As you know in linear theory of elasticity equations for strains are given as (according to picture):
But what will be if an element will just rotate (strain equation is on the picture):
Intuitively i think there should be no strains. The main question is how to deal with it?
The answer to this lies in the defining equation for the strains that you have supplied. The full displacement gradient $u_{i,j}$ (where the $\bullet_{,j}$ represents the $j^{\rm th}$ derivative) can be linearly decomposed into a symmetric and an anti-symmetric component: $u_{i,j} = \epsilon_{i,j} + \omega_{i,j}$, in the usual way as for all rank-2 tensors.
The symmetric part, $\epsilon_{i,j} = \frac{u_{i,j}+u_{j,i}}{2}$ represents the strain, the only thing that counts for elastic energy, whereas the anti-symmetric part $\omega_{i,j} = \frac{u_{i,j}-u_{j,i}}{2}$ represents rigid body rotation. And as you remarked, this costs no energy to the system. Hence, for elasticity, we only ever talk about $\epsilon_{i,j}$.
From your picture, your deformation map would be something like
$$ \varphi(x) = \left[\begin{array}{c} xcos(\varphi) - ysin(\varphi) \\ xsin(\varphi) + ycos(\varphi) \\ z \\ \end{array}\right] $$ Deformation gradient $$ F = \left[\begin{array}{ccc} cos(\varphi) & -sin(\varphi) & 0 \\ sin(\varphi) & cos(\varphi) & 0 \\ 0 & 0 & 1\\ \end{array}\right] $$ Green deformation tensor $$ G = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{array}\right] $$ $$ \therefore E = \frac{1}{2} [G - I] = 0 $$ Which is what we would expect, but in order to calculate this we needed to use higher order terms. Since: $$ G = FF^T = (\nabla u + I)(\nabla u + I)^T = \nabla u + \nabla u^T + \nabla u \nabla u ^T + I $$ If each component of: $$ \nabla u < 1 $$ then each component of: $$ \nabla u \nabla u ^T \ll 1 $$ We expect our $\varphi$ to be very small, so we can ignore our higher order terms. Let's linearize the deformation gradient: $$ cos(\varphi) \approx 1, sin(\varphi) \approx \varphi \\ F = \left[\begin{array}{ccc} 1 & -\varphi & 0 \\ \varphi & 1 & 0 \\ 0 & 0 & 1\\ \end{array}\right] \\ G = \left[\begin{array}{ccc} 1 + \varphi ^2 & 0 & 0 \\ 0 & 1 + \varphi ^2 & 0 \\ 0 & 0 & 1\\ \end{array}\right] \\ \therefore E = \frac{1}{2} \left[\begin{array}{ccc} \varphi ^2 & 0 & 0 \\ 0 & \varphi ^2 & 0 \\ 0 & 0 & 0\\ \end{array}\right] \approx 0 $$ This makes since, because if $\varphi$ is small, $\varphi^2$ is much smaller and we can accept the error assuming geometrically linear behavior. We just need to watch out for non-small rotations that will cause our linearization to have a larger error.
Resource:
Fundamentals of Structural Mechanics 2nd, Keith D. Hjelmstad
