## PHY454H1S Continuum Mechanics. Lecture 2. Introduction and strain tensor. Taught by Prof. K. Das.

Posted by peeterjoot on January 21, 2012

# 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. components. Examples: density, Temperature, …

\item Vector fields. components. Examples: Force, velocity.

\item Tensor fields. 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 to the coordinates so that the transformation is related by

\begin{figure}[htp]

\centering

\includegraphics[totalheight=0.2\textheight]{continuumL2fig4}

\caption{Differential change to the object.}

\end{figure}

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

or

with

we have

We write

where we define the \emph{strain tensor} as

Here is a matrix in Cartesian coordinates

We see from 3.11 that is symmetric, so we have

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

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

We are also free to define a volume element

So the change of volume is given by the trace

## 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

where

Since and are functions of position, we will need their differentials

but these are just scaled basis vectors

So for our and differentials we find

and

Putting these together we have

For the squared magnitude’s difference from we have

Expanding this out, but dropping all the terms that are quadratic in the components of or its differentials, we have

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

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

Observe that we have to introduce factors of along with all the ‘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

As with the first order case, we can read off the tensor coordinates by inspection (once we factor out the various factors of and ). 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 , and its reciprocal frame , as defined by

Our coordinate representation, with summation and dimensionality implied, is

Our differentials are

and

Summing these we have

Taking dot products to form the squares we have

and

Taking the difference we find

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

Here is used to denote summation over the pairs just once, not neccessarily any numeric ordering. For example with , this could be the set .

## 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

However, since we also have , this is

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 derivatives, and for those we have

If we write

We have for all the partials

latex \alpha = x^\mu = r$ or } \\ 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 the value of

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

and

For the term we have

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

Now we have the two messy mixed terms. For the , term we get

Finally for the , term we have

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

Factors of 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.

## Spherical tensor.

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.

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