Eq. (1) is proposed for the relation between the tensile force and elongation of a cross-linked polymer which shows the rubbery region based upon the ideal Affine deformation.
f=ν(0)RT(α-α
-2) (1)
Eq. (2) or (3) which is a corrected eq. (1) holds, however, for a tightly cross-linked polymer because of its abnormal structure.
f=r
2(network chain)/r
20(free chain)•ν(0)RT(α-α
-2) (2)
or f=Φ•ν(0)RT(α-α
-2) (3)
Here, Φ=
r2/
r02.
When the shear modulus G is used, eq. (3) becomes,
G=Φ•ν(0)RT (4)
Taking elongation into consideration,
G=Φ•ν(0)RTΓ(λ
m) (5)
Here, λ
m=
rm/ri,
rm is the maximum chain length and
ri is the average chain length without elongation.
Moreover,
Γ(λ
m)=1+6/5λ
m2+297/175λ
m4+……(6)
Assuming polymerization degree of
n, and segment length of
l, most probable average square length
r02, and maximum chain length
rm are respectively expressed by eqs. (7) and (8).
r
02=C
n•nl
2 (7)
r
m=qnl (8)
Here,
Cn and
q are constants regarding the Polymer structure, and eq. (9) is obtained.
λ
m=(q
2•n/Φ
n•C
n)
1/2 (9)
Eq. (6) is approximated by eq. (10) in the range of 0<1/λ
m<0.88.
Γ
app(λ
m)=1/(1-6/5λ
m2)=5λ
m2/(5λ
m2-6) (10)
The next equation (11) is derived from eqs. (5), (6), (9) and (10).
ν(0)RT/G=1/Φ
n-6•C
n/5q
2•n (11)
Assuming the real cross-linking density is ν′(0), and the apparent one is ν(0) for a tightly cross-linked polymer, eq. (12) is derived from eq. (11).
ν′(0)/ν(0)=1/Φ
n-6/5(0.83)
2•n
0.43 (12)
The reasonable relation of ν
M(0)>ν
S(0) holds in the smaller range of ν(0), while the reciprocal relation of ν
M(0)<ν
S(0) which is not understandable does in the higher range of ν(0). The latter has been made understandable by using eqs. (12).
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