CEE 220 - Test 2

1.  (15%)

The tube shown has a shear modulus G = 10,000 ksi. What is the allowable torque if the allowable shear stress is t_a = 10 ksi, and the allowable angle of twist is 0.025 radians?

J = p(r_o⁴ - r_i⁴)/2 = p(2.5⁴ - 2⁴)/2 = 36.226 in⁴

t = Tr/J

T₁ = t_aJ/r_o = 10(36.226)/2.5 = 144.9 k-in

j = TL/GJ

T₂ = GJj_a/L = 10000(36.226)(0.025)/(6)(12) = 125.79 k-in

Allowable T = 125.79 k-in

2.  (35%)

For the beam and loading shown, determine the required depth h of a rectangular cross section of width b = 1.5 inch, assuming an allowable bending stress s_a = 1200 psi.

R_A = (2/3)P + (1/3)(12w) = 400 lb

R_B = (1/3)P + (2/3)(12w) = 500 lb

Point of zero shear at x_b from B,

x_b = R_B/w = 10 ft

M_max = R_Bx_b/2 = 2500 lb-ft

s_a = M_maxh/2I   or s_a = M_max/S

I = M_maxh/2s_a = 1.5h³/12  or S = M_max/s_a = 1.5h²/6

h² = 6M_max/1.5s_a  = 6(2500)(12)/1.5(1200) = 100 in²

h = 10 in.

3.  (25%)

A shear force V =  1000 lb acts on the cross-section.  What is the location and magnitude of the maximum shear stress ?

Neutral Axis

Section is made of 3 parts having equal areas of 8 in².  Choosing y origin for centroid at bottom:

(8 + 8 + 8)y_c = 8(4) + 8(4) + 8(8.5)

y_c = 5.5 in

I = [1(8)³/12 + 8(1.5)²](2) + 8(1)³/12 + 8(3)²  = 194 in⁴

t = VQ/It

Q_max = moment of area below cut at neutral axis.

Q_max = 1(5.5)(5.5/2) = 15.125 in³

t_max = VQ_max/It = 1000(15.125)/194(1) = 77.96 psi  at neutral axis.

4.  25%

Obtain the rotation at A by integration of the second-order differential equation of the elastic line.  Partial credit is given for:

a) the expressions of the bending moment

b) integration of the differential equation

c) equations for determining the integration constants

a) R_A = (1/3)(12w) = 8 k

0 < x < 6:  M = 8x

6 < x < 18: M = 8x - 2(x-6)²/2 = 8x - (x-6)²

b)  EIv'' = M

0 < x < 6

EIv' = 4x² + A

EIv = 4x³/3 + Ax + B

6< x < 18

EIv' = 4x² - (x-6)³/3 + C

EIv = 4x³/3 - (x-6)⁴/12 + Cx + D

c) at x = 0, EIv =  B = 0

at x = 6, EIv' = 4(6)² + A = 4(6)² +0  + C,  A = C

              EIv = 4(6)³/3 + A(6) =  4(6)³/3 + 0  + C(6) + D,   D = 0

at x = 18, EIv = 4(18)³/3 - (12)⁴/12 + C(18) + 0 = 0, C = -336 k-ft²

Rotation at A: q_A = (v')_x=0 = A/EI = C/EI 

q_A = -336 k-ft²/EI