categories: physics, automation & robotics, mechanical engineering
A shaft in a real drivetrain (a motor coupling, a gearbox output shaft, a robot wheel axle) almost never sees just one type of load — it's usually being twisted by a driving torque and bent by a side load at the same time. Treating those separately and just picking a diameter that survives each one individually isn't safe, because both stresses act on the same material simultaneously. This post covers the standard way to combine them into one number you can actually design against.
Task: minimum shaft diameter under combined torsion and bending
A shaft carries a torsional moment $$M_s=150\text{ Nm}$$ and a bending moment $$M_g=200\text{ Nm}$$ simultaneously. Find the minimum shaft diameter.
Step 1 — reduce the combined load to a single equivalent moment (Huber-Mises-Hencky). Bending and torsion stress the material's cross-section in different ways — one stretches/compresses it, the other shears it — so you can't just add $$M_g$$ and $$M_s$$ directly. The HMH (distortion-energy) failure criterion accounts for this by weighting the torsional term, since shear stress contributes to yielding differently than normal stress does: $$M_{zred}=\sqrt{M_g^2+0.75M_s^2}=\sqrt{200^2+0.75\cdot150^2}=\sqrt{40000+16875}=\sqrt{56875}\approx238.48\text{ Nm}$$
Step 2 — allowable stress from the material. For structural steel S355 (yield strength $$R_e=355\text{ MPa}$$) with safety factor $$k=2.0$$ — the safety factor exists precisely to absorb everything the idealized calculation doesn't capture: fatigue, stress concentrations at fillets/keyways, manufacturing tolerances: $$\sigma_{adm}=\frac{R_e}{k}=\frac{355\text{ MPa}}{2.0}=177.5\text{ MPa}=177.5\times10^6\text{ Pa}$$
Step 3 — section modulus and minimum diameter. The strength condition requires the actual stress to stay under the allowable: $$\sigma_{zred}=\dfrac{M_{zred}}{W_g}\leq\sigma_{adm}$$, where a circular cross-section has section modulus $$W_g=\dfrac{\pi d^3}{32}$$. Solve for the minimum $$W_g$$, then invert the cube to get $$d$$: $$W_g\geq\frac{M_{zred}}{\sigma_{adm}}=\frac{238.48\text{ Nm}}{177.5\times10^6\text{ Pa}}=1.3435\times10^{-6}\text{ m}^3$$ $$d\geq\sqrt[3]{\frac{32\cdot W_g}{\pi}}=\sqrt[3]{\frac{32\cdot1.3435\times10^{-6}}{\pi}}=\sqrt[3]{1.368\times10^{-5}}\approx0.0239\text{ m}=23.9\text{ mm}$$ Rounding up to the nearest standardized size: shaft diameter d = 25 mm — same rule as the battery post, always round up when a physical safety margin is on the line, never to the nearest value.
The pattern generalizes past shafts: whenever a part sees more than one type of load at once, look for the right combined-stress theory (HMH here; there's also the maximum-shear-stress theory, more conservative but simpler) instead of checking each load in isolation. Thank you for reading! Next post: leaving solid mechanics for fluid mechanics — deriving a velocity profile straight from the Navier-Stokes equations.
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