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 $$\nabla_j = \pdv*{x_j}$$ 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} %= (\nabla \vec{u})^\top \cdot \dd{\va{l}} } \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.