WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A compound shaft is attached to a fixed wall as shown below. Section AB is made of brass. Sections BC and CD are made of steel. If the allowable shear stress for steel and brass are:
t _steel = 18,000 psi t _brass = 12,000 psi
A. Determine the maximum torques which could be applied at points D, C, & B (T_D,T_C, & T_B) without exceeding the allowable shear stress in any section of the shaft.
B. Using the torques found in part A, 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 = T_CD - T_D = 0; So T_D = T_CD
Then: t_steel = T_CD r / J
18,000 lb/in² = T_CD(.375in) /[3.1416 * (.75 in)⁴ / 32]
Solving: T_CD = T_D = 1491 in-lb = 124 ft-lb

Section II:
From equilibrium condtions:
Sum of Torque = T_BC - T_CD - 124 ft-lb = 0; So T_C = (T_BC - 124 ft-lb)
Then t_steel = T_BCr / J
18,000 lb/in² = T_BC(.5in) /[3.1416 * (1 in)⁴ / 32]
Solving: T_BC = 3534 in-lb = 295 ft-lb
T_C = (295 ft-lb - 124 ft-lb) = 171 ft-lb

Section III:
From equilibrium condtions:
Sum of Torque = -T_AB + T_B - 171 ft-lb - 124 ft-lb = 0; So T_B = (T_AB + 171 ft-lb + 124 ft-lb)
Then t_brass = T_ABr / J
12,000 lb/in² = T_AB(1 in) /[3.1416 * (2 in)⁴ / 32]
Solving: T_AB = 18,850 in-lb = 1570 ft-lb.
T_B = (1570 ft-lb + 171 ft-lb + 124 ft-lb) = 1865 ft-lb

Part B:
Resultant angle of twist:
f_AB = TL/JG = (1,570 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.1416 * (2 in)⁴ / 32)(6x10⁶ lb/in²)] = .024 radians (ccw)
f_BC = TL/JG = (245 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.1416 * (1 in)⁴ / 32)(12x10⁶ lb/in²)] = .0361 radians (cw)
f_CD = TL/JG = (124 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.1416 * (.75 in)⁴ / 32)(12x10⁶ lb/in²)] = .0479 radians (cw)
f_Total = + f_AB - f_BC -f_CD = .06 radians (cw)