Transcript Slide 1
Design of Concrete Structure I
Dr. Ali
Tayeh
First
Semester
2009
1
Lecture 8
DESIGN OF
T-Section BEAMS FOR
MOMENTS
2
Analysis of Flanged Section
• Floor systems with slabs and beams are placed in
monolithic pour.
• Slab acts as a top flange to the beam; T-beams, and
Inverted L(Spandrel) Beams.
3
Analysis of Flanged Sections
Positive and Negative Moment Regions in a T-beam
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Analysis of Flanged Sections
If the neutral axis falls
within the slab depth
analyze the beam as a
rectangular beam,
otherwise as a T-beam.
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Analysis of Flanged Sections
Effective Flange Width
Portions near the webs are more highly stressed than
areas away from the web.
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Analysis of Flanged Sections
Effective width (beff)
beff is width that is stressed uniformly to give the same
compression force actually developed in compression
zone of width b(actual)
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ACI Code Provisions for Estimating beff
From ACI 318, Section 8.10.2
T Beam Flange:
L
beff
4
16hf bw
bactual
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ACI Code Provisions for Estimating beff
From ACI 318, Section 8.10.3
Inverted L Shape Flange
L
beff
bw
12
6hf bw
bactual bw 0.5* clear distance to next web
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ACI Code Provisions for Estimating beff
From ACI 318, Section 8.10
Isolated T-Beams
bw
hf
2
beff 4bw
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Various Possible Geometries of T-Beams
Single Tee
Twin Tee
Box
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Analysis of T-Beam
Case 1:
a hf
Equilibrium
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As f y
T Ca
0.85fc beff
Analysis of T-Beam
Case 1:
a hf
Confirm
s y
c
a
1
d c
s
cu 0.005
c
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Analysis of T-Beam
Case 1:
a hf
Calculate Mn
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a
M n As f y d
2
Analysis of T-Beam
Case 2:
a hf
Assume steel yields
Cf 0.85 f cb bw hf
Cw 0.85 f c bw a
T As f y
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Analysis of T-Beam
Case 2:
a hf
Equilibrium
Assume steel yields
0.85 f c b bw hf
Asf
fy
The flanges are considered to be equivalent compression steel.
T Cf Cw a
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As Asf f y
0.85 fcbw
Analysis of T-Beam
Case 2:
a hf
Confirm
a hf
c
a
1
d c
s
cu 0.005
c
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Analysis of T-Beam
Case 2:
a hf
Calculate nominal
moments
M n M n1 M n2
a
M n1 As Asf f y d
2
hf
M n2 Asf f y d
2
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Analysis of T-Beams
The definition of Mn1 and Mn2 for the T-Beam are
given as:
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Limitations on Reinforcement for
Flange Beams
• Lower Limits
– Positive Reinforcement
min
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f c
As
4f y
larger of
b wd
1.4
fy
Limitations on Reinforcement for
Flange Beams
• Lower Limits
– For negative reinforcement and T sections
with flanges in tension
(min)
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f c
2f y
larger of
1.4
f
y
Example - T-Beam
Find Mn and Mu for T-Beam.
hf = 15 cm
d = 40cm
As = 50cm2
fy = 420Mpa fc = 25Mpa
bw= 30cm
S=2.15m
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L = 5.5m
Example of T-Beam
Confirm beff
beff
L 550
137cm .
4
4
16hf b w 16 15 30=270cm.
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b 2.15 m 215cm .
Compute the equivalent c value and check the strain
in the steel, s
Assume a<t.
2
50
cm
420mpa
As f y
a
7.21cm t 15cm ok
0.85f c b eff 0.85 25Mpa 137cm
a
7.21
c
8.5 cm
1 0.85
d
40
s 1 0.003
1 0.003 0.011 0.005
c
8.5
Steel will yield in the tension zone.
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Compute the reinforcement and check to make sure it
is greater than min
2
50
cm
As
0.042
b w d 30 cm 40 cm
min
1.4 1.4
f 420 0.0033
y
min 0.0033
25
fc
0.003
4f
y 4 420
0.042 0.0033
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Section works for minimum
reinforcement.
Compute nominal moment components
a
M n As f y d
2
0.0721
50 10 420 10 0.40
2
764295 N .m
4
6
M u M n
0.9 764295 =688 kN .m
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