Prob. 2. 22 Line rotation from uniaxial extensi Consider a line inclined at an angle p from the vertical, with a slope yo/xo. After stretching by an amount A,=1, we have φ 元,=4→Ax= √元 tan o y Ayyo Orientation function Fraction of segments in range d(o f(phi): =2 Pir2*sin(phi)/(2*Pi*rA2); f(o):=sin(o Segment orientation after stretching(from previous problem) phi 2: = arctan((1/lambda (3/2))tan(phi)); 2 Integrate over sphere to obtain mean square orientation angle phi avg: simplify( int( cos(phi 2)2 f(phi), phi=0. P1/2)); 入3-1- arcta arctan 少 Evaluate for 2=3 Digits: =4; evalf( subs(lambda=3, phi avg)); Digits: =4 Denote the Herrman orientation parameter as ff: 1/2)*(3° phi avg-1);⎜ ⎜ ⎟⎟ Prob. 2.22 Line rotation from uniaxial extension: Consider a line inclined at an angle p from the vertical, with a slope y0/x0. After stretching by an amount λy = λ, we have: φ ' φ0 x x0 y0 1 λ y x z = λ ⇒ λ = λ = λ φ x x λ x λ ′ = I K J 0 F G H 1 x0 1 == = tan φ 0 tan 3 2/ y λ y y 0 0 λy λ 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 := 3 2/ λ ⎛⎜ ⎜ ⎜ ⎝ ⎛⎜ ⎜ ⎜ ⎝ arctan ⎞⎟ ⎟ ⎟ ⎠ ⎞⎟ ⎟ ⎟ ⎠ λ3 3 ⎛⎜ 1 λ − 2 ⎜ ⎜⎝ arctan ⎞⎟ ⎟ ⎟⎠ 3 1 λ− − − 1 3 2 λ − 3 1 1 λ − phi_avg := 3 2/ ⎛ ⎝ 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) ;