The basic 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 2.1a

Show that when the only nonzero applied stress is ${\sigma }_{ii}$, Hooke’s law requires that the normal strains ${\varepsilon }_{yy}={\varepsilon }_{zz}$, and that Poisson’s ratio ${\sigma }$, defined as ${\sigma }=-{\varepsilon }_{yy}/{\varepsilon }_{xx}=-{\varepsilon }_{zz}/{\varepsilon }_{xx}$, satisfies the equation

${\begin{aligned}{\sigma ={\frac {\lambda }{{2}\left({\lambda }+{\mu }\right)}}}.\end{aligned}}$

(2.1a)

Background

Stress is force/unit area and is denoted by ${\mathrm {\sigma } }_{xy}$, etc., where a force in the $x$-direction acts upon a surface perpendicular to the $y$-axis. The stresses ${\mathrm {\sigma } }_{xx}$ and ${\mathrm {\sigma } }_{xy}$ are, respectively, a normal stress and a shearing stress.

Stresses produce strains (changes in size and/or shape). If the stresses cause a point $P\left(x,\;y,\;z\right)$ to have displacements $\left(u,\;v,\;w\right)$ along the coordinate axes, the basic strains are given by derivatives of these displacements as follows:

${\begin{aligned}{Normal}\ {strains}:\ \,\ {\mathrm {\varepsilon } }_{xx}={\frac {{\mathrm {\partial } }u}{{\mathrm {\partial } }x}},{\quad }{\mathrm {\varepsilon } }_{yy}={\frac {{\mathrm {\partial } }v}{{\mathrm {\partial } }y}},{\quad }{\mathrm {\varepsilon } }_{zz}={\frac {{\mathrm {\partial } }w}{{\mathrm {\partial } }z}}.\end{aligned}}$

(2.1b)

${\begin{aligned}{Shearing}\ {strains}:\ {\mathrm {\varepsilon } }_{xy}={\mathrm {\varepsilon } }_{yx}={\frac {{\mathrm {\partial } }v}{{\mathrm {\partial } }x}}+{\frac {{\mathrm {\partial } }u}{{\mathrm {\partial } }y}},{\quad }{\hbox{and so on for }}{\mathrm {\varepsilon } }_{yz}{\hbox{ and }}{\mathrm {\varepsilon } }_{zx}.\end{aligned}}$

(2.1c)

The vector displacement ${\mathrm {\zeta } }$ is

${\begin{aligned}{\mathrm {\zeta } }=ui+vj+wk,\end{aligned}}$

(2.1d)

where ${\textbf {i}},{\textbf {j}},{\textbf {k}}$ are unit vectors in the ${\textit {x}}-,{\textit {y}}-,{\textit {z}}$-directions (see Sheriff and Geldart, 1995, problem 15.3). The dilatation $\Delta$ is the change in volume per unit volume, i.e.,

${\begin{aligned}\Delta \approx \left(1+{\mathrm {\varepsilon } }_{xx}\right)\left(1+{\mathrm {\varepsilon } }_{yy}\right)\left(1+{\mathrm {\varepsilon } }_{zz}\right)-1\approx {\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}={\nabla }\cdot {\mathrm {\zeta } }.\end{aligned}}$

(2.1e)

A pressure ${\mathcal {P}}$ produces a decrease in volume, the proportionality constant being the bulk modulus $k$:

${\begin{aligned}k=-{\mathcal {P}}/\Delta .\end{aligned}}$

(2.1f)

Sometimes the compressibility, $1/k$, is used instead of $k$.

In addition to creating strains, stresses cause rotation of the medium, the vector rotation ${\theta }$ being equal to

${\begin{aligned}{\theta }={\mathrm {\theta } }_{x}i+{\mathrm {\theta } }_{y}j+{\mathrm {\theta } }_{z}k={\frac {1}{2}}{\nabla }\times {\mathrm {\zeta } },\end{aligned}}$

(2.1g)

where ${\mathrm {\theta } }_{x}=\left(1/2\right)\left({\mathrm {\partial } }w/{\mathrm {\partial } }y-{\mathrm {\partial } }v/{\mathrm {\partial } }z\right)$, ${\mathrm {\theta } }_{y}=\left(1/2\right)\left({\mathrm {\partial } }u/{\mathrm {\partial } }z-{\mathrm {\partial } }w/{\mathrm {\partial } }x\right)$, ${\mathrm {\theta } }_{z}=\left(1/2\right)\left({\mathrm {\partial } }v/{\mathrm {\partial } }x-{\mathrm {\partial } }u/{\mathrm {\partial } }y\right).$

For small strains and an isotropic medium (where properties are the same regard-less of the direction of measurement), Hooke’s law relates the stresses to the strains as follows:

${\begin{aligned}{\mathrm {\sigma } }_{ii}={\mathrm {\lambda } }{\mathrm {\Delta } }+2{\mu }{\mathrm {\varepsilon } }_{ii},{\qquad }i=x,\;y,\;z;\end{aligned}}$

(2.1h)

${\begin{aligned}{\mathrm {\sigma } }_{ij}={\mathrm {\mu } }{\mathrm {\varepsilon } }_{ij},{\qquad }{\qquad }{\quad }i\neq j.\end{aligned}}$

(2.1i)

where ${\mathrm {\lambda } }$ and ${\mathrm {\mu } }$ are Lamé’s constants (${\mathrm {\mu } }$ is usually called the modulus of rigidity or the shear modulus).

Solution

Subtracting equation (2.1h) for $i=y$ from the same equation for $i=z$ gives ${\mathrm {\varepsilon } }_{yy}={\mathrm {\varepsilon } }_{zz}$.

Dividing equation (2.1h) for $i=y$ by ${\mathrm {\varepsilon } }_{xx}$ gives

${\begin{aligned}0={\mathrm {\lambda } }\left(1+2{\frac {{\mathrm {\varepsilon } }_{yy}}{{\mathrm {\varepsilon } }_{xx}}}\right)+2{\mathrm {\mu } }{\frac {{\mathrm {\varepsilon } }_{yy}}{{\mathrm {\varepsilon } }_{xx}}}={\mathrm {\lambda } }-2\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)\sigma ,\end{aligned}}$

so

${\begin{aligned}{\mathrm {\sigma } }={\mathrm {\lambda } }/2\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right).\end{aligned}}$

Problem 2.1b

2.1b Show that Young’s modulus $E$, defined as ${\mathrm {\sigma } _{xx}/{\mathrm {\varepsilon } }_{xx}}$, is given by the equation

${\begin{aligned}{E={\frac {{\mu }\left({3}{\lambda }+{2}{\mu }\right)}{\left({\lambda }+{\mu }\right)}}.}\end{aligned}}$

(2.1j)

Solution

Adding the three equations (2.1h) for $i=x$, $y$, and $z$, and recalling that ${\mathrm {\sigma } }_{yy}=0={\mathrm {\sigma } }_{zz}$, we get

${\begin{aligned}{\mathrm {\sigma } }_{xx}=\left[3{\mathrm {\lambda } }{\mathrm {\Delta } }+2{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}\right)\right]=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\Delta .\end{aligned}}$

Dividing both sides by ${\mathrm {\varepsilon } }_{xx}$ gives

${\begin{aligned}E={\mathrm {\sigma } }_{xx}/{\mathrm {\varepsilon } }_{xx}=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\left(1+2{\mathrm {\varepsilon } }_{yy}/{\mathrm {\varepsilon } }_{xx}\right)=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\left(1-2{\mathrm {\sigma } }\right).\end{aligned}}$

Using equation (2.1a) we get

${\begin{aligned}E={\frac {{\mathrm {\mu } }\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)}{\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)}}.\end{aligned}}$

Problem 2.1c

A pressure $\mathbf {\mathcal {P}}$ is equivalent to stresses ${\sigma }_{xx}={\sigma }_{yy}={\sigma }_{zz}=-{\mathcal {P}}$. Derive the following result for the bulk modulus $k$:

${\begin{aligned}k{\mathbf {=} \left({\lambda }\mathbf {+{\frac {2}{3}}} {\mu }\right)}.\end{aligned}}$

(2.1k)

Solution

Since ${\mathcal {P}}=-{\mathrm {\sigma } }_{xx}=-{\mathrm {\sigma } }_{yy}=-{\mathrm {\sigma } }_{zz}$, we add equation (2.1h) for each of the three values $i=x,y,z$, obtaining $-3{\mathcal {P}}=\left[3{\mathrm {\lambda } }{\mathrm {\Delta } }+2{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}\right)\right]=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\Delta$, so from equation (2.1f),

${\begin{aligned}k=-{\mathcal {P}}/{\mathrm {\Delta } }=\left({\mathrm {\lambda } }+{\frac {2}{3}}{\mathrm {\mu } }\right).\end{aligned}}$

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Introduction	Theory of seismic waves
Table of Contents (book)

Also in this chapter

• Interrelationships among elastic constants
• Magnitude of disturbance from a seismic source
• Magnitudes of elastic constants
• 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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