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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.











Figure 1



Assuming beam weight=0.55k/ft

ultimate dead load=1.2*(0.55+2)=3.06 k/ft

ultimate live load=1.6*20=32k

the ultimate moment from figure 2

Mu=604.32k-ft=7,251,840 lb-in
Figure 2

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=20x22.5

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





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