Stress-Strain Relation

Recall

Recall from Cauchy’s Stress Theory that the Cauchy-Green tensor is given by $\mathbf{C}={\mathbf{F}}^{\top }\mathbf{F}$, or more preicsely $\mathbf{C}=F^{\ast }{♭}_{W}F$.

Here, we postulate that energy depends on the frame indifferent induced metric and the potential energy takes the form $\mathcal{U}=U(C)$.

The 2nd Piola-Kirchhoff Stress is given by $S=2\frac{\partial U}{\partial \mathbb{C}}$.

How do we design how the system stores energy when stretched?

Postulate (Existence of Rest Metric)

We postulate there exists a time-independent metric ${♭}_M$ on the material space. Then, we can define $\hat{C}={♯}_{M}C$ where there is “no deformation” iff $\hat{C}=\mathrm{id}$, as seen in Actual Model. We write energy as $U(C)=\Psi (\hat{C}){\rho }_M$ and $\Psi$ is the specific internal energy or Helmholtz free energy.

2nd Piola-Kirchoff Stress is

$$S=2\frac{\partial U}{\partial C}=2{♯}_M\frac{\partial \Psi }{\partial \hat{C}}{\rho }_M$$

A strain is an expression of $\hat{C}$ that measures its deviation from $\text{id}$. For large deformations, Hencky strain is usually considered as the natural true strain. For small deformations, $\frac{\partial \Psi }{\partial \hat{C}}\approx \frac{\partial \Psi }{\partial E}$.

Small Deformations

An example energy is the St Venant-Kirchhoff energy, which is given by

$${\Psi }_{\text{StVK}}=\frac{\lambda }{2}\mathrm{tr}(E{)}^2+\mu \mathrm{tr}(E^2)$$

with the corresponding stress

$$S={♯}_M\left(\lambda \mathrm{tr}(E)\mathrm{id}+2\mu E\right){\rho }_M$$

The Lame constants $\lambda ,\mu$ describe how stress relate to the isotropic part of deformation and anisotropic part of deformation, respectively.

We can measure Young’s modulus $E$ and Poisson’s ratio $\nu$:

$$\{\begin{array}{rl}E& =\frac{F/A}{\Delta \ell /\ell }\\ \nu & =-\frac{\Delta L^{\prime }}{\Delta L}\end{array}$$

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• $E$ measures how much force is needed to stretch a material linearly (think of it like a spring constant). It is defined as stress over strain.
• $\nu$ measures how much a material contracts in the directions perpendicular to the direction of stretching (how much a material thins out when you stretch it).

We get the Lame constants from $E$ and $\nu$:

$$\{\begin{array}{rl}\lambda & =\frac{E\nu }{(1+\nu )(1-2\nu )}\\ \mu & =\frac{E}{2(1+\nu )}\end{array}$$

Postulate (General Isotropic Material)

A material is called isotropic if

$$\Psi (\hat{C})=\Psi (R^{-1}\hat{C}R)$$

for all “rotation” operators $R$ characterized by $R^{\ast }{♭}_{M}R={♭}_M$. In ${\mathbb{R}}^3$, this means $R$ is in Special Orthogonal Group $\mathrm{SO}(3)$.

The energy is only a function of the eigenvalues (modulo permutations)

$$\text{eigenvalues}(\hat{C})=\{{\lambda }_1,{\lambda }_2,{\lambda }_3\}$$

These eigenvalues are the square of the eigenvalues of $\mathbf{Y}$ in polar decomposition $\mathbf{F}=\mathbf{RY}$. Equivalently, they are the square of the singular values of $\mathbf{F}$ or the square of “principal stretching”.

How do we model $U$ such that $U(\mathbf{C})=u({\lambda }_1,{\lambda }_2,{\lambda }_3)$? We can view the eigenvalues as the roots of a polynomial, and use the coefficient of the polynomial as our parameters:

$$\{{\lambda }_1,{\lambda }_2,{\lambda }_3\}=\text{roots}\left(t^3-I_1{t}^2+I_{2}t-I_3;t\right)$$

These coefficents are called the principal invariants:

$$\begin{array}{rlrl}{I}_1& ={\lambda }_1+{\lambda }_2+{\lambda }_3& & =\mathrm{tr}(\mathbf{C})\\ I_2& ={\lambda }_1{\lambda }_2+{\lambda }_2{\lambda }_3+{\lambda }_3{\lambda }_1& & =\frac{1}{2}\left(\mathrm{tr}(\mathbf{C}{)}^2-\mathrm{tr}({\mathbf{C}}^2)\right)\\ I_3& ={\lambda }_1{\lambda }_2{\lambda }_3& & =\det (\mathbf{C})\end{array}$$

This is the Characteristic Polynomial:

$$t^3-I_1{t}^2+I_{2}t-I_3=\det (t\mathbf{I}-\mathbf{C})$$

which allows us to express the energy as a function of the principal invariants:

$$U(\mathbf{C})=w(I_1,I_2,I_3)$$

Why do we care about principal invariants?

We can express the energy as a function of the principal invariants, which are easier to compute than the eigenvalues.

Since an isotropic material has no preferred direction, its stored energy $\Psi$ can only depend on the principal stretches (the eigenvalues of $\hat{C}$). However, eigenvalues do not have a fixed order (they can swap), so a valid energy function must mod out permutation, meaning it must yield the same energy regardless of how the eigenvalues are ordered.

The principal invariants guarantee this symmetry, as the coefficients of the characteristic polynomial. Thus, we can model the energy as $U(\mathbf{C})=w(I_1,I_2,I_3)$.

Theorem (No Fourth Invariants)

We only need three trace invariants to compute the system, no matter the material.

So $I_1=\text{tr}(\mathbf{C})$. $\text{tr}({\mathbf{C}}^2)=I_1^2-2I_2$. We also get $\text{tr}({\mathbf{C}}^3)=I_1^3-3I_1{I}_2+3I_3$. How do we find $\text{tr}({\mathbf{C}}^k)$?

We use the Cayley-Hamilton Theorem:

$${\mathbf{C}}^3-I_1{\mathbf{C}}^2+I_2\mathbf{C}-I_3=0$$

Mutiply by ${\mathbf{C}}^k$ and take its trace:

$$\begin{array}{rl}0& =\mathrm{tr}({\mathbf{C}}^{k+3})-I_1\mathrm{tr}({\mathbf{C}}^{k+2})+I_2\mathrm{tr}({\mathbf{C}}^{k+1})-I_3\mathrm{tr}({\mathbf{C}}^k)\\ \mathrm{tr}({\mathbf{C}}^{k+3})& =I_1\mathrm{tr}({\mathbf{C}}^{k+2})-I_2\mathrm{tr}({\mathbf{C}}^{k+1})+I_3\mathrm{tr}({\mathbf{C}}^k)\end{array}$$

giving us a recursive formula to compute $\mathrm{tr}({\mathbf{C}}^k)$ for all $k$.

Chain Rule for Stress

We still need to calculate stress, and so we need to use the chain rule. Recall the 2nd Piola-Kirchhoff Stress. Using the chain rule with $w$,

$$\frac{\partial U}{\partial \mathbf{C}}=\frac{\partial w}{\partial I_1}\frac{\partial I_1}{\partial \mathbf{C}}+\frac{\partial w}{\partial I_2}\frac{\partial I_2}{\partial \mathbf{C}}+\frac{\partial w}{\partial I_3}\frac{\partial I_3}{\partial \mathbf{C}}$$

The scalar terms are easy to compute. However, the tensor terms are more involved. In particular, we need to compute $\frac{\partial I_k}{\partial \mathbf{C}}$.

Instead, define the characteristic polynomial as a function of a dummy variable $z$:

$$P_{\mathbf{C}}(z)=\det (z\mathbf{I}-\mathbf{C})=z^n-I_1{z}^{n-1}+I_2{z}^{n-2}-\cdots +(-1{)}^n{I}_n$$

We can use Jacobi’s Formula to compute the derivative of the determinant:

$$\frac{\partial \det (A)}{\partial A}=\det (A)A^{-\top }$$

By applying Jacobi’s formula to $A=z\mathbf{I}-\mathbf{C}$, we get:

$$\begin{array}{rl}\frac{\partial P_{\mathbf{C}}(z)}{\partial \mathbf{C}}& =\frac{\partial \det (z\mathbf{I}-\mathbf{C})}{\partial \mathbf{C}}\\ & =-\det (z\mathbf{I}-\mathbf{C})(z\mathbf{I}-\mathbf{C}{)}^{-\top }\\ & =-P_{\mathbf{C}}(z)(z\mathbf{I}-\mathbf{C}{)}^{-\top }\end{array}$$

Note that applying the chain rule to the argument $z\mathbf{I}-\mathbf{C}$ with respect to $\mathbf{C}$ introduces a negative sign to the Jacobi formula result.

We can also compute the derivative of $P_{\mathbf{C}}(z)$ by differentiating its expanded scalar polynomial form term-by-term with respect to $\mathbf{C}$:

$$\frac{\partial P_{\mathbf{C}}(z)}{\partial \mathbf{C}}=-\frac{\partial I_1}{\partial \mathbf{C}}{z}^{n-1}+\frac{\partial I_2}{\partial \mathbf{C}}{z}^{n-2}-\cdots +(-1{)}^n\frac{\partial I_n}{\partial \mathbf{C}}$$

Because both methods describe the same derivative $\frac{\partial P_{\mathbf{C}}(z)}{\partial \mathbf{C}}$, we can equate them.

$$-\frac{\partial I_1}{\partial \mathbf{C}}{z}^{n-1}+\frac{\partial I_2}{\partial \mathbf{C}}{z}^{n-2}-\cdots +(-1{)}^n\frac{\partial I_n}{\partial \mathbf{C}}=-P_{\mathbf{C}}(z)(z\mathbf{I}-\mathbf{C}{)}^{-\top }$$

Notice that $\frac{\partial I_k}{\partial \mathbf{C}}$ is isolated as the coefficient for the $z^{n-k}$ term. To extract this specific coefficient from the right-hand side equation, we can take the $(n-k)$-th derivative with respect to $z$ and evaluate it at $z=0$ (similar to finding coefficients in a Maclaurin series).

This provides the generalized formula for the derivative of the $k$-th principal invariant:

$$\frac{\partial I_k}{\partial \mathbf{C}}=(-1{)}^{n-k}\frac{1}{(n-k)!}\frac{d^{n-k}}{d{z}^{n-k}}{|}_{z=0}\left(P_{\mathbf{C}}(z)(z\mathbf{I}-\mathbf{C}{)}^{-\top }\right)$$

Example: Fluid Potential

Consider a fluid where the energy potential $\Psi$ only depends on volume deformation, entirely ignoring the shearing part. This means the energy depends exclusively on the third invariant, $I_3=\det (\mathbf{C})$:

$$\Psi (\mathbf{C})=w(I_3)$$

To find the 2nd Piola-Kirchoff Stress $\mathbf{S}$, we apply the chain rule using the derivative for $I_3$ and Jacobi’s Formula:

$$\mathbf{S}=2\frac{\partial \Psi (\mathbf{C})}{\partial \mathbf{C}}=2w^{\prime }\frac{\partial \det (\mathbf{C})}{\partial \mathbf{C}}=2w^{\prime }\det (\mathbf{C}){\mathbf{C}}^{-\top }$$

Since $\mathbf{C}={\mathbf{F}}^{\top }\mathbf{F}$, we know $\det (\mathbf{C})=J^2$ (where $J=\det (\mathbf{F})$). We can then compute the 1st Piola-Kirchhoff Stress $\mathbf{P}$:

$$\begin{array}{rl}\mathbf{P}& =\mathbf{FS}\\ & =2w^{\prime }{J}^2\mathbf{F}{\mathbf{C}}^{-\top }\\ & =2w^{\prime }{J}^2\mathbf{F}({\mathbf{F}}^{\top }\mathbf{F}{)}^{-\top }\\ & =2w^{\prime }{J}^2\mathbf{F}{\mathbf{F}}^{-1}{\mathbf{F}}^{-\top }\\ & =2w^{\prime }{J}^2{\mathbf{F}}^{-\top }\end{array}$$

Finally, transforming this into the Cauchy stress $\mathbf{\sigma }$ yields a purely hydrostatic pressure relationship $p\mathbf{I}$:

$$\mathbf{\sigma }=\frac{1}{J}\mathbf{P}{\mathbf{F}}^{\top }=\frac{2w^{\prime }{J}^2}{J}\mathbf{I}=p\mathbf{I}$$

Example: Neo-Hookean Model

We can have $w$ represent

$$w(I_1,I_2,I_3)=\frac{\mu }{2}(I_1-3-\ln (I_3))+\frac{\lambda }{2}{\left(\sqrt{I_3}-1\right)}^2$$

or approximately

$$U(\mathbf{C})=\left(\frac{\lambda }{2}\mathrm{tr}(E{)}^2+\mu \mathrm{tr}(E^2)\right)d{V}_M$$

Example: Degenerate

Consider

$$w(I_1,I_2,I_3)=\frac{\lambda }{2}{\left(\sqrt{I_3}-1\right)}^2$$

then we get a fluid body!. If $\lambda ,\mu \to \infty$ and if $w$ is nondegenerate, then we obtain a rigid body. If $w=w(I_3)$, we obtain an incompressible fluid.