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Example 1:design of rectangular beam

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Design a rectangular beam for a 22-ft simple span if a dead load of 1 k/ft (not including the beam weight) and a live load of 2 k/ft are to be supported. Use f'c = 4000psi and fy = 60,000 psi.(assume h=0.1*L)





h=0.1*L=0.1*22=2.2ft say 27 in and d=24.5in
assuming b=0.5*h=0.5*2.2=1.1ft say 14in
Beam weight=27*14*150/144=393lb/ft3=0.393k/ft3

 Wu=1.2*DL+1.6L.L
Wu=1.2*(1+.393)+1.6*2
Wu=4.87k/ft2

 Mu for simply supported beam=(Wu*(L^2))/8=(4.87*(22^2))/8=292.215k/ft2
  ρ=(0.85*fc'/fy)*(1-√(1-(2*Rn/0.85*fc'))
Rn=ɸ*Mu/(b*d)

assuming ɸ=0.90

Rn=Mu/(ɸ*b*d^2)
Rn=292.21/(0.90*1.1*2.2^2)
Rn=60.98 k/ft2=423.47psi

  ρ=(0.85*4000/60000)*(1-√(1-(2*423.47/0.85*4000))
ρ=0.00756

selecting reinforcement

As=ρ*b*d
As=0.00756*2.2*1.1*12*12=2.63in2
use 3#9 bars (As=3in2)

ρ=As/b*d=3/(14*24.5)=0.00875
ρmin=(3√fc'/fy) and not less than 200/fy
ρmin=((3√4000)/60000) and not less than 200/60000

 ρmin=0.0033
so ρ>ρmin then ok

 now we will calculate εt

 0.003/C=(εt+.003)/d
C=a/β1

 a=As*fy/0.85*fc'*b
a=3*60000/0.85*4000*14
a=3.78
β1=0.85 for fc'=4000psi
C=3.78/.85=4.44in


0.003/4.44=(εt+.003)/24.5


εt=0.01>0.005, therefore, the section is tension control and our assumption is ok
Figure 1

checking if ɸ*Mn>Mu

ɸ*Mn=ɸ*As*fy*(d-a/2)
ɸ*Mn=0.9*3*60000*(24.5-3.78/2)
ɸ*Mn=3662in-k=305.2ft-k>292.12k-ft

Figure 2 (final section)

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