Prof. G. V. TRAPEZNIKOV

On Stresses

Under Tension.

Tomsk — 1929.

Prof. G. V. Trapeznikov.

On Stresses Under Tension.

I. Residual Cell-Deformations.

The present work was conceived by me in the winter of 1924–25, when some preliminary experiments were carried out, and was completed in the winter of the following year in the Mechanical Laboratory of the STI.

The thought that prompted me to recognize the necessity of such work was the doubt as to whether a cylindrical or, in general, a bar of constant cross-section, under elastic deformation, preserves the same form of the contour of its axial section, with the corresponding, of course, change in its linear dimensions, as it had before deformation. In other words, the question arose for me whether the transverse dimensions of an elastically stretched bar change equally in different transverse sections.

The relative transverse deformation, expressed through the relative longitudinal deformation and Poisson’s ratio, is written as follows:

\[ \varepsilon = m i, \]

where it is assumed that in all sections the normal stresses are distributed quite uniformly, if, of course, the tensile force is applied strictly along the axis of the bar.

Hence follows the condition of uniformity, for all sections, of the transverse deformation, and consequently also of Poisson’s ratio. But everyday laboratory experience and the numerical data that we have for the value of Poisson’s ratio among various experimenters indicate its extreme variability, even for one and the same material. For greater clarity I give a table of values of the coefficient for several metals.

Metal	m	Experimenter
Hardened steel	0.294	Kirchhoff
Annealed steel	0.273	Kirchhoff
″ ″	0.304	Okatov
″ ″	0.333	Young
Iron	0.243	Young
″	0.310	Bach
Brass	0.226	Young
″	0.469	Okatov
Red copper	0.250	Voigt
″ ″	0.348	Bach
Lead	0.375	Voigt
″	0.456	Okatov

Such contradictory data indicate that, despite a carefully and precisely arranged experiment, each of the experimenters had certain particular conditions of his own experiment. Leaving aside the question of the purity of the metals—the elements—and also the errors that could have occurred when different measuring instruments were used (for this we have too great a discrepancy in the values of \(m\) for one and the same material), we must acknowledge that the cause of such a discrepancy in the figures for \(m\) lies in the distribution of stresses in the bar being tested.

In ordinary laboratory tensile tests of metallic specimens, a series of thin transverse scratches, spaced 1 cm apart, is first applied to them. In Fig. 1 such a specimen with the scratches, blackened

Fig. 1. Diagrams of residual elongations.
A—large divisions.
B—small divisions.

in ink, for better viewing in the photograph, is shown between diagrams A and B. After rupture of the specimen, the elongation of each centimeter segment is found and, for them, a diagram of relative elongations is constructed. The diagram is constructed in such a way that the ordinate expressing the elongation of each centimeter is laid off from its middle, along the length of the specimen with unchanged deformation. One such diagram is given in Fig. 1, under the letter A. It was constructed for a specimen stretched until the appearance of a neck; the photograph of the specimen is denoted by C. In this diagram the greatest ordinate corresponds to the neck. On the diagram, besides one sharply expressed maximum, there are usually also others, of considerably smaller size (see the right-hand part of diagram A). The well-known angularity of the curve is explained by the relative coarseness of such divisions (the normal diameter of the specimen is 20 mm). On this same stretched specimen C, before its drawing, marks were made on the side opposite to the centimeter segments at intervals of 2.5 mm. The lower diagram—B—was constructed for these shorter segments. Besides much greater smoothness, in comparison

Fig. 2. Stretched specimens with double and triple necks.

with curve A, we see here necks that have appeared quite definitely and sharply, distributed over the entire length of the specimen; together with the maximum there are seven of them. This circumstance indicates that along the length of a specimen stretched beyond the elastic limit of the material, not one but several necks are established, with rupture occurring along one of them. In most cases all the necks, except the principal (maximum) one, are small and cannot easily be noticed by eye upon direct examination of the specimen. It is true that we can see all the necks on it if we set the specimen so that some point source of light (a lamp, a candle), the eye, and the axis of the specimen form a plane, and so that the ray of light falling on the generatrix of the bar before being reflected into the eye makes a “grazing angle” with it. Quite often, some necks become so large that they are noticeable even upon superficial examination. In Fig. 2 a photograph is given of several specimens possessing double and triple necks.

Under compression we likewise observe the appearance not of one, but of two, sometimes three, bulges in cylindrical specimens, in cases where the ratio of length to diameter is not especially small. Fig. 3 is a photograph of cylinders of iron, red copper, and lead compressed until the appearance of sharply noticeable alternating bulges and

Fig. 3. Compressed specimens with single and double transition.

depressions. In a tube being flattened along its axis, collars appear and the tube is transformed into a harmonica, whose waves have the same size. Figs. 4 and 5 give photographs of such tubes at various stages of compression.

Fig. 4. Compressed tubes at various stages of compression.

In some tubes having relatively thin walls, these waves are not annular, but form a harmonic pattern like the bellows of a camera; that is, the circumference is transformed into a regular polygon with an even number of sides, half of the latter corresponding to depressions and half to bulges.

A normal cylindrical round tensile specimen has a length-to-diameter ratio of 10; it is therefore rather difficult to trace the appearance of many necks, when the supporting heads are very close to one another, owing to the strong restraining action of the heads on the occurrence of transverse deformations. It is necessary to have specimens with a much greater length-to-diameter ratio.

Such long specimens, in the form of simple round rods or bars, with heads and without them, were made by me from various metals. Copper, iron, steel, soft brass, and aluminum were taken from round rolled stock; tin and lead—in cast form, and nickel in the form of thick rheostat wire.

Fig. 5. Compressed thin-walled tubes.
The central tube in the background had been filled with cement before compression.

Before tension, marks were made on the specimens with a comparator of the firm “Société Genevoise pour la construction d’Instruments de Physique,” at whole numbers of millimeters, with an accuracy of up to 0.001 mm. The length of the interval between marks was chosen so that it lay within the limits of \(0.1 d\).

After the specimen had been stretched to the moment when the principal neck, which upon further stretching would give the place of rupture, had already become sharply outlined and the load of the machine began to fall, it was unloaded. On the comparator, with the aid of a microscope, the length of each interval was determined with the same accuracy as in making the marks. Fig. 6 gives an idea of the specimen and of the diagram constructed for it of the elongations of the individual sections. Here we see an extremely regular alternation of necks, with which the entire length of the specimen is filled. Determination of the wavelength for all the metals tested, with various specimen diameters (from 6 mm to 65 mm), gives a constant ratio—

between the wavelength—\(L\) and the diameter—\(d\): as the average of observations on 73 specimens:

\[ \frac{L}{d}=1.50 . \]

Such regularity and uniformity (monotonicity) in the alternation of crests and hollows on the generatrix of a cylindrical bar during its tension must, without doubt, also occur under compression, but cannot be verified for solid long cylinders, since it is impossible to compress a sufficiently long bar without bending its axis. The constancy of the ratio \(L:d\) indicates the non-random character of the phenomenon of wave formation.

Experiment establishes two propositions:

1 — under deformation beyond the elastic limit, a stretched (compressed) bar of constant cross-section has transverse deformation of different magnitude at different places along its length.

2 — the residual transverse deformation, as it varies along the length of the bar, follows a certain law of periodicity.

The latter proposition indicates the presence of a certain foundation, within the limits of elasticity, for the formation of such regular waves. In other words, we arrive at the idea of the necessity of having some stress distribution, obligatory for a round cylindrical bar, which must produce periodic elastic necks. If experiment confirms these conclusions, then it must be recognized that the cylindrical form of a stretched or compressed bar is not a stable form.

Fig. 6. Diagram of residual elongations.
(Stretched along the axis)

Such a definitive conclusion is far from paradoxical. We know, in the theory of elasticity, two simplest problems where a form that appears at first glance to be an equilibrium-stable one, upon analysis of the problems and upon experimental verification, proves not to be so. Consideration of the phenomenon of longitudinal bending and torsion of a long shaft shows that, beyond a known limit of force and twisting moment, the stable form of the axes of the bars is not straight lines, which, as

as would be expected, while the curves, respectively, would be plane and spatial.

On the other hand, we have certain indications of the necessity of the appearance of regularly alternating elastic necks in Rejtö’s work: “Die innere Reibung der Festen Körper” (1897).

The scheme of tensile (compressive) deformation, according to Rejtö, is as follows (Fig. 7). A tensile force, uniformly distributed over the end sections \(ab\) and \(a'b'\), acts on a round cylindrical bar, the longitudinal section of which is shown in the drawing, in such a way that

Fig. 7. Distribution of internal forces according to Rejtö.

each molecule of these sections receives the same elementary force. Rejtö assumes that molecules can transmit the action of forces only along lines connecting the centers of neighboring molecules. Therefore each end elementary force must be resolved into two components along the line of centers. In view of the regularity of the structural lattice, the lines of centers, or the lines of action of the forces, also form a regular lattice of parallel lines, along which force is transmitted from molecule to molecule: \(a—g\), \(g—m\), \(b—e\), \(e—n\), etc. These lines form, with the axis of the bar, an angle that is constant for the given material, which Rejtö calls the “angle of action” and denotes by \(\beta\). In such a force lattice, the molecules located in the conical portions of the bar: \(a—c—b\), \(c—e—h—g\), etc., are subjected to the action of identical elementary forces, symmetrically arranged with respect to the axis of the bar, and therefore the resultants of these forces will be directed along the same axis. In other words, these molecules will be in a zone of pure uniform tension. In the annular portions of the triangular section, as for example: \(c—a—e—c—g—b—c\), only some components act, in the direction of the angle of action.

The result of such an “oblique” action of the elementary forces must be a tendency of the molecules of these portions to approach the axis of the bar.

In Fig. 8 the zones of pure tension are shown by longitudinal hatching, and the zones of “oblique” tension by transverse hatching. Leaving aside

Fig. 8. Division of the body of a deformed bar into zones of longitudinal and transverse deformation (according to Rejtö).

the question of Rejtö’s division of all materials into groups and subgroups according to the law of relation between the coefficient of interparticle friction and the force of cohesion of the particles, we arrive at the conclusion that correctly alternating necks must necessarily appear within the limits of elasticity.

The angle of action is found by Rejtö experimentally, as the result of processing diagrams obtained in testing bars in tension. According to Rejtö, it varies within rather narrow limits, from \(48^\circ\) to \(61^\circ\), and depends on the kind of metal. For the majority of metals it remains almost unchanged, staying within the limits from \(53^\circ\) to \(56^\circ\).

Taking \(55^\circ\) as the mean value, let us compute the mean distance between necks (the wave length).

\[ \operatorname{tg}\beta=\frac{2r}{L} \]

hence:

\[ L=\frac{2r}{\operatorname{tg}\beta}=\frac{2r}{\operatorname{tg}55^\circ}\simeq 1.4r . \]

The wave length differs from that found by me. The explanation of this discrepancy, in my opinion, should be sought in the fact that Rejtö’s angle of action is found by means of a whole series of graphical transformations of the tensile diagram of a bar, taken by an ordinary indicator. The small scale of the diagram and the inevitable errors in graphical constructions must undoubtedly give a large error, as compared with a simple, direct measurement of the length of the necks. Moreover, and this is much more important, Rejtö assumes without proof the constancy of the angle of action from the moment the material passes beyond the elastic limit up to its rupture. Fig. 9 gives an idea of how Rejtö understands the transmission of elementary forces by the molecules to one another in the neck that has formed. As is evident from the figure, the angle of action remains

Fig. 9. Constancy of the angle of action according to Rejtö.

unchanged both in the slightly and in the strongly deformed part of the bar. In this case necks must form within the necks themselves, i.e. we should observe a shortening of the wave length at the places of greatest deformation. Experiment does not show this. Since Rejtö determines the angle of action not directly but indirectly, considering it constant, the calculation of the magnitude of the angle must always give a value greater than the actual one. In other words, the ratio \(\frac{r}{L}\), according to Rejtö, must turn out to be smaller than the actual one, which is what we observe. The constancy of the angle of action under residual deformation is contradicted by the very fact of the change in the size of the cells of the structural network beyond the elastic limit.

It must be pointed out that Rejtö gives no indication of experiments measuring the necks either before or beyond the elastic limit.

Having come to the conclusion that it is necessary to have an idea of whether correctly alternating waves of necks appear before the limit of…

elasticity, and what the amplitude of such waves is; I determined the magnitude of the minimum deformation with which one would have to deal when setting up experiments by the method of necks.

II. Determination of Elastic Deformations.

To determine the elastic deformations of the cross sections of the specimens being tested, an instrument was needed that would make it possible to measure the amplitude of oscillation of the size of the cross section, in various sections, if such occurred within the limits of elasticity of the material. Let us take, as the limiting stress beyond which we shall not go in the experiment, two thirds of the elastic limit of the given metal, in order to avoid the phenomenon of elastic aftereffect.

The absolute transverse deformation of the diameter will be expressed as follows:

\[ \varepsilon = m \frac{2k}{3E} d \]

where \(k\) is the elastic limit
and \(E\) is Young’s modulus.

Let the amplitude of the waves amount to the \(n\)-th part of the mean change in the diameter.

\[ \Delta \varepsilon = \frac{\varepsilon}{n}; \]

then

\[ \Delta \varepsilon = \frac{2mkd}{3En} \]

Let us impose the requirement that the accuracy of measuring this quantity be up to \(5\%\). We obtain the minimum reading of deformation:

\[ \Delta \Delta \varepsilon = \frac{\Delta \varepsilon}{20} = \frac{mkd}{30En}. \]

In order to have an idea of the minimum magnitude of deformation that must still be recorded by our instrument, we shall substitute numbers. We take an iron specimen, of diameter: \(d = 30\) mm (as an average); then, for

\[ m = \frac{1}{3} \]

\[ E = 20000\ \mathrm{kg}/\mathrm{mm}^{2} \]

\[ k = 20\ \mathrm{kg}/\mathrm{mm}^{2}; \]

we have

\[ \Delta \Delta \varepsilon = \frac{1}{3000 n}\ \mathrm{mm} \]

If the wave amplitude amounts to only \(5\%\) of the mean deformation of the diameter, then: \((n = 20)\)

\[ \Delta \Delta \varepsilon = \frac{1}{60000}\ \mathrm{mm} \]

The instrument had to record such a fraction of a millimeter. In the Mechanical Laboratory of the STI there was no instrument of such sensitivity, and therefore I had to design and make an instrument giving a reading accuracy no less than that required by the conditions of the experiment. (For a description of the instrument and of work with it, see Izv. STI, 1928, vol. 49, no. 2.)

For determining elastic deformations, bars of circular cross section were taken, with diameters from 15 to 65 mm and lengths from 400 to

1500 mm. The specimens were loaded as follows: the smaller ones—in Amsler’s press on columns, and the larger ones—in Olsen’s 100 t machine. Fig. 10 shows the installation of the apparatus during the test.

Fig. 10. Installation of the instruments in determining elastic deformations.

The procedure was as follows. Transverse marks were made on the specimen, as before, at an integral number of millimeters, with the length of the interval between them approximately one tenth of the diameter. The specimen was placed in the testing machine and, in the middle of the first outer interval between the marks, an instrument for determining transverse deformations was fastened.

In addition to this instrument, on the same section, a Martens apparatus with short strips was installed for the simultaneous determination of longitudinal deformations; the latter was necessary, since there was reason to suppose that if the elastic necks along the axis

of the specimen, then it should be expected that the longitudinal deformations of the individual sections will also be different.

The load, up to its maximum, was applied in 12 equal intervals, the first 2 of which were passed through without taking readings on the instruments, so that it would be possible to guarantee that the entire installation system (instrument–specimen–machine) had acquired a definite operating regime. In the remaining ten intervals readings were taken (11 readings) and diagrams of the course of deformations were plotted: \(P-\lambda\) and \(P-\varepsilon\), in order to be certain of the linear dependence between the force and the deformations and, consequently, of the correctness of the installation and of the readings.

As the recorded figures for a given section, the arithmetic means were taken from the 10 differences obtained for \(\lambda\) and \(\varepsilon\) under the successively increasing load. After unloading the machine, the instruments were set on the neighboring section and the entire installation “rested” for at least half an hour, in order to equalize the temperature differences of its parts; then the experiment was repeated. From the obtained mean differences of readings on the scales, two diagrams were constructed: of transverse contractions and longitudinal elongations, in exactly the same way as the diagram of residual elongations had previously been constructed.

Fig. 11. Complex diagram of deformations of an iron bar.
Owing to the small scale of the drawing, every other point is plotted.

After measuring the longitudinal and transverse elastic deformations in all sections of the specimen, the latter was stretched, now without instruments, almost to rupture, i.e. until a neck visible to the eye was formed. For the residual elongations of the intervals, a diagram was constructed on the former axis.

The very first experiment gave a picture which confirmed the supposition I had made concerning the presence of elastic waves on the generatrix of the bar. On the other hand, a result was obtained which proved unexpected to me, namely: the elastic elongations, in all sections along the length—

of the specimen were perfectly identical. In Fig. 11 we have one of the diagrams constructed for a complete experiment with an iron specimen. The wavy line C represents, by its ordinates, the residual elongations. Line B represents the elastic changes of diameter. Line A represents the elastic changes in the length of the segment. The complete coincidence of the maxima and minima in diagrams C and B is clearly visible, as is the constancy along the length of the bar of the fractional elastic deformation.

Thus the experiment confirmed the assumptions concerning the regularity of the distribution of necks, on the one hand, and, on the other, revealed a new datum, a new fact, in the form of the constancy of the fractional deformation. Both the former and the latter required a theoretical analysis of the phenomenon and its substantiation.

III. Theoretical substantiations.

Let us consider the equilibrium of an elementary volume (Fig. 12) inside a cylindrical bar subjected to the action of a central tensile force. In the present case it is most convenient to write the equations of equilibrium in cylindrical coordinates. We direct the \(Z\)-axis along the axis of the bar. We shall denote:

\[ \sigma_\rho \text{— normal stress in the radial direction,} \]

\[ \sigma_\varphi \text{— normal stress in the tangential direction,} \]

\[ \sigma_z \text{— normal stress along the axis,} \]

\[ t_\varphi \text{— shear stress perpendicular to the edge of the element} \]

coinciding with the tangent.

The two remaining shear stresses, \(t_\rho\) and \(t_z\), will be absent here, since the deformation has an axis of symmetry coinciding with \(Z\). The first of these stresses would give torsion of the bar, and the second would transform a circular cross-section into a non-circular one.

Projecting the elementary forces onto the \(Z\)-axis, we obtain:

\[ (\sigma_z+d\sigma_z)\rho\,d\varphi\,d\rho-\sigma_z\rho\,d\varphi\,d\rho+ \left(t_\varphi+\frac{\partial t_\varphi}{\partial \rho}\,d\rho\right)\rho\,d\varphi\,dz -t_\varphi\rho\,d\varphi\,dz=0; \]

Projecting them onto the \(\rho\)-axis, we obtain:

\[ (\sigma_\rho+d\sigma_\rho)\rho\,dz\,d\varphi-\sigma_\rho\rho\,dz\,d\varphi+ \left[\left(t_\varphi+\frac{\partial t_\varphi}{\partial z}\,dz\right)-t_\varphi\right]\rho\,d\rho\,d\varphi=0; \]

In final form, neglecting infinitesimals of the second order, the equations are represented as follows:

\[ \frac{\partial \sigma_\rho}{\partial \rho} +\frac{\partial t_\varphi}{\partial z} +\frac{\sigma_z-\sigma_\rho}{\rho}=0 \ldots\ldots\ldots\ldots\ldots (*) \]

\[ \frac{\partial t_\varphi}{\partial \rho} +\frac{\partial \sigma_z}{\partial z} +\frac{t_\varphi}{\rho}=0 \ldots\ldots\ldots\ldots\ldots (**) \]

In addition to these two equations of equilibrium, in the case of symmetric deformation two further surface conditions must be satisfied:

\[ p_\rho=\sigma_\rho\cos(\rho\nu)+t_\varphi\cos(z\nu) \]

\[ p_z=t_\varphi\cos(\rho\nu)+\sigma_z\cos(z\nu), \]

where \(\nu\) is the direction of the normal to the surface at the given point.

1 — \(\sigma_\rho\)
2 — \(\sigma_\rho + d\sigma_\rho\)
3 — \(\sigma_z\)
4 — \(\sigma_z + d\sigma_z\)
5 — \(\sigma_\varphi\)
6 — \(t_\varphi\)
7 — \(t_\varphi + \dfrac{\partial t_\varphi}{\partial \rho}\,d\rho\)
8 — \(t_\varphi + \dfrac{\partial t_\varphi}{\partial z}\,dz\)

Fig. 12. Distribution of stresses in an element of a stretched cylindrical bar.

But since we are dealing with a cylindrical bar, then:

for the end face — \((\rho \nu)=-\dfrac{\pi}{2}\), \((z\nu)=0\);

\[ p_{\rho}=t_{\rho} \]

\[ p_{z}=\sigma_{z} \]

for the lateral surface — \((\rho \nu)=0\); \((z\nu)=-\dfrac{\pi}{2}\)

\[ p_{\rho}=\sigma_{\rho} \]

\[ p_{z}=t_{\rho} \]

We shall satisfy these conditions when determining the arbitrary constant coefficients that will occur in the equations of stresses and strains. As for the stress \(p_{\varphi}\), in the present case it will be absent, since we are considering a bar with the normal gripping of the ends, i.e. by means of a head, thread, or wedges.

The four equations written down, however, are insufficient for solving the problem. It is necessary to have still a number of equations that would give the connection between the strains of the individual elements of the body, which, after deformation, must remain as continuous as it was before it.

Let us derive such equations. We shall denote the reduced relative linear strain by \(e\), indicating by a subscript the direction of the axis along which it is taken. We shall denote the displacements of an element under deformation by \(u\), \(v\), and \(w\).

Thus, in rectangular coordinates:

Along the \(X\)-axis — displacement — \(u\) and strain — \(e_x\)
“ “ \(Y\) “ — \(v\) “ — \(e_y\)
“ “ \(Z\) “ — \(w\) “ — \(e_z\)

In accordance with these notations the angles of shear will also be written:

In the plane \(XOY\) — \(e_{xy}\)
“ \(YOZ\) — \(e_{yz}\)
“ \(ZOX\) — \(e_{zx}\)

We have:

\[ e_x=\frac{\partial u}{\partial x};\quad e_y=\frac{\partial v}{\partial y};\quad e_z=\frac{\partial w}{\partial z} \]

\[ e_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x};\quad e_{yz}=\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y};\quad e_{zx}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}; \]

Let us form derivatives of the following kind:

\[ \frac{\partial^{2}e_x}{\partial y^{2}} = \frac{\partial^{3}u}{\partial x\,\partial y^{2}}; \quad \frac{\partial^{2}e_y}{\partial x^{2}} = \frac{\partial^{3}v}{\partial y\,\partial x^{2}}; \quad \frac{\partial^{2}e_{xy}}{\partial x\,\partial y} = \frac{\partial^{3}u}{\partial x\,\partial y^{2}} + \frac{\partial^{3}v}{\partial y\,\partial x^{2}} \]

Comparing the expressions obtained with one another, we get:

\[ \frac{\partial^{2}e_x}{\partial y^{2}} + \frac{\partial^{2}e_y}{\partial x^{2}} = \frac{\partial^{2}e_{xy}}{\partial x\,\partial y} \tag{1} \]

By cyclic permutation of the letters, we shall obtain two more such equations.

Then let us form derivatives of the following kind:

\[ \frac{\partial^2 e_{xy}}{\partial y\,\partial z} = \frac{\partial^3 u}{\partial x\,\partial y\,\partial z}; \qquad \frac{\partial e_{yz}}{\partial x} = \frac{\partial^2 v}{\partial x\,\partial z} + \frac{\partial^2 w}{\partial x\,\partial y}; \]

\[ \frac{\partial e_{zx}}{\partial y} = \frac{\partial^2 w}{\partial y\,\partial x} + \frac{\partial^2 u}{\partial y\,\partial z}; \qquad \frac{\partial e_{xy}}{\partial z} = \frac{\partial^2 u}{\partial z\,\partial y} + \frac{\partial^2 v}{\partial z\,\partial x}. \]

Comparing them, we obtain:

\[ 2\,\frac{\partial^2 e_x}{\partial y\,\partial z} = \frac{\partial}{\partial x} \left[ -\frac{\partial e_{yz}}{\partial x} + \frac{\partial e_{zx}}{\partial y} + \frac{\partial e_{xy}}{\partial z} \right] \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (2) \]

A cyclic permutation of the letters here gives two more equations of the same type.

We now have 6 additional equations for solving the problem. In order that these equations may be solved jointly with those obtained earlier, it is necessary, first, to express in them the strains in terms of the stresses, and, second, to replace the Cartesian coordinates by cylindrical ones. To replace the strains by stresses we shall make use of Cauchy’s equations:

\[ \frac{\partial \sigma_x}{\partial x} + \frac{\partial t_y}{\partial z} + \frac{\partial t_z}{\partial y} = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ a) \]

\[ \frac{\partial \sigma_y}{\partial y} + \frac{\partial t_z}{\partial x} + \frac{\partial t_x}{\partial z} = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ b) \]

\[ \frac{\partial \sigma_z}{\partial z} + \frac{\partial t_x}{\partial x} + \frac{\partial t_y}{\partial y} = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ c) \]

The reduced strain, expressed in terms of the actual stresses, will be:

\[ e_x=\frac{1}{E}\left(\sigma_x-m\sigma_y-m\sigma_z\right) \]

Let us introduce, for convenience in the transformations, the following new quantity:

\[ \sigma_x+\sigma_y+\sigma_z=\Theta; \]

then:

\[ e_x=\frac{1}{E}\left[(1+m)\sigma_x-m\Theta\right] \]

\[ e_y=\frac{1}{E}\left[(1+m)\sigma_y-m\Theta\right] \]

\[ e_{zy}=\frac{2(1+m)}{E}\,t_z \]

Substitution into equation (1) gives:

\[ (1+m)\left[\frac{\partial^2 \sigma_y}{\partial x^2}+\frac{\partial^2 \sigma_x}{\partial y^2}\right] -m\left[\frac{\partial^2 \Theta}{\partial x^2}+\frac{\partial^2 \Theta}{\partial y^2}\right] = 2(1+m)\frac{\partial^2 t_z}{\partial x\,\partial y}. \tag{A} \]

Let us determine from equations (a) and (b):

\[ \frac{\partial t_z}{\partial x} = -\frac{\partial \sigma_y}{\partial y} -\frac{\partial t_x}{\partial z} \]

\[ \frac{\partial t_z}{\partial y} = -\frac{\partial \sigma_x}{\partial x} -\frac{\partial t_y}{\partial z} \]

Differentiating the first with respect to \(y\), the second with respect to \(x\), and adding the resulting expressions, we shall have:

\[ 2\frac{\partial^2 t_z}{\partial x\,\partial y} = -\frac{\partial^2 \sigma_x}{\partial x^2} -\frac{\partial^2 \sigma_y}{\partial y^2} -\frac{\partial}{\partial z} \left[ \frac{\partial t_x}{\partial y} + \frac{\partial t_y}{\partial x} \right] \]

Having from equation (c):

\[ \frac{\partial t_x}{\partial y} + \frac{\partial t_y}{\partial x} = -\frac{\partial \sigma_z}{\partial z} \]

we may write:

\[ 2\frac{\partial^2 t_z}{\partial x\,\partial y} = \frac{\partial^2 \sigma_z}{\partial z^2} -\frac{\partial^2 \sigma_x}{\partial x^2} -\frac{\partial^2 \sigma_y}{\partial y^2}. \]

We make the substitution in (A):

\[ (1+m)\left[\frac{\partial^2 \sigma_y}{\partial x^2}+\frac{\partial^2 \sigma_x}{\partial y^2}\right] -m\left[\frac{\partial^2 \Theta}{\partial x^2}+\frac{\partial^2 \Theta}{\partial y^2}\right] \]

\[ = (1+m)\left[ \frac{\partial^2 \sigma_z}{\partial z^2} -\frac{\partial^2 \sigma_x}{\partial x^2} -\frac{\partial^2 \sigma_y}{\partial y^2} \right] \]

We introduce the operator:

\[ \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}. \]

The last equation takes the form:

\[ (1+m)\left[ \nabla^2\Theta - \nabla^2\sigma_x - \frac{\partial^2\Theta}{\partial x^2} \right] - m\left[ \nabla^2\Theta - \frac{\partial^2\Theta}{\partial x^2} \right] =0; \]

Carrying out the operations indicated in the equation, we shall see that it assumes the very simple form:

\[ (1+m)\nabla^2\sigma_x + \frac{\partial^2\Theta}{\partial x^2} =0; \]

Two other equations of the same type are obtained by a cyclic permutation of the indices.

The last three equations, with the aid of the transformation just given, will be of the following form:

\[ (1+m)\nabla^{2} t_z+\frac{\partial^{2}\Theta}{\partial x\,\partial y}=0 \]

Thus the last six equations, relating the deformations of individual elements of the body, will be written in terms of stresses as follows:

\[ \left. \begin{aligned} (1+m)\nabla^{2}\sigma_x+\frac{\partial^{2}\Theta}{\partial x^{2}}&=0\\[1.2em] (1+m)\nabla^{2}\sigma_y+\frac{\partial^{2}\Theta}{\partial y^{2}}&=0\\[1.2em] (1+m)\nabla^{2}\sigma_z+\frac{\partial^{2}\Theta}{\partial z^{2}}&=0\\[1.2em] (1+m)\nabla^{2}t_x+\frac{\partial^{2}\Theta}{\partial y\,\partial z}&=0\\[1.2em] (1+m)\nabla^{2}t_y+\frac{\partial^{2}\Theta}{\partial z\,\partial x}&=0\\[1.2em] (1+m)\nabla^{2}t_z+\frac{\partial^{2}\Theta}{\partial x\,\partial y}&=0 \end{aligned} \right\} \qquad . . . . . . . . (A') \]

Let us write these equations in cylindrical coordinates, so that, first, they may be solved jointly with the equilibrium equations, and, second, the solution itself may be simplified, since in the presence of an axis of symmetry of the deformation some components of the complete stresses must vanish.

To pass to cylindrical coordinates, the independent variables \(x\), \(y\), and \(z\) must be replaced by the new ones: \(\rho\), \(\varphi\), and \(z\).

The relation between the old and the new coordinates is established by the equations:

\[ \rho^{2}=x^{2}+y^{2} \]

\[ \operatorname{tg}\varphi=\frac{y}{x} \]

\[ z=z \]

we write:

\[ \frac{\partial}{\partial x} = \frac{\partial}{\partial \rho}\cdot \cos\varphi - \frac{\partial}{\rho\,\partial\varphi}\cdot \sin\varphi \]

\[ \frac{\partial}{\partial y} = \frac{\partial}{\partial \rho}\cdot \sin\varphi + \frac{\partial}{\rho\,\partial\varphi}\cdot \cos\varphi \]

The operation of obtaining the second derivatives will be given by the equations:

\[ \frac{\partial^2}{\partial x^2} = \frac{\partial^2}{\partial \rho^2}\cdot \cos^2\varphi - \frac{\partial^2}{\rho\,\partial\rho\,\partial\varphi}\cdot \operatorname{sn}2\varphi + \frac{\partial}{\rho\,\partial\rho}\cdot \operatorname{sn}^2\varphi + \]

\[ + \frac{\partial}{\rho^2\,\partial\varphi}\cdot \operatorname{sn}2\varphi + \frac{\partial^2}{\rho^2\,\partial^2\varphi}\operatorname{sn}^2\varphi; \]

\[ \frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial\rho^2}\cdot \operatorname{sn}^2\varphi + \frac{\partial^2}{\rho\,\partial\rho\,\partial\varphi}\cdot \operatorname{sn}2\varphi + \frac{\partial}{\rho\,\partial\rho}\cdot \cos^2\varphi - \]

\[ - \frac{\partial}{\rho^2\,\partial\varphi}\cdot \operatorname{sn}2\varphi + \frac{\partial^2}{\rho^2\,\partial\varphi^2}\cdot \cos^2\varphi; \]

\[ \frac{\partial^2}{\partial x\,\partial y} = \frac{\partial^2}{\partial\rho^2}\cdot \frac{\operatorname{sn}2\varphi}{2} + \frac{\partial^2}{\rho\,\partial\rho\,\partial\varphi}\cdot \cos 2\varphi - \frac{\partial}{\rho^2\,\partial\varphi}\cdot \cos 2\varphi - \]

\[ - \frac{\partial}{\rho\,\partial\rho}\cdot \frac{\operatorname{sn}2\varphi}{2} - \frac{\partial^2}{\rho^2\,\partial\varphi^2}\cdot \frac{\operatorname{sn}2\varphi}{2}; \]

The stresses are related by the equations:

\[ \sigma_x=\sigma_\rho\cos^2\varphi+\sigma_\varphi\operatorname{sn}^2\varphi \qquad t_y=t_\varphi\cos\varphi \]

\[ \sigma_y=\sigma_\rho\operatorname{sn}^2\varphi+\sigma_\varphi\cos^2\varphi \]

\[ \sigma_z=\sigma_z \]

\[ \nabla^2\sigma_x = \left( \frac{\partial^2}{\partial\rho^2} + \frac{\partial}{\rho\,\partial\rho} + \frac{\partial^2}{\rho^2\,\partial\varphi^2} + \frac{\partial^2}{\partial z^2} \right) (\sigma_\rho\cos^2\varphi+\sigma_\varphi\operatorname{sn}^2\varphi) = \]

\[ = \nabla^2(\sigma_\rho\cos^2\varphi+\sigma_\varphi\operatorname{sn}^2\varphi) - \frac{2}{\rho^2}(\sigma_\rho-\sigma_\varphi)\cos 2\varphi; \]

\[ \nabla^2\sigma_y = \left( \frac{\partial^2}{\partial\rho^2} + \frac{\partial}{\rho\,\partial\rho} + \frac{d^2}{\rho^2\,d\varphi^2} + \frac{\partial^2}{\partial z^2} \right) (\sigma_\rho\operatorname{sn}^2\varphi+\sigma_\varphi\cos^2\varphi) = \]

\[ = \nabla^2(\sigma_\rho\operatorname{sn}^2\varphi+\sigma_\varphi\cos^2\varphi) + \frac{2}{\rho^2}(\sigma_\rho-\sigma_\varphi)\cos 2\varphi; \]

\[ \nabla^2 t_\varphi = \left( \frac{\partial^2}{\partial\rho^2} + \frac{\partial}{\rho\,\partial\rho} + \frac{\partial^2}{\rho^2\,\partial\varphi^2} + \frac{\partial^2}{\partial z^2} \right) t_\varphi\cos\varphi = \]

\[ = \left( \nabla^2 t_\varphi-\frac{t_\varphi}{\rho^2} \right)\cos\varphi \]

Having:

\[ \frac{\partial^2\Theta}{\partial x^2} = \frac{\partial^2\Theta}{\partial\rho^2}\cos^2\varphi + \frac{\partial\Theta}{\rho\,\partial\rho}\operatorname{sn}^2\varphi \]

\[ \frac{\partial^2\Theta}{\partial y^2} = \frac{\partial^2\Theta}{\partial\rho^2}\operatorname{sn}^2\varphi + \frac{\partial\Theta}{\rho\,\partial\rho}\cos^2\varphi \]

\[ \frac{\partial^2\Theta}{\partial x\,\partial z} = \frac{\partial^2\Theta}{\partial\rho\,\partial z}\cos\varphi \]

We shall substitute the obtained quantities into the equations of system \((A')\). Since these equations must be satisfied for any value of the angle \(\varphi\), the latter will be absent from the transformed equations.

The third equation of system \((A')\) is rewritten without change, and instead of the last three we shall have only one. Thus, in cylindrical coordinates, the compatibility equations will be:

\[ \left. \begin{aligned} &(1+m)\left[\nabla^2 \sigma_\rho-\frac{2}{\rho^2}(\sigma_\rho-\sigma_\varphi)\right] +\frac{\partial^2 \Theta}{\partial \rho^2}=0,\\[4pt] &(1+m)\left[\nabla^2 \sigma_\varphi+\frac{2}{\rho^2}(\sigma_\rho-\sigma_\varphi)\right] +\frac{\partial \Theta}{\rho\,\partial \rho}=0,\\[4pt] &(1+m)\nabla^2 \sigma_z+\frac{\partial^2 \Theta}{\partial z^2}=0,\\[4pt] &(1+m)\left[\nabla^2 t_\varphi-\frac{t_\varphi}{\rho^2}\right] +\frac{\partial^2 \Theta}{\partial \rho\,\partial z}=0 \end{aligned} \right\} \qquad\ldots\ldots (B) \]

For the solution of the problem in the general form, we have obtained six equations. We shall use the method of A. Love. Let us choose such a function \(\varphi\) through which it would be possible to express all the stresses entering into the equations, with the obligatory condition that, in the general form, both the equilibrium equations \((*)\) and \((**)\) and the compatibility equations \((B)\) be satisfied. According to Love, the functional composition of the stresses must consist of first derivatives with respect to \(z\) of functions of the following kind:

\[ \frac{\partial^2 \varphi}{\partial \rho^2},\quad \frac{\partial \varphi}{\rho\,\partial \rho},\quad \frac{\partial^2 \varphi}{\partial z^2},\quad \frac{\partial^2 \varphi}{\partial \rho\,\partial z}. \]

We write, with undetermined constant coefficients, the following equations:

\[ \left. \begin{aligned} \sigma_\rho &= \frac{\partial}{\partial z}\left( a\,\frac{\partial^2\varphi}{\partial \rho^2} +b\,\frac{\partial\varphi}{\rho\,\partial \rho} +c\,\frac{\partial^2\varphi}{\partial z^2} +d\,\frac{\partial^2\varphi}{\partial \rho\,\partial z} \right),\\[4pt] \sigma_\varphi &= \frac{\partial}{\partial z}\left( e\,\frac{\partial^2\varphi}{\partial \rho^2} +f\,\frac{\partial\varphi}{\rho\,\partial \rho} +g\,\frac{\partial^2\varphi}{\partial z^2} +h\,\frac{\partial^2\varphi}{\partial \rho\,\partial z} \right),\\[4pt] \sigma_z &= \frac{\partial}{\partial z}\left( i\,\frac{\partial^2\varphi}{\partial \rho^2} +k\,\frac{\partial\varphi}{\rho\,\partial \rho} +l\,\frac{\partial^2\varphi}{\partial z^2} +m\,\frac{\partial^2\varphi}{\partial \rho\,\partial z} \right),\\[4pt] t_\varphi &= \frac{\partial}{\partial z}\left( n\,\frac{\partial^2\varphi}{\partial \rho^2} +o\,\frac{\partial\varphi}{\rho\,\partial \rho} +p\,\frac{\partial^2\varphi}{\partial z^2} +r\,\frac{\partial^2\varphi}{\partial \rho\,\partial z} \right) \end{aligned} \right\} \qquad\ldots\ldots (C) \]

Having substituted these stress values into both equilibrium equations \((*)\) and \((**)\), we obtain the values of part of these coefficients and establish the relation between the remaining ones. The substitution gives:

\[ a=d=f=h=l=m=n=o=p=0; \]

\[ b=e=A \]

\[ c=g=i=k=-r=B \]

The equations will take the form:

\[ \left. \begin{aligned} \sigma_\rho &= \frac{\partial}{\partial z} \left( A\,\frac{\partial \varphi}{\rho\,\partial \rho} + B\,\frac{\partial^2 \varphi}{\partial z^2} \right), \\[0.4em] \sigma_\varphi &= \frac{\partial}{\partial z} \left( A\,\frac{\partial^2 \varphi}{\partial \rho^2} + B\,\frac{\partial^2 \varphi}{\partial z^2} \right), \\[0.4em] \sigma_z &= B\,\frac{\partial}{\partial z} \left( \frac{\partial^2 \varphi}{\partial \rho^2} + \frac{\partial \varphi}{\rho\,\partial \rho} \right), \\[0.4em] t_\varphi &= -\,B\,\frac{\partial}{\partial z} \left( \frac{\partial^2 \varphi}{\partial \rho\,\partial z} \right). \end{aligned} \right\} \qquad . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (D) \]

We shall substitute the expressions obtained for the stresses into the equation of continuity (B). The latter is then written as:

\[ \frac{\partial}{\partial z} \left[ (1+m) \left\{ A\,\frac{\partial}{\rho\,\partial \rho}\nabla^2\varphi + B\,\frac{\partial^2}{\partial z^2}\nabla^2\varphi \right\} + (A+B)\frac{\partial^2}{\partial \rho^2}\nabla^2\varphi + (B-A)\frac{\partial^4 \varphi}{\partial \rho^2\,\partial z^2} \right]=0; \]

\[ \frac{\partial}{\partial z} \left[ (1+m) \left\{ A\,\frac{\partial^2}{\partial \rho^2}\nabla^2\varphi + B\,\frac{\partial^2}{\partial z^2}\nabla^2\varphi \right\} + (A+B)\frac{\partial}{\rho\,\partial \rho}\nabla^2\varphi + (B-A)\frac{\partial^3 \varphi}{\rho\,\partial \rho\,\partial z^2} \right]=0; \]

\[ \frac{\partial}{\partial z} \left[ (1+m)B \left\{ \frac{\partial^2}{\partial \rho^2}\nabla^2\varphi + \frac{\partial}{\rho\,\partial \rho}\nabla^2\varphi \right\} + (A+B)\frac{\partial^2}{\partial z^2}\nabla^2\varphi + (B-A)\frac{\partial^4 \varphi}{\partial z^4} \right]=0; \]

\[ \frac{\partial}{\partial z} \left[ (A-mB)\frac{\partial^2}{\partial \rho\,\partial z}\nabla^2\varphi + (B-A)\frac{\partial^4\varphi}{\partial \rho\,\partial z^3} \right]=0; \]

Solving jointly the four equations obtained, we shall have:

\[ \frac{\partial^4}{\partial z^4} \left[ \frac{\partial^2\varphi}{\partial \rho^2} + M\,\frac{\partial\varphi}{\rho\,\partial \rho} + N\,\frac{\partial^2\varphi}{\partial z^2} \right] = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (3) \]

where the coefficients \(M\) and \(N\) have the values:

\[ M = - \frac{(c-B)(ac-B^2)} {(c^2-B^2)(c-B)-(c-a)(ac-2B^2+c^2)} \]

\[ N = - \frac{(c-B)(c-a)} {(c^2-B^2)(c-B)-(c-a)(ac-2B^2+c^2)} \]

Here, for brevity of notation, the following have been denoted:

\[ a=(1+m)A \]

\[ b=(1+m)B \]

\[ c=A+B \]

Let us first find a particular integral of this equation, i.e., solve the equation:

\[ \frac{\partial^{2}\varphi}{\partial \rho^{2}}+M\frac{\partial\varphi}{\rho\,\partial \rho}+N\frac{\partial^{2}\varphi}{\partial z^{2}}=0 \ldots \ldots \ldots \ldots \ldots (4) \]

We seek its solution in the form:

\[ \varphi=R\sin \beta z \]

where \(R\) is a function of \(\rho\) alone and \(\beta\) is a constant coefficient.

Carrying out the differentiation of \(\varphi\) and substituting the values of the derivatives into (4), we obtain:

\[ \frac{\partial^{2}R}{\partial \rho^{2}}+M\frac{\partial R}{\rho\,\partial \rho}-\beta^{2}NR=0 \ldots \ldots \ldots (5) \]

This is Bessel’s equation. Its solution cannot be found in the form of a finite algebraic or simple transcendental function [A. R. Forsyth, of Differential Equations, 1902].

The function \(R\) may be represented by an infinite series:

\[ R=a_{1}+a_{2}\rho^{2}+a_{3}\rho^{4}+a_{4}\rho^{6}+\ldots \ldots \]

After differentiating \(R\) and substituting the derivatives into equation (5), we obtain:

\[ \begin{aligned} &2a_{2}+12a_{3}\rho^{2}+30a_{4}\rho^{4}+\ldots \ldots \ldots\\ &+2Ma_{2}+4Ma_{3}\rho^{2}+6Ma_{4}\rho^{4}+\ldots \ldots\\ &-\beta^{2}Na_{1}-\beta^{2}Na_{2}\rho^{2}-\beta^{2}Na_{3}\rho^{4}-\ldots \ldots=0; \end{aligned} \]

Equating the coefficients of like powers of \(\rho\), we have:

\[ \begin{aligned} &2(1+M)a_{2}-\beta^{2}Na_{1}=0,\\ &4(3+M)a_{3}-\beta^{2}Na_{2}=0,\\ &6(5+M)a_{4}-\beta^{2}Na_{3}=0, \end{aligned} \]

\[ \begin{array}{cccc} \cdot & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot \end{array} \]

The coefficients turn out to be related to one another and can all be expressed in terms of one, namely \(a_{1}\).

\[ a_{2}=\frac{\beta^{2}N}{2(1+M)}\,a_{1} \]

\[ a_{3}=\frac{\beta^{2}N}{4(3+M)}\,a_{2} = \frac{\beta^{4}N^{2}}{2\cdot4(1+M)(3+M)}\,a_{1} \]

\[ a_{4}=\frac{\beta^{4}N}{6(5+M)}\,a_{3} = \frac{\beta^{6}N^{3}}{2\cdot4\cdot6(1+M)(3+M)(5+M)}\,a_{1} \]

\[ \begin{array}{ccccc} \cdot & \cdot & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot & \cdot \end{array} \]

\[ a_{n}= \frac{\beta^{2(n-1)}N^{\,n-1}} {2(n-1)!\,(1+M)(3+M)\ldots[(2n-3)+M]}\,a_{1} \]

The coefficients decrease rapidly as the series is lengthened; therefore, in an approximate solution, one may confine oneself to only the first few terms of the series.

Let us take the first three terms. The stress function will be expressed as:

\[ \varphi=(a_1+a_2\rho^2+a_3\rho^4)\,\operatorname{sn}\beta z \]

But this is only a particular integral of equation (3). The complete integral may be represented by the function:

\[ \varphi=(a_1+a_2\rho^2+a_3\rho^4)\,\operatorname{sn}\beta z+C\rho^2 z+D z^3+E \]

Performing the differentiation of \(\varphi\), we substitute the derivatives into equations (D):

\[ \sigma_\rho=\beta\left[(2Aa_2-\beta^2Ba_1)+(4Aa_3-\beta^2Ba_2)\rho^2-\beta^2Ba_3\rho^4\right]\cos\beta z+ \]

\[ +2(AC+2BD) \]

\[ \sigma_\varphi=\beta\left[(2Aa_2-\beta^2Ba_1)+(12Aa_3-\beta^2Ba_2)\rho^2-\beta^2Ba_3\rho^4\right]\cos\beta z+ \]

\[ +2(AC+3BD) \]

\[ \sigma_z=4\beta B\left[a_2+4a_3\rho^2\right]\cos\beta z+4C \]

\[ t_\rho=2\beta^2B\left[a_2\rho+2a_3\rho^3\right]\operatorname{Sn}\beta z \]

To determine the constant coefficients, let us write a number of conditions.

On the lateral surface of the cylindrical bar there are no radial normal or tangential stresses, i.e.

\[ \text{for } \rho=r \quad \sigma_\rho=0 \ldots\ldots\ldots\ldots\ldots a) \]

\[ \sigma_\varphi=0 \ldots\ldots\ldots\ldots\ldots\ldots c) \]

The sum of the elementary normal forces in any transverse section must be equal to the tensile force.

\[ 2\pi\int_F \rho\sigma_z\,d\rho=P=\pi r^2 k_0 \ldots\ldots\ldots\ldots c) \]

Here, \(k_0\) denotes the mean tensile stress.

Owing to the presence of an axis of symmetry of the deformation and its coincidence with the axis of the bar, we must have, at the center of any transverse section, equality of the normal stresses along the radius and tangentially. Moreover, the tangential stress must be equal to zero, since at the center the shear angle changes its direction to the opposite one:

\[ \text{for } \rho=0 \quad \sigma_\rho=\sigma_\varphi \ldots\ldots\ldots\ldots d) \]

\[ t_\rho=0 \ldots\ldots\ldots\ldots\ldots\ldots l) \]

The generator of the cylinder, under deformation, becomes a wave-like line, which, in view of its periodicity, we take to be a sinusoid (cosinusoid). Therefore the deformation of an arbitrary radius may be expressed as:

\[ u=u_0+u_1\cos\beta z \ldots\ldots\ldots\ldots f) \]

where \(u_0\) is the mean linear deformation of the radius, and \(u_1\) is the half-amplitude of its variation along the length of the cylinder.

Experiment gives for the magnitude of the longitudinal deformation of the outer fibers the constant value

\[ \text{for } \rho=r \quad e_z=\text{const}=i_0=\frac{k_0}{E}\ . . . . . \quad (g) \]

Let us carry out all the conditions written above.

Condition (a) gives:

\[ \beta\left[(2Aa_2-\beta^2Ba_1)+(4Aa_3-\beta^2Ba_2)r^2-\beta^2Ba_3r^4\right]\cos\beta z+ \]

\[ +\,2(AC+3BD)=0. \]

Since this condition must be satisfied for any value of \(z\), we have, separately:

\[ (2Aa_2-\beta^2Ba_1)+(4Aa_3-\beta^2Ba_2)r^2-\beta^2Ba_3r^4=0 \ . . . \quad (1) \]

and

\[ AC+BD \ . . . . . . . . . . . . . . . . . \quad (2) \]

Equation (b) gives:

\[ 2\beta^2B\left[a_2r+2a_3r^3\right]\operatorname{Sn}\beta z=0 \]

i.e.

\[ a_2+2a_3r^2=0 \]

Condition (c) gives:

\[ 8\pi\beta B\left[\frac{a_2r^2}{2}+a_3r^4\right]\cos\beta z+4\pi Cr^2=\pi r^2 k^0, \]

or, by the condition of independence from the position of the cross-section, separately,

\[ a_2+2a_2r^2=0 \ . . . . . . . . . . . . \quad (4) \]

i.e. the condition obtained earlier—(3)!,

\[ \text{and}\quad 4C=k_0 \ . . . . . . . . . . . . \quad (5) \]

Conditions (d) and (e), as is seen from the equations, are satisfied in the general form.

Condition (f) gives:

\[ u=u_0+u_1\cos\beta z=(\rho e_\varphi)_{\rho=r}= \]

\[ =\frac{r}{E}\left[\sigma_\varphi-m\sigma_\rho-m\sigma_z\right]_{\rho=r} =\frac{r\beta}{E}\left[(1-m)(2Aa_2-\beta^2Ba_1)-4mBa_2+\right. \]

\[ \left.+\left\{(12Aa_3-\beta^2Ba_2)-m(4Aa_3-\beta^2Ba^2-16mBa_3)r^2-\right.\right. \]

\[ \left.\left.-(1-m)\beta^2Ba_3r^4\right\}\cos\beta z\right] +\frac{2r}{E}(1-m)(AC+2BD)-4\frac{mr}{E}C. \]

Hence, paying attention to condition (2), we have:

\[ u_0=-4\frac{mr}{E}C=-mr\frac{k^0}{E}\ . . . . . . . . \quad (6) \]

i.e. that linear deformation of the radius which occurs under the condition of invariability of the cylindrical form of the bar during deformation.

The compound coefficient of \(\cos\beta z\) is equal to \(u_1\) . . . . . . \((7)\)

Condition (g) gives:

\[ i_0=e_z=\frac{1}{E}\left[\tau_z-m\sigma_r-m\sigma_\varphi\right] =\frac{\beta}{E}\left[4Ba_2-\right. \]

\[ \left.-2m(2Aa_2-\beta^2Ba_1)+\{16Ba_3-2m(9Aa_3-\beta^2Ba_1)\}r^2+\right. \]

\[ \left.+2m\beta^2Ba_3r^4\right]\cos\beta z +\frac{4}{E}\left[C-m(AC+3BD)\right]=\mathrm{const} \]

\(i_0\) does not depend on \(z\); therefore, taking (1), we have:

\[ 4Ba_1-2m(2Aa_2-\beta^2Ba_1)+\{16Ba_3-2m(8Aa_3-\beta^2Ba_2)\}r^2+ \]

\[ +2m\beta^2Ba_3r^4=\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (8) \]

and

\[ i_0=\frac{4C}{E}=\frac{k_0}{E}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (9) \]

which already occurred in (5).

To simplify the equations obtained, let us determine the value of the coefficient \(\beta\).

We have taken the deformed generatrix of the cylinder to be a sinusoid. For \(\cos\), a complete period is equal to \(2\pi\); therefore:

\[ \cos \beta z=\cos(2\pi+\beta z)=\cos\beta(L+z) \]

where \(L\) is the wavelength.

Hence

\[ 2\pi=\beta L \]

or

\[ \beta=\frac{2\pi}{L}. \]

From the experiments, as was already mentioned above, the wavelength was found to be:

\[ L=1.59d=3.18r. \]

Obviously, the experimental coefficient relating the wavelength to the radius is:

\[ \pi. \]

Hence:

\[ \beta=\frac{2}{r}. \]

Substituting the obtained value of \(\beta\) and the values of the integral coefficients \(a_2\) and \(a_3\) into the equations obtained, from (1) to (9), we determine the constants.

Substitution into (3) and (1), respectively, gives:

\[ 3+M+2N=0 \]

\[ 1+M+N=0 \]

Solving them simultaneously, we find:

\[ M=1\quad\text{and}\quad N=-2. \]

Substituting their values into the expressions for the integral coefficients:

\[ a_2=-\frac{2}{r^2}a_1 \]

\[ a_3=\frac{1}{r^4}a_1 \]

from (7) it is determined that:

\[ u_1=\frac{16a_1}{Er^2}(A-mB) \]

Knowing the literal expression (in \(A\) and \(B\)) of the coefficients \(M\) and \(N\), and now knowing their numerical value, we shall determine from equations \((****)\) and \((****)\) the dependence between \(A\) and \(B\).

We obtain:

\[ m A = B \]

This relation makes it possible to determine, from (2):

\[ D = -\frac{1}{3m} = C \frac{k_0}{12m} \]

Finally, carrying out the substitution in (1) and (8), we shall see that these equations are identically satisfied.

In its final form the semi-amplitude of the vibration of the wave (neck) is expressed as:

\[ u_1 = \frac{16 A a_1}{E r^2}(1-m^2)\ldots \ldots \ldots \ldots \ldots \ldots (10) \]

Upon substituting the constants into all the stress equations \((D)\), we obtain the same, in all equations, complex coefficient

\[ A a_1, \]

which can be determined only experimentally.

Calling:

\[ A a_1=\alpha, \]

we shall determine it through \(u_1\) and \(u_0\).

From (10) we have:

\[ \alpha=\frac{E r^2 u_1}{(16\,1-m^2)} = -\frac{m u_1}{16(1-m^2)n_{\xi}}\, k_0 r^3 \]

Calling the ratio:

\[ \frac{u_1}{u_0}=\gamma, \]

we shall now rewrite the expressions for the stresses and the deformation of the radius in their final form.

\[ \sigma_{\rho} = \frac{\gamma m k_0}{2(1-m^2)r^4} \left[(1+m)r^4-(1+2m)r^2\rho^2+m\rho^4\right] \cos\frac{2z}{r} \]

\[ \sigma_{\varphi} = \frac{\gamma m k_0}{2(1-m^2)r^4} \left[(1+m)r^4-(3+2m)r^2\rho^2+m\rho^4\right] \cos\frac{2z}{r} \]

\[ \sigma_z = k_0+ \frac{\gamma m^2 k_0}{(1-m^2)r^2} \left(r^2-2\rho^2\right) \cos\frac{2z}{r} \]

\[ \tau_{\varphi} = \frac{\gamma m^2 k_0}{(1-m^2)r^3} \left(r^2\rho-\rho^3\right) \operatorname{Sn}\frac{2z}{r} \]

\[ u = -\,m\frac{k_0}{E}\rho -\gamma m\frac{k_0}{E}\rho\cos\frac{2z}{r} \]

In Fig. 13 the distribution of stresses in various transverse sections of the bar is shown; moreover, for \(\sigma_z\) not the full value of \(\sigma_z\) is shown, but only:

\[ \Delta\sigma_z = \frac{\gamma m^2 k_0}{(2-m^2)r^2} \left(r^2-2\rho^2\right) \cos\frac{2z}{r} \]

Labels in the figure: \(\sigma_\rho\), \(\sigma_\varphi\), \(\Delta\sigma_z\), \(\tau_\varphi\), \(\Delta u\).

Fig. 13. Distribution of stresses and radial displacements in various sections of a round specimen. The scale of \(\Delta\sigma_z\) is enlarged.

In Fig. 14 are shown those changes in form of the internal fibers which occur under tension (the scale of the changes has been taken large).

Fig. 14. Exaggerated change in form of individual fibers of a stretched circular specimen.

The experimental coefficient \(\gamma\) is a constant quantity. From the above-mentioned experiments its mean value was obtained:

\[ \gamma = 0.0829 \simeq \frac{1}{12}. \]

For estimating the maximum values of stresses and deformations we give a table, taking for \(m\), as the mean value: \(m = \frac{1}{3}\).

Ratio	Place in the section \(\rho\)	Place along the length of the bar \(Z\)	Value of the ratio	Note: (where the cross-section occurs)
\(\dfrac{s_z}{k_0}\)	0	0	\(\dfrac{1}{48}\)	Trough of the wave.
\(\dfrac{s_z}{k_0}\)	0	\(\dfrac{L}{2}\)	\(-\dfrac{1}{48}\)	Crest of the wave.
\(\dfrac{s_\varphi}{k_0}\)	0	0	\(\dfrac{1}{48}\)	t. w.
\(\dfrac{s_\varphi}{k_0}\)	0	\(\dfrac{L}{2}\)	\(-\dfrac{1}{48}\)	c. w.
\(\dfrac{s_\varphi}{k_0}\)	\(r\)	0	\(-\dfrac{1}{24}\)	t. w.
\(\dfrac{s_\varphi}{k_0}\)	\(r\)	\(\dfrac{L}{2}\)	\(\dfrac{1}{24}\)	c. w.
\(\dfrac{\nabla z_z}{k_0}\)	0	0	\(\dfrac{1}{96}\)	t. w.
\(\dfrac{\nabla z_z}{k_0}\)	0	\(\dfrac{L}{2}\)	\(-\dfrac{1}{96}\)	c. w.
\(\dfrac{\nabla z_z}{k_0}\)	\(r\)	0	\(-\dfrac{1}{96}\)	t. w.
\(\dfrac{\nabla z_z}{k_0}\)	\(r\)	\(\dfrac{L}{2}\)	\(\dfrac{1}{96}\)	c. w.
\(\dfrac{r_\varphi}{k_0}\)	\(0.577\,r\)	\(\dfrac{L}{4}\)	\(\dfrac{1}{250}\)	On the rise of the wave.
\(\dfrac{r_\varphi}{k_0}\)	\(0.577\,r\)	\(\dfrac{3L}{4}\)	\(-\dfrac{1}{250}\)	On the fall of the wave.
\(\dfrac{u_0}{u'}\)	\(r\)	0	\(-\dfrac{1}{12}\)	t. w.
\(\dfrac{u_0}{u'}\)	\(r\)	\(\dfrac{L}{2}\)	\(\dfrac{1}{12}\)	c. w.

In addition to experiments with cylindrical bars, I carried out a series of experiments with bars of rectangular cross-section. The bars were stretched beyond the elastic limit (elastic deformations were not measured), and the entire experiment was reduced to constructing a diagram of residual deformations and to determining the wavelength. In such bars the distribution of stresses must have an even more complicated character than in cylindrical ones. This is indicated, if only, by the shape of the cross-sections in the necks, a photograph of which is given in Fig. 15. Setting aside the problem of the distribution of stresses in bars of non-circular cross-section as a completely independent one, I cannot pass over in silence the interesting fact that the wavelength in bars of rectangular cross-section, which occurred in my experiments (14 bars), is found to be:

Fig. 15. Change in the rectangular cross-section in the necks of specimens at rupture.

\[ l = \frac{\pi}{2} d, \]

where \(d\) is the diameter of a certain circular cross-section equal, in area, to the area of the specimen with rectangular transverse cross-section.

In the experiments indicated, the ratio of the sides of the cross-section reached \(1:4\).

Summary

1. The hypothesis that the cylindrical form of a body, under tensile (compressive) deformation, is not a stable form is confirmed by indirect experiment with residual deformations.

Experiment indicates the periodicity of the distribution of necks under tension (compression).

1. The same is also confirmed by direct experiment with elastic deformations.

2. The theoretical development of the problem leads to the conclusion that the periodicity of necks is necessary, within the limits of elasticity of the material, thereby confirming both the initial hypothesis and experiment.

3. The distribution of stresses in a normally stretched (compressed) cylindrical bar has a complex character; moreover, tangential stresses are present in the transverse sections.

4. The waves formed by regularly alternating necks have an amplitude proportional to the Poisson coefficient of the given material and equal to

\[ \frac{1}{12} \]

of the mean change in the diameter of the bar, for all the metals tested.

Mechanical Lab., STI.