# Strain rate tensor derivation

The strain (deformation) tensor $\varepsilon_{ij}$ is given by \begin{align} \varepsilon_{ij}&=\left[\begin{array}{ccc}\varepsilon_{xx} & \varepsilon_{xy} &\varepsilon_{xz} \\ \varepsilon_{yx} & \varepsilon_{yy} &\varepsilon_{yz} \\ \varepsilon_{zx} & \varepsilon_{zy} & \varepsilon_{zz} \end{array}\right]\\ &= \left[\begin{array}{ccc} \partial_xu & \frac12\left(\partial_yu+\partial_x v\right) & \frac12\left(\partial_zu+\partial_xw\right)\\ \frac12\left(\partial_xv+\partial yu\right)&\partial_yv & \frac12\left(\partial_zv+\partial_yw\right)\\ \frac12\left(\partial_xw+\partial_zu\right)&\frac12\left(\partial_yw+\partial_zv\right)&\partial_zw \end{array} \right] \end{align}

Does anyone know how to derive this tensor from physical laws, such as Newton's laws?

## 2 Answers

Mike Stone is correct. There is no derivation from Newton's laws, and it is just geometry, but I will present it a little differently. Strain angles and rotation angles are how we parameterize all the 3x3 matrices that strain and rotate 3-vectors. Rotations and strains form the group GL(3,R). This is the group of all invertible 3x3 matrices M of real numbers.

We can describe what these transformations do by just talking about the matrices $M$ that are very close to the identity matrix, where all elements in the matrix $\Theta$ are <<1. All these elements are in radians. $$M=I+\Theta$$ $$\Theta = \begin{bmatrix} 0 & \theta^{12} &-\theta^{13} \\ -\theta^{12} & 0 & \theta^{23} \\ \theta^{13} &-\theta^{23} & 0 \\ \end{bmatrix}_{Asymmetric} \ + \begin{bmatrix} \epsilon^{11} & \epsilon^{12} & \epsilon^{13} \\ \epsilon^{12} & \epsilon^{22} & \epsilon^{23} \\ \epsilon^{13} & \epsilon^{23} & \epsilon^{33} \\ \end{bmatrix}_{Symmetric}$$

Now apply M to a vector x to get X. We have moved a piece of a body from x to X. $$X^i=M^{ij}x^j=(\delta^{ij}+\Theta^{ij})x^j$$ $$u^i=(X^i-x^i)=\Theta^{ij} x^j$$ Where u is the “displacement” of the point. As we move around to different points x in the body, we will get different u ‘s. Differentiating the last equation gives $$\frac{\partial u^i}{\partial x^j}=\Theta^{ij}$$ Therefore we can express the elements of $\Theta$ also in terms the derivatives of the displacements, remembering the strains $\epsilon$ are the symmetric part of $\Theta$ and the rotations $\theta$ are the antisymmetric part of $\Theta$. $$\epsilon^{ij}=\frac{1}{2}[\frac{\partial u^i}{\partial x^j}+\frac{\partial u^j}{\partial x^i}]$$ $$\theta^{ij}=\frac{1}{2}[\frac{\partial u^i}{\partial x^j}-\frac{\partial u^j}{\partial x^i}]$$ So far, we have been talking about doing strains and rotations by small angles. Now, by taking products of these small angle transformations, we build all the matrices $M$ in the group for any size elements in $\Theta$. This shows us the matrices $M$ which do macroscopic rotations and strains. $$M=\lim_{n\to \infty}(1+\frac{\Theta}{n})^n=e^{\Theta}=I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +…$$

The $\theta$ are antisymmetric, make M orthogonal ($M^T =M^{-1}$), and leave the length ${x_1}^2 + {x_2}^2+{x_3}^2$ invariant. Because lengths are invariant, the transformations are called rotations. The $\epsilon$ do not leave lengths invariant and are called strains.

The parameter $\theta^{12}$ means rotate the object about axis1 X axis2. That is, put your right thumb perpendicular to the plane formed by axis1 and axis2, such that you fingers would push axis1 into axis 2. Then in this same rotation direction about your thumb rotate the object by $\theta^{12}$ radians.

The parameter $\epsilon^{11}$ means stretch the object by $(1+\epsilon^{11})$ along axis1.

The parameter $\epsilon^{12}$ means parallelepiped the object in the plane containing axis1 and axis2. For example, a square box with its sides initially along axis1 and axis2, becomes a parallelepiped with its sides tilted inward from axis1 and axis2 and its diagonal from the origin stretched. Both sides now make inward angles of $\epsilon^{12}$ radians with their respective axis1 or axis2.

Examples of large rotation and strain matrices are:

Rotate about axis3 by $\theta^{12}$ radians. $$M(\theta^{12})=\begin{bmatrix} cos(\theta^{12}) & -sin(\theta^{12}) & 0 \\ sin(\theta^{12}) & cos(\theta^{12}) & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ Strain (parallelepiped) about the axis3 by $\epsilon^{12}$ radians. $$M(\epsilon^{12})=\begin{bmatrix} cosh(\epsilon^{12}) & sinh(\epsilon^{12}) & 0 \\ sinh(\epsilon^{12}) & cosh(\epsilon^{12}) & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ Strain (stretch) along the axis1 by $\epsilon^{11}$ radians. $$M(\epsilon^{11})=\begin{bmatrix} e^{\epsilon^{11} } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ Notice that for $\epsilon^{11}<<1$, the fractional change of the length of an object in the 1 direction is $\epsilon^{11}$.

The purpose of this rather too long answer was to show the strains and rotations are intimately related and are a group of transformations we can do with our fingers to any piece of material. How the displacements u change as you move around in the body is just what the transformations cause, but is not the fundamental concept of strain or rotation.

There is no derivation from Newton, because strain is purely geometric concept. It is measuring the deformation (the change in the length and angles of the spacing between the atome) of the body. If you take an orthonormal basis of vectors ${\bf e}_1$, ${\bf e}_2$, ${\bf e}_3$ at a point ${\bf r}_0$ and regard them as painted on the atoms in the body. Now displace each atom (and hence the stretch and rotate the basis vectors) by a small position dependent amount ${\bf u}=(u_1,u_2,u_3)$, the first-order change in the induced metric tensor $g_{ij}\equiv {\bf e}_i\cdot {\bf e}_j=\delta_{ij}$ is easily seen to be $$\delta_{ij}\to g_{ij}= \delta_{ij}+ \left(\frac{\partial u_i}{\partial x_j}+ \frac{\partial u_j}{\partial x_i}\right)$$ The strain tensor is, by definition, one half of this change.

• can you explain why it is easily seen that the first-order change in the induced metric tensor is as described as above? And relate it to one of the elements in the strain tensor above? – torgny Feb 12 '17 at 20:13
• It's straightforward plug-and-chug. You have ${\bf e}_1\mapsto {\bf e}_1+ \partial_{x_1}{\bf u}_1$ etc. Then take the dot product of the new ${\bf e}_i$. – mike stone Feb 12 '17 at 21:55