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# Posts Tagged ‘strain 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.

\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.

\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.

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.

## 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

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}

\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.

## Strain tensor in spherical coordinates

Posted by peeterjoot on January 23, 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.)]

## Spherical tensor.

To perform the derivation in spherical coordinates we have some setup to do first, since we need explicit representations of all three unit vectors. The radial vector we can get easily by geometry and find the usual

\begin{aligned}\hat{\mathbf{r}} =\begin{bmatrix}\sin\theta \cos\phi \\ \sin\theta \sin\phi \\ \cos\theta\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.61)

We can get $\hat{\boldsymbol{\phi}}$ by geometrical intuition since it the plane unit vector at angle $\phi$ rotated by $\pi/2$. That is

\begin{aligned}\hat{\boldsymbol{\phi}} =\begin{bmatrix}-\sin\phi \\ \cos\phi \\ 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.62)

We can get $\hat{\boldsymbol{\theta}}$ by utilizing the right handedness of the coordinates since

\begin{aligned}\hat{\boldsymbol{\phi}} \times \hat{\mathbf{r}} = \hat{\boldsymbol{\theta}}\end{aligned} \hspace{\stretch{1}}(3.63)

and find

\begin{aligned}\hat{\boldsymbol{\theta}} =\begin{bmatrix}\cos\theta \cos\phi \\ \cos\theta \sin\phi \\ -\sin\theta\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.64)

\begin{aligned}\begin{aligned}&d\mathbf{l}'^2 - d\mathbf{x}^2 \\ &=2 (dr)^2 \biggl(\frac{\partial u_r}{\partial r}+ \frac{1}{{2}}\frac{\partial u_m}{\partial r} \frac{\partial u_m}{\partial r}\biggr) \\ & + 2 r^2 (d\theta )^2 \biggl(\frac{1}{{r}} u_r + \frac{1}{{2r^2}}(u_r^2 + u_{\theta }^2) - \frac{1}{{r^2}} u_{\theta } \frac{\partial u_r}{\partial \theta }+ \left(\frac{1}{{r}} + \frac{1}{{r^2}}u_r\right) \frac{\partial u_{\theta }}{\partial \theta }+ \frac{1}{{2 r^2}} \frac{\partial u_m}{\partial \theta } \frac{\partial u_m}{\partial \theta }\biggr) \\ &+ 2 r^2 \sin^2\theta (d\phi )^2 \biggl( \frac{1}{{2 r^2 \sin^2\theta}} u_\phi^2+ \frac{1}{{2 r^2 }} u_{\theta }^2 \cot^2\theta+ \frac{1}{{r}} u_r+ \frac{1}{{2 r^2}} u_r^2+ \left(\frac{1}{{r}} + \frac{1}{{r^2}}u_r\right) u_{\theta } \cot\theta \\ &\qquad- \frac{1}{{r^2 \sin\theta}} u_{\phi } \frac{\partial u_r}{\partial \phi }- \frac{1}{{r^2 }} u_{\phi } \frac{\cos\theta}{\sin^2\theta} \frac{\partial u_{\theta }}{\partial \phi }+ \frac{1}{{r^2 }} \frac{\partial u_{\phi }}{\partial \phi } \left(u_{\theta } \frac{\cos\theta}{\sin^2\theta} + \left(r + u_r\right) \frac{1}{{\sin\theta}} \right)+ \frac{1}{{2 r^2 \sin^2\theta}} \frac{\partial u_m}{\partial \phi } \frac{\partial u_m}{\partial \phi }\biggr) \\ & + 2 dr r d\theta \biggl(- \frac{1}{{r}} u_{\theta }+ \frac{1}{{r}} \frac{\partial u_r}{\partial \theta }- \frac{1}{{r}} u_{\theta } \frac{\partial u_r}{\partial r}+ \frac{\partial u_{\theta }}{\partial r} \left(1 + \frac{u_r}{r} \right)+ \frac{1}{{r}} \frac{\partial u_m}{\partial r} \frac{\partial u_m}{\partial \theta }\biggr) \\ & + 2 r^2 \sin\theta d\theta d\phi \biggl(\frac{1}{{r^2 }} u_{\theta } u_{\phi }- \frac{1}{{r^2 \sin\theta}} u_{\theta } \frac{\partial u_r}{\partial \phi }- \frac{1}{{r^2 }} u_{\phi } \frac{\partial u_r}{\partial \theta }- \frac{1}{{r^2 }} u_{\phi } \cot\theta \left(r + u_r + \frac{\partial u_{\theta }}{\partial \theta }\right) \\ &\qquad+ \frac{1}{{r^2 \sin\theta}} \left(r + u_r \right) \frac{\partial u_{\theta }}{\partial \phi }+ \frac{\partial u_{\phi }}{\partial \theta } \left(\frac{u_{\theta }}{r^2} \cot\theta + \frac{1}{{r}} + \frac{u_r}{r^2} \right)+ \frac{1}{{r^2 \sin\theta}} \frac{\partial u_m}{\partial \theta } \frac{\partial u_m}{\partial \phi }\biggr) \\ & + 2 r \sin\theta d\phi dr \biggl(- \frac{1}{{r }} u_{\phi }+ \frac{1}{{r \sin\theta}} \frac{\partial u_r}{\partial \phi }- u_{\phi } \frac{1}{{r }} \frac{\partial u_r}{\partial r}- u_{\phi } \cot\theta \frac{1}{{r }} \frac{\partial u_{\theta }}{\partial r}+ \frac{1}{{r }} \frac{\partial u_{\phi }}{\partial r} \left( u_{\theta } \cot\theta + r + u_r \right)+ \frac{1}{{r \sin\theta}} \frac{\partial u_m}{\partial \phi } \frac{\partial u_m}{\partial r}\biggr)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.65)

### A manual derivation.

Doing the calculation pretty much completely with Mathematica is rather unsatisfying. To set up for it let’s first compute the unit vectors from scratch. I’ll use geometric algebra to do this calculation. Consider figure (\ref{fig:qmTwoExamReflection:continuumL2fig5})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig5}
\caption{Composite rotations for spherical polar unit vectors.}
\end{figure}

We have two sets of rotations, the first is a rotation about the $z$ axis by $\phi$. Writing $i = \mathbf{e}_1 \mathbf{e}_2$ for the unit bivector in the $x,y$ plane, we rotate

\begin{aligned}\mathbf{e}_1' &= \mathbf{e}_1 e^{i\phi} = \mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi \\ \mathbf{e}_2' &= \mathbf{e}_2 e^{i\phi} = \mathbf{e}_2 \cos\phi - \mathbf{e}_1 \sin\phi \\ \mathbf{e}_3' &= \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(3.66)

Now we rotate in the plane spanned by $\mathbf{e}_3$ and $\mathbf{e}_1'$ by $\theta$. With $j = \mathbf{e}_3 \mathbf{e}_1'$, our vectors in the plane rotate as

\begin{aligned}\mathbf{e}_1'' &= \mathbf{e}_1' e^{j\phi} = \mathbf{e}_1 e^{i\phi} e^{j\theta} \\ \mathbf{e}_3'' &= \mathbf{e}_3' e^{j\theta} = \mathbf{e}_3 e^{j\theta},\end{aligned} \hspace{\stretch{1}}(3.69)

(with $\mathbf{e}_2'' = \mathbf{e}_2$ since $\mathbf{e}_2 \cdot j = 0$).

\begin{aligned}\hat{\boldsymbol{\theta}} = \mathbf{e}_1''&= \mathbf{e}_1 e^{i\phi} e^{j\theta} \\ &= \mathbf{e}_1 e^{i\phi} (\cos\theta + \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} \sin\theta) \\ &= \mathbf{e}_1 e^{i\phi} \cos\theta -\mathbf{e}_3 \sin\theta \\ &= (\mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi) \cos\theta -\mathbf{e}_3 \sin\theta \\ \end{aligned}

\begin{aligned}\hat{\mathbf{r}} = \mathbf{e}_3''&= \mathbf{e}_3 e^{j\theta} \\ &= \mathbf{e}_3 (\cos\theta + \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} \sin\theta) \\ &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 e^{i\phi} \sin\theta \\ &= \mathbf{e}_3 \cos\theta + (\mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi) \sin\theta \\ \end{aligned}

Now, these are all the same relations that we could find with coordinate algebra

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_1 \cos\phi \sin\theta +\mathbf{e}_2 \sin\phi \sin\theta +\mathbf{e}_3 \cos\theta \\ \hat{\boldsymbol{\theta}} &= \mathbf{e}_1 \cos\phi \cos\theta +\mathbf{e}_2 \sin\phi \cos\theta -\mathbf{e}_3 \sin\theta \\ \hat{\boldsymbol{\phi}} &= -\mathbf{e}_1 \sin\phi + \mathbf{e}_2 \cos\phi\end{aligned} \hspace{\stretch{1}}(3.71)

There’s nothing special in this approach if that is as far as we go, but we can put things in a nice tidy form for computation of the differentials of the unit vectors. Introducing the unit pseudoscalar $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$ we can write these in a compact exponential form.

\begin{aligned}\hat{\mathbf{r}}&= (\mathbf{e}_1 \cos\phi +\mathbf{e}_2 \sin\phi ) \sin\theta +\mathbf{e}_3 \cos\theta \\ &= \mathbf{e}_1 e^{i\phi} \sin\theta +\mathbf{e}_3 \cos\theta \\ &= \mathbf{e}_3 ( \cos\theta + \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} \sin\theta ) \\ &= \mathbf{e}_3 ( \cos\theta + \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_2 e^{i\phi} \sin\theta ) \\ &= \mathbf{e}_3 ( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta ) \\ &= \mathbf{e}_3 e^{ I \hat{\boldsymbol{\phi}} \theta }\end{aligned}

\begin{aligned}\hat{\boldsymbol{\theta}}&=\mathbf{e}_1 \cos\phi \cos\theta +\mathbf{e}_2 \sin\phi \cos\theta -\mathbf{e}_3 \sin\theta \\ &=(\mathbf{e}_1 \cos\phi +\mathbf{e}_2 \sin\phi ) \cos\theta -\mathbf{e}_3 \sin\theta \\ &=\mathbf{e}_1 e^{i\phi} \cos\theta -\mathbf{e}_3 \sin\theta \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta - e^{-i\phi} \mathbf{e}_1 \mathbf{e}_3 \sin\theta ) \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta - \mathbf{e}_1 \mathbf{e}_3 e^{i\phi} \sin\theta ) \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta - \mathbf{e}_1 \mathbf{e}_3 \mathbf{e}_2 \mathbf{e}_2 e^{i\phi} \sin\theta ) \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta ) \\ &=\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_2 e^{i\phi} ( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta ) \\ &=i \hat{\boldsymbol{\phi}} e^{I \hat{\boldsymbol{\phi}} \theta}.\end{aligned}

To summarize we have

\begin{aligned}\hat{\boldsymbol{\phi}} &= \mathbf{e}_2 e^{i\phi} \\ \hat{\mathbf{r}} &= \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} \\ \hat{\boldsymbol{\theta}} &= i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta}.\end{aligned} \hspace{\stretch{1}}(3.74)

Taking differentials we find first

\begin{aligned}d\hat{\boldsymbol{\phi}} = \mathbf{e}_2 e^{i\phi} i d\phi = \hat{\boldsymbol{\phi}} i d\phi\end{aligned}

\begin{aligned}d\hat{\boldsymbol{\theta}}&= d \left( i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} \right) \\ &= i d \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} + i \hat{\boldsymbol{\phi}} d \left( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta \right) \\ &= i d \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta}+ i \hat{\boldsymbol{\phi}} I (d \hat{\boldsymbol{\phi}}) \sin\theta+ i \hat{\boldsymbol{\phi}} I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= i \hat{\boldsymbol{\phi}} i e^{I\hat{\boldsymbol{\phi}} \theta} d\phi+ i \hat{\boldsymbol{\phi}} I \hat{\boldsymbol{\phi}} i \sin\theta d\phi+ i \hat{\boldsymbol{\phi}} I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\phi- I \sin\theta d\phi- \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= \hat{\boldsymbol{\phi}} (\cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta) d\phi- I \sin\theta d\phi- \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta\end{aligned}

\begin{aligned}d \hat{\mathbf{r}}&=\mathbf{e}_3 d \left( e^{I\hat{\boldsymbol{\phi}} \theta} \right) \\ &=\mathbf{e}_3 d \left( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta \right) \\ &=\mathbf{e}_3 \left( I (d \hat{\boldsymbol{\phi}}) \sin\theta + I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \right) \\ &=\mathbf{e}_3 \left( I \hat{\boldsymbol{\phi}} i \sin\theta d\phi + I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \right) \\ &=i \hat{\boldsymbol{\phi}} i \sin\theta d\phi + i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &=\hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta\end{aligned}

Summarizing these differentials we have

\begin{aligned}d\hat{\mathbf{r}} &= \hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta \\ d\hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta \\ d\hat{\boldsymbol{\phi}} &= \hat{\boldsymbol{\phi}} i d\phi\end{aligned} \hspace{\stretch{1}}(3.77)

A final cleanup is required. While $\hat{\boldsymbol{\phi}} i$ is a vector and has a nicely compact form, we need to decompose this into components in the $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$ and $\hat{\boldsymbol{\phi}}$ directions. Taking scalar products we have

\begin{aligned}\hat{\boldsymbol{\phi}} \cdot (\hat{\boldsymbol{\phi}} i) = 0\end{aligned}

\begin{aligned}\hat{\mathbf{r}} \cdot (\hat{\boldsymbol{\phi}} i)&=\left\langle{{ \hat{\mathbf{r}} \hat{\boldsymbol{\phi}} i}}\right\rangle \\ &=\left\langle{{ \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} \mathbf{e}_2 e^{i\phi} i}}\right\rangle \\ &=\left\langle{{ \mathbf{e}_3 (\cos\theta + I \mathbf{e}_2 e^{i\phi} \sin\theta) \mathbf{e}_2 e^{i\phi} i}}\right\rangle \\ &=\left\langle{{ I (\cos\theta e^{-i\phi} + I \mathbf{e}_2 \sin\theta) \mathbf{e}_2 }}\right\rangle \\ &=-\sin\theta\end{aligned}

\begin{aligned}\hat{\boldsymbol{\theta}} \cdot (\hat{\boldsymbol{\phi}} i)&=\left\langle{{ \hat{\boldsymbol{\theta}} \hat{\boldsymbol{\phi}} i }}\right\rangle \\ &=\left\langle{{ i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} \hat{\boldsymbol{\phi}} i }}\right\rangle \\ &=-\left\langle{{ \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} \hat{\boldsymbol{\phi}} }}\right\rangle \\ &=-\left\langle{{ e^{I\hat{\boldsymbol{\phi}} \theta} }}\right\rangle \\ &=- \cos\theta.\end{aligned}

Summarizing once again, but this time in terms of $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$ and $\hat{\boldsymbol{\phi}}$ we have

\begin{aligned}d\hat{\mathbf{r}} &= \hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta \\ d\hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta \\ d\hat{\boldsymbol{\phi}} &= -(\hat{\mathbf{r}} \sin\theta + \hat{\boldsymbol{\theta}} \cos\theta) d\phi\end{aligned} \hspace{\stretch{1}}(3.80)

Now we are set to take differentials. With

\begin{aligned}\mathbf{x} = r \hat{\mathbf{r}},\end{aligned} \hspace{\stretch{1}}(3.83)

we have

\begin{aligned}d\mathbf{x} =dr \hat{\mathbf{r}}+ r d\hat{\mathbf{r}}=dr \hat{\mathbf{r}} + \hat{\boldsymbol{\phi}} r \sin\theta d\phi + r \hat{\boldsymbol{\theta}} d\theta.\end{aligned} \hspace{\stretch{1}}(3.84)

Squaring this we get the usual spherical polar line scalar line element

\begin{aligned}d\mathbf{x}^2 = dr^2 + r^2 \sin^2\theta d\phi^2 + r^2 d\theta^2.\end{aligned} \hspace{\stretch{1}}(3.85)

With

\begin{aligned}\mathbf{u} = u_r \hat{\mathbf{r}} + u_\theta \hat{\boldsymbol{\theta}} + u_\phi \hat{\boldsymbol{\phi}},\end{aligned} \hspace{\stretch{1}}(3.86)

our differential is

\begin{aligned}d\mathbf{u}&=du_r \hat{\mathbf{r}} + du_\theta \hat{\boldsymbol{\theta}} + du_\phi \hat{\boldsymbol{\phi}}+ u_r d\hat{\mathbf{r}} + u_\theta d\hat{\boldsymbol{\theta}} + u_\phi d \hat{\boldsymbol{\phi}} \\ &=du_r \hat{\mathbf{r}} + du_\theta \hat{\boldsymbol{\theta}} + du_\phi \hat{\boldsymbol{\phi}}+ u_r \left(\hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta \right)+ u_\theta \left( \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta \right)- u_\phi (\hat{\mathbf{r}} \sin\theta + \hat{\boldsymbol{\theta}} \cos\theta) d\phi\\ &=\hat{\mathbf{r}} \left( du_r - u_\theta d\theta - u_\phi \sin\theta d\phi \right) \\ &+\hat{\boldsymbol{\theta}} \left( du_\theta + u_r d\theta - u_\phi \cos\theta d\phi \right) \\ &+\hat{\boldsymbol{\phi}} \left( du_\phi + u_r \sin\theta d\phi + u_\theta \cos\theta d\phi \right).\end{aligned}

We can add $d\mathbf{x}$ to this and take differences

\begin{aligned}\begin{aligned}(d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2&=\left( du_r - u_\theta d\theta - u_\phi \sin\theta d\phi + dr \right)^2 \\ &+\left( du_\theta + u_r d\theta - u_\phi \cos\theta d\phi + r d\theta \right)^2 \\ &+\left( du_\phi + u_r \sin\theta d\phi + u_\theta \cos\theta d\phi + r \sin\theta d\phi \right)^2\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.87)

For each $m = r,\theta,\phi$ we have

\begin{aligned}du_m=\frac{\partial {u_m}}{\partial {r}} dr +\frac{\partial {u_m}}{\partial {\theta}} d\theta +\frac{\partial {u_m}}{\partial {\phi}} d\phi,\end{aligned} \hspace{\stretch{1}}(3.88)

and plugging through that calculation is really all it takes to derive the textbook result. To do this to first order in $u_m$, we find

\begin{aligned}\frac{1}{{2}} \left((d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2\right)&=du_r dr- u_\theta d\theta dr- u_\phi \sin\theta d\phi dr \\ &+ du_\theta r d\theta+ u_r r d\theta^2- u_\phi r \cos\theta d\phi d\theta \\ &+ r \sin\theta du_\phi d\phi+ r \sin^2\theta u_r d\phi^2+ r \sin\theta \cos\theta u_\theta d\phi^2 \\ &=\left( \frac{\partial {u_r}}{\partial {r}} dr + \frac{\partial {u_r}}{\partial {\theta}} d\theta + \frac{\partial {u_r}}{\partial {\phi}} d\phi \right)dr- u_\theta d\theta dr- u_\phi \sin\theta d\phi dr \\ &+\left( \frac{\partial {u_\theta}}{\partial {r}} dr + \frac{\partial {u_\theta}}{\partial {\theta}} d\theta + \frac{\partial {u_\theta}}{\partial {\phi}} d\phi \right) r d\theta+ u_r r d\theta^2- u_\phi r \cos\theta d\phi d\theta \\ &+\left( \frac{\partial {u_\phi}}{\partial {r}} dr + \frac{\partial {u_\phi}}{\partial {\theta}} d\theta + \frac{\partial {u_\phi}}{\partial {\phi}} d\phi \right)r \sin\theta d\phi+ r \sin^2\theta u_r d\phi^2+ r \sin\theta \cos\theta u_\theta d\phi^2\end{aligned}

Collecting terms we have the result of the text in the braces

\begin{aligned}\begin{aligned}\left((d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2\right)&=2 dr^2 \left(\frac{\partial {u_r}}{\partial {r}}\right) \\ &+2 r^2 d\theta^2 \left(\frac{1}{{r}} \frac{\partial {u_\theta}}{\partial {\theta}} + u_r \frac{1}{{r}}\right) \\ &+2 r^2 \sin^2\theta d\phi^2 \left(\frac{\partial {u_\phi}}{\partial {\phi}} \frac{1}{{r \sin\theta}} + \frac{1}{{r}} u_r + \frac{1}{{r}} \cot\theta u_\theta\right) \\ &+2 dr r d\theta \left(\frac{1}{{r}} \frac{\partial {u_r}}{\partial {\theta}} - \frac{1}{{r}} u_\theta +\frac{\partial {u_\theta}}{\partial {r}}\right) \\ &+2 r^2 \sin\theta d\theta d\phi \left(\frac{\partial {u_\theta}}{\partial {\phi}} \frac{1}{{r \sin\theta}} - \frac{1}{{r}} u_\phi \cot\theta +\frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\theta}}\right) \\ &+2 r \sin\theta d\phi dr \left(\frac{1}{{r \sin\theta}} \frac{\partial {u_r}}{\partial {\phi}} - \frac{1}{{r}} u_\phi + \frac{\partial {u_\phi}}{\partial {r}}\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.89)

It should be possible to do the calculation to second order too, but to include all the quadratic terms in $u_m$ is again really messy. Trying that with mathematica gives the same results as above using the strictly coordinate algebra approach.

# 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 2. Introduction and strain tensor. 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.

# Introduction.

Mechanics could be defined as the study of effects of forces and displacements on a physical body

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig1}
\caption{Physical body.}
\end{figure}

In continuum mechanics we have a physical body and we are interested in the internal motions in the object.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig2}
\caption{Control volume elements.}
\end{figure}

For the first time considering mechanics we have to introduce the concepts of fields to make progress tackling these problems.

We will have use of the following types of fields

\begin{itemize}
\item Scalar fields. $3^0$ components. Examples: density, Temperature, …
\item Vector fields. $3^1$ components. Examples: Force, velocity.
\item Tensor fields. $3^2$ components. Examples: stress, strain.
\end{itemize}

We have to consider objects (a control volume) that is small enough that we can consider that we have a point in space limit for the quantities of density and velocity. At the same time we cannot take this limiting process to the extreme, since if we use a control volume that is sufficiently small, quantum and inter-atomic effects would have to be considered.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig3}
\caption{Mass and volume ratios at different scales.}
\end{figure}

## Stress and Strain definitions.

\begin{definition}
\emph{(Stress)}

Measure of the Internal force on the surfaces.
\end{definition}

\begin{definition}
\emph{(Strain)}

Measure of the deformation of the body.
\end{definition}

# Strain Tensor.

This follows [1] section 1 very closely.

Utilizing summation convention consider a set of small internal displacements $u_1, u_2, u_3$ to the $x, y, z$ coordinates so that the transformation $x_i \rightarrow x_i'$ is related by

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig4}
\caption{Differential change to the object.}
\end{figure}

\begin{aligned}u_i &= x_i' - x_i \\ x_i' &= g(x_i) \\ u_i &= f(x_i)\end{aligned} \hspace{\stretch{1}}(3.1)

(ie: $x_i'$ is a function of all the initial coordinates, as are the displacements $u_i$).

\begin{aligned}dx_i' = dx_i + du_i\end{aligned} \hspace{\stretch{1}}(3.4)

\begin{aligned}dl &= \sqrt{dx_k dx_k} \\ dl' &= \sqrt{d{x'}_k d{x'}_k}\end{aligned} \hspace{\stretch{1}}(3.5)

or

\begin{aligned}{dl'}^2 = (dx_k + du_k)(dx_k + du_k)= dl^2 + 2 dx_k du_k + du_k du_k\end{aligned} \hspace{\stretch{1}}(3.7)

with

\begin{aligned}du_i = \frac{\partial {u_i}}{\partial {x_k}} dx_k\end{aligned} \hspace{\stretch{1}}(3.8)

we have

\begin{aligned}du_i^2 = \frac{\partial {u_i}}{\partial {x_k}} dx_k\frac{\partial {u_i}}{\partial {x_l}} dx_l\end{aligned} \hspace{\stretch{1}}(3.9)

\begin{aligned}{dl'}^2 &= dl^2 + 2 \frac{\partial {u_i}}{\partial {x_k}} dx_k dx_i + \frac{\partial {u_l}}{\partial {x_i}} \frac{\partial {u_l}}{\partial {x_k}} dx_i dx_k \\ &= dl^2 + \left(\frac{\partial {u_i}}{\partial {x_k}} +\frac{\partial {u_k}}{\partial {x_i}} \right)dx_k dx_i + \frac{\partial {u_l}}{\partial {x_i}} \frac{\partial {u_l}}{\partial {x_k}} dx_i dx_k \\ &=dl^2 + 2 e_{ik} dx_i dx_k\end{aligned}

We write

\begin{aligned}{dl'}^2 - dl^2 = 2 e_{ik} dx_i dx_k\end{aligned} \hspace{\stretch{1}}(3.10)

where we define the \emph{strain tensor} as

\begin{aligned}e_{ik} = \frac{1}{{2}} \left(\left(\frac{\partial {u_i}}{\partial {x_k}} +\frac{\partial {u_k}}{\partial {x_i}} \right)+ \frac{\partial {u_l}}{\partial {x_i}} \frac{\partial {u_l}}{\partial {x_k}} \right)\end{aligned} \hspace{\stretch{1}}(3.11)

Here $e_{ik}$ is a $3 \times 3$ matrix in Cartesian coordinates

\begin{aligned}\begin{bmatrix}e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.12)

We see from 3.11 that $e_{ik}$ is symmetric, so we have

\begin{aligned}e_{21} &= e_{12} \\ e_{31} &= e_{13} \\ e_{32} &= e_{23}\end{aligned} \hspace{\stretch{1}}(3.13)

Because any real symmetric matrix can be diagonalized we can write in some coordinate system

\begin{aligned}\bar{e}_{ik} = \begin{bmatrix}\bar{e}_{11} & 0 & 0 \\ 0 & \bar{e}_{22} & 0 \\ 0 & 0 & \bar{e}_{33} \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.16)

\begin{aligned}{dx_1'}^2 &= (1 + 2 \bar{e}_{11}) dx_1^2 \\ {dx_2'}^2 &= (1 + 2 \bar{e}_{22}) dx_2^2 \\ {dx_3'}^2 &= (1 + 2 \bar{e}_{33}) dx_3^2\end{aligned} \hspace{\stretch{1}}(3.17)

If our changes are small enough we can also write approximately, taking the first order term in the square root evaluation

\begin{aligned}dx_1' &\approx (1 + \bar{e}_{11}) dx_1 \\ dx_2' &\approx (1 + \bar{e}_{22}) dx_2 \\ dx_3' &\approx (1 + \bar{e}_{33}) dx_3\end{aligned} \hspace{\stretch{1}}(3.20)

We are also free to define a volume element

\begin{aligned}dV' = dx_1'dx_2'dx_3'\approx(1 + e_{11})(1 + e_{22})(1 + e_{33})dx_1 dx_2 dx_3\end{aligned} \hspace{\stretch{1}}(3.23)

\begin{aligned}dV' = (1 + e_{11} +e_{22} +e_{33} ) dV\end{aligned} \hspace{\stretch{1}}(3.24)

So the change of volume is given by the trace

\begin{aligned}dV' = ( 1 + e_{ii} )^2 dV\end{aligned} \hspace{\stretch{1}}(3.25)

## Strain Tensor in cylindrical coordinates.

At the end of the section in the text, the formulas for the spherical and cylindrical versions (to first order) of the strain tensor is given without derivation. Let’s do that derivation for the cylindrical case, which is simpler. It appears that use of explicit vector notation is helpful here, so we write

\begin{aligned}\mathbf{x} &= r \hat{\mathbf{r}} + z \hat{\mathbf{z}} \\ \mathbf{u} & u_r \hat{\mathbf{r}} + u_\phi \hat{\boldsymbol{\phi}} + u_z \hat{\mathbf{z}}\end{aligned} \hspace{\stretch{1}}(3.26)

where

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_1 e^{i\phi} \\ \hat{\boldsymbol{\phi}} &= \mathbf{e}_2 e^{i\phi} \\ i &= \mathbf{e}_1 \mathbf{e}_2\end{aligned} \hspace{\stretch{1}}(3.28)

Since $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\phi}}$ are functions of position, we will need their differentials

\begin{aligned}d\hat{\mathbf{r}} &= \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 e^{i\phi} d\phi = \mathbf{e}_2 e^{i \phi} d\phi \\ d\hat{\boldsymbol{\phi}} &= \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2 e^{i\phi} d\phi = -\mathbf{e}_2 e^{i \phi} d\phi,\end{aligned} \hspace{\stretch{1}}(3.31)

but these are just scaled basis vectors

\begin{aligned}d\hat{\mathbf{r}} &= \hat{\boldsymbol{\phi}} d\phi \\ d\hat{\boldsymbol{\phi}} &= -\hat{\mathbf{r}} d\phi.\end{aligned} \hspace{\stretch{1}}(3.33)

So for our $\mathbf{x}$ and $\mathbf{u}$ differentials we find

\begin{aligned}d\mathbf{x} &= dr \hat{\mathbf{r}} + r d\hat{\mathbf{r}} + dz \hat{\mathbf{z}} \\ &= dr \hat{\mathbf{r}} + r \hat{\boldsymbol{\phi}} d\phi + dz \hat{\mathbf{z}},\end{aligned}

and

\begin{aligned}d\mathbf{u} &= du_r \hat{\mathbf{r}} + du_\phi \hat{\boldsymbol{\phi}} + du_z \hat{\mathbf{z}} + u_r \hat{\boldsymbol{\phi}} d\phi - u_\phi \hat{\mathbf{r}} d\phi \\ &= \hat{\mathbf{r}}( du_r - u_\phi d\phi )+ \hat{\boldsymbol{\phi}} ( du_\phi + u_r d\phi )+ \hat{\mathbf{z}} ( du_z ).\end{aligned}

Putting these together we have

\begin{aligned}d\mathbf{l}' &= d\mathbf{u} + d\mathbf{x} \\ &= \hat{\mathbf{r}}( du_r - u_\phi d\phi + dr )+ \hat{\boldsymbol{\phi}} ( du_\phi + u_r d\phi + r d\phi )+ \hat{\mathbf{z}} ( du_z + dz ).\end{aligned}

For the squared magnitude’s difference from $d\mathbf{x}^2$ we have

\begin{aligned}(d\mathbf{l}')^2 - d\mathbf{x}^2&= ( du_r - u_\phi d\phi + dr )^2+ ( du_\phi + u_r d\phi + r d\phi )^2+ ( du_z + dz )^2-dr^2 - r^2 d\phi^2 - dz^2 \\ &=( du_r - u_\phi d\phi )^2 + 2 dr ( du_r - u_\phi d\phi )+ ( du_\phi + u_r d\phi )^2+ 2 r d\phi ( du_\phi + u_r d\phi )+ du_z^2 + 2 du_z dz \\ \end{aligned}

Expanding this out, but dropping all the terms that are quadratic in the components of $\mathbf{u}$ or its differentials, we have

\begin{aligned}(d\mathbf{l}')^2 - d\mathbf{x}^2&\approx 2 dr ( du_r - u_\phi d\phi )+ 2 r d\phi ( du_\phi + u_r d\phi )+ 2 du_z dz \\ &= 2 dr du_r - 2 dr u_\phi d\phi + 2 r d\phi du_\phi + 2 r d\phi u_r d\phi + 2 du_z dz \\ &= 2 dr \left( \frac{\partial {u_r}}{\partial {r}} dr+\frac{\partial {u_r}}{\partial {\phi}} d\phi+\frac{\partial {u_r}}{\partial {z}} dz\right) \\ &- 2 dr d\phi u_\phi \\ &+ 2 r d\phi \left( \frac{\partial {u_\phi}}{\partial {r}} dr+\frac{\partial {u_\phi}}{\partial {\phi}} d\phi+\frac{\partial {u_\phi}}{\partial {z}} dz\right) \\ &+ 2 r d\phi d\phi u_r \\ &+ 2 dz \left( \frac{\partial {u_z}}{\partial {r}} dr+\frac{\partial {u_z}}{\partial {\phi}} d\phi+\frac{\partial {u_z}}{\partial {z}} dz\right) \\ \end{aligned}

Grouping all terms, with all the second order terms neglected, we have

\begin{aligned}\begin{aligned}(d\mathbf{l}')^2 - d\mathbf{x}^2&=2 dr dr \frac{\partial {u_r}}{\partial {r}} + 2 r^2 d\phi d\phi \left( \frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\phi}} +\frac{1}{{r}} u_r \right)+ 2 dz dz \frac{\partial {u_z}}{\partial {z}} \\ &+ 2 dz dr \left( \frac{\partial {u_r}}{\partial {z}} + \frac{\partial {u_z}}{\partial {r}} \right)+ 2 dr r d\phi \left( \frac{\partial {u_\phi}}{\partial {r}} - \frac{1}{{r}} u_\phi + \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}} \right)+ 2 dz r d\phi \left( \frac{\partial {u_\phi}}{\partial {z}} +\frac{1}{{r}} \frac{\partial {u_z}}{\partial {\phi}} \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.35)

From this we can read off the result quoted in the text

\begin{aligned}2 e_{rr} &= \frac{\partial {u_r}}{\partial {r}} \\ 2 e_{\phi\phi} &= \frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\phi}} +\frac{1}{{r}} u_r \\ 2 e_{zz} &= \frac{\partial {u_z}}{\partial {z}} \\ 2 e_{zr} &= \frac{\partial {u_r}}{\partial {z}} + \frac{\partial {u_z}}{\partial {r}} \\ 2 e_{r\phi} &= \frac{\partial {u_\phi}}{\partial {r}} - \frac{1}{{r}} u_\phi + \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}} \\ 2 e_{\phi z} &= \frac{\partial {u_\phi}}{\partial {z}} +\frac{1}{{r}} \frac{\partial {u_z}}{\partial {\phi}}.\end{aligned} \hspace{\stretch{1}}(3.36)

Observe that we have to introduce factors of $r$ along with all the $d\phi$‘s, when we factored out the tensor components. That’s an important looking detail, which isn’t obvious unless one works through the derivation.

Note that in class we retained the second order terms. That becomes a messier calculation and I’ve cheated using the symbolic capabilities of mathematica to do it

\begin{aligned}\begin{aligned}&(d\mathbf{l}')^2 - d\mathbf{x}^2 \\ &= (dr)^2 \left(2 \frac{\partial u_r}{\partial r}+\left(\frac{\partial u_r}{\partial r}\right)^2+\left(\frac{\partial u_z}{\partial r}\right)^2+\left(\frac{\partial u_{\phi }}{\partial r}\right)^2\right) \\ &+(d\phi )^2 \left(2 r u_r+u_r^2+u_{\phi }^2-2 u_{\phi } \frac{\partial u_r}{\partial \phi }+\left(\frac{\partial u_r}{\partial \phi }\right)^2+\left(\frac{\partial u_z}{\partial \phi }\right)^2+2 r \frac{\partial u_{\phi }}{\partial \phi }+2 u_r \frac{\partial u_{\phi }}{\partial \phi }+\left(\frac{\partial u_{\phi }}{\partial \phi }\right)^2\right) \\ &+(dz)^2 \left(\left(\frac{\partial u_r}{\partial z}\right)^2+2 \frac{\partial u_z}{\partial z}+\left(\frac{\partial u_z}{\partial z}\right)^2+\left(\frac{\partial u_{\phi }}{\partial z}\right)^2\right) \\ &+dr d\phi \left(-2 u_{\phi }-2 u_{\phi } \frac{\partial u_r}{\partial r}+2 \frac{\partial u_r}{\partial \phi }+2 \frac{\partial u_r}{\partial r} \frac{\partial u_r}{\partial \phi }+2 \frac{\partial u_z}{\partial r} \frac{\partial u_z}{\partial \phi }+2 r \frac{\partial u_{\phi }}{\partial r}+2 u_r \frac{\partial u_{\phi }}{\partial r}+2 \frac{\partial u_{\phi }}{\partial r} \frac{\partial u_{\phi }}{\partial \phi }\right) \\ &+dz d\phi \left(-2 u_{\phi } \frac{\partial u_r}{\partial z}+2 \frac{\partial u_r}{\partial z} \frac{\partial u_r}{\partial \phi }+2 \frac{\partial u_z}{\partial \phi }+2 \frac{\partial u_z}{\partial z} \frac{\partial u_z}{\partial \phi }+2 r \frac{\partial u_{\phi }}{\partial z}+2 u_r \frac{\partial u_{\phi }}{\partial z}+2 \frac{\partial u_{\phi }}{\partial z} \frac{\partial u_{\phi }}{\partial \phi }\right) \\ &+dr dz \left(2 \frac{\partial u_r}{\partial z}+2 \frac{\partial u_r}{\partial r} \frac{\partial u_r}{\partial z}+2 \frac{\partial u_z}{\partial r}+2 \frac{\partial u_z}{\partial r} \frac{\partial u_z}{\partial z}+2 \frac{\partial u_{\phi }}{\partial r} \frac{\partial u_{\phi }}{\partial z}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.42)

As with the first order case, we can read off the tensor coordinates by inspection (once we factor out the various factors of $2$ and $r$). The next logical step would be to do the spherical tensor calculation. That would likely be particularily messy if we attempted it in the brute force fashion. Let’s step back and look at the general case, before tackling there sphereical polar form explicitly.

## Strain Tensor for general coordinate representation.

Now let’s dispense with the assumption that we have an orthonormal frame. Given an arbitrary, not neccessarily orthonormal, position dependent frame $\{e_\mu\}$, and its reciprocal frame $\{e^\mu\}$, as defined by

\begin{aligned}e_\mu \cdot e^\nu = {\delta_\mu}^\nu.\end{aligned} \hspace{\stretch{1}}(3.43)

Our coordinate representation, with summation and dimensionality implied, is

\begin{aligned}\mathbf{x} &= x^\mu e_\mu = x_\nu e^\nu \\ \mathbf{u} &= u^\mu e_\mu = u_\nu e^\nu.\end{aligned} \hspace{\stretch{1}}(3.44)

Our differentials are

\begin{aligned}\begin{aligned}d\mathbf{x} &= dx^\mu e_\mu + x^\mu d e_\mu \\ &= \sum_\alpha d\alpha \left( \frac{\partial {x^\mu}}{\partial {\alpha}} e_\mu+x^\mu\frac{\partial {e_\mu}}{\partial {\alpha}} \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.46)

and

\begin{aligned}\begin{aligned}d\mathbf{u} &= du^\mu e_\mu + u^\mu d e_\mu \\ &= \sum_\alpha d\alpha \left( \frac{\partial {u^\mu}}{\partial {\alpha}} e_\mu+u^\mu\frac{\partial {e_\mu}}{\partial {\alpha}} \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.47)

Summing these we have

\begin{aligned}d\mathbf{u} + d\mathbf{u} = \sum_\alpha d\alpha \left( \left(\frac{\partial {x^\mu}}{\partial {\alpha}} +\frac{\partial {u^\mu}}{\partial {\alpha}} \right)e_\mu+\left(x^\mu+u^\mu\right)\frac{\partial {e_\mu}}{\partial {\alpha}} \right).\end{aligned} \hspace{\stretch{1}}(3.48)

Taking dot products to form the squares we have

\begin{aligned}d\mathbf{x}^2 &= \sum_{\alpha, \beta} d\alpha d\beta \left( \frac{\partial {x^\mu}}{\partial {\alpha}} e_\mu+x^\mu\frac{\partial {e_\mu}}{\partial {\alpha}} \right)\cdot\left( \frac{\partial {x_\nu}}{\partial {\beta}} e^\nu+x_\nu\frac{\partial {e^\nu}}{\partial {\beta}} \right) \\ &=\sum_{\alpha, \beta} d\alpha d\beta \left( \frac{\partial {x^\mu}}{\partial {\alpha}} \frac{\partial {x_\mu}}{\partial {\beta}} +x^\mu x_\nu\frac{\partial {e_\mu}}{\partial {\alpha}} \cdot\frac{\partial {e^\nu}}{\partial {\beta}} + 2 \frac{\partial {x^\mu}}{\partial {\alpha}} x_\nu e_\mu \cdot\frac{\partial {e^\nu}}{\partial {\beta}} \right),\end{aligned}

and

\begin{aligned}&(d\mathbf{u} + d\mathbf{x})^2 \\ &= \sum_{\alpha, \beta}d\alpha d\beta \left( \left(\frac{\partial {x^\mu}}{\partial {\alpha}} +\frac{\partial {u^\mu}}{\partial {\alpha}} \right)e_\mu+\left(x^\mu+u^\mu\right)\frac{\partial {e_\mu}}{\partial {\alpha}} \right)\cdot\left( \left(\frac{\partial {x_\nu}}{\partial {\beta}} +\frac{\partial {u_\nu}}{\partial {\beta}} \right)e^\nu+\left(x_\nu+u_\nu\right)\frac{\partial {e^\nu}}{\partial {\beta}} \right) \\ &= \sum_{\alpha, \beta}d\alpha d\beta \left(\left(\frac{\partial {x^\mu}}{\partial {\alpha}} +\frac{\partial {u^\mu}}{\partial {\alpha}} \right)\left(\frac{\partial {x_\mu}}{\partial {\beta}} +\frac{\partial {u_\mu}}{\partial {\beta}} \right)+\left(x^\mu+u^\mu\right)\left(x_\nu+u_\nu\right)\frac{\partial {e_\mu}}{\partial {\alpha}} \cdot\frac{\partial {e^\nu}}{\partial {\beta}} +2\left(x^\mu+u^\mu\right)e^\nu\cdot\frac{\partial {e_\mu}}{\partial {\alpha}} \left(\frac{\partial {x_\nu}}{\partial {\beta}} +\frac{\partial {u_\nu}}{\partial {\beta}} \right)\right).\end{aligned}

Taking the difference we find

\begin{aligned}\begin{aligned}&(d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2 \\ &=\sum_{\alpha, \beta}d\alpha d\beta \left( \frac{\partial {u^\mu}}{\partial {\alpha}} \frac{\partial {u_\mu}}{\partial {\beta}} +2\frac{\partial {u^\mu}}{\partial {\alpha}} \frac{\partial {x_\mu}}{\partial {\beta}} + \left(u^\mu u_\nu +x^\mu u_\nu +u^\mu x_\nu \right)\frac{\partial {e_\mu}}{\partial {\alpha}}\cdot\frac{\partial {e^\nu}}{\partial {\beta}} +2 \left(\frac{\partial {x^\mu}}{\partial {\alpha}}u_\nu+\frac{\partial {u^\mu}}{\partial {\alpha}}(x_\nu+u_\nu)\right)e_\mu \cdot \frac{\partial {e^\nu}}{\partial {\beta}}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.49)

To evaluate this, it is useful, albeit messier, to group terms a bit

\begin{aligned}\begin{aligned}&(d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2 \\ &=\sum_{\alpha}2 d\alpha d\alpha \left( \frac{1}{{2}}\frac{\partial {u^\mu}}{\partial {\alpha}} \frac{\partial {u_\mu}}{\partial {\alpha}} +\frac{\partial {u^\mu}}{\partial {\alpha}} \frac{\partial {x_\mu}}{\partial {\alpha}} + \frac{1}{{2}}\left(u^\mu u_\nu +x^\mu u_\nu +u^\mu x_\nu \right)\frac{\partial {e_\mu}}{\partial {\alpha}}\cdot\frac{\partial {e^\nu}}{\partial {\alpha}} +\left(\frac{\partial {x^\mu}}{\partial {\alpha}}u_\nu+\frac{\partial {u^\mu}}{\partial {\alpha}}(x_\nu+u_\nu)\right)e_\mu \cdot \frac{\partial {e^\nu}}{\partial {\alpha}}\right) \\ &+\sum_{\alpha < \beta}2 d\alpha d\beta \left( \frac{\partial {u^\mu}}{\partial {\alpha}} \frac{\partial {u_\mu}}{\partial {\beta}} +\frac{\partial {u^\mu}}{\partial {\alpha}} \frac{\partial {x_\mu}}{\partial {\beta}} +\frac{\partial {u^\mu}}{\partial {\beta}} \frac{\partial {x_\mu}}{\partial {\alpha}} + \frac{1}{{2}}\left(u^\mu u_\nu +x^\mu u_\nu +u^\mu x_\nu \right)\left(\frac{\partial {e_\mu}}{\partial {\alpha}}\cdot\frac{\partial {e^\nu}}{\partial {\beta}} +\frac{\partial {e_\mu}}{\partial {\beta}}\cdot\frac{\partial {e^\nu}}{\partial {\alpha}} \right) \right) \\ &+\sum_{\alpha < \beta}2 d\alpha d\beta \left( \left(\frac{\partial {x^\mu}}{\partial {\alpha}}u_\nu+\frac{\partial {u^\mu}}{\partial {\alpha}}(x_\nu+u_\nu)\right)e_\mu \cdot \frac{\partial {e^\nu}}{\partial {\beta}}+\left(\frac{\partial {x^\mu}}{\partial {\beta}}u_\nu+\frac{\partial {u^\mu}}{\partial {\beta}}(x_\nu+u_\nu)\right)e_\mu \cdot \frac{\partial {e^\nu}}{\partial {\alpha}}\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.50)

Here $\alpha < \beta$ is used to denote summation over the pairs $\alpha \ne \beta$ just once, not neccessarily any numeric ordering. For example with $\alpha, \beta \in \{r, \phi, z\}$, this could be the set $\{\alpha, \beta\} \in \{r \phi, \phi z, z r\}$.

## Cartesian tensor.

In the Cartesian case all the partials of the unit vectors are zero, and we also have no need of upper or lower indexes. We are left with just

\begin{aligned}(d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2 =\sum_{i, j, k}dx^idx^j\left( \frac{\partial {u^k}}{\partial {x^i}} \frac{\partial {u^k}}{\partial {x^j}} +2\frac{\partial {u^k}}{\partial {x^i}} \frac{\partial {x^k}}{\partial {x^j}} \right)\end{aligned} \hspace{\stretch{1}}(3.51)

However, since we also have ${\partial {x^k}}/{\partial {x^j}} = \delta_{jk}$, this is

\begin{aligned}(d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2 =\sum_{i, j}2dx^idx^j\left( \frac{1}{{2}}\sum_k\frac{\partial {u^k}}{\partial {x^i}} \frac{\partial {u^k}}{\partial {x^j}} +\frac{\partial {u^j}}{\partial {x^i}} \right).\end{aligned} \hspace{\stretch{1}}(3.52)

This essentially recovers the result 3.11 derived in class.

## Cylindrial tensor.

Now lets do the cylindrical tensor again, but this time without resorting mathematica brute force.

First we recall that all our basis vector derivatives are zero except for the $\phi$ derivatives, and for those we have

\begin{aligned}\frac{\partial {\hat{\mathbf{r}}}}{\partial {\phi}} &= \hat{\boldsymbol{\phi}} \\ \frac{\partial {\hat{\boldsymbol{\theta}}}}{\partial {\phi}} &= -\hat{\mathbf{r}}.\end{aligned} \hspace{\stretch{1}}(3.53)

If we write

\begin{aligned}\mathbf{x} = r \hat{\mathbf{r}} + z \hat{\mathbf{z}} = x_r \hat{\mathbf{r}} + x_\phi \hat{\boldsymbol{\phi}} + x_z \hat{\mathbf{z}}\end{aligned} \hspace{\stretch{1}}(3.55)

We have for all the $x^\mu$ partials

\begin{aligned}\frac{\partial {x^\mu}}{\partial {\alpha}} = \left\{\begin{array}{l l}1 & \quad \mbox{iflatex \alpha = x^\mu = ror $\alpha = x^\mu = z$} \\ 0 & \quad \mbox{otherwise}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(3.56)

We are now set to evaluate the terms in the sum of 3.50 for the cylindrical coordinate system and shouldn’t need Mathematica to do it. Let’s do this one at a time, starting with all the squared differential pairs. Those are, for $\alpha \in \{r, \phi, z\}$ the value of

\begin{aligned}2 d\alpha d\alpha \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {\alpha}} \frac{\partial {u_m}}{\partial {\alpha}} +\frac{\partial {u_m}}{\partial {\alpha}} \frac{\partial {x_m}}{\partial {\alpha}} + \frac{1}{{2}}\left(u_m u_n +x_m u_n +u_m x_n \right)\frac{\partial {e_m}}{\partial {\alpha}}\cdot\frac{\partial {e_n}}{\partial {\alpha}} +\left(\frac{\partial {x_m}}{\partial {\alpha}}u_n+\frac{\partial {u_m}}{\partial {\alpha}}(x_n+u_n)\right)e_m \cdot \frac{\partial {e_n}}{\partial {\alpha}}\right)\end{aligned} \hspace{\stretch{1}}(3.60)

For both $r$ and $z$ all our unit vectors have zero derivatives so we are left respectively with

\begin{aligned}2 dr dr \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {r}} \frac{\partial {u_m}}{\partial {r}} +\frac{\partial {u_r}}{\partial {r}} \right),\end{aligned} \hspace{\stretch{1}}(3.60)

and

\begin{aligned}2 dz dz \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {z}} +\frac{\partial {u_z}}{\partial {z}} \right).\end{aligned} \hspace{\stretch{1}}(3.60)

For the $\alpha = \phi$ term we have

\begin{aligned}&2 d\phi d\phi \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {\phi}} \frac{\partial {u_m}}{\partial {\phi}} + \frac{1}{{2}}\sum_{m = r, \phi}\left(u_m u_m +2 x_m u_m \right)+\sum_{m n \in \{r \phi, \phi r\}}\left(\frac{\partial {x_m}}{\partial {\phi}}u_n+\frac{\partial {u_m}}{\partial {\phi}}(x_n+u_n)\right)e_m \cdot \frac{\partial {e_n}}{\partial {\phi}}\right) \\ &=2 d\phi d\phi \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {\phi}} \frac{\partial {u_m}}{\partial {\phi}} + \frac{1}{{2}} \left( u_r^2 + u_\phi^2 \right) + r u_r-\frac{\partial {u_r}}{\partial {\phi}}u_\phi+\frac{\partial {u_\phi}}{\partial {\phi}}(r+u_r)\right)\end{aligned}

Now, on to the mixed terms. The easiest is the $dz dr$ term, for which all the unit vector derivatives are zero, and we are left with just

\begin{aligned}2 dz dr \left( \frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {r}} +\frac{\partial {u_m}}{\partial {z}} \frac{\partial {x_m}}{\partial {r}} +\frac{\partial {u_m}}{\partial {r}} \frac{\partial {x_m}}{\partial {z}} \right)=2 dz dr \left( \frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {r}} +\frac{\partial {u_r}}{\partial {z}} +\frac{\partial {u_z}}{\partial {r}} \right)\end{aligned}

Now we have the two messy mixed terms. For the $r$, $\phi$ term we get

\begin{aligned}&2 dr d\phi \left( \frac{\partial {u_m}}{\partial {r}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_m}}{\partial {r}} \not{{\frac{\partial {x_m}}{\partial {\phi}}}}+\frac{\partial {u_m}}{\partial {\phi}} \frac{\partial {x_m}}{\partial {r}} + \frac{1}{{2}}\left(u_m u_n +x_m u_n +u_m x_n \right)\left(\not{{\frac{\partial {e_m}}{\partial {r}}}}\cdot\frac{\partial {e_n}}{\partial {\phi}} +\frac{\partial {e_m}}{\partial {\phi}}\cdot\not{{\frac{\partial {e_n}}{\partial {r}} }}\right) \right) \\ &+2 dr d\phi \left( \left(\frac{\partial {x_m}}{\partial {r}}u_n+\frac{\partial {u_m}}{\partial {r}}(x_n+u_n)\right)e_m \cdot \frac{\partial {e_n}}{\partial {\phi}}+\left(\frac{\partial {x_m}}{\partial {\phi}}u_n+\frac{\partial {u_m}}{\partial {\phi}}(x_n+u_n)\right)e_m \cdot \not{{\frac{\partial {e_n}}{\partial {r}}}}\right) \\ &=2 dr d\phi \left( \frac{\partial {u_m}}{\partial {r}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_r}}{\partial {\phi}} +u_n\hat{\mathbf{r}} \cdot \frac{\partial {e_n}}{\partial {\phi}}+\frac{\partial {u_m}}{\partial {r}}(x_n+u_n)e_m \cdot \frac{\partial {e_n}}{\partial {\phi}}\right) \\ &=2 dr d\phi \left( \frac{\partial {u_m}}{\partial {r}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_r}}{\partial {\phi}} -u_\phi+\frac{\partial {u_r}}{\partial {r}}(x_n+u_n)\hat{\mathbf{r}} \cdot \frac{\partial {e_n}}{\partial {\phi}}+\frac{\partial {u_\phi}}{\partial {r}}(x_n+u_n)\hat{\boldsymbol{\phi}} \cdot \frac{\partial {e_n}}{\partial {\phi}}\right) \\ &=2 dr d\phi \left( \frac{\partial {u_m}}{\partial {r}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_r}}{\partial {\phi}} -u_\phi-\frac{\partial {u_r}}{\partial {r}}u_\phi+\frac{\partial {u_\phi}}{\partial {r}}(r +u_r)\right) \\ \end{aligned}

Finally for the $z$, $\phi$ term we have

\begin{aligned}&2 dz d\phi \left( \frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_m}}{\partial {z}} \not{{\frac{\partial {x_m}}{\partial {\phi}} }}+\frac{\partial {u_m}}{\partial {\phi}} \frac{\partial {x_m}}{\partial {z}} + \frac{1}{{2}}\left(u_m u_n +x_m u_n +u_m x_n \right)\left(\not{{\frac{\partial {e_m}}{\partial {z}}}}\cdot\frac{\partial {e_n}}{\partial {\phi}} +\frac{\partial {e_m}}{\partial {\phi}}\cdot\not{{\frac{\partial {e_n}}{\partial {z}} }}\right) \right) \\ &+2 d\phi dz \left( \left(\frac{\partial {x_m}}{\partial {z}}u_n+\frac{\partial {u_m}}{\partial {z}}(x_n+u_n)\right)e_m \cdot \frac{\partial {e_n}}{\partial {\phi}}+\left(\frac{\partial {x_m}}{\partial {\phi}}u_n+\frac{\partial {u_m}}{\partial {\phi}}(x_n+u_n)\right)e_m \cdot \not{{\frac{\partial {e_n}}{\partial {z}}}}\right) \\ &=2 dz d\phi \left( \frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_m}}{\partial {\phi}} \frac{\partial {x_m}}{\partial {z}} +\not{{u_n\hat{\mathbf{z}} \cdot \frac{\partial {e_n}}{\partial {\phi}}}}+\frac{\partial {u_m}}{\partial {z}}(x_n+u_n)e_m \cdot \frac{\partial {e_n}}{\partial {\phi}}\right) \\ &=2 dz d\phi \left( \frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {\phi}} +\frac{\partial {u_z}}{\partial {\phi}} -\frac{\partial {u_r}}{\partial {z}}u_\phi+\frac{\partial {u_\phi}}{\partial {z}}(r+u_r)\right) \\ \end{aligned}

To summarize we have, including both first and second order terms,

\begin{aligned}\begin{aligned}{d\mathbf{l}'}^2 - d\mathbf{x}^2&=2 dr dr \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {r}} \frac{\partial {u_m}}{\partial {r}} +\frac{\partial {u_r}}{\partial {r}} \right) \\ &+2 r^2 d\phi d\phi \left( \frac{1}{{2 r^2}}\frac{\partial {u_m}}{\partial {\phi}} \frac{\partial {u_m}}{\partial {\phi}} + \frac{1}{{2 r^2}} \left( u_r^2 + u_\phi^2 \right) + \frac{u_r}{r}-\frac{1}{{r}}\frac{\partial {u_r}}{\partial {\phi}}\frac{u_\phi}{r}+\frac{1}{{r}}\frac{\partial {u_\phi}}{\partial {\phi}}\left(1+\frac{u_r}{r}\right)\right) \\ &+2 dz dz \left( \frac{1}{{2}}\frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {z}} +\frac{\partial {u_z}}{\partial {z}} \right) \\ &+2 dr r d\phi \left( \frac{\partial {u_m}}{\partial {r}} \frac{1}{{r}}\frac{\partial {u_m}}{\partial {\phi}} +\frac{1}{{r}}\frac{\partial {u_r}}{\partial {\phi}} -\frac{u_\phi}{r}-\frac{\partial {u_r}}{\partial {r}}\frac{u_\phi}{r}+\frac{\partial {u_\phi}}{\partial {r}}\left(1 +\frac{u_r}{r}\right)\right) \\ &+2 r d\phi dz \left( \frac{\partial {u_m}}{\partial {z}} \frac{1}{{r}}\frac{\partial {u_m}}{\partial {\phi}} +\frac{1}{{r}}\frac{\partial {u_z}}{\partial {\phi}} -\frac{\partial {u_r}}{\partial {z}}\frac{u_\phi}{r}+\frac{\partial {u_\phi}}{\partial {z}}\left(1+\frac{u_r}{r}\right)\right) \\ &+2 dz dr \left( \frac{\partial {u_m}}{\partial {z}} \frac{\partial {u_m}}{\partial {r}} +\frac{\partial {u_r}}{\partial {z}} +\frac{\partial {u_z}}{\partial {r}} \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.60)

Factors of $r$ have been pulled out so that the portions remaining in the braces are exactly the cylindrical tensor elements as given in the text (except also with the second order terms here). Observe that the pre-calculation of the general formula has allowed an on paper expansion of the cylindrical tensor without too much pain, and this time without requiring Mathematica.

FIXME: TODO.

# 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.

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}