Design of shafts for combined bending and torsion

Sizing solid and hollow shafts under combined bending and torsion with equivalent torque and equivalent bending moment, ASME shock factors and rigidity checks.

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Why it matters

Shafts carry every motor's torque to pulleys, gears, couplings and rollers, and the radial forces from those elements bend the shaft at the same time. A shaft that is too thin breaks at a keyway or shoulder or twists and deflects enough to ruin gear meshing and bearing life; one that is too thick wastes mass and inertia. Combined bending and torsion is the core shaft-sizing calculation in every design course and exam.

Key ideas

Loads on a shaft. Torque T comes from the power transmitted. Bending moment M comes from gear tooth forces, belt tensions, pulley and gear weights acting between bearings. Draw bending-moment diagrams in the horizontal and vertical planes and combine them at each section: M = √(M_H² + M_V²). The critical section is where the combination of M and T (and any stress raiser) is worst - usually under a gear or pulley.

Stresses at the surface of a solid shaft: bending σb = 32M/(π·d³) and torsional shear τ = 16T/(π·d³). Both are maximum at the outer fibre, so the surface point is in plane stress with σ = σb and τ.

Failure theories turned into equivalent moments:

  • Maximum shear stress theory (ductile materials): τmax = √((σb/2)² + τ²) = 16/(π·d³) · √(M² + T²). The quantity Te = √(M² + T²) is the equivalent torque - the pure torque that gives the same τmax.
  • Maximum principal stress theory (brittle materials): σmax = σb/2 + √((σb/2)² + τ²) = 32/(π·d³) · ½[M + √(M² + T²)]. The quantity Me = ½[M + √(M² + T²)] is the equivalent bending moment. Many textbook problems compute d from both and take the larger.

ASME code approach (widely used in Indian syllabi): multiply M and T by combined shock and fatigue factors Kb (often written Km) and Kt, taken from a table depending on whether the load is gradually applied, sudden-minor shock or heavy shock, and whether the shaft rotates. Permissible shear stress is the smaller of 0.30·σyt and 0.18·σut, reduced by 25% if a keyway is present. These values come from the code - take them from your data book.

Hollow shafts. With k = di/do, the section modulus is multiplied by (1 − k⁴). A hollow shaft is lighter for the same strength and stiffness because the material near the axis carries little stress; used in robot joints (cables pass through) and long drive shafts.

Rigidity. Strength is not enough: limit the angle of twist (commonly about 0.25° per metre for line shafts - check your data book) and lateral deflection at gears and bearings. θ = T·L/(G·J).

Fatigue. A rotating shaft under steady M sees fully reversed bending stress at every point, so for accurate design the stresses are combined with fatigue criteria (endurance limit, Kf at keyways and shoulders, Goodman/Soderberg). The ASME factors above are the simplified version of this.

Formulas

T = 60·P / (2π·N)

  • T: torque (N·m); P: power (W); N: speed (rpm).

σb = 32·M / (π·d³) · τ = 16·T / (π·d³)

  • M: bending moment (N·m); d: diameter (m).

Te = √(M² + T²) · d³ = 16·Te / (π·τ_allow) (max shear stress theory) Me = ½·[M + √(M² + T²)] · d³ = 32·Me / (π·σ_allow) (max principal stress theory)

Te = √((Kb·M)² + (Kt·T)²) (ASME code)

  • Kb, Kt: combined shock and fatigue factors for bending and torsion (from the code table).

Hollow: τ = 16·T / (π·do³·(1 − k⁴)), k = di / do

θ = T·L / (G·J), J = π·d⁴ / 32

  • θ: angle of twist (rad); L: length (m); G: shear modulus (Pa, about 80 GPa for steel).

M = √(M_H² + M_V²) (resultant of moments in two perpendicular planes)

Worked examples

Example 1 (standard). A shaft transmits 20 kW at 300 rpm. At the critical section the bending moment is 800 N·m. τ_allow = 50 MPa, σ_allow = 100 MPa. Find the diameter by both theories.

  1. Torque: T = 60P/(2πN) = 60 × 20,000 / (2π × 300) = 636.6 N·m.
  2. Equivalent torque: Te = √(800² + 636.6²) = 1022.4 N·m.
  3. Max shear theory: d = (16 × 1022.4 × 10³ / (π × 50))^(1/3) = 47.05 mm.
  4. Equivalent bending moment: Me = ½(800 + 1022.4) = 911.2 N·m.
  5. Max principal stress theory: d = (32 × 911.2 × 10³ / (π × 100))^(1/3) = 45.28 mm.
  6. Larger value governs: 47.05 mm → d = 50 mm (next standard size).

Example 2 (GATE level). A rotating shaft is supported on bearings 600 mm apart and carries a pulley at mid-span with a total downward load of 3 kN. It transmits a torque of 400 N·m. Using ASME factors Kb = 1.5 and Kt = 1.0 and τ_allow = 40 MPa, find (a) the solid shaft diameter, (b) the outer diameter of a hollow shaft with k = 0.6.

  1. Bending moment at mid-span: M = W·L/4 = 3000 × 0.6 / 4 = 450 N·m.
  2. Te = √((1.5 × 450)² + (1.0 × 400)²) = √(675² + 400²) = 784.6 N·m.
  3. Solid: d = (16 × 784.6 × 10³ / (π × 40))^(1/3) = 46.40 mm → 50 mm.
  4. Hollow: do = d_solid / (1 − k⁴)^(1/3) = 46.40 / (1 − 0.1296)^(1/3) = 46.40 / 0.9548 = 48.60 mm, di = 0.6 × 48.60 = 29.16 mm.
  5. Mass ratio (hollow/solid, same length) = (48.60² − 29.16²)/46.40² = 0.70, i.e. about 30% lighter before rounding to standard sizes.

Common mistakes

  • Using T from power with N in rpm but forgetting the factor 60/(2π).
  • Adding horizontal and vertical bending moments arithmetically instead of as √(M_H² + M_V²).
  • Using 32 for torsion or 16 for bending (σb uses 32, τ uses 16).
  • Applying Te with σ_allow, or Me with τ_allow.
  • Forgetting shock/fatigue factors or the 25% keyway reduction when the problem calls for the ASME code.
  • Choosing the diameter on strength alone and never checking twist or deflection.

For GATE ME

Common question types: diameter of a solid or hollow shaft from Te or Me, the ratio of diameters or weights of hollow and solid shafts of equal strength, equivalent torque given M and T, and torque from power and speed. Practise drawing bending-moment diagrams for gears and pulleys between bearings, since finding M at the critical section is usually the longest step.

Quick check

  1. M = 300 N·m, T = 400 N·m. Find Te and Me.
  2. Which equivalent moment goes with brittle materials?
  3. Torque transmitted by 15 kW at 1440 rpm?
  4. For equal strength, is a hollow shaft lighter or heavier than a solid one? Answers: 1. Te = 500 N·m; Me = 400 N·m. 2. Equivalent bending moment Me (max principal stress theory). 3. 99.5 N·m. 4. Lighter.

Interview questions

All Machine Design interview questions

Try answering each one aloud before you open it.

  1. 1.What is the significance of designing shafts for combined bending and torsion?Concept

    Designing shafts for combined bending and torsion is crucial because shafts in machinery often experience both types of loads simultaneously. This ensures that the shaft can withstand the complex stress state without failure, providing reliability and safety in mechanical systems.

  2. 2.Explain the concept of equivalent bending moment in the context of shaft design.Concept

    The equivalent bending moment is a hypothetical bending moment that would produce the same maximum normal stress as the combined effect of actual bending and torsional moments. It simplifies the analysis by allowing engineers to use standard bending stress formulas to evaluate the shaft's strength.

  3. 3.How do you calculate the equivalent torque and equivalent bending moment for a shaft subjected to both bending and torsion?Concept

    For ductile shafts, the maximum shear stress theory gives the equivalent torque Te = √(M² + T²), and d³ = 16·Te/(π·τ_allow). For brittle shafts, the maximum principal stress theory gives the equivalent bending moment Me = ½[M + √(M² + T²)], and d³ = 32·Me/(π·σ_allow). In the ASME code approach, M and T are first multiplied by shock and fatigue factors Kb and Kt taken from a table.

  4. 4.Why is it important to consider both bending and torsion in shaft design?Application

    Considering both bending and torsion is important because shafts in real-world applications are rarely subjected to pure bending or pure torsion. Ignoring one of these stresses can lead to underestimating the actual stress state, potentially causing premature failure.

  5. 5.What happens if a shaft is designed only for bending and not for torsion?Application

    If a shaft is designed only for bending, it may not have sufficient strength to withstand torsional loads. This oversight can lead to excessive twisting, increased wear, or even catastrophic failure due to torsional stresses that were not accounted for.

  6. 6.In what scenarios would a hollow shaft be preferred over a solid shaft for combined loading?Application

    A hollow shaft is often preferred when weight reduction is critical, as it provides a better strength-to-weight ratio. Hollow shafts can also have higher torsional stiffness compared to solid shafts of the same weight, making them suitable for applications where both bending and torsion are significant.

  7. 7.How does the presence of keyways affect the design of shafts under combined loading?Application

    A keyway removes material and creates a sharp stress raiser at its root corners, which matters most in fatigue because a rotating shaft sees fully reversed bending. In simplified design the ASME code reduces the permissible shear stress by 25% when a keyway is present; in fatigue design a fatigue stress concentration factor Kf for the keyway is applied. Sled-runner keyways and generous root fillets give lower stress concentration than profile (end-milled) keyways.

  8. 8.A shaft carries a bending moment of 500 N·m and a torque of 300 N·m. Find the equivalent torque and the equivalent bending moment.Numerical

    Equivalent torque Te = √(M² + T²) = √(500² + 300²) = √340,000 = 583.1 N·m, used with the maximum shear stress theory. Equivalent bending moment Me = ½[M + Te] = ½(500 + 583.1) = 541.5 N·m, used with the maximum principal stress theory. The diameter then follows from d³ = 16·Te/(π·τ_allow) or d³ = 32·Me/(π·σ_allow).

  9. 9.A rotating shaft carries a bending moment of 200 N·m and a torque of 150 N·m. With ASME factors Kb = 1.5 and Kt = 1.0, find the equivalent torque.Numerical

    Te = √((Kb·M)² + (Kt·T)²) = √((1.5 × 200)² + (1.0 × 150)²) = √(300² + 150²) = √112,500 = 335.4 N·m. The diameter then follows from d³ = 16·Te/(π·τ_allow), with τ_allow reduced by 25% if the shaft has a keyway.

  10. 10.What are the typical materials used for shafts designed for combined bending and torsion, and why?Application

    Typical materials include alloy steels, carbon steels, and stainless steels due to their high strength, toughness, and fatigue resistance. These properties are essential for withstanding the complex stress states encountered in combined loading conditions.

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