Torsion of Circular Shafts — the Torsion Formula

Why twisting a shaft creates shear stress that is zero at the centre and maximum at the surface — and why hollow shafts win on efficiency.

Mechanics of MaterialsMechanical Engineering Year 1Free preview
⏱️ About 18 min

Turn a screwdriver and the shaft twists. You cannot see it, but the material is being sheared — stressed parallel to the cross-section. That shear stress is not uniform: it is zero exactly at the centre of the shaft and largest at the outer surface. Engineers size drive shafts, propellers, and drill bits using one formula that captures this entire behaviour: the torsion formula.

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The big idea: For a circular shaft in elastic torsion, the shear stress at any radius is τ = Tρ / J, where T is the applied torque, ρ is the radial distance from the centre, and J is the polar moment of inertia of the cross-section. The stress varies linearly from zero at the centre to a maximum at the outer radius. Hollow shafts concentrate material where the stress is highest, giving a larger J per unit mass and making them more efficient than solid shafts.
🎯 By the end, you'll be able to
  • State the torsion formula τ = Tρ / J and identify each symbol
  • Compute J for solid and hollow circular cross-sections
  • Calculate maximum shear stress in a shaft under a given torque
  • Explain why a hollow shaft can be lighter and stronger than a solid shaft of the same outer diameter
📎 Helpful to know first

From shear strain linearity to the torsion formula

When a torque is applied to a circular shaft, every cross-section rotates slightly relative to its neighbour. Imagine drawing a grid of longitudinal and circumferential lines on the surface: after twisting, the longitudinal lines become helices. The angle change between a longitudinal line and a circumferential line is the shear strain γ at the surface.

For a small twist angle, the arc length at radius ρ is proportional to ρ. That means the shear strain — and therefore the shear stress in the elastic range — grows linearly from the centre outward:

τ(ρ) ∝ ρ

To turn this proportionality into an equation, we use the fact that the sum of the moment produced by shear stresses over the entire cross-section must equal the applied torque T. That integration yields the torsion formula:

\[ \tau = \frac{T\,\rho}{J} \]
Torsion formula: shear stress at radius ρ in a circular shaft subjected to torque T. J is the polar moment of inertia.
\[ J_{\text{solid}} = \frac{\pi}{32}\,d^{4} \qquad J_{\text{hollow}} = \frac{\pi}{32}\left(D^{4} - d^{4}\right) \]
Polar moment of inertia for solid (diameter d) and hollow (outer D, inner d) circular cross-sections.
\[ \tau_{\max} = \frac{T\,R}{J} = \frac{16\,T}{\pi\,d^{3}} \quad\text{(solid shaft)} \]
Maximum shear stress occurs at the outer radius R = d/2. For a solid circular shaft, substituting J gives the compact form on the right.
Circular shaft cross-section showing linear shear stress distribution: zero at the centre, maximum at the outer radius. ρ R τ = 0 τ_max τ(ρ) = T·ρ / J Linear distribution Zero at centre Maximum at surface Valid only for circular shafts in elastic range
Shear stress varies linearly with radius. The formula is valid only for circular shafts behaving elastically.
⚠️ The torsion formula is valid only for circular shafts in the elastic range

Two common misconceptions cause real design errors:

  • Non-circular sections warp. A square bar under torsion does not remain plane; its cross-sections distort out of plane. The simple linear τ = Tρ/J relationship breaks down completely. Square shafts need St. Venant torsion theory — a different, more advanced analysis.
  • Plastic deformation changes the distribution. Once the outer fibres yield, the stress profile flattens and spreads inward. The elastic torsion formula overestimates the torque capacity if the material has yielded.

Whenever you reach for τ = Tρ/J, confirm both that the shaft is circular and that the maximum stress is below the shear yield strength.

🎮 Interactive: shaft twist and shear stress LIVE
Predict first: Set the torque to 2.0 kN·m and diameter to 50 mm. Switch to the cross-section view. Where is the shear stress zero, and where is it maximum? Now increase the diameter to 80 mm — how does tau_max change?

An interactive shaft-twist simulator with sliders for torque, diameter, and length, showing a 3D shaft and cross-section stress distribution.

Explore how torque, diameter, and length affect twist angle and maximum shear stress in a circular shaft. Toggle between the 3D shaft view and the cross-section stress distribution.

Why hollow shafts are more efficient

Material near the centre of a solid shaft carries almost no stress — the torsion formula shows τ is proportional to ρ, so the core is essentially along for the ride. By removing that low-stress material and relocating it to a larger outer radius, a hollow shaft achieves a much larger polar moment of inertia J without a proportional increase in mass.

The result is a higher strength-to-weight ratio. Aircraft propeller shafts, race-car drive shafts, and bicycle bottom-bracket axles are often hollow for exactly this reason. The design trade-off is that very thin walls can buckle locally, so wall thickness is chosen to resist both shear stress and local instability.

📝 Worked example: A solid shaft has diameter 50 mm. A hollow shaft has outer diameter 60 mm and inner diameter 40 mm. Both are made of steel and carry the same torque of 2.5 kN·m. Compare their polar moments of inertia, maximum shear stresses, and cross-sectional areas (mass is proportional to area).
  1. Solid shaft: J = π/32 × 50⁴ = 6.136 × 10⁵ mm⁴ = 6.136 × 10⁻⁷ m⁴.
  2. Solid shaft: τ_max = 2500 × 0.025 / 6.136×10⁻⁷ = 101.9 MPa.
  3. Hollow shaft: J = π/32 × (60⁴ − 40⁴) = 1.021 × 10⁶ mm⁴ = 1.021 × 10⁻⁶ m⁴.
  4. Hollow shaft: τ_max = 2500 × 0.030 / 1.021×10⁻⁶ = 73.5 MPa.
  5. Area solid = π/4 × 50² = 1964 mm². Area hollow = π/4 × (60² − 40²) = 1571 mm².
  6. Comparison: the hollow shaft has 20% less mass but 66% higher J, giving 28% lower peak stress. This is why hollow shafts are more efficient for torsion.
✓ Solid: J = 6.14×10⁵ mm⁴, τ_max = 102 MPa. Hollow: J = 1.02×10⁶ mm⁴, τ_max = 73.5 MPa. Hollow is lighter and stronger in torsion.
✏️ Practice: A solid circular shaft of diameter 40 mm carries a torque of 1.5 kN·m. What is the maximum shear stress, in MPa?
MPa
Solution
  1. τ_max = 16T / (πd³) with T = 1500 N·m, d = 0.040 m.
  2. τ_max = 16 × 1500 / (π × 0.040³) = 24 000 / (2.011×10⁻⁴) = 1.194×10⁸ Pa = 119.4 MPa.
✏️ Practice: A hollow shaft has outer diameter 80 mm and inner diameter 50 mm. What is its polar moment of inertia J, in mm⁴?
mm⁴
Solution
  1. J = π/32 × (D⁴ − d⁴) = π/32 × (80⁴ − 50⁴) = π/32 × (40 960 000 − 6 250 000).
  2. J = π/32 × 34 710 000 = 3 406 400 mm⁴.

Check your understanding

1. In the elastic torsion of a circular shaft, the shear stress at the centre is:
τ = Tρ/J. At ρ = 0 (the centre), the shear stress is exactly zero.
2. The torsion formula τ = Tρ / J is valid for:
The derivation assumes plane sections remain plane and circular — true for circular shafts in elastic torsion, false for non-circular or plastic behaviour.
3. Compared to a solid shaft of the same outer diameter, a hollow shaft of similar mass typically has:
Relocating material to a larger radius increases J more than mass increases, reducing peak stress for the same torque.
🔎 Related lessons
Angle of Twist & Power Transmission What Is Stress? Normal & Shear Stress
✅ Key takeaways
  • The torsion formula τ = Tρ/J gives shear stress at any radius in a circular elastic shaft.
  • J_solid = πd⁴/32; J_hollow = π(D⁴ − d⁴)/32.
  • τ_max occurs at the outer radius and is zero at the centre.
  • Hollow shafts are more efficient because they place material where stress is highest.
  • The formula is valid only for circular shafts in the elastic range.
➡️ Knowing the stress is only half the story. The next question is: how much does the shaft actually twist under load? That angle of twist governs alignment, vibration, and drivetrain behaviour.
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