Search Finite Elements Website

Hooke's Law

 home > basic mechanics > hooke's law

Introduction

We have talked about Hooke's Law some already, and used it for tensor notation exercises and examples. Hooke's Law is linear and isotropic. (Isotropic means that it has equal stiffness in every direction.) It is in fact the 1st order linearization of any hyperelastic material Law, including nonlinear ones, as long as the Law is also isotropic. So it can be applied to rubber as long as the strains are small. And it is the standard for metals in the elastic range.

---

Normal Components

The first step in getting to the full model is to start with simple uniaxial tension/compression. In this case, the relationship is

\[ \sigma = E \, \epsilon \]
That's simple enough. It is linear because the stress and strain terms are all first-order. Keep in mind that \(E\) has units of stress just like \(\sigma\) because strain is unitless.

The next step is to generalize this for the case of multiple dimensions. For starters, decide that the tension/compression direction is \(x\). So the stress and strain in the above equation will be in the \(x\) direction. Also write the above equation a little differently as

\[ \epsilon_{xx} = {1 \over E} \sigma_{xx} \]
So far, so good. Now add some complexity. Add an additional normal stress in the \(y\) direction. Any tensile stress in the \(y\) direction will cause the strain in the \(x\) direction to decrease. So modify the above equation as follows.

\[ \epsilon_{xx} = {1 \over E} \sigma_{xx} - B \, \sigma_{yy} \]
In this case, \(B\) is a material property just like \(E\) and requires a lab test to determine. Nevertheless, using \(B\) in this form is not very practical. It is much more useful to express \(B\) as a fraction of \((1 / E)\). Something like

\[ \epsilon_{xx} = {1 \over E} \sigma_{xx} - f {1 \over E} \sigma_{yy} \]
where \(f\) is the "fraction". It is a number between 0 and 1 because material behavior dictates it to be so. It is less than 1 because strains in the \(x\) direction, \(\epsilon_{xx}\), are less sensitive to stresses in the \(y\) direction, \(\sigma_{yy}\), than they are to stresses in the \(x\) direction, \(\sigma_{xx}\).

Except that instead of using \(f\) as the symbol, use \(\nu\) instead. This is Poisson's ratio. It is the ratio of the sensitivity of strain to two stresses, one in line with the strain, and the other perpendicular to it.

We will see that Poisson's ratio is 1/2 for incompressible materials such as rubber. It is pretty consistently 1/3 for face-centered-cubic (FCC) materials like aluminum, nickel, and copper. And it is about 0.3 for steel.

So the above equation becomes

\[ \epsilon_{xx} = {1 \over E} \big[ \sigma_{xx} - \nu \, \sigma_{yy} \big] \]
And now add in a \(\sigma_{zz}\) to the mix. If the material is isotropic, then the response of \(\epsilon_{xx}\) to \(\sigma_{zz}\) will be exactly the same as \(\sigma_{yy}\). So the equation can be written as

\[ \epsilon_{xx} = {1 \over E} \big[ \sigma_{xx} - \nu \left( \sigma_{yy} + \sigma_{zz} \big) \right] \]
This is the complete Hooke's Law, at least for \(\epsilon_{xx}\). It is also the defining equation for Poisson's ratio (not \(-\epsilon_{yy} / \epsilon_{xx}\) by the way).

POISSON RATIOS
Material	Poisson's Ratio	Comments
Foams	~0	Negligible effect
Steels	~0.3	Varies by a few % among steels
Aluminum	0.33	Face centered cubic (FCC) metals
Copper	0.33
Nickel	0.33
Rubber	0.50	Incompressible

The complete set of equations are

\[ \epsilon_{xx} = {1 \over E} \big[ \sigma_{xx} - \nu \, ( \sigma_{yy} + \sigma_{zz} ) \big] \] \[ \epsilon_{yy} = {1 \over E} \big[ \sigma_{yy} - \nu \, ( \sigma_{xx} + \sigma_{zz} ) \big] \] \[ \epsilon_{zz} = {1 \over E} \big[ \sigma_{zz} - \nu \, ( \sigma_{xx} + \sigma_{yy} ) \big] \]

---

Shear Components

We've worked out Hooke's Law for the normal components. We will now use them to develop the relationship between the shear stresses and strains. Start with a shear stress, \(\tau_{xy}\), and shear strain, \(\gamma_{xy}\), and rotate them 45° to get the principal values.

For the shear stress

\[ \left[ \matrix { 0 & \tau_{xy} \\ \tau_{xy} & 0 } \right] \qquad \matrix { 45^\circ \\ \text{transform} \\ \longrightarrow \; } \qquad \left[ \matrix { \tau_{xy} & 0 \\ 0 & -\tau_{xy} } \right] \]
and the shear strain

\[ \left[ \matrix { 0 & \gamma_{xy} / 2 \\ \gamma_{xy} / 2 & 0 } \right] \qquad \matrix { 45^\circ \\ \text{transform} \\ \longrightarrow \; } \qquad \left[ \matrix { \gamma_{xy} / 2 & 0 \\ 0 & -\gamma_{xy} / 2 } \right] \]
So the principal values are

\[ \sigma_1 = \tau_{xy} \qquad \qquad \sigma_2 = -\tau_{xy} \qquad \qquad \epsilon_1 = \gamma_{xy} / 2 \qquad \qquad \epsilon_2 = -\gamma_{xy} / 2 \]
Plugging the positive principal strain into Hooke's Law gives

\[ \epsilon_1 = {1 \over E} \big[ \sigma_1 - \nu \left( \sigma_2 + \sigma_3 \right) \big] \]

\[ {\gamma_{xy} \over 2} = {1 \over E} \big[ \tau_{xy} - \nu \left( -\tau_{xy} + 0 \right) \big] \]
Rearrange to get

\[ \tau_{xy} = {E \over 2 (1 + \nu) } \gamma_{xy} \]
And this leads to the expression for the shear modulus.

\[ G = {\tau_{xy} \over \gamma_{xy}} = {E \over 2 (1 + \nu) } \]

---

Bulk Modulus

Return to the three equations for normal strain in terms of normal stress.

\[ \epsilon_{xx} = {1 \over E} \big[ \sigma_{xx} - \nu \, ( \sigma_{yy} + \sigma_{zz} ) \big] \] \[ \epsilon_{yy} = {1 \over E} \big[ \sigma_{yy} - \nu \, ( \sigma_{xx} + \sigma_{zz} ) \big] \] \[ \epsilon_{zz} = {1 \over E} \big[ \sigma_{zz} - \nu \, ( \sigma_{xx} + \sigma_{yy} ) \big] \]
Add the three equations together.

\[ \epsilon_{xx} + \epsilon_{yy} + \epsilon_{zz} = { (1 - 2 \nu) \over E} ( \sigma_{xx} + \sigma_{yy} + \sigma_{zz} ) \]
But \((\epsilon_{xx} + \epsilon_{yy} + \epsilon_{zz})\) is the volumetric strain, \(\epsilon_\text{Vol} = \Delta V / V\). We will discuss this at length in the next chapter on strain and deformations.

And \((\sigma_{xx} + \sigma_{yy} + \sigma_{zz})\) is three times the hydrostatic stress, \(3 \, \sigma_\text{Hyd}\). Note that hydrostatic stress is simply the average of the diagonal components and is the pressure your ears feel when you go deep under water.

So the equation becomes

\[ \epsilon_\text{Vol} = { 3 (1 - 2 \nu) \over E} \sigma_\text{Hyd} \]
And the ratio of the \(\sigma_\text{Hyd}\) to \(\epsilon_\text{Vol}\) is the bulk modulus, \(K\).

\[ K = { \sigma_\text{Hyd} \over \epsilon_\text{Vol} } = { E \over 3 (1 - 2 \nu) } \]
And a little more rearrangement gives

\[ {E \over K} = 3 (1 - 2 \nu) \]
This is the important relationship between tensile modulus and bulk modulus. Although the bulk modulus for rubber is not large, it is nevertheless much larger than the tensile modulus. So the ratio of the two is very small. This dictates that the Poisson Ratio for rubber must be 0.5. This is so that the RHS term, \(3 ( 1 - 2 \nu )\), equals zero.

---

Matrix & Tensor Notation

Hooke's Law can be written in matrix notation as

\[ \boldsymbol{\epsilon} = {1 \over E} \left[ (1 + \nu) \boldsymbol{\sigma} - \nu \; {\bf I} \; \text{tr}(\boldsymbol{\sigma}) \right] \]
and in tensor notation as

\[ \epsilon_{ij} = {1 \over E} \left[ (1 + \nu) \sigma_{ij} - \nu \; \delta_{ij} \; \sigma_{kk} \right] \]
Writing all the matrices out gives

\[ \left[ \matrix { \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} } \right] = {1 \over E} \left\{ (1 + \nu) \left[ \matrix { \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} } \right] - \nu \, ( \sigma_{11} + \sigma_{22} + \sigma_{33} ) \left[ \matrix { 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \right] \right\} \]
The individual component equations all together are

\[ \epsilon_{11} = {1 \over E} \left[ (1 + \nu) \sigma_{11} - \nu \; (\sigma_{11} + \sigma_{22} + \sigma_{33}) \right] \qquad \qquad \epsilon_{12} = {1 \over E} \left[ (1 + \nu) \sigma_{12} \right] \] \[ \epsilon_{22} = {1 \over E} \left[ (1 + \nu) \sigma_{22} - \nu \; (\sigma_{11} + \sigma_{22} + \sigma_{33}) \right] \qquad \qquad \epsilon_{13} = {1 \over E} \left[ (1 + \nu) \sigma_{13} \right] \] \[ \epsilon_{33} = {1 \over E} \left[ (1 + \nu) \sigma_{33} - \nu \; (\sigma_{11} + \sigma_{22} + \sigma_{33}) \right] \qquad \qquad \epsilon_{23} = {1 \over E} \left[ (1 + \nu) \sigma_{23} \right] \]
but then clean up into the familiar forms

\[ \epsilon_{11} = {1 \over E} \left[ \sigma_{11} - \nu \; (\sigma_{22} + \sigma_{33}) \right] \qquad \qquad \epsilon_{12} = {(1 + \nu) \over E} \sigma_{12} \] \[ \epsilon_{22} = {1 \over E} \left[ \sigma_{22} - \nu \; (\sigma_{11} + \sigma_{33}) \right] \qquad \qquad \epsilon_{13} = {(1 + \nu) \over E} \sigma_{13} \] \[ \epsilon_{33} = {1 \over E} \left[ \sigma_{33} - \nu \; (\sigma_{11} + \sigma_{22}) \right] \qquad \qquad \epsilon_{23} = {(1 + \nu) \over E} \sigma_{23} \]

---

Inverting Hooke's Law

This section will simply give the results of inverting Hooke's Law to obtain the stress as a function of strain. Detailed steps of the inversion can be found at http://www.continuummechanics.org/hookeslaw.html#inverting.

The inversion process involves manipulation of the equation in tensor notation format. A complete review of tensor notation can be found on the following pages: http://www.continuummechanics.org/tensornotationbasic.html and http://www.continuummechanics.org/tensornotationadvanced.html.

Recall Hooke's Law in matrix and tensor notation...

\[ \boldsymbol{\epsilon} = {1 \over E} \left[ (1 + \nu) \boldsymbol{\sigma} - \nu \; {\bf I} \; \text{tr}(\boldsymbol{\sigma}) \right] \qquad \text{and} \qquad \epsilon_{ij} = {1 \over E} \left[ (1 + \nu) \sigma_{ij} - \nu \; \delta_{ij} \; \sigma_{kk} \right] \]
The inverted forms are:

\[ \boldsymbol{\sigma} = {E \over (1 + \nu)} \left[ \boldsymbol{\epsilon} + { \nu \over (1 - 2 \nu)} \; {\bf I} \; \text{tr}(\boldsymbol{\epsilon}) \right] \qquad \text{and} \qquad \sigma_{ij} = {E \over (1 + \nu)} \left[ \epsilon_{ij} + { \nu \over (1 - 2 \nu)} \; \delta_{ij} \; \epsilon_{kk} \right] \]
Note that in this form, it is impossible to calculate the stress tensor given a strain tensor if the material is incompressible because \(\nu = 0.5\).

---

Stiffness Tensor

The stiffness tensor is a 4th rank quantity (3x3x3x3) that relates stress and strain.

\[ \boldsymbol{\sigma} = {\bf C} : \boldsymbol{\epsilon} \qquad \qquad \sigma_{ij} = C_{ijkl} \epsilon_{kl} \]
It can be determined from Hooke's Law by taking the derivative of the stress tensor with respect to the strain tensor.

\[ C_{ijkl} \; = \; {\partial \sigma_{ij} \over \partial \epsilon_{kl} } \; = \; {\partial \over \partial \epsilon_{kl} } \left\{ {E \over (1 + \nu)} \left[ \epsilon_{ij} + { \nu \over (1 - 2 \nu)} \delta_{ij} \, \epsilon_{mm} \right] \right\} \\ \]
The result, after quite a bit of work, is

\[ C_{ijkl} = {E \over (1 + \nu)} \left[ {1 \over 2} ( \delta_{ik} \, \delta_{jl} + \delta_{jk} \, \delta_{il} ) + { \nu \over (1 - 2 \nu)} \delta_{ij} \, \delta_{kl} \right] \]
Note how the stiffness tensor is symmetric with respect to \(i\) and \(j\) and with respect to \(k\) and \(l\).

Details of the derivation can be found on this page.
In general, \({\bf C}\) can relate every strain component to every stress component. The practical challenge is that one cannot write a 4th rank 3x3x3x3 tensor on 2-D paper. However, this can be overcome by using so-called Voigt notation. This amounts to writing

\[ \left\{ \matrix{ \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \tau_{12} \\ \tau_{23} \\ \tau_{13} } \right\} = \left[ \matrix{ C_{11} & C_{12} & C_{13} & C_{14} & C_{15} & C_{16} \\ & C_{22} & C_{23} & C_{24} & C_{25} & C_{26} \\ & & C_{33} & C_{34} & C_{35} & C_{36} \\ & & & C_{44} & C_{45} & C_{46} \\ & sym & & & C_{55} & C_{56} \\ & & & & & C_{66} } \right] \left\{ \matrix{ \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \gamma_{12} \\ \gamma_{23} \\ \gamma_{13} } \right\} \]
Note that this is not conventional matrix forms that tensor notation can be applied to. Second, the full shear strain values, \(\gamma_{ij}\), are used here, not the half values, \(\epsilon_{ij}\). Finally, the order that shear stresses and strains are listed is not universal. Here they are listed as \(\gamma_{12}, \gamma_{23}, \gamma_{13}\), but other sources may also list them as \(\gamma_{12}, \gamma_{13}, \gamma_{23}\).

---

Plane Stress

The state of plane stress is a loading condition found in thin sheets and plates. In such cases, the stresses (which come from loads) are high in the plane of the part, and negligible normal to it. The normal direction is selected as \(z\), leaving the \(x\) and \(y\) directions to be the loaded ones.

This situation is represented in Hooke's Law by simply setting \(\sigma_{zz} = 0\). Doing so gives

\[ \begin{eqnarray} \epsilon_{xx} & = & {1 \over E} \left[ \sigma_{xx} - \nu \; \sigma_{yy} \right] \\ \\ \epsilon_{yy} & = & {1 \over E} \left[ \sigma_{yy} - \nu \; \sigma_{xx} \right] \\ \\ \epsilon_{zz} & = & {-\nu \over E} \left[ \sigma_{xx} + \sigma_{yy} \right] \\ \\ \gamma_{xy} & = & \tau_{xy} \; / \; G \end{eqnarray} \]
with all other terms equal zero.

Note that \(\epsilon_{zz}\) is not zero even tho \(\sigma_{zz}\) is. This is correct, if not intuitive.

Keep in mind that there is no new physics here. It's simply a special case of Hooke's Law when \(\sigma_{zz} = 0\). Furthermore, the shear term is no different than in the general case. The interesting relationships are found in the equations for normal stress and strain.

Inverting the equations gives stress in terms of strain.

\[ \begin{eqnarray} \sigma_{xx} & = & { E \over 1 - \nu^2 } \left[ \epsilon_{xx} + \nu \, \epsilon_{yy} \right] \\ \\ \sigma_{yy} & = & { E \over 1 - \nu^2 } \left[ \epsilon_{yy} + \nu \, \epsilon_{xx} \right] \\ \\ \tau_{xy} & = & G \; \gamma_{xy} \end{eqnarray} \]
There is no need to include \(\epsilon_{zz}\) in the equations, even though it is not zero.

---

Plane Strain

The plane strain condition occurs in very thick plates. In this case, \(\epsilon_{zz}\) is negligible and assumed to be zero. And \(\sigma_{zz}\) will not be zero. Taking the equation for \(\epsilon_{zz}\) and setting it equal to zero gives

\[ \epsilon_{zz} = {1 \over E} \left[ \sigma_{zz} - \nu \, ( \sigma_{xx} + \sigma_{yy} ) \right] = 0 \]
Solving for \(\sigma_{zz}\) gives

\[ \sigma_{zz} = \nu \, ( \sigma_{xx} + \sigma_{yy} ) \]