When bending, the rods are subjected to shear force or bending moment. Bending is called pure if only a bending moment acts, and transverse if a load perpendicular to the axis of the rod acts. A beam (rod) that bends is usually called a beam. Beams are the most common elements of structures and machines that receive loads from other structural elements and transfer them to those parts that support the beam (most often supports).
In building structures and machine-building structures, you can most often find the following cases of fastening beams: cantilever - with one pinched end (with a rigid seal), two-support beams - with one hinged-fixed support and with one hinged-movable support, and multi-supported beams. If the support reactions can be found from static equations alone, then the beams are called statically determinate. If the number of unknown support reactions is greater than the number of static equations, then such beams are called statically indeterminate. To determine reactions in such beams, additional equations have to be drawn up - displacement equations. With flat transverse bending everything external loads perpendicular to the axis of the beam.
Determining the internal force factors acting in the cross sections of the beam should begin with determining the support reactions. After this, we use the method of sections, mentally cut the beam into two parts and consider the equilibrium of one part. We replace the interaction of beam parts internal factors: bending moment and shear force.
The transverse force in a section is equal to the algebraic sum of the projections of all forces, and the bending moment is equal to the algebraic sum of the moments of all forces located on one side of the section. The signs of the acting forces and moments should be determined in accordance with accepted rules. It is necessary to learn how to correctly determine the resultant force and bending moment from a load uniformly distributed along the length of the beam.
It should be borne in mind that when determining the stresses arising during bending, the following assumptions are made: flat sections before bending remain flat after bending (hypothesis of flat sections); longitudinal adjacent fibers do not press one another; the relationship between stress and strain is linear.
When studying bending, attention should be paid to the uneven distribution of normal stresses in the cross section of the beam. Normal stresses vary with height cross section proportional to the distance from the neutral axis. You should be able to determine bending stresses, which depend on the magnitude of the effective bending moment M I and moment of resistance of the section during bending W O(axial moment of resistance of the section).
Bending strength condition: σ = M I / W O £ [σ]. Meaning W O depends on the size, shape and location of the cross section relative to the axis.
The presence of a transverse force acting on a beam is associated with the occurrence of tangential stresses in transverse sections, and, according to the law of pairing of tangential stresses, in longitudinal sections. Tangential stresses are determined using the formula of D.I. Zhuravsky.
The transverse force shifts the section under consideration relative to the adjacent one. The bending moment, which consists of elementary normal forces arising in the cross section of the beam, rotates the section relative to the adjacent one, which causes the curvature of the axis of the beam, i.e. its bending.
When a beam experiences pure bending, a bending moment of constant magnitude acts along the entire length of the beam or on a separate section of it in each section, and the transverse force in any section of this section is zero. In this case, only normal stresses arise in the cross sections of the beam.
In order to understand more deeply physical phenomena bending and in the methodology for solving problems when calculating strength and stiffness, it is necessary to thoroughly understand the geometric characteristics of flat sections, namely: static moments of sections, moments of inertia of sections simplest form and complex sections, determination of the center of gravity of figures, main moments of inertia of sections and main axes of inertia, centrifugal moment of inertia, change in moments of inertia when turning axes, theorems on the transfer of axes.
When studying this section, you should learn how to correctly construct diagrams of bending moments and shear forces, determine dangerous sections and the stresses acting in them. In addition to determining stresses, you should learn to determine displacements (beam deflections) during bending. For this purpose it is used differential equation curved axis of the beam (elastic line), written in general form.
To determine deflections, the elastic line equation is integrated. In this case, it is necessary to correctly determine the constants of integration WITH And D based on the beam support conditions (boundary conditions). Knowing the quantities WITH And D, you can determine the rotation angle and deflection of any beam section. The study of complex resistance usually begins with oblique bending.
The phenomenon of oblique bending is especially dangerous for sections with significantly different main moments of inertia; beams with such a cross-section work well for bending in the plane of greatest rigidity, but even at small angles of inclination of the plane external forces Towards the plane of greatest rigidity, significant additional stresses and deformations arise in the beams. For a beam of circular cross-section, oblique bending is impossible, since all the central axes of such a section are the main ones and the neutral layer will always be perpendicular to the plane of external forces. Oblique bending is also impossible for a square beam.
When determining stresses in the case of eccentric tension or compression, it is necessary to know the position of the main central axes of the section; It is from these axes that the distances of the point of application of force and the point at which stress is determined are measured.
An eccentrically applied compressive force can cause tensile stresses in the cross section of the rod. In this regard, eccentric compression is especially dangerous for rods made of brittle materials that weakly resist tensile forces.
In conclusion, we should study the case of complex resistance, when the body experiences several deformations simultaneously: for example, bending together with torsion, tension-compression together with bending, etc. It should be borne in mind that bending moments acting in different planes can add up like vectors.
Bend is called deformation in which the axis of the rod and all its fibers, i.e. longitudinal lines parallel to the axis of the rod, are bent under the action of external forces. The simplest case of bending occurs when external forces lie in a plane passing through the central axis of the rod and do not produce projections onto this axis. This type of bending is called transverse bending. There are flat bends and oblique bends.
Flat bend- such a case when the curved axis of the rod is located in the same plane in which external forces act.
Oblique (complex) bend– a case of bending when the bent axis of the rod does not lie in the plane of action of external forces.
A bending rod is usually called beam.
During flat transverse bending of beams in a section with the coordinate system y0x, two internal forces can arise - transverse force Q y and bending moment M x; in what follows we introduce the notation for them Q And M. If there is no transverse force in a section or section of a beam (Q = 0), and the bending moment is not zero or M is const, then such a bend is usually called clean.
Lateral force in any section of the beam is numerically equal to the algebraic sum of the projections onto the axis of all forces (including support reactions) located on one side (either) of the drawn section.
Bending moment in a beam section is numerically equal to the algebraic sum of the moments of all forces (including support reactions) located on one side (any) of the drawn section relative to the center of gravity of this section, more precisely, relative to the axis passing perpendicular to the drawing plane through the center of gravity of the drawn section.
Force Q is resultant distributed over the cross-section of internal shear stress, A moment M– sum of moments around the central axis of section X internal normal stress.
There is a differential relationship between internal forces
which is used in constructing and checking Q and M diagrams.
Since some of the fibers of the beam are stretched, and some are compressed, and the transition from tension to compression occurs smoothly, without jumps, in the middle part of the beam there is a layer whose fibers only bend, but do not experience either tension or compression. This layer is called neutral layer. The line along which the neutral layer intersects the cross section of the beam is called neutral line th or neutral axis sections. Neutral lines are strung on the axis of the beam.
Lines drawn on the side surface of the beam perpendicular to the axis remain flat when bending. These experimental data make it possible to base the conclusions of the formulas on the hypothesis of plane sections. According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and turn out to be perpendicular to the curved axis of the beam when it is bent. The cross section of the beam is distorted when bending. Due to transverse deformation The cross-sectional dimensions in the compressed zone of the beam increase, and in the tension zone they compress.
Assumptions for deriving formulas. Normal voltages
1) The hypothesis of plane sections is fulfilled.
2) Longitudinal fibers do not press on each other and, therefore, under the influence of normal stresses, linear tension or compression operates.
3) Deformations of fibers do not depend on their position along the cross-sectional width. Consequently, normal stresses, changing along the height of the section, remain the same along the width.
4) The beam has at least one plane of symmetry, and all external forces lie in this plane.
5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same.
6) The relationship between the dimensions of the beam is such that it operates under plane bending conditions without warping or twisting.
In case of pure bending of a beam, only normal stress, determined by the formula:
where y is the coordinate of an arbitrary section point, measured from the neutral line - the main central axis x.
Normal bending stresses along the height of the section are distributed over linear law. On the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero.
The nature of normal stress diagrams for symmetrical sections relative to the neutral line
The nature of normal stress diagrams for sections that do not have symmetry with respect to the neutral line
Dangerous points are the points furthest from the neutral line.
Let's choose some section
For any point of the section, let's call it a point TO, the beam strength condition for normal stresses has the form:
, where n.o. - This neutral axis
This axial section modulus relative to the neutral axis. Its dimension is cm 3, m 3. The moment of resistance characterizes the influence of the shape and dimensions of the cross section on the magnitude of the stresses.
Normal stress strength condition:
The normal stress is equal to the ratio of the maximum bending moment to the axial moment of resistance of the section relative to the neutral axis.
If the material does not equally resist tension and compression, then two strength conditions must be used: for the tensile zone with the permissible tensile stress; for a compression zone with permissible compressive stress.
During transverse bending, the beams on the platforms in its cross-section act as normal, so tangents voltage.
We will start with the simplest case, the so-called pure bend.
Clean bend There is special case bending, in which the transverse force in the sections of the beam is zero. Pure bending can only occur when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads causing pure
bending, shown in Fig. 88. In sections of these beams, where Q = 0 and, therefore, M = const; pure bending takes place.
The forces in any section of the beam during pure bending are reduced to a pair of forces, the plane of action of which passes through the axis of the beam, and the moment is constant.
Voltages can be determined based on the following considerations.
1. The tangential components of forces along elementary areas in the cross section of a beam cannot be reduced to a pair of forces, the plane of action of which is perpendicular to the section plane. It follows that the bending force in the section is the result of action along elementary areas
only normal forces, and therefore with pure bending the stresses are reduced only to normal.
2. In order for efforts on elementary sites to be reduced to only a couple of forces, among them there must be both positive and negative. Therefore, both tension and compression fibers of the beam must exist.
3. Due to the fact that the forces in different sections are the same, the stresses at the corresponding points of the sections are the same.
Let's consider some element near the surface (Fig. 89, a). Since no forces are applied along its lower edge, which coincides with the surface of the beam, there are no stresses on it. Therefore, there are no stresses on the upper edge of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we arrive at
The same conclusion, etc. It follows that there are no stresses along the horizontal edges of any element. Considering the elements that make up the horizontal layer, starting with the element near the surface of the beam (Fig. 90), we come to the conclusion that there are no stresses along the lateral vertical edges of any element. Thus, the stress state of any element (Fig. 91, a), and in the limit, fibers, should be represented as shown in Fig. 91,b, i.e. it can be either axial tension or axial compression.
4. Due to the symmetry of the application of external forces, the section along the middle of the length of the beam after deformation should remain flat and normal to the axis of the beam (Fig. 92, a). For the same reason, sections in quarters of the length of the beam also remain flat and normal to the axis of the beam (Fig. 92, b), unless the extreme sections of the beam during deformation remain flat and normal to the axis of the beam. A similar conclusion is valid for sections in eighths of the length of the beam (Fig. 92, c), etc. Consequently, if during bending the outer sections of the beam remain flat, then for any section it remains 
It is a fair statement that after deformation it remains flat and normal to the axis of the curved beam. But in this case, it is obvious that the change in elongation of the fibers of the beam along its height should occur not only continuously, but also monotonically. If we call a layer a set of fibers that have the same elongations, then it follows from what has been said that the stretched and compressed fibers of the beam should be located on opposite sides of the layer in which the elongations of the fibers are equal to zero. We will call fibers whose elongations are zero neutral; a layer consisting of neutral fibers is a neutral layer; the line of intersection of the neutral layer with the cross-sectional plane of the beam - the neutral line of this section. Then, based on the previous reasoning, it can be argued that with pure bending of a beam, in each section there is a neutral line that divides this section into two parts (zones): a zone of stretched fibers (stretched zone) and a zone of compressed fibers (compressed zone). ). Accordingly, at the points of the stretched zone of the section, normal tensile stresses should act, at the points of the compressed zone - compressive stresses, and at the points of the neutral line the stresses are equal to zero.
Thus, with pure bending of a beam of constant cross-section:
1) only normal stresses act in sections;
2) the entire section can be divided into two parts (zones) - stretched and compressed; the boundary of the zones is the neutral section line, at the points of which the normal stresses are equal to zero;
3) any longitudinal element of the beam (in the limit, any fiber) is subjected to axial tension or compression, so that adjacent fibers do not interact with each other;
4) if the extreme sections of the beam during deformation remain flat and normal to the axis, then all its cross sections remain flat and normal to the axis of the curved beam.
Stress state of a beam under pure bending
Let us consider an element of a beam subject to pure bending, concluding 
located between sections m-m and n-n, which are spaced one from the other at an infinitesimal distance dx (Fig. 93). Due to position (4) of the previous paragraph, sections m- m and n - n, which were parallel before deformation, after bending, remaining flat, will form an angle dQ and intersect along a straight line passing through point C, which is the center of curvature neutral fiber NN. Then the part AB of the fiber enclosed between them, located at a distance z from the neutral fiber (the positive direction of the z axis is taken towards the convexity of the beam during bending), will turn after deformation into an arc AB. A piece of neutral fiber O1O2, having turned into an arc, O1O2 will not change its length, while fiber AB will receive an elongation:
before deformation
after deformation
where p is the radius of curvature of the neutral fiber.
Therefore, the absolute lengthening of segment AB is equal to
and relative elongation
Since, according to position (3), fiber AB is subjected to axial tension, then during elastic deformation
This shows that normal stresses along the height of the beam are distributed according to a linear law (Fig. 94). Since the equal force of all forces over all elementary sections of the section must be equal to zero, then
from where, substituting the value from (5.8), we find
But the last integral is a static moment about the Oy axis, perpendicular to the plane of action of the bending forces.
Due to its equality to zero, this axis must pass through the center of gravity O of the section. Thus, the neutral line of the section of the beam is a straight line y, perpendicular to the plane of action of bending forces. It is called the neutral axis of the beam section. Then from (5.8) it follows that the stresses at points lying at the same distance from the neutral axis are the same.
The case of pure bending, in which the bending forces act in only one plane, causing bending only in that plane, is planar pure bending. If the said plane passes through the Oz axis, then the moment of elementary forces relative to this axis should be equal to zero, i.e.
Substituting here the value of σ from (5.8), we find
The integral on the left side of this equality, as is known, is the centrifugal moment of inertia of the section relative to the y and z axes, so
The axes about which the centrifugal moment of inertia of the section is zero are called the main axes of inertia of this section. If they, in addition, pass through the center of gravity of the section, then they can be called the main central axes of inertia of the section. Thus, with flat pure bending, the direction of the plane of action of bending forces and the neutral axis of the section are the main central axes of inertia of the latter. In other words, to obtain a flat, pure bend of a beam, a load cannot be applied to it arbitrarily: it must be reduced to forces acting in a plane that passes through one of the main central axes of inertia of the sections of the beam; in this case, the other main central axis of inertia will be the neutral axis of the section.
As is known, in the case of a section that is symmetrical about any axis, the axis of symmetry is one of its main central axes of inertia. Consequently, in this particular case we will certainly obtain pure bending by applying appropriate loads in a plane passing through the longitudinal axis of the beam and the axis of symmetry of its section. A straight line perpendicular to the axis of symmetry and passing through the center of gravity of the section is the neutral axis of this section.
Having established the position of the neutral axis, it is not difficult to find the magnitude of the stress at any point in the section. In fact, since the sum of the moments of elementary forces relative to the neutral axis yy must be equal to the bending moment, then
whence, substituting the value of σ from (5.8), we find
Since the integral 
is. moment of inertia of the section relative to the yy axis, then
and from expression (5.8) we obtain
The product EI Y is called the bending stiffness of the beam.
The greatest tensile and greatest compressive stresses in absolute value act at the points of the section for which absolute value z is greatest, i.e., at points farthest from the neutral axis. With the notation, Fig. 95 we have
The value Jy/h1 is called the moment of resistance of the section to tension and is designated Wyr; similarly, Jy/h2 is called the moment of resistance of the section to compression
and denote Wyc, so
and therefore
If the neutral axis is the axis of symmetry of the section, then h1 = h2 = h/2 and, therefore, Wyp = Wyc, so there is no need to distinguish them, and they use the same notation:
calling W y simply the moment of resistance of the section. Consequently, in the case of a section symmetrical about the neutral axis,
All the above conclusions were obtained on the basis of the assumption that the cross sections of the beam, when bent, remain flat and normal to its axis (hypothesis of flat sections). As has been shown, this assumption is valid only in the case when the extreme (end) sections of the beam remain flat during bending. On the other hand, from the hypothesis of plane sections it follows that elementary forces in such sections should be distributed according to a linear law. Therefore, for the validity of the resulting theory of flat pure bending, it is necessary that the bending moments at the ends of the beam be applied in the form of elementary forces distributed along the height of the section according to a linear law (Fig. 96), coinciding with the law of stress distribution along the height of the section beams. However, based on the Saint-Venant principle, it can be argued that changing the method of applying bending moments at the ends of the beam will cause only local deformations, the influence of which will affect only a certain distance from these ends (approximately equal to the height of the section). The sections located throughout the rest of the length of the beam will remain flat. Consequently, the stated theory of flat pure bending for any method of applying bending moments is valid only within the middle part of the length of the beam, located from its ends at distances approximately equal to the height of the section. From here it is clear that this theory is obviously inapplicable if the height of the section exceeds half the length or span of the beam.
As in § 17, we assume that the cross section of the rod has two axes of symmetry, one of which lies in the bending plane.
When transverse bending tangential stresses arise in the cross section of the rod, and when the rod is deformed, it does not remain flat, as in the case of pure bending. However, for a beam of solid cross-section, the influence of tangential stresses during transverse bending can be neglected and it can be approximately assumed that, just as in the case of pure bending, the cross-section of the rod remains flat during its deformation. Then the formulas for stress and curvature derived in § 17 remain approximately valid. They are accurate for the special case of a constant shear force along the length of the rod 1102).
Unlike pure bending, in transverse bending the bending moment and curvature do not remain constant along the length of the rod. The main task in the case of transverse bending is to determine the deflections. To determine small deflections, you can use the known approximate dependence of the curvature of a bent rod on deflection 11021. Based on this dependence, the curvature of a bent rod x c and deflection V e, resulting from the creep of the material, are related by the relation x c = = dV
Substituting curvature into this relation according to formula (4.16), we establish that
Integrating the last equation makes it possible to obtain the deflection resulting from the creep of the beam material.
Analyzing the above solution to the problem of creep of a bent rod, we can conclude that it is completely equivalent to the solution to the problem of bending a rod made of a material for which the tension-compression diagrams can be approximated power function. Therefore, the determination of deflections arising due to creep, in the case under consideration, can also be made using the Mohr integral to determine the movement of rods made of material that does not obey Hooke’s law)