Example 8: design of rectangular beam

Design rectangular sections for the beams and loads shown. Beam weights are not included in the given loads. fy= 60,000 psi and fc'= 4000 psi. Live loads and dead load(except weight) are to be placed where they will cause the most severe conditions at the sections being considered. Select beam size for the largest moment (positive or negative), and then select the steel required for maximum positive and negative moment. Finally, sketch the beam and show approximate bar locations. (One ans. 12 in. × 28 in. with 3 #10 bars negative reinforcement and 3 #9 bars positive reinforcement)

Figure 1

placing the loads as shown in figure 2 and 3 to generate max positive and negative moment

assuming beam weight=0.35 k/ft
Figure 2(max.positive moment)
Figure 3(max.negative moment)

Mu/b*d^2=Ф*ρ*fy(1-ρ*fy/1.7*fc')
Mu=374 k-ft=4,488,000lb-in
ρ=0.18fc'/fy=0.012
assuming Ф=0.90
b*d^2=4,488,000/(0.9*60,000*0.012*(1-.012*60,000/1.7*4000))

b*d^2=7,746.1
selecting bxd=12inx25.5in
bxh=12inx28in
checking the weight of beam=12*28*150/(144*1000)=0.35k    ok

As=ρ*b*d
As=0.012*12*25.5=3.67in2
selecting 3#10, As=3*1.27=3.81in2

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=3.81*60,000/(0.85*4000*25)=2.69in

C=a/β1=2.69/0.85=3.16in

εt=0.003*(25.5)/3.16-0.003=0.021>0.005 so section is tension control Ф=0.9  is ok

checking if ФMn>Mu

ФMn=0.9*3.81*60,000*(25.5-2.69/2)=4,969,649.70 lb-in>Mu   ok

for positive moment Mu=364.9 k-ft=4,378,800lb-in
ρ=(0.85*fc'/fy)*(1-√(1-(2*Rn/0.85*fc'))
Rn=Mu/(Ф*b*d^2)
assuming  Ф=0.90
Rn=4,378,800/(0.90*12*25.5^2)=623.52

ρ=(0.85*fc'/fy)*(1-√(1-(2*Rn/0.85*fc'))

ρ=(0.85*4000/60000)*(1-√(1-(2*623.52/0.85*4000))

ρ=0.0116
As=  ρ*b*d
As=0.0116*12*25.5=3.54in2

using 3#10, As=3*1.27=3.81in2

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=3.81*60,000/(0.85*4000*25)=2.69in

C=a/β1=2.69/0.85=3.16in

εt=0.003*(25.5)/3.16-0.003=0.021>0.005 so section is tension control Ф=0.9  is ok

checking if ФMn>Mu

ФMn=0.9*3.81*60,000*(25.5-2.69/2)=4,969,649.70 lb-in>Mu   ok