# How to deal with rotation of an element in theory of elasticity?

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?

• On mobile so can't give full answer, but look up Mohr's circle. You'll have individual strains in both x and y, but you will find one axis x' and y' where you can define no strain. – Mark Jun 10 '17 at 14:59
• I'm guessing you're looking for torsion. – JMac Jun 10 '17 at 15:47
• In the first picture, the equations for the strain use engineering strain which is only valid for small rotations. So in the second picture, $\cos \varphi$ is small (approximately $\cos\varphi = 1 - \frac 1 2 \varphi^2$) and the second-order term $\varphi^2$ is ignored. So $\epsilon_x = 0$. If you want to model large rotations, use Green strain, not engineering strain, and you will get $\epsilon_x = 0$ exactly, for any value of $\varphi$. – alephzero Jun 10 '17 at 20:11

$$\varphi(x) = \left[\begin{array}{c} xcos(\varphi) - ysin(\varphi) \\ xsin(\varphi) + ycos(\varphi) \\ z \\ \end{array}\right]$$ 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.
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 $\epsilon_{i,j}$.