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With the following quantities defined as follows:

Normal stress along x, $\sigma_x = \frac{F_{nx}}{S}$

Strain along x, $\epsilon_x= \frac{\Delta L_x}{L_x}$

and Poisson's Law: $\epsilon_y=\epsilon_z=-\nu \epsilon_x= -\nu\frac{\sigma_x}{E}$, as well as Hooke's Law: $\epsilon_x=\frac{1}{E}\sigma_x$,

with $F_{nx}, L_x,S, E, \nu$ being the normal force applied, length along x, area, Young's modulus and Poisson's coefficient respectively, we are trying to find the change of volume of a uniformly compressed parallelepiped $(\sigma_x=\sigma_y=\sigma_z=\sigma=-p)$.

Parallelepiped

I do not understand the passage: $\epsilon_x=\frac{\Delta L_x}{L_x}=\frac{\sigma_x}{E}-\frac{\nu}{E}(\sigma_y+\sigma_z)$ which is supposedly validated by the superposition principle. This already makes little sense to me because of Hooke's law implying $\frac{\nu}{E}(\sigma_y+\sigma_z)=0$. In this context, I interpret the superposition principle as the sum of the inputs(stresses) equalling the sum of the outputs (extensions). What's happening here? As you might guess, this is a new topic for me.

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  • $\begingroup$ I think I finally understand what you are asking. See the ADDENDUM to my answer below. $\endgroup$ Aug 30, 2017 at 12:27

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The general equations for the three strains are: $$\epsilon_x=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}{E}$$ $$\epsilon_y=\frac{\sigma_y-\nu(\sigma_x+\sigma_z)}{E}$$ $$\epsilon_z=\frac{\sigma_z-\nu(\sigma_x+\sigma_y)}{E}$$ For the case of uniaxial loading $\sigma_x=\sigma$ in the x-direction, while the stresses in the y and z directions are zero ($\sigma_y=\sigma_z=0$),we get from the above three equations:$$\epsilon_x=\frac{\sigma}{E}$$ $$\epsilon_y=-\nu\frac{\sigma}{E}=-\nu\epsilon_x$$ $$\epsilon_z=-\nu\frac{\sigma}{E}=-\nu\epsilon_x$$

Now let's consider a different kind of loading where, instead of the load being only in the x direction, there are also equal stresses in the y and z directions, such that $$\sigma_x=\sigma_y=\sigma_z=\sigma$$If we substitute these into the three general equations for the strains in terms of the stresses, we obtain:$$\epsilon_x=\epsilon_y=\epsilon_z=(1-2\nu)\frac{\sigma}{E}$$This equation can also be obtained by starting with a uniaxial stress in the x-direction, then superimposing an equal stress in the y-direction, and then superimposing an equal stress in the z direction.

If $\sigma=-p$, the linear strains in the three directions are $$\epsilon=-(1-2\nu)\frac{p}{E}$$The volumetric strain $\epsilon_v$ is three times the linear strain, so,$$\epsilon_v=-3(1-2\nu)\frac{p}{E}$$ Since the original volume is $L_xL_yL_z$, the change in volume is $$\Delta V=-3(1-2\nu)\frac{p}{E}L_xL_yL_z$$ ADDENDUM

Here's another way of looking at it. Suppose you start out with a uniaxial stress of $\sigma_x$ on the body. Then, the strains in the three directions are $$\epsilon_x=\frac{\sigma_x}{E}$$ $$\epsilon_y=-\nu\epsilon_x=-\nu\frac{\sigma_x}{E}$$and$$\epsilon_z=-\nu\epsilon_x=-\nu\frac{\sigma_x}{E}$$ If, instead, you start out with a unixial stress of $\sigma_y$ on the body, then the strains in the three directions are $$\epsilon_y=\frac{\sigma_y}{E}$$ $$\epsilon_x=-\nu\epsilon_y=-\nu\frac{\sigma_y}{E}$$and$$\epsilon_z=-\nu\epsilon_y=-\nu\frac{\sigma_y}{E}$$

If, instead, you start out with a unixial stress of $\sigma_z$ on the body, then the strains in the three directions are $$\epsilon_z=\frac{\sigma_z}{E}$$ $$\epsilon_x=-\nu\epsilon_z=-\nu\frac{\sigma_z}{E}$$and$$\epsilon_y=-\nu\epsilon_z=-\nu\frac{\sigma_z}{E}$$

If, instead, you impose all three of these stresses simultaneously on the body, the strains you get are obtained by linearly superimposing (i.e., adding together) the three strains from the uniaxial stress cases: $$\epsilon_x=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}{E}$$ $$\epsilon_y=\frac{\sigma_y-\nu(\sigma_x+\sigma_z)}{E}$$ $$\epsilon_z=\frac{\sigma_z-\nu(\sigma_x+\sigma_y)}{E}$$

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  • $\begingroup$ Yes, the addendum clarified everything! $\endgroup$ Aug 30, 2017 at 19:17

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