Irodov 1.298: A Rod Stretched by Its Own Weight – Amit & Deepali Sinha Foundation

Problem Irodov 1.298

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

(a)
the elongation $\Delta l$ this self-weight produces;
(b)
the resulting fractional change in the rod’s volume, $\Delta V/V$ .

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

copper rod of length $l$ , fixed at one end

Find

(a)

\Delta l

, (b)

\Delta V/V

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. — An element $dx$ at distance $x$ below the fixed end carries the weight of the length $(l-x)$ still hanging beneath it.

Let $A$ be the cross-sectional area and $\rho$ the density of copper. Measure $x$ 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 2Elongation, by integrating the local strain

Each element’s local strain is $T(x)/(EA)$ with $T$ from (1), and the total elongation sums this over the rod:

\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 \tag{2}

\Delta l = \frac{1}{2}\frac{\rho g l^2}{E}

(3)

Step 3The volume change, part (b)

A rod element under axial strain $\varepsilon(x) = T(x)/(EA)$ also contracts sideways by Poisson’s ratio $\mu$ : the two transverse strains are each $-\mu\varepsilon(x)$ . To first order, the local fractional volume change of that element is the sum of all three strains,

\frac{dV}{dV_0} = \varepsilon(x) - 2\mu\,\varepsilon(x) = (1-2\mu)\,\varepsilon(x).

Since $dV_0 = A\,dx$ , the total volume change is

\Delta V = \int_0^l A(1-2\mu)\varepsilon(x)\,dx = A(1-2\mu)\int_0^l \varepsilon(x)\,dx = A(1-2\mu)\,\Delta l,

using the same integral (2) that produced $\Delta l$ . Dividing by $V=Al$ gives the relative volume change.

\frac{\Delta V}{V} = (1-2\mu)\frac{\Delta l}{l}

(4)

Check

For an incompressible material, $\mu=1/2$ , the formula gives $\Delta V/V=0$ : stretching such a material changes its shape but never its volume, which is the defining property of $\mu=1/2$ .

Remark

The relation $\Delta V/V=(1-2\mu)\,\Delta l/l$ never used gravity, length, or Young’s modulus directly, only the fractional stretch $\Delta l/l$ itself. It therefore holds for any axial strain, however produced, self-weight here, a hanging mass elsewhere, or a testing machine’s grip: two rods of the same material carrying the same fractional elongation always carry the same fractional volume change, regardless of what stretched them.