Categories: Continuum physics, Physics.

# Cauchy strain tensor

Strain quantifies the deformation of a solid object. If the body has been deformed, e.g. by pulling or bending, its constituent particles have moved a bit. Let $\va{X}$ be the original location of a particle, and $\va{x}$ its new location after the deformation. We can thus define the displacement field $\va{u}$:

\begin{aligned} \va{u} \equiv \va{x} - \va{X} \end{aligned}

We restrict ourselves to infinitesimal strain, where $\va{u}$ is so tiny that the material’s properties are unchanged, and a slowly-varying strain, where the particle’s neighbourhood has been distorted, but not completely changed.

A key challenge when quantifying deformation is that we need to somehow exclude movements of the entire body: for example, you can bend a twig in your hands while walking or dancing, but we are only interested in the twig’s shape change, not in your movements. The above definition of $\vu{u}$ includes both, so we should be careful how we extract the strain from it.

## Definition

We use the Eulerian description of deformation, where the new position $\va{x}$ is the reference, and the old position $\va{X}$ is expressed as a function of $\va{x}$:

\begin{aligned} \va{u}(\va{x}) \equiv \va{x} - \va{X}(\va{x}) \end{aligned}

Let us choose two nearby points in the deformed solid, and call them $\va{x}$ and $\va{x} + \va{a}$, where $\va{a}$ is a tiny vector pointing from one to the other. Before the displacement, those points respectively had these positions, where we define $\va{A}$ as the “old” version of $\va{a}$:

\begin{aligned} \va{X} = \va{X}(\va{x}) \qquad \va{X} + \va{A} = \va{X}(\va{x} + \va{a}) \end{aligned}

Because the new positions $\va{x}$ are our reference, we would like to write $\va{A}$ without $\va{X}$. To do so, we use the definition of $\va{u}(\va{x})$, yielding:

\begin{aligned} \va{A} &= \va{X}(\va{x} + \va{a}) - \va{X}(\va{x}) \\ &= \big( \va{x} + \va{a} - \va{u}(\va{x} + \va{a}) \big) - \big( \va{x} - \va{u}(\va{x}) \big) \\ &= \va{a} - \va{u}(\va{x} + \va{a}) - \va{u}(\va{x}) \end{aligned}

Using the fact that $\va{a}$ is tiny by definition, we expand the middle term to first order in $\va{a}$:

\begin{aligned} \va{u}(\va{x} + \va{a}) \approx \va{u}(\va{x}) + a_x \pdv{\va{u}}{x} + a_y \pdv{\va{u}}{y} + a_z \pdv{\va{u}}{z} = \va{u}(\va{x}) + (\va{a} \cdot \nabla) \va{u}(\va{x}) \end{aligned}

With this, we can now define the “shift” $\delta\va{a}$ as the difference between $\va{a}$ and $\va{A}$ like so:

\begin{aligned} \delta{\va{a}} \equiv \va{a} - \va{A} = (\va{a} \cdot \nabla) \va{u}(\va{x}) \end{aligned}

In index notation, we write this expression as follows, with $\nabla_j \equiv \ipdv{}{x_j}$ simply being the partial derivative with respect to the $j$th coordinate:

\begin{aligned} \delta a_i = \sum_{j} a_j \nabla_j u_i \end{aligned}

Where $\nabla_j u_i$ are called the displacement gradients, and are just one step away from the desired definition of strain. Note that these gradients are dimensionless, so we can more formally define a slowly-varying displacement $\va{u}(\va{x})$ as one where $|\nabla_j u_i| \ll 1$.

Now, to solve the problem of macroscopic movements, we take another tiny vector $\va{b}$ starting in the same point $\va{x}$ as $\va{a}$. Here is the trick: if the whole body is uniformly translated or rotated, the scalar product $\va{a} \cdot \va{b}$ is unchanged, but if there is a non-uniform distortion, it changes. We thus define the scalar product’s difference like so:

\begin{aligned} \delta(\va{a} \cdot \va{b}) \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B} \end{aligned}

Where $\va{B}$ is the old version of $\va{b}$. Since these vectors are all tiny, we apply the product rule:

\begin{aligned} \delta(\va{a} \cdot \va{b}) &= \delta\va{a} \cdot \va{b} + \va{a} \cdot \delta\va{b} \end{aligned}

It is more informative to switch to index notation here. Inserting $\delta\va{a}$ and $\delta\va{b}$ yields:

\begin{aligned} \delta(\va{a} \cdot \va{b}) &= \sum_{i} \delta{a}_i \: b_i + \sum_{i} \delta{b}_i \: a_i \\ &= \sum_{ij} \nabla_j u_i \: a_j b_i + \sum_{ij} \nabla_j u_i \: a_i b_j \\ &= \sum_{ij} \big( \nabla_i u_j + \nabla_j u_i \big) \: a_i b_j \end{aligned}

At last, we define the Cauchy infinitesimal strain tensor $\hat{u}$ such that it has $u_{ij}$ as components:

\begin{aligned} \boxed{ u_{ij} \equiv \frac{1}{2} \big( \nabla_j u_i + \nabla_i u_j \big) } \end{aligned}

Which allows us to rewrite the shift of the scalar product in the following compact way:

\begin{aligned} \delta(\va{a} \cdot \va{b}) &= 2 \sum_{ij} u_{ij} a_i b_j = 2 \va{a} \cdot \hat{u} \cdot \va{b} \end{aligned}

The Cauchy strain tensor $\hat{u}$ is a second-rank tensor, and can alternatively be expressed like so:

\begin{aligned} \boxed{ \hat{u} \equiv \frac{1}{2} \big( \nabla \va{u} + (\nabla \va{u})^\top \big) } \end{aligned}

Where $\top$ is the transpose. Being defined from the scalar product, all macroscopic movements of the body are removed from the tensor, which turns out to make it symmetric, i.e. $u_{ij} = u_{ji}$.

## Geometry

So far we have used Cartesian coordinates, but we can choose any three vectors $\va{a}$, $\va{b}$ and $\va{c}$, and project $\hat{u}$ onto this basis. For example, the component $u_{ab}$ then becomes:

\begin{aligned} \boxed{ u_{ab} = \frac{\va{a} \cdot \hat{u} \cdot \va{b}}{\big|\va{a}\big| \big|\va{b}\big|} } \end{aligned}

And so forth, for the other eight components. The basis in which $\hat{u}$ is diagonal is the one formed by its eigenvectors, and their directions are the principal axes of strain at that point in the solid. Because $\hat{u}$ is symmetric, such a basis always exists.

Given a vector $\va{a}$, its relative length change due to the deformation is simply given by:

\begin{aligned} \boxed{ \frac{\delta|\va{a}|}{|\va{a}|} = u_{aa} } \end{aligned}

To find the angle change $\delta\theta$ between two vectors $\va{a}$ and $\va{b}$, we start with the product rule:

\begin{aligned} \delta(\va{a} \cdot \va{b}) = \delta(\big|\va{a}\big| \big|\va{b}\big| \cos\theta) = \delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta - \big|\va{a}\big| \big|\va{b}\big| \sin\theta \: \delta\theta \end{aligned}

We isolate this for $\delta\theta$, using the fact that $\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$ thanks to the projection $u_{ab}$:

\begin{aligned} \delta\theta = \frac{\delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta - 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}} {\big|\va{a}\big| \big|\va{b}\big| \sin\theta} \end{aligned}

By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$, we arrive at the following expression:

\begin{aligned} \boxed{ \delta\theta = \frac{(u_{aa} + u_{bb}) \cos\theta - u_{ab}}{\sin\theta} } \end{aligned}

Now, everything so far has been about tiny vectors, so the change of the line element $\dd{\va{l}}$ is easy to express using the displacement field $\va{u}$:

\begin{aligned} \boxed{ \delta(\dd{\va{l}}) = (\dd{\va{l}} \cdot \nabla) \va{u} } \end{aligned}

Next, we calculate the change of the differential volume element $\dd{V}$ by treating it as the volume of a tiny parallelepiped spanned by $\va{a}$, $\va{b}$ and $\va{c}$:

\begin{aligned} \delta(\dd{V}) = \delta(\va{a} \cross \va{b} \cdot \va{c}) &= \delta\va{a} \cross \va{b} \cdot \va{c} + \va{a} \cross \delta\va{b} \cdot \va{c} + \va{a} \cross \va{b} \cdot \delta\va{c} \\ &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c} + \va{a} \cross (\va{b} \cdot \nabla )\va{u} \cdot \va{c} + \va{a} \cross \va{b} \cdot (\va{c} \cdot \nabla) \va{u} \end{aligned}

We can reorder the factors like so (write it out in index notation if you are not convinced):

\begin{aligned} \delta(\dd{V}) &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c} + (\va{b} \cdot \nabla) \va{a} \cross \va{u} \cdot \va{c} + (\va{c} \cdot \nabla) \va{a} \cross \va{b} \cdot \va{u} \end{aligned}

By applying a couple of vector identities, we can rewrite this more compactly as follows:

\begin{aligned} \delta(\dd{V}) &= \Big( \va{b} \cross \va{c} (\va{a} \cdot \nabla) \cross \va{b} + \va{c} \cross \va{a} (\va{b} \cdot \nabla) + \va{a} \cross \va{b} (\va{c} \cdot \nabla) \Big) \cdot \va{u} \\ &= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u}) \end{aligned}

Here, we recognize the definition of $\dd{V}$, leading to the following infinitesimal volume change:

\begin{aligned} \boxed{ \delta(\dd{V}) = \nabla \cdot \va{u} \dd{V} } \end{aligned}

Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$, we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$:

\begin{aligned} \delta(\dd{V}) = \delta(\va{c} \cdot \dd{\va{S}}) = \delta\va{c} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) \end{aligned}

By comparing this to the previous result for $\delta(\dd{V})$, we arrive at the following equation:

\begin{aligned} \nabla \cdot \va{u} (\va{c} \cdot \dd{\va{S}}) = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) \end{aligned}

Since $\va{c}$ is dot-multiplied at the front of each term, we remove it, and isolate the rest for $\delta(\dd{\va{S}})$:

\begin{aligned} \boxed{ \delta(\dd{\va{S}}) = \big( (\nabla \cdot \va{u}) \hat{1} - \nabla \va{u} \big) \cdot \dd{\va{S}} } \end{aligned}

## References

1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.