Irodov 1.310: The Same Stretch Hides More Energy – Amit & Deepali Sinha Foundation

Problem Irodov 1.310

A steel cylindrical rod of length $l$ and radius $r$ hangs vertically, fixed to the ceiling at its upper end, and stretches elastically under its own weight. Find:

the total elastic deformation energy $U$ stored in the rod;
$U$ expressed in terms of the rod’s own overall tensile strain $\Delta l/l$ , its fractional elongation.

The rod of length l hanging under its own weight from its fixed upper end. — The rod of length $l$ hanging under its own weight from its fixed upper end.

Given

steel rod, length $l$ , radius $r$ , fixed at one end

Find

(a)

U

, (b)

U

in terms of

\Delta l/l

Solution

Step 1The tension at a cross-section

An element dx at distance x below the fixed end carries the weight of the length (l-x) still hanging beneath it, and so is less stretched, and stores less energy, than material nearer the top. — An element $dx$ at distance $x$ below the fixed end carries the weight of the length $(l-x)$ still hanging beneath it, and so is less stretched, and stores less energy, than material nearer the top.

Let $A=\pi r^2$ be the rod’s cross-sectional area and $\rho$ its density. Measuring $x$ from the fixed (upper) end, $0\le x\le l$ , the cross-section at $x$ carries the weight of everything hanging below it, a length $(l-x)$ .

T(x) = \rho A g\,(l-x). \tag{1}

Step 2Elastic energy stored, part (a)

An element $dx$ under tension $T(x)$ stores energy equal to half the tension times its own stretch, the same pattern as a spring’s $\frac12$ (force)(displacement), with $E$ the steel’s Young’s modulus.

dU = \frac12 T(x)\,d(\delta x) = \frac12 T(x)\cdot\frac{T(x)}{EA}\,dx = \frac{T(x)^2}{2EA}\,dx.

Integrating that stored energy along the whole rod, with $T(x)$ from (1), gives the total.

U = \int_0^l \frac{T(x)^2}{2EA}\,dx = \frac{\rho^2 A g^2}{2E}\int_0^l (l-x)^2\,dx = \frac{\rho^2 A g^2}{2E}\left[-\frac{(l-x)^3}{3}\right]_0^l = \frac{\rho^2 A g^2 l^3}{6E}.

Substituting $A=\pi r^2$ :

U = \frac{1}{6}\pi r^2 l^3\rho^2 g^2/E

(2)

Step 3The rod’s overall strain, to set up part (b)

Each element’s local strain is $T(x)/(EA)$ , so the elongation $\Delta l$ is that strain integrated along the rod; dividing by $l$ gives the strain $\Delta l/l$ part (b) asks for.

\Delta l = \int_0^l \frac{T(x)}{EA}\,dx = \frac{\rho g}{E}\int_0^l (l-x)\,dx = \frac{\rho g}{E}\left[lx-\frac{x^2}{2}\right]_0^l = \frac{\rho g l^2}{2E}.

\frac{\Delta l}{l} = \frac{\rho g l}{2E}

(3)

Step 4Elastic energy in terms of the strain, part (b)

Solving (3) for $\rho g$ and squaring it gives $\rho^2 g^2$ purely in terms of $\Delta l/l$ .

\rho g = \frac{2E}{l}\cdot\frac{\Delta l}{l} \qquad\Longrightarrow\qquad \rho^2 g^2 = \frac{4E^2}{l^2}\left(\frac{\Delta l}{l}\right)^2.

Putting that into the part (a) energy (2) replaces the density and $g$ with the strain.

U = \frac{\pi r^2 l^3}{6E}\cdot\frac{4E^2}{l^2}\left(\frac{\Delta l}{l}\right)^2 = \frac{4}{6}\pi r^2 l E\left(\frac{\Delta l}{l}\right)^2.

U = \frac{2}{3}\pi r^2 l E\left(\Delta l/l\right)^2

(4)

Check

Substituting the strain (3) back into part (b)’s boxed result reproduces part (a) exactly: $\frac23\pi r^2 l E\cdot \rho^2g^2l^2/(4E^2) = \pi r^2 l^3\rho^2g^2/(6E)$ , confirming the two forms describe the same energy. Dimensionally, $[\pi r^2 l E]=\text{m}^2\cdot\text{m}\cdot\text{Pa}=\text{N}\cdot\text{m}=\text{J}$ , correct for an energy, since $\Delta l/l$ is dimensionless.

Remark

A rod stretched uniformly by the same overall fraction $\Delta l/l$ , one constant tension throughout its length, would store $U_{\mathrm{unif}} = EAl(\Delta l/l)^2/2$ , a spring of stiffness $EA/l$ extended by $\Delta l$ . Part (b)’s result is $U = 2EAl(\Delta l/l)^2/3$ , a full third larger, because the energy density $T^2/(2EA)$ is quadratic in the local tension: concentrating more of the load near the fixed end, where $T$ is largest, raises the average of the square above the square of the average. The same overall stretch hides more stored energy when the strain producing it is spread unevenly.