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 X⃗\va{X} be the original location of a particle, and x⃗\va{x} its new location after the deformation. We can thus define the displacement field u⃗\va{u}:

u⃗≡x⃗−X⃗\begin{aligned} \va{u} \equiv \va{x} - \va{X} \end{aligned}

We restrict ourselves to infinitesimal strain, where u⃗\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 u^\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 x⃗\va{x} is the reference, and the old position X⃗\va{X} is expressed as a function of x⃗\va{x}:

u⃗(x⃗)≡x⃗−X⃗(x⃗)\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 x⃗\va{x} and x⃗+a⃗\va{x} + \va{a}, where a⃗\va{a} is a tiny vector pointing from one to the other. Before the displacement, those points respectively had these positions, where we define A⃗\va{A} as the “old” version of a⃗\va{a}:

X⃗=X⃗(x⃗)X⃗+A⃗=X⃗(x⃗+a⃗)\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 x⃗\va{x} are our reference, we would like to write A⃗\va{A} without X⃗\va{X}. To do so, we use the definition of u⃗(x⃗)\va{u}(\va{x}), yielding:

A⃗=X⃗(x⃗+a⃗)−X⃗(x⃗)=(x⃗+a⃗−u⃗(x⃗+a⃗))−(x⃗−u⃗(x⃗))=a⃗−u⃗(x⃗+a⃗)−u⃗(x⃗)\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 a⃗\va{a} is tiny by definition, we expand the middle term to first order in a⃗\va{a}:

u⃗(x⃗+a⃗)≈u⃗(x⃗)+ax∂u⃗∂x+ay∂u⃗∂y+az∂u⃗∂z=u⃗(x⃗)+(a⃗⋅∇)u⃗(x⃗)\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” δa⃗\delta\va{a} as the difference between a⃗\va{a} and A⃗\va{A} like so:

δa⃗≡a⃗−A⃗=(a⃗⋅∇)u⃗(x⃗)\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 ∇j≡∂/∂xj\nabla_j \equiv \ipdv{}{x_j} simply being the partial derivative with respect to the jjth coordinate:

δai=∑jaj∇jui\begin{aligned} \delta a_i = \sum_{j} a_j \nabla_j u_i \end{aligned}

Where ∇jui\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 u⃗(x⃗)\va{u}(\va{x}) as one where ∣∇jui∣≪1|\nabla_j u_i| \ll 1.

Now, to solve the problem of macroscopic movements, we take another tiny vector b⃗\va{b} starting in the same point x⃗\va{x} as a⃗\va{a}. Here is the trick: if the whole body is uniformly translated or rotated, the scalar product a⃗⋅b⃗\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:

δ(a⃗⋅b⃗)≡a⃗⋅b⃗−A⃗⋅B⃗\begin{aligned} \delta(\va{a} \cdot \va{b}) \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B} \end{aligned}

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

δ(a⃗⋅b⃗)=δa⃗⋅b⃗+a⃗⋅δb⃗\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 δa⃗\delta\va{a} and δb⃗\delta\va{b} yields:

δ(a⃗⋅b⃗)=∑iδai bi+∑iδbi ai=∑ij∇jui ajbi+∑ij∇jui aibj=∑ij(∇iuj+∇jui) aibj\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 u^\hat{u} such that it has uiju_{ij} as components:

uij≡12(∇jui+∇iuj)\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:

δ(a⃗⋅b⃗)=2∑ijuijaibj=2a⃗⋅u^⋅b⃗\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 u^\hat{u} is a second-rank tensor, and can alternatively be expressed like so:

u^≡12(∇u⃗+(∇u⃗)⊤)\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. uij=ujiu_{ij} = u_{ji}.

Geometry

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

uab=a⃗⋅u^⋅b⃗∣a⃗∣∣b⃗∣\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 u^\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 u^\hat{u} is symmetric, such a basis always exists.

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

δ∣a⃗∣∣a⃗∣=uaa\begin{aligned} \boxed{ \frac{\delta|\va{a}|}{|\va{a}|} = u_{aa} } \end{aligned}

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

δ(a⃗⋅b⃗)=δ(∣a⃗∣∣b⃗∣cos⁡θ)=δ∣a⃗∣∣b⃗∣cos⁡θ+∣a⃗∣δ∣b⃗∣cos⁡θ−∣a⃗∣∣b⃗∣sin⁡θ δθ\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 δθ\delta\theta, using the fact that δ(a⃗⋅b⃗)=2∣a⃗∣∣b⃗∣uab\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab} thanks to the projection uabu_{ab}:

δθ=δ∣a⃗∣∣b⃗∣cos⁡θ+∣a⃗∣δ∣b⃗∣cos⁡θ−2∣a⃗∣∣b⃗∣uab∣a⃗∣∣b⃗∣sin⁡θ\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 δ∣a⃗∣/∣a⃗∣=uaa\delta|\va{a}|/|\va{a}| = u_{aa}, we arrive at the following expression:

δθ=(uaa+ubb)cos⁡θ−uabsin⁡θ\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 dl⃗\dd{\va{l}} is easy to express using the displacement field u⃗\va{u}:

δ(dl⃗)=(dl⃗⋅∇)u⃗\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 dV\dd{V} by treating it as the volume of a tiny parallelepiped spanned by a⃗\va{a}, b⃗\va{b} and c⃗\va{c}:

δ(dV)=δ(a⃗×b⃗⋅c⃗)=δa⃗×b⃗⋅c⃗+a⃗×δb⃗⋅c⃗+a⃗×b⃗⋅δc⃗=(a⃗⋅∇)u⃗×b⃗⋅c⃗+a⃗×(b⃗⋅∇)u⃗⋅c⃗+a⃗×b⃗⋅(c⃗⋅∇)u⃗\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):

δ(dV)=(a⃗⋅∇)u⃗×b⃗⋅c⃗+(b⃗⋅∇)a⃗×u⃗⋅c⃗+(c⃗⋅∇)a⃗×b⃗⋅u⃗\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:

δ(dV)=(b⃗×c⃗(a⃗⋅∇)×b⃗+c⃗×a⃗(b⃗⋅∇)+a⃗×b⃗(c⃗⋅∇))⋅u⃗=(a⃗×b⃗⋅c⃗)(∇⋅u⃗)\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 dV\dd{V}, leading to the following infinitesimal volume change:

δ(dV)=∇⋅u⃗dV\begin{aligned} \boxed{ \delta(\dd{V}) = \nabla \cdot \va{u} \dd{V} } \end{aligned}

Finally, for the surface element dS⃗=a⃗×b⃗\dd{\va{S}} = \va{a} \cross \va{b}, we use that the volume element dV=c⃗⋅dS⃗\dd{V} = \va{c} \cdot \dd{\va{S}}:

δ(dV)=δ(c⃗⋅dS⃗)=δc⃗⋅dS⃗+c⃗⋅δ(dS⃗)=(c⃗⋅∇)u⃗⋅dS⃗+c⃗⋅δ(dS⃗)\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 δ(dV)\delta(\dd{V}), we arrive at the following equation:

∇⋅u⃗(c⃗⋅dS⃗)=(c⃗⋅∇)u⃗⋅dS⃗+c⃗⋅δ(dS⃗)\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 c⃗\va{c} is dot-multiplied at the front of each term, we remove it, and isolate the rest for δ(dS⃗)\delta(\dd{\va{S}}):

δ(dS⃗)=((∇⋅u⃗)1^−∇u⃗)⋅dS⃗\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.