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Derivation of relationships
The most general linear relationship between two symmetric tensors $$\sigma_{ij}\,$$ and $$\varepsilon_{ij}$$ which is homogeneous (i.e. independent of direction) is:



\sigma_{ij}=A\delta_{ij}\varepsilon_{kk}+B(\epsilon_{ij}-\textstyle{\frac{1}{3}}\delta_{ij}\varepsilon_{kk}) $$

Where A and B are constants, and $$\delta_{ij}\,$$ is the Kroneker delta tensor. When $$\sigma_{ij}\,$$ is the stress and $$\varepsilon_{ij}$$ is the strain, this is an expression of Hooke's law, also known as the constitutive equations of linear elasticity.

Bulk modulus
Consider a cube of linearly elastic material with each side having length L. A purely compressive force consists of a force $$\sigma L^2\,$$ directed normal to each face, causing each face to move a distance $$\Delta L\,$$. The stress and strain may be written as:



\sigma_{ij}=\begin{bmatrix}\sigma & 0 & 0\\ 0 & \sigma & 0 \\ 0 & 0 & \sigma\end{bmatrix} \qquad \varepsilon_{ij}=\begin{bmatrix}\ \frac{\Delta L}{L} & 0 & 0\\ 0 & \frac{\Delta L}{L} & 0 \\ 0 & 0 & \frac{\Delta L}{L}\end{bmatrix} $$

Thus:

\varepsilon_{kk}=3\frac{\Delta L}{L} $$

and the constitutive equations become:



\sigma=3A\frac{\Delta L}{L} $$

The incompressibility or bulk modulus K is defined as



K=V \frac{\sigma}{\Delta V} $$

where $$V=L^3\,$$. Taking the derivative, it follows that $$\Delta V\approx 3L^2\Delta L\,$$, thus:



K=L^3\, \frac{\sigma}{3L^2\Delta L} $$

and it is clear that the constant A is simply the bulk modulus K.

Shear modulus
For a pure shear force $$\tau L^2\,$$ applied only to the z-face of the cube:



\sigma_{ij}=\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0 \\ \tau & \tau & 0\end{bmatrix} \qquad \varepsilon_{ij}=\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ \frac{\Delta L}{L} & \frac{\Delta L}{L} & 0\end{bmatrix} $$

Since only off-diagonal elements are non-zero, the constitutive equations become:



\tau=B\frac{\Delta L}{L} $$

The shear modulus is defined as



G= \frac{F L}{2 A \Delta L} $$

where $$F=\tau L^2\,$$ and $$A=L^2\,$$ thus:



G= \frac{\tau L}{2 \Delta L} $$

and it is clear that the constant B is simply twice the shear modulus G. The constitutive equations may now be written:



\sigma_{ij}=K\delta_{ij}\varepsilon_{kk}+2G(\epsilon_{ij}-\textstyle{\frac{1}{3}}\delta_{ij}\varepsilon_{kk}) $$

Young's modulus and Poisson's ratio
By the definition of Poisson's ratio, if a positive force is applied only to the x-faces of the cube, they will move a distance of $$\Delta L\,$$, and the other faces will move a distance of $$-\nu\Delta L\,$$. The stress and strain are written:



\sigma_{ij}=\begin{bmatrix}\sigma & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} \qquad \varepsilon_{ij}=\begin{bmatrix}\ \frac{\Delta L}{L} & 0 & 0\\ 0 & -\frac{\nu\Delta L}{L} & 0 \\ 0 & 0 & -\frac{\nu\Delta L}{L}\end{bmatrix} $$

Thus:

\varepsilon_{kk}=(1-2\nu)\frac{\Delta L}{L} $$

The constitutive equations become:



\sigma=K(1-2\nu)\frac{\Delta L}{L} + \frac{4}{3}G(1-\nu)\frac{\Delta L}{L} $$



0=K(1-2\nu)\frac{\Delta L}{L} - \frac{2}{3}G(1+\nu)\frac{\Delta L}{L} $$

Young's modulus is defined as:



E= \frac{\sigma L}{ \Delta L} $$

thus:



E=K(1-2\nu) + \textstyle{\frac{4}{3}}G(1-\nu) $$

and



0=K(1-2\nu) - \textstyle{\frac{2}{3}}G(1+\nu) $$

These two equations may be solved for any one of the variables in terms of the other two, yielding the relationships:


 * $$E=2G(1+\nu)=3K(1-2\nu)\,$$


 * $$K=\frac{2G(1+\nu)}{3(1-2\nu)}=\frac{E}{3(1-2\nu)}$$


 * $$G=\frac{3K(1-2\nu)}{2(1+\nu)}=\frac{E}{2(1+\nu)}$$

Lame's first parameter and the p-wave modulus
If equal and opposite forces $$\sigma L^2\,$$ are applied to the x faces of the cube, and a force $$\sigma_0 L^2\,$$ is applied to the other faces such that these other faces do not move, then:



\sigma_{ij}=\begin{bmatrix}\sigma & 0 & 0\\ 0 & \sigma_0 & 0 \\ 0 & 0 & \sigma_0\end{bmatrix} \qquad \varepsilon_{ij}=\begin{bmatrix}\ \frac{\Delta L}{L} & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} $$

Thus:

\varepsilon_{kk}=\frac{\Delta L}{L} $$

The constitutive equations become:



\sigma=K\frac{\Delta L}{L} + \textstyle{\frac{4}{3}}G\frac{\Delta L}{L} $$

and



\sigma_0=K\frac{\Delta L}{L} - \textstyle{\frac{2}{3}}G\frac{\Delta L}{L} $$

The p-wave modulus is defined as:



M= \frac{\sigma L}{ \Delta L} $$

and Lame's first parameter is defined as:



\lambda= \frac{\sigma_0 L}{ \Delta L} $$

Thus:



M=K + \textstyle{\frac{4}{3}}G $$



\lambda=K - \textstyle{\frac{2}{3}}G $$

These two equations, along with the relationships derived above may be used to express any two elastic moduli in terms of any other two.

Volumetric change
The relative change of volume ΔV/V due to the stretch of the material can be calculated using a simplified formula:


 * $$\frac {\Delta V} {V} = \left(1-\frac{\Delta L}{L}\right)^{1-2\nu}-1$$

which, for small deformations reduces to:


 * $$\frac {\Delta V} {V} = (1-2\nu)\frac {\Delta L} {L}$$

where
 * $$ V $$ is material volume
 * $$ \Delta V $$ is change in material volume
 * $$ L $$ is original length, before stretch
 * $$ \Delta L $$ is the change of length along the direction of compression: $$ \Delta L = L_\mathrm{new} - L$$

Note that for an incompressible material, $$\Delta V=0$$ which implies that $$\nu=1/2$$. For a material which does not have any transverse expansion or contraction, the volume change will be simply $$\Delta V=L^2\Delta L$$, which implies that $$\nu=0$$.

Derivation A cube with sides of length $$L$$, made of an isotropic material, has volume $$V=L^3$$. An axial stress gives new dimensions to the cube: $$L_a$$ axially and $$L_t$$ transversely.

For small changes in dimension, Poisson's ratio gives the relationship between these new dimensions:


 * $$\nu\dfrac{L_a-L}{L}=-\dfrac{L_t-L}{L} \Rightarrow L_t=L-\nu\Delta L$$

where $$\Delta L=L_a-L\,$$ is the change in length due to the axial stress. The new volume of the cube is:


 * $$V_{new}=L_a L_t^2=(L+\Delta L)(L-\nu\Delta L)^2$$

To the first order in $$\Delta L\,$$ we have:


 * $$V_{new}=L^3+L^2(1-2\nu)\Delta L\,$$

Defining $$\Delta V=V_{new}-V$$ yields:


 * $$\dfrac{\Delta V}{V} = (1 - 2\nu)\dfrac{\Delta L}{L}$$