Torsion of circular shafts

Why it matters

Helicopter main- and tail-rotor drive shafts, turbine engine spools, propeller shafts and control torque tubes all carry torque. Their design must keep the shear stress below the allowable value and the angle of twist small enough for the drive or control system to work accurately, at the lowest possible weight. That last requirement is why aerospace shafts are almost always hollow.

Key ideas

Assumptions of elementary torsion theory (circular shafts only):

• The shaft is straight with a uniform circular (solid or hollow) cross-section, and the material is homogeneous, isotropic and linear elastic.
• Plane cross-sections remain plane and do not warp; radii remain straight.
• The torque is constant along the length (or piecewise constant) and the twist is small.

Non-circular sections (rectangles, open thin-walled sections) warp and need different formulas.

Strain and stress. A cylinder twisted by an angle θ over length L shears each element at radius r by γ = r·θ/L. With τ = G·γ, the shear stress is zero at the axis and grows linearly to a maximum at the outer surface. Integrating the torque of these stresses gives T = G·J·θ/L, where J is the polar moment of inertia of the section.

The torsion equation. T/J = τ/r = G·θ/L. τ_max = T·R/J = T/Z_p, with Z_p = J/R the polar section modulus.

Torsional stiffness. k = T/θ = G·J/L. G·J is the torsional rigidity. Shafts in series (stepped shafts) add twists; shafts in parallel (a tube shrunk over a rod) share torque in proportion to their G·J.

Hollow shafts. The material near the axis carries little stress, so removing it costs little strength but saves a lot of weight. For the same weight, a hollow shaft is both stronger and stiffer than a solid one.

Power transmission. A shaft turning at N rpm and carrying torque T transmits P = 2π·N·T/60. Higher speed means lower torque for the same power, which is why gas-turbine shafts are slender.

Pure shear and failure. The surface element is in pure shear, so the principal stresses are ±τ on planes at 45° to the axis. Ductile shafts fail by shear on a transverse plane; brittle shafts (chalk, cast iron) fail along a 45° helix, perpendicular to the maximum tensile stress.

Statically indeterminate shafts. A shaft fixed at both ends with a torque applied at an intermediate point is solved like the axial case: equilibrium T_A + T_B = T, plus compatibility (the twists of the two parts are equal and opposite).

Formulas

T / J = τ / r = G·θ / L T = torque (N·m), J = polar moment of inertia (m⁴), τ = shear stress at radius r (Pa), r = radial distance (m), G = shear modulus (Pa), θ = angle of twist (rad), L = length (m).

J = π·d⁴ / 32 Solid shaft of diameter d (m).

J = π·(D⁴ − d⁴) / 32 Hollow shaft, outer diameter D and inner diameter d (m).

τ_max = 16·T / (π·d³) Solid shaft, at the surface.

τ_max = 16·T·D / [π·(D⁴ − d⁴)] Hollow shaft, at the outer surface.

θ = T·L / (G·J) Angle of twist (rad); multiply by 180/π for degrees. For stepped shafts use θ = Σ Tᵢ·Lᵢ/(Gᵢ·Jᵢ).

P = 2·π·N·T / 60 Power (W) with N in rev/min and T in N·m; equivalently P = T·ω with ω in rad/s.

Worked examples

Example 1 (standard). A solid shaft must transmit 100 kW at 1500 rpm. The allowable shear stress is 60 MPa and G = 80 GPa. Find the required diameter, then the stress and twist over 1.5 m if a 40 mm shaft is used.

1. T = 60·P/(2π·N) = 60 × 100 000 /(2π × 1500) = 636.6 N·m = 636 620 N·mm.
2. d³ = 16·T/(π·τ) = 16 × 636 620/(π × 60) = 54 038 mm³, so d = 37.8 mm. Choose 40 mm.
3. τ_max = 16·T/(π·d³) = 16 × 636 620/(π × 40³) = 50.7 MPa.
4. J = π·d⁴/32 = π × 40⁴/32 = 2.513 × 10⁵ mm⁴.
5. θ = T·L/(G·J) = 636 620 × 1500/(80 000 × 2.513 × 10⁵) = 0.0475 rad = 2.72°.

Answer: d_required ≈ 37.8 mm; with d = 40 mm, τ_max ≈ 50.7 MPa and θ ≈ 2.72°.

Example 2 (GATE level). The 40 mm solid shaft of Example 1 is replaced by a hollow shaft of outer diameter 50 mm and inner diameter 40 mm, same material and length. Compare stress, twist and weight.

1. J = π·(D⁴ − d⁴)/32 = π × (50⁴ − 40⁴)/32 = 3.623 × 10⁵ mm⁴.
2. τ_max = T·R/J = 636 620 × 25/3.623 × 10⁵ = 43.9 MPa.
3. θ = T·L/(G·J) = 636 620 × 1500/(80 000 × 3.623 × 10⁵) = 0.0329 rad = 1.89°.
4. Weight ratio = area ratio = (50² − 40²)/40² = 900/1600 = 0.5625.

Answer: the hollow shaft has τ ≈ 43.9 MPa and θ ≈ 1.89°, so it is stronger and stiffer, yet weighs only about 56 % as much.

Common mistakes

• Using J = π·d⁴/64 (that is the bending I) instead of π·d⁴/32.
• Using the radius in place of the diameter in J = π·d⁴/32.
• Leaving θ in radians when degrees are asked, or using degrees inside the formula.
• Forgetting to convert rpm and kW correctly in P = 2π·N·T/60.
• Assuming shear stress is uniform over the section, or maximum at the centre.
• Applying T/J = τ/r to rectangular or open sections.

For GATE AE

Expect numericals on diameter from power and speed, maximum shear stress and twist in solid and hollow shafts, strength and stiffness comparisons of hollow and solid shafts of equal weight, stepped shafts, and shafts fixed at both ends with an intermediate torque. Conceptual questions test the stress distribution and the brittle versus ductile failure plane. Practise converting units quickly and working in N·mm and mm.

Quick check

1. Where is the shear stress zero in a solid circular shaft under torsion?
2. If the diameter of a solid shaft is doubled, by what factor does its torque capacity rise?
3. By what factor does the twist fall for the same torque?
4. On what plane does a cast-iron shaft fail in torsion?

Answers: 1. At the axis. 2. 8 (τ ∝ T/d³). 3. 16 (θ ∝ 1/d⁴). 4. A 45° helical surface, normal to the maximum tensile stress.