Step strain deformation

Immediately following a step strain deformation, all of the segments of a fully entangled melt can be assumed to have the same degree of orientation. In other words, both the short and the long chains will be characterized by identical functions (uu)f Q+, where... 

Figure 8 Relaxation of a polymeric chain after a step-strain deformation[6] process A (8c) reequilibration of chain segments process B (8d) reequilibration across slip links process C (8e) reptation.

D.C.Venerus, C.M.Vrentas, J.S.Vrentas, Step strain deformations for viscoelastic fluids experiment, J. Rheol. 34 (1990), 657-682. 

The relaxation of the segmental orientation and the chain extension in polycarbonate (Makrolon 1143 from Bayer AG) were studied by IR dichroism and heat shrinkage after a step strain deformation to a draw ratio of 1.75. This makes it possible to distinguish between local relaxation mechanisms (Rouse) and large-scale relaxation... 

Fig. 7. (a) General step-strain deformation history relevant to Boltzmann-type linear superposition, (b) Schematic of stress additivity of responses for a Maxwell model in a two-step strain history (see text for discussion). 

D. C. Venerus, E. F. Brown, and W. R. Burghardt, The Nonlinear Response ofa Polydis-perse Polymer Solution to Step Strain Deformations Macromolecules 31, 9206-9212... 

This function can most readily be obtained by performing step strain deformations if a shear strain y is imposed on the material at time zero, eq. 4.4.4 gives... 

Venerus, D. C., Brown, E. R, Burghardt, W. R. The nonlinear response of a polydisperse polymer solution to step strain deformations. Macromol (1998) 31, pp. 9206-9212... 

These two experiments are fundamentally different in the nature of the applied deformation. In the case of the relaxation experiment a step strain is applied whereas the modulus is measured by an applied oscillating strain. Thus we are transforming between the time and frequency domains. In fact during the derivation of the storage and loss moduli these transforms have already been defined by Equation (4.53). In complex number form this becomes... 

Each submolecule will experience a frictional drag with the solvent represented by the frictional coefficient /0. This drag is related to the frictional coefficient of the monomer unit (0- If there are x monomer units per link then the frictional coefficient of a link is x(0- If we aPply a step strain to the polymer chain it will deform and its entropy will fall. In order to attain its equilibrium conformation and maximum entropy the chain will rearrange itself by diffusion. The instantaneous elastic response can be thought of as being due to an entropic spring . The drag on each submolecule can be treated in terms of the motion of the N+ 1 ends of the submolecules. We can think of these as beads linked... 

It became clear in the early development of the tube model that it provided a means of calculating the response of entangled polymers to large deformations as well as small ones [2]. Some predictions, especially in steady shear flow, lead to strange anomaUes as we shall see, but others met with surprising success. In particular the same step-strain experiment used to determine G(t) directly in shear is straightforward to extend to large shear strains y. In many cases of such experiments on polymer melts both Hnear and branched, monodisperse and polydisperse,the experimental strain-dependent relaxation function G(t,Y) may be written... 

Reasons have been advanced for both an increase and a decrease of the tube diameter with strain. A justification of the former view might be the retraction process itself [38]. If it acts in a similar way to the dynamic dilution and the effective concentration of entanglement network follows the retraction then Cgjy < E.u > so that a < E.u On the other hand one might guess that at large strains the tube deforms at constant tube volume La. The tube length must increase as < E.u >,so from this effect a < E.u > . Indeed, Marrucci has recently proposed that both these effects exist and remain unnoticed in step strain because they cancel [69] Of course this is far from idle speculation because there is another situation in which such effects would have important consequences. This is in conditions of continuous deformation, to which we now turn. 

When reptation is used to develop a description of the linear viscoelasticity of polymer melts [5, 6], the same underlying hypothesis ismade, and the same phenomenological parameter Ng appears. Basically, to describe the relaxation after a step strain, for example, each chain is assumed to first reorganise inside its deformed tube, with a Rouse-like dynamics, and then to slowly return to isotropy, relaxing the deformed tube by reptation (see the paper by Montfort et al in this book). Along these lines, the plateau relaxation modulus, the steady state compliance and the zero shear viscosity should be respectively ... 

Transient shear is defined as when a material is subject to an instantaneous change in deformation and the response as a function of time is measured. For example. Figure 3.72 shows an instantaneously applied step-strain test. 

A step shear deformation A can be regarded as equivalent to a constant strain rate X/5t being applied over a very short time interval St (cf. Eq. (4.27)). The velocity gradient tensor K = (VV) is expressed by... 

For URPs, the emphasis is somewhat different. Due to their relatively low stiffness, component deformations under load may be much higher than for metals and the design criteria in step (b) are often defined in terms of maximum acceptable deflections. Thus, for example, a metal panel subjected to a transverse load may be limited by the stresses leading to yield and to a permanent dent. Whereas a URPs panel may be limited by a maximum acceptable transverse deflection even though the panel may recover without permanent damage upon removal of the loads. Even when the design is limited by material failure it is usual to specify the materials criterion in terms of a critical failure strain rather than a failure stress. Thus, it is evident that strain and deformation play a much more important role for URP than they do for metals. As a consequence, step (a) is usually required to provide a full stress/strain/ deformation analysis and, because of the viscoelastic nature of plastics, this can pose a more difficult problem than for metals. 

Deformation and rearrangement of the morphological structure have been described by Peterlin [201-203] and Samuels [183]. During the initial step of deformation, a crystalline fiber undergoes an affine transformation, i.e., the strains are uniform throughout the material. However, even at still smaller deformations the morphological inhomogeneity of the... 

Next we consider a polymer melt of high molecular weight in which entanglement is very important. To calculate G(t)> it is convenient to consider the stress relaxation after a step strain. Suppose at t = 0 a shear strain y is applied to the system in equilibrium. The strain causes the deformation of the molecular conformation, and creates the stress, which relaxes with time as the conformation of polymers goes back to... 

Fig. 7.2. Explanation of the stress relaxation after small step strain, (a) Before deformation, the conformation of the tube is in equilibrium, (b) Immediately after the deformation, the whole tube is deformed. The deformed part is indicated by the oblique lines. For small strain, the contour length of die tube is unchanged, (c) At a later time t, the chain is pardy confined in a deformed tube.

Ilg. 7J2. Explanation of the stress relaxation after large step strain, (a) Before deformation the conformatian of the fnimitive chain is in equilibrium (r = —0). (b) Immediately after deformation, the primitive chain is in the afiindy deformed conformation (t = -1-0). (c) After time Tj, the primitive chain contracts along the tube and recovers the eqi brium contour length (t Tj,). (d) After the time Xj, the primitive chain leaves the deformed tube by reptation (t Xa). The oblique lines indicates the deformed part of the tube. Reproduced from ref. 107. 

We consider the inextensible chain model. Figure 7.24 eiqilains the change of polymer conformation under the double step strain. Figure 7.24a shows the undeformed state just before the first deformation. Figure 7.24h represents the state immediately after the deformation the primitive chain is deformed by the shear Yi. Figure 7.24c indicates the state just before the second deformation the inner part AB still remains in the deformed tube, while the outer parts are in the undeformed tube. Now when the second deformation is applied, the inner part AB is deformed by the shear Yi + Yi from the equilibrium state, while the outer part is deformed by the shear Yi- It is important to note that the second shear stretches the contour length of the outer part by the factor < (72), but that of the inner part by the factor... 

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