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

Posted by peeterjoot on April 8, 2012

# 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

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

In either case we have

(where the primes indicate the scope of the gradient, showing here that we are operating only on , and not ).

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

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

# Verifying the relationship.

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

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

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

so we find

# 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

and our cylindrical velocity is

Our curl is then

Since and , we have only one cross term and our curl is

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 .

Our directional derivative component for a normal direction doesn’t have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

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

\begin{subequations}

\end{subequations}

## With .

Our directional derivative component for a normal direction will have some cross terms since both and are functions of

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

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

\begin{subequations}

\end{subequations}

## With .

Like the normal direction, our directional derivative component for a normal direction will not have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

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

\begin{subequations}

\end{subequations}

## Summary.

\begin{subequations}

\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

and our gradient mirrors this structure

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

\begin{subequations}

\end{subequations}

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

Finally, our velocity in spherical coordinates is just

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

So we have

## With .

The directional derivative portion of our strain is

The other portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

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

## With .

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

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

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

## With .

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

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

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

## Summary

\begin{subequations}

\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.

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