For a cantilever beam loaded with a distributed load of intensity kN/m and a concentrated moment of kN m (Fig. 3.12), it is required to: construct diagrams of shear forces and bending moments, select a beam of circular cross-section with an allowable normal stress kN/cm2 and check the strength of the beam according to tangential stresses with permissible tangential stress kN/cm2. Beam dimensions m; m; m.
Calculation scheme for the problem of direct transverse bending
Rice. 3.12
Solution of the problem "straight transverse bending"
Determining support reactions
The horizontal reaction in the embedment is zero, since external loads in the z-axis direction do not act on the beam.
We choose the directions of the remaining reactive forces arising in the embedment: we will direct the vertical reaction, for example, downward, and the moment – clockwise. Their values are determined from the static equations:
When composing these equations, we consider the moment to be positive when rotating counterclockwise, and the projection of the force to be positive if its direction coincides with the positive direction of the y-axis.
From the first equation we find the moment at the seal:
From the second equation - vertical reaction:
Received by us positive values for the moment and vertical reaction in the embedment indicate that we guessed their directions.
In accordance with the nature of the fastening and loading of the beam, we divide its length into two sections. Along the boundaries of each of these sections we will outline four cross sections (see Fig. 3.12), in which we will use the method of sections (ROZU) to calculate the values of shearing forces and bending moments.
Section 1. Let's mentally discard the right side of the beam. Let's replace its action on the remaining left side with a cutting force and a bending moment. For the convenience of calculating their values, let’s cover the discarded right side of the beam with a piece of paper, aligning the left edge of the sheet with the section under consideration.
Let us recall that the shear force arising in any cross section, must balance all external forces (active and reactive) that act on the part of the beam we are considering (that is, visible). Therefore, the shearing force must be equal to the algebraic sum of all the forces that we see.
Let us also present the rule of signs for the shearing force: an external force acting on the part of the beam under consideration and tending to “rotate” this part relative to the section in a clockwise direction causes a positive shearing force in the section. Such an external force is included in the algebraic sum for the definition with a plus sign.
In our case, we see only the reaction of the support, which rotates the part of the beam visible to us relative to the first section (relative to the edge of the piece of paper) counterclockwise. That's why
kN.
The bending moment in any section must balance the moment created by the external forces visible to us relative to the section in question. Consequently, it is equal to the algebraic sum of the moments of all forces that act on the part of the beam we are considering, relative to the section under consideration (in other words, relative to the edge of the piece of paper). Wherein external load, bending the part of the beam under consideration with a convex downward direction, causes a positive bending moment in the section. And the moment created by such a load is included in the algebraic sum for determination with a “plus” sign.
We see two efforts: reaction and closing moment. However, the force's leverage relative to section 1 is zero. That's why
kNm.
We took the “plus” sign because the reactive moment bends the part of the beam visible to us with a convex downward.
Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now, unlike the first section, the force has a shoulder: m. Therefore
kN; kNm.
Section 3. Closing the right side of the beam, we find
kN;
Section 4. Cover the left side of the beam with a sheet. Then
kNm.
kNm.
.
Using the found values, we construct diagrams of shearing forces (Fig. 3.12, b) and bending moments (Fig. 3.12, c).
Under unloaded areas, the diagram of shearing forces goes parallel to the axis of the beam, and under a distributed load q - along an inclined straight line upward. Under the support reaction in the diagram there is a jump down by the value of this reaction, that is, by 40 kN.
In the diagram of bending moments we see a break under the support reaction. The bend angle is directed towards the support reaction. Under a distributed load q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6 on the diagram there is an extremum, since the diagram of the shearing force in this place passes through the zero value.
Determine the required cross-sectional diameter of the beam
The normal stress strength condition has the form:
,
where is the moment of resistance of the beam during bending. For a beam of circular cross-section it is equal to:
.
The largest absolute value of the bending moment occurs in the third section of the beam: 
kN cm
Then the required beam diameter is determined by the formula
cm.
We accept mm. Then
kN/cm2 kN/cm2.
"Overvoltage" is
,
what is allowed.
We check the strength of the beam by the highest shear stresses
The greatest tangential stresses arising in the cross section of a beam of circular cross-section are calculated by the formula
,
where is the cross-sectional area.
According to the diagram, the largest algebraic value of the shearing force is equal to 
kN. Then
kN/cm2 kN/cm2,
that is, the strength condition for tangential stresses is also satisfied, and with a large margin.
An example of solving the problem "straight transverse bending" No. 2
Condition of an example problem on straight transverse bending
For a simply supported beam loaded with a distributed load of intensity kN/m, concentrated force kN and concentrated moment kN m (Fig. 3.13), it is necessary to construct diagrams of shear forces and bending moments and select a beam of I-beam cross-section with an allowable normal stress kN/cm2 and permissible tangential stress kN/cm2. Beam span m.
An example of a straight bending problem - calculation diagram
 |
Rice. 3.13
Solution of an example problem on straight bending
Determining support reactions
For a given simply supported beam, it is necessary to find three support reactions: , and . Since only vertical loads perpendicular to its axis act on the beam, the horizontal reaction of the fixed hinged support A is zero: .
The directions of vertical reactions are chosen arbitrarily. Let us direct, for example, both vertical reactions upward. To calculate their values, let’s create two static equations:
Let us recall that the resultant of the linear load , uniformly distributed over a section of length l, is equal to , that is, equal to the area of the diagram of this load and it is applied at the center of gravity of this diagram, that is, in the middle of the length.
;
kN.
Let's check: .
Recall that forces whose direction coincides with the positive direction of the y-axis are projected (projected) onto this axis with a plus sign:
that is true.
We construct diagrams of shearing forces and bending moments
We divide the length of the beam into separate sections. The boundaries of these sections are the points of application of concentrated forces (active and/or reactive), as well as points corresponding to the beginning and end of the distributed load. There are three such sections in our problem. Along the boundaries of these sections, we will outline six cross sections, in which we will calculate the values of shear forces and bending moments (Fig. 3.13, a).
Section 1. Let's mentally discard the right side of the beam. For the convenience of calculating the shearing force and bending moment arising in this section, we will cover the part of the beam we discarded with a piece of paper, aligning the left edge of the sheet of paper with the section itself.
The shearing force in the beam section is equal to the algebraic sum of all external forces(active and reactive) that we see. IN in this case we see the reaction of the support and the linear load q distributed over an infinitesimal length. The resultant linear load is zero. That's why
kN.
The plus sign is taken because the force rotates the part of the beam visible to us relative to the first section (the edge of a piece of paper) clockwise.
The bending moment in the beam section is equal to the algebraic sum of the moments of all the forces that we see relative to the section under consideration (that is, relative to the edge of the piece of paper). We see the support reaction and linear load q distributed over an infinitesimal length. However, the force has a leverage of zero. The resultant linear load is also zero. That's why
Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now we see the reaction and load q acting on a section of length . The resultant linear load is equal to . It is attached in the middle of a section of length . That's why
Let us recall that when determining the sign of the bending moment, we mentally free the part of the beam we see from all the actual supporting fastenings and imagine it as if pinched in the section under consideration (that is, we mentally imagine the left edge of the piece of paper as a rigid embedment).
Section 3. Let's close the right side. We get
Section 4. Cover the right side of the beam with a sheet. Then
Now, to check the correctness of the calculations, let’s cover the left side of the beam with a piece of paper. We see the concentrated force P, the reaction of the right support and the linear load q distributed over an infinitesimal length. The resultant linear load is zero. That's why
kNm.
That is, everything is correct.
Section 5. As before, close the left side of the beam. Will have
kN;
kNm.
Section 6. Let's close the left side of the beam again. We get
kN;
Using the found values, we construct diagrams of shearing forces (Fig. 3.13, b) and bending moments (Fig. 3.13, c).
We make sure that under the unloaded area the diagram of shearing forces runs parallel to the axis of the beam, and under a distributed load q - along a straight line sloping downwards. There are three jumps in the diagram: under the reaction - up by 37.5 kN, under the reaction - up by 132.5 kN and under the force P - down by 50 kN.
In the diagram of bending moments we see kinks under concentrated force P and under support reactions. The fracture angles are directed towards these forces. Under a distributed load of intensity q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. Under the concentrated moment there is a jump of 60 kN m, that is, by the magnitude of the moment itself. In section 7 on the diagram there is an extremum, since the diagram of the shearing force for this section passes through the zero value (). Let us determine the distance from section 7 to the left support.
Forces acting perpendicular to the axis of the beam and located in a plane passing through this axis cause deformation called transverse bending. If the plane of action of the mentioned forces – main plane, then a straight (flat) transverse bend occurs. Otherwise, the bend is called oblique transverse. A beam that is subject to predominantly bending is called beam 1 .
Essentially, transverse bending is a combination of pure bending and shear. In connection with the curvature of cross sections due to the uneven distribution of shears along the height, the question arises about the possibility of using the normal stress formula σ X, derived for pure bending based on the hypothesis of plane sections.
1 A single-span beam, having at the ends, respectively, one cylindrical fixed support and one cylindrical movable one in the direction of the beam axis, is called simple. A beam with one end clamped and the other free is called console. A simple beam having one or two parts hanging over a support is called console.
If, in addition, the sections are taken far from the places where the load is applied (at a distance not less than half the height of the section of the beam), then it can be assumed, as in the case of pure bending, that the fibers do not exert pressure on each other. This means that each fiber experiences uniaxial tension or compression.
Under the action of a distributed load, the transverse forces in two adjacent sections will differ by an amount equal to qdx. Therefore, the curvature of the sections will also be slightly different. In addition, the fibers will exert pressure on each other. A thorough study of the issue shows that if the length of the beam l quite large compared to its height h (l/ h> 5), then even with a distributed load, these factors do not have a significant effect on the normal stresses in the cross section and therefore may not be taken into account in practical calculations.
a B C
Rice. 10.5 Fig. 10.6
In sections under concentrated loads and near them, the distribution of σ X deviates from the linear law. This deviation, which is local in nature and is not accompanied by an increase in the highest stresses (in the outermost fibers), is usually not taken into account in practice.
Thus, with transverse bending (in the plane xy) normal stresses are calculated using the formula
σ X= – [M z(x)/Iz]y.
If we draw two adjacent sections on a section of the beam that is free from load, then the transverse force in both sections will be the same, and therefore the curvature of the sections will be the same. In this case, any piece of fiber ab(Fig. 10.5) will move to a new position a"b", without undergoing additional elongation, and therefore, without changing the value of the normal stress.
Let us determine the tangential stresses in the cross section through their paired stresses acting in the longitudinal section of the beam.
Select an element of length from the timber dx(Fig. 10.7 a). Let's draw a horizontal section at a distance at from neutral axis z, dividing the element into two parts (Fig. 10.7) and consider the equilibrium of the upper part, which has a base
width b. In accordance with the law of pairing of tangential stresses, the stresses acting in the longitudinal section are equal to the stresses acting in the cross section. Taking this into account, under the assumption that the shear stresses in the site b distributed uniformly, using the condition ΣХ = 0, we obtain:
N * - (N * +dN *)+
where: N * is the resultant of normal forces σ in the left cross section of the element dx within the “cut off” area A * (Fig. 10.7 d):
where: S = - static moment of the “cut off” part of the cross section (shaded area in Fig. 10.7 c). Therefore, we can write:
Then we can write:
This formula was obtained in the 19th century by the Russian scientist and engineer D.I. Zhuravsky and bears his name. And although this formula is approximate, since it averages the stress over the width of the section, the calculation results obtained from it are in good agreement with the experimental data.
In order to determine the shear stresses at an arbitrary cross-section point located at a distance y from the z axis, you should:
Determine from the diagram the magnitude of the transverse force Q acting in the section;
Calculate the moment of inertia I z of the entire section;
Draw a plane parallel to the plane through this point xz and determine the section width b;
Calculate the static moment of the clipped area S relative to the main central axis z and substitute the found values into the Zhuravsky formula.
Let us determine, as an example, tangential stresses in a rectangular cross section (Fig. 10.6, c). Static moment about the axis z parts of the section above line 1-1, on which the stress is determined, will be written in the form:
It varies according to the law of a square parabola. Section width V for a rectangular beam is constant, then the law of change in tangential stresses in the section will also be parabolic (Fig. 10.6, c). At y = and y = − the tangential stresses are zero, and on the neutral axis z they reach their greatest value.
For a beam of circular cross section on the neutral axis we have.
Bend is the type of loading of a beam in which a moment is applied to it lying in a plane passing through the longitudinal axis. Bending moments occur in the cross sections of the beam. When bending, deformation occurs in which the axis of a straight beam bends or the curvature of a curved beam changes.
A beam that bends is called beam . A structure consisting of several bendable rods, most often connected to each other at an angle of 90°, is called frame .
The bend is called flat or straight , if the load plane passes through the main central axis of inertia of the section (Fig. 6.1).
Fig.6.1
When plane transverse bending occurs in a beam, two types of internal forces arise: transverse force Q and bending moment M. In a frame with flat transverse bending, three forces arise: longitudinal N, transverse Q forces and bending moment M.
If the bending moment is the only internal force factor, then such bending is called clean (Fig. 6.2). When there is a shear force, bending is called transverse . Strictly speaking, to simple types resistance relates only to pure bending; transverse bending is conventionally classified as a simple type of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected when calculating strength.
22.Flat transverse bend. Differential dependencies between internal forces and external load. Between the bending moment, shear force and the intensity of the distributed load, there are differential dependencies based on the Zhuravsky theorem, named after the Russian bridge engineer D.I. Zhuravsky (1821-1891).
This theorem is formulated as follows:
The transverse force is equal to the first derivative of the bending moment along the abscissa of the beam section.
23. Flat transverse bend. Plotting diagrams of shear forces and bending moments. Determination of shear forces and bending moments - section 1
Let's discard the right side of the beam and replace its action on the left side with a transverse force and a bending moment. For ease of calculation, let’s cover the discarded right side of the beam with a piece of paper, aligning the left edge of the sheet with the section 1 under consideration.
The transverse force in section 1 of the beam is equal to the algebraic sum of all external forces that are visible after closure
We see only the reaction of the support directed downward. Thus, the shear force is:
kN.
We took the “minus” sign because the force rotates the part of the beam visible to us relative to the first section counterclockwise (or because it is in the same direction as the direction of the transverse force according to the sign rule)
The bending moment in section 1 of the beam is equal to the algebraic sum of the moments of all the forces that we see after closing the discarded part of the beam, relative to the section 1 under consideration.
We see two forces: the reaction of the support and the moment M. However, the force has a shoulder that is practically equal to zero. Therefore, the bending moment is equal to: 
kNm.
Here we took the “plus” sign because the external moment M bends the part of the beam visible to us with a convex downward. (or because it is opposite to the direction of the bending moment according to the sign rule)
Determination of shear forces and bending moments - section 2
Unlike the first section, the reaction force now has a shoulder equal to a.
shear force:
kN;
bending moment:
Determination of shear forces and bending moments - section 3
shear force:
bending moment:
Determination of shear forces and bending moments - section 4
Now it's more convenient cover the left side of the beam with a sheet.
shear force:
bending moment:
Determination of shear forces and bending moments - section 5
shear force:
bending moment:
Determination of shear forces and bending moments - section 1
shear force and bending moment:
.
Using the found values, we construct a diagram of transverse forces (Fig. 7.7, b) and bending moments (Fig. 7.7, c).
CONTROL OF THE CORRECTNESS OF CONSTRUCTION OF DIAGRAMS
Let's make sure that diagrams are constructed correctly based on external features, using the rules for constructing diagrams.
Checking the shear force diagram
We are convinced: under unloaded areas the diagram of transverse forces runs parallel to the axis of the beam, and under a distributed load q - along a downward inclined straight line. On the diagram longitudinal force three jumps: under reaction – down by 15 kN, under force P – down by 20 kN and under reaction – up by 75 kN.
Checking the bending moment diagram
In the diagram of bending moments we see kinks under the concentrated force P and under the support reactions. The fracture angles are directed towards these forces. Under a distributed load q, the diagram of bending moments changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6 on the diagram of the bending moment there is an extremum, since the diagram of the transverse force in this place passes through the zero value.
Bending deformation consists in curvature of the axis of a straight rod or in a change in the initial curvature of a straight rod (Fig. 6.1). Let's get acquainted with the basic concepts that are used when considering bending deformation.
Rods that bend are called beams.
Clean called bending, in which the bending moment is the only internal force factor arising in the cross section of the beam.
More often, in the cross section of the rod, along with the bending moment, a transverse force also arises. This bending is called transverse.
Flat (straight) called bending when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section.
At oblique bend the plane of action of the bending moment intersects the cross section of the beam along a line that does not coincide with any of the main central axes of the cross section.
We begin our study of bending deformation with the case of pure plane bending.
Normal stresses and strains during pure bending.
As already mentioned, with pure plane bending in the cross section, of the six internal force factors, only the bending moment is nonzero (Fig. 6.1, c):
Experiments carried out on elastic models show that if a grid of lines is applied to the surface of the model (Fig. 6.1, a), then with pure bending it deforms as follows (Fig. 6.1, b):
a) longitudinal lines are curved along the circumference;
b) the contours of the cross sections remain flat;
c) the contour lines of the sections intersect everywhere with the longitudinal fibers at right angles.
Based on this, it can be assumed that in pure bending, the cross sections of the beam remain flat and rotate so that they remain normal to the curved axis of the beam (flat sections in bending hypothesis).
Rice. 6.1
By measuring the length of the longitudinal lines (Fig. 6.1, b), you can find that the upper fibers lengthen when the beam bends, and the lower ones shorten. Obviously, it is possible to find fibers whose length remains unchanged. A set of fibers that do not change their length when a beam is bent is called neutral layer (n.s.). The neutral layer intersects the cross section of the beam in a straight line, which is called neutral line (n.l.) section.
To derive a formula that determines the magnitude of normal stresses arising in the cross section, consider a section of the beam in a deformed and undeformed state (Fig. 6.2).
Rice. 6.2
Using two infinitesimal cross sections, we select an element of length 
. Before deformation, sections bounding the element 
, were parallel to each other (Fig. 6.2, a), and after deformation they bent slightly, forming an angle 
. The length of the fibers lying in the neutral layer does not change when bending 
. Let us denote the radius of curvature of the trace of the neutral layer on the drawing plane by the letter 
. Let us determine the linear deformation of an arbitrary fiber 
, located at a distance 
from the neutral layer.
The length of this fiber after deformation (arc length 
) is equal to 
. Considering that before deformation all fibers had the same length 
, we find that the absolute elongation of the fiber under consideration
It's obvious that 
, since the length of the fiber lying in the neutral layer has not changed. Then after substitution 
we get
(6.2)
Therefore, the relative longitudinal strain is proportional to the distance of the fiber from the neutral axis.
Let us introduce the assumption that when bending, the longitudinal fibers do not press on each other. Under this assumption, each fiber is deformed in isolation, experiencing simple tension or compression, in which 
. Taking into account (6.2)
, (6.3)
that is, normal stresses are directly proportional to the distances of the cross-section points under consideration from the neutral axis.
Let us substitute dependence (6.3) into the expression for the bending moment 
in cross section (6.1)
.
Recall that the integral 
represents the moment of inertia of the section relative to the axis 
.
(6.4)
Dependence (6.4) represents Hooke's law for bending, since it relates the deformation (curvature of the neutral layer 
) with a moment acting in the section. Work 
is called the section stiffness during bending, N m 2.
Let's substitute (6.4) into (6.3)
(6.5)
This is the required formula for determining normal stresses during pure bending of a beam at any point in its cross-section.
In order to establish where the neutral line is located in the cross section, we substitute the value of normal stresses into the expression for the longitudinal force 
and bending moment 
Because the 
,
;
(6.6)
(6.7)
Equality (6.6) indicates that the axis 
– neutral axis of the section – passes through the center of gravity of the cross section.
Equality (6.7) shows that 
And 
- the main central axes of the section.
According to (6.5), the highest voltage is achieved in the fibers furthest from the neutral line
Attitude 
represents the axial moment of resistance of the section 
relative to its central axis 
, Means
Meaning 
for the simplest cross sections the following:
For rectangular cross section
, (6.8)
Where 
- side of the section perpendicular to the axis 
;
- side of the section parallel to the axis 
;
For round cross section
, (6.9)
Where 
- diameter of the circular cross-section.
The strength condition for normal bending stresses can be written in the form
(6.10)
All formulas obtained were obtained for the case of pure bending of a straight rod. The action of the transverse force leads to the fact that the hypotheses underlying the conclusions lose their strength. However, calculation practice shows that even during transverse bending of beams and frames, when in the section, in addition to the bending moment 
there is also a longitudinal force 
and shear force 
, you can use the formulas given for pure bending. The error is insignificant.
To visually represent the nature of the deformation of beams (rods) during bending, the following experiment is carried out. On side faces rubber beam rectangular section a grid of lines is applied parallel and perpendicular to the axis of the beam (Fig. 30.7, a). Then moments are applied to the beam at its ends (Fig. 30.7, b), acting in the plane of symmetry of the beam, intersecting each of its cross sections along one of the main central axes of inertia. The plane passing through the axis of the beam and one of the main central axes of inertia of each of its cross sections will be called the main plane.
Under the influence of moments, the beam experiences a straight pure bend. As a result of deformation, as experience shows, the grid lines parallel to the axis of the beam are bent, maintaining the same distances between them. When indicated in Fig. 30.7, b in the direction of the moments, these lines in the upper part of the beam are lengthened, and in the lower part they are shortened.
Each grid line perpendicular to the axis of the beam can be considered as a trace of the plane of some cross section of the beam. Since these lines remain straight, it can be assumed that the cross sections of the beam, flat before deformation, remain flat during deformation.
This assumption, based on experience, is known as the hypothesis of plane sections, or Bernoulli's hypothesis (see § 6.1).
The hypothesis of plane sections applies not only to pure bending, but also to transverse bending. For transverse bending it is approximate, and for pure bending it is strict, which is confirmed by theoretical studies carried out using methods of elasticity theory.
Let us now consider a straight beam with a cross section symmetrical about the vertical axis, embedded at the right end and loaded at the left end with an external moment acting in one of the main planes of the beam (Fig. 31.7). In each cross section of this beam, only bending moments occur acting in the same plane as the moment
Thus, the beam is in a state of straight, pure bend throughout its entire length. Individual sections of the beam may be in a state of pure bending even if it is subject to transverse loads; for example, section 11 of the beam shown in Fig. experiences pure bending. 32.7; in the sections of this section the shear force
From the beam under consideration (see Fig. 31.7) we select an element of length . As a result of deformation, as follows from Bernoulli's hypothesis, the sections will remain flat, but will tilt relative to each other by a certain angle. Let us take the left section conditionally as stationary. Then, as a result of rotating the right section through an angle, it will take the position (Fig. 33.7).
The straight lines will intersect at a certain point A, which is the center of curvature (or, more precisely, the trace of the axis of curvature) of the longitudinal fibers of the element. The upper fibers of the element in question when shown in Fig. 31.7 in the direction of the moment are lengthened, and the lower ones are shortened. The fibers of some intermediate layer perpendicular to the plane of action of the moment retain their length. This layer is called the neutral layer.
Let us denote the radius of curvature of the neutral layer, i.e., the distance from this layer to the center of curvature A (see Fig. 33.7). Let's consider a certain layer located at a distance y from the neutral layer. The absolute elongation of the fibers of this layer is equal to and the relative elongation
Considering similar triangles we establish that Therefore,
In the theory of bending, it is assumed that the longitudinal fibers of the beam do not press on each other. Experimental and theoretical research show that this assumption does not significantly affect the calculation results.
With pure bending, shear stresses do not arise in the cross sections of the beam. Thus, all fibers in pure bending are under conditions of uniaxial tension or compression.
According to Hooke's law, for the case of uniaxial tension or compression, the normal stress o and the corresponding relative deformation are related by the dependence
or based on formula (11.7)
From formula (12.7) it follows that the normal stresses in the longitudinal fibers of the beam are directly proportional to their distances y from the neutral layer. Consequently, in the cross section of the beam at each point, the normal stresses are proportional to the distance y from this point to the neutral axis, which is the line of intersection of the neutral layer with the cross section (Fig.
34.7, a). From the symmetry of the beam and the load it follows that the neutral axis is horizontal.
At the points of the neutral axis, normal stresses are zero; on one side of the neutral axis they are tensile, and on the other they are compressive.
The stress diagram o is a graph bounded by a straight line, with the largest absolute values of stress for the points furthest from the neutral axis (Fig. 34.7b).
Let us now consider the equilibrium conditions of the selected beam element. Let us represent the action of the left part of the beam on the section of the element (see Fig. 31.7) in the form of a bending moment; the remaining internal forces in this section with pure bending are equal to zero. Let us imagine the action of the right side of the beam on the cross-section of the element in the form of elementary forces applied to each elementary area of the cross-section (Fig. 35.7) and parallel to the axis of the beam.
Let's create six equilibrium conditions for an element
Here are the sums of the projections of all forces acting on the element, respectively, on the axes - the sums of the moments of all forces relative to the axes (Fig. 35.7).
The axis coincides with the neutral axis of the section and the y-axis is perpendicular to it; both of these axes are located in the cross-sectional plane
An elementary force does not produce projections on the y-axis and does not cause a moment about the axis. Therefore, the equilibrium equations are satisfied for any values of o.
The equilibrium equation has the form
Let us substitute the value of a into equation (13.7) according to formula (12.7):
Since (a curved beam element is considered, for which), then
The integral represents the static moment of the cross section of the beam about the neutral axis. Its equality to zero means that the neutral axis (i.e., the axis) passes through the center of gravity of the cross section. Thus, the center of gravity of all cross sections of the beam, and therefore the axis of the beam, which is the geometric location of the centers of gravity, are located in the neutral layer. Therefore, the radius of curvature of the neutral layer is the radius of curvature of the curved axis of the beam.
Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element relative to the neutral axis:
Here represents a moment of elementary inner strength relative to the axis.
Let us denote the area of the cross-section of the beam located above the neutral axis - below the neutral axis.
Then it will represent the resultant of elementary forces applied above the neutral axis, below the neutral axis (Fig. 36.7).
Both of these resultants are equal to each other in absolute value, since their algebraic sum, based on condition (13.7), is equal to zero. These resultants form an internal pair of forces acting in the cross section of the beam. The moment of this pair of forces, equal to the product of the magnitude of one of them and the distance between them (Fig. 36.7), is a bending moment in the cross section of the beam.
Let us substitute the value of a into equation (15.7) according to formula (12.7):
Here represents the axial moment of inertia, i.e., the axis passing through the center of gravity of the section. Hence,
Let's substitute the value from formula (16.7) into formula (12.7):
When deriving formula (17.7), it was not taken into account that with an external torque directed, as shown in Fig. 31.7, according to the accepted sign rule, the bending moment is negative. If we take this into account, then we must put a minus sign in front of the right side of formula (17.7). Then, with a positive bending moment in the upper zone of the beam (i.e., at ), the values of a will turn out to be negative, which will indicate the presence of compressive stresses in this zone. However, usually the minus sign is not placed on the right side of formula (17.7), and this formula is used only to determine the absolute values of stresses a. Therefore, the absolute values of the bending moment and the ordinate y should be substituted into formula (17.7). The sign of the stresses is always easily determined by the sign of the moment or by the nature of the deformation of the beam.
Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element relative to the y-axis:
Here it represents the moment of the elementary internal force about the y-axis (see Fig. 35.7).
Let us substitute the value of a into expression (18.7) according to formula (12.7):
Here the integral represents the centrifugal moment of inertia of the cross section of the beam relative to the y and axis. Hence,
But since
As is known (see § 7.5), the centrifugal moment of inertia of the section is equal to zero relative to the main axes of inertia.
In the case under consideration, the y-axis is the axis of symmetry of the cross-section of the beam and, therefore, the y-axes and are the main central axes of inertia of this section. Therefore, condition (19.7) is satisfied here.
In the case when the cross section of the bent beam does not have any axis of symmetry, condition (19.7) is satisfied if the plane of action of the bending moment passes through one of the main central axes of inertia of the section or is parallel to this axis.
If the plane of action of the bending moment does not pass through any of the main central axes of inertia of the cross section of the beam and is not parallel to it, then condition (19.7) is not satisfied and, therefore, there is no direct bending - the beam experiences oblique bending.
Formula (17.7), which determines the normal stress at an arbitrary point of the beam section under consideration, is applicable provided that the plane of action of the bending moment passes through one of the main axes of inertia of this section or is parallel to it. In this case, the neutral axis of the cross section is its main central axis of inertia, perpendicular to the plane of action of the bending moment.
Formula (16.7) shows that during direct pure bending, the curvature of the curved axis of the beam is directly proportional to the product of the elastic modulus E and the moment of inertia. We will call the product the stiffness of the section during bending; it is expressed in, etc.
In pure bending of a beam of constant cross-section, the bending moments and section stiffnesses are constant along its length. In this case, the radius of curvature of the curved axis of the beam has a constant value [see. expression (16.7)], that is, the beam bends along a circular arc.
From formula (17.7) it follows that the largest (positive - tensile) and smallest (negative - compressive) normal stresses in the cross section of the beam arise at the points furthest from the neutral axis, located on both sides of it. For a cross section symmetrical about the neutral axis, the absolute values of the greatest tensile and compressive stresses are the same and can be determined by the formula
For sections that are not symmetrical about the neutral axis, for example, for a triangle, tee, etc., the distances from the neutral axis to the most distant stretched and compressed fibers are different; Therefore, for such sections there are two moments of resistance:
where are the distances from the neutral axis to the most distant stretched and compressed fibers.
