Stress

What is the force experienced by each point? This is called stress inside the material.

There are a couple types of stress:

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Cauchy’s Stress Theory

Consider an infinitesimal plane within a material. The force acting on that surface is called the traction force denoted as ${\mathbf{T}}^{(\mathbf{n})}$, dependent on the orientation of the plane defined by the normal vector $\mathbf{n}$. It satisfies the reciprocity property, meaning the force on the opposite side of the plane is equal in magnitude, but opposite in direction: ${\mathbf{T}}^{(-\mathbf{n})}=-{\mathbf{T}}^{(\mathbf{n})}$.

The total force experienced by a volume is given by the total traction force

\iiint_{V} \bf dV = \oiint_{\del V} \bT^{(\bn)} dA

Cauchy’s Stress Theorem is defined as: ${\mathbf{T}}^{(\mathbf{n})}$ is linear in $\mathbf{n}$ (i.e. there is a matrix $\sigma$ called the stress tensor) such that ${\mathbf{T}}^{(\mathbf{n})}=\sigma \mathbf{n}$.

The net force at each point is defined as the divergence of the stress tensor: $\mathbf{f}=\nabla \cdot \sigma$ (or written in index notation as $f^i={\partial }_j{\sigma }^{ij}$). For a material in equilibrium, the sum of these internal foces any any external forces must be zero:

$$\nabla \cdot \sigma +{\mathbf{f}}_{ext}=0$$

Furthermore, at equilibrium, the Cauchy stress tensor $\sigma$ must be symmetric.

• This can be shown by requesting zero net torque.
• When it’s not in equilibrium, the Cauchy stress tensor can be non-symmetric, such as viscoelastic fluids.
• For a different that we will see next, in pure elastic material, Cauchy stress $\sigma$ is always symmetric even at non-equilibrium.

Alternative Stress Tensors

The Cauchy Stress Tensor is not the only way to represent stress. Consider the deformation gradient:

$$\mathbf{F}=\left[\begin{array}{ccc}\frac{\partial {\varphi }^1}{\partial X}& \frac{\partial {\varphi }^1}{\partial Y}& \frac{\partial {\varphi }^1}{\partial Z}\\ \frac{\partial {\varphi }^2}{\partial X}& \frac{\partial {\varphi }^2}{\partial Y}& \frac{\partial {\varphi }^2}{\partial Z}\\ \frac{\partial {\varphi }^3}{\partial X}& \frac{\partial {\varphi }^3}{\partial Y}& \frac{\partial {\varphi }^3}{\partial Z}\end{array}\right]$$

And let $J=\det (\mathbf{F})$.

The First Piola-Kirchhoff Stress Tensor is defined as

$$\mathbf{P}=J\sigma {\mathbf{F}}^{-\top }$$

allowing us to calculate the net “world-force” at a point using the original coordinate (Lagrangian) grid $\mathbf{f}\circ \varphi =\nabla \cdot \mathbf{P}$. This tensor is generally non-symmetric.

The Second Piola-Kirchhoff Stress Tensor is defined as

$$\mathbf{S}={\mathbf{F}}^{-1}\sigma$$

It is symmetric iff Cauchy is symmetric.

Equation of Motion

We first need to define the strain. Strain is a measure of how much a material has deformed from its rest state. The right Cauchy-Green deformation tensor is

$$\mathbf{C}={\mathbf{F}}^{\top }\mathbf{F}$$

And then the Green-St Venant strain

$$E=\frac{1}{2}(\mathbf{C}-\mathbf{I})$$

Next, we need to model the stress-strain relationship (deformation-”internal forces”). To calculate the 2nd Piola-Kirchhoff stress $\mathbf{S}$,

$$\mathbf{S}=\lambda \text{tr}(E)\mathbf{I}+2\mu E$$

where $\lambda ,\mu$ are the Lamé parameters, dictating the specific material properties.

To easily compute the forces, we convert it to the 1st Piola-Kirchhoff stress $\mathbf{P}$:

$$\mathbf{P}=\mathbf{FS}$$

Now we want to find the actual force vector $\mathbf{f}$ acting on each point in the original (Lagrangian) coordinate system by taking its divergence (like before!):

$$\mathbf{f}=\nabla \cdot \mathbf{P}\text{or in index notation}{f}^i=\frac{\partial }{\partial X^i}{P}_j^i$$

Fianlly, applying Newton’s second law to get the equation of motion:

$${\rho }_M\ddot{\varphi }=\mathbf{f}$$

Type-Sensitive Tensor Analysis

Having a type-sensitive tensor analysis will make it more transparent. In particular, theorems like Cauchy stress symmetry is just the result of type checking.