Irodov 1.307: How Thick Buys More Power Than Fast – Amit & Deepali Sinha Foundation

Problem Irodov 1.307

A solid steel shaft of length $l = 200$ cm and radius $r = 1.50$ cm rotates about its own axis at angular velocity $\omega = 120$ rad/s. The shaft may not be twisted, along its own length, by more than a permissible angle $\varphi = 2.5^\circ$ . Find the greatest power the shaft can transmit without exceeding this limiting twist.

The shaft's solid circular cross-section, radius r. — The shaft’s solid circular cross-section, radius $r$ .

Given

$\omega = 120$ rad/s
$l = 200$ cm
$r = 1.50$ cm
$\varphi = 2.5^\circ$

Find

P

, the maximum power the shaft can transmit

Solution

Step 1The polar moment of the solid circular section

A ring element dρ at radius ρ contributes 2πρ^3 dρ to the polar moment of the section. — A ring element $d\rho$ at radius $\rho$ contributes $2\pi\rho^3\,d\rho$ to the polar moment of the section.

A thin ring at radius $\rho$ , width $d\rho$ , has area $dS=2\pi\rho\,d\rho$ , so it contributes $\rho^2\,dS=2\pi\rho^3\,d\rho$ to the polar moment $I_p=\int\rho^2\,dS$ . Here the section runs from the axis to the shaft’s own radius $r$ :

I_p = \int_0^r 2\pi\rho^3\,d\rho = \frac{\pi r^4}{2}. \tag{1}

Step 2The torque the shaft can carry

The torsion relation $\varphi = Nl/(GI_p)$ gives the torque corresponding to the permissible twist angle $\varphi$ . Solving for $N$ and substituting $I_p$ from (1):

N = \frac{GI_p\varphi}{l}

N = \frac{\pi r^4 G\varphi}{2l}

(2)

Step 3Power at that torque

A shaft turning at angular velocity $\omega$ under torque $N$ transmits power $P$ .

P = N\omega

Step 4Numbers

Steel: $G = 8.1\times10^{10}$ Pa. Converting $\varphi$ to radians,

\varphi = 2.5^\circ \times \frac{\pi}{180^\circ} = 0.0436\ \text{rad}.

With $r = 0.0150$ m, $l = 2.00$ m,

I_p = \frac{\pi(0.0150)^4}{2} = 7.95\times10^{-8}\ \text{m}^4,

P = \frac{7.95\times10^{-8}\times8.1\times10^{10}\times0.0436\times120}{2.00} \approx 1.7\times10^4\ \text{W}.

P = \frac{\pi r^4 G\varphi\omega}{2l} \approx 17\ \text{kW}

(3)

Check

Dimensions: $[r^4 G\varphi\omega/l] = \text{m}^4\cdot\text{Pa}\cdot 1\cdot\text{s}^{-1}/\text{m} = \text{m}^3\cdot(\text{N/m}^2)/\text{s} = \text{N}\cdot\text{m/s} = \text{W}$ , correct for a power. Doubling the shaft’s length halves the permissible power: the same total twist $\varphi$ spread over twice the length means half the twist per unit length, so the allowable torque $N=GI_p\varphi/l$ halves, and at the same $\omega$ the power $P=N\omega$ halves with it.

Remark

A solid shaft of this radius transmits more power than a tube of the same outer radius, since none of its cross-section sits idle, but it does so wastefully in terms of material: the innermost material contributes almost nothing to $I_p$ , which grows as $\rho^4$ under the integral, yet it still carries its full share of the shaft’s mass. That is exactly why real drive shafts are so often built hollow: for a fixed mass, moving material outward from the axis buys far more torsional stiffness than the removed core was worth.