Question

A rectangular beam has f'c = 4000 pai and steel with fy = 60ksi. The moment due to load is 42 kip-feet in the moment due to live load is 35 kip-feet. If the width of the beam is required to be 12 inches, what is the beams depth, the total area of steel reinforcing required, and an acceptable number of steel bars in their spacing?

Answer 1

Step 1:

D.L moment=42 ft-kips

L.L moment=35 ft-kips

Mu=1.2(42)+1.6(35)=106.4 ft-kips

Step 2:

Assume =0.012

Mu/(bd2)=Fy(1-Fy/Fc)

Mu/(0.9x12xd2)=0.012x60000(1-0.012x60000/4000)

106.4x12x1000/0.9x12xd2=640.8

d=13.5 in, is minimum required, but d=1.5b is usually assumed as safe

Step 3:

d=18 in, is taken

Now take the area of steel

As=bd=0.012x12x18=2.5 in2

Now calculate the strength of the beam:

Provide 3#9 bars

a=AsFy/(0.85fcb)

a=3x60/(0.85x4x12)=4.41 in

c=a/0.85=5.19 in

Step 4:

t=d-c(0.003)/c=(18-5.19)x(0.003)/5.19=0.007>0.005

Mn=0.9AsFy(d-a/2)

=0.9x3x60x(18-4.41/2)

=2558 in-kips

=213 ft-kips>106 ft-kips

Provide a beam of 12inx18 in with 3#9 bars,

Step 5:

Spacing of bars:

As beam width is 12 in, 1.5 in taken as cover on both sides, so remaining=9 in

Number of bars=3

Provide spacing of bars=9/3=3 in