When a wire is stretched by a load, is there any change in its volume? [closed]

Question

When a wire, or a rod(whose diameter is not negligible) is subjected to a tensile stress, is there any change in its volume?

If yes, does the volume increase or decrease? Take into consideration both the longitudinal elongation and the lateral contraction.

If no, then what is wrong with the following derivation?

$V=\pi d^2 l/4$$\ \ \ \ \ \ \ \ \ \ \ (V$ is the volume of the wire)

$\implies\Delta V/V=2\Delta d/d +\Delta l/l$

$\implies\Delta V/V=(1-2\sigma)\ \Delta l/l$ $\ \ \ \ \ \ \ (\Delta d/d=-\sigma \Delta l/l)$

where $\sigma$ is Poisson's ratio.

Answer 1

The volume changes. In tension, the chemical bonds all get a little bigger, resulting in a larger volume. To a first approximation, the change of volume is linear, as you have shown. A more accurate analysis shows that the wire gets a little narrower, so the increase in volume is a little less than your result predicts.

Things are entirely different with a rubber band. When stretched, the volume remains nearly the same. The elasticity of a polymer is due to a completely different mechanism than that of a wire.

Answer 2

Poisson's ration for most metals is close to 1/3. That means for a stretch of 1%, you get a reduction in the diameter of 0.33% Adding up the changes in all three dimensions gives you a volume increase of 0.33%. (The "rubber-band" deformation with no volume change corresponds to a Poisson's Ration of 1/2.)