Torsion

When an object, like a cylindrical rod or shaft, experiences torsion, shear stress is produced inside the material. It is the internal force that resists the twisting motion. For an object under torsion, the shear stress at a point within the material depends on how far that point is from the center of the object, which is the axis of rotation. Its formula is:

\[ \tau = \frac{T r}{J} \]

Where:

– τ: Shear stress

– T: Applied torque

– r: Radial distance from the center

– J: Polar moment of inertia 

The polar moment of inertia depends on the shape and size of the object and indicates resistance to twisting. The shear stress is greatest at the outer surface of the object because 

𝑟 is maximum at the outer surface. At the center, where r = 0, there is no shear stress.

As the object twists, it not only resists the force but deforms as well. The amount of deformation is called shear strain, which measures how much the material is stretched or compressed at different points. The formula for shear strain is:

\[ \gamma = \frac{r \theta}{L} \]

Where:

– γ: Shear strain

– r: Radial distance

– θ: Angle of twist

– L: Length of the object

The farther a point is from the center, the greater the strain. This means the outer regions of the object experience more twisting than those near the center.

To derive the expression for the angle of twist (θ), we use Hooke’s Law, which states that stress is proportional to strain.

\[ \text{Stress } \propto \text{Strain} \\

=> \tau = G \cdot \gamma \]

Where G is the shear modulus of the material

From the equations above:

\[ \frac{T r}{J} = G \cdot \frac{r \theta}{L} \]

Canceling r from both sides:

\[ \frac{T}{J} = G \frac{\theta}{L} \]

Solving for the angle of twist:

\[ \theta = \frac{T L}{G J} \]

This equation shows that the angle of twist (θ) is:

• Directly proportional to the applied torque (T) and the object’s length (L). It means that increasing the applied torque or the length of the shaft results in greater twisting.
• Inversely proportional to the shear modulus (G) and the polar moment of inertia (J), meaning that stiffer materials and larger cross-sections resist twisting more.

• Automotive & Machinery: Drive shafts in cars and industrial machines transmit power through rotational motion, experiencing torsion.
• Hand Tools: Screwdrivers and wrenches rely on torsion to turn screws and bolts efficiently.
• Structural Engineering: Bridges and tall buildings are designed to withstand torsional forces from wind and earthquakes.
• Springs & Suspension Systems: Torsion springs in clocks and vehicle suspension systems store and release energy through twisting.
• Biological Systems: DNA molecules experience torsion when they unwind and replicate during cell processes.
• Material Testing & Physics Experiments: Torsion pendulums help measure material properties like rigidity and shear modulus.
• Aerospace Engineering: Helicopter rotor blades and propeller shafts experience torsion during operation.
• Seismic Studies: Engineers study torsional waves to understand how buildings respond to twisting forces during earthquakes.

Problem 1: A solid steel shaft with a radius of 20 mm is subjected to a torque of 500 N·m. Calculate the shear stress at the outer surface of the shaft. (Take the polar moment of inertia for a solid circular shaft as \( J = \frac{\pi r^4}{2} \)​).

Solution

Given:

– Torque: T = 500 N·m
– Radius: r = 20 mm = 0.02 m

The polar moment of inertia is: 

\[ J = \frac{\pi r^4}{2} = \frac{\pi (0.02)^{4}}{2} = 2.51 \times 10^{-7} \text{ m}^4 \]

The shear stress is:

\[ \tau = \frac{T r}{J} = \frac{(500)(0.02)}{2.51 \times 10^{-7}} = 3.98 \times 10^7 \text{ Pa} = 39.8 \text{ MPa}

The shear stress at the outer surface is 39.8 MPa.

Problem 2: A solid brass rod with a radius of 15 mm can withstand a maximum shear stress of 50 MPa before failing. What is the maximum torque it can handle before failing? (Use \( J = \frac{\pi r^4}{2} \)​).

Solution

Given:

– Radius: r =15 mm = 0.015 m

– Maximum shear stress: τ_max = 50 x 10⁶ Pa

The formula for torque is:

\[ T = \frac{\tau_{\max} J}{r} \]

First, calculate J:

\[ J = \frac{\pi r^4}{2} = \frac{\pi (0.015)^4}{2} = 7.95 \times 10^{-8} \text{ m}^4 \]

Now, find T:

\[ T = \frac{(50 \times 10^6)(7.95 \times 10^{-8})}{0.015} = 265 \text{ N}\cdot \text{m} \]

The maximum torque before failure is 265 N·m.

Problem 3: A hollow aluminum shaft of 1.5 m in length has an inner radius of 10 mm and an outer radius of 20 mm. It is subjected to a torque of 600 N·m. Find the angle of twist if the shear modulus of aluminum is 25 x 10⁹ Pa. (For a hollow shaft, \( J = \frac{\pi (r_o^4 – r_i^4)}{2} \)).

Solution

Given:

– T = 600 N·m

– L = 1.5 m

– r_o ​= 20 mm = 0.02 m

– r_i ​= 10 mm = 0.01 m

– G = 25 x 10⁹  Pa

First, calculate J:

\[ J = \frac{\pi (r_o^4 – r_i^4)}{2} = \frac{\pi ((0.02)^4 – (0.01)^4)}{2} = 2.36 \times 10^{-7} \text{ m}^4 \]

The formula for the angle of twist (θ) is:

\[ \theta = \frac{600 \times 1.5 }{25 \times 10^9 \times 2.36 \times 10^{-7}} = 0.153 \text{ radians} = 0.153 \times \frac{180}{\pi} = 8.74 \circ \]

Thus, the angle of twist is 8.74°.