We will start with the simplest case, the so-called pure bend.
Clean bend There is special case bending, at which in the sections of the beam shear force equal to 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; line of intersection of the neutral layer with the plane cross section beams - 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 largest compressive stresses in absolute value act at the points of the section for which the absolute value of z is greatest, i.e., at the points furthest 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.
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 influence 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 relationships between the dimensions of the beam are such that it works under conditions flat bend no warping or curling.
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.
1. Straight pure bending Transverse bending is the deformation of a rod by forces perpendicular to the axis (transverse) and in pairs, the planes of action of which are perpendicular to the normal sections. A bending rod is called a beam. With direct pure bending in the cross section of the rod, only one force factor arises - the bending moment Mz. Since Qy=d. Mz/dx=0, then Mz=const and pure straight bending can be realized when the rod is loaded with pairs of forces applied in the end sections of the rod. σ Since the bending moment Mz, by definition, is equal to the sum of the moments of internal forces relative to the Oz axis with normal stresses, it is connected by the static equation that follows from this definition:
Analysis of the stress state during pure bending Let us analyze the deformations of the rod model on the side surface of which a grid of longitudinal and transverse marks is applied: Since the transverse marks when the rod is bent by pairs of forces applied in the end sections remain straight and perpendicular to the curved longitudinal marks, this allows us to conclude that hypotheses of flat sections, and therefore By measuring the change in the distances between the longitudinal risks, we come to the conclusion that the hypothesis about the non-pressure of the longitudinal fibers is valid, that is, that is, of all the components of the stress tensor during pure bending, only stress σx=σ and pure straight bending of the prismatic rod are not equal to zero reduces to uniaxial tension or compression of longitudinal fibers by stresses σ. In this case, part of the fibers is in the tension zone (in the figure these are the lower fibers), and the other part is in the compression zone (upper fibers). These zones are separated by a neutral layer (n-n), which does not change its length, and the stresses in which are zero.
Rule of signs of bending moments Rules of signs of moments in problems of theoretical mechanics and strength of materials do not coincide. The reason for this is the difference in the processes under consideration. In theoretical mechanics, the process under consideration is motion or equilibrium solids, therefore, the two moments in the figure tending to rotate Mz the rod in different directions (the right moment clockwise, and the left moment counterclockwise) have problems in theoretical mechanics different sign. In strength problems, stresses and deformations occurring in the body are considered. From this point of view, both moments cause compressive stresses in the upper fibers and tensile stresses in the lower fibers, so the moments have the same sign. Rules for signs of bending moments regarding sections C-C are presented in the diagram:
Calculation of stress values for pure bending Let us derive formulas for calculating the radius of curvature of the neutral layer and normal stresses in the rod. Let us consider a prismatic rod under conditions of direct pure bending with a cross section symmetrical with respect to vertical axis Oy. We place the Ox axis on a neutral layer, the position of which is unknown in advance. Note that the constancy of the cross section of the prismatic rod and the bending moment (Mz=const) ensures the constancy of the radius of curvature of the neutral layer along the length of the rod. When bending with constant curvature, the neutral layer of the rod becomes a circular arc limited by the angle φ. Let us consider an infinitesimal element of length dx cut from a rod. When bending, it will turn into an infinitesimal arc element limited by an infinitesimal angle dφ. φ ρ dφ Taking into account the relationships between the radius of the circle, angle and arc length:
Since the deformations of the element, determined by the relative displacement of its points, are of interest, one of the end sections of the element can be considered stationary. Due to the smallness of dφ, we assume that the cross-sectional points, when rotated by this angle, move not along arcs, but along corresponding tangents. Let's calculate relative deformation longitudinal fiber AB, spaced from the neutral layer at y: From the similarity of triangles COO 1 and O 1 BB 1, it follows that: The longitudinal deformation turned out to be a linear function of the distance from the neutral layer, which is a direct consequence of the law of plane sections. Then the normal tensile stress of fiber AB, based on Hooke’s law, will be equal to:
The resulting formula is not suitable for practical use, since it contains two unknowns: the curvature of the neutral layer 1/ρ and the position of the neutral axis Ox, from which the y coordinate is measured. To determine these unknowns, we will use the equilibrium equations of statics. The first expresses the requirement that the longitudinal force be equal to zero. Substituting the expression for σ into this equation: and taking into account that, we obtain that: The integral on the left side of this equation represents the static moment of the cross section of the rod relative to the neutral axis Ox, which can be equal to zero only relative to the central axis (axis passing through the center of gravity of the section). Therefore, the neutral axis Ox passes through the center of gravity of the cross section. The second static equilibrium equation is one that relates normal stresses to bending moment. Substituting the expression for stresses into this equation, we obtain:
The integral in the resulting equation was previously studied: Jz is the moment of inertia relative to the Oz axis. In accordance with the selected position of the coordinate axes, it is also the main central moment of inertia of the section. We obtain the formula for the curvature of the neutral layer: The curvature of the neutral layer 1/ρ is a measure of the deformation of the rod during straight pure bending. The greater the value of EJz, called the cross-sectional stiffness during bending, the smaller the curvature. Substituting the expression into the formula for σ, we obtain: Thus, the normal stresses during pure bending of a prismatic rod are a linear function of the y coordinate and reach highest values in the fibers furthest from the neutral axis. geometric characteristic, having a dimension m 3 is called the moment of resistance during bending.
Determination of the moments of resistance Wz of cross sections - For the simplest figures in the reference book (Lecture 4) or calculate it yourself - For standard profiles in the GOST range
Calculation of strength during pure bending Design calculation The strength condition when calculating pure bending will have the form: From this condition, Wz is determined, and then either the required profile is selected from the range of standard rolled products, or the section dimensions are calculated using geometric dependencies. When calculating beams made of brittle materials, it is necessary to distinguish between the highest tensile and maximum compressive stresses, which are compared, respectively, with the permissible tensile and compressive stresses. In this case, there will be two strength conditions, separately for tension and compression: Here are the permissible tensile and compressive stresses, respectively.
2. Straight transverse bendingτxy τxz σ In case of direct transverse bending, a bending moment Mz and transverse force Qy arise in the sections of the rod, which are associated with normal and tangential stresses. The formula derived in the case of pure bending of the rod for calculating normal stresses in the case of direct transverse bending, strictly speaking, is not applicable, since from - due to shears caused by tangential stresses, deplanation (curvature) of the cross sections occurs, that is, the hypothesis of flat sections is violated. However, for beams with section height h
When deriving the strength condition for pure bending, the hypothesis of the absence of transverse interaction of longitudinal fibers was used. During transverse bending, deviations from this hypothesis are observed: a) in places where concentrated forces are applied. Under a concentrated force, the transverse interaction stresses σy can be quite large and many times higher than the longitudinal stresses, decreasing, in accordance with the Saint-Venant principle, with distance from the point of application of the force; b) in places where distributed loads are applied. So, in the case shown in Fig., the stress is due to pressure on the upper fibers of the beam. Comparing them with longitudinal stresses σz, which are of the order of magnitude: we come to the conclusion that stresses σy
Calculation of tangential stresses during direct transverse bending Let us assume that the tangential stresses are evenly distributed across the width of the cross section. Direct determination of stresses τyx is difficult, therefore we find equal tangential stresses τxy that arise on the longitudinal area with coordinate y of an element of length dx cut from the beam z x Mz
From this element with a longitudinal section spaced from the neutral layer by y, we cut off top part, replacing the action of the rejected lower part with tangential stresses τ. We will also replace the normal stresses σ and σ+dσ acting on the end areas of the element with the resultant stresses y Mz τ Mz+d. Mz by ω y z Qy Qy +d. Qy dx Nω+d Nω d. T is the static moment of the cut-off part of the cross-sectional area ω relative to the Oz axis. Let us consider the equilibrium condition of the cut element by composing for it the static equation Nω dx b
from where, after simple transformations, taking into account that we obtain the Zhuravsky Formula Tangential stresses along the height of the section change according to the law of a quadratic parabola, reaching a maximum on the neutral axis Mz z Considering that the highest normal stresses arise in the outermost fibers, where there are no tangential stresses, and the highest tangential stresses in In many cases, they take place in the neutral layer, where normal stresses are equal to zero, the strength conditions in these cases are formulated separately for normal and shear stresses
3. Composite beams in bending Shear stresses in longitudinal sections are an expression of the existing connection between the layers of the rod in transverse bending. If this connection is broken in some layers, the nature of the bending of the rod changes. In a rod made up of sheets, each sheet bends independently in the absence of friction forces. The bending moment is evenly distributed between the composite sheets. Maximum value the bending moment will be in the middle of the beam and will be equal. Mz=P l. The greatest normal stress in the cross section of the sheet is equal to:
If the sheets are tightened tightly with sufficiently rigid bolts, the rod will bend as a whole. In this case, the maximum normal stress turns out to be n times less, i.e. In the cross sections of the bolts, when the rod is bent, transverse forces arise. The greatest shear force will be in the section coinciding with the neutral plane of the curved rod.
This force can be determined from the equality of the sums of transverse forces in the cross sections of the bolts and the longitudinal resultant tangential stresses in the case of a whole rod: where m is the number of bolts. Let's compare the change in the curvature of the rod in the seal in the case of connected and unconnected packages. For a connected package: For an unconnected package: Proportional to changes in curvature, deflections also change. Thus, compared to a whole rod, a set of loosely folded sheets turns out to be n 2 times more flexible and only n times less strong. This difference in the coefficients of reduction in rigidity and strength when moving to a sheet package is used in practice when creating flexible spring suspensions. The frictional forces between the sheets increase the rigidity of the package, since they partially restore the tangential forces between the layers of the rod, which were eliminated when moving to the sheet package. Springs therefore require leaf lubrication and should be protected from contamination.
4. Rational forms of cross sections during bending The most rational is the section that has the minimum area for a given load on the beam. In this case, the consumption of material for the manufacture of the beam will be minimal. To obtain a beam with minimal material consumption, one must strive to ensure that the largest possible volume of material works at stresses equal to or close to the permissible ones. First of all, the rational section of the beam during bending must satisfy the condition of equal strength of the tensile and compressed zones of the beam. To do this, it is necessary that the highest tensile stresses and the highest compressive stresses simultaneously reach the permissible stresses. We arrive at a section that is rational for a plastic material in the form of a symmetrical I-beam, in which the largest possible part of the material is concentrated on the flanges connected by a wall, the thickness of which is determined from the conditions of the strength of the wall in terms of tangential stresses. . According to the criterion of rationality, the so-called box section is close to the I-section
For beams made of brittle material, the most rational section will be in the form of an asymmetrical I-beam, satisfying the condition of equal strength in tension and compression, which follows from the requirement. The idea of rationality of the cross-section of rods during bending is implemented in standard thin-walled profiles obtained by hot pressing or rolling from ordinary and alloyed high-quality structural steels, as well as aluminum and aluminum alloys. a-I-beam, b-channel, c-unequal angle, cold-formed closed d-equilateral angle. welded profiles
Clean bend This type of bending is called in which the action takes place bending moment only(Fig. 3.5, A). Let us mentally draw the section plane I-I perpendicular to the longitudinal axis of the beam at a distance * from the free end of the beam to which the external moment is applied m z . Let us carry out actions similar to those that we performed when determining stresses and strains during torsion, namely:
- 1) let’s draw up equilibrium equations for the mentally cut-off part of the part;
- 2) we determine the deformation of the material of the part based on the conditions of compatibility of deformations of elementary volumes of a given section;
- 3) solve the equations of equilibrium and compatibility of deformations.
From the equilibrium condition of the cut off section of the beam (Fig. 3.5, b)
we find that the moment of internal forces Mz equal to the moment of external forces t: M = t.
Rice. 3.5.
The moment of internal forces is created by normal stresses o v directed along the x axis. With pure bending there are no external forces, therefore the sum of the projections of internal forces onto any coordinate axis is zero. On this basis, we write the equilibrium conditions in the form of equalities
Where A- cross-sectional area of the beam (rod).
In pure bending, external forces Fx, F, Fv as well as moments of external forces t x, t y are equal to zero. Therefore, the remaining equilibrium equations are identically equal to zero.
From the equilibrium condition 
when o^O it follows that
normal voltage c x in the cross section, both positive and negative values. (Experience shows that when bending, the material of the lower side of the beam in Fig. 3.5, A stretched, and the upper one is compressed.) Consequently, in the cross section during bending there are such elementary volumes (of the transition layer from compression to tension) in which there is no elongation or compression. This - neutral layer. The line of intersection of the neutral layer with the cross-sectional plane is called neutral line.
The conditions for the compatibility of deformations of elementary volumes during bending are formed on the basis of the hypothesis of flat sections: the cross sections of the beam are flat before bending (see Fig. 3.5, b) will remain flat even after bending (Fig. 3.6).
As a result of the action of an external moment, the beam bends, and the planes sections I-I and II-II rotate relative to each other by an angle dy(Fig. 3.6, b). In pure bending, the deformation of all sections along the beam axis is the same, therefore the radius pk of curvature of the neutral layer of the beam along the x axis is the same. Because dx= p K dip, then the curvature of the neutral layer is equal to 1 / p k = dip / dx and is constant along the length of the beam.
The neutral layer is not deformed; its length before and after deformation is equal to dx. Below this layer the material is stretched, above it is compressed.
Rice. 3.6.
The elongation value of the stretched layer located at a distance y from the neutral one is equal to ydq. Relative elongation of this layer:
Thus, in the adopted model, a linear distribution of deformations is obtained depending on the distance of a given elementary volume to the neutral layer, i.e. along the height of the beam section. Assuming that there is no mutual pressure of parallel layers of material on each other (o y = 0, a, = 0), we write Hooke’s law for linear stretching:
According to (3.13), normal stresses in the cross section of the beam are distributed according to a linear law. The stress of the elementary volume of the material farthest from the neutral layer (Fig. 3.6, V), maximum and equal 
? Problem 3.6
Determine the elastic limit of a steel blade with a thickness / = 4 mm and a length / = 80 cm, if its bending into a semicircle does not cause residual deformation.
Solution
Bending stress o v = Ey/ r k. Let's take y max = t/ 2i r k = / / To.
The elastic limit must correspond to the condition with уп > c v = 1 / 2 kE t /1.
Answer: o = ] / 2 to 2 10 11 4 10 _3 / 0.8 = 1570 MPa; The yield strength of this steel is a t > 1800 MPa, which exceeds the a t of the strongest spring steels. ?
? Problem 3.7
Determine the minimum radius of the drum for winding tape with a thickness of / = 0.1 mm heating element made of a nickel alloy, in which the tape material is not plastically deformed. Module E= 1.6 10 5 MPa, elastic limit about yp = 200 MPa.
Answer: minimum radius р = V 2 ?ir/a yM = У? 1.6-10 11 0.1 10 -3 / (200 10 6) = = 0.04 m?
1. When solving the first equilibrium equation (3.12) and the deformation compatibility equation (3.13) together, we obtain 
Meaning E/ r k φ 0 and the same for all elements dA integration areas. Consequently, this equality is satisfied only under the condition
This integral is called static moment of cross-sectional area about the axisz? What physical meaning this integral?
Let's take the record constant thickness/, but of an arbitrary profile (Fig. 3.7). Let's hang this plate at a point WITH so that it is in a horizontal position. Let us denote by the symbol y m specific gravity material of the plate, then the weight of the elementary volume with area dA equals dq= y JdA. Since the plate is in a state of equilibrium, then from the equality to zero of the projections of forces on the axis at we get
Where G= y M tA- weight of the record.
Rice. 3.7.
The sum of the moments of forces of all forces about the axis z passing through any section of the plate is also zero:
Considering that Y c = G, let's write down
Thus, if an integral of the form J xdA by area A equals
zero, then x c = 0. This means that point C coincides with the center of gravity of the plate. Therefore, from the equality S z = J ydA = 0 when due
bending it follows that the center of gravity of the cross section of the beam is on the neutral line.
Therefore, the value y s the cross section of the beam is zero.
- 1. The neutral line during bending passes through the center of gravity of the cross section of the beam.
- 2. The center of gravity of the cross section is the center of reduction of the moments of external and internal forces.
Problem 3.8
Problem 3.9
2. When solving the second equilibrium equation (3.12) and the deformation compatibility equation (3.13) together, we obtain
Integral J z= J y 2 dA called moment of inertia of the transverse
section of the beam (rod) relative to the z axis, passing through the center of gravity of the cross section.
Thus, M z = E J z / r k. Considering that c x = Ee x = Ey/ r k i E/ r k = a x / y, we obtain the dependence of normal stresses Oh when bending:
1. The bending stress at a given point of the section does not depend on the normal elastic modulus E, but depends on the geometric parameter of the cross section J z and distances at from a given point to the center of gravity of the cross section.
2. The maximum bending stress occurs in the elementary volumes furthest from the neutral line (see Fig. 3.6, V):
Where W z- moment of resistance of the cross section relative to the axis Z-
The condition for strength under pure bending is similar to the condition for strength under linear tension:
where [a m | - permissible bending stress.
It is obvious that the internal volumes of the material, especially near the neutral axis, are practically not loaded (see Fig. 3.6, V). This contradicts the requirement to minimize the material consumption of the structure. Below we will show some ways to overcome this contradiction.
Straight bend- this is a type of deformation in which two internal force factors arise in the cross sections of the rod: bending moment and transverse force.
Clean bend- this is a special case of direct bending, in which only a bending moment occurs in the cross sections of the rod, and the transverse force is zero.
An example of a pure bend - a section CD on the rod AB. Bending moment is the quantity Pa a pair of external forces causing bending. From the equilibrium of the part of the rod to the left of the cross section mn it follows that the internal forces distributed over this section are statically equivalent to the moment M, equal and opposite to the bending moment Pa.
To find the distribution of these internal forces over the cross section, it is necessary to consider the deformation of the rod.
In the simplest case, the rod has a longitudinal plane of symmetry and is subject to the action of external bending pairs of forces located in this plane. Then the bending will occur in the same plane.
Rod axis nn 1 is a line passing through the centers of gravity of its cross sections.
Let the cross section of the rod be a rectangle. Let's draw two vertical lines on its edges mm And pp. When bending, these lines remain straight and rotate so that they remain perpendicular to the longitudinal fibers of the rod.
Further theory of bending is based on the assumption that not only lines mm And pp, but the entire flat cross-section of the rod remains, after bending, flat and normal to the longitudinal fibers of the rod. Therefore, during bending, the cross sections mm And pp rotate relative to each other around axes perpendicular to the bending plane (drawing plane). In this case, the longitudinal fibers on the convex side experience tension, and the fibers on the concave side experience compression.
Neutral surface- This is a surface that does not experience deformation when bending. (Now it is located perpendicular to the drawing, the deformed axis of the rod nn 1 belongs to this surface).
Neutral axis of section- this is the intersection of a neutral surface with any cross-section (now also located perpendicular to the drawing).
Let an arbitrary fiber be at a distance y from a neutral surface. ρ – radius of curvature of the curved axis. Dot O– center of curvature. Let's draw a line n 1 s 1 parallel mm.ss 1– absolute fiber elongation.
Relative extension εx fibers
It follows that deformation of longitudinal fibers proportional to distance y from the neutral surface and inversely proportional to the radius of curvature ρ .
Longitudinal elongation of the fibers of the convex side of the rod is accompanied by lateral narrowing, and the longitudinal shortening of the concave side is lateral expansion, as in the case of simple stretching and compression. Because of this, the appearance of all cross sections changes, the vertical sides of the rectangle become inclined. Lateral deformation z:
μ - Poisson's ratio.
Due to this distortion, all straight cross-sectional lines parallel to the axis z, are bent so as to remain normal to the lateral sides of the section. The radius of curvature of this curve R will be more than ρ in the same respect as ε x in absolute value is greater than ε z and we get
These deformations of longitudinal fibers correspond to stresses
The voltage in any fiber is proportional to its distance from the neutral axis n 1 n 2. Neutral axis position and radius of curvature ρ – two unknowns in the equation for σ x – can be determined from the condition that forces distributed over any cross section form a pair of forces that balances the external moment M.
All of the above is also true if the rod does not have a longitudinal plane of symmetry in which the bending moment acts, as long as the bending moment acts in the axial plane, which contains one of the two main axes cross section. These planes are called main bending planes.
When there is a plane of symmetry and the bending moment acts in this plane, deflection occurs precisely in it. Moments of internal forces relative to the axis z balance the external moment M. Moments of effort about the axis y are mutually destroyed.