A rectangular beam is to be sized with fy= 60,000 psi, fc’= 3000psi, and a ρ approximately equal to 0.18fc’ /fy. It is to have a 25-ft simple span and to support a dead load, in addition to its own weight, equal to 2 k/ft and a live load equal to 3 k/ft.
Assume beam weight=400 Lb/ft
Mu for simply supported beam=(Wu*L^2)/8=600k-ft
Mu=Ф*As*fy*(d-As*fy/(0.85*fc’*b*2))——–divide by b*d^2
16 in. × 32.19 in.
18 in. × 30.35 in.——– let try this one
20 in. × 28.79 in.
try 18inx33in (d=30.5in)
beam weight is bigger than assumed
now let assume beam weight bigger than 618.75lb/ft
Mu for simply supported beam=(Wu*L^2)/8=623.4k-ft
16 in. × 32.81 in.
18 in. × 30.93 in.———–let try this
20 in. × 29.35 in.
Try 18in x 34in(d=31in)
Beam weight=18*34*150/144=637.5lb/ft ok
try 5#9=5in2(the used area here is smaller but very close to calculated one, so will check whether it sufficient or now)
ρmin=3*√fc’/fy and not smaller than 200/fy
ρmin=0.0027 and not smaller than 0.0033
now check if Ф=0.90, to calculate Ф we need to calculate εt
β1=0.85 for fc’=4000 psi or less
εt=.009>0.005, section id tension control and assumption is ok