Library / Mechanical / MEC-34
Mechanical
Torsional shear stress in a solid shaft
Maximum shear stress at the surface of a solid round shaft under torque.
- Ref
- MEC-34
- Rev
- v1
- Field
- Mechanical
- Result
- τ [MPa]
T is the torque, in N·m, and d the shaft diameter in mm. The answer is the largest shear stress, which occurs at the outer surface; it falls to zero at the centre. The 16/(πd³) factor comes from J = πd⁴/32 and the outer radius d/2.
Maximum shear stress at the surface of a solid round shaft under torque. Diameter enters as a cube, so it is the most powerful design lever.
Calculator
- Polar second moment J = πd⁴/32251,300 mm⁴
- Stress if the diameter is 20 % larger23.03 MPa
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 26 % bigger diameter halves the stress (1.26³ ≈ 2). Because the diameter is cubed, a small increase in size buys a lot of margin.
- The centre of a solid shaft carries almost no stress, so a hollow shaft of the same strength is much lighter.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (MPa) | Local step | Change in result |
|---|---|---|---|---|---|---|
| d Shaft diameter | 1 to 500 mm | 0.02037 | 2,546,000 | 2,546,000 | +10% | −24.9 % |
| T Torque | 1 to 1,000,000 N·m | 0.07958 | 79,580 | 79,580 | +10% | +10 % |
Sweep of Shaft diameter
| Shaft diameter (mm) | Shear stress (MPa) |
|---|---|
| 1 | 2,546,000 |
| 63.38 | 10 |
| 125.8 | 1.281 |
| 188.1 | 0.3825 |
| 250.5 | 0.162 |
| 312.9 | 0.08314 |
| 375.3 | 0.04819 |
| 437.6 | 0.03038 |
| 500 | 0.02037 |
Sweep one variable
Variables and how each one contributes
Shear stress
What the equation gives you. Everything else feeds it.
Torque
The torque carried by the shaft, in N·m. Directly proportional: doubling the torque doubles the surface shear stress.
Shaft diameter
The diameter of the solid round shaft, in mm. Inverse cube, the strongest lever in the formula: a 10 % rise in d lowers τ by about 25 %.
Worked example with the values above
- Convert torque: T = 500 N·m = 500,000 N·mm.
- Numerator: 16T = 8,000,000; denominator: πd³ = π × 64,000 = 201,100.
- τ = 8,000,000 / 201,100 = 39.79 MPa.
Assumptions and limits
- A solid circular shaft of uniform diameter, loaded in pure torsion.
- Elastic behaviour, and a section away from keyways, shoulders and splines where stress concentrates.
Common mistakes
- Putting the radius in place of the diameter, which gives a result 8 times too high.
- Using the formula for a hollow shaft. It needs the inner diameter as well.
Statistics and uncertainty
τ ∝ T/d³, so a 1 % error in the diameter gives a 3 % error in the stress. A machining tolerance of ±0.1 mm on 40 mm gives a ±0.75 % change in the stress, which is small beside the uncertainty in the torque itself, especially with shock loads.
Where it is used
- Sizing a drive shaft, propeller shaft or torsion bar.
- Checking the shear stress against the allowable value.
Maths you need
- Polar second moment of area: J = πd⁴/32
- Cubes: d³