⎜⎟⎜⎟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 y01λ=λ⇒λ=λ=y x z λφλλx x x′=IKJ0 FGH 1 x0 1 tan φ0 == =tan 32/y λyy λy λ00 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 φSegment 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 := /32λ⎛⎜⎜⎜⎝ ⎛⎜⎜⎜⎝arctan ⎞⎟⎟⎟⎠⎞⎟⎟⎟⎠λ 3 31 −λ 2⎛⎜⎜⎜⎝arctan ⎞⎟⎟⎟⎠ 1 3−λ− −1 23−λ3−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 31 −λ 2 3 ⎛⎜⎜⎜⎝ ⎛⎜⎜⎜⎝arctan ⎛⎜⎜⎜⎝arctan ⎞⎟⎟⎟⎠ 3−λ− − λ1 23−λ3−λ3 1 1 −1 ff := /322⎛⎝ 3−λ 1⎞⎠ Evaluate for λ=3: evalf( subs(lambda=3,ff) ); .6370 Plot orientation function versus extension ratio: plot(ff, lambda=.1..5); 543210.80.60.40.20-0.2-0.4λf
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