Torsion of solid and hollow circular shafts

Torsion of solid and hollow circular shafts: T/J = τ/r = Gθ/L, power transmission, strength and stiffness design, solid versus hollow comparisons, and shafts in series and parallel.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Every powertrain transmits torque through shafts: the crankshaft, gearbox shafts, the propeller shaft and the half-shafts to the wheels. A shaft must be strong enough not to yield or fatigue in shear and stiff enough not to twist too much (which upsets timing, causes vibration or lets gears mis-mesh). The torsion equation gives both checks, and it explains why propeller shafts are tubes.

Key ideas

Assumptions of the elementary theory.

  • The shaft is straight, circular (solid or hollow) and of constant section over the length considered.
  • The material is homogeneous, isotropic and linear elastic.
  • Plane cross-sections remain plane and radii remain straight after twisting (true only for circular sections).
  • The twist is small and the torque is constant along the length (or piecewise constant).

Strain and stress distribution. A line on the surface that was parallel to the axis becomes a helix. The shear strain at radius r is γ = r·θ/L, proportional to r. With τ = G·γ, the shear stress is also proportional to r: zero at the centre, maximum at the outer surface. In a hollow shaft it is not zero at the bore: τ_inner = τ_outer·(d_i/d_o).

Torsion equation. Moment equilibrium of the shear stresses over the section gives T = (τ/r)·J, where J is the polar second moment of area. Combined: T/J = τ/r = G·θ/L.

Complementary shear and failure planes. Shear on cross-sections is accompanied by equal shear on longitudinal planes, and by principal stresses ±τ on 45° planes. Ductile shafts fail by shear on the cross-section (a flat break); brittle shafts fail in tension along a 45° helix.

Torsional rigidity and stiffness. G·J is the torsional rigidity; the torsional stiffness of a length L is k_t = G·J/L (torque per radian). Design often limits the twist, for example to a fraction of a degree per metre; the actual limit comes from the application or a design code.

Power transmission. P = T·ω = 2π·N·T/60. For a given power, torque is inversely proportional to speed, so slow shafts (after a reduction gear, or the final drive) are thicker.

Solid versus hollow.

  • For the same outer diameter, a solid shaft is stronger and stiffer, but a hollow one loses little strength because the core carries little stress.
  • For the same weight (same area), a hollow shaft is considerably stronger and stiffer, since its material sits at large radius where it does the most work.
  • Hollow shafts also have higher bending stiffness per kilogram, which raises the whirling speed, important for long propeller shafts.
  • Very thin walls can buckle locally under torsion, so wall thickness has a practical lower limit.

Shafts in series and parallel.

  • Series (stepped shaft, same torque in each part): total twist = sum of twists.
  • Parallel (composite shaft, a tube shrunk on a core, or a shaft fixed at both ends with torque applied between): the parts share the torque so that their twists are equal.

Non-circular sections (rectangles, open thin sections) warp and do not follow these formulas; use the appropriate tables from a data book.

Formulas

T / J = τ / r = G·θ / L

  • T: torque (N·m); J: polar second moment of area (m⁴); τ: shear stress at radius r (Pa); G: modulus of rigidity (Pa); θ: angle of twist over length L (rad).

J = π·d⁴ / 32 (solid), J = π·(d_o⁴ − d_i⁴) / 32 (hollow)

τ_max = 16·T / (π·d³) (solid), τ_max = 16·T·d_o / (π·(d_o⁴ − d_i⁴)) (hollow)

  • Polar section modulus Z_p = J / r_max.

θ = T·L / (G·J), series shaft θ = Σ Tᵢ·Lᵢ / (G·Jᵢ)

P = 2π·N·T / 60

  • P: power (W); N: speed (rev/min).

T = T₁ + T₂, T₁·L₁ / (G₁·J₁) = T₂·L₂ / (G₂·J₂)

  • Composite (parallel) shafts sharing a torque with equal twist.

Worked examples

Example 1 (standard): sizing a solid shaft for strength and stiffness. Given: a solid steel shaft transmits 30 kW at 1500 rev/min. Allowable shear stress 40 MPa, allowable twist 1° per metre, G = 80 GPa. Find the diameter.

  1. Torque: T = P·60 / (2π·N) = 30 000 × 60 / (2π × 1500) = 190.99 N·m.
  2. Strength: d³ = 16·T / (π·τ) = 16 × 190 990 / (π × 40) = 24 318 mm³, so d = 28.97 mm.
  3. Stiffness: θ/L = 1° = 0.017 45 rad/m, so J = T / (G·θ/L) = 190.99 / (80 × 10⁹ × 0.017 45) = 1.368 × 10⁻⁷ m⁴.
  4. d = (32·J/π)^(1/4) = 0.03436 m = 34.36 mm.
  5. Take the larger value. Answer: d ≈ 34.4 mm (stiffness governs); round up to the next standard size.

Example 2 (GATE level): replacing a solid shaft by a hollow one. Given: a 60 mm solid shaft is to be replaced by a hollow shaft of the same material with d_i = 0.6·d_o, carrying the same torque at the same maximum shear stress. Find d_o, d_i, the weight saving and the change in twist.

  1. Equal strength needs equal polar section modulus: d_o³·(1 − k⁴) = d³ with k = 0.6, so d_o = 60 / (1 − 0.1296)^(1/3) = 60 / 0.9548 = 62.84 mm and d_i = 37.70 mm.
  2. Area ratio (hollow/solid) = d_o²·(1 − k²) / d² = 62.84² × 0.64 / 60² = 0.702, a weight saving of about 30 %.
  3. Twist ratio: θ ∝ 1/J. J_hollow / J_solid = [d_o⁴·(1 − k⁴)] / d⁴ = (d_o/d) × 1 = 1.047, so the hollow shaft twists about 4.5 % less. Answer: d_o ≈ 62.8 mm, d_i ≈ 37.7 mm, about 30 % lighter and slightly stiffer.

Common mistakes

  • Using rev/min directly in P = T·ω; convert to rad/s or use P = 2πNT/60.
  • Mixing π·d⁴/32 (polar, torsion) with π·d⁴/64 (bending).
  • Saying the shear stress is zero at the inner surface of a hollow shaft.
  • Using degrees in θ = T·L/(G·J), which gives radians.
  • Checking only strength and forgetting the stiffness limit (or the reverse).
  • Adding twists for parallel shafts or sharing torque in series shafts.

For GATE ME

Expect maximum shear stress and twist for solid and hollow shafts, power–torque–speed conversions, strength and weight comparison of solid and hollow shafts, stepped and composite shafts, and shafts fixed at both ends with an intermediate torque. Practise ratio questions (for example, how τ changes when d doubles) without full calculation.

Quick check

  1. If the diameter of a solid shaft doubles at the same torque, how does τ_max change?
  2. Find τ_max in a 50 mm solid shaft carrying 200 N·m.
  3. A hollow shaft has d_o = 60 mm and d_i = 40 mm. Find J.
  4. What is the twist of a 1 m length of a 120 mm solid shaft carrying 600 N·m (G = 80 GPa)?

Answers: 1. It falls to one-eighth. 2. 16 × 200 000 / (π × 50³) = 8.15 MPa. 3. 1.021 × 10⁶ mm⁴. 4. 3.68 × 10⁻⁴ rad.

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is torsion in the context of solid and hollow circular shafts?Concept

    Torsion refers to the twisting of an object due to an applied torque. In the context of solid and hollow circular shafts, it involves the rotation of the shaft about its longitudinal axis, causing shear stress over the cross-section. The amount of twist depends on the material properties, the geometry of the shaft, and the magnitude of the applied torque.

  2. 2.Explain the difference between solid and hollow circular shafts in terms of torsional strength.Concept

    For the same outer diameter a solid shaft is slightly stronger and stiffer, because J = π(d_o⁴ − d_i⁴)/32 falls when the core is removed, but the loss is small since the core is lightly stressed. For the same weight (same cross-sectional area) a hollow shaft is much stronger and stiffer, because its material sits at a larger radius where shear stress and its contribution to J are highest. For example, a hollow shaft with d_i = 0.6·d_o can match a solid shaft's torque capacity with about 30 % less material.

  3. 3.What is the polar moment of inertia, and why is it important in the analysis of torsion?Concept

    The polar moment of inertia is a measure of an object's ability to resist torsion and is calculated based on the geometry of the cross-section. It is crucial in torsion analysis because it directly affects the shear stress and angle of twist in the shaft. A higher polar moment of inertia indicates a greater resistance to twisting.

  4. 4.Why are hollow shafts often preferred over solid shafts in automotive applications?Application

    In a shaft under torsion the material near the centre carries little stress, so removing it saves weight with little loss of strength; for equal weight a tube is both stronger and stiffer than a solid bar. Propeller (drive) shafts are long, so their tubular form also gives high bending stiffness per kilogram, which raises the critical whirling speed. Lower rotating mass also reduces inertia and improves response. The limits are cost, a minimum wall thickness to avoid local buckling, and space at joints and splines.

  5. 5.What happens to the shear stress distribution in a shaft when it is subjected to torsion?Application

    In a circular shaft the shear stress on the cross-section is proportional to the radius, τ = T·r/J, so it rises linearly from zero at the centre of a solid shaft to its maximum at the outer surface. In a hollow shaft it rises linearly from τ_max·(d_i/d_o) at the bore, not zero, to τ_max at the outside. Equal complementary shear acts on longitudinal planes, and principal stresses of ±τ act on 45° planes.

  6. 6.How does the angle of twist differ between solid and hollow shafts under the same torque?Application

    The twist is θ = T·L/(G·J), so it depends inversely on J. For the same outer diameter, a hollow shaft has a smaller J and twists slightly more than a solid one. For the same weight, the hollow shaft has a larger outer diameter and a much larger J, so it twists less. That is why tubes are used where torsional stiffness per kilogram matters.

  7. 7.Calculate the maximum shear stress in a solid circular shaft with a diameter of 50 mm subjected to a torque of 200 Nm.Numerical
    1. J = π·d⁴/32 = π × 0.05⁴/32 = 6.14 × 10⁻⁷ m⁴.
    2. τ_max = T·r/J = 200 × 0.025 / 6.14 × 10⁻⁷ = 8.15 × 10⁶ Pa.
    3. Equivalently τ_max = 16·T/(π·d³) = 16 × 200 / (π × 0.05³) = 8.15 MPa.
  8. 8.Determine the angle of twist for a hollow shaft with an outer diameter of 60 mm, inner diameter of 40 mm, length of 2 m, and subjected to a torque of 300 Nm. Assume the material's shear modulus is 80 GPa.Numerical
    1. J = π(d_o⁴ − d_i⁴)/32 = π × (0.06⁴ − 0.04⁴)/32 = π × 1.04 × 10⁻⁵ / 32 = 1.021 × 10⁻⁶ m⁴.
    2. θ = T·L/(G·J) = 300 × 2 / (80 × 10⁹ × 1.021 × 10⁻⁶) = 600 / 81 680 = 7.35 × 10⁻³ rad.
    3. In degrees: 7.35 × 10⁻³ × 180/π ≈ 0.42°.
  9. 9.What are the limitations of using hollow shafts in engineering applications?Application

    Hollow shafts cost more to make (tube stock, boring or friction-welded ends) and need a larger outer diameter for the same torque, which can clash with bearing, seal and spline sizes. If the wall is made very thin to save weight, it can buckle locally under torsion or be dented in service. They also need end fittings, such as welded yokes, that introduce stress concentrations and must be balanced carefully to avoid vibration.

  10. 10.Explain how the material properties affect the torsional behavior of a shaft.Concept

    Material properties such as the shear modulus (G) and yield strength influence the torsional behavior of a shaft. A higher shear modulus indicates that the material is more rigid and will experience less deformation under the same torque. The yield strength determines the maximum stress the material can withstand before permanent deformation occurs. Selecting the appropriate material is crucial for ensuring that the shaft can handle the expected loads without failure.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousShear stress distribution in beamsNext Deflection of beams: double integration and Macaulay's method