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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}

\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.