Irodov 1.297: The Squeeze That Still Loses Volume – Amit & Deepali Sinha Foundation

Problem Irodov 1.297

A solid copper cylinder of length $l = 65$ cm stands upright on a rigid horizontal surface. A vertical force $F = 1000$ N, spread uniformly over the cylinder’s top end face, presses straight down on it. Find the resulting change in the cylinder’s volume, in cubic millimeters.

The cylinder of length l standing on the surface. The applied force F loads the top face uniformly, and the surface supplies an equal reaction N at the base. — The cylinder of length $l$ standing on the surface. The applied force $F$ loads the top face uniformly, and the surface supplies an equal reaction $N$ at the base.

Given

$l = 65$ cm $= 0.65$ m
$F = 1000$ N
copper: $E$ , $\mu$ known

Find

\Delta V

, the change in volume, in mm

^3

Solution

Step 1The axial and transverse strains

The force $F$ , spread uniformly over the end face of cross-sectional area $A$ , loads the cylinder in pure axial compression. Hooke’s law gives the fractional shortening directly, and Poisson’s ratio gives the fractional expansion each transverse direction acquires in response.

|\varepsilon_\parallel| = \frac{F}{AE}, \qquad |\varepsilon_\perp| = \mu\,|\varepsilon_\parallel| = \mu\,\frac{F}{AE}.

Step 2The fractional volume change

To first order, the fractional volume change is the sum of the strain along all three principal directions: one axial contraction and two equal transverse expansions.

\frac{|\Delta V|}{V} = |\varepsilon_\parallel| - 2|\varepsilon_\perp| = \frac{F}{AE} - 2\mu\,\frac{F}{AE}.

\frac{|\Delta V|}{V} = (1-2\mu)\frac{F}{AE}

(1)

Step 3Eliminating the cross-section

The cross-sectional area does not appear in the given data. Multiplying (1) by the volume $V = Al$ replaces $A$ with the length $l$ instead.

|\Delta V| = V\cdot\frac{|\Delta V|}{V} = (Al)\cdot(1-2\mu)\frac{F}{AE}.

The area cancels between $V$ and the strain, leaving the length in its place. Since $\mu<1/2$ for any real material, the axial contraction always outweighs the transverse bulge, so the volume decreases.

Step 4Numbers

Copper: $E = 1.3\times10^{11}$ Pa, $\mu = 0.34$ . With $F = 1000$ N and $l = 0.65$ m,

1-2\mu = 1-2(0.34) = 0.32, \qquad \frac{Fl}{E} = \frac{1000\times0.65}{1.3\times10^{11}} = 5.00\times10^{-9}\ \text{m}^3.

Multiplying the two factors together gives the volume change itself.

|\Delta V| = 0.32\times5.00\times10^{-9}\ \text{m}^3 = 1.6\times10^{-9}\ \text{m}^3 = 1.6\ \text{mm}^3.

\Delta V = (1-2\mu)\frac{Fl}{E} \approx 1.6\ \text{mm}^3\ \text{(a decrease)}

(2)

Check

Dimensions: $[Fl/E] = (\text{N})(\text{m})/(\text{N/m}^2) = \text{m}^3$ , a volume, as required. And at $\mu=1/2$ , (1) gives zero: an incompressible material would not change volume under any load, consistent with the same $(1-2\mu)$ factor vanishing at that limit.

Remark

The combination $(1-2\mu)F/(AE)$ here is the same one that set the volume change of Problem 1.298’s copper rod stretched by its own weight, $(1-2\mu)\,\Delta l/l$ , since $F/(AE)$ is exactly this cylinder’s axial strain and $\Delta l/l$ is that rod’s mean axial strain (its local strain varies along its length, while this cylinder’s does not). Whether the load pulls the material apart or pushes it together, a single elastic combination, the same $(1-2\mu)$ weighted by the axial strain, governs how much volume a uniaxial load buys or costs; only the sign of the loading changes which way the volume moves.