## 1. Properties of rubber-like materials (amorphous crosslinked polymer above Tg)

### 1.1. High extensibility (minimum 100% strain)

### 1.2. Complete recovery after mechanical deformation

### 1.3. Engineering rubber

1.3.1. Natural rubber is weak

1.3.1.1. Crosslinking with vulcanization (commonly sulphur) strengthens natural rubber

1.3.1.1.1. Usually one crosslink per 500-1000 isoprene repeat unit

1.3.1.1.2. Chemical crosslinking vs physical crosslinking

1.3.1.2. Carbon black is also commonly added for mechanical reinforcement, UV and ozone stabilization to help with degradation

### 1.4. Ideal network

1.4.1. Flexible chains

1.4.2. Liquid-like motion between crosslinks

1.4.3. Chain between tie-points is a sub molecule

1.4.4. Randomness of chain is represented by the Gaussian theory

1.4.4.1. Concept Connection: if a chain is stretched too much (beyond an extension ratio of 1.6), the chain alignment becomes too great and the stress-strain relationship derived in the other section of the chapter no longer holds

### 1.5. Several properties (especially extensibility and complete recovery) are due to deformation induced entropy changes

1.5.1. Term: conformational entropy (probability substance is in a specific conformation)

1.5.1.1. Rubber can have a huge number of conformations because it's mobile

1.5.2. Rubber chains want to attain and maintain the highest conformational entropy. For rubbers, highest entropy is achieved when the chains are as tightly and highly coiled as possible

1.5.2.1. Rubbers arrange themselves to achieve the highest entropic state. The more coiled the better

1.5.2.1.1. Concept Connection: relate the force equation for rubbers with properties of rubber-like materials

## 2. Derivation of stress-strain relationship for an amorphous crosslinked network

### 2.1. Model: elongate a material with an external force F

2.1.1. Objective: derive the force equation for elastic solids and rubbers from the Helmholtz free energy equation

2.1.1.1. What does the force equation tell us?

2.1.1.1.1. Entropy decreases as deformation increases

2.1.1.2. Equations and assumptions

2.1.1.2.1. W = Fd

2.1.1.2.2. work = energy

2.1.1.2.3. Constant volume and temperature

2.1.1.3. F = W/d

2.1.1.3.1. F = A/d = dA/dL = d(E-TS)/dL

2.1.1.4. Force equation for elastic solids (metals and ceramics) vs rubbers

2.1.1.4.1. When elastic solids are pulled the force goes to changing the internal energy so the change in entropy is 0. In other words, the bonds themselves are deformed (pulled) instead of the molecules moving around

2.1.1.4.2. When rubbers are pulled the force goes to changing the conformation (entropy) so the change in internal energy is 0. In other words, the molecules move around instead of stretching their bonds.

2.1.2. Helmholtz free energy (A = E-TS)

2.1.2.1. A: Helmholtz free energy E: internal energy T: temperature S: entropy

2.1.2.1.1. Internal energy: energy stored in bonds

2.1.2.1.2. Temperature: equivalent to thermal energy

2.1.2.1.3. Entropy: conformational entropy

### 2.2. Entropy change of a single molecule

2.2.1. S = So + k*lnW

2.2.1.1. S: final entropy So: initial entropy k: Boltzman's constant W: probability of state

2.2.2. Objective: derive the force of one chain equation

2.2.2.1. rho (characteristic length) is designated with a p

2.2.2.2. S = So + k*lnW = So + k*lnP(r) = So - k[3ln(p*sqrt(pi)) + (r/p)^2] = dS/dr = -(2k/p^2)r

2.2.2.2.1. Concept Connection: use the derivation of the force of rubbers to find the force of one chain

### 2.3. Elasticity of a network

2.3.1. 1) Assumptions and setup

2.3.1.1. Work done internally = work done externally

2.3.1.2. No volume change

2.3.1.3. Isotropic (behaves and looks the same in all directions)

2.3.1.4. Affine deformation (forces exerted on the macro level are reflected on the micro scale)

2.3.1.5. In network vs out of network <r^2>

2.3.1.5.1. Means squared distance between crosslinks vs between ends

2.3.1.6. Scenario: pulling on a rubber band cube with a force F. Initial dimensions of xi, yi, and zi and final dimensions of x, y, and z

2.3.2. 2) Extension ratios: ERx = x/xi and ERy = y/yi and ERz = z/zi note that ER will be used in place of lambda

2.3.2.1. Volume for relaxed state, deformed state, and in terms of ER

2.3.2.1.1. Relaxed: Vi = xi*yi*zi

2.3.2.1.2. Deformed: V = x*y*z

2.3.2.1.3. In terms of the extension ratio: Vi/V = ERx*ERy*ERz = 1

2.3.3. 3) Define the distance between crosslinks recalling f = Kr where K = 2kT/p^2

2.3.3.1. ri = <xi, yi, zi>

2.3.3.2. r = <xi*ERx , yi*ERy , zi*ERz>

2.3.3.3. fi = <K*xi*ERx , K*yi*ERy , K*zi*ERz>

2.3.4. 4) Determine the work in the z-direction for one chain segment between crosslinks

2.3.4.1. To find Wz take the integral from (zi) to (ERz*zi) of fz*dz

2.3.4.1.1. Wz = (kT/p^2)*[(zi*ERz)^2 - (zi)^2]

2.3.5. 5) Determine the total work done in the z direction, assuming there are u segments between crosslinks

2.3.5.1. Start with the integral from 1 to u of Wz*dz

2.3.5.1.1. Summation of Wz from 1 to u equals (kT/p^2)*[ERz^2 - 1]*u(<r^2>i / 3)

2.3.6. 6) Determine the total work done in all directions

2.3.6.1. Take the summation from 1 to u of Wx and add it to the summation from 1 to u of Wy and the summation from 1 to u of Wz

2.3.6.1.1. Wt = (ukT/3p^2)<r^2>i*[ERx^2 + ERy^2 + ERz^2 - 3]

### 2.4. In and out of network chains

2.4.1. out of network is an intrinsic system property and defines the end to end distance as <r^2>o = (3/2)p^2

2.4.1.1. In the total work done in all directions equation, replace p^2 with (2/3)<r^2>o

2.4.1.1.1. Introduce the Shear Modulus (G) where G = NkT(<r^2>i)/(<r^2>o) where N = [number of chain segments between crosslinks]/[volume]. In other words, N is the crosslink density

2.4.2. in network <r^2>i is a function of volume

### 2.5. Deriving stress from the work equation

2.5.1. Recall that W = Fd and ER = L/Li where L is the final length and Li is the initial length

2.5.1.1. Take the partial derivative of work with respect to L to obtain F = (dW/dER) * (dER/dL) = VG/2(2ER - 2/ER^2)(1/Li) = VG/Li(ER - 1/ER^2)

2.5.1.1.1. F = AiG*(ER - 1/ER^2)

### 2.6. Using the theory correctly

2.6.1. All the equations derived in this chapter are for natural rubbers that aren't stretched beyond an extension ratio of 1.6

2.6.2. Effects of stretching the rubber too far (beyond ER = 1.6)

2.6.2.1. Chain alignment becomes so great that the material is no longer isotropic which means gaussian distribution no longer holds

2.6.2.1.1. Concept Connection: isotropy was one of the major assumptions made when deriving these equations. Isotropy allowed us to convert all three extension ratios into one extension ration. Isotropy also allowed us to define <r^2>i as one term instead of having to split it into three separate terms for each direction

2.6.2.2. High levels of orientation can result in stress-induced crystallization and the crystals change the stress-strain properties