Plane stressed state of strength of materials. stressed and deformed state. Torsion of rectangular beam
Fundamentals of the theory of elasticity
Lecture 4
Plane problem of elasticity theory
slide 2
In the theory of elasticity, there is a large class of problems that are important in the sense of practical applications and, at the same time, allow significant simplifications of the mathematical side of the solution. The simplification lies in the fact that in these problems one of the coordinate axes of the body, for example, the z axis, can be discarded and all phenomena can be considered as occurring in the same coordinate plane x0y of the loaded body. In this case, stresses, strains and displacements will be functions of two coordinates - x and y.
A problem considered in two coordinates is called plane problem of elasticity theory.
under the term " plane problem of elasticity theory» combine two physically different problems, leading to very similar mathematical relationships:
1) the problem of a plane deformed state (plane deformation);
2) the problem of a plane stress state.
These problems are most often characterized by a significant difference between one geometric dimension and the other two dimensions of the bodies under consideration: a large length in the first case and a small thickness in the second case.
Plane deformation
The deformation is called flat if the displacements of all points of the body can occur only in two directions in one plane and do not depend on the coordinate normal to this plane, i.e.
u=u(x,y); v=v(x,y); w=0 (4.1)
Plane deformation occurs in long prismatic or cylindrical bodies with an axis parallel to the z axis, along which a load acts on the lateral surface, perpendicular to this axis and not changing in magnitude along it.
An example of plane deformation is the stress-strain state that occurs in a long straight dam and a long arch of an underground tunnel (Fig. 4.1).
Figure - 4.1. Plane deformation occurs in the body of the dam and the vault of the underground tunnel
slide 3
Substituting the components of the displacement vector (4.1) into the Cauchy formulas (2.14), (2.15), we obtain:
(4.2)
The absence of linear deformations in the direction of the z axis leads to the appearance of normal stresses σ z . From the Hooke's law formula (3.2) for the deformation ε z it follows that
whence the expression for the stress σ z is obtained:
(4.3)
Substituting this ratio into the first two formulas of Hooke's law, we find:
(4.4)
slide 4
From the analysis of formulas (4.2) − (4.4) and (3.2) it also follows that
Thus, the basic equations of the three-dimensional theory of elasticity in the case of plane deformation are greatly simplified.
Of the three Navier differential equilibrium equations (2.2), only two equations remain:
(4.5)
and the third turns into an identity.
Since the direction cosine is everywhere on the lateral surface n=cos(v,z)=cos90 0 =0, Z v =0, only two equations remain of the three conditions on the surface (2.4):
(4.6)
where l, m are the direction cosines of the outer normal v to the contour surface;
X, Y, X v, Y v are the components of body forces and the intensity of external surface loads on the x and y axes, respectively.
slide 5
The six Cauchy equations (2.14), (2.15) are reduced to three:
(4.7)
Of the six Saint-Venant deformation continuity equations (2.17), (2.18), one equation remains:
(4.8)
and the rest turn into identities.
Of the six formulas of Hooke's law (3.2), taking into account (4.2), (4.4), three formulas remain:
In these relations, for the type of record traditional in the theory of elasticity, new elastic constants are introduced:
slide 6
Plane stress state
A plane stress state occurs when the length of the same prismatic body is small compared to the other two dimensions. In this case, it is called thickness. Stresses in the body act only in two directions in the xOy coordinate plane and do not depend on the z coordinate. An example of such a body is a thin plate of thickness h, loaded along the side surface (rib) with forces parallel to the plane of the plate and uniformly distributed over its thickness (Fig. 4.2).
Figure 4.2 - Thin plate and loads applied to it
In this case, simplifications similar to those in the plane strain problem are also possible. The stress tensor components σ z , τ xz , τ yz on both planes of the plate are equal to zero. Since the plate is thin, we can assume that they are equal to zero inside the plate as well. Then the stress state will be determined only by the components σ x , σ y , τ xy which do not depend on the z coordinate, i.e. do not change along the thickness of the plate, but are functions of only x and y.
Thus, the following stress state occurs in a thin plate:
Slide 7
With respect to stresses, the plane stress state differs from the plane strain by the condition
In addition, from the formula of Hooke's law (3.2), taking into account (4.10), for the linear deformation ε z we obtain that it is not equal to zero:
Consequently, the bases of the plate will be curved, as there will be displacements 
along the z-axis.
Under these assumptions, the basic plane strain equations: differential equilibrium equations (4.5), surface conditions (4.6), Cauchy equations (4.7), and strain continuity equations (4.8) retain the same form in the plane stress problem.
The formulas of Hooke's law will take the following form:
Formulas (4.11) differ from formulas (4.9) of Hooke's law for plane deformation only by the values of elastic constants: E and E 1 , v And v 1 .
Slide 8
In reverse form, Hooke's law can be written as follows:
(4.12)
Thus, when solving these two problems (plane deformation and plane stress state), one can use the same equations and combine the problems into one plane problem of elasticity theory.
There are eight unknowns in the plane problem of elasticity theory:
are two components of the displacement vector u and v;
– three components of the stress tensor σ x , σ y , τ xy ;
are three components of the strain tensor ε x , ε y , γ xy .
Eight equations are used to solve the problem:
– two differential equilibrium equations (4.5);
– three Cauchy equations (4.7);
are three formulas of Hooke's law (4.9), or (4.11).
In addition, the strains obtained must obey the strain continuity equation (4.8), and the equilibrium conditions (4.6) between the internal stresses and the intensities of the external surface load X v, Y v.
Stressed and deformed state
There are three types of stress state:
1) linear stress state - tension (compression) in one direction;
2) plane stress state - tension (compression) in two directions;
3) volumetric stress state - tension (compression) in three mutually perpendicular directions.
Consider an infinitesimal parallelepiped (cube). On its faces there can be normal s and tangential stresses t. When the position of the "cube" is changed, the voltages change. You can find a position in which there are no shear stresses, see fig.
https://pandia.ru/text/78/374/images/image002_227.gif" align="left" width="337" height="217 src="> Let's cut an elementary parallelepiped (Fig. a) with an oblique section. only one plane.We consider an elementary triangular prism (Fig. b).The position of the inclined area is determined by the angle a.If the rotation from the x-axis is counterclockwise (see Fig.b), then a>0.
Normal stresses have an index corresponding to the axis of their direction. shear stresses, usually, have two indices: the first corresponds to the direction of the normal to the site, the second to the direction of the stress itself (unfortunately, there are other designations and a different choice of coordinate axes, which leads to a change in signs in some formulas).
Normal stress is positive if it is tensile, shear stress is positive if it tends to rotate the considered part of the element clockwise about the internal point. pp (for shear stress in some textbooks and universities, the opposite is accepted).
Stresses on an inclined platform:
The law of pairing of shear stresses: if a tangential stress acts on the site, then a tangential stress equal in magnitude and opposite in sign will act on the site perpendicular to it. (txz=-tzx)
There are two main tasks in the theory of stress state.
Direct problem . Based on the known principal stresses: s1= smax, s2= smin, it is required to determine for a site inclined at a given angle (a) to the main sites, normal and shear stresses:
https://pandia.ru/text/78/374/images/image007_125.gif" width="219" height="33"> 
or 
.
For a perpendicular platform:
.
From where it can be seen that sa + sb = s1 + s2 is the sum of normal stresses over two mutually perpendicular areas of the invariant (independent) with respect to the slope of these areas.
As in the linear stress state, the maximum shear stresses occur at a=±45o, i.e..gif" align="left" width="240" height="227">.gif" width="154" height= "55 src=">.gif" align="left" width="253" height="176 src=">If one of the main stresses turns out to be negative, then they should be denoted s1, s3, if both are negative, then s2, s3.
Volume stress state
Stresses in any site with known principal stresses s1, s2, s3:
where a1, a2, a3 are the angles between the normal to the area under consideration and the directions of principal stresses.
Maximum shear stress: 
.
It acts on a platform parallel to the main stress s2 and inclined at an angle of 45o to the main stresses s1 and s3.
https://pandia.ru/text/78/374/images/image023_60.gif" width="171" height="48 src=">
https://pandia.ru/text/78/374/images/image025_53.gif" width="115" height="48 src="> (sometimes called principal shear stresses).
A plane stress state is a special case of a three-dimensional one and can also be represented by three Mohr circles, while one of the main stresses must be equal to 0. For shear stresses, as well as in a plane stress state, pairing law: components of shear stresses along mutually perpendicular areas, perpendicular to the line of intersection of these areas, are equal in magnitude and opposite in direction.
https://pandia.ru/text/78/374/images/image027_53.gif" width="166" height="51 src=">;
The octahedral normal stress is equal to the average of the three principal stresses.
https://pandia.ru/text/78/374/images/image029_49.gif" width="199" height="50">, The octahedral shear stress is proportional to the geometric sum of the principal shear stresses. Stress intensity:
DIV_ADBLOCK135">
https://pandia.ru/text/78/374/images/image032_47.gif" width="177" height="49"> 
The change in volume does not depend on the ratio between the principal stresses, but depends on the sum of the principal stresses. That is, an elementary cube will receive the same change in volume if the same average stresses are applied to its faces: 
, then 
, where K= - bulk modulus. When a body is deformed, the material of which has a Poisson's ratio m = 0.5 (for example, rubber), the volume of the body does not change.
Potential strain energy
With simple tension (compression), the potential energy is U=https://pandia.ru/text/78/374/images/image038_46.gif" width="95" height="47 src=">.gif" width="234 "height="50 src="> or
The total strain energy accumulated per unit volume can be considered as consisting of two parts: 1) the energy uo accumulated due to a change in volume (i.e., the same change in all dimensions of the cube without changing the cubic shape) and 2) the energy uf associated with changing the shape of the cube (i.e., the energy expended on turning the cube into a parallelepiped). u = uo + uf.
https://pandia.ru/text/78/374/images/image043_42.gif" width="389" height="50 src=">
https://pandia.ru/text/78/374/images/image045_41.gif" width="160" height="84 src=">. When you rotate the coordinate system, the tensor coefficients change, the tensor itself remains constant.
Three stress state invariants:
https://pandia.ru/text/78/374/images/image047_39.gif" width="249" height="48"> 
ea - relative strain, ga - shear angle.
The same analogy holds for the bulk state. Therefore, we have the invariants of the deformed state:
J1 = ex + ey + ez;
J2= exey +eyez + ezex - https://pandia.ru/text/78/374/images/image051_31.gif height="140 src="> - strain tensor.
ex, ey, ez, gxy, gyz, gzx are the components of the deformed state.
For axes coinciding with the directions of principal strains e1, e2, e3, the strain tensor takes the form: 
.
Strength theories
In the general case, the dangerous stress state of a structural element depends on the ratio between the three principal stresses (s1,s2,s3). That is, strictly speaking, for each ratio it is necessary to experimentally determine the magnitude of the limiting stress, which is unrealistic. Therefore, such methods for calculating the strength were adopted that would make it possible to assess the degree of danger of any stress state from the tensile-compression stress. They are called strength theories (theories of limit stress states).
1st strength theory(the theory of the greatest normal stresses): the cause of the onset of the limiting stress state is the greatest normal stresses. smax= s1£ [s]. Main disadvantage: two other main stresses are not taken into account. It is confirmed by experience only when stretching very brittle materials (glass, gypsum). Currently, it is practically not used.
2nd strength theory(the theory of the largest relative deformations): the cause of the onset of the limit stress state is the greatest elongation. emax= e1£ [e]..gif" width="63 height=47" height="47">, strength condition: sequiIII= s1 - s3£ [s]. The main drawback is that it does not take into account the influence of s2.
In plane stress state: sequivIII= 
£[s]. For sy=0 we get 
Widely used for plastic materials.
4th strength theory(energy theory): the cause of the onset of the limit stress state is the value of the specific potential energy of shape change. uf£..gif" width="367" height="55 src=">..gif" width="166" height="57">. It is used in the calculations of brittle materials, in which the allowable tensile and compressive stresses are not the same (cast iron).
For plastic materials = Mohr's theory turns into the 3rd theory.
Mohr's Circle (stress circle). The coordinates of the points of the circle correspond to the normal and shear stresses at different sites. We postpone the beam from the s axis from the center C at an angle 2a (a> 0, then counterclockwise page), we find the point D,
whose coordinates are: sa, ta. You can graphically solve both direct and inverse problems.
Pure shift
https://pandia.ru/text/78/374/images/image063_27.gif" width="48 height=47" height="47">, where Q is the force acting along the face, F is the face area. , on which only shear stresses act, are called areas of pure shear. Shear stresses on them are the largest. Pure shear can be represented as simultaneous compression and tension occurring in two mutually perpendicular directions. That is, this is a special case of a plane stress state, in which principal stresses: s1= - s3 = t, s2= 0. The main areas make an angle of 45° with the pure shear areas.
https://pandia.ru/text/78/374/images/image065_26.gif" width="16" height="48 src="> - relative shift or shear angle.
Hooke's law in shear : g = t/G or t = G×g.
G- shear modulus or modulus of elasticity of the second kind [MPa] - a material constant that characterizes the ability to resist shear deformations. (E - modulus of elasticity, m - Poisson's ratio).
Potential energy in shear: 
.
Specific potential energy of shear strain: https://pandia.ru/text/78/374/images/image069_26.gif" width="63" height="53">.
All potential energy in pure shear is spent only on the change in shape, the change in volume during shear deformation is zero.
Mohr's circle in pure shift.
Torsion
https://pandia.ru/text/78/374/images/image072_23.gif" align="left" width="175" height="125 src=">This type of deformation, in which only one torques - Mk It is convenient to determine the sign of the torque Mk in the direction of the external moment If, when viewed from the side of the section, the external moment is directed counterclockwise, then Mk> 0 (there is also an inverse rule).During torsion, one section rotates relative to another on the twist angle- j. When a round bar (shaft) is twisted, a pure shear stress state arises (there are no normal stresses), only tangential stresses arise. It is assumed that the plane sections before twisting remain flat and after twisting - law of plane sections. Shear stresses at the points of the section change in proportion to the distance of the points from the axis. ..gif" width="103" height="57 src="> - relative twist angle..gif" width="127 height=57" height="57">, [t] =, for a plastic material, tlim is taken to be the shear yield strength tm, for a brittle material, tv is the ultimate strength, [n] is the coefficient torsional stiffness condition: qmax£[q] – allowable angle of twist.
Torsion of rectangular beam
https://pandia.ru/text/78/374/images/image081_17.gif" width="46" height="46">Shear stress diagrams of a rectangular section.
; , Jk and Wk - conditionally called the moment of inertia and the moment of resistance during torsion. Wk=ahb2,
Jk= bhb3, Maximum shear stresses tmax will be in the middle of the long side, stresses in the middle of the short side: t= g×tmax, coefficients: a, b, g are given in reference books depending on the ratio h/b (for example, when h/b= 2, a=0.246, b=0.229, g=0.795.
bend
https://pandia.ru/text/78/374/images/image085_18.gif" width="270" height="45">.
https://pandia.ru/text/78/374/images/image087_16.gif" width="71" height="53">, r - radius of curvature of the neutral layer, y - distance from some fiber to the neutral layer. Hooke's law in bending: , whence (Navier formula): , Jx - moment of inertia of the section relative to the main central axis perpendicular to the plane of the bending moment, EJx - bending stiffness, https://pandia.ru/text/78/374/images/image091_15.gif" width="126" height="54">, Jx/ymax=Wx-section modulus in bending, .
https://pandia.ru/text/78/374/images/image094_14.gif" width="103 height=54" height="54">, where Sx(y) is the static moment relative to the neutral axis of that part of the area, which is located below or above the layer spaced at a distance "y" from the neutral axis; Jx - moment of inertia Total cross section relative to the neutral axis, b(y) is the width of the section in the layer on which shear stresses are determined.
https://pandia.ru/text/78/374/images/image096_14.gif" width="89" height="49 src=">, F=b×h, for circular section:, F=p×R2 , for a section of any shape ,
k- coefficient depending on the shape of the section (rectangle: k= 1.5; circle - k= 1.33).
https://pandia.ru/text/78/374/images/image100_12.gif" align="left" width="244" height="85 src=">The action of the discarded part is replaced by internal force factors M and Q, which are determined from the equilibrium equations.In some universities, the moment M>0 is laid down, i.e., the diagram of moments is built on stretched fibers.When Q= 0, we have an extremum of the diagram of moments. Differential dependencies between M,QAndq: https://pandia.ru/text/78/374/images/image102_10.gif" width="187" height="54">.
Flexural strength calculation
:
two strength conditions related to different points of the beam: a) by normal stresses 
, (points furthest from C); b) by shear stresses https://pandia.ru/text/78/374/images/image105_10.gif "width="96" height="51">, which are checked according to b). There may be points in the sections of the beams, where both normal and large tangential stresses are found.For these points, equivalent stresses are found, which should not exceed the allowable ones.Strength conditions are checked according to various strength theories
I-I: 
; II-I: (with Poisson's ratio m=0.3); - rarely used.
III-I: 
, IV-I: 
,
Mohr's theory: , (used for cast iron, in which the allowable tensile stress ¹ - compressive).
Determination of displacements in beams during bending
https://pandia.ru/text/78/374/images/image113_9.gif" width="104" height="52 src=">, where r(x) is the radius of curvature of the bent axis of the beam in section x, M (x) - bending moment in the same section, EJ - stiffness of the beam It is known from higher mathematics: - tangent of the angle between the x-axis and the tangent to the curved axis. This value is very small (beam deflections are small) Þ its square is neglected and the angle of rotation of the section is equated to the tangent. approximate differential equation for curved beam axis: 
. If the y-axis is pointing up, then the sign (+). In some universities, the y-axis goes down Þ(-). Integrating diff..gif" width="226" height="50 src="> - we get deflection level. The integration constants C and D are found from the boundary conditions, which depend on the methods of fixing the beam.
a" from the origin, it is multiplied by the factor (x - a) 0, which is equal to 1. Any distributed load is extended to the end of the beam, and a load in the opposite direction is applied to compensate for it.
EJ= M(x) = RA×x – https://pandia.ru/text/78/374/images/image122_8.gif" width="79 height=49" height="49"> - P(x - a – b); we integrate:
EJ = EJq0 + RA× – – M(x – a) + – P;
EJy =EJy0 + EJq0x + RA× – – M + https://pandia.ru/text/78/374/images/image132_8.gif" width="93" height="51 src=">.
The initial parameters are what we have at the origin, i.e. for the figure: M0=0, Q0=RA, deflection y0=0, angle of rotation q0¹0. q0 we find from the substitution into the second equation the conditions for fixing the right support: x=a+b+c; y(x)=0.
Differential dependencies in bending :
; 
; https://pandia.ru/text/78/374/images/image136_6.gif" width="56" height="48 src=">.
Definition of displacements by the method of fictitious load. Matching the equations:
https://pandia.ru/text/78/374/images/image138_5.gif" align="left" width="203" height="120 src="> and we have an analogy, Þ the definition of deflections can be reduced to the definition of moments from some fictitious (conditional) load in a fictitious beam: The moment from a fictitious load Mf after dividing by EJ is equal to the deflection "y" in a given beam from a given load Considering that and , we obtain that the angle of rotation in a given beam is numerically equal to the fictitious transverse force in a fictitious beam.. In this case, there should be a complete analogy in the boundary conditions of two beams.Each given beam corresponds to its own fictitious beam.
The fixing of fictitious beams is chosen from the condition that at the ends of the beam and on the supports there is a complete correspondence between "y" and "q" in a given beam and Mf and Qf in a fictitious beam. If the diagrams of the moments in both the real and the fictitious beams are built from the side of the stretched fiber (i.e., the positive moment is laid down), then the deflection lines in the given beam coincide with the diagram of the moments in the fictitious beam.
Statically indeterminate beams.
Systems are called statically indeterminate if the reactions in which cannot be determined from the equations of equilibrium of a solid body. In such systems, there are more bonds than is necessary for equilibrium. The degree of static indeterminacy of the beam(having no intermediate hinges - continuous beams) is equal to the excess (extra) number of external links (more than three).
https://pandia.ru/text/78/374/images/image120_7.gif" width="21" height="25 src=">.gif" width="20" height="25 src=">. gif" width="39" height="51 src="> + C;
EJy = RВ×https://pandia.ru/text/78/374/images/image129_6.gif" width="40" height="49 src="> + С×х + D..gif" width=" 39" height="49 src=">+ MA=0; are RA and MA.
extra "fixing" is called main system. For the "extra" unknown, you can take any of the reactions. Having applied the given loads to the main system, we add a condition that ensures the coincidence of the given beam and the main one - the displacement compatibility equation. For Fig.: yB=0, i.e. deflection at point B = 0. The solution to this equation is possible in different ways.
Way to compare displacements
.
The deflection of point B (Fig.) is determined in the main system under the action of a given load (q): yВq = "extra" unknown RB, and the deflection from the action of RB is found: 
. Substitute in the displacement compatibility equation: yB= yВq += 0, i.e. += 0, whence RB=https://pandia.ru/text/78/374/images/image153_4.gif" align="left" width ="371" height="300 src="> Three moment theorem
. Used in the calculation continuous beams- beams on many supports, one of which is fixed, the rest are movable. To move from a statically indeterminate beam to a statically determinate basic system, hinges are inserted above the extra supports. Extra unknowns: moments Mn applied to the ends of spans over extra supports.
Plots of moments are built for each beam span from a given load, considering each span as a simple beam on two supports. For each intermediate support "n" is compiled equation of three moments:
wn, wn+1 – plot areas, an – distance from the center of gravity of the left diagram to the left support, bn+1 – distance from the center of gravity of the right diagram to the right support. The number of moment equations is equal to the number of intermediate supports. Their joint solution makes it possible to find unknown support moments. Knowing the support moments, individual spans are considered and unknown support reactions are found from the static equations. If there are only two spans, then the left and right moments are known, since these are either given moments, or they are equal to zero. As a result, we obtain one equation with one unknown М1.
General methods for determining displacements
m" , which is caused by the action of the force of the generalized "n". Total displacement caused by several force factors: DР = DРP + DРQ + DРM. Displacements caused by a single force or a single moment: d - specific displacement. If a single force P=1 caused a displacement dP, then the total displacement caused by the force P will be: DP=P×dP. If the force factors acting on the system are designated X1, X2, X3, etc., then the movement in the direction of each of them:
where Х1d11=+D11; X2d12=+D12; Хidmi=+Dmi. Dimension of specific displacements: 
, J - joules, the dimension of work is 1J = 1Nm.
The work of external forces acting on an elastic system: 
.
https://pandia.ru/text/78/374/images/image160_3.gif" width="307" height="57">,
k - coefficient taking into account the uneven distribution of shear stresses over the cross-sectional area, depends on the shape of the section.
Based on the law of conservation of energy: potential energy U=A.
D 
11 - movement in the direction. force P1 from the action of force P1;
D12 - movement in the direction. force P1 from the action of force P2;
D21 - movement in the direction. force P2 from the action of force P1;
D22 - movement in the direction. force P2 from the action of force P2.
А12=Р1×D12 is the work of the force Р1 of the first state on the movement in its direction, caused by the force Р2 of the second state. Similarly: A21=P2×D21 is the work of the force P2 of the second state on the movement in its direction, caused by the force P1 of the first state. A12=A21. The same result is obtained for any number of forces and moments. Work reciprocity theorem: Р1×D12=Р2×D21.
The work of the forces of the first state on displacements in their directions, caused by the forces of the second state, is equal to the work of the forces of the second state on displacements in their directions, caused by the forces of the first state.
Theorem on the reciprocity of displacements (Maxwell's theorem) If P1=1 and P2=1, then P1d12=P2d21, i.e. d12=d21, in general dmn=dnm.
For two unit states of an elastic system, the movement in the direction of the first unit force caused by the second unit force is equal to the movement in the direction of the second unit force caused by the first force.
https://pandia.ru/text/78/374/images/image163_4.gif" width="104" height="27 src="> from the action of a unit force; 4) the found expressions are substituted into the Mohr integral and integrated according to the given If the resulting Dmn>0, then the displacement coincides with the chosen direction of the unit force, if<0, то противоположно.
For flat design:
https://pandia.ru/text/78/374/images/image165_3.gif" width="155" height="58">.
https://pandia.ru/text/78/374/images/image167_4.gif" width="81 height=43" height="43"> for the case when the diagram from a given load has an arbitrary shape, and from a single load - rectilinear is conveniently determined by the graph-analytical method proposed by Vereshchagin. 
, where W is the area of the diagram Мр from an external load, yc is the ordinate of the diagram from a unit load under the center of gravity of the diagram Мр. The result of multiplication of diagrams is equal to the product of the area of one of the diagrams by the ordinate of the other diagram, taken under the center of gravity of the area of the first diagram. The ordinate must be taken from a straight line plot. If both diagrams are rectilinear, then the ordinate can be taken from any one.
https://pandia.ru/text/78/374/images/image170_3.gif" width="119" height="50 src=">. This formula is calculated by sections, each of which should have a straight-line diagram without fractures.The complex diagram Mp is divided into simple geometric shapes, for which it is easier to determine the coordinates of the centers of gravity.When multiplying two diagrams that look like trapezoids, it is convenient to use the formula: 
. The same formula is also suitable for triangular diagrams, if we substitute the corresponding ordinate = 0.
https://pandia.ru/text/78/374/images/image173_3.gif" width="71" height="48"> (for fig., i.e. 
, xC=L/2).
blind "embedding with a uniformly distributed load, we have a concave quadratic parabola, for which =3L/4. It can also be obtained if the diagram is represented by the difference between the area of a triangle and the area of a convex quadratic parabola: 
. The "missing" area is considered negative.
Castigliano's theorem.
– the displacement of the point of application of the generalized force in the direction of its action is equal to the partial derivative of the potential energy with respect to this force. Neglecting the influence of axial and transverse forces on the movement, we have the potential energy: 
, where 
.
Statically indeterminate systems- systems, the force factors in the elements of which cannot be determined only from the equations of equilibrium of a rigid body. In such systems, the number of bonds is greater than necessary for equilibrium. Degree of static indeterminacy: S = 3n - m, n - the number of closed loops in the structure, m - the number of single hinges (a hinge connecting two rods is counted as one, connecting three rods - as two, etc.). force method force factors are taken as unknowns. The sequence of calculation: 1) set the degree of static. indefinability; 2) by removing unnecessary connections, the original system is replaced by a statically determinate one - the main system (there may be several such systems, but when removing unnecessary connections, the geometric invariability of the structure should not be violated); 3) the main system is loaded with given forces and unnecessary unknowns; 4) unknown forces should be selected so that the deformations of the original and main systems do not differ. That is, the reactions of the rejected bonds should have such values at which the displacements in their directions = 0. The canonical equations of the method of forces:
These equations are additional ur-strains that allow you to open static. indefinability. The number of ur-s = the number of discarded connections, i.e., the degree of indeterminacy of the system.
dik is the movement in direction i, caused by a unit force acting in direction k. dii - main, dik - side movements. According to the reciprocity theorem: dik=dki. Dip - movement in the direction of the i-th connection, caused by the action of a given load (load members). The displacements included in the canonical equations are conveniently determined by the Mohr method.
To do this, single loads X1=1, X2=1, Xn=1, external load are applied to the main system and curves of bending moments are plotted. The Mohr integral is used to find: 
; 
; ….; 
;
; 
; ….; 
;
; 
; ….; 
.
The line over M indicates that these internal forces are caused by the action of a unit force.
For systems consisting of rectilinear elements, it is convenient to multiply diagrams using the Vereshchagin method. 
; 
etc. WP is the area of the Mp diagram from an external load, yСр is the ordinate of the diagram from a single load under the center of gravity of the Мр diagram, W1 is the area of the M1 diagram from a single load. The result of multiplication of diagrams is equal to the product of the area of one of the diagrams by the ordinate of the other diagram, taken under the center of gravity of the area of the first diagram.
Calculation of flat curved bars (rods)
Curved beams include hooks, chain links, arches, etc. Limitations: the cross section has an axis of symmetry, the axis of the beam is a flat curve, the load acts in the same plane. There are bars of small curvature: h / R<1/5, большой кривизны: h/R³1/5. При изгибе брусьев малой кривизны нормальные напряжения рассчитывают по формуле Навье, как для балок с прямой осью: https://pandia.ru/text/78/374/images/image198_3.gif" width="115" height="55">,
rН is the radius of the neutral layer, e=R – rН, R is the radius of the layer in which the centers of gravity of the section are located. The neutral axis of the curved beam does not pass through the center of gravity of section C. It is always located closer to the center of curvature than the center of gravity of the section. , r=rН – y. Knowing the radius of the neutral layer, you can determine the distance "e" from the neutral layer to the center of gravity. For a rectangular section with height h, with outer radius R2 and inner R1: ; for different sections, the formulas are given in the reference literature. For h/R<1/2 независимо от формы сечения можно определять "е" по приближенной формуле: , где Jx – момент инерции сечения относительно оси, проходящей через его центр тяжести перпендикулярно плоскости кривизны бруса.
Normal stresses in the section are distributed according to the hyperbolic law (less at the outer edge of the section, more at the inner edge). Under the action of a normal force N: 
(here rН is the radius of the neutral layer, which would be under the action of only the moment M, i.e., at N=0, but in reality, in the presence of a longitudinal force, this layer is no longer neutral). Strength condition: 
, while considering the extreme points at which the total stresses from bending and tension-compression will be the largest, i.e. y= – h2 or y= h1. Displacements are conveniently determined by Mohr's method.
Stability of compressed rods. Longitudinal bend
The destruction of the rod can occur not only because the strength will be broken, but also because the rod does not retain the desired shape. For example, bending under longitudinal compression of a thin ruler. The loss of stability of a rectilinear form of equilibrium of a centrally compressed rod is called buckling. Elastic balance steadily, if the deformed body, with any small deviation from the equilibrium state, tends to return to its original state and returns to it when the external influence is removed. The load, the excess of which causes loss of stability, is called critical load Rcr (critical force). Permissible load [P]=Pkr/nу, nу – normative stability factor..gif" width="111" height="51 src=">.gif" width="115 height=54" height="54"> - the formula gives the value of the critical force for a rod with hinged ends. With various fixings: 
, m is the length reduction factor.
With hinged fastening of both ends of the rod m=1; for a rod with closed ends m=0.5; for a rod with one closed and other free end m=2; for a rod with one end fixed and the other end hinged, m=0.7.
Critical compressive stress.: 
, – rod flexibility, 
is the smallest principal radius of inertia of the cross-sectional area of the rod. These formulas are valid only when the voltages skr £ spts are the limit of proportionality, i.e., within the limits of applicability of Hooke's law. The Euler formula is applicable when the rod is flexible: 
, for example, for steel St3 (C235) lkr "100. For the case l
Sufficiently short rods for which l
(Fnet = Fgross-Fweak – the area of the weakened section, taking into account the area of holes in the section Fweak, for example, from rivets). \u003d scr / nу, nу - standard coefficient. margin of stability. The allowable stress is expressed in terms of the main allowable stress [s] used in strength calculations: =j×[s], j - allowable stress reduction factor for compressed rods (buckling coefficient). The values of j are given in Table. in textbooks and depend on the material of the rod and its flexibility (for example, for steel St3 at l=120 j=0.45).
In the design calculation of the required cross-sectional area, j1 = 0.5–0.6 is taken at the first step; find: 
. Further, knowing Fgross, select the section, determine Jmin, imin and l, set according to Table. the actual j1I, if it differs significantly from j1, the calculation is repeated with the average j2= (j1+j1I)/2. As a result of the second attempt, j2I is found, compared with the previous value, and so on, until a close enough match is achieved. It usually takes 2-3 tries..
Relationship between moments of inertia when turning the axes:
https://pandia.ru/text/78/374/images/image249_2.gif" width="17" height="47 src=">(Jx - Jy)sin2a + Jxycos2a ;
Angle a>0, if the transition from the old coordinate system to the new one occurs counterclockwise. p. Jy1 + Jx1= Jy + Jx
Extreme (maximum and minimum) values of moments of inertia are called main moments of inertia. The axes with respect to which the axial moments of inertia have extreme values are called main axes of inertia. The principal axes of inertia are mutually perpendicular. Centrifugal moments of inertia about the main axes \u003d 0, i.e., the main axes of inertia are the axes relative to which the centrifugal moment of inertia \u003d 0. If one of the axes coincides or both coincide with the axis of symmetry, then they are principal. Angle defining the position of the main axes: 
, if a0>0 Þ the axes are rotated counterclockwise. p. The axis of maximum always makes a smaller angle with that of the axes, relative to which the moment of inertia has a greater value. Principal axes passing through the center of gravity are called main central axes of inertia. Moments of inertia about these axes: 
Jmax + Jmin = Jx + Jy. The centrifugal moment of inertia about the main central axes of inertia is 0. If the main moments of inertia are known, then the formulas for the transition to rotated axes are:
Jx1=Jmaxcos2a + Jminsin2a; Jy1=Jmaxcos2a + Jminsin2a; Jx1y1=(Jmax - Jmin)sin2a;
The ultimate goal of calculating the geometric characteristics of the section is to determine the main central moments of inertia and the position of the main central axes of inertia. 
Radius of inertia- https://pandia.ru/text/78/374/images/image254_3.gif" width="85" height="32 src=">. For sections with more than two axes of symmetry (for example: circle, square, ring, etc.) axial moments of inertia relative to all central axes are equal to each other, Jxy=0, the ellipse of inertia turns into a circle of inertia.
s- normal voltage[Pa], 1Pa (pascal) = 1 N/m2,
106Pa = 1 MPa (megapascal) = 1 N/mm2
N - longitudinal (normal) force [N] (newton); F - cross-sectional area [m2]
e - relative deformation [dimensionless value];
DL - longitudinal deformation [m] (absolute elongation), L - bar length [m].
Hooke's law - s = E×e
E - tensile modulus (modulus of elasticity of the 1st kind or Young's modulus) [MPa]. For steel E = 2×105MPa = 2×106 kg/cm2 (in the "old" system of units).
(the more E, the less extensible the material)
; - Hooke's law
EF - rod stiffness in tension (compression).
When the rod is stretched, it "thinners", its width - a decreases by transverse deformation - Da.
Relative transverse deformation.
 |
Basic mechanical characteristics of materials
sp - limit of proportionality, st - yield strength, sВ- strength limit or temporary resistance, sk is the voltage at the moment of rupture.
Brittle materials, such as cast iron, break at low elongations and do not have a yield plateau, resisting compression better than stretching.
Allowable voltage https://pandia.ru/text/78/374/images/image276_3.gif" align="left" width="173" height="264"> stresses along the slope:
Direct task…………………………………………………..3
Inverse problem……………………………………………………3
Volume stress state……………………………4
Stresses along the octahedral site…………………..5
Deformations under volumetric stress state.
Generalized Hooke's law …………………………………………6
Potential strain energy…………………………7
Strength theories…………………………………………………………………9
Mohr's strength theory …………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
Mohr Circle ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
Net shift…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
Hooke's law in shear………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
Torsion………………………………………………………..13
Torsion of a rectangular bar…………………….14
Bend………………………………………………………………15
Zhuravsky’s formula…………………………………………………………………16
Calculation for bending strength…………………………………………………………………18
Determination of displacements in beams during bending……………19
Differential dependencies in bending……………….20
Displacement compatibility equation……………………..22
Way to compare displacements……………………………..22
The three-moment theorem………………………………………..22
General methods for determining displacements………………….24
Work reciprocity theorem (Betley's theorem)……………….25
The theorem on the reciprocity of displacements (Maxwell's theorem).. 26
Calculation of the Mohr integral by the Vereshchagin method……….27
Castigliano's theorem…………………………………………..28
Statically indeterminate systems………………………..29
Calculation of flat curved bars (rods)………………...31
Stability of compressed rods. Longitudinal bend………33
Geometric characteristics of flat sections…………36
Moments of inertia of the section…………………………………..37
Centrifugal moment of inertia of the section …………………..37
Moments of inertia of simple shape sections………………..38
Moments of inertia about parallel axes……..39
The relationship between the moments of inertia when turning
axes……………………………………………………………40
Moments of resistance…………………………………….42
Tension and compression……………………………………………………43
Basic mechanical characteristics of materials…….45
Biaxial or flat called such a stress state of the body, in which at all its points one of the main stresses is equal to zero. It can be shown * that a plane stress state occurs in a prismatic or cylindrical body (Fig. 17.1) with loose and unloaded ends, if a system of external forces normal to the axis is applied to the side surface of the body Oz and changing depending on z according to the quadratic law, it is symmetrical with respect to the mean section. It turns out that in all cross sections of the body
and voltage a x, a y, x change depending on z also, according to the quadratic law, it is symmetrical with respect to the mean section. The introduction of these assumptions makes it possible to obtain a solution to the problem that satisfies conditions (17.13) and all equations of the theory of elasticity.
Of interest is the special case when the stresses do not depend on the variable z‘-
Such a stressed state is possible only under the action of a load uniformly distributed along the length. It follows from the formulas of Hooke's law (16.3) that the deformations e x, e y, e z , y also do not depend on z, and deformations y and y zx taking into account (17.13) are equal to zero. In this case, the fourth and fifth of the deformation continuity equations (16.4), (16.5) are identically satisfied, and the second, third, and sixth equations take the form
Integrating these equations and taking into account the third formula of Hooke's law (16.3) with az = 0, we get
Cm.: Timoshenko S. P., Goodyear J. Theory of elasticity. Moscow: Nauka, 1975.
Thus, a plane stress state in a prismatic or cylindrical body with free ends loaded with a surface load constant along the length of the body is possible only in the particular case when the sum of the stresses a x + a y varies depending on the variables x and at linear or constant.
If the distance between the end planes of the body (Fig. 7.1) is small compared to the dimensions of the sections, then we have the case of a thin plate (Fig. 17.5) loaded along the outer contour with forces symmetrically distributed relative to the middle plane of the plate according to a quadratic law. Since the plate thickness h is small, then with a slight error it can be assumed that for any symmetrical with respect to the median plane loading of the stress plate a x, a v , txv are uniformly distributed over its thickness.
In this case, stresses should be understood as their average values over the thickness, for example
It should also be noted that when the assumption (17.14) is introduced, the condition (17.13) of zero stresses
The considered case of the stress state of a thin plate with assumptions (17.13) and (17.14) is often called generalized plane stress state.
Let us consider the basic equations of elasticity theory for this case.
Taking into account (17.13), the formulas of Hooke's law (16.3) can be written in the form
The corresponding inverse relations have the form
Formulas (17.17) and (17.18) differ from formulas (17.7) and (17.9) of Hooke's law for plane deformation only in that in the latter, instead of the elastic modulus E and Poisson's ratio v includes the reduced quantities E ( and vr
Equilibrium equations, Cauchy relations, strain continuity equation and static boundary conditions do not differ from the corresponding equations (17.10), (17.3), (17.11), (17.12) for plane strain.
Plane deformation and generalized plane stress state are essentially described by the same equations. The only difference is in the values of the elasticity constants in the formulas of Hooke's law. Therefore, both tasks are combined by a common name: plane problem of the theory of elasticity.
The complete system of equations of the plane problem consists of two equilibrium equations (17.10), three geometric Cauchy relations (17.3) and three formulas of Hooke's law (17.7) or (17.17). They contain eight unknown functions: three voltages a x, a y, x xy, three strains e x, e y, y xy and two moves And And And.
If when solving the problem it is not required to determine the displacements, then the number of unknowns is reduced to six. To determine them, there are six equations: two equilibrium equations, three formulas of Hooke's law and the equation of continuity of deformations (17.11).
The main difference between the two types of plane problem considered is as follows. For plane deformation ? z = 0,oz * 0, and the value c z can be found by formula (17.6) after the stresses o x io, have been determined. For a generalized plane stress state a z = 0, ? z F 0, and warp ? z can be expressed in terms of stresses o x and OU according to the formula (17.16). moving w can be found by integrating the Cauchy equation
DEFORMED STATES ("FLAT PROBLEM")
Plane stress and plane strain states are characterized by the following features.
1. All stress components do not depend on one of the coordinates common to all components, and remain constant when it changes.
2. In planes normal to the axis of this coordinate:
a) shear stress components are equal to zero;
b) normal stress is either equal to zero (plane stress state), or equal to half the sum of two other normal stresses (plane strain state).
Let us take for the axis, which was mentioned earlier, the y-axis. It is clear from the foregoing that this axis will be principal, i.e., it can also be denoted by the index 2. Moreover, , and do not depend on y; at the same time, and , and hence, and and are equal to zero.
For a plane stressed state = 0. For a plane deformed state (this feature of a plane deformed state will be proved below).
One should always take into account the significant difference between the plane stress and plane strain states.
In the first, in the direction of the third axis, there is no normal stress, but there is deformation, in the second there is normal stress, but no deformation.
A plane stress state can be, for example, in a plate subject to the action of forces applied to its contour parallel to the plane of the plate and distributed evenly over its thickness (Fig. 3.16). The change in the thickness of the plate in this case does not matter, and its thickness can be taken as unity. With sufficient accuracy, the stress state of the flange can be considered flat when drawing a cylindrical billet from sheet material.
A plane deformed state can be accepted for sections of a cylindrical or prismatic body of great length, remote from its ends, if the body is loaded with forces that do not change along its length and are directed perpendicular to the generators. In a flat deformed state, for example, a beam can be considered to be subject to upset in its thickness direction, when the deformation along the length can be neglected.
All stress state equations for a plane problem are greatly simplified and the number of variables is reduced.
The equations for the plane problem can be easily obtained from those derived earlier for the bulk stress state, taking into account that 
\u003d 0 and taking \u003d 0, since inclined areas should only be considered parallel to the y axis, i.e., normal to areas that are free from stresses in a plane stress state or free from deformations in a plane deformed state (Fig. 3.17).
In the case under consideration
Denoting the angle (see Fig. 3.17) between the normal to the inclined area and the axis (or the axis, if the stress state is given in the main axes 1 and 2) through , we get , from where .
Considering the above, by direct substitutions in the corresponding expressions (3.10) and (3.11) for the volumetric stress state, we obtain the normal and shear stresses in the inclined area (see Fig. 3.17).
Fig.3.15. Plane stress state (a), stress on an inclined platform (b)
normal voltage
shear stress
. (3.41)
From expression (3.41) it is easy to see that it has a maximum at sin 2 \u003d 1, i.e. at \u003d 45 °:
. (3.42)
The magnitude of the principal stresses can be expressed in terms of components in arbitrary axes using equation (3.13), from which we obtain
. (3.43)
In this case, for a plane stress state = 0; for flat strained state
Knowing the stress state in the main axes, it is easy to switch to any arbitrary coordinate axes (Fig. 3.18). Let the new coordinate axis x make an angle with the axis, then, considering it as a normal to the inclined area, we have for the latter according to equation (3.40)
but for the axis, the voltage is the voltage, hence
this expression can be converted as follows:
(3.44)
The new axis will be tilted to axis 1 by an angle (+90°); therefore, replacing in the previous equation by ( + 90°), we obtain
We determine the voltage from expression (3.41):
. (3.46)
Denoting the average voltage through, i.e., taking
, (3.47)
and taking into account equation (3.42), we obtain the so-called transformation formulas, which express the stress components as a function of the angle:
(3.48)
When constructing the Mohr diagram, we take into account that since we are considering areas parallel to the y-axis (i.e., axis 2), the direction cosine is always zero, i.e., angle = 90 °. Therefore, all corresponding values and will be located on the circle defined by equation (3.36 b) when substituting = 0 into it, namely:
, (3.49)
or taking into account expressions (3.47) and (3.42)
. (3.49a)
This circle is shown in Fig. 3.19 and is a Mohr diagram. The coordinates of some point P, located on the circle, determine the corresponding values and Let's connect the point P with the point . It is easy to see that the segments 0 2 P = ;
Рр= , Ор= , and, consequently, sin 
= 
.
Comparing the obtained expressions with equations (3.48), we can establish that
P0 2 A \u003d 2, P0 2 A \u003d.
Thus, knowing the position of the inclined area, determined by the angle , one can find the values of the stresses and acting in this area.
Fig.3.17. Mohr diagram
,
then the segment OP expresses the total stress S.
If an element of a stressed body, in the inclined face of which stresses are considered, is drawn so that the main stress is directed parallel to the axis, then the normal N drawn to this inclined face, and hence the direction of the stress, will be parallel to the segment СР.
Continuing the line P0 2 to the intersection with the circle, at the point P "we get the second pair of values and for another inclined area, in which " = + 90 °, i.e. for the area perpendicular to the first, with the direction of the normal ". Directions of the normals N and N" can be taken respectively as the directions of the new axes: and , and the stresses and " - respectively for the coordinate stresses and. Thus, it is possible to determine the stress state in arbitrary axes without using formulas (3.44) - (3.46). are equal to each other according to the law of pairing.
It is not difficult to solve the inverse problem: for given stresses in two mutually perpendicular areas , and , t "(where t" = t) find the main stresses.
We draw coordinate axes n and (Fig. 3.19). We plot points P and P "with coordinates corresponding to the given stresses , and ,. The intersection of the segment PP" with the axis will determine the center of the Mohr circle 0 2 with a diameter PP "= 2 31. Further, if we build the axes N, N" (or, something same, , ) and rotate the figure so that the directions of these axes are parallel to the directions of stresses and at the considered point of the given body, then the directions of the axes and the diagram will be parallel to the direction of the main axes 1 and 2.
We obtain the differential equilibrium equation for a plane problem from equations (3.38), taking into account that all derivatives with respect to y are equal to zero, and are also equal to zero and :
(3.50)
When solving some problems related to flat ones, it is sometimes convenient to use polar coordinates instead of rectangular coordinates, determining the position of a point by the radius vector and the polar angle, i.e., the angle that the radius vector makes with the axis.
The equilibrium conditions in polar coordinates can be easily obtained from the same conditions in cylindrical coordinates by equating
And given that the derivatives are equal
(3.51)
A special case of a plane problem is when the stresses do not also depend on the coordinate (stress distribution symmetrical about the axis). In this case, the derivatives with respect to and stresses and will vanish, and the equilibrium conditions are determined by one differential equation
. (3.52)
It is clear that the stresses are the main ones here as well.
Such a stressed state can be taken for the flange of a round billet during drawing without pressing the cylindrical cup.
Type of stress state
The stress state at any point of the deformable body is characterized by three main normal stresses and directions of the main axes.
There are three main types of stress state: volume (triaxial), in which all three principal stresses are not equal to zero, flat (biaxial), in which one of the principal stresses is zero, and linear (uniaxial), in which only one principal stress is different from zero.
If all normal stresses have the same sign, then the stress state is called of the same name, and for stresses of different sign - opposite.
Thus, there are nine types of stress state: four volumetric, three flat and two linear (Fig. 3.18).
The stress state is called homogeneous when at any point of the deformable body the directions of the main axes and the magnitude of the main normal stresses remain unchanged.
The type of stress state affects the ability of the metal to deform plastically without collapsing and the amount of external force that must be applied to achieve a deformation of a given value.
So, for example, deformation under conditions of the same volumetric stress state requires more effort than under the opposite stress state, all other things being equal.
test questions
1.What is voltage? What characterizes the stress state of a point, of a body as a whole?
2. What do the indices express in the notation of the stress tensor components?
3. Give the sign rule for the stress tensor components.
4. Write down Cauchy's formulas for stresses on inclined platforms. What is the basis for their conclusion?
5. What is a stress tensor? What are the components of the stress tensor?
6. What are the eigenvectors and eigenvalues of the stress tensor called?
7. What are principal stresses? How many?
8. Give the rule for assigning indices to the main normal stresses.
9. Give a physical interpretation of the main normal stresses and the main axes of the stress tensor.
10. Show the diagrams of the main normal stresses for the main processes of OMD - rolling, drawing, pressing.
11. What are stress tensor invariants? How many?
12. What is the mechanical meaning of the first stress tensor invariant?
13. What is called the intensity of shear stresses?
14..What are the main shear stresses? Find their platforms
15.. How many areas of the main shear stresses can be indicated at some point of the deformable body?
16. What is the maximum shear stress, the normal stress on the site on which it acts?
17. What is an axisymmetric stress state? Give examples.
18. Show the diagrams of the main normal stresses for the main OMD processes - rolling, drawing, pressing.
19. What is common between a plane stressed and a plane deformed state and what is the difference between them? To which of these states does a simple shift refer?
20. Give the formulas of stress theory known to you in the main coordinate system
21. What is a stress ellipsoid? Write down its equation and indicate the construction order. What is the form of the stress ellipsoid for hydrostatic pressure, plane and linear stress states?
22. Write down an equation for finding the main normal stresses and three systems of equations for finding the main axes T a.
23..What is a spherical tensor and a stress deviator? What quantities are used to calculate the second and third invariants of the stress deviator?
24. Show that the main coordinate systems of the stress tensor and the stress deviator coincide.
25. Why are stress intensity and shear stress intensity introduced into consideration? Explain their physical meaning and give geometric interpretations.
26. What is a Mohr diagram? What are the radii of the principal circles?
27. How will the Mohr diagram change when the average voltage changes?
28. What are octahedral stresses?
29. How many characteristic areas can be drawn through a point of a body in a stressed state?
30. Equilibrium conditions for the volumetric stress state in rectangular coordinates, in cylindrical and spherical coordinates.
31. Equilibrium equations for a plane problem.
BIBLIOGRAPHY
1. Ilyushin A. A. Plasticity. Ch. I. M.-L., GTI, 1948. 346 p. (33)
2. I. M. Pavlov, “On the physical nature of tensor representations in the theory of plasticity,” Izvestiya vuzov. Ferrous metallurgy”, 1965, No. 6, p. 100–104.
3. V. V. Sokolovsky, Theory of Plasticity. M., Higher School, 1969. 608 p. (91)
4. M. V. Storozhev and E. A. Popov, Theory of metal pressure treatment. M., "Engineering", 1971. 323 p. (99)
5. S. P. Timoshenko, Theory of Elasticity. Gostekhizdat, 1934. 451 p. (104)
6. Shofman L. A. Fundamentals of calculation of the stamping and pressing process. Mashgiz, 1961. (68)
Let us consider the case of a plane stress state, which is important for applications and is realized, for example, in the plane Oyz. The stress tensor in this case has the form
The geometric illustration is shown in Fig.1. At the same time, the sites x= const are principal with corresponding zero principal voltages. The stress tensor invariants are , and the characteristic equation takes the form
The roots of this equation are
|
The numbering of the roots is made for the case 
Fig.1. Initial plane stress state. 
Fig.2. Position of principal stresses
An arbitrary site is characterized by an angle in Fig. 1, while the vector P has components: , , n x \u003d 0. Normal and shear stresses on an inclined site are expressed in terms of the angle as follows:
The smallest positive root of equation (4) will be denoted by . Since tg( X) is a periodic function with period , then we have two mutually orthogonal directions that make up the angles and 
with axle OU. These directions correspond to mutually perpendicular main areas (Fig. 2).
If we differentiate relation (2) with respect to and equate the derivative to zero, then we arrive at equation (4), which proves that the principal stresses are extremal.
To find the orientation of the areas with extreme shear stresses, we equate to zero the derivative of the expression
from where we get
|
Comparing relations (4) and (5), we find that
This equality is possible if the angles and differ by the angle . Consequently, the directions of areas with extreme shear stresses differ from the directions of the main areas by an angle (Fig. 3).
Fig.3. Extreme shear stress
The values of extreme shear stresses are obtained after substituting (5) into relation (3) using the formulas
.
After some transformations, we get
Comparing this expression with the values of the principal stresses (2.21) obtained earlier, we express the extreme shear stresses in terms of the principal stresses
A similar substitution into (2) leads to an expression for normal stresses on areas with
The relations obtained allow us to carry out a directionally oriented strength analysis of structures in the case of a plane stress state.
STRAIN TENSOR
Let us first consider the case of plane deformation (Fig. 4). Let the flat element MNPQ moves within the plane and deforms (changes shape and size). The coordinates of the points of the element before and after deformation are marked in the figure.
Fig.4. Flat deformation.
By definition, relative linear strain at a point M in axis direction Oh is equal to
From fig. 4 follows
Given that MN=dx, we get
In the case of small deformations, when , , we can neglect the quadratic terms. Taking into account the approximate ratio
fair at x<<1, окончательно для малой деформации получим
Angular deformation is defined as the sum of angles and (4). In the case of small deformations
For the angular deformation we have
Carrying out similar calculations in the general case of three-dimensional deformation, we have nine relations
This tensor completely determines the deformed state of the solid. It has the same properties as the stress tensor. The property of symmetry follows directly from the definition of angular deformations. The principal values and principal directions, as well as the extreme values of the angular strains and their corresponding directions, are found by the same methods as for the stress tensor.
The strain tensor invariants are defined by analogous formulas, and the first invariant of the small strain tensor has a clear physical meaning. Before deformation, its volume is equal to dV 0 =dxdydz. If we neglect the shear deformations, which change the shape, not the volume, then after the deformation, the ribs will have dimensions
(Fig. 4), and its volume will be equal to
Relative volume change
within small deformations will be
which coincides with the definition of the first invariant. Obviously, the change in volume is a physical quantity that does not depend on the choice of the coordinate system.
Just like the stress tensor, the strain tensor can be decomposed into a spherical tensor and a deviator. In this case, the first invariant of the deviator is equal to zero, i.e. the deviator characterizes the deformation of the body without changing its volume.
