Torsion of Shafts
Shear Stress, Angle of Twist, Power Transmission & Solid vs Hollow Shafts
Last Updated: March 2026
📌 Key Takeaways
- Torsion equation: T/J = τ/r = Gθ/L — connects torque, shear stress, and angle of twist.
- Maximum shear stress: τmax = TR/J (at the outer surface).
- Angle of twist: θ = TL/(GJ) radians.
- Power-torque: P = 2πNT/60 (N in rpm, P in watts, T in N·m).
- Polar moment of inertia: Solid shaft J = πD⁴/32. Hollow shaft J = π(Do⁴ − Di⁴)/32.
- Hollow shafts are lighter and more efficient in torsion than solid shafts of the same outer diameter.
1. What is Torsion?
Torsion is the twisting of a structural member when it is subjected to a torque (twisting moment) about its longitudinal axis. When you twist a screwdriver, wring a towel, or turn a shaft in a machine — you are applying torsion.
In engineering, torsion is most important in rotating shafts that transmit power from engines and motors to machines. Every drive shaft, propeller shaft, axle, and spindle in mechanical systems experiences torsion. The analysis determines the shear stress developed in the shaft and the angle through which it twists — both critical for safe design.
2. Assumptions
The standard torsion theory applies to circular cross-sections (solid or hollow) and assumes:
- The shaft material is homogeneous, isotropic, and linear elastic.
- Plane cross-sections remain plane after twisting (no warping).
- The shaft is straight and of uniform cross-section.
- Radial lines remain straight after twisting — they simply rotate.
- The angle of twist is small.
- The applied torque is constant along the length (or varies in a known way).
Important: This theory is exact only for circular sections. Non-circular sections (square, rectangular, elliptical) warp when twisted and require more advanced analysis.
3. The Torsion Equation
Torsion Equation
T/J = τ/r = Gθ/L
Where:
- T = applied torque (N·m)
- J = polar moment of inertia of the cross-section (m⁴ or mm⁴)
- τ = shear stress at radial distance r from centre (Pa or MPa)
- r = radial distance from the centre of the shaft (m or mm)
- G = shear modulus (Pa or GPa)
- θ = angle of twist (radians)
- L = length of shaft (m or mm)
Maximum Shear Stress
τmax = TR / J (at the outer surface, r = R)
For solid shaft: τmax = 16T / (πD³)
The shear stress varies linearly from zero at the centre of the shaft to maximum at the outer surface. This is analogous to how bending stress varies linearly from the neutral axis in bending.
4. Polar Moment of Inertia
| Cross-Section | J (Polar Moment of Inertia) | Zp = J/R (Polar Section Modulus) |
|---|---|---|
| Solid circle (diameter D) | πD⁴/32 | πD³/16 |
| Hollow circle (Do, Di) | π(Do⁴ − Di⁴)/32 | π(Do⁴ − Di⁴)/(16Do) |
Note: The polar moment of inertia J = Ix + Iy (sum of the two rectangular moments of inertia). For a circle: Ix = Iy = πD⁴/64, so J = 2 × πD⁴/64 = πD⁴/32.
5. Angle of Twist
Angle of Twist
θ = TL / (GJ) (radians)
Convert to degrees: θ° = θ × 180/π
For stepped shafts (different diameters or materials along the length):
θtotal = Σ (TiLi) / (GiJi)
Add the twist of each segment, taking care with the sign of the torque in each segment.
6. Power Transmission
Shafts exist to transmit power from a source (engine, motor) to a load (machine, wheel). The relationship between power, torque, and speed is:
Power-Torque-Speed
P = 2πNT / 60
Where: P = power (W), N = rotational speed (rpm), T = torque (N·m)
Rearranging: T = 60P / (2πN) = 9549P/N (P in watts, T in N·m)
This formula is the starting point for most shaft design problems: given the power and speed, find the torque, then calculate the required shaft diameter from the allowable shear stress.
Shaft Design Equation
From τmax = 16T/(πD³) ≤ τallow:
D ≥ ∛(16T / (πτallow))
7. Hollow vs Solid Shafts
Hollow shafts are more efficient than solid shafts because the core material in a solid shaft contributes very little to torsional resistance (shear stress is proportional to r, so it is nearly zero near the centre).
| Comparison (same outer diameter Do) | Solid Shaft | Hollow Shaft (Di/Do = 0.6) |
|---|---|---|
| J | πDo⁴/32 | 0.8704 × πDo⁴/32 (87% of solid) |
| Weight (per unit length) | ρ × πDo²/4 | 0.64 × ρ × πDo²/4 (64% of solid) |
| Torque capacity | τallow × πDo³/16 | 87% of solid shaft capacity |
| Strength-to-weight ratio | Baseline | 36% higher |
A hollow shaft with Di/Do = 0.6 retains 87% of the torque capacity but weighs only 64% as much. This is why hollow shafts are preferred in aircraft, automotive, and bicycle applications where weight matters.
Strength Equivalence — Same Torque Capacity
For a hollow shaft to carry the same torque as a solid shaft at the same maximum stress:
Do,hollow = Dsolid × [1/(1 − k⁴)]1/3
Where k = Di/Do
8. Worked Numerical Examples
Example 1: Maximum Shear Stress
Problem: A solid steel shaft of 60 mm diameter transmits a torque of 1.5 kN·m. Find the maximum shear stress.
Solution
τmax = 16T/(πD³) = 16 × 1,500,000 / (π × 60³)
= 24,000,000 / (π × 216,000) = 24,000,000 / 678,584
τmax = 35.4 MPa
Example 2: Shaft Design for Power Transmission
Problem: A motor shaft transmits 50 kW at 1,500 rpm. Allowable shear stress is 60 MPa. Find the minimum shaft diameter.
Solution
T = 60P/(2πN) = 60 × 50,000 / (2π × 1500) = 3,000,000 / 9,424.8 = 318.3 N·m
D ≥ ∛(16T/(πτ)) = ∛(16 × 318,300 / (π × 60))
= ∛(5,092,800 / 188.5) = ∛(27,019) = 30.0 mm
Use D = 32 mm or next standard shaft size.
Example 3: Angle of Twist
Problem: A solid shaft (D = 50 mm, L = 2 m, G = 80 GPa) is subjected to T = 800 N·m. Find the angle of twist.
Solution
J = πD⁴/32 = π × 50⁴/32 = π × 6,250,000/32 = 613,592 mm⁴
θ = TL/(GJ) = 800,000 × 2,000 / (80,000 × 613,592)
= 1.6 × 10⁹ / 4.909 × 10¹⁰ = 0.0326 rad = 1.87°
9. Common Mistakes Students Make
- Using I instead of J: Torsion uses the POLAR moment of inertia J = πD⁴/32, NOT the rectangular moment I = πD⁴/64. Using I gives answers that are exactly half the correct value.
- Forgetting unit conversions for power: P = 2πNT/60 requires P in watts, N in rpm, T in N·m. If power is given in kW, convert to watts first. If torque is wanted in N·mm, multiply by 1000.
- Not converting angle to degrees: θ = TL/(GJ) gives the answer in radians. Many problems ask for the answer in degrees — multiply by 180/π.
- Applying the formula to non-circular sections: The torsion equation T/J = τ/r is valid only for circular (solid or hollow) cross-sections. Square, rectangular, and other shapes warp during torsion and require different formulas.
- Confusing D and R: τmax = TR/J uses the RADIUS (R = D/2), not the diameter. Using diameter gives double the correct stress.
10. Frequently Asked Questions
What is the torsion formula?
The torsion equation T/J = τ/r = Gθ/L relates the applied torque T to the shear stress τ, angle of twist θ, shaft geometry (J, r, L), and material property (G). The maximum shear stress τmax = TR/J occurs at the outer surface of the shaft.
Why are hollow shafts more efficient than solid shafts?
In a solid shaft, material near the centre carries almost no shear stress (since τ ∝ r). Removing this core (making the shaft hollow) eliminates dead weight without significantly reducing torsional strength. A hollow shaft with Di/Do = 0.6 weighs only 64% of a solid shaft but retains 87% of its torque capacity — a 36% improvement in strength-to-weight ratio.