Bending examples. Simple types of resistance. flat bend. Normal and shear stresses

Bending deformation consists in bending the axis of a straight bar or in changing the initial curvature of a straight bar (Fig. 6.1). Let's get acquainted with the basic concepts that are used when considering bending deformation.

Bending rods are called beams.

Clean bending is called, in which the bending moment is the only internal force factor that occurs in the cross-section of the beam.

More often, in the cross-section of the bar, along with the bending moment, a transverse force also arises. This bend is called transverse.

Flat (straight) bending is called when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section.

At oblique bend the plane of action of the bending moment intersects the beam cross-section along a line that does not coincide with any of the main central axes of the cross-section.

We begin our study of bending deformation with the case of pure plane bending.

Normal stresses and strains in pure bending.

As already mentioned, with a pure plane bending in the cross section of six internal force factors, only the bending moment is not zero (Fig. 6.1, c):

Experiments performed on elastic models show that if a grid of lines is applied to the surface of the model (Figure 6.1, a), then with pure bending it deforms as follows (Figure 6.1, b):

a) longitudinal lines are curved along the circumference;

b) the contours of the cross-sections remain flat;

c) the lines of the contours of the sections intersect everywhere with the longitudinal fibers at right angles.

Based on this, it can be assumed that in pure bending, the cross-sections of the beam remain flat and rotate so that they remain normal to the curved axis of the beam (hypothesis of flat sections during bending).

Rice. 6.1

By measuring the length of the longitudinal lines (Fig. 6.1, b), it can be found that the upper fibers lengthen when the beam is deformed, and the lower ones shorten. Obviously, it is possible to find such fibers, the length of which remains unchanged. A set of fibers that do not change their length when the beam is bent is called neutral layer (n. s.)... The neutral layer crosses the cross-section of the beam in a straight line, which is called neutral line (n. l.) of the section.

To derive a formula that determines the magnitude of the normal stresses arising in the cross section, consider a section of the beam in a deformed and non-deformed state (Fig. 6.2).

Rice. 6.2

With two infinitesimal cross-sections, select an element of length
... Before deformation, the sections bounding the element
, were parallel to each other (Fig. 6.2, a), and after deformation they slightly tilted, forming an angle
... The length of the fibers lying in the neutral layer does not change when bent
... Let us denote the radius of curvature of the trace of the neutral layer on the plane of the drawing by the letter ... Define the linear deformation of an arbitrary fiber
at a distance from the neutral layer.

The length of this fiber after deformation (arc length
) is equal to
... Considering that before deformation all fibers had the same length
, we obtain that the absolute elongation of the considered fiber

Its relative deformation

It's obvious that
, since the length of the fiber lying in the neutral layer has not changed. Then after substitution
get

(6.2)

Therefore, the relative longitudinal deformation is proportional to the distance of the fiber from the neutral axis.

Let us introduce the assumption that longitudinal fibers do not press against each other during bending. Under this assumption, each fiber is deformed in isolation, undergoing simple tension or compression, in which
... Taking into account (6.2)

, (6.3)

that is, normal stresses are directly proportional to the distances of the considered section points from the neutral axis.

Let us substitute dependence (6.3) into the expression for the bending moment
in cross section (6.1)

Recall that the integral
represents the moment of inertia of the section about the axis

(6.4)

Dependence (6.4) is Hooke's law in bending, since it relates the deformation (curvature of the neutral layer
) with the moment acting in the section. Work
is called the stiffness of the section in bending, Nm 2.

Substitute (6.4) into (6.3)

(6.5)

This is the sought-for formula for determining the normal stresses during pure bending of a beam at any point of its section.

In order to establish where the neutral line is in the cross-section, we substitute the value of normal stresses in the expression for the longitudinal force
and bending moment

Insofar as
,

;

(6.6)

(6.7)

Equality (6.6) indicates that the axis - the neutral axis of the section - passes through the center of gravity of the cross section.

Equality (6.7) shows that and - the main central axes of the section.

According to (6.5), the highest stress is reached in the fibers farthest from the neutral line

Attitude represents the axial moment of resistance of the section about its central axis , means

Meaning for the simplest cross-sections, the following:

For rectangular cross section

, (6.8)

where - the side of the section perpendicular to the axis ;

- the side of the section is parallel to the axis ;

For round cross section

, (6.9)

where is the diameter of the circular cross-section.

The condition for the strength under normal bending stresses can be written as

(6.10)

All the formulas obtained are obtained for the case of pure bending of a straight bar. The action of the transverse force leads to the fact that the hypotheses underlying the conclusions lose their validity. However, the practice of calculations shows that in the case of transverse bending of beams and frames, when in the section, in addition to the bending moment
the longitudinal force still acts
and lateral force , you can use the formulas given for pure bend. In this case, the error turns out to be insignificant.

General concepts.

Bending deformationconsists in bending the axis of a straight bar or in changing the initial curvature of a straight bar(fig. 6.1) ... Let's get acquainted with the basic concepts that are used when considering bending deformation.

Bending rods are called beams.

Clean bending is called, in which the bending moment is the only internal force factor that occurs in the cross-section of the beam.

More often, in the cross-section of the bar, along with the bending moment, a transverse force also arises. This bend is called transverse.

Flat (straight) bending is called when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section.

Oblique bending the plane of action of the bending moment intersects the beam cross-section along a line that does not coincide with any of the main central axes of the cross-section.

We begin our study of bending deformation with the case of pure plane bending.

Normal stresses and strains in pure bending.

As already mentioned, with a pure plane bending in the cross section of six internal force factors, only the bending moment is not zero (Fig. 6.1, c):

; (6.1)

Experiments performed on elastic models show that if a grid of lines is applied to the surface of the model(Fig. 6.1, a) , then in pure bending it deforms as follows(Figure 6.1, b):

a) longitudinal lines are curved along the circumference;

b) the contours of the cross-sections remain flat;

c) the lines of the contours of the sections intersect everywhere with the longitudinal fibers at right angles.

Rice. ...

By measuring the length of the longitudinal lines (Fig. 6.1, b), it can be found that the upper fibers lengthen when the beam is deformed, and the lower ones shorten. Obviously, it is possible to find such fibers, the length of which remains unchanged. A set of fibers that do not change their length when the beam is bent is calledneutral layer (n. s.)... The neutral layer crosses the cross-section of the beam in a straight line, which is calledneutral line (n. l.) of the section.

To derive a formula that determines the magnitude of the normal stresses arising in the cross section, consider a section of the beam in a deformed and non-deformed state (Fig. 6.2).

Rice. ...

Select the element with two infinitesimal cross-sections. Before deformation, the sections bounding the element were parallel to each other (Fig. 6.2, a), and after deformation they slightly tilted, forming an angle. The length of the fibers lying in the neutral layer does not change upon bending. Let us designate the radius of curvature of the trace of the neutral layer on the plane of the drawing by a letter. Let us determine the linear deformation of an arbitrary fiber spaced at a distance from the neutral layer.

The length of this fiber after deformation (arc length) is equal to. Taking into account that before deformation all fibers had the same length, we obtain that the absolute elongation of the considered fiber

Its relative deformation

Obviously, since the length of the fiber lying in the neutral layer has not changed. Then after substitution we get

(6.2)

Therefore, the relative longitudinal deformation is proportional to the distance of the fiber from the neutral axis.

, (6.3)

that is, normal stresses are directly proportional to the distances of the considered section points from the neutral axis.

Substitute dependence (6.3) into the expression for the bending moment in the cross section (6.1)

Recall that the integral is the moment of inertia of the section about the axis

(6.4)

Dependence (6.4) is Hooke's law in bending, since it connects the deformation (curvature of the neutral layer) with the moment acting in the section. The product is called the stiffness of the section in bending, N m 2.

Substitute (6.4) into (6.3)

(6.5)

This is the sought-for formula for determining the normal stresses during pure bending of a beam at any point of its section.

For in order to establish where the neutral line is in the cross-section, we substitute the value of normal stresses in the expression of the longitudinal force and bending moment

Insofar as,

then

(6.6)

(6.7)

Equality (6.6) indicates that the axis - the neutral axis of the section - passes through the center of gravity of the cross section.

Equality (6.7) shows that and are the main central axes of the section.

According to (6.5), the highest stress is reached in the fibers farthest from the neutral line

The ratio is the axial moment of resistance of the section relative to its central axis, which means

The meaning for the simplest cross-sections is as follows:

For rectangular cross section

, (6.8)

where is the side of the section perpendicular to the axis;

The side of the section is parallel to the axis;

For round cross section

, (6.9)

where is the diameter of the circular cross-section.

The condition for the strength under normal bending stresses can be written as

(6.10)

All the formulas obtained are obtained for the case of pure bending of a straight bar. The action of the transverse force leads to the fact that the hypotheses underlying the conclusions lose their validity. However, the practice of calculations shows that in the case of transverse bending of beams and frames, when, in addition to the bending moment, a longitudinal force and a transverse force also act in the section, one can use the formulas given for pure bending. In this case, the error turns out to be insignificant.

Determination of shear forces and bending moments.

As already mentioned, in the case of plane transverse bending, two internal force factors and arise in the cross section of the beam.

Before determining and determine the reactions of the beam supports (Fig. 6.3, a), making up the equations of equilibrium of the statics.

To determine and apply the method of sections. In the place of interest to us, we will make a mental section of the beam, for example, at a distance from the left support. Let's discard one of the parts of the beam, for example, the right one, and consider the equilibrium of the left side (Figure 6.3, b). We replace the interaction of the beam parts with internal forces and.

Let us establish the following sign rules for and:

The transverse force in the section is positive if its vectors tend to rotate the considered section clockwise;
The bending moment in the section is positive if it causes compression of the upper fibers.

Rice. ...

To determine these efforts, we use two equilibrium equations:

1. ; ; .

2. ;

In this way,

a) the transverse force in the cross section of the beam is numerically equal to the algebraic sum of the projections onto the transverse axis of the section of all external forces acting on one side of the section;

b) the bending moment in the cross section of the beam is numerically equal to the algebraic sum of the moments (calculated relative to the center of gravity of the section) of external forces acting on one side of the given section.

In practical calculations, they are usually guided by the following:

If the external load tends to turn the beam clockwise relative to the considered section (Fig. 6.4, b), then in the expression for it gives a positive term.
If the external load creates a moment relative to the considered section, causing compression of the upper fibers of the beam (Fig. 6.4, a), then in the expression for in this section it gives a positive term.

Rice. ...

Construction of diagrams in beams.

Consider a two-support beam(fig. 6.5, a) ... A concentrated moment acts on the beam at a point, a concentrated force at a point, and a uniformly distributed intensity load on a section.

We define support reactions and(Fig. 6.5, b) ... The resultant distributed load is equal, and its line of action passes through the center of the section. Let us compose the equations of the moments with respect to the points and.

We define the shear force and bending moment in an arbitrary section located on a section at a distance from point A(Fig. 6.5, c) .

(Fig. 6.5, d). The distance can vary within ().

The value of the shear force does not depend on the coordinate of the section, therefore, in all sections of the section, the shear forces are the same and the diagram has the form of a rectangle. Bending moment

The bending moment changes linearly. Let's define the ordinates of the plot for the boundaries of the plot.

We define the shear force and bending moment in an arbitrary section located on a section at a distance from the point(Fig. 6.5, d). The distance can vary within ().

The transverse force changes linearly. Define for the boundaries of the site.

Bending moment

The diagram of bending moments in this section will be parabolic.

To determine the extreme value of the bending moment, we equate to zero the derivative of the bending moment along the abscissa of the section:

From here

For a section with a coordinate, the value of the bending moment will be

As a result, we obtain diagrams of shear forces(Figure 6.5, e) and bending moments (Figure 6.5, g).

Differential Bending Dependencies.

(6.11)

(6.12)

(6.13)

These dependencies make it possible to establish some features of the diagrams of bending moments and shear forces:

N in areas where there is no distributed load, the diagrams are limited to straight lines parallel to the zero line of the diagram, and diagrams in the general case are limited by inclined straight lines.

N in areas where a uniformly distributed load is applied to the beam, the diagram is limited by inclined straight lines, and the diagram is limited by quadratic parabolas with a convexity facing in the direction opposite to the direction of the load action.

V sections, where, the tangent to the plot is parallel to the zero line of the plot.

N and in areas where, the moment increases; in areas where, the moment decreases.

V sections where concentrated forces are applied to the beam, there will be jumps on the diagram by the magnitude of the applied forces, and on the diagram there will be fractures.

In sections where concentrated moments are applied to the beam, there will be jumps on the diagram by the magnitude of these moments.

The plot ordinates are proportional to the tangent of the angle of inclination of the tangent to the plot.

Straight bend. Plane transverse bending Plotting internal force factors for beams Plotting Q and M plots using equations Plotting Q and M plots from characteristic sections (points) Strength calculations for direct bending of beams Principal bending stresses. Full check of the strength of beams. Concept of bending center. Determination of displacements in beams during bending. Concepts of deformation of beams and conditions of their rigidity Differential equation of a curved axis of a beam Direct integration method Examples of determining displacements in beams by the method of direct integration Physical meaning of integration constants Method of initial parameters (universal equation of a curved axis of a beam). Examples of determining displacements in a beam by the method of initial parameters. Determining displacements by Mohr's method. Rule A.K. Vereshchagin. Calculation of the Mohr integral according to A.K. Vereshchagin Examples of determining displacements by means of Mohr's integral Bibliography Direct bend. Flat lateral bend. 1.1. Plotting internal force factors for beams Direct bending is a type of deformation in which two internal force factors arise in the cross-sections of the bar: bending moment and shear force. In a particular case, the shear force can be equal to zero, then the bend is called pure. In a plane transverse bending, all forces are located in one of the main planes of inertia of the rod and are perpendicular to its longitudinal axis, moments are located in the same plane (Fig. 1.1, a, b). Rice. 1.1 The transverse force in an arbitrary cross-section of the beam is numerically equal to the algebraic sum of projections onto the normal to the axis of the beam of all external forces acting on one side of the considered section. The transverse force in the section of the m-n beam (Fig. 1.2, a) is considered positive if the resultant of external forces to the left of the section is directed upwards, and on the right - downwards, and negative - in the opposite case (Fig. 1.2, b). Rice. 1.2 When calculating the shear force in a given section, the external forces lying to the left of the section are taken with a plus sign if they are directed upwards, and with a minus sign if downwards. The opposite is true for the right side of the beam. 5 The bending moment in an arbitrary cross-section of the beam is numerically equal to the algebraic sum of moments about the central z-axis of the section of all external forces acting on one side of the section under consideration. The bending moment in the section of the m-n beam (Fig. 1.3, a) is considered positive if the resultant moment of external forces to the left of the section is directed clockwise, and on the right - counterclockwise, and negative - in the opposite case (Fig. 1.3, b). Rice. 1.3 When calculating the bending moment in a given section, the moments of external forces lying to the left of the section are considered positive if they are directed clockwise. The opposite is true for the right side of the beam. It is convenient to determine the sign of the bending moment by the nature of the deformation of the beam. The bending moment is considered positive if, in the section under consideration, the cut-off part of the beam is bent downward, i.e., the lower fibers are stretched. Otherwise, the bending moment in the section is negative. Differential relationships exist between the bending moment M, the shear force Q and the load intensity q. 1. The first derivative of the shear force along the abscissa of the section is equal to the intensity of the distributed load, i.e. ... (1.1) 2. The first derivative of the bending moment along the abscissa of the section is equal to the transverse force, ie. (1.2) 3. The second derivative with respect to the abscissa of the section is equal to the intensity of the distributed load, ie. (1.3) The distributed load directed upwards is considered positive. A number of important conclusions follow from the differential dependencies between M, Q, q: 1. If on a section of the beam: a) the transverse force is positive, then the bending moment increases; b) the transverse force is negative, then the bending moment decreases; c) the shear force is zero, then the bending moment has a constant value (pure bending); 6 d) the transverse force passes through zero, changing sign from plus to minus, max M M, in the opposite case M Mmin. 2. If there is no distributed load on the section of the beam, then the shear force is constant, and the bending moment changes linearly. 3. If there is a uniformly distributed load on a section of the beam, then the shear force changes according to a linear law, and the bending moment - according to the law of a square parabola, convex facing towards the load (in the case of plotting an M diagram from the side of stretched fibers). 4. In the section under the concentrated force, the Q diagram has a jump (by the value of the force), the M diagram has a kink in the direction of the force. 5. In the section where the concentrated moment is applied, the diagram M has a jump equal to the value of this moment. This is not reflected in the Q plot. With complex loading of the beam, diagrams of shear forces Q and bending moments M are plotted. Diagram Q (M) is a graph showing the law of change of the shear force (bending moment) along the length of the beam. Based on the analysis of the M and Q diagrams, dangerous sections of the beam are established. Positive ordinates of the Q plot are plotted upward, and negative ordinates are plotted downward from the baseline drawn parallel to the longitudinal axis of the beam. Positive ordinates of the M plot are laid down, and negative ordinates - up, that is, the M plot is built from the side of the stretched fibers. The construction of plots Q and M for beams should begin with the definition of support reactions. For a beam with one restrained and the other free ends, the construction of the Q and M diagrams can be started from the free end without defining the reactions in the embedment. 1.2. Plotting Q and M diagrams according to the equations The beam is divided into sections, within which the functions for the bending moment and shear force remain constant (do not have discontinuities). The boundaries of the sections are the points of application of concentrated forces, pairs of forces and places of change in the intensity of the distributed load. At each section, an arbitrary section is taken at a distance x from the origin of coordinates, and for this section, equations for Q and M are compiled. These equations are used to construct diagrams Q and M. Example 1.1 Construct diagrams of shear forces Q and bending moments M for a given beam (Fig. 1.4, a). Solution: 1. Determination of support reactions. We compose the equilibrium equations: from which we obtain The reactions of the supports are defined correctly. The beam has four sections Fig. 1.4 loads: CA, AD, DB, BE. 2. Plotting Q. Plot CA. On the CA 1 section, we draw an arbitrary section 1-1 at a distance x1 from the left end of the beam. We define Q as the algebraic sum of all external forces acting to the left of the section 1-1: The minus sign is taken because the force acting to the left of the section is directed downward. The expression for Q is independent of the variable x1. Diagram Q in this area will be depicted as a straight line parallel to the abscissa axis. Plot AD. On the site, we draw an arbitrary section 2-2 at a distance x2 from the left end of the beam. We define Q2 as the algebraic sum of all external forces acting to the left of the section 2-2: 8. The value of Q is constant in the section (does not depend on the variable x2). The plot Q on the site is a straight line parallel to the abscissa axis. Plot DB. On the site, we make an arbitrary section 3-3 at a distance x3 from the right end of the beam. We define Q3 as the algebraic sum of all external forces acting to the right of section 3-3: The resulting expression is the equation of an inclined straight line. Plot BE. On the site, we make a section 4-4 at a distance x4 from the right end of the beam. We define Q as the algebraic sum of all external forces acting to the right of section 4-4: 4 Here the plus sign is taken because the resultant load to the right of section 4-4 is directed downward. Based on the obtained values, we plot the diagrams Q (Fig. 1.4, b). 3. Plotting M. Plot m1. We define the bending moment in section 1-1 as the algebraic sum of the moments of forces acting to the left of section 1-1. - equation of a straight line. Section A 3 Define the bending moment in section 2-2 as the algebraic sum of the moments of forces acting to the left of section 2-2. - equation of a straight line. Section DB 4 Define the bending moment in section 3-3 as the algebraic sum of the moments of forces acting to the right of section 3-3. - the equation of a square parabola. 9 Find three values at the ends of the section and at a point with coordinate xk, where Section BE 1 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: Using the obtained values, we construct a diagram of M (Fig. 1.4, c). In sections CA and AD, the Q plot is limited by straight lines parallel to the abscissa axis, and in sections DB and BE - by inclined straight lines. In sections C, A and B on the plot Q, there are jumps by the value of the corresponding forces, which serves as a check of the correctness of plotting the plot Q. In the sections where Q  0, the moments increase from left to right. On the sections where Q  0, the moments decrease. Under the concentrated forces there are kinks towards the action of the forces. Under the concentrated moment, there is a jump by the magnitude of the moment. This indicates the correctness of plotting M. Example 1.2 Construct diagrams Q and M for a beam on two supports, loaded with a distributed load, the intensity of which varies linearly (Fig. 1.5, a). Solution Determination of support reactions. The resultant of the distributed load is equal to the area of the triangle, which is a diagram of the load and is applied at the center of gravity of this triangle. We compose the sums of the moments of all forces relative to points A and B: Plotting a diagram Q. Let's draw an arbitrary section at a distance x from the left support. The ordinate of the load diagram corresponding to the section is determined from the similarity of the triangles The resultant of that part of the load that is located to the left of the section The transverse force in the section is equal to The transverse force varies according to the law of a square parabola Equating the equation of the transverse force to zero, we find the abscissa of the section in which the diagram Q passes through zero: Diagram Q is shown in Fig. 1.5, b. The bending moment in an arbitrary section is equal to The bending moment changes according to the law of a cubic parabola: The bending moment has a maximum value in the section, where 0, i.e. at Diagram M is shown in Fig. 1.5, c. 1.3. Plotting Q and M diagrams by characteristic sections (points) Using the differential dependencies between M, Q, q and the conclusions arising from them, it is advisable to plot Q and M diagrams by characteristic sections (without drawing up equations). Using this method, the values of Q and M are calculated in characteristic sections. Typical sections are the boundary sections of the sections, as well as sections where the given internal force factor is of extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of the differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct plots Q and M for the beam shown in fig. 1.6, a. Rice. 1.6. Solution: We start plotting the Q and M diagrams from the free end of the beam, while the reactions in the embedding can be omitted. The beam has three loading areas: AB, BC, CD. There is no distributed load on sections AB and BC. The lateral forces are constant. Diagram Q is limited by straight lines parallel to the abscissa axis. Bending moments change linearly. Diagram M is limited by straight lines inclined to the abscissa axis. There is a uniformly distributed load on the CD section. Transverse forces change linearly, and bending moments - according to the law of a square parabola with a bulge in the direction of a distributed load. On the border of sections AB and BC, the lateral force changes abruptly. At the boundary of the sections BC and CD, the bending moment changes abruptly. 1. Plotting Q. We calculate the values of the shear forces Q in the boundary sections of the sections: Based on the results of the calculations, we plot the Q plot for the beam (Fig. 1, b). From the diagram Q it follows that the transverse force on the section CD is equal to zero in the section located at a distance qa a q from the beginning of this section. In this section, the bending moment has a maximum value. 2. Construction of the M diagram. We calculate the values of bending moments in the boundary sections of the sections: At the maximum moment in the section. Based on the results of the calculations, we construct the M diagram (Fig. 5.6, c). Example 1.4 Using a given diagram of bending moments (Fig. 1.7, a) for a beam (Fig. 1.7, b), determine the acting loads and build a diagram Q. The circle denotes the vertex of a square parabola. Solution: Determine the loads acting on the beam. The AC section is loaded with a uniformly distributed load, since the M diagram in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting clockwise, since on the M diagram we have a jump upward by the magnitude of the moment. In the NE section, the beam is not loaded, since the M diagram in this section is bounded by an inclined straight line. The reaction of support B is determined from the condition that the bending moment in section C is zero, i.e., to determine the intensity of the distributed load, we compose an expression for the bending moment in section A as the sum of the moments of forces on the right and equate to zero. Now we define the reaction of support A. To do this, we compose an expression for the bending moments in the section as the sum of the moments of forces on the left.The design diagram of a beam with a load is shown in Fig. 1.7, c. Starting from the left end of the beam, we calculate the values of the shear forces in the boundary sections of the sections: Diagram Q is shown in Fig. 1.7, d. The considered problem can be solved by drawing up functional dependencies for M, Q at each site. Select the origin at the left end of the beam. On the section AC, the diagram M is expressed by a square parabola, the equation of which has the form Constants a, b, c are found from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the equation of the parabola, we obtain: The expression for the bending moment will be Differentiating the function M1 , we obtain the dependence for the transverse force After differentiating the function Q, we obtain the expression for the intensity of the distributed load On the section CB, the expression for the bending moment is represented as a linear function To determine the constants a and b, we use the conditions that this straight line passes through two points whose coordinates are known. We obtain two equations:, b from which we have a 20. The equation for the bending moment on the section CB will be After two-fold differentiation of M2 we will find By the found values of M and Q we plot the diagrams of bending moments and shear forces for the beam. In addition to the distributed load, concentrated forces are applied to the beam in three sections, where there are jumps on the Q diagram and concentrated moments in the section where there is a jump on the M diagram. Example 1.5 For a beam (Fig. 1.8, a), determine the rational position of the hinge C, at which the greatest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Build Q and M diagrams. Solution Determination of support reactions. Although the total number of support ties is four, the beam is statically definable. The bending moment in the joint C is equal to zero, which allows us to draw up an additional equation: the sum of the moments relative to the joint of all external forces acting on one side of this joint is equal to zero. Let us compose the sum of the moments of all forces to the right of the hinge C. Diagram Q for the beam is bounded by an inclined straight line, since q = const. We determine the values of the shear forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation whence Diagram M for the beam is bounded by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are written accordingly as follows: From the condition of equality of moments, we obtain a quadratic equation for the sought parameter x: Real value x2x 1, 029 m. Determine the numerical values of the shear forces and bending moments in the characteristic sections of the beam Figure 1.8, b shows the diagram Q, and in Fig. 1.8, c - diagram M. The considered problem could be solved by dividing the hinged beam into its constituent elements, as shown in Fig. 1.8, d. At the beginning, the reactions of the supports VC and VB are determined. Diagrams Q and M are plotted for the suspended beam CB from the action of the load applied to it. Then they go to the main beam of the AC, loading it with an additional force VC, which is the force of pressure of the CB beam on the AC beam. Then the diagrams Q and M are plotted for the AC beam. 1.4. Strength calculations for direct bending of beams Strength calculations for normal and shear stresses. With direct bending of the beam, normal and tangential stresses arise in its cross-sections (Fig. 1.9). Fig. 18 1.9 Normal stresses are associated with a bending moment, shear stresses are associated with a shear force. In straight pure bending, the shear stresses are zero. Normal stresses at an arbitrary point of the cross-section of the beam are determined by the formula (1.4) where M is the bending moment in the given section; Iz is the moment of inertia of the section relative to the neutral z-axis; 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 vary linearly and reach the greatest value at the points farthest from the neutral axis If the section is symmetrical about the neutral axis (Fig. 1.11), then Fig. 1.11 the largest tensile and compressive stresses are the same and are determined by the formula,  is the axial moment of resistance of the section in bending. For a rectangular section of width b and height h: (1.7) For a circular section of diameter d: (1.8) For an annular section   - the inner and outer diameters of the ring, respectively. For beams made of plastic materials, the most rational are symmetrical 20 sectional shapes (I-beams, box-shaped, annular). For beams made of brittle materials that are not equally resistant to tension and compression, sections that are asymmetric with respect to the neutral z-axis (T, U-shaped, asymmetrical I-beam) are rational. For beams of constant cross-section made of plastic materials with symmetric cross-sectional shapes, the strength condition is written as follows: (1.10) where Mmax is the maximum bending moment modulo; - allowable stress for the material. For beams of constant cross-section made of plastic materials with asymmetric cross-sectional shapes, the strength condition is written in the following form: (1. 11) For beams made of brittle materials with sections that are asymmetric about the neutral axis, if the M diagram is unambiguous (Fig. 1.12), you need to write down two strength conditions - the distance from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; P - allowable stresses in tension and compression, respectively. Figure 1.12. 21 If the diagram of bending moments has sections of different signs (Fig. 1.13), then in addition to checking section 1-1, where Mmax acts, it is necessary to calculate the largest tensile stresses for section 2-2 (with the largest moment of the opposite sign). Rice. 1.13 Along with the basic calculation for normal stresses, in some cases it is necessary to check the strength of the beam in terms of shear stresses. Shear stresses in the beams are calculated by the formula of DI Zhuravsky (1.13) where Q is the shear force in the considered cross-section of the beam; Szotc - static moment relative to the neutral axis of the area of a part of the section located on one side of a straight line drawn through a given point and parallel to the z axis; b is the width of the section at the level of the point in question; Iz is the moment of inertia of the entire section relative to the neutral z-axis. In many cases, the maximum shear stresses occur at the level of the neutral layer of the beam (rectangle, I-beam, circle). In such cases, the shear stress strength condition is written in the form, (1.14) where Qmax is the largest shear force in modulus; Is the permissible shear stress for the material. For a rectangular section of a beam, the strength condition has the form (1.15) A - the cross-sectional area of the beam. For a circular section, the strength condition is represented in the form (1.16) For an I-section, the strength condition is written as follows: (1.17) where Szо, тmсax is the static half-section moment relative to the neutral axis; d - wall thickness of the I-beam. Usually, the dimensions of the cross-section of the beam are determined from the condition of the strength with respect to normal stresses. Checking the strength of beams for shear stresses is mandatory for short beams and beams of any length, if there are large concentrated forces near the supports, as well as for wooden, riveted and welded beams. Example 1.6 Check the strength of a box-section beam (Fig. 1.14) for normal and shear stresses, if MPa. Plot the dangerous section of the beam. Rice. 1.14 Solution 23 1. Construction of Q and M diagrams by characteristic sections. Considering the left side of the beam, we obtain the Diagram of transverse forces is shown in Fig. 1.14, c. A diagram of bending moments is shown in Fig. 5.14, g. 2. Geometric characteristics of the cross section 3. The highest normal stresses in the section C, where Mmax acts (modulo): MPa. The maximum normal stresses in the beam are practically equal to the permissible ones. 4. The largest shear stresses in section C (or A), where max Q acts (modulo): Here is the static moment of the half-section area relative to the neutral axis; b2 cm - section width at the level of the neutral axis. 5. Shear stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 cm3 is the static moment of the area of the part of the section located above the line passing through the point K1; b2 cm - wall thickness at the level of point K1. Diagrams  and  for section C of the beam are shown in Fig. 1.15. Example 1.7 For the beam shown in fig. 1.16, a, it is required: 1. Construct diagrams of shear forces and bending moments by characteristic sections (points). 2. Determine the dimensions of the cross-section in the form of a circle, rectangle and I-beam from the condition of strength with respect to normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of the cross-sections of the beams in terms of shear stress. Given: Solution: 1. Determine the reactions of the beam supports Check: 2. Plotting the diagrams Q and M. The values of the shear forces in the characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, the intensity of the load is q = const. Consequently, in these areas, the Q diagram is limited by straight lines inclined to the axis. In the section DB, the intensity of the distributed load q = 0, therefore, in this section of the diagram Q is limited by a straight line parallel to the x axis. The Q plot for the beam is shown in Fig. 1.16, b. The values of the bending moments in the characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section, in which Q = 0: The maximum moment in the second section The diagram M for the beam is shown in Fig. 1.16, c. 2. We formulate the strength condition for normal stresses from where we determine the required axial moment of resistance of the section from the expression the required diameter d of the circular section area The area of the circular section For the rectangular section The required section height The area of the rectangular section Define the required number of the I-beam. According to the tables of GOST 8239-89, we find the nearest higher value of the axial moment of resistance 597 cm3, which corresponds to I-beam No. 33 with the following characteristics: A z 9840 cm4. Check for tolerance: (underloading by 1% of the permissible 5%) the nearest I-beam No. 30 (W 2 cm3) leads to a significant overload (more than 5%). Finally, we accept I-beam No. 33. We compare the areas of circular and rectangular sections with the smallest area A of the I-beam: Of the three considered sections, the I-section is the most economical. 3. We calculate the highest normal stresses in the dangerous section of the 27 I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-section of the beam The diagram of normal stresses in the dangerous section of the beam is shown in Fig. 1.17, b. 5. Determine the highest shear stresses for the selected sections of the beam. a) rectangular section 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 (right) (at point 2): The diagram of shear stresses in the dangerous sections of the I-beam is shown in Fig. 1.17, c. The maximum shear stresses in the beam do not exceed the permissible stresses Example 1.8 Determine the permissible load on the beam (Figure 1.18, a), if 60 MPa, the cross-sectional dimensions are given (Figure 1.19, a). Construct a diagram of normal stresses in the dangerous section of the beam at the permissible load. Figure 1.18 1. Determination of the reactions of the beam supports. Due to the symmetry of the system 2. Construction of diagrams Q and M on characteristic sections. Shear forces in characteristic sections of the beam: Diagram Q for the beam is shown in Fig. 5.18, b. Bending moments in characteristic sections of the beam For the second half of the beam, the ordinates M are along the axes of symmetry. Diagram M for a beam is shown in Fig. 1.18, b. 3. Geometric characteristics of the section (Fig. 1.19). We break the figure into two simplest elements: an I-beam - 1 and a rectangle - 2. Fig. 1.19 According to the assortment for I-beam No. 20, we have For a rectangle: Static moment of the section area relative to the z1 axis Distance from the z1 axis to the center of gravity of the section. dangerous point "a" (Fig. 1.19) in the dangerous section I (Fig. 1.18): After substitution of the numerical data 5. Under the permissible load in the dangerous section, the normal stresses at points "a" and "b" will be equal: Diagram of normal stresses for dangerous section 1-1 is shown in Fig. 1.19, b.

10.1. General concepts and definitions

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

A bending rod is called a beam (or bar). In what follows, we will consider rectilinear beams, the cross-section of which has at least one axis of symmetry.

In the strength of materials, a distinction is made between flat, oblique and complex bending.

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

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

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

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

10.2. Determination of internal bending forces

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

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

The rest of the equilibrium equations are obviously identically equal to zero.

Thus, in the general case of flat bending in the beam section, out of six internal forces, two arise - bending momentМz and lateral force Qy (or when bending about another main axis - bending moment My and shear force Qz).

In this case, in accordance with the two considered loading cases, plane bending can be subdivided into pure and transverse.

Pure bend- plane bending, in which only one out of six internal forces arises in the bar sections - a bending moment (see the first case).

Transverse bending- bending, in which, in addition to the internal bending moment, a transverse force arises in the cross-sections of the bar (see the second case).

Strictly speaking, only pure bending belongs to simple types of resistance; transverse bending is conventionally referred to as simple types of resistance, since in most cases (for sufficiently long beams) the effect of transverse force in strength calculations can be neglected.

When determining internal efforts, we will adhere to the following rule of signs:

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

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

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

10.3. Differential Bending Constraints

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

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

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

The first of the two written equations gives the condition

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

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

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

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

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

10.4. Normal stresses at pure bending of a straight bar

Consider the case of pure flat bending of the beam and derive a formula for determining the normal stresses for this case.

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

There are three such hypotheses in bending:

a - the hypothesis of flat sections (Bernoulli's hypothesis) - sections that are flat before deformation remain flat and after deformation, but only rotate about a certain line, which is called the neutral axis of the beam section. In this case, the fibers of the beam lying on one side of the neutral axis will stretch, and on the other, they will be compressed; fibers lying on the neutral axis do not change their length;

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

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

The static side of the problem

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

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

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

Geometric side of the problem

Consider the deformation of a beam element of length dx, selected from a bending bar at an arbitrary point with the x coordinate. Taking into account the previously accepted hypothesis of plane sections, after bending the section of the beam, turn about the neutral axis (n.d.) by an angle dϕ, while the fiber ab, spaced from the neutral axis at a distance y, will turn into a circular arc a1b1, and its length will change by some magnitude. Here we recall that the length of the fibers lying on the neutral axis does not change, and therefore the arc a0b0 (the radius of curvature of which we denote by ρ) has the same length as the segment a0b0 before deformation a0b0 = dx.

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

Pure bend is called the kind of bend in which the action takes place bending moment only(fig. 3.5, a). Let us mentally draw the plane of section 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 m z. We will carry out actions similar to those that were performed by us in determining the stresses and strains during torsion, namely:

1) compose the equilibrium equations for the mentally cut off part of the part;
2) we determine the deformation of the material of the part based on the conditions of compatibility of deformations of elementary volumes of a given section;
3) we will solve the equations of equilibrium and compatibility of deformations.

From the equilibrium condition of the cut-off section of the beam (Fig. 3.5, b)

we get that the moment of internal forces M z 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 A- 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 stress with x in the cross section take on both positive and negative values. (Experience shows that when bending, the material of the lower side of the bar in Fig.3.5, a stretched, and the upper one - compressed.) Consequently, 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. This - neutral layer. The line of intersection of the neutral layer with the cross-sectional plane is called neutral line.

The compatibility conditions for the deformations of elementary volumes during bending are formed on the basis of the hypothesis of flat sections: the cross-sections of the beam are flat before bending (see Fig. 3.5, b) remain flat even after bending (fig. 3.6).

As a result of the action of the external moment, the beam bends, and the planes of sections I-I and II-II rotate relative to each other at 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 p to the 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 1 / p k = dip / dx and is constant along the length of the beam.

The neutral layer is not deformed, 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 elongation value of a stretched layer located at a distance y from neutral is ydq. Elongation of this layer:

Thus, in the adopted model, a linear distribution of deformations was obtained depending on the distance of a given elementary volume to the neutral layer, i.e. along the height of the section of the beam. Assuming that there is no mutual pressure of parallel layers of material on each other (about y = 0, a, = 0), we write Hooke's law for linear tension:

According to (3.13), normal stresses in the cross-section of the beam are linearly distributed. The stress of the elementary volume of the material farthest from the neutral layer (Fig. 3.6, v), 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 = Ey/ p k. Let us take y max = t/ 2 and p k = / / To.

The elastic limit must correspond to the condition with yn> c v = 1/2 kE t / 1.

Answer: oh = ] / 2 to 2 10 11 4 10 _3 / 0.8 = 1570 MPa; the yield point of this steel is at> 1800 MPa, which is higher than at the most durable spring steels. ?

? Problem 3.7

Determine the minimum radius of the drum for winding a tape with a thickness / = 0.1 mm of a heating element made of nickel alloy, in which the tape material does not deform plastically. 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) = = 0.04 m.?

1. With a joint solution of the first equilibrium equation (3.12) and the compatibility equation of deformations (3.13), we obtain

Meaning E/ p k φ 0 and is the same for all elements dA integration area. 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). Let's hang this plate at the point WITH so that it is in a horizontal position. Let us denote by the symbol y m the specific gravity of the plate material, then the weight of an elementary volume with an area dA is equal to 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 get

where G= y M tA is the weight of the plate.

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 zero:

Considering that Y c = G, write down

Thus, if an integral of the form J xdA by area A is equal to

zero then x c = 0. This means that point C coincides with the center of gravity of the plate. Therefore, from the equality S z = J ydA = 0 when due

bending, it follows that the center of gravity of the cross-section of the beam is on the neutral line.

Hence the value with the cross-section of the beam is zero.

1. The bending neutral line passes through the center of gravity of the cross-section of the beam.
2. The center of gravity of the cross section is the center of reduction of the moments of external and internal forces.

Target 3.8

Task 3.9

2. With the joint solution of the second equilibrium equation (3.12) and the compatibility equation of deformations (3.13), we obtain

Integral J z= J y 2 dA called moment of inertia of the transverse

section of the beam (rod) relative to the z-axis, passing through the center of gravity of the cross section.

In this way, M z = Е J z / p k. Considering that c x = Her x = Ey/ p to and E/ p k = a x / y, we obtain the dependence of the normal stresses Oh when bending:

1. The bending stress at a given point of the section does not depend on the modulus of normal elasticity E, but depends on the geometric parameter of the cross section J z and distance at from this point to the center of gravity of the cross section.

2. The maximum bending stress occurs in the elementary volumes farthest from the neutral line (see Fig. 3.6, v):

where W z- the moment of resistance of the cross section relative to the axis Z-

The condition for pure bending strength is similar to the condition for linear tensile strength:

where [a m | - allowable bending stress.

It is obvious that the internal volumes of the material, especially near the neutral axis, are practically unloaded (see Fig. 3.6, v). This contradicts the requirement to minimize the material consumption of the structure. Some ways to overcome this contradiction will be shown below.