# Peeter Joot's Blog.

• ## Archives

 ivor on Just Energy Canada nasty busin… A final pre-exam upd… on An updated compilation of note… Anon on About peeterjoot on About Anon on About
• ## People not reading this blog: 6,973,738,433 minus:

• 132,875 hits

## 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}