3.2 Linear elastic material behavior

 

Linear elastic stress-strain laws are used to describe the behavior of materials that experience small, reversible strains when subjected to modest stress. You are probably familiar with the behavior of a linear elastic material from introductory materials courses. Their main features are reviewed briefly below.

 

 

 

3.2.1 Isotropic, linear elastic material behavior

If you conduct a uniaxial tensile test on almost any material, and keep the stress levels sufficiently low, you will observe the following behavior:

 

· The specimen deforms reversibly:  if you remove the loads, the solid returns to its original shape.

 

· The strain in the specimen depends only on the stress applied to it $–$ it doesn’t depend on the rate of loading, or the history of loading.

 

· For most materials, the stress is a linear function of strain, as shown in the figure. Because the strains are small, this is true whatever stress measure is adopted (Cauchy stress or nominal stress), and is true whatever strain measure is adopted (Lagrange strain or infinitesimal strain).

 

· For most, but not all, materials, the material has no characteristic orientation.  Thus, if you cut a tensile specimen out of a block of material, as shown in the figure, the stress$—$strain curve will be independent of the orientation of the specimen relative to the block of material.  Such materials are said to be isotropic.

 

· If you heat a specimen of the material, increasing its temperature uniformly, it will generally change its shape slightly.  If the material is isotropic (no preferred material orientation) and homogeneous, then the specimen will simply increase in size, without shape change.

 

 

 

3.2.2 Stress$—$strain relations for isotropic, linear elastic materials. Young’s Modulus, Poissons ratio and the Thermal Expansion Coefficient.

 

Before writing down stress$—$strain relations, we need to decide what strain and stress measures we want to use.  Because the model only works for small shape changes

 

· Deformation is characterized using the infinitesimal strain tensor ${\epsilon }_{ij}=\left(\partial u_i/\partial x_j+\partial u_j/\partial x_i\right)/2$ defined in Section 2.1.7.  This is convenient for calculations, but has the disadvantage that linear elastic constitutive equations can only be used if the solid experiences small rotations, as well as small shape changes. 

 

· All stress measures are taken to be equal.  We can use the Cauchy stress ${\sigma }_{ij}$ as the stress measure.

 

You probably already know the stress$—$strain relations for an isotropic, linear elastic solid.  They are repeated below for convenience.

 

 

Here, E and $\nu$ are Young’s modulus and Poisson’s ratio, $\alpha$ is the coefficient of thermal expansion, and $\Delta T$ is the increase in temperature of the solid.  The remaining relations can be deduced from the fact that both ${\sigma }_{ij}$ and ${\epsilon }_{ij}$ are symmetric. 

 

The inverse relationship can be expressed as

 

HEALTH WARNING: Note the factor of 2 in the strain vector.  Most texts, and most finite element codes use this factor of two, but not all.  In addition, shear strains and stresses are often listed in a different order in the strain and stress vectors.  For isotropic materials this makes no difference, but you need to be careful when listing material constants for anisotropic materials.

 

We can write this expression in a much more convenient form using index notation.  Verify for yourself that the matrix expression above is equivalent to

${\epsilon }_{ij}=\frac{1+\nu }{E}{\sigma }_{ij}-\frac{\nu }{E}{\sigma }_{kk}{\delta }_{ij}+\alpha \Delta T{\delta }_{ij}$

 

The inverse relation is

${\sigma }_{ij}=\frac{E}{1+\nu }\left\{{\epsilon }_{ij}+\frac{\nu }{1-2\nu }{\epsilon }_{kk}{\delta }_{ij}\right\}-\frac{E\alpha \Delta T}{1-2\nu }{\delta }_{ij}$

 

The stress-strain relations are often expressed using the elastic modulus tensor $C_{ijkl}$ or the elastic compliance tensor $S_{ijkl}$ as

${\sigma }_{ij}=C_{ijkl}\left({\epsilon }_{kl}-\alpha \Delta T{\delta }_{kl}\right){\epsilon }_{ij}=S_{ijkl}{\sigma }_{kl}+\alpha \Delta T{\delta }_{ij}$

In terms of elastic constants, $C_{ijkl}$ and $S_{ijkl}$ are

$\begin{array}{l}{C}_{ijkl}=\frac{E}{2\left(1+\nu \right)}\left({\delta }_{il}{\delta }_{jk}+{\delta }_{ik}{\delta }_{jl}\right)+\frac{E\nu }{\left(1+\nu \right)\left(1-2\nu \right)}{\delta }_{ij}{\delta }_{kl}\\ S_{ijkl}=\frac{1+\nu }{2E}\left({\delta }_{il}{\delta }_{jk}+{\delta }_{ik}{\delta }_{jl}\right)-\frac{\nu }{E}{\delta }_{ij}{\delta }_{kl}\end{array}$

 

 

 

3.2.3 Reduced stress-strain equations for plane deformation of isotropic solids

 

For plane strain or plane stress deformations, some strain or stress components are always zero (by definition) so the stress-strain laws can be simplified. 

 

· For a plane strain deformation ${\epsilon }_{33}={\epsilon }_{23}={\epsilon }_{13}=0$.  The stress strain laws are therefore

$\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right]=\frac{(1+\nu )}{E}\left[\begin{array}{ccc}1-\nu & -\nu & 0\\ -\nu & 1-\nu & 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right]+\left(1+\nu \right)\alpha \Delta T\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$

$\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right]=\frac{E}{(1+\nu )(1-2\nu )}\left[\begin{array}{ccc}1-\nu & \nu & 0\\ \nu & 1-\nu & 0\\ 0& 0& \frac{1-2\nu }{2}\end{array}\right]\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right]-\frac{E\alpha \Delta T}{1-2\nu }\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$

${\sigma }_{33}=\frac{E\nu \left({\epsilon }_{11}+{\epsilon }_{22}\right)}{\left(1-2\nu \right)\left(1+\nu \right)}+\frac{E\alpha \Delta T}{1-2\nu },{\sigma }_{13}={\sigma }_{23}=0$

In index notation

$\begin{array}{l}{\epsilon }_{\alpha \beta }=\frac{1+\nu }{E}\left\{{\sigma }_{\alpha \beta }-\nu {\sigma }_{\gamma \gamma }{\delta }_{\alpha \beta }\right\}+\left(1+\nu \right)\alpha \Delta T{\delta }_{\alpha \beta }\\ {\sigma }_{\alpha \beta }=\frac{E}{1+\nu }\left\{{\epsilon }_{\alpha \beta }+\frac{\nu }{1-2\nu }{\epsilon }_{\gamma \gamma }{\delta }_{\alpha \beta }\right\}-\frac{E\alpha \Delta T}{1-2\nu }{\delta }_{\alpha \beta }\end{array}$

where Greek subscripts $\alpha ,\beta$ can have values 1 or 2.

 

· For a plane stress deformation ${\sigma }_{33}={\sigma }_{23}={\sigma }_{13}=0$, and the stres-strain relations are

$\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right]=\frac{1}{E}\left[\begin{array}{ccc}1& -\nu & 0\\ -\nu & 1& 0\\ 0& 0& 2(1+\nu )\end{array}\right]\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right]+\alpha \Delta T\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$

$\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right]=\frac{E}{(1-{\nu }^2)}\left[\begin{array}{ccc}1& \nu & 0\\ \nu & 1& 0\\ 0& 0& (1-\nu )/2\end{array}\right]\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right]-\frac{E\alpha \Delta T}{\left(1-\nu \right)}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$

${\epsilon }_{33}=-\frac{\nu }{E}\left({\sigma }_{11}+{\sigma }_{22}\right)+\alpha \Delta T$

$\begin{array}{l}{\epsilon }_{\alpha \beta }=\frac{1+\nu }{E}\left({\sigma }_{\alpha \beta }-\frac{\nu }{1+\nu }{\sigma }_{\gamma \gamma }{\delta }_{\alpha \beta }\right)+\alpha \Delta T{\delta }_{\alpha \beta }\\ {\sigma }_{\alpha \beta }=\frac{E}{1+\nu }\left\{{\epsilon }_{\alpha \beta }+\frac{\nu }{1-\nu }{\epsilon }_{\gamma \gamma }{\delta }_{\alpha \beta }\right\}-\frac{E\alpha \Delta T}{1-\nu }{\delta }_{\alpha \beta }\end{array}$

 

 

 

3.2.4 Representative values for density, and elastic constants of isotropic solids

 

The table below shows representative elastic constants for a range of different materials. The data are partly from Jones and Ashby (2019), and partly from manufacturers data sheets.

 

Note the units $–$ values of E are given in $GN/m^2$; the G stands for Giga, and is short for ${10}^9$.  The units for density are in $Mg{m}^{-3}$ - that’s Mega grams.  One mega gram is 1000 kg.

 

 

 

 

3.2.5 Other Elastic Constants $–$ bulk, shear and Lame modulus.

 

Young’s modulus and Poisson’s ratio are the most common properties used to characterize elastic solids, but other measures are also used.  For example, we define the shear modulus,  bulk modulus and Lame modulus of an elastic solid as follows:

$\begin{array}{l}\text{Bulk Modulus }K=\frac{E}{3\left(1-2\nu \right)}\\ \text{Shear Modulus }\mu \text{=}\frac{E}{2\left(1+\nu \right)}\\ \text{Lame Modulus }\lambda =\frac{\nu E}{\left(1+\nu \right)\left(1-2\nu \right)}\end{array}$

The table below relates all the possible combinations of moduli to all other possible combinations.  Enjoy!

 

 

 

 

3.2.6 Physical interpretation of elastic constants for isotropic solids

 

It is important to have a feel for the physical significance of the various elastic constants.  They can be interpreted as follows

 

· Young’s modulus E is the slope of the stress$—$strain curve in uniaxial tension.  It has dimensions of stress ( $N/m^2$ ) and is usually large $–$ for steel, $E=210\times {10}^9{\text{N/m}}^{\text{2}}$. You can think of E as a measure of the stiffness of the solid. The larger the value of E, the stiffer the solid.  For a stable material, the Young’s modulus must satisfy E>0.

 

· Poisson’s ratio $\nu$ is the ratio of lateral to longitudinal strain in uniaxial tensile stress. It is dimensionless and typically ranges from 0.2$—$0.49, and is around 0.3 for most metals.  For a stable material, Poisson’s ratio is in the range $-1<\nu <0.5$. It is a measure of the compressibility of the solid.  If $\nu =0.5$, the solid is incompressible $–$ its volume remains constant, no matter how it is deformed.  If $\nu =0$, then stretching a specimen causes no lateral contraction.  Some bizarre materials have $\nu <0$ --  if you stretch a round bar of such a material, the bar increases in diameter!!

 

· Thermal expansion coefficient quantifies the change in volume of a material if it is heated in the absence of stress.  It has dimensions of (degrees Kelvin)^-1 and is usually very small.  For steel, $\alpha \approx 6-10\times {10}^{-6}{\text{K}}^{\text{-1}}$

 

· The bulk modulus quantifies the resistance of the solid to volume changes.  It has a large value (usually bigger than E).

 

· The shear modulus quantifies its resistance to volume preserving shear deformations.  Its value is usually somewhat smaller than E. 

 

 

 

3.2.7 Strain Energy Density for Isotropic Solids

 

The following observations are the basis for defining the strain energy density of an elastic material

 

· If you deform a block of material, you do work on it (or, in some cases, it may do work on you…)

 

· In an elastic material, the work done during loading is stored as recoverable strain energy in the solid.  If you unload the material, the specimen does work on you, and when it reaches its initial configuration you come out even.

 

· The work done to deform a specimen depends only on the state of strain at the end of the test.  It is independent of the history of loading. 

 

Based on these observations, we define the strain energy density of a solid as the work done per unit volume to deform a material from a stress free reference state to a loaded state.

 

To write down an expression for the strain energy density, it is convenient to separate the strain into two parts

${\epsilon }_{ij}={\epsilon }_{ij}^e+{\epsilon }_{ij}^T$

where, for an isotropic solid,

${\epsilon }_{ij}^T=\alpha \Delta T{\delta }_{ij}$

represents the strain due to thermal expansion (known as thermal strain), and

${\epsilon }_{ij}^e=\frac{1+\nu }{E}{\sigma }_{ij}-\frac{\nu }{E}{\sigma }_{kk}{\delta }_{ij}$

is the strain due to mechanical loading (known as elastic strain).

 

Work is done on the specimen only during mechanical loading.  It is straightforward to show that the strain energy density is

$U=\frac{1}{2}{\sigma }_{ij}{\epsilon }_{ij}^e$

You can also re-write this as

$\begin{array}{l}U=\frac{1+\nu }{2E}{\sigma }_{ij}{\sigma }_{ij}-\frac{\nu }{2E}{\sigma }_{kk}{\sigma }_{jj}\\ U=\frac{E}{2\left(1+\nu \right)}{\epsilon }_{ij}^e{\epsilon }_{ij}^e+\frac{E\nu }{2\left(1+\nu \right)\left(1-2\nu \right)}{\epsilon }_{jj}^e{\epsilon }_{kk}^e\end{array}$

Observe that

${\epsilon }_{ij}^e=\frac{\partial U}{\partial {\sigma }_{ij}}{\sigma }_{ij}=\frac{\partial U}{\partial {\epsilon }_{ij}^e}$

 

 

 

3.2.8 Stress-strain relation for a general anisotropic linear elastic material $–$ the elastic stiffness and compliance tensors

 

The simple isotropic model described in the preceding section is unable to describe the response of some materials accurately, even though the material may deform elastically.  This is because some materials do have a characteristic orientation.  For example, in a block of wood, the grain is oriented in a particular direction in the specimen.  The block will be stiffer if it is loaded parallel to the grain than if it is loaded perpendicular to the grain.  The same observation applies to fiber reinforced composite materials. Generally, single crystal specimens of a material will also be anisotropic $–$ this is important when modeling stress effects in small structures such as microelectronic circuits. Even polycrystalline metals may be anisotropic, because a preferred texture may form in the specimen during processing.

 

A more general stress$—$strain relation is needed to describe anisotropic solids. 

 

The most general linear stress$—$strain relation has the form

${\sigma }_{ij}=C_{ijkl}\left({\epsilon }_{kl}-{\alpha }_{kl}\Delta T\right)$

Here, $C_{ijkl}$ is a fourth order tensor (horrors!), known as the elastic stiffness tensor, and ${\alpha }_{kl}={\alpha }_{lk}$ is the thermal expansion coefficient tensor. The stress strain relation is invertible:

${\epsilon }_{ij}=S_{ijkl}{\sigma }_{kl}+{\alpha }_{ij}\Delta T$

where $S_{ijkl}$ is known as the elastic compliance tensor

 

At first sight it appears that the stiffness tensor has 81 components.  Imagine having to measure and keep track of 81 material properties!  Fortunately, $C_{ijkl}$ must have the following symmetries

$C_{ijkl}=C_{klij}=C_{jikl}=C_{ijlk}$

This reduces the number of material constants to 21. The compliance tensor has the same symmetries as $C_{ijkl}$.

 

To see the origin of the symmetries of $C_{ijkl}$, note that

 

· The stress tensor is symmetric, which is only possible if $C_{ijkl}=C_{jikl}.$

 

· If a strain energy density exists for the material, the elastic stiffness tensor must satisfy $C_{ijkl}=C_{klij}$

 

· The previous two symmetries imply $C_{ijkl}=C_{ijlk}$, since $C_{ijkl}=C_{jikl}=C_{klji}$ and . $C_{ijkl}=C_{lkij}$

 

 

To see that $C_{ijkl}=C_{klij}$, note that by definition

$C_{ijkl}=\frac{\partial {\sigma }_{ij}}{\partial {\epsilon }_{kl}}$

and recall further that the stress is the derivative of the strain energy density with respect to strain

${\sigma }_{ij}=\frac{\partial U}{\partial {\epsilon }_{ij}}$

Combining these,

$C_{ijkl}=\frac{{\partial }^{2}U}{\partial {\epsilon }_{ij}\partial {\epsilon }_{kl}}$

Now, note that

$\frac{{\partial }^{2}U}{\partial {\epsilon }_{ij}\partial {\epsilon }_{kl}}=\frac{{\partial }^{2}U}{\partial {\epsilon }_{kl}\partial {\epsilon }_{ij}}$

so that

$C_{ijkl}=C_{klij}$

These symmetries allow us to write the stress-strain relations in a more compact matrix form as

$\begin{array}{c}\sigma =C(\epsilon -\alpha \Delta T)\\ \sigma =\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{33}\\ {\sigma }_{23}\\ {\sigma }_{13}\\ {\sigma }_{12}\end{array}\right]\epsilon =\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ {\epsilon }_{33}\\ 2{\epsilon }_{23}\\ 2{\epsilon }_{13}\\ 2{\epsilon }_{12}\end{array}\right]\alpha =\left[\begin{array}{c}{\alpha }_{11}\\ {\alpha }_{22}\\ {\alpha }_{33}\\ 2{\alpha }_{23}\\ 2{\alpha }_{13}\\ 2{\alpha }_{12}\end{array}\right]\\ C=\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& c_{14}& c_{15}& c_{16}\\ c_{12}& c_{22}& c_{23}& c_{24}& c_{25}& c_{26}\\ c_{13}& c_{23}& c_{33}& c_{34}& c_{35}& c_{36}\\ c_{14}& c_{24}& c_{34}& c_{44}& c_{45}& c_{46}\\ c_{15}& c_{25}& c_{35}& c_{45}& c_{55}& c_{56}\\ c_{16}& c_{26}& c_{36}& c_{46}& c_{56}& c_{66}\end{array}\right]\end{array}$

where $c_{11}\equiv C_{1111}{c}_{12}\equiv C_{1122}=C_{2211}$, etc are the elastic stiffnesses of the material.  The inverse has the form

$\begin{array}{c}\epsilon =S\sigma +\alpha \Delta T\\ S=\left[\begin{array}{cccccc}{s}_{11}& s_{12}& s_{13}& s_{14}& s_{15}& s_{16}\\ s_{12}& s_{22}& s_{23}& s_{24}& s_{25}& s_{26}\\ s_{13}& s_{23}& s_{33}& s_{34}& s_{35}& s_{36}\\ s_{14}& s_{24}& s_{34}& s_{44}& s_{45}& s_{46}\\ s_{15}& s_{25}& s_{35}& s_{45}& s_{55}& s_{56}\\ s_{16}& s_{26}& s_{36}& s_{46}& s_{56}& s_{66}\end{array}\right]\end{array}$

where $s_{11}=S_{1111},s_{12}=S_{1122}$, etc are the elastic compliances of the material.

 

To satisfy Drucker stability, the eigenvalues of the elastic stiffness and compliance matrices must all be greater than zero.

 

HEALTH WARNING: The shear strain and shear stress components are not always listed in the order given when defining the elastic and compliance matrices.  The conventions used here are common and are particularly convenient in analytical calculations involving anisotropic solids.  But many sources use other conventions.  Be careful to enter material data in the correct order when specifying properties for anisotropic solids.

 

 

 

3.2.9 Physical Interpretation of the Anisotropic Elastic Constants.

 

It is easiest to interpret $s_{11},s_{12}\mathrm{....}{s}_{66}$, rather than $c_{11},c_{`12},\mathrm{...}{c}_{66}$.  Imagine applying a uniaxial stress, say ${\sigma }_{11}$, to an anisotropic specimen.  In general, this would induce both extensional and shear deformation in the solid, as shown in the figure.

 

The strain induced by  the uniaxial stress would be

$\begin{array}{l}{\epsilon }_{11}=s_{11}{\sigma }_{11},{\epsilon }_{22}=s_{12}{\sigma }_{11},{\epsilon }_{33}=s_{13}{\sigma }_{11}\\ {\epsilon }_{23}=s_{14}{\sigma }_{11},{\epsilon }_{13}=s_{15}{\sigma }_{11},{\epsilon }_{12}=s_{16}{\sigma }_{11}\end{array}$

All the constants have dimensions $m^2/N$.  The constant $s_{11}$ looks like a uniaxial compliance, (like $1/E$ ), while the ratios $s_{12}/s_{11},s_{13}/s_{11}$ are generalized versions of Poisson’s ratio: they quantify the lateral contraction of a uniaxial tensile specimen.   The shear terms are new $–$ in an isotropic material, no shear strain is induced by uniaxial tension.

 

 

 

3.2.10 Strain energy density for anisotropic, linear elastic solids

 

The strain energy density of an anisotropic material is

$\begin{array}{l}U=\frac{1}{2}{C}_{ijkl}\left({\epsilon }_{ij}-{\alpha }_{ij}\Delta T\right)\left({\epsilon }_{kl}-{\alpha }_{kl}\Delta T\right)\\ =\frac{1}{2}{S}_{ijkl}{\sigma }_{ij}{\sigma }_{kl}\end{array}$

 

 

 

3.2.11 Basis change formulas for anisotropic elastic constants

 

The material constants $c_{ij}$ or $s_{ij}$ for a particular material are usually specified in a basis with coordinate axes aligned with particular symmetry planes (if any) in the material.  When solving problems involving anisotropic materials it is frequently necessary to transform these values to a coordinate system that is oriented in some convenient way relative to the boundaries of the solid.  Since $C_{ijkl}$ is a fourth rank tensor, the basis change formulas are highly tedious, unfortunately. 

 

Suppose that the components of the stiffness tensor are given in a basis $\left\{e_1,e_2,e_3\right\}$, and we wish to determine its components in a second basis, $\left\{m_1,m_2,m_3\right\}$, as shown in the figure. We define the usual transformation tensor ${\Omega }_{ij}$ with components ${\Omega }_{ij}=m_i\cdot e_j$, or in matrix form

$\left[\Omega \right]=\left[\begin{array}{l}{m}_1\cdot e_1{m}_1\cdot e_2{m}_1\cdot e_3\\ m_2\cdot e_1{m}_2\cdot e_2{m}_2\cdot e_3\\ m_3\cdot e_1{m}_3\cdot e_2{m}_3\cdot e_3\end{array}\right]$

This is an orthogonal matrix satisfying $\Omega {\Omega }^T={\Omega }^T\Omega =I$. In practice, the matrix can be computed in terms of the angles between the basis vectors. It is straightforward to show that stress, strain, thermal expansion and elasticity tensors transform as

$\begin{array}{l}{\sigma }_{ij}^{(m)}={\Omega }_{ik}{\sigma }_{kl}^{(e)}{\Omega }_{jl}{\epsilon }_{ij}^{(m)}={\Omega }_{ik}{\epsilon }_{kl}^{(e)}{\Omega }_{jl}{\alpha }_{ij}^{(m)}={\Omega }_{ik}{\alpha }_{kl}^{(e)}{\Omega }_{jl}\\ C_{ijkl}^{(m)}={\Omega }_{ip}{\Omega }_{jq}{C}_{pqrs}^{(e)}{\Omega }_{kr}{\Omega }_{ls}\end{array}$

 

The basis change formula for the elasticity tensor in matrix form can be expressed as

$C^{(m)}=K{C}^{(e)}{K}^T$

where the basis change matrix K is computed as

$\begin{array}{l}K=\left[\begin{array}{cc}{K}^{(1)}& 2K^{(2)}\\ K^{(3)}& K^{(4)}\end{array}\right]\\ \left.\begin{array}{l}{K}_{ij}^{(1)}={\Omega }_{ij}^2{K}_{ij}^{(2)}={\Omega }_{i\mathrm{mod}(j+1,3)}{\Omega }_{i\mathrm{mod}(j+2,3)}\\ K_{ij}^{(3)}={\Omega }_{\mathrm{mod}(i+1,3)j}{\Omega }_{\mathrm{mod}(i+2,3)j}\\ K_{ij}^{(4)}={\Omega }_{\mathrm{mod}(i+1,3)\mathrm{mod}(j+1,3)}{\Omega }_{\mathrm{mod}(i+2,3)\mathrm{mod}(j+2,3)}\\ +{\Omega }_{\mathrm{mod}(i+1,3)\mathrm{mod}(j+2,3)}{\Omega }_{\mathrm{mod}(i+2,3)\mathrm{mod}(j+1,3)}\end{array}\right\}i,j=\mathrm{1..3}\end{array}$

and the modulo function satisfies

$\mathrm{mod}(i,3)=\left\{\begin{array}{c}ii\le 3\\ i-3i>3\end{array}\right.$

Although these expressions look cumbersome they are quite easy to code in a computer program.

 

The basis change for the compliance tensor follows as

$S^{(m)}=K^{-T}{S}^{(e)}{K}^{-1}$

where

$K^{-T}=\left[\begin{array}{cc}{K}^{(1)}& K^{(2)}\\ 2K^{(3)}& K^{(4)}\end{array}\right]$

 

The proof of these expressions is merely tiresome algebra and will not be given here.  Ting (1996) has a nice clear discussion.

 

For the particular case of rotation through an angle $\theta$ with right hand screw convention about the $e_1,e_2,e_3$ axes, respectively, the rotation matrix for the elasticity tensor reduces to

 

Rotation about $e_1$

 

$\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& c^2& s^2& 2cs& 0& 0\\ 0& s^2& c^2& -2cs& 0& 0\\ 0& -cs& cs& c^2-s^2& 0& 0\\ 0& 0& 0& 0& c& -s\\ 0& 0& 0& 0& s& c\end{array}\right]$

Rotation about $e_2$

 

   $\left[\begin{array}{cccccc}{c}^2& 0& s^2& 0& -2cs& 0\\ 0& 1& 0& 0& 0& 0\\ s^2& 0& c^2& 0& 2cs& 0\\ 0& 0& 0& c& 0& s\\ cs& 0& -cs& 0& c^2-s^2& 0\\ 0& 0& 0& -s& 0& c\end{array}\right]$

Rotation about $e_3$

 

     $\left[\begin{array}{cccccc}{c}^2& s^2& 0& 0& 0& 2cs\\ s^2& c^2& 0& 0& 0& -2cs\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& c& s& 0\\ 0& 0& 0& -s& c& 0\\ -cs& cs& 0& 0& 0& c^2-s^2\end{array}\right]$

where $c=\cos \theta s=\sin \theta$. The inverse matrix $K^{-1}$ can be obtained simply by changing the sign of the angle $\theta$ in each rotation matrix.  Applying the three rotations successively can produce an arbitrary orientation change.

 

For an isotropic material, the elastic stress-strain relations, the elasticity matrices and thermal expansion coefficient are unaffected by basis changes.

 

 

 

3.2.12 The effect of material symmetry on stress-strain relations for anisotropic materials

 

A general anisotropic solid has 21 independent elastic constants. Note that, in general, tensile stress may induce shear strain, and shear stress may cause extension.

 

 

If a material has a symmetry plane, then applying stress normal or parallel to this plane, as shown in the figure,  induces only extension in direction normal and parallel to the plane.

 

For example, suppose the material contains a single symmetry plane, and let $e_1$  be normal to this plane. Then the components of the elastic stiffnes matrix

$c_{15}=c_{16}=c_{25}=c_{26}=c_{35}=c_{36}=0$

Or equivalently

$C_{1112}=C_{1113}=C_{2212}=C_{2213}=C_{3312}=C_{3313}=0$

 (symmetrical terms also vanish, of course). This leaves 13 independent constants.

 

Similar restrictions on the thermal expansion coefficient can be determined using symmetry conditions.  Details are left as an exercise.

 

In the following sections, we list the stress-strain relations for anisotropic materials with various numbers of symmetry planes.

 

 

 

3.2.13 Stress-strain relations for linear elastic orthotropic materials

 

An orthotropic material has three mutually perpendicular symmetry planes. This type of material has 9 independent material constants.  With basis vectors perpendicular to the symmetry planes, as shown in the figure,  the elastic stiffness matrix has the form

$C=\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& 0& 0& 0\\ & c_{22}& c_{23}& 0& 0& 0\\ & & c_{33}& 0& 0& 0\\ & sym& & c_{44}& 0& 0\\ & & & 0& c_{55}& 0\\ & & & 0& 0& c_{66}\end{array}\right]$

This relationship is sometimes expressed in inverse form, in terms of generalized Young’s moduli and Poisson’s ratios (which have the same significance as Young’s modulus and Poisson’s ratio for uniaxial loading along the three basis vectors) as follows                                                                    

$S=\left[\begin{array}{cccccc}1/E_1& -{\nu }_{21}/E_2& -{\nu }_{31}/E_3& 0& 0& 0\\ -{\nu }_{12}/E_1& 1/E_2& -{\nu }_{32}/E_3& 0& 0& 0\\ -{\nu }_{13}/E_1& -{\nu }_{23}/E_2& 1/E{}_3& 0& 0& 0\\ 0& 0& 0& 1/{\mu }_{23}& 0& 0\\ 0& 0& 0& 0& 1/{\mu }_{13}& 0\\ 0& 0& 0& 0& 0& 1/{\mu }_{12}\end{array}\right]$

Here the generalized Poisson’s ratios are not symmetric but instead satisfy ${\nu }_{ij}/E_i={\nu }_{ji}/E_j$ (no sums). This ensures that the stiffness matrix is symmetric.

 

The engineering constants are related to the components of the compliance tensor by

$\begin{array}{l}{c}_{11}=E_1\left(1-{\nu }_{23}{\nu }_{32}\right)\Upsilon c_{22}=E_2\left(1-{\nu }_{13}{\nu }_{31}\right)\Upsilon c_{33}=E_3\left(1-{\nu }_{12}{\nu }_{21}\right)\Upsilon \\ c_{12}=E_1\left({\nu }_{21}+{\nu }_{31}{\nu }_{23}\right)\Upsilon =E_2\left({\nu }_{12}+{\nu }_{32}{\nu }_{13}\right)\Upsilon \\ c_{13}=E_1\left({\nu }_{31}+{\nu }_{21}{\nu }_{32}\right)\Upsilon =E_3\left({\nu }_{13}+{\nu }_{12}{\nu }_{23}\right)\Upsilon \\ c_{23}=E_2\left({\nu }_{32}+{\nu }_{12}{\nu }_{31}\right)\Upsilon =E_3\left({\nu }_{23}+{\nu }_{21}{\nu }_{13}\right)\Upsilon \\ c_{44}={\mu }_{23}{c}_{55}={\mu }_{13}{c}_{66}={\mu }_{12}\\ \Upsilon =\frac{1}{1-{\nu }_{12}{\nu }_{21}-{\nu }_{23}{\nu }_{32}-{\nu }_{31}{\nu }_{13}-2{\nu }_{21}{\nu }_{32}{\nu }_{13}}\end{array}$

or in inverse form

$\begin{array}{l}{E}_1=\left(c_{11}{c}_{22}{c}_{33}+2c_{23}{c}_{12}{c}_{13}-c_{11}{c}_{23}^2-c_{22}{c}_{13}^2-c_{33}{c}_{12}^2\right)/\left(c_{22}{c}_{33}-c_{23}^2\right)\\ E_2=\left(c_{11}{c}_{22}{c}_{33}+2c_{23}{c}_{12}{c}_{13}-c_{11}{c}_{23}^2-c_{22}{c}_{13}^2-c_{33}{c}_{12}^2\right)/\left(c_{11}{c}_{33}-c_{13}^2\right)\\ E_3=\left(c_{11}{c}_{22}{c}_{33}+2c_{23}{c}_{12}{c}_{13}-c_{11}{c}_{23}^2-c_{22}{c}_{13}^2-c_{33}{c}_{12}^2\right)/\left(c_{11}{c}_{22}-c_{12}^2\right)\\ {\nu }_{21}=(c_{12}{c}_{33}-c_{13}{c}_{23})/(c_{11}{c}_{33}-c_{13}^2)\\ {\nu }_{12}=(c_{12}{c}_{33}-c_{13}{c}_{23})/(c_{22}{c}_{33}-c_{23}^2)\\ {\nu }_{31}=(c_{13}{c}_{22}-c_{12}{c}_{23})/(c_{11}{c}_{22}-c_{12}^2)\\ {\nu }_{13}=(c_{22}{c}_{13}-c_{12}{c}_{23})/(c_{22}{c}_{33}-c_{23}^2)\\ {\nu }_{23}=(c_{11}{c}_{23}-c_{12}{c}_{13})/(c_{11}{c}_{33}-c_{13}^2)\\ {\nu }_{32}=(c_{11}{c}_{23}-c_{12}{c}_{13})/(c_{11}{c}_{22}-c_{12}^2)\\ {\mu }_{23}=c_{44},{\mu }_{13}=c_{55}{\mu }_{12}=c_{66}\end{array}$

 

For an orthotropic material thermal expansion cannot induce shear (in this basis) but the expansion in the three directions need not be equal.  Consequently the thermal expansion coefficient tensor has the form

$\left[\begin{array}{ccc}{\alpha }_1& 0& 0\\ 0& {\alpha }_2& 0\\ 0& 0& {\alpha }_3\end{array}\right]$

 

 

 

3.2.14 Stress-strain relations for linear elastic Transversely Isotropic Material

 

A special case of an orthotropic solid is one that contains a plane of isotropy (this implies that the solid can be rotated with respect to the loading direction about one axis without measurable effect on the solid’s response).  Choose $e_3$ perpendicular to this symmetry plane.  Then, transverse isotropy requires that $c_{22}=c_{11}$, $c_{23}=c_{13}$, $c_{55}=c_{44}$, $c_{66}=(c_{11}-c_{12})/2$, so that the stiffness matrix has the form

 

The engineering constants must satisfy

$\begin{array}{l}{E}_1=E_2=E_p{E}_3=E_t\\ {\nu }_{12}={\nu }_{21}={\nu }_p{\nu }_{31}={\nu }_{32}={\nu }_{tp}\\ {\nu }_{13}={\nu }_{23}={\nu }_{pt}\end{array}$

and the compliance matrix has the form

$S=\left[\begin{array}{cccccc}1/E_p& -{\nu }_p/E_p& -{\nu }_{tp}/E_t& 0& 0& 0\\ -{\nu }_p/E_p& 1/E_p& -{\nu }_{tp}/E_t& 0& 0& 0\\ -{\nu }_{pt}/E_p& -{\nu }_{pt}/E_p& 1/E_t& 0& 0& 0\\ 0& 0& 0& 1/{\mu }_t& 0& 0\\ 0& 0& 0& 0& 1/{\mu }_t& 0\\ 0& 0& 0& 0& 0& 1/{\mu }_p\end{array}\right]$

where ${\mu }_p=E_p/2(1+{\nu }_p)$.  As before the Poisson’s ratios are not symmetric, but satisfy ${\nu }_{tp}/E_t={\nu }_{pt}/E_p$

 

The engineering constants and stiffnesses are related by

$\begin{array}{l}{c}_{11}=c_{22}=E_p\left(1-{\nu }_{pt}{\nu }_{tp}\right)\Upsilon c_{33}=E_t\left(1-{\nu }_p^2\right)\Upsilon c_{12}=E_p\left({\nu }_p+{\nu }_{pt}{\nu }_{tp}\right)\Upsilon \\ c_{13}=c_{23}=E_p\left({\nu }_{tp}+{\nu }_p{\nu }_{tp}\right)\Upsilon =E_t\left({\nu }_{pt}+{\nu }_p{\nu }_{pt}\right)\Upsilon c_{44}={\mu }_t{c}_{66}={\mu }_p\\ \Upsilon =\frac{1}{1-{\nu }_p^2-2{\nu }_{pt}{\nu }_{tp}-2{\nu }_p{\nu }_{pt}{\nu }_{tp}}\end{array}$

$\begin{array}{l}{E}_p=\left(c_{11}^2{c}_{33}+2c_{13}^2{c}_{12}-2c_{11}{c}_{13}^2-c_{33}{c}_{12}^2\right)/\left(c_{11}{c}_{33}-c_{13}^2\right)\\ E_t=\left(c_{11}^2{c}_{33}+2c_{13}^2{c}_{12}-2c_{11}{c}_{13}^2-c_{33}{c}_{12}^2\right)/\left(c_{11}^2-c_{12}^2\right)\\ {\nu }_p=(c_{12}{c}_{33}-c_{13}^2)/(c_{11}{c}_{33}-c_{13}^2)\\ {\nu }_{tp}=(c_{13}{c}_{11}-c_{12}{c}_{13})/(c_{11}^2-c_{12}^2)\\ {\nu }_{pt}=(c_{11}{c}_{13}-c_{12}{c}_{13})/(c_{11}{c}_{33}-c_{13}^2)\\ {\mu }_t=c_{44}{\mu }_p=c_{66}\end{array}$

 

For this material the two thermal expansion coefficients in the symmetry plane must be equal, so the thermal expansion coefficient tensor has the form

$\left[\begin{array}{ccc}{\alpha }_1& 0& 0\\ 0& {\alpha }_1& 0\\ 0& 0& {\alpha }_3\end{array}\right]$

 

 

 

3.2.15 Representative values for elastic constants of transversely isotropic hexagonal close packed crystals

 

Hexagonal close-packed crystals are an example of transversely isotropic materials.  The $e_3$ axis must be taken to be perpendicular to the basal (0001) plane of the crystal, as shown in the figure.  Since the plane perpendicular to $e_3$ is isotropic the orientation of $e_1$ and $e_2$ is arbitrary.

 

 

A table of values of stiffnesses for a few hexagonal materials is listed in the table below

 

 

The engineering constants can be calculated and are listed in the table below

 

 

 

3.2.16 Linear elastic stress-strain relations for cubic materials

 

A huge number of materials have cubic symmetry $–$ all the FCC and BCC metals, for example.  The constitutive law for such a material is particularly simple, and can be parameterized by only 3 material constants.  Pick basis vectors perpendicular to the symmetry planes, as shown in Figure 3.12. <Figure 3.12 near here>

 

 

Then

$\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{33}\\ {\sigma }_{23}\\ {\sigma }_{13}\\ {\sigma }_{11}\end{array}\right]=\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{12}& 0& 0& 0\\ & c_{11}& c_{12}& 0& 0& 0\\ & & c_{11}& 0& 0& 0\\ & sym& & c_{44}& 0& 0\\ & & & 0& c_{44}& 0\\ & & & 0& 0& c_{44}\end{array}\right]\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ {\epsilon }_{33}\\ 2{\epsilon }_{23}\\ 2{\epsilon }_{13}\\ 2{\epsilon }_{12}\end{array}\right]$

or in terms of engineering constants

$\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ {\epsilon }_{33}\\ 2{\epsilon }_{23}\\ 2{\epsilon }_{13}\\ 2{\epsilon }_{12}\end{array}\right]=\left[\begin{array}{cccccc}1/E& -\nu /E& -\nu /E& 0& 0& 0\\ -\nu /E& 1/E& -\nu /E& 0& 0& 0\\ -\nu /E& -\nu /E& 1/E& 0& 0& 0\\ 0& 0& 0& 1/\mu & 0& 0\\ 0& 0& 0& 0& 1/\mu & 0\\ 0& 0& 0& 0& 0& 1/\mu \end{array}\right]\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{33}\\ {\sigma }_{23}\\ {\sigma }_{13}\\ {\sigma }_{12}\end{array}\right]$

This is virtually identical to the constitutive law for an isotropic solid, except that the shear modulus $\mu$ is not related to the Poisson’s ratio and Young’s modulus through the usual relation given in Section 3.1.6.   In fact, the ratio

$A=\frac{2\mu (1+\nu )}{E}=\frac{2c_{44}}{c_{11}-c_{12}}$

provides a convenient measure of anisotropy.  For $A=1$ the material is isotropic. 

 

The thermal expansion coefficient matrix must be isotropic for materials with cubic symmetry.

 

The relationships between the elastic constants are

$\begin{array}{l}E=\left(c_{11}^2+c_{12}{c}_{11}-2c_{12}^2\right)/\left(c_{11}+c_{12}\right)\nu =c_{12}/(c_{11}+c_{12})\mu =c_{44}\\ c_{11}=E(1-\nu )/(1-\nu -2{\nu }^2)c_{12}=E\nu /(1-\nu -2{\nu }^2)\end{array}$

 

 

 

3.2.17 Representative values for elastic properties of cubic crystals and compounds

 

The table below lists values of elastic constants for various cubic crystals and compounds. The elastic constants  $E,\nu$ were calculated using the formulas in the preceding section.