WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A compound shaft with applied torques is shown below. Section AB is made of steel. Section BC is made of brass. Section CD is made of steel. A driving torque of 2000 ft-lb is applied as shown. Load torques of 600 ft-lbs, 1000 ft-lb, and 400 ft-lbs are also shown. The shaft is rotating at 2400 rev/minute. The allowable shear stress of steel is 20,000 lb/sq.in., and the allowable shear stress for brass in 18,000 lb/sq.in. For this shaft:
A. Determine the horsepower being transfer through each shaft.
B. Determine the minimum diameter for each shaft, such that it can safely carry the transmitted power.

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 = 600 ft-lb - T_AB = 0; So T_AB = 600 ft-lb
Then: HP = 2phT / 550 ft-lb/s / hp = 2p (40 Rev/s )( 600 ft-lb) / 550 ft-lb/s / hp = 274 hp

Section II:
From equilibrium condtions:
Sum of Torque = 600 ft-lb - 2000 ft-lb + T_BC = 0; So T_BC = 1400 ft-lb
Then: HP = 2phT / 550 ft-lb/s / hp = 2p (40 Rev/s)( 1,400 ft-lb) / 550 ft-lb/s / hp = 640 hp

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: HP = 2phT / 550 ft-lb/s / hp = 2p (40 Rev/s)( 400 ft-lb) / 550 ft-lb/s / hp = 183 hp

Part B:
Section I:
t_AB = Tr / J => 20,000 = (600 ft-lb)(12 in/ft)( d/2 ) /[3.1416 * (d)⁴ / 32]
Solve for d³_AB = (600 ft-lb)(12 in/ft)( 1/2 ) /[(3.1416 / 32) (20,000 1b/in²)] = 1.8335 in³
so: d_AB = 1.22 inches

Section II:
t_BC = Tr / J => 18,000 = (1,400 ft-lb)(12 in/ft)( d/2 ) /[3.1416 * (d)⁴ / 32]
Solve for d³_BC = (1,400 ft-lb)(12 in/ft)( 1/2 ) /[(3.1416 / 32) (18,000 1b/in²)] = 4.7534 in³
so: d_BC = 1.68 inches

Section III:
t_CD = Tr / J => 20,000 = (400 ft-lb)(12 in/ft)( d/2 ) /[3.1416 * (d)⁴ / 32]
Solve for d³_CD = (400 ft-lb)(12 in/ft)( 1/2 ) /[(3.1416 / 32) (20,000 1b/in²)] = 1.2223 in³
so: d_CD = 1.07 inches