Shaft Design Calculation Explained: Bending + Torsion

A power-transmission shaft is one of the most common load-carrying members in machine design, and one of the most commonly mis-sized. It looks like a simple round bar, but it almost never sees a single, simple load. A shaft carrying a gear or a pulley feels torsion from the torque it transmits and bending from the radial reactions of the very same gear teeth or belt, all while spinning — which turns that steady bending into a fully reversed fatigue load. Get the combination wrong and the shaft either snaps at a keyway after a few million cycles or is grotesquely over-built. This article walks through the standard ASME approach to sizing a solid or hollow shaft for combined bending and torsion, how to choose the allowable shear stress, where stress concentrations bite, and a fully worked numeric example you can check against the shaft design calculator.

What loads does a shaft actually see?

Before reaching for a formula, it pays to enumerate the loads, because the governing equation only handles two of them directly. A typical line shaft in a gearbox, pump, or conveyor experiences:

• Torque (T). The whole point of the shaft — the twisting moment transmitted from the prime mover to the driven element. Torque produces a torsional shear stress τ = T·r/J. The torsion calculator handles this pure-torsion component on its own.
• Bending moment (M).Gears push on each other with a radial separating force; belts and chains pull with tension; the shaft’s own weight and any overhung mass add more. These transverse forces, reacted at the bearings, create a bending-moment diagram along the shaft. Bending produces a normal stress σ = M·c/I.
• Axial load (F). Helical and bevel gears, and any thrust from the driven machine, add an axial force. On most general shafts this term is small and is often neglected or rolled into the factor of safety; on screw and propeller shafts it dominates and must be carried explicitly.
• Transverse shear (V). Real but usually negligible at the critical section, because the maximum bending moment and the maximum transverse shear almost never coincide on a beam.

The crucial physical fact for a rotating shaft is that a constant bending moment is felt by the material as a completely reversed alternating stress: a given fiber on the surface is in tension at the top of its rotation and in compression half a turn later, swinging from +σ to −σ every revolution. Torque, by contrast, is usually steady or only mildly fluctuating. That asymmetry is why bending and torsion are weighted differently in a full fatigue analysis, and why shafts are designed against fatigue rather than just static yield.

The governing equation

For a ductile shaft under combined bending and torsion, the standard sizing equation comes from the maximum-shear-stress (Tresca) or distortion-energy (von Mises) failure theory applied to the combined stress state at the surface. The ASME design code for transmission shafting (ASME B106.1M) packages it into a single, memorable form. For a solid circular shaft the minimum diameter is:

• d³ = (16 / (π · τ_allow)) · √(M² + T²)

where d is the shaft diameter, τ_allow is the allowable shear stress, M is the bending moment, and T is the torque (all in consistent units — meters, newton-meters, and pascals, or millimeters, newton-millimeters, and MPa). The term √(M² + T²) is the heart of the method. Bending and torsion do notadd linearly; they combine like the legs of a right triangle, because they act on the cross-section in perpendicular “directions” in stress space. Adding them as M + T overstates the load and oversizes the shaft; using just the larger of the two understates it.

A more complete ASME form inserts shock-and-fatigue factors that amplify each moment before combining them:

• d³ = (16 / (π · τ_allow)) · √((K_b·M)² + (K_t·T)²)

Here K_b and K_t are combined shock-and-fatigue factors applied to the bending and torsional moments respectively. Representative values are 1.5 for K_b and 1.0 for K_tunder gradually applied steady loads, rising to 2.0–3.0 for shafts subjected to suddenly applied or heavy shock loads such as those in crushers, presses, and reciprocating machinery. Setting both factors to 1.0 recovers the basic combined-load formula above.

For a hollow circular shaft with outer diameter d_o and inner diameter d_i, define the bore ratio c = d_i/d_o. The bore removes material near the neutral axis (which carries little stress) but also near the surface (which carries the most), and the net effect is a (1 − c⁴) factor in the section modulus. The sizing equation becomes:

• d_o³ = (16 / (π · τ_allow· (1 − c⁴))) · √(M² + T²)

Because (1 − c⁴) is less than one, a hollow shaft of a given outer diameter is weaker than a solid one of the same outer diameter, so the outer diameter must grow to compensate. The payoff is mass: the material removed from the bore contributed little strength but plenty of weight, so a hollow shaft delivers nearly the same strength and torsional stiffness at a substantially lower mass. That is why aerospace, automotive, and high-speed drivelines favor hollow shafting despite the higher manufacturing cost.

Choosing the allowable shear stress

The formula is only as good as the τ_allowyou feed it. The allowable shear stress is derived from the material’s strength and a factor of safety, and the ASME shafting code gives explicit guidance. For a shaft with no keyway, the allowable shear stress is taken as the smaller of:

• 0.30 × S_y (30% of the yield strength), or
• 0.18 × S_ut (18% of the ultimate tensile strength).

The smaller of the two governs, which protects both ductile (yield-led) and harder, higher-strength (ultimate-led) steels. For a typical carbon steel shaft with S_y = 400 MPa, the yield-based limit is 0.30 × 400 = 120 MPa; if S_ut were 600 MPa the ultimate-based limit would be 0.18 × 600 = 108 MPa, so 108 MPa would govern. Many designers round to a round 100 MPa for a conventional medium-carbon shaft, which is the value used in the worked example below. Whatever value you adopt, be aware that allowable shear is expressed in megapascals — see the megapascal unit reference for the N/mm² identity that makes these numbers fall out cleanly when you work in millimeters and newtons.

Keyways and the 25% penalty. A keyway is a stress raiser, and the ASME code accounts for it bluntly: reduce the allowable shear stress by 25% wherever the shaft is keyed. The 120 MPa unkeyed allowable above drops to 0.75 × 120 = 90 MPa at a keyseat. Since the critical section of a shaft is very often exactly where a gear or coupling is keyed on, this penalty frequently governs the diameter.

The factor of safety is already embedded in those coefficients (roughly a factor of 2 to 3 against yield), but you can and should layer an explicit factor of safety on top for uncertain loads, critical service, or poor inspection access. Relating the computed working stress back to the material’s yield or ultimate strength — the basis of every safety-factor decision — is exactly what the factor of safety calculator is for.

Stress concentrations: where shafts really fail

The √(M² + T²) formula computes a nominal stress on a smooth, prismatic section. Real shafts are full of geometric features that amplify the local stress well above nominal, and fatigue cracks nucleate at exactly these points:

• Shoulder fillets. A step change in diameter (needed to locate bearings and gears axially) concentrates stress at the fillet radius. A sharp fillet can raise the local stress by a factor K_t of 2 to 3; a generous radius brings it down toward 1.5. The single cheapest fatigue improvement you can make to a shaft is to increase the fillet radius at every shoulder.
• Keyways. The sharp corners of a profiled (end-milled) keyway are severe raisers, with bending K_t around 1.6 and torsional K_taround 1.4 or higher. This is the physical reason behind the code’s 25% allowable-stress reduction.
• Grooves and undercuts. Retaining-ring grooves, thread relief undercuts, and oil holes all concentrate stress and should be radiused and located away from high-moment sections where possible.
• Press-fit and bearing seats. The edge of an interference fit creates a fretting-fatigue hotspot even with no geometric notch.

In a rigorous fatigue analysis these are handled with theoretical concentration factors K_t(geometric, from Peterson’s charts) reduced by a notch-sensitivity factor q to a fatigue concentration factor K_f, which then de-rates the material’s endurance limit. For a first-pass size, the shock-and-fatigue factors K_b and K_t in the ASME formula and the keyway de-rating capture the dominant effects.

A note on fatigue

Because the bending stress on a rotating shaft is fully reversed, shaft design is fundamentally a fatigueproblem, not a static one. A complete design follows a fatigue criterion — most commonly the ASME-elliptic, Goodman, or Soderberg line — that plots the alternating stress (from bending) against the mean stress (from steady torque) and checks the operating point against a failure envelope built from the material’s corrected endurance limit S_e. The endurance limit itself is the textbook S′_e ≈ 0.5·S_ut for steel, knocked down by Marin factors for surface finish, size, loading type, temperature, and reliability, and then again by the stress-concentration factor K_f. The simplified ASME diameter formula used here is a sound and widely taught starting point that bakes these effects into the allowable shear and the K factors; for critical or weight-sensitive shafts, follow it with a full ASME-elliptic fatigue check at every notch.

Worked example: sizing a solid transmission shaft

A solid power-transmission shaft must carry a bending moment M = 500 N·m and a torque T = 800 N·m. The allowable shear stress is τ_allow = 100 MPa. Find the minimum diameter, then check the actual stresses and factor of safety if a 50 mm stock shaft is selected. (These are the exact inputs verified in the shaft design calculator, so you can reproduce every number.)

Step 1 — combine the loads. Form the resultant of bending and torsion:

• √(M² + T²) = √(500² + 800²) = √(250,000 + 640,000) = √890,000 = 943.40 N·m

Note this is well below the linear sum M + T = 1,300 N·m and above the larger single load of 800 N·m, exactly as the right-triangle picture predicts.

Step 2 — solve the ASME diameter formula. With K_b = K_t = 1 and τ_allow = 100 MPa = 100 × 10⁶ Pa:

• 16 / (π · τ_allow) = 16 / (π × 100×10⁶) = 5.0930×10⁻⁸
• d³ = 5.0930×10⁻⁸ × 943.40 = 4.8047×10⁻⁵ m³
• d_min = (4.8047×10⁻⁵)^1/3 = 0.03635 m = 36.35 mm

The minimum required diameter is 36.35 mm, so a 50 mm stock shaft comfortably clears the requirement. Now confirm the margin at the chosen size.

Step 3 — section properties at d = 50 mm. For a solid circle of diameter d = 0.05 m, with outer radius c = d/2 = 0.025 m:

• Polar second moment J = π·d⁴ / 32 = 6.1359×10⁻⁷ m⁴
• Bending second moment I = π·d⁴ / 64 = 3.0680×10⁻⁷ m⁴

Step 4 — actual stresses. The torsional shear stress and bending stress at the surface are:

• τ = T·c / J = 800 × 0.025 / 6.1359×10⁻⁷ = 32.59×10⁶ Pa = 32.59 MPa
• σ_b = M·c / I = 500 × 0.025 / 3.0680×10⁻⁷ = 40.74 MPa

Step 5 — von Mises check and factor of safety. Combine the bending and shear into the von Mises equivalent stress for a yield check, then take the shear-based factor of safety against the allowable:

• σ_VM = √(σ_b² + 3τ²) = √(40.74² + 3×32.59²) = √(1659.7 + 3186.3) = √4846.0 = 69.62 MPa
• FOS = τ_allow / τ = 100 / 32.59 = 3.07

The 50 mm shaft carries an actual torsional shear of 32.59 MPa against a 100 MPa allowable — a factor of safety of about 3.07 — and a von Mises equivalent stress of 69.62 MPa, comfortably below the yield strength of any reasonable shaft steel. This is a healthy, well-proportioned design: a factor of safety of 2 or more is the conventional green band for general machinery. Had a keyway been present at this section, the allowable would drop to 75 MPa and the margin would shrink accordingly, which is why the keyseat is usually the section you check first.

For readers working in US customary units: the 800 N·m torque is about 590 ft·lbf, and the 100 MPa allowable shear is roughly 14,500 psi. The diameter comes out the same 36.35 mm (≈ 1.43 in) either way — the physics does not care about the unit system, but the arithmetic is far cleaner in SI.

A quick word on hollow shafts

Suppose the same combined load had to be carried by a hollow shaft with a fixed bore ratio c = d_i/d_o= 0.6. The hollow factor is 1 − c⁴ = 1 − 0.6⁴ = 1 − 0.1296 = 0.8704, so the outer diameter is scaled up by (1/0.8704)^1/3 ≈ 1.047 relative to the equivalent solid shaft. The outer diameter grows by under 5%, the inner bore removes a large fraction of the cross-sectional area, and the result is a markedly lighter shaft of essentially the same strength. That is the entire case for hollow shafting — a few percent more outer diameter buys a large reduction in mass.

Design checklist beyond strength

Sizing for combined bending and torsion gives you the strength-driven diameter, but a shaft is rarely governed by stress alone. Before finalizing, also check:

• Lateral deflection and slope. Excessive deflection misaligns gears and overloads bearings. Typical limits are L/1000 to L/500 of the span for deflection, and around 0.5° of slope at a bearing. Stiffness frequently governs over strength, especially for long, lightly loaded shafts.
• Critical (whirling) speed. Keep the operating speed comfortably below the first lateral natural frequency — a common rule is below 75% of the first critical speed — to avoid resonant whirl.
• Torsional rigidity. Angle of twist θ = T·L / (G·J) must stay within limits for positioning and timing accuracy; the torsion calculator gives this directly.
• Bearing and seal seats, fits, and assembly. Standard stock diameters, bearing bore sizes, retaining-ring grooves, and key stock all constrain the final dimension. Round the calculated minimum up to the next convenient stock size — as we did from 36.35 mm to 50 mm — and re-check the stresses at that size.

The takeaway

Shaft design under combined bending and torsion reduces to one durable idea: the two moments combine as √(M² + T²), and that resultant divided by an allowable shear stress sets the cube of the diameter via d³ = (16 / (π·τ_allow)) · √(M² + T²), with a (1 − c⁴) factor for a bore. Choose τ_allow as the smaller of 0.30·S_y and 0.18·S_ut, drop it 25% at any keyway, layer on shock and fatigue factors for rough service, and remember that the rotating shaft is really a fatigue problem with the bending stress fully reversed. Stress concentrations at fillets, keyways, and grooves — not the smooth nominal section — are where shafts actually crack, so radius them generously. Size for strength first, then verify deflection, critical speed, and torsional rigidity. Run the worked example above through the shaft design calculator and you should land on the same 36.35 mm minimum and 3.07 factor of safety — a sign the method, and your arithmetic, are sound.