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
[tex]V^\prime=V(1+\epsilon_1)(1+\epsilon_2)(1+\epsilon_3)[/tex]
where [itex]V^\prime[/itex] and [itex]V[/itex] are the new and old volumes, respectively, and [itex]\epsilon_i[/itex] ([itex]i=1,2,3[/itex]) are the normal strains. For the small strains we see in a metal, the normalized change in volume can be simplified to
[tex]\frac{V^\prime-V}{V}=\epsilon_1+\epsilon_2+\epsilon_3[/tex]
Continuing, the normal strain at any plane of an isotropic, homogeneous solid is
[tex]\epsilon_1=\frac{1}{E}\sigma_1-\frac{\nu}{E}\sigma_2-\frac{\nu}{E}\sigma_3[/tex]
etc., [itex]\nu[/itex] and [itex]E[/itex] are the Poisson's ratio and Young's elastic modulus, respectively, and [itex]\sigma_1[/itex], [itex]\sigma_2[/itex], and [itex]\sigma_3[/itex] 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
[tex]\frac{\Delta V}{V}=\frac{1-2\nu}{E}(\sigma_1+\sigma_2+\sigma_3)[/tex]
(note the correspondence to the bulk modulus [itex]K=E/3(1-2\nu)[/itex], 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 [itex]\sigma/E[/itex], as the Poisson's ratio [itex]\nu[/itex] 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.