Irodov 1.313: Where Torsional Energy Really Lives – Amit & Deepali Sinha Foundation

Problem Irodov 1.313

A steel rod of length $l$ is twisted about its own axis through an angle $\varphi$ . Find how the volume density $u$ of the elastic (torsional) deformation energy stored in the rod depends on the distance $r$ from the rod’s own axis.

A thin cylindrical shell of radius r and thickness dr, running the rod's full length l, seen end on. Every fiber in it lies the same distance from the axis, so shares one energy density u(r). — A thin cylindrical shell of radius $r$ and thickness $dr$ , running the rod’s full length $l$ , seen end on. Every fiber in it lies the same distance from the axis, so shares one energy density $u(r)$ .

Given

steel rod, length $l$ , twisted through angle $\varphi$

Find

u(r)

, the volume density of elastic energy

Solution

Step 1The local shear strain at radius

r

A prismatic rod under a constant torque twists at a uniform rate along its own length, so the twist per unit length is $\varphi/l$ everywhere. A fiber running parallel to the axis at radius $r$ is carried sideways by $r\,d\varphi'$ over an element $dx$ of length, so its shear angle is that sideways shift divided by $dx$ .

\gamma(r) = r\cdot\frac{d\varphi}{dx} = r\cdot\frac{\varphi}{l} = \frac{r\varphi}{l}. \tag{1}

Step 2The shear stress, by Hooke’s law

Hooke’s law for shear relates stress to strain through the shear modulus $G$ ; substituting the strain (1) gives the stress as a function of $r$ .

\tau(r) = G\gamma(r) = \frac{Gr\varphi}{l}. \tag{2}

Step 3The volume density of elastic energy

The elastic energy density in a state of pure shear is one half the shear stress times the shear strain, the same $\frac12$ (force)(displacement) pattern behind every elastic energy in this section. Substituting (1) and (2),

u(r) = \frac12\tau(r)\gamma(r) = \frac12\cdot\frac{Gr\varphi}{l}\cdot\frac{r\varphi}{l} = \frac{1}{2}\frac{G\varphi^2 r^2}{l^2}.

u = \frac{1}{2}G\varphi^2 r^2/l^2

(3)

Check

Integrating $u(r)$ over the rod’s own volume reproduces the total torsional energy: for a solid rod of outer radius $R$ , $\int_0^R u(r)(2\pi r l)\,dr = (\pi G\varphi^2/l)\int_0^R r^3\,dr = \pi R^4 G\varphi^2/(4l)$ , exactly problem 1.312’s boxed total energy with $R$ standing in for that problem’s rod radius $r$ . Also $u(0)=0$ : a fiber sitting exactly on the rotation axis undergoes no shear at all, however large $\varphi$ is.

Remark

Because $u\propto r^2$ , the outer material of the rod does almost all the work of resisting the twist while the material near the axis does almost none, the same weighting by the square of the radial coordinate under the integral that made the polar moment $I_p=\int r^2\,dS$ grow so steeply with radius. It is also the reason a hollow shaft is such an efficient one: all of its material sits out where $u$ is large, and the core it leaves out was barely resisting the twist in the first place.