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## PHY454H1S Continuum Mechanics. Lecture 6: Compatibility condition and elastostatics. Taught by Prof. K. Das.

Posted by peeterjoot on January 29, 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: Elastostatics

We’ve defined the strain tensor, where assuming the second order terms are ignored, was

\begin{aligned}e_{ij} = \frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+ \frac{\partial {e_j}}{\partial {x_i}} \right).\end{aligned} \hspace{\stretch{1}}(2.1)

We’ve also defined a stress tensor defined implicitly as a divergence relationship using the force per unit volume $F_i$ in direction $i$

\begin{aligned}\sigma_{ij} \leftrightarrow F_i = \frac{\partial {\sigma_{ij}}}{\partial {x_j}}.\end{aligned} \hspace{\stretch{1}}(2.2)

We’ve also discussed the constitutive relation, relating stress $\sigma_{ij}$ and strain $e_{ij}$.

We’ve also discussed linear constitutive relationships (Hooke’s law).

# 2D strain.

\begin{aligned}e_{ij} = \begin{bmatrix}e_{11} & e_{12} \\ e_{21} & e_{22}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.3)

From 2.1 we see that we have

\begin{aligned}e_{11} &= \frac{\partial {e_1}}{\partial {x_1}} \\ e_{22} &= \frac{\partial {e_2}}{\partial {x_2}} \\ e_{12} = e_{21} &= \frac{1}{{2}} \left( \frac{\partial {e_2}}{\partial {x_1}}+ \frac{\partial {e_1}}{\partial {x_2}} \right).\end{aligned} \hspace{\stretch{1}}(3.4)

We have a relationship between these displacements (called the compatibility relationship), which is

\begin{aligned}\boxed{\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2} +\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} = 2\frac{\partial^2 e_{12}}{\partial x_1 \partial x_2}.}\end{aligned} \hspace{\stretch{1}}(3.7)

We find this by straight computation

\begin{aligned}\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2} &= \frac{\partial^2 {{}}}{\partial {{x_2}}^2}\left( \frac{\partial {e_1}}{\partial {x_1}}\right) \\ &=\frac{\partial^3 e_1}{\partial x_1 \partial x_2^2},\end{aligned}

and

\begin{aligned}\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} &= \frac{\partial^2 {{}}}{\partial {{x_1}}^2}\left( \frac{\partial {e_2}}{\partial {x_2}}\right) \\ &= \frac{\partial^3 e_2}{\partial x_2 \partial x_1^2},\end{aligned}

Now, looking at the cross term we find

\begin{aligned}2 \frac{\partial^2 e_{12}}{\partial x_1 \partial x_2} &= \frac{\partial^2 e_{12}}{\partial x_1 \partial x_2} \left(\frac{\partial {e_2}}{\partial {x_1}}+ \frac{\partial {e_1}}{\partial {x_2}} \right) \\ &=\left(\frac{\partial^3 e_1}{\partial x_1 \partial x_2^2} +\frac{\partial^3 e_2}{\partial x_2 \partial x_1^2} \right) \\ &=\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} +\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2} \end{aligned}

This is called the compatibility condition, and ensures that we don’t have a disjoint deformation of the form in figure (\ref{fig:continuumL6:continuumL6fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL6fig1}
\caption{disjoint deformation illustrated.}
\end{figure}

# 3D strain.

While we have 9 components in the tensor, not all of these are independent. The sets above and below the diagonal can be related, as illustrated in figure (\ref{fig:continuumL6:continuumL6fig2}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL6fig2}
\caption{continuumL6fig2}
\end{figure}

Here we have 6 relationships between the components of the strain tensor $e_{ij}$. Deriving these will be assigned in the homework.

# Elastodynamics. Elastic waves.

Reading: Chapter III (section 22 – section 26) of the text [1].

Example: sound or water waves (i.e. waves in a solid or liquid material that comes back to its original position.)

\begin{definition}
\emph{(Elastic Wave)}

An elastic wave is a type of mechanical wave that propagates through or on the surface of a medium. The elasticity of the material provides the restoring force (that returns the material to its original state). The displacement and the restoring force are assumed to be linearly related.
\end{definition}

In symbols we say

\begin{aligned}e_i(x_j, t) \quad \mbox{related to force},\end{aligned} \hspace{\stretch{1}}(5.8)

and specifically

\begin{aligned}\rho \frac{\partial^2 {{e_i}}}{\partial {{t}}^2} = F_i = \frac{\partial {\sigma_{ij}}}{\partial {x_j}}.\end{aligned} \hspace{\stretch{1}}(5.9)

This is just Newton’s second law, $F = ma$, but expressed in terms of a unit volume.

Should we have an external body force (per unit volume) $f_i$ acting on the body then we must modify this, writing

\begin{aligned}\boxed{\rho \frac{\partial^2 {{e_i}}}{\partial {{t}}^2} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + f_i}\end{aligned} \hspace{\stretch{1}}(5.10)

Note that we are separating out the “original” forces that produced the stress and strain on the object from any constant external forces that act on the body (i.e. a gravitational field).

With

\begin{aligned}e_{ij} = \frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+ \frac{\partial {e_j}}{\partial {x_i}} \right),\end{aligned} \hspace{\stretch{1}}(5.11)

we can expand the stress divergence, for the case of homogeneous deformation, in terms of the Lam\’e parameters

\begin{aligned}\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2 \mu e_{ij}.\end{aligned} \hspace{\stretch{1}}(5.12)

We compute

\begin{aligned}\frac{\partial {\sigma_{ij}}}{\partial {x_j}}&=\lambda \frac{\partial {e_{kk}}}{\partial {x_j}}\delta_{ij} + 2 \mu \frac{\partial {}}{\partial {x_j}}\frac{1}{{2}} \left( \frac{\partial {e_i}}{\partial {x_j}}+ \frac{\partial {e_j}}{\partial {x_i}} \right), \\ &=\lambda \frac{\partial {e_{kk}}}{\partial {x_i}}+ \mu \left(\frac{\partial^2 {{e_{i}}}}{\partial {{x_j}}^2}+\frac{\partial^2 e_{j} }{ \partial x_j \partial x_i}\right) \\ &=\lambda \frac{\partial {}}{\partial {x_i}}\frac{\partial {e_k}}{\partial {x_k}}+ \mu \left(\frac{\partial^2 {{e_{i}}}}{\partial {{x_j}}^2}+\frac{\partial^2 e_{k} }{ \partial x_k \partial x_i}\right) \\ &=(\lambda + \mu)\frac{\partial {}}{\partial {x_i}}\frac{\partial {e_k}}{\partial {x_k}}+ \mu \frac{\partial^2 {{e_{i}}}}{\partial {{x_j}}^2}\end{aligned}

We find, for homogeneous deformations, that the force per unit volume on our element of mass, in the absence of external forces (the body forces), takes the form

\begin{aligned}\rho \frac{\partial^2 {{e_i}}}{\partial {{t}}^2} = (\lambda + \mu) \frac{\partial^2 e_i}{\partial x_i \partial x_j}+ \mu\frac{\partial^2 e_i}{\partial x_j^2}.\end{aligned} \hspace{\stretch{1}}(5.13)

This can be seen to be equivalent to the vector relationship

\begin{aligned}\boxed{\rho \frac{\partial^2 {\mathbf{e}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{e}) + \mu \boldsymbol{\nabla}^2 \mathbf{e}.}\end{aligned} \hspace{\stretch{1}}(5.14)

TODO: What form do the stress and strain tensors take in vector form?

# 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.