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# Posts Tagged ‘Cauchy tetrahedron’

## Continuum mechanics elasticity review.

Posted by peeterjoot on April 23, 2012

[Click here for a PDF of this post with nicer formatting]

# Motivation.

Review of key ideas and equations from the theory of elasticity portion of the class.

# Strain Tensor

Identifying a point in a solid with coordinates $x_i$ and the coordinates of that portion of the solid after displacement, we formed the difference as a measure of the displacement

\begin{aligned}u_i = x_i' - x_i.\end{aligned} \hspace{\stretch{1}}(2.1)

With $du_i = {\partial {u_i}}/{\partial {x_j}} dx_j$, we computed the difference in length (squared) for an element of the displaced solid and found

\begin{aligned}dx_k' dx_k' - dx_k dx_k = \left( \frac{\partial {u_j}}{\partial {x_i}} + \frac{\partial {u_i}}{\partial {x_j}} + \frac{\partial {u_k}}{\partial {x_i}} \frac{\partial {u_k}}{\partial {x_j}} \right) dx_i dx_j,\end{aligned} \hspace{\stretch{1}}(2.2)

or defining the \textit{strain tensor} $e_{ij}$, we have

\begin{aligned}(d\mathbf{x}')^2 - (d\mathbf{x})^2= 2 e_{ij} dx_i dx_j\end{aligned} \hspace{\stretch{1}}(2.3a)

\begin{aligned}e_{ij}=\frac{1}{{2}}\left( \frac{\partial {u_j}}{\partial {x_i}} + \frac{\partial {u_i}}{\partial {x_j}} + \frac{\partial {u_k}}{\partial {x_i}} \frac{\partial {u_k}}{\partial {x_j}} \right).\end{aligned} \hspace{\stretch{1}}(2.3b)

In this course we use only the linear terms and write

\begin{aligned}e_{ij}=\frac{1}{{2}}\left( \frac{\partial {u_j}}{\partial {x_i}} + \frac{\partial {u_i}}{\partial {x_j}} \right).\end{aligned} \hspace{\stretch{1}}(2.4)

## Unresolved: Relating displacement and position by strain

In [1] it is pointed out that this strain tensor simply relates the displacement vector coordinates $u_i$ to the coordinates at the point at which it is measured

\begin{aligned}u_i = e_{ij} x_j.\end{aligned} \hspace{\stretch{1}}(2.5)

When we get to fluid dynamics we perform a linear expansion of $du_i$ and find something similar

\begin{aligned}dx_i' - dx_i = du_i = \frac{\partial {u_i}}{\partial {x_k}} dx_k = e_{ij} dx_k + \omega_{ij} dx_k\end{aligned} \hspace{\stretch{1}}(2.6)

where

\begin{aligned}\omega_{ij} = \frac{1}{{2}} \left( \frac{\partial {u_j}}{\partial {x_i}} +\frac{\partial {u_i}}{\partial {x_j}} \right).\end{aligned} \hspace{\stretch{1}}(2.7)

Except for the antisymmetric term, note the structural similarity of 2.5 and 2.6. Why is it that we neglect the vorticity tensor in statics?

## Diagonal strain representation.

In a basis for which the strain tensor is diagonal, it was pointed out that we can write our difference in squared displacement as (for $k = 1, 2, 3$, no summation convention)

\begin{aligned}(dx_k')^2 - (dx_k)^2 = 2 e_{kk} dx_k dx_k\end{aligned} \hspace{\stretch{1}}(2.8)

from which we can rearrange, take roots, and apply a first order Taylor expansion to find (again no summation convention)

\begin{aligned}dx_k' \approx (1 + e_{kk}) dx_k.\end{aligned} \hspace{\stretch{1}}(2.9)

An approximation of the displaced volume was then found in terms of the strain tensor trace (summation convention back again)

\begin{aligned}dV' \approx (1 + e_{kk}) dV,\end{aligned} \hspace{\stretch{1}}(2.10)

allowing us to identify this trace as a relative difference in displaced volume

\begin{aligned}e_{kk} \approx \frac{dV' - dV}{dV}.\end{aligned} \hspace{\stretch{1}}(2.11)

## Strain in cylindrical coordinates.

Useful in many practice problems are the cylindrical coordinate representation of the strain tensor

\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}}(2.12)

This can be found in [2]. It was not derived there or in class, but is not too hard, even using the second order methods we used for the Cartesian form of the tensor.

An easier way to do this derivation (and understand what the coordinates represent) follows from the relation found in section 6 of [3]

\begin{aligned}2 \mathbf{e}_i e_{ij} n_j = 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u} + \hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u}),\end{aligned} \hspace{\stretch{1}}(2.18)

where $\hat{\mathbf{n}}$ is the normal to the surface at which we are measuring a force applied to the solid (our Cauchy tetrahedron).

The cylindrical tensor coordinates of 2.12 follow from
2.18 nicely taking $\hat{\mathbf{n}} = \hat{\mathbf{r}}, \hat{\boldsymbol{\phi}}, \hat{\mathbf{z}}$ in turn.

## Compatibility condition.

For a 2D strain tensor we found an interrelationship between the components of the strain tensor

\begin{aligned}2 \frac{\partial^2 e_{12}}{\partial x_1 \partial x_2} =\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} +\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2},\end{aligned} \hspace{\stretch{1}}(2.19)

and called this the compatibility condition. It was claimed, but not demonstrated that this is what is required to ensure a deformation maintained a coherent solid geometry.

I wasn’t able to find any references to this compatibility condition in any of the texts I have, but found [4], [5], and [6]. It’s not terribly surprising to see Christoffel symbol and differential forms references on those pages, since one can imagine that we’d wish to look at the mappings of all the points in the object as it undergoes the transformation from the original to the deformed state.

Even with just three points in a plane, say $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, the general deformation of an object doesn’t seem like it’s the easiest thing to describe. We can imagine that these have trajectories in the deformation process $\mathbf{a} = \mathbf{a}(\alpha$, $\mathbf{b} = \mathbf{b}(\beta)$, $\mathbf{c} = \mathbf{c}(\gamma)$, with $\mathbf{a}', \mathbf{b}', \mathbf{c}'$ at the end points of the trajectories. We’d want to look at displacement vectors $\mathbf{u}_a, \mathbf{u}_b, \mathbf{u}_c$ along each of these trajectories, and then see how they must be related. Doing that carefully must result in this compatibility condition.

# Stress tensor.

By sought and found a representation of the force per unit area acting on a body by expressing the components of that force as a set of divergence relations

\begin{aligned}f_i = \partial_k \sigma_{i k},\end{aligned} \hspace{\stretch{1}}(3.20)

and call the associated tensor $\sigma_{ij}$ the \textit{stress}.

Unlike the strain, we don’t have any expectation that this tensor is symmetric, and identify the diagonal components (no sum) $\sigma_{i i}$ as quantifying the amount of compressive or contractive force per unit area, whereas the cross terms of the stress tensor introduce shearing deformations in the solid.

With force balance arguments (the Cauchy tetrahedron) we found that the force per unit area on the solid, for a surface with unit normal pointing into the solid, was

\begin{aligned}\mathbf{t} = \mathbf{e}_i t_i = \mathbf{e}_i \sigma_{ij} n_j.\end{aligned} \hspace{\stretch{1}}(3.21)

## Constitutive relation.

In the scope of this course we considered only Newtonian materials, those for which the stress and strain tensors are linearly related

\begin{aligned}\sigma_{ij} = c_{ijkl} e_{kl},\end{aligned} \hspace{\stretch{1}}(3.22)

and further restricted our attention to isotropic materials, which can be shown to have the form

\begin{aligned}\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2 \mu e_{ij},\end{aligned} \hspace{\stretch{1}}(3.23)

where $\lambda$ and $\mu$ are the Lame parameters and $\mu$ is called the shear modulus (and viscosity in the context of fluids).

By computing the trace of the stress $\sigma_{ii}$ we can invert this to find

\begin{aligned}2 \mu e_{ij} = \sigma_{ij} - \frac{\lambda}{3 \lambda + 2 \mu} \sigma_{kk} \delta_{ij}.\end{aligned} \hspace{\stretch{1}}(3.24)

## Uniform hydrostatic compression.

With only normal components of the stress (no shear), and the stress having the same value in all directions, we find

\begin{aligned}\sigma_{ij} = ( 3 \lambda + 2 \mu ) e_{ij},\end{aligned} \hspace{\stretch{1}}(3.25)

and identify this combination $-3 \lambda - 2 \mu$ as the pressure, linearly relating the stress and strain tensors

\begin{aligned}\sigma_{ij} = -p e_{ij}.\end{aligned} \hspace{\stretch{1}}(3.26)

With $e_{ii} = (dV' - dV)/dV = \Delta V/V$, we formed the Bulk modulus $K$ with the value

\begin{aligned}K = \left( \lambda + \frac{2 \mu}{3} \right) = -\frac{p V}{\Delta V}.\end{aligned} \hspace{\stretch{1}}(3.27)

## Uniaxial stress. Young’s modulus. Poisson’s ratio.

For the special case with only one non-zero stress component (we used $\sigma_{11}$) we were able to compute Young’s modulus $E$, the ratio between stress and strain in that direction

\begin{aligned}E = \frac{\sigma_{11}}{e_{11}} = \frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu } = \frac{3 K \mu}{K + \mu/3}.\end{aligned} \hspace{\stretch{1}}(3.28)

Just because only one component of the stress is non-zero, does not mean that we have no deformation in any other directions. Introducing Poisson’s ratio $\nu$ in terms of the ratio of the strains relative to the strain in the direction of the force we write and then subsequently found

\begin{aligned}\nu = -\frac{e_{22}}{e_{11}} = -\frac{e_{33}}{e_{11}} = \frac{\lambda}{2(\lambda + \mu)}.\end{aligned} \hspace{\stretch{1}}(3.29)

We were also able to find

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}}(3.30)

\begin{aligned}\lambda = \frac{E \nu}{(1 - 2 \nu)(1 + \nu)}\end{aligned} \hspace{\stretch{1}}(3.31)

\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}}(3.32)

# Displacement propagation

It was argued that the equation relating the time evolution of a one of the vector displacement coordinates was given by

\begin{aligned}\rho \frac{\partial^2 {{u_i}}}{\partial {{t}}^2} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + f_i,\end{aligned} \hspace{\stretch{1}}(4.35)

where the divergence term ${\partial {\sigma_{ij}}}/{\partial {x_j}}$ is the internal force per unit volume on the object and $f_i$ is the external force. Employing the constitutive relation we showed that this can be expanded as

\begin{aligned}\rho \frac{\partial^2 {{u_i}}}{\partial {{t}}^2} = (\lambda + \mu) \frac{\partial^2 u_k}{\partial x_i \partial x_k}+ \mu\frac{\partial^2 u_i}{\partial x_j^2},\end{aligned} \hspace{\stretch{1}}(4.36)

or in vector form

\begin{aligned}\rho \frac{\partial^2 {\mathbf{u}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \mu \boldsymbol{\nabla}^2 \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(4.37)

## P-waves

Operating on 4.37 with the divergence operator, and writing $\Theta = \boldsymbol{\nabla} \cdot \mathbf{u}$, a quantity that was our relative change in volume in the diagonal strain basis, we were able to find this divergence obeys a wave equation

\begin{aligned}\frac{\partial^2 {{\Theta}}}{\partial {{t}}^2} = \frac{\lambda + 2 \mu}{\rho} \boldsymbol{\nabla}^2 \Theta.\end{aligned} \hspace{\stretch{1}}(4.38)

We called these P-waves.

## S-waves

Similarly, operating on 4.37 with the curl operator, and writing $\boldsymbol{\omega} = \boldsymbol{\nabla} \times \mathbf{u}$, we were able to find this curl also obeys a wave equation

\begin{aligned}\rho \frac{\partial^2 {{\boldsymbol{\omega}}}}{\partial {{t}}^2} = \mu \boldsymbol{\nabla}^2 \boldsymbol{\omega}.\end{aligned} \hspace{\stretch{1}}(4.39)

These we called S-waves. We also noted that the (transverse) compression waves (P-waves) with speed $C_T = \sqrt{\mu/\rho}$, traveled faster than the (longitudinal) vorticity (S) waves with speed $C_L = \sqrt{(\lambda + 2 \mu)/\rho}$ since $\lambda > 0$ and $\mu > 0$, and

\begin{aligned}\frac{C_L}{C_T} = \sqrt{\frac{ \lambda + 2 \mu}{\mu}} = \sqrt{ \frac{\lambda}{\mu} + 2}.\end{aligned} \hspace{\stretch{1}}(4.40)

## Scalar and vector potential representation.

Assuming a vector displacement representation with gradient and curl components

\begin{aligned}\mathbf{u} = \boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H},\end{aligned} \hspace{\stretch{1}}(4.41)

We found that the displacement time evolution equation split nicely into curl free and divergence free terms

\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}}(4.42)

When neglecting boundary value effects this could be written as a pair of independent equations

\begin{aligned}\rho \frac{\partial^2 {{\phi}}}{\partial {{t}}^2} - (\lambda + 2\mu) \boldsymbol{\nabla}^2 \phi = 0\end{aligned} \hspace{\stretch{1}}(4.43a)

\begin{aligned}\rho \frac{\partial^2 {\mathbf{H}}}{\partial {{t}}^2} - \mu \boldsymbol{\nabla}^2 \mathbf{H}= 0.\end{aligned} \hspace{\stretch{1}}(4.43b)

This are the irrotational (curl free) P-wave and solenoidal (divergence free) S-wave equations respectively.

## Phasor description.

It was mentioned that we could assume a phasor representation for our potentials, writing

\begin{aligned}\phi = A \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \end{aligned} \hspace{\stretch{1}}(4.44a)

\begin{aligned}\mathbf{H} = \mathbf{B} \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right)\end{aligned} \hspace{\stretch{1}}(4.44b)

finding

\begin{aligned}\mathbf{u} = i \mathbf{k} \phi + i \mathbf{k} \times \mathbf{H}.\end{aligned} \hspace{\stretch{1}}(4.45)

We did nothing with neither the potential nor the phasor theory for solid displacement time evolution, and presumably won’t on the exam either.

# Some wave types

Some time was spent on non-qualitative descriptions and review of descriptions for solutions to the time evolution equations we did not attempt

1. P-waves [7]. Irrotational, non volume preserving body wave.
2. S-waves [8]. Divergence free body wave. Shearing forces are present and volume is preserved (slower than S-waves)
3. Rayleigh wave [9]. A surface wave that propagates near the surface of a body without penetrating into it.
4. Love wave [10]. A polarized shear surface wave with the shear displacements moving perpendicular to the direction of propagation.

For reasons that aren’t clear both the midterm and last years final ask us to spew this sort of stuff (instead of actually trying to do something analytic associated with them).

# References

[1] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics.[Lectures on physics], chapter Elastic Materials. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963.

[2] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of Elasticity: Vol. 7 of Course of Theoretical Physics. 1960.

[3] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

[4] Wikipedia. Compatibility (mechanics) — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 23-April-2012]. http://en.wikipedia.org/w/index.php?title=Compatibility_(mechanics)&oldid=463812965.

[5] Wikipedia. Infinitesimal strain theory — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 23-April-2012]. http://en.wikipedia.org/w/index.php?title=Infinitesimal_strain_theory&oldid=478640283.

[6] Wikipedia. Saint-venant’s compatibility condition — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 23-April-2012]. http://en.wikipedia.org/w/index.php?title=Saint-Venant\%27s_compatibility_condition&oldid=436103127.

[7] 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.

[8] 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.

[9] Wikipedia. Rayleigh wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 4-February-2012]. http://en.wikipedia.org/w/index.php?title=Rayleigh_wave&oldid=473693354.

[10] Wikipedia. Love wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 4-February-2012]. http://en.wikipedia.org/w/index.php?title=Love_wave&oldid=474355253.

## 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}

Recall (3.20) that we had

\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.

## PHY454H1S Continuum Mechanics. Lecture 4: Strain tensor components. 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.

# Stress tensor.

Reading: Portions of this lecture cover section 2 from the text [1].

For the stress tensor

\begin{aligned}\sigma_{ij},\end{aligned} \hspace{\stretch{1}}(2.1)

a second rank tensor, the first index $i$ defines the direction of the force, and the second index $j$ defines the surface.

Observe that the dimensions of $\sigma_{ij}$ is force per unit area, just like pressure. We will in fact show that this tensor is akin to the pressure, and the diagonalized components of this tensor represent the pressure.

We’ve illustrated the stress tensor in a couple of 2D examples. The first we call uniaxial stress, having just the $1,1$ element of the matrix as illustrated in figure (\ref{fig:continuumL4:continuumL4fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig1}
\caption{Uniaxial stress}
\end{figure}

\begin{aligned}\sigma = \begin{bmatrix}\sigma_{11} & 0 \\ 0 & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.2)

A biaxial stress is illustrated in figure (\ref{fig:continuumL4:continuumL4fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig2}
\caption{Biaxial stress.}
\end{figure}

where for $\sigma_{11} \ne \sigma_{22}$ our tensor takes the form

\begin{aligned}\sigma = \begin{bmatrix}\sigma_{11} & 0 \\ 0 & \sigma_{22}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.3)

In the general case we have

\begin{aligned}\sigma = \begin{bmatrix}\sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.4)

We can attempt to illustrate this, but it becomes much harder to visualize as shown in figure (\ref{fig:continuumL4:continuumL4fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig3}
\caption{General strain}
\end{figure}

In equilibrium we must have

\begin{aligned}\sigma_{12} = \sigma_{21}.\end{aligned} \hspace{\stretch{1}}(2.5)

We can use similar arguments to show that the stress tensor is symmetric.

In 3D we have three components of the stress tensor acting on each surface, as illustrated in figure (\ref{fig:continuumL4:continuumL4fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig5}
\caption{Strain components on a 3D volume.}
\end{figure}

We have three unique surface orientations and three components of the force for each of these, resulting in nine components, but these are not all independent. For an object in equilibrium we must have $\sigma_{ij} = \sigma_{ji}$ (FIXME: justify?). Explicitly, that is

\begin{aligned}\sigma_{12} &= \sigma_{21} \\ \sigma_{23} &= \sigma_{32} \\ \sigma_{31} &= \sigma_{13}\end{aligned} \hspace{\stretch{1}}(2.6)

## Diagonalization

We’ll look at the two dimensional case in some detail, as in figure (\ref{fig:continuumL4:continuumL4fig6})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig6}
\caption{Area element under strain with and without rotation.}
\end{figure}

Under this coordinate transformation, a rotation, the diagonal stress tensor is taken to a non-diagonal form

\begin{aligned}\begin{bmatrix}\sigma_{11} & 0 \\ 0 & \sigma_{22} \end{bmatrix}\leftrightarrow\begin{bmatrix}\sigma_{11}' & \sigma_{12}' \\ \sigma_{21}' & \sigma_{22}' \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.9)

## How do the stress tensor and the force relate

We form a Cauchy tetrahedron as in figure (\ref{fig:continuumL4:continuumL4fig7})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig7}
\caption{Cauchy tetrahedron}
\end{figure}

\begin{aligned}\mathbf{f} = \frac{\text{external force}}{\text{unit area}} = f_j \mathbf{e}_j\end{aligned} \hspace{\stretch{1}}(2.10)

\begin{aligned}\text{internal stress} = \text{external force}\end{aligned} \hspace{\stretch{1}}(2.11)

We write $\hat{\mathbf{n}}$ in terms of the direction cosines

\begin{aligned}\hat{\mathbf{n}} = n_1 \mathbf{e}_1 + n_2 \mathbf{e}_2 + n_3 \mathbf{e}_3 \end{aligned} \hspace{\stretch{1}}(2.12)

Here

\begin{aligned}n_1 &= \hat{\mathbf{n}} \cdot \mathbf{e}_1 \\ n_2 &= \hat{\mathbf{n}} \cdot \mathbf{e}_2 \\ n_3 &= \hat{\mathbf{n}} \cdot \mathbf{e}_3,\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}n_j = \hat{\mathbf{n}} \cdot \mathbf{e}_j = \cos\phi_j\end{aligned} \hspace{\stretch{1}}(2.16)

Force balance on $x_1$ direction, matching total external force in this direction to the total internal force ($\sigma_{ij}'s$) as follows

\begin{aligned}\begin{aligned}f_1 \times \text{area ABC} &= \sigma_{11} \times \text{area BOC} \\ &+\sigma_{12} \times \text{area AOC} \\ &+\sigma_{13} \times \text{area AOB}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.17)

Similarily

\begin{aligned}\begin{aligned}f_2 \times \text{area ABC} &= \sigma_{21} \times \text{area BOC} \\ &+\sigma_{22} \times \text{area AOC} \\ &+\sigma_{23} \times \text{area AOB},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.18)

and

\begin{aligned}\begin{aligned}f_3 \times \text{area ABC} &= \sigma_{31} \times \text{area BOC} \\ &+\sigma_{32} \times \text{area AOC} \\ &+\sigma_{33} \times \text{area AOB},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.19)

We can therefore write these force components like

\begin{aligned}f_1 = \sigma_{11} \frac{BOC}{ABC} + \sigma_{12} \frac{AOC}{ABC} + \sigma_{13} \frac{AOB}{ABC} \end{aligned} \hspace{\stretch{1}}(2.20)

but these ratios are really just the projections of the areas as illustrated in figure (\ref{fig:continuumL4:continuumL4fig8})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig8}
\caption{Area projection.}
\end{figure}

where an arbitrary surface with area $\Delta S$ can be decomposed into projections

\begin{aligned}\Delta S \cos\phi_j,\end{aligned} \hspace{\stretch{1}}(2.21)

utilizing the direction cosines. We can therefore write

\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}}(2.22)

or in matrix notation

\begin{aligned}\begin{bmatrix}f_1 \\ f_2 \\ f_3 \end{bmatrix}=\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}}(2.25)

This is just

\begin{aligned}\boxed{f_i = \sigma_{ij} n_j.}\end{aligned} \hspace{\stretch{1}}(2.26)

This force with components $f_i$ is also called the traction vector

\begin{aligned}T_i = \sigma_{ij} n_j.\end{aligned} \hspace{\stretch{1}}(2.27)

# 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.