Chapter 6: Anisotropic Linear Elasticity

Anisotropic Linear Elasticity

Thus far we have been working with a very specific subset of materials—those being isotropic, linear, elastic, cubic materials. Whew... that is a lot of caveats! And as you know, many materials that we work with do not fall under this category, most notably composite materials, which are becoming increasingly critical and ubiquitous in both the aerospace and automotive industries due to the push toward materials with a high stiffness or strength-to-density ratio, which saves on fuel costs.

But before we get into cool composite mechanics, let's start and look at some very general expressions for how to still utilize our elastic constitutive relationships to determine stress from strain or vice versa. This is a critical point: we are still in the linear elastic regime. We have not left this regime and will not do so until our plasticity lecture.

Now we have hinted at a very complex relationship and have introduced it in a slightly different form previously, but again I want to introduce our simple version of Hooke's Law for uniaxial tension:

\[
\sigma = E \epsilon
\]

But now, for complex stress states, we treat stress and strain as second rank tensor properties, \( \overline{\overline{\sigma}} \) and \( \overline{\overline{\epsilon}} \). And if we want a relationship between two second rank tensors we have to introduce another tensor to be consistent with linear algebra to relate the two, and that tensor essentially stands in for our Young's modulus. It is referred to as our stiffness tensor \( C \) and our compliance tensor \( S \), and we have the relationship that \( S = C^{-1} \).

We can see our general relationships below:

\[
\sigma_{ij} = C_{ijkl} \epsilon_{kl}
\]
\[
\epsilon_{ij} = S_{ijkl} \sigma_{kl}
\]

As you can see, this will get very messy quickly. Instead of our nice and simple stress tensor that we were working with previously that we assumed to be isotropic and cubic and having only 6 independent components, the full anisotropic fourth rank stiffness and compliance tensors have 81 independent elastic components. Now, typically for almost every material this will not be the case—there will be symmetries that reduce this number drastically.

We can also reduce it further right now by remembering the convenient relationships that reduced the number of independent components in our stress and strain second rank tensors to 6. Remembering this, we can make a quick change in notation where we have:

We can make a quick change in notation where we have:

\[
\sigma_{11} = \sigma_{1}
\]
\[
\sigma_{22} = \sigma_{2}
\]
\[
\sigma_{33} = \sigma_{3}
\]
\[
\sigma_{23} = \sigma_{4}
\]
\[
\sigma_{13} = \sigma_{5}
\]
\[
\sigma_{12} = \sigma_{6}
\]

Before we move on to strain, it is critical at this point that we stop and say that here \( \sigma_1 \) is not the maximum principal stress—it is literally just \( \sigma_{11} \). So again, be very careful with notation here, develop an intuition and understanding for why we are using these equations, and do not plug and chug. That will be critical for your future success as an engineer well beyond this course.

We can also do something similar for strain as well:

\[
\epsilon_{11} = \epsilon_{1}
\]
\[
\epsilon_{22} = \epsilon_{2}
\]
\[
\epsilon_{33} = \epsilon_{3}
\]
\[
\gamma_{23} = 2\epsilon_{23} = \epsilon_{4}
\]
\[
\gamma_{13} = 2\epsilon_{13} = \epsilon_{5}
\]
\[
\gamma_{12} = 2\epsilon_{12} = \epsilon_{6}
\]

With this we can now write out the most general relationship, and it should be noted that once we make this change we are no longer working with a tensor but a matrix, since we changed our directional notation. So now our relationship becomes:

\[
\epsilon_{i} = S_{ij} \sigma_{j}
\]

or equivalently:

\[
[\epsilon] = [S][\sigma]
\]

Now, just by making this change from tensor to matrix we have gone from 81 components in our compliance tensor to a 36-component matrix, so our math has become much simpler.

Now if we expand this expression for the most general anisotropic material, we would have:

\[
\begin{bmatrix}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\epsilon_4\\
\epsilon_5\\
\epsilon_6
\end{bmatrix}
=
\begin{bmatrix}
S_{11} & S_{12} & S_{13} & S_{14} & S_{15} & S_{16} \\
S_{21} & S_{22} & S_{23} & S_{24} & S_{25} & S_{26} \\
S_{31} & S_{32} & S_{33} & S_{34} & S_{35} & S_{36} \\
S_{41} & S_{42} & S_{43} & S_{44} & S_{45} & S_{46} \\
S_{51} & S_{52} & S_{53} & S_{54} & S_{55} & S_{56} \\
S_{61} & S_{62} & S_{63} & S_{64} & S_{65} & S_{66}
\end{bmatrix}
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3\\
\sigma_4\\
\sigma_5\\
\sigma_6
\end{bmatrix}
\]

However, we must remember that we had previously shown that our stress and strain tensors were symmetric due to our static equilibrium conditions (i.e., the sum of forces and moments must equal zero). With this symmetry in our stress and strain tensors—and the fact that we can have no hysteresis in our strain energy density (see full proof in Roylance, *Constitutive Equations*)—we can write that \( S_{ij} = S_{ji} \).

\[
\begin{bmatrix}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\epsilon_4\\
\epsilon_5\\
\epsilon_6
\end{bmatrix}
=
\begin{bmatrix}
S_{11} & S_{12} & S_{13} & S_{14} & S_{15} & S_{16} \\
S_{12} & S_{22} & S_{23} & S_{24} & S_{25} & S_{26} \\
S_{13} & S_{23} & S_{33} & S_{34} & S_{35} & S_{36} \\
S_{14} & S_{24} & S_{34} & S_{44} & S_{45} & S_{46} \\
S_{15} & S_{25} & S_{35} & S_{45} & S_{55} & S_{56} \\
S_{16} & S_{26} & S_{36} & S_{46} & S_{56} & S_{66}
\end{bmatrix}
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3\\
\sigma_4\\
\sigma_5\\
\sigma_6
\end{bmatrix}
\]

So we have reduced from 81 to 21 independent components, and this is the compliance matrix that will hold true for the most anisotropic material possible—i.e., a triclinic structure.

But we can reduce this even more by invoking Onsager’s theorem, which states that materials cannot exhibit properties of higher symmetry than that of the material itself. In other words, depending on the symmetry of the material, some of these constants could disappear. Note: for a more complete discussion of Onsager, read Sam Allen’s *Kinetics of Materials*—the best description that I have come across.

For example:

- Triclinic materials: have all 21 independent coefficients
- Monoclinic materials: have 13 independent coefficients
- Orthorhombic/orthotropic materials: have 9 independent coefficients (examples include wood, composite laminate, and semi-crystalline polyethylene)
- Tetragonal materials: have 7 independent coefficients
- Cubic materials: have 3 independent coefficients, which can be reduced to 2 if the material is isotropic

Before going into this let's prove that we will only have 2 independent components for isotropic cubic linear elastic materials. Remember for cubic materials (a=b=c and \(\alpha = \beta = \gamma = 90^{o}\) for cubic) you will find that there are only \textbf{2 independent elastic constants} \(E\) and \(\nu\) as we have worked with previously.

But let's break this down piece by piece to determine what each of these compliance components are. Let's start with assuming that we only apply \(\sigma_{1}\) then we know that we should obtain the following expression our Hooke's Law

\[
\epsilon_{1} = S_{11} \sigma_{1}
\]

thus we find that

\[
S_{11} = \frac{1}{E_1}
\]

You can do a similar proof to find that

\[
S_{22} = \frac{1}{E_{2}}, \quad
S_{33} = \frac{1}{E_{3}}
\]

If we were also to apply pure shear for example only \(\sigma_{4}\) then we would get the following expression

\[
\epsilon_4 = S_{44} \sigma_4
\]

and thus

\[
S_{44} = \frac{1}{G_{23}}
\]

and similarly

\[
S_{55} = \frac{1}{G_{13}}, \quad
S_{66} = \frac{1}{G_{12}}
\]

However we also know previously that if we apply a tension stress in one direction the material will contract in transverse directions due to volume conservation and Poisson's ratio thus we also know that if we only apply \(\sigma_1\) then for the strain in the 2 direction

\[
\epsilon_2 = S_{21} \sigma_1
\]

and in general to define Poisson's ratio we have

\[
\nu_{ij} = -\frac{\epsilon_j}{\epsilon_i}
\]

where \(\epsilon_j\) is the transverse strain and \(\epsilon_i\) is the direction of the applied stress so with this we have that

\[
S_{21} = -\frac{\nu_{12}}{E_1}
\]

and similarly we will find that

\[
S_{31} = -\frac{\nu_{13}}{E_1}, \quad
S_{32} = -\frac{\nu_{23}}{E_2}
\]

and we also know that \(S_{ij} = S_{ji}\) so we know that

\[
S_{12} = S_{21}, \quad
-\frac{\nu_{12}}{E_1} = \frac{\nu_{21}}{E_2}
\]

and you can derive the others similarly. Now the question becomes what about \(S_{14}\), \(S_{15}\), and \(S_{16}\)?. These components attempt to\textbf{ relate normal strain to shear stress} and we know previously that \textbf{all such stresses should be zero}, but note that there are times for woven fabrics they may have non-zero values so it can depend on the material but \textbf{typically these will be zero as shear stresses shouldn't generate normal strains}.

Additionally for components like \(S_{45}\), \(S_{46}\), and \(S_{56}\) which attempt to relate \textbf{shear strain in one plain to shear stress in another plain these will also typically be zero}. So in general we should expect a matrix to look as follows

\[
\begin{bmatrix}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\epsilon_4\\
\epsilon_5\\
\epsilon_6
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E_{1}} & -\frac{\nu_{21}}{E_{2}} & -\frac{\nu_{31}}{E_{3}} & 0 & 0 & 0 \\
-\frac{\nu_{12}}{E_{1}} & \frac{1}{E_{2}} & -\frac{\nu_{32}}{E_{3}} & 0 & 0 & 0 \\
-\frac{\nu_{13}}{E_{1}} & -\frac{\nu_{23}}{E} & \frac{1}{E_{3}} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{G_{23}} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{G_{13}} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{G_{12}}
\end{bmatrix}
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3\\
\sigma_4\\
\sigma_5\\
\sigma_6
\end{bmatrix}
\]

Now for cubic materials the Young's modulus does not vary as a function of direction and the same holds for the shear modulus, as well as the Poisson's ratio so we get for cubic materials that

\[
\begin{bmatrix}
S_{11} & S_{12} & S_{12} & 0 & 0 & 0 \\
S_{12} & S_{11} & S_{12} & 0 & 0 & 0 \\
S_{12} & S_{12} & S_{11} & 0 & 0 & 0 \\
0 & 0 & 0 & S_{44} & 0 & 0 \\
0 & 0 & 0 & 0 & S_{44} & 0 \\
0 & 0 & 0 & 0 & 0 & S_{44}
\end{bmatrix}
\]

and for isotropic materials we remember that we have the following relationship between Young's Modulus and the shear modulus which is

\[
G = \frac{E}{2(1+\nu)} = \frac{1}{2(S_{11} -S_{12})}
\]

So we can re-write the matrix to be a function of two constant for linear elastic isotropic cubic materials

\[
\begin{bmatrix}
S_{11} & S_{12} & S_{12} & 0 & 0 & 0 \\
S_{12} & S_{11} & S_{12} & 0 & 0 & 0 \\
S_{12} & S_{12} & S_{11} & 0 & 0 & 0 \\
0 & 0 & 0 & 2(S_{11} -S_{12}) & 0 & 0 \\
0 & 0 & 0 & 0 & 2(S_{11} -S_{12}) & 0 \\
0 & 0 & 0 & 0 & 0 & 2(S_{11} -S_{12})
\end{bmatrix}
\]

Well that is great but what about for materials with lower symmetries? Let's dive into how our compliance matrix looks like for orthorhromic/orthotropic materials, just a reminder you can look up orthorhombic crystal structure or you can see that orthotropic materials will have \textbf{9 independent elastic constants} as seen below

\[
\begin{bmatrix}
S_{11} & S_{12} & S_{13} & 0 & 0 & 0 \\
S_{12} & S_{22} & S_{23} & 0 & 0 & 0 \\
S_{13} & S_{23} & S_{33} & 0 & 0 & 0 \\
0 & 0 & 0 & S_{44} & 0 & 0 \\
0 & 0 & 0 & 0 & S_{55} & 0 \\
0 & 0 & 0 & 0 & 0 & S_{66}
\end{bmatrix}
\]

There is also an interesting subset of materials that also exhibit transversely isotropic symmetry, i.e. the materiel will be isotropic in one plane (for this example assume the 1-2 plane is isotropic) and will have \textbf{5 independent elastic constants}, one such of example of these materials are extruded materials.

\[
\begin{bmatrix}
S_{11} & S_{12} & S_{13} & 0 & 0 & 0 \\
S_{12} & S_{11} & S_{13} & 0 & 0 & 0 \\
S_{13} & S_{13} & S_{33} & 0 & 0 & 0 \\
0 & 0 & 0 & S_{44} & 0 & 0 \\
0 & 0 & 0 & 0 & S_{44} & 0 \\
0 & 0 & 0 & 0 & 0 & 2(S_{11} - S_{12})
\end{bmatrix}
\]

We can re-write this here because we know that

\[
\frac{1}{G_{12}} = \frac{2(1+\nu_{12})}{E_1} = 2(S_{11} - S_{12})
\]

We have been very general thus far but one of the most common anisotropic materials that you will encounter are biomaterials and more importantly composites. Let's take a quick detour to discuss these increasingly important materials.

Rotation of Compliance Tensor

We can also develop some rotation matrices to rotate our compliance matrix as well as follows for rotations around the 1 axis for stress and strain respectively to be \(T^{\sigma}_{1}\) and \(T^{\epsilon}_{1}\)

\[
T^{\sigma}_{1} =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & c^2 & s^2 & 2sc & 0 & 0 \\
0 & s^2 & c^2 & -2sc & 0 & 0 \\
0 & -sc & sc & c^2-s^2 & 0 & 0 \\
0 & 0 & 0 & 0 & c & -s \\
0 & 0 & 0 & 0 & s & c \\
\end{bmatrix}
\]

and

\[
T^{\epsilon}_{1} =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & c^2 & s^2 & sc & 0 & 0 \\
0 & s^2 & c^2 & -sc & 0 & 0 \\
0 & -2sc & 2sc & c^2-s^2 & 0 & 0 \\
0 & 0 & 0 & 0 & c & -s \\
0 & 0 & 0 & 0 & s & c \\
\end{bmatrix}
\]

Also we now that to transform to a new coordinate system

\[
\sigma^{'} = T^{\sigma}_{1} \sigma^{o}
\]

and

\[
\epsilon^{'} = T^{\epsilon}_{1} \epsilon^{o}
\]

so we get that

\[
\epsilon^{o} = S^{o} \sigma^{o}
\]

and

\[
\epsilon^{'} = S^{'} \sigma^{'}
\]

Now we get that

\[
S^{'} = T^{\epsilon}_{1} \, S^{o} \, \left(T^{\sigma}_{1}\right)^{-1}
\]

And for the 2 axis:

\[
T^{\sigma}_{2} =
\begin{bmatrix}
c^2 & 0 & s^2 & 0 & -2sc & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
s^2 & 0 & c^2 & 0 & 2sc & 0 \\
0 & 0 & 0 & c & 0 & s \\
sc & 0 & -sc & 0 & c^2-s^2 & 0 \\
0 & 0 & 0 & -s & 0 & c \\
\end{bmatrix}
\]

and

\[
T^{\epsilon}_{2} =
\begin{bmatrix}
c^2 & 0 & s^2 & 0 & -sc & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
s^2 & 0 & c^2 & 0 & sc & 0 \\
0 & 0 & 0 & c & 0 & s \\
2sc & 0 & -2sc & 0 & c^2-s^2 & 0 \\
0 & 0 & 0 & -s & 0 & c \\
\end{bmatrix}
\]

And for the 3 axis:

\[
T^{\sigma}_{3} =
\begin{bmatrix}
c^2 & s^2 & 0 & 0 & 0 & 2sc \\
s^2 & c^2 & 0 & 0 & 0 & -2sc \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & c & -s & 0 \\
0 & 0 & 0 & s & c & 0 \\
-sc & sc & 0 & 0 & 0 & c^2-s^2 \\
\end{bmatrix}
\]

and

\[
T^{\epsilon}_{3} =
\begin{bmatrix}
c^2 & s^2 & 0 & 0 & 0 & sc \\
s^2 & c^2 & 0 & 0 & 0 & -sc \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & c & -s & 0 \\
0 & 0 & 0 & s & c & 0 \\
-2sc & 2sc & 0 & 0 & 0 & c^2-s^2 \\
\end{bmatrix}
\]

Composite Mechanics

Composite materials are materials that contain typically more than one type of material combined. This typically involves a material matrix which is the major component in terms of volume fraction. This matrix material is reinforced by an additional material typically one that is stiffer or tougher than the matrix material. The material that reinforces the matrix can be in particle form, fibers, or precipitates. A composite can also be a porous material like metal foams, concrete, ECM, etc.

Let's take a look at a particle reinforced composite that is composed of a matrix which has some elastic modulus, \(E_{m}\), and volume fraction, \(f_{m}\). There is also the particle reinforcement in this case which again has an associated elastic modulus, \(E_{p}\), and volume fraction \(f_{p}\). Thus the total composite modulus, \(E_{composite}\), is:

\[
E_{composite} = E_{m}f_{m} + E_{p}f_{p}
\]

Empirically, there is typically a constant that is less than 1 for the particle reinforcement contribution, but again that constant can only be found empirically.

[Figure: Composites pulled in the transverse or longitudinal/axial direction.]

Let's now look at a fiber reinforced composite that is pulled or stressed in the transverse direction with respect to the longitudinal or axial direction of the fibers. The stresses are the same on the matrix and the fibers. However, the strains of the fiber and matrix will be different. With these constraints:

\[
\sigma_{m}=\sigma_{f}=\sigma
\]
\[
\epsilon_{c} = \epsilon_{m}f_{m} + \epsilon_{f}f_{f}
\]
\[
\epsilon_{c} = \sigma\left(\frac{f_{m}}{E_{m}} + \frac{f_{f}}{E_{f}}\right)
\]
\[
\frac{1}{E_{c}} = \frac{f_{f}}{E_{f}} + \frac{f_{m}}{E_{m}}
\]

If we instead pull parallel to the fiber longitudinal or axial direction, the strain is the same but now the stresses will be different for the fibers and the matrix. With those constraints we get:

\[
\sigma_{c}=\sigma_{f}f_{f} + \sigma_{m}f_{m}
\]
\[
\epsilon_{c}=\epsilon_{m}=\epsilon_{f}
\]
\[
\sigma_{c} = E_{f}\epsilon f_{f} + E_{m}\epsilon f_{m}
\]
\[
E_{c} = E_{f}f_{f} + E_{m}f_{m}
\]

We can see the difference in the Young's modulus in both of these directions as a function of volume fraction in the plot below:

Transversely Isotropic Composite

Let's go a bit further and do an example of a composite or laminate that is transversely isotropic (i.e. \(E_{33} = E_{22} \neq E_{11}\)).

In this scenario our compliance and stiffness matrices would look like this:

\[
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3\\
\sigma_4\\
\sigma_5\\
\sigma_6
\end{bmatrix}
=
\begin{bmatrix}
C_{11} & C_{12} & C_{13} & 0 & 0 & 0 \\
C_{12} & C_{22} & C_{13} & 0 & 0 & 0 \\
C_{13} & C_{13} & C_{33} & 0 & 0 & 0 \\
0 & 0 & 0 & C_{44} & 0 & 0 \\
0 & 0 & 0 & 0 & C_{44} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{C_{11} - C_{12}}{2}
\end{bmatrix}
\begin{bmatrix}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\epsilon_4\\
\epsilon_5\\
\epsilon_6
\end{bmatrix}
\]

or equivalently:

\[
\begin{bmatrix}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\epsilon_4\\
\epsilon_5\\
\epsilon_6
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E_{1}} & -\frac{\nu_{21}}{E_{2}} & -\frac{\nu_{31}}{E_2} & 0 & 0 & 0 \\
-\frac{\nu_{12}}{E_{1}} & \frac{1}{E_{2}} & -\frac{\nu_{32}}{E_2} & 0 & 0 & 0 \\
-\frac{\nu_{13}}{E_{1}} & -\frac{\nu_{23}}{E_{2}} & \frac{1}{E_{2}} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{G_{12}} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{G_{23}} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{G_{13}}
\end{bmatrix}
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3\\
\sigma_4\\
\sigma_5\\
\sigma_6
\end{bmatrix}
\]

We can see from this that as a result we will have that \( \epsilon_4 = \epsilon_5 = 0 \). We also know that from our conditions of tensor symmetry:

\[
\frac{\nu_{21}}{E_{2}} = \frac{\nu_{12}}{E_{1}}
\]
\[
\frac{\nu_{13}}{E_{1}} = \frac{\nu_{31}}{E_2}
\]
\[
\frac{\nu_{23}}{E_{2}} = \frac{\nu_{32}}{E_2}
\]

Now a typical scenario for composites will be when they are placed under plane stress conditions, and we can then simplify our matrix and only focus on the in-plane stress and strains where we now have:

\[
\begin{bmatrix}
\epsilon_1\\
\epsilon_2\\
0\\
0\\
0\\
\epsilon_6
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E_{1}} & -\frac{\nu_{21}}{E_{2}} & -\frac{\nu_{31}}{E_2} & 0 & 0 & 0 \\
-\frac{\nu_{12}}{E_{1}} & \frac{1}{E_{2}} & -\frac{\nu_{32}}{E_2} & 0 & 0 & 0 \\
-\frac{\nu_{13}}{E_{1}} & -\frac{\nu_{23}}{E_{2}} & \frac{1}{E_{2}} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{G_{12}} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{G_{23}} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{G_{13}}
\end{bmatrix}
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3\\
\sigma_4\\
\sigma_5\\
\sigma_6
\end{bmatrix}
\]

or more visually appealing:

\[
\begin{bmatrix}
\epsilon_{11}\\
\epsilon_{22}\\
\gamma_{12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E_{1}} & -\frac{\nu_{21}}{E_{2}} & 0 \\
-\frac{\nu_{12}}{E_{1}} & \frac{1}{E_{2}} & 0 \\
0 & 0 & \frac{1}{G_{12}}
\end{bmatrix}
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_{12}
\end{bmatrix}
\]

This is great, but this only applies if we are pulling along the principal axes. How can we express the properties of this material when stressed along arbitrary directions with respect to the principal axes? Well, we will use linear algebra!

We have previously seen in the elasticity lecture how to rotate stresses and strains in relation to Mohr's circle, and we can do the same thing here. We will use our same rotation matrices, transformation matrix \( T \), as previously defined, and we can obtain stress after an arbitrary rotation:

\[
\begin{bmatrix}
\sigma_{x'}\\
\sigma_{y'}\\
\sigma_{xy'}
\end{bmatrix}
=
\begin{bmatrix}
T
\end{bmatrix}
\begin{bmatrix}
\sigma_x\\
\sigma_y\\
\sigma_{xy}
\end{bmatrix}
\]

For strain, again we have to deal with this factor of 2 in relating shear strain, so first we have to account for \( \epsilon_{12} = \frac{\gamma_{12}}{2} \). Therefore, we have to introduce another matrix, the Reuter Matrix (\( R \)) as previously defined. So to convert to a rotated coordinate system:

\[
\begin{bmatrix}
\epsilon_{x'}\\
\epsilon_{y'}\\
\gamma_{xy'}
\end{bmatrix}
=
R T R^{-1}
\begin{bmatrix}
\epsilon_x\\
\epsilon_y\\
\gamma_{xy}
\end{bmatrix}
\]

Now we can finally determine the strain at an arbitrary rotation for a transverse isotropic polymer composite or laminate, which can be seen below when we combine these expressions:

\[
\begin{bmatrix}
\epsilon_{x'}\\
\epsilon_{y'}\\
\gamma_{xy'}
\end{bmatrix}
=
R T R^{-1} S T^{-1}
\begin{bmatrix}
\sigma_{x'}\\
\sigma_{y'}\\
\sigma_{xy'}
\end{bmatrix}
\]

You can take this composite mechanics a bit further and discuss layered composites. A great discussion—but one that involves a considerable amount of linear algebra knowledge—is given by Roylance, so please feel free to look this up and perhaps it may inspire a unique project... Mathematica anyone?