Abstract

AbstractThe stress‐strain relations for transversely isotropic deformations of linear and branched polymer melts as well as of (crosslinked) rubbers are discussed in terms of the orientation tensor. It is shown that orientation and network strand extension are decoupled, and that the relative tube diameter and its inverse, the molecular stress function f, can be extracted directly from experimental data, if the effect of network orientation is accounted for by the order parameter. The tension of the average network strand increases with increasing deformation. This is caused by an increasing restriction of lateral movement of polymer chains due to deformation. At small strains, f2 is found to be linear in the average stretch for melts as well as for rubbers, which corresponds to a constant volume of the tube. At large strains, melts show maximum molecular tension, depending on the degree of longchain branching, while rubbers show maximum extensibility.

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