• 346,727

# Posts Tagged ‘stress tensor’

## Putting the stress tensor (and traction vector) into explicit vector form.

Posted by peeterjoot on April 8, 2012

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

Exersize 6.1 from [1] is to show that the traction vector can be written in vector form (a rather curious thing to have to say) as

\begin{aligned}\mathbf{t} = -p \hat{\mathbf{n}} + \mu ( 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})\mathbf{u} + \hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})).\end{aligned} \hspace{\stretch{1}}(1.1)

Note that the text uses a wedge symbol for the cross product, and I’ve switched to standard notation. I’ve done so because the use of a Geometric-Algebra wedge product also can be used to express this relationship, in which case we would write

\begin{aligned}\mathbf{t} = -p \hat{\mathbf{n}} + \mu ( 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}).\end{aligned} \hspace{\stretch{1}}(1.2)

In either case we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}=\hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})=\boldsymbol{\nabla}' (\hat{\mathbf{n}} \cdot \mathbf{u}') - (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u}\end{aligned} \hspace{\stretch{1}}(1.3)

(where the primes indicate the scope of the gradient, showing here that we are operating only on $\mathbf{u}$, and not $\hat{\mathbf{n}}$).

After computing this, lets also compute the stress tensor in cylindrical and spherical coordinates (a portion of that is also problem 6.10), something that this allows us to do fairly easily without having to deal with the second order terms that we encountered doing this by computing the difference of squared displacements.

We’ll work primarily with just the strain tensor portion of the traction vector expressions above, calculating

\begin{aligned}2 {\mathbf{e}}_{\hat{\mathbf{n}}}=2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})\mathbf{u} + \hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})=2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}.\end{aligned} \hspace{\stretch{1}}(1.4)

We’ll see that this gives us a nice way to interpret these tensor relationships. The interpretation was less clear when we computed this from the second order difference method, but here we see that we are just looking at the components of the force in each of the respective directions, dependent on which way our normal is specified.

# Verifying the relationship.

Let’s start with the the plain old cross product version

\begin{aligned}(\hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u}) + 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u})_i&=n_a (\boldsymbol{\nabla} \times \mathbf{u})_b \epsilon_{a b i} + 2 n_a \partial_a u_i \\ &=n_a \partial_r u_s \epsilon_{r s b} \epsilon_{a b i} + 2 n_a \partial_a u_i \\ &=n_a \partial_r u_s \delta_{ia}^{[rs]} + 2 n_a \partial_a u_i \\ &=n_a ( \partial_i u_a -\partial_a u_i ) + 2 n_a \partial_a u_i \\ &=n_a \partial_i u_a + n_a \partial_a u_i \\ &=n_a (\partial_i u_a + \partial_a u_i) \\ &=\sigma_{i a } n_a\end{aligned}

We can also put the double cross product in wedge product form

\begin{aligned}\hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})&=-I \hat{\mathbf{n}} \wedge (\boldsymbol{\nabla} \times \mathbf{u}) \\ &=-\frac{I}{2}\left(\hat{\mathbf{n}} (\boldsymbol{\nabla} \times \mathbf{u})- (\boldsymbol{\nabla} \times \mathbf{u}) \hat{\mathbf{n}}\right) \\ &=-\frac{I}{2}\left(-I \hat{\mathbf{n}} (\boldsymbol{\nabla} \wedge \mathbf{u})+ I (\boldsymbol{\nabla} \wedge \mathbf{u}) \hat{\mathbf{n}}\right) \\ &=-\frac{I^2}{2}\left(- \hat{\mathbf{n}} (\boldsymbol{\nabla} \wedge \mathbf{u})+ (\boldsymbol{\nabla} \wedge \mathbf{u}) \hat{\mathbf{n}}\right) \\ &=(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}\end{aligned}

Equivalently (and easier) we can just expand the dot product of the wedge and the vector using the relationship

\begin{aligned}\mathbf{a} \cdot (\mathbf{c} \wedge \mathbf{d} \wedge \mathbf{e} \wedge \cdots )=(\mathbf{a} \cdot \mathbf{c}) (\mathbf{d} \wedge \mathbf{e} \wedge \cdots ) - (\mathbf{a} \cdot \mathbf{d}) (\mathbf{c} \wedge \mathbf{e} \wedge \cdots ) +\end{aligned} \hspace{\stretch{1}}(2.5)

so we find

\begin{aligned}((\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}} + 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u})_i&=(\boldsymbol{\nabla}' (\mathbf{u}' \cdot \hat{\mathbf{n}})-(\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u}+ 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u})_i \\ &=\partial_i u_a n_a+n_a \partial_a u_i \\ &=\sigma_{ia} n_a.\end{aligned}

# Cylindrical strain tensor.

Let’s now compute the strain tensor (and implicitly the traction vector) in cylindrical coordinates.

Our gradient in cylindrical coordinates is the familiar

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\phi}} \frac{1}{{r }}\frac{\partial {}}{\partial {\phi}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}},\end{aligned} \hspace{\stretch{1}}(3.6)

and our cylindrical velocity is

\begin{aligned}\mathbf{u} = \hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z.\end{aligned} \hspace{\stretch{1}}(3.7)

Our curl is then

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}&=\left(\hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\phi}} \frac{1}{{r }}\frac{\partial {}}{\partial {\phi}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\right)\wedge\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}\left(\partial_r u_\phi -\frac{1}{{r}} \partial_\phi u_r\right)+\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right)+\frac{1}{{r}} \hat{\boldsymbol{\phi}} \wedge \left((\partial_\phi \hat{\mathbf{r}}) u_r+(\partial_\phi \hat{\boldsymbol{\phi}}) u_\phi\right)\end{aligned}

Since $\partial_\phi \hat{\mathbf{r}} = \hat{\boldsymbol{\theta}}$ and $\partial_\phi \hat{\boldsymbol{\phi}} = -\hat{\mathbf{r}}$, we have only one cross term and our curl is

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right).\end{aligned} \hspace{\stretch{1}}(3.8)

We can now move on to compute the directional derivatives and complete the strain calculation in cylindrical coordinates. Let’s consider this computation of the stress for normals in each direction in term.

## With $\hat{\mathbf{n}} = \hat{\mathbf{r}}$.

Our directional derivative component for a $\hat{\mathbf{r}}$ normal direction doesn’t have any cross terms

\begin{aligned}2 (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=2 \partial_r\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=2\left(\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\phi}} \partial_r u_\phi + \hat{\mathbf{z}} \partial_r u_z\right).\end{aligned}

Projecting our curl bivector onto the $\hat{\mathbf{r}}$ direction we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{r}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\mathbf{r}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}) \cdot \hat{\mathbf{r}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+(\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}) \cdot \hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=-\hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+\hat{\mathbf{z}}\left(\partial_z u_r - \partial_r u_z\right).\end{aligned}

Putting things together we have

\begin{aligned}2 \mathbf{e}_{\hat{\mathbf{r}}}&=2\left(\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\phi}} \partial_r u_\phi + \hat{\mathbf{z}} \partial_r u_z\right)-\hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+\hat{\mathbf{z}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\mathbf{r}}\left(2 \partial_r u_r\right)+\hat{\boldsymbol{\phi}}\left(2 \partial_r u_\phi-\partial_r u_\phi+\frac{1}{{r}} \partial_\phi u_r- \frac{u_\phi}{r}\right)+\hat{\mathbf{z}}\left(2 \partial_r u_z+\partial_z u_r - \partial_r u_z\right).\end{aligned}

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\mathbf{r}}} = - p \hat{\mathbf{r}} + 2 \mu e_{\hat{\mathbf{r}}},\end{aligned} \hspace{\stretch{1}}(3.9)

we can now read off our components by taking dot products to yield

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(3.10a)

\begin{aligned}\sigma_{r \phi}=\mu \left( \frac{\partial {u_\phi}}{\partial {r}}+\frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.10b)

\begin{aligned}\sigma_{r z}=\mu \left( \frac{\partial {u_z}}{\partial {r}}+\frac{\partial {u_r}}{\partial {z}}\right).\end{aligned} \hspace{\stretch{1}}(3.10c)

\end{subequations}

## With $\hat{\mathbf{n}} = \hat{\boldsymbol{\phi}}$.

Our directional derivative component for a $\hat{\boldsymbol{\phi}}$ normal direction will have some cross terms since both $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\phi}}$ are functions of $\phi$

\begin{aligned}2 (\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\frac{2}{r}\partial_\phi\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=\frac{2}{r}\left(\hat{\mathbf{r}} \partial_\phi u_r + \hat{\boldsymbol{\phi}} \partial_\phi u_\phi + \hat{\mathbf{z}} \partial_\phi u_z+(\partial_\phi \hat{\mathbf{r}}) u_r + (\partial_\phi \hat{\boldsymbol{\phi}}) u_\phi\right) \\ &=\frac{2}{r}\left(\hat{\mathbf{r}} (\partial_\phi u_r - u_\phi) + \hat{\boldsymbol{\phi}} (\partial_\phi u_\phi + u_r )+ \hat{\mathbf{z}} \partial_\phi u_z\right) \\ \end{aligned}

Projecting our curl bivector onto the $\hat{\boldsymbol{\phi}}$ direction we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}) \cdot \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+(\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}) \cdot \hat{\boldsymbol{\phi}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\mathbf{r}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)-\hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)\end{aligned}

Putting things together we have

\begin{aligned}2 \mathbf{e}_{\hat{\boldsymbol{\phi}}}&=\frac{2}{r}\left(\hat{\mathbf{r}} (\partial_\phi u_r - u_\phi) + \hat{\boldsymbol{\phi}} (\partial_\phi u_\phi + u_r )+ \hat{\mathbf{z}} \partial_\phi u_z\right)+\hat{\mathbf{r}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)-\hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right) \\ &=\hat{\mathbf{r}}\left(\frac{1}{r}\partial_\phi u_r-\frac{u_\phi}{r}+\partial_r u_\phi\right)+\frac{2}{r} \hat{\boldsymbol{\phi}}\left(\partial_\phi u_\phi + u_r\right)+\hat{\mathbf{z}}\left(\frac{1}{r} \partial_\phi u_z + \partial_z u_\phi\right).\end{aligned}

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\boldsymbol{\phi}}} = - p \hat{\boldsymbol{\phi}} + 2 \mu e_{\hat{\boldsymbol{\phi}}},\end{aligned} \hspace{\stretch{1}}(3.11)

we can now read off our components by taking dot products to yield

\begin{subequations}

\begin{aligned}\sigma_{\phi \phi}=-p + 2 \mu \left(\frac{1}{{r}}\frac{\partial {u_\phi}}{\partial {\phi}} + \frac{u_r}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.12a)

\begin{aligned}\sigma_{\phi z}=\mu \left(\frac{1}{r} \frac{\partial {u_z}}{\partial {\phi}} + \frac{\partial {u_\phi}}{\partial {z}}\right)\end{aligned} \hspace{\stretch{1}}(3.12b)

\begin{aligned}\sigma_{\phi r}=\mu \left(\frac{1}{r}\frac{\partial {u_r}}{\partial {\phi}}-\frac{u_\phi}{r}+\frac{\partial {u_\phi}}{\partial {r}}\right).\end{aligned} \hspace{\stretch{1}}(3.12c)

\end{subequations}

## With $\hat{\mathbf{n}} = \hat{\mathbf{z}}$.

Like the $\hat{\mathbf{r}}$ normal direction, our directional derivative component for a $\hat{\mathbf{z}}$ normal direction will not have any cross terms

\begin{aligned}2 (\hat{\mathbf{z}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\partial_z\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=\hat{\mathbf{r}} \partial_z u_r + \hat{\boldsymbol{\phi}} \partial_z u_\phi + \hat{\mathbf{z}} \partial_z u_z\end{aligned}

Projecting our curl bivector onto the $\hat{\mathbf{z}}$ direction we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\mathbf{z}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}) \cdot \hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+(\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}) \cdot \hat{\mathbf{z}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)-\hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right)\end{aligned}

Putting things together we have

\begin{aligned}2 \mathbf{e}_{\hat{\mathbf{z}}}&=2 \hat{\mathbf{r}} \partial_z u_r + 2 \hat{\boldsymbol{\phi}} \partial_z u_\phi + 2 \hat{\mathbf{z}} \partial_z u_z+\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)-\hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\mathbf{r}}\left(2 \partial_z u_r -\partial_z u_r + \partial_r u_z\right)+\hat{\boldsymbol{\phi}}\left(2 \partial_z u_\phi +\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+\hat{\mathbf{z}}\left(2 \partial_z u_z\right) \\ &=\hat{\mathbf{r}}\left(\partial_z u_r + \partial_r u_z\right)+\hat{\boldsymbol{\phi}}\left(\partial_z u_\phi +\frac{1}{{r}} \partial_\phi u_z\right)+\hat{\mathbf{z}}\left(2 \partial_z u_z\right).\end{aligned}

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\mathbf{z}}} = - p \hat{\mathbf{z}} + 2 \mu e_{\hat{\mathbf{z}}},\end{aligned} \hspace{\stretch{1}}(3.13)

we can now read off our components by taking dot products to yield

\begin{subequations}

\begin{aligned}\sigma_{z z}=-p + 2 \mu \frac{\partial {u_z}}{\partial {z}}\end{aligned} \hspace{\stretch{1}}(3.14a)

\begin{aligned}\sigma_{z r}=\mu \left(\frac{\partial {u_r}}{\partial {z}}+ \frac{\partial {u_z}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(3.14b)

\begin{aligned}\sigma_{z \phi}=\mu \left(\frac{\partial {u_\phi}}{\partial {z}}+\frac{1}{{r}} \frac{\partial {u_z}}{\partial {\phi}}\right).\end{aligned} \hspace{\stretch{1}}(3.14c)

\end{subequations}

## Summary.

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(3.15a)

\begin{aligned}\sigma_{\phi \phi}=-p + 2 \mu \left(\frac{1}{{r}}\frac{\partial {u_\phi}}{\partial {\phi}} + \frac{u_r}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.15b)

\begin{aligned}\sigma_{z z}=-p + 2 \mu \frac{\partial {u_z}}{\partial {z}}\end{aligned} \hspace{\stretch{1}}(3.15c)

\begin{aligned}\sigma_{r \phi}=\mu \left( \frac{\partial {u_\phi}}{\partial {r}}+\frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.15d)

\begin{aligned}\sigma_{\phi z}=\mu \left(\frac{1}{r} \frac{\partial {u_z}}{\partial {\phi}} + \frac{\partial {u_\phi}}{\partial {z}}\right)\end{aligned} \hspace{\stretch{1}}(3.15e)

\begin{aligned}\sigma_{z r}=\mu \left(\frac{\partial {u_r}}{\partial {z}}+ \frac{\partial {u_z}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(3.15f)

\end{subequations}

# Spherical strain tensor.

Having done a first order cylindrical derivation of the strain tensor, let’s also do the spherical case for completeness. Would this have much utility in fluids? Perhaps for flow over a spherical barrier?

We need the gradient in spherical coordinates. Recall that our spherical coordinate velocity was

\begin{aligned}\frac{d\mathbf{r}}{dt} = \hat{\mathbf{r}} \dot{r} + \hat{\boldsymbol{\theta}} (r \dot{\theta}) + \hat{\boldsymbol{\phi}} ( r \sin\theta \dot{\phi} ),\end{aligned} \hspace{\stretch{1}}(4.16)

and our gradient mirrors this structure

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\theta}} \frac{1}{{r }}\frac{\partial {}}{\partial {\theta}} + \hat{\boldsymbol{\phi}} \frac{1}{{r \sin\theta}} \frac{\partial {}}{\partial {\phi}}.\end{aligned} \hspace{\stretch{1}}(4.17)

We also previously calculated \inbookref{phy454:continuumL2}{eqn:continuumL2:1010} the unit vector differentials

\begin{subequations}

\begin{aligned}d\hat{\mathbf{r}} = \hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta\end{aligned} \hspace{\stretch{1}}(4.18a)

\begin{aligned}d\hat{\boldsymbol{\theta}} = \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta\end{aligned} \hspace{\stretch{1}}(4.18b)

\begin{aligned}d\hat{\boldsymbol{\phi}} = -(\hat{\mathbf{r}} \sin\theta + \hat{\boldsymbol{\theta}} \cos\theta) d\phi,\end{aligned} \hspace{\stretch{1}}(4.18c)

\end{subequations}

and can use those to read off the partials of all the unit vectors

\begin{aligned}\frac{\partial \hat{\mathbf{r}}}{\partial \{r,\theta, \phi\}} &= \{0, \hat{\boldsymbol{\theta}}, \hat{\boldsymbol{\phi}} \sin\theta \} \\ \frac{\partial \hat{\boldsymbol{\theta}}}{\partial \{r,\theta, \phi\}} &= \{0, -\hat{\mathbf{r}}, \hat{\boldsymbol{\phi}} \cos\theta \} \\ \frac{\partial \hat{\boldsymbol{\phi}}}{\partial \{r,\theta, \phi\}} &= \{0, 0, -\hat{\mathbf{r}} \sin\theta -\hat{\boldsymbol{\theta}} \cos\theta \}.\end{aligned} \hspace{\stretch{1}}(4.19)

Finally, our velocity in spherical coordinates is just

\begin{aligned}\mathbf{u} = \hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi,\end{aligned} \hspace{\stretch{1}}(4.22)

from which we can now compute the curl, and the directional derivative. Starting with the curl we have

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}&=\left( \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\theta}} \frac{1}{{r }}\frac{\partial {}}{\partial {\theta}} + \hat{\boldsymbol{\phi}} \frac{1}{{r \sin\theta}} \frac{\partial {}}{\partial {\phi}} \right) \wedge\left( \hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi \right) \\ &=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r\right)\\ & +\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta\right)\\ & +\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi\right)\\ & +\frac{1}{{r}} \hat{\boldsymbol{\theta}} \wedge \left(u_\theta \underbrace{\partial_\theta \hat{\boldsymbol{\theta}}}_{-\hat{\mathbf{r}}}+u_\phi \underbrace{\partial_\theta \hat{\boldsymbol{\phi}}}_{0}\right)\\ & +\frac{1}{{r \sin\theta}} \hat{\boldsymbol{\phi}} \wedge \left(u_\theta \underbrace{\partial_\phi \hat{\boldsymbol{\theta}}}_{\hat{\boldsymbol{\phi}} \cos\theta}+u_\phi \underbrace{\partial_\phi \hat{\boldsymbol{\phi}}}_{-\hat{\mathbf{r}} \sin\theta - \hat{\boldsymbol{\theta}} \cos\theta}\right).\end{aligned}

So we have

\begin{aligned}\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}&=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.23)

## With $\hat{\mathbf{n}} = \hat{\mathbf{r}}$.

The directional derivative portion of our strain is

\begin{aligned}2 (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=2 \partial_r (\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi ) \\ &=2 (\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\theta}} \partial_r u_\theta + \hat{\boldsymbol{\phi}} \partial_r u_\phi ).\end{aligned}

The other portion of our strain tensor is

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{r}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}) \cdot \hat{\mathbf{r}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +(\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\mathbf{r}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}) \cdot \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=-\hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +\hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Putting these together we find

\begin{aligned}2 {\mathbf{e}}_{\hat{\mathbf{r}}}&=2 (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{r}} \\ &=2 (\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\theta}} \partial_r u_\theta + \hat{\boldsymbol{\phi}} \partial_r u_\phi )-\hat{\boldsymbol{\theta}}\left(\partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=\hat{\mathbf{r}}\left(2 \partial_r u_r\right)+\hat{\boldsymbol{\theta}}\left(2 \partial_r u_\theta-\partial_r u_\theta + \frac{1}{{r}} \partial_\theta u_r - \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\phi}}\left(2 \partial_r u_\phi+ \frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Which gives

\begin{aligned}2 {\mathbf{e}}_{\hat{\mathbf{r}}}=\hat{\mathbf{r}}\left(2 \partial_r u_r\right)+\hat{\boldsymbol{\theta}}\left(\partial_r u_\theta+ \frac{1}{{r}} \partial_\theta u_r - \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\phi}}\left(\partial_r u_\phi+ \frac{1}{{r \sin\theta}} \partial_\phi u_r- \frac{u_\phi}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.24)

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\mathbf{r}}} = - p \hat{\mathbf{r}} + 2 \mu e_{\hat{\mathbf{r}}},\end{aligned} \hspace{\stretch{1}}(4.25)

we can now read off our components by taking dot products

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(4.26a)

\begin{aligned}\sigma_{r \theta}=\mu \left(\frac{\partial {u_\theta}}{\partial {r}}+ \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\theta}} - \frac{u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.26b)

\begin{aligned}\sigma_{r \phi}=\mu \left(\frac{\partial {u_\phi}}{\partial {r}}+ \frac{1}{{r \sin\theta}} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.26c)

\end{subequations}

This is consistent with (15.20) from [3] (after adjusting for minor notational differences).

## With $\hat{\mathbf{n}} = \hat{\boldsymbol{\theta}}$.

Now let’s do the $\hat{\boldsymbol{\theta}}$ direction. The directional derivative portion of our strain will be a bit more work to compute because we have $\theta$ variation of the unit vectors

\begin{aligned}(\hat{\boldsymbol{\theta}} \cdot \boldsymbol{\nabla}) \mathbf{u} &= \frac{1}{r} \partial_\theta (\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi ) \\ &= \frac{1}{r} \left( \hat{\mathbf{r}} \partial_\theta u_r + \hat{\boldsymbol{\theta}} \partial_\theta u_\theta + \hat{\boldsymbol{\phi}} \partial_\theta u_\phi \right)+\frac{1}{r} \left( (\partial_\theta \hat{\mathbf{r}}) u_r + (\partial_\theta \hat{\boldsymbol{\theta}}) u_\theta + (\partial_\theta \hat{\boldsymbol{\phi}}) u_\phi \right) \\ &= \frac{1}{r}\left(\hat{\mathbf{r}} \partial_\theta u_r + \hat{\boldsymbol{\theta}} \partial_\theta u_\theta + \hat{\boldsymbol{\phi}} \partial_\theta u_\phi \right)+\frac{1}{r} \left( \hat{\boldsymbol{\theta}} u_r - \hat{\mathbf{r}} u_\theta \right).\end{aligned}

So we have

\begin{aligned}2 (\hat{\boldsymbol{\theta}} \cdot \boldsymbol{\nabla}) \mathbf{u}=\frac{2}{r} \hat{\mathbf{r}} (\partial_\theta u_r- u_\theta)+ \frac{2}{r} \hat{\boldsymbol{\theta}} (\partial_\theta u_\theta+ u_r) + \frac{2}{r} \hat{\boldsymbol{\phi}} \partial_\theta u_\phi,\end{aligned} \hspace{\stretch{1}}(4.27)

and can move on to projecting our curl bivector onto the $\hat{\boldsymbol{\theta}}$ direction. That portion of our strain tensor is

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\theta}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}) \cdot \hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +(\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\boldsymbol{\theta}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}) \cdot \hat{\boldsymbol{\theta}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=\hat{\mathbf{r}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)-\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right).\end{aligned}

Putting these together we find

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\theta}}}&=2 (\hat{\boldsymbol{\theta}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\theta}} \\ &= \frac{2}{r} \hat{\mathbf{r}} (\partial_\theta u_r - u_\theta )+ \frac{2}{r} \hat{\boldsymbol{\theta}} (\partial_\theta u_\theta + u_r )+ \frac{2}{r} \hat{\boldsymbol{\phi}} \partial_\theta u_\phi \\ &+\hat{\mathbf{r}}\left(\partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)-\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta + \frac{u_\phi \cot\theta}{r}\right).\end{aligned}

Which gives

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\theta}}}=\hat{\mathbf{r}} \left( \frac{1}{r} \partial_\theta u_r + \partial_r u_\theta- \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\theta}} \left( \frac{2}{r} \partial_\theta u_\theta+ \frac{2}{r} u_r\right)+\hat{\boldsymbol{\phi}} \left(\frac{1}{r} \partial_\theta u_\phi+ \frac{1}{{r \sin\theta}} \partial_\phi u_\theta- \frac{u_\phi \cot\theta}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.28)

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\boldsymbol{\theta}}} = - p \hat{\boldsymbol{\theta}} + 2 \mu e_{\hat{\boldsymbol{\theta}}},\end{aligned} \hspace{\stretch{1}}(4.29)

we can now read off our components by taking dot products

\begin{subequations}

\begin{aligned}\sigma_{\theta \theta}=-p+\mu \left( \frac{2}{r} \frac{\partial {u_\theta}}{\partial {\theta}}+ \frac{2}{r} u_r\right)\end{aligned} \hspace{\stretch{1}}(4.30a)

\begin{aligned}\sigma_{\theta \phi}=\mu \left(\frac{1}{r} \frac{\partial {u_\phi}}{\partial {\theta}}+ \frac{1}{{r \sin\theta}} \frac{\partial {u_\theta}}{\partial {\phi}}- \frac{u_\phi \cot\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.30b)

\begin{aligned}\sigma_{\theta r}= \mu \left(\frac{1}{r} \frac{\partial {u_r}}{\partial {\theta}} + \frac{\partial {u_\theta}}{\partial {r}}- \frac{u_\theta}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.30c)

\end{subequations}

This again is consistent with (15.20) from [3].

## With $\hat{\mathbf{n}} = \hat{\boldsymbol{\phi}}$.

Finally, let’s do the $\hat{\boldsymbol{\phi}}$ direction. This directional derivative portion of our strain will also be a bit more work to compute because we have $\hat{\boldsymbol{\phi}}$ variation of the unit vectors

\begin{aligned}(\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\frac{1}{r \sin\theta} \partial_\phi (\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi) \\ &=\frac{1}{r \sin\theta}(\hat{\mathbf{r}} \partial_\phi u_r+\hat{\boldsymbol{\theta}} \partial_\phi u_\theta+\hat{\boldsymbol{\phi}} \partial_\phi u_\phi+(\partial_\phi \hat{\mathbf{r}} )u_r+(\partial_\phi \hat{\boldsymbol{\theta}} )u_\theta+(\partial_\phi \hat{\boldsymbol{\phi}} )u_\phi) \\ &=\frac{1}{r \sin\theta}(\hat{\mathbf{r}} \partial_\phi u_r+\hat{\boldsymbol{\theta}} \partial_\phi u_\theta+\hat{\boldsymbol{\phi}} \partial_\phi u_\phi+\hat{\boldsymbol{\phi}} \sin\thetau_r+\hat{\boldsymbol{\phi}} \cos\thetau_\theta-(\hat{\mathbf{r}} \sin\theta+ \hat{\boldsymbol{\theta}} \cos\theta)u_\phi)\end{aligned}

So we have

\begin{aligned}2 (\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla}) \mathbf{u}=2 \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \frac{u_\phi}{r}\right)+2 \hat{\boldsymbol{\theta}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\theta-\frac{1}{{r}} \cot\theta u_\phi\right)+2 \hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\phi+ \frac{1}{{r}} u_r+ \frac{1}{{r}} \cot\theta u_\theta\right),\end{aligned} \hspace{\stretch{1}}(4.31)

and can move on to projecting our curl bivector onto the $\hat{\boldsymbol{\phi}}$ direction. That portion of our strain tensor is

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}) \cdot \hat{\boldsymbol{\phi}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +(\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}) \cdot \hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=\hat{\boldsymbol{\theta}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ &-\hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Putting these together we find

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\theta}}}&=2 (\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}} \\ &=2 \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \frac{u_\phi}{r}\right)+2 \hat{\boldsymbol{\theta}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\theta-\frac{1}{{r}} \cot\theta u_\phi\right)+2 \hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\phi+ \frac{1}{{r}} u_r+ \frac{1}{{r}} \cot\theta u_\theta\right) \\ &+\hat{\boldsymbol{\theta}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)-\hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Which gives

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\phi}}}=\hat{\mathbf{r}} \left( \frac{ \partial_\phi u_r }{r \sin\theta}- \frac{u_\phi}{r}+ \partial_r u_\phi\right)+\hat{\boldsymbol{\theta}} \left(\frac{\partial_\phi u_\theta}{r \sin\theta}- \frac{u_\phi \cot\theta}{r}+\frac{\partial_\theta u_\phi}{r}\right)+2 \hat{\boldsymbol{\phi}} \left(\frac{\partial_\phi u_\phi}{r \sin\theta}+ \frac{u_r}{r}+ \frac{\cot\theta u_\theta}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.32)

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\boldsymbol{\phi}}} = - p \hat{\boldsymbol{\phi}} + 2 \mu e_{\hat{\boldsymbol{\phi}}},\end{aligned} \hspace{\stretch{1}}(4.33)

we can now read off our components by taking dot products

\begin{subequations}

\begin{aligned}\sigma_{\phi \phi}=-p+2 \mu \left(\frac{1}{{r \sin\theta}} \frac{\partial {u_\phi}}{\partial {\phi}}+ \frac{u_r}{r}+ \frac{\cot\theta u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.34a)

\begin{aligned}\sigma_{\phi r}=\mu \left( \frac{1}{r \sin\theta} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}+ \frac{\partial {u_\phi}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(4.34b)

\begin{aligned}\sigma_{\phi \theta}= \mu \left(\frac{1}{r \sin\theta} \frac{\partial {u_\theta}}{\partial {\phi}}- \frac{u_\phi \cot\theta}{r}+\frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\theta}}\right).\end{aligned} \hspace{\stretch{1}}(4.34c)

\end{subequations}

This again is consistent with (15.20) from [3].

## Summary

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(4.35a)

\begin{aligned}\sigma_{\theta \theta}=-p+2 \mu \left( \frac{1}{r} \frac{\partial {u_\theta}}{\partial {\theta}}+ \frac{ u_r }{r}\right)\end{aligned} \hspace{\stretch{1}}(4.35b)

\begin{aligned}\sigma_{\phi \phi}=-p+2 \mu \left(\frac{1}{{r \sin\theta}} \frac{\partial {u_\phi}}{\partial {\phi}}+ \frac{u_r}{r}+ \frac{\cot\theta u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.35c)

\begin{aligned}\sigma_{r \theta}=\mu \left(\frac{\partial {u_\theta}}{\partial {r}}+ \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\theta}} - \frac{u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.35d)

\begin{aligned}\sigma_{\theta \phi}= \mu \left(\frac{1}{r \sin\theta} \frac{\partial {u_\theta}}{\partial {\phi}}- \frac{u_\phi \cot\theta}{r}+\frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\theta}}\right).\end{aligned} \hspace{\stretch{1}}(4.35e)

\begin{aligned}\sigma_{\phi r}=\mu \left( \frac{1}{r \sin\theta} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}+ \frac{\partial {u_\phi}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(4.35f)

\end{subequations}

# References

[1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

[2] Peeter Joot. Continuum mechanics., chapter {Introduction and strain tensor.} http://sites.google.com/site/peeterjoot2/math2012/phy454.pdf.

[3] L.D. Landau and E.M. Lifshitz. A Course in Theoretical Physics-Fluid Mechanics. Pergamon Press Ltd., 1987.

## Compilation of class notes for phy454h1s, continuum mechanics (so far).

Posted by peeterjoot on February 17, 2012

Have collected all my pre-midterm continuum mechanics notes into a single document. The individual pdfs below are still available, but won’t be updated further.

Feb 17, 2012 Flow in a pipe. Gravity driven flow of a film.

Feb 10, 2012 Navier-Stokes equation.

Feb 1, 2012 P-waves and S-waves.

Jan 27, 2012 Compatibility condition and elastostatics.

Jan 25, 2012 Constitutive relationship.

Jan 20, 2012 Strain tensor components.

Jan 18, 2012 Strain tensor review. Stress tensor.

Jan 13, 2012 Introduction and strain tensor.

Jan 11, 2012 Overview.

## PHY454H1S Continuum Mechanics. Lecture 12: Flow in a pipe. Gravity driven flow of a film. Taught by Prof. K. Das.

Posted by peeterjoot on February 17, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Steady rectilinear flow.

\begin{aligned}\frac{\partial {}}{\partial {t}} = 0\end{aligned} \hspace{\stretch{1}}(2.1)

Rectilinear is a unidirectional flow such as

\begin{aligned}\mathbf{u} = \hat{\mathbf{x}} u( x, y, z ),\end{aligned} \hspace{\stretch{1}}(2.2)

\begin{enumerate}
\item
Utilizing an incompressibility assumption $\boldsymbol{\nabla} \cdot \mathbf{u} = 0$, so for this case we have

\begin{aligned}\frac{\partial {u}}{\partial {x}} = 0\end{aligned}

or

\begin{aligned}u = u(y, z)\end{aligned}

Note that Prof. Das called this a continuity requirement, and justified this label with the relation

\begin{aligned}\frac{d\rho}{dt} = \rho (\boldsymbol{\nabla} \cdot \mathbf{u}),\end{aligned} \hspace{\stretch{1}}(2.3)

which was a consequence of mass conservation. It’s still not clear to me why he would call this a continuity requirement.

\item Nonlinear term is zero. $(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = 0$
\item $p = p(x)$. Since $\frac{d^2 p}{dx^2} = 0$ we also have $\frac{dp}{dx} = -G$, a constant.

\item $\mu \left( \frac{\partial^2 {{u}}}{\partial {{y}}^2} + \frac{\partial^2 {{u}}}{\partial {{z}}^2} \right) = G$

\end{enumerate}

# Solution by intuition.

Two examples that we have solved analytically are illustrated in figure (\ref{fig:continuumL12:continuumL12fig1}) and figure (\ref{fig:continuumL12:continuumL12fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig1}
\caption{Simple shear flow}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig2}
\caption{Channel flow}
\end{figure}

Sometimes we can utilize solutions already found to understand the behaviour of more complex systems. Combining the two we can look at flow over a plate as in figure (\ref{fig:continuumL12:continuumL12fig3})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig3}
\caption{Flow on a plate}
\end{figure}

Example 2. Fluid in a container. If the surface tension is altered on one side, we induce a flow on the surface, leading to a circulation flow. This can be done for example, by introducing a heat source or addition of surfactant.

This is illustrated in figure (\ref{fig:continuumL12:continuumL12fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig4}
\caption{Circulation flow induced by altering surface tension.}
\end{figure}

This sort of flow is hard to analyze, only first done by Steve Davis in the 1980’s. The point here is that we can use some level of intuition to guide our attempts at solution.

# Flow down a pipe.

Reading: section 2 from [1].

Recall that the Navier-Stokes equation is

\begin{aligned}\boxed{\rho \frac{\partial {\mathbf{u}}}{\partial {t}} + \rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = - \boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \rho \mathbf{f}.}\end{aligned} \hspace{\stretch{1}}(4.4)

We need to express this in cylindrical coordinates $(r, \theta, z)$ as in figure (\ref{fig:continuumL12:continuumL12fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig5}
\caption{Flow through a pipe.}
\end{figure}

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \frac{\hat{\boldsymbol{\theta}}}{r} \frac{\partial {}}{\partial {\theta}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}},\end{aligned} \hspace{\stretch{1}}(4.5)

For our Laplacian we find

\begin{aligned}\boldsymbol{\nabla}^2 &= \left(\hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \frac{\hat{\boldsymbol{\theta}}}{r} \frac{\partial {}}{\partial {\theta}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\right) \cdot\left(\hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \frac{\hat{\boldsymbol{\theta}}}{r} \frac{\partial {}}{\partial {\theta}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\right) \\ &=\partial_{rr} + \frac{\hat{\boldsymbol{\theta}}}{r} \cdot (\partial_\theta \hat{\mathbf{r}}) \partial_r+ \frac{1}{{r}} \partial_\theta \left( \frac{1}{{r}} \partial_\theta \right)+ \partial_{zz} \\ &=\partial_{rr} + \frac{1}{{r}} \partial_r + \frac{1}{{r^2}} \partial_{\theta\theta} + \partial_{zz},\end{aligned}

which we can write as

\begin{aligned}\boldsymbol{\nabla}^2 = \frac{1}{{r}} \frac{\partial {}}{\partial {r}} \left( r \frac{\partial {}}{\partial {r}} \right) + \frac{1}{{r^2}} \frac{\partial^2 {{}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2}.\end{aligned} \hspace{\stretch{1}}(4.6)

NS takes the form

\begin{aligned}\boxed{\begin{aligned}\rho \frac{\partial {\mathbf{u}}}{\partial {t}} &+ \rho \left(u_r \frac{\partial {}}{\partial {r}} + \frac{u_\theta}{r} \frac{\partial {}}{\partial {\theta}} + u_z \frac{\partial {}}{\partial {z}} \right) \mathbf{u} = \\ &- \left(\hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \frac{\hat{\boldsymbol{\theta}}}{r} \frac{\partial {}}{\partial {\theta}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\right)p + \mu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}} \left( r \frac{\partial {}}{\partial {r}} \right) + \frac{1}{{r^2}} \frac{\partial^2 {{}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2} \right)\mathbf{u} + \rho \mathbf{f}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(4.7)

For steady state and incompressible fluids in the absence of body forces we have

\begin{aligned}\left(\hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \frac{\hat{\boldsymbol{\theta}}}{r} \frac{\partial {}}{\partial {\theta}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\right)p = \mu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}} \left( r \frac{\partial {}}{\partial {r}} \right) + \frac{1}{{r^2}} \frac{\partial^2 {{}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2} \right)\mathbf{u},\end{aligned} \hspace{\stretch{1}}(4.8)

or, in coordinates

\begin{aligned}\frac{\partial {p}}{\partial {r}} &= \mu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}} \left( r \frac{\partial {}}{\partial {r}} \right) + \frac{1}{{r^2}} \frac{\partial^2 {{}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2} \right)u_r \\ \frac{1}{r} \frac{\partial {p}}{\partial {\theta}}&= \mu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}} \left( r \frac{\partial {}}{\partial {r}} \right) + \frac{1}{{r^2}} \frac{\partial^2 {{}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2} \right)u_\theta \\ \frac{\partial {p}}{\partial {z}}&= \mu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}} \left( r \frac{\partial {}}{\partial {r}} \right) + \frac{1}{{r^2}} \frac{\partial^2 {{}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2} \right)u_z\end{aligned} \hspace{\stretch{1}}(4.9)

With an assumption that we have no radial or circulatory flows ($u_r = u_\theta = 0$), and with $u_z = w$ assumed to only have a radial dependence, our velocity is

\begin{aligned}\mathbf{u} = \hat{\mathbf{z}} w(r),\end{aligned} \hspace{\stretch{1}}(4.12)

and an assumption of linear pressure dependence

\begin{aligned}\frac{dp}{dz} = -G,\end{aligned} \hspace{\stretch{1}}(4.13)

then NS takes the final simple form

\begin{aligned}\frac{1}{{r}} \frac{d}{dr} \left( r \frac{dw}{dr} \right) = - \frac{G}{\mu}.\end{aligned} \hspace{\stretch{1}}(4.14)

Solving this we have

\begin{aligned}r \frac{dw}{dr} = - \frac{G r^2}{2\mu} + A\end{aligned} \hspace{\stretch{1}}(4.15)

\begin{aligned}w = -\frac{G r^2}{4 \mu} + A \ln(r) + B\end{aligned} \hspace{\stretch{1}}(4.16)

Requiring finite solutions for $r = 0$ means that we must have $A = 0$. Also $w(a) = 0$, we have $B = G a^2/4 \mu$ so we must have

\begin{aligned}w(r) = \frac{G}{4 \mu}( a^2 - r^2 )\end{aligned} \hspace{\stretch{1}}(4.17)

# Example: Gravity driven flow of a liquid film

(This is one of our Professor’s favorite problems).

Coordinates as in figure (\ref{fig:continuumL12:continuumL12fig6})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig6}
\caption{Gravity driven flow down an inclined plane.}
\end{figure}

\begin{aligned}\mathbf{u} = \hat{\mathbf{x}} u(y)\end{aligned} \hspace{\stretch{1}}(5.18)

Boundary conditions

\begin{enumerate}
\item $u(y = 0) = 0$
\item Tangential stress at the air-liquid interface $y = h$ is equal.

\begin{aligned}\boldsymbol{\tau} \cdot (\boldsymbol{\sigma}_l \cdot \hat{\mathbf{n}}) = \boldsymbol{\tau} \cdot (\boldsymbol{\sigma}_a \cdot \hat{\mathbf{n}}),\end{aligned} \hspace{\stretch{1}}(5.19)

\end{enumerate}

We write

\begin{aligned}\boldsymbol{\tau} &= \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} \\ \hat{\mathbf{n}} &= \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} \\ \end{aligned} \hspace{\stretch{1}}(5.20)

and seek simultaneous solutions to the pair of stress tensor equations

\begin{aligned}\sigma_{ij}^l &= - p \delta_{ij} + \mu^l \left( \frac{\partial {u_i}}{\partial {x_j}} +\frac{\partial {u_j}}{\partial {x_i}}\right) \\ \sigma_{ij}^a &= - p \delta_{ij} + \mu^a \left( \frac{\partial {u_i}}{\partial {x_j}} +\frac{\partial {u_j}}{\partial {x_i}}\right).\end{aligned} \hspace{\stretch{1}}(5.23)

In general this requires an iterated approach, solving for one with an initial approximation of the other, then switching and tuning the numerical method carefully for convergence.

We expect that the flow of liquid will induce a flow of air at the interface, but may be able to make a one-sided approximation. Let’s see how far we get before we have to introduce any approximations and compute the traction vector for the liquid

\begin{aligned}\boldsymbol{\sigma}^l \cdot \hat{\mathbf{n}} &= \begin{bmatrix}-p & \mu^l {\partial {u}}/{\partial {y}} & 0 \\ \mu^l {\partial {u}}/{\partial {y}} & -p & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} \\ &=\begin{bmatrix}\mu^l {\partial {u}}/{\partial {y}} \\ -p \\ 0\end{bmatrix}\end{aligned}

So

\begin{aligned}\boldsymbol{\tau} \cdot (\boldsymbol{\sigma}^l \cdot \hat{\mathbf{n}})=\begin{bmatrix}1 & 0 & 0\end{bmatrix}\begin{bmatrix}\mu^l {\partial {u}}/{\partial {y}} \\ -p \\ 0\end{bmatrix}=\mu^l \frac{\partial {u}}{\partial {y}}\end{aligned} \hspace{\stretch{1}}(5.25)

Our boundary value condition is therefore

\begin{aligned}{\left.{{\mu^l \frac{\partial {u^l}}{\partial {y}}}}\right\vert}_{{y = h}} ={\left.{{\mu^a \frac{\partial {u^a}}{\partial {y}}}}\right\vert}_{{y = h}}\end{aligned} \hspace{\stretch{1}}(5.26)

When can we decouple this, treating only the liquid? Observe that we have

\begin{aligned}{\left.{{\frac{\partial {u^l}}{\partial {y}}}}\right\vert}_{{y = h}} ={\left.{{\frac{\mu^a}{\mu^l} \frac{\partial {u^a}}{\partial {y}}}}\right\vert}_{{y = h}}\end{aligned} \hspace{\stretch{1}}(5.27)

so if

\begin{aligned}\frac{\mu_a}{\mu_l} \ll 1\end{aligned} \hspace{\stretch{1}}(5.28)

we can treat only the liquid portion of the problem, with a boundary value condition

\begin{aligned}{\left.{{\frac{\partial {u^l}}{\partial {y}}}}\right\vert}_{{y = h}} = 0.\end{aligned} \hspace{\stretch{1}}(5.29)

Let’s look at the component of the traction vector in the direction of the normal (liquid pressure acting on the air)

\begin{aligned}\hat{\mathbf{n}} \cdot (\boldsymbol{\sigma}^l \cdot \hat{\mathbf{n}}) = \hat{\mathbf{n}} \cdot (\boldsymbol{\sigma}^a \cdot \hat{\mathbf{n}}) \end{aligned} \hspace{\stretch{1}}(5.30)

or

\begin{aligned}\begin{bmatrix}0 & 1 & 0\end{bmatrix}\begin{bmatrix}\mu^l \frac{\partial {u}}{\partial {y}} \\ -p^l \\ 0\end{bmatrix}= -{\left.{{p^l}}\right\vert}_{{y = h}} = -{\left.{{p^a}}\right\vert}_{{y = h}}\end{aligned} \hspace{\stretch{1}}(5.31)

i.e. We have pressure matching at the interface.

Our body force is

\begin{aligned}\mathbf{f} = \begin{bmatrix}g \sin\alpha \\ -g \cos\alpha \\ 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.32)

Referring to the Navier-Stokes equation 4.4, we see that our only surviving parts are

\begin{subequations}

\begin{aligned}0 = -\frac{\partial {p}}{\partial {x}} + \mu \frac{\partial^2 {{u}}}{\partial {{y}}^2} + \rho g \sin\alpha \end{aligned} \hspace{\stretch{1}}(5.33a)

\begin{aligned}0 = -\frac{\partial {p}}{\partial {y}} - \rho g \cos\alpha \end{aligned} \hspace{\stretch{1}}(5.33b)

\begin{aligned}0 = -\frac{\partial {p}}{\partial {z}} \end{aligned} \hspace{\stretch{1}}(5.33c)

\end{subequations}

The last gives us $p \ne p(z)$. Integrating the second we have

\begin{aligned}p = \rho g y \cos\alpha + p_1\end{aligned} \hspace{\stretch{1}}(5.34)

Since $p = p_{\text{atm}}$ at $y = h$, we have

\begin{aligned}p_{\text{atm}} = \rho g h \cos\alpha + p_1\end{aligned} \hspace{\stretch{1}}(5.35)

Our first NS equation 5.33a becomes

\begin{aligned}0 = \mu \frac{\partial^2 {{u}}}{\partial {{y}}^2} + g \sin\alpha,\end{aligned} \hspace{\stretch{1}}(5.36)

or

\begin{aligned}\frac{\partial^2 {{u}}}{\partial {{y}}^2} = -\frac{g}{\mu} \sin\alpha\end{aligned} \hspace{\stretch{1}}(5.37)

Solving we have

\begin{aligned}u = - \rho g \frac{\sin\alpha}{2 \mu} y^2 + A y + B\end{aligned} \hspace{\stretch{1}}(5.38)

With

\begin{aligned}u(0) &= 0 \\ {\left.{{\frac{\partial {u}}{\partial {y}}}}\right\vert}_{{y = h}} &= 0\end{aligned} \hspace{\stretch{1}}(5.39)

\begin{aligned}u = \rho g \frac{\sin\alpha}{2 \mu} \left( 2 h y - y^2 \right) .\end{aligned} \hspace{\stretch{1}}(5.41)

This velocity distribution is illustrated figure (\ref{fig:continuumL12:continuumL12fig7}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig7}
\caption{Velocity streamlines for flow down a plane.}
\end{figure}

It’s important to note that in these problems we have to derive our boundary value conditions! They are not given.

In this discussion, the height $h$ was assumed to be constant, with the tangential direction constant and parallel to the surface that the liquid is flowing on. It’s claimed in class that this is actually a consequence of surface tension only! That’s not at all intuitive, but will be covered when we learn about “stability conditions”.

# Study note.

Memorizing the NS equation is required for midterm, but more complex stuff (like cylindrical forms of the strain tensor if required) will be given.

# References

[1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

## PHY454H1S Continuum Mechanics. Lecture 10: Navier-Stokes equation. Taught by Prof. K. Das.

Posted by peeterjoot on February 11, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Newtonian fluid.

Reading: section 6.* from [1].

We stated the model for a newtonian fluid

\begin{aligned}\sigma_{ij} = -p \delta_{ij} + 2 \mu e_{ij}\end{aligned} \hspace{\stretch{1}}(2.1)

and started considering conservation of mass with a volume $dV$ through an area element $d\mathbf{s}$. For the rate of change of mass flowing out of the volume $V$ is

\begin{aligned}\oint \rho \mathbf{u} \cdot d\mathbf{s} = - \frac{\partial {}}{\partial {t}} \int_V \rho dV.\end{aligned} \hspace{\stretch{1}}(2.2)

Application of Green’s theorem, for a fixed (in time) volume $V$ produces

\begin{aligned}0 = \int_V \left( \boldsymbol{\nabla} \cdot (\rho \mathbf{u}) + \frac{\partial {\rho}}{\partial {t}} \right) dV,\end{aligned} \hspace{\stretch{1}}(2.3)

or in differential form for an infinitesimal volume

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot (\rho \mathbf{u}).\end{aligned} \hspace{\stretch{1}}(2.4)

Expanding out the divergence term using

\begin{aligned}\boldsymbol{\nabla} \cdot (a \mathbf{b})&=\partial_i (a b_i) \\ &=b_i \partial_i a +a \partial_i b_i \\ &=\mathbf{b} \cdot \boldsymbol{\nabla} a+ a \boldsymbol{\nabla} \cdot \mathbf{b}\end{aligned}

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \rho \boldsymbol{\nabla} \cdot \mathbf{u}+ \mathbf{u} \cdot \boldsymbol{\nabla} \rho.\end{aligned} \hspace{\stretch{1}}(2.7)

For an incompressible fluid

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u} = 0\end{aligned} \hspace{\stretch{1}}(2.6)

so the conservation of mass equality relation takes the form

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \mathbf{u} \cdot \boldsymbol{\nabla} \rho.\end{aligned} \hspace{\stretch{1}}(2.7)

# Conservation of momentum.

In classical mechanics we have

\begin{aligned}\mathbf{f} = m \mathbf{a},\end{aligned} \hspace{\stretch{1}}(3.8)

our analogue here is found in terms of the stress tensor

\begin{aligned}\int_V F_i dV = \int_V \frac{\partial {\sigma_{ij}}}{\partial {x_j}} dV\end{aligned} \hspace{\stretch{1}}(3.9)

Here $F_i$ is the force per unit volume. With body forces we have

\begin{aligned}F_i = \rho \frac{du_i}{dt} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + \rho f_i\end{aligned} \hspace{\stretch{1}}(3.10)

where $f_i$ is an external force per unit volume. Observe that $\sigma_{ij}$, through the constituative relation, includes both contributions of linear displacement and the vorticity component.

From the constitutive relation 2.1, we have

\begin{aligned}\frac{\partial {\sigma_{ij}}}{\partial {x_j}} &= - \frac{\partial {p}}{\partial {x_j}} \delta_{ij} + 2 \mu \frac{\partial {e_{ij}}}{\partial {x_j}} \\ &= - \frac{\partial {p}}{\partial {x_i}} + 2 \mu \frac{\partial {}}{\partial {x_j}} \left( \frac{1}{{2}} \left( \frac{\partial {u_i}}{\partial {x_j}}+ \frac{\partial {u_j}}{\partial {x_i}}\right)\right) \\ &= - \frac{\partial {p}}{\partial {x_i}} + \mu \left(\frac{\partial^2 u_i}{\partial x_j \partial x_j}+\frac{\partial^2 u_j}{\partial x_i \partial x_j}\right) \end{aligned}

Observe that the term

\begin{aligned}\frac{\partial^2 u_i}{\partial x_j \partial x_j}\end{aligned} \hspace{\stretch{1}}(3.11)

is the $i^{\text{th}}$ component of $\boldsymbol{\nabla}^2 \mathbf{u}$, whereas

\begin{aligned}\frac{\partial^2 u_j}{\partial x_i \partial x_j} &= \frac{\partial {}}{\partial {x_i}} \left( \frac{\partial {u_j}}{\partial {x_j}} \right) \\ &= \frac{\partial {}}{\partial {x_i}} (\boldsymbol{\nabla} \cdot \mathbf{u})\end{aligned}

is the $i^{\text{th}}$ component of $\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u})$.

We have therefore that

\begin{aligned}\rho \frac{du_i}{dt} = \left( -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}\right)_i,\end{aligned} \hspace{\stretch{1}}(3.12)

or in vector notation

\begin{aligned}\rho \frac{d\mathbf{u}}{dt} = -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(3.13)

We can expand this a bit more writing our velocity $\mathbf{u} = \mathbf{u}(x, y, z, t)$ differential

\begin{aligned}du_i = \frac{\partial {u_i}}{\partial {x_j}} \delta x_j + \frac{\partial {u_i}}{\partial {t}} \delta t.\end{aligned} \hspace{\stretch{1}}(3.14)

Considering rates

\begin{aligned}\frac{du_i}{dt} = \frac{\partial {u_i}}{\partial {x_j}} \frac{dx_j}{dt} + \frac{\partial {u_i}}{\partial {t}} .\end{aligned} \hspace{\stretch{1}}(3.15)

In vector notation we have

\begin{aligned}\frac{d\mathbf{u}}{dt} = (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} + \frac{\partial {\mathbf{u}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(3.16)

Newton’s second law 3.13 now becomes

\begin{aligned}\boxed{\rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} + \rho \frac{\partial {\mathbf{u}}}{\partial {t}} = -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}.}\end{aligned} \hspace{\stretch{1}}(3.20)

This is the Navier-Stokes equation. Observe that we have an explicitly non-linear term

\begin{aligned}(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} ,\end{aligned} \hspace{\stretch{1}}(3.18)

something we don’t encounter in most classical mechanics. The impacts of this non-linear term are very significant and produce some interesting effects.

## Incompressible fluids.

Incompressibility was the condition

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u} = 0,\end{aligned} \hspace{\stretch{1}}(3.19)

so the Navier-Stokes equation takes the form

\begin{aligned}\begin{aligned}\rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} + \rho \frac{\partial {\mathbf{u}}}{\partial {t}} &= -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u} + \rho \mathbf{f} \\ \boldsymbol{\nabla} \cdot \mathbf{u} &= 0\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.20)

## Boundary value conditions.

In order to solve any sort of PDE we need to consider the boundary value conditions. Consider the interface between two layers of liquids as in figure (\ref{fig:continuumL9:continuumL10fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL10fig1}
\caption{Rocker tank with two viscosity fluids.}
\end{figure}

Also found an illustration of this in fig 1.13 of white’s text online.

We see the fluids sticking together at the boundary. This is due to matching of the tangential velocities at the interface.

# References

[1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

## PHY456H1S Continuum mechanics. Problem Set 1. Stress, Strain, Traction vector. Force free equilibrium.

Posted by peeterjoot on February 9, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

This problem set is as yet ungraded.

# Problem Q1.

## Statement

For the stress tensor

\begin{aligned}\sigma =\begin{bmatrix}6 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 3\end{bmatrix}\text{M Pa}\end{aligned} \hspace{\stretch{1}}(2.1)

Find the corresponding strain tensor, assuming an isotropic solid with Young’s modulus $E = 200 \times 10^9 \text{N}/\text{m}^2$ and Poisson’s ration $\nu = 0.35$.

## Solution

We need to express the relation between stress and strain in terms of Young’s modulus and Poisson’s ratio. In terms of Lam\’e parameters our model for the relations between stress and strain for an isotropic solid was given as

\begin{aligned}\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2 \mu e_{ij}.\end{aligned} \hspace{\stretch{1}}(2.2)

Computing the trace

\begin{aligned}\sigma_{kk} = (3 \lambda + 2 \mu) e_{kk},\end{aligned} \hspace{\stretch{1}}(2.3)

allows us to invert the relationship

\begin{aligned}2 \mu e_{ij} = \sigma_{ij} - \lambda \frac{\sigma_{kk}}{3 \lambda + 2 \mu} \delta_{ij}.\end{aligned} \hspace{\stretch{1}}(2.4)

In terms of Poisson’s ratio $\nu$ and Young’s modulus $E$, our Lam\’e parameters were found to be

\begin{aligned}\lambda &= \frac{ E \nu }{(1 - 2 \nu)(1 + \nu)} \\ \mu &= \frac{E}{2(1 + \nu)},\end{aligned} \hspace{\stretch{1}}(2.5)

and

\begin{aligned}3 \lambda + 2 \mu&= \frac{ 3 E \nu }{(1 - 2 \nu)(1 + \nu)} + \frac{E}{1 + \nu} \\ &= \frac{E}{1 + \nu} \left( \frac{3 \nu}{1 - 2 \nu} + 1\right) \\ &= \frac{E}{1 + \nu} \frac{1 + \nu}{1 - 2 \nu} \\ &= \frac{E}{1 - 2 \nu}.\end{aligned}

Our stress strain model for the relationship for an isotropic solid becomes
we find

\begin{aligned}\frac{E}{1 + \nu} e_{ij}&=\sigma_{ij}-\frac{ E \nu }{(1 - 2 \nu)(1 + \nu)} \frac{1 - 2 \nu}{E}\sigma_{kk} \delta_{ij} \\ &=\sigma_{ij}-\frac{ \nu }{1 + \nu}\sigma_{kk} \delta_{ij} \\ \end{aligned}

or

\begin{aligned}e_{ij}=\frac{1}{{E}}\left((1 + \nu)\sigma_{ij}-\nu\sigma_{kk} \delta_{ij}\right).\end{aligned} \hspace{\stretch{1}}(2.7)

As a sanity check note that this matches (5.12) of [1], although they use a notation of $\sigma$ instead of $\nu$ for Poisson’s ratio. We are now ready to tackle the problem. First we need the trace of the stress tensor

\begin{aligned}\sigma_{kk} = (6 + 1 + 3) \text{M Pa} = 10 \text{M Pa},\end{aligned} \hspace{\stretch{1}}(2.8)

\begin{aligned}e_{ij}&=\frac{1}{{E}}\left((1 + \nu)\begin{bmatrix}6 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 3\end{bmatrix}-10 \nu\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\right)\text{M Pa} \\ &=\frac{1}{{E}}\left(\begin{bmatrix}6 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 3\end{bmatrix}+ 0.35\begin{bmatrix}-4 & 0 & 2 \\ 0 & -9 & 1 \\ 2 & 1 & -7\end{bmatrix}\right)\text{M Pa} \\ &=\frac{1}{{2 \times 10^{5}}}\left(\begin{bmatrix}6 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 3\end{bmatrix}+ 0.35\begin{bmatrix}-4 & 0 & 2 \\ 0 & -9 & 1 \\ 2 & 1 & -7\end{bmatrix}\right)\end{aligned}

Expanding out the last bits of arithmetic the strain tensor is found to have the form

\begin{aligned}e_{ij}=\begin{bmatrix} 23 & 0 & 13.5 \\ 0 & -10.75 & 6.75 \\ 13.5 & 6.75 & 2.75\end{bmatrix} 10^{-6}.\end{aligned} \hspace{\stretch{1}}(2.9)

Note that this is dimensionless, unlike the stress.

# Problem Q2.

## Statement

Small displacement field in a material is given by

\begin{aligned}e_1 &= 2 x_1 x_2 \\ e_2 &= x_3^2 \\ e_3 &= x_1^2 - x_3\end{aligned} \hspace{\stretch{1}}(3.10)

Find

\begin{enumerate}
\item the infinitesimal strain tensor $e_{ij}$,
\item the principal strains and the corresponding principal axes at $(x_1, x_2, x_3) = (1, 2, 4)$,
\item Is the body under compression or expansion?
\end{enumerate}

## Solution. infinitesimal strain tensor $e_{ij}$

Diving right in, we have

\begin{aligned}e_{11}&= \frac{\partial {e_1}}{\partial {x_1}} \\ &= \frac{\partial {}}{\partial {x_1}}2 x_1 x_2 \\ &= 2 x_2\end{aligned}

\begin{aligned}e_{22}&= \frac{\partial {e_2}}{\partial {x_2}} \\ &= \frac{\partial {}}{\partial {x_2}} x_3^2 \\ &= 0\end{aligned}

\begin{aligned}e_{33}&= \frac{\partial {e_3}}{\partial {x_3}} \\ &= \frac{\partial {}}{\partial {x_3}} ( x_1^2 - x_3 ) \\ &= -1\end{aligned}

\begin{aligned}e_{12}&=\frac{1}{{2}} \left(\frac{\partial {e_2}}{\partial {x_1}}+\frac{\partial {e_1}}{\partial {x_2}}\right) \\ &=\frac{1}{{2}}\left(\not{{\frac{\partial {}}{\partial {x_1}} x_3^2 }}+\frac{\partial {}}{\partial {x_2}} 2 x_1 x_2\right) \\ &=x_1\end{aligned}

\begin{aligned}e_{23}&=\frac{1}{{2}} \left(\frac{\partial {e_3}}{\partial {x_2}}+\frac{\partial {e_2}}{\partial {x_3}}\right) \\ &=\frac{1}{{2}}\left(\not{{\frac{\partial {}}{\partial {x_2}} (x_1^2 - x_3 )}}+\frac{\partial {}}{\partial {x_3}} x_3^2\right) \\ &=x_3\end{aligned}

\begin{aligned}e_{31}&=\frac{1}{{2}} \left(\frac{\partial {e_1}}{\partial {x_3}}+\frac{\partial {e_3}}{\partial {x_1}}\right) \\ &=\frac{1}{{2}}\left(\not{{\frac{\partial {}}{\partial {x_3}} 2 x_1 x_2 }}+\frac{\partial {}}{\partial {x_1}} (x_1^2 - x_3 )\right) \\ &=x_1\end{aligned}

In matrix form we have

\begin{aligned}\mathbf{e} =\begin{bmatrix}2 x_2 & x_1 & x_1 \\ x_1 & 0 & x_3 \\ x_1 & x_3 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.13)

## Solution. principle strains and axes

At the point $(1, 2, 4)$ the strain tensor has the value

\begin{aligned}\mathbf{e} =\begin{bmatrix}4 & 1 & 1 \\ 1 & 0 & 4 \\ 1 & 4 & -1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.14)

We wish to diagonalize this, solving the characteristic equation for the eigenvalues $\lambda$

\begin{aligned}0 &=\begin{vmatrix}4 -\lambda & 1 & 1 \\ 1 & -\lambda & 4 \\ 1 & 4 & -1 -\lambda\end{vmatrix} \\ &=(4 -\lambda )\begin{vmatrix} -\lambda & 4 \\ 4 & -1 -\lambda\end{vmatrix}-\begin{vmatrix}1 & 1 \\ 4 & -1 -\lambda\end{vmatrix}+\begin{vmatrix}1 & 1 \\ -\lambda & 4 \\ \end{vmatrix} \\ &=(4 - \lambda)(\lambda^2 + \lambda - 16)-(-1 -\lambda - 4)+(4 + \lambda) \\ \end{aligned}

We find the characteristic equation to be

\begin{aligned}0 = -\lambda^3 + 3 \lambda^2 + 22\lambda - 55.\end{aligned} \hspace{\stretch{1}}(3.15)

This doesn’t appear to lend itself easily to manual solution (there are no obvious roots to factor out). As expected, since the matrix is symmetric, a plot (\ref{fig:continuumL8:continuumProblemSet1Q2fig1}) shows that all our roots are real

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumProblemSet1Q2fig1}
\caption{Q2. Characteristic equation.}
\end{figure}

Numerically, we determine these roots to be

\begin{aligned}\{5.19684, -4.53206, 2.33522\}\end{aligned} \hspace{\stretch{1}}(3.16)

with the corresponding basis (orthonormal eigenvectors), the principle axes are

\begin{aligned}\left\{\hat{\mathbf{p}}_1,\hat{\mathbf{p}}_2,\hat{\mathbf{p}}_3\right\}=\left\{\begin{bmatrix}0.76291 \\ 0.480082 \\ 0.433001\end{bmatrix},\begin{bmatrix}-0.010606 \\ -0.660372 \\ 0.750863\end{bmatrix},\begin{bmatrix}-0.646418 \\ 0.577433 \\ 0.498713\end{bmatrix}\right\}.\end{aligned} \hspace{\stretch{1}}(3.17)

## Solution. Is body under compression or expansion?

To consider this question, suppose that as in the previous part, we determine a basis for which our strain tensor $e_{ij} = p_i \delta_{ij}$ is diagonal with respect to that basis at a given point $\mathbf{x}_0$. We can then simplify the form of the stress tensor at that point in the object

\begin{aligned}\sigma_{ij}&=\frac{E}{1 + \nu} \left(e_{ij} + \frac{\nu}{1 - 2 \nu} e_{mm} \delta_{ij}\right) \\ &=\frac{E}{1 + \nu} \left(p_i + \frac{\nu}{1 - 2 \nu} e_{mm}\right)\delta_{ij}.\end{aligned}

We see that the stress tensor at this point is also necessarily diagonal if the strain is diagonal in that basis (with the implicit assumption here that we are talking about an isotropic material). Noting that the Poisson ratio is bounded according to

\begin{aligned}-1 \le \nu \le \frac{1}{{2}},\end{aligned} \hspace{\stretch{1}}(3.18)

so if our trace is positive (as it is in this problem for all points $x_2 > 1/2$), then any positive principle strain value will result in a positive stress along that direction). For example at the point $(1,2,4)$ of the previous part of this problem (for which $x_2 > 1/2$), we have

\begin{aligned}\sigma_{ij}=\frac{E}{1 + \nu}\begin{bmatrix}5.19684+ \frac{3 \nu}{1 - 2 \nu} & 0 & 0 \\ 0 & -4.53206+ \frac{3 \nu}{1 - 2 \nu} & 0 \\ 0 & 0 & 2.33522+ \frac{3 \nu}{1 - 2 \nu}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.19)

We see that at this point the $(1,1)$ and $(3,3)$ components of stress is positive (expansion in those directions) regardless of the material, and provided that

\begin{aligned}\frac{3 \nu}{1 - 2 \nu} > 4.53206\end{aligned} \hspace{\stretch{1}}(3.20)

(i.e. $\nu > 0.375664$) the material is under expansion in all directions. For $\nu < 0.375664$ the material at that point is expanding in the $\hat{\mathbf{p}}_1$ and $\hat{\mathbf{p}}_3$ directions, but under compression in the $\hat{\mathbf{p}}_2$ directions.

(save to disk and run with either Mathematica or the free Wolfram CDF player ( http://www.wolfram.com/cdf-player/ ) )

For a Mathematica notebook that visualizes this part of this problem see https://raw.github.com/peeterjoot/physicsplay/master/notes/phy454/continuumProblemSet1Q2animated.cdf. This animates the stress tensor associated with the problem, for different points $(x,y,z)$ and values of Poisson’s ratio $\nu$, with Mathematica manipulate sliders available to alter these (as well as a zoom control to scale the graphic, keeping the orientation and scale fixed with any variation of the other parameters). This generalizes the solution of the problem (assuming I got it right for the specific $(1,2,4)$ point of the problem). The vectors are the orthonormal eigenvectors of the tensor, scaled by the magnitude of the eigenvectors of the stress tensor (also diagonal in the basis of the diagonalized strain tensor at the point in question). For those directions that are under expansive stress, I’ve colored the vectors blue, and for compressive directions, I’ve colored the vectors red.

This requires either a Mathematica client or the free Wolfram CDF player, either of which can run the notebook after it is saved to your computer’s hard drive.

# Problem Q3.

## Statement

The stress tensor at a point has components given by

\begin{aligned}\sigma =\begin{bmatrix}1 & -2 & 2 \\ -2 & 3 & 1 \\ 2 & 1 & -1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(4.21)

Find the traction vector across an area normal to the unit vector

\begin{aligned}\hat{\mathbf{n}} = ( \sqrt{2} \mathbf{e}_1 - \mathbf{e}_2 + \mathbf{e}_3)/2\end{aligned} \hspace{\stretch{1}}(4.22)

Can you construct a tangent vector $\boldsymbol{\tau}$ on this plane by inspection? What are the components of the force per unit area along the normal $\hat{\mathbf{n}}$ and tangent $\boldsymbol{\tau}$ on that surface? (hint: projection of the traction vector.)

## Solution

The traction vector, the force per unit volume that holds a body in equilibrium, in coordinate form was

\begin{aligned}P_i = \sigma_{ik} n_k\end{aligned} \hspace{\stretch{1}}(4.23)

where $n_k$ was the coordinates of the normal to the surface with area $df_k$. In matrix form, this is just

\begin{aligned}\mathbf{P} = \sigma \hat{\mathbf{n}},\end{aligned} \hspace{\stretch{1}}(4.24)

so our traction vector for this stress tensor and surface normal is just

\begin{aligned}\mathbf{P} &=\frac{1}{{2}}\begin{bmatrix}1 & -2 & 2 \\ -2 & 3 & 1 \\ 2 & 1 & -1\end{bmatrix}\begin{bmatrix}\sqrt{2} \\ -1 \\ 1\end{bmatrix} \\ &=\frac{1}{{2}}\begin{bmatrix}\sqrt{2} + 2 + 2 \\ -2\sqrt{2} - 3 + 1 \\ 2\sqrt{2} - 1 -1\end{bmatrix} \\ &=\begin{bmatrix}\sqrt{2}/2 + 2 \\ -\sqrt{2} -1 \\ \sqrt{2} - 1\end{bmatrix}\end{aligned}

We also want a vector in the plane, and can pick

\begin{aligned}\boldsymbol{\tau} = \frac{1}{{\sqrt{2}}}\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.25)

or

\begin{aligned}\boldsymbol{\tau}' = \begin{bmatrix}\frac{1}{{\sqrt{2}}} \\ \frac{1}{{2}} \\ -\frac{1}{{2}}\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.26)

It’s clear that either of these is normal to $\hat{\mathbf{n}}$ (the first can also be computed by normalizing $\hat{\mathbf{n}} \times \mathbf{e}_1$, and the second with one round of Gram-Schmidt). However, neither of these vectors in the plane are particularly interesting since they are completely arbitrary. Let’s instead compute the projection and rejection of the traction vector with respect to the normal. We find for the projection

\begin{aligned}(\mathbf{P} \cdot \hat{\mathbf{n}}) \hat{\mathbf{n}}&=\frac{1}{{4}}\left(\begin{bmatrix}\sqrt{2}/2 + 2 \\ -\sqrt{2} -1 \\ \sqrt{2} - 1\end{bmatrix}\cdot \begin{bmatrix}\sqrt{2} \\ -1 \\ 1\end{bmatrix} \right)\begin{bmatrix}\sqrt{2} \\ -1 \\ 1\end{bmatrix} \\ &=\frac{1}{{4}}\left( 1 + 2\sqrt{2}+\sqrt{2} +1 +\sqrt{2} - 1\right)\begin{bmatrix}\sqrt{2} \\ -1 \\ 1\end{bmatrix} \\ &=\frac{1}{{2}}\left( 1 + 4\sqrt{2}\right)\hat{\mathbf{n}}\end{aligned}

Our rejection, the component of the traction vector in the plane, is

\begin{aligned}(\mathbf{P} \wedge \hat{\mathbf{n}}) \hat{\mathbf{n}} &=\mathbf{P} - (\mathbf{P} \cdot \hat{\mathbf{n}})\hat{\mathbf{n}} \\ &=\frac{1}{{2}}\begin{bmatrix}\sqrt{2}/2 + 2 \\ -\sqrt{2} -1 \\ \sqrt{2} - 1\end{bmatrix}-\frac{1}{{4}}(1 + r \sqrt{2})\begin{bmatrix}\sqrt{2} \\ -1 \\ 1\end{bmatrix} \\ &=\frac{1}{{4}}\begin{bmatrix}\sqrt{2} \\ -3 \\ -5\end{bmatrix}\end{aligned}

This gives us a another vector perpendicular to the normal $\hat{\mathbf{n}}$

\begin{aligned}\hat{\boldsymbol{\tau}} = \frac{1}{{6}}\begin{bmatrix}\sqrt{2} \\ -3 \\ -5\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(4.27)

Wrapping up, we find the decomposition of the traction vector in the direction of the normal and its projection onto the plane to be

\begin{aligned}\mathbf{P} = \frac{1}{{2}}(1 + 4\sqrt{2}) \hat{\mathbf{n}}+\frac{3}{2} \hat{\boldsymbol{\tau}}.\end{aligned} \hspace{\stretch{1}}(4.28)

The components we can read off by inspection.

# Problem Q4.

## Statement

The stress tensor of a body is given by

\begin{aligned}\sigma =\begin{bmatrix}A \cos x & y^2 & C x \\ y^2 & B \sin y & z \\ C x & z & z^3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.29)

Determine the constant $A$, $B$, and $C$ if the body is in equilibrium.

## Solution

In the absence of external forces our equilibrium condition was

\begin{aligned}\partial_k \sigma_{ik} = 0.\end{aligned} \hspace{\stretch{1}}(5.30)

In matrix form we wish to operate (to the left) with the gradient coordinate vector

\begin{aligned}0 &= \sigma \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \\ &=\begin{bmatrix}A \cos x & y^2 & C x \\ y^2 & B \sin y & z \\ C x & z & z^3\end{bmatrix}\begin{bmatrix}\stackrel{ \leftarrow }{\partial}_x \\ \stackrel{ \leftarrow }{\partial}_y \\ \stackrel{ \leftarrow }{\partial}_z \\ \end{bmatrix} \\ &=\begin{bmatrix}\partial_x (A \cos x) + \partial_y(y^2) + \not{{\partial_z(C x)}} \\ \not{{\partial_x (y^2)}} + \partial_y(B \sin y) + \partial_z(z) \\ \partial_x (C x) + \not{{\partial_y(z)}} + \partial_z(z^3)\end{bmatrix} \\ &=\begin{bmatrix}-A \sin x + 2 y \\ B \cos y + 1 \\ C + 3 z^2 \end{bmatrix} \\ \end{aligned}

So, our conditions for equilibrium will be satisfied when we have

\begin{aligned}A &= \frac{2 y }{\sin x} \\ B &= -\frac{1}{\cos y} \\ C &= -3 z^2,\end{aligned} \hspace{\stretch{1}}(5.31)

provided $y \ne 0$, and $y \ne \pi/2 + n\pi$ for integer $n$. If equilibrium is to hold along the $y = 0$ plane, then we must either also have $A = 0$ or also impose the restriction $x = m \pi$ (for integer $m$).

# A couple other mathematica notebooks

Some of the hand calculations done in this problem set I’ve confirmed using Mathematica. Those notebooks are available here

These all require either a Mathematica client or the free Wolfram CDF player. Note that I haven’t figured out a way to get a browser based CDF player to play these without explicit download.

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

## PHY454H1S Continuum Mechanics. Lecture 7: P-waves and S-waves. Taught by Prof. K. Das.

Posted by peeterjoot on February 1, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Setup

Reading: section 22 from [1].

We got as far as expressing the vector displacement $\mathbf{e}$ for an isotropic material at a given point in terms of the Lam\’e parameters

\begin{aligned}\rho \frac{\partial^2 {\mathbf{e}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{e}) + \mu \boldsymbol{\nabla}^2 \mathbf{e}.\end{aligned} \hspace{\stretch{1}}(2.1)

## P-waves.

Operating on this with the divergence once more, and writing $\theta = \boldsymbol{\nabla} \cdot \mathbf{e}$, we have

\begin{aligned}\rho \frac{\partial^2 {{\boldsymbol{\nabla} \cdot \mathbf{e}}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{e}) + \mu \boldsymbol{\nabla}^2 (\boldsymbol{\nabla} \cdot \mathbf{e})\end{aligned} \hspace{\stretch{1}}(2.2)

or

\begin{aligned}\frac{\partial^2 {{\theta}}}{\partial {{t}}^2} = \frac{\lambda + 2 \mu}{\rho} \boldsymbol{\nabla}^2 \theta.\end{aligned} \hspace{\stretch{1}}(2.3)

We see that our divergence is governed by a wave equation where the speed of the wave $C_L$ is specified by

\begin{aligned}C_L^2 = \frac{\lambda + 2 \mu}{\rho},\end{aligned} \hspace{\stretch{1}}(2.4)

so the displacement wave equation is given by

\begin{aligned}\frac{\partial^2 {{\theta}}}{\partial {{t}}^2} = C_L^2 \boldsymbol{\nabla}^2 \theta.\end{aligned} \hspace{\stretch{1}}(2.5)

Let’s look at the divergence of the displacement vector in some more detail. By definition this is just

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{e} = \frac{\partial {e_1}}{\partial {x_1}}+\frac{\partial {e_2}}{\partial {x_2}}+\frac{\partial {e_3}}{\partial {x_3}}.\end{aligned} \hspace{\stretch{1}}(2.6)

Recall that the strain tensor $e_{ij}$ was defined as

\begin{aligned}e_{ij} = \frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+\frac{\partial {e_j}}{\partial {x_i}}\right),\end{aligned} \hspace{\stretch{1}}(2.7)

so we have

\begin{aligned}\frac{\partial {e_1}}{\partial {x_1}} &= e_{11} \\ \frac{\partial {e_2}}{\partial {x_2}} &= e_{22} \\ \frac{\partial {e_3}}{\partial {x_3}} &= e_{33}.\end{aligned} \hspace{\stretch{1}}(2.8)

So the divergence in question can be written in terms of the strain tensor

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{e} = e_{11}+e_{22}+e_{33} = e_{ii}.\end{aligned} \hspace{\stretch{1}}(2.11)

We also found that the trace of the strain tensor was the relative change in volume. We call this the dilatation. A measure of change in volume as illustrated (badly) in figure (\ref{fig:continuumL7:continuumL7fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL7fig1}
\caption{Illustrating changes in a control volume.}
\end{figure}

This idea can be found nicely animated in the wikipedia page [2].

## S-waves.

Now let’s operate on our equation 2.1 with the curl operator

\begin{aligned}\rho \frac{\partial^2 {{\boldsymbol{\nabla} \times \mathbf{e}}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} \times (\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{e})) + \mu \boldsymbol{\nabla}^2 (\boldsymbol{\nabla} \times \mathbf{e}).\end{aligned} \hspace{\stretch{1}}(2.12)

Writing

\begin{aligned}\boldsymbol{\omega} = \boldsymbol{\nabla} \times \mathbf{e},\end{aligned} \hspace{\stretch{1}}(2.13)

and observing that $\boldsymbol{\nabla} \times \boldsymbol{\nabla} f = 0$ (with $f = \boldsymbol{\nabla} \cdot \mathbf{e}$), we find

\begin{aligned}\rho \frac{\partial^2 {{\boldsymbol{\omega}}}}{\partial {{t}}^2} = \mu \boldsymbol{\nabla}^2 \boldsymbol{\omega}.\end{aligned} \hspace{\stretch{1}}(2.14)

We call this the S-wave equation, and write $C_T$ for the speed of this wave

\begin{aligned}C_T^2 = \mu,\end{aligned} \hspace{\stretch{1}}(2.15)

so that we have

\begin{aligned}\frac{\partial^2 {{\boldsymbol{\omega}}}}{\partial {{t}}^2} = C_T^2 \boldsymbol{\nabla}^2 \boldsymbol{\omega}.\end{aligned} \hspace{\stretch{1}}(2.16)

Again, we can find nice animations of this on wikipedia [3].

## Relative speeds of the p-waves and s-waves.

Taking ratios of the wave speeds we find

\begin{aligned}\frac{C_L}{C_T} = \sqrt{\frac{ \lambda + 2 \mu}{\mu}} = \sqrt{ \frac{\lambda}{\mu} + 2}.\end{aligned} \hspace{\stretch{1}}(2.17)

Since both $\lambda > 0$ and $\mu > 0$, we have

\begin{aligned}C_L > C_T.\end{aligned} \hspace{\stretch{1}}(2.18)

Divergence (p-waves) are faster than rotational (s-waves) waves.

In terms of the Poisson ratio $\nu = \lambda/(2(\lambda + \mu))$, we find

\begin{aligned}\frac{\mu}{\lambda} = \frac{1}{{2 \nu}} - 1.\end{aligned} \hspace{\stretch{1}}(2.19)

we see that Poisson’s ratio characterizes the speeds of the waves for the medium

\begin{aligned}\frac{C_L}{C_T} = \sqrt{\frac{2(1-\nu)}{1 - 2\nu}}\end{aligned} \hspace{\stretch{1}}(2.20)

## Assuming a gradient plus curl representation.

Let’s assume that our displacement can be written in terms of a gradient and curl as we do for the electric field

\begin{aligned}\mathbf{e} = \boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H},\end{aligned} \hspace{\stretch{1}}(2.21)

Inserting this into 2.1 we find

\begin{aligned}\rho \frac{\partial^2 {{(\boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H})}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H})) + \mu \boldsymbol{\nabla}^2 (\boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H}).\end{aligned} \hspace{\stretch{1}}(2.22)

using

\begin{aligned}\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} \phi = \boldsymbol{\nabla}^2 \phi.\end{aligned} \hspace{\stretch{1}}(2.23)

Observe that

\begin{aligned}\boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \times \mathbf{H}) &=\partial_k (\partial_a H_b \epsilon_{abk})&=0\end{aligned}

Here we make use of the fact that an antisymmetric sum of symmetric partials is zero assuming sufficient continuity. Grouping terms we have

\begin{aligned}\boldsymbol{\nabla} \left(\rho \frac{\partial^2 {{\phi}}}{\partial {{t}}^2} - (\lambda + 2\mu) \boldsymbol{\nabla}^2 \phi\right)+ \boldsymbol{\nabla} \times \left(\rho \frac{\partial^2 {\mathbf{H}}}{\partial {{t}}^2} - \mu \boldsymbol{\nabla}^2 \mathbf{H}\right)= 0.\end{aligned} \hspace{\stretch{1}}(2.24)

When the material is infinite in scope, so that boundary value coupling is not a factor, we can write this as a set of independent P-wave and S-wave equations

\begin{aligned}\rho \frac{\partial^2 {{\phi}}}{\partial {{t}}^2} - (\lambda + 2\mu) \boldsymbol{\nabla}^2 \phi = 0\end{aligned} \hspace{\stretch{1}}(2.25)

The P-wave is irrotational (curl free).

\begin{aligned}\rho \frac{\partial^2 {\mathbf{H}}}{\partial {{t}}^2} - \mu \boldsymbol{\nabla}^2 \mathbf{H} = 0\end{aligned} \hspace{\stretch{1}}(2.26)

The S-wave is solenoidal (divergence free).

## A couple summarizing statements.

\begin{itemize}
\item
P-waves: irrotational. Volume not preserved.
\item
S-waves: divergence freee. Shearing forces are present and volume is preserved.
\item
P-waves are faster than S-waves.
\end{itemize}

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

[2] Wikipedia. P-wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 1-February-2012]. http://en.wikipedia.org/w/index.php?title=P-wave&oldid=474119033.

[3] Wikipedia. S-wave — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 1-February-2012]. http://en.wikipedia.org/w/index.php?title=S-wave&oldid=468110825.

## PHY454H1S Continuum Mechanics. Lecture 6: Compatibility condition and elastostatics. Taught by Prof. K. Das.

Posted by peeterjoot on January 29, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review: Elastostatics

We’ve defined the strain tensor, where assuming the second order terms are ignored, was

\begin{aligned}e_{ij} = \frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+ \frac{\partial {e_j}}{\partial {x_i}} \right).\end{aligned} \hspace{\stretch{1}}(2.1)

We’ve also defined a stress tensor defined implicitly as a divergence relationship using the force per unit volume $F_i$ in direction $i$

\begin{aligned}\sigma_{ij} \leftrightarrow F_i = \frac{\partial {\sigma_{ij}}}{\partial {x_j}}.\end{aligned} \hspace{\stretch{1}}(2.2)

We’ve also discussed the constitutive relation, relating stress $\sigma_{ij}$ and strain $e_{ij}$.

We’ve also discussed linear constitutive relationships (Hooke’s law).

# 2D strain.

\begin{aligned}e_{ij} = \begin{bmatrix}e_{11} & e_{12} \\ e_{21} & e_{22}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.3)

From 2.1 we see that we have

\begin{aligned}e_{11} &= \frac{\partial {e_1}}{\partial {x_1}} \\ e_{22} &= \frac{\partial {e_2}}{\partial {x_2}} \\ e_{12} = e_{21} &= \frac{1}{{2}} \left( \frac{\partial {e_2}}{\partial {x_1}}+ \frac{\partial {e_1}}{\partial {x_2}} \right).\end{aligned} \hspace{\stretch{1}}(3.4)

We have a relationship between these displacements (called the compatibility relationship), which is

\begin{aligned}\boxed{\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2} +\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} = 2\frac{\partial^2 e_{12}}{\partial x_1 \partial x_2}.}\end{aligned} \hspace{\stretch{1}}(3.7)

We find this by straight computation

\begin{aligned}\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2} &= \frac{\partial^2 {{}}}{\partial {{x_2}}^2}\left( \frac{\partial {e_1}}{\partial {x_1}}\right) \\ &=\frac{\partial^3 e_1}{\partial x_1 \partial x_2^2},\end{aligned}

and

\begin{aligned}\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} &= \frac{\partial^2 {{}}}{\partial {{x_1}}^2}\left( \frac{\partial {e_2}}{\partial {x_2}}\right) \\ &= \frac{\partial^3 e_2}{\partial x_2 \partial x_1^2},\end{aligned}

Now, looking at the cross term we find

\begin{aligned}2 \frac{\partial^2 e_{12}}{\partial x_1 \partial x_2} &= \frac{\partial^2 e_{12}}{\partial x_1 \partial x_2} \left(\frac{\partial {e_2}}{\partial {x_1}}+ \frac{\partial {e_1}}{\partial {x_2}} \right) \\ &=\left(\frac{\partial^3 e_1}{\partial x_1 \partial x_2^2} +\frac{\partial^3 e_2}{\partial x_2 \partial x_1^2} \right) \\ &=\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} +\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2} \end{aligned}

This is called the compatibility condition, and ensures that we don’t have a disjoint deformation of the form in figure (\ref{fig:continuumL6:continuumL6fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL6fig1}
\caption{disjoint deformation illustrated.}
\end{figure}

# 3D strain.

While we have 9 components in the tensor, not all of these are independent. The sets above and below the diagonal can be related, as illustrated in figure (\ref{fig:continuumL6:continuumL6fig2}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL6fig2}
\caption{continuumL6fig2}
\end{figure}

Here we have 6 relationships between the components of the strain tensor $e_{ij}$. Deriving these will be assigned in the homework.

# Elastodynamics. Elastic waves.

Reading: Chapter III (section 22 – section 26) of the text [1].

Example: sound or water waves (i.e. waves in a solid or liquid material that comes back to its original position.)

\begin{definition}
\emph{(Elastic Wave)}

An elastic wave is a type of mechanical wave that propagates through or on the surface of a medium. The elasticity of the material provides the restoring force (that returns the material to its original state). The displacement and the restoring force are assumed to be linearly related.
\end{definition}

In symbols we say

\begin{aligned}e_i(x_j, t) \quad \mbox{related to force},\end{aligned} \hspace{\stretch{1}}(5.8)

and specifically

\begin{aligned}\rho \frac{\partial^2 {{e_i}}}{\partial {{t}}^2} = F_i = \frac{\partial {\sigma_{ij}}}{\partial {x_j}}.\end{aligned} \hspace{\stretch{1}}(5.9)

This is just Newton’s second law, $F = ma$, but expressed in terms of a unit volume.

Should we have an external body force (per unit volume) $f_i$ acting on the body then we must modify this, writing

\begin{aligned}\boxed{\rho \frac{\partial^2 {{e_i}}}{\partial {{t}}^2} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + f_i}\end{aligned} \hspace{\stretch{1}}(5.10)

Note that we are separating out the “original” forces that produced the stress and strain on the object from any constant external forces that act on the body (i.e. a gravitational field).

With

\begin{aligned}e_{ij} = \frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+ \frac{\partial {e_j}}{\partial {x_i}} \right),\end{aligned} \hspace{\stretch{1}}(5.11)

we can expand the stress divergence, for the case of homogeneous deformation, in terms of the Lam\’e parameters

\begin{aligned}\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2 \mu e_{ij}.\end{aligned} \hspace{\stretch{1}}(5.12)

We compute

\begin{aligned}\frac{\partial {\sigma_{ij}}}{\partial {x_j}}&=\lambda \frac{\partial {e_{kk}}}{\partial {x_j}}\delta_{ij} + 2 \mu \frac{\partial {}}{\partial {x_j}}\frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+ \frac{\partial {e_j}}{\partial {x_i}} \right), \\ &=\lambda \frac{\partial {e_{kk}}}{\partial {x_i}}+ \mu \left(\frac{\partial^2 {{e_{i}}}}{\partial {{x_j}}^2}+\frac{\partial^2 e_{j} }{ \partial x_j \partial x_i}\right) \\ &=\lambda \frac{\partial {}}{\partial {x_i}}\frac{\partial {e_k}}{\partial {x_k}}+ \mu \left(\frac{\partial^2 {{e_{i}}}}{\partial {{x_j}}^2}+\frac{\partial^2 e_{k} }{ \partial x_k \partial x_i}\right) \\ &=(\lambda + \mu)\frac{\partial {}}{\partial {x_i}}\frac{\partial {e_k}}{\partial {x_k}}+ \mu \frac{\partial^2 {{e_{i}}}}{\partial {{x_j}}^2}\end{aligned}

We find, for homogeneous deformations, that the force per unit volume on our element of mass, in the absence of external forces (the body forces), takes the form

\begin{aligned}\rho \frac{\partial^2 {{e_i}}}{\partial {{t}}^2} = (\lambda + \mu) \frac{\partial^2 e_i}{\partial x_i \partial x_j}+ \mu\frac{\partial^2 e_i}{\partial x_j^2}.\end{aligned} \hspace{\stretch{1}}(5.13)

This can be seen to be equivalent to the vector relationship

\begin{aligned}\boxed{\rho \frac{\partial^2 {\mathbf{e}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{e}) + \mu \boldsymbol{\nabla}^2 \mathbf{e}.}\end{aligned} \hspace{\stretch{1}}(5.14)

TODO: What form do the stress and strain tensors take in vector form?

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

## PHY454H1S Continuum Mechanics. Lecture 5: Constitutive relationship. Taught by Prof. K. Das.

Posted by peeterjoot on January 28, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review: Cauchy Tetrahedron.

Referring to figure (\ref{fig:continuumL5:continuumL5fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig1}
\caption{Cauchy tetrahedron direction cosines.}
\end{figure}

recall that we can decompose our force into components that refer to our direction cosines $n_i = \cos\phi_i$

\begin{aligned}f_1 &= \sigma_{11} n_1 + \sigma_{12} n_2 + \sigma_{13} n_3 \\ f_2 &= \sigma_{21} n_1 + \sigma_{22} n_2 + \sigma_{23} n_3 \\ f_3 &= \sigma_{31} n_1 + \sigma_{32} n_2 + \sigma_{33} n_3\end{aligned} \hspace{\stretch{1}}(2.1)

Or in tensor form

\begin{aligned}f_i = \sigma_{ij} n_j.\end{aligned} \hspace{\stretch{1}}(2.4)

We call this the traction vector and denote it in vector form as

\begin{aligned}\mathbf{T} = \boldsymbol{\sigma} \cdot \hat{\mathbf{n}}=\begin{bmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33}\end{bmatrix}\begin{bmatrix}n_1 \\ n_2 \\ n_3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.5)

# Constitutive relation.

Reading: section 2, section 4 and section 5 from the text [1].

We can find the relationship between stress and strain, both analytically and experimentally, and call this the Constitutive relation. We prefer to deal with ranges of distortion that are small enough that we can make a linear approximation for this relation. In general such a linear relationship takes the form

\begin{aligned}\sigma_{ij} = c_{ijkl} e_{kl}.\end{aligned} \hspace{\stretch{1}}(3.6)

Consider the number of components that we are talking about for various rank tensors

\begin{aligned}\begin{array}{l l}\mbox{latex 0^\text{th}rank tensor} & \mbox{$3^0 = 1$ components} \\ \mbox{$1^\text{st}$ rank tensor} & \mbox{$3^1 = 3$ components} \\ \mbox{$2^\text{nd}$ rank tensor} & \mbox{$3^2 = 9$ components} \\ \mbox{$3^\text{rd}$ rank tensor} & \mbox{$3^3 = 81$ components}\end{array}\end{aligned} \hspace{\stretch{1}}(3.7)

We have a lot of components, even for a linear relation between stress and strain. For isotropic materials we model the constitutive relation instead as

\begin{aligned}\boxed{\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2 \mu e_{ij}.}\end{aligned} \hspace{\stretch{1}}(3.8)

For such a modeling of the material the (measured) values $\lambda$ and $\mu$ (shear modulus or modulus of rigidity) are called the Lam\’e parameters.

It will be useful to compute the trace of the stress tensor in the form of the constitutive relation for the isotropic model. We find

\begin{aligned}\sigma_{ii}&= \lambda e_{kk} \delta_{ii} + 2 \mu e_{ii} \\ &= 3 \lambda e_{kk} + 2 \mu e_{jj},\end{aligned}

or

\begin{aligned}\sigma_{ii} = (3 \lambda + 2 \mu) e_{kk}.\end{aligned} \hspace{\stretch{1}}(3.9)

We can now also invert this, to find the trace of the strain tensor in terms of the stress tensor

\begin{aligned}e_{ii} = \frac{\sigma_{kk}}{3 \lambda + 2 \mu}\end{aligned} \hspace{\stretch{1}}(3.10)

Substituting back into our original relationship 3.8, and find

\begin{aligned}\sigma_{ij} = \lambda \frac{\sigma_{kk}}{3 \lambda + 2 \mu} \delta_{ij} + 2 \mu e_{ij},\end{aligned} \hspace{\stretch{1}}(3.12)

which finally provides an inverted expression with the strain tensor expressed in terms of the stress tensor

\begin{aligned}\boxed{2 \mu e_{ij} =\sigma_{ij} - \lambda \frac{\sigma_{kk}}{3 \lambda + 2 \mu} \delta_{ij}.}\end{aligned} \hspace{\stretch{1}}(3.12)

## Special cases.

### Hydrostatic compression

Hydrostatic compression is when we have no shear stress, only normal components of the stress matrix $\sigma_{ij}$ is nonzero. Strictly speaking we define Hydrostatic compression as

\begin{aligned}\sigma_{ij} = -p \delta_{ij},\end{aligned} \hspace{\stretch{1}}(3.13)

i.e. not only diagonal, but with all the components of the stress tensor equal.

We can write the trace of the stress tensor as

\begin{aligned}\sigma_{ii} = - 3 p = (3 \lambda + 2 \mu) e_{kk}.\end{aligned} \hspace{\stretch{1}}(3.14)

Now, from our discussion of the strain tensor $e_{ij}$ recall that we found in the limit

\begin{aligned}dV' = (1 + e_{ii}) dV,\end{aligned} \hspace{\stretch{1}}(3.15)

allowing us to express the change in volume relative to the original volume in terms of the strain trace

\begin{aligned}e_{ii} = \frac{dV' - dV}{dV}.\end{aligned} \hspace{\stretch{1}}(3.16)

Writing that relative volume difference as $\Delta V/V$ we find

\begin{aligned}- 3 p = (3 \lambda + 2 \mu) \frac{\Delta V}{V},\end{aligned} \hspace{\stretch{1}}(3.17)

or

\begin{aligned}- \frac{ p V}{\Delta V} = \left( \lambda + \frac{2}{3} \mu \right) = K,\end{aligned} \hspace{\stretch{1}}(3.18)

where $K$ is called the Bulk modulus.

### Uniaxial stress

Again illustrated in the plane as in figure (\ref{fig:continuumL5:continuumL5fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig2}
\caption{Uniaxial stress.}
\end{figure}

Expanding out 3.12 we have for the $1,1$ element of the strain tensor

\begin{aligned}\boldsymbol{\sigma} =\begin{bmatrix}\sigma_{11} & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19)

\begin{aligned}2 \mu e_{11}&= \sigma_{11} - \frac{\lambda ( \sigma_{11} + \not{{\sigma_{22}}} ) }{3 \lambda + 2 \mu} \\ &= \sigma_{11} \frac{3 \lambda + 2 \mu - \lambda }{3 \lambda + 2 \mu} \\ &= 2 \sigma_{11} \frac{\lambda + \mu }{3 \lambda + 2 \mu}\end{aligned}

or

\begin{aligned}\frac{\sigma_{11}}{e_{11}} = \frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu } = E\end{aligned} \hspace{\stretch{1}}(3.20)

where $E$ is Young’s modulus. Young’s modulus in the text (5.3) is given in terms of the bulk modulus $K$. Using $\lambda = K - 2\mu/3$ we find

\begin{aligned}E &=\frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu } \\ &=\frac{\mu(3 (K - 2\mu/3)+ 2 \mu)}{K - 2\mu/3 + \mu } \\ &=\frac{3 K \mu}{ K + \mu/3 } \end{aligned}

\begin{aligned}\boxed{E =\frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu } =\frac{9 K \mu}{ 3 K + \mu } }\end{aligned} \hspace{\stretch{1}}(3.21)

FIXME: figure (\ref{fig:continuumL5:continuumL5fig3}) reference?

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig3}
\caption{stress associated with Young’s modulus}
\end{figure}

We define Poisson’s ratio $\nu$ as the quantity

\begin{aligned}\frac{e_{22}}{e_{11}} = \frac{e_{33}}{e_{11}} = - \nu.\end{aligned} \hspace{\stretch{1}}(3.22)

Note that we are still talking about uniaxial stress here. Referring back to 3.12 we have

\begin{aligned}2 \mu e_{2 2}&= \sigma_{2 2} - \lambda \frac{\sigma_{k k}}{3 \lambda + 2 \mu} \delta_{2 2} \\ &= \sigma_{2 2} - \lambda \frac{\sigma_{k k}}{3 \lambda + 2 \mu} \\ &= - \frac{\lambda \sigma_{11}}{3 \lambda + 2 \mu}\end{aligned}

Recall (3.20) that we had

\begin{aligned}\sigma_{11} = \frac{\mu (3 \lambda + 2 \mu)}{\lambda + \mu} e_{11}.\end{aligned} \hspace{\stretch{1}}(3.23)

Inserting this gives us

\begin{aligned}2 \mu e_{22} = - \frac{\lambda}{\not{{3 \lambda + 2 \mu}}} \frac{ \mu (\not{{3 \lambda + 2\mu}})}{\lambda + \mu} e_{11}\end{aligned}

so

\begin{aligned}\boxed{\nu = -\frac{e_{22}}{e_{11}} = \frac{\lambda}{2 (\lambda + \mu)}.}\end{aligned} \hspace{\stretch{1}}(3.24)

We can also relate the Poisson’s ratio $\nu$ to the shear modulus $\mu$

\begin{aligned}\mu = \frac{E}{2(1 + \nu)}\end{aligned} \hspace{\stretch{1}}(3.25)

\begin{aligned}\lambda = \frac{E \nu}{(1 - 2 \nu)(1 + \mu)}\end{aligned} \hspace{\stretch{1}}(3.26)

\begin{aligned}e_{11} &= \frac{1}{{E}}\left( \sigma_{11} - \nu(\sigma_{22} + \sigma_{33}) \right) \\ e_{22} &= \frac{1}{{E}}\left( \sigma_{22} - \nu(\sigma_{11} + \sigma_{33}) \right) \\ e_{33} &= \frac{1}{{E}}\left( \sigma_{33} - \nu(\sigma_{11} + \sigma_{22}) \right)\end{aligned} \hspace{\stretch{1}}(3.27)

These ones are (5.14) in the text, and are easy enough to verify (not done here).

### Appendix. Computing the relation between Poisson’s ratio and shear modulus.

Young’s modulus is given in 3.21 (equation (43) in the Professor’s notes) as

\begin{aligned}E = \frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu },\end{aligned} \hspace{\stretch{1}}(3.30)

and for Poisson’s ratio 3.24 (equation (46) in the Professor’s notes) we have

\begin{aligned}\nu = -\frac{e_{22}}{e_{11}} = \frac{\lambda}{2 (\lambda + \mu)}.\end{aligned} \hspace{\stretch{1}}(3.31)

Let’s derive the other stated relationships (equation (47) in the Professor’s notes). I get

\begin{aligned}2 (\lambda + \mu) \nu = \lambda \\ \implies \\ \lambda ( 2 \nu - 1 ) = - 2\mu\nu\end{aligned}

or

\begin{aligned}\lambda = \frac{ 2 \mu \nu} { 1 - 2 \nu }\end{aligned}

For substitution into the Young’s modulus equation calculate

\begin{aligned}\lambda + \mu &= \frac{ 2 \mu \nu} { 1 - 2 \nu } + \mu \\ &= \mu \left( \frac{ 2 \nu} { 1 - 2 \nu } + 1 \right) \\ &= \mu \frac{ 2 \nu + 1 - 2 \nu} { 1 - 2 \nu } \\ &= \frac{ \mu} { 1 - 2 \nu } \\ \end{aligned}

and

\begin{aligned}3 \lambda + 2 \mu &= 3 \frac{ \mu} { 1 - 2 \nu } - \mu \\ &= \mu \frac{ 3 - (1 - 2 \nu)} { 1 - 2 \nu } \\ &= \mu \frac{ 2 + 2 \nu} { 1 - 2 \nu } \\ &= 2 \mu \frac{ 1 + \nu} { 1 - 2 \nu } \\ \end{aligned}

Putting these together we find

\begin{aligned}E &= \frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu } \\ &= \mu 2 \mu \frac{ 1 + \nu} { 1 - 2 \nu } \frac{ 1 - 2 \nu}{\mu} \\ &= 2 \mu ( 1 + \nu ) \\ \end{aligned}

Rearranging we have

\begin{aligned}\mu = \frac{E}{2 (1 + \nu)}.\end{aligned} \hspace{\stretch{1}}(3.32)

This matches (5.9) in the text (where $\sigma$ is used instead of $\nu$).

We also find

\begin{aligned}\lambda &= \frac{ 2 \mu \nu} { 1 - 2 \nu } \\ &= \frac{ \nu} { 1 - 2 \nu } \frac{E }{1 + \nu}.\end{aligned}

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

## PHY454H1S Continuum Mechanics. Lecture 4: Strain tensor components. Taught by Prof. K. Das.

Posted by peeterjoot on January 21, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Stress tensor.

Reading: Portions of this lecture cover section 2 from the text [1].

For the stress tensor

\begin{aligned}\sigma_{ij},\end{aligned} \hspace{\stretch{1}}(2.1)

a second rank tensor, the first index $i$ defines the direction of the force, and the second index $j$ defines the surface.

Observe that the dimensions of $\sigma_{ij}$ is force per unit area, just like pressure. We will in fact show that this tensor is akin to the pressure, and the diagonalized components of this tensor represent the pressure.

We’ve illustrated the stress tensor in a couple of 2D examples. The first we call uniaxial stress, having just the $1,1$ element of the matrix as illustrated in figure (\ref{fig:continuumL4:continuumL4fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig1}
\caption{Uniaxial stress}
\end{figure}

\begin{aligned}\sigma = \begin{bmatrix}\sigma_{11} & 0 \\ 0 & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.2)

A biaxial stress is illustrated in figure (\ref{fig:continuumL4:continuumL4fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig2}
\caption{Biaxial stress.}
\end{figure}

where for $\sigma_{11} \ne \sigma_{22}$ our tensor takes the form

\begin{aligned}\sigma = \begin{bmatrix}\sigma_{11} & 0 \\ 0 & \sigma_{22}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.3)

In the general case we have

\begin{aligned}\sigma = \begin{bmatrix}\sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.4)

We can attempt to illustrate this, but it becomes much harder to visualize as shown in figure (\ref{fig:continuumL4:continuumL4fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig3}
\caption{General strain}
\end{figure}

In equilibrium we must have

\begin{aligned}\sigma_{12} = \sigma_{21}.\end{aligned} \hspace{\stretch{1}}(2.5)

We can use similar arguments to show that the stress tensor is symmetric.

In 3D we have three components of the stress tensor acting on each surface, as illustrated in figure (\ref{fig:continuumL4:continuumL4fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig5}
\caption{Strain components on a 3D volume.}
\end{figure}

We have three unique surface orientations and three components of the force for each of these, resulting in nine components, but these are not all independent. For an object in equilibrium we must have $\sigma_{ij} = \sigma_{ji}$ (FIXME: justify?). Explicitly, that is

\begin{aligned}\sigma_{12} &= \sigma_{21} \\ \sigma_{23} &= \sigma_{32} \\ \sigma_{31} &= \sigma_{13}\end{aligned} \hspace{\stretch{1}}(2.6)

## Diagonalization

We’ll look at the two dimensional case in some detail, as in figure (\ref{fig:continuumL4:continuumL4fig6})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig6}
\caption{Area element under strain with and without rotation.}
\end{figure}

Under this coordinate transformation, a rotation, the diagonal stress tensor is taken to a non-diagonal form

\begin{aligned}\begin{bmatrix}\sigma_{11} & 0 \\ 0 & \sigma_{22} \end{bmatrix}\leftrightarrow\begin{bmatrix}\sigma_{11}' & \sigma_{12}' \\ \sigma_{21}' & \sigma_{22}' \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.9)

## How do the stress tensor and the force relate

We form a Cauchy tetrahedron as in figure (\ref{fig:continuumL4:continuumL4fig7})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig7}
\caption{Cauchy tetrahedron}
\end{figure}

\begin{aligned}\mathbf{f} = \frac{\text{external force}}{\text{unit area}} = f_j \mathbf{e}_j\end{aligned} \hspace{\stretch{1}}(2.10)

\begin{aligned}\text{internal stress} = \text{external force}\end{aligned} \hspace{\stretch{1}}(2.11)

We write $\hat{\mathbf{n}}$ in terms of the direction cosines

\begin{aligned}\hat{\mathbf{n}} = n_1 \mathbf{e}_1 + n_2 \mathbf{e}_2 + n_3 \mathbf{e}_3 \end{aligned} \hspace{\stretch{1}}(2.12)

Here

\begin{aligned}n_1 &= \hat{\mathbf{n}} \cdot \mathbf{e}_1 \\ n_2 &= \hat{\mathbf{n}} \cdot \mathbf{e}_2 \\ n_3 &= \hat{\mathbf{n}} \cdot \mathbf{e}_3,\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}n_j = \hat{\mathbf{n}} \cdot \mathbf{e}_j = \cos\phi_j\end{aligned} \hspace{\stretch{1}}(2.16)

Force balance on $x_1$ direction, matching total external force in this direction to the total internal force ($\sigma_{ij}'s$) as follows

\begin{aligned}\begin{aligned}f_1 \times \text{area ABC} &= \sigma_{11} \times \text{area BOC} \\ &+\sigma_{12} \times \text{area AOC} \\ &+\sigma_{13} \times \text{area AOB}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.17)

Similarily

\begin{aligned}\begin{aligned}f_2 \times \text{area ABC} &= \sigma_{21} \times \text{area BOC} \\ &+\sigma_{22} \times \text{area AOC} \\ &+\sigma_{23} \times \text{area AOB},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.18)

and

\begin{aligned}\begin{aligned}f_3 \times \text{area ABC} &= \sigma_{31} \times \text{area BOC} \\ &+\sigma_{32} \times \text{area AOC} \\ &+\sigma_{33} \times \text{area AOB},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.19)

We can therefore write these force components like

\begin{aligned}f_1 = \sigma_{11} \frac{BOC}{ABC} + \sigma_{12} \frac{AOC}{ABC} + \sigma_{13} \frac{AOB}{ABC} \end{aligned} \hspace{\stretch{1}}(2.20)

but these ratios are really just the projections of the areas as illustrated in figure (\ref{fig:continuumL4:continuumL4fig8})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig8}
\caption{Area projection.}
\end{figure}

where an arbitrary surface with area $\Delta S$ can be decomposed into projections

\begin{aligned}\Delta S \cos\phi_j,\end{aligned} \hspace{\stretch{1}}(2.21)

utilizing the direction cosines. We can therefore write

\begin{aligned}f_1 &= \sigma_{11} n_1 + \sigma_{12} n_2 + \sigma_{13} n_3 \\ f_2 &= \sigma_{21} n_1 + \sigma_{22} n_2 + \sigma_{23} n_3 \\ f_3 &= \sigma_{31} n_1 + \sigma_{32} n_2 + \sigma_{33} n_3,\end{aligned} \hspace{\stretch{1}}(2.22)

or in matrix notation

\begin{aligned}\begin{bmatrix}f_1 \\ f_2 \\ f_3 \end{bmatrix}=\begin{bmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix}\begin{bmatrix}n_1 \\ n_2 \\ n_3 \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.25)

This is just

\begin{aligned}\boxed{f_i = \sigma_{ij} n_j.}\end{aligned} \hspace{\stretch{1}}(2.26)

This force with components $f_i$ is also called the traction vector

\begin{aligned}T_i = \sigma_{ij} n_j.\end{aligned} \hspace{\stretch{1}}(2.27)

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. Physics Today, 13:44, 1960.

## PHY454H1S Continuum Mechanics. Lecture 3. Strain tensor review. Stress tensor. Taught by Prof. K. Das.

Posted by peeterjoot on January 20, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Strain.

Strain is the measure of stretching. This is illustrated pictorially in figure (\ref{fig:continuumL3:continuumL3fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig1}
\caption{Stretched line elements.}
\end{figure}

\begin{aligned}{ds'}^2 - ds^2 = 2 e_{ik} dx_i dx_k,\end{aligned} \hspace{\stretch{1}}(1.1)

where $e_{ik}$ is the strain tensor. We found

\begin{aligned}e_{ik} = \frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_k}} +\frac{\partial {e_k}}{\partial {x_i}} +\frac{\partial {e_l}}{\partial {x_i}} \frac{\partial {e_l}}{\partial {x_k}} \right)\end{aligned} \hspace{\stretch{1}}(1.2)

Why do we have a factor two? Observe that if the deformation is small we can write

\begin{aligned}{ds'}^2 - ds^2 &= (ds' - ds)(ds' + ds) \\ &\approx (ds' - ds) 2 ds\end{aligned}

so that we find

\begin{aligned}\frac{{ds'}^2 - ds^2 }{ds^2}\approx\frac{ds' - ds }{ds}\end{aligned} \hspace{\stretch{1}}(1.3)

Suppose for example, that we have a diagonalized strain tensor, then we find

\begin{aligned}{ds'}^2 - ds^2 = 2 e_{ii} \left(\frac{dx_i}{ds}\right)^2\end{aligned} \hspace{\stretch{1}}(1.4)

so that

\begin{aligned}\frac{{ds'}^2 - ds^2 }{ds^2}= 2 e_{ii} dx_i^2\end{aligned} \hspace{\stretch{1}}(1.5)

Observe that here again we see this factor of two.

If we have a diagonalized strain tensor, the tensor is of the form

\begin{aligned}\begin{bmatrix}e_{11} & 0 & 0 \\ 0 & e_{22} & 0 \\ 0 & 0 & e_{33} \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.6)

we have

\begin{aligned}{dx_i'}^2 - dx_i^2 = 2 e_{ii} dx_i^2\end{aligned} \hspace{\stretch{1}}(1.7)

\begin{aligned}{ds'}^2 = (1 + 2 e_{11}) dx_1^2+(1 + 2 e_{22}) dx_2^2+(1 + 2 e_{33}) dx_3^2\end{aligned} \hspace{\stretch{1}}(1.8)

\begin{aligned}ds^2 = dx_1^2+dx_2^2+dx_3^2\end{aligned} \hspace{\stretch{1}}(1.9)

so

\begin{aligned}dx_1' &= \sqrt{1 + 2 e_{11}} dx_1 \sim ( 1 + e_{11}) dx_1 \\ dx_2' &= \sqrt{1 + 2 e_{22}} dx_2 \sim ( 1 + e_{22}) dx_2 \\ dx_3' &= \sqrt{1 + 2 e_{33}} dx_3 \sim ( 1 + e_{33}) dx_3\end{aligned} \hspace{\stretch{1}}(1.10)

Observe that the change in the volume element becomes the trace

\begin{aligned}dV' = dx_1'dx_2'dx_3'= dV(1 + e_{ii})\end{aligned} \hspace{\stretch{1}}(1.13)

How do we use this? Suppose that you are given a strain tensor. This should allow you to compute the stretch in any given direction.

FIXME: find problem and try this.

# Stress tensor.

Reading for this section is section 2 from the text associated with the prepared notes [1].

We’d like to consider a macroscopic model that contains the net effects of all the internal forces in the object as depicted in figure (\ref{fig:continuumL3:continuumL3fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig2}
\caption{Internal forces.}
\end{figure}

We will consider a volume big enough that we won’t have to consider the individual atomic interactions, only the average effects of those interactions. Will will look at the force per unit volume on a differential volume element

The total force on the body is

\begin{aligned}\iiint \mathbf{F} dV,\end{aligned} \hspace{\stretch{1}}(2.14)

where $\mathbf{F}$ is the force per unit volume. We will evaluate this by utilizing the divergence theorem. Recall that this was

\begin{aligned}\iiint (\boldsymbol{\nabla} \cdot \mathbf{A}) dV= \iint \mathbf{A} \cdot d\mathbf{s}\end{aligned} \hspace{\stretch{1}}(2.15)

We have a small problem, since we have a non-divergence expression of the force here, and it is not immediately obvious that we can apply the divergence theorem. We can deal with this by assuming that we can find a vector valued tensor, so that if we take the divergence of this tensor, we end up with the force. We introduce the quantity

\begin{aligned}\mathbf{F} = \frac{\partial {\sigma_{ik}}}{\partial {x_k}},\end{aligned} \hspace{\stretch{1}}(2.16)

and require this to be a vector. We can then apply the divergence theorem

\begin{aligned}\iiint \mathbf{F} dV = \iiint \frac{\partial {\sigma_{ik}}}{\partial {x_k}} d\mathbf{x}^3 \iint \sigma_{ik} ds_k,\end{aligned} \hspace{\stretch{1}}(2.17)

where $ds_k$ is a surface element. We identify this tensor

\begin{aligned}\sigma_{ik} = \frac{\text{Force}}{\text{Unit Area}}\end{aligned} \hspace{\stretch{1}}(2.18)

and

\begin{aligned}f_i = \sigma_{ik} ds_k,\end{aligned} \hspace{\stretch{1}}(2.19)

as the force on the surface element $ds_k$. In two dimensions this is illustrated in the following figures (\ref{fig:continuumL3:continuumL3fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig3}
\caption{2D strain tensor.}
\end{figure}

Observe that we use the index $i$ above as the direction of the force, and index $k$ as the direction normal to the surface.

Note that the strain tensor has the matrix form

\begin{aligned}\begin{bmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.20)

We will show later that this tensor is in fact symmetric.

FIXME: given some 3D forces, compute the stress tensor that is associated with it.

## Examples of the stress tensor

### Example 1. stretch in two opposing directions.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig4}
\caption{Opposing stresses in one direction.}
\end{figure}

Here, as illustrated in figure (\ref{fig:continuumL3:continuumL3fig4}), the associated (2D) stress tensor takes the simple form

\begin{aligned}\begin{bmatrix}\sigma_{11} & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.21)

### Example 2. stretch in a pair of mutually perpendicular directions

For a pair of perpendicular forces applied in two dimensions, as illustrated in figure (\ref{fig:continuumL3:continuumL3fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig5}
\caption{Mutually perpendicular forces}
\end{figure}

our stress tensor now just takes the form

\begin{aligned}\begin{bmatrix}\sigma_{11} & 0 \\ 0 & \sigma_{22}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.22)

It’s easy to imagine now how to get some more general stress tensors, should we make a change of basis that rotates our frame.

### Example 3. radial stretch

Suppose we have a fire fighter’s safety net, used to catch somebody jumping from a burning building (do they ever do that outside of movies?), as in figure (\ref{fig:continuumL3:continuumL3fig6}). Each of the firefighters contributes to the stretch.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig6}