Question

Select the lightest W-shape beam to support a uniformly distributed lead of 1 kip/ft on a simple span of 30 ft. Lateral support exists at the ends and mid-span. Assume A-36 steel. Consider moment and shear.

Answer 1

Step 1:

Fy=36 ksi

L=30 ft

D.L=1 kip/ft

Step 2:

Using LRFD method, the ultimate bending moment will be

Pu=1.4kip/ft

M=1.4x30x30/8

M=157.5 ft-kips

Step 3:

Zx=Mu/(0.9Fy)

Zx=157.5x12/(0.9x36)

Zx=58.33 in³

Select A W section from Zx tables and also check for self weight

Step 4:

Consider W14X38, also cheking for self weight with Zx=61.55 in³

W_d=1.4(1+0.038)=1.45 kip/ft

M=1.4504x30x30/8=163.485 ft-kips

Zx=M/(0.9Fy)=163.485x12/(0.9x36)=60.55 in³

Step 5:

Mp=ZxFy

Mp=61.55x36/12

Mp=2216 ft-kips

As there is a bracing in between Cb should be considered and also checked for moment capacity

Lb=15 ft

Lr=20 ft

Lp=6.45 ft

M_n=Cb[Mp-(Mp-0.7FySx)(Lb-Lp)/(Lr-Lp)]<Mp

Mn=1.30[2216-(2216-0.7x36x54.6)(15-6.45)/(20-6.45)]

Mn=184.55 ft-kips

Capacity=0.9x184.55=166.095 ft-kips, this capacity is greate than bending moment

Step 6:

S.F=1.4x30/12=3.5 kips

Check for shear:

V=0.6Fydt_w

V=0.6x36x14.1x0.31

V=94.41 kips

The beam W14X38 is enough to resist the bending moment and shear for this particular section