A shaft must transmit 50 N·m; the material allows 40 N/mm² shear stress. What minimum diameter is safe?
Stress lives at the outer edge
Under torsion on a solid round shaft, the shear stress is τ = T·r/J, with J = π·d⁴/32 and r = d/2. Stress is zero at the core and maximum at the outer edge. Solving for d gives d = ∛(16T/(π·τ_allow)) — the smallest diameter that doesn't exceed the allowed stress.
Worked example: 50 N·m at 40 N/mm²
With T = 50 N·m (50,000 N·mm) and τ_allow = 40 N/mm², d = ∛(16 · 50,000 / (π · 40)) ≈ 18.53 mm. Working backwards confirms it: at Ø18.53 mm, 50 N·m produces approximately 40.02 N/mm². The unrounded minimum diameter is approximately 18.5336 mm; round a specified diameter upwards. A smaller shaft exceeds the allowed stress; a larger one has margin to spare.
Torsion is rarely the only load
A gear, pulley or sprocket also introduces bending and transverse shear on the shaft — for most shafts that dominates, not pure torsion. Stress concentration at keyways, shoulders and holes raises the local stress further. A rotating shaft under alternating bending is also a fatigue problem, not a static one. Mechify's safety band (below 1.0 is unsafe, 1.0-1.2 is a caution zone) is a screening threshold, not a value from a design code.
Assumptions and applicability limits
The model calculates pure torsion on a solid round shaft only. Combined loading, stress concentration, fatigue, stiffness (twist angle, deflection, critical speed) and hollow shafts fall outside it. For a real shaft with keyways and alternating loads, use a full strength and fatigue analysis.