FLEXURE AND SHEAR

Problem A box beam carries a distributed load of 200 lb/ft and a concentrated load P as shown in Fig. P-

590. Determine the maximum value of P if fb 1200 psi and fv 150 psi.

Solution 590

MR2=0

10R1=5P+3(2800)

R1=0.5P+840

MR1=0

10R2=5P+7(2800)

R2=0.5P+1960

MB=12[(0.5P+840)+(0.5P160)](5)

MB=2.5P+1700 lbft

MC=MB12[(0.5P+160)+(0.5P+1160)](5)

MC=MB(2.5P+1320)

MC=(2.5P+1700)(2.5P+3300)

MC=1600 lbft

Check MC from the overhang segment

MC=12(4)(800)

MC=1600 lbft (okay!)

Based on allowable bending stress

fb=McI

Where

M = 2.5P + 1700 lbft

c = 12/2 = 6 in

I = 10(123)/12 - 8(10

3)/12 = 773.33 in

4

Thus,

1200=(2.5P+1700)(12)(6)773.33

P=4475.56 lb

Based on allowable shear stress

fv=VQIb

Where

V = 0.5P + 1160 lb

Q = 10(1)(5.5) + 2 [ 5(1)(2.5) ] = 80 in3

b = 2 in

Thus,

150=(0.5P+1160)(80)773.33(2)

P=3480 lb

For safe value of P, use P = 3480 lb answer