Bending stress σb
-Torsional shear τ
-Von Mises stress σ'
-Safety factor (von Mises)
-Sy / σ'Safety factor (max shear)
-Sy / (2 τmax)Minimum outer diameter
-All results
| Axial stress σa | - |
| Maximum principal stress σ1 | - |
| Minimum principal stress σ2 | - |
| Maximum shear stress τmax | - |
| Yield strength used | - |
| Second moment of area I | - |
| Polar constant J | - |
| Angle of twist | - |
| Twist per unit length | - |
Method and formulas
The shaft is a round bar of outer diameter d and bore di (zero for a solid shaft). Stresses are evaluated at the outer surface, where bending and torsion are largest. These are standard strength-of-materials results that you can check by hand.
| Quantity | Formula |
|---|---|
| Second moment of area | I = π (d⁴ − di⁴) / 64 |
| Polar constant | J = π (d⁴ − di⁴) / 32 |
| Area | A = π (d² − di²) / 4 |
| Bending stress | σb = M (d/2) / I |
| Axial stress | σa = F / A |
| Torsional shear | τ = T (d/2) / J |
| Normal stress at the critical point | σ = σb + σa (bending and axial tension add on the tension side) |
| Principal stresses | σ1,2 = σ/2 ± √((σ/2)² + τ²) |
| Maximum shear | τmax = √((σ/2)² + τ²) |
| Von Mises | σ' = √(σ² + 3 τ²) |
| Safety factors | n = Sy / σ' (von Mises, distortion energy); n = Sy / (2 τmax) (Tresca) |
| Angle of twist | θ = T L / (G J) |
The minimum outer diameter is found by increasing d, keeping the bore-to-outer ratio you entered, until the von Mises safety factor equals your target. For a solid shaft without axial load this gives the closed form d = [32 n √(M² + 0.75 T²) / (π Sy)]1/3. Yield strengths are typical handbook values; use the certified value for your bar.
Worked example: 40 mm shaft, 500 N·m torque, 400 N·m bending
Solid 40 mm shaft. I = π 40⁴ / 64 = 125 664 mm⁴ and J = 251 327 mm⁴. σb = 400 000 N·mm × 20 / 125 664 = 63.7 MPa. τ = 500 000 × 20 / 251 327 = 39.8 MPa. σ' = √(63.7² + 3 × 39.8²) = 93.8 MPa. For AISI 1045 (Sy 530 MPa) the von Mises safety factor is 530 / 93.8 = 5.65. Over 0.5 m, with G = 79.3 GPa, the twist is 500 × 0.5 / (79.3×10⁹ × 2.513×10−7) = 0.0125 rad = 0.72°.
Questions
How do you calculate the stress in a shaft under torque and bending?
For a round shaft of outer diameter d and inner diameter di, the second moment of area is I = π(d⁴ − di⁴)/64 and the polar constant is J = π(d⁴ − di⁴)/32. Bending stress is σ = M·(d/2)/I, torsional shear stress is τ = T·(d/2)/J and axial stress is σa = F/A. The von Mises equivalent stress is σ' = √((σ + σa)² + 3τ²). Example: a 40 mm solid shaft carrying 500 N·m torque and 400 N·m bending gives σ = 63.7 MPa, τ = 39.8 MPa and σ' = 93.8 MPa, so a 1045 steel shaft (yield 530 MPa) has a static safety factor of 5.65.
What is the von Mises safety factor for a shaft?
It is the material yield strength divided by the von Mises equivalent stress at the outer surface, n = Sy/σ'. A value above 1 means the shaft does not yield under that static load. The calculator also gives the maximum shear (Tresca) safety factor, which is slightly more conservative, n = Sy/(2·τmax).
What diameter shaft do I need for a given torque and bending moment?
For a solid shaft with no axial load, d = [32·n·√(M² + 0.75·T²) / (π·Sy)]^(1/3), where n is the safety factor. The calculator solves this for your safety factor, including axial load and the bore ratio you entered, and shows the minimum outer diameter.
Does this check fatigue?
No. This is a static yield check at the surface. Rotating shafts see fully reversed bending, so fatigue (endurance limit, stress concentration at shoulders and keyways, surface finish and size factors) usually governs and needs a separate check such as the Soderberg or ASME elliptic criterion. Use this result as a first sizing step and for slowly rotating or occasional peak loads.
How is the angle of twist calculated?
θ = T·L/(G·J) in radians, where L is the length under torque and G the shear modulus (about 79 GPa for carbon steel, 26 GPa for aluminium). Many drive shafts are limited to roughly 0.25° per metre of length for stiffness, which often governs before strength does.
Does the calculator include stress concentrations?
No. Shoulders, keyways, cross-holes and grooves raise the local stress by a factor Kt of typically 1.5 to 3. Multiply the stresses (or divide the safety factor) by Kt for a conservative local check.
Engineering aid for guidance. Static yield check only: it does not cover fatigue, stress concentrations, buckling of long compressed shafts, dynamic loads or deflection under bending. Verify critical designs; you remain responsible for the design. Nothing you enter leaves your browser except through a share link you copy yourself.