Quiz 4 - Solution

1. Draw the bending moment diagram by the key-points approach.  Show the values at the key-points.  Indicate the scale on the M axis.

M_A = 0

M_B- = -5(4) = -20 k-ft

M_B+ = 20 + 30 = 10

M_C = -2(4)(4/2) = -16

Mid-point of BC:

M = (10 - 16)/2 + wL²/8 = =3 + 2(100)/8 = 22 k-ft

Point of zero shear in BC:

R_b = 2(10)/2 - (16 + 10)/10 = 7.4 k

x_b = R_b/w = 3.7 ft

M_max = 10 + R_bx_b/2 = 23.69 k-ft

2.

a) What is the allowable bending moment in k-ft for the cross section shown if the allowable bending stress is 1000 psi?

Neutral axis.

Using y-axis origin at bottom:

(24 + 24)y_c = 24(6) + 24(13)

y_c = 9.5 in

Moment of inertia.

I_f = 12(2)³/12 + 24(3.5)² = 302 in⁴

I_w = 2(12)³/12 + 24(3.5)² = 582 in⁴

I = I_f + I_w = 884 in4

Allowable bending moment:

M_a = s_aI/c = 1000(884)/9.5 = 93,053 lb-in  or 7.654 k-ft

b) What is the allowable shear force if the allowable shear stress is 80 psi?

The maximum shear stress occurs at the neutral axis.

t = VQ/It

Isolate the area below the neutral axis.

Q = 2(9.5)(9.5/2 = 90.25 in³

Allowable shear force

V_a = t_aIt/Q = 80(884)(2)/90.25 = 1567.2 lb