Question

2. Design a reinforced concrete beam of 20 feet span carrying a uniformly distributed load of 1200 lbs/ft. Use Fc = 3,000 psi and steel stress of 20,000 psi and assume appropriate modulus of elasticity and other required material properties.

Answer 1

Step 1:

Wu=1200 lb/ft

Mu=WuL2/8

Mu=1.2x20x20/8=60 ft-kips

Step 2:

Assume h=0.1L

h=20x12x0.1=24 in

Assume effective depth=22.5 in, b=12 in

Step 3:

Rn=Mu/(0.9bd2)

Rn=60x12x1000/(0.9x12x22.5x22.5)=131.68

=0.85fc/fy(1-(1-2Rn/0.85fc))0.5

=0.85x3/60(1-(1-2x131.68/0.85x3000))0.5=0.06375

Ast=0.06375x12x22.5=1.7 in2

Step 4:

Provide 2#9 bars

As=2 in2

a=AsFy/(0.85fcb)

a=2x20/(0.85x3x12)=1.3 in

c=a/0.85=1.53 in

t=d-c(0.003)/c=22.5-1.53x0.003/1.53=0.04

Mn=AsFy(d-a/2)

Mn=2x20x(22.5-1.3/2)

Mn=874 in-kips

Mn=874x0.9/12=65.5 ft-kips

So provide a beam of 12 inx24in (full depth) and 22.5 in effective depth with 2#9 bars provided