Overview of Elasticity of Materials/Anisotropic Response

This section covers linear anisotropic response.

At this point, we have developed our notation for describing stress and strain, and have written expressions relating the two in terms of material parameters such as $E$, $G$, and $\nu$. This is all within the assumption of a homogeneous, isotropic solid. This is often times a fairly good approximation, and you are likely to recognize the expressions derived here. Hopefully, this section will provide an insight to mechanics of solids as well as other materials-focused topics.

We will begin by looking at an example. Consider a brittle material that is hot at first and then suddenly quenched to low temperature. This can cause thermal shock and breakage. So the part is quenched from $T_{max}$ to $T_{min}$, where $\Delta T=T_{max}-T_{min}$.

It is known that, without proof, the thermal stress is

$\sigma _{TH}=-{\frac {\alpha \Delta TE}{(1-2\nu )}}$           [1]

Where $\alpha$ is the coefficient of thermal expansion.

Where is this from? We know that this is the hydrostatic stress:

$\sigma _{ii}={\frac {E}{1-2\nu }}\varepsilon _{kk}$

Which can be rewritten as

$\sigma _{11}+\sigma _{22}+\sigma _{33}={\frac {E}{1-2\nu }}(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})$

Where

$\sigma _{m}={\frac {1}{3}}\sigma _{ii}$

Then it must be approximated that

${\frac {1}{3}}\Delta =\alpha \Delta T$

For spherical, circular flaws, or cracks with radius $r$, the energy is

$U_{TOT}=U_{o}-U_{STRAIN}+U_{SURF}$           [2]

Where $U_{TOT}$ is the total energy of the system and $U_{o}$ is the energy of the stress-free and crack-free system with volume $V_{o}$.

The strain energy is

$U_{STRAIN}={\frac {V_{o}\sigma _{TH}^{2}}{2E}}-{\frac {N\sigma _{TH}^{2}}{2E}}{\frac {4}{3}}\pi r^{3}$           [3]

Where $N$ is the number of cracks.

Where is this from? Our expression for strain energy when normal stress is applied is

$U_{o}^{NORMAL}={\frac {1}{2}}{\frac {\sigma _{11}^{2}}{E}}$

Where $U_{o}^{NORMAL}$ is the energy per unit volume.

Therefore, the first term in Equation 3 is simply the energy due to the thermal strain. The second term is the strain relieved when $N$ number of cracks of volume ${\frac {4}{3}}\pi r^{3}$ open (we are subtracting this term because the strain is being relieved):

$N\left({\frac {4}{3}}\pi r^{3}\right){\frac {\sigma _{TH}^{2}}{2E}}$

Is this reasonable? This is probably slightly underestimated. Finally, the last term is

$U_{SURF}=NG_{c}\pi r^{2}$           [4]

Where $G_{c}$ is the toughness energy required to create new surface (units of $J/m^{2}$). Note for ideal brittle materials, $G_{c}=2\gamma$.

In this example, isotropic elasticity is used, and in all likelihood is a reasonable approximation for bulk polycrystalline or amorphous solids. Sometimes we cannot assume an isotropic solid. This is most common when dealing with systems where single crystals are studied such as the one in Figure 1. Looking at Figure 1, even if $|P_{1}|=|P_{3}|$, pulling in direction $|P_{1}|$ will encounter different resistance (constant of proportionality in Hooke's Law) than pulling in direction $|P_{2}|$.

Figure 1: Example of a single crystal system where bonds behave in spring-like fashion.

The anisotropic expression of Hooke's Law is a tensor relation:

$\sigma _{ij}=C_{ijkl}\varepsilon _{kl}$           [5]

$\varepsilon _{ij}=S_{ijkl}\sigma _{kl}$           [6]

Where $C_{ijkl}$ is the stiffness or elastic constant and $S_{ijkl}$ is the elastic compliance.

Here, the primary focus will be on $C_{ijkl}$, but the two behave similarly. The elastic constant tensor is a fourth rank tensor that connects two second rank tensors. In Equation 5, the right hand side is a double sum over $k$ and $l$, resulting in 9 terms in the sum. Since there are 9 expressions for $\sigma _{ij}$, there is a total of 81 elements in ${\boldsymbol {C}}$. This seems like a large number, but are they unique? We know ${\boldsymbol {\sigma }}$ and ${\boldsymbol {\varepsilon }}$ are symmetric, therefore

$\sigma _{ij}=\sigma _{ji}\rightarrow C_{ijkl}=C_{jikl}$

And

$\varepsilon _{kl}=\varepsilon _{lk}\rightarrow C_{ijkl}=C_{ijlk}$

This reduces the number of elastic constants from 81 to 36 unique values. To further simplify, we must consider the elastic energy. We know that the energy of the system is

$U_{o}={\frac {1}{2}}\sigma _{ij}\varepsilon _{ij}$           [7]

Figure 2: Example of a homogeneous elastic stress-strain curve. The area underneath the curve is highlighted to show the total elastic energy.

As depicted in Figure 2.

For homogeneous elastic loading, we can write

$dU_{o}=\sigma _{ij}d\varepsilon _{ij}$           [8]

Which is the superposition of

${\begin{aligned}dU_{o}&=\sigma _{11}d\varepsilon _{11}+\sigma _{12}d\varepsilon _{12}+\sigma _{13}d\varepsilon _{13}\\&+\sigma _{21}d\varepsilon _{21}+\sigma _{22}d\varepsilon _{22}+\sigma _{23}d\varepsilon _{23}\\&+\sigma _{31}d\varepsilon _{31}+\sigma _{32}d\varepsilon _{32}+\sigma _{33}d\varepsilon _{33}\end{aligned}}$           [9]

Now we will consider starting in an initial state $\sigma =\varepsilon =0$ and straining $\varepsilon _{11}$. This increases the internal energy, the energy stored in bonds, by

$dw_{1}=\sigma _{11}d\varepsilon _{11}=C_{1111}\varepsilon _{11}d\varepsilon _{11}$           [10]

Where $\sigma _{11}=C_{1111}\varepsilon _{11}$ from Equation 5 (all strain except $\varepsilon _{11}$ are zero).

Integrating this

$\int _{0}^{w_{1}}dw_{1}=\int _{0}^{\varepsilon _{11}}C_{1111}\varepsilon _{11}d\varepsilon _{11}$

Becomes

$w_{1}={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}$           [11]

Now apply a second strain deformation $\varepsilon _{22}$ and integrate:

$\int _{0}^{w_{2}}dw_{2}=\int _{0}^{\varepsilon _{22}}\sigma _{22}d\varepsilon _{22}=\int _{0}^{\varepsilon _{22}}(C_{2222}\varepsilon _{22}+C_{2211}\varepsilon _{11})d\varepsilon _{22}$

Using Equation 5 and the fact that all $\varepsilon =0$ except $\varepsilon _{11}$ and $\varepsilon _{22}$. This becomes

$w_{2}={\frac {1}{2}}C_{2222}\varepsilon _{22}^{2}+C_{2211}\varepsilon _{11}\varepsilon _{22}$           [12]

Then the total work is

$w_{TOT}=w_{1}+w_{2}={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}+{\frac {1}{2}}C_{2222}\varepsilon _{22}^{2}+C_{2211}\varepsilon _{11}\varepsilon _{22}$           [13]

Imagine now we reverse the order; first applying $\varepsilon _{22}$ then $\varepsilon _{11}$.

$w_{1}'={\frac {1}{2}}C_{2222}\varepsilon _{22}$

$w_{2}'={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}+C_{1122}\varepsilon _{22}\varepsilon _{11}$

$w_{TOT}'={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}+{\frac {1}{2}}C_{2222}\varepsilon _{22}^{2}+C_{1122}\varepsilon _{11}\varepsilon _{22}$           [14]

In the linear regime superposition holds, therefore the order is not important and

${\begin{aligned}w_{TOT}=w_{TOT}'\\\therefore C_{2211}=C_{1122}\end{aligned}}$

This is generalized to

$C_{ijkl}=C_{klij}$

Which reduces the number of unique values to 21. This is the fewest number of elastic constants that must be specified for an arbitrary crystal.

Crystal Symmetry Simplifications

Fortunately, tensor relations are tied to the symmetry of a crystal. For high symmetry crystals, such as cubic, only 3 unique values are needed and further many zeroes exist. The following rules can be utilized to simplify the notation:

1. All $C_{iiii}=C_{1111}$
2. All $C_{iijj}=C_{1122}$
3. All $C_{ijij}=C_{2323}$
4. All other $C_{ijkl}=0$ (not in Equations 1-3).

This makes the situation much simpler. Next, we will simplify the notation using Voigt notation as described in the table below.

${\boldsymbol {ij}}$ pair	11	22	33	23	31	12
m	1	2	3	4	5	6

This is often (not always) found in textbooks and manuscripts. In this notation, work is

${\begin{aligned}dw&=\sigma _{1}d\varepsilon _{1}+\sigma _{6}d\varepsilon _{6}+\sigma _{5}d\varepsilon _{5}\\&+\sigma _{6}d\varepsilon _{6}+\sigma _{2}d\varepsilon _{2}+\sigma _{4}d\varepsilon _{4}\\&+\sigma _{5}d\varepsilon _{5}+\sigma _{4}d\varepsilon _{4}+\sigma _{3}d\varepsilon _{3}\\dw&=\sigma _{1}d\varepsilon _{1}+\sigma _{2}d\varepsilon _{2}+\sigma _{3}d\varepsilon _{3}+2\sigma _{4}d\varepsilon _{4}+2\sigma _{5}d\varepsilon _{5}+2\sigma _{6}d\varepsilon _{6}\end{aligned}}$

Using Voigt notation, we can now write the stress-strain relation

${\boldsymbol {\sigma }}={\boldsymbol {C}}{\boldsymbol {\varepsilon }}$

In 2D instead of the original 4D format.

$\left[{\begin{matrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{matrix}}\right]=\left[{\begin{matrix}{\text{C}}_{11}&{\text{C}}_{12}&{\text{C}}_{13}&{\text{C}}_{14}&{\text{C}}_{15}&{\text{C}}_{16}&{\text{C}}_{14}&{\text{C}}_{15}&{\text{C}}_{16}\\{\text{C}}_{12}&{\text{C}}_{22}&{\text{C}}_{23}&{\text{C}}_{24}&{\text{C}}_{25}&{\text{C}}_{26}&{\text{C}}_{24}&{\text{C}}_{25}&{\text{C}}_{26}\\{\text{C}}_{13}&{\text{C}}_{23}&{\text{C}}_{33}&{\text{C}}_{34}&{\text{C}}_{35}&{\text{C}}_{36}&{\text{C}}_{34}&{\text{C}}_{35}&{\text{C}}_{36}\\{\text{C}}_{14}&{\text{C}}_{24}&{\text{C}}_{34}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}\\{\text{C}}_{15}&{\text{C}}_{25}&{\text{C}}_{35}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}\\{\text{C}}_{16}&{\text{C}}_{26}&{\text{C}}_{36}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}\\{\text{C}}_{14}&{\text{C}}_{24}&{\text{C}}_{34}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}\\{\text{C}}_{15}&{\text{C}}_{25}&{\text{C}}_{35}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}\\{\text{C}}_{16}&{\text{C}}_{26}&{\text{C}}_{36}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}\end{matrix}}\right]\left[{\begin{matrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{matrix}}\right]$           [15]

In this format, it is still possible to transform the vectors and tensor ${\boldsymbol {C}}$ together. This can further be shortened to

$\left[{\begin{matrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{matrix}}\right]=\left[{\begin{matrix}{\text{C}}_{11}&{\text{C}}_{12}&{\text{C}}_{13}&{\text{C}}_{14}&{\text{C}}_{15}&{\text{C}}_{16}\\{\text{C}}_{12}&{\text{C}}_{22}&{\text{C}}_{23}&{\text{C}}_{24}&{\text{C}}_{25}&{\text{C}}_{26}\\{\text{C}}_{13}&{\text{C}}_{23}&{\text{C}}_{33}&{\text{C}}_{34}&{\text{C}}_{35}&{\text{C}}_{36}\\{\text{C}}_{14}&{\text{C}}_{24}&{\text{C}}_{34}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}\\{\text{C}}_{15}&{\text{C}}_{25}&{\text{C}}_{35}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}\\{\text{C}}_{16}&{\text{C}}_{26}&{\text{C}}_{36}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}\end{matrix}}\right]\left[{\begin{matrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\gamma _{4}\\\gamma _{5}\\\gamma _{6}\end{matrix}}\right]$           [16]

However, this has shear strain components ($\gamma$), which means that in this representation, $C$ is not a tensor but a matrix. This is because the rotational properties of tensors are not satisfied. We can see how these two are related by writing out one of the stresses from Equation 15:

$\sigma _{1}=C_{11}\varepsilon _{1}+C_{12}\varepsilon _{2}+C_{13}\varepsilon _{3}+C_{14}\varepsilon _{4}+C_{15}\varepsilon _{5}+C_{16}\varepsilon _{6}+C_{14}\varepsilon _{4}+C_{15}\varepsilon _{5}+C_{16}\varepsilon _{6}$

$\sigma _{1}=C_{11}\varepsilon _{1}+C_{12}\varepsilon _{2}+C_{13}\varepsilon _{3}+C_{14}2\varepsilon _{4}+C_{15}2\varepsilon _{5}+C_{16}2\varepsilon _{6}$

Where $\gamma _{4}=2\varepsilon _{4}$, $\gamma _{5}=2\varepsilon _{5}$, and $\gamma _{6}=2\varepsilon _{6}$.

Applying our simplifications due to crystalline symmetry we can write for a cubic crystal:

$C_{ij}^{CUBIC}=\left[{\begin{matrix}{\text{C}}_{11}&{\text{C}}_{12}&{\text{C}}_{12}&0&0&0\\{\text{C}}_{12}&{\text{C}}_{11}&{\text{C}}_{12}&0&0&0\\{\text{C}}_{12}&{\text{C}}_{12}&{\text{C}}_{11}&0&0&0\\0&0&0&{\text{C}}_{44}&0&0\\0&0&0&0&{\text{C}}_{44}&0\\0&0&0&0&0&{\text{C}}_{44}\end{matrix}}\right]$           [17]

Returning to our definition of strain energy (Equation 7) we can write this in Voigt notation:

$U_{o}={\frac {1}{2}}\sigma _{i}\varepsilon _{i}$           [18]

And we can write the stress-strain relation (Equation 5) as

$\sigma _{j}=C_{jk}\varepsilon _{k}$           [19]

Which allows us to write

$U_{o}={\frac {1}{2}}C_{ij}\varepsilon _{i}\varepsilon _{j}$           [20]

Which is a double sum over $i$ and $j$. Putting $C_{ij}^{CUBIC}$ from Equation 17 into Equation 20 yields

$U_{o}^{CUBIC}={\frac {1}{2}}C_{11}(\varepsilon _{1}^{2}+\varepsilon _{2}^{2}+\varepsilon _{3}^{2})+2C_{44}(\varepsilon _{4}^{2}+\varepsilon _{5}^{2}+\varepsilon _{6}^{2})+C_{12}(\varepsilon _{1}\varepsilon _{2}+\varepsilon _{2}\varepsilon _{3}+\varepsilon _{3}\varepsilon _{1})$           [21]

Here are a few useful relationships for cubic crystals which are given without proof:

$C_{11}={\frac {S_{11}+S_{12}}{(S_{11}-S_{12})(S_{11}+2S_{12})}}$           [22]

$C_{12}={\frac {-S_{12}}{(S_{11}-S_{12})(S_{11}+2S_{12})}}$           [23]

$C_{44}={\frac {1}{S_{44}}}$           [24]

${\frac {1}{E}}=S_{11}-2\left[(S_{11}-S_{12})-{\frac {1}{2}}S_{44}\right](l^{2}m^{2}+m^{2}n^{2}+l^{2}n^{2})$           [25]

Here $l$, $m$, and $n$ are direction cosine from $x_{1}$, $x_{2}$, and $x_{3}$ and are the direction of applied uniaxial loading. Mapping the anisotropic elastic properties to the isotropic limit, we have

$S_{11}={\frac {1}{E}}$           [26]

$S_{12}={\frac {-\nu }{E}}$           [27]

$S_{44}={\frac {1}{G}}$           [28]

Yet for a truly isotropic medium, only 2 material parameters are needed. We find

$S_{44}=2(S_{11}-S_{12})$           [29]

And

$C_{12}=\lambda$           [30]

$C_{11}=2G+\lambda$           [31]

$C_{44}={\frac {1}{2}}(C_{11}-C_{12})$           [32]

$A={\frac {2C_{44}}{C_{11}-C_{12}}}$           [33]

Where $A$ is the Zener anisotropy ratio ($A=1$ for isotropic material).