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

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

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

We have:

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

{\begin{aligned}{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{aligned}} Equation (6,3) gives

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

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

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

{\begin{aligned}\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{aligned}} {\begin{aligned}\sigma =0.38=\lambda /2\left(\lambda +\mu \right),\quad \mathrm {i.e.} ,0.76\mu =0.24\lambda .\end{aligned}} Solving the two equations gives ${\mathrm {\lambda } }=11{\times }10^{9}$ Pa, ${\mathrm {\mu } }=3.6{\times }10^{9}$ Pa.

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

{\begin{aligned}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{aligned}} Finally,

{\begin{aligned}\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{aligned}} #### Sandstone

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

{\begin{aligned}{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{aligned}} [We could also have obtained ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$ 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:

{\begin{aligned}{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{aligned}} #### Granite

Again the solution is the same as for sandstone.

{\begin{aligned}{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{aligned}} Table 2.4b summarizes the results.

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