**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 **; whereT = 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 10

^{6}lb/in

^{2}, G

_{brass}= 6 x 10

^{6}lb/in

^{2}.

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) d^{4} for a solid shaft = (3.1416/32) (1.5^{4}in^{4}) = .5 in^{4}.

G _{steel} = 12 x 10^{6} lb/in^{2}

Then,**= T L / J G = (12,000 in-lb.* 24 in) / (.5 in ^{4} * 12 x 10^{6} lb/in^{2}) = .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 in ^{4}. = 18,000 lb/in^{2}.**

**Part I. **In Diagram 1a we have shown a solid shaft with what we will call a driving external torque of 1000 ft-lb. at end A, and a load torque of 1000 ft-lb. at end B. The shaft is in equilibrium. We would like to determine the maximum transverse shear stress in the shaft due to the applied torque.

To solve, we first need to determine the internal torque in the shaft. We cut the shaft a distance x from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 1b. Where we cut the shaft there is an internal torque, which in this case must be equal and opposite to the torque at end A for equilibrium. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.

We next simply apply the torsion formula for the shear stress:** ****= T r / J**; where:**T** is the internal torque in that section of the shaft **= 1000 ft-lb = 12,000 in-lb.****r** = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum;** r = 1 in.****J** = polar moment of inertia = **(****/32) d ^{4} for a solid shaft = (3.1416/32) (2^{4}in^{4}) = 1.57 in^{4}**.

So

**,**

**= T r / J = 12,000 in-lb. * 1 in./1.57 in**

This is the Maximum Transverse (and longitudinal) Shear Stress in the shaft.

^{4}. = 7,640 lb/in^{2}.**Part II**We now would like to consider the case where the shaft is not solid, but a hollow shaft with an outer diameter of 2″ and an inner diameter of 1″, as shown in Diagram 2a. We still apply the same driving and load torque, and still have the same value of the internal torque, as is shown in Diagram 2b.

We next apply the torsion formula for the shear stress for the hollow shaft:**= T r / J**; where we observe that all the values are the same as in part one, except for the value of J, the polar moment of inertia.**T** is the internal torque in that section of the shaft** = 1000 ft-lb = 12,000 in-lb.r** = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum;

**r = 1 in.**

J= polar moment of inertia = (3.1416/32) [d

J

_{o}

^{4}– d

_{i}

^{4}] for a hollow shaft = (3.1416/32) [(2

^{4}in

^{4}) – (1.0

^{4}in

^{4})] =

**1.47 in**

So

^{4}.**,**

**= T r / J = 12,000 in-lb. * 1 in./1.47 in**

^{4}. = 8,150 lb/in^{2}.This then is the Maximum Transverse (and longitudinal) Shear Stress in the hollow shaft.

**Shear Stress – Example 2**

In Diagram 1 we have shown a solid compound shaft with what we will call the driving external torque of 1600 ft-lb. acting at point B, and load torque of 400 ft-lb. at end A, 900 ft-lb. at point C, and 300 ft-lb. at end D. Notice that the shaft is in rotational equilibrium. We would like to determine the maximum transverse shear stress in each section of the shaft due to the applied torque.

To solve, we first need to determine the internal torque in each section of the shaft. We cut the shaft a distance 0′ < x < 1′ from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 2. Where we cut the shaft there is an internal torque, which in this case must be equal and opposite to the torque at end A for equilibrium. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.

We next apply the torsion formula for the shear stress:** ****= T r / J**; where:**T** is the internal torque in that section of the shaft** = 400 ft-lb. = 4,800 in-lb**.**r** = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem r is to the outer edge of the shaft since that is where the transverse shear stress is a maximum; **r = .5 in**.**J** = polar moment of inertia** = (**** /32) d^{4} for a solid shaft = (3.1416/32) (1^{4}in^{4}) = .098 in^{4}**.

So

**,**

**= T r / J = 4,800 in-lb. * .5 in./.098 in**

^{4}. = 24,500 lb./in^{2}.This value is the Maximum Transverse Shear Stress in the shaft section AB, and it falls in a reasonable range for allowable shear stresses for metals.

We now determine the internal torque in the next section of the shaft. We cut the shaft a distance 1′ < x < 3′ from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 3. Where we cut the shaft there is an internal torque, and by mentally summing torque, we see that in order to have rotational equilibrium we must have an internal torque in section BC of 1200 ft-lb. acting in the direction shown.

We apply the torsion formula for the shear stress once again:** ****= T r / J**; where:**T** is the internal torque in that section of the shaft **= 1,200 ft-lb. = 14,400 in-lb**.**r** = the radial distance to the outer edge of the shaft since that is where the transverse shear stress is a maximum; **r = 1 in.****J** = polar moment of inertia** = (**** /32) d^{4} for a solid shaft = (3.1416/32) (2^{4}in^{4}) = 1.57 in^{4}.**So

**,**

**= T r / J = 14,400 in-lb. * 1 in./ 1.57 in**

^{4}. = 9,170 lb./in^{2}.This then is the Maximum Shear Stress in shaft section BC. We note that even though the internal torque is much larger in section BC as compared to section AB, because of the size of the shaft in section BC, the shear stress is much lower in BC.

We now determine the internal torque in the next section of the shaft. We cut the shaft a distance 3′ < x < 4′ from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 4. Where we cut the shaft there is an internal torque. From the Free Body Diagram we see that in order to have rotational equilibrium we must have an internal torque in section CD of 300 ft-lb. acting in the direction shown.

We apply the torsion formula for the shear stress:** ****= T r / J**; where:**T** is the internal torque in that section of the shaft** = 300 ft-lb. = 3,600 in-lb**.**r** = the radial distance from the center of the shaft to the outer edge of the shaft since that is where the transverse shear stress is a maximum; **r = .25 in**.**J** = polar moment of inertia = (/32) d^{4} for a solid shaft **= (3.1416/32) (.5 ^{4}in^{4}) = .0061 in^{4}**.

So

**,**

**= T r / J = 3,600 in-lb. * .25 in./ .0061 in**

^{4}. = 147,500 lb./in^{2}.This is the Maximum Shear Stress in shaft section CD. We note that this is much larger than the ultimate shear stress most metals, thus this section of the shaft would fail – a larger diameter is needed to carry the torque.