ( a ) +% 04
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1681
PROBLEM 9.186 (Continued)
Then
-
I y = dI y =
-
a
0
a
"
1 5/2 #
1
2 7/ 2 !
b & x2 −
x ' dx = b $ x3 −
x %
3
7 a
a
(
)
*
+0
or
and
k y2
=
Iy
A
=
Iy =
1 3
ab
21
1 3
ab
21
1 ab
3
or
ky =
a
7
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1682
PROBLEM 9.187
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
At x = a,
y1 = y2 = b :
y1 : b = ka 2
or k =
y2 : b = 2b − ca 2
Then
Now
b
a2
or c =
b 2
x
a2
"
x2
y2 = b && 2 − 2
a
(
b
a2
y1 =
#
''
)
dA = ( y2 − y1 ) dx
!
"
x2 # b
= $b && 2 − 2 '' − 2 x 2 % dx
a ) a
$* (
%+
2b
= 2 (a 2 − x 2 ) dx
a
Then
-
A = dA =
-
a
0
2b 2
(a − x 2 )dx
a2
a
2b 2
1 !
a x − x3 %
2 $
3 +0
a *
4
= ab
3
=
Now
2b
!
dI y = x 2 dA = x 2 $ 2 (a 2 − x 2 ) dx %
a
*
+
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1683
PROBLEM 9.187 (Continued)
Then
-
I y = dI y =
-
a
0
2b 2 2
x (a − x 2 ) dx
a2
a
=
2b 1 2 3 1 5 !
a x − x %
5 +0
a 2 $* 3
or
and
k y2 =
Iy
A
=
4 3
ab
15
4 ab
3
=
Iy =
4 3
ab
15
1 2
a
5
or
ky =
a
5
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1684
PROBLEM 9.188
Determine the moments of inertia of the shaded area shown with respect to the
x and y axes.
SOLUTION
We have
where
I x = ( I x )1 + ( I x )2 − ( I x )3
( I x )1 =
1
4
(2a )(2a)3 = a 4
12
3
π
a4
Now
( I AA ) 2 = ( I BB )3 =
and
( I AA ) 2 = ( I x2 ) 2 + Ad 2
or
( I x2 ) 2 = ( I x3 )3 =
Then
8
2
8
" π #" 4a # " π
a 4 − & a 2 '&
' =& −
8
( 2 )( 3π ) ( 8 9π
π
8
"π
( I x )2 = ( I x2 )2 + Ad 22 = & −
( 8 9π
" 4 5π # 4
=& +
'a
(3 8 )
# 4
' a1
)
4a #
# 4 " π 2 #"
' a + & a '& a +
'
3π )
)
( 2 )(
8 # 4 " π 2 #"
4a #
"π
( I x )3 = ( I x3 )3 + Ad32 = & −
' a + & a '& a −
'
3π )
( 8 9π )
( 2 )(
" 4 5π # 4
= &− +
'a
( 3 8 )
4 4 " 4 5π # 4 ! " 4 5π # 4 !
a + $& +
' a % − $& − +
'a %
3
* ( 3 8 ) + *( 3 8 ) +
Finally
Ix =
Also
I y = ( I y )1 + ( I y ) 2 − ( I y )3
Where
( I y )1 =
2
2
I x = 4a 4
1
16
(2a)(2a)3 + (2a) 2 (a) 2 = a 4
12
3
!
π
"π #
( I y ) 2 = ( I y )3 $ = a 4 + & a 2 ' ( a ) 2 %
8
2
(
)
*
+
Iy =
16 4
a
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1685
PROBLEM 9.189
Determine the polar moment of inertia of the area shown with
respect to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the area.
A, mm 2
1
2
Σ
π
2
−
48
(54)(36) = 3053.6
π
2
π
24
(18) 2 = −508.9
π
y , mm
yA, mm3
= 15.2789
46,656
= 7.6394
−3888
42,768
2544.7
Y Σ A = Σ y A: Y (2544.7 mm 2 ) = 42, 768 mm3
Then
Y = 16.8067 mm "
or
J O = ( J O )1 − ( J O ) 2
(a)
π
π
(54 mm)(36 mm)[(54 mm)2 + (36 mm)2 ] − (18 mm) 4
8
4
6
6
4
= (3.2155 × 10 − 0.0824 × 10 ) mm
=
= 3.1331 × 106 mm 4
or
J O = 3.13 × 106 mm 4
J O = J C + A(Y )2
(b)
or
J C = 3.1331 × 106 mm 4 − (2544.7 mm 2 )(16.8067 mm) 2
or
J C = 2.41 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1686
PROBLEM 9.190
To form an unsymmetrical girder, two L76 × 76 × 6.4-mm angles
and two L152 × 102 × 12.7-mm angles are welded to a 16-mm steel
plate as shown. Determine the moments of inertia of the combined
section with respect to its centroidal x and y axes.
SOLUTION
A = 929 mm 2
L76 × 76 × 6.4:
I x × I y = 0.512 × 106 mm 4
A = 3060 mm 2
L152 × 102 × 12.7:
I x = 2.59 × 106 mm 4
I y = 7.20 × 106 mm 4
First locate centroid C of the section:
Then
or
A, mm 2
y
yA, mm3
1
2(3060) = 6120
24.9
152,388
2
(16)(540) = 8640
270
2,332,800
3
2(929) = 1858
518.8
963,930
Σ
16,618
3,449,118
YΣA = Σy A
Y (16, 618 mm 2 ) = 3, 449,118
or
Y = 207.553 mm
Now
I x = 2( I x )1 + ( I x ) 2 + 2( I x )3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1687
PROBLEM 9.190 (Continued)
where
( I x )1 = I x + Ad 2 = 2.59 × 106 mm 4 + (3060 mm 2 )[(207.553 − 24.9) mm]2
= 104.678 × 106 mm 4
1
( I x )2 = I x + Ad 2 = (16 mm)(540 mm)3 + (16 mm)(540 mm)[(270 − 207.553) mm]2
12
= 243.645 × 106 mm 4
( I x )3 = I x + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 [(518.8 − 207.553) mm]2
= 90.5086 × 106 mm 4
Then
I x = [2(104.678) + 243.645 + 2(90.5086)] × 106 mm 4
or
Also
where
I x = 634 × 106 mm 4
I y = 2( I y )1 + ( I y ) 2 + 2( I y )3
( I y )1 = I y + Ad 2 = 7.20 × 106 mm 4 + (3060 mm 2 )[(8 + 50.3) mm]2
= 17.6006 × 106 mm 4
1
( I y ) 2 = (540 mm)(16 mm)3
12
= 0.1843 × 106 mm 4
( I y )3 = I y + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(8 + 21.2) mm]2
= 1.3041 × 106 mm 4
Then
I y = [2(17.6006) + 0.1843 + 2(1.3041)] × 106 mm 4
or
I y = 38.0 × 106 mm 4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1688
PROBLEM 9.191
Using the parallel-axis theorem, determine the product of inertia of
the L5 × 3 × 12 -in. angle cross section shown with respect to the
centroidal x and y axes.
SOLUTION
We have
I xy = ( I xy )1 + ( I xy ) 2
For each rectangle:
I xy = I x′y′ + x yA
I x′y′ = 0 (symmetry)
and
I xy = Σ x yA
Thus
A, in.2
1
3×
2
4.5 ×
1
= 1.5
2
1
= 2.25
2
x , in.
y , in.
x yA, in.4
0.754
−1.49
−1.68519
− 0.496
1.01
−1.12716
− 2.81235
Σ
I xy = − 2.81 in.4
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1689
PROBLEM 9.192
For the L5 × 3 × 12 -in. angle cross section shown, use Mohr’s circle
to determine (a) the moments of inertia and the product of inertia
with respect to new centroidal axes obtained by rotating the x and y
axes 30° clockwise, (b) the orientation of the principal axes through
the centroid and the corresponding values of the moments of inertia.
SOLUTION
From Figure 9.13a:
I x = 9.43 in.4
I y = 2.55 in.4
From the solution to Problem 9.191
I xy = −2.81235 in.4
The Mohr’s circle is defined by the diameter XY, where
X (9.43 − 2.81235),
Now
I ave =
Y (2.55, 2.81235)
1
1
( I x + I y ) = (9.43 + 2.55) = 5.99 in.4
2
2
2
and
2
1
1
!
!
R = $ ( I x − I y ) % + I xy2 = $ (9.43 − 2.55) % + (−2.81235) 2
2
2
*
+
*
+
= 4.4433 in.4
The Mohr’s circle is then drawn as shown.
tan 2θ m = −
2 I xy
Ix − Iy
2( −2.81235)
9.43 − 2.55
= 0.81754
=−
or
and
Then