MechanicalMachine DesignhardCourseworkInterviewPE / Advanced

Shaft Sizing from Transmitted Power

A solid circular shaft transmits $P = 15\ \text{kW}$ at a rotational speed of $N = 1500\ \text{rpm}$ .

The shaft material has an allowable shear stress of $\tau_{allow} = 60\ \text{MPa}$ .

Determine the minimum required shaft diameter.

Given

Hint 1

Three chained steps: rotational speed to angular velocity, angular velocity plus power to torque, torque to diameter.

Hint 2

ω = 2πN/60 converts rpm to rad/s — both factors matter.

Hint 3

For a solid shaft in torsion, τ_max = 16T/(πd³). Rearranging for d needs a cube root, not a square root.

Worked solution — try the problem first

Step 1 — Convert speed to angular velocity.

$\omega = \frac{2\pi N}{60} = \frac{2\pi (1500)}{60} = 157.08\ \text{rad/s}$

The factor of 60 converts rev/min to rev/s; the $2\pi$ converts rev to radians. Both are needed.

Step 2 — Torque from power. For rotating machinery, $P = T\omega$ :

$T = \frac{P}{\omega} = \frac{15{,}000}{157.08} = 95.49\ \text{N·m}$

Step 3 — Size the shaft from the torsion formula. For a solid shaft,

$\tau_{max} = \frac{16T}{\pi d^3} \quad \Longrightarrow \quad d^3 = \frac{16T}{\pi \tau_{allow}}$

$d^3 = \frac{16 \times 95.49}{\pi \times 60\times10^{6}} = \frac{1527.9}{1.885\times10^{8}} = 8.106\times10^{-6}\ \text{m}^3$

$d = \left(8.106\times10^{-6}\right)^{1/3} = 0.02009\ \text{m}$

$\boxed{d = 20.1\ \text{mm minimum}}$

In practice you would round up to the next standard stock size (say 22 or 25 mm) — rounding down would exceed the allowable stress.

Concepts:Power transmissionTorsionShaft designAngular velocity

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