Question

1) Select the lightest section that can be used for the beam shown below if lateral bracing is provided only at the ends of the beam. Given loads are service loads (dead load given includes beam dead weight). Use Fy= 50ksi. The point load is applied at midspan.

4) Select the lightest section that can be used for the beam shown below if lateral bracing is provided only at the ends of the beam. Given loads are service loads (dead load given includes beam dead weight). Use Fy 50 ksi. The point load is applied at midspan Concentrated Dead Load 20 k Wi 2.0 klf wd 1.5 klf 15'

Answer 1

Given:

L=15 ft

P=20 kips

w_dl=1.5 klf

w_LL=2.0 klf

Step 1:

Calculate B.M for point load and UDL separately

P_u=1.2(D.L)=1.2(20)=24 kips

M=PL/4=24x15/4=90 ft-kips (For point load)

W_u=1.2(D.L)+1.6(L.L)

W_u=1.2(1.5)+1.6(2)

W_u=5 klf

M=5x15²/8=140.625 ft-kips

Total B.M=140.625+90=230.625 ft-kips

Step 2:

F_y=50 ksi

Calculate the Z_x value

Z_x=M/ $\phi$ F_y

Z_x=230.625x12/(0.9x50)

Z_x=61.5 in³

Step 3:

Select a W section having Zx>61.5 in³

Since we have W14X38 having Zx exactly equal to 61.5 in³, it is better to prefer next higher section

Select W16X36

From the Zx tables, Moment capacity=240 kips

Step 4:

Calculate shear

V=w_uL/2+W/2

V=5x15/2+20/2

V=47.5 kips

Shear capacity=340 kips from Zx tables

Step 5:

Check live load deflection:

$\delta$ =5wL⁴/384EI

$\delta$ =5x(2/12)x(15x12)⁴/384x29000x448

$\delta$ =0.175 in

Permissible deflection=L/240=5*12/240=0.75 in

So W16X36 can be used in bending, shear and deflection safely