Irodov 1.309: Strain Energy Scales with Mass – Amit & Deepali Sinha Foundation

Problem Irodov 1.309

A steel rod of mass $m = 3.1$ kg is stretched elastically to a tensile strain $\varepsilon = 1.0\times10^{-3}$ . Find the total elastic deformation energy stored in the rod, using standard tabulated values for steel.

The rod pulled by an axial force F at each end, stretched uniformly from its natural length l_0; the tensile strain ε is the same at every cross-section since the rod is uniform. — The rod pulled by an axial force $F$ at each end, stretched uniformly from its natural length $l_0$ ; the tensile strain $\varepsilon$ is the same at every cross-section since the rod is uniform.

Given

$m = 3.1$ kg
$\varepsilon = 1.0\times10^{-3}$
steel: $E$ , $\rho$ known

Find

U

, the total elastic deformation energy stored

Solution

Step 1Energy density of a strained rod

Hooke’s law gives the tensile stress at strain $\varepsilon'$ as $\sigma = E\varepsilon'$ ; the elastic energy stored per unit volume is the work done straining an element from zero up to $\varepsilon$ , the area under the linear stress-strain line.

u = \int_0^\varepsilon \sigma(\varepsilon')\,d\varepsilon' = \int_0^\varepsilon E\varepsilon'\,d\varepsilon'

u = \dfrac{E\varepsilon^2}{2}

(1)

Step 2Total energy from the rod’s volume

The rod’s volume follows from its mass and the steel density; multiplying the energy density (1) by that volume gives the total stored energy.

V = \frac{m}{\rho}, \qquad U = uV = \frac{E\varepsilon^2}{2}\cdot\frac{m}{\rho}

Step 3Evaluating

Using the steel values $E = 2.0\times10^{11}$ Pa and $\rho = 7.8\times10^{3}\ \text{kg/m}^3$ with the given $m$ and $\varepsilon$ .

U = \frac{(3.1\ \text{kg})(2.0\times10^{11}\ \text{Pa})(1.0\times10^{-3})^2}{2(7.8\times10^{3}\ \text{kg/m}^3)} = \frac{6.2\times10^{5}}{1.56\times10^{4}}\ \text{J} = 39.7\ \text{J}

U = \dfrac{mE\varepsilon^2}{2\rho} \approx 0.04\ \text{kJ}

(2)

Check

Units: $[m][E]/[\rho] = \text{kg}\cdot(\text{N/m}^2)/(\text{kg/m}^3) = \text{N}\cdot\text{m} = \text{J}$ , correct for an energy, and $\varepsilon^2$ is dimensionless. A second, independent route reproduces the box: the rod is a spring of stiffness $k = EA/l_0$ stretched by $\Delta l = \varepsilon l_0$ , so $U = \frac12 k(\Delta l)^2 = \frac12 EAl_0\varepsilon^2$ , and $Al_0$ is the rod’s volume $m/\rho$ , which returns $mE\varepsilon^2/(2\rho)$ without ever integrating an energy density.

Remark

The specific energy $U/m = E\varepsilon^2/(2\rho)$ depends only on the strain and on the material’s $E/\rho$ , never on the rod’s length or cross-section: two rods of the same steel and the same strain store the same energy per unit mass however differently shaped they are. That ratio $E/\rho$ is the material’s specific stiffness. It ranks materials only at a fixed strain, though, and a real energy-storing element is limited instead by the stress its material can carry: at a given working stress $\sigma$ the energy stored per unit mass is $\sigma^2/(2E\rho)$ , which rewards a low modulus rather than a high one. That is why a rubber cord outstores a steel spring of the same mass, even though steel’s $E/\rho$ is larger by a factor of tens of thousands.