Chapter 4: Tensor Notation of Stress and Strain

We have introduced some general notions of stress and strain definitions and you have likely encountered them in past courses; however, at this point in this class we have been focusing on relatively simple stress states, uniaxial tension. However, there are many other complex stress states that are more commonly found in real life applications both in industry and academia. The other important thing to clarify is that all the expressions that we will work with here only apply to the first regime of our stress-strain curve, i.e. linear elastic, and we will also assume the materials we work with are cubic and behave isotropically.

Tensorial Definition of Stress

We have previously defined normal stress as

\[
\sigma = \frac{F}{A}
\]

and we define positive stress if the material is under tension (material extends) or negative if the material compresses.

In addition to normal stress one will often encounter shear stresses as well. Shear stress occurs when the force is applied parallel to the area. For shear stresses we will define positive faces of elements as those with normal in positive 1, 2, or 3 directions. Positive shear will act in the positive direction on positive faces or act in the negative direction on the negative faces.

Shear stresses are also defined as force per unit area as seen below

\[
\tau = \frac{F}{A}
\]

where \( \tau \) is shear stress. In this class we will instead utilize the tensorial definition of both stress and strain, but it should be noted that both are equivalent (unlike engineering and tensorial shear strain... more on this later).

Stress is a second rank tensor property. You have most likely encountered several first rank tensor properties, i.e. vector quantities. The difference between a matrix and a tensor is that a tensor indicates a coordinate system while a matrix is just a matrix of numbers. The second rank nature of stress \( \overline{\overline{\sigma}} \) or \( \sigma_{ij} \), has indices \( i \) and \( j \) that indicate directions in our coordinate system. In this class we will use a 1–2–3 coordinate system. For stress, \( i \) denotes the normal to the plane on which the force is acting, and \( j \) is the direction of the force:

\[
\sigma_{ij} = \frac{F_j}{A_i}
\]

So our full stress tensor or our most generic stress state would be:

\[
\sigma =
\begin{bmatrix}
\sigma_{11} & \sigma_{12} & \sigma_{13} \\
\sigma_{21} & \sigma_{22} & \sigma_{23} \\
\sigma_{31} & \sigma_{32} & \sigma_{33}
\end{bmatrix}
\]

Now this looks like a complex matrix with 9 independent components; however, they are not completely independent. We need to ensure that our representative volume element (RVE) is in equilibrium, i.e. \( \Sigma F = 0 \) and \( \Sigma M = 0 \).

To ensure that we are at equilibrium, let's look at the sum of the forces in the 2 direction:

\[
\Sigma F_2 = \sigma_{32} \delta_1 \delta_2 - \sigma_{32} \delta_1 \delta_2 = 0
\]

This is a somewhat trivial result but the cool finding comes when we set the moments about the 1 axis to be equal to zero

\[
\Sigma M_{1} = \sigma_{23} \delta_3 \delta_1 \delta_2 - \sigma_{32} \delta_1 \delta_2 \delta_3 = 0
\]

We then find the following relationships that

\[
\sigma_{23} = \sigma_{32}
\]

Moreover we can extend this proof and then we will find that

\[
\sigma_{12} = \sigma_{21}
\]
\[
\sigma_{23} = \sigma_{32}
\]
\[
\sigma_{31} = \sigma_{13}
\]

Our second rank tensor reduces from 9 independent components to 6 independent components as seen below.

\[
\sigma =
\begin{bmatrix}
\sigma_{11} & \sigma_{12} & \sigma_{13} \\
\sigma_{12} & \sigma_{22} & \sigma_{23} \\
\sigma_{13} & \sigma_{23} & \sigma_{33}
\end{bmatrix}
\]

Now this ends our initial discussion of the tensorial definition of stress.

How about an inverse problem, draw the following RVE:

\[
\sigma =
\begin{bmatrix}
86 & 0 & -8 \\
0 & -37 & 3 \\
-8 & 3 & 14
\end{bmatrix}
\]

Now you are experts but are you ready for strain? Of course you are—let's go.

Engineering Definition of Strain

We have previously defined normal strain as well as true strain but forget about true stress and strain for a little bit. Normal strain was defined as

\[
\epsilon = \frac{\Delta l}{l_o}
\]

and this was engineering strain. We also have a definition of engineering shear strain as seen below.

\[
\gamma = \frac{dL}{L} = \tan \theta
\]

This particular case is for simple shear and pure shear as well. Unfortunately, this definition is going to cause an issue in just a second. For a pure stress state we observe a proportional relationship between shear stress and shear strain in the elastic regime which is defined as:

\[
\tau = G \gamma
\]

where \( G \) is the shear modulus. Remember this only holds for pure shear just like this \( \sigma = E \epsilon \) holds for uniaxial tension only.

Now I mentioned a hiccup previously. When we talked about the engineering definition of normal and shear stress those definitions map directly to the tensorial definition of stress for both normal and shear stresses. For strain, engineering normal strain is the exact same as the tensorial definition but there is a difference for the shear strain. Thus we must introduce infinitesimal strain theory to reconcile this so we can move on to some very fun and complex problems.

Infinitesimal Strain Theory

Now more formally we can define strain using infinitesimal strain theory (if you are interested please read in Chapter 5 of Young and Lovell Introduction to Polymers). The essential idea is that you take an infinitesimally small representative volume element and displace the points. So the definition of strain becomes as follows:

\[
\epsilon_{ij} = \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)
\]

where \( x \) is an axis dimension and \( u \) is the displacement. We also have from this theory that:

Again, here we can see that our engineering definition of strain and our tensorial definition match perfectly. However, looking below for the case of strain we have an issue.

We can see here that when applying the infinitesimal strain theory definition of tensorial strain we do not see this equivalence for engineering shear strain and tensorial shear strain. In fact, we see that the tensorial definition of shear strain is half the engineering shear strain. Thus, we must account for this and be careful when working with problems, i.e. is this the tensorial or engineering definition of shear strain.

\[
\epsilon_{11} = \frac{\partial u_1}{\partial x_1}
\]
\[
\epsilon_{22} = \frac{\partial u_2}{\partial x_2}
\]
\[
\epsilon_{33} = \frac{\partial u_3}{\partial x_3}
\]
\[
\epsilon_{12} = \epsilon_{21} = \frac{1}{2} \left( \frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1} \right)
\]
\[
\epsilon_{23} = \epsilon_{32} = \frac{1}{2} \left( \frac{\partial u_2}{\partial x_3} + \frac{\partial u_3}{\partial x_2} \right)
\]
\[
\epsilon_{31} = \epsilon_{13} = \frac{1}{2} \left( \frac{\partial u_3}{\partial x_1} + \frac{\partial u_1}{\partial x_3} \right)
\]
\[
\gamma_{12} = \gamma_{21} = \left( \frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1} \right)
\]
\[
\gamma_{23} = \gamma_{32} = \left( \frac{\partial u_2}{\partial x_3} + \frac{\partial u_3}{\partial x_2} \right)
\]
\[
\gamma_{31} = \gamma_{13} = \left( \frac{\partial u_3}{\partial x_1} + \frac{\partial u_1}{\partial x_3} \right)
\]

With all this we can now create our strain matrix, and seeing from the definitions above we know that \( \epsilon_{12} = \epsilon_{21} \), \( \epsilon_{23} = \epsilon_{32} \), and \( \epsilon_{13} = \epsilon_{31} \), so that now our strain matrix has 6 independent components and can be written as:

\[
\epsilon =
\begin{bmatrix}
\epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\
\epsilon_{12} & \epsilon_{22} & \epsilon_{23} \\
\epsilon_{13} & \epsilon_{23} & \epsilon_{33}
\end{bmatrix}
\]

or equivalently:

\[
\epsilon =
\begin{bmatrix}
\epsilon_{11} & \frac{\gamma_{12}}{2} & \frac{\gamma_{13}}{2} \\
\frac{\gamma_{12}}{2} & \epsilon_{22} & \frac{\gamma_{23}}{2} \\
\frac{\gamma_{13}}{2} & \frac{\gamma_{23}}{2} & \epsilon_{33}
\end{bmatrix}
\]

This will become increasingly important when we begin to rotate our coordinate systems in problems to come.

Now at this point we want to start to use these expressions to relate stress and strain using our series of constitutive equations, but one question at this point you may be asking is what is the origin of linear elasticity at the atomistic level. Well, I'm so glad you asked...

Atomistic Basis of Linear Elasticity and the Young’s Modulus

In the previous lecture we learned about the Young's modulus and \( \sigma_f \), but these aren't just values that we only find on a stress-strain curve. These are values that have an atomistic origin in nature. To understand these properties we have to remember that in the elastic regime we are simply stretching bonds and that the distance between bonds and the equilibrium distance between bonds is governed by an interatomic potential. Also, whenever you have a potential or a distribution you can relate the width of the distribution or any potential to the stiffness of that potential or distribution—see my paper about quantum harmonic oscillators and chemotactic traps, riveting read...

But we can relate the chemical bonding to how stiff a material will be or how compliant they are. This is very, very unique and something that we cannot do in the plasticity regime or fracture due to the fact that we have to deal with inhomogeneities in the system and other atomistic details, i.e. energy of dislocations, defects, etc. We will also see soon that for certain materials the stiffness is derived from entropic considerations.

Let's approach this from a thermodynamic perspective and consider a system under uniaxial tension in the 1 direction. Now the key thing to remember is that when you stretch bonds this changes the internal energy of the system. Looking at the Lennard-Jones (LJ) potential, if we increase the distance between atoms the internal energy of the system increases.

Now we know that the change in the internal energy of the system is equal to the sum of the work done on the material by the applied stress and the heat absorbed by the material (aren’t you glad we had that reminder lecture for thermodynamics!!!)

\[
dU = dW + dq
\]

where work will simply be

\[
dW = \sigma_{11} \delta_2 \delta_3 d\epsilon_{11} \delta_1 = \sigma_{11} d\epsilon_{11} V
\]

We also know from thermodynamics that:

\[
dq = T dS
\]

and thus:

\[
dU = V \sigma d\epsilon + T dS
\]

and rearranging in terms of stress:

\[
\sigma = \frac{1}{V} \frac{dU}{d\epsilon} - \frac{T dS}{V d\epsilon}
\]

Right here it is important to note that for most crystalline materials (metals and ceramics) as well as amorphous materials that are well below the \( T_g \), the value of \( \frac{dS}{d\epsilon} \) will be very small to the point where the values may be negligible.

Now for polymers above the \( T_g \) and for rubbers or other elastomers we will see very soon that \( \frac{dS}{d\epsilon} \) is very significant—a fantastic derivation of rubber elasticity is coming soon, don't worry it is riveting!

So for crystalline materials and for amorphous glassy materials we can simplify the expression and look at the internal energy per atom and instead at the atomic volume \( \Omega \), so our previous expression becomes:

\[
\sigma = \frac{1}{\Omega} \frac{dU}{d\epsilon}
\]

and thus for our Young's modulus:

\[
E = \frac{d\sigma}{d\epsilon} = \frac{1}{\Omega} \frac{d^2U}{d\epsilon^2}
\]

In general many potentials will have the following form:

\[
U(r) = -\frac{A}{r^m} + \frac{B}{r^n}
\]

where the first term is the attractive term and \( A \) is a constant while the \( m \) exponent is typically 1 for ionic crystals, and for covalent or metallic bonding it can vary between 4–6, whereas the repulsive term is the second term, \( B \) is a constant, and the \( n \) exponent can vary between 9–12.

We can then start to write some components of the previous equation in terms of our potential. For example, we see that:

\[
\epsilon = \frac{r - r_o}{r_o} = \frac{r}{r_o} - 1
\]

And if you plot the force you will also see that it is linear or Hookean near the equilibrium distance, and finally if you also plot the Young's modulus you will see that as \( r \) increases the Young's modulus decreases. This is important a little later but let's do some more math quickly...

So back to our derivation, now we know that:

\[
E = \left( \frac{d\sigma}{d\epsilon} \right)_{\epsilon = 0}
\]

Also we know that:

\[
d\sigma = \frac{dF}{r_o^2}
\]

And we have that:

\[
\frac{d\epsilon}{dr} = \frac{1}{r_o}
\]

And now finally we can write:

\[
E = \left( \frac{d\sigma}{dr} \frac{dr}{d\epsilon} \right)_{\epsilon = 0} = \frac{1}{r_o} \left( \frac{d^2U}{dr^2} \right)_{\epsilon = 0}
\]

You can arrive at a similar expression to find an estimate of the Young's modulus using the elastic strain energy which we previously defined:

\[
U_{ESE} = \frac{E \epsilon^2}{2}
\]

And remember the elastic strain energy is simply a potential energy of our system so we can write:

\[
E = \frac{d^2U_{ESE}}{d\epsilon^2}
\]

and again we can relate the change in the distance to strain via this relationship \( r = r_o (1 + \epsilon) \) and then:

\[
E = \frac{d}{d\epsilon} \left( \frac{dU_{ESE}}{dr} \frac{dr}{d\epsilon} \right)
= \frac{d^2U_{ESE}}{dr^2} \left( \frac{dr}{d\epsilon} \right)^2
+ \frac{dU_{ESE}}{dr} \frac{d^2r}{d\epsilon^2}
= \frac{d^2U_{ESE}}{dr^2} r_o^2
\]

Since \( r \) is linear in strain. We also know that the atomic volume \( V = V_o = r_o^3 \), so our strain energy per unit volume will simply be the energy per atom divided by the volume, and here the energy per atom will simply be:

\[
E_{atom} = \frac{Z U(r)}{2}
\]

where \( Z \) is the number of nearest neighbors. So we find:

\[
U_{ESE} = \frac{E_{atom}(r) - E_{atom}(r_o)}{V}
\]

And finally we arrive at the expression that:

\[
E = \frac{Z}{r_o} \frac{d^2U}{dr^2}
\]

Now you may find that this calculation may be very different from known values; however, it is important to note that there are much more complex potentials that are typically used, in particular embedded atom potentials (EAM), which take into account more complex and long-range next-neighbor interactions.

Now one question that we hinted at is what happens when temperature changes? Right now we have been assuming we are working at essentially 0 K. Well, let's figure this out together.

Thermal Expansion Coefficient

Let's go back once again to our good old friend the Lennard-Jones potential. When temperature increases, we know that the thermal energy increases, causing the atoms to vibrate and oscillate around their equilibrium distance \( r_o \). Now if the potential was symmetric or harmonic (like at the minimum of a quadratic function), then the atoms would be equally distributed at distances less than or greater than the equilibrium distance. However, as you can see, the Lennard-Jones and most atomic potentials are asymmetric or anharmonic, and thus when the atoms fluctuate they will tend to occupy positions with the lowest energy. One can see that as temperature increases and as the molecules deviate further and further from their equilibrium position, they will tend to adopt positions with lower energies. Thus, as you can see, on average the molecules will tend to expand because the larger interatomic distances have lower energies, and thus the materials will tend to expand.

So to put it simply, at \( T = 0 \, K \) the atoms will be at their equilibrium distance \( r_o \), and then as temperature increases, the atoms will essentially move up the anharmonic potential to lower their energy. The equilibrium distance thus shifts to the right depending on the thermal expansion coefficient \( \alpha \), and thus we can write this thermally induced strain as:

\[
\epsilon_{T} = \alpha \Delta T
\]

Now for linear elastic isotropic materials, the material will expand uniformly in all directions and there will be no shear strain from thermal expansion. Some typical thermal expansion coefficients for some materials can be seen below, all multiplied by \( 10^{-6} \, K^{-1} \):

- Polymers: 50–500
- Metals: 5–50
- Ceramics: 1–10
- Glasses: 1–2

Taking into account thermal strain is incredibly important in many materials processes. When you heat a material that is a composite and cool it, this can induce residual stresses due to the incompatible thermal strains.

Rubber Elasticity

In previous courses you have hopefully come across the notion that for polymers there is a characteristic relaxation time where, when a mechanical load is applied, the polymer can behave like a viscous liquid or an elastic solid. We will talk more about this in this class as well. Many polymers are also viscoelastic in mechanical nature and we will discuss this as well, but there are also even other polymers (this is why polymers are the best) that exhibit elastic behavior for very long timescales, and these polymers are elastomers, and this property is rubber elasticity. We will describe the molecular characteristics that give rise to rubber elasticity, the physical model used to describe the rubbery response, and finally how we can use this model to find the stress-strain behavior of an elastomeric polymer.

Rubber Elasticity:

Typically when we discuss rubbers and the mechanics of rubbers or elastomers we will talk about networks. And when we deform those networks we describe this behavior as rubber elasticity. Microscopically, we envision a rubber network as a large, entwined set of polymer chains that are attached together by crosslinks, referring to points that join otherwise distinct chains. Crosslinks can be chemical or physical in nature — chemical crosslinks refer to actual covalent bonds between chains (vulcanization), while physical crosslinks refer to non-specific interactions between chains. Examples of physical crosslinks include entanglements in polymer melts, glassy phase-separated regions in a rubbery matrix (as in high-impact polystyrene or HIPS), block copolymers, or possibly hydrogen bonding or other types of strong non-covalent bonds between groups that are highly separated along a chain.

The presence of crosslinks joins otherwise disparate chains together such that the entire rubber is connected and individual chains are no longer capable of diffusing completely independently in response to a perturbation. As a result, stressing a crosslinked network leads to a mechanical response that is partially characteristic of an elastic material and partially characteristic of a viscous material, with the important caveat that under a mechanical stress the overall network maintains mechanical robustness.

We will quantitatively describe how rubbery networks respond to stress using basic principles of mechanics and a thermodynamic description as well. It is critical to keep in mind the molecular structure that gives rise to rubbery behavior. The key concept to recognize is that a rubber is still composed of flexible polymer chains that can easily rearrange in the presence of a stress, allowing large amounts of strain when stressed. However, the presence of crosslinks prevents the chains from completely flowing, and as a result some solid-like mechanical behavior is observed. In the context of the discussion of relaxation time scales we could say that the rubber exhibits a longest relaxation time characteristic of a solid (the relaxation time associated with removing crosslinks) while having shorter relaxation times characteristic of liquids (the relaxation time associated with diffusing the polymer chains between crosslinks). We will have multiple relaxation times and see behaviors characteristic of both liquid and solid behavior.

Before describing this behavior and creating a theoretical framework we will be making several key assumptions:

1. Gaussian subchains between permanent crosslinks — that is, you can envision the part of a chain between two crosslinked points as a chain itself, which can be described by a Gaussian distribution. This assumption breaks down if the distance between crosslinks is very small.
2. Temperature well above \( T_g \), so that the system has sufficient thermal energy for chains to rearrange.
3. Flexible chains with relatively easy backbone bond rotation potentials, again reflecting the ability of chains to rearrange.
4. Chain deformation occurs by conformational changes.
5. No relaxation by chain slip — the network appears as fixed on the time scale of experiment, i.e. physical crosslinks can be assumed to be permanent. This means that as we pull the rubber, the chains are not able to slide past each other and thus fall apart.
6. Affine deformation — microscopic deformation by same amount as macroscopic deformation; that is, any given chain's dimensions change by the same ratio as the change in dimensions of the overall network.
7. No crystallization of the chains at large strains.
8. No change in volume of the rubber upon deformation, i.e. incompressibility assumption.

Now there are two distinct types of behavior that are well described by our rubber elasticity framework:

- Mechanical Deformation: we apply some stress and measure the strain response for a fixed volume.
- Gel Swelling: a polymer network swells due to mixing a solvent and the volume increases.

We have hinted at energetic competition for mechanically deforming a rubber network previously in other courses, specifically that stretching polymer chains decreases the conformational entropy of the chain and we had an expression for the energy of a stretched chain based on the change in the number of microstates:

\[
G(r)_{stretch} = \frac{3kT}{2} \frac{r^2}{r_0^2}
\]

Here energy increases if the chain extends, and thus we have an entropic spring restoring force that opposes pulling.

Let's start with the mechanical deformation first.

Rubber Elasticity System

As usual we want to think about the energetics and thermodynamics associated with changing the state of our system when we apply mechanical stress to the crosslinked polymer network. The initial state is the unstressed, relaxed polymer network. We assume that we have a large number of subchains which are the chain segments between crosslinked points. We assume that these subchains can be defined as Gaussian with unperturbed end-to-end distance \( \langle r_0^2 \rangle \). We also assume that the molecular weight of each subchain, \( M_x \), is the same. Given an initial set of chains, the molecular weight of each subchain will thus decrease as the degree of crosslinking is increased.

Upon applying a force (stress) to the network, we assume that the dimensions extend to a new end-to-end distance \( \langle r^2 \rangle \). Here, we can invoke the assumption of an affine deformation — the deformation experienced by each individual subchain is equivalent to the deformation experienced by the whole network. Using this assumption, we can relate the microscopic perturbed dimensions to unperturbed dimensions via an extension ratio in each direction 1, 2, and 3, which is defined as the ratio between the macroscopic deformed length \( l_i \) in that direction and the undeformed length \( l_0 \) (assuming a sample with the same \( l_0 \) in all 3 directions). These extension ratios are:

\[
\alpha_1 = \frac{l_1}{l_0}
\]
\[
\alpha_2 = \frac{l_2}{l_0}
\]
\[
\alpha_3 = \frac{l_3}{l_0}
\]

Because of the affine deformation assumption, we can also write coordinates of an individual subchain in terms of the extension ratios, if we imagine deforming a subchain between one crosslink fixed at (0, 0, 0) and one crosslink fixed at (1, 2, 3), with deformed coordinates (1′, 2′, 3′):

\[
1' = \alpha_1 (1), \quad 2' = \alpha_2 (2), \quad 3' = \alpha_3 (3)
\]

We also then see that the end-to-end distances are defined as:

\[
r^2 = 1'^2 + 2'^2 + 3'^2
\]
\[
r_0^2 = 1^2 + 2^2 + 3^2
\]

Finally, since we assume that the network is incompressible, the total volume change has to be equal to 0, which means that the product of the deformed lengths has to be equal to the product of the undeformed lengths:

\[
\alpha_1 \alpha_2 \alpha_3 = 1
\]

Thermodynamic Perspective of Rubber Elasticity

Now we need a thermodynamic relationship to derive an expression between force and extension for a general material, which we can then apply to our polymer network. As usual, let's start with Gibbs:

\[
G = H - TS
\]

Finding the differential form of \( G \), and including a work term for extension in \( U \), gives:

\[
dG = -S dT + V dP + F d l
\]

where \( G \) is the Gibbs free energy, \( l \) is the extension of the chain, and \( F \) is the force. By rearranging this equation we can see that \( F \) is:

\[
F = \left( \frac{\partial G}{\partial l} \right)_{T,P}
\]

We can thus take the derivative of our first expression for \( G \) with respect to \( l \) at constant \( T \) and \( P \) to get:

\[
F = \left( \frac{\partial H}{\partial l} \right)_{T,P} - T \left( \frac{\partial S}{\partial l} \right)_{T,P}
\]

Remember the physical interpretation: the equation above tells us that the force associated with a change in chain extension is related to the derivative of the enthalpy, which measures bond interactions, and entropy. For ideal rubbers, we assume that the change in enthalpy is 0 as the rubber network is stretched. So then:

\[
F = - T \left( \frac{\partial S}{\partial l} \right)_{T,P}
\]

Here we see that the restoring force of rubber elasticity is driven by entropy. We assume there is no change in enthalpy when a rubber is stretched, in part because of the fixed volume assumption — since the volume is not changing, the net number of interactions is assumed to be the same on average, so there is no net enthalpy change. This assumption of no net enthalpy change is one of the unique characteristics of polymers — for example, we could not apply the same reasoning to metals because of strong enthalpic changes associated with stretching bonds, etc. Even if we relax the assumption of a zero enthalpy change, we would still find that for polymers the change in entropy is the dominant term, allowing us to refer to the corresponding force as an entropic spring force.

Since the elastic force that arises from stressing a rubber network is related to the derivative of the entropy, we need to calculate the entropy based on the extension ratios defined previously. Entropy is related to the number of microstates available to the polymer, which is related to the probability distribution of the Gaussian subchains. So we can define the entropy change for a single chain as:

\[
S_2 - S_1 = \Delta S = k \ln \frac{\Omega_2}{\Omega_1}
\]
\[
\Omega_1 = \text{const} \times \exp \left( \frac{-3 r_0^2}{2 n l^2} \right)
\]
\[
\Omega_2 = \text{const} \times \exp \left( \frac{-3 (\alpha_1^2 1^2 + \alpha_2^2 2^2 + \alpha_3^2 3^2)}{2 n l^2} \right)
\]

Here we assume that \( \Omega_1 \propto P(r_0) \) and \( \Omega_2 \propto P(r) \) where \( P(r) \) is the Gaussian distribution, and \( r_0 \) and \( r \) are the relaxed and stressed dimensions defined above. This then gives:

\[
\Delta S = k \ln \frac{\Omega_2}{\Omega_1} = \frac{-3k}{2 n l^2} \left[ (\alpha_1^2 - 1)1^2 + (\alpha_2^2 - 1)2^2 + (\alpha_3^2 - 1)3^2 \right]
\]

Again, we are able to state that the entropy of a single chain is related to the extension ratios associated with the macroscopic sample because of our assumption of affine/homogeneous deformation. Right now we have the entropy in terms of the relaxed coordinates 1, 2, 3. In this state, where the overall chain dimensions can be described by the Gaussian distribution, then we can expect that there is no distinction between a random walk in either of the 3 directions, so the average contribution to \( \langle r_0^2 \rangle \) from \( 1^2 \), \( 2^2 \), and \( 3^2 \) is the same so:

\[
\langle 1^2 \rangle = \langle 2^2 \rangle = \langle 3^2 \rangle = \frac{\langle r_0^2 \rangle}{3}
\]

Substituting in these values and knowing that \( n l^2 = r_0 \) gives:

\[
\Delta S = \frac{-k}{2} \left( \alpha_1^2 + \alpha_2^2 + \alpha_3^2 - 3 \right)
\]

Finally, \( F \) is:

\[
\Delta F_i = -T \left( \frac{\partial \Delta S(\alpha_i)}{\partial l_i} \right)_{T,P}
\]

where \( l_i \) denotes the force in a given direction \( i = 1, 2, 3 \).

Uniaxial Deformation of a Rubber Network

Now that we have an expression for the change in entropy of a chain given some deformation and have a thermodynamic relation that yields the force associated with that deformation, we can start analyzing different scenarios. Let's start simple with uniaxial deformation and find the resulting elastic force. Let's take deformation along the \( x \)-axis so that:

\[
\alpha_1 = \frac{l_1}{l_0}
\]
\[
\alpha_2 = \frac{1}{\sqrt{\alpha_1}}
\]
\[
\alpha_3 = \frac{1}{\sqrt{\alpha_1}}
\]

Here, we calculate \( \alpha_2 \) and \( \alpha_3 \) from the incompressibility condition \( \alpha_1 \alpha_2 \alpha_3 = 1 \). We can then write \( \Delta S \) in terms of just \( \alpha_1 \):

\[
\Delta S (\alpha_1) = -\frac{k}{2} \left( \alpha_1^2 + \frac{2}{\alpha_1} - 3 \right)
\]

The 3-dimensional problem has now broken down into only knowing the extension in a single dimension. We can take the derivative with respect to \( l_1 \) to find \( F \), using the chain rule:

\[
F_1 = -T \frac{\partial}{\partial l_1} \Delta S (\alpha_1)
\]
\[
= -T \frac{\partial}{\partial \alpha_1} \left( \frac{\partial \alpha_1}{\partial l_1} \right) \Delta S (\alpha_1)
\]
\[
= \frac{kT}{2 l_0} \frac{\partial}{\partial \alpha_1} \left( \alpha_1^2 + \frac{2}{\alpha_1} - 3 \right)
\]
\[
F_1 = \frac{kT}{l_0} \left( \alpha_1 - \frac{1}{\alpha_1^2} \right)
\]

To put this in more familiar terms, we can find the stress by dividing by the cross-sectional area (in the 2–3 plane), which we'll call \( A_0 \). So far we've derived \( F_1 \) for a single chain only, and would ideally like to know the stress required to deform a network of \( z \) total subchains per volume \( V \):

\[
\sigma_1 = \frac{z F_1}{A_0} = \frac{z k T}{A_0 l_0} \left( \alpha_1 - \frac{1}{\alpha_1^2} \right) = N_{xlinks} k T \left( \alpha_1 - \frac{1}{\alpha_1^2} \right)
\]

Here, we simplify by letting \( A_0 l_0 = V \) and \( \frac{z}{V} = N_{xlinks} \), where \( N_{xlinks} \) is the number of crosslinks per unit volume. Typically the restoring force is written in terms of the molecular mass of crosslinks instead of their number, so we can rearrange to find the crosslink molecular mass in terms of the density as:

\[
M_x = \frac{\text{mass of polymer/volume}}{\text{number of subchains per volume}} = \frac{\rho}{N_{xlinks}/N_A}
\]
\[
\sigma_x = \frac{\rho N_A k T}{M_x} \left( \alpha_1 - \frac{1}{\alpha_1^2} \right)
\]

where \( N_A \) is Avogadro's number. We can write the stress as a function of strain (the typical measurement of material deformation) using the relation:

\[
\alpha_1 = \frac{l_1}{l_0} = \frac{l_0 + \Delta l}{l_0} = 1 + \frac{\Delta l}{l_0} = 1 + \epsilon_1
\]
\[
\sigma_1 = \frac{\rho N_A k T}{M_x} \left( 1 + \epsilon_1 - \frac{1}{(1 + \epsilon_1)^2} \right)
\]

In the limit of small strains, we see that the stress scales linearly with the strain, and hence we can define the Young's modulus:

\[
E = \frac{\rho N_A k T}{M_x}
\]

Let's stop here for a moment and appreciate this result. We see here that for a rubber network the Young's modulus increases with temperature and decreases with the molecular weight of crosslinks. This implies that if we increase the density of crosslinks, then the molecular weight of the crosslinks decreases and the Young's modulus goes up, leading to a greater elastic restoring force. However, do recall that if the density gets too high, the assumption of Gaussian subchains will eventually break down, and the derivation will no longer be valid.

Since we have shown that the Young's modulus depends on crosslink molecular mass (and hence crosslink density), and we have stated that entanglements can act as physical crosslinks at short times (since at longer times the entanglements will eventually diffuse away), then we can measure the elastic response of an entangled melt at some known temperature to approximate the number of statistical segments between entanglements \( N_E \). In principle, this type of experiment would also allow us to identify the critical molecular weight for crossing over between the Rouse and reptation regimes, since the onset of the reptation regime would be associated with the onset of elastic behavior at small timescales.

Limits of Rubber Elasticity Model

This model works very well at applied strains that are relatively low; however, in regimes where the chain extension is high, the model begins to break down.

- Strain-induced Crystallization: at large extensions the polymer chains can pack tightly and form crystalline regions (shish-kebob spherulites), and thus the Young's modulus will be larger than predicted by theory.
- Subchain Length Reaches Contour Length: at this point we are pulling on covalent bonds and the Young's modulus will approach that of metals.
- Bond Rotation Away from Trans State at High Extension: (see discussion below).
- Dangling Ends: experimental Young's modulus is lower than theory because there are regions between the end of the chain and crosslink that can form entanglements. This increases the crosslink density of the network somewhat artificially, as the physical crosslinks disappear when deformation occurs.
- System Temperature Drops Below \( T_g \): recall that we are implicitly assuming that the rubbery network is above its glass transition temperature, and thus chains are freely able to deform and hence explore a full conformational space. If the network is cooled below the glass transition temperature, the entropic spring arguments will fail because the network no longer acts like a Gaussian chain, but instead will give an elastic response consistent with the kinetic barriers to chain rearrangement.
- Too Many Crosslinks: a major assumption of this model is that every subchain in the rubbery network can be treated as a Gaussian chain, and thus loses a significant amount of conformational entropy upon stretching. If the crosslink density is high, however, the number of segments between crosslinks will correspondingly become very low and the subchains may no longer be considered Gaussian.
- Too Few Crosslinks: if the chains are too long, entanglements can occur between subchains and the entanglements can act like crosslinks on short timescales, so we could imagine the effective crosslink density being much higher than anticipated due to the presence of crosslinks. Note that even in the case of moderate crosslinking density, there will typically be chain ends which are capable of forming crosslinks and may result in spurious results.

Now that was quite a deviation from where we were headed in terms of relating stress to strain via constitutive equations for linear elastic isotropic cubic materials. So let's get back on track, and before we get into some really nice linear algebra we have to define one more very important material property, and that is the Poisson ratio.

Poisson’s Ratio

Now what happens to the dimensions perpendicular to the direction of applied force when the material is subjected to uniaxial tension? We know from experience that a body being pulled in tension will contract laterally. This lateral strain is described by the Poisson's Ratio:

\[
\nu = -\frac{\epsilon_L}{\epsilon_A}
\]

[Figure: General Poisson’s Ratio]

where \( \epsilon_L \) is the lateral strain and \( \epsilon_A \) is the axial strain. Note the negative sign which signifies a decrease in length. Some typical examples for Poisson's ratio are:

- Metals: 0.3
- Ceramics: 0.2
- Polymers/Biomaterials: 0.4
- Regular Hexagonal Honeycomb: 1
- Cork: 0
- Isotropic solids: \( -1 < \nu < 0.5 \)

Now that we know the Poisson ratio, we can start to develop our constitutive equations for complex stress states.

Biaxial and Complex Stress States

What happens when we have a more complex stress state and we are interested in the stress state of a particular plane of orientation angle \( \theta \)? Well, we do this the same way we resolve force vectors onto a new axis of interest. Let's look at a case of plane stress, which is a case when all the stress is contained in one plane.

So far we have been dealing with simple uniaxial stress along the principal testing direction; however, often materials can also be subjected to shear stress or strain.

With this information let us move on to more complex stress states, specifically one that typically exists on a free, not constrained, surface of a stressed material.

Consider the element which is initially stressed in the 1 direction by applying \( \sigma_{11} \). There will be a resultant strain:

\[
\epsilon_{11} = \frac{\sigma_{11}}{E}
\]
\[
\epsilon_{22} = \frac{-\nu \sigma_{11}}{E}
\]

The element is then stressed in the 2 direction by applying \( \sigma_{22} \). The resultant strain will be:

\[
\epsilon_{22} = \frac{\sigma_{22}}{E}
\]
\[
\epsilon_{11} = \frac{-\nu \sigma_{22}}{E}
\]

Therefore, the total strain in the x and y directions will be:

\[
\epsilon_{11} = \frac{\sigma_{11} - \nu \sigma_{22}}{E}
\]
\[
\epsilon_{22} = \frac{\sigma_{22} - \nu \sigma_{11}}{E}
\]

We can solve these two equations (two unknowns and two equations) to find the total stress in the x and y directions:

\[
\sigma_{11} = \frac{E (\epsilon_{11} + \nu \epsilon_{22})}{1 - \nu^2}
\]
\[
\sigma_{22} = \frac{E (\epsilon_{22} + \nu \epsilon_{11})}{1 - \nu^2}
\]

3D Stress State

We can extend this even further to find the total strain if there is another stress applied, \( \sigma_{z} \), in the z direction:

\[
\epsilon_{11} = \frac{1}{E} [\sigma_{11} - \nu(\sigma_{22} + \sigma_{33})]
\]
\[
\epsilon_{22} = \frac{1}{E} [\sigma_{22} - \nu(\sigma_{33} + \sigma_{11})]
\]
\[
\epsilon_{33} = \frac{1}{E} [\sigma_{33} - \nu(\sigma_{11} + \sigma_{22})]
\]

We can similarly write the stress as well:

\[
\sigma_{11} = \frac{E}{(1+\nu)(1-2\nu)} \bigg[ (1-\nu)\epsilon_{11} + \nu(\epsilon_{22} + \epsilon_{33})\bigg]
\]
\[
\sigma_{22} = \frac{E}{(1+\nu)(1-2\nu)} \bigg[ (1-\nu)\epsilon_{22} + \nu(\epsilon_{33} + \epsilon_{11})\bigg]
\]
\[
\sigma_{33} = \frac{E}{(1+\nu)(1-2\nu)} \bigg[ (1-\nu)\epsilon_{33} + \nu(\epsilon_{11} + \epsilon_{22})\bigg]
\]

General Expression for Strain in Complex Stress-State

We can also introduce here a more general way to write strain where:

\[
\epsilon_{ij} = \frac{1}{E} \bigg[ (1+\nu)\sigma_{ij} - \nu \sigma_{kk} \delta_{ij} \bigg]
\]

where \( \delta_{ij} \) is the Kronecker Delta, which will be 1 when \( i = j \) and 0 when \( i \neq j \).

Or we can describe this in linear algebra form as follows, where that large matrix relating our strain and stress tensors is known as our compliance matrix \( S \). If you flip and relate stress to strain you will work with a stiffness matrix \( C \).

\[
\begin{bmatrix}
\epsilon_{11}\\
\epsilon_{22}\\
\epsilon_{33}\\
\epsilon_{23}\\
\epsilon_{13}\\
\epsilon_{12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E} & -\frac{\nu}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & \frac{1}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & -\frac{\nu}{E} & \frac{1}{E} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1+\nu}{E} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1+\nu}{E} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1+\nu}{E}
\end{bmatrix}
\begin{bmatrix}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{33}\\
\sigma_{23}\\
\sigma_{13}\\
\sigma_{12}
\end{bmatrix}
\]

We can also relate this to engineering shear strain and the shear modulus \( G \):

\[
\begin{bmatrix}
\epsilon_{11}\\
\epsilon_{22}\\
\epsilon_{33}\\
\gamma_{23}\\
\gamma_{13}\\
\gamma_{12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E} & -\frac{\nu}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & \frac{1}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & -\frac{\nu}{E} & \frac{1}{E} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{G} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{G} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{G}
\end{bmatrix}
\begin{bmatrix}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{33}\\
\sigma_{23}\\
\sigma_{13}\\
\sigma_{12}
\end{bmatrix}
\]

Where we can write the shear modulus in terms of the Young's modulus as:

\[
G = \frac{E}{2(1+\nu)}
\]

We can also define the bulk modulus, \( \kappa \), in the case of hydrostatic stress, i.e. all shear stresses are 0 and all normal stresses are equal, i.e. \( \sigma_{11} = \sigma_{22} = \sigma_{33} = \sigma_{hydro} \):

\[
\kappa = \frac{\sigma_{hydro}}{\frac{\Delta V}{V_0}} = \frac{E}{3(1-2\nu)}
\]

Example: Jewelry Maker — Strain Without Stress?

A jewelry maker creates a die to form a new cufflink made of steel. We need to know how much stress to apply to the metal slab to reduce the thickness to 3 mm. The die is a simple channel that does not constrain the metal in the 1-direction, and the channel is well lubricated so that all frictional forces and stresses along the channel walls can be ignored.

a.) State all the components of the stress tensor under the applied stress.

Well, there is clearly a stress in the 3 direction. What about 1 and 2?

b.) Which normal strain components are zero and why?

c.) Use your answers in part a) and b) to express the relationship between the non-zero stresses in this system.

Finally, I won't leave until we complete thin-walled pressure vessels!

Pressure Vessels: A Special Stress State

Finally, let's consider a very common scenario, a thin-walled pressure vessel. The thin-walled vessel has an applied pressure difference \( \Delta P \) between the internal pressure and the environment. The wall thickness is sufficiently small where we can say \( t \ll R \) or that \( \frac{R}{t} \geq 10 \).

What is the stress state? It is a plane stress state due to symmetry considerations. Therefore, we need to find \( \sigma_{1} \) and \( \sigma_{2} \), which we will accomplish by drawing free body diagrams.

Let us first consider the longitudinal direction. Remember that the forces must sum to zero.

\[
\sum F_{1} = 0 = \Delta P \pi r^{2} - \sigma_{11} 2 \pi r t
\]
\[
\sigma_{11} = \sigma_{Longitudinal} = \frac{\Delta P r}{2 t}
\]

We have the longitudinal stress, so now let us look at the other direction, circumferential or hoop, and again the forces must sum to equal zero:

\[
\sum F_{2} = 0 = -\Delta P \Delta x 2 R + \sigma_{22} \Delta x t 2
\]
\[
\sigma_{22} = \sigma_{Hoop} = \frac{\Delta P R}{t}
\]

As you can see, the hoop stress is twice as large as the longitudinal stress, so you should expect the material to fail in such a manner, unless there are extenuating conditions, i.e. defects, corrosion, different processing, etc. The strain ratio for hoop to longitudinal strain is nearly 4:1.