Does Young's Modulus Affect Volume Changes in Solid Materials?

AI Thread Summary
When stress is applied to a solid, such as an iron rod, the volume can change depending on the material's properties, particularly Poisson's ratio. Conservation of mass dictates that as the rod's length changes, the cross-sectional area must adjust to maintain a constant volume. For isotropic linear elastic materials, the volume change can be expressed mathematically, indicating that a Poisson's ratio of 0.5 results in no volume change, classifying the material as incompressible. Most metals have Poisson's ratios between 0.3 and 0.35, leading to some volume change under low strains, while rubbers, with ratios close to 0.5, exhibit minimal volume change. Understanding these relationships is crucial in material science and engineering applications.
Sakha
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After a tough discussion with my friends, I opt to ask here because we aren't coming to an answer.
When stress is applied to a solid, let's say an iron rod, will the volume change, or when the length of the rod shrinks or stretches, the area of the cross sectional area will reduce or increase to maintain a constant volume?
 
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Conservation of mass... the volume is the same and poisson's ratio helps tell how much the cross sectional area will change.
 
That was exactly what I was looking for, thanks!
 
Dilatation or volume change can be given for small strains in an isotropic linear elastic material by:

\frac{dV}{V}=\frac{1-2\nu}{E}(\sigma_{x}+\sigma_{y}+\sigma_{z})

In a uniaxial tensile test at small strains, \sigma_{y}=\sigma{z}=0. You can see that for there to be no volume change, \nu (Poisson's ratio) must be equal to 0.5 at which point the material is said to be incompressible, as the bulk modulus tends to infinity.

Typically, most metals exhibit Poisson's ratios of between 0.3 and 0.35, meaning you will observe some volume change at low strains. Rubbers have Poisson's ratios tending to 0.5, meaning you will exhibit little to no volume change.
 
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