Design of shafts for combined bending and torsion

Why it matters

Gearbox shafts, axle shafts, camshafts and countershafts carry torque from the engine and, at the same time, are bent by gear tooth forces, belt pulls and their own supports. A shaft sized for torque alone will be undersized, often by a wide margin. Combining bending and torsion correctly is one of the most frequently examined design calculations.

Key ideas

Loads on a shaft: torque T from the power transmitted, bending moment M from transverse forces (gear tangential and radial forces, belt tensions, weights) found by drawing bending moment diagrams in two planes and combining them as M = √(M_H² + M_V²), and sometimes an axial force (helical gears). The critical section is where the combination is largest, usually under a gear or pulley or at a shoulder.

Torque from power: T = 60P/(2πN), with P in W and N in rpm.

Stresses at the surface of a solid shaft: bending σ = 32M/(πd³), torsion τ = 16T/(πd³). For a hollow shaft with k = d_i/d_o, multiply the denominators by (1 − k⁴).

Combining them (static or steady loading):

• Maximum shear stress theory (ductile, most common in Indian syllabi): equivalent torque T_e = √(M² + T²), and τ_max = 16T_e/(πd³).
• Maximum principal stress theory (brittle): equivalent bending moment M_e = ½[M + √(M² + T²)], and σ_max = 32M_e/(πd³).
• Distortion energy theory: σ_e = (32/(πd³))·√(M² + 0.75·T²). Design by both T_e and M_e and take the larger diameter when the material is not clearly ductile or brittle.

Shock and fatigue factors: for suddenly applied or fluctuating loads, the ASME/textbook method multiplies the moments by combined shock and fatigue factors: T_e = √((K_m·M)² + (K_t·T)²). K_m and K_t depend on whether the load is gradual, sudden with minor shock, or heavy shock, and on whether the shaft rotates; take them from your design data book. A full fatigue check uses the Goodman or Soderberg approach of the fatigue topic.

Stiffness: shafts are also checked for angle of twist θ = TL/(GJ) (often limited to about 0.25° per metre of length for line shafts, but take the limit from the application) and for lateral deflection at gears.

Hollow shafts are lighter for the same strength because material near the axis carries little stress; propeller shafts are a typical example.

Formulas

T = 60·P / (2π·N) σ_b = 32M / (π·d³), τ = 16T / (π·d³) T_e = √(M² + T²), τ_max = 16·T_e / (π·d³) M_e = ½·[M + √(M² + T²)], σ_max = 32·M_e / (π·d³) σ_e = (32 / (π·d³))·√(M² + 0.75·T²) T_e = √((K_m·M)² + (K_t·T)²) Hollow: τ_max = 16·T_e / (π·d_o³·(1 − k⁴)), k = d_i / d_o θ = T·L / (G·J), J = π·d⁴ / 32

Symbols: P = power (W); N = speed (rpm); T = torque (N·mm); M = resultant bending moment (N·mm); T_e = equivalent twisting moment, M_e = equivalent bending moment (N·mm); d, d_o, d_i = diameters (mm); σ_b, σ_max, σ_e = normal and equivalent stresses (MPa); τ, τ_max = shear stresses (MPa); K_m, K_t = combined shock and fatigue factors for bending and torsion (–); G = shear modulus (MPa); L = length (mm); J = polar second moment of area (mm⁴); θ = angle of twist (rad).

Worked examples

Example 1 (standard). A countershaft transmits 20 kW at 500 rpm. The resultant bending moment at the critical section is 300 N·m. The material has permissible shear stress 50 MPa and permissible tensile stress 80 MPa. Find a suitable diameter.

1. T = 60 × 20 000 / (2π × 500) = 381.97 N·m.
2. T_e = √(300² + 381.97²) = 485.70 N·m.
3. By shear: d³ = 16T_e/(πτ) = 16 × 485 700 / (π × 50), d = 36.71 mm.
4. M_e = ½ × (300 + 485.70) = 392.85 N·m.
5. By normal stress: d³ = 32M_e/(πσ) = 32 × 392 850 / (π × 80), d = 36.85 mm.
6. Take the larger and round up to a standard size: d = 40 mm.

Example 2 (GATE level). A hollow shaft with d_i = 0.6·d_o carries M = 1.2 kN·m and T = 1.8 kN·m. The load is suddenly applied with minor shock, for which the data book gives K_m = 1.5 and K_t = 1.0. Permissible shear stress = 60 MPa. Find d_o and compare with a solid shaft.

1. T_e = √((1.5 × 1200)² + (1.0 × 1800)²) = 2545.6 N·m.
2. 1 − k⁴ = 1 − 0.6⁴ = 0.8704.
3. d_o³ = 16 × 2 545 600 / (π × 60 × 0.8704), so d_o = 62.85 mm. Choose d_o = 65 mm, d_i = 39 mm.
4. A solid shaft needs d³ = 16 × 2 545 600 / (π × 60), d = 60.0 mm.
5. Comparing the minimum sections, the hollow shaft area is (1 − 0.36) × 62.85² / 60.0² = 0.70 of the solid one, so it is about 30 % lighter for the same stress.

Common mistakes

• Calling √(M² + T²) the equivalent bending moment. It is the equivalent twisting moment T_e; M_e = ½[M + T_e].
• Adding horizontal and vertical bending moments algebraically instead of as √(M_H² + M_V²).
• Using N·m with d in mm. Convert to N·mm.
• Forgetting (1 − k⁴) for hollow shafts, or using (1 − k²).
• Using the radius in place of the diameter in 16T/(πd³).
• Designing for strength only and never checking twist or deflection.

For GATE ME

Expect: shaft diameter from power, speed and bending moment using T_e or M_e; comparison of the theories; ratio of strengths or weights of hollow and solid shafts; and the equivalent stress for a given M and T. Practise drawing bending moment diagrams in two planes for a shaft with a gear and a pulley.

Quick check

1. M = 3 kN·m, T = 4 kN·m. Find T_e and M_e.
2. How does the shear stress in a solid shaft change if the diameter is doubled for the same T?
3. Torque transmitted by 15 kW at 1000 rpm?
4. For k = 0.5, by what factor is the hollow shaft's torsional strength reduced compared with a solid shaft of the same outer diameter?
5. Which theory gives M_e?

Answers: 1. 5 kN·m and 4 kN·m. 2. It falls to one-eighth. 3. 143.2 N·m. 4. 1 − 0.5⁴ = 0.9375. 5. Maximum principal (normal) stress theory.