10.1. General concepts and definitions
Bend- this is a type of loading in which the rod is loaded with moments in planes passing through the longitudinal axis of the rod.
A rod that bends is called a beam (or timber). In the future, we will consider rectilinear beams, the cross section of which has at least one axis of symmetry.
The resistance of materials is divided into flat, oblique and complex bending.
Flat bend– bending, in which all the forces bending the beam lie in one of the planes of symmetry of the beam (in one of the main planes).
The main planes of inertia of a beam are the planes passing through the main axes cross sections and the geometric axis of the beam (x-axis).
Oblique bend– bending, in which the loads act in one plane that does not coincide with the main planes of inertia.
Complex bend – bending, in which loads act in different (arbitrary) planes.
10.2. Determination of internal bending forces
Let us consider two typical cases of bending: in the first, the cantilever beam is bent by a concentrated moment Mo; in the second - concentrated force F.
Using the method of mental sections and composing equilibrium equations for the cut off parts of the beam, we determine the internal forces in both cases:
The remaining equilibrium equations are obviously identically equal to zero.
Thus, in the general case of plane bending in the section of a beam, out of six internal forces, two arise - bending moment Mz and shear force Qy (or when bending relative to another main axis - bending moment My and shear force Qz).
Moreover, in accordance with the two loading cases considered, flat bend can be divided into pure and transverse.
Clean bend– flat bending, in which in the sections of the rod, out of six internal forces, only one arises – a bending moment (see the first case).
Transverse bend– bending, in which in the sections of the rod, in addition to the internal bending moment, a transverse force also arises (see the second case).
Strictly speaking, to simple types resistance only applies pure bend; 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.
When determining internal efforts, we will adhere to next rule signs:
1) the transverse force Qy is considered positive if it tends to rotate the beam element in question clockwise;
2) bending moment Mz is considered positive if, when bending a beam element, the upper fibers of the element are compressed and the lower fibers are stretched (umbrella rule).
Thus, we will build the solution to the problem of determining the internal forces during bending according to the following plan: 1) at the first stage, considering the equilibrium conditions of the structure as a whole, we determine, if necessary, the unknown reactions of the supports (note that for a cantilever beam the reactions in the embedment can be and not found if we consider the beam from the free end); 2) at the second stage, we select characteristic sections of the beam, taking as the boundaries of the sections the points of application of forces, points of change in the shape or size of the beam, points of fastening of the beam; 3) at the third stage, we determine the internal forces in the sections of the beam, considering the conditions of equilibrium of the beam elements in each section.
10.3. Differential dependencies during bending
Let us establish some relationships between internal forces and external bending loads, as well as characteristics diagrams Q and M, knowledge of which will facilitate the construction of diagrams and allow you to control their correctness. For convenience of notation, we will denote: M≡Mz, Q≡Qy.
Let us select a small element dx in a section of a beam with an arbitrary load in a place where there are no concentrated forces and moments. Since the entire beam is in equilibrium, the element dx will also be in equilibrium under the action of the forces applied to it shear forces, bending moments and external load. Since Q and M generally vary along
axis of the beam, then transverse forces Q and Q+dQ, as well as bending moments M and M+dM, will arise in the sections of element dx. From the equilibrium condition of the selected element we obtain
The first of the two equations written gives the condition
From the second equation, neglecting the term q dx (dx/2) as an infinitesimal quantity of the second order, we find
Considering expressions (10.1) and (10.2) together we can obtain
Relations (10.1), (10.2) and (10.3) are called differential dependences of D.I. Zhuravsky during bending.
Analysis of the above differential dependencies during bending allows us to establish some features (rules) for constructing diagrams of bending moments and transverse forces: a - in areas where there is no distributed load q, diagrams Q are limited to straight lines parallel to the base, and diagrams M are limited to inclined straight lines; b – in areas where a distributed load q is applied to the beam, Q diagrams are limited by inclined straight lines, and M diagrams are limited by quadratic parabolas.
Moreover, if we construct diagram M “on a stretched fiber,” then the convexity of the parabola will be directed in the direction of action q, and the extremum will be located in the section where diagram Q intersects the base line; c – in sections where a concentrated force is applied to the beam, on the diagram Q there will be jumps by the magnitude and in the direction of this force, and on the diagram M there will be kinks, the tip directed in the direction of action of this force; d – in sections where a concentrated moment is applied to the beam, there will be no changes on diagram Q, and on diagram M there will be jumps in the magnitude of this moment; d – in areas where Q>0, the moment M increases, and in areas where Q<0, момент М убывает (см. рисунки а–г).
10.4. Normal stresses during pure bending of a straight beam
Let us consider the case of pure plane bending of a beam and derive a formula for determining normal stresses for this case.
Note that in the theory of elasticity it is possible to obtain an exact dependence for normal stresses during pure bending, but if this problem is solved using methods of strength of materials, it is necessary to introduce some assumptions.
There are three such hypotheses for bending:
a – hypothesis of flat sections (Bernoulli hypothesis) – flat sections before deformation remain flat after deformation, but only rotate relative to a certain line, which is called the neutral axis of the beam section. In this case, the fibers of the beam lying on one side of the neutral axis will stretch, and on the other, compress; fibers lying on the neutral axis do not change their length;
b – hypothesis about the constancy of normal stresses - stresses acting at the same distance y from the neutral axis are constant across the width of the beam;
c – hypothesis about the absence of lateral pressures – adjacent longitudinal fibers do not press on each other.
Static side of the problem
To determine the stresses in the cross sections of the beam, we consider, first of all, the static sides of the problem. Using the method of mental sections and composing equilibrium equations for the cut-off part of the beam, we will find the internal forces during bending. As was shown earlier, the only internal force acting in the beam section during pure bending is the internal bending moment, which means that normal stresses associated with it will arise here.
We will find the relationship between internal forces and normal stresses in the beam section by considering the stresses on the elementary area dA, selected in the cross section A of the beam at the point with coordinates y and z (the y axis is directed downward for convenience of analysis):
As we see, the problem is internally statically indeterminate, since the nature of the distribution of normal stresses over the section is unknown. To solve the problem, consider the geometric picture of deformations.
Geometric side of the problem
Let us consider the deformation of a beam element of length dx, separated from a bending rod at an arbitrary point with coordinate x. Taking into account the previously accepted hypothesis of flat sections, after bending the beam section, rotate relative to the neutral axis (n.o.) by an angle dϕ, while the fiber ab, spaced from the neutral axis at a distance y, will turn into an arc of a circle a1b1, and its length will change by some size. Let us recall here that the length of the fibers lying on the neutral axis does not change, and therefore the arc a0b0 (the radius of curvature of which is denoted by ρ) has the same length as the segment a0b0 before the deformation a0b0=dx.
Let us find the relative linear deformation εx of the fiber ab of the curved beam.
The hypothesis of plane sections during bending can be explained with an example: let us apply a grid consisting of longitudinal and transverse (perpendicular to the axis) straight lines on the side surface of an undeformed beam. As a result of bending the beam, the longitudinal lines will take on a curved outline, while the transverse lines will practically remain straight and perpendicular to the curved axis of the beam.
Formulation of the plane section hypothesis: cross sections that are flat and perpendicular to the axis of the beam before , remain flat and perpendicular to the curved axis after it is deformed.
This circumstance indicates: when fulfilled plane section hypothesis, as with and
In addition to the hypothesis of flat sections, the assumption is accepted: the longitudinal fibers of the beam do not press on each other when it bends.
The plane section hypothesis and assumption are called Bernoulli's hypothesis.
Consider a beam of rectangular cross-section undergoing pure bending (). Let's select a beam element with a length (Fig. 7.8. a). As a result of bending, the cross sections of the beam will rotate, forming an angle. The upper fibers experience compression, and the lower fibers experience tension. We denote the radius of curvature of the neutral fiber as .
Conventionally, we assume that the fibers change their length while remaining straight (Fig. 7.8. b). Then the absolute and relative elongations of the fiber located at a distance y from the neutral fiber:
Let us show that longitudinal fibers, which do not experience either tension or compression when the beam bends, pass through the main central axis x.
Since the length of the beam does not change during bending, the longitudinal force (N) arising in the cross section must be zero. Elementary longitudinal force.
Given the expression :
The factor can be taken out of the integral sign (does not depend on the integration variable).
The expression represents the cross section of the beam about the neutral x-axis. It is zero when the neutral axis passes through the center of gravity of the cross section. Consequently, the neutral axis (zero line) when the beam bends passes through the center of gravity of the cross section.
Obviously: the bending moment is associated with normal stresses arising at points in the cross section of the rod. Elementary bending moment created by an elementary force:
,
where is the axial moment of inertia of the cross section relative to the neutral x-axis, and the ratio is the curvature of the beam axis.
Rigidity beams in bending(the larger, the smaller the radius of curvature).
The resulting formula represents Hooke's law of bending for a rod: The bending moment occurring in the cross section is proportional to the curvature of the beam axis.
Expressing the radius of curvature () from the formula of Hooke’s law for a rod during bending and substituting its value into the formula , we obtain a formula for normal stresses () at an arbitrary point in the cross section of the beam located at a distance y from the neutral axis x: .
In the formula for normal stresses () at an arbitrary point in the cross section of the beam, the absolute values of the bending moment () and the distance from the point to the neutral axis (y coordinates) should be substituted. Whether the stress at a given point will be tensile or compressive can be easily determined by the nature of the deformation of the beam or by the diagram of bending moments, the ordinates of which are plotted on the side of the compressed fibers of the beam.
From the formula it is clear: normal stresses () change along the height of the cross section of the beam according to a linear law. In Fig. 7.8, shows the diagram. The greatest stresses during beam bending occur at points furthest from the neutral axis. If a line is drawn in the cross section of the beam parallel to the neutral x axis, then equal normal stresses arise at all its points.
Simple analysis normal stress diagrams shows that when a beam bends, the material located near the neutral axis practically does not work. Therefore, in order to reduce the weight of the beam, it is recommended to choose cross-sectional shapes in which most of the material is removed from the neutral axis, such as an I-section.
Bend is called deformation in which the axis of the rod and all its fibers, i.e. longitudinal lines parallel to the axis of the rod, are bent under the influence of external forces. The simplest case of bending occurs when external forces lie in a plane passing through the central axis of the rod and do not produce projections onto this axis. This type of bending is called transverse bending. There are flat bends and oblique bends.
Flat bend- such a case when the curved axis of the rod is located in the same plane in which external forces act.
Oblique (complex) bend– a case of bending when the bent axis of the rod does not lie in the plane of action of external forces.
A bending rod is usually called beam.
During flat transverse bending of beams in a section with the coordinate system y0x, two internal forces can arise - transverse force Q y and bending moment M x; in what follows we introduce the notation for them Q And M. If there is no transverse force in a section or section of a beam (Q = 0), and the bending moment is not zero or M is const, then such a bend is usually called clean.
Lateral force in any section of the beam is numerically equal to the algebraic sum of the projections onto the axis of all forces (including support reactions) located on one side (either) of the drawn section.
Bending moment in a beam section is numerically equal to the algebraic sum of the moments of all forces (including support reactions) located on one side (any) of the drawn section relative to the center of gravity of this section, more precisely, relative to the axis passing perpendicular to the drawing plane through the center of gravity of the drawn section.
Force Q is resultant distributed over the cross-section of internal shear stress, A moment M– sum of moments around the central axis of section X internal normal stress.
There is a differential relationship between internal forces
which is used in constructing and checking Q and M diagrams.
Since some of the fibers of the beam are stretched, and some are compressed, and the transition from tension to compression occurs smoothly, without jumps, in the middle part of the beam there is a layer whose fibers only bend, but do not experience either tension or compression. This layer is called neutral layer. The line along which the neutral layer intersects the cross section of the beam is called neutral line th or neutral axis sections. Neutral lines are strung on the axis of the beam.
Lines drawn on the side surface of the beam perpendicular to the axis remain flat when bending. These experimental data make it possible to base the conclusions of the formulas on the hypothesis of plane sections. According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and turn out to be perpendicular to the curved axis of the beam when it is bent. The cross section of the beam is distorted when bending. Due to transverse deformation, the cross-sectional dimensions in the compressed zone of the beam increase, and in the tensile zone they are compressed.
Assumptions for deriving formulas. Normal voltages
1) The hypothesis of plane sections is fulfilled.
2) Longitudinal fibers do not press on each other and, therefore, under the influence of normal stresses, linear tension or compression operates.
3) Deformations of fibers do not depend on their position along the cross-sectional width. Consequently, normal stresses, changing along the height of the section, remain the same along the width.
4) The beam has at least one plane of symmetry, and all external forces lie in this plane.
5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same.
6) The relationship between the dimensions of the beam is such that it operates under plane bending conditions without warping or twisting.
In case of pure bending of a beam, only normal stress, determined by the formula:
where y is the coordinate of an arbitrary section point, measured from the neutral line - the main central axis x.
Normal bending stresses along the height of the section are distributed over linear law. On the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero.
The nature of normal stress diagrams for symmetrical sections relative to the neutral line
The nature of normal stress diagrams for sections that do not have symmetry with respect to the neutral line
Dangerous points are the points furthest from the neutral line.
Let's choose some section
For any point of the section, let's call it a point TO, the beam strength condition for normal stresses has the form:
, where n.o. - This neutral axis
This axial section modulus relative to the neutral axis. Its dimension is cm 3, m 3. The moment of resistance characterizes the influence of the shape and dimensions of the cross section on the magnitude of the stresses.
Normal stress strength condition:
The normal stress is equal to the ratio of the maximum bending moment to the axial moment of resistance of the section relative to the neutral axis.
If the material does not equally resist tension and compression, then two strength conditions must be used: for the tensile zone with the permissible tensile stress; for a compression zone with permissible compressive stress.
During transverse bending, the beams on the platforms in its cross-section act as normal, so tangents voltage.
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.
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
Its relative deformation 
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.
We will start with the simplest case, the so-called pure bend.
Pure bending is a special case of bending in which the transverse force in the sections of the beam is zero. Pure bending can only occur when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads causing pure
bending, shown in Fig. 88. In sections of these beams, where Q = 0 and, therefore, M = const; pure bending takes place.
The forces in any section of the beam during pure bending are reduced to a pair of forces, the plane of action of which passes through the axis of the beam, and the moment is constant.
Voltages can be determined based on the following considerations.
1. The tangential components of forces along elementary areas in the cross section of a beam cannot be reduced to a pair of forces whose plane of action is perpendicular to the section plane. It follows that the bending force in the section is the result of action along elementary areas
only normal forces, and therefore with pure bending the stresses are reduced only to normal.
2. In order for efforts on elementary sites to be reduced to only a couple of forces, among them there must be both positive and negative. Therefore, both tension and compression fibers of the beam must exist.
3. Due to the fact that the forces in different sections are the same, the stresses at the corresponding points of the sections are the same.
Let's consider some element near the surface (Fig. 89, a). Since no forces are applied along its lower edge, which coincides with the surface of the beam, there are no stresses on it. Therefore, there are no stresses on the upper edge of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we arrive at
The same conclusion, etc. It follows that there are no stresses along the horizontal edges of any element. Considering the elements that make up the horizontal layer, starting with the element near the surface of the beam (Fig. 90), we come to the conclusion that there are no stresses along the lateral vertical edges of any element. Thus, the stress state of any element (Fig. 91, a), and in the limit, fibers, should be represented as shown in Fig. 91,b, i.e. it can be either axial tension or axial compression.
4. Due to the symmetry of the application of external forces, the section along the middle of the length of the beam after deformation should remain flat and normal to the axis of the beam (Fig. 92, a). For the same reason, sections in quarters of the length of the beam also remain flat and normal to the axis of the beam (Fig. 92, b), unless the extreme sections of the beam during deformation remain flat and normal to the axis of the beam. A similar conclusion is valid for sections in eighths of the length of the beam (Fig. 92, c), etc. Consequently, if during bending the outer sections of the beam remain flat, then for any section it remains 
It is a fair statement that after deformation it remains flat and normal to the axis of the curved beam. But in this case, it is obvious that the change in elongation of the fibers of the beam along its height should occur not only continuously, but also monotonically. If we call a layer a set of fibers that have the same elongations, then it follows from what has been said that the stretched and compressed fibers of the beam should be located on opposite sides of the layer in which the elongations of the fibers are equal to zero. We will call fibers whose elongations are zero neutral; a layer consisting of neutral fibers is a neutral layer; the line of intersection of the neutral layer with the cross-sectional plane of the beam - the neutral line of this section. Then, based on the previous reasoning, it can be argued that with pure bending of a beam, in each section there is a neutral line that divides this section into two parts (zones): a zone of stretched fibers (stretched zone) and a zone of compressed fibers (compressed zone). ). Accordingly, at the points of the stretched zone of the section, normal tensile stresses should act, at the points of the compressed zone - compressive stresses, and at the points of the neutral line the stresses are equal to zero.
Thus, with pure bending of a beam of constant cross-section:
1) only normal stresses act in sections;
2) the entire section can be divided into two parts (zones) - stretched and compressed; the boundary of the zones is the neutral section line, at the points of which the normal stresses are equal to zero;
3) any longitudinal element of the beam (in the limit, any fiber) is subjected to axial tension or compression, so that adjacent fibers do not interact with each other;
4) if the extreme sections of the beam during deformation remain flat and normal to the axis, then all its cross sections remain flat and normal to the axis of the curved beam.
Let us consider an element of a beam subject to pure bending, concluding 
located between sections m-m and n-n, which are spaced one from the other at an infinitesimal distance dx (Fig. 93). Due to position (4) of the previous paragraph, sections m- m and n - n, which were parallel before deformation, after bending, remaining flat, will form an angle dQ and intersect along a straight line passing through point C, which is the center of curvature neutral fiber NN. Then the part AB of the fiber enclosed between them, located at a distance z from the neutral fiber (the positive direction of the z axis is taken towards the convexity of the beam during bending), will turn after deformation into an arc AB. A piece of neutral fiber O1O2, having turned into an arc, O1O2 will not change its length, while fiber AB will receive an elongation:
before deformation
after deformation
where p is the radius of curvature of the neutral fiber.
Therefore, the absolute lengthening of segment AB is equal to
and relative elongation
Since, according to position (3), fiber AB is subjected to axial tension, then during elastic deformation
This shows that normal stresses along the height of the beam are distributed according to a linear law (Fig. 94). Since the equal force of all forces over all elementary cross-sectional areas must be equal to zero, then
from where, substituting the value from (5.8), we find
But the last integral is a static moment about the Oy axis, perpendicular to the plane of action of the bending forces.
Due to its equality to zero, this axis must pass through the center of gravity O of the section. Thus, the neutral line of the section of the beam is a straight line y, perpendicular to the plane of action of bending forces. It is called the neutral axis of the beam section. Then from (5.8) it follows that the stresses at points lying at the same distance from the neutral axis are the same.
The case of pure bending, in which the bending forces act in only one plane, causing bending only in that plane, is planar pure bending. If the said plane passes through the Oz axis, then the moment of elementary forces relative to this axis should be equal to zero, i.e.
Substituting here the value of σ from (5.8), we find
The integral on the left side of this equality, as is known, is the centrifugal moment of inertia of the section relative to the y and z axes, so
The axes about which the centrifugal moment of inertia of the section is zero are called the main axes of inertia of this section. If they, in addition, pass through the center of gravity of the section, then they can be called the main central axes of inertia of the section. Thus, with flat pure bending, the direction of the plane of action of bending forces and the neutral axis of the section are the main central axes of inertia of the latter. In other words, to obtain a flat, pure bend of a beam, a load cannot be applied to it arbitrarily: it must be reduced to forces acting in a plane that passes through one of the main central axes of inertia of the sections of the beam; in this case, the other main central axis of inertia will be the neutral axis of the section.
As is known, in the case of a section that is symmetrical about any axis, the axis of symmetry is one of its main central axes of inertia. Consequently, in this particular case we will certainly obtain pure bending by applying appropriate loads in a plane passing through the longitudinal axis of the beam and the axis of symmetry of its section. A straight line perpendicular to the axis of symmetry and passing through the center of gravity of the section is the neutral axis of this section.
Having established the position of the neutral axis, it is not difficult to find the magnitude of the stress at any point in the section. In fact, since the sum of the moments of elementary forces relative to the neutral axis yy must be equal to the bending moment, then
whence, substituting the value of σ from (5.8), we find
Since the integral 
is. moment of inertia of the section relative to the yy axis, then
and from expression (5.8) we obtain
The product EI Y is called the bending stiffness of the beam.
The greatest tensile and largest compressive stresses in absolute value act at the points of the section for which the absolute value of z is greatest, i.e., at the points furthest from the neutral axis. With the notation, Fig. 95 we have
The value Jy/h1 is called the moment of resistance of the section to tension and is designated Wyr; similarly, Jy/h2 is called the moment of resistance of the section to compression
and denote Wyc, so
and therefore
If the neutral axis is the axis of symmetry of the section, then h1 = h2 = h/2 and, therefore, Wyp = Wyc, so there is no need to distinguish them, and they use the same notation:
calling W y simply the moment of resistance of the section. Consequently, in the case of a section symmetrical about the neutral axis,
All the above conclusions were obtained on the basis of the assumption that the cross sections of the beam, when bent, remain flat and normal to its axis (hypothesis of flat sections). As has been shown, this assumption is valid only in the case when the extreme (end) sections of the beam remain flat during bending. On the other hand, from the hypothesis of plane sections it follows that elementary forces in such sections should be distributed according to a linear law. Therefore, for the validity of the resulting theory of flat pure bending, it is necessary that the bending moments at the ends of the beam be applied in the form of elementary forces distributed along the height of the section according to a linear law (Fig. 96), coinciding with the law of stress distribution along the height of the section beams. However, based on the Saint-Venant principle, it can be argued that changing the method of applying bending moments at the ends of the beam will cause only local deformations, the effect of which will affect only a certain distance from these ends (approximately equal to the height of the section). The sections located throughout the rest of the length of the beam will remain flat. Consequently, the stated theory of flat pure bending for any method of applying bending moments is valid only within the middle part of the length of the beam, located from its ends at distances approximately equal to the height of the section. From here it is clear that this theory is obviously inapplicable if the height of the section exceeds half the length or span of the beam.
