Irodov 1.312: The Energy Cost of a Twist – Amit & Deepali Sinha Foundation

Problem Irodov 1.312

A steel rod of length $l = 1.0$ m and radius $r = 10$ mm has one end fixed and the other end twisted, about the rod’s own axis, through an angle $\varphi = 6.0^\circ$ . Find the elastic deformation energy stored in the rod.

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

Given

$l = 1.0$ m
$r = 10$ mm
$\varphi = 6.0^\circ$
steel rod

Find

U

, the elastic deformation energy

Solution

Step 1The polar moment of the solid circular cross-section

A ring element dρ at radius ρ contributes 2πρ^3 dρ to the polar moment of the section, the quantity that sets how much torque, and so how much elastic energy, a given twist angle stores. — A ring element $d\rho$ at radius $\rho$ contributes $2\pi\rho^3\,d\rho$ to the polar moment of the section, the quantity that sets how much torque, and so how much elastic energy, a given twist angle stores.

A thin ring at radius $\rho$ , width $d\rho$ , contributes $\rho^2\,dS = 2\pi\rho^3\,d\rho$ to the polar moment; the section runs from the axis to the rod’s own radius $r$ .

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

I_p = \frac{\pi r^4}{2}

(1)

Step 2Elastic energy stored, twisting quasi-statically to

\varphi

Twisting the rod slowly from zero angle up to the final angle $\varphi$ , the torque needed at each instant obeys the linear torsion law $N(\varphi') = GI_p\varphi'/l$ , growing in step with the angle already twisted. The stored elastic energy is the work done against this torque, its integral from $0$ to $\varphi$ .

U = \int_0^\varphi N(\varphi')\,d\varphi' = \int_0^\varphi \frac{GI_p}{l}\varphi'\,d\varphi' = \frac{GI_p}{l}\left[\frac{\varphi'^2}{2}\right]_0^\varphi = \frac{GI_p\varphi^2}{2l}. \tag{2}

Step 3Substituting the polar moment and evaluating

Substituting the polar moment (1) gives $U$ in terms of the rod’s own geometry and material.

\begin{aligned} U &= \frac{G\varphi^2}{2l}\cdot I_p\\ &= \frac{G\varphi^2}{2l}\cdot\frac{\pi r^4}{2} \end{aligned}

The twist is given in degrees, so convert it to radians first.

\varphi = 6.0^\circ\times\frac{\pi}{180^\circ} = 0.10472\ \text{rad}

Now evaluate with steel’s shear modulus $G = 8.1\times10^{10}$ Pa, $r = 10\times10^{-3}$ m and $l = 1.0$ m.

U = \frac{\pi\times(0.010)^4\times8.1\times10^{10}\times(0.10472)^2}{4\times1.0} \approx 6.98\ \text{J}.

U = \frac{\pi r^4 G\varphi^2}{4l} \approx 7\ \text{J}

(3)

Check

Differentiating the boxed $U$ with respect to $\varphi$ gives $dU/d\varphi = \pi r^4 G\varphi/(2l) = GI_p\varphi/l = N$ , exactly the torque of the linear torsion law integrated in (2), confirming $U$ is the correct potential whose derivative is the restoring torque.

Remark

For a fixed twist angle $\varphi$ , the stored energy falls off as $1/l$ : a longer rod spreads the same total twist over more length, so the local shear strain $\rho\varphi/l$ at any given radius is smaller everywhere along it, and it takes less work to reach $\varphi$ . This is the opposite sense from problem 1.310’s hanging rod, where a longer rod stores more self-weight energy; there the load itself, the weight below each cross-section, grows with the rod’s own length, while here the imposed angle $\varphi$ is held fixed regardless of how long the rod is.