Low chain extension regime, Ia

The Molecular Kink Paradigm envisions a representative network chain as a series of vectors that follow the chain contour within its tube. Each vector represents the equilibrium end-to-end distance of a kink. The actual 3-dimensional path of the chain is not pertinent, since all elastic forces are assumed to operate along the chain contour. In addition to the chain's contour length, the only other important parameter is its tortuosity, the ratio of its contour length to its end-to-end distance. As the chain is extended, in response to an applied strain, the induced elastic force is assumed to propagate uniformly along its contour. Consider a network chain whose end points (network nodes) are more or less aligned with the tensile strain axis. As the initial strain is applied to the rubber sample, the network nodes at the ends of the chain begin to move apart and all of the kink vectors along the contour are stretched simultaneously. Physically, the applied strain forces the kinks to stretch beyond their thermal equilibrium end-to-end distances, causing a decrease in their entropy. The increase in free energy associated with this change in entropy, gives rise to a (linear) elastic force that opposes the strain. The force constant for the low strain regime can be estimated by sampling molecular dynamics (MD) trajectories of a kink, i.e. short chains, composed of 2–3 isoprene units, at relevant temperatures, e.g. 300K.^[5] By taking many samples of the coordinates over the course of the simulations, the probability distributions of end-to-end distance for a kink can be obtained. Since these distributions (which turn out to be approximately Gaussian) are directly related to the number of states, we may associate them with the entropy of the kink at any end-to-end distance. By numerically differentiating the probability distribution, the change in entropy, and hence free energy, with respect to the kink end-to-end distance can be found. The force model for this regime is found to be linear and proportional to the temperature divided by the chain tortuosity.

Fig. 2 Isoprene backbone unit. composed of carbon atoms (dark grey) and hydrogen atoms (white). Carbon atoms labeled '1' and '6' are in adjacent units on the polymer chain.

Moderate chain extension regime, Ib

At some point in the low extension regime, i.e. as all of the kinks along the chain are being extended simultaneously, it becomes energetically more favorable to have one kink transition to an extended conformation in order to stretch the chain further. The applied strain can force a single isoprene unit within a kink into an extended conformation, slightly increasing the end-to-end distance of the chain, and the energy required to do this is less than that needed to continue extending all of the kinks simultaneously. Numerous experiments^[13] strongly suggest that stretching a rubber network is accompanied by a decrease in entropy. As shown in Fig. 2, an isoprene unit has three single C-C bonds and there are two or three preferred rotational angles (orientations) about these bonds that have energy minima. Of the18 allowed^[6] rotational conformations, only 6 have extended end-to-end distances and forcing the isoprene units in a chain to reside in some subset of the extended states must reduce the number of rotational conformations available for thermal motion. It is this reduction in the number of available states that causes the entropy to decrease. As the chain continues to straighten, all of the isoprene units in the chain are eventually forced into extended conformations and the chain is considered to be ‘taut’. A force constant for chain extension can be estimated from the resulting change in free energy associated with this entropy change.^[6] As with regime Ia, the force model for this regime is linear and proportional to the temperature divided by the chain tortuosity.

High chain extension regime, II

When all of the isoprene units in a network chain have been forced to reside in just a few extended rotational conformations, the chain becomes taut. It may be regarded as sensibly straight, except for the zigzag path that the C-C bonds make along the chain contour. However, further extension is still possible by bond distortions, e.g., bond angle increases, bond stretches and dihedral angle rotations. These forces are spring-like and are not associated with entropy changes. A taut chain can be extended by only about 40%. At this point the force along the chain is sufficient to mechanically rupture the C-C covalent bond. This tensile force limit has been calculated^[7] via quantum chemistry simulations and it is approximately 7 nN, about a factor of a thousand greater than the entropic chain forces at low strain. The angles between adjacent backbone C-C bonds in an isoprene unit vary between about 115–120 degrees and the forces associated with maintaining these angles are quite large, so within each unit, the chain backbone always follows a zigzag path, even at bond rupture. This mechanism accounts for the steep upturn in the elastic stress, observed at high strains (Fig. 1).

Network morphology

Although the network is completely described by only two parameters (the number of network nodes per unit volume and the statistical de-correlation length of the polymer, the Kuhn length), the way in which the chains are connected is actually quite complicated. There is a wide variation in the lengths of the chains and most of them are not connected to the nearest neighbor network node. Both the chain length and its end-to-end distance are described by probability distributions. The term ‘morphology’ refers to this complexity. If the cross-linking agent is thoroughly mixed, there is an equal probability for any isoprene unit to become a network node. For dicumyl peroxide, the cross linking efficiency in natural rubber is unity,^[14] but this is not the case for sulfur.^[15] The initial morphology of the network is dictated by two random processes: the probability for a cross-link to occur at any isoprene unit and the Markov random walk nature of a chain conformation.^[8][9] The probability distribution function for how far one end of a chain end can ‘wander’ from the other is generated by a Markov sequence.^[16] This conditional probability density function relates the chain length ${\displaystyle n}$ in units of the Kuhn length ${\displaystyle b}$ to the end-to-end distance ${\displaystyle r}$:

$${\displaystyle P(r|n)=4\pi r^{2}\left({\frac {2nb^{2}\pi }{3}}\right)^{-{3}/{2}}\exp \left(-{\frac {3r^{2}}{2nb^{2}}}\right)\,}$$

(1)

The probability that any isoprene unit becomes part of a cross-link node is proportional to the ratio of the concentrations of the cross-linker molecules (e.g., dicumyl-peroxide) to the isoprene units:

$${\displaystyle p_{x}=2{\frac {\text{[cross-link]}}{\text{[isoprene]}}}}$$

The factor of two comes about because two isoprene units (one from each chain) participate in the cross-link. The probability for finding a chain containing ${\displaystyle N}$ isoprene units is given by:

$${\displaystyle P(N)=p_{x}{\left(1-p_{x}\right)}^{N-1}\,,}$$

(3)

where ${\displaystyle N\geq 1}$. The equation can be understood as simply the probability that an isoprene unit is NOT a cross-link (1−p_x) in N−1 successive units along a chain. Since P(N) decreases with N, shorter chains are more probable than longer ones. Note that the number of statistically independent backbone segments is not the same as the number of isoprene units. For natural rubber networks, the Kuhn length contains about 2.2 isoprene units, so ${\displaystyle N\sim 2.2n}$. It is the product of equations (1) and (3) (the joint probability distribution) that relates the network chain length (${\displaystyle N}$) and end-to-end distance (${\displaystyle r}$) between its terminating cross-link nodes:

$${\displaystyle P(r,N)\;=\;P(N)P(r|N)\;=\;p_{x}{\left(1-p_{x}\right)}^{N-1}\,4\pi r^{2}\left({\frac {2nb^{2}\pi }{3}}\right)^{-{3}/{2}}\exp \left(-{\frac {3r^{2}}{2nb^{2}}}\right)}$$

(4)

Fig. 3 Probability density for an average network chain vs. end-to-end distance in units of mean cross-link node spacing (2.9 nm); n= 52, b= 0.96 nm.

The complex morphology of a natural rubber network can be seen in Fig. 3, which shows the probability density vs. end-to-end distance (in units of mean node spacing) for an ‘average’ chain. For the common experimental cross-link density of 4x10¹⁹ cm⁻³, an average chain contains about 116 isoprene units (52 Kuhn lengths) and has a contour length of about 50 nm. Fig. 3 shows that a significant fraction of chains span several node spacings, i.e., the chain ends overlap other network chains. Natural rubber, cross-linked with dicumyl peroxide, has tetra-functional cross-links, i.e. each cross-link node has 4 network chains emanating from it. Depending on their initial tortuosity and the orientation of their endpoints with respect to the strain axis, each chain associated with an active cross-link node can have a different elastic force constant as it resists the applied strain. To preserve force equilibrium (zero net force) on each cross-link node, a node may be forced to move in tandem with the chain having the highest force constant for chain extension. It is this complex node motion, arising from the random nature of the network morphology, that makes the study of the mechanical properties of rubber networks so difficult. As the network is strained, paths composed of these more extended chains emerge that span the entire sample, and it is these paths that carry most of the stress at high strains.

Numerical network simulation model

To calculate the elastic response of a rubber sample, the three chain force models (regimes Ia, Ib and II) and the network morphology must be combined in a micro-mechanical network model.^[10][11][12] Using the joint probability distribution in equation (4) and the force extension models, it is possible to devise numerical algorithms to both construct a faithful representative volume element of a network and to simulate the resulting mechanical stress as it is subjected to strain. An iterative relaxation algorithm is used to maintain approximate force equilibrium at each network node as strain is imposed. When the force constant obtained for kinks having 2 or 3 isoprene units (approximately one Kuhn length) is used in numerical simulations, the predicted stress is found to be consistent with experiments. The results of such a calculation^[15] are shown in Fig. 1 (dashed red line) for sulfur cross-linked natural rubber and compared with experimental data ^[17] (solid blue line). These simulations also predict a steep upturn in the stress as network chains become taut and, ultimately, material failure due to bond rupture. In the case of sulfur cross-linked natural rubber, the S-S bonds in the cross-link are much weaker than the C-C bonds on the chain backbone and are the network failure points. The plateau in the simulated stress, starting at a strain of about 7, is the limiting value for the network. Stresses greater than about 7 MPa cannot be supported and the network fails. Near this stress limit, the simulations predict^[12] that less than 10% of the chains are taut, i.e. in the high chain extension regime and less than 0.1% of the chains have ruptured. While the very low rupture fraction may seem surprising, it is not inconsistent with our experience of stretching a rubber band until it breaks. The elastic response of the rubber after breaking is not noticeably different from the original.

Freely-jointed chain model

The freely joined chain, also called an ideal chain, follows the random walk model. Microscopically, the 3-D random walk of a polymer chain assumes the overall end-to-end distance is expressed in terms of the x, y and z directions:

Model of the freely jointed chain

$${\displaystyle {\vec {R}}=R_{x}{\hat {x}}+R_{y}{\hat {y}}+R_{z}{\hat {z}}}$$

In the model, ${\displaystyle b}$ is the length of a rigid segment, ${\displaystyle N}$ is the number of segments of length ${\displaystyle b}$, ${\displaystyle R}$ is the distance between the fixed and free ends, and ${\displaystyle L_{\text{c}}}$ is the "contour length" or ${\displaystyle Nb}$. Above the glass transition temperature, the polymer chain oscillates and ${\displaystyle r}$ changes over time. The probability distribution of the chain is the product of the probability distributions of the individual components, given by the following Gaussian distribution:

$${\displaystyle P({\vec {R}})=P(R_{x})P(R_{y})P(R_{z})=\left({\frac {2nb^{2}\pi }{3}}\right)^{-{3}/{2}}\exp \left({\frac {-3R^{2}}{2Nb^{2}}}\right)}$$

Therefore, the ensemble average end-to-end distance is simply the standard integral of the probability distribution over all space. Note that the movement could be backwards or forwards, so the net average ${\displaystyle \langle R\rangle }$ will be zero. However, one can use the root mean square as a useful measure of the distance.

$${\displaystyle {\begin{aligned}\langle R\rangle &=0\\\langle R^{2}\rangle &=\int _{0}^{\infty }R^{2}4\pi R^{2}P({\vec {R}})dR=Nb^{2}\\\langle R^{2}\rangle ^{\frac {1}{2}}&={\sqrt {N}}b\end{aligned}}}$$

The Flory theory of rubber elasticity has pointed out the rubber elasticity has primarily entropic origins. By using the following basic equations for Helmholtz free energy and its discussion about entropy, the force generated from the deformation of a rubber chain from its original un-stretched conformation can be derived. The ${\displaystyle \Omega }$ is the number of conformations of the polymer chain. Since the deformation does not involve enthalpy change, the change in free energy can just be calculated as the change in entropy${\displaystyle -T\Delta S}$. It can be observed that the force equation resembles the behavior of a spring and follows Hooke's law: ${\displaystyle F=kx}$, where F is the force, k is the spring constant and x is the distance. Usually, neo-Hookean model can be used on cross-linked polymers to predict their stress-strain relations:

$${\displaystyle \Omega =C\exp \left({\frac {-3{\vec {R}}^{2}}{2Nb^{2}}}\right)}$$

$${\displaystyle S=k_{\text{B}}\ln \Omega \,\approx {\frac {-3k_{\text{B}}{\vec {R}}^{2}}{2Nb^{2}}}}$$

$${\displaystyle \Delta F({\vec {R}})\approx -T\Delta S_{d}({\vec {R}}^{2})=C+{\frac {3k_{\text{B}}T}{Nb^{2}}}{\vec {R}}^{2}}$$

$${\displaystyle f={\frac {dF({\vec {R}})}{d{\vec {R}}}}={\frac {d}{d{\vec {R}}}}\left({\frac {3k_{\text{B}}T{\vec {R}}^{2}}{2Nb^{2}}}\right)={\frac {3k_{\text{B}}T}{Nb^{2}}}{\vec {R}}}$$

Note that the elastic coefficient ${\displaystyle 3k_{\text{B}}T/Nb}$ is temperature dependent. If we increase the rubber temperature, the elastic coefficient also rises. This is the reason why rubber under constant stress shrinks when its temperature increases.

We can further expand the Flory theory into a macroscopic view, where bulk rubber material is discussed. Assume the original dimension of the rubber material is ${\displaystyle L_{x}}$, ${\displaystyle L_{y}}$ and ${\displaystyle L_{z}}$, a deformed shape can then be expressed by applying an individual extension ratio ${\displaystyle \lambda _{i}}$ to the length (${\displaystyle \lambda _{x}L_{x}}$, ${\displaystyle \lambda _{y}L_{y}}$, ${\displaystyle \lambda _{z}L_{z}}$). So microscopically, the deformed polymer chain can also be expressed with the extension ratio: ${\displaystyle \lambda _{x}R_{x}}$, ${\displaystyle \lambda _{y}R_{y}}$, ${\displaystyle \lambda _{z}R_{z}}$. The free energy change due to deformation can then be expressed as follows:

$${\displaystyle {\begin{aligned}\Delta F_{\text{def}}({\vec {R}})&=-{\frac {3k_{\text{B}}T{\vec {R}}^{2}}{2Nb^{2}}}=-{\frac {3k_{\text{B}}T\left(\left(R_{x}^{2}-R_{x0}^{2}\right)+\left(R_{y}^{2}-R_{y0}^{2}\right)+\left(R_{z}^{2}-R_{z0}^{2}\right)\right)}{2Nb^{2}}}\\&=-{\frac {3k_{\text{B}}T\left(\left(\lambda _{x}^{2}-1\right)R_{x0}^{2}+\left(\lambda _{y}^{2}-1\right)R_{y0}^{2}+\left(\lambda _{z}^{2}-1\right)R_{z0}^{2}\right)}{2Nb^{2}}}\end{aligned}}}$$

Assume that the rubber is cross-linked and isotropic, the random walk model gives ${\displaystyle R_{x}}$, ${\displaystyle R_{y}}$ and ${\displaystyle R_{z}}$ are distributed according to a normal distribution. Therefore, they are equal in space, and all of them are 1/3 of the overall end-to-end distance of the chain: ${\displaystyle \langle R_{x0}^{2}\rangle =\langle R_{y0}^{2}\rangle =\langle R_{z0}^{2}\rangle =\langle R^{2}\rangle /3}$. Plugging in the change of free energy equation above, it is easy to get:

$${\displaystyle {\begin{aligned}\Delta F_{\text{def}}({\vec {R}})&=-{\frac {k_{\text{B}}Tn_{s}\langle R^{2}\rangle \left(\lambda _{x}^{2}+\lambda _{y}^{2}+\lambda _{z}^{2}-3\right)}{2Nb^{2}}}\\&=-{\frac {k_{\text{B}}Tn_{s}\langle R^{2}\rangle \left(\lambda _{x}^{2}+\lambda _{y}^{2}+\lambda _{z}^{2}-3\right)}{2R_{0}^{2}}}\end{aligned}}}$$

The free energy change per volume is just:

$${\displaystyle \Delta f_{\text{def}}={\frac {\Delta F_{\text{def}}({\vec {R}})}{V}}=-{\frac {k_{\text{B}}Tv_{s}\beta \left(\lambda _{x}^{2}+\lambda _{y}^{2}+\lambda _{z}^{2}-3\right)}{2}}}$$

where ${\displaystyle n_{s}}$ is the number of strands in network, the subscript "def" means "deformation", ${\displaystyle v_{s}=n_{s}/V}$, which is the number density per volume of polymer chains, ${\displaystyle \beta =\langle R^{2}\rangle /R_{0}^{2}}$ which is the ratio between the end-to-end distance of the chain and the theoretical distance that obey random walk statistics. If we assume incompressibility, the product of extension ratios is 1, implying no change in the volume: ${\displaystyle \lambda _{x}\lambda _{y}\lambda _{z}=1}$.

Case study: Uniaxial deformation:

In a uniaxial deformed rubber, because ${\displaystyle \lambda _{x}\lambda _{y}\lambda _{z}=1}$ we assume ${\displaystyle \lambda _{x}=\lambda _{y}=\lambda _{z}^{-1/2}}$. So the previous free energy per volume equation is:

$${\displaystyle \Delta f_{\text{def}}={\frac {\Delta F_{\text{def}}({\vec {R}})}{V}}=-{\frac {k_{\text{B}}Tv_{s}\beta \left(\lambda _{x}^{2}+\lambda _{y}^{2}+\lambda _{z}^{2}-3\right)}{2}}={\frac {k_{\text{B}}Tv_{s}\beta }{2}}\left(\lambda _{z}^{2}+{\frac {2}{\lambda _{z}}}-3\right)}$$

The engineering stress (by definition) is the first derivative of the energy in terms of the extension ratio, which is equivalent to the concept of strain:

$${\displaystyle \sigma _{\text{eng}}={\frac {d(\Delta f_{\text{def}})}{\lambda _{z}}}=k_{\text{B}}Tv_{s}\beta \left(\lambda _{z}-{\frac {1}{\lambda _{z}^{2}}}\right)}$$

and the Young's Modulus ${\displaystyle E}$ is defined as derivative of the stress with respect to strain, which measures the stiffness of the rubber in laboratory experiments.

$${\displaystyle E={\frac {d(\sigma _{\text{eng}})}{d\lambda _{z}}}=k_{\text{B}}Tv_{s}\beta \left.\left(1+{\frac {2}{\lambda _{z}^{3}}}\right)\right|_{\lambda _{z}=1}=3k_{\text{B}}Tv_{s}\beta ={\frac {3\rho \beta RT}{M_{s}}}}$$

where ${\displaystyle v_{s}=\rho N_{a}/M_{s}}$, ${\displaystyle \rho }$ is the mass density of the chain, ${\displaystyle M_{s}}$ is the number average molecular weight of a network strand between crosslinks. Here, this type of analysis^[27] links the thermodynamic theory of rubber elasticity to experimentally measurable parameters. In addition, it gives in sights into the cross-linking condition of the materials.

Worm-like chain model

The worm-like chain model (WLC) takes the energy required to bend a molecule into account. The variables are the same except that ${\displaystyle L_{\text{p}}}$, the persistence length, replaces ${\displaystyle b}$. Then, the force follows this equation:

$${\displaystyle F\approx {\frac {k_{\text{B}}T}{L_{\text{p}}}}\left({\frac {1}{4\left(1-{\frac {r}{L_{\rm {c}}}}\right)^{2}}}-{\frac {1}{4}}+{\frac {r}{L_{\text{c}}}}\right)}$$

Therefore, when there is no distance between chain ends (r=0), the force required to do so is zero, and to fully extend the polymer chain (${\displaystyle r=L_{\text{c}}}$), an infinite force is required, which is intuitive. Graphically, the force begins at the origin and initially increases linearly with ${\displaystyle r}$. The force then plateaus but eventually increases again and approaches infinity as the chain length approaches ${\displaystyle L_{\text{c}}}$.