# Example 6: Design of Rectangular Beam

design rectangular sections for the beams, loads, and ρ=0.01425. 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.

Assuming beam weight=0.55k/ft

the ultimate moment from figure 2

Mu=604.32k-ft=7,251,840 lb-in

Mu/bd^2=Ф*ρ*fy*(1-ρ*fy/(1.7fc’))

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

bd^2=7,251,840/0.9*0.01425*60,000*(1-0.01425*60,000/(1.7*4000))

bd^2=10779.45

trail and error selecting   bxd=20×22.5

beam dimension=20inx26in

checking beam weight=20*26*150/(144*1000)=0.541 k/ft    ok

As=ρ*b*d=0.01425*20*22.5=6.40 in2

using 7#9 placed at top because the moment is positive

As=7*1=7 in2

checking if ρ>ρmin

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

ρmin=3*√ 4000/60000 and not less than 200/60000
ρmin=0.00316 and 0.00333     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=7*60,000/(0.85*4000*20)=6.17in
C=a/β1=6.17/0.85=7.27in
εt=0.003*(22.5)/7.27-0.003=0.0063>0.005 so section is tension control Ф=0.9  is ok

checking if ФMn>Mu
ФMn=0.9*7.0*60,000*(22.5-6.17/2)=7,338,870 lb-in>Mu   ok