To visually represent the nature of the deformation of beams (rods) during bending, the following experiment is carried out. On the side edges of the rubber beam rectangular section a grid of lines is applied parallel and perpendicular to the axis of the beam (Fig. 30.7, a). Then moments are applied to the beam at its ends (Fig. 30.7, b), acting in the plane of symmetry of the beam, intersecting each of its cross sections along one of the main central axes of inertia. The plane passing through the axis of the beam and one of the main central axes of inertia of each of its cross sections will be called the main plane.

Under the influence of moments, the beam experiences a straight pure bend. As a result of deformation, as experience shows, the grid lines parallel to the axis of the beam are bent, maintaining the same distances between them. When indicated in Fig. 30.7, b in the direction of the moments, these lines in the upper part of the beam are lengthened, and in the lower part they are shortened.

Each grid line perpendicular to the axis of the beam can be considered as a trace of the plane of some cross section of the beam. Since these lines remain straight, it can be assumed that the cross sections of the beam, flat before deformation, remain flat during deformation.

This assumption, based on experience, is known as the hypothesis of plane sections, or Bernoulli's hypothesis (see § 6.1).

The hypothesis of plane sections applies not only to pure but also to transverse bending. For transverse bending it is approximate, and for pure bending it is strict, which is confirmed by theoretical studies carried out using methods of elasticity theory.

Let us now consider a straight beam with a cross section symmetrical with respect to vertical axis, embedded at the right end and loaded at the left end with an external moment acting in one of the main planes of the beam (Fig. 31.7). In each cross section of this beam, only bending moments occur acting in the same plane as the moment

Thus, the beam is in a state of straight, pure bend throughout its entire length. Individual sections of the beam may be in a state of pure bending even if it is subject to transverse loads; for example, section 11 of the beam shown in Fig. experiences pure bending. 32.7; in the sections of this section the shear force

From the beam under consideration (see Fig. 31.7) we select an element of length . As a result of deformation, as follows from Bernoulli's hypothesis, the sections will remain flat, but will tilt relative to each other by a certain angle. Let us take the left section conditionally as stationary. Then, as a result of rotating the right section through an angle, it will take the position (Fig. 33.7).

The straight lines will intersect at a certain point A, which is the center of curvature (or, more precisely, the trace of the axis of curvature) of the longitudinal fibers of the element. The upper fibers of the element in question when shown in Fig. 31.7 in the direction of the moment are lengthened, and the lower ones are shortened. The fibers of some intermediate layer perpendicular to the plane of action of the moment retain their length. This layer is called the neutral layer.

Let us denote the radius of curvature of the neutral layer, i.e., the distance from this layer to the center of curvature A (see Fig. 33.7). Let's consider a certain layer located at a distance y from the neutral layer. The absolute elongation of the fibers of this layer is equal to and the relative elongation

Considering similar triangles we establish that Therefore,

In the theory of bending, it is assumed that the longitudinal fibers of the beam do not press on each other. Experimental and theoretical studies show that this assumption does not significantly affect the calculation results.

With pure bending, shear stresses do not arise in the cross sections of the beam. Thus, all fibers in pure bending are under conditions of uniaxial tension or compression.

According to Hooke's law, for the case of uniaxial tension or compression, the normal stress o and the corresponding relative deformation are related by the dependence

or based on formula (11.7)

From formula (12.7) it follows that the normal stresses in the longitudinal fibers of the beam are directly proportional to their distances y from the neutral layer. Consequently, in the cross section of the beam at each point, the normal stresses are proportional to the distance y from this point to the neutral axis, which is the line of intersection of the neutral layer with the cross section (Fig.

34.7, a). From the symmetry of the beam and the load it follows that the neutral axis is horizontal.

At the points of the neutral axis, normal stresses are zero; on one side of the neutral axis they are tensile, and on the other they are compressive.

The stress diagram o is a graph bounded by a straight line, with the largest absolute values ​​of stress for the points furthest from the neutral axis (Fig. 34.7b).

Let us now consider the equilibrium conditions of the selected beam element. Let us represent the action of the left part of the beam on the section of the element (see Fig. 31.7) in the form of a bending moment; the remaining internal forces in this section with pure bending are equal to zero. Let us imagine the action of the right side of the beam on the cross-section of the element in the form of elementary forces applied to each elementary area of ​​the cross-section (Fig. 35.7) and parallel to the axis of the beam.

Let's create six equilibrium conditions for an element

Here are the sums of the projections of all forces acting on the element, respectively, on the axes - the sums of the moments of all forces relative to the axes (Fig. 35.7).

The axis coincides with the neutral axis of the section and the y-axis is perpendicular to it; both of these axes are located in the cross-sectional plane

An elementary force does not produce projections on the y-axis and does not cause a moment about the axis. Therefore, the equilibrium equations are satisfied for any values ​​of o.

The equilibrium equation has the form

Let us substitute the value of a into equation (13.7) according to formula (12.7):

Since (a curved beam element is considered, for which), then

The integral represents the static moment of the cross section of the beam about the neutral axis. Its equality to zero means that the neutral axis (i.e., the axis) passes through the center of gravity of the cross section. Thus, the center of gravity of all cross sections beam, and consequently the axis of the beam, which is the geometric location of the centers of gravity, are located in the neutral layer. Therefore, the radius of curvature of the neutral layer is the radius of curvature of the curved axis of the beam.

Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element relative to the neutral axis:

Here represents the moment of the elementary internal force relative to the axis.

Let us denote the area of ​​the cross-section of the beam located above the neutral axis - below the neutral axis.

Then it will represent the resultant of elementary forces applied above the neutral axis, below the neutral axis (Fig. 36.7).

Both of these resultants are equal to each other in absolute value, since their algebraic sum, based on condition (13.7), is equal to zero. These resultants form an internal pair of forces acting in the cross section of the beam. The moment of this pair of forces, equal to the product of the magnitude of one of them and the distance between them (Fig. 36.7), is a bending moment in the cross section of the beam.

Let us substitute the value of a into equation (15.7) according to formula (12.7):

Here represents the axial moment of inertia, i.e., the axis passing through the center of gravity of the section. Hence,

Let's substitute the value from formula (16.7) into formula (12.7):

When deriving formula (17.7), it was not taken into account that with an external torque directed, as shown in Fig. 31.7, according to the accepted sign rule, the bending moment is negative. If we take this into account, then we must put a minus sign in front of the right side of formula (17.7). Then, with a positive bending moment in the upper zone of the beam (i.e., at ), the values ​​of a will turn out to be negative, which will indicate the presence of compressive stresses in this zone. However, usually the minus sign is not placed on the right side of formula (17.7), and this formula is used only to determine the absolute values ​​of stresses a. Therefore, the absolute values ​​of the bending moment and the ordinate y should be substituted into formula (17.7). The sign of the stresses is always easily determined by the sign of the moment or by the nature of the deformation of the beam.

Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element relative to the y-axis:

Here it represents the moment of the elementary internal force about the y-axis (see Fig. 35.7).

Let us substitute the value of a into expression (18.7) according to formula (12.7):

Here the integral represents the centrifugal moment of inertia of the cross section of the beam relative to the y and axis. Hence,

But since

As is known (see § 7.5), the centrifugal moment of inertia of the section is equal to zero relative to the main axes of inertia.

In the case under consideration, the y-axis is the axis of symmetry of the cross-section of the beam and, therefore, the y-axes and are the main central axes of inertia of this section. Therefore, condition (19.7) is satisfied here.

In the case when the cross section of the bent beam does not have any axis of symmetry, condition (19.7) is satisfied if the plane of action of the bending moment passes through one of the main central axes of inertia of the section or is parallel to this axis.

If the plane of action of the bending moment does not pass through any of the main central axes of inertia of the cross section of the beam and is not parallel to it, then condition (19.7) is not satisfied and, therefore, there is no direct bending - the beam experiences oblique bending.

Formula (17.7), which determines the normal stress at an arbitrary point of the beam section under consideration, is applicable provided that the plane of action of the bending moment passes through one of the main axes of inertia of this section or is parallel to it. In this case, the neutral axis of the cross section is its main central axis of inertia, perpendicular to the plane of action of the bending moment.

Formula (16.7) shows that during direct pure bending, the curvature of the curved axis of the beam is directly proportional to the product of the elastic modulus E and the moment of inertia. We will call the product the stiffness of the section during bending; it is expressed in, etc.

In pure bending of a beam of constant cross-section, the bending moments and section stiffnesses are constant along its length. In this case, the radius of curvature of the curved axis of the beam has a constant value [see. expression (16.7)], that is, the beam bends along a circular arc.

From formula (17.7) it follows that the largest (positive - tensile) and smallest (negative - compressive) normal stresses in the cross section of the beam arise at the points furthest from the neutral axis, located on both sides of it. For a cross section symmetrical about the neutral axis, the absolute values ​​of the greatest tensile and compressive stresses are the same and can be determined by the formula

For sections that are not symmetrical about the neutral axis, for example, for a triangle, tee, etc., the distances from the neutral axis to the most distant stretched and compressed fibers are different; Therefore, for such sections there are two moments of resistance:

where are the distances from the neutral axis to the most distant stretched and compressed fibers.

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 the 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 that is: 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 of equality to zero longitudinal force 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 (the axis passing through the center of gravity sections). 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, eliminated during the transition 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

Flat transverse bending of beams. Internal bending forces. Differential dependencies of internal forces. Rules for checking diagrams of internal forces during bending. Normal and shear stresses during bending. Strength calculation based on normal and tangential stresses.

10. SIMPLE TYPES OF RESISTANCE. FLAT BEND

10.1. General concepts and definitions

Bending 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.

Plane bending is a 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 of the cross sections and the geometric axis of the beam (x axis).

Oblique bending is a bending in which the loads act in one plane that does not coincide with the main planes of inertia.

Complex bending is a bending in which loads act in different (arbitrary) planes.

10.2. Determination of internal bending forces

Let's consider two typical cases of bending: in the first, the cantilever beam is bent by a concentrated moment M o ; 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 general case of flat bending in the section of a beam, out of six internal forces, two arise - bending moment M z and shear force Q y (or when bending relative to another main axis - bending moment M y and shear force Q z).

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

Pure bending is a flat bending in which only one out of six internal forces occurs in the sections of the rod - 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 relates only to pure bending; 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 the following rule of signs:

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

2) bending moment M z 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, we will build the solution to the problem of determining the internal forces during bending 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 characteristics 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 ≡ M z, Q ≡ Q y.

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 element dx will also be in equilibrium under the action of shear forces, bending moments and external load. Since Q and M generally change along the axis of the beam, transverse forces Q and Q +dQ, as well as bending moments M and M +dM will appear in sections of the element dx. From the equilibrium condition of the selected element we obtain

∑ F y = 0 Q + q dx − (Q + dQ) = 0;

∑ M 0 = 0 M + Q dx + q dx dx 2 − (M + dM ) = 0.

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

Relations (10.1), (10.2) and (10.3) are called differential dependencies 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 shear 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, diagrams Q are limited by inclined straight lines, and diagrams M are limited by quadratic parabolas. Moreover, if we construct diagram M “on a stretched fiber,” then the convexity of the pa-

the work will be directed in the direction of action q, and the extremum will be located in the section where the diagram Q intersects the base line;

c – in sections where a concentrated force is applied to the beam, on diagram Q there will be jumps by the magnitude and in the direction of this force, and on 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 on the epi-

there will be no changes in re Q, and on the diagram M there will be jumps by the value 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 by methods of resistance of materials, it is necessary to introduce some assumptions.

There are three such hypotheses for bending:

a – hypothesis of plane sections (Bernoulli hypothesis)

– sections that are flat 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

niy – 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 – co-

The gray longitudinal fibers do not press on each other.

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 located upper extreme fibers and equals zero on the lower outermost fibers. Main stress σ 3 It 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 both 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, which 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 us 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 before 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, VC 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

On 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, I x– 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):

For determining 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 shear 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 pretend 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 verification 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 limiting 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 it 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. On 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 dimensions of the cross section on the magnitude of the stresses.

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 straight 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

Task. Construct diagrams Q and M for a statically indeterminate beam. Let's calculate the beams using the formula:

n= Σ R- Sh— 3 = 4 — 0 — 3 = 1

Beam once is statically indeterminate, which means one of the reactions is "extra" unknown. Let us take the support reaction as the “extra” unknown IN — R B.

A statically determinate beam, which is obtained from a given one by removing the “extra” connection, is called the main system (b).

Now this system should be presented equivalent given. To do this, load the main system given load, and at the point IN let's apply "extra" reaction R B(rice. V).

However for equivalence this not enough, since in such a beam the point IN Maybe move vertically, and in a given beam (Fig. A ) this cannot happen. Therefore we add condition, What deflection t. IN in the main system should be equal to 0. Deflection t. IN consists of deflection from the active load Δ F and from deflection from the “extra” reaction Δ R.

Then we make up condition for compatibility of movements:

Δ F + Δ R=0 (1)

Now it remains to calculate these movements (deflections).

Loading main system given load(rice .G) and we'll build load diagramM F (rice. d ).

IN T. IN Let's apply and build an ep. (rice. hedgehog ).

Using Simpson's formula we determine deflection due to active load.

Now let's define deflection from the action of “extra” reaction R B , for this we load the main system R B (rice. h ) and build a diagram of the moments from its action M R (rice. And ).

We compose and solve equation (1):

Let's build ep. Q And M (rice. k, l ).

Building a diagram Q.

Let's build a diagram M method characteristic points. We place points on the beam - these are the points of the beginning and end of the beam ( D,A ), concentrated moment ( B ), and also mark the middle of a uniformly distributed load as a characteristic point ( K ) is an additional point for constructing a parabolic curve.

We determine bending moments at points. Rule of signs cm. - .

The moment in IN we will define it as follows. First let's define:

Full stop TO let's take in middle area with a uniformly distributed load.

Building a diagram M . Plot AB – parabolic curve(umbrella rule), area ВD – straight slanted line.

For a beam, determine the support reactions and construct diagrams of bending moments ( M) and shear forces ( Q).

1. We designate supports letters A And IN and direct support reactions R A And R B .

Compiling equilibrium equations.

Examination

Write down the values R A And R B on design scheme.

2. Constructing a diagram shear forces method sections. We arrange the sections on characteristic areas(between changes). According to the dimensional thread - 4 sections, 4 sections.

sec. 1-1 move left.

The section passes through the area with evenly distributed load, mark the size z 1 to the left of the section before the start of the section. The length of the section is 2 m. Rule of signs For Q - cm.

We build according to the found value diagramQ.

sec. 2-2 move on the right.

The section again passes through the area with a uniformly distributed load, mark the size z 2 to the right from the section to the beginning of the section. The length of the section is 6 m.

Building a diagram Q.

sec. 3-3 move on the right.

sec. 4-4 move on the right.

We are building diagramQ.

3. Construction diagrams M method characteristic points.

Feature point- a point that is somewhat noticeable on the beam. These are the points A, IN, WITH, D , and also a point TO , wherein Q=0 And bending moment has an extremum. also in middle console we will put an additional point E, since in this section under a uniformly distributed load the diagram M described crooked line, and it is built at least according to 3 points.

So, the points are placed, let's start determining the values ​​​​in them bending moments. Rule of signs - see.

Sites NA, AD – parabolic curve(the “umbrella” rule for mechanical specialties or the “sail rule” for construction specialties), sections DC, SV – straight slanted lines.

Moment at a point D should be determined both left and right from point D . The very moment in these expressions Excluded. At the point D we get two values ​​with difference by the amount m – leap by its size.

Now we need to determine the moment at the point TO (Q=0). However, first we define point position TO , designating the distance from it to the beginning of the section as unknown X .

T. TO belongs second characteristic area, its equation for shear force(see above)

But the shear force incl. TO equal to 0 , A z 2 equals unknown X .

We get the equation:

Now knowing X, let's determine the moment at the point TO on the right side.

Building a diagram M . The construction can be carried out for mechanical specialties, putting aside positive values up from the zero line and using the “umbrella” rule.

For a given design of a cantilever beam, it is necessary to construct diagrams of the transverse force Q and the bending moment M, and perform a design calculation by selecting a circular section.

Material - wood, design resistance of the material R=10MPa, M=14kN m, q=8kN/m

There are two ways to construct diagrams in a cantilever beam with a rigid embedment - the usual way, having previously determined the support reactions, and without determining the support reactions, if you consider the sections, going from the free end of the beam and discarding the left part with the embedding. Let's build diagrams ordinary way.

1. Let's define support reactions.

Evenly distributed load q replace with conditional force Q= q·0.84=6.72 kN

In a rigid embedment there are three support reactions - vertical, horizontal and moment; in our case, the horizontal reaction is 0.

We'll find vertical ground reaction R A And supporting moment M A from equilibrium equations.

In the first two sections on the right there is no shear force. At the beginning of a section with a uniformly distributed load (right) Q=0, in the background - the magnitude of the reaction R A.
3. To construct, we will compose expressions for their determination in sections. Let's construct a diagram of moments on fibers, i.e. down.

(the diagram of individual moments has already been constructed earlier)

We solve equation (1), reduce by EI

Static indetermination revealed, the value of the “extra” reaction has been found. You can start constructing diagrams of Q and M for a statically indeterminate beam... We sketch the given diagram of the beam and indicate the magnitude of the reaction Rb. In this beam, reactions in the embedment can not be determined if you move from the right.

Construction Q plots for a statically indeterminate beam

Let's plot Q.

Construction of diagram M

Let us define M at the extremum point - at the point TO. First, let's determine its position. Let us denote the distance to it as unknown “ X" Then

We are building a diagram of M.

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 us calculate the static moment for top shelf:

Now let's calculate shear stress:

We are building shear stress diagram:

Design and verification calculations. For a beam with constructed diagrams of internal forces, select a section in the form of two channels from the condition of strength under normal stresses. Check the strength of the beam using the shear stress strength condition and the energy strength criterion. Given:

Let's show a beam with constructed diagrams Q and M

According to the diagram of bending moments, dangerous is section C, in which M C = M max = 48.3 kNm.

Normal stress strength condition for this beam has the form σ max =M C /W X ≤σ adm . It is necessary to select a section from two channels.

Let's determine the required calculated value axial moment of resistance of the section:

For a section in the form of two channels, we accept according to two channels No. 20a, moment of inertia of each channel I x =1670cm 4, Then axial moment of resistance of the entire section:

Overvoltage (undervoltage) at dangerous points we calculate using the formula: Then we get undervoltage:

Now let's check the strength of the beam based on strength conditions for tangential stresses. According to shear force diagram dangerous are sections on section BC and section D. As can be seen from the diagram, Q max =48.9 kN.

Strength condition for tangential stresses has the form:

For channel No. 20 a: static moment of area S x 1 = 95.9 cm 3, moment of inertia of the section I x 1 = 1670 cm 4, wall thickness d 1 = 5.2 mm, average flange thickness t 1 = 9.7 mm , channel height h 1 =20 cm, shelf width b 1 =8 cm.

For transverse sections of two channels:

S x = 2S x 1 =2 95.9 = 191.8 cm 3,

I x =2I x 1 =2·1670=3340 cm 4,

b=2d 1 =2·0.52=1.04 cm.

Determining the value maximum shear stress:

τ max =48.9 10 3 191.8 10 −6 /3340 10 −8 1.04 10 −2 =27 MPa.

As seen, τ max<τ adm (27MPa<75МПа).

Hence, the strength condition is satisfied.

We check the strength of the beam according to the energy criterion.

From consideration diagrams Q and M follows that section C is dangerous, in which they operate M C =M max =48.3 kNm and Q C =Q max =48.9 kN.

Let's carry out analysis of the stress state at the points of section C

Let's define normal and shear stresses at several levels (marked on the section diagram)

Level 1-1: y 1-1 =h 1 /2=20/2=10cm.

Normal and tangent voltage:

Main voltage:

Level 2−2: y 2-2 =h 1 /2−t 1 =20/2−0.97=9.03 cm.

Main voltages:

Level 3−3: y 3-3 =h 1 /2−t 1 =20/2−0.97=9.03 cm.

Normal and shear stresses:

Main voltages:

Extreme shear stress:

Level 4−4: y 4-4 =0.

(in the middle the normal stresses are zero, the tangential stresses are maximum, they were found in the strength test using tangential stresses)

Main voltages:

Extreme shear stress:

Level 5−5:

Normal and shear stresses:

Main voltages:

Extreme shear stress:

Level 6−6:

Normal and shear stresses:

Main voltages:

Extreme shear stress:

Level 7−7:

Normal and shear stresses:

Main voltages:

Extreme shear stress:

In accordance with the calculations performed stress diagrams σ, τ, σ 1, σ 3, τ max and τ min are presented in Fig.

Analysis these diagram shows, which is in the section of the beam dangerous points are at level 3-3 (or 5-5), in which:

Using energy criterion of strength, we get

From a comparison of equivalent and permissible stresses it follows that the strength condition is also satisfied

(135.3 MPa<150 МПа).

The continuous beam is loaded in all spans. Construct diagrams Q and M for a continuous beam.

1. Define degree of static indetermination beams according to the formula:

n= Sop -3= 5-3 =2, Where Sop – number of unknown reactions, 3 – number of static equations. To solve this beam it is required two additional equations.

2. Let us denote numbers supports from zero in order ( 0,1,2,3 )

3. Let us denote span numbers from the first in order ( ι 1, ι 2, ι 3)

4. We consider each span as simple beam and build diagrams for each simple beam Q and M. What pertains to simple beam, we will denote with index "0", that which relates to continuous beam, we will denote without this index. Thus, is the shear force and bending moment for a simple beam.