10.1. General concepts and definitions

Bend- this is a type of loading in which the rod is loaded with moments in planes passing through the longitudinal axis of the rod.

A rod that bends is called a beam (or timber). In the future, we will consider rectilinear beams, the cross section of which has at least one axis of symmetry.

The resistance of materials is divided into flat, oblique and complex bending.

Flat bend– bending, in which all the forces bending the beam lie in one of the planes of symmetry of the beam (in one of the main planes).

The main planes of inertia of a beam are the planes passing through the main axes cross sections and the geometric axis of the beam (x-axis).

Oblique bend– bending, in which the loads act in one plane that does not coincide with the main planes of inertia.

Complex bend – bending, in which loads act in different (arbitrary) planes.

10.2. Determination of internal bending forces

Let us consider two typical cases of bending: in the first, the cantilever beam is bent by a concentrated moment Mo; in the second - concentrated force F.

Using the method of mental sections and composing equilibrium equations for the cut off parts of the beam, we determine the internal forces in both cases:

The remaining equilibrium equations are obviously identically equal to zero.

Thus, in the general case of plane bending in the section of a beam, out of six internal forces, two arise - bending moment Mz and shear force Qy (or when bending relative to another main axis - bending moment My and shear force Qz).

Moreover, in accordance with the two loading cases considered, plane bending can be divided into pure and transverse.

Clean bend– flat bending, in which in the sections of the rod, out of six internal forces, only one arises – a bending moment (see the first case).

Transverse bend– bending, in which in the sections of the rod, in addition to the internal bending moment, a transverse force also arises (see the second case).

Strictly speaking, to simple types resistance only applies pure bend; transverse bending is conventionally classified as a simple type of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected when calculating strength.

When determining internal efforts, we will adhere to next rule signs:

1) the transverse force Qy is considered positive if it tends to rotate the beam element in question clockwise;

2) bending moment Mz is considered positive if, when bending a beam element, the upper fibers of the element are compressed and the lower fibers are stretched (umbrella rule).

Thus, the solution to the problem of determining the internal forces during bending will be built according to the following plan: 1) at the first stage, considering the equilibrium conditions of the structure as a whole, we determine, if necessary, the unknown reactions of the supports (note that for a cantilever beam the reactions in the embedment can be and not found if we consider the beam from the free end); 2) at the second stage, we select characteristic sections of the beam, taking as the boundaries of the sections the points of application of forces, points of change in the shape or size of the beam, points of fastening of the beam; 3) at the third stage, we determine the internal forces in the sections of the beam, considering the conditions of equilibrium of the beam elements in each section.

10.3. Differential dependencies during bending

Let us establish some relationships between internal forces and external bending loads, as well as characteristic features diagrams Q and M, knowledge of which will facilitate the construction of diagrams and allow you to control their correctness. For convenience of notation, we will denote: M≡Mz, Q≡Qy.

Let us select a small element dx in a section of a beam with an arbitrary load in a place where there are no concentrated forces and moments. Since the entire beam is in equilibrium, the dx element will also be in equilibrium under the action of the forces applied to it shear forces, bending moments and external load. Since Q and M generally vary along

axis of the beam, then transverse forces Q and Q+dQ, as well as bending moments M and M+dM, will appear in the sections of element dx. From the equilibrium condition of the selected element we obtain

The first of the two equations written gives the condition

From the second equation, neglecting the term q·dx·(dx/2) as an infinitesimal quantity of the second order, we find

Considering expressions (10.1) and (10.2) together we can obtain

Relations (10.1), (10.2) and (10.3) are called differential dependences of D.I. Zhuravsky during bending.

Analysis of the above differential dependencies during bending allows us to establish some features (rules) for constructing diagrams of bending moments and transverse forces: a - in areas where there is no distributed load q, diagrams Q are limited to straight lines parallel to the base, and diagrams M are limited to inclined straight lines; b – in areas where a distributed load q is applied to the beam, Q diagrams are limited by inclined straight lines, and M diagrams are limited by quadratic parabolas.

Moreover, if we construct diagram M “on a stretched fiber,” then the convexity of the parabola will be directed in the direction of action q, and the extremum will be located in the section where diagram Q intersects the base line; c – in sections where a concentrated force is applied to the beam, on the diagram Q there will be jumps by the magnitude and in the direction of this force, and on the diagram M there will be kinks, the tip directed in the direction of action of this force; d – in sections where a concentrated moment is applied to the beam, there will be no changes on diagram Q, and on diagram M there will be jumps in the magnitude of this moment; d – in areas where Q>0, the moment M increases, and in areas where Q<0, момент М убывает (см. рисунки а–г).

10.4. Normal stresses during pure bending of a straight beam

Let us consider the case of pure plane bending of a beam and derive a formula for determining normal stresses for this case.

Note that in the theory of elasticity it is possible to obtain an exact dependence for normal stresses during pure bending, but if this problem is solved using methods of strength of materials, it is necessary to introduce some assumptions.

There are three such hypotheses for bending:

a – hypothesis of flat sections (Bernoulli hypothesis) – flat sections before deformation remain flat after deformation, but only rotate relative to a certain line, which is called the neutral axis of the beam section. In this case, the fibers of the beam lying on one side of the neutral axis will stretch, and on the other, compress; fibers lying on the neutral axis do not change their length;

b – hypothesis about the constancy of normal stresses - stresses acting at the same distance y from the neutral axis are constant across the width of the beam;

c – hypothesis about the absence of lateral pressures – adjacent longitudinal fibers do not press on each other.

Static side of the problem

To determine the stresses in the cross sections of the beam, we consider, first of all, the static sides of the problem. Using the method of mental sections and composing equilibrium equations for the cut-off part of the beam, we will find the internal forces during bending. As was shown earlier, the only internal force acting in the beam section during pure bending is the internal bending moment, which means that normal stresses associated with it will arise here.

We will find the relationship between internal forces and normal stresses in the beam section by considering the stresses on the elementary area dA, selected in the cross section A of the beam at the point with coordinates y and z (the y axis is directed downward for convenience of analysis):

As we see, the problem is internally statically indeterminate, since the nature of the distribution of normal stresses over the section is unknown. To solve the problem, consider the geometric picture of deformations.

Geometric side of the problem

Let us consider the deformation of a beam element of length dx, separated from a bending rod at an arbitrary point with coordinate x. Taking into account the previously accepted hypothesis of flat sections, after bending the beam section, rotate relative to the neutral axis (n.o.) by an angle dϕ, while the fiber ab, spaced from the neutral axis at a distance y, will turn into an arc of a circle a1b1, and its length will change by some size. Let us recall here that the length of the fibers lying on the neutral axis does not change, and therefore the arc a0b0 (the radius of curvature of which is denoted by ρ) has the same length as the segment a0b0 before the deformation a0b0=dx.

Let us find the relative linear deformation εx of the fiber ab of the curved beam.

When building diagrams of bending momentsM at builders accepted: ordinates expressing on a certain scale positive values ​​of bending moments, set aside stretched fibers, i.e. - down, A negative - up from the beam axis. Therefore, they say that builders construct diagrams on stretched fibers. At the mechanics positive values ​​of both shear force and bending moment are postponed up. Mechanics draw diagrams on compressed fibers.

Principal stresses when bending. Equivalent voltages.

In the general case of direct bending in the cross sections of a beam, normal And tangentsvoltage. These voltages vary both along the length and height of the beam.

Thus, in the case of bending, there is plane stress state.

Let's consider a diagram where the beam is loaded with force P

Largest normal tensions arise in extreme, points most distant from the neutral line, and There are no shear stresses in them. Thus, for extreme fibers non-zero principal stresses are normal stresses in cross section.

At the neutral line level in the cross section of the beam there are highest shear stress, A normal stresses are zero. means in the fibers neutral layer the principal stresses are determined by the values ​​of the tangential stresses.

In this design scheme, the upper fibers of the beam will be stretched, and the lower ones will be compressed. To determine the principal stresses we use the well-known expression:

Full stress analysis Let's imagine it in the picture.

Bending Stress Analysis

Maximum principal stress σ 1 is on upper extreme fibers and equals zero on the lower outermost fibers. Main stress σ 3 has the largest absolute value is on the lower fibers.

Trajectory of principal stresses depends on load type And method of securing the beam.

When solving problems it is enough separately check normal And separately tangential stresses. However sometimes the most stressful turn out to be intermediate fibers in which there are both normal and shear stresses. This happens in sections where At the same time, both the bending moment and the shear force reach large values- this can be in the embedding of a cantilever beam, on the support of a beam with a cantilever, in sections under concentrated force, or in sections with sharply changing widths. For example, in an I-section the most dangerous the junction of the wall and the shelf- there are significant normal and shear stresses.

The material is in a plane stress state and is required check for equivalent voltages.

Strength conditions for beams made of plastic materials By third(theory of maximum tangential stresses) And fourth(theory of energy of shape changes) theories of strength.

As a rule, in rolled beams the equivalent stresses do not exceed the normal stresses in the outermost fibers and no special testing is required. Another thing - composite metal beams, who have the wall is thinner than for rolled profiles at the same height. Welded composite beams made of steel sheets are more often used. Calculation of such beams for strength: a) selection of the section - height, thickness, width and thickness of the beam chords; b) checking strength by normal and tangential stresses; c) checking strength using equivalent stresses.

Determination of shear stresses in an I-section. Let's consider the section I-beam S x =96.9 cm 3 ; Yх=2030 cm 4 ; Q=200 kN

To determine the shear stress, it is used formula,where Q is the shear force in the section, S x 0 is the static moment of the part of the cross section located on one side of the layer in which the tangential stresses are determined, I x is the moment of inertia of the entire cross section, b is the width of the section in the place where shear stress is determined

Let's calculate maximum shear stress:

Let's calculate the static moment for top shelf:

Now let's calculate shear stress:

We are building shear stress diagram:

Let us consider the cross section of a standard profile in the form I-beam and define shear stress, acting parallel to the shear force:

Let's calculate static moments simple figures:

This value can be calculated and otherwise, using the fact that for the I-beam and trough sections the static moment of half the section is given. To do this, it is necessary to subtract from the known value of the static moment the value of the static moment to the line A 1 B 1:

The tangential stresses at the junction of the flange and the wall change spasmodically, because sharp wall thickness varies from t st to b.

Diagrams of tangential stresses in the walls of trough, hollow rectangular and other sections have the same form as in the case of an I-section. The formula includes the static moment of the shaded part of the section relative to the X axis, and the denominator includes the width of the section (net) in the layer where the shear stress is determined.

Let us determine the tangential stresses for a circular section.

Since the shear stresses at the section contour must be directed tangent to the contour, then at points A And IN at the ends of any chord parallel to the diameter AB, shear stresses are directed perpendicular to the radii OA And OV. Hence, directions tangential stresses at points A, V, K converge at some point N on the Y axis.

Static moment of the cut-off part:

That is, the shear stresses change according to parabolic law and will be maximum at the level of the neutral line, when y 0 =0

Formula for determining shear stress (formula)

Consider a rectangular section

At a distance y 0 from the central axis we draw section 1-1 and determine the tangential stresses. Static moment area cut off part:

It should be borne in mind that it is fundamental indifferent, take the static moment of area shaded or remaining part cross section. Both static moments equal and opposite in sign, so their sum, which represents static moment of area of ​​the entire section relative to the neutral line, namely the central x axis, will be equal to zero.

Moment of inertia of a rectangular section:

Then shear stress according to the formula

The variable y 0 is included in the formula in second degrees, i.e. tangential stresses in a rectangular section vary according to law of a square parabola.

Shear stress reached maximum at the level of the neutral line, i.e. When y 0 =0:

, Where A is the area of ​​the entire section.

Strength condition for tangential stresses has the form:

, Where S x 0– static moment of the part of the cross section located on one side of the layer in which the shear stresses are determined, Ix– moment of inertia of the entire cross section, b– section width in the place where the shear stress is determined, Q-lateral force, τ - shear stress, [τ] — permissible tangential stress.

This strength condition allows us to produce three type of calculation (three types of problems when calculating strength):

1. Verification calculation or strength test based on tangential stresses:

2. Selection of section width (for a rectangular section):

3. Determination of permissible lateral force (for a rectangular section):

To determine tangents stresses, consider a beam loaded with forces.

The task of determining stresses is always statically indeterminate and requires involvement geometric And physical equations. However, it is possible to accept such hypotheses about the nature of stress distribution that the task will become statically definable.

By two infinitely close cross sections 1-1 and 2-2 we select dz element, Let's depict it on a large scale, then draw a longitudinal section 3-3.

In sections 1–1 and 2–2, normal σ 1, σ 2 stresses, which are determined by the well-known formulas:

Where M - bending moment in cross section, dM - increment bending moment at length dz

Lateral force in sections 1–1 and 2–2 is directed along the main central axis Y and, obviously, represents the sum of the vertical components of internal tangential stresses distributed over the section. In strength of materials it is usually taken assumption of their uniform distribution across the width of the section.

To determine the magnitude of tangential stresses at any point in the cross section located at a distance y 0 from the neutral X axis, draw a plane parallel to the neutral layer (3-3) through this point and remove the clipped element. We will determine the voltage acting across the ABCD area.

Let's project all the forces onto the Z axis

The resultant of the internal longitudinal forces along the right side will be equal to:

Where A 0 – area of ​​the façade edge, S x 0 – static moment of the cut-off part relative to the X axis. Similarly on the left side:

Both resultants directed towards each other, since the element is in compressed beam area. Their difference is balanced by the tangential forces on the lower edge of 3-3.

Let's assume that shear stress τ distributed across the width of the beam cross section b evenly. This assumption is the more likely the smaller the width compared to the height of the section. Then resultant of tangential forces dT equal to the stress value multiplied by the area of ​​the face:

Let's compose now equilibrium equation Σz=0:

or where from

Let's remember differential dependencies, according to which Then we get the formula:

This formula is called formulas. This formula was obtained in 1855. Here S x 0 – static moment of part of the cross section, located on one side of the layer in which the shear stresses are determined, I x – moment of inertia the entire cross section, b – section width in the place where the shear stress is determined, Q - shear force in cross section.

— bending strength condition, Where

- maximum moment (modulo) from the diagram of bending moments; - axial moment of resistance of the section, geometric characteristic; - permissible stress (σ adm)

- maximum normal voltage.

If the calculation is carried out according to limit state method, then instead of the permissible voltage, we enter into the calculation design resistance of the material R.

Types of flexural strength calculations

1. Check calculation or testing of strength using normal stresses

2. Design calculation or selection of section

3. Definition permissible load (definition lifting capacity and or operational carrier capabilities)

When deriving the formula for calculating normal stresses, we consider the case of bending, when the internal forces in the sections of the beam are reduced only to bending moment, A the shear force turns out to be zero. This case of bending is called pure bending. Consider the middle section of the beam, which is subject to pure bending.

When loaded, the beam bends so that it The lower fibers lengthen and the upper ones shorten.

Since part of the fibers of the beam is stretched, and part is compressed, and the transition from tension to compression occurs smoothly, without jumps, V average part of the beam is located 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 or neutral axis sections. Neutral lines are strung on the axis of the beam. Neutral line is the line in which normal stresses are zero.

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 hypothesis of plane sections (conjecture). 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.

Assumptions for deriving normal stress formulas: 1) The hypothesis of plane sections is fulfilled. 2) Longitudinal fibers do not press on each other (non-pressure hypothesis) and, therefore, each of the fibers is in a state of uniaxial tension or compression. 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 in plane bending conditions without warping or twisting.

Let's consider a beam of arbitrary cross-section, but having an axis of symmetry. Bending moment represents resultant moment of internal normal forces, arising on infinitely small areas and can be expressed in integral form: (1), where y is the arm of the elementary force relative to the x axis

Formula (1) expresses static side of the problem of bending a straight beam, but along it at a known bending moment It is impossible to determine normal stresses until the law of their distribution is established.

Let us select the beams in the middle section and consider section of length dz, subject to bending. Let's depict it on an enlarged scale.

Sections bounding the area dz, parallel to each other until deformed, and after applying the load rotate around their neutral lines by an angle . The length of the neutral layer fiber segment will not change. and will be equal to: , where is this radius of curvature the curved axis of the beam. But any other fiber lying lower or higher neutral layer, will change its length. Let's calculate relative elongation of fibers located at a distance y from the neutral layer. Relative elongation is the ratio of absolute deformation to the original length, then:

Let's reduce by and bring similar terms, then we get: (2) This formula expresses geometric side of the pure bending problem: The deformations of the fibers are directly proportional to their distances to the neutral layer.

Now let's move on to stresses, i.e. we will consider physical side of the task. in accordance with non-pressure assumption we use fibers under axial tension-compression: then, taking into account the formula (2) we have (3), those. normal stress when bending along the section height linearly distributed. At the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero. Let's substitute (3) into the equation (1) and take the fraction out of the integral sign as a constant value, then we have . But the expression is axial moment of inertia of the section relative to the x axis - I x. Its dimension cm 4, m 4

Then ,where (4) ,where is the curvature of the curved axis of the beam, and is the rigidity of the beam section during bending.

Let's substitute the resulting expression curvature (4) into expression (3) and we get formula for calculating normal stresses at any point in the cross section: (5)

That. maximum tensions arise at points furthest from the neutral line. Attitude (6) called axial moment of section resistance. Its dimension cm 3, m 3. The moment of resistance characterizes the influence of the shape and size of the cross section on the magnitude of stress.

Then maximum voltages: (7)

Bending strength condition: (8)

When transverse bending occurs not only normal, but also shear stresses, because available shear force. Shear stress complicate the picture of deformation, they lead to curvature cross sections of the beam, resulting in the hypothesis of plane sections is violated. However, research shows that distortions introduced by shear stresses slightly affect normal stresses calculated by the formula (5) . Thus, when determining normal stresses in the case of transverse bending The theory of pure bending is quite applicable.

Neutral line. Question about the position of the neutral line.

During bending there is no longitudinal force, so we can write Let us substitute here the formula for normal stresses (3) and we get Since the modulus of longitudinal elasticity of the beam material is not equal to zero and the curved axis of the beam has a finite radius of curvature, it remains to assume that this integral is static moment of area cross section of the beam relative to the neutral line-axis x , and, since it is equal to zero, then the neutral line passes through the center of gravity of the section.

The condition (absence of moment of internal forces relative to the field line) will give or taking into account (3) . For the same reasons (see above) . In integrand - the centrifugal moment of inertia of the section relative to the x and y axes is zero, which means these axes are main and central and make up direct corner. Hence, The force and neutral lines in a straight bend are mutually perpendicular.

Having installed neutral line position, easy to build normal stress diagram along the section height. Her linear character is determined equation of the first degree.

The nature of the diagram σ for symmetrical sections relative to the neutral line, M<0

Forces acting perpendicular to the axis of the beam and located in a plane passing through this axis cause deformation called transverse bending. If the plane of action of the mentioned forces – main plane, then a straight (flat) transverse bend occurs. Otherwise, the bend is called oblique transverse. A beam that is subject to predominantly bending is called beam 1 .

Essentially, transverse bending is a combination of pure bending and shear. In connection with the curvature of cross sections due to the uneven distribution of shears along the height, the question arises about the possibility of using the normal stress formula σ X, derived for pure bending based on the hypothesis of plane sections.

1 A single-span beam, having at the ends, respectively, one cylindrical fixed support and one cylindrical movable one in the direction of the beam axis, is called simple. A beam with one end clamped and the other free is called console. A simple beam having one or two parts hanging over a support is called console.

If, in addition, the sections are taken far from the places where the load is applied (at a distance not less than half the height of the section of the beam), then it can be assumed, as in the case of pure bending, that the fibers do not exert pressure on each other. This means that each fiber experiences uniaxial tension or compression.

Under the action of a distributed load, the transverse forces in two adjacent sections will differ by an amount equal to qdx. Therefore, the curvature of the sections will also be slightly different. In addition, the fibers will exert pressure on each other. A thorough study of the issue shows that if the length of the beam l quite large compared to its height h (l/ h> 5), then even with a distributed load, these factors do not have a significant effect on the normal stresses in the cross section and therefore may not be taken into account in practical calculations.

a b c

Rice. 10.5 Fig. 10.6

In sections under concentrated loads and near them, the distribution of σ X deviates from the linear law. This deviation, which is local in nature and is not accompanied by an increase in the highest stresses (in the outermost fibers), is usually not taken into account in practice.

Thus, with transverse bending (in the plane xy) normal stresses are calculated using the formula

σ X= – [M z(x)/I z]y.

If we draw two adjacent sections on a section of the beam that is free from load, then the transverse force in both sections will be the same, and therefore the curvature of the sections will be the same. In this case, any piece of fiber ab(Fig. 10.5) will move to a new position a"b", without undergoing additional elongation, and therefore, without changing the value of the normal stress.

Let us determine the tangential stresses in the cross section through their paired stresses acting in the longitudinal section of the beam.

Select an element of length from the timber dx(Fig. 10.7 a). Let's draw a horizontal section at a distance at from neutral axis z, dividing the element into two parts (Fig. 10.7) and consider the equilibrium of the upper part, which has a base

width b. In accordance with the law of pairing of tangential stresses, the stresses acting in the longitudinal section are equal to the stresses acting in the cross section. Taking this into account, under the assumption that the shear stresses in the site b distributed uniformly, using the condition ΣХ = 0, we obtain:

N * - (N * +dN *)+

where: N * is the resultant of normal forces σ in the left cross section of the element dx within the “cut off” area A * (Fig. 10.7 d):

where: S = - static moment of the “cut off” part of the cross section (shaded area in Fig. 10.7 c). Therefore, we can write:

Then we can write:

This formula was obtained in the 19th century by the Russian scientist and engineer D.I. Zhuravsky and bears his name. And although this formula is approximate, since it averages the stress over the width of the section, the calculation results obtained from it are in good agreement with the experimental data.

In order to determine the shear stresses at an arbitrary cross-section point located at a distance y from the z axis, you should:

Determine from the diagram the magnitude of the transverse force Q acting in the section;

Calculate the moment of inertia I z of the entire section;

Draw a plane parallel to the plane through this point xz and determine the section width b;

Calculate the static moment of the clipped area S relative to the main central axis z and substitute the found values ​​into the Zhuravsky formula.

Let us determine, as an example, tangential stresses in a rectangular cross section (Fig. 10.6, c). Static moment about the axis z parts of the section above line 1-1, on which the stress is determined, will be written in the form:

It changes according to the law of a square parabola. Section width V for a rectangular beam is constant, then the law of change in tangential stresses in the section will also be parabolic (Fig. 10.6, c). At y = and y = − the tangential stresses are zero, and on the neutral axis z they reach their greatest value.

For a beam of circular cross section on the neutral axis we have.

Classification of types of rod bending

Bend This type of deformation is called in which bending moments occur in the cross sections of the rod. A rod that bends is usually called beam. If bending moments are the only internal force factors in the cross sections, then the rod experiences clean bend. If bending moments occur together with transverse forces, then such bending is called transverse.

Beams, axles, shafts and other structural parts work for bending.

Let's introduce some concepts. The plane passing through one of the main central axes of the section and the geometric axis of the rod is called main plane. The plane in which external loads act, causing the beam to bend, is called force plane. The line of intersection of the force plane with the cross-sectional plane of the rod is called power line. Depending on the relative position of the force and main planes of the beam, straight or oblique bending is distinguished. If the force plane coincides with one of the main planes, then the rod experiences straight bend(Fig. 5.1, A), if it does not match - oblique(Fig. 5.1, b).

Rice. 5.1. Rod bend: A- straight; b- oblique

From a geometric point of view, the bending of the rod is accompanied by a change in the curvature of the rod axis. The initially straight axis of the rod becomes curved when it is bent. In case of direct bending, the curved axis of the rod lies in the plane of force; in case of oblique bending, it lies in a plane different from the force plane.

Observing the bending of a rubber rod, you can notice that part of its longitudinal fibers are stretched, and the other part is compressed. Obviously, between the stretched and compressed fibers of the rod there is a layer of fibers that experience neither tension nor compression - the so-called neutral layer. The line of intersection of the neutral layer of the rod with the plane of its cross section is called neutral section line.

As a rule, loads acting on a beam can be classified into one of three types: concentrated forces R, concentrated moments M distributed loads of intensity ts(Fig. 5.2). Part I of the beam located between the supports is called in flight, part II of the beam located on one side of the support - console.

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.

Elongation εx fibers

It follows from this 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.