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E. for uc=ua ΄ ΅ 2Vct0 tanh(aua/u0) Smax = 9923 1 – 994 m tanh(auc/u0) aua/u0 (83) For a long average chain length ua , Eq. 83 can be approximated by 2Vct0 Smax = 9 m (84) The Tensile Strength of Polymer Fibres 55 The values of Smax given by Eqs. 83 and 84 are the same as those given by Yoon, because h(u) equals f(u) for the monodisperse function [11]. ) and f(z) is the chain length distribution. The effect of the width of the distribution on the ultimate fibre strength, sL, is calculated for the case in which the molecular weight distribution fw(z)= zf(z)zn–1 is a uniform distribution.

35. For partially bonded chains the tensile and shear stresses become ΄ ΅ cosh(2ay/u0) s (y) = ece 1 – 9922 cosh(auc/u0) for 0 ≤ |y| ≤ –2 uc (74a) s (y) = 0 for –12 uc ≤ |y| ≤ –12 u (74b) sinh(2ay/u0) 1 t (y) = –2 mece 9922 sinh(auc/u0) for 0 ≤ |y| ≤ –2 uc (75a) t (y) = 0 for –12 uc ≤ |y| ≤ –12 u (75b) 1 and 1 In Eqs. 74a, 74b, 75a and 75b the quantity uc equals the bonded length of the chain. These relations are almost equal to Eqs. 69 and 70, except for the bonded length uc which replaces the total length u.

Energy in a sphere with radius q around the crack has been released. If w gS is the surface energy of the crack, the energy to create the crack is given by W1 = 2pq2 wgS (34) In the case of a crack parallel to the chains the surface energy relates to the cleavage of secondary bonds between the chains by the shear stress ssinqcosq. The shear strain energy released by the sphere is equal to the product of the volume of the sphere multiplied by the shear strain energy given by the second term in Eq.

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Topologie- und Formoptimierung von duennwandigen Tragwerken by Maute K.

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