On bending on them constantly. Solving typical problems on strength of materials. An example of a task for a straight bend - a design scheme
The hypothesis of flat sections in bending can be explained by an example: let's apply a grid on the side surface of an undeformed beam, consisting of longitudinal and transverse (perpendicular to the axis) straight lines. As a result of the bending of the beam, the longitudinal lines will take on a curvilinear shape, while the transverse lines will practically remain straight and perpendicular to the bent axis of the beam.
Formulation of the planar section hypothesis: cross sections, flat and perpendicular to the axis of the beam before , remain flat and perpendicular to the curved axis after its deformation.
This circumstance indicates that when flat section hypothesis, as with and
In addition to the hypothesis of flat sections, an assumption is made: the longitudinal fibers of the beam do not press each other when it is bent.
The hypothesis of flat sections and the assumption are called Bernoulli's conjecture.
Consider a beam of rectangular cross section experiencing pure bending (). Let's select a beam element with a length (Fig. 7.8. a). As a result of bending, the cross sections of the beam will rotate, forming an angle. The top fibers are in compression and the bottom fibers are in tension. The radius of curvature of the neutral fiber is denoted by .
We conditionally consider that the fibers change their length, while remaining straight (Fig. 7.8. b). Then the absolute and relative elongation of the fiber, spaced at a distance y from the neutral fiber:
Let us show that the longitudinal fibers, which do not experience either tension or compression during beam bending, pass through the main central axis x.
Since the length of the beam does not change during bending, the longitudinal force (N) arising in the cross section must be zero. Elementary longitudinal force.
Given the expression :
The multiplier can be taken out of the integral sign (does not depend on the integration variable).
The expression represents the cross section of the beam with respect to the neutral x-axis. It is zero when the neutral axis passes through the center of gravity of the cross section. Consequently, the neutral axis (zero line) when the beam is bent passes through the center of gravity of the cross section.
Obviously: the bending moment is associated with normal stresses that occur at the points of the cross section of the rod. Elementary bending moment created by elemental force:
,
where is the axial moment of inertia of the cross section about the neutral axis x, and the ratio is the curvature of the beam axis.
Rigidity beams in bending(the larger, the smaller the radius of curvature).
The resulting formula represents Hooke's law in bending for a rod: the bending moment occurring in the cross section is proportional to the curvature of the beam axis.
Expressing from the formula of Hooke's law for a rod when bending the radius of curvature () and substituting its value in the formula , we obtain the formula for normal stresses () at an arbitrary point of the cross section of the beam, spaced at a distance y from the neutral axis x: .
In the formula for normal stresses () at an arbitrary point of the cross section of the beam, the absolute values of the bending moment () and the distance from the point to the neutral axis (y coordinates) should be substituted. Whether the stress at a given point will be tensile or compressive is easy to establish by the nature of the deformation of the beam or by the diagram of bending moments, the ordinates of which are plotted from the side of the compressed fibers of the beam.
From the formula you can see: normal stresses() change along the height of the cross section of the beam according to a linear law. On fig. 7.8, the plot is shown. The greatest stresses during beam bending occur at points furthest from the neutral axis. If a line is drawn in the cross section of the beam parallel to the neutral axis x, then the same normal stresses arise at all its points.
Simple analysis normal stress diagrams shows that when the beam is bent, the material located near the neutral axis practically does not work. Therefore, in order to reduce the weight of the beam, it is recommended to choose cross-sectional shapes in which most of the material is removed from the neutral axis, such as, for example, an I-profile.
Forces acting perpendicular to the axis of the beam and located in a plane passing through this axis cause a deformation called transverse bend. If the plane of action of the mentioned forces – main plane, then there is a straight (flat) transverse bend. Otherwise, the bend is called oblique transverse. A beam that is predominantly subject to bending is called beam 1 .
Essentially transverse bending is a combination of pure bending 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 normal stress formula σ X derived for pure bending 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 axis of the beam, is called simple. A beam with one fixed end and the other free end is called console. A simple beam having one or two parts hanging over a support is called console.
If, in addition, the sections are taken far from the points of application of the load (at a distance not less than half the height of the beam section), 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 experiences uniaxial tension or compression.
Under the action of a distributed load, the transverse forces in two adjacent sections will differ by an amount equal to qdx. Therefore, the curvature of the sections will also be slightly different. In addition, the fibers will exert pressure on each other. A careful study of the issue shows that if the length of the beam l quite large compared to its height h (l/ h> 5), then even with a distributed load, these factors do not have a significant effect on the normal stresses in the cross section and, therefore, may not be taken into account in practical calculations.
a B C
Rice. 10.5 Fig. 10.6
In sections under concentrated loads and near them, the distribution σ X deviates from the linear law. This deviation, which is of a local nature and is not accompanied by an increase in the greatest 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 X= – [Mz(x)/Iz]y.
If we draw two adjacent sections on a section of the bar that is free from 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 beam.
Select from the bar an element with 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 balance of the upper part, which has a base
width b. In accordance with the law of pairing of shear 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 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 \u003d - 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 calculation 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, one should:
Determine from the diagram the magnitude of the transverse force Q acting in the section;
Calculate the moment of inertia I z of the entire section;
Draw through this point a plane parallel to the plane xz and determine the section width b;
Calculate the static moment of the cut-off area S with respect to the main central axis z and substitute the found values into Zhuravsky's formula.
Let us define, as an example, shear stresses in a rectangular cross section (Fig. 10.6, c). Static moment about the axis z parts of the section above the line 1-1, on which the stress is determined, we write in the form:
It changes according to the law of a square parabola. Section width V for a rectangular beam is constant, then the law of change in shear stresses in the section will also be parabolic (Fig. 10.6, c). For y = and y = − tangential stresses are equal to zero, and on the neutral axis z they reach their highest point.
For a beam with a circular cross section on the neutral axis, we have
With direct pure bending, only one force factor arises in the cross section of the rod bending moment M x(Fig. 1). Because Q y \u003d dM x / dz \u003d 0, That M x=const and pure direct bending can be realized when the bar is loaded with pairs of forces applied in the end sections of the bar. Since the bending moment M x a-priory is equal to the sum moments internal forces about the axis Oh it is connected with normal stresses by the equation of statics that follows from this definition
Let us formulate the premises of the theory of pure direct bending of a prismatic rod. To do this, we analyze the deformations of a model of a rod made of a low-modulus material, on the side surface of which a grid of longitudinal and transverse scratches is applied (Fig. 2). Since the transverse risks, when the rod is bent by pairs of forces applied in the end sections, remain straight and perpendicular to the curved longitudinal risks, this allows us to conclude that plane section hypotheses, which, as the solution of this problem by the methods of the theory of elasticity shows, ceases to be a hypothesis, becoming an exact fact the law of plane sections. Measuring the change in the distances between the longitudinal risks, we come to the conclusion about the validity of the hypothesis of non-pressure of the longitudinal fibers.
Orthogonality of longitudinal and transverse scratches before and after deformation (as a reflection of the action of the law of flat sections) also indicates the absence of shifts, shear stresses in the transverse and longitudinal sections of the rod.
Fig.1. Relationship between internal effort and stress 
Fig.2. Pure bending model
Thus, pure direct bending of a prismatic rod is reduced to uniaxial tension or compression of longitudinal fibers by stresses (index G omitted later). In this case, part of the fibers is in the tension zone (in Fig. 2, these are the lower fibers), and the other part is in the compression zone (upper fibers). These zones are separated by a neutral layer (np), not changing its length, the stresses in which are equal to zero. Taking into account the prerequisites formulated above and assuming that the material of the rod is linearly elastic, i.e. Hooke's law in this case has the form: , we derive formulas for the curvature of the neutral layer (radius of curvature) and normal stresses . We first note that the constancy of the cross section of the prismatic rod and the bending moment (M x = const), ensures the constancy of the radius of curvature of the neutral layer along the length of the rod (Fig. 3, A), neutral layer (np) described by an arc of a circle.
Consider a prismatic rod under conditions of direct pure bending (Fig. 3, a) with a cross section symmetrical about the vertical axis OU. This condition will not affect the final result (in order for a straight bend to be possible, the coincidence of the axis Oh with main axis of inertia of the cross section, which is the axis of symmetry). Axis Ox put on the neutral layer, position whom not known in advance.
A) calculation scheme, b) strains and stresses
Fig.3. Fragment of a pure bend of a beamConsider an element cut from a rod with length dz, which is shown on a scale with proportions distorted in the interests of clarity in Fig. 3, b. Since the deformations of the element, determined by the relative displacement of its points, are of interest, one of the end sections of the element can be considered fixed. In view of the smallness, we assume that the points of the cross section, when rotated through this angle, move not along arcs, but along the corresponding tangents.
Let us calculate the relative deformation of the longitudinal fiber AB, separated from the neutral layer by at:
From the similarity of triangles C00 1 And 0 1 BB 1 follows that
Longitudinal deformation turned out to be a linear function of the distance from the neutral layer, which is a direct consequence of the law of plane sections
This formula is not suitable for practical use, since it contains two unknowns: the curvature of the neutral layer and the position of the neutral axis Oh, from which the coordinate is counted y. To determine these unknowns, we use the equilibrium equations of statics. The first expresses the requirement that the longitudinal force be equal to zero
|
Substituting expression (2) into this equation
and taking into account that , we get that
The integral on the left side of this equation is the static moment of the rod cross section about the neutral axis Oh, which can be equal to zero only relative to the central axis. Therefore, the neutral axis Oh passes through the center of gravity of the cross section.
The second static equilibrium equation is that relating normal stresses to the bending moment (which can easily be expressed in terms of external forces and is therefore considered a given value). Substituting the expression for into the bundle equation. voltage, we get:
and given that 
Where J x principal central moment of inertia about the axis Oh, for the curvature of the neutral layer, we obtain the formula
Fig.4. Normal stress distribution
which was first obtained by S. Coulomb in 1773. To match the signs of the bending moment M x and normal stresses, the minus sign is put on the right side of formula (5), since at M x >0 normal stresses at y>0 turn out to be contractive. However, in practical calculations, it is more convenient, without adhering to the formal rule of signs, to determine the stresses modulo, and put the sign according to the meaning. Normal stresses in pure bending of a prismatic bar are a linear function of the coordinate at and reach the highest values in the fibers most distant from the neutral axis (Fig. 4), i.e.
Here the geometric characteristic is introduced 
, which has the dimension m 3 and is called moment of resistance in bending. Since for a given M x voltage max? the less the more W x , moment of resistance is geometric characteristic of the strength of the cross-sectional bending. Let us give examples of calculating the moments of resistance for the simplest forms of cross sections. For a rectangular cross section (Fig. 5, A) we have J x \u003d bh 3 / 12, y max = h/2 And W x = J x /y max = bh 2 /6. Similarly for a circle (Fig. 5 ,a J x =d4 /64, ymax=d/2) we get W x =d3/32, for a circular annular section (Fig. 5, V), which one
Straight bend. Flat transverse bending Plotting diagrams of internal force factors for beams Plotting Q and M diagrams according to equations Plotting Q and M diagrams using characteristic sections (points) Calculations for strength in direct bending of beams Principal stresses in bending. Complete verification of the strength of beams Understanding the center of bending Determination of displacements in beams during bending. Concepts of deformation of beams and conditions of their rigidity Differential equation of the bent axis of the beam Method of direct integration Examples of determining displacements in beams by the method of direct integration Physical meaning of the constants of integration Method of initial parameters (universal equation of the bent axis of the beam). Examples of determining displacements in a beam using the method of initial parameters Determination of displacements using the Mohr method. A.K.'s rule Vereshchagin. Calculation of the Mohr integral according to A.K. Vereshchagin Examples of determination of displacements by means of Mohr's integral Bibliography Direct bending. Flat transverse bend. 1.1. Plotting diagrams of 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: a bending moment and a transverse force. In a particular case, the transverse force can be equal to zero, then the bend is called pure. With a flat transverse bending, all forces are located in one of the main planes of inertia of the rod and are perpendicular to its longitudinal axis, the 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 the projections onto the normal to the axis of the beam 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 upwards, and to the right - downwards, and negative - in the opposite case (Fig. 1.2, b). Rice. 1.2 When calculating the transverse 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. For the right side of the beam - vice versa. 5 The bending moment in an arbitrary cross section of the beam is numerically equal to the algebraic sum of the moments about the central axis z of the section of all external forces acting on one side of the section under consideration. The bending moment in the m-n section of the beam (Fig. 1.3, a) is considered positive if the resultant moment of external forces is directed clockwise from the section to the left of the section, and counterclockwise to the right, 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. For the right side of the beam - vice versa. 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 bends with a convexity downward, i.e., the lower fibers are stretched. Otherwise, the bending moment in the section is negative. Between the bending moment M, the transverse force Q and the intensity of the load q, there are differential dependencies. 1. The first derivative of the transverse 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, i.e. . (1.2) 3. The second derivative with respect to the abscissa of the section is equal to the intensity of the distributed load, i.e. . (1.3) We consider the distributed load directed upwards to be positive. A number of important conclusions follow from the differential dependencies between M, Q, q: 1. If on the beam section: a) the transverse force is positive, then the bending moment increases; b) the transverse force is negative, then the bending moment decreases; c) the transverse 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, otherwise M Mmin. 2. If there is no distributed load on the beam section, then the transverse force is constant, and the bending moment changes linearly. 3. If there is a uniformly distributed load on the section of the beam, then the transverse force changes according to a linear law, and the bending moment - according to the law of a square parabola, convex in the direction of the load (in the case of plotting M from the side of the 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 break in the direction of the force. 5. In the section where a 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. Under complex loading, beams build diagrams of transverse forces Q and bending moments M. Plot Q (M) is a graph showing the law of change in the transverse force (bending moment) along the length of the beam. Based on the analysis of diagrams M and Q, dangerous sections of the beam are established. The positive ordinates of the Q diagram are plotted upwards, and the negative ordinates are plotted downwards from the base line drawn parallel to the longitudinal axis of the beam. The positive ordinates of the diagram M are laid down, and the negative ordinates are plotted upwards, i.e., the diagram M is built from the side of the stretched fibers. The construction of diagrams Q and M for beams should begin with the definition of support reactions. For a beam with one fixed end and the other free end, plotting Q and M can be started from the free end without defining reactions in the embedment. 1.2. The construction of diagrams Q and M according to the Balk equations is divided into sections, within which the functions for the bending moment and the shear force remain constant (have no 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. On each section, an arbitrary section is taken at a distance x from the origin, and equations for Q and M are drawn up for this section. Plots Q and M are built using these equations. Example 1.1 Construct plots of shear forces Q and bending moments M for a given beam (Fig. 1.4a). Solution: 1. Determination of reactions of supports. 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 loadings: CA, AD, DB, BE. 2. Plotting Q. Plot SA. On section CA 1, 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 does not depend on the variable x1. Plot Q in this section will be depicted as a straight line parallel to the x-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 section 2-2: 8 The value of Q is constant on the section (does not depend on the variable x2). Plot Q on the plot is a straight line parallel to the x-axis. DB site. 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 B.E. On the site, we draw 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 build 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. is the 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. is the equation of a straight line. Plot DB 4 We 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. is the equation of a square parabola. 9 Find three values at the ends of the section and at the point with coordinate xk , where Section BE 1 Define 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: Based on the obtained values, we build a plot M (Fig. 1.4, c). In sections CA and AD, plot Q is limited by straight lines parallel to the abscissa axis, and in sections DB and BE, by oblique straight lines. In the sections C, A and B on the diagram Q there are jumps by the magnitude of the corresponding forces, which serves as a check of the correctness of the construction of the diagram Q. In sections where Q 0, the moments increase from left to right. In sections where Q 0, the moments decrease. Under the concentrated forces there are kinks in the direction of the action of the forces. Under the concentrated moment, there is a jump by the moment value. This indicates the correctness of plotting M. Example 1.2 Construct plots 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 make up the sums of the moments of all forces relative to points A and B: Plotting 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 triangles The resultant of that part of the load that is located to the left of the section The shear force in the section is equal to zero: Plot 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 maximum value of the bending moment is in the section, where 0, i.e. at. 1.5, c. 1.3. Construction of diagrams Q and M by characteristic sections (points) Using the differential relationships between M, Q, q and the conclusions arising from them, it is advisable to build diagrams Q and M by characteristic sections (without formulating equations). Using this method, the values of Q and M are calculated in characteristic sections. The characteristic sections are the boundary sections of the sections, as well as the sections where the given internal force factor has an extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct diagrams 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 embedment can be omitted. The beam has three loading areas: AB, BC, CD. There is no distributed load in sections AB and BC. The transverse forces are constant. Plot Q is limited by straight lines parallel to the x-axis. Bending moments change linearly. Plot M is limited to straight lines inclined to the x-axis. On section CD there is a uniformly distributed load. The transverse forces change linearly, and the bending moments change according to the law of a square parabola with a convexity in the direction of the distributed load. At the boundary of sections AB and BC, the transverse force changes abruptly. At the boundary of sections BC and CD, the bending moment changes abruptly. 1. Plotting Q. We calculate the values of transverse forces Q in the boundary sections of the sections: Based on the results of calculations, we build a diagram Q for the beam (Fig. 1, b). It follows from the diagram Q that the transverse force in 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 diagram M. We calculate the values of bending moments in the boundary sections of the sections: Example 1.4 According to the given diagram of bending moments (Fig. 1.7, a) for the beam (Fig. 1.7, b), determine the acting loads and plot Q. The circle indicates the vertex of the square parabola. Solution: Determine the loads acting on the beam. Section AC is loaded with a uniformly distributed load, since the diagram M in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting in a clockwise direction, since on the diagram M we have an upward jump by the magnitude of the moment. In the NE section, the beam is not loaded, since the diagram M in this section is limited 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 determine the reaction of support A. To do this we will compose an expression for bending moments in the section as the sum of the moments of forces on the left. The calculation scheme 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 transverse forces in the boundary sections of the sections: Plot Q is shown in fig. 1.7, d. The considered problem can be solved by compiling functional dependencies for M, Q in each section. Let's choose the origin of coordinates at the left end of the beam. On the section AC, the plot M is expressed by a square parabola, the equation of which is of the form Constants a, b, c, we find 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 get: The expression for the bending moment will be , we obtain the dependence for the transverse force After differentiating the function Q, we obtain an expression for the intensity of the distributed load In the section NE, 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 line passes through two points whose coordinates are known We obtain two equations: ,b of which we have a 20. The equation for the bending moment in the section NE will be After a twofold differentiation of M2, we will find. Based on the found values of M and Q, we build 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 largest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Build diagrams Q and M. Solution Determination of reactions of supports. Despite the fact that the total number of support links is four, the beam is statically determinate. The bending moment in hinge C is equal to zero, which allows us to make an additional equation: the sum of the moments about the hinge of all external forces acting on one side of this hinge is equal to zero. Compose the sum of the moments of all forces to the right of the hinge C. Diagram Q for the beam is limited by an inclined straight line, since q = const. We determine the values of transverse forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation whence Plot M for the beam is limited by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are written respectively as follows: From the condition of equality of moments, we obtain a quadratic equation with respect to the desired parameter x: Real value x2x 1.029 m. in the characteristic sections of the beam Figure 1.8, b shows the diagram Q, and in Figure. 1.8, c - plot 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. Plots Q and M are constructed for the suspension beam SV from the action of the load applied to it. Then they move to the main beam AC, loading it with an additional force VC, which is the pressure force of the beam CB on the beam AC. After that, diagrams Q and M are built for the AC beam. 1.4. Strength calculations for direct bending of beams Strength calculation for normal and shear stresses. With a direct bending of a beam, normal and shear stresses arise in its cross sections (Fig. 1.9). 18 Fig. 1.9 Normal stresses are related to the bending moment, shear stresses are related to the transverse force. In direct pure bending, shear stresses are equal to zero. Normal stresses at an arbitrary point of the beam cross section 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 most distant from the neutral axis If the section is symmetrical about the neutral axis (Fig. 1.11), then Fig. 1.11 the greatest tensile and compressive stresses are the same and are determined by the formula, - axial moment of section resistance in bending. For a rectangular section with a width b and a height h: (1.7) For a circular section with a diameter d: (1.8) For an annular section are the inner and outer diameters of the ring, respectively. For beams made of plastic materials, the most rational are symmetrical 20 section shapes (I-beam, box-shaped, annular). For beams made of brittle materials that do not equally resist tension and compression, sections that are asymmetrical about the neutral axis z (ta-br., U-shaped, asymmetrical I-beam) are rational. For beams of constant section made of plastic materials with symmetrical section 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 section made of ductile materials with asymmetric cross-sectional shapes, the strength condition is written in the following form: (1.11) strength conditions - distances from the neutral axis to the most remote points of the stretched and compressed zones of the dangerous section, respectively; P - allowable stresses, respectively, in tension and compression. Fig.1.12. 21 If the bending moment diagram has sections of different signs (Fig. 1.13), then in addition to checking the section 1-1, where Mmax acts, it is necessary to calculate the maximum tensile stresses for the 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 beam strength for shear stresses. Shear stresses in beams are calculated by the formula of D. I. Zhuravsky (1.13) where Q is the transverse force in the considered cross section of the beam; Szots is the static moment about the neutral axis of the area of the part of the section located on one side of the straight line drawn through the given point and parallel to the z axis; b is the width of the section at the level of the considered point; Iz is the moment of inertia of the entire section about the neutral axis z. 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 as, (1. 14) where Qmax is the transverse force with the highest modulus; - allowable shear stress for the material. For a rectangular beam section, 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 as (1.16) For an I-section, the strength condition is written as follows: (1.17) d is the wall thickness of the I-beam. Usually, the dimensions of the cross section of the beam are determined from the condition of strength for normal stresses. Checking the strength of beams for shear stresses is mandatory for short beams and beams of any length, if there are concentrated forces of large magnitude 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. Build diagrams in the dangerous section of the beam. Rice. 1.14 Decision 23 1. Plot Q and M plots from characteristic sections. Considering the left side of the beam, we obtain. The diagram of the transverse forces is shown in fig. 1.14, c. The plot 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 allowable ones. 4. The greatest tangential 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 is the width of the section at the level of the neutral axis. Fig. 5. Tangential stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 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 is the wall thickness at the level of point K1. Plots 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 transverse forces and bending moments along 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 for normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of the beam sections for shear stresses. Given: Solution: 1. Determine the reactions of the beam supports Check: 2. Plot Q and M diagrams. Values of transverse forces in characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, the load intensity q = const. Therefore, in these sections, the diagram Q is limited to straight lines inclined to the axis. In the section DB, the intensity of the distributed load q \u003d 0, therefore, in this section, the diagram Q is limited to a straight line parallel to the x axis. Diagram Q for the beam is shown in fig. 1.16b. Values of 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 Diagram M for the beam is shown in fig. 1.16, c. 2. We compose the strength condition for normal stresses from which we determine the required axial section modulus from the expression determined the required diameter d of a circular section beam Circular section area For a rectangular beam Required section height Rectangular section area According to the tables of GOST 8239-89, we find the nearest greater value of the axial moment of resistance 597 cm3, which corresponds to the I-beam No. 33 with the characteristics: A z 9840 cm4. Tolerance check: (underload by 1% of the allowable 5%) the nearest I-beam No. 30 (W 2 cm3) leads to a significant overload (more than 5%). We finally accept the 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 largest normal stresses in the dangerous section 27 of the I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-beam section. 1.17b. 5. We determine the largest shear stresses for the selected sections of the beam. a) rectangular section of the beam: b) round section beams: c) I-beam section: Shear stresses in the wall near the flange of the I-beam in the dangerous section A (on the right) (at point 2): The diagram of shear stresses in the dangerous sections of the I-beam is shown in fig. 1.17, in. The maximum shear stresses in the beam do not exceed the allowable stresses Example 1.8 Determine the allowable load on the beam (Fig. 1.18, a), if 60MPa, the cross-sectional dimensions are given (Fig. 1.19, a). Construct a diagram of normal stresses in the dangerous section of the beam under the allowable load. Fig 1.18 1. Determination of the reactions of the beam supports. In view of the symmetry of the system 2. Construction of diagrams Q and M from characteristic sections. Shear forces in the characteristic sections of the beam: Diagram Q for the beam is shown in fig. 5.18b. Bending moments in the characteristic sections of the beam For the second half of the beam, the ordinates M are along the axes of symmetry. Diagram M for the beam is shown in fig. 1.18b. 3. Geometric characteristics of the section (Fig. 1.19). We divide the figure into two simple 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 sectional 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 axis z of the entire section according to the formulas for the transition to parallel axes dangerous point "a" (Fig. 1.19) in the dangerous section I (Fig. 1.18): After substituting numerical data 5. With a permissible load in the dangerous section, the normal stresses at points "a" and "b" will be equal: dangerous section 1-1 is shown in fig. 1.19b.
We start with the simplest case, the so-called pure bending.
There is a clean bend special case bending, at which the transverse force in the sections of the beam is zero. Pure bending can only take place when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads that cause net
bend, shown in Fig. 88. On sections of these beams, where Q \u003d 0 and, therefore, M \u003d const; there is a pure bend.
The forces in any section of the beam with pure bending are reduced to a pair of forces, the plane of action of which passes through the axis of the beam, and the moment is constant.
Stresses can be determined based on the following considerations.
1. The tangential components of the forces on the elementary areas in the cross section of the beam cannot be reduced to a pair of forces, the plane of action of which is perpendicular to the plane of the section. It follows that the bending force in the section is the result of action on elementary areas
only normal forces, and therefore, with pure bending, stresses are reduced only to normal ones.
2. In order for efforts on elementary platforms to be reduced to only a couple of forces, there must be both positive and negative ones among them. Therefore, both tensioned and compressed beam fibers must exist.
3. Due to the fact that the forces in different sections are the same, the stresses at the corresponding points of the sections are the same.
Consider any element near the surface (Fig. 89, a). Since no forces are applied along its lower face, which coincides with the surface of the beam, there are no stresses on it either. Therefore, there are no stresses on the upper face of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we arrive at
The same conclusion, etc. It follows that there are no stresses along the horizontal faces of any element. Considering the elements that make up the horizontal layer, starting with the element near the surface of the beam (Fig. 90), we come to the conclusion that there are no stresses along the lateral vertical faces of any element. Thus, the stress state of any element (Fig. 91, a), and in the limit of the fiber, must be represented as shown in Fig. 91b, i.e., it can be either axial tension or axial compression.
4. Due to the symmetry of the application of external forces, the section along the middle of the beam length after deformation should remain flat and normal to the beam axis (Fig. 92, a). For the same reason, sections in quarters of the beam length also remain flat and normal to the beam axis (Fig. 92, b), if only the extreme sections of the beam remain flat and normal to the beam axis during deformation. A similar conclusion is also valid for sections in eighths of the length of the beam (Fig. 92, c), etc. Therefore, if the extreme sections of the beam remain flat during bending, then for any section it remains 
it is fair to say that after deformation it remains flat and normal to the axis of the curved beam. But in this case, it is obvious that the change in the elongation of the fibers of the beam along its height should occur not only continuously, but also monotonously. If we call a layer a set of fibers having the same elongations, then it follows from what has been said that the stretched and compressed fibers of the beam should be located on opposite sides of the layer in which the fiber elongations are equal to zero. We will call fibers whose elongations are equal to zero, neutral; a layer consisting of neutral fibers - a neutral layer; the line of intersection of the neutral layer with the plane of the cross section of the beam - the neutral line of this section. Then, based on the previous considerations, it can be argued that with a pure bending of the beam in each of its sections there is a neutral line that divides this section into two parts (zones): the zone of stretched fibers (tensioned zone) and the zone of compressed fibers (compressed zone ). Accordingly, normal tensile stresses should act at the points of the stretched zone of the cross-section, compressive stresses at the points of the compressed zone, and at the points of the neutral line the stresses are equal to zero.
Thus, with a pure bending of a beam of constant cross-section:
1) only normal stresses act in the sections;
2) the entire section can be divided into two parts (zones) - stretched and compressed; the boundary of the zones is the neutral line of the section, at the points of which the normal stresses are equal to zero;
3) any longitudinal element of the beam (in the limit, any fiber) is subjected to axial tension or compression, so that adjacent fibers do not interact with each other;
4) if the extreme sections of the beam during deformation remain flat and normal to the axis, then all its cross sections remain flat and normal to the axis of the curved beam.
Stress state of a beam in pure bending
Consider an element of a beam subject to pure bending, concluding 
measured between sections m-m and n-n, which are spaced one from the other at an infinitely small distance dx (Fig. 93). Due to the provision (4) of the previous paragraph, the sections m-m and n-n, which were parallel before deformation, after bending, remaining flat, will form an angle dQ and intersect along a straight line passing through point C, which is the center of curvature neutral fiber NN. Then the part of the AB fiber enclosed between them, located at a distance z from the neutral fiber (the positive direction of the z axis is taken towards the convexity of the beam during bending), will turn into an arc A "B" after deformation. A segment of the neutral fiber O1O2, turning into an O1O2 arc, it will not change its length, while the AB fiber will receive an elongation:
before deformation
after deformation
where p is the radius of curvature of the neutral fiber.
Therefore, the absolute elongation of the segment AB is
and elongation
Since, according to position (3), the fiber AB is subjected to axial tension, then with elastic deformation
From this it can be seen that the normal stresses along the height of the beam are distributed according to a linear law (Fig. 94). Since the equal force of all efforts on all elementary sections of the section must be equal to zero, then
whence, substituting the value from (5.8), we find
But the last integral is a static moment about the Oy axis, which is perpendicular to the plane of action of the bending forces.
Due to its equality to zero, this axis must pass through the center of gravity O of the section. Thus, the neutral line of the beam section is a straight line yy, perpendicular to the plane of action of the bending forces. It is called the neutral axis of the beam section. Then from (5.8) it follows that the stresses at points lying at the same distance from the neutral axis are the same.
The case of pure bending, in which the bending forces act in only one plane, causing bending in that plane only, is a planar pure bending. If the named plane passes through the Oz axis, then the moment of elementary efforts relative to this axis must be equal to zero, i.e.
Substituting here the value of σ from (5.8), we find
The integral on the left side of this equality, as is known, is the centrifugal moment of inertia of the section about the y and z axes, so that
The axes with respect to which the centrifugal moment of inertia of the section is equal to zero are called the main axes of inertia of this section. If, in addition, they pass through the center of gravity of the section, then they can be called the main central axes of inertia of the section. Thus, with a flat pure bending, the direction of the plane of action of the bending forces and the neutral axis of the section are the main central axes of inertia of the latter. In other words, to obtain a flat pure bending of a beam, a load cannot be applied to it arbitrarily: it must be reduced to forces acting in a plane that passes through one of the main central axes of inertia of the beam sections; in this case, the other main central axis of inertia will be the neutral axis of the section.
As is known, in the case of a section that is symmetrical about any axis, the axis of symmetry is one of its main central axes of inertia. Consequently, in this particular case, we will certainly obtain a pure bending by applying the appropriate analoads in the plane passing through the longitudinal axis of the beam and the axis of symmetry of its section. The straight line, perpendicular to the axis of symmetry and passing through the center of gravity of the section, is the neutral axis of this section.
Having established the position of the neutral axis, it is not difficult to find the magnitude of the stress at any point in the section. Indeed, since the sum of the moments of elementary forces relative to the neutral axis yy must be equal to the bending moment, then
whence, substituting the value of σ from (5.8), we find
Since the integral 
is. moment of inertia of the section about the y-axis, then
and from expression (5.8) we obtain
The product EI Y is called the bending stiffness of the beam.
The largest tensile and largest compressive stresses in absolute value act at the points of the section for which the absolute value of z is the largest, i.e., at the points furthest from the neutral axis. With the designations, Fig. 95 have
The value of Jy / h1 is called the moment of resistance of the section to stretching and is denoted by Wyr; similarly, Jy/h2 is called the moment of resistance of the section to compression
and denote Wyc, so
and therefore
If the neutral axis is the axis of symmetry of the section, then h1 = h2 = h/2 and, consequently, Wyp = Wyc, so there is no need to distinguish between them, and they use the same designation:
calling W y simply the section modulus. Therefore, in the case of a section symmetrical about the neutral axis,
All the above conclusions are obtained on the basis of the assumption that the cross sections of the beam, when bent, remain flat and normal to its axis (the hypothesis of flat sections). As shown, this assumption is valid only if the extreme (end) sections of the beam remain flat during bending. On the other hand, it follows from the hypothesis of flat sections that elementary forces in such sections should be distributed according to a linear law. Therefore, for the validity of the obtained theory of flat pure bending, it is necessary that the bending moments at the ends of the beam be applied in the form of elementary forces distributed along the height of the section according to a linear law (Fig. 96), which coincides with the law of stress distribution along the height of the section beams. However, based on the Saint-Venant principle, it can be argued that a change in the method of application of bending moments at the ends of the beam will cause only local deformations, the effect of which will affect only at a certain distance from these ends (approximately equal to the height of the section). The sections located in the rest of the length of the beam will remain flat. Consequently, the stated theory of flat pure bending, with any method of applying bending moments, is valid only within the middle part of the length of the beam, located at distances from its ends approximately equal to the height of the section. From this it is clear that this theory is obviously inapplicable if the height of the section exceeds half the length or span of the beam.
