Pure bending definition. Simple types of resistance. flat bend. We check the strength of the beam for the highest tangential stresses
bend the type of loading of a bar is called, in which a moment is applied to it, lying in a plane passing through the longitudinal axis. Bending moments occur in the cross sections of the beam. When bending, deformation occurs, in which the axis of the straight beam is bent or the curvature of the curved beam changes.
A bar that works in bending 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 (Fig. 6.1).
Fig.6.1
With a flat transverse bending in the beam, two types of internal forces arise: the 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 bend is called clean (fig.6.2). In the presence of a transverse force, a bend is called transverse . Strictly speaking, only pure bending belongs to the simple types of resistance; transverse bending is conditionally referred to as simple types of resistance, since in most cases (for sufficiently long beams) the action of a transverse force can be neglected in strength calculations.
22.Flat transverse bend. Differential dependencies between internal forces and external load. Between the bending moment, the transverse force and the intensity of the distributed load, there are differential relationships 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 transverse forces and bending moments. Determination of shear forces and bending moments - section 1
We discard the right side of the beam and replace its action on the left side with a transverse force and a bending moment. For the convenience of calculations, we close the discarded right side of the beam with a sheet of paper, aligning the left edge of the sheet with the considered section 1.
The transverse force in section 1 of the beam is equal to the algebraic sum of all external forces that are visible after closing
We see only the downward reaction of the support. Thus, the transverse force is:
kN.
We took the minus sign because the force rotates the visible part of the beam 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 efforts 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 arm of the force is almost zero. So the bending moment is: 
kN m
Here the plus sign is taken by us because the external moment M bends the visible part of the beam with a convexity downwards. (or because it is opposite to the direction of the bending moment according to the rule of signs)
Determination of shear forces and bending moments - section 2
In contrast to the first section, the reaction force has a shoulder equal to a.
transverse force:
kN;
bending moment:
Determination of shear forces and bending moments - section 3
transverse force:
bending moment:
Determination of shear forces and bending moments - section 4
Now more comfortable cover the left side of the beam with a leaf.
transverse force:
bending moment:
Determination of transverse forces and bending moments - section 5
transverse force:
bending moment:
Determination of shear forces and bending moments - section 1
transverse force and bending moment:
.
Based on the values found, we construct a diagram of transverse forces (Fig. 7.7, b) and bending moments (Fig. 7.7, c).
CONTROL OF CORRECT CONSTRUCTION OF PHYSICS
We will verify the correctness of the construction of diagrams according to external features, using the rules for constructing diagrams.
Checking the Shear Force Plot
We are convinced: under unloaded sections, the diagram of transverse forces runs parallel to the axis of the beam, and under a distributed load q, along a straight line inclined downward. There are three jumps on the longitudinal force diagram: under the reaction - down by 15 kN, under the force P - 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 breaks under the concentrated force P and under the support reactions. The fracture angles are directed towards these forces. Under a distributed load q, the diagram of bending moments changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6, there is an extremum on the diagram of the bending moment, since the diagram of the transverse force in this place passes through zero.
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. The transverse force in the m-n section of the beam (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. 1.3b). 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 beam section, then the transverse force changes according to a linear law, and the bending moment - according to the law of a square parabola, convex inverted towards the load (in the case of plotting M from the side of tensioned 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. An arbitrary section is taken at each section 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. Plot A 3 We 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 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 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 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 AC section, 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 parabola equation, we get: 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 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 termination are written respectively as follows: From the condition of equality of the moments, we obtain a quadratic equation for the desired parameter x: The real value is x2x 1.029 m. We determine the numerical values of the transverse forces and bending moments in the characteristic sections of the beam. 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 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 with respect to the neutral z axis (T-shaped, 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 plastic materials with asymmetric section 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 diagram M is unambiguous (Fig. 1.12), two strength conditions must be written - 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, 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 strength condition for shear stresses 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) where Szo,tmsax is the static half-section moment relative to the neutral axis; 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 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. 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 highest 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 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 round beam. Round 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 highest 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 highest shear stresses for the selected sections of the beam. a) rectangular section of the beam: 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 (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 60 MPa, 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.
Pure bend called this type of bending, in which the action takes place only bending moment(Fig. 3.5, a). Let's mentally draw the section plane I-I perpendicular to the longitudinal axis of the beam at a distance * from the free end of the beam, to which the external moment is applied mz . Let's carry out actions similar to those that were performed by us when determining stresses and strains during torsion, namely:
- 1) compose the equilibrium equations of the mentally cut off part of the part;
- 2) we determine the deformation of the material of the part based on the conditions for the compatibility of deformations of elementary volumes of a given section;
- 3) solve the equations of equilibrium and compatibility of deformations.
From the condition of equilibrium of the cut-off section of the beam (Fig. 3.5, b)
we get that the moment of internal forces Mz equal to the moment of external forces t: M = t.
Rice. 3.5.
The moment of internal forces is created by normal stresses o v directed along the x axis. With pure bending, there are no external forces, so the sum of the projections of internal forces on any coordinate axis is zero. On this basis, we write the equilibrium conditions in the form of equalities
where BUT- cross-sectional area of the beam (rod).
In pure bending, external forces F x , F, F v as well as moments of external forces t x, t y are equal to zero. Therefore, the rest of the equilibrium equations are identically equal to zero.
From the equilibrium condition 
for o > 0 it follows that
normal voltage with x in cross section take both positive and negative values. (Experience shows that when bending, the material of the underside of the beam in Fig. 3.5, a stretched, and the upper one is compressed.) Therefore, in the cross section during bending there are such elementary volumes (of the transition layer from compression to tension) in which there is no elongation or compression. It - neutral layer. The line of intersection of the neutral layer with the plane of the cross section is called neutral line.
The conditions for the compatibility of deformations of elementary volumes during bending are formed on the basis of the hypothesis of flat sections: flat cross sections of the beam before bending (see Fig. 3.5, b) will remain flat even after bending (Fig. 3.6).
As a result of the action of an external moment, the beam bends, and the planes of sections I-I and II-II rotate relative to each other by an angle dy(Fig. 3.6, b). With pure bending, the deformation of all sections along the axis of the beam is the same, therefore, the radius pk of curvature of the neutral layer of the beam along the x axis is the same. Because dx= p k dip, then the curvature of the neutral layer is equal to 1 / p k = dip / dx and is constant along the length of the beam.
The neutral layer does not deform, its length before and after deformation is equal to dx. Below this layer the material is stretched, above it is compressed.
Rice. 3.6.
The value of the elongation of the stretched layer, located at a distance y from the neutral one, is equal to ydq. Relative elongation of this layer:
Thus, in the adopted model, a linear distribution of strains is obtained depending on the distance of a given elementary volume to the neutral layer, i.e. along the height of the beam section. Assuming that there is no mutual pressing of parallel layers of material on each other (o y \u003d 0, a, \u003d 0), we write Hooke's law for linear tension:
According to (3.13), the normal stresses in the cross section of the beam are distributed according to a linear law. The stress of the elementary volume of the material, the most distant from the neutral layer (Fig. 3.6, in), maximum and equal to 
? Task 3.6
Determine the elastic limit of a steel blade with a thickness / = 4 mm and a length / = 80 cm, if its bending into a semicircle does not cause permanent deformation.
Solution
Bending stress o v = eu/ p k. Let's take y max = t/ 2i p k = / / to.
The elastic limit must correspond to the condition with yn > c v = 1/2 kE t /1.
Answer: about = ] / 2 to 2 10 11 4 10 _3 / 0.8 = 1570 MPa; the yield strength of this steel is a m > 1800 MPa, which exceeds a m of the strongest spring steels. ?
? Task 3.7
Determine the minimum drum radius for winding a tape with a thickness / = 0.1 mm of a heating element made of nickel alloy, at which the tape material does not plastically deform. Module E= 1.6 10 5 MPa, elastic limit o yn = 200 MPa.
Answer: minimum radius р = V 2 ?ir/a yM = У? 1.6-10 11 0.1 10 -3 / (200 10 6) \u003d \u003d 0.04 m.?
1. By jointly solving the first equilibrium equation (3.12) and the strain compatibility equation (3.13), we obtain 
Meaning E/ r k f 0 and the same for all elements dA area of integration. Therefore, this equality is satisfied only under the condition
This integral is called static moment of the cross-sectional area about the axisz? What is the physical meaning of this integral?
Let us take a plate of constant thickness /, but of an arbitrary profile (Fig. 3.7). Hang this plate at the point FROM so that it is in a horizontal position. We denote by the symbol y m the specific gravity of the material of the plate, then the weight of an elementary volume with an area dA equals dq= y JdA. Since the plate is in a state of equilibrium, then from the equality to zero of the projections of forces on the axis at we get
where G= y MtA- plate weight.
Rice. 3.7.
The sum of the moments of forces of all forces about the axis z passing in any section of the plate is also equal to zero:
Given that Y c = g, write down
Thus, if an integral of the form J xdA by area BUT equals
zero, then x c = 0. This means that point C coincides with the center of gravity of the plate. Therefore, from the equality Sz = J ydA= 0 at
bend it follows that the center of gravity of the cross section of the beam is on the neutral line.
Therefore, the value u s cross section of the beam is zero.
- 1. The neutral line during bending passes through the center of gravity of the beam cross section.
- 2. The center of gravity of the cross section is the center of reduction of the moments of external and internal forces.
Task 3.8
Task 3.9
2. By jointly solving the second equilibrium equation (3.12) and the strain compatibility equation (3.13), we obtain
Integral Jz= J y2dA called moment of inertia of the transverse
section of a beam (rod) relative to the z-axis, passing through the center of gravity of the cross section.
In this way, M z \u003d E J z / p k. Considering that c x = ee x = ey/ p k i E/ p k = a x / y, we obtain the dependence of normal stresses oh when bending:
1. The bending stress at a given section point does not depend on the modulus of normal elasticity E, but depends on the geometric parameter of the cross section Jz and distance at from this point to the center of gravity of the cross section.
2. The maximum bending stress occurs in elementary volumes, the most distant from the neutral line (see Fig. 3.6, in):
where Wz- moment of resistance of the cross section about the axis Z-
The condition of strength in pure bending is similar to the condition of strength in linear tension:
where [a m | - allowable bending stress.
Obviously, the internal volumes of the material, especially near the neutral axis, are practically not loaded (see Fig. 3.6, in). This contradicts the requirement to minimize the material consumption of the structure. Some ways of overcoming this contradiction will be shown below.
We start with the simplest case, the so-called pure bending.
Pure bending is a special case of bending, in which the transverse force in the beam sections is zero. Pure bending can only take place when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads that cause net
bend, shown in Fig. 88. On sections of these beams, where Q \u003d 0 and, therefore, M \u003d const; there is a pure bend.
The forces in any section of the beam with pure bending are reduced to a pair of forces, the plane of action of which passes through the axis of the beam, and the moment is constant.
Stresses can be determined based on the following considerations.
1. The tangential components of the forces on the elementary areas in the cross section of the beam cannot be reduced to a pair of forces, the plane of action of which is perpendicular to the plane of the section. It follows that the bending force in the section is the result of action on elementary areas
only normal forces, and therefore, with pure bending, stresses are reduced only to normal ones.
2. In order for efforts on elementary platforms to be reduced to only a couple of forces, there must be both positive and negative ones among them. Therefore, both tensioned and compressed beam fibers must exist.
3. Due to the fact that the forces in different sections are the same, the stresses at the corresponding points of the sections are the same.
Consider any element near the surface (Fig. 89, a). Since no forces are applied along its lower face, which coincides with the surface of the beam, there are no stresses on it either. Therefore, there are no stresses on the upper face of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we 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 beam length (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 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 only in 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. Therefore, 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 over the height of the section according to a linear law (Fig. 96), which coincides with the law of stress distribution over 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 influence 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.
Straight transverse bend occurs when all loads are applied perpendicular to the axis of the rod, 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. Direct transverse bending refers to a simple form of resistance and is plane stress state, i.e. the two principal stresses are different from zero. With this type of deformation, internal forces arise: a transverse force and a bending moment. A special case of a direct transverse bend is pure bend, with such resistance there are cargo sections, within which the transverse force vanishes, and the bending moment is nonzero. In the cross sections of the rods with a direct transverse bending, normal and shear stresses arise. Stresses are a function of the internal force, in this case normal stresses are a function of the bending moment, and tangential stresses are a function of the transverse force. For direct transverse bending, several hypotheses are introduced:
1) The cross sections of the beam, which are flat before deformation, remain flat and orthogonal to the neutral layer after deformation (the hypothesis of flat sections or J. Bernoulli's hypothesis). This hypothesis holds for pure bending and is violated when a shear force, shear stresses, and angular deformation appear.
2) There is no mutual pressure between the longitudinal layers (hypothesis about non-pressure of the fibers). From this hypothesis it follows that the longitudinal fibers experience uniaxial tension or compression, therefore, with pure bending, Hooke's law is valid.
A bar 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 introduced hypotheses for pure bending, a formula for determining normal stresses is obtained, which is also used 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 axis 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 during bending in 1856. Russian engineer-builder of bridges D.I. Zhuravsky obtained the dependence
.
(2)
The shear stress in a particular section does not depend on the ratio of the transverse 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, there are movements: 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 bent axis of the beam ( 
).
Here v 0 , Θ 0 ,M 0 , Q 0 – initial parameters, x – distance from the origin of coordinates to the section in which the displacement is defined , a is the distance from the origin of coordinates to the place of application or the beginning of the load.
The calculation for strength and stiffness is carried out using the conditions of strength and stiffness. Using these conditions, you can solve verification problems (perform verification of the condition), determine the size of the cross section, or select the allowable value of the load parameter. There are several strength conditions, some of them are given below. Strength condition for normal stresses looks like:
,
(4)
here 
–
section modulus relative to the z-axis, R is the design resistance for normal stresses.
Strength condition for shear stresses 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 greatest shear stresses can be written in the following form:
.
(6)
Stiffness conditions can be written for deflections (v ) and rotation angles (Θ ) :
where displacement values in square brackets are valid.
An example of completing an individual task No. 4 (term 2-8 weeks)
