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Spatial bend is called a type of complex resistance in which only bending moments act in the cross-section of the bar and
... In this case, the total bending moment acts in any of the principal planes of inertia. There is no longitudinal force. A spatial or complex bend is often referred to as nonplanar bend since the curved axis of the bar is not a flat curve. This bending is caused by forces acting in different planes perpendicular to the beam axis (Figure 12.4).

Following the order of solving problems with complex resistance, described above, we expand the spatial system of forces presented in Fig. 12.4, into two such that each of them acts in one of the principal 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 this case in the cross-section of the beam
, we will take into account the influence of only bending moments
... We build diagrams
caused respectively by forces
(Figure 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 arise
and
... Now it is necessary to set the dangerous points of section A. To do this, draw the zero line. The zero line equation, taking into account the rule of signs 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.

Determine the angle of inclination of the zero line with positive direction of the axis (Figure 12.6):

. (12.8)

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

From fig.12.5 it can be seen that the greatest stresses will occur at the points of section No. 2 and No. 4 farthest from the zero line. The magnitude of the normal stresses at these points will be the same, but they differ in sign: at point No. 4, the stresses will be positive, i.e. stretching, at point # 2 - negative, i.e. compressing. The signs of these stresses were established from physical considerations.

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

. (12.9)

The strength condition (12.9) makes it possible 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

Oblique is called a type of complex resistance in which only bending moments appear in the cross-sections of the beam
and
, but unlike spatial bending, all forces applied to the beam act in one (force) plane that does not coincide with any of the principal planes of inertia. This type of bend is most often encountered in practice, so we will investigate it in more detail.

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

As in the case of spatial bending, at oblique bending there is no longitudinal force... We will neglect the influence of transverse forces when calculating the strength of the beam.

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

Expand the force to the vertical and horizontal components and from each of these components we construct diagrams of bending moments
and
.

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

;
.

Total bending moment in section is equal to

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 can be seen that with oblique bending there is no need to decompose the system of external forces into components, since these components of the total bending moment are related 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
full bending moment. It is enough to plot 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 of interest to us. This conclusion greatly simplifies the solution of problems with oblique bending.

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 near the total bending moment is specially placed for the purpose of automatically obtaining the correct sign of the normal stress at the considered point of the cross-section. Full bending moment
and point coordinates and are taken with their signs, provided that in the first quadrant the signs of the coordinates of the point are taken positive.

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

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

(12.12)

. (12.13)

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

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

From Fig. 12.8 it follows that the highest stresses in magnitude arise at the points of the section farthest from the zero line. In this case, such points are points # 1 and # 3. Thus, in oblique bending, the strength condition is as follows:

. (12.14)

Here:
;
.

If the moments of resistance of the section relative to the main axes of inertia can be expressed through 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 set by the ratio ... Knowing
,
and angle , by successive attempts determine the values
and satisfying the strength condition

. (12.16)

For asymmetric sections without protruding corners, the strength condition is used in the form (12.14). In this case, with each new attempt to select a section, you must first re-find the position of the zero line and the coordinates of the most distant point (
). For rectangular section
... Given the ratio, from the strength condition (12.16), one can easily find the value
and cross-sectional dimensions.

Consider the definition of displacements in oblique bending. Find the deflection in the section cantilever beam (Figure 12.9). To do this, we will depict a beam in a single state and plot a diagram of unit bending moments in one of the principal planes. We will determine the total deflection in the section , having previously defined the projections of the displacement vector on the axis and ... The projection of the full deflection vector onto the axis find using Mohr's formula:

The projection of the full deflection vector onto the axis find in a similar way:

The total deflection is determined by the formula:

. (12.19)

It should be noted that in the case of 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, we multiply the unit diagram
for freight
(Fig. 12.9) plotted in the force plane, and then the result obtained is multiplied sequentially by constant coefficients, respectively, and ... As a result, we get the projection of the full deflection and on the coordinate axis and ... Expressions for the projected deflection for the general loading case, when the beam has plots will look like:

; (12.20)

. (12.21)

We postpone the found values ​​for ,and (Figure 12.8). Full deflection vector makes up with axle sharp corner , the values ​​of which can be found by 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 perpedicular. Injection complements the corner up to 90 0. This condition can be used to check when solving oblique bending problems:

. (12.24)

Thus, the direction of the deflections during oblique bending is perpendicular to the zero line. This implies the important condition that the direction of the deflections does not coincide with the direction of the acting force(Figure 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 is skewed in relation to the load-bearing 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 Figure 12.10.

1. Angle to the trace of the force plane we will postpone from the positive direction of the axis ... Injection 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 set aside 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 at an oblique bend in the form, whence we find the angle :

;
.

4. Angle turned out to be positive, so we postpone it from the positive direction of the axis counterclockwise to the zero line (Figure 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 Beam cross-section dimensions and load plane slope are shown in Figure 12.11.

1. Let's preliminarily calculate the moments of inertia of the section about the axes and :

cm 4;
cm 4.

2. Let us write down the formula (12.11) for determining the 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 is positive by the condition of the problem.

7.78 MPa.

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

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

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

,

Where do we find it, given that the ratio
:

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, whence

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 force plane trace we use the expression (12.22):

, where
.

Moment of inertia ratio
(see example 12.1). Then

.

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

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

.

The zero line is shown in Figure 12.13 with a dash-dotted line. The zero line should be perpendicular to the deflection line. Let's check it out:

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

1. Let's define the projection of the vector of the total deflection in section A and ... To do this, construct a load diagram of bending moments
(Fig. 12.14, c), a single plot
(Fig.12.14, d).

2. Applying the Mohr-Simpson method, we multiply the load
and a single
bending moment diagrams using expressions (12.20) and (12.21):

m
mm.

m
mm.

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

3. Determine the total deflection of the section B:

.

The found values ​​of the projections of the full deflection and the full deflection itself are put aside in the drawing (Fig. 12.14b). Since the projections of the total deflection turned out to be positive when solving the problem, we postpone them in the direction of the action of a 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 :

Add up the modules of the angles of the direction of the total deflection and :

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

This combination of internal force factors is typical when calculating shafts. The problem is flat, since the concept of "oblique bend" for a bar of circular cross-section, in which any central axis is the main one, is inapplicable. In the general case, the action of external forces, such a bar experiences a combination of the following types of deformation: direct lateral bending, torsion and central tension (compression). In fig. 11.5 shows a beam loaded with external forces that cause all four types of deformation.

Internal force plots allow you to identify dangerous sections, and stress plots - dangerous points in these sections. Shear stresses from shear forces reach their maximum on the beam axis and are insignificant for a solid bar and can be neglected, compared to shear stresses from torsion, which reach their maximum at peripheral points (point B).

A dangerous section is in the termination, where at the same time they have great importance longitudinal and transverse forces, bending and torques.

The dangerous point in this section will be the point where σ x and τ xy reach a significant value (point B). At this point, the greatest normal bending stress and shear stress from torsion act, as well as normal tensile stress

Having determined the main stresses by the formula:

we find σ red =

(when using the criterion of the highest shear stresses m = 4, when using the criterion for the specific energy of the shape change m = 3).

Substituting the expressions σ α and τ xy, we obtain:

or taking into account that W р = 2 W z, A = (see 10.4),

If the shaft is bent in two mutually perpendicular planes, then instead of M z it is necessary to substitute M tot =

The reduced stress σ red should not exceed the permissible stress σ adm, determined during tests at a linear stress state, taking into account the safety factor. For given dimensions and permissible stresses, a verification calculation is performed, the dimensions necessary to ensure safe strength are found from the condition

11.5. Calculation of momentless shells of revolution

In engineering, structural elements are widely used, which, from the point of view of strength and stiffness calculations, can be attributed to thin shells. It is customary to consider the shell thin if the ratio of its thickness to the overall size is less than 1/20. For thin shells, the hypothesis of straight normals is applicable: the segments of the normal to the median surface remain straight and inextensible after deformation. In this case, there is a linear distribution of deformations and, consequently, normal stresses (at small elastic deformations) over the shell thickness.

The surface of the shell is obtained by rotating a flat curve about an axis lying in the plane of the curve. If the curve is replaced by a straight line, then when it rotates parallel to the axis, a circular cylindrical shell is obtained, and when it rotates at an angle to the axis, a conical one.

In design schemes, the shell is represented by its median surface (equidistant from the front ones). The median surface is usually associated with a curvilinear orthogonal coordinate system Ө and φ. The angle θ () determines the position of the parallel to the line of intersection of the mid-surface with a plane passing normally to the axis of rotation.

Fig. 11.6 Fig. 11.7

A set of planes can be drawn through the normal with the middle of the surface, which will be normal to it and, in sections with it, form lines with different radii of curvature. Two of these radii are extreme. The lines to which they correspond are called lines of principal curvatures. One of the lines is a meridian, its radius of curvature is denoted r 1... The radius of curvature of the second curve is r 2(the center of curvature lies on the axis of rotation). Radius centers r 1 and r 2 can coincide (spherical shell), lie on one or on opposite sides of the middle surface, one of the centers can go to infinity (cylindrical and conical shells).

When compiling the basic equations, the forces and displacements are referred to the normal sections of the shell in the planes of the main curvatures. Let's compose ura-vnenie for internal efforts. Consider an infinitesimal element of the shell (Fig.11.6), cut out by two adjacent meridional planes (with angles θ and θ + dθ) and two adjacent parallel circles normal to the axis of rotation (with angles φ and φ + dφ). As a system of axes of projections and moments, we select a rectangular system of axes x, y, z... Axis y directed tangentially to the meridian, axis z- along the normal.

By virtue of axial symmetry(load P = 0) only normal forces will act on the element. N φ - linear meridian force directed tangentially to the meridian: N θ - linear annular force directed tangentially to the circle. The equation ΣX = 0 becomes an identity. We project all the forces onto the axis z:

2N θ r 1 dφsinφ + r o dθdφ + P z r 1 dφr o dθ = 0.

If we neglect the infinitesimal higher-order value () r o dθ dφ and divide the equation by r 1 r o dφ dθ, then taking into account that we will obtain the equation due to P. Laplace:

Instead of the equation ΣY = 0 for the element under consideration, we compose the equilibrium equation for the upper part of the shell (Fig. 11.6). Let's project all forces onto the axis of rotation:

ude: R v - vertical projection of the resultant external forces applied to the cut off part of the shell. So,

Substituting the values ​​of N φ into the Laplace equation, we find N θ. Determination of the forces in the shell of rotation according to the momentless theory is a statically definable problem. This became possible as a result of the fact that we immediately postulated the law of stress variation across the shell thickness - we considered them constant.

In the case of a spherical dome, we have r 1 = r 2 = r and r о = r. If the load is specified in the form of intensity P on the horizontal projection of the shell, then

Thus, in the meridional direction, the dome is uniformly compressed. Components of the surface load along the normal z is equal to P z = P. We substitute the values ​​of N φ and P z into the Laplace equation and find from it:

Annular compressive forces reach a maximum at the top of the dome at φ = 0. At φ = 45 º - N θ = 0; at φ> 45- N θ = 0 becomes tensile and reaches a maximum at φ = 90.

The horizontal component of the meridional effort is equal to:

Consider an example of calculating a momentless shell. The main pipeline is filled with gas, the pressure of which is R.

Here r 1 = R, r 2 = a in accordance with the previously accepted assumption that stresses are distributed uniformly throughout the thickness δ shell

where: σ m - normal meridional stresses, and

σ t - circumferential (latitudinal, ring) normal stresses.

The combination of bending and torsion of beams with circular cross-section is most often considered in the design of shafts. Cases of bending with torsion of non-circular beams are much less common.

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

Let us consider such a special case of calculating a round bar, when the longitudinal force in its cross sections is equal to zero. In this case, the beam works for the combined action of bending and torsion. To find the dangerous point of the bar, it is necessary to establish how the values ​​of bending and torque moments change along the length of the bar, that is, to construct diagrams of the total bending moments M and torques. specific example the shaft shown in Fig. 22.9, a. The shaft is supported by bearings A and B and is driven by motor C.

Pulleys E and F are mounted on the shaft, through which the tensioning drive belts are thrown. Suppose the shaft rotates in bearings without friction; we neglect the own weight of the shaft and pulleys (in the case when their own weight is significant, it should be taken into account). Let us direct the y-axis of the shaft cross-section 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 ratio 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, twisting moments and equal, respectively, are applied to the shaft in the sections in which the pulleys E and F are located. These moments are balanced by the torque transmitted from the engine (Fig. 22.9, b). Then the forces are decomposed into vertical and horizontal components. Vertical forces will cause vertical reactions in bearings, and horizontal forces - horizontal reactions. The magnitudes of these reactions are determined as for a beam lying on two supports.

The diagram of bending moments acting in a vertical plane is constructed 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 by the formula

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

(see scan)

For example, in the section of the considered shaft, the length of the diagram M is limited by 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.

The point O of the intersection of the straight line with the axis of the diagram is also located on the same vertical. A similar position is typical for a shaft section with a length

The diagram of the 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 are conditionally aligned for all sections with the plane of the drawing.

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

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 and the greatest torque acts, then this section is dangerous. In particular, for the shaft under consideration, this is the section located to the right of the pulley F at an infinitely small distance from it.

If the greatest bending moment M and the greatest torque act in different cross-sections, then the section in which neither the value nor the greatest can be dangerous. With bars of variable diameter, the most dangerous section may be in which significantly lower bending and twisting moments act than in other sections.

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

After a dangerous section of the bar has been established (or several sections have been outlined, one of which may turn out to be dangerous), it is necessary to find dangerous points in it. For this, we will consider the stresses arising in the cross-section of the bar, when the bending moment M and the torque moment act simultaneously in it

In the beams of a circular cross-section, the length of which is many times greater than the diameter, the values ​​of the greatest shear stresses from the shear force are small and are not taken into account when calculating the strength of the beams for the combined action of bending and torsion.

In fig. 23.9 shows a cross-section of a round bar. A bending moment M and a torque act in this section. The y-axis is taken as the axis perpendicular to the plane of action of the bending moment, the y-axis is thus the neutral axis of the section.

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

Normal stresses a are determined by the formula Diagram of these stresses is shown in Fig. 23.9. The highest in 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 bar.

Shear 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 along the normal to the radius connecting this point with the center of the section. The greatest shear stresses arise 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 bar.

With a plastic material, points A and B of the cross-section, in which both normal and tangential stresses reach the highest value, are dangerous. In the case of brittle material, the dangerous point is that of these points at which tensile stresses arise from the bending moment M.

The stress state of an elementary parallelepiped selected 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 bar, normal stresses and tangents act. Based on the law of pairing of shear stresses, stresses also arise on the upper and lower faces of the parallelepiped. The other two faces are stress-free. Thus, in this case, there is private view plane stress state, discussed in detail in Ch. 3. The main stresses atax and are determined by formulas (12.3).

After substituting the values ​​into them, we obtain

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

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

Introduction.

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

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

In engineering practice, it is also considered a special case bending - longitudinal I. ( rice. one, c), characterized by the 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 a longitudinal-transverse bending ( rice. one, G).

Rice. 1. Bend of the bar: a - clean: b - transverse; в - longitudinal; d - longitudinal-transverse.

A beam that works in bending is called a beam. A bend is called flat if the beam axis remains a flat line after deformation. The plane where the beam's curved axis is located is called the bending plane. The plane of action of the load forces is called the force plane. If the force plane coincides with one of the principal planes of inertia of the cross section, the bend is called straight. (Otherwise, oblique bending occurs). The main plane of inertia of the cross-section is the plane formed by one of the main axes of the cross-section with the longitudinal axis of the bar. When flat straight bend the bending plane and the force plane coincide.

The problem of torsion and bending of a bar (Saint-Venant problem) is of great practical interest. The application of the bending theory established by Navier makes up an extensive department of 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 give a general summary of the basic equations for the problems of equilibrium of an elastic body, which form the content of a section of the theory of elasticity, which is usually called the statics of an elastic body.

The deformed state of the body is completely determined by the tensor of the deformation field or the displacement field Components of the deformation tensor related to displacements by differential Cauchy dependencies:

(1)

The strain tensor components must satisfy the Saint-Venant differential relationships:

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

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

Stress tensor components and displacement are related by six equations of Hooke's law:

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

, (5)

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

where are the components of the vector t surface strength, - components of the unit vector P, directed along the outer normal to the surface at its considered point.

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

where - functions specified on the surface.

Boundary conditions can also be mixed, when on one part body surfaces are given external surface forces and on the other side body surface, displacements are specified:

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

§ 2.the main problems of statics of an elastic body

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

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 displacement of points inside the region and surface points bodies for 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 displacements points inside the area and the stress field tensor component for given mass forces and for given displacements on the surface of the body.

Functions sought and must satisfy the basic equations (3) and (4) and boundary conditions (7).

Note that boundary conditions (7) reflect the requirement of continuity of the defined functions on the border body, i.e. when the interior 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 the 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) under the mixed boundary conditions (8).

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

By lenght timber, then bar deformed. Deformation timber accompanied at the same time ... wood, polymer, etc. When bend timber lying on two supports, ... bend will be characterized by a deflection arrow. In this case, the compressive stresses in the concave part timber ...

In the case of calculating a round bar under the action of bending and torsion (Fig. 34.3), it is necessary to take into account normal and shear stresses, since the maximum values ​​of stresses in both cases arise on the surface. The calculation should be carried out according to the theory of strength, replacing a complex stress state with an equally dangerous simple one.

Maximum voltage torsion in section

Maximum bending stress in the section

According to one of the theories of strength, depending on the material of the bar, they calculate equivalent voltage for a dangerous section and check the bar for strength using the allowable bending stress for the bar material.

For a round bar, the moments of resistance of the section are as follows:

When calculated according to the third theory of strength, the theory of maximum shear stresses, the equivalent stress is calculated by the formula

The theory is applicable to plastic materials.

When calculated according to the theory of energy of shape change, the equivalent stress is calculated by the formula

The theory is applicable to ductile and brittle materials.

Equivalent stress when calculated by the theory of energy of shape change:

where is the equivalent moment.

Strength condition

Examples of problem solving

Example 1. For a given stress state (Fig. 34.4), using the hypothesis of maximum shear stresses, calculate the safety factor if σ T = 360 N / mm 2.

1. How is the stress state at a point characterized and how is it depicted?

2. What sites and what voltages are called the main ones?

3. List the types of stress conditions.

4. What characterizes the deformed state at a point?

5. In what cases do limiting stress states arise in ductile and brittle materials?

6. What is the equivalent voltage?

7. Explain the purpose of the theory of strength.

8. Write formulas for calculating equivalent stresses in calculations according to the theory of maximum tangential stresses and the theory of energy of shape change. Explain how to use them.

LECTURE 35

Topic 2.7. Calculation of a round bar with a combination of basic deformations

Know the formulas for equivalent stresses based on the hypotheses of the greatest shear stresses and energy of shape change.

To be able to calculate a circular cross-section beam for strength with a combination of basic deformations.

Equivalent stress formulas

Equivalent stress on the hypothesis of maximum shear stresses

Equivalent stress on the hypothesis of energy of shape change

Strength condition under the combined action of bending and torsion

where M EKV- equivalent moment.

Equivalent moment according to the hypothesis of maximum shear stresses

Equivalent moment according to the hypothesis of energy of shape change

The peculiarity of the calculation of shafts

Most shafts experience a combination of bending and torsional deformations. Usually the shafts are straight beams with a round or annular section. When calculating shafts, shear stresses from the action of transverse forces are not taken into account due to their insignificance.

Calculations are carried out for hazardous cross-sections. In the case of spatial loading of the shaft, the hypothesis of the independence of the action of forces is used and the bending moments are considered in two mutually perpendicular planes, and the total bending moment is determined by geometric summation.

Examples of problem solving

Example 1. In the dangerous cross-section of a round bar, internal force factors arise (Fig.35.1) M x; M y; M z.

M x and M y- bending moments in planes wooh and zOx respectively; M z- torque. Check the strength on the hypothesis of the highest shear stresses, if [ σ ] = 120 MPa. Initial data: M x= 0.9 kN m; M y = 0.8 kN m; M z = 2.2 kN * m; d= 60 mm.

Solution

We build diagrams of normal stresses from the action of bending moments relative to the axes Oh and OU and a diagram of shear stresses from torsion (Fig. 35.2).

The maximum shear stress occurs at the surface. Maximum normal stress from torque M x arise at the point A, maximum normal stress from the moment M y at the point V. Normal stresses add up because bending moments in mutually perpendicular planes add up geometrically.

Total bending moment:

We calculate the equivalent moment according to the theory of maximum tangential stresses:

Strength condition:

Section resistance moment: W oce in oe = 0.1 60 3 = 21600mm 3.

We check the strength:

Durability is assured.

Example 2. Calculate the required shaft diameter from the strength condition. There are two wheels mounted on the shaft. Two circumferential forces act on the wheels F t 1 = 1.2kN; F t 2= 2kN and two radial forces in the vertical plane F r 1= 0.43kN; F r 2 = 0.72kN (fig. 35.3). The diameters of the wheels are respectively equal d 1= 0.1m; d 2= 0.06 m.

Accept for shaft material [ σ ] = 50MPa.

The calculation is carried out on the hypothesis of maximum shear stresses. Disregard the weight of the shaft and wheels.

Solution

Indication. We use the principle of independence of the action of forces, we draw up the calculation schemes of the shaft in the vertical and horizontal planes. We determine the reactions in the supports in the horizontal and vertical planes separately. We build diagrams of bending moments (Fig. 35.4). Under the action of circumferential forces, the shaft is twisted. Determine the torque acting on the shaft.

Let's make a design diagram of the shaft (Fig. 35.4).

1. Torque on the shaft:

2. The bend is considered in two planes: horizontal (square H) and vertical (square V).

In the horizontal plane, we determine the reactions in the support:

WITH and V:

In the vertical plane, we determine the reactions in the support:

Determine bending moments at points C and B:

Total bending moments at points C and B:

At the point V maximum bending moment, torque also acts here.

The calculation of the shaft diameter is carried out according to the most loaded section.

3. Equivalent moment at point V according to the third theory of strength

4. Determine the diameter of the shaft of a circular cross-section from the strength condition

Round off the resulting value: d= 36 mm.

Note. When choosing shaft diameters, use the standard range of diameters (Appendix 2).

5. Determine the required dimensions of the shaft of the annular section at c = 0.8, where d is the outer diameter of the shaft.

The diameter of the annular shaft can be determined by the formula

We will accept d = 42 mm.

The overload is negligible. d BH = 0.8d = 0.8 42 = 33.6mm.

Round up to value d BH= 33 mm.

6. Let us compare the cost of metal by the cross-sectional areas of the shaft in both cases.

Solid shaft cross-sectional area

Cross-sectional area of ​​a hollow shaft

The cross-sectional area of ​​a solid shaft is almost twice the annular shaft:

Example 3... Determine the dimensions of the cross-section of the shaft (Fig. 2.70, a) control drive. Pedal thrust P 3, forces transmitted by the mechanism P 1, P 2, P 4... Shaft material - steel StZ with a yield point σ t = 240 N / mm 2, the required safety factor [ n] = 2.5. The calculation should be carried out according to the hypothesis of the energy of shape change.

Solution

Consider the balance of the shaft, preliminarily bringing the forces R 1, R 2, R 3, R 4 to the points lying on its axis.

Transferring forces R 1 parallel to ourselves at points TO and E, it is necessary to add pairs of forces with moments equal to the moments of forces R 1 regarding points TO and E, i.e.

These pairs of forces (moments) are conventionally shown in Fig. 2.70 , b in the form of arched lines with arrows. Similarly, when transferring forces R 2, R 3, R 4 to points K, E, L, H you need to add a couple of forces with moments

The support of the shaft shown in fig. 2.70, a, should be considered as spatial hinged supports that prevent movement in the direction of the axes X and at(the selected coordinate system is shown in Fig. 2.70, b).

Taking advantage of design scheme shown in Fig. 2.70, v, we compose the equilibrium equations:

therefore, support reactions ON THE and N B defined correctly.

Torque plots M z and bending moments M y are shown in Fig. 2.70, G... The section to the left of point L is dangerous.

The strength condition is as follows:

where the equivalent moment according to the hypothesis of the energy of shape change

Required shaft outer diameter

We accept d = 45 mm, then d 0 = 0.8 * 45 = 36 mm.

Example 4. Check the strength of the intermediate shaft (Fig. 2.71) of the helical spur gear, if the shaft transmits power N= 12.2 kW at rotational speed P= 355 rpm. The shaft is made of steel St5 with a yield point σ t = 280 N / mm 2. Required safety factor [ n] = 4. When calculating, apply the hypothesis of the highest shear stresses.

Indication. District efforts R 1 and R 2 lie in a horizontal plane and are directed along tangents to the circles gear wheels... Radial forces T 1 and T 2 lie in a vertical plane and are expressed in terms of the corresponding circumferential force as follows: T = 0,364R.

Solution

In fig. 2.71, a a schematic drawing of the shaft is presented; in fig. 2.71, b shows a diagram of the shaft and the forces arising in the gearing.

Determine the moment transmitted by the shaft:

Obviously, m = m 1 = m 2(the torsional moments applied to the shaft with uniform rotation are equal in magnitude and opposite in direction).

Let's define the forces acting on the gear wheels.

District efforts:

Radial forces:

Consider the equilibrium of the shaft AB preliminarily bringing forces R 1 and R 2 to the points lying on the axis of the shaft.

Transferring power R 1 parallel to itself to a point L, you need to add a couple of forces with a moment equal to the moment of force R 1 relative to point L, i.e.

This pair of forces (moment) is conventionally shown in Fig. 2.71, v in the form of an arched line with an arrow. Similarly, when transferring force R 2 exactly TO it is necessary to attach (add) a couple of forces with the moment

The support of the shaft shown in fig. 2.71, a, should be considered as spatial hinged supports that prevent linear movements in the directions of the axes X and at(the selected coordinate system is shown in Fig, 2.71, b).

Using the design scheme shown in Fig. 2.71, G, we compose the equations of equilibrium of the shaft in the vertical plane:

Let's compose a test equation:

therefore, the support reactions in the vertical plane are determined correctly.

Consider the balance of the shaft in the horizontal plane:

Let's compose a test equation:

therefore, the support reactions in the horizontal plane are determined correctly.

Torque plots M z and bending moments M x and M y are shown in Fig. 2.71, d.

The section is dangerous TO(see fig. 2.71, G,d). Equivalent moment according to the hypothesis of the greatest shear stresses

Equivalent stress on the hypothesis of the highest shear stresses for the dangerous point of the shaft

Safety factor

which is much more [ n] = 4, therefore, the strength of the shaft is ensured.

When calculating the strength of the shaft, the change in stresses over time was not taken into account, which is why such a significant safety factor was obtained.

Example 5. Determine the dimensions of the cross-section of the timber (Fig. 2.72, a). The material of the bar is steel 30KhGS with conditional yield points in tension and compression σ о, 2р = σ tr = 850 N / mm 2, σ 0.2 c = σ Tc = 965 N / mm 2. Safety factor [ n] = 1,6.

Solution

The bar works on the combined action of tension (compression) and torsion. With this loading, two internal force factors arise in the cross sections: longitudinal force and torque.

Longitudinal force diagrams N and torques M z are shown in Fig. 2.72, b, c. In this case, determine the position of the dangerous section using the diagrams N and M z impossible, since the dimensions of the cross-sections of the sections of the timber are different. To find out the position of the dangerous section, it is necessary to construct diagrams of normal and maximum shear stresses along the length of the bar.

According to the formula

we calculate the normal stresses in the cross-sections of the timber and build a diagram o (Fig. 2.72, G).

According to the formula

we calculate the maximum shear stresses in the cross-sections of the beam and build a diagram t tah(rice * 2.72, e).

Points of the contour of the cross-sections of the sections are probably dangerous. AB and CD(see fig. 2.72, a).

In fig. 2.72, e plots shown σ and τ for section cross-sections AB.

Recall that in this case (a bar of a circular cross-section works for the combined action of tension - compression and torsion), all points of the cross-section contour are equally dangerous.

In fig. 2.72, f

In fig. 2.72, s shown diagrams a and t for cross-sections of the site CD.

In fig. 2.72, and shows the voltages at the source sites at the hazardous point.

Main stresses at the dangerous point of the site CD:

According to Mohr's hypothesis of strength, the equivalent stress for the dangerous point of the section under consideration is

The points of the contour of the cross-sections of the section AB turned out to be dangerous.

The strength condition is as follows:

Example 2.76. Determine the allowable force value R from the condition of the strength of the bar Sun(Figure 2.73) The material of the bar is cast iron with a tensile strength σ bp = 150 N / mm 2 and a compressive strength σ bc = 450 N / mm 2. Required safety factor [ n] = 5.

Indication. Broken timber ABS is located in the horizontal plane, and the bar AB perpendicular to Sun. Forces R, 2P, 8P lie in a vertical plane; strength 0.5 R, 1.6 R- horizontal and perpendicular to the bar Sun; strength 10P, 16P coincide with the axis of the bar Sun; a pair of forces with a moment m = 25Pd is located in a vertical plane perpendicular to the axis of the rod Sun.

Solution

Let's bring forces R and 0.5P to the center of gravity of cross-section B.

Transferring force P parallel to itself to point B, you need to add a couple of forces with a moment equal to the moment of force R relative to point V, i.e., a pair with the moment m 1 = 10 Pd.

Strength 0.5R we transfer along its line of action to point B.

Loads acting on the bar Sun, are shown in Fig. 2.74, a.

We build diagrams of internal force factors for a bar Sun. Under the indicated loading of the bar, six of them arise in its cross sections: the longitudinal force N, lateral forces Qx and Qy, torque Mz bending moments Mx and Mu.

Diagrams N, Mz, Mx, Mu are shown in Fig. 2.74, b(the ordinates of the diagrams are expressed through R and d).

Diagrams Qy and Qx we do not construct, since the shear stresses corresponding to lateral forces are small.

In the example under consideration, the position of the dangerous section is not obvious, Presumably, the dangerous sections K (the end of the section I) and S.

According to Mohr's hypothesis of strength, the equivalent stress for point L

Let us determine the magnitude and plane of action of the bending moment Mi in section C, shown separately in Fig. 2.74, d... The same figure shows the diagrams σ И, σ N, τ for section C.

Source pads stresses at a point N(Fig. 2.74, e)

Principal stresses at a point N:

According to Mohr's hypothesis of strength, the equivalent stress for the point N

Stresses at the original sites at point E (Fig. 2.74, g):

The main stresses at point E:

According to Mohr's hypothesis of strength, the equivalent stress for point E

The point turned out to be dangerous L, for which

The strength condition is as follows:

Test questions and tasks

1. What stress state occurs in the cross section of the shaft under the combined action of bending and torsion?

2. Write the strength condition for calculating the shaft.

3. Write the formulas for calculating the equivalent moment when calculating the hypothesis of maximum shear stresses and the hypothesis of the energy of shaping.

4. How is the dangerous section selected when calculating the shaft?