Prob 2,19
Estimate for the theoretical stiffness of a polymer molecule,
Consider half of a polymer link,
T
F
A
φ
l/2
φ
φ
I
Kf
The force along the link direction is F
l
= F sin θ,and the
torque around point A is T = F (l/2) cos φ,The strain energy is
then
L O
F
l
2
T
φ
2
U ++
M
M
N
P
P
Q
Un= U
φ
= n
l
2k 2k
φl
2
L O
F
H
l
F cos
M P
2
F sin a
2
M P += n
M P
2k 2k
φl F
P
Q
M
N
Castigliano's Theorem then gives the deflection as
2
L O
2F
F
H
l
2
I
K
cosφ
M P
U
2
φ2F sin
δ M P +==n
F
M P
2k 2k
φl
P
Q
M
N
L O
P
P
Q
φ
2
φ l
22
cos sin
+
M
M
N
= nF
k 4k
φl
The effective spring stiffness k from F = k δ is then
k[eff]:= (n*( sin(phi)^2/(k[l]) + l^2*(cos(phi)^2/(4*k[phi]))))^(-1);
1
k,=
n
()sin
2
φ
+
2
1 l
2
()cos φ
k 4 k
l φ
eff
T modulus is then E = k
eff
L/A,The extended chain length is L = nl sin φ and the effective chain area from
2
crystallographic measurements is 0.181 nm,
L:=n*l*sin(phi); l:=153e-12;phi:=(56*Pi/180);A:=.181e-18;
k[l]:=435;k[phi]:=35;
The modulus (in Pa) is then,
Digits:=4;'E'=evalf(k[eff]*L/A);
E =,4435 10
12
This is more than twice the stiffness of steel,at a fraction of the weight,
