Combined Loading Stress⚠ unverified
Mechanical / Stress Analysis · Compute the von Mises stress for a solid circular shaft under bending and torsion
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| M | M | N*m | 1.0 | Applied bending moment |
| T | T | N*m | 1.0 | Applied torsional moment |
| d | d | m | 1.0 | Shaft diameter |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | σ' | Pa | Von Mises equivalent stress, in pascals (Pa) |
The science & history
Understanding the Parameters
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Bending moment $M$ — produces the surface bending stress $\sigma = 32M/(\pi d^3)$ (from $\sigma = Mc/I$ with $c = d/2$, $I = \pi d^4/64$). It is the normal-stress contribution.
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Torque $T$ — produces the surface shear stress $\tau = 16T/(\pi d^3)$ (from $\tau = Tr/J$ with $J = \pi d^4/32$). It is the shear contribution.
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Diameter $d$ — appears as $d^3$: both stresses scale as $1/d^3$, so a modest diameter increase sharply cuts the equivalent stress. Sizing a shaft means solving this for $d$.
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Equivalent stress $\sigma'$ — a single number comparable to the yield strength (or, with an appropriate allowable, to a fatigue limit). The $\sqrt{M^2+T^2}$ grouping is the maximum-shear result; swapping to $\sqrt{M^2+\tfrac34 T^2}$ gives von Mises.
Derivation (Approaching a Proof)
On the outer surface the shaft is in a plane stress state: a bending normal stress $\sigma$ and a torsional shear $\tau$, with no second normal stress. From beam and torsion theory,
$$\sigma = \frac{32M}{\pi d^3}, \qquad \tau = \frac{16T}{\pi d^3}.$$
The maximum shear stress for this state (Mohr's circle radius with $\sigma_y = 0$) is
$$\tau_{max} = \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2} = \sqrt{\left(\frac{16M}{\pi d^3}\right)^2 + \left(\frac{16T}{\pi d^3}\right)^2} = \frac{16}{\pi d^3}\sqrt{M^2 + T^2}.$$
The Tresca equivalent stress is $\sigma' = 2\tau_{max}$:
$$\sigma' = \frac{32}{\pi d^3}\sqrt{M^2 + T^2}.$$
(Had we used the von Mises combination $\sigma' = \sqrt{\sigma^2 + 3\tau^2}$ instead, the torsion term would carry weight $\tfrac34$: $\sigma' = \tfrac{32}{\pi d^3}\sqrt{M^2 + \tfrac34 T^2}$.)
Dimensional check. $\sigma' = \dfrac{32}{\pi d^3}\sqrt{M^2+T^2} = \dfrac{\text{N}\cdot\text{m}}{\text{m}^3} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$ — a stress, as required ($\sqrt{M^2+T^2}$ carries the units of a moment).
History and Development
Combining bending and torsion into an equivalent shaft stress is a founding application of the yield criteria to machine design. The maximum-shear ($\sqrt{M^2+T^2}$) and distortion-energy ($\sqrt{M^2+\tfrac34 T^2}$) forms both appear in Shigley and in the ANSI/ASME B106 shaft standard; the choice sets how conservatively torsion is treated. Adding a fatigue treatment (applying $K_f$ and a Goodman/elliptic criterion to the alternating and mean components) turns this static check into the full shaft-fatigue design equation.
Related Concepts: Combined Stress Shaft, Shaft Diameter Combined, Von Mises Stress, Tresca Stress, Combined Loading Equivalent Stress, Shaft Fatigue Factor
Notes: $\sqrt{M^2+T^2}$ is the maximum-shear (Tresca) form (mislabeled "von Mises" in the registry); von Mises uses $\sqrt{M^2+\tfrac34 T^2}$. Solid shaft; surface stress. $\sigma=32M/\pi d^3$, $\tau=16T/\pi d^3$. Solve for $d$ to size a shaft.