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By Luqman Ismat © 2025

Engineering API Solutions • Hydraulics Calculations • Thermal Systems • Pump Design

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Shaft Stress, Twist and Power

Compare circular sections without confusing a lighter shaft with a stronger one.

Source review: September 8, 2026

Define the section and load

The shaft torsion calculator accepts outside and inside diameters, length, shear modulus, torque and rotational speed. Use metres for dimensions and pascals for shear modulus: 50 mm is 0.05 m; 80 GPa is 80,000,000,000 Pa. A zero bore selects a solid section.

For a uniform circular shaft, J = π(Do⁴ − Di⁴)/32, outer stress magnitude = |T|Do/(2J), and relative twist = TL/(GJ). J measures area distribution, in m⁴. The derivation assumes a concentric section, isotropic linear elasticity and small shear strains. See MIT, equations 12–14.

Two different hollow-shaft comparisons

Original calculated example: hold length at 1 m, shear modulus at 80 GPa and torque at 500 N·m. These are illustrative inputs, not a material grade or design approval. All three sections use the same assumed material.

Dimensions are outside / inside diameter. Values are calculated from the live tool's model.
SectionArea (mm²)Stress (MPa)Twist (°)
Solid, 40 mm1256.6439.7891.4248
Hollow, 40 / 24 mm804.2545.7131.6370
Hollow, 50 / 30 mm1256.6423.4050.6705

Fixed outside diameter: boring the 40 mm shaft to 24 mm removes 36% of its area. Stress and twist both rise by 14.89%. Removing material has not made this shaft stiffer.

Equal material area: the 50 / 30 mm tube has the same area as the 40 mm solid shaft because 50² − 30² = 40². At equal length and density their masses match. Its larger envelope reduces stress by 41.18% and twist by 52.94%. This comparison requires space for a larger diameter.

Try doubling only the length in the calculator: twist doubles and nominal stress stays fixed. Doubling only shear modulus halves twist and leaves stress fixed. These checks help expose an incorrect input or unit conversion.

Power and elastic energy are different outputs

Rotational power is P = Tω, with ω = 2πn/60 for speed n in rpm. OpenStax, section 10.8 derives this from the rate of rotational work. Choose one positive axis for both torque and speed; opposite signs give negative power.

At 1500 rpm and 500 N·m, all three examples transmit 78.540 kW. Changing geometry alone does not change this product. At zero speed the model reports zero power even with nonzero static torque, stress and twist.

The 50 / 30 mm shaft stores 2.926 J of elastic energy, calculated as Tθ/2 for gradual linear elastic loading. This is not its rotational kinetic energy or energy transmitted per second. The reported 0.6705° is the relative rotation between its ends, not its accumulated operating rotation. See MIT's elastic-energy derivation, equation 16.

Where this calculation stops

Nominal stress is not a safe torque rating. The tool omits keyways, shoulders, fatigue, yielding, bending, instability and machinery losses. Noncircular sections require a different torsion solution. Material allowables and the actual loading history remain separate inputs to a design assessment.

For axial loading use the axial bar calculator; for its implemented rectangular beam cases use the beam calculator. Their individual results do not form a combined-stress or fatigue assessment.

References

  • MIT OpenCourseWare: Shear and Torsion, circular-section derivation and elastic energyAccessed 2026-09-08
  • OpenStax University Physics: Work and Power for Rotational MotionAccessed 2026-09-08