### Example 4: design of rectangular beam

design rectangular sections for the beams, loads are shown in figure below. Beam weights are not included in the loads shown. Assume concrete weighs 150 lb/ft3 , fy= 60,000 psi, and fc'= 3000 psi, ρ=0.0107 unless given otherwise.

Figure 1

assuming the weight of beam=0.75k/ft

Mu=876.74k-ft=10,520,880lb-in

Figure 2

Mu/(b*d^2)=Ф*ρ*fy(1-fy*ρ/1.7*fc')

ρ=0.0107

bd^2=Mu/(Ф*ρ*fy(1-fy*ρ/1.7*fc')   assuming Ф=0.90

Mu=10,520,880lb-in

bd^2=10,520,880/(0.9*.0107*60,000*(1-60,000*0.0107/1.7*3000))=20,830.7

selecting bxd=    18x33.5

beam dimension bxh=18x37

checking beam weight=18x37x150=0.693k/ft so it is ok
As=ρxbxd=0.0107x18x37=7.126in2
Take 5#11=5x1.56=7.8in2

checking ρ,min

ρ,min  =3*√fc'/fy and not less than 200/fy

ρ,min=0.0027 and not less than 0.0033    ok

checking for Ф=0.9 or no

(εt+0.003)/(d)=0.003/c

C=a/β1

a=As*fy/(0.85*fc'*b)
a=7.8*60,000/(0.85*3000*18)=10.196in
β1=0.85 for fc'=3000psi
C=10.196/0.85=11.99in

εt=0.003*(33.5)/11.99-0.003=0.0054>0.005 so section is tension control Ф=0.9  is ok

checking Ф*Mn>Mu

Ф*Mn=0.9*6.24*60,000(d-a/2)

Ф*Mn=0.9*7.8*60,000(33.5-10.196/2)=11,836,562.4lb-in> Mu ok