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# t beam design

Anuncio ```Chapter 5 Analysis and Design of T Beams
and Doubly Reinforced Beams
• T Beams
– Analysis
– Design
• Doubly reinforced concrete beams
– Compression steel
T Beams
• Monolithically constructed floor system
resulting in T-shaped beams (or L-shaped
beams)
• Consists of flange and web
• Effective only when the flange is subjected to
flexural compression
S1
S2
Effective Flange Width
• For T-shaped beams, lesser of
– L/4
– bw+2&times;8hf
– bw+(S1+S2)/2
• For L-shaped beams, lesser of
– L/12
– bw+6hf
– bw+S/2
S1
S2
Isolated T Beams
8.10.4 ― Isolated beams, in which the T-shape is
used to provide a flange for additional
compression area, shall have a flange
thickness not less than one-half the width of
web and an effective flange width not more
than four times the width of web.
ACI 318M-05
Location of neutral axis determines
whether the rectangular beam formulas
apply
Rectangular
beam
formulas can
be applied
Rectangular
beam formulas
cannot be
applied
Analysis of T Beams
• Check As,min as per ACI Section 10.5.1
using bw as the web width
• Compute T=Asfy
• Determine Ac stressed to 0.85f'c
T
Ac =
0.85 f c′
• Calculate a, c, εt
• Calculate φMn
Example 5.1 Determine the design strength of
the T beam shown below with f'c=27.6 MPa and
fy=414 MPa.
b=1524 mm
hf=100
d=610 mm
6-#29
(As=3870 mm2)
bw=254 mm
Minimum reinforcement: As ,min =
(ACI 318M-05 10.5.1)
As ,min =
f c′
27.6
bw d =
&times; 254 &times; 610 = 491.5 mm 2
4 fy
4 &times; 414
1.4bw d 1.4 &times; 254 &times; 610
=
= 524 mm 2 ←
fy
414
As , provided &gt; As ,min O.K .
Example 5.1 Determine the design strength of
the T beam shown below with f'c=27.6 MPa and
b=1524 mm
fy=414 MPa.
hf=100
d=610 mm
6-#29
(As=3870 mm2)
Assuming steel yielding,
Ac =
bw=254 mm
T
3870 &times; 414
=
= 68294.1 mm 2
0.85 f c′ 0.85 &times; 27.6
a=
Ac 68294.1
a 44.8
=
= 44.8 mm → c =
=
= 52.7 mm (neutral axis within the slab)
b
1524
β1 0.85
εt =
0.003
(d − c ) = 0.003 &times; (610 − 52.7 ) = 0.0317 &gt; 0.005 → φ = 0.9
c
52.7
Design strength:
φM n = φAs f y (d − a 2 ) = 0.9 &times; 3871&times; 414 &times; (610 − 44.8 2 )&times;10 −6 = 847.5 kN − m
Method for Analyzing T Beams
When a&gt;hf, slab contribution to compression can
be divided into web portion and flange portion.
Cw = 0.85 f c′abw
C f = 0.85 f c′(b − bw )h f
M n = Cw (d − a 2 ) + C f (d − h f 2 )
Example 5.2 Compute the design strength for
the T beam shown below in which f'c=27.6 MPa
and fy=414 MPa
b=762 mm
hf=102 mm
Minimum reinforcement:
As ,min =
f c′
27.6
bw d =
&times; 254 &times; 610 = 491.5 mm 2
4 fy
4 &times; 414
1.4bw d 1.4 &times; 356 &times; 762
d=762 mm
As ,min =
=
= 917.3 mm 2 ←
660 mm
414
fy
8-#32
As=6552 mm2
As , provided &gt; As ,min O.K .
Assuming steel yielding:
bw=356 mm
Ac =
a=
T
6552 &times; 414
=
= 115623.5 mm 2
0.85 f c′ 0.85 &times; 27.6
Ac 115623.5
=
= 151.7 mm &gt; h f
b
762
→ neutral axis is located out of slab
Example 5.2 Compute the design strength for
the T beam shown below in which f'c=27.6 MPa
and fy=414 MPa
Calculation of stress block depth:
As f y − 0.85 f c′h f (b − bw )
a=
0.85 f c′bw
b=762 mm
hf=102 mm
6552 &times; 414 − 0.85 &times; 27.6 &times;102 &times; (762 − 356 )
0.85 &times; 27.6 &times; 356
= 208.5 mm
a 208.5
=
= 245.2 mm
c=
β1 0.85
=
d=762 mm
As=6552 mm2
bw=356 mm
εt =
0.003
(d − c ) = 0.003 (762 − 245.2) = 0.0063 &gt; 0.005 → φ = 0.9
c
245.2
Design strength:
φM n = φ 0.85 f c′[abw (d − a 2 ) + h f (b − bw )(d − h f 2 )]
= 0.9 &times; 0.85 &times; 27.6 &times; [208.5 &times; 356 &times; (762 − 208.5 2 ) + 102 &times; (762 − 356 )(762 − 102 2 )]&times;10 −6
= 1652.5 kN − m
Design of T Beams
• Follow the procedure similar to the
design of rectangular beams
Example 5.4 Design a T Beam for the floor system shown
below for which bw and d are given. MD=108.5 kN-m,
ML=135.6 kN-m, f'c=27.6 MPa, fy=414 MPa and simple
100 mm
span=6.1 m
d=457 mm
305 mm
3048 mm
3048 mm
3048 mm
3048 mm
Effective flange width: b = L 4 = 6100 4 = 1525 mm ← smallest
b = bw + 2 &times; 8h f = 305 + 16 &times;100 = 1905 mm
b = bw + (S1 + S 2 ) 2 = 3048 mm
Factored moment: M u = 1.2 M D + 1.6 M L = 1.2 &times;108.5 + 1.6 &times;135.6 = 347.2 kN − m
Assuming steel yielding and φ=0.9, jd=0.9d,
Mu
347.2 &times;106
=
= 2265.6 mm 2
As ,required =
φf y jd 0.9 &times; 414 &times; 0.9 &times; 457
→ Use 4-#25 (As,provided=2040 mm2)
Beam width: bw = 2&times;40+2&times;10+4&times;25+3S = 305 mm → S = 35mm &gt; 25mm O.K.
Example 5.4 Design a T Beam for the floor system shown
below for which bw and d are given . MD=108.5 kN-m,
ML=135.6 kN-m, f'c=27.6 MPa, fy=414 MPa and simple
100 mm
span=6.1 m
d=457 mm
305 mm
3048 mm
3048 mm
As ,min =
f c′
bw d
4 fy
3048 mm
As ,min =
1.4bw d
fy
3048 mm
As ,max = 0.0181bw d
= 0.0181&times; 305 &times; 457
1.4 &times; 305 &times; 457
27.6
= 2522.9 mm 2 &gt; As , provided
=
&times; 305 &times; 457
414
4 &times; 414
= 471.4 mm 2 &lt; As , provided
= 442.2 mm 2 &lt; As , provided
As f y
a 24.4
2040 &times; 414
a=
=
= 24.4 mm &lt; h f → c =
=
= 28.7 mm → φ = 0.9
β1 0.85
0.85 f c′b 0.85 &times; 26.7 &times;1525
=
φM n = φAs f y (d − a 2) = 0.9 &times; 2040 &times; 414 &times; (457 − 24.4 2 )&times;10 −6 = 338 kN − m ≈ M u
(2.7% under, say O.K.)
Example 5.5 Design T beam for the floor system below for which bw
and d are given MD=271.3 kN-m, ML=576.5 kN-m, f'c=20.7 MPa,
fy=414 MPa, and sipmle span=5.5 m
76
533 mm
381
1829 mm
381
1829 mm
381
1829 mm
1829 mm
Effective flange width: b = L 4 = 5500 4 = 1375 mm ← smallest
b = bw + 2 &times; 8h f = 381 + 16 &times; 76 = 1597 mm
b = bw + (S1 + S 2 ) 2 = 1829 mm
Factored moment: M u = 1.2M D + 1.6M L = 1.2 &times; 271.3 + 1.6 &times; 576.5 = 1248 kN − m
610
Example 5.5 Design T beam for the floor system below for which bw
and d are given MD=271.3 kN-m, ML=576.5 kN-m, f'c=20.7 MPa,
fy=414 MPa, and sipmle span=5.5 m
76
533 mm
381
1829 mm
381
1829 mm
610
381
1829 mm
1829 mm
Assuming steel yielding and φ=0.9, jd=0.9d,
Mu
1248 &times;106
=
= 6101 mm 2
As ,required =
φ f y jd 0.9 &times; 414 &times; 0.9 &times; 610
double
layer
→ Use 10-#29 double layer (As,provided=6450 mm2)
Beam width: bw = 2&times;40+2&times;10+5&times;29+4S = 381 mm → S = 34mm &gt; 25mm O.K.
Example 5.5 Design T beam for the floor system below for which bw
and d are given MD=271.3 kN-m, ML=576.5 kN-m, f'c=20.7 MPa,
fy=414 MPa, and sipmle span=5.5 m
76
533 mm
381
381
381
f c′
20.7
bw d =
&times; 381&times; 610 = 638.5 mm 2 &lt; (As , provided )mod ified
18294mm
1829 mm
fy
4 &times; 414 1829 mm
As ,min =
As ,min =
1829 mm
1.4bw d 1.4 &times; 381&times; 610
=
= 786 mm 2 &lt; (As , provided )mod ified
fy
414
As ,max = 0.0135bw d = 0.0135 &times; 381&times; 610 = 3138 mm 2 &lt; As , provided
(A
610
)
s , provided mod ified
N .G.
= As , provided − 0.85( f c′ f y )h f (b − bw ) = 6450 − 0.85 &times; (20.7 414 )&times; 76 &times; (1375 − 381) = 3239 mm 2
Since still (As , provided )mod ified &gt; As ,max , slab thickness should be increased. Let' s use h f = 80 mm
Example 5.5 Design T beam for the floor system below for which bw
and d are given MD=271.3 kN-m, ML=576.5 kN-m, f'c=20.7 MPa,
fy=414 MPa, and sipmle span=5.5 m
80
533 mm
381
1829 mm
(A
)
s , provided mod ified
381
1829 mm
381
1829 mm
1829 mm
= As , provided − 0.85( f c′ f y )h f (b − bw ) = 6450 − 0.85 &times; (20.7 414)&times; 80 &times; (1375 − 381) = 3070 mm 2
As ,max = 0.0135bw d = 0.0135 &times; 381&times; 610 = 3138 mm 2 &gt; (As , provided )mod ified O.K .
6450 &times; 414
a 110.4
= 110.4 mm &gt; h f → c =
=
= 129.9 mm
0.85 f c′b 0.85 &times; 20.7 &times;1375
β1 0.85
Since the neutral axis is located out of flange, take the secton into web and flange parts!
a=
As f y
610
=
Example 5.5 Design T beam for the floor system below for which bw
and d are given MD=271.3 kN-m, ML=576.5 kN-m, f'c=20.7 MPa,
fy=414 MPa, and sipmle span=5.5 m
80
533 mm
381
1829 mm
381
1829 mm
610
381
1829 mm
1829 mm
As f y
a 189.6
6450 &times; 414 − 0.85 &times; 20.7 &times; (1375 − 381)&times; 80
=
= 189.6 mm → c =
=
= 223.1 mm
′
β1 0.85
0.85 f c bw
0.85 &times; 20.7 &times; 381
0.003
(d − c ) = 0.003 (610 − 223.1) = 0.0052 &gt; 0.005 → φ = 0.9
εt =
c
223.1
a=
Example 5.5 Design T beam for the floor system below for which bw
and d are given MD=271.3 kN-m, ML=576.5 kN-m, f'c=20.7 MPa,
fy=414 MPa, and sipmle span=5.5 m
80
533 mm
381
1829 mm
381
1829 mm
381
1829 mm
1829 mm
φM n = φ [C f (d − h f 2) + Cw (d − a 2)]
= 0.9 &times; [0.85 f c′(b − bw )h f (d − h f 2 ) + 0.85 f c′abw (d − a 2 )]
= 0.9 &times; 0.85 &times; 20.7 &times; (1375 − 381)&times; 80 &times; (610 − 80 2 )&times;10 −6
+ 0.9 &times; 0.85 &times; 20.7 &times;189.6 &times; 381&times; (610 − 189.6 2 )&times;10 −6
= 1307.1 kN − m &gt; M u
610
Design of T Beams for Negative Moments
• Part of beams under negative moment
causing the convex shape of beam deflection
• In this case, flanges in tension and lower part
of web in compression
• No beneficial effect from flange concrete
cannot be expected
• Therefore, beams in this category are rather
the rectangular beam than the T beam
tension
T
compression
C
Compression Steel
• Results in doubly reinforced concrete beams
– Different from the double layered steel
• Increases moment capacity by increasing
tension steel without violating the maximum
steel limit
• Makes the beam ductile and tough to be
earthquake resistant
• Reduces long-term deflection due to
shrinkage and plastic flow
• Makes the fabrication of transverse bar right
and easier by providing the supports
Design of Doubly Reinforced
Concrete Beams
Compression steel reduces the contribution of concrete
stress block. However, the existence of compression steel
does not significantly enhance the moment strength.
C = 0.85 f c′ab + As′ f s′
T = As f y
From force equilibrium and assumption that tensile steel
yields,
T = C = Cc + C s
T = As f y
Cc = 0.85 f c′ab
Cs = As′ f s′ = As′ε s′ Es ≤ As′ f y
As′
d d′
As
b
Design of Doubly Reinforced
Concrete Beams
From strain profile,
ε s′
c − d′
0.003
c
ε s′ = 0.003(1 − d ′ c ) = 0.003(1 − β1d ′ a )
=
Substituting this into the equilibrium relation,
Cs = As′ε s′ Es = As′ ⋅ 0.003(1 − β1d ′ a )Es = 600 As′ (1 − β1d ′ a )
As f y = 0.85 f c′ab + 600 As′ (1 − β1d ′ a )
Rearranging the equilibrium equation
(0.85 f c′b )a 2 − (As f y − 600 As′ )a − 600 As′ β1d ′ = 0
Solving this quadratic equation gives the value of &quot;a&quot;.
Make sure the tensile steel strain greater than or equal
to 0.005.
Alternative Analysis of Doubly
Reinforced Concrete Beams
Various T-Beams
Various Doubly Reinforced
Concrete Beams
```