A comparatively simple repeating element whose relative deformation corresponds to the deformation of the entire sample can be distinguished in many materials made of chemical (including rigid-chain) fibres. A simple program based on the MATLAB mathematics package that allows calculating the relative deformation of each element and consequently also the deformation of the sample is described. The result of calculating the deformation of a repeating element consisting of two circles of different radius twisted to opposite sides is reported as an example of running the program.Fibres and thread made from chemical, including rigid-chain, polymers are used for manufacturing fabrics, knits, and other textile materials for different applications. Since such fibres and thread stretch very little on deformation of the sample, deformation primarily takes place due to bending of the fibres and thread. In most textile materials except for nonwovens, a comparatively simple repeating element whose relative deformation corresponds to the relative deformation of the entire sample can be distinguished. A photomicrograph of one type of knit is shown in Fig. 1a and a photomicrograph of a column separated (prepared) from this knit is shown in Fig. 1b. A comparison of photomicrographs 1a and b shows that the shape and size of the loops changes little when the fibre is free from interacting with the other fibres. This indicates that the fibre used in manufacturing the knit is plastically deformed, as otherwise it would be in the form of a straight line. After calculating the relative deformation of such a repeating element, we obtain the relative deformation of the entire sample.It is convenient to use the two-dimensional nonlinear theory of elasticity for these calculations. We will assume that only concentrated, and not distributed, forces act on the examined element (henceforth the elastic line).Consider a two-dimensional elastic line obtained as a result of the fact that a fibre of arbitrary initial curvature (for example, plastically deformed) is loaded with some finite number of concentrated forces and external moments. We will use the approach proposed by Popov.*The elastic line can always be divided into segments in the general case so that concentrated forces F 0 and F 1 and external bending moments M 0 , M 1 were applied only on ends 0 and 1 (Fig. 2) of the examined line segment (they are taken with consideration of the effect of the cut parts of the fibre on segment 0-1). If there are no distributed forces on segment 0-1, then from the condition of equilibrium of the segment, we have F 0 + F 1 = 0 or F 0 = -F 1 . We stipulate the direction of force F 0 applied at the initial point of segment 0-1 of the elastic line as the basic direction.Let us introduce angle δ read counterclockwise from direction F 0 to axis X (or clockwise from axis X to the direction of force F 0 ) (i.e., the slope of axis X to force vector F 0 ). We will consider the initial curvature of the rod (fibre) (the curvature is a quantity inverse to the radi...

677.023 L. A. Kudryavin, and M. V. ShablyginThis article examines certain aspects of the unwinding of thin filaments of chemical fibers from smalldiameter bobbins. The aspects discussed are related either to the penetration of turns in lower-lying layers of the winding by turns in higher layers or to the "sticking" of turns (if the filament is coated) deep in the winding. Both of these phenomena lead to rupture of the filament as it is being unwound from the bobbin. A method is described for calculating the optimum parameters of a flanged cylindrical bobbin in order to prevent filament rupture due to penetration and sticking of the turns. The optimum parameters that are calculated are for filaments of specified lengths.Thin filaments being unwound from cylindrical bobbins will rupture if their length on the bobbin exceeds a certain value L. This phenomenon is usually related to the fact that the upper layers of the filament press on the lower layers and this pressure causes the former to penetrate the latter. During the unwinding operation, additional force is necessary to remove the upper layers from the lower layers that they penetrated. If a filament is not strong enough, it will rupture. This limits the length of the filament that can be wound on the feed bobbin and complicates the production process.We are proposing a theoretical method that makes it possible to determine the parameters of a cylindrical bobbin in such a way as to prevent the rupture of a filament of a prescribed length as it is unwound from the bobbin. To do this, first we calculate the change in the pressure between the layers of the microwire as the radius of a layer ρ decreases (Fig. 1). We introduce the following notation: r -the internal winding radius; R -the external radius; ρ -the radius of the layer being unwound; B -the distance between the flanges of the bobbin.We will assume that during the winding operation the plane of any turn of the microwire is nearly perpendicular to the axis of the bobbin. This assumption is valid in the case when the diameter of the microwire is small compared to the radius of the bobbin and the outermost turns are wound almost flush with the flanges of the bobbin. The subsequent turns are then wound adjacent to one another. These are the conditions that usually exist when a thin filament is wound on a bobbin. Then the length of one turn of the radius ρ will be equal to 2πρ.We theoretically dissect the bobbin with a plane that passes through its axis. This section is shown in Fig. 1a. We then use the letter n to represent the number of turns that intersect a unit area of this plane. The value of n determines the winding density. Let us examine a layer of very small thickness dρ in the winding (Fig. 1b). Since the layer is very thin, all of the turns that end up in this layer will have nearly the same radius ρ and, thus, the same length 2πρ. The area of the cross section of this layer which coincides with the plane passing through the axis of the bobbin (Fig. 1a) is equal to Bdρ, while the number of turns ...

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678.01:548The problem of fabricating high-strength and high-modulus fibres from flexible-chain crystalline polymers was investigated. These polymers include polyethylene, polypropylene, and polyacrylonitrile, used to fabricate high-strength fibres with gel technology.Many mechanical and physical properties of fibres fabricated from crystallizable polymers, for example, the modulus of elasticity, strength, thermal stability, and others, are greatly determined by the degree of folding of the macromolecules in the crystalline phase --the less folding there is, the better these indexes are. We can intuitively predict that stretching of macromolecules during crystallization should suppress folding and it should disappear with a high enough degree of stretching of the macromolecules. However, experiments show that even with a high degree of stretching of the macromolecules, folding cannot be totally suppressed. In order to explain the causes of the appearance of folds in even strongly oriented polymers, let us consider the change in the free energy of an oriented melt of a polymer during its crystallization.Let all of the polymer molecules be in a force field, that is, let each macromolecule be stretched the same with respect to the direction and magnitude of forces f applied to its ends. We will only consider crystals whose formation is most thermodynamically advantageous, i.e., crystals with the axes of the molecules directed along the direction of stretching.Using the data in [I, 2], we can show that the change in the free energy of the system in formation of one such crystal will be determined with the equationfor AFIc = 2Ao~ef -2Ao(o~f -o) I t o -AolZ3d-/~(l -T I T~) -k/~b I t o I I T s = I I T o + k In(shb / b) / (Aolz~r'l) b = fl / (kT),where F, F are the free energy of a system without and with a crystal; I is the length of the static fold of the molecule; A 0 is the area of the cross section of a segment with the plane perpendicular to its axis; t is the total number of segments in the cross section of the crystal (tA o is the area of the cross section of the crystal); AFlk is the average change in the free energy of the system when t increases by 1; t o is the average number of segments in the cross section of the crystal belonging to one molecule; t 0-1 is the average number of folds belonging to one molecule in the crystal (t0A 0 is the average area of the end face of the cross section of the crystal per molecule in the crystal); AH is the specific heat of fusion of the crystal; T O is its equilibrium melting point in the absence of forces that stretch the macromolecules; T is the crystallization temperature; ~ is the number of segments over the thickness of the crystal (~l is the thickness of the crystal); k is the Boltzmann constant;Oef is the end face energy of a crystal from a folded chain on conversion to the unit of surface area; o is the free surface energy at the interface of crystalline and amorphous phases, also on conversion to the unit of surface area; in a first approximation, can be set equal ...

The effect of the cross-section of crystals and thickness of the amorphous phase in fibre-forming oriented polymers on transfer chains, kinks, and "cilia" in Both oriented and unoriented crystalline polymers consist of alternating crystalline and amorphous regions. The molecules coming out in one crystalline region can terminate in an amorphous region, forming kinks (pos. 2) and can pass into a neighboring crystalline region, forming transfer chains (pos. 3). As previously established in [1,2], the mechanical properties of polymers (for example, the modulus of elasticity and strength of fibres) are greatly determined by the proportion of transfer chains --the higher it is, the higher the indexes of these properties. It can be intuitively predicted that when the cross-sections of the crystals and the thickness of the amorphous region change, the proportion of transfer chains should also change. However, it is somewhat difficult to verify this with the usual experimental methods. We used the method of the computer experiment in a cubic lattice based on the widely used Monte Carlo method for this purpose [3][4][5][6]. It was hypothesized that the cross-section of the crystals was in the shape of a square, and the polymer segments in both the crystalline and in the amorphous phases can only be positioned along the sides of the cubic lattice cells.We will select a rectangular system of coordinates and place the origin of the coordinates in the center of the end of one of the crystals. We will position axis z along the axis of the molecules in the crystal in the direction of the nearest amorphous phase and axes x and y along the sides of the cubic lattice cell perpendicular to the side faces of the crystal (Fig. 1).The segments of the molecules can come out into the amorphous region from any cubic lattice site on the end face of the crystal. If the site examined is not on the side of the crystal, then the first amorphous segment can only be positioned in one way --along axis z in the direction of its enlargement, since the neighboring segments will interfere in positioning it in other directions. The second amorphous segment can be positioned in five different ways: four ways --perpendicular to axis z (one along axes x and y and one opposite them) and one way --along axis z. This segment cannot be positioned opposite axis z, since the first segment has already been placed there.If the second segment occupies a position along axis z, then the third segment will be positioned similar to the second segment. If the second segment is located along axisx, for example, then the third segment can be positioned in the following five ways: one along axisx, one along axis y, one opposite axisy, one along axis z, and one opposite axis z. The third segment cain be positioned opposite axis x, since the second segment is located there. If the third segment is positioned opposite axis 2000.Moscow State Textile University.