### Example 5: design of rectangular beam

design rectangular sections for the beams, loads, and ρ values shown. Beam weights are not included in the loads shown.  Assume concrete weighs 150 lb/ft3 , fy= 60,000 psi, and fc'= 4000 psi, unless given otherwise.

Figure 1

assuming beam weight=0.50 lb/in

ultimate moment from figure 2

Mu=532.8k-ft=6,393,600 lb-in

Figure 2

Mu/(b*d^2)=Ф*fy*ρ (1-ρfy/1.7*fc')

b*d^2=Mu/(Ф*fy*ρ (1-ρfy/1.7*fc'))

assuming Ф=0.9

ρ=0.18*4000/60,000

b*d^2=6,393,600/(0.90*60,000*0.012 (1-0.012*60,000/1.7*4000))

b*d^2=11,035.08

trial and error we cant take bxd=14x27.5

bxh=14inx31in

weight of the beam=14*31*150/144=452.08lb/ft=0.452k/ft so our assumpiation is valid

As=ρ*b*d=0.012*14*27.5=4.62 in2

using 4#10 at the top, As=4*1.27=5.08 in2

checking for ρmin

ρmin=3√fc'/fy and not less than 200/fy

ρmin=3√4000/60,000 and not less than 200/60,000

ρmin=0.0031and not less than 0.0033    ok

checking if Ф=0.9 valid or no

(εt+0.003)/(d)=0.003/c

C=a/β1

β1=0.85 for fc'=4000 psi or less

a=Asfy/(0.85*fc'*b)

a=5.08*60,000/(0.85*4000*14)=6.4in

C=a/β1=6.4/0.85=7.53in

εt=0.003*(27.5)/7.53-0.003=0.008>0.005 so section is tension control Ф=0.9  is ok

checking if ФMn>Mu
ФMn=0.9*5.08*60,000*(27.5-6.4/2)=6,665,976 lb-in>Mu   ok