Simple types of resistance. flat bend

When building diagrams of bending momentsM at builders accepted: ordinates expressing on a certain scale positive values of bending moments, set aside stretched fibers, i.e. - down, A negative - up from the beam axis. Therefore, they say that builders construct diagrams on stretched fibers. At the mechanics positive values of both shear force and bending moment are postponed up. Mechanics draw diagrams on compressed fibers.

Principal stresses when bending. Equivalent voltages.

In the general case of direct bending in the cross sections of a beam, normal And tangentsvoltage. These voltages vary both along the length and height of the beam.

Thus, in the case of bending, there is plane stress state.

Let's consider a diagram where the beam is loaded with force P

Largest normal tensions arise in extreme, points most distant from the neutral line, and There are no shear stresses in them. Thus, for extreme fibers non-zero principal stresses are normal stresses in cross section.

At the neutral line level in the cross section of the beam there are highest shear stress, A normal stresses are zero. means in the fibers neutral layer the principal stresses are determined by the values of the tangential stresses.

In this design scheme, the upper fibers of the beam will be stretched, and the lower ones will be compressed. To determine the principal stresses we use the well-known expression:

Full stress analysis Let's imagine it in the picture.

Bending Stress Analysis

Maximum principal stress σ 1 is on upper extreme fibers and equals zero on the lower outermost fibers. Main stress σ 3 has the largest absolute value is on the lower fibers.

Trajectory of principal stresses depends on load type And method of securing the beam.

When solving problems it is enough separately check normal And separately tangential stresses. However sometimes the most stressful turn out to be intermediate fibers in which there are both normal and shear stresses. This happens in sections where simultaneously, both the bending moment and the shear force reach large values- this can be in the embedding of a cantilever beam, on the support of a beam with a cantilever, in sections under concentrated force, or in sections with sharply changing widths. For example, in an I-section the most dangerous the junction of the wall and the shelf- there are significant both normal and shear stresses.

The material is in a plane stress state and is required check for equivalent voltages.

Strength conditions for beams made of plastic materials By third(theory of maximum tangential stresses) And fourth(theory of energy of shape changes) theories of strength.

As a rule, in rolled beams the equivalent stresses do not exceed the normal stresses in the outermost fibers and special check not required. Another thing - composite metal beams, who have the wall is thinner than for rolled profiles at the same height. Welded composite beams made of steel sheets are more often used. Calculation of such beams for strength: a) selection of the section - height, thickness, width and thickness of the beam chords; b) checking strength by normal and tangential stresses; c) checking strength using equivalent stresses.

Determination of shear stresses in an I-section. Let's consider the section I-beam S x =96.9 cm 3 ; Yх=2030 cm 4 ; Q=200 kN

To determine the shear stress, it is used formula,where Q is the shear force in the section, S x 0 is the static moment of the part cross section, located on one side of the layer in which the shear stress is determined, I x is the moment of inertia of the entire cross section, b is the width of the section in the place where the shear stress is determined

Let's calculate maximum shear stress:

Let's calculate the static moment for top shelf:

Now let's calculate shear stress:

We are building shear stress diagram:

Let us consider the cross section of a standard profile in the form I-beam and define shear stress, acting parallel to the shear force:

Let's calculate static moments simple figures:

This value can be calculated and otherwise, using the fact that for the I-beam and trough sections the static moment of half the section is given. To do this, it is necessary to subtract from the known value of the static moment the value of the static moment to the line A 1 B 1:

The tangential stresses at the junction of the flange and the wall change spasmodically, because sharp wall thickness varies from t st to b.

Diagrams of tangential stresses in the walls of trough, hollow rectangular and other sections have the same form as in the case of an I-section. The formula includes the static moment of the shaded part of the section relative to the X axis, and the denominator includes the width of the section (net) in the layer where the shear stress is determined.

Let us determine the tangential stresses for a circular section.

Since the shear stresses at the section contour must be directed tangent to the contour, then at points A And IN at the ends of any chord parallel to the diameter AB, shear stresses are directed perpendicular to the radii OA And OV. Hence, directions tangential stresses at points A, V, K converge at some point N on the Y axis.

Static moment of the cut-off part:

That is, the shear stresses change according to parabolic law and will be maximum at the level of the neutral line, when y 0 =0

Formula for determining shear stress (formula)

Consider a rectangular section

At a distance y 0 from the central axis we draw section 1-1 and determine the tangential stresses. Static moment area cut off part:

It should be borne in mind that it is fundamental indifferent, take the static moment of area shaded or remaining part cross section. Both static moments equal and opposite in sign, so their sum, which represents static moment of area of the entire section relative to the neutral line, namely the central x axis, will be equal to zero.

Moment of inertia of a rectangular section:

Then shear stress according to the formula

The variable y 0 is included in the formula in second degrees, i.e. tangential stresses in a rectangular section vary according to law of a square parabola.

Shear stress reached maximum at the level of the neutral line, i.e. When y 0 =0:

, Where A is the area of the entire section.

Strength condition for tangential stresses has the form:

, Where S x 0– static moment of the part of the cross section located on one side of the layer in which the tangential stresses are determined, Ix– moment of inertia of the entire cross section, b– section width in the place where the shear stress is determined, Q-lateral force, τ - shear stress, [τ] — permissible tangential stress.

This strength condition allows us to produce three type of calculation (three types of problems when calculating strength):

1. Verification calculation or strength test based on tangential stresses:

2. Selection of section width (for a rectangular section):

3. Determination of permissible lateral force (for a rectangular section):

To determine tangents stresses, consider a beam loaded with forces.

The task of determining stresses is always statically indeterminate and requires involvement geometric And physical equations. However, it is possible to accept such hypotheses about the nature of stress distribution that the task will become statically definable.

By two infinitely close cross sections 1-1 and 2-2 we select dz element, Let's depict it on a large scale, then draw a longitudinal section 3-3.

In sections 1–1 and 2–2, normal σ 1, σ 2 stresses, which are determined by the well-known formulas:

Where M - bending moment in cross section, dM - increment bending moment at length dz

Lateral force in sections 1–1 and 2–2 is directed along the main central axis Y and, obviously, represents the sum of the vertical components of internal tangential stresses distributed over the section. In strength of materials it is usually taken assumption of their uniform distribution across the width of the section.

To determine the magnitude of shear stresses at any point in the cross section located at a distance y 0 from the neutral X axis, draw a plane parallel to the neutral layer (3-3) through this point and remove the clipped element. We will determine the voltage acting across the ABCD area.

Let's project all the forces onto the Z axis

The resultant of the internal longitudinal forces along the right side will be equal to:

Where A 0 – area of the façade edge, S x 0 – static moment of the cut-off part relative to the X axis. Similarly on the left side:

Both resultants directed towards to each other, since the element is in compressed beam area. Their difference is balanced by the tangential forces on the lower edge of 3-3.

Let's assume that shear stress τ distributed across the width of the beam cross section b evenly. This assumption is the more likely the smaller the width compared to the height of the section. Then resultant of tangential forces dT equal to the stress value multiplied by the area of the face:

Let's compose now equilibrium equation Σz=0:

or where from

Let's remember differential dependencies , according to which Then we get the formula:

This formula is called formulas. This formula was obtained in 1855. Here S x 0 – static moment of part of the cross section, located on one side of the layer in which the shear stresses are determined, I x – moment of inertia the entire cross section, b – section width in the place where the shear stress is determined, Q - shear force in cross section.

— bending strength condition, Where

- maximum moment (modulo) from the diagram of bending moments; - axial moment of resistance of the section, geometric characteristic; - permissible stress (σ adm)

- maximum normal voltage.

If the calculation is carried out according to limit state method, then instead of the permissible voltage, we enter into the calculation design resistance of the material R.

Types of flexural strength calculations

1. Check calculation or testing of strength using normal stresses

2. Design calculation or selection of section

3. Definition permissible load (definition lifting capacity and or operational carrier capabilities)

When deriving the formula for calculating normal stresses, we consider the case of bending, when the internal forces in the sections of the beam are reduced only to bending moment, A the shear force turns out to be zero. This case of bending is called pure bending . Consider the middle section of the beam, which is subject to pure bending.

When loaded, the beam bends so that it The lower fibers lengthen and the upper fibers shorten.

Since part of the fibers of the beam is stretched, and part is compressed, and the transition from tension to compression occurs smoothly, without jumps, V average part of the beam is located 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 or neutral axis sections. Neutral lines are strung on the axis of the beam. Neutral line is the line in which normal stresses are zero.

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 hypothesis of plane sections (conjecture). 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.

Assumptions for deriving normal stress formulas: 1) The hypothesis of plane sections is fulfilled. 2) Longitudinal fibers do not press on each other (non-pressure hypothesis) and, therefore, each of the fibers is in a state of uniaxial tension or compression. 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.

Let's consider a beam of arbitrary cross-section, but having an axis of symmetry. Bending moment represents resultant moment of internal normal forces, arising on infinitely small areas and can be expressed in integral form: (1), where y is the arm of the elementary force relative to the x axis

Formula (1) expresses static side of the problem of bending a straight beam, but along it at a known bending moment It is impossible to determine normal stresses until the law of their distribution is established.

Let us select the beams in the middle section and consider section of length dz, subject to bending. Let's depict it on an enlarged scale.

Sections limiting the area dz, parallel to each other until deformed, and after applying the load rotate around their neutral lines by an angle . The length of the neutral layer fiber segment will not change. and will be equal to: , where is this radius of curvature the curved axis of the beam. But any other fiber lying lower or higher neutral layer, will change its length. Let's calculate relative elongation of fibers located at a distance y from the neutral layer. Relative elongation is the ratio of absolute deformation to the original length, then:

Let's reduce by and bring similar terms, then we get: (2) This formula expresses geometric side of the pure bending problem: The deformations of the fibers are directly proportional to their distances to the neutral layer.

Now let's move on to stresses, i.e. we will consider physical side of the task. in accordance with non-pressure assumption we use fibers under axial tension-compression: then, taking into account the formula (2) we have (3), those. normal stress when bending along the section height linearly distributed. On the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero. Let's substitute (3) into the equation (1) and take the fraction out of the integral sign as a constant value, then we have . But the expression is axial moment of inertia of the section relative to the x axis - I x. Its dimension cm 4, m 4

Then ,where (4) ,where is the curvature of the curved axis of the beam, and is the rigidity of the beam section during bending.

Let's substitute the resulting expression curvature (4) into expression (3) and we get formula for calculating normal stresses at any point in the cross section: (5)

That. maximum tensions arise at points furthest from the neutral line. Attitude (6) called axial moment of section resistance. Its dimension cm 3, m 3. The moment of resistance characterizes the influence of the shape and size of the cross section on the magnitude of stress.

Then maximum voltages: (7)

Bending strength condition: (8)

When transverse bending occurs not only normal, but also shear stresses, because available shear force. Shear stress complicate the picture of deformation, they lead to curvature cross sections of the beam, resulting in the hypothesis of plane sections is violated. However, research shows that distortions introduced by shear stresses slightly affect normal stresses calculated by the formula (5) . Thus, when determining normal stresses in the case transverse bending The theory of pure bending is quite applicable.

Neutral line. Question about the position of the neutral line.

During bending there is no longitudinal force, so we can write Let us substitute here the formula for normal stresses (3) and we get Since the modulus of longitudinal elasticity of the beam material is not equal to zero and the curved axis of the beam has a finite radius of curvature, it remains to assume that this integral is static moment of area cross section of the beam relative to the neutral line-axis x , and, since it is equal to zero, then the neutral line passes through the center of gravity of the section.

The condition (absence of moment of internal forces relative to the field line) will give or taking into account (3) . For the same reasons (see above) . In integrand - the centrifugal moment of inertia of the section relative to the x and y axes is zero, which means these axes are main and central and make up direct corner. Hence, The force and neutral lines in a straight bend are mutually perpendicular.

Having installed neutral line position, easy to build normal stress diagram along the section height. Her linear character is determined equation of the first degree.

The nature of the diagram σ for symmetrical sections relative to the neutral line, M<0

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 not equal to zero (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 bending stiffness of the section, 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
,

;

(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 - 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.

Flat transverse bending of beams. Internal bending forces. Differential dependencies of internal forces. Rules for checking diagrams of internal bending forces. Normal and shear stresses during bending. Strength calculation based on normal and tangential stresses.

10. SIMPLE TYPES OF RESISTANCE. FLAT BEND

10.1. General concepts and definitions

Bending 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.

Plane bending is a 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 of the cross sections and the geometric axis of the beam (x-axis).

Oblique bending is a bending in which the loads act in one plane that does not coincide with the main planes of inertia.

Complex bending is a bending in which loads act in different (arbitrary) planes.

10.2. Determination of internal bending forces

Let's consider two typical cases of bending: in the first, the cantilever beam is bent by a concentrated moment M o ; 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 M z and shear force Q y (or when bending relative to another main axis - bending moment M y and shear force Q z).

Moreover, in accordance with the two loading cases considered, plane bending can be divided into pure and transverse.

Pure bending is a flat bending in which only one out of six internal forces occurs in the sections of the rod - 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, simple types of resistance include only 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.

When determining internal efforts, we will adhere to the following rule of signs:

1) the transverse force Q y is considered positive if it tends to rotate the beam element in question clockwise;

2) bending moment M z 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 loads during bending, as well as characteristic features of diagrams Q and M, knowledge of which will facilitate the construction of diagrams and allow us to control their correctness. For convenience of notation, we will denote: M ≡ M z, Q ≡ Q y.

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, element dx will also be in equilibrium under the action of shear forces, bending moments and external load applied to it. Since Q and M generally change along the axis of the beam, transverse forces Q and Q +dQ, as well as bending moments M and M +dM will appear in sections of the element dx. From the equilibrium condition of the selected element we obtain

∑ F y = 0 Q + q dx − (Q + dQ) = 0;

∑ M 0 = 0 M + Q dx + q dx dx 2 − (M + dM ) = 0.

From the second equation, neglecting the term q dx (dx /2) as an infinitesimal quantity of the second order, we find

Relations (10.1), (10.2) and (10.3) are called differential dependencies 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, diagrams Q are limited by inclined straight lines, and diagrams M are limited by quadratic parabolas. Moreover, if we construct diagram M “on a stretched fiber,” then the convexity of the pa-

the work will be directed in the direction of action q, and the extremum will be located in the section where the diagram Q intersects the base line;

c – in sections where a concentrated force is applied to the beam, on diagram Q there will be jumps by the magnitude and in the direction of this force, and on 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 on the epi-

there will be no changes in re Q, and on the diagram M there will be jumps by the value 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 by methods of resistance of materials, it is necessary to introduce some assumptions.

There are three such hypotheses for bending:

a – hypothesis of plane sections (Bernoulli hypothesis)

– sections that are flat 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

niy – 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 – co-

The gray longitudinal fibers do not press on each other.

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 of the 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, plane bending 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).

When determining internal efforts, we will adhere to the following rule of 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).

10.3. Differential dependencies during bending

Let us establish some relationships between internal forces and external loads during bending, as well as the characteristic features of the Q and M diagrams, knowledge of which will facilitate the construction of diagrams and allow us to control their correctness. For convenience of notation, we will denote: M≡Mz, Q≡Qy.

axis of the beam, then transverse forces Q and Q+dQ, as well as bending moments M and M+dM, will appear 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, diagrams Q are limited by inclined straight lines, and diagrams M 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 diagram Q there will be jumps by the magnitude and in the direction of this force, and on 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.

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, simple types of resistance include only 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. There are differential relationships between the bending moment, shear force and the intensity of the distributed load, 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 downwards. 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 of the longitudinal force there are three jumps: under the reaction - down by 15 kN, under the force P - down by 20 kN and under the 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.