### Example 3: 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 the weight of beam=0.50k/ft

Mu from moment diagram in figure 1=657.5k-ft=7,890,000lb-in

Figure 1

Mu=Ф*Mn=Ф*Asfy(d-a/2)

Mu/(b*d^2)=Ф*ρ*fy(1-fy*ρ/1.7*fc')

ρ=0.18*4,000/60,000=0.012

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

Mu=7,890,000lb-in

bd^2=7,890,000/(0.9*.012*60,000*(1-60,000*0.012/1.7*4000))=13,617.88

taking bd=    16x29.5

using 16*33 as beam dimension (bxh)

checking beam weight=16*33*150/144=0.55k/ft    not ok

Mu=664.25 k-ft=7,971,000lb-in

bd^2=7,971,000/(0.9*.012*60,000*(1-60,000*0.012/1.7*4000))=13,757.61

bd=    16x29.5 is ok

As=ρ*b*d=0.012*16*29.5=5.664in2

using 4#11=4*1.56=6.24 in2   ok

checking for Ф=0.9 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=6.24*60,000/(0.85*4000*16)=6.88in

C=a/β1=6.88/0.85=8.1in

εt=0.003*(29.5)/8.1-0.003=0.0079>0.005 so section is tension control Ф=0.9  is ok

checking for ρmin

As,min=(3√fc'/fy)*bw*d and not less than 200bw*d/fy

As,min=3*√4000*16*29.5/60,000  and not less than 200*29.5*16/60,000

As,min=1.49 in2 and not less than 1.57in2 ok, our As is larger than the minimum

checking Ф*Mn>Mu

Ф*Mn=0.9*6.24*60,000(d-a/2)

Ф*Mn=0.9*6.24*60,000(29.5-6.88/2)=8,781,177.6lb-in> Mu ok