# Magnitudes of elastic constants

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 2.4

To illustrate the relationships and magnitudes of the elastic constants, complete Table 2.4a. Note that these values apply to specific specimens; the elastic constants for rocks range considerably, especially as porosity and pressure change.

### Solution

We use the row-column notation to designate equations from Table 2.2a.

#### Water

Since water is a fluid we know that $\displaystyle {\mathrm{\mu}} =0=\beta$ . Equation (4,1) shows that $\displaystyle E=0$ also. From equation (4,5) we get $\displaystyle {\mathrm{\lambda}} =k=2.1\times 10^{9}$ Pa.

Table 2.4a. Magnitudes of elastic constants and velocities.
Constant Water Stiff mud Shale Sandstone Limestone Granite
$\displaystyle E\,\,\,({\times}10^9 \hbox{Pa})$ 16 54 50
$\displaystyle k\,\,\,({\times}10^9 \hbox{Pa})$ 2.1
$\displaystyle {\mathrm{\mu}}\,\,\,({\times}10^9 \hbox{Pa})$
$\displaystyle {\mathrm{\lambda}}\,\,\,({\times}10^9 \hbox{Pa})$
$\displaystyle {\mathrm{\sigma}}$ 0.50 0.43 0.38 0.34 0.25 0.20
$\displaystyle {\mathrm{\rho}}\,\,\,(\hbox{g/cm}^{3})$ 1.0 1.5 1.8 1.9 2.5 2.7
$\displaystyle {\alpha}\,\,\,(\hbox{km/s})$ 1.5 1.6 3.2
$\displaystyle {\beta}\,\,\,(\hbox{km/s})$

#### Stiff mud

Because $\displaystyle {\mathrm{\sigma}}<0.5$ , stiff mud is equivalent to a solid, hence $\displaystyle {\mathrm{\mu}} \ne 0$ . From $\displaystyle {\mathrm{\rho}}$ and $\displaystyle {\mathrm{\alpha}}$ we get $\displaystyle \left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)$ using equation (9,6), while equation (9,2) expresses $\displaystyle {\mathrm{\sigma}}$ in terms of $\displaystyle {\mathrm{\lambda}}$ and $\displaystyle {\mathrm{\mu}}$ , thus enabling us to find both $\displaystyle {\mathrm{\lambda}}$ and $\displaystyle {\mathrm{\mu}}$ .

We have:

\displaystyle \begin{align} {{\rho} {{\alpha} ^2} = \left( {{\lambda} + 2{\mu} } \right) = \left( {1.5\;{\rm{g/c}}{{\rm{m}}^3}} \right) {\times} {{(1.6 {\times} {{10}^5}\;{\rm{cm/s}})}^2}}\\ {{\qquad} = 3.8 {\times} {{10}^{10}}\;{\rm{dynes/c}}{{\rm{m}}^2} = 3.8 {\times} {{10}^9}\;{\rm{Pa}},}\\ {{\quad}{\sigma} = 0.43 = {\lambda} /2\left( {{\lambda} + {\mu} } \right),\;{\rm{i}}.{\rm{e}}.,\;0.86{\mu} = 0.14{\lambda} .} \end{align}

Solving the two equations, we get $\displaystyle {\mathrm{\lambda}} =2.9{\times} 10^{9}$ Pa, $\displaystyle {\mathrm{\mu}} =0.47{\times} 10^{9}$ Pa. Using equation (6,1),

\displaystyle \begin{align} {E={\mathrm{\lambda}} \frac{\left(1+{\mathrm{\sigma}} \right)\left(1-2{\mathrm{\sigma}} \right)}{{\mathrm{\sigma}} } =2.9{\times} 10^{9} {\times} 1.43{\times} 0.14/0.43}\\ {{\qquad}{\qquad}{\qquad}{\qquad}{\quad}=1.4{\times} 10^{9}\ \mathrm{Pa}.} \end{align}

Equation (6,3) gives

\displaystyle \begin{align} k={\mathrm{\lambda}} \frac{\left(1+{\mathrm{\sigma}} \right)}{3{\mathrm{\sigma}} } =2.9{\times} 10^{9} {\times} 1.43/1.29=3.2{\times} 10^{9}\ \mathrm{Pa}. \end{align}

Finally, to get $\displaystyle {\mathrm{\beta}}$ we note that $\displaystyle {\mathrm{\lambda}}$ , $\displaystyle {\mathrm{\mu}}$ , and $\displaystyle k$ are in $\displaystyle N/m^{2}$ , and so we must express $\displaystyle {\mathrm{\rho}}$ in appropriate units of $\displaystyle kg/m^{3}$ , i.e., $\displaystyle {\mathrm{\rho}}=1.5{\times} 10^{3}$ $\displaystyle kg/m^{3}$ . We now have

$\displaystyle \beta =({\mathrm{\mu}} /{\mathrm{\rho}})^{1/2} =\left(0.47{\times} 10^{9} /1.5{\times} 10^{3} \right)=0.56\ \mathrm{\hbox{km/s}}.$

#### Shale

As with stiff mud, we have been given $\displaystyle {\mathrm{\rho}}$ , $\displaystyle \alpha$ , and $\displaystyle {\mathrm{\sigma}}$ , so that again $\displaystyle {\mathrm{\rho}}$ and $\displaystyle {\mathrm{\alpha}}$ give us $\displaystyle \left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)$ and $\displaystyle {\mathrm{\sigma}}$ gives us $\displaystyle {\mathrm{\lambda}} /2\left({\mathrm{\lambda}} +{\mathrm{\mu}} \right)$ so that we can solve these equations for $\displaystyle {\mathrm{\lambda}}$ and $\displaystyle {\mathrm{\mu}}$ , then find $\displaystyle k$ and $\displaystyle E$ using equations (9,3) and (9,1). Thus,

\displaystyle \begin{align} \left(\lambda +2\mu \right)=\rho\alpha^{2} =\left(1.8\ \mathrm{g/cm}^{3} \right)\times (3.2\times 10^{5}\ \mathrm{cm/s})^{2} =18.4\times 10^{9}\ \mathrm{Pa}, \end{align}

\displaystyle \begin{align} \sigma =0.38=\lambda /2\left(\lambda +\mu \right),\quad \mathrm{i.e.}, 0.76\mu =0.24\lambda. \end{align}

Solving the two equations gives $\displaystyle {\mathrm{\lambda}} =11{\times} 10^{9}$ Pa, $\displaystyle {\mathrm{\mu}} =3.6{\times} 10^{9}$ Pa.

From equation (9,3), $\displaystyle k={\mathrm{\lambda}} +\frac{2}{3} {\mathrm{\mu}} =14{\times} 10^{9}$ Pa, and equation (9,1) gives

\displaystyle \begin{align} E={\mathrm{\mu}} \frac{\left(3{\mathrm{\lambda}} +2{\mathrm{\mu}} \right)}{\left({\mathrm{\lambda}} +{\mathrm{\mu}} \right)} =9.9{\times} 10^{9}\ \mathrm{Pa}. \end{align}

Finally,

\displaystyle \begin{align} \beta =({\mathrm{\mu}} /{\mathrm{\rho}})^{1/2} =(3.6{\times} 10^{9} /1.8{\times} 10^{3} )^{1/2} =1.4\ \mathrm{\hbox{km/s}}. \end{align}

#### Sandstone

We are given the elastic constants $\displaystyle E$ and $\displaystyle {\mathrm{\sigma}}$ (plus $\displaystyle {\mathrm{\rho}}$ ), so we get $\displaystyle k$ , $\displaystyle {\mathrm{\mu}}$ , $\displaystyle {\mathrm{\lambda}}$ , $\displaystyle \alpha$ , $\displaystyle {\mathrm{\beta}}$ using equations (1,3) to (1,7) in Table 2.2a. Thus

\displaystyle \begin{align} {k=16{\times} 10^{9} /3{\times} 0.32=17{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\mu}} =16{\times} 10^{9} /2{\times} 1.34=6.0{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\lambda}} =16{\times} 10^{9} {\times} 0.34/1.34{\times} 0.32=13{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\alpha}} =[16{\times} 10^{9} {\times} 0.66/1.34{\times} 0.32{\times} 1.9{\times} 10^{3} ]^{1/2} =3.6\ \mathrm{\hbox{km/s}},}\\ {{\mathrm{\beta}} =[16{\times} 10^{9} /2{\times} 1.34{\times} 1.9{\times} 10^{3} ]^{1/2} =1.8\ \mathrm{\hbox{km/s}}.} \end{align}

[We could also have obtained $\displaystyle {\mathrm{\alpha}}$ and $\displaystyle {\mathrm{\beta}}$ by using equations (9,6) and (9,7).]

#### Limestone

We solve in the same way as with sandstone since we are given the same constants:

\displaystyle \begin{align} {k =54{\times} 10^{9} /3{\times} 0.50=36{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\mu}} =54{\times} 10^{9} /2{\times} 1.25=22{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\lambda}} =54{\times} 10^{9} {\times} 0.25/1.25{\times} 0.50=22{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\alpha}} =[54{\times} 10^{9} {\times} 0.75/1.25{\times} 0.50{\times} 2.5{\times} 10^{3} ]^{1/2} =5.1\ \mathrm{\hbox{km/s}},}\\ {{\mathrm{\beta}} =[54{\times} 10^{9} /2{\times} 1.25{\times} 2.5{\times} 10^{3} ]^{1/2} =2.9\ \mathrm{\hbox{km/s}}.} \end{align}

#### Granite

Again the solution is the same as for sandstone.

\displaystyle \begin{align} {k=50{\times} 10^{9} /3{\times} 0.60=28{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\mu}} =50{\times} 10^{9} /2{\times} 1.2=21{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\lambda}} =50{\times} 10^{9} {\times} 0.20/1.2{\times} 0.60=14{\times} 10^{9}\ \mathrm{Pa},}\\ {{\mathrm{\alpha}} =[50{\times} 10^{9} {\times} 0.80/1.2{\times} 0.60{\times} 2.7{\times} 10^{3} ]^{1/2} =4.5\ \mathrm{\hbox{km/s}},}\\ {{\mathrm{\beta}} =[50{\times} 10^{9} /2{\times} 1.2{\times} 2.7{\times} 10^{3} ]^{1/2} =2.8\ \mathrm{\hbox{km/s}}.} \end{align}

Table 2.4b summarizes the results.

Table 2.4b Magnitudes of elastic constants.
Constant Water Stiffmud Shale Sandstone Limestone Granite
$\displaystyle E\,\,\,({\times} 10^{9} \hbox{Pa})$ 0 1.4 9.9 16 54 50
$\displaystyle K\,\,\,({\times} 10^{9} \hbox{Pa})$ 2.1 3.2 13 17 36 28
$\displaystyle {\mathrm{\mu}}\,\,\,({\times} 10^{9} \hbox{Pa})$ 0 0.47 3.6 6.0 22 21
$\displaystyle {\mathrm{\lambda}}\,\,\,({\times} 10^{9} \hbox{Pa})$ 2.1 2.9 11 13 22 14
$\displaystyle {\mathrm{\sigma}}$ 0.50 0.43 0.38 0.34 0.25 0.20
$\displaystyle {\mathrm{\mu}}\,\,\,(\hbox{g/cm}^{3})$ 1.0 1.5 1.8 1.9 2.5 2.7
$\displaystyle {\mathrm{\alpha}}\,\,\,(\hbox{km/s})$ 1.5 1.6 3.2 3.6 5.1 4.5
$\displaystyle {\mathrm{\beta}}\,\,\,(\hbox{km/s})$ 0 0.56 1.4 1.8 2.9 2.8