Estimate for the theoretical stiffness of a polymer molecule Consider half of a polymer link: The force along the link direction is F= F sin 0, and the torque around point A is T= F(/2)cos o. The strain energy is U=+U引 F2,76 2k12k =刀( Fsin g Castigliano's Theorem then gives the deflection as cos sin o /coS k, The effective spring stiffness k from F=k8 is then kefn:=(n°(sin(phi)^2/(k)+1^2(cos(phi)~2/(4*klphi)^(-1); 〃/sim(9)21Pcos()2 k+4 T modulus is then e= ketL/A. The extended chain length is L=nI sin o and the effective chain area fre crystallographic measurements is 0. 181 nm": L:= n*sin(phi);l:=153e-12;phi:=(56*Pi/180);A:=18le-18; k:=435; kephir:=35; The modulus (in Pa)is then: Digits: 4; ' E=evalf(kleff]"L/A) This is more than twice the stiffness of steel, at a fraction of the weight
⎜ ⎟ 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 f K The force along the link direction is Fl = F sin θ , and the torque around point A is T = F (l/2) cos φ. The strain energy is then L F O l 2 Tφ 2 U + + M MN P PQ U n = Uφ = n l 2k 2k l φ 2 L O F H l F cos M P 2 aF sin 2 = n M + P M 2k l 2kφ P F P Q M N Castigliano's Theorem then gives the deflection as 2 L O 2FF H l 2 I K M cosφ P ∂U 2 2F sin φ δ = = nM + P ∂F M 2k l 2kφ P P Q M N L O P PQ φ 2 φ l 2 2 sin cos M + MN = nF k 4k l φ 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 = keff 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 1012 This is more than twice the stiffness of steel, at a fraction of the weight
