If the plate with cylindrical anisotropy is at the same time also ortho- tropic, i. Equation The problem of plane strain is completely analogous to the problem of plane stress of the plate. If a body represented in Fig. Zhitkov studied the generalized plane stress of orthotropic body with cylindrical anisotropy, the moduli of elasticity of which are func- tions of coordinates r and 6.
In this case instead of Prikladnaya matematika i mekhanika, v. Some plane problems of the theory of elasticity of anisotropic body. Experi- mental methods for the determination of stresses and deformations in elastic and plastic zones, Collection of works, ONTI, Ibid, p.
See Ref. Mikhlin, S. Savin, G. Sherman, D. VI, No. I'ya Vekua, Application of N. I, No. Kupradze, V. XV, No. Coker, E. ONTI, Krasnoy, V. Uchenye zapiski Leningrad, Gosudar. Krasnov, V. Leningrad, Zapiski Leningrad. Nauk, No. Leipzig-Berlin Teubner , Auerbach, F. III, Leipzig, Geckeler, J. Handbook of Physics, vol. II, Berlin, Mitinskiy, A. Trudy Lesotekhnich. Akademii im. Kitova, No. Handbook of Airplane Design, vol. III, Strength of Airplane.
TsAGI, , p. Zapiski Saratov. I XIV , No. The Simplest Cases In this section we will consider some cases of stress distribution caused by bending forces in a rectangular flat plate, in a wedge-shaped canti- lever with rectangular cross-section, and in a curved beam in the shape of a segment of a flat circular ring. In all cases it is assumed that each point of the body possesses a plane of elastic symmetry which is parallel to its middle surface which is taken as plane xy or r0.
We consider at first the simplest cases of the equilibrium of a homo- geneous anisotropic rectangular plate of thickness h which is subjected to generalized plane stress caused by forces distributed on its edges. A rectangular plate is subjected to tension by normal forces p dustributed uniformly over two sides Fig. The stress distribution is identical to the case of an isotropic plate subjected to tension, and the strains are determined from The dotted line shows the stress distribution for an isotropic beam.
For an orthotropic beam, in which the direction of axis x coincides with one of the principal directions, we obtain from It is necessary to add a constant corrective term to the expression determined by the elementary theory, which depends on the elastic constants and the dimensions of the cross section. Beam on two supports. For a beam simply supported at the ends and bent by a uniformly distributed load see Fig.
After solving this problem we reach the follow- ing conclusion. The nor- mal stresses o, on the longitudinal cross sections, which express the effect of longitudinal layers on each other, are distributed in both cases according to the formula: 1 -F Bending of a Beam by a Linearly Distributed Load With the use as a stress function a polynominal of the sixth order it is easy to obtain the solution for a beam loaded by normal forces distri- buted lengthwise in a linear manner.
We consider two cases of a beam bent by a linearly distributep load. For simplicity we consider that the beam is orthotropic direction of the axis coincides with the principal direction. A beam is fixed at one end and is bent by a normal load distributed in a linear manner. Coordinate directions are shown in Fig. A beam on two supports. Stresses which are equal to zero corre- spond, evidently, to terms of the zero and first power.
They can be dis- regarded. Terms of the second and third order satisfy the equation for the stress function 7. The poly- nomial terms of higher orders, fourth, fifth, etc.
One can be easily convinced of this by making the function P, the solution of 7. The arbitrary constants of Generally speaking, it is impossible to satisfy conditions on short sides of the beam i. We can only provide that those stresses at the ends considering the method of clamping be equi- valent to forces and moments which would balance the external load.
In order, for example, to obtain the stress distribution in a beam with a load given as a quadratic function in x according to a parabolic relation , it is necessary to take the stress function as a polynomial of the seventh power, or as a sum of homogeneous polynomial from the second to seventh power inclusively.
The solution obtained in this man- ner is cumbersome, and we are not going to present it here. It need only be noted that in all cases when load is given by The first terms are the stresses which are obtained by the elementary theory of bending, and the second, the corrective terms, not considered by the fundamental theory. The expression for the curvature Eu" 2 es The second term is not accounted for by the elementary theory.
A solution for the more general case of elastic equilibrium of an anisotropic rectangular plate was obtained by A.
We will consider the application of this method to an orthotropic beam supported at ends and bent by a normal load distributed symmetrically relative to the middle according to an arbitrary law. Other cases of Fig. Let the plane of elastic symmetry be parallel to the beam edges, and, consequently, the axial direction is the principal one. By arranging the coordinate axes as shown in Fig.
Case I. The roots are real and not equal. Case II. The roots are real and equal. Case IIL. The roots are complex. Next, according to 5.
The boundary condi- tions provide a system of equations for the determination of constants Ams Bms Cn; Dm for each ma new system is obtained. By solving these equations we find all the constants.
As a result we obtain formulas for stresses in the form of series of rather complicated structure. The stresses o, at the beam ends can be reduced to moments, which can be found using the solution for pure bending. This type of problem for an anisotropic beam is, evidentialy, seldom solved.
Kufarev and V. Bending of a Composite Multi-layered Beam It is possible to obtain the stress distribution for certain cases of bend- ing of a beam composed of an arbitrary number of anisotropic layers of uniform thickness by a polynomial stress function. We will consider here only the case of an orthotropic cantilever which is bent by a force and a moment.!! A beam is given which is made up of any arbitrary number of strips of the same thickness but with different elastic properties.
One of its ends is rigidly fixed and the other is subjected to load which leads to a moment M and transverse force P. It is necessary to determine the stress in each layer, as well as equation of the neutral axis and the bending ridigity. The x coordinate is measured from the free end, and the x axis runs along the upper edge, as shown in Fig. The final results are the following. The equation of the bent axis i. All formulas can be slightly simplified when layers are of the same thickness equal to bin.
In particular, the formula for rigidity Using the formulas In a homogeneous cantilever the highest normal stress is 6m, and the highest tangential stress is p. The stress distribution o, and t,, at a cross section is shown in Fig.
The broken line represents stresses in a homogeneous canti- lever. Case 2. Comparing Fig. The highest normal stress in the composite cantilever in both cases exceeds considerably the highest stress in the identical homogeneous cantilever. Bending of a Beam with Variable Elastic Moduli In a preceeding section we considered a beam, the elastic moduli of which changed jumpwise relative to its height, i. It is also easy to obtain the solution for a beam with elastic moduli which change continuously with respect to Fig.
We consider here only an orthotropic cantilever, the prin- cipal directions of which are parallel to the sides and which is bent by force P and moment M Fig. Imo We assume that the general nature of the stress distribution in a given beam is the same as in the case of a homogeneous cantilever, i. The un- known function f y is determined from They will be expressed by f y.
By insuring that w and v also satisfy the third equation of All arbitrary constants are deter- mined from No other moduli influence the values of stresses and rigidity. Displacements u and v of any point not located on the neutral axis will depend on G and », and modulus E; is not present in any formulas.
The ratio of its largest value to its smallest is equal to 5. Two case are possible here. The normal stress at each cross section reaches its largest value at the beam edges, and the tangential, at the neutral axis. It is shown in Fig. Deformation of a Wedge-shaped Body by a Force Applied at the Apex We consider here the elastic equilibrium of a rectangular cantilever, the wide end of which is rigidly fixed and which is subjected to force P applied at the apex. It is assumed that the material of the cantilever is homogeneous and anisotropic.
The existence of a plane of clastic symmetry parallel to the middle plane is assumed to be present always. We consider the cantilever as an infinite wedge, i. The apex of the wedge is the origin of coordinates and the x-axis runs arbitrarily in the middle Plane.
We also will use polar coordinates and measure the polar angle 6 from the x-axis. The problem is confined to the selection of a solution for 5. In order to determine this function we must present 5. Arbitrary constants A and B are determined from equilibrium condi- tions for a part of the wedge cut off by a circle of radius r Fig.
The calculations are simplified when the cantilever is orthotropic and the directions of axes coincide with the principal directions. Calculation of the integrals which are present in Regarding the general nature of the stress distribution, the following should be noted. The radial stress o, at any point is at the same time a principal stress. Also at the same time, the other principal stress og, acting on the radial plane is equal to zero.
The material is compressed on one side of this line, and under tension on the other side Fig. On one side of the neutral line there are curves which correspond to compressive stresses negative 0 ; on other side are the curves which correspond to tensile stresses positive oo. Constants A, B, C, Dare found from boundary conditions The final expressions are complicated and we are not going to present them here. The stress in a wedge bent by a uniform load does not depend on distance r. The stress distribution in an anisotropic body in this case is completely identical with that of an isotropic body.
When the loaded side is horizontal, then in any homogeneous wedge-shaped body, both anisotropic and iso- tropic Fig. The solution of the general case of bending of an anisotropic wedge by a load distributed along the edge according to any law or given in the form of a concentrated force was obtained by V. Abramov's, Fig.
Still another solution of this problem was suggested by P. There are no numerical results as yet for these solutions. What is presented above with appropriate changes can be applied to a wedge with cylindrical anisotropy, the anisotropy pole of which coin- cides with the apex. For example, in the case of bending by uniform load Fig. Pure Bending of a Curved Beam with Cylindrical Anisotropy We will consider the elastic equilibrium of a curved beam as a seqment ofa plane circular ring subjected to forces which are applied at the ends and produce moments.
One pole of anisotropy is located at the common center of the circular arcs which form the outside and inside edges of the beam. Aside from planes of elastic symmetry, which are parallel to the middle plane, there are no other elements of elastic symmetry. Function F satisfies The solution is found with a function F which does not depend on polar angle The normal stress o in the cross section does not follow either the linear or the hyperbolic Jaw.
The displacements u,, ue in the radial and tangential direc- tions can be easily determined from Bending of a Linearly Anisotropic Curved Beam by End Force A curved beam in the form of a segment of a circular ring is fixed at one end and deformed by forces distributed at the other end which produce a force P applied at the center of the cross section.
It is assumed that the beam possesses cylindrical anisotropy and its pole is located Fig. We consider the anisotropy pole as the origin of the coordinates, the x-axis runs along the radius at the loaded end. We designate by w the angle between the force and the x-axis Fig. The magnitude of the angle between the edge sections is arbitrary, but not larger than We will use equation The solution is obtained with the use of a stress function of the type!
The shear stresses in these cross sections are equal to zero. The shear stresses reach their highest numerical values in sections where the force acts; the normal stresses are equal to zero in these sections.
Kosmodamianskiy studied a more general case when a beam shown in Fig. The solution for this case was found with a stress function? Generally speaking they are complex and are determined from boundary conditions and conditions at the free end. A, B, C, D are conjugate values. Kosmodamianskiy also studied numerical examples and con- structed diagrams of stress distribution in cross sections of the aniso- tropic beam with given elastic constants for various values of c.
All constants will be found from boundary conditions which can always be completely satisfied at the curvilinear edges, and only approximately at the ends. As an example, we will consider a beam with cylindrical anisotropy, simply supported at the ends and bent by a normal load uniformly distributed along the outside edge Fig.
The common center of circles delimiting the beam, which is at the same time the anisotropy Pole, serves as the orgin of coordinates. We assume that both supports are hinged-type and are construc- ted in such a manner that supporting reactions form identical angles y with the symmetry axis. The stress components will not depend on angle 6 and will be found by formulas G ". It is also comparatively easy to obtain the solution for the general case, when a curved beam in a form of a segment of a plane ring is deformed by normal and tangential forces distributed uniformly along the curvilinear sides.
Each of the given forces should be expanded into Fourier series, i. All constants present in The former are always satisfied, and the latter, only approximately.
Stress Distribution in an Annular Plate with Cylindrical Anisotropy Let us consider the elastic equilibrium of a plate having the shape of a complete circular concentric ring with cylindrical anisotropy and compressed along the external and internal contours by a uniformly distributed normal force.
We consider that the anisotropy pole coin- cides with the ring center and there are no elements of elasticsymmetry besides the planes which are parallel to the middle plane.
By solving this problem, we obtain at the same time the solution of an analogous problem regarding the stress distribution in a pipe made of material with cylindrical anisotropy subjected to internal and external pressures. Saint-Venant and Voigt solved the problem for pipe with cylindrical anisotropy of a special type??. We consider here a more general case, the nonorthotropic ring, which corresponds to the nonorthotropic pipe.
We de- signate p and q the magnitudes of the internal and external pressures per unit area, and a and 5, the internal and external ring radii. The stress distribution is identical in both the orthotropic and nonorthotropic rings, for which 46 and a 25 are not equal to zero.
Where there are such planes, radial cross sections remain plane; and when they are absent, the radial cross sections become warped. Displacements of points in the plate in the radial and tangential directions u, and u, can be found from But when E, 1andk , it is necessary to know how E,, Ey and vy depend on r.
In addition, we will study the problem of the elastic equilibrium of an infinite plane medium subjected to concen- trated force and moment. To be specific, we study here generalized plane stress.
All the results obtained will be applicable to plane strain. Let us consider an elastic homogeneous anisotropic plate with a rectilinear edge, along which forces are so distributed that they act on the middle plane.
If the dimensions of the plate along the rectilinear edge, as well as in the other directions are large in comparison with the length of the loaded edge, the stress distribution can be obtained in a simplified manner. Namely, we can consider the plate as an infinite plane elastic medium with a rectilinear border, or in other words, as an elastic half-plane. There are several methods for finding the stress distribution in an elastic half-plane. The first method is based on the use of a Fourier integral and is of advantage in the case of an orthotropic half-plane.
We will briefly summarize the essentials of the method. The problem can be solved without these limitations, but the solution will be considerably more complicated. The center of the loaded area is the origin of coordinates, the y-axis runs along the border, and the x-axis into the half-plane Fig.
G E we exclude the body forces. The solution is obtained with the help of a stress function Fel P x, x cos ay dev. Zhitkov obtained equation Zhitkov indicated that such dependencies of elastic moduli on distance r is true for pressed wood page 20 in his work. In both cases equation References 1, Lekhnitskiy, S. Timoshenko, S. ONTI, , p. Reissner, E. Orlova, E. Kurdyumov, A. Kufarev, P. Vestnik inzhen. Abramov, V.
Procedings of the conference on photoelastic studies of stresses. XIV , ser. Kosmodamianskiy, A. XVI, No. Liouville , t. Gesellschaft der Wissenschaften und d. Georg-August- Univ. Zhitkov, P. XXVII, fiz-matem. The following three cases are possible: Case I. The roots of Case III. By satisfying the bound- ary conditions In the case of a simple load distribution, the determination of integrals presents no difficulties. Formulas for the two other cases of roots can be obtained from The second method for solving the problem of the half-plane is based on the use of certain properties of Cauchy integrals and is known as the N.
By using this method we are able to find the stress distribution in an elastic anisotropic half-plane subjected to so-called generalized plane strain when the planes of elastic symmetry parallel to xy are absent.
The solution of the plane problem is obtained automatically from the solution for generalized plane strain, when we assume in the latter that the deformation coefficients a14, 15, , G25, G35, Gao and ds, are equal to zero.
In the following we will give this solution without derivation. The force distribution in all other aspects can be arbitrary. Savin suggested one additional method for solving this problem. The method is based on a known Schwartz formula which furnishes the analytical function whose real part is given at the contour. Procedures in general remain the same as in the case of forces given. The problem becomes more complicated when partially forces and partially displacements are given at the edge problem of the mixed type.
These problems include, for example, the effect of rigid punches on an elastic half-plane. Savin and D. Sokolovskiy gave the solution of the contact of an in- finitely long elastic beam and an elastic anisotropic half-plane infinite beam located on elastic base and bent by normal forces. It is assumed that friction forces are absent on the contact surface and the beam can- not be separated from the half-plane.
The solution in this case was ob- tained with Fourier integrals. The Action of a Concentrated Force and Moment Applied to an Edge In order to obtain the stress distribution caused by a normal concen- trated force P applied at point 0 on the rectilinear edge of an infinite elastic medium Fig.
We take only the case of an orthotropic half-plane, the principal directions of which are parallel and perpendicular to the edge. By using Fig. Psinw G22 thi the complex parameters 4, and j:, are roots of equation 7.
The complex parameters ; and jz. Note the very simple law of distribution of oy and t,», and the very complicated dependence of a, an angle 9. Similarly to the case of the half-plane, the moment is con- sidered as the limiting case of two equal but opposing forces. GL where L is the value which is opposite to E, see A very simple stress distribution is obtained for an isotropic medium M1 2ah 7? On the basis of formulas given in this and the preceding sections, it is not difficult to obtain by superposition the stress distribution in a half- plane subjected to force and moment, the points of application of which are located not at the boundary but inside the half-plane.
Conway and M. Sokolovskiy solved the problem of an ortho- tropic half-plane loaded by a force applied at some distance from the boundary. A second method for the same problem was suggested by P. Stress Distribution in a Plane Medium with a Parabolic or Hyperbolic Boundary Let us consider the elastic equilibrium of an anisotropic plate with a concave edge subjected to load applied to this edge. If the length of the loaded edge is small relative to the plate size, then the latter can be considered as an infinite plane elastic medium with a parabolic bound- ary.
It is simpler in this case to use the second method for the elastic half-plane presented in Section Let the origin of coordinates be at a point of the concave edge where the curvature is greatest; the x-axis runs along the tangent, and the y- axis runs outwardly Fig. We will consider, in regard to these forces, that their resultant force for any section of the body, finite or infinite, has a finite value or is equal to zero. The material in the general case is non- orthotropic.
No special cases of the elastic equilibrium of plates with parabolic boundary have been studied as yet. The solution is known for only one special case of an anisotropic plate with a hyperbolic boundary, which is discussed below. An elastic anisotropic plate is given which is bounded by two hyper- bolas and two equal rectilinear sections Fig.
Smith and Okubo. The results of Smith and Okubo are easily generalized to include an- isotropic plates. A plate is infinite and bounded by two hyperbolas. By arranging the axes as shown in Fig. By solving them we find that 4-d Multiplier K is the concentration factor.
It shows by how many times the maximal stress in a plate with hyperbolic boundary is larger than the nominal stress. The formulas and Table above can be used for approximate cal- culations of stresses in rectangular plates subjected to tension and weakened by two identical side grooves. References 1. Smirnov, V. Moscow-Leningrad, , pp. Gostekhizdat, Moscow-Leningrad, , pp. XXII, No. Vestnik inzhenerov i tekhnikov, No. Franko, v. V, seriya fiz-matem. Galin, L. Grilitskiy, D.
Sokolowski, M. Wolf, K. Okubo, H. Tohoku Univ. Conway, H. Zagubizhenko, P. Neyber, G. Smith, Bassel C. Quarterly of Appl. Philosophical Magazine, vo. Stress Distribution in an Elliptic Plate Loaded along the Edge We will discuss in this chapter the stress distribution in a homogeneous elliptic plate loaded along the edge, as well as in an elliptic plate ro- tating around its axis with a constant angular velocity. We will also Fig. Let us consider first the elliptic plate which is made of a homogeneous anisotropic material and is in equilibrium with forces arbitrarily distrib- uted along the edge.
For generality we will consider the plate as an- isotropic. Let X, and Y, be the compo- nents of the external force relative to a unit area and a and b the semi- axes of the ellipse.
We assume that body forces are absent and the surface forces X,, Y, are in equilibrium so that the resultant force and the resultant moment are equal to zero. This problem for an isotropic plate was solved by N. Muskhelish- vili!. The general solution for the anisotropic plate presented hereis ob- tained from the method by the author?
By expressing the components of stresses and displacements in terms of the functions of complex vatiables , z, and , z, , the relation of the boundary conditions to these functions can be written in the form 8. The coefficients «,, 6,, Z, and A, should satisfy the equilibrium condition the resultant moment must equal zero ; Zi —o, Pith po Hee eee BuPanl 2. At the edge of the plate Thus it is possible to derive formulas for stresses for any external load distribution. After determination of the A,, and B,, from equations The problems of the elliptic and circular plates are approximately of equal difficulty.
The matter is quite differ- ent in the case of an isotropic plate where the problem of stress distrib- ution in the circular plate is much simpler in comparison with the elliptic plate.
All formulas above make it possible to calculate the average stresses across the thickness. In reality the stresses change across the thickness, i. Stress Distribution in a Rotating Curvilinearly Anisotropic Disc It is easy to obtain the stress distribution in a rotating circular disc with cylindrical anisotropy.
We assume that the disc shown in Figure 65 possesses cylindrical anisotropy with anisotropy pole at the center and, in addition, is ortho- tropic, so that any radial plane is an elastic symmetry plane. The stress distribution in such a disc, either solid or weakened by a circular open- ing at the center, is obtained by a stress function wich depends only on distance r. This function is a solution of the non-homogeneous equa- tion Alll rigid displace- ments are also disregarded.
Constants C and D can also be used in such a way as to satisfy the required conditions at the plate and opening edges. Uniform pressure applied at the opening contour. Normal forces q per unit area; see Fig. In the case of Fig. Tangential forces uniformly distributed along the opening contour Fig. The stress distribution , along the opening edge in anisotropic or non-orthotropic plates follows more complex laws than in the iso- tropic plate.
The stress distribution is symmetrical in relation to the opening cen- ter, but is in general unsymmetrical in relation to its axes. Stress Distribution in an Orthotropic Plate with a Circular Opening The most interesting cases of stress distribution in an orthotropic plate with a circular opening are considered here and in the next sections. The origin of the coordinates in all cases is at the opening center and the principal directions of elasticity are assumed as axes x and y.
We used different elastic constants and complex parameters in the first edition of this book. In the diagrams, values of stresses are measured from the circumfer- ence radially outward. Positive values are shown by arrows directed away from the center, and negative, by arrows pointing toward the center.
The load scale is shown at the right-hand side of each figure and the dotted line shows the stress distribution og in an isotropic plate subjected to identical load. Normal pressure distributed uniformly along the opening edge Fig.
In an orthotropic plate stress a is distributed nonuniformly along the contour, according to a complicated law. Lekhnitskii, S. Tsai, T. Unlike the former results, which have different orders of approximation for different openings, the solutions presented here have only one simple unified expression for various openings such as the ellipse, circle, crack, triangle, oval, and square. Related Content Customize your page view by dragging and repositioning the boxes below.
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