Irodov 1.306: Twisting a Hollow Steel Tube – Amit & Deepali Sinha Foundation

Problem Irodov 1.306

A hollow steel tube of length $l = 3.0$ m, with inside diameter $d_1 = 30$ mm and outside diameter $d_2 = 50$ mm, is twisted about its own axis through an angle $\varphi = 2.0^\circ$ . Find the torque $N$ required to produce this twist.

The tube's cross-section, an annulus of inside diameter d_1 and outside diameter d_2. — The tube’s cross-section, an annulus of inside diameter $d_1$ and outside diameter $d_2$ .

Given

$l = 3.0$ m
$\varphi = 2.0^\circ$
$d_1 = 30$ mm
$d_2 = 50$ mm

Find

N

, the twisting torque

Solution

Step 1The polar moment of the annular cross-section

A ring element dr at radius r, between d_1/2 and d_2/2, contributes 2π r^3 dr to the polar moment of the section. — A ring element $dr$ at radius $r$ , between $d_1/2$ and $d_2/2$ , contributes $2\pi r^3\,dr$ to the polar moment of the section.

For a shaft under torsion, the relevant section property is the polar moment $I_p = \int r^2\,dS$ , taken over the cross-section. A thin ring of radius $r$ and width $dr$ has area $dS = 2\pi r\,dr$ , so

I_p = \int_{d_1/2}^{d_2/2} r^2\cdot 2\pi r\,dr = 2\pi\int_{d_1/2}^{d_2/2} r^3\,dr = \frac{\pi}{2}\left[\left(\frac{d_2}{2}\right)^4-\left(\frac{d_1}{2}\right)^4\right].

I_p = \frac{\pi}{32}\left(d_2^4-d_1^4\right)

(1)

Step 2Relating torque to twist angle

For a shaft of length $l$ under torque $N$ , the standard torsion relation is $\varphi = Nl/(GI_p)$ , where $G$ is the shear modulus. Solving for $N$ , with $I_p$ from (1), gives the torque in terms of the tube’s geometry.

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

Step 3Numbers

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

\varphi = 2.0^\circ \times \frac{\pi}{180^\circ} = 0.0349\ \text{rad}.

With $d_1=0.030$ m, $d_2=0.050$ m,

d_2^4-d_1^4 = (0.050)^4-(0.030)^4 = 6.25\times10^{-6}-0.81\times10^{-6}=5.44\times10^{-6}\ \text{m}^4,

N = \frac{\pi\times5.44\times10^{-6}\times8.1\times10^{10}\times0.0349}{32\times3.0} \approx 5.0\times10^2\ \text{N}\cdot\text{m}.

N = \frac{\pi\left(d_2^4-d_1^4\right)G\varphi}{32\,l} \approx 0.5\ \text{kN}\cdot\text{m}

(2)

Check

Letting $d_1\to0$ collapses the tube to a solid shaft, and the boxed formula reduces to $N=\pi d_2^4 G\varphi/(32l)$ , the familiar solid-shaft torsion formula, exactly as it should.

Remark

The $r^4$ weighting inside $I_p=\int r^2\,dS$ means a tube built from the same amount of material as a solid shaft, but pushed farther from the axis, resists twisting far more than the solid shaft does: torsional stiffness depends on where the material sits, not merely how much of it there is. That is the opposite of problem 1.292’s rupturing rod, where the cross-sectional area canceled out of the answer entirely and placement did not matter at all.