WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A compound shaft with applied torques and dimensions is shown below. Section AB is made of brass. 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/sq. in.
The modulus of rigidity for brass = 6 x 10⁶ lb/sq. in.

Part A:
Section I:
From equilibrium condtions:
Sum of Torque = 400 ft-lb - T_AB = 0; So T_AB = 400 ft-lb
Then t_AB = Tr / J = (400 ft-lb)(12 in/ft)(.375in) /[3.1416 * (.75 in)⁴ / 32] = 57,900 psi

Section II:
From equilibrium condtions:
Sum of Torque = 400 ft-lb - 1,400 ft-lb + T_BC = 0; So T_BC = 1,000 ft-lb
Then t_BC= Tr / J = (1,000 ft-lb)(12 in/ft)(.75 in) /[3.1416 * (1.5 in)⁴ / 32] = 18,100 psi

Section III:
From equilibrium condtions:
Sum of Torque = 400 ft-lb - 1,400 ft-lb + 800 ft-lb + T_CD = 0; So T_CD = 200 ft-lb
Then t_CD = Tr / J = (200 ft-lb) (12 in/ft)(.25 in) /[3.1416 * (.5 in)⁴ / 32] = 97,800 psi

Part B:
Resultant angle of twist:
f_AB = TL/JG = (400 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.1416 * (.75 in)⁴ / 32)(6x10⁶ lb/in²)] = .309 radians (cw)
f_BC = TL/JG = (1,000 ft-lb)(12 in./ft)(2 ft)(12 in/ft)/[(3.1416 * (1.5 in)⁴ / 32)(6x10⁶ lb/in²)] = .0966 radians (ccw)
f_CD = TL/JG = (200 ft-lb)(12 in./ft)(1.5 ft)(12 in/ft)/[(3.1416 * (.5 in)⁴ / 32)(12x10⁶ lb/in²)] = .587 radians (ccw)
f_Total = - f_AB + f_BC + f_CD = .375 radians (ccw)