The basic elastic constants

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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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