By bending we mean a type of loading in which bending moments occur in the cross sections of the beam. If the bending moment in the section is the only force factor, then the bend is called pure. If, along with the bending moment, transverse forces also arise in the cross sections of the beam, then the bending is called transverse.
It is assumed that the bending moment and shear force lie in one of the main planes of the beam (let us assume that this plane is ZOY). This type of bend is called flat.
In all cases considered below, there is a flat transverse bending of the beams.
To calculate a beam for strength or rigidity, it is necessary to know the internal force factors that arise in its sections. For this purpose, diagrams of transverse forces (diagram Q) and bending moments (M) are constructed.
When bending, the straight axis of the beam is bent; the neutral axis passes through the center of gravity of the section. For certainty, when constructing diagrams of transverse forces and bending moments, we will establish sign rules for them. Let us assume that the bending moment will be considered positive if the beam element bends convexly downward, i.e. in such a way that its compressed fibers are in the upper part.
If the moment bends the beam with a convex upwards, then this moment will be considered negative.
When constructing a diagram, positive values of bending moments are plotted, as usual, in the direction of the Y axis, which corresponds to constructing a diagram on a compressed fiber.
Therefore, the rule of signs for the diagram of bending moments can be formulated as follows: the ordinates of the moments are plotted from the side of the layers of the beam.
Bending moment in section equal to the sum moments relative to this section of all forces located on one side (either) of the section.
To determine the transverse forces (Q), we establish a sign rule: the transverse force is considered positive if the external force tends to rotate the cut-off part of the beam hourly. arrow relative to the axis point that corresponds to the drawn section.
The transverse force (Q) in an arbitrary cross section of a beam is numerically equal to the sum of the projections onto the OU axis external forces, attached to its truncated part.
Let's consider several examples of constructing diagrams of transverse forces and bending moments. All forces are perpendicular to the axis of the beams, so the horizontal component of the reaction is zero. The deformed axis of the beam and the forces lie in the main plane ZOY.
A beam of length is clamped at its left end and loaded with a concentrated force F and a moment m=2F.
Let's construct diagrams of transverse forces Q and bending moments M from.
In our case, on a beam with right side no connections are made. Therefore, in order not to determine support reactions, it is advisable to consider the equilibrium of the right cut-off part of the beam. The given beam has two loading sections. Boundaries of section sections in which external forces are applied. 1st section - NE, 2nd - VA.
We carry out an arbitrary section in section 1 and consider the equilibrium of the right cut-off part of length Z 1.
From the equilibrium condition it follows:
Q=F ; M out = -FZ 1 ()
The shear force is positive because external force F tends to rotate the cut-off part clockwise. The bending moment is considered negative, because it bends the part of the beam in question with its convex upward.
When drawing up equilibrium equations, we mentally fix the location of the section; from equations () it follows that the transverse force in section I does not depend on Z 1 and is a constant value. Positive force Q=F is plotted on a scale upward from the centerline of the beam, perpendicular to it.
The bending moment depends on Z 1.
When Z 1 =O M from =O when Z 1 = M from =
We put the resulting value () down, i.e. diagram M from is built on a compressed fiber.
Let's move on to the second section
We cut section II at an arbitrary distance Z 2 from the free right end of the beam and consider the equilibrium of the cut part of length Z 2 . The change in shear force and bending moment based on equilibrium conditions can be expressed by the following equations:
Q=FM from = - FZ 2 +2F
The magnitude and sign of the shear force have not changed.
The magnitude of the bending moment depends on Z 2 .
When Z 2 = M from =, when Z 2 =
The bending moment turned out to be positive, both at the beginning of section II and at the end of it. In section II, the beam bends convexly downwards.
We plot on a scale the magnitude of the moments up along the centerline of the beam (i.e., the diagram is built on a compressed fiber). The largest bending moment occurs in the section where an external moment m is applied and its absolute value is equal to
Note that over the length of the beam, where Q remains constant, the bending moment M changes linearly and is represented on the diagram by inclined straight lines. From the diagrams Q and M from it is clear that in the section where an external transverse force is applied, the diagram Q has a jump by the magnitude of this force, and the diagram M from has a kink. In the section where an external bending moment is applied, the Miz diagram has a jump by the value of this moment. This is not reflected in the Q diagram. From diagram M we see that
max M from =
hence, dangerous section extremely close on the left side to the so-called.
For the beam shown in Fig. 13, a, construct diagrams of transverse forces and bending moments. Along its length, the beam is loaded with a uniformly distributed load with intensity q(KN/cm).
At support A (fixed hinge), a vertical reaction R a will occur (the horizontal reaction is zero), and at support B (a movable hinge), a vertical reaction R v will occur.
Let us determine the vertical reactions of the supports by composing an equation of moments relative to supports A and B.
Let's check the correctness of the reaction definition:
those. the support reactions are determined correctly.
The given beam has two loading sections: Section I - AC.
Section II - NE.
In the first section a, in the current section Z 1, from the equilibrium condition of the cut-off part we have
Equation of bending moments on 1 section of the beam:
The moment from the reaction R a bends the beam in section 1, with the convex side down, so the bending moment from the reaction Ra is entered into the equation with a plus sign. The load qZ 1 bends the beam with its convexity upward, so the moment from it is entered into the equation with a minus sign. The bending moment varies according to the law of a square parabola.
Therefore, it is necessary to find out whether there is an extremum. There is a differential relationship between the transverse force Q and the bending moment, which we will analyze further.
As you know, a function has an extremum where the derivative is zero. Therefore, in order to determine at what value of Z 1 the bending moment will be extreme, it is necessary to equate the transverse force equation to zero.
Since the transverse force in this section changes sign from plus to minus, the bending moment in this section will be maximum. If Q changes sign from minus to plus, then the bending moment in this section will be minimal.
So, the bending moment at
is the maximum.
Therefore, we build a parabola using three points
When Z 1 =0 M from =0
We cut the second section at a distance Z 2 from support B. From the equilibrium condition of the right cut off part of the beam we have:
When the value Q=const,
the bending moment will be:
at, at, i.e. M FROM
varies according to a linear law.
A beam on two supports, having a span of 2 and a left console of length, is loaded as shown in Fig. 14, a., where q(KN/cm) is the linear load. Support A is hingedly stationary, support B is a movable roller. Construct diagrams of Q and M from.
Solving the problem should begin with determining the reactions of the supports. From the condition that the sum of the projections of all forces on the Z axis is equal to zero, it follows that the horizontal component of the reaction at support A is equal to 0.
To check we use the equation
The equilibrium equation is satisfied, therefore, the reactions are calculated correctly. Let's move on to defining internal power factors. A given beam has three loading sections:
- 1st section - SA,
- Section 2 - AD,
- Section 3 - Far East.
Let's cut 1 section at a distance Z 1 from the left end of the beam.
at Z 1 =0 Q=0 M IZ =0
at Z 1 = Q= -q M FROM =
Thus, on the diagram of transverse forces, an inclined straight line is obtained, and on the diagram of bending moments, a parabola is obtained, the vertex of which is located at the left end of the beam.
In section II (a Z 2 2a), to determine the internal force factors, we consider the equilibrium of the left cut-off part of the beam with length Z 2. From the equilibrium condition we have:
The shear force in this area is constant.
In section III()
From the diagram we see that the largest bending moment occurs in the section under force F and is equal to. This section will be the most dangerous.
In diagram M from there is a shock at support B, equal to the external moment applied in this section.
Looking at the diagrams constructed above, it is easy to notice a certain natural connection between the diagrams of bending moments and the diagrams of transverse forces. Let's prove it.
The derivative of the shear force along the length of the beam is equal to the modulus of the load intensity.
Discarding the value higher order we get a little:
those. the shear force is the derivative of the bending moment along the length of the beam.
Taking into account the obtained differential dependencies, general conclusions can be drawn. If the beam is loaded with a uniformly distributed load of intensity q=const, obviously, the function Q will be linear, and M will be quadratic.
If the beam is loaded with concentrated forces or moments, then in the intervals between the points of their application the intensity q=0. Consequently, Q = const, and M from is a linear function of Z. At the points of application of concentrated forces, the diagram Q undergoes a jump by the magnitude of the external force, and in the diagram M from a corresponding kink (discontinuity in the derivative) appears.
At the point where the external bending moment is applied, a gap in the moment diagram is observed, equal in magnitude to the applied moment.
If Q>0, then M grows, and if Q<0, то М из убывает.
Differential dependencies are used to check the equations compiled to construct diagrams Q and M from, as well as to clarify the type of these diagrams.
The bending moment changes according to the law of a parabola, the convexity of which is always directed towards the external load.
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. There is no longitudinal force. 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. By size normal stress at these points will be the same, but different in sign: at point No. 4 the voltages 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.
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). 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 beam. In flat straight bending, 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, constitutes an extensive branch 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 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 given to 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
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Brief information from the theory
The timber is subject to conditions of complex resistance if several internal force factors in the cross sections are not equal to zero at the same time.
The following cases of complex loading are of greatest practical interest:
1. Oblique bend.
2. Bending with tension or compression when in transverse
cross-sections arise longitudinal force and bending moments such as
for example, during eccentric compression of a beam.
3. Bend with torsion, characterized by the presence in the butt
river sections of bending (or two bending) and torsional
moments.
Oblique bend.
Oblique bending is a case of beam bending in which the plane of action of the total bending moment in the section does not coincide with any of the main axes of inertia. It is most convenient to consider oblique bending as the simultaneous bending of a beam in two main planes zoy and zox, where the z axis is the axis of the beam, and the x and y axes are the main central axes of the cross section.
Let us consider a cantilever beam of rectangular cross-section loaded with force P (Fig. 1).
Having expanded the force P along the main central axes of the cross section, we obtain:
P y =Pcos φ, P x =Psin φ
Bending moments occur in the current section of the beam
M x = - P y z = -P z cos φ,
M y = P x z = P z sin φ.
The sign of the bending moment M x is determined in the same way as in the case straight bend. We will consider the moment M y positive if at points with positive value coordinate x this moment causes tensile stresses. By the way, the sign of the moment M y can be easily established by analogy with the determination of the sign of the bending moment M x, if you mentally rotate the section so that the x axis coincides with the original direction of the y axis.
The stress at an arbitrary point in the cross section of a beam can be determined using formulas for determining stress for the case of plane bending. Based on the principle of independent action of forces, we summarize the stresses caused by each of the bending moments
(1)
The values of bending moments (with their own signs) and the coordinates of the point at which the stress is calculated are substituted into this expression.
To determine the dangerous points of the section, it is necessary to determine the position of the zero or neutral line (the geometric location of the points of the section at which stresses σ = 0). Maximum stresses occur at points furthest from the zero line.
The zero line equation is obtained from equation (1) at =0:
whence it follows that the zero line passes through the center of gravity of the cross section.
The tangential stresses arising in the sections of the beam (at Q x ≠0 and Q y ≠0), as a rule, can be neglected. If there is a need to determine them, then first the components of the total shear stress τ x and τ y are calculated according to the formula of D.Ya. Zhuravsky, and then the latter are geometrically summed up:
To assess the strength of a beam, it is necessary to determine the maximum normal stresses in the dangerous section. Since at the most loaded points the stress state is uniaxial, the strength condition when calculating using the permissible stress method takes the form
For plastic materials,
For fragile materials,
n - safety factor.
If you calculate using the method limit states, then the strength condition has the form:
where R is the design resistance,
m – coefficient of working conditions.
In cases where the beam material has different resistance to tension and compression, it is necessary to determine both the maximum tensile and maximum compressive stresses, and a conclusion about the strength of the beam is made from the relationships:
where R p and R c - respectively calculated resistances material under tension and compression.
To determine the deflections of a beam, it is convenient to first find the displacements of the section in the main planes in the direction of the x and y axes.
The calculation of these displacements ƒ x and ƒ y can be done by constructing a universal equation for the curved axis of the beam or by energy methods.
The total deflection can be found as a geometric sum:
the beam rigidity condition has the form:
where - is the permissible deflection of the beam.
Eccentric compression
In this case, the compressive force P on the beam is directed parallel to the axis of the beam and is applied at a point that does not coincide with the center of gravity of the section. Let X p and Y p be the coordinates of the point of application of force P, measured relative to the main central axes (Fig. 2).
Effective load causes the appearance of the following internal force factors in cross sections: N= -P, Mx= -Py p, My=-Px p
The signs of the bending moments are negative, since the latter cause compression at points belonging to the first quarter. The stress at an arbitrary point of the section is determined by the expression
(9)
Substituting the values of N, Mx and Mu, we get
(10)
Since Ух= F, Уу= F (where i x and i y are the main radii of inertia), the last expression can be reduced to the form
(11)
We obtain the zero line equation by setting =0
1+ 
(12)
The segments and cut off by the zero line on the coordinate axes are expressed as follows:
Using dependencies (13), you can easily find the position of the zero line in the section (Fig. 3), after which the points most distant from this line are determined, which are dangerous, since maximum stresses arise in them.
The stressed state at the points of the section is uniaxial, therefore the condition for the strength of the beam is similar to the previously considered case of oblique bending of the beam - formulas (5), (6).
During eccentric compression of beams, the material of which weakly resists tension, it is desirable to prevent the appearance of tensile stresses in the cross-section. Stresses of the same sign will arise in the section if the zero line passes outside the section or, in extreme cases, touches it.
This condition is satisfied when the compressive force is applied inside a region called the core of the section. The core of the section is an area covering the center of gravity of the section and is characterized by the fact that any longitudinal force applied inside this zone causes stresses of the same sign at all points of the beam.
To construct the core of the section, it is necessary to set the position of the zero line so that it touches the section without intersecting it anywhere, and find the corresponding point of application of the force P. By drawing a family of tangents to the section, we obtain a set of poles corresponding to them, the geometric location of which will give the outline (contour) of the core sections.
Let, for example, be given the section shown in Fig. 4, with the main central axes x and y.
To construct the core of the section, we present five tangents, four of which coincide with the sides AB, DE, EF and FA, and the fifth connects points B and D. By measuring or calculating from the cutting, cut off by the indicated tangents I-I, . . . ., 5-5 on the x, y axes and substituting these values in dependence (13), we determine the coordinates x p, y p for the five poles 1, 2....5, corresponding to the five positions of the zero line. Tangent I-I can be moved to position 2-2 by rotating around point A, while pole I must move in a straight line and, as a result of rotating the tangent, move to point 2. Consequently, all poles corresponding to intermediate positions of the tangent between I-I and 2-2 will be located on straight 1-2. Similarly, it can be proven that the remaining sides of the core of the section will also be rectangular, i.e. the core of the section is a polygon, to construct which it is enough to connect poles 1, 2, ... 5 with straight lines.
Bending with torsion round timber.
When bending with torsion in the cross section of a beam, in the general case, five internal force factors are not equal to zero: M x, M y, M k, Q x and Q y. However, in most cases, the influence of shear forces Q x and Q y can be neglected if the section is not thin-walled.
Normal stresses in a cross section can be determined from the magnitude of the resulting bending moment
because the neutral axis is perpendicular to the cavity of action of the moment M u.
In Fig. Figure 5 shows the bending moments M x and M y in the form of vectors (the directions M x and M y are chosen positive, i.e., such that at the points of the first quadrant the stress sections are tensile).
The direction of the vectors M x and M y is chosen in such a way that an observer, looking from the end of the vector, sees them directed counterclockwise. In this case, the neutral line coincides with the direction of the resulting moment vector M u, and the most loaded points of the section A and B lie in the plane of action of this moment.
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 of the following types of deformation: 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 under the combined action of 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. The largest normal stresses in absolute value occur 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. At fragile 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 coinciding with cross sections timber, 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
