Polymers- Rubber Elasticity Theory Flashcards
Elastic properties of rubber
Extremely high extensibility, up to x10, generated by low mechanical stress.
Complete recovery after mechanical deformation.
These are both due to deformation-induced changes in entropy
What is the origin of polymer elasticity at molecular level due to?
The fact that the molecules prefer disordered, coiled conformations. This is entropy driven
Formula for change in internal energy of a system
dU=dQ-dW
dQ is heat absorbed by the system
dW is the work done by the system
Formula for work done by the system
dW=-fdl
f is tensile force
dl is change in length of elastomer
Deriving formula for tensile force in terms of energy changes for a reversible process (entropic elasticity formula)
For reversible process: dQ=TdS Using previous formulae: fdl=dU-TdS Divide through by dl f=(δU/δl)T,V - T(δS/δl)T,V Where δs are curly ds. T,V is subscript for constant T and V
Which terms are dominant for metals, ceramics and elastomers in entropic elasticity formula?
For metals and ceramics the internal energy term is dominant but for elastomers the change in entropy gives the largest contribution to the force. So can be approximated as:
f=-T(δS/δl)T,V
How do pressure, volume, internal energy and entropy of ideal gases change when they are compressed?
Pressure increases
Volume decreases
Internal energy constant
Entropy decreases
Difference between natural rubber and vulcanised natural rubber
Natural rubber just has lots of the polymer chains mixed around together.
Vulcanised has some sulfur crosslinks between the rubber molecules.
Repeating unit [CH2-C(CH3)=CH-CH2]n (has wings)
Structural requirement for a polymer to be elastic
Amorphous
Above Tg
Lightly cross-linked
Ideal rubber
Consists of flexible cross-linked polymer chains undergoing violent liquid-like motions.
The chains are joined to the network at both ends (sub-molecules).
The chains or sub-molecules have equal contour length
Freely jointed chain model
One rubber molecule (the chain) has lots of zigzags (bonds). Two bonds have an angle between them and form two equal length sides of a triangle. These have length l. The length of the open side is rf (f sub). Over the whole molecule:
rf^2=nl^2
n is number of bonds of length l
Statistical problem of where the two chain ends of a rubber molecule will be
If one end is the origin, find the probability that the other chain end lies in a certain volume, dV, whose centre is a distance r from the origin.
How does the number of conformations that a single chain can take up depend on end-to-end separation?
Number of conformations possible decreases as end-to-end separation increases (eventually only a straight line possible)
Gaussian chain model
P(x,y,z)dxdydz=(exp(-(x2+y2+z2)/ρ2))/(rt(π)ρ)^3 dxdydz
Where =3/2 ρ2
Graph of P(x,y,z) vs r starts high, curves down to diagonal then starts to level off near x axis
Entropy of a Gaussian chain
s=s0+klnP(x,y,x), k is Boltzmann
Sub in formula for Gaussian chain model
s=s0-k[3ln(rt(π)ρ)+(x2+y2+z2)/ρ2]
So entropy goes down with increasing end-to-end distance as the number of conformations it can adopt decreases. This is the molecular origin of rubber elasticity
Elastic behaviour of Gaussian chain in different directions formulae for tensile force
fx=-T(δs/δx)=2kTx/ρ2 fy= 2kTy/ρ2 fz= 2kTz/ρ2 f bar = (2kT/ρ2)r bar So Gaussian chain is linearly elastic and the force constant is linear in T
How can elastic properties of molecular network in a rubber be represented?
By interconnected linearly elastic springs
How do extension ratios work?
When something stretched there is no change in volume so
Xi,Yi,Zi=XYZ (initial dimensions to new dimensions)
Extension ratios: λx=X/Xi, λy=Y/Yi, etc
λxλyλz=1
This applies to each sub-molecule
Work done by chain during deformation
Integrate fx from Xi to λxXi with respect to x (capitals actually LC).
wx=(kT/ρ2)(λx^2-1)xi^2
Same for wy and wz just replace x
Total work done by external force
W=Σ(wx+wy+wz) from1 to v (number of chains (sub-molecules))
Sum of one component of work from 1 to v
Σwx=(kT/ρ2)(λx^2-1)Σxi^2
Σwx= (kT/ρ2)(λx^2-1)vx^2 i (i sub outside arrows)
x^2 i is same for y and z so equals r^2 i/3
Where i is mean square end-to-end distance of chains in the unreformed state. For Gaussian chain:
r^2 0=3/2 ρ^2 (arrows)
Formula for shear modulus
G=NkT x (r^2 i/r^2 0)
Where N is v/V (number of sub-molecules per unit volume)
The r^2 terms have arrows around them and sub i or 0.
k is Boltzmann constant
T is temperature
r^2 in arrows sub i is mean square end-to end distance of the chains in the undeformed state
r^2 in arrows sub 0 is mean square end-to-end distance of the chains for a Gaussian state
Nominal stress for simple stretching along z direction
σn=f/XiYi=G(λ-1/λ2)
True stress for simple stretching along z direction
σt=f/XY=G(λ2-1/λ)
How well do theoretical results for relationship between stress and extension ratio compare to experimental results?
At low values of λ (0-1.5, theory matches well with curve up with decreasing gradient. Les than 1 has negative stress and above 1 positive stress. Theory then largely a straight diagonal line as λ increases whereas experimental goes below and then crosses as it goes to exponential increase.
Reasons for discrepancies at large extension ratios
The theory neglects network defects, chain entanglements, etc.
The chains are no longer Gaussian at high extensions.
At high extensions vulcanised natural rubber crystallises
Mechanical hysteresis graph
Stress vs %extension
Loading curve always above unloading curve apart from the joins at both ends. This is for a strain-crystallising elastomer
