WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A compound shaft with applied torques and dimensions is shown below. Section AB is made of steel. Section BC is made of brass. Section CD is made of steel. For this shaft:
A. Determine the maximum shear stress in each of the sections of the shaft.
B. Determine the resultant angle of twist of a point on end D with respect to a point on end A.
The modulus of rigidity for steel = 12 x 106 lb/in².
The modulus of rigidity for brass = 6 x 106 lb/in².

Part A:
Section I:
From equilibrium condtions:
Sum of Torque = 600 ft-lb - T_AB = 0; So T_AB = 600 ft-lb
Then t_AB = Tr / J = (600 ft-lb)(12 in/ft)(.5 in) /[3.1416 * (1 in)⁴ / 32] = 36,700 psi

Section II:
From equilibrium condtions:
Sum of Torque = 600 ft-lb - 2000 ft-lb + T_BC = 0; So T_BC = 1400 ft-lb
Then t_BC = Tr / J = (1400 ft-lb)(12 in/ft)(1 in) /[3.1416 * (2 in)⁴ / 32] = 10,700 psi

Section III:

From equilibrium condtions:
Sum of Torque = 600 ft-lb - 2000 ft-lb + 1000 ft-lb + T_CD = 0; So T_CD = 400 ft-lb
Then t_CD = Tr / J = (400 ft-lb) (12 in/ft)(.375 in) /[3.1416 * (.75 in)⁴ / 32] = 57,900 psi

Part B:
Resultant angle of twist:
f_AB = TL/JG = (600 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.1416 * (1 in)⁴ / 32)(12x10⁶ lb/in²)] = .0733 radians (cw)
f_BC = TL/JG = (1,400 ft-lb)(12 in./ft)(2 ft)(12 in/ft)/[(3.1416 * (2 in)⁴ / 32)(6x10⁶ lb/in²)] = .0428 radians (ccw)
f_CD = TL/JG = (400 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.1416 * (.75 in)⁴ / 32)(12x10⁶ lb/in²)] = .155 radians (ccw)
f_Total = - f_AB + f_BC + f_CD = .1245 radians (ccw)