Question

Use A992 and select a W shape for the following beam:

- Simply supported with a span length of 30 ft

- Braced at the ends of the beam only

- Uniform dead load of .75 kips

- A 34 kip point load at midspan

Use ASD

Answer 1

ASD solution:

Step 1:

Fy=50 ksi for A992 steel

Load=0.75 kips

L=30 ft

M=WL²/8

M=0.75x30x30/8

M_d=84.375 ft-kips, for dead load

M=34x30/4

M_L=255 ft-kips

M=M_d+M_L=339.375 ft-kips

Step 2:

Zx=M/0.6Fy

Zx=339.375x12/(0.6x50)

Zx=135.75 in³

Select a W shape having Zx>135 in³

Step 3:

Accounting for self weight, consider W21x62

Now check for self weight, D.L=0.062 kips/ft is self weight

D.L=0.062+0.75=0.812

L.L=34 kips

M_D=0.812x30x30/8=91.35 ft-kips

M_L=255 ft-kips

M=346.35 ft-kips

Zx=346.35x12/(0.6x50)

Zx=138.54 is the required section modulus, provided is 144 in³

Step 4:

ASD strength=359 ft-kips

B.M=339 ft-kips

It is sufficient for Bending moment

Step 5:

Shear=0.812x30/2+34/2

Shear=29.18 kips

Capacity=168 kips

This is safe in shear capacity also

Step 6:

Check for deflection:

Allowable deflection=L/240=30x12/240=1.5 in

Actual Live load deflection=WL³/48EI=34x(360)³/(48x29000x1330)=0.85 in

This beam is safe in bending, shear and deflection