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## 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.

1. ### Nasser M. Abbasisaid

When I download your CDF file for solution under probem Q2 above, which is:

https://raw.github.com/peeterjoot/physicsplay/master/notes/phy454/continuumProblemSet1Q2animated.cdf

and I run it on windows, using Mathematica 8.04, I get these errors when I open the CDF file

Part::partd: Part specification standardBasis[[1]] is longer than depth of object. >>
Part::partd: Part specification standardBasis[[1]] is longer than depth of object. >>
Part::partd: Part specification standardBasis[[2]] is longer than depth of object. >>

thank you,
–Nasser

• ### peeterjootsaid

Are these errors, or warnings?

I can’t reproduce the behavior that you see. I’m also running mathematica 8.0.4, and either don’t see these messages, or am not looking in the right place.

• ### Nasser M. Abbasisaid

I do not know how you are trying to reproduce it, but the steps are:
2. open mathematica
3. open the cdf file
4, Enable dynamics is not already enabled

now you will see the error. Clearly this is an error. trying to access out-of-bound index.

I am using windows 7, using M V 8.04

It happens all the time when I open the file.

–Nasser

• ### peeterjootsaid

That’s exactly what I did, and didn’t see the error, but if I restart Mathematica I see the error. I don’t understand the source of the error, do you? The odd thing is that a second shift-enter after the fact appears to clear the error.

• ### peeterjootsaid

http://mathematica.stackexchange.com/questions/1669/table-function-with-part-call-misbehaving-but-only-after-initial-startup-of-m

I’ll update the online source for this .cdf, presuming that I find an answer to my question. Until then, you can use the workaround of re-evaluating the expression.

• ### peeterjootsaid

should be fixed now. I needed selected Dynamic and DynamicModule calls to make things work properly on initial load.

2. ### Nasser M. Abbasisaid

btw, there is no need to put the CDF’s as links in the way you did. Instead, you can now embed the CDF directly in the web page, so that one can run them on the web, just like an applet. This makes it easier for someone to run them. Information how to do this is here