What is transverse bending of a beam. bend. Differential dependencies in bending
With direct pure bending, only one force factor arises in the cross section of the rod bending moment M x(Fig. 1). Because Q y \u003d dM x / dz \u003d 0, then M x=const and pure direct bending can be realized when the bar is loaded with pairs of forces applied in the end sections of the bar. Since the bending moment M x by definition is equal to the sum of the moments of internal forces about the axis Oh it is connected with normal stresses by the equation of statics that follows from this definition
Let us formulate the premises of the theory of pure direct bending of a prismatic rod. To do this, we analyze the deformations of a model of a rod made of a low-modulus material, on the side surface of which a grid of longitudinal and transverse scratches is applied (Fig. 2). Since the transverse risks, when the rod is bent by pairs of forces applied in the end sections, remain straight and perpendicular to the curved longitudinal risks, this allows us to conclude that plane section hypotheses, which, as the solution of this problem by the methods of the theory of elasticity shows, ceases to be a hypothesis, becoming an exact fact the law of plane sections. Measuring the change in the distances between the longitudinal risks, we come to the conclusion about the validity of the hypothesis of non-pressure of the longitudinal fibers.
Orthogonality of longitudinal and transverse scratches before and after deformation (as a reflection of the action of the law of flat sections) also indicates the absence of shifts, shear stresses in the transverse and longitudinal sections of the rod.
Fig.1. Relationship between internal effort and stress 
Fig.2. Pure bending model
Thus, pure direct bending of a prismatic rod is reduced to uniaxial tension or compression of longitudinal fibers by stresses (index G omitted later). In this case, part of the fibers is in the tension zone (in Fig. 2, these are the lower fibers), and the other part is in the compression zone (upper fibers). These zones are separated by a neutral layer (np), not changing its length, the stresses in which are equal to zero. Taking into account the prerequisites formulated above and assuming that the material of the rod is linearly elastic, i.e. Hooke's law in this case has the form: , we derive formulas for the curvature of the neutral layer (radius of curvature) and normal stresses . We first note that the constancy of the cross section of the prismatic rod and the bending moment (M x = const), ensures the constancy of the radius of curvature of the neutral layer along the length of the rod (Fig. 3, a), neutral layer (np) described by an arc of a circle.
Consider a prismatic rod under conditions of direct pure bending (Fig. 3, a) with a cross section symmetrical about the vertical axis OU. This condition will not affect the final result (in order for a straight bend to be possible, the coincidence of the axis Oh with main axis of inertia of the cross section, which is the axis of symmetry). Axis Ox put on the neutral layer, position whom not known in advance.
a) calculation scheme, b) strains and stresses
Fig.3. Fragment of a pure bend of a beamConsider an element cut from a rod with length dz, which is shown on a scale with proportions distorted in the interests of clarity in Fig. 3, b. 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 fixed. In view of the smallness, we assume that the points of the cross section, when rotated through this angle, move not along arcs, but along the corresponding tangents.
Let us calculate the relative deformation of the longitudinal fiber AB, separated from the neutral layer by at:
From the similarity of triangles C00 1 and 0 1 BB 1 follows that
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
This formula is not suitable for practical use, since it contains two unknowns: the curvature of the neutral layer and the position of the neutral axis Oh, from which the coordinate is counted y. To determine these unknowns, we use the equilibrium equations of statics. The first expresses the requirement that the longitudinal force be equal to zero
Substituting expression (2) into this equation
and taking into account that , we get that
The integral on the left side of this equation is the static moment of the rod cross section about the neutral axis Oh, which can be equal to zero only relative to the central axis. Therefore, the neutral axis Oh passes through the center of gravity of the cross section.
The second static equilibrium equation is that relating normal stresses to the bending moment (which can easily be expressed in terms of external forces and is therefore considered a given value). Substituting the expression for into the bundle equation. voltage, we get:
and given that 
where J x principal central moment of inertia about the axis Oh, for the curvature of the neutral layer, we obtain the formula
Fig.4. Normal stress distribution
which was first obtained by S. Coulomb in 1773. To match the signs of the bending moment M x and normal stresses, the minus sign is put on the right side of formula (5), since at M x >0 normal stresses at y>0 turn out to be contractive. However, in practical calculations, it is more convenient, without adhering to the formal rule of signs, to determine the stresses modulo, and put the sign according to the meaning. Normal stresses in pure bending of a prismatic bar are a linear function of the coordinate at and reach the highest values in the fibers most distant from the neutral axis (Fig. 4), i.e.
Here a geometric characteristic is introduced, which has the dimension m 3 and is called moment of resistance in bending. Since for a given M x voltage max? the less the more W x , moment of resistance is geometric characteristic of the strength of the cross-sectional bending. Let us give examples of calculating the moments of resistance for the simplest forms of cross sections. For a rectangular cross section (Fig. 5, a) we have J x \u003d bh 3 / 12, y max = h/2 and W x = J x /y max = bh 2 /6. Similarly for a circle (Fig. 5 ,a J x =d4 /64, ymax=d/2) we get W x =d3/32, for a circular annular section (Fig. 5, in), which one
bend
Basic concepts about bending
Bending deformation is characterized by the loss of straightness or original shape by the beam line (its axis) when an external load is applied. In this case, in contrast to the shear deformation, the beam line changes its shape smoothly.
It is easy to see that the resistance to bending is affected not only by the cross-sectional area of the beam (beam, rod, etc.), but also by the geometric shape of this section.
Since the body (beam, beam, etc.) is bent relative to any axis, the bending resistance is affected by the magnitude of the axial moment of inertia of the body section relative to this axis.
For comparison, during torsion deformation, the section of the body is subjected to twisting relative to the pole (point), therefore, the polar moment of inertia of this section affects the torsion resistance.
Many structural elements can work on bending - axles, shafts, beams, gear teeth, levers, rods, etc.
In the resistance of materials, several types of bends are considered:
- depending on the nature of the external load applied to the beam, they distinguish pure bend and transverse bend;
- depending on the location of the plane of action of the bending load relative to the axis of the beam - straight bend and oblique bend.
Pure and transverse beam bending
A pure bend is a type of deformation in which only a bending moment occurs in any cross section of the beam ( rice. 2).
The deformation of pure bending will, for example, take place if two pairs of forces equal in magnitude and opposite in sign are applied to a straight beam in a plane passing through the axis. Then only bending moments will act in each section of the beam.
If the bend takes place as a result of applying a transverse force to the bar ( rice. 3), then such a bend is called transverse. In this case, both the transverse force and the bending moment act in each section of the beam (except for the section to which an external load is applied).
If the beam has at least one axis of symmetry, and the plane of action of the loads coincides with it, then direct bending takes place, if this condition is not met, then oblique bending takes place.
When studying bending deformation, we will mentally imagine that a beam (beam) consists of an innumerable number of longitudinal fibers parallel to the axis.
In order to visualize the deformation of a direct bend, we will conduct an experiment with a rubber bar, on which a grid of longitudinal and transverse lines is applied. 
Subjecting such a bar to a direct bend, one can notice that ( rice. one):
The transverse lines will remain straight when deformed, but will turn at an angle to each other;
- the beam sections will expand in the transverse direction on the concave side and narrow on the convex side;
- longitudinal straight lines will be curved.
From this experience it can be concluded that:
For pure bending, the hypothesis of flat sections is valid;
- the fibers lying on the convex side are stretched, on the concave side they are compressed, and on the border between them lies a neutral layer of fibers that only bend without changing their length.
Assuming the hypothesis of non-pressure of the fibers to be fair, it can be argued that with pure bending in the cross section of the beam, only normal tensile and compressive stresses arise, which are unevenly distributed over the section.
The line of intersection of the neutral layer with the plane of the cross section is called neutral axis. It is obvious that the normal stresses on the neutral axis are equal to zero.
Bending moment and shear force
As is known from theoretical mechanics, the support reactions of beams are determined by compiling and solving the static equilibrium equations for the entire beam. When solving the problems of resistance of materials, and determining the internal force factors in the bars, we took into account the reactions of the bonds along with the external loads acting on the bars.
To determine the internal force factors, we use the section method, and we will depict the beam with only one line - the axis to which active and reactive forces are applied (loads and reactions of bonds).
Consider two cases:
1. Two equal and opposite pairs of forces are applied to the beam.
Considering the balance of the part of the beam located to the left or right of section 1-1 (Fig. 2), we see that in all cross sections there is only a bending moment M and equal to the external moment. Thus, this is a case of pure bending.
The bending moment is the resulting moment about the neutral axis of the internal normal forces acting in the cross section of the beam.
Let's pay attention to the fact that the bending moment has a different direction for the left and right parts of the beam. This indicates the unsuitability of the rule of signs of statics in determining the sign of the bending moment.
2. Active and reactive forces (loads and reactions of bonds) perpendicular to the axis are applied to the beam (rice. 3). Considering the balance of the beam parts located on the left and right, we see that the bending moment M should act in the cross sections and
and shear force Q.
From this it follows that in the case under consideration, not only normal stresses corresponding to the bending moment, but also tangential stresses corresponding to the transverse force act at the points of the cross sections.
The transverse force is the resultant of the internal tangential forces in the cross section of the beam.
Let us pay attention to the fact that the shear force has the opposite direction for the left and right parts of the beam, which indicates the unsuitability of the rule of static signs when determining the sign of the shear force.
Bending, in which a bending moment and a transverse force act in the cross section of the beam, is called transverse.
For a beam in equilibrium with the action of a flat system of forces, the algebraic sum of the moments of all active and reactive forces relative to any point is equal to zero; therefore, the sum of the moments of external forces acting on the beam to the left of the section is numerically equal to the sum of the moments of all external forces acting on the beam to the right of the section.
In this way, the bending moment in the beam section is numerically equal to the algebraic sum of the moments about the center of gravity of the section of all external forces acting on the beam to the right or left of the section.
For a beam in equilibrium under the action of a plane system of forces perpendicular to the axis (i.e., a system of parallel forces), the algebraic sum of all external forces is zero; therefore, the sum of external forces acting on the beam to the left of the section is numerically equal to the algebraic sum of the forces acting on the beam to the right of the section.
In this way, the transverse force in the beam section is numerically equal to the algebraic sum of all external forces acting to the right or left of the section.
Since the rules of signs of statics are unacceptable for establishing the signs of the bending moment and the transverse force, we will establish other rules of signs for them, namely: beam convex upwards, then the bending moment in the section is considered negative ( Figure 4a).
If the sum of external forces lying on the left side of the section gives a resultant directed upwards, then the shear force in the section is considered positive, if the resultant is directed downward, then the shear force in the section is considered negative; for the part of the beam located to the right of the section, the signs of the transverse force will be opposite ( rice. 4b). Using these rules, one should mentally imagine the section of the beam as rigidly clamped, and the connections as discarded and replaced by reactions.
Once again, we note that to determine the reactions of bonds, the rules of the signs of statics are used, and to determine the signs of the bending moment and the transverse force, the rules of the signs of the resistance of materials are used.
The rule of signs for bending moments is sometimes called the "rain rule", meaning that in the case of a downward bulge, a funnel is formed in which rain water is retained (the sign is positive), and vice versa - if under the action of loads the beam bends upward in an arc, the water on it does not delayed (the sign of the bending moments is negative).
Materials of the section "Bending":
The hypothesis of flat sections in bending can be explained by an example: let's apply a grid on the side surface of an undeformed beam, consisting of longitudinal and transverse (perpendicular to the axis) straight lines. As a result of the bending of the beam, the longitudinal lines will take on a curvilinear shape, while the transverse lines will practically remain straight and perpendicular to the bent axis of the beam.
Formulation of the planar section hypothesis: cross-sections that are flat and perpendicular to the axis of the beam before , remain flat and perpendicular to the curved axis after it has been deformed.
This circumstance indicates that when flat section hypothesis, as with and
In addition to the hypothesis of flat sections, an assumption is made: the longitudinal fibers of the beam do not press each other when it is bent.
The hypothesis of flat sections and the assumption are called Bernoulli's conjecture.
Consider a beam of rectangular cross section experiencing pure bending (). Let's select a beam element with a length (Fig. 7.8. a). As a result of bending, the cross sections of the beam will rotate, forming an angle. The top fibers are in compression and the bottom fibers are in tension. The radius of curvature of the neutral fiber is denoted by .
We conditionally consider that the fibers change their length, while remaining straight (Fig. 7.8. b). Then the absolute and relative elongation of the fiber, spaced at a distance y from the neutral fiber:
Let us show that the longitudinal fibers, which do not experience either tension or compression during beam bending, pass through the main central axis x.
Since the length of the beam does not change during bending, the longitudinal force (N) arising in the cross section must be zero. Elementary longitudinal force.
Given the expression :
The multiplier can be taken out of the integral sign (does not depend on the integration variable).
The expression represents the cross section of the beam with respect to the neutral x-axis. It is zero when the neutral axis passes through the center of gravity of the cross section. Consequently, the neutral axis (zero line) when the beam is bent passes through the center of gravity of the cross section.
Obviously: the bending moment is associated with normal stresses that occur at the points of the cross section of the rod. Elementary bending moment created by elemental force:
,
where is the axial moment of inertia of the cross section about the neutral axis x, and the ratio is the curvature of the beam axis.
Rigidity beams in bending(the larger, the smaller the radius of curvature).
The resulting formula represents Hooke's law in bending for a rod: the bending moment occurring in the cross section is proportional to the curvature of the beam axis.
Expressing from the formula of Hooke's law for a rod when bending the radius of curvature () and substituting its value in the formula , we obtain the formula for normal stresses () at an arbitrary point of the cross section of the beam, spaced at a distance y from the neutral axis x: .
In the formula for normal stresses () at an arbitrary point of the cross section of the beam, the absolute values of the bending moment () and the distance from the point to the neutral axis (y coordinates) should be substituted. Whether the stress at a given point will be tensile or compressive is easy to establish by the nature of the deformation of the beam or by the diagram of bending moments, the ordinates of which are plotted from the side of the compressed fibers of the beam.
It can be seen from the formula: normal stresses () change along the height of the cross section of the beam according to a linear law. On fig. 7.8, the plot is shown. The greatest stresses during beam bending occur at points furthest from the neutral axis. If a line is drawn in the cross section of the beam parallel to the neutral axis x, then the same normal stresses arise at all its points.
Simple analysis normal stress diagrams shows that when the beam is bent, the material located near the neutral axis practically does not work. Therefore, in order to reduce the weight of the beam, it is recommended to choose cross-sectional shapes in which most of the material is removed from the neutral axis, such as, for example, an I-profile.
We start with the simplest case, the so-called pure bending.
Pure bending is a special case of bending, in which the transverse force in the beam sections is zero. Pure bending can only take place when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads that cause net
bend, shown in Fig. 88. On sections of these beams, where Q \u003d 0 and, therefore, M \u003d const; there is a pure bend.
The forces in any section of the beam with 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.
Stresses can be determined based on the following considerations.
1. The tangential components of the forces on the elementary areas in the cross section of the beam cannot be reduced to a pair of forces, the plane of action of which is perpendicular to the plane of the section. It follows that the bending force in the section is the result of action on elementary areas
only normal forces, and therefore, with pure bending, stresses are reduced only to normal ones.
2. In order for efforts on elementary platforms to be reduced to only a couple of forces, there must be both positive and negative ones among them. Therefore, both tensioned and compressed beam fibers 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.
Consider any element near the surface (Fig. 89, a). Since no forces are applied along its lower face, which coincides with the surface of the beam, there are no stresses on it either. Therefore, there are no stresses on the upper face of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we come to
The same conclusion, etc. It follows that there are no stresses along the horizontal faces 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 faces of any element. Thus, the stress state of any element (Fig. 91, a), and in the limit of the fiber, must be represented as shown in Fig. 91b, 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 beam length after deformation should remain flat and normal to the beam axis (Fig. 92, a). For the same reason, sections in quarters of the beam length also remain flat and normal to the beam axis (Fig. 92, b), if only the extreme sections of the beam remain flat and normal to the beam axis during deformation. A similar conclusion is also valid for sections in eighths of the length of the beam (Fig. 92, c), etc. Therefore, if the extreme sections of the beam remain flat during bending, then for any section it remains 
it is fair to say 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 the elongation of the fibers of the beam along its height should occur not only continuously, but also monotonously. If we call a layer a set of fibers having 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 fiber elongations are equal to zero. We will call fibers whose elongations are equal to zero, neutral; a layer consisting of neutral fibers - a neutral layer; the line of intersection of the neutral layer with the plane of the cross section of the beam - the neutral line of this section. Then, based on the previous considerations, it can be argued that with a pure bending of the beam in each of its sections there is a neutral line that divides this section into two parts (zones): the zone of stretched fibers (tensioned zone) and the zone of compressed fibers (compressed zone ). Accordingly, normal tensile stresses should act at the points of the stretched zone of the cross-section, compressive stresses at the points of the compressed zone, and at the points of the neutral line the stresses are equal to zero.
Thus, with a pure bending of a beam of constant cross-section:
1) only normal stresses act in the sections;
2) the entire section can be divided into two parts (zones) - stretched and compressed; the boundary of the zones is the neutral line of the section, 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 in pure bending
Consider an element of a beam subject to pure bending, concluding 
measured between sections m-m and n-n, which are spaced one from the other at an infinitely small distance dx (Fig. 93). Due to the provision (4) of the previous paragraph, the 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 of the AB 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 into an arc A "B" after deformation. A segment of the neutral fiber O1O2, turning into an O1O2 arc, it will not change its length, while the AB fiber will receive an elongation:
before deformation
after deformation
where p is the radius of curvature of the neutral fiber.
Therefore, the absolute elongation of the segment AB is
and elongation
Since, according to position (3), the fiber AB is subjected to axial tension, then with elastic deformation
From this it can be seen that the normal stresses along the height of the beam are distributed according to a linear law (Fig. 94). Since the equal force of all efforts on all elementary sections of the section must be equal to zero, then
whence, substituting the value from (5.8), we find
But the last integral is a static moment about the Oy axis, which is 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 beam section is a straight line yy, perpendicular to the plane of action of the 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 in that plane only, is a planar pure bending. If the named plane passes through the Oz axis, then the moment of elementary efforts relative to this axis must 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 about the y and z axes, so that
The axes with respect to which the centrifugal moment of inertia of the section is equal to zero are called the main axes of inertia of this section. If, in addition, they 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 a flat pure bending, the direction of the plane of action of the 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 bending 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 beam sections; 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 a pure bending by applying the appropriate analoads in the plane passing through the longitudinal axis of the beam and the axis of symmetry of its section. The 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. Indeed, 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 about the y-axis, then
and from expression (5.8) we obtain
The product EI Y is called the bending stiffness of the beam.
The largest tensile and largest compressive stresses in absolute value act at the points of the section for which the absolute value of z is the largest, i.e., at the points furthest from the neutral axis. With the designations, Fig. 95 have
The value of Jy / h1 is called the moment of resistance of the section to stretching and is denoted by 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, consequently, Wyp = Wyc, so there is no need to distinguish between them, and they use the same designation:
calling W y simply the section modulus. Therefore, in the case of a section symmetrical about the neutral axis,
All the above conclusions are obtained on the basis of the assumption that the cross sections of the beam, when bent, remain flat and normal to its axis (the hypothesis of flat sections). As shown, this assumption is valid only if the extreme (end) sections of the beam remain flat during bending. On the other hand, it follows from the hypothesis of flat sections that elementary forces in such sections should be distributed according to a linear law. Therefore, for the validity of the obtained 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), which coincides 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 a change in the method of application of bending moments at the ends of the beam will cause only local deformations, the effect of which will affect only at a certain distance from these ends (approximately equal to the height of the section). The sections located in the rest of the length of the beam will remain flat. Consequently, the stated theory of flat pure bending, with any method of applying bending moments, is valid only within the middle part of the length of the beam, located at distances from its ends approximately equal to the height of the section. From this it is clear that this theory is obviously inapplicable if the height of the section exceeds half the length or span of the beam.
Pure bend called this type of bending, in which the action takes place only bending moment(Fig. 3.5, a). Let's 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 mz . Let's carry out actions similar to those that were performed by us when determining stresses and strains during torsion, namely:
- 1) compose the equilibrium equations of the mentally cut off part of the part;
- 2) we determine the deformation of the material of the part based on the conditions for the compatibility of deformations of elementary volumes of a given section;
- 3) solve the equations of equilibrium and compatibility of deformations.
From the condition of equilibrium of the cut-off section of the beam (Fig. 3.5, b)
we get 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, so the sum of the projections of internal forces on any coordinate axis is zero. On this basis, we write the equilibrium conditions in the form of equalities
where BUT- cross-sectional area of the beam (rod).
In pure bending, external forces F x , F, F v as well as moments of external forces t x, t y are equal to zero. Therefore, the rest of the equilibrium equations are identically equal to zero.
It follows from the equilibrium condition for o > 0 that
normal voltage with x in cross section take both positive and negative values. (Experience shows that when bending, the material of the underside 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. It - neutral layer. The line of intersection of the neutral layer with the plane of the cross section 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: flat cross sections of the beam 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 of sections I-I and II-II rotate relative to each other by an angle dy(Fig. 3.6, b). With pure bending, the deformation of all sections along the axis of the beam 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 does not deform, 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 value of the elongation 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 strains 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 pressing of parallel layers of material on each other (o y \u003d 0, a, \u003d 0), we write Hooke's law for linear tension:
According to (3.13), the 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, the most distant from the neutral layer (Fig. 3.6, in), maximum and equal to 
? Task 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 permanent deformation.
Solution
Bending stress o v = eu/ p k. Let's take y max = t/ 2i p k = / / to.
The elastic limit must correspond to the condition with yn > c v = 1/2 kE t /1.
Answer: about = ] / 2 to 2 10 11 4 10 _3 / 0.8 = 1570 MPa; the yield strength of this steel is a m > 1800 MPa, which exceeds a m of the strongest spring steels. ?
? Task 3.7
Determine the minimum drum radius for winding a tape with a thickness / = 0.1 mm of a heating element made of nickel alloy, at which the tape material does not plastically deform. Module E= 1.6 10 5 MPa, elastic limit o yn = 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. By jointly solving the first equilibrium equation (3.12) and the strain compatibility equation (3.13), we obtain 
Meaning E/ r k f 0 and the same for all elements dA area of integration. Therefore, this equality is satisfied only under the condition
This integral is called static moment of the cross-sectional area about the axisz? What is the physical meaning of this integral?
Let us take a plate of constant thickness /, but of an arbitrary profile (Fig. 3.7). Hang this plate at the point FROM so that it is in a horizontal position. We denote by the symbol y m the specific gravity of the material of the plate, then the weight of an elementary volume with an 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 MtA- plate weight.
Rice. 3.7.
The sum of the moments of forces of all forces about the axis z passing in any section of the plate is also equal to zero:
Given that Y c = g, write down
Thus, if an integral of the form J xdA by area BUT equals
zero, then x c = 0. This means that point C coincides with the center of gravity of the plate. Therefore, from the equality Sz = J ydA= 0 at
bend it follows that the center of gravity of the cross section of the beam is on the neutral line.
Therefore, the value u s cross section of the beam is zero.
- 1. The neutral line during bending passes through the center of gravity of the beam cross section.
- 2. The center of gravity of the cross section is the center of reduction of the moments of external and internal forces.
Task 3.8
Task 3.9
2. By jointly solving the second equilibrium equation (3.12) and the strain compatibility equation (3.13), we obtain
Integral Jz= J y2dA called moment of inertia of the transverse
section of a beam (rod) relative to the z-axis, passing through the center of gravity of the cross section.
In this way, M z \u003d E J z / p k. Considering that c x = Ee x = Ey/ p k and E/ p k = a x / y, we obtain the dependence of normal stresses oh when bending:
1. The bending stress at a given section point does not depend on the modulus of normal elasticity E, but depends on the geometric parameter of the cross section Jz and distance at from this point to the center of gravity of the cross section.
2. The maximum bending stress occurs in elementary volumes, the most distant from the neutral line (see Fig. 3.6, in):
where Wz- moment of resistance of the cross section about the axis Z-
The condition of strength in pure bending is similar to the condition of strength in linear tension:
where [a m | - allowable bending stress.
Obviously, the internal volumes of the material, especially near the neutral axis, are practically not loaded (see Fig. 3.6, in). This contradicts the requirement to minimize the material consumption of the structure. Some ways of overcoming this contradiction will be shown below.
