HW#6 solution

HW #6 solution
!
.m
m패
"
=→
mB
세니
.
.
"
5
도
Find the
.
"며
I
먹
학차
=
"=
=
voy
시다이
=
.
다괴
=
qc
x-
3 78 KW 1 명
VQy
-
IEZ
=
0
20 .
t
11111
=
8694 ×
(
5
10
-
5
m
.
k
151003 t
4
…
y
:
대에
f,
m
+ zl
2 3000
065
…
개
ydy
n on 5
+
. i
5
f
.
.
m 에
86939050 mmy
식울에더째 "
KNI 에
=6.
2로7
12
시여 -
.
=
%
VQy
2162HS
094
=
=
qA
1 au
.
(
- 230153
IE =
,
I
*
IEZ
1090
=
shear flow
ad
GXTS
3003
S 15 ydy s
+
SdAsydA y
o
d
J .200 5 ydyt
=
.
"
meatra 에
nk ,
….
01
/
/ - 11
) 시에 ,
111
11,
10015232 )
+
k
153003
ti
5
z
omm -
=
sydA
*1
고
전
oun
-
= san -
( f xosoydytfiyioogoydy
)
o
9
00
.
5
= 20
W
mm
mmA
150
오
s
ㅡ
IEz
150452
150 30 45 + 21 .
15030 + 3015630
.
=
R
100
고
=
shear flow
q
F
=-
6 5 kW 1 에
-
mm
F
4
27
=
10
0
4
m
VQy
-
=
IEE
6
5 4
+
.
로
x
F
F
⇒
0
.
+
.
슈
A
Ʃ
'
B
수
MB
8 F 1 - 262
=
⇒
0
=
F=
6
F2
261 원
c
=
kW
. 5
F= 19 5 kN
.
< singularity function >
FE
8
V
6 5 < x3 '
=
-
. Ke -430 + 19 . 5 Kx -8
6
.
↓
↓ d
-
=
-
6 5 <c30
.
>
-
1
5
+
6
. <
5
e-4 1 - 19 . EpK
)
5
-
870
MV
1 일
"
g
4
:
6 5
-
⇒ Vma =
19
.
5
kN
2 x
,
shear
Uma
flow
qmax
에
-
=
IEE
QY
19 5 103
.
27 104
=
omax
=
38
max
-
3010
=
4 875 MPa
.
×
=
146250 N 서
6360150
.
y
dy
-
10
9
15 t
.
!
a
Neufrol
*
방
Y
axis
sydA
=
f tzaydy
=
-
sto
+
.
SaA
mat
ㅡ
있다
(a
+
t)
-
(
1
iytfydy
(
)
. 5t
황높냐
*
까
는
더
3 a
4
lat 23 )
Maximym bendi 서 gmorment
.
dddldddddw
xva )
UC=w
(
1
KCO
C= ( Oa
M ( x)
M (o )
-
=
.
.<
ω
K>
<
'
a
+
)
o
-
C
,
.<
←
-
G ,
)
w x 르
M
0
=
CF -
10 a
CZ
5 wba
⇒
(
=
)
x
tCG
=
o
O
Zw <x ) zt 5 w
( xF
ax
.
:
M
.
-
∞
=
K
Zw
w
=
.
고
2 + 5 w 0 ac
( c-
5
a
)
'
+
225
w0
az
고5
C= 50
(
Maximym
-
.
stress
at
ascars
compressive
aximum
y
iax
- 서바
작
ta
IEZ
-
mn
3
y
=
↑
a
t
1 ocl
=
Mo
:
IEZ
= 3
서
그
=
310 T 1
-
y
이에
o
3 a
*
=
4
:
IEE
12 11
=
- 5t
5
≈
-
16
=
w
apmmaxt
z
yoccarsat
-
stresr
tensile
~
a
3
)(
2
a
23
=
-t
3
( atz )
)
t(
3
1 .5
t ) a(
t
) zatt - la pt *
4
(
t
w
25 afic
IEzMmaxy
tow
165
-
그
.
=
at
∴
w=
otoT
=
)
(
t
=
3
)
a
tz
[
f
z
3 a ) t( la
- zy
(
if
t < ca
)
2
4
.
61km
↑
-
고서
ㄴ
1
5
59 09 mm
=
.
Bending
m
at
moment
a
area
-
a
C
-
(2
=
6 kN
-
2
)X B
1
( 고 6 km ]
+
X 2
5
X
.
×
-
1B
( 59 09 ×
-
10
.
33
a
4
=
85909 kNn
.
AXIl force
25
Z
1
=
D -2
:
Mby
oz. c, top =
IEZ -
Mb
-
bxk bottom
. 085909
A
-
16 19
422576
-
2576
-
X
(o
t
152003
6
4
0
.
.
59 wg
×
mmlkm
11000
.
091
l 61
=
KN
o
≈
15 . 200 40
.
.
91
)
p
4
( Ho
6
-
.
P
3
=
- 161
어 MPa
P+ 3
=
paXimom
y
shear force
bar AB
Tn
:
shear flow
endure
should
Nails
↑1111111Qp
a1 o
oa 1 MP에
F
고
%
)
.
mpa
fr
2
- m← →
8y
= kN
5
3
1ㆍ
X 10
.
0
m
-
13
G
LO
42 2576
=
-
kN ) X
6
103 [ 5909603 = 6
naMpa
×
5
d
m
-
( 200 ) =
.
3
←
(ㅁ
2
.
-
.
RF
09 t
.
2
)
4
IKN
aXTl =
42
석4
IEZ
lomax =
5
=
=
,
.
2502049
422576 X 10 bm
=
OKK
3 t
-
IEE
o
(
=
Vmax
>
xd
Z
IEE
=
=
-
+
72
S yn
x
1 B ) ( 1503=
100 mm
4
212250303
-
=
72 ×
No
bm
5
Q
-
4
25 oydy
=
450000
+
25030
.
602 )
4
mmz
=
4 5
시
:
θ
04
spacing
Anol
IE
'
m
.
-
ABpITed
Shear force
for
one
nail
P
. 1xqty
3
= 1
λ2
:
P
드
6 6 kN
.
+
P
+ 3
( KN
)
at the
interfacey
ny
…
=
× 보 KN
q