Displacements Against Randomly Distributed Forces

Chain Displacements Against Randomly Distributed Forces 

The chain stress is derived from Eq. (5.24) replacing the chain elasticity constant K by the elastic constant of a series of elastic elements 

The above equation shows quantitatively that large chain stresses will only be obtained if the interaction force constants Wj are of about the same order they are in a crystal and if the elastic moduli of the chain sections are large throughout. One weak section will greatly enhance average displacement and reduce the stress. It should also be mentioned that it is not the chain length L and the absolute number n of interaction sites which raise the chain stress but the intensity of the interaction forces per unit chain length. 

As an example of the static excitation of a chain under these conditions a nearly extended chain of end-to-end distance L embedded in a glassy matrix may be considered. If the matrix (index m) and the chain (no index) were homogeneously strained in chain axis direction they would experience the following stresses  

Due to the different elastic response the free chain would have to act against the lattice potential until chain stress and lattice interaction were in equilibrium at any point of the chain. This interaction can be described by identically the same mathematical formalism developed above. Since the chain ends are assumed to be free they cannot bear any axial stresses. This condition will exactly be met if one superimposes the hypothetical stress o (Eq. 5.35) by a compressive stress according to Eq. (5.33) where 

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D. Chain Displacements Against Randomly Distributed Forces.107... 

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