WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

The diagram below represents two hollow brass shafts attached to a solid wall at end A. The torque acting at point B is 600 ft-lbs, and the torque acting at end C is 200 ft-lbs. For this shaft:
A. Determine the maximum shear stress in each shaft.
B. Determine the angle of twist of end C with respect to 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.

The diameters of each shaft are as follows:
outer diameter of AB = 3.5 in, inner diameter of AB = 2.8 in
outer diameter of BC = 2 in , inner diameter of BC = 1.6 in

Part A:
Section I:
From equilibrium condtions:
Sum of Torque = T_BC - 200 ft-lb = 0; So T_BC = 200 ft-lb
Then t_AB = Tr / J = (200 ft-lb)(12 in/ft)(1 in) /[3.1416 * [(2 in)⁴ - (1.6 in)⁴] / 32] = 2,600 psi

Section II:
From equilibrium condtions:
Sum of Torque = T_AB - 600 ft-lb - 200 ft-lb = 0; So T_AB = 800 ft-lb
Then t_AB = Tr / J = (800 ft-lb)(12 in/ft)(1.75 in) /[3.1416 * [(3.5 in)⁴ - (2.8 in)⁴] / 32] = 1,150 psi

Part B:
Resultant angle of twist:
f_AB = TL/JG = (800 ft-lb)(12 in./ft)(2 ft)(12 in/ft)/[(3.1416 * [(3.5 in)⁴ - (2.8 in)⁴] / 32)(6x10⁶ lb/in²)] = .0044 radians (cw)
f_BC = TL/JG = (200 ft-lb)(12 in./ft)(3 ft)(12 in/ft)/[(3.1416 * [(2 in)⁴ - (1.6 in)⁴] / 32)(6x10⁶ lb/in²)] = .0155 radians (cw)
f_Total = - f_AB - f_BC = -.0199 radians (cw)