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 action 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 tension zone they compress.
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 is a type of deformation in which the longitudinal axis of the beam is bent. Straight beams that bend are called beams. Direct bending is a bend in which the external forces acting on the beam lie in one plane (force plane) passing through the longitudinal axis of the beam and the main central axis of inertia of the cross section.
The bend is called pure, if in any cross section beam there is only one bending moment.
Bending, in which a bending moment and a transverse force simultaneously act in the cross section of a beam, is called transverse. The line of intersection of the force plane and the cross-sectional plane is called the force line.
Internal force factors during beam bending.
During plane transverse bending, two internal force factors arise in the beam sections: transverse force Q and bending moment M. To determine them, the method of sections is used (see lecture 1). The transverse force Q in the beam section is equal to the algebraic sum of the projections onto the section plane of all external forces acting on one side of the section under consideration.
Sign rule for shear forces Q:
The bending moment M in a beam section is equal to the algebraic sum of the moments relative to the center of gravity of this section of all external forces acting on one side of the section under consideration.
Sign rule for bending moments M:
Zhuravsky's differential dependencies.
Differential relationships have been established between the intensity q of the distributed load, the expressions for the transverse force Q and the bending moment M:
Based on these dependencies, the following can be distinguished: general patterns diagrams of transverse forces Q and bending moments M:
Features of diagrams of internal force factors during bending.
1. In the section of the beam where there is no distributed load, the Q diagram is presented straight line , parallel to the base of the diagram, and diagram M - an inclined straight line (Fig. a).
2. In the section where a concentrated force is applied, Q should be on the diagram leap , equal to value this force, and on the diagram M - breaking point (Fig. a).
3. In the section where a concentrated moment is applied, the value of Q does not change, and the diagram M has leap , equal to the value of this moment (Fig. 26, b).
4. On the beam section with distributed load intensity q, the diagram Q changes according to a linear law, and the diagram M - according to a parabolic law, and the convexity of the parabola is directed towards the direction of the distributed load (Fig. c, d).
5. If, within a characteristic section, the diagram Q intersects the base of the diagram, then in the section where Q = 0, the bending moment has an extreme value M max or M min (Fig. d).
Normal bending stresses.
Determined by the formula:
The moment of resistance of a section to bending is the quantity:
Dangerous cross section during bending, the cross section of the beam in which the maximum normal stress occurs is called.
Shear stresses during straight bending.
Determined by Zhuravsky's formula for shear stresses during straight beam bending:
where S ots is the static moment of the transverse area of the cut-off layer of longitudinal fibers relative to the neutral line.
Calculations of bending strength.
1. At verification calculation The maximum design stress is determined and compared with the permissible stress:
2. At design calculation the selection of the beam section is made from the condition:
3. When determining the permissible load, the permissible bending moment is determined from the condition:
Bending movements.
Under the influence of bending load, the axis of the beam bends. In this case, tension of the fibers is observed on the convex part and compression on the concave part of the beam. In addition, there is a vertical movement of the centers of gravity of the cross sections and their rotation relative to the neutral axis. To characterize bending deformation, the following concepts are used:
Beam deflection Y- movement of the center of gravity of the cross section of the beam in the direction perpendicular to its axis.
Deflection is considered positive if the center of gravity moves upward. The amount of deflection varies along the length of the beam, i.e. y = y(z)
Section rotation angle- angle θ through which each section rotates relative to its original position. The rotation angle is considered positive when the section is rotated counterclockwise. The magnitude of the rotation angle varies along the length of the beam, being a function of θ = θ (z).
The most common methods for determining displacements is the method Mora And Vereshchagin's rule.
Mohr's method.
The procedure for determining displacements using Mohr's method:
1. An “auxiliary system” is built and loaded with a unit load at the point where the displacement is required to be determined. If linear displacement is determined, then a unit force is applied in its direction; when determining angular displacements, a unit moment is applied.
2. For each section of the system, expressions for bending moments M f from the applied load and M 1 from the unit load are written down.
3. Over all sections of the system, Mohr’s integrals are calculated and summed, resulting in the desired displacement:
4. If the calculated displacement has a positive sign, this means that its direction coincides with the direction of the unit force. A negative sign indicates that the actual displacement is opposite to the direction of the unit force.
Vereshchagin's rule.
For the case when the diagram of bending moments from a given load has an arbitrary outline, and from a unit load – a rectilinear outline, it is convenient to use the graphic-analytical method, or Vereshchagin’s rule.
where A f is the area of the diagram of the bending moment M f from a given load; y c – ordinate of the diagram from a unit load under the center of gravity of the diagram M f; EI x is the section stiffness of the beam section. Calculations using this formula are made in sections, in each of which the straight-line diagram should be without fractures. The value (A f *y c) is considered positive if both diagrams are located on the same side of the beam, negative if they are located on different sides. A positive result of multiplying diagrams means that the direction of movement coincides with the direction of a unit force (or moment). A complex diagram M f should be divided into simple figures (the so-called “plot stratification” is used), for each of which it is easy to determine the ordinate of the center of gravity. In this case, the area of each figure is multiplied by the ordinate under its center of gravity.
Straight bend- this is a type of deformation in which two internal force factors arise in the cross sections of the rod: bending moment and transverse force.
Clean bend - This special case direct bending, in which only a bending moment occurs in the cross sections of the rod, and the transverse force is zero.
An example of a pure bend - a section CD on the rod AB. Bending moment is the quantity Pa pairs of external forces, bending. From the equilibrium of the part of the rod to the left of the cross section mn it follows that the internal forces distributed over this section are statically equivalent to the moment M, equal and opposite to the bending moment Pa.
To find the distribution of these internal forces over the cross section, it is necessary to consider the deformation of the rod.
In the simplest case, the rod has a longitudinal plane of symmetry and is subject to the action of external bending pairs of forces located in this plane. Then the bending will occur in the same plane.
Rod axis nn 1 is a line passing through the centers of gravity of its cross sections.
Let the cross section of the rod be a rectangle. Let's draw two vertical lines on its edges mm And pp. When bending, these lines remain straight and rotate so that they remain perpendicular to the longitudinal fibers of the rod.
Further theory of bending is based on the assumption that not only lines mm And pp, but the entire flat cross-section of the rod remains, after bending, flat and normal to the longitudinal fibers of the rod. Therefore, during bending, the cross sections mm And pp rotate relative to each other around axes, perpendicular to the plane bending (drawing plane). In this case, the longitudinal fibers on the convex side experience tension, and the fibers on the concave side experience compression.
Neutral surface- This is a surface that does not experience deformation when bending. (Now it is located perpendicular to the drawing, the deformed axis of the rod nn 1 belongs to this surface).
Neutral axis of section- this is the intersection of a neutral surface with any cross-section (now also located perpendicular to the drawing).
Let an arbitrary fiber be at a distance y from a neutral surface. ρ
– radius of curvature of the curved axis. Dot O– center of curvature. Let's draw a line n 1 s 1 parallel mm.ss 1– absolute fiber elongation.
Relative extension ε x fibers
It follows that deformation of longitudinal fibers proportional to distance y from the neutral surface and inversely proportional to the radius of curvature ρ .
Longitudinal elongation of the fibers of the convex side of the rod is accompanied by lateral narrowing, and the longitudinal shortening of the concave side is lateral expansion, as in the case of simple stretching and compression. Because of this, the appearance of all cross sections changes, the vertical sides of the rectangle become inclined. Lateral deformation z:
μ - Poisson's ratio.
Due to this distortion, all straight cross-sectional lines parallel to the axis z, are bent so as to remain normal to the lateral sides of the section. The radius of curvature of this curve R will be more than ρ in the same respect as ε x by absolute value more than ε z and we get
These deformations of longitudinal fibers correspond to stresses
The voltage in any fiber is proportional to its distance from the neutral axis n 1 n 2. Neutral axis position and radius of curvature ρ
– two unknowns in the equation for σ
x – can be determined from the condition that forces distributed over any cross section form a pair of forces that balances the external moment M.
All of the above is also true if the rod does not have a longitudinal plane of symmetry in which the bending moment acts, as long as the bending moment acts in the axial plane, which contains one of the two main axes cross section. These planes are called main bending planes.
When there is a plane of symmetry and the bending moment acts in this plane, deflection occurs precisely in it. Moments of internal forces relative to the axis z balance the external moment M. Moments of effort about the axis y are mutually destroyed.
Bend
Basic concepts about bending
Bending deformation is characterized by the loss of straightness or original shape by the beam line (its axis) when an external load is applied. In this case, unlike shear deformation, the beam line changes its shape smoothly.
It is easy to see that bending resistance is affected not only by the cross-sectional area of the beam (beam, rod, etc.), but also geometric shape this section.
Since the bending of a body (beam, timber, etc.) is carried out relative to any axis, the resistance to bending is affected by the value of the axial moment of inertia of the body’s section relative to this axis.
For comparison, during torsional deformation, the section of the body is subject to twisting relative to the pole (point), therefore, the resistance to torsion is influenced by the polar moment of inertia of this section.
Many structural elements can bend - axles, shafts, beams, gear teeth, levers, rods, etc.
In the strength of materials, several types of bends are considered:
- depending on the nature of the external load applied to the beam, there are pure bend And transverse bending;
- depending on the location of the plane of action of the bending load relative to the axis of the beam - straight bend And oblique bend.
Pure and transverse beam bending
Pure bending is a type of deformation in which only a bending moment occurs in any cross section of the beam ( rice. 2).
Pure bending deformation will, for example, occur if two pairs of forces equal in magnitude and opposite in sign are applied to a straight beam in a plane passing through the axis. Then in each section of the beam only bending moments will act.
If bending occurs as a result of applying a transverse force to the beam ( rice. 3), then such a bend is called transverse. In this case, in each section of the beam there is both a transverse force and a bending moment (except for the section to which the external load).
If the beam has at least one axis of symmetry, and the plane of action of the loads coincides with it, then direct bending occurs, but if this condition is not met, then oblique bending occurs.
When studying bending deformation, we will mentally imagine that the beam (timber) consists of an innumerable number of longitudinal fibers parallel to the axis.
To visualize the deformation of a straight bend, we will conduct an experiment with a rubber bar on which a grid of longitudinal and transverse lines is applied. 
Having subjected such a beam to straight bending, one can notice that ( rice. 1):
The transverse lines will remain straight during deformation, but will turn at an angle to each other;
- the sections of the beam will expand in the transverse direction on the concave side and narrow on the convex side;
- longitudinal straight lines will bend.
From this experience we can conclude that:
For pure bending, the hypothesis of plane sections is valid;
- fibers lying on the convex side are stretched, on the concave side they are compressed, and on the border between them there is a neutral layer of fibers that only bend without changing their length.
Assuming that the hypothesis that there is no pressure on the fibers is valid, it can be argued that with pure bending in the cross section of the beam, only normal tensile and compressive stresses arise, unevenly distributed over the cross section.
The line of intersection of the neutral layer with the cross-sectional plane is called neutral axis. It is obvious that on the neutral axis the normal stresses are zero.
Bending moment and shear force
As is known from theoretical mechanics, the support reactions of beams are determined by composing and solving static equilibrium equations for the entire beam. When solving problems of resistance of materials, and determining the internal force factors in the beams, we took into account the reactions of the connections along with the external loads acting on the beams.
To determine the internal force factors, we will use the section method, and we will depict the beam with only one line - the axis to which active and reactive forces are applied (loads and reaction reactions).
Let's consider two cases:
1. Two pairs of forces of equal and opposite sign are applied to a beam.
Considering the equilibrium of the part of the beam located to the left or right of section 1-1 (Fig. 2), we see that in all cross sections only a bending moment M and equal to the external moment occurs. Thus, this is a case of pure bending.
The bending moment is the resulting moment about the neutral axis of the internal normal forces acting in the cross section of the beam.
Let us pay attention to the fact that the bending moment has different direction for left and right parts beams. This indicates the unsuitability of the static sign rule when determining the sign of the bending moment.
2. Active and reactive forces (loads and reaction reactions) perpendicular to the axis are applied to the beam (rice. 3). Considering the equilibrium of the parts of the beam located on the left and right, we see that the bending moment M must act in the cross sections And
and shear force Q.
It follows from this that in the case under consideration, at the points of the cross sections there are not only normal stresses corresponding to the bending moment, but also tangent stresses corresponding to the transverse force.
The transverse force is the resultant of the internal tangential forces in the cross section of the beam.
Let us pay attention to the fact that the transverse force has the opposite direction for the left and right parts of the beam, which indicates that the rule of static signs is unsuitable when determining the sign of the transverse force.
Bending, in which a bending moment and shear force act in the cross section of the beam, is called transverse.
For a beam that is in water equilibrium under the action of a plane system of forces, the algebraic sum of the moments of all active and reactive forces relative to any point is equal to zero; therefore, the sum of the moments of external forces acting on the beam to the left of the section is numerically equal to the sum of the moments of all external forces acting on the beam to the right of the section.
Thus, the bending moment in the beam section is numerically equal to the algebraic sum of the moments relative to the center of gravity of the section of all external forces acting on the beam to the right or left of the section.
For a beam in equilibrium under the action of a plane system of forces perpendicular to the axis (i.e., a system of parallel forces), the algebraic sum of all external forces is equal to zero; therefore, the sum of external forces acting on the beam to the left of the section is numerically equal to the algebraic sum of the forces acting on the beam to the right of the section.
Thus, the transverse force in the beam section is numerically equal to the algebraic sum of all external forces acting to the right or left of the section.
Since the rules of static signs are unacceptable for establishing the signs of bending moment and shear force, we will establish other sign rules for them, namely: If an external load tends to bend the beam with its convexity downward, then the bending moment in the section is considered positive, and vice versa, if the external load tends to bend beam with a convex upwards, then the bending moment in the section is considered negative ( Fig 4,a).
If the sum of external forces lying on the left side of the section gives a resultant directed upward, then the transverse force in the section is considered positive; if the resultant is directed downward, then the transverse force in the section is considered negative; for the part of the beam located to the right of the section, the signs of the shear force will be opposite ( rice. 4,b). Using these rules, you should mentally imagine the section of the beam as rigidly clamped, and the connections as discarded and replaced by reactions.
Let us note once again that to determine the reactions of bonds, the rules of signs of statics are used, and to determine the signs of bending moment and transverse force, the rules of signs of resistance of materials are used.
The sign rule for bending moments is sometimes called the "rain rule", meaning that in the case of a downward convexity, a funnel is formed in which the rainwater(the sign is positive), and vice versa - if under the influence of loads the beam bends upward in an arc, water does not linger on it (the sign of bending moments is negative).
Materials from the "Bending" section:
Straight transverse bend occurs when all loads are applied perpendicular to the axis of the rod, lie in the same plane and, in addition, the plane of their action coincides with one of the main central axes of inertia of the section. Straight transverse bending refers to simple view resistance is flat stress state, i.e. two principal stresses are non-zero. With this type of deformation, internal forces arise: shear force and bending moment. A special case of direct transverse bending is pure bend, with such resistance there are load areas within which the transverse force becomes zero and the bending moment is non-zero. In the cross sections of the rods during direct transverse bending, normal and tangential stresses arise. Stresses are a function of internal force, in this case normal stresses are a function of bending moment, and tangential stresses are a function of shear force. For direct transverse bending, several hypotheses are introduced:
1) The cross sections of the beam, flat before deformation, remain flat and orthogonal to the neutral layer after deformation (hypothesis of plane sections or J. Bernoulli’s hypothesis). This hypothesis is satisfied under pure bending and is violated when shear forces, shear stresses, and angular deformation occur.
2) There is no mutual pressure between the longitudinal layers (hypothesis of non-pressure of fibers). From this hypothesis it follows that longitudinal fibers experience uniaxial tension or compression, therefore, with pure bending, Hooke's law is valid.
A rod undergoing bending is called beam. When bending, one part of the fibers stretches, the other part contracts. The layer of fibers located between the stretched and compressed fibers is called neutral layer, it passes through the center of gravity of the sections. The line of its intersection with the cross section of the beam is called neutral axis. Based on the introduced hypotheses for pure bending, a formula was obtained for determining normal stresses, which is also used for direct transverse bending. The normal stress can be found using the linear relationship (1), in which the ratio of the bending moment to the axial moment of inertia ( 
) in a particular section is a constant value, and the distance ( y) along the ordinate axis from the center of gravity of the section to the point at which the stress is determined varies from 0 to
.
.
(1)
To determine the shear stress during bending in 1856. Russian engineer and bridge builder D.I. Zhuravsky became addicted
.
(2)
The shear stress in a particular section does not depend on the ratio of the transverse force to the axial moment of inertia ( 
), because this value does not change within one section, but depends on the ratio of the static moment of the area of the cut-off part to the width of the section at the level of the cut-off part ( 
).
When straight transverse bending occurs movements: deflections (v
) and rotation angles (Θ
)
. To determine them, use the equations of the initial parameters method (3), which are obtained by integrating the differential equation of the curved axis of the beam ( 
).
Here v 0 , Θ 0 ,M 0 , Q 0 – initial parameters, x – distance from the origin to the section in which the displacement is determined , a– the distance from the origin of coordinates to the place of application or the beginning of the load.
Strength and stiffness calculations are made using the strength and stiffness conditions. Using these conditions, you can solve verification problems (check the fulfillment of a condition), determine the size of the cross section, or select the permissible value of the load parameter. There are several strength conditions, some of which are given below. Normal stress strength condition has the form:
,
(4)
Here 
–
moment of resistance of the section relative to the z axis, R – design resistance based on normal stresses.
Strength condition for tangential stresses looks like:
,
(5)
here the notations are the same as in Zhuravsky’s formula, and R s – calculated shear resistance or calculated resistance to tangential stresses.
Strength condition according to the third strength hypothesis or the hypothesis of the greatest tangential stresses can be written in the following form:
.
(6)
Severity conditions can be written for deflections (v ) And rotation angles (Θ ) :
where the displacement values in square brackets are valid.
Example of completing individual task No. 4 (term 2-8 weeks)