path: root/source/know/concept/cauchy-strain-tensor/index.md

---
title: "Cauchy strain tensor"
sort_title: "Cauchy strain tensor"
date: 2021-03-31
categories:
- Physics
- Continuum physics
layout: "concept"
---

**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.