Magnitudes of elastic constants

Problems in Exploration Seismology and their Solutions

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

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Introduction	Partitioning at an interface
Table of Contents (book)

Also in this chapter

• The basic elastic constants
• Interrelationships among elastic constants
• Magnitude of disturbance from a seismic source
• General solutions of the wave equation
• Wave equation in cylindrical and spherical coordinates
• Sum of waves of different frequencies and group velocity
• Magnitudes of seismic wave parameters
• Potential functions used to solve wave equations
• Boundary conditions at different types of interfaces
• Boundary conditions in terms of potential functions
• Disturbance produced by a point source
• Far- and near-field effects for a point source
• Rayleigh-wave relationships
• Directional geophone responses to different waves
• Tube-wave relationships
• Relation between nepers and decibels
• Attenuation calculations
• Diffraction from a half-plane

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