### Example 3:design of rectangular beam

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

Wu=1.2*(0.4+2)+1.6*3=7.68k/ft

Mu for simply supported beam=(Wu*L^2)/8=600k-ft

Ф*Mn=Mu
Mu=Ф*As*fy*(d-a/2)
a=As*fy/(0.85*gc'*b)
Mu=Ф*As*fy*(d-As*fy/(0.85*fc'*b*2))--------divide by b*d^2

ρ=As/(b*d)=0.18fc'/fy=0.009

Mu/(b*d^2)=Ф*ρ*fy*(1-ρ*fy/(1.7*fc'))

assuming Ф=0.90

b*d^2=Mu/(Ф*ρ*fy*(1-ρ*fy/(1.7*fc')))
b*d^2=7,200,000/434.54 =16,569.24in3

bxd

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=18*33*150/144=618.75lb/ft

beam weight is bigger than assumed

now let assume beam weight bigger than 618.75lb/ft
Beam weight=650lb/ft

Wu=1.2*(0.65+2)+1.6*3=7.98k/ft

Mu for simply supported beam=(Wu*L^2)/8=623.4k-ft

b*d^2=Mu/(Ф*ρ*fy*(1-ρ*fy/(1.7*fc')))

b*d^2=7,481,250/434.54 =17,216.48in3

bxd

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

As=ρ*b*d=0.009*18*31=5.022in2.
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
ρ=As/(b*d)=5/(18*31)=0.009>ρmin                ok

now check if Ф=0.90, to calculate Ф we need to calculate εt

t+0.003)/d=0.003/C

C=a/β1

a=As*fy/(0.85fc'*b)=6.53in

β1=0.85 for fc'=4000 psi or less

C=6.53/0.853=7.68in

t+0.003)/31=0.003/7.68

εt=.009>0.005, section id tension control and assumption is ok

Ф*Mn=Ф*As*fy*(d-a/2)

Ф*Mn=0.9*5*60,000*(31-6.53/2)=7,488,450lb-in=624.1k-ft
Mu=(w*L^2)/8=(7.96*25^2)/8=622.21k-ft

Ф*Mn>Mu      ok