It's interesting that you chose a spring as an example. Since the torsional load on a compression spring is essentially all shear stress, relatively little volume change is expected.
For elastic loading, we want to start with
V^\prime=V(1+\epsilon_1)(1+\epsilon_2)(1+\epsilon_3)
where V^\prime and V are the new and old volumes, respectively, and \epsilon_i (i=1,2,3) are the normal strains. For the small strains we see in a metal, the normalized change in volume can be simplified to
\frac{V^\prime-V}{V}=\epsilon_1+\epsilon_2+\epsilon_3
Continuing, the normal strain at any plane of an isotropic, homogeneous solid is
\epsilon_1=\frac{1}{E}\sigma_1-\frac{\nu}{E}\sigma_2-\frac{\nu}{E}\sigma_3
etc., \nu and E are the Poisson's ratio and Young's elastic modulus, respectively, and \sigma_1, \sigma_2, and \sigma_3 are the normal stresses at any plane. Again, shear stresses aren't included (as modeled in mechanics of materials, shear causes a change in shape, not a change in size).
So the normalized change in volume is
\frac{\Delta V}{V}=\frac{1-2\nu}{E}(\sigma_1+\sigma_2+\sigma_3)
(note the correspondence to the bulk modulus K=E/3(1-2\nu), which is the triaxial pressure needed to achieve unit volumetric compression). Thus, normal stresses will generally cause a normalized volume change in metals of order \sigma/E, as the Poisson's ratio \nu is often near 0.3.
For plastic loading, the volume is usually assumed to be constant, but the problem becomes more complicated here. It sounds like you're interested specifically in elastic loading.
