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
4
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.
5
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 Ca
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 cb  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 fcbw
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
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 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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