Abstract
Elastic moduli of the crystalline region in the direction perpendicular to the chain axis (Et) for isotactic polypropylene (Isot. PP) were determined by means of an X-ray diffraction method with the same experimental technique described in Part I. Based on a simple series model the Et-values for Isot. PP were calculated for the following two defferent equatorial planes.
Isot. PP:(110)…Et=2.9×104kg/cm2
(040)…Et=3.2×104kg/cm2 (28°C)
As the modulus parallel to the chain direction was found in our previous paper to be 42×104kg/cm2, there is a remarkable anisotropy of the modulus in the parallel and perpendicular directions to the chain axis. This may reasonably be attributed to the different extension mechanisms of the polymer crystal in the two directions. In view of the experimental error importance may not be placed on the difference between the Et-values obtained for two different planes. It was shown that the crystal moduli in the perpendicular direction are of the same order of magnitude as for the specimen modulus (Yt=2.3×104kg/cm2). This seems to mean that the extension of crystalline and amorphous regions in the direction perpendicular to the chain axis occurs by a similar mechanism. It should further be noted that the Et-values for Isot. PP have a magnitude just comparable to those for polyethylene (PE)(3-4×104kg/cm2) described in Part I. Since, in an essential meaning, the crystals of both polymers are composed of saturated paraffine chains, the interchain cohesion forces may be compared by comparing the closeness of the packing of the atoms in the crystal. Thus the similarity of the Et-values for the two polymers seems to be explained from the similar values of the mean effective volume per carbon atom calculated for their crystal structures (23.1 Å3 and 24.8 Å3 for PE and Isot. PP, respectively, neglecting the hdrogen atoms).
