Torsion, Shear and Combined Loads
Torsion of shafts and tubes, direct and bending shear, combining stresses and the von Mises criterion.
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Torsion of round shafts
When a torque T twists a round bar, every cross-section rotates relative to its neighbor and the material is sheared. The shear stress grows linearly from zero on the axis to a maximum at the surface: τ = T r ÷ J, and for a solid shaft of diameter d, τmax = 16 T ÷ (π d³), with the polar second moment of area J = π d⁴ ÷ 32. This holds for round sections (solid or hollow) in the elastic range; for other shapes the formulas differ (below).
Angle of twist and stiffness
θ = T L ÷ (G J) kt = T ÷ θ = G J ÷ L
The twist angle θ (in radians) of a shaft of length L is proportional to the torque and the length and inversely proportional to the shear modulus G (about 80 GPa for steel, 27 for aluminum) and to J. The torsional stiffness kt = G J ÷ L matters for timing and positioning drives and for the torsional vibration of a system (see Shafts, Keys and Couplings).
Worked example: a 30 mm shaft
- A 30 mm steel shaft, 600 mm long, transmits 150 N·m. J = π × 30⁴ ÷ 32 = 79 522 mm⁴.
- Stress: τ = 16 × 150 000 ÷ (π × 30³) = 28.3 MPa. The shear yield of steel with Sy = 350 MPa is about 0.577 × 350 = 202 MPa, so n = 7.1 for static torque (fatigue and keyways need more: see Fatigue and Stress Concentration).
- Twist: θ = T L ÷ (G J) = 150 000 × 600 ÷ (80 000 × 79 522) = 0.014 rad = 0.81° over 600 mm. A twist of this size spoils a precision positioning axis; the rule of thumb for machine shafts is 0.25° per meter or less (a limit of about 1° in 20 diameters for general use).
From power to torque
Shafts are specified by the power they carry: T = 9.549 × P ÷ n (N·m, P in watts, n in rpm). At 5 000 W and 1 000 rpm the torque is 47.75 N·m. With an allowable shear stress of 60 MPa the required diameter is d = (16 T ÷ (π τ))1/3 = (16 × 47 745 ÷ (π × 60))1/3 = 15.9 mm (round up to 25 or 30 mm to allow for keyways, bending and shock). The same shaft at half the speed has twice the torque and needs 26 percent more diameter.
Hollow shafts and non-circular sections
Worked example: solid against hollow
A hollow shaft of outside diameter 40 mm and bore 28 mm has J = π (40⁴ − 28⁴) ÷ 32 = 190 984 mm⁴, more than the solid 30 mm shaft (79 522 mm⁴), with a similar cross-section area (641 mm² against 707 mm²). Under the same torque: τ = T (D ÷ 2) ÷ J = 150 000 × 20 ÷ 190 984 = 15.7 MPa (against 28.3 MPa) and the twist is 0.34° (against 0.81°): 44 percent lower stress for the same weight, because material is moved to where the shear stress is greatest.
- Square and rectangular solid bars are much less efficient in torsion: the shear stress is highest at the middle of the sides and zero at the corners. Use handbook formulas (for a square of side a: τmax ≈ 4.8 T ÷ a³ and θ = 7.1 T L ÷ (G a⁴)) with their coefficients.
- Keyways and splines raise the local stress (see Shafts, Keys and Couplings).
- Stepped shafts have stress concentrations at the shoulders; fillet radii reduce them.
Thin-walled tubes: closed against open sections
A thin-walled closed tube carries torque by a shear flow around the loop: τ = T ÷ (2 Am t) and Jeff = 4 Am² t ÷ p, with Am the area enclosed by the wall mid-line and p the mid-line perimeter. An open thin-walled section (a channel, an angle, a slit tube) carries torque far less efficiently: J ≈ ⅓ Σ b t³, which collapses with the cube of the wall thickness.
Worked example: a tube against the same tube slit open
A 40 × 40 × 3 mm square tube: the mid-line is 37 mm square, so Am = 1 369 mm² and p = 148 mm. Closed: J = 4 × 1 369² × 3 ÷ 148 = 151 959 mm⁴. Slit along one side (open), J = ⅓ × 148 × 3³ = 1 332 mm⁴. The closed tube is 114 times stiffer in torsion, which is why closed sections are used for anything that twists (see Section Properties and Structural Profiles).
Direct shear and bearing
Direct (transverse) shear stress appears in bolts, pins, rivets, welds, keys and the webs of beams (see Stress, Strain and Safety Factors and Bolted and Welded Joints). It is usually taken as τ = V ÷ A over the shear area, a good average. The strength in shear of a ductile metal is about 0.577 × Sy (von Mises) or 0.5 × Sy (Tresca) and the ultimate shear strength about 0.6 to 0.75 × Su. Bearing (contact) stress on holes and pins adds a check of its own.
Combined loads and the von Mises criterion
Most real parts see several loads at once: a shaft in bending and torsion, a bolt in tension and shear, a pressure vessel in hoop and axial stress. Their stresses do not simply add: the material yields according to a combination. The von Mises (distortion energy) criterion, which fits ductile metals best, says it yields when σ′ = √(σ² + 3 τ²) ≥ Sy for a normal stress σ and a shear stress τ (more generally σ′ = √(σ1² − σ1 σ2 + σ2²) for principal stresses). The maximum shear (Tresca) criterion, σ′ = √(σ² + 4 τ²), is slightly more conservative. Compare the equivalent stress, not the individual stresses, with the yield strength divided by the safety factor.
Worked example: a shaft in bending and torsion
- A 32 mm shaft carries M = 120 N·m and T = 90 N·m. Bending stress σ = 32 M ÷ (π d³) = 37.3 MPa; torsional shear τ = 16 T ÷ (π d³) = 14.0 MPa.
- Von Mises: σ′ = √(37.3² + 3 × 14.0²) = 44.5 MPa; Tresca: √(37.3² + 4 × 14.0²) = 46.6 MPa.
- For Sy = 350 MPa the safety factor is 7.87 (von Mises) or 7.51 (Tresca). The simple sum σ + τ would be 51.3 MPa: wrong (too conservative for the torsion).
- The static check is the first step: a rotating shaft must also be checked for fatigue (see Fatigue and Stress Concentration).
A worked application: the helical spring
A coil spring is a torsion problem: when it is compressed the wire is twisted. The shear stress in the wire is τ = Kw × 8 F D ÷ (π d³), with F the force, D the mean coil diameter, d the wire diameter and Kw = (4C − 1) ÷ (4C − 4) + 0.615 ÷ C the Wahl factor (C = D ÷ d, the spring index) for the curvature and direct shear. The spring rate is k = G d⁴ ÷ (8 D³ Na) for Na active coils.
- A spring of mean diameter 20 mm and wire 2 mm (C = 10) carries 50 N: Kw = 1.145 and τ = 1.145 × 8 × 50 × 20 ÷ (π × 2³) = 364 MPa: within the range for music wire (shear yield about 800 to 1 000 MPa for static loads).
- With 8 active coils the rate is k = 80 000 × 2⁴ ÷ (8 × 20³ × 8) = 2.50 N/mm, so 50 N compresses it by 20.0 mm.
Design checklist
- Find the torque from the power and the speed, with the peak and shock torque (see Sizing a Motion Axis).
- Size for strength: d from τ = 16 T ÷ (π d³) (or the hollow formula), with the safety factor.
- Check the stiffness: twist angle against the limit; torsional vibration for a drive with large inertias.
- Combine with bending, axial and shear with the von Mises equivalent stress.
- Allow for stress concentrations at keyways, holes and shoulders, and check fatigue.
- Prefer closed, round and hollow sections for torsion; avoid open thin sections.
- Use the right modulus: G for torsion, E for bending; plastics and aluminum twist several times as much as steel.
Related
- Structures and Strength of Materials: the overview of this guide.
- Stress, Strain and Safety Factors and Beam Bending and Deflection: shear, normal stress and bending.
- Fatigue and Stress Concentration: shafts under cyclic loads.
- Shafts, Keys and Couplings: the shaft design procedure and keys.
- Worked Design Checks: combined-load checks.
FAQ
Why is the shear stress zero at the center of a shaft?
By symmetry the material at the axis does not move relative to its neighbors when the shaft twists: it only rotates. The relative displacement and therefore the strain and the shear stress rise in proportion to the distance from the axis.
Why do hollow shafts transmit more torque per weight?
The material near the axis is under little stress and so wasted; removing it costs little strength and saves weight, while the outer material has a larger lever arm for the same shear stress.
What is the difference between von Mises and Tresca?
Both predict yielding under combined stress. Tresca uses the maximum shear stress and is simpler and slightly conservative; von Mises uses the distortion energy and matches tests of ductile metals better. For pure shear they differ by 15 percent in the predicted yield (0.577 against 0.5 of Sy).
Can I just add bending and torsion stresses?
No. The normal stress from bending and the shear stress from torsion act in different directions and combine as σ′ = √(σ² + 3τ²) (von Mises). Adding them directly is too conservative.
References
- Beer, F. P., Johnston, E. R. and DeWolf, J. T., Mechanics of Materials, McGraw-Hill: torsion, thin-walled tubes, combined loading.
- Shigley, J. E. and Mischke, C. R., Mechanical Engineering Design, McGraw-Hill: shafts, springs and failure theories (von Mises).
- Young, W. C. and Budynas, R. G., Roark’s Formulas for Stress and Strain: torsion of non-circular sections.
Last reviewed: October 2026. Formulas assume linear elastic behavior; check the code and use the real stress concentrations for critical parts.