3.2.2 Strain Tensor

Under applied forces solids are strained resulting in a change of volume and shape. In the approximation of the elastic continuum, the position of each point of a solid is described by the vector $\vec{r}$ which in some Cartesian coordinate system has the components $x_{1}$, $x_{2}$, $x_{3}$. Under strain all points of a solid are in general shifted. If the position of a given point before strain was $\vec{r}$, then after strain it is $\vec{r}^{'}$ with components $x^{'}_{i}$, $i=1,2,3$. The displacement of the point is characterized by the displacement vector defined as

$$\vec{u}=\vec{r}^{'}-\vec{r}.$$ (3.9)

The coordinates $x_{i}^{'}$ of a shifted point are functions of the coordinates $x_{i}$ of the same point before strain. This means that the displacement vector $\vec{u}$ is also a function of $x_{i}^{'}$. This function completely determines the strained state of a solid.

When a solid is strained, the distances between points change. If before strain the distance between two infinitely close points was $dx_{i}$, then after strain it is equal to $dx_{i}^{'}=dx_{i}+du_{i}$. The distance between these two points before strain is

$$dl=\sqrt{dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}}$$ (3.10)

and after strain:

$$dl^{'}=\sqrt{dx_{1}^{'2}+dx_{2}^{'2}+dx_{3}^{'2}}.$$ (3.11)

Substituting the expressions for $dx_{i}^{'}$ through $du_{i}$ the following expression for $dl^{'2}$ is obtained:

$$dl^{'2}=dl^{2}+2\frac{\partial u_{i}}{\partial x_{k}}dx_{i}dx_{k}... ...artial u_{i}}{\partial x_{k}}\frac{\partial u_{i}}{\partial x_{l}}dx_{k}dx_{l}.$$ (3.12)

Since indices in the double sum can be exchanged, the last expression can be rearranged and rewritten as:

$$dl^{'2}=dl^{2}+2\varepsilon_{ik}dx_{i}dx_{k},$$ (3.13)

where a tensor of the second rank has been introduced:

$$\varepsilon_{ik}=\frac{1}{2}\biggl(\frac{\partial u_{i}}{\partial... ...ac{\partial u_{l}}{\partial x_{i}}\frac{\partial u_{l}}{\partial x_{k}}\biggr).$$ (3.14)

The second rank tensor $\varepsilon_{ik}$ is called the strain tensor. As can be seen from the definition (3.14), it represents a symmetric tensor:

$$\varepsilon_{ik}=\varepsilon_{ki},\thickspace\forall\thickspace i,k = 1,2,3.$$ (3.15)

Any symmetric tensor can be reduced to the principle axes. This means that at each given point the coordinate system can be chosen in such a way that only diagonal elements $\varepsilon_{11}$, $\varepsilon_{22}$ and $\varepsilon_{33}$ will be non-zero and all non-diagonal elements vanish. It should be noted that if a tensor is reduced to the diagonal form at a given point, it will be in general non-diagonal at all other points of a given solid considered as continuum.

If the strain tensor is reduced at a given point to its principle axes, in the elementary volume built around this point the element of the length (3.13) takes the form:

$$dl^{'2}=(1+2\varepsilon_{11})dx_{1}^{2}+(1+2\varepsilon_{22})dx_{2}^{2}+(1+2\varepsilon_{33})dx_{3}^{2}.$$ (3.16)

This expression is decomposed into three independent terms. This means that at any given elementary volume of a solid the strain can be considered as a set of three independent deformations along three relatively orthogonal directions - the principal axes of the tensor. Each of these deformations represents a simple stretching or compressing along the corresponding direction: the length $dx_{i}$ along the $i$-th principle axis turns into the length $dx_{i}^{'}$:

$$dx_{i}^{'}=\sqrt{1+2\varepsilon_{ii}}dx_{i}.$$ (3.17)

The relative elongation along the $i$-th axis is thus given as:

$$\frac{dx_{i}^{'}-dx_{i}}{dx_{i}}=\sqrt{1+2\varepsilon_{ii}}-1.$$ (3.18)

A deformation is considered small if the change of any distance in a solid turns out to be much less than the distance itself. In other words all relative elongations are much less than unity. In this work only strain of this kind is considered.

When strain is weak in the sense mentioned above, the displacements $u_{i}$ and their derivatives are small. Thus, in the general expression (3.14) the last term is negligible and can be omitted. Therefore, in the case of weak strain, the components $\varepsilon_{ik}$ of the strain tensor are determined by the following expression:

$$\varepsilon_{ik}=\frac{1}{2}\biggl(\frac{\partial u_{i}}{\partial x_{k}}+\frac{\partial u_{k}}{\partial x_{i}}\biggr).$$ (3.19)

In this case the relative elongations are thus equal:

$$\sqrt{1+2\varepsilon_{ii}}-1\approx\varepsilon_{ii},$$ (3.20)

and given by the eigen values of the strain tensor. S. Smirnov: