Straight bend- this is a type of deformation in which two internal force factors arise in the cross-sections of the bar: bending moment and shear force.

Pure bend - this is special case direct bending, in which only a bending moment occurs in the cross-sections of the bar, and the shear force is zero.

An example of a pure bend - a stretch CD on the rod AB. Bending moment Is the value Pa a pair of external forces causing bending. From the equilibrium of the part of the bar 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 to and opposite to the directional bending moment Pa.

To find the distribution of these internal forces over the cross section, it is necessary to consider the deformation of the bar.

In the simplest case, the bar has a longitudinal plane of symmetry and is subjected to the action of external bending pairs of forces located in this plane. Then the bending will occur in the same plane.

Bar axis nn 1 Is a line passing through the centers of gravity of its cross-sections.

Let the cross-section of the bar be a rectangle. Let's draw two vertical lines on its edges. mm and pp... When bent, these lines remain straight and rotate so that they remain perpendicular to the longitudinal fibers of the bar.

Further bending theory is based on the assumption that not only lines mm and pp, but the entire flat cross-section of the bar remains flat after bending and normal to the longitudinal fibers of the bar. Therefore, when 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 undergo tension, and the fibers on the concave side undergo compression.

Neutral surface Is a surface that does not undergo bending deformation. (Now it is perpendicular to the drawing, the deformed axis of the bar nn 1 belongs to this surface).

Neutral section axis- this is the intersection of a neutral surface with any one with any cross section (now it is also located perpendicular to the drawing).

Let an arbitrary fiber be at a distance y from a neutral surface. ρ Is the radius of curvature of the curved axis. Point O Is the center of curvature. Let's draw a line n 1 s 1 parallel mm.ss 1- absolute fiber elongation.

Relative extension ε x fiber

It follows that deformation of longitudinal fibers proportional to distance y from the neutral surface and inversely proportional to the radius of curvature ρ .

Longitudinal elongation of the fibers of the convex side of the rod is accompanied by lateral constriction, 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 is greater than in absolute value ε z, and we get

These strains 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 the 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 bar does not have a longitudinal plane of symmetry, in which the bending moment acts, as long as only the bending moment acts in the axial plane, which contains one of the two main axes cross section. These planes are called principal bending planes.

When there is a plane of symmetry and the bending moment acts in this plane, the deflection occurs precisely in it. Moments of internal forces relative to the axis z balance the external moment M... Moments of forces relative to the axis y mutually destroyed.

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

Essentially, lateral bend is a combination of pure bend and shear. In connection with the curvature of the cross-sections due to the uneven distribution of shears along the height, the question arises of the possibility of applying the formula for the normal stress σ NS derived for pure bend based on the hypothesis of flat sections.

1 A single-span beam, having at the ends, respectively, one cylindrical fixed support and one cylindrical movable in the direction of the beam axis, is called simple... A beam with one restrained and the other free end is called console... A simple beam with one or two parts hanging over the 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 bar), then, as in the case of pure bending, it can be assumed that the fibers do not exert pressure on each other. This means that each fiber undergoes uniaxial tension or compression.

Under the action of a distributed load, the lateral 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 put pressure on each other. A careful study of the question shows that if the length of the bar l is large enough compared to its height h (l/ h> 5), then even with a distributed load, these factors do not significantly affect 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 the sections under concentrated loads and near them, the distribution of σ NS deviates from the linear law. This deviation, which is of a local nature and is not accompanied by an increase in the highest stresses (in the extreme fibers), is usually not taken into account in practice.

Thus, with transverse bending (in the plane hu) normal stresses are calculated by the formula

σ NS= – [M z(x)/I z]y.

If we draw two adjacent sections on a section of the beam free from the load, then the transverse force in both sections will be the same, which means that 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 magnitude of the normal stress.

Let us determine the shear stresses in the cross section through their paired stresses acting in the longitudinal section of the bar.

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

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. With this in mind, under the assumption that the shear stresses in the site b distributed uniformly, we use the condition ΣX = 0, we get:

N * - (N * + dN *) +

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

Then you 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 calculated results obtained using it are in good agreement with the experimental data.

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

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

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

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

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

Let us define, as an example, shear stresses in a rectangular cross section (Fig. 10.6, c). Static moment about the axis z part of the section above the line 1-1, on which the voltage is determined, we will write in the form:

It changes according to the law of a square parabola. Section width v for a rectangular bar is constant, then the law of change in shear stresses in the section will also be parabolic (Figure 10.6, c). At y = and y = - the tangential stresses are equal to 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.

Straight transverse bend arises in the case when all loads are applied perpendicular to the axis of the bar, lie in the same plane and, in addition, the plane of their action coincides with one of the main central axes of inertia of the section. Straight transverse bending refers to a simple type of resistance and is flat stress state, i.e. the two main voltages are nonzero. With this type of deformation, internal forces arise: shear force and bending moment. A special case of direct transverse bending is clean bend, with such a resistance, there are load sections, within which the lateral force becomes zero, and the bending moment is nonzero. Normal and tangential stresses arise in the cross-sections of the rods under direct transverse bending. Stresses are a function of internal stress, in this case normal stresses are a function of bending moment, and tangential stresses are a function of shear force. For direct transverse bending, several hypotheses are introduced:

1) The cross-sections of the beam, flat before deformation, remain flat and orthogonal to the neutral layer after deformation (the hypothesis of flat sections or the hypothesis of J. Bernoulli). This hypothesis holds for pure bending and is violated when shear forces, shear stresses, and angular deformation occur.

2) There is no mutual pressure between the longitudinal layers (hypothesis of non-compression of fibers). It follows from this hypothesis that longitudinal fibers experience uniaxial tension or compression; therefore, Hooke's law is valid in pure bending.

The rod undergoing bending is called beam... When bending, one part of the fibers is stretched, the other part is compressed. The layer of fibers between the stretched and compressed fibers is called neutral layer, it passes through the center of gravity of the sections. The line of its intersection with the cross-section of the beam is called neutral axis... On the basis of the hypotheses introduced for pure bending, a formula for determining normal stresses is obtained, which is also applied for direct transverse bending. The normal stress can be found using the linear relationship (1), in which the ratio of the bending moment to the axial moment of inertia (
) in a particular section is a constant value, and the distance ( y) along the ordinate from the center of gravity of the section to the point at which the stress is determined varies from 0 to
.

. (1)

To determine the shear stress in bending in 1856. Russian engineer - bridge builder D.I. Zhuravsky received the dependence

. (2)

The tangential stress in a particular section does not depend on the ratio of the shear force to the axial moment of inertia (
), because this value does not change within one section, but depends on the ratio of the static moment of the area of ​​the cut-off part to the width of the section at the level of the cut-off part (
).

In direct transverse bending, displacement: deflections (v ) and rotation angles (Θ ) ... To determine them, the equations of the method of initial parameters (3) are used, which are obtained by integrating the differential equation of the curved axis of the beam (
).

Here v 0 , Θ 0 ,M 0 , Q 0 - initial parameters, x – distance from the origin to the section in which the displacement is defined , a- the distance from the origin of coordinates to the place of application or the beginning of the action of the load.

Strength and stiffness calculations are made using strength and stiffness conditions. Using these conditions, you can solve verification tasks (check whether the condition is fulfilled), determine the size of the cross-section, or select the permissible value of the load parameter. Several strength conditions are distinguished, some of which are given below. Strength condition for normal stresses looks like:

, (4)

here
– moment of resistance of the section relative to the z-axis, R - design resistance for normal voltages.

Tensile strength condition looks like:

, (5)

here the notation is the same as in the Zhuravsky formula, and R s - design shear resistance or design shear stress resistance.

Strength condition according to the third strength hypothesis or the hypothesis of the highest shear stresses can be written in the following form:

. (6)

Stiffness conditions can be written for deflections (v ) and rotation angles (Θ ) :

where the displacement values ​​in square brackets are valid.

An example of an individual task number 4 (term 2-8 weeks)

Bending the type of loading of the bar is called, in which a moment is applied to it, lying in the plane passing through the longitudinal axis. Bending moments arise in the cross-sections of the beam. When bending, deformation occurs, in which the axis of the straight bar bends or the curvature of the curved bar changes.

Bending beam is called beam ... A structure consisting of several bending rods connected to each other most often at an angle of 90 ° is called frame .

The bend is called flat or straight if the plane of action of the load passes through the main central axis of inertia of the section (Figure 6.1).

Figure 6.1

With flat transverse bending, two types of internal forces arise in the beam: transverse force Q and bending moment M... In the frame with a flat transverse bending, three forces arise: longitudinal N, transverse Q forces and bending moment M.

If the bending moment is the only internal force factor, then such a bending is called clean (Figure 6.2). In the presence of a lateral force, the bend is called transverse ... Strictly speaking, to simple types resistance applies only to pure bending; transverse bending is conventionally referred to as simple types of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected in strength calculations.

22.Flat lateral bend. Differential relationships between internal forces and external load. There are differential dependences between the bending moment, shear force and the intensity of the distributed load, based on the Zhuravsky theorem, named after the Russian bridge engineer D.I. Zhuravsky (1821-1891).

This theorem is formulated as follows:

The transverse force is equal to the first derivative of the bending moment along the abscissa of the beam section.

23. Flat transverse bend. Plotting shear forces and bending moments. Determination of Shear Forces and Bending Moments - Section 1

Let's discard right side of the beam and replace its action with the left-hand side with a shear force and a bending moment. For the convenience of the calculation, we cover the discarded right part of the beam with a sheet of paper, aligning the left edge of the sheet with the section under consideration 1.

The transverse force in section 1 of the beam is equal to the algebraic sum of all external forces that we see after closing

We see only the reaction of the support directed downward. Thus, the shear force is:

kN.

We took the "minus" sign because the force rotates the part of the beam we see relative to the first section counterclockwise (or because it is equally directed with the direction of the transverse force according to the rule of signs)

The bending moment in section 1 of the beam is equal to the algebraic sum of the moments of all the forces that we see after closing the discarded part of the beam, relative to the considered section 1.

We see two efforts: the reaction of the support and the moment M. However, the shoulder of the force is practically equal to zero. Therefore, the bending moment is:

kN m.

Here we have taken the plus sign because the external moment M bends the visible part of the beam with a convexity downward. (or because it is oppositely directed to the direction of the bending moment according to the rule of signs)

Determination of Shear Forces and Bending Moments - Section 2

Unlike the first section, the reaction force has a shoulder equal to a.

lateral force:

kN;

bending moment:

Determination of Shear Forces and Bending Moments - Section 3

lateral force:

bending moment:

Determination of Shear Forces and Bending Moments - Section 4

More convenient now cover the left side of the beam with a leaf.

lateral force:

bending moment:

Determination of Shear Forces and Bending Moments - Section 5

lateral force:

bending moment:

Determination of Shear Forces and Bending Moments - Section 1

shear force and bending moment:

.

Using the found values, we construct a diagram of transverse forces (Fig. 7.7, b) and bending moments (Fig. 7.7, c).

CONTROL OF THE CORRECT CONSTRUCTION OF EPURES

Let us make sure of the correctness of plotting diagrams by external signs, using the rules for constructing diagrams.

Checking the Shear Force Plot

We are convinced: under the unloaded sections, the shear force diagram runs parallel to the beam axis, and under the distributed load q - along a straight line inclined downward. On the longitudinal force diagram there are three jumps: under the reaction — down by 15 kN, under the P force — down by 20 kN, and under the reaction — up by 75 kN.

Checking the Bending Moment Plot

On the diagram of bending moments, we see kinks under the concentrated force P and under the support reactions. The angles of the kinks are directed towards these forces. Under a distributed load q, the bending moment diagram changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6, there is an extremum on the bending moment diagram, since the shear force diagram at this point passes through a zero value.

Straight bend. Plane transverse bending Plotting internal force factors for beams Plotting Q and M plots using equations Plotting Q and M plots from characteristic sections (points) Strength calculations for direct bending of beams Principal bending stresses. Full check of the strength of beams Understand the center of bend Determination of displacements in beams during bending. Concepts of deformation of beams and conditions of their rigidity Differential equation of the curved axis of the beam Direct integration method Examples of determining displacements in beams by the direct integration method Physical meaning of the integration constants Method of initial parameters (universal equation of the curved axis of the beam). Examples of defining displacements in a beam by the method of initial parameters Determining displacements by Mohr's method. Rule A.K. Vereshchagin. Calculation of the Mohr integral according to the rule of A.K. Vereshchagin Examples of determining displacements by means of Mohr's integral Bibliography Direct bend. Flat lateral bend. 1.1. Plotting internal force factors for beams Direct bending is a type of deformation in which two internal force factors arise in the cross-sections of the bar: bending moment and shear force. In a particular case, the shear force can be equal to zero, then the bend is called pure. With plane transverse bending, all forces are located in one of the main planes of inertia of the rod and are perpendicular to its longitudinal axis, moments are located in the same plane (Fig. 1.1, a, b). Rice. 1.1 The transverse force in an arbitrary cross-section of the beam is numerically equal to the algebraic sum of projections onto the normal to the beam axis of all external forces acting on one side of the section under consideration. Shear force in section m-n beams (Fig. 1.2, a) is considered positive if the resultant of external forces to the left of the section is directed upward, and on the right - downward, and negative - in the opposite case (Fig. 1.2, b). Rice. 1.2 When calculating the shear force in a given section, the external forces lying to the left of the section are taken with a plus sign if they are directed upwards, and with a minus sign if downwards. The opposite is true for the right side of the beam. 5 The bending moment in an arbitrary cross-section of the beam is numerically equal to the algebraic sum of moments about the central z-axis of the section of all external forces acting on one side of the section under consideration. Bending moment in section m-n beams (Fig. 1.3, a) is considered positive if the resultant moment of external forces to the left of the section is directed along the clock hand, and on the right - counterclockwise, and negative - in the opposite case (Fig. 1.3, b). Rice. 1.3 When calculating the bending moment in a given section, the moments of external forces lying to the left of the section are considered positive if they are directed clockwise. The opposite is true for the right side of the beam. It is convenient to determine the sign of the bending moment by the nature of the deformation of the beam. The bending moment is considered positive if, in the section under consideration, the cut-off part of the beam is bent downward, i.e., the lower fibers are stretched. Otherwise, the bending moment in the section is negative. Differential relationships exist between the bending moment M, the shear force Q and the load intensity q. 1. The first derivative of the shear force along the abscissa of the section is equal to the intensity of the distributed load, i.e. ... (1.1) 2. The first derivative of the bending moment along the abscissa of the section is equal to the transverse force, ie. (1.2) 3. The second derivative with respect to the abscissa of the section is equal to the intensity of the distributed load, ie. (1.3) The distributed load directed upwards is considered positive. A number of important conclusions follow from the differential dependencies between M, Q, q: 1. If on a section of the beam: a) the transverse force is positive, then the bending moment increases; b) the transverse force is negative, then the bending moment decreases; c) the shear force is zero, then the bending moment has a constant value (pure bending); 6 d) the transverse force passes through zero, changing sign from plus to minus, max M M, in the opposite case M Mmin. 2. If there is no distributed load on the section of the beam, then the shear force is constant, and the bending moment changes linearly. 3. If there is a uniformly distributed load on a section of the beam, then the shear force changes according to a linear law, and the bending moment - according to the law of a square parabola facing the convexity towards the load (in the case of plotting an M diagram from the side of stretched fibers). 4. In the section under the concentrated force, the diagram Q has a jump (by the magnitude of the force), the diagram M has a kink in the direction of the force. 5. In the section where the concentrated moment is applied, the diagram M has a jump equal to the value of this moment. This is not reflected in the Q plot. With complex loading of the beam, diagrams of shear forces Q and bending moments M are plotted. Diagram Q (M) is a graph showing the law of change of the shear force (bending moment) along the length of the beam. Based on the analysis of the M and Q diagrams, dangerous sections of the beam are established. Positive ordinates of the Q plot are plotted upward, and negative ordinates are plotted downward from the baseline drawn parallel to the longitudinal axis of the beam. The positive ordinates of the M plot are laid down, and the negative ones - up, that is, the M plot is built from the side of the stretched fibers. The construction of Q and M diagrams for beams should begin with defining the support reactions. For a beam with one restrained and the other free ends, the construction of Q and M diagrams can be started from the free end without defining the reactions in the embedment. 1.2. Plotting Q and M diagrams according to the equations The beam is divided into sections, within which the functions for the bending moment and shear force remain constant (do not have discontinuities). The boundaries of the sections are the points of application of concentrated forces, pairs of forces and places of change in the intensity of the distributed load. At each site, an arbitrary section is taken at a distance x from the origin, and equations for Q and M are drawn up for this section. These equations are used to construct diagrams Q and M. Example 1.1 Construct diagrams of shear forces Q and bending moments M for a given beam (Fig. 1.4, a). Solution: 1. Determination of support reactions. We compose the equilibrium equations: from which we obtain The reactions of the supports are defined correctly. The beam has four sections Fig. 1.4 loads: CA, AD, DB, BE. 2. Plotting Q. Plot CA. On the CA 1 section, we draw an arbitrary section 1-1 at a distance x1 from the left end of the beam. We define Q as the algebraic sum of all external forces acting to the left of the section 1-1: The minus sign is taken because the force acting to the left of the section is directed downward. The expression for Q is independent of the variable x1. Diagram Q in this area will be depicted as a straight line parallel to the abscissa axis. Plot AD. On the site, we draw an arbitrary section 2-2 at a distance x2 from the left end of the beam. We define Q2 as the algebraic sum of all external forces acting to the left of the section 2-2: 8. The value of Q is constant in the section (does not depend on the variable x2). The plot Q on the site is a straight line parallel to the abscissa axis. Plot DB. On the site, we draw an arbitrary section 3-3 at a distance x3 from the right end of the beam. We define Q3 as the algebraic sum of all external forces acting to the right of section 3-3: The resulting expression is the equation of an inclined straight line. Plot BE. On the site, we make a section 4-4 at a distance x4 from the right end of the beam. We define Q as the algebraic sum of all external forces acting to the right of section 4-4: 4 Here the plus sign is taken because the resultant load to the right of section 4-4 is directed downward. Based on the obtained values, we plot the diagrams Q (Fig. 1.4, b). 3. Plotting M. Plot m1. We define the bending moment in section 1-1 as the algebraic sum of the moments of forces acting to the left of section 1-1. - equation of a straight line. Section A 3 Define the bending moment in section 2-2 as the algebraic sum of the moments of forces acting to the left of section 2-2. - equation of a straight line. Section DB 4 Define the bending moment in section 3-3 as the algebraic sum of the moments of forces acting to the right of section 3-3. - the equation of a square parabola. 9 Find three values ​​at the ends of the section and at a point with coordinate xk, where Section BE 1 Determine the bending moment in section 4-4 as the algebraic sum of the moments of forces acting to the right of section 4-4. - the equation of a square parabola, we find three values ​​of M4: Using the obtained values, we construct a diagram of M (Fig. 1.4, c). In sections CA and AD, the Q plot is limited by straight lines parallel to the abscissa axis, and in sections DB and BE - by inclined straight lines. In sections C, A and B on the plot Q, there are jumps by the value of the corresponding forces, which serves as a check of the correctness of plotting the plot Q. In the sections where Q  0, the moments increase from left to right. On the sections where Q  0, the moments decrease. Under the concentrated forces there are kinks towards the action of the forces. Under the concentrated moment, there is a jump by the magnitude of the moment. This indicates the correctness of plotting M. Example 1.2 Construct diagrams Q and M for a beam on two supports, loaded with a distributed load, the intensity of which varies linearly (Fig. 1.5, a). Solution Determination of support reactions. The resultant of the distributed load is equal to the area of ​​the triangle representing the load diagram and is applied at the center of gravity of this triangle. We compose the sums of the moments of all forces relative to points A and B: Plotting a diagram Q. Let's draw an arbitrary section at a distance x from the left support. The ordinate of the load diagram corresponding to the section is determined from the similarity of the triangles The resultant of that part of the load that is located to the left of the section The transverse force in the section is equal to The transverse force varies according to the law of a square parabola Equating the equation of the transverse force to zero, we find the abscissa of the section in which the diagram Q passes through zero: Diagram Q is shown in Fig. 1.5, b. The bending moment in an arbitrary section is equal to The bending moment changes according to the law of a cubic parabola: The bending moment has a maximum value in the section, where 0, i.e. at Diagram M is shown in Fig. 1.5, c. 1.3. Plotting Q and M diagrams by characteristic sections (points) Using the differential dependencies between M, Q, q and the conclusions arising from them, it is advisable to plot Q and M diagrams by characteristic sections (without drawing up equations). Using this method, the values ​​of Q and M are calculated in the characteristic sections. Typical sections are the boundary sections of the sections, as well as sections where the given internal force factor is of extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of the differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct plots Q and M for the beam shown in fig. 1.6, a. Rice. 1.6. Solution: We start plotting Q and M diagrams from the free end of the beam, while the reactions in the embedding can be omitted. The beam has three loading areas: AB, BC, CD. There is no distributed load on sections AB and BC. The lateral forces are constant. Plot Q is limited by straight lines parallel to the abscissa axis. Bending moments change linearly. Diagram M is limited by straight lines inclined to the abscissa axis. There is a uniformly distributed load on the CD section. Transverse forces change linearly, and bending moments - according to the law of a square parabola with a bulge in the direction of a distributed load. On the border of sections AB and BC, the lateral force changes abruptly. At the boundary of the sections BC and CD, the bending moment changes abruptly. 1. Plotting Q. We calculate the values ​​of the shear forces Q in the boundary sections of the sections: Based on the results of the calculations, we plot the Q plot for the beam (Fig. 1, b). From the diagram Q it follows that the transverse force on the section CD is equal to zero in the section spaced at a distance qa a q from the beginning of this section. In this section, the bending moment has a maximum value. 2. Construction of the M diagram. We calculate the values ​​of the bending moments in the boundary sections of the sections: At the maximum moment in the section. Based on the results of the calculations, we construct the M diagram (Fig. 5.6, c). Example 1.4 For a given diagram of bending moments (Fig. 1.7, a) for a beam (Fig. 1.7, b), determine the acting loads and build a diagram Q. The circle denotes the vertex of a square parabola. Solution: Determine the loads acting on the beam. The AC section is loaded with a uniformly distributed load, since the M diagram in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting clockwise, since on the diagram M we have a jump upward by the magnitude of the moment. On the NE section, the beam is not loaded, since the M diagram in this section is bounded by an inclined straight line. The reaction of support B is determined from the condition that the bending moment in section C is equal to zero, i.e., to determine the intensity of the distributed load, we compose an expression for the bending moment in section A as the sum of the moments of forces on the right and equate to zero Now we define the reaction of support A. For this Let us compose an expression for the bending moments in the section as the sum of the moments of forces on the left.The design diagram of a beam with a load is shown in Fig. 1.7, c. Starting from the left end of the beam, we calculate the values ​​of the shear forces in the boundary sections of the sections: Diagram Q is shown in Fig. 1.7, d. The considered problem can be solved by drawing up functional dependencies for M, Q at each site. Select the origin at the left end of the beam. On the section AC, the diagram M is expressed by a square parabola, the equation of which has the form Constants a, b, c are found from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the equation of the parabola, we obtain: The expression for the bending moment will be Differentiating the function M1 , we obtain the dependence for the transverse force After differentiating the function Q, we obtain the expression for the intensity of the distributed load On the section CB, the expression for the bending moment is represented as a linear function To determine the constants a and b, we use the conditions that this straight line passes through two points whose coordinates are known. We obtain two equations:, b from which we have a 20. The equation for the bending moment on the section CB will be After two-fold differentiation of M2, we will find By the found values ​​of M and Q, we plot the diagrams of bending moments and shear forces for the beam. In addition to the distributed load, concentrated forces are applied to the beam in three sections, where there are jumps on the Q diagram and concentrated moments in the section where there is a jump on the M diagram. Example 1.5 For a beam (Fig. 1.8, a), determine the rational position of the hinge C, at which the greatest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Build Q and M diagrams. Solution Determination of support reactions. Although total number the support ties are equal to four, the beam is statically definable. The bending moment in the hinge C is equal to zero, which allows us to draw up an additional equation: the sum of the moments relative to the hinge of all external forces acting on one side of this hinge is zero. Let us compose the sum of the moments of all forces to the right of the hinge C. Diagram Q for the beam is bounded by an inclined straight line, since q = const. We determine the values ​​of the shear forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation whence Diagram M for the beam is bounded by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are written accordingly as follows: From the condition of equality of moments, we obtain quadratic equation relative to the sought parameter x: The real value x2x 1, 029 m. Determine the numerical values ​​of the shear forces and bending moments in the characteristic sections of the beam Figure 1.8, b shows the diagram Q, and Fig. 1.8, c - diagram M. The considered problem could be solved by dividing the hinged beam into its constituent elements, as shown in Fig. 1.8, d. At the beginning, the reactions of the supports VC and VB are determined. Diagrams Q and M are plotted for the suspended beam CB from the action of the load applied to it. Then they go to the main beam of the AC, loading it with an additional force VC, which is the force of pressure of the CB beam on the AC beam. Then the diagrams Q and M are plotted for the AC beam. 1.4. Strength calculations for direct bending of beams Strength calculations for normal and shear stresses. With direct bending of the beam, normal and tangential stresses arise in its cross sections (Fig. 1.9). Fig. 18 1.9 Normal stresses are associated with a bending moment, shear stresses are associated with a shear force. In straight pure bending, the shear stresses are zero. Normal stresses at an arbitrary point of the cross-section of the beam are determined by the formula (1.4) where M is the bending moment in the given section; Iz is the moment of inertia of the section relative to the neutral axis z; y is the distance from the point where the normal stress is determined to the neutral z-axis. Normal stresses along the height of the section change linearly and reach the greatest value at the points farthest from the neutral axis If the section is symmetrical about the neutral axis (Fig. 1.11), then Fig. 1.11 the largest tensile and compressive stresses are the same and are determined by the formula,  is the axial moment of resistance of the section in bending. For a rectangular section of width b and height h: (1.7) For a circular section of diameter d: (1.8) For an annular section   - respectively internal and outer diameters rings. For beams made of plastic materials, the most rational are symmetrical 20 sectional shapes (I-beams, box-shaped, annular). For beams made of brittle materials that are not equally resistant to tension and compression, sections that are asymmetric with respect to the neutral z-axis (T, U-shaped, asymmetrical I-beam) are rational. For beams of constant cross-section made of plastic materials with symmetric cross-sectional shapes, the strength condition is written as follows: (1.10) where Mmax is the maximum bending moment modulo; - allowable stress for the material. For beams of constant cross-section made of plastic materials with asymmetric sectional shapes, the strength condition is written in the following form: (1.11) For beams made of brittle materials with sections that are asymmetric about the neutral axis, if the M diagram is unambiguous (Figure 1.12), you need to write two strength conditions - the distance from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; P - allowable stresses in tension and compression, respectively. Figure 1.12. 21 If the diagram of bending moments has sections of different signs (Fig. 1.13), then in addition to checking section 1-1, where Mmax acts, it is necessary to calculate the largest tensile stresses for section 2-2 (with the largest moment of the opposite sign). Rice. 1.13 Along with the basic calculation for normal stresses, in some cases it is necessary to check the strength of the beam in terms of shear stresses. Shear stresses in the beams are calculated by the formula of DI Zhuravsky (1.13) where Q is the shear force in the considered cross-section of the beam; Szotc - static moment relative to the neutral axis of the area of ​​a part of the section located on one side of a straight line drawn through a given point and parallel to the z axis; b is the width of the section at the level of the point under consideration; Iz is the moment of inertia of the entire section relative to the neutral z axis. In many cases, the maximum shear stresses occur at the level of the neutral layer of the beam (rectangle, I-beam, circle). In such cases, the shear stress strength condition is written in the form, (1.14) where Qmax is the largest shear force in modulus; Is the permissible shear stress for the material. For a rectangular section of a beam, the strength condition has the form (1.15) A is the cross-sectional area of ​​the beam. For a circular section, the strength condition is represented in the form (1.16) For an I-section, the strength condition is written as follows: (1. 17) where Szо, тmсax is the static half-section moment relative to the neutral axis; d - wall thickness of the I-beam. Usually, the dimensions of the cross-section of a beam are determined from the condition of strength with respect to normal stresses. Checking the strength of beams for shear stresses is mandatory for short beams and beams of any length, if there are large concentrated forces near the supports, as well as for wooden, riveted and welded beams. Example 1.6 Check the strength of a box-section beam (Fig. 1.14) for normal and shear stresses, if MPa. Plot the dangerous section of the beam. Rice. 1.14 Solution 23 1. Plotting Q and M diagrams using characteristic sections. Considering the left side of the beam, we obtain the Diagram of transverse forces is shown in Fig. 1.14, c. The diagram of bending moments is shown in Fig. 5.14, g. 2. Geometric characteristics of the cross-section 3. The highest normal stresses in the section C, where Mmax acts (modulo): MPa. The maximum normal stresses in the beam are practically equal to the permissible ones. 4. The largest shear stresses in section C (or A), where max Q acts (modulo): Here is the static moment of the half-section area relative to the neutral axis; b2 cm - section width at the level of the neutral axis. 5. Shear stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 cm3 is the static moment of the area of ​​the part of the section located above the line passing through the point K1; b2 cm - wall thickness at the level of point K1. Diagrams  and  for section C of the beam are shown in Fig. 1.15. Example 1.7 For the beam shown in fig. 1.16, a, it is required: 1. Construct diagrams of shear forces and bending moments by characteristic sections (points). 2. Determine the dimensions of the cross-section in the form of a circle, rectangle and I-beam from the condition of strength with respect to normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of the cross-sections of the beams in terms of shear stress. Given: Solution: 1. Determine the reactions of the beam supports Check: 2. Plotting the diagrams Q and M. The values ​​of the shear forces in the characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, the intensity of the load is q = const. Consequently, in these areas, the Q diagram is limited by straight lines inclined to the axis. In the section DB, the intensity of the distributed load q = 0, therefore, in this section of the diagram Q is limited by a straight line parallel to the x axis. The Q plot for the beam is shown in Fig. 1.16, b. The values ​​of the bending moments in the characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section, in which Q = 0: The maximum moment in the second section The diagram M for the beam is shown in Fig. 1.16, c. 2. We formulate the strength condition for normal stresses from where we determine the required axial moment of resistance of the section from the expression determined required diameter d of the circular section Area of ​​the circular section For the rectangular section The required section height Rectangular section Determine the required number of the I-beam. According to the tables of GOST 8239-89, we find the nearest greater importance axial moment of resistance 597 cm3, which corresponds to I-beam No. 33 with characteristics: A z 9840 cm4. Check for tolerance: (underloading by 1% of the permissible 5%) the nearest I-beam No. 30 (W 2 cm3) leads to a significant overload (more than 5%). Finally, we accept I-beam No. 33. We compare the areas of circular and rectangular sections with the smallest area A of the I-beam: Of the three considered sections, the I-section is the most economical. 3. We calculate the highest normal stresses in the dangerous section of the 27 I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-beam section The diagram of normal stresses in the dangerous section of the beam is shown in Fig. 1.17, b. 5. Determine the highest shear stresses for the selected sections of the beam. a) rectangular section beams: b) circular section of the beam: c) I-section of the beam: Shear stresses in the wall near the flange of the I-beam in the dangerous section A (right) (at point 2): The diagram of shear stresses in the dangerous sections of the I-beam is shown in Fig. 1.17, c. The maximum shear stresses in the beam do not exceed the permissible stresses Example 1.8 Determine the permissible load on the beam (Figure 1.18, a), if 60 MPa, the cross-sectional dimensions are given (Figure 1.19, a). Construct a diagram of normal stresses in the dangerous section of the beam at the permissible load. Figure 1.18 1. Determination of the reactions of the beam supports. Due to the symmetry of the system 2. Construction of diagrams Q and M on characteristic sections. Shear forces in characteristic sections of the beam: Diagram Q for the beam is shown in Fig. 5.18, b. Bending moments in characteristic sections of the beam For the second half of the beam, the ordinates M are along the axes of symmetry. Diagram M for a beam is shown in Fig. 1.18, b. 3. Geometric characteristics of the section (Fig. 1.19). We divide the figure into two simplest elements: an I-beam - 1 and a rectangle - 2. Fig. 1.19 According to the assortment for I-beam No. 20, we have For a rectangle: Static moment of the section area relative to the z1 axis Distance from the z1 axis to the center of gravity of the section Moment of inertia of the section relative to the main central z axis of the entire section according to the formulas for the transition to parallel axes 4. The condition of strength under normal stresses for dangerous point "a" (Fig. 1.19) in the dangerous section I (Fig. 1.18): After substitution of the numerical data 5. With the permissible load in the dangerous section, the normal stresses at points "a" and "b" will be equal: Diagram of normal stresses for dangerous section 1-1 is shown in Fig. 1.19, b.