Prob,2.22
Line rotation from uniaxial extension,
Consider a line inclined at an angle p from the vertical,with a slope y
0
/x
0
,After
stretching by an amount λ
y
= λ,we have,
φ
'
φ
0
x
x
0
y
0
1
λ = λ?λ = λ =
y x z
λ
φ
λ λ
x
x x
′ =
I
K
J
0
F
G
H
1
x
0
1
tan φ
0
== =tan
32/
y λ
y
y λy λ
00
Prob,2.23
Orientation function,
Fraction of sigments in range d(φ),
f(phi):=2*Pi*r^2*sin(phi)/(2*Pi*r^2);
,= ()f φ ()sin φ
Segment orientation after stretching (from previous problem);
phi_2:=arctan((1/lambda^(3/2))*tan(phi));
()tan φ?
arctan
Integrate over sphere to obtain mean square orientation angle,
phi_avg:= simplify( int( cos(phi_2)^2*f(phi),phi=0..Pi/2) );
phi_2,=
/32
λ
arctan
λ
3
3
1?λ 2?
arctan
1
3
λ1
23
λ
3
λ1 1
phi_avg,=
/32
3
λ 1
Evaluate for λ=3,
Digits:=4;evalf( subs(lambda=3,phi_avg) );
Digits,= 4
.7581
Denote the Herrman orientation parameter as ff,
ff:= (1/2)*(3*phi_avg-1) ;
2
1
3
1?λ 2
3
arctan
arctan
3
λ λ1
23
λ
3
λ3 1 1
1
ff,=
/32
2
3
λ 1
Evaluate for λ=3,
evalf( subs(lambda=3,ff) );
.6370
Plot orientation function versus extension ratio,
plot(ff,lambda=.1..5);
54321
0.8
0.6
0.4
0.2
0
-0.2
-0.4
λ
f
