Is shear strain an additive quantity? Consider a box with rigid walls containing an elasic medium, subject possibly to some body forces or tractions.
The volume is an additive quantity, in the sense that the total volume change of the system may be written as the sum of volume changes of subsystems. Therefore, if I define the strain tensor $\epsilon_{ij}$, then for any forcing carried out while holding the walls fixed, $$ \Delta V = \int \mathrm{d}{V} \, \mathrm{Tr\,}{\epsilon_{ij}} = 0.$$
Is there any analogous additivity for shear strain? Is it true that if the boundaries of the medium are fixed (say, cubical) then  $$  \int \mathrm{d}{V} \, \epsilon_{xy} = 0?$$
Is it true more generally that, if the box is deformed to a parallelopiped with shear strain $\gamma$, then $$  \int \mathrm{d}{V} \, \epsilon_{xy} = V \gamma?$$
Perhaps there is some theorem about integrating the gradient of a vector field which will help?
 A: Calling the box size $L$, it seems like we can say
$$\int \mathrm{d}{V} \, \epsilon_{xy} = \frac{1}{2}\int \mathrm{d}{x} \,\mathrm{d}{y} \,\mathrm{d}{z}\, \left[ \frac{\partial u_x}{\partial y} + \frac{\partial u_y}{\partial x} \right],$$
and thus 
$$\int \mathrm{d}{V} \, \epsilon_{xy} = \frac{1}{2} L^2 \left[ u_x{\left(y = L\right)} - u_x{\left(y = 0\right)} + u_y{\left(x = L\right)} - u_y{\left(x = 0\right)}\right]$$
$$\int \mathrm{d}{V} \, \epsilon_{xy} = \frac{1}{2} \gamma V.$$
I guess the factor of $1/2$ is the usual factor between the "engineering strain" and $\epsilon_{xy}$. But this still almost seems too simple to be true.
A: *

*Is shear strain an additive quantity?____Yes, if in the same plane.

If two "true"(ie: non-infinitesimal) strains are done sequentially in the same plane, then you can just do the sum of the two strain angles as a single transformation.  This is because "true" strain angles and rotation angles are how we parameterize all the 3x3 matrices that strain and rotate 3-vectors. Rotations and strains form the Lie group GL(3,R). This is the group of all invertible 3x3 matrices M of real numbers.  Two shear strains in the same plane commute.


*How do strains change a 3-volume?____Shear strains do not change the volume.
$$
M=e^{\Theta}
$$
$$
\Theta = \begin{bmatrix}
           0 & \theta^{12} &-\theta^{13} \\
-\theta^{12} &           0 & \theta^{23} \\ 
 \theta^{13} &-\theta^{23} &           0  \\
\end{bmatrix}_{Asymmetric}
+ \begin{bmatrix} 
\epsilon^{11} & \epsilon^{12} & \epsilon^{13} \\
\epsilon^{12} & \epsilon^{22} & \epsilon^{23} \\
\epsilon^{13} & \epsilon^{23} & \epsilon^{33} \\ 
\end{bmatrix}_{Symmetric}
$$
$$
dV'=det[M]dV
$$
where $det[M]$ is the Jacobian of the transformation of coordinates from (x,y,z) to (x',y',z').
$$
=det[e^{\Theta}]dV
$$
$$
=e^{Trace[\Theta]}dV
$$
$$
=e^{Trace[\begin{bmatrix} 
\epsilon^{11} & \epsilon^{12} & \epsilon^{13} \\
\epsilon^{12} & \epsilon^{22} & \epsilon^{23} \\
\epsilon^{13} & \epsilon^{23} & \epsilon^{33} \\ 
\end{bmatrix}]}dV
$$
$$
=e^{(\epsilon^{11}+\epsilon^{22}+\epsilon^{33})}dV
$$
So, if $(\epsilon^{11}+\epsilon^{22}+\epsilon^{33})=0$, then the volume remains unchanged.  Therefore, the shear strains (ie: the off diagonal elements $\epsilon^{12},\ \epsilon^{13},\ \epsilon^{23}$) do not change the volume.
Please see my answer to this question for more of an explanation as to how all this group stuff applies to the usual explanation of strain in an engineering class.
