WORLD FOR CIVIL

opic 6.9a Torsion, Rivets & Welds - Topic Examination

Torsion Exam Solutions

Modulus of Rigidity for several materials:
Steel = 12 x 10⁶ lb/in².; Brass = 6 x 10⁶ lb/in².; Aluminum = 4 x 10⁶ lb/in².

1. A compound shaft is attached to a fixed wall as shown in Diagram 1. Section AB brass, section BC is steel, and section CD is brass. If the allowable shear stress for steel is 18,000 lb/in², and for brass is 16,000 lb/in².

A. Determine the maximum torque 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 torque found in part A, determine the resultant angle of twist of end D with respect to end A.

Solution: We start by determining the maximum allowable internal torque in each section. We begin from the right end and work to the left, cutting shaft CD and drawing the section shown in Diagram 2.

We next applied the torsion formula and solve for the internal torque.
= T r / J ; then solving for the torque: T = J / r ; where
J = (3.1416)(.75)⁴ / 32 = .031 in⁴; r = .375 inch; and = 16,000 lb/in², (We use the allowable shear stress as the maximum stress in the shaft.) Putting values into the equation and solving:
T = 16,000 lb/in² * .031 in⁴ / .375 in = 1325 in-lb = 110 ft-lb.
This is the internal torque in section CD, and is equal to the externally applied T_D, as is clear from rotational equilibrium conditions and Diagram 2.

We next cut the shaft in section BC, and look at section to the right, as shown in Diagram 3.

We next applied the torsion formula and solve for the internal torque in section BC.
= T r / J ; then solving for the torque: T = J / r ; where
J = (3.1416)(1")⁴ / 32 = .098 in⁴; r = .5 inch; and = 18,000 lb/in², (We use the allowable shear stress as the maximum stress in the shaft.) Putting values into the equation and solving:
T = 18,000 lb/in² * .098 in⁴ / .5 in = 3534 in-lb = 295 ft-lb.
This is the internal torque in section BC. From Diagram 3 we see that the internal toque is equal to the sum of the external torque T_D(110 ft-lb) and T_C. We can solve for the value of T_C, finding T_C = 185 ft-lb

We next applied the torsion formula and solve for the internal torque in section AB.
= T r / J ; then solving for the torque: T = J / r ; where
J = (3.1416)(2")⁴ / 32 = 1.57 in⁴; r = 1 inch; and = 16,000 lb/in², (We use the allowable shear stress as the maximum stress in the shaft.) Putting values into the equation and solving:
T = 16,000 lb/in² * 1.57 in⁴ / 1 in = 25,132 in-lb = 2094 ft-lb.
This is the internal torque in section AB. From Diagram 4 we see that T_B is equal to the sum of the internal toque (2094 ft-lb) + T_D(110 ft-lb) + T_C(185 ft-lb). We can solve for the value of T_B, finding T_B = 2389 ft-lb

Part B: The angle of twist of end D with respect to end A can be found by determining the angle of twist in each section and then finding the net angle of twist.
Apply the Angle of Twist relationship to each section of the shaft.
_AB =T L/J G= (25,132 in-lb.* 12 in)/(1.57 in⁴ * 6 x 10⁶ lb/in²)=.032 radians =1.83^o(ccw).
_BC =T L/J G= (3,534 in-lb.* 12 in)/(.098 in⁴ * 12 x 10⁶ lb/in²)=.036 radians = 2.06^o(cw).
_CD =T L/J G= (1,325 in-lb.* 12 in)/(.031 in⁴ * 6 x 10⁶ lb/in²)=.085 radians = 4.87^o(cw).
Then the total twist of end D with respect to end A will be:
_total = + 1.83 ^o - 2.06^o - 4.87^o = - 5.1^o (clockwise); where our signs are taken from the direction of the internal torque in each section. In this example, AB - counter clockwise, BC - clockwise, and CD - clockwise, resulting in the (+, -, -) signs of the angles of twist.

2. A 2 foot long hollow steel shaft with an outer diameter of 2 inches and an inner diameter of 1.5 inches is to transmit power while being driven a 3000 rpm.

A.) If the allowable shear stress in the shaft is 15,000 lb/in²., what is the maximum horsepower which can be transmitted down the shaft ?
B.) If we were not given the outer diameter of the shaft, but were told that the inner diameter was to be four-tenths of the outer diameter, what would be the minimum outer diameter of the shaft which could safely transmit the horse power found in part A? (The allowable shear stress in the shaft is 15,000 lb/in² )

Solution:
Part A We use the shear stress relationship to first find the maximum torque which may be applied to the shaft: = T r / J ; then solving for the torque: T = J / r ; where
J = (3.1416)(2"⁴ - 1.5"⁴)/ 32 = 1.07 in⁴; r = 1 inch; and = 15,000 lb/in², (We use the allowable shear stress as the maximum stress in the shaft.) Putting values into the equation and solving:
T = 15,000 lb/in² * 1.07 in⁴ / 1 in = 16,107 in-lb = 1342 ft-lb.
Then we apply the power equation to determine the maximum power which may be transmitted.
P_AB=2 pi Tn/(550 ft-lb/s/hp)=2*3.1416*1342 ft-lb.*(50 rev/s) /(550 ft-lb/s/hp)= 767 hp

Part B: Using the torque found in part A, and the maximum allowable shear stress, we can now solve of the outer diameter. = T r / J ; where
J = (3.1416)[D_o⁴ - (.4D_o)⁴]/ 32; r = D_o /2, = 15,000 lb/in²,and T = 16107 in-lb.
Inserting values and solve for D_o we find:
15,000 lb/in² = (16107 in-lb * D_o /2)/(3.1416/32)[D_o⁴ - (.4D_o)⁴]
so, D_o= 1.78 inches