Introduction.

Bending is a type of deformation characterized by curvature (change in curvature) of the axis or middle surface of a deformable object (beam, beam, slab, shell, etc.) under the influence of external forces or temperature. Bending is associated with the occurrence of bending moments in the cross sections of the beam. If out of the six internal force factors in the cross-section of a beam, only one bending moment is non-zero, the bending is called pure:

If in the cross sections of a beam, in addition to the bending moment, there is also a transverse force, the bending is called transverse:

In engineering practice, a special case of bending is also considered - longitudinal I. ( rice. 1, c), characterized by buckling of the rod under the action of longitudinal compressive forces. The simultaneous action of forces directed along the axis of the rod and perpendicular to it causes longitudinal-transverse bending ( rice. 1, G).

Rice. 1. Bending of the beam: a - clean: b - transverse; c - longitudinal; g - longitudinal-transverse.

A beam that bends is called a beam. The bend is called flat if the axis of the beam remains a flat line after deformation. The plane of location of the curved axis of the beam is called the bending plane. The plane of action of load forces is called the force plane. If the force plane coincides with one of the main planes of inertia of the cross section, the bend is called straight. (Otherwise, oblique bending occurs). Main plane of inertia cross section- this is a plane formed by one of the main cross-sectional axes with the longitudinal axis of the beam. When flat straight bend the bending plane and the force plane coincide.

The problem of torsion and bending of a beam (Saint-Venant problem) is of great practical interest. The application of the theory of bending established by Navier forms a vast branch structural mechanics and is of enormous practical importance, since it serves as the basis for calculating the dimensions and checking the strength of various parts of structures: beams, bridges, machine elements, etc.

BASIC EQUATIONS AND PROBLEMS OF ELASTICITY THEORY

§ 1. basic equations

First, we will give a general summary of the basic equations for equilibrium problems of an elastic body, which form the content of the section of the theory of elasticity, usually called the statics of an elastic body.

The deformed state of a body is completely determined by the deformation field tensor or the displacement field Components of the deformation tensor are associated with displacements by differential Cauchy dependencies:

(1)

The components of the deformation tensor must satisfy the Saint-Venant differential dependencies:

which are necessary and sufficient conditions for the integrability of equations (1).

The stressed state of the body is determined by the stress field tensor Six independent components of a symmetric tensor () must satisfy three differential equilibrium equations:

Components of the stress tensor And movements connected by six equations of Hooke's law:

In some cases, the equations of Hooke's law have to be used in the form of a formula

, (5)

Equations (1)-(5) are the basic equations of static problems in the theory of elasticity. Sometimes equations (1) and (2) are called geometric equations, equations ( 3) are static equations, and equations (4) or (5) are physical equations. To the basic equations that determine the state of a linearly elastic body at its internal points of volume, it is necessary to add conditions on its surface. These conditions are called boundary conditions. They are determined either by given external surface forces or specified movements points on the body surface. In the first case, the boundary conditions are expressed by the equality:

where are the vector components t surface force, - components of the unit vector P, directed along the outer normal to the surface at the point in question.

In the second case, the boundary conditions are expressed by the equality

Where - functions specified on the surface.

Boundary conditions can also be of a mixed nature, when on one part external surface forces are specified on the surface of the body and on the other part the surface of the body is given displacements:

Other types of boundary conditions are also possible. For example, on a certain area of ​​the body surface, only some components of the displacement vector are specified and, in addition, not all components of the surface force vector are specified.

§ 2. main problems of statics of an elastic body

Depending on the type of boundary conditions, three types of basic static problems in the theory of elasticity are distinguished.

The main task of the first type is to determine the components of the stress field tensor within the area , occupied by the body, and the component of the vector of movement of points inside the area and surface points bodies according to given mass forces and surface forces

The required nine functions must satisfy the basic equations (3) and (4), as well as the boundary conditions (6).

The main task of the second type is to determine the movements points inside the area and the stress field tensor component according to given mass forces and according to specified movements on the body surface.

Features you're looking for And must satisfy the basic equations (3) and (4) and the boundary conditions (7).

Note that the boundary conditions (7) reflect the requirement for the continuity of the defined functions on the border body, i.e. when the internal point tends to some point on the surface, the function should tend to a given value at a given point on the surface.

The main problem of the third type or mixed problem is that given surface forces on one part of the body surface and according to given displacements on another part of the body surface and also, generally speaking, according to given mass forces it is required to determine the components of the stress and displacement tensor , satisfying the basic equations (3) and (4) when mixed boundary conditions (8) are met.

Having obtained the solution to this problem, it is possible to determine, in particular, the forces of connections on , which must be applied at points of the surface in order to realize specified displacements on this surface, and it is also possible to calculate the displacements of surface points . Coursework >> Industry, production

By lenght timber, That timber deformed. Deformation timber accompanied simultaneously... wood, polymer, etc. When bend timber lying on two supports... bend will be characterized by an arrow of deflection. In this case, the compressive stress in the concave part timber ...

The combination of bending and torsion of beams of circular cross-section is most often considered when calculating shafts. Cases of bending with torsion of beams are much less common. round section.

In § 1.9 it is established that in the case when the moments of inertia of the section relative to the main axes are equal to each other, oblique bending of the beam is impossible. In this regard, oblique bending of round beams is impossible. Therefore, in the general case of external forces, a round beam experiences a combination the following types deformations: direct transverse bending, torsion and central tension (or compression).

Let's consider this special case calculation of a round beam when the longitudinal force in its cross sections is zero. In this case, the beam works at joint action bending and torsion. To find the dangerous point of the beam, it is necessary to establish how the values ​​of bending and torque moments change along the length of the beam, i.e., construct diagrams of the total bending moments M and torques. We will consider the construction of these diagrams at specific example shaft shown in Fig. 22.9, a. The shaft rests on bearings A and B and is driven by motor C.

Pulleys E and F are mounted on the shaft, through which drive belts with tension are thrown. Let us assume that the shaft rotates in bearings without friction; we neglect the own weight of the shaft and pulleys (in the case where their own weight is significant, it should be taken into account). Let's direct the axis of the cross section of the shaft vertically, and the axis horizontally.

The magnitudes of the forces can be determined using formulas (1.6) and (2.6), if, for example, the power transmitted by each pulley, the angular velocity of the shaft and the ratios are known. After determining the magnitudes of the forces, these forces are transferred parallel to themselves to the longitudinal axis of the shaft. In this case, torsional moments are applied to the shaft in the sections in which pulleys E and F are located and are equal to, respectively. These moments are balanced by the moment transmitted from the engine (Fig. 22.9, b). The forces are then decomposed into vertical and horizontal components. Vertical forces will cause vertical reactions in the bearings, and horizontal forces will cause horizontal reactions. The magnitudes of these reactions are determined as for a beam lying on two supports.

Diagram of bending moments acting in vertical plane, is built from vertical forces (Fig. 22.9, c). It is shown in Fig. 22.9, d. Similarly, from horizontal forces (Fig. 22.9, e), a diagram of bending moments acting in the horizontal plane is constructed (Fig. 22.9, f).

From the diagrams you can determine (in any cross section) the total bending moment M using the formula

Using the values ​​of M obtained using this formula, a diagram of the total bending moments is constructed (Fig. 22.9, g). In those sections of the shaft in which straight, limiting diagrams intersect the axes of the diagrams at points located on the same vertical, diagram M is limited by straight lines, and in other areas it is limited by curves.

(see scan)

For example, in the section of the shaft in question, the length of the diagram M is limited to a straight line (Fig. 22.9, g), since the diagrams in this section are limited by straight lines and intersecting the axes of the diagrams at points located on the same vertical.

Point O of the intersection of the straight line with the axis of the diagram is located on the same vertical. A similar situation is typical for a shaft section with a length

The diagram of total (total) bending moments M characterizes the magnitude of these moments in each section of the shaft. The planes of action of these moments in different sections of the shaft are different, but the ordinates of the diagram for all sections are conventionally aligned with the plane of the drawing.

The torque diagram is constructed in the same way as for pure torsion(see § 1.6). For the shaft in question, it is shown in Fig. 22.9, z.

The dangerous section of the shaft is established using diagrams of the total bending moments M and torques. If in the section of a beam of constant diameter with the greatest bending moment M the greatest torque also acts, then this section is dangerous. In particular, the shaft under consideration has such a section located to the right of pulley F at an infinitesimal distance from it.

If the maximum bending moment M and the maximum torque act in different cross sections, then a section in which neither the value is the greatest may turn out to be dangerous. With beams of variable diameter, the most dangerous section may be the one in which significantly lower bending and torsional moments act than in other sections.

In cases where the dangerous section cannot be determined directly from the diagrams M and it is necessary to check the strength of the beam in several of its sections and in this way establish dangerous stresses.

Once a dangerous section of the beam has been established (or several sections have been identified, one of which may turn out to be dangerous), it is necessary to find dangerous points in it. To do this, let us consider the stresses arising in the cross section of the beam when a bending moment M and a torque are simultaneously acting in it

In beams of round cross-section, the length of which is many times greater than the diameter, the values ​​of the highest tangential stresses from transverse force are small and are not taken into account when calculating the strength of beams under the combined action of bending and torsion.

In Fig. Figure 23.9 shows the cross section of a round beam. In this section, a bending moment M and a torque act. The axis y is taken to be perpendicular to the plane of action of the bending moment. The y axis is therefore the neutral axis of the section.

In the cross section of the beam, normal stresses arise from bending and shear stresses from torsion.

Normal stresses a are determined by the formula. The diagram of these stresses is shown in Fig. 23.9. Largest by absolute value normal stresses arise at points A and B. These stresses are equal

where is the axial moment of resistance of the cross section of the beam.

Tangential stresses are determined by the formula. The diagram of these stresses is shown in Fig. 23.9.

At each point of the section they are directed normal to the radius connecting this point with the center of the section. The highest shear stresses occur at points located along the perimeter of the section; they are equal

where is the polar moment of resistance of the cross section of the beam.

For a plastic material, points A and B of the cross section, at which both normal and shear stresses simultaneously reach highest value, are dangerous. For a brittle material, the dangerous point is the one at which tensile stresses arise from the bending moment M.

The stressed state of an elementary parallelepiped isolated in the vicinity of point A is shown in Fig. 24.9, a. Along the faces of the parallelepiped, which coincide with the cross sections of the beam, normal stresses and tangential stresses act. Based on the law of pairing of tangential stresses, stresses also arise on the upper and lower faces of the parallelepiped. Its remaining two faces are stress-free. Thus, in in this case available private view plane stress state, discussed in detail in Chap. 3. The main stresses amax and are determined by formulas (12.3).

After substituting the values ​​into them we get

Voltages have different signs and therefore

An elementary parallelepiped, highlighted in the vicinity of point A by the main areas, is shown in Fig. 24.9, b.

The calculation of beams for strength during bending with torsion, as already noted (see the beginning of § 1.9), is carried out using strength theories. In this case, the calculation of beams from plastic materials is usually carried out on the basis of the third or fourth theory of strength, and from brittle ones - according to Mohr’s theory.

According to the third theory of strength [see. formula (6.8)], substituting the expressions into this inequality [see. formula (23.9)], we obtain

Spatial (complex) bending

Spatial bending is a type of complex resistance in which only bending moments and act in the cross section of the beam. The full bending moment acts in none of the main planes of inertia. Longitudinal force absent. Spatial or complex bending is often called non-planar bending because the bent axis of the rod is not a plane curve. This bending is caused by forces acting in different planes perpendicular to the axis of the beam (Fig. 1.2.1).

Fig.1.2.1

Following the order of solving problems with complex resistance outlined above, we lay out the spatial system of forces presented in Fig. 1.2.1, into two such that each of them acts in one of the main planes. As a result, we get two flat transverse bends - in the vertical and horizontal planes. Of the four internal force factors that arise in the cross section of the beam, we will take into account the influence of only bending moments. We construct diagrams caused by the corresponding forces (Fig. 1.2.1).

Analyzing the diagrams of bending moments, we come to the conclusion that section A is dangerous, since it is in this section that the largest bending moments and occur. Now it is necessary to establish the dangerous points of section A. To do this, we will construct a zero line. The zero line equation, taking into account the sign rule for the terms included in this equation, has the form:

Here the “” sign is adopted near the second term of the equation, since the stresses in the first quarter caused by the moment will be negative.

Let us determine the angle of inclination of the zero line with the positive direction of the axis (Fig. 12.6):

Rice. 1.2.2

From equation (8) it follows that the zero line for spatial bending is a straight line and passes through the center of gravity of the section.

From Fig. 1.2.2 it is clear that the greatest stresses will arise at the points of section No. 2 and No. 4 furthest from the zero line. The normal stresses at these points will be the same in magnitude, but different in sign: at point No. 4 the stresses will be positive, i.e. tensile, at point No. 2 - negative, i.e. compressive. The signs of these stresses were established from physical considerations.

Now that the dangerous points have been established, let's calculate the maximum stresses in section A and check the strength of the beam using the expression:

The strength condition (10) allows not only to check the strength of the beam, but also to select the dimensions of its cross section if the aspect ratio of the cross section is specified.

Spatial bending This type of complex resistance is called in which only bending moments and
. The full bending moment acts in none of the main planes of inertia. There is no longitudinal force. Spatial or complex bending is often called non-planar bend, since the curved axis of the rod is not a flat curve. This bending is caused by forces acting in different planes perpendicular to the axis of the beam (Fig. 12.4).

Following the order of solving problems with complex resistance outlined above, we lay out the spatial system of forces shown in Fig. 12.4, into two such that each of them acts in one of the main planes. As a result, we get two flat transverse bends - in the vertical and horizontal plane. Of the four internal force factors that arise in the cross section of the beam
, we will take into account the influence of only bending moments
. We build diagrams
, caused respectively by the forces
(Fig. 12.4).

Analyzing the diagrams of bending moments, we come to the conclusion that section A is dangerous, since it is in this section that the largest bending moments occur
And
. Now it is necessary to establish the dangerous points of section A. To do this, we will construct a zero line. The zero line equation, taking into account the sign rule for the terms included in this equation, has the form:

. (12.7)

Here the “” sign is adopted near the second term of the equation, since the stresses in the first quarter caused by the moment
, will be negative.

Let's determine the angle of inclination of the zero line with positive axis direction (Fig.12.6):

. (12.8)

From equation (12.7) it follows that the zero line for spatial bending is a straight line and passes through the center of gravity of the section.

From Fig. 12.5 it is clear that the greatest stresses will arise at the points of section No. 2 and No. 4 furthest from the zero line. The normal stresses at these points will be the same in magnitude, but different in sign: at point No. 4 the stresses will be positive, i.e. tensile, at point No. 2 – negative, i.e. compressive. The signs of these stresses were established from physical considerations.

Now that the dangerous points have been established, let's calculate the maximum stresses in section A and check the strength of the beam using the expression:

. (12.9)

The strength condition (12.9) allows not only to check the strength of the beam, but also to select the dimensions of its cross section if the aspect ratio of the cross section is specified.

12.4. Oblique bend

Obliquely This type of complex resistance is called in which only bending moments occur in the cross sections of the beam
And
, but unlike spatial bending, all forces applied to the beam act in one (force) plane, which does not coincide with any of the main planes of inertia. This type of bending is most often encountered in practice, so we will study it in more detail.

Consider a cantilever beam loaded with a force , as shown in Fig. 12.6, and made of isotropic material.

Just as with spatial bending, with oblique bending there is no longitudinal force. We will neglect the influence of transverse forces on the strength of the beam when calculating it.

The design diagram of the beam shown in Fig. 12.6 is shown in Fig. 12.7.

Let's break down the force to vertical and horizontal components and from each of these components we will construct diagrams of bending moments
And
.

Let us calculate the components of the total bending moment in the section :

;
.

Total bending moment in section equals

Thus, the components of the total bending moment can be expressed in terms of the total moment as follows:

;
. (12.10)

From expression (12.10) it is clear that during oblique bending there is no need to decompose the system of external forces into components, since these components of the total bending moment are connected to each other using the angle of inclination of the trace of the force plane . As a result, there is no need to construct diagrams of the components
And
total bending moment. It is enough to plot a diagram of the total bending moment
in the force plane, and then, using expression (12.10), determine the components of the total bending moment in any section of the beam that interests us. The obtained conclusion significantly simplifies the solution of problems with oblique bending.

Let us substitute the values ​​of the components of the total bending moment (12.10) into the formula for normal stresses (12.2) at
. We get:

. (12.11)

Here, the “” sign next to the total bending moment is placed specifically for the purpose of automatically obtaining the correct sign of the normal stress at the cross-section point under consideration. Total bending moment
and point coordinates And are taken with their signs, provided that in the first quadrant the signs of the point coordinates are taken positive.

Formula (12.11) was obtained from considering the special case of oblique bending of a beam, clamped at one end and loaded at the other with a concentrated force. However, this formula is a general formula for calculating stresses in oblique bending.

The dangerous section, as with spatial bending in the case under consideration (Fig. 12.6), will be section A, since in this section the largest total bending moment occurs. We will determine the dangerous points of section A by constructing a zero line. We obtain the zero line equation by calculating, using formula (12.11), the normal stresses at the point with coordinates And , belonging to the zero line and equate the found voltages to zero. After simple transformations we get:

(12.12)

. (12.13)

Here angle of inclination of the zero line to the axis (Fig. 12.8).

By examining equations (12.12) and (12.13), we can draw some conclusions about the behavior of the zero line during oblique bending:

From Fig. 12.8 it follows that the highest stresses occur at the cross-section points furthest from the zero line. In the case under consideration, such points are points No. 1 and No. 3. Thus, with oblique bending, the strength condition has the form:

. (12.14)

Here:
;
.

If the moments of resistance of the section relative to the main axes of inertia can be expressed in terms of the dimensions of the section, it is convenient to use the strength condition in this form:

. (12.15)

When selecting sections, one of the axial moments of resistance is taken out of the bracket and specified by the relation . Knowing
,
and angle , through successive attempts, determine the values
And , satisfying the strength condition

. (12.16)

For asymmetrical sections that do not have protruding corners, the strength condition in the form (12.14) is used. In this case, with each new attempt to select a section, it is necessary to first again find the position of the zero line and the coordinates of the most distant point (
). For rectangular section
. Given the relation, from the strength condition (12.16) one can easily find the quantity
and cross-sectional dimensions.

Let us consider the determination of displacements during oblique bending. Let's find the deflection in the section cantilever beam (Fig. 12.9). To do this, we will depict the beam in a single state and construct a diagram of single bending moments in one of the main planes. We will determine the total deflection in the section , having previously determined the projections of the displacement vector on the axis And . Projection of the total deflection vector onto the axis we find using Mohr's formula:

Projection of the total deflection vector onto the axis let's find it in a similar way:

The total deflection is determined by the formula:

. (12.19)

It should be noted that with oblique bending in formulas (12.17) and (12.18), when determining the projections of the deflection on the coordinate axes, only the constant terms in front of the integral sign change. The integral itself remains constant. When solving practical problems, we will calculate this integral using the Mohr-Simpson method. To do this, multiply the unit diagram
for cargo
(Fig. 12.9), constructed in the force plane, and then multiply the resulting result sequentially by constant coefficients, respectively, And . As a result, we obtain projections of the total deflection And on the coordinate axis And . Expressions for deflection projections for the general case of loading, when the beam has plots will look like:

; (12.20)

. (12.21)

Let us set aside the found values ​​for ,And (Fig. 12.8). Total deflection vector is with the axis sharp corner , the values ​​of which can be found using the formula:

, (12.22)

. (12.23)

Comparing equation (12.22) with the zero line equation (12.13), we come to the conclusion that

or
,

whence it follows that the zero line and the vector of total deflection mutually peredicular. Corner is the complement of an angle up to 90 0. This condition can be used to check when solving oblique bending problems:

. (12.24)

Thus, the direction of deflections during oblique bending is perpendicular to the zero line. It follows from this important condition, What the direction of deflections does not coincide with the direction of the acting force(Fig. 12.8). If the load is a plane system of forces, then the axis of the curved beam lies in a plane that does not coincide with the plane of action of the forces. The beam skews relative to the force plane. This circumstance served as the basis for the fact that such a bend began to be called oblique.

Example 12.1. Determine the position of the zero line (find the angle ) for the cross section of the beam shown in Fig. 12.10.

1. Angle to the force plane trace we will plot from the positive direction of the axis . Corner We will always take it sharp, but taking into account the sign. Any angle is considered positive if in the right coordinate system it is plotted from the positive direction of the axis counterclockwise, and negative if the angle is laid clockwise. In this case the angle is considered negative (
).

2. Determine the ratio of axial moments of inertia:

.

3. We write the equation of the zero line for oblique bending in the form from which we find the angle :

;
.

4. Angle turned out to be positive, so we set it aside from the positive direction of the axis counterclockwise to the zero line (Fig. 12.10).

Example 12.2. Determine the magnitude of the normal stress at point A of the cross section of the beam during oblique bending, if the bending moment
kNm, point coordinates
cm,
see Dimensions of the cross section of the beam and the angle of inclination of the force plane are shown in Fig. 12.11.

1. Let us first calculate the moments of inertia of the section relative to the axes And :

cm 4;
cm 4.

2. Let us write formula (12.11) to determine normal stresses at an arbitrary point of the cross section during oblique bending. When substituting the value of the bending moment into formula (12.11), it should be taken into account that the bending moment according to the conditions of the problem is positive.

7.78 MPa.

Example 12.3. Determine the dimensions of the cross section of the beam shown in Fig. 12.12a. Beam material – steel with permissible stress
MPa. The aspect ratio is given
. Loads and angle of inclination of the force plane are shown in Fig. 12.12c.

1. To determine the position of the dangerous section, we construct a diagram of bending moments (Fig. 12.12b). Section A is dangerous. Maximum bending moment in dangerous section
kNm.

2. The dangerous point in section A will be one of the corner points. We write the strength condition in the form

,

Where can we find it, given that the relation
:

3. Determine the dimensions of the cross section. Axial moment of resistance
taking into account the relationship of the parties
is equal to:

cm 3, from where

cm;
cm.

Example 12.4. As a result of the bending of the beam, the center of gravity of the section moved in the direction determined by the angle with axle (Fig. 12.13, a). Determine the angle of inclination force plane. The shape and dimensions of the cross section of the beam are shown in the figure.

1. To determine the angle of inclination of the trace of the force plane Let's use expression (12.22):

, where
.

Ratio of moments of inertia
(see example 12.1). Then

.

Let's set aside this angle value from positive axis direction (Fig. 12.13, b). The trace of the force plane in Fig. 12.13b is shown as a dashed line.

2. Let's check the resulting solution. To do this, with the found value of the angle Let's determine the position of the zero line. Let's use expression (12.13):

.

The zero line is shown in Fig. 12.13 as a dotted line. The zero line must be perpendicular to the deflection line. Let's check this:

Example 12.5. Determine the total deflection of the beam in section B during oblique bending (Fig. 12.14a). Beam material – steel with elastic modulus
MPa. Cross-sectional dimensions and inclination angle of the force plane are shown in Fig. 12.14b.

1. Determine the projections of the total deflection vector in section A And . To do this, we will construct a load diagram of bending moments
(Fig. 12.14, c), single diagram
(Fig. 12.14, d).

2. Using the Mohr-Simpson method, we multiply the cargo
and single
diagrams of bending moments using expressions (12.20) and (12.21):

m
mm.

m
mm.

Axial moments of inertia of the section
cm 4 and
We take cm 4 from example 12.1.

3. Determine the total deflection of section B:

.

The found values ​​of the projections of the total deflection and the full deflection itself are plotted in the drawing (Fig. 12.14b). Since the projections of the total deflection turned out to be positive when solving the problem, we put them aside in the direction of action of the unit force, i.e. down ( ) and left ( ).

5. To check the correctness of the solution, we determine the angle of inclination of the zero line to the axis :

Let's add up the modules of the angles of the direction of the total deflection And :

This means that the full deflection is perpendicular to the zero line. Thus, the problem was solved correctly.