Torsion of circular shafts

Why it matters

Drive shafts, axles, spindles, propeller shafts and torsion bars transmit power by twisting. A shaft must be strong enough not to yield in shear and stiff enough not to twist so much that gears mis-mesh or vibrations set in. The torsion formula sizes almost every rotating machine element, and the same theory gives the stiffness of close-coiled springs.

Key ideas

Assumptions (circular shafts).

1. The shaft is straight, circular (solid or hollow) and of uniform section over the length considered.
2. The material is homogeneous, isotropic and linearly elastic (Hooke's law in shear).
3. Plane cross-sections remain plane and do not warp; radii remain straight.
4. The twist is small, and the torque is applied about the shaft axis. These hold exactly only for circular sections; non-circular sections warp and need different formulas.

Shear strain and stress. A line on the surface parallel to the axis becomes a helix. The shear strain at radius r is γ = r θ / L, so it grows linearly from zero at the centre to a maximum at the surface. With τ = Gγ, shear stress also varies linearly with r: zero at the axis, maximum at the outer surface. By complementary shear, equal shear stresses act on longitudinal planes (this is why wooden shafts split lengthwise).

Torsion equation. T/J = τ/r = Gθ/L. J is the polar moment of inertia of the section. Torsional stiffness is T/θ = GJ/L; GJ is called torsional rigidity.

Hollow vs solid. Material near the axis is lightly stressed, so removing it costs little strength. A hollow shaft is lighter for the same torque capacity, or stronger and stiffer for the same weight, at the cost of a larger outside diameter and more expensive manufacture.

Power transmission. P = T ω = 2πNT/60. Lower speed means higher torque for the same power, so slow shafts are thicker.

Shafts in series and parallel.

• Series (stepped shaft, segments end to end): same torque in each part; angles of twist add.
• Parallel (composite shaft, tube shrunk on a core, or a shaft fixed at both ends with torque applied between): same angle of twist; torques add. These are statically indeterminate; use compatibility of twist.

Design. Choose d to satisfy both conditions: τmax ≤ τallow (strength) and θ/L ≤ allowable twist (stiffness, often 0.25–1° per metre for machine shafts; take the value from the design data book). The larger diameter governs. With bending as well, use the equivalent torque or moment from the failure-theories topic.

Limits. Stress concentration at keyways, shoulders and holes raises local stress (use factors from a data book). Beyond yield, the linear distribution no longer applies.

Formulas

T / J = τ / r = G θ / L

• T = torque (N·mm), J = polar moment of inertia (mm⁴), τ = shear stress at radius r (MPa), G = modulus of rigidity (MPa), θ = angle of twist (rad), L = length (mm).

J = π d⁴ / 32 (solid); J = π (D⁴ − d⁴) / 32 (hollow, outer D, inner d)

τmax = 16 T / (π d³) (solid); τmax = 16 T D / [π (D⁴ − d⁴)] (hollow)

Zp = J / r_max (polar section modulus): π d³ / 16 (solid)

θ = T L / (G J); for segments in series θ = Σ Ti Li / (Gi Ji)

P = 2π N T / 60

• P = power (W), N = speed (rpm), T = torque (N·m).

U = T² L / (2 G J) (strain energy stored in torsion)

Worked examples

Example 1 (standard, design). A solid steel shaft transmits 30 kW at 300 rpm. Allowable shear stress 60 MPa, allowable twist 1° per metre, G = 80 GPa. Find the minimum diameter.

1. Torque: T = 60 P / (2π N) = 60 × 30 000 / (2π × 300) = 954.9 N·m = 954 930 N·mm.
2. Strength: d³ = 16 T / (π τ) = 16 × 954 930 / (π × 60) = 81 057 mm³, so d = 43.3 mm.
3. Stiffness: θ/L = 1° per m = 0.017 45 rad per 1000 mm. From θ = TL/(GJ): J = T L / (G θ) = 954 930 × 1000 / (80 000 × 0.017 45) = 6.84 × 10⁵ mm⁴.
4. d⁴ = 32 J / π = 6.97 × 10⁶ mm⁴, so d = 51.4 mm.
5. Stiffness governs.

Answer: d ≥ 51.4 mm (stiffness governs); choose the next standard size from the data book.

Example 2 (GATE level, hollow shaft). For the same torque and allowable stress (954.9 N·m, 60 MPa), design a hollow shaft with inner diameter = 0.6 × outer diameter, and find the weight saving compared with the strength-based solid shaft (d = 43.3 mm).

1. τ = 16 T / [π D³ (1 − k⁴)] with k = 0.6, 1 − k⁴ = 1 − 0.1296 = 0.8704.
2. D³ = 16 × 954 930 / (π × 60 × 0.8704) = 93 126 mm³, so D = 45.3 mm, d = 0.6 × 45.3 = 27.2 mm.
3. Weight ratio (same material and length) = area ratio: D²(1 − k²) / d_solid² = 45.33² × 0.64 / 43.28² = 0.702.
4. Saving = 1 − 0.702 = 29.8%.

Answer: D ≈ 45.3 mm, d ≈ 27.2 mm; about 30% lighter than the solid shaft.

Common mistakes

• Using π d⁴/64 (that is I for bending) instead of J = π d⁴/32.
• Mixing N·m and N·mm, or kW and W, in the power formula.
• Using degrees for θ in TL/(GJ); it gives radians.
• Applying the torsion formula to square or rectangular sections.
• Forgetting the stiffness check, which often governs long shafts.
• For stepped shafts, using one J for all segments.

For GATE PI

Expect shaft diameter from power and speed, ratios of strength and weight for hollow vs solid shafts, angle of twist of stepped or composite shafts, shafts fixed at both ends with an intermediate torque, and combined bending and torsion. Practise converting power, speed and torque in one line.

Quick check

1. Where is shear stress zero in a solid shaft under torsion?
2. If the diameter of a solid shaft is doubled, how does its torque capacity change at the same allowable stress?
3. What is J for a solid shaft of 100 mm diameter?
4. Two segments in series carry torque: what is common to them, and what adds?

Answers: 1. At the axis. 2. It becomes 8 times larger (∝ d³). 3. 9.82 × 10⁶ mm⁴. 4. Same torque; angles of twist add.