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Example 1: Design of T shape beam

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Design a T beam for the floor system shown in Figure 1, for which bw and d are given.MD = 80 ft-k, ML = 100 ft-k, fc' = 4000 psi, fy = 60,000 psi, and simple span = 20 ft.
Figure 1
Effective Flange Width

  • 1/4ft × 20 ft = 5 ft = 60 in.
  • 12 in. + (2) (8) (4 in.) = 76 in.
  • 10 ft = 120 in.
bf=60 in, ACI Code (8.12.2) calls for a smaller width with an assumed uniform stress distribution for design purposes.

Mu=1.2MD+1.6ML=256ft-K

assuming ϕ=0.9
Mn=Mu/ϕ=284.4ft-K=3,412,800in-lb
Assuming a Lever Arm z Equal to the Larger of 0.9d or d − (hf/2)

z=0.9*d=0.9*18=16.2in


Steel trial area

As*fy*Z=Mn
As=3,412,800/(60,000*16.2)
As=3.51in2

computing Z and a

0.85*fc'*bf*a=As*fy
a=(As*fy)/(0.85*fc'*bf)
a=1.0326in     then neutral axis in the flange

z=d-a/2
z=18-1.03/2=17.48in

computing As using revised Z value

As=Mn/(fy*Z)
As=3,412,800/(60000*17.48)=3.25in2 

compute revised a and Z

a=(As*fy)/(0.85*fc'*bf)
a=(3.25*60,000)/(0.85*4000*60)
a=0.96in

Z=18-0.95/2=17.52in


computing As using revised Z value

As=Mn/(fy*Z)
As=3,412,800/(60000*17.52)=3.24 in2     ok

Checking Minimum Reinforcing
As, min=3*√fc'*bw*d/fy and not less than 200*bw*d/fy
As, min=3*√fc'*bw*d/fy=0.68in2 and not less than 200*bw*d/fy=.72in2<3.25in2 ok


checking if ϕ=0.9
c=a/β=0.96/0.85=1.13 in

εt=((d-c)/c)*0.003=0.045>0.005 assumption is ok






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