Shear Strain and no change in volume of the element

Question

Consider that at a particular point of the body the state of stress was that of a pure shear one, with $\tau_{xy}$ and $\gamma_{xy}$ as the shear stress and strain. I read two statements in two different books in this regard-

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In 1) it is stated that there will be no change in volume of the element. In 2) it is stated that the z face will be a rhombus. I'm thinking of these statements to be corollaries of each other.

So if I start off by saying that volume of the element doesn't change how does that makes me conclude that the x, y and z dimensions won't change as well?

OR

If the x, y , z dimensions do not change how can I conclude that the volume doesn't change?

Answer 1

Let's say that a cube under horizontal shear, $ \ \tau \ $deforms to a parallelogram as shown.

As we see a small triangular wedge with the base $\ W= shear\ strain$ is subtracted from the left side but the same volume is added to the right side.

So under small angle (linear) shear strain, the volume remains constant while the coordinates of x or Y change.

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Answer 2

The mass and mass density of elastic material will hold constant, they will not change (since no gain nor loss) due to deformation. The shape can change, but the volume maintains the same. It can be observed by applying moderate pressure on the dole, so as not to change the mass density, as a result, it deforms but weights the same ($V = \dfrac{m}{\rho}$).

Answer 3

kamran's answer above is a very intuitive and elegant way to look at this. However, I feel it falls a little bit short when looking at the general phenomenon of shear deformation.

First, notice that both pictures you posted, reproduced below for clarity, as well as kamran's drawing, are bodies subjected to simple shear, meaning that there is only one of the 3 possible shear being applied.

So it's natural to ask, ¿are all shear deformations volume-preserving? To answer this we can turn to continuum mechanics and introduce the concept of motion (in the material description) and the deformation gradient.

Let there be a body in a space described by the coordinates $X_1,X_2$ and $X_3$ as shown in the figure below .

The motion (or deformation) of the body is a function $\mathbf{x}(\mathbf{X},t)$ that describes the position of the particle originally at position $\mathbf(X)$ at time $t$. For example, simple shear is described by $$ x_1=X_1+kX_3 \qquad x_2=X_2 \qquad x_3=X_3, $$ which results in the following configuration

To know how much the volume changed, first we look at a line element in the reference configuration, call it $d\mathbf{X}$, and we want to see how it transforms under the motion $\mathbf{x}(\mathbf{X},t)$. The next figure shows the line element $d\mathbf{X}$ and its image on the deformed configuration, $d\mathbf{x}$.

Now we note that $$ \frac{d\mathbf{x}}{d\mathbf{X}}=Grad(\mathbf{x}) \\ d\mathbf{x}=Grad(\mathbf{x})d\mathbf{X} $$

But looking at the previous image, $d\mathbf{x}$ is equal to $$ d\mathbf{x}=\mathbf{x}(\mathbf{X}+d\mathbf{x})-\mathbf{x}(\mathbf{X}) $$

Then, the gradient of the motion, call it $\mathbf{F}$, relates how a line segment transforms under a motion. Now we can apply this to a volume element. Recall that the volume of a parallelepiped is given by the triple scalar product, then in the reference configuration the volume of an infinitesimal volume element is $$ dV=d\mathbf{X}_1\cdot(d\mathbf{X}_2\times d\mathbf{X}_3) $$
and, due to the property we just established for the gradient, the volume of the volume element on the deformed configuration is $$ \begin{aligned} dv&=d\mathbf{x}_1\cdot(d\mathbf{x}_2\times d\mathbf{x}_3) \\ &=\mathbf{F}d\mathbf{X}_1\cdot(\mathbf{F}d\mathbf{X}_2\times \mathbf{F}d\mathbf{X}_3) \\ &=det\, \mathbf{F}\,\left(d\mathbf{X}_1\cdot(d\mathbf{X}_2\times d\mathbf{X}_3)\right)\\ &=det\,\mathbf{F}\, dV \end{aligned} $$

This gives us a way to evaluate the volume change for a body subjected to a given motion.

Going back to our simple shear motion $\mathbf{x}$, it's deformation gradient is given by

$$ \mathbf{F}=\frac{d\mathbf{x}}{d\mathbf{X}}=\left[ \begin{matrix} 1 & 0 & k\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix} \right]. $$ The determinant of $\mathbf{F}$ is $det\,\mathbf{F}=1$, meaning a volume element will retain its volume under this transformation (because $dV=1\cdot dv$).

If you had pure (but not simple) shear, i.e. a motion given by, for example, $$ x_1=X_1+kX_3 \qquad x_2=X_2 \qquad x_3=X_3+kX_1, $$ then deformation gradient is $$ \mathbf{F}=\left[ \begin{matrix} 1 & 0 & k\\ 0 & 1 & 0\\ k & 0 & 1 \end{matrix} \right]. $$ Then the determinant of the deformation gradient is $det\,\mathbf{F}=1-k^2$, so pure shear is not a volume-preserving transformation.