### Example 2:design of rectangular beam

Design a beam using  ρ=0.0120, Mu = 600 ft-k, fy = 60,000 psi, and fc' = 4000psi.

φ*Mn=Mu=φAs*fy*(d-a/2)
a=(As*fy)/(0.85*fc'*b)

φ*Mn=Mu=φAs*fy*(d-(As*fy)/(0.85*fc'*b*2))----divide by bd^2

Mu/(b*d^2)=φρ*fy*(1-ρ*fy/1.7*fc')
b*d^2=Mu/(φρ*fy*(1-ρ*fy/1.7*fc'))

600 K-ft=7,200,000Lb-in
assuming φ=0.90
b*d^2=7200000/(0.90*0.012*60000*(1-0.012*60000/1.7*4000))
b*d^2=12,426.9in2

bd2 = 12,427in.
b × d
12 in. × 32.18 in.
14 in. × 29.79 in.-------selecting this one
16 in. × 27.87 in.

so we obtain b=14in , h=33in(d=30in)

As=ρ*b*d
As=0.012*14*30=5.04in2

select 4#10 (As=5.06in2)

checking solution

ρ=As/b*d=0.012

ρmin=(3√fc'/fy) and not less than 200/fy
ρmin=(3√4000/60000) and not less than 200/6000
ρmin=0.0033

ρ>ρmin  ok

checking for the reduction factor φ
0.003/C=(εt+.003)/d
C=a/β1

a=As*fy/0.85*fc'*b
a=5.06*60000/0.85*4000*14
a=6.378
β1=0.85 for fc'=4000psi
C=6.378/.85=7.5in

0.003/7.5=(εt+.003)/30

εt=0.009>0.005, therefore, the section is tension control and our assumption is ok

Checking φ*Mn

φ*Mn=φ*As*fy*(d-a/2)
φ*Mn=0.9*5.06*60000*(30-6.78/2)
φ*Mn=7,325,837.64>Mu    ok