Quiz 3 - Solution

The figure shows an assemblage of a steel bar and an aluminum tube.

The shear moduli of steel and aluminum are, respectively,

G_s = 11,000 ksi

G_a = 4,000 ksi

a) Determine the torque T required to cause a twist of  j = 0.5 deg.  to the assemblage.

T = T_s + T_a = (G_sJ_s/L)j + (G_aJ_a/L)j

j = 0.5(p/180)  =  8.7266(10-³) rad

Steel:

J_s = pc⁴/2 = p(0.5)⁴/2 = 0.098175 in⁴

k_s = G_sJ_s/L  = 11,000(0.098175)/12 = 89.994 k-in/rad

T_s = k_sj = 89.9935*8.7266(10-³) = 0.785 k-in

Aluminum:

J_a = p(c_o⁴ - c_i4)/2 = p[(1.5)⁴ - (1)⁴]/2 = 6.3814 in⁴

k_a = G_aJ_a/L  = 4,000(6.38136)/12 = 2127.1  k-in/rad

T_a = k_aj = 2127.1*8.7266(10-³) = 18.563 k-in

T = T_s + T_a = 19.348 k-in

b) Assuming allowable stresses

(t_s)_a = 10 ksi  for steel

(t_a)_a = 3 ksi   for aluminum

find out whether T exceeds the allowable torque.

t_s = T_sc/J_s = 0.785(0.5)/0.098175 = 4. < 10 ksi,  O.K.

t_a = T_ac_o/J_a = 18.563(1.5)/6.38136  = 4.363 > 4 ksi

T exceeds allowable torque.