Example 4: design of rectangular beam

design rectangular sections for the beams, loads are shown in figure below. Beam weights are not included in the loads shown. Assume concrete weighs 150 lb/ft3 , fy= 60,000 psi, and fc’= 3000 psi, ρ=0.0107 unless given otherwise.

example2b4-3611056 construction management: concrete construction
bridge construction:How to become a bridge engineer first, we will calculate the ultimate load, we have point live load and distributed dead load, therefore will take each load separately ultimate live load=1.6*PL=1.6*40=64k assuming the weight of beam=0.75k/ft ultimate dead load=1.2*(0.75+3)=4.5k/ft Mu=876.74k-ft=10,520,880lb-in example2b5-2541095 Mu/(b*d^2)=Ф*ρ*fy(1-fy*ρ/1.7*fc’)     bd^2=Mu/(Ф*ρ*fy(1-fy*ρ/1.7*fc’)   assuming Ф=0.90 bd^2=10,520,880/(0.9*.0107*60,000*(1-60,000*0.0107/1.7*3000))=20,830.7 checking beam weight=18x37x150=0.693k/ft so it is ok As=ρxbxd=0.0107x18x37=7.126in2 ρ,min  =3*√fc’/fy and not less than 200/fy ρ,min=0.0027 and not less than 0.0033    ok a=7.8*60,000/(0.85*3000*18)=10.196in εt=0.003*(33.5)/11.99-0.003=0.0054>0.005 so section is tension control Ф=0.9  is ok Ф*Mn=0.9*6.24*60,000(d-a/2) Ф*Mn=0.9*7.8*60,000(33.5-10.196/2)=11,836,562.4lb-in> Mu ok