Torsion of solid and hollow circular shafts

Why it matters

Every motor, gearbox and coupling in a mechatronic drive transmits power through a shaft in torsion. The shaft must be strong enough not to yield or fatigue, and in a servo system it must also be stiff enough: torsional wind-up between motor and load adds position error and lowers the resonant frequency that limits control bandwidth. Torque sensors are themselves short shafts with strain gauges at 45°, read through the same torsion equations.

Key ideas

Assumptions of elementary torsion theory.

• The shaft is straight with a uniform circular (solid or hollow) cross-section.
• Material is homogeneous, isotropic and linear-elastic.
• Plane cross-sections remain plane and radii remain straight as the shaft twists (true only for circular sections).
• The twist is small, and the torque is constant along the length considered.

Shear strain and stress vary linearly with radius. 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, where θ is the angle of twist over length L. With τ = G·γ, the shear stress also grows linearly from zero at the centre to a maximum at the outer surface. Complementary shear stresses of the same size act along the shaft's length.

The torsion equation. Summing the moment of the shear stresses over the section gives T = (G·θ/L)·J, where J is the polar second moment of area. Together: T/J = τ/r = G·θ/L.

Strength and stiffness. The design must satisfy two limits:

• strength: τ_max = T·R/J ≤ allowable shear stress (from the material's shear yield strength and a factor of safety; take values from your design data book);
• stiffness: angle of twist θ = T·L/(G·J) ≤ allowable twist (often specified per metre of length). For precision drives the stiffness limit often governs. Torsional stiffness is k_t = T/θ = G·J/L.

Hollow shafts. Material near the centre carries little stress and contributes little to J, so removing it saves weight with little loss of strength. For the same weight, a hollow shaft is stronger and stiffer than a solid one; for the same outer diameter, it is weaker but much lighter.

Power transmission. P = T·ω = 2πN·T/60, with N in rpm. Low speed means high torque for the same power, so gearbox output shafts are larger than input shafts.

Shafts in series and in parallel. In series (stepped shaft or two shafts coupled end to end), the torque is the same in each segment and the twists add. In parallel (a composite shaft with a sleeve bonded to a core, or a shaft fixed at both ends with a torque applied in between), the twists are equal and the torques add; this is statically indeterminate and needs compatibility.

Torsion and principal stresses. Pure shear τ at the surface is equivalent to principal stresses +τ and −τ at 45° to the axis. Ductile shafts fail in shear on a transverse plane; brittle ones crack along a 45° helix. A strain gauge at 45° reads ε = τ/(2G), which is how torque sensors work.

Limits. These equations do not apply to non-circular sections (rectangles warp), to thin-walled open sections, or near keyways, shoulders and splines where stress concentration factors from your data book must be applied.

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); r: radius (m); G: shear modulus (Pa); θ: angle of twist (rad); L: length (m).

J = π·d⁴/32 (solid), J = π·(D⁴ − d⁴)/32 (hollow)

• d: diameter of a solid shaft, or inner diameter of a hollow one (m); D: outer diameter (m).

τ_max = 16·T / (π·d³) (solid), τ_max = 16·T·D / [π·(D⁴ − d⁴)] (hollow)

• Maximum shear stress at the outer surface (Pa).

θ = T·L / (G·J), k_t = G·J / L

• Angle of twist (rad) and torsional stiffness (N·m/rad).

P = T·ω = 2π·N·T / 60

• P: power (W); ω: angular speed (rad/s); N: speed (rpm).

ε₄₅ = τ / (2G)

• Strain read by a gauge at 45° to the axis on the surface of a shaft in pure torsion.

Worked examples

Example 1 (standard): servo output shaft by strength and stiffness. Given: a servo motor delivers 3 kW at 1500 rpm through a solid steel shaft. Allowable shear stress 40 MPa; allowable twist 0.25° per metre; G = 80 GPa. Find the minimum diameter.

1. Torque: T = 60P/(2πN) = 60 × 3000/(2π × 1500) = 19.10 N·m = 19 100 N·mm.
2. Strength: τ_max = 16T/(πd³) ≤ 40 gives d³ = 16 × 19 100/(π × 40) = 2432 mm³, so d = 13.4 mm.
3. Stiffness: allowable θ/L = 0.25 × π/180 per 1000 mm = 4.363 × 10⁻⁶ rad/mm.
4. θ/L = T/(G·J) gives J = 19 100/(80 000 × 4.363 × 10⁻⁶) = 54 710 mm⁴.
5. d⁴ = 32J/π = 557 300 mm⁴, so d = 27.3 mm. Answer: d ≈ 27.3 mm (stiffness governs; strength alone needs only 13.4 mm). Choose 28 mm: the actual twist is then 0.23° per metre.

Example 2 (GATE level): hollow shaft to replace a solid one, and a torque sensor. Given: a solid steel shaft of 60 mm diameter works at τ_max = 60 MPa. It is to be replaced by a hollow shaft of the same material with d/D = 0.6, transmitting the same torque at the same maximum stress. Find the torque, the hollow shaft's dimensions, the weight ratio, and the reading of a 45° strain gauge on the surface (G = 80 GPa).

1. Torque: T = π·d³·τ/16 = π × 60³ × 60/16 = 2.545 × 10⁶ N·mm = 2.545 kN·m.
2. Same T and τ_max means the same J/R: D³·(1 − 0.6⁴) = 60³, so D = 60/(0.8704)^(1/3) = 62.84 mm; d = 0.6 × 62.84 = 37.70 mm.
3. Weight ratio (same length and material) = area ratio = (62.84² − 37.70²)/60² = 0.702.
4. Gauge: ε₄₅ = τ/(2G) = 60/(2 × 80 000) = 375 × 10⁻⁶. Answer: T ≈ 2.55 kN·m; D ≈ 62.8 mm, d ≈ 37.7 mm; the hollow shaft weighs about 70 % of the solid one; the gauge reads 375 με.

Common mistakes

• Using the radius where the formula needs diameter (or vice versa) in J = π·d⁴/32.
• Using degrees in θ = T·L/(G·J); the result is in radians.
• Mixing up J (polar, torsion) with I (bending): for a circle J = 2I.
• Forgetting the stiffness check; precision drives often need larger shafts than strength suggests.
• Saying a hollow shaft is stronger than a solid one of the same outer diameter; it is stronger for the same weight.
• Applying these formulas to square or rectangular shafts.
• Converting power with N in rpm but forgetting the 2π/60.

For GATE ME

Expect maximum shear stress and twist of solid and hollow shafts, power-torque-speed conversions, strength and weight comparisons of solid versus hollow shafts, stepped shafts and shafts fixed at both ends, composite shafts, and the 45° principal stresses in torsion. Practise ratio problems (how τ or θ changes when d is doubled: τ falls by 8, θ by 16).

Quick check

1. A solid 100 mm shaft carries 500 N·m. What is τ_max?
2. Find J for a 50 mm solid shaft.
3. If the diameter of a solid shaft is doubled at the same torque, by what factor does the twist change?
4. What power does a shaft carry at 300 N·m and 1000 rpm?
5. Where is the shear stress zero in a solid shaft in torsion?

Answers: 1. 16 × 500/(π × 0.1³) = 2.55 MPa. 2. 6.14 × 10⁻⁷ m⁴. 3. It becomes 1/16. 4. 2π × 1000 × 300/60 = 31.4 kW. 5. At the axis (centre).