WORLD FOR CIVIL

Topic 6.2: Torsion: Deformation - Angle of Twist

Another effect of applying an external torque to a shaft is a resulting deformation or twist as the material is stressed. The resulting shaft deformation is expressed as an Angle of Twist of one end of the shaft with respect to the other.

In Diagram 1 we have shown a section of a solid shaft. An external torque T is applied to the left end of the shaft, and an equal internal torque T develops inside the shaft. Additionally there is a corresponding deformation (angle of twist) which results from the applied torque and the resisting internal torque causing the shaft to twist through an angle, phi, shown in Diagram 1.

The angle of twist may be calculated from:
= T L / J G ; where
T = the internal torque in the shaft
L = the length of shaft being "twisted"
J = the polar moment of inertia of the shaft
G = the Modulus of Rigidity (Shear Modulus) for the material, for example for steel and brass we have, G _steel = 12 x 10⁶ lb/in², G _brass = 6 x 10⁶ lb/in².

We will now look at an example of determining the angle of twist in a shaft. In Diagram 2a we have shown a solid steel circular shaft with an external torque of 1000 ft-lb. being applied at each end of the shaft, in opposite directions.

The shaft has a diameter of 1.5 inch. We would like to determine the angle of twist of end B with respect to end A.

To find the angle of twist we first determine the internal torque in the shaft. We cut the shaft a distance x feet from the left end, and make a free body diagram of the left section of the shaft - shown in Diagram 2b. From the free body diagram, we see that the internal torque must be 1000 ft-lb. to satisfy rotational equilibrium.

We next apply the Angle of Twist formula: = T L / J G ; where
T = 1000 ft-lb. = 12,000 in-lb.
L = 2 ft. = 24 inches
J = polar moment of inertia = (/32) d⁴ for a solid shaft = (3.1416/32) (1.5⁴in⁴) = .5 in⁴.
G _steel = 12 x 10⁶ lb/in²
Then,
= T L / J G = (12,000 in-lb.* 24 in) / (.5 in⁴ * 12 x 10⁶ lb/in²) = .048 radians = 2.75^o.

The angle of twist will have units of radians, and in this problem is clockwise with respect to end A as shown in Diagram 3.

We also take a moment to calculate the maximum shear stress in the shaft, just out of interest in it's value.
= T r / J = 12,000 in-lb. * .75 in./.5 in⁴. = 18,000 lb/in².