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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.)]


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}


\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})

\caption{disjoint deformation illustrated.}

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}).


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.)

\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.

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).


\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?


[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.


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