Calculate bending beam There are several options:
1. Calculation maximum load which she will endure
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle selection of beam section on two supports loaded with a uniformly distributed load or concentrated force.
To begin with, you will need to find the point (section) at which there will be a maximum moment. This depends on whether the beam is supported or embedded. Below are diagrams of bending moments for the most common schemes.

After finding the bending moment, we must find the moment of resistance Wx of this section using the formula given in the table:

Further, when dividing the maximum bending moment by the moment of resistance in a given section, we get maximum voltage in the beam and we must compare this stress with the stress that our beam of a given material can generally withstand.

For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, A for fragile(cast iron) – tensile strength. We can find the yield strength and tensile strength from the tables below.

Let's look at a couple of examples:
1. [i]You want to check whether an I-beam No. 10 (steel St3sp5) 2 meters long, rigidly embedded in the wall, will support you if you hang on it. Let your mass be 90 kg.
First we need to choose design scheme.

This diagram shows that the maximum moment will be at the seal, and since our I-beam has equal section along the entire length, then the maximum voltage will be in the termination. Let's find it:

P = m * g = 90 * 10 = 900 N = 0.9 kN

M = P * l = 0.9 kN * 2 m = 1.8 kN * m

Using the I-beam assortment table, we find the moment of resistance of I-beam No. 10.

It will be equal to 39.7 cm3. Let's convert to Cubic Meters and we get 0.0000397 m3.
Next, using the formula, we find the maximum stresses that arise in the beam.

b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa

After we have found the maximum stress that occurs in the beam, we can compare it with the maximum permissible stress equal to the limit the fluidity of steel St3sp5 is 245 MPa.

45.34 MPa is correct, which means this I-beam will withstand a mass of 90 kg.

2. [i] Since we have quite a large supply, we will solve the second problem, in which we will find the maximum possible mass that the same I-beam No. 10, 2 meters long, will support.
If we want to find the maximum mass, then we must equate the values ​​of the yield strength and the stress that will arise in the beam (b = 245 MPa = 245,000 kN*m2).

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, the solution to the problem of determining internal bending forces will be built according to next 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 may not be 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 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 the beam is attached distributed load q, 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 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.

Task. Construct diagrams Q and M for a statically indeterminate beam. Let's calculate the beams using the formula:

n= Σ R- Sh— 3 = 4 — 0 — 3 = 1

Beam once is statically indeterminate, which means one of the reactions is "extra" unknown. Let us take the support reaction as the “extra” unknown IN — R B.

A statically determinate beam, which is obtained from a given one by removing the “extra” connection, is called the main system (b).

Now this system should be presented equivalent given. To do this, load the main system given load, and at the point IN let's apply "extra" reaction R B(rice. V).

However for equivalence this not enough, since in such a beam the point IN Maybe move vertically, and in a given beam (Fig. A ) this cannot happen. Therefore we add condition, What deflection t. IN in the main system should be equal to 0. Deflection t. IN consists of deflection from the active load Δ F and from deflection from the “extra” reaction Δ R.

Then we make up condition for compatibility of movements:

Δ F + Δ R=0 (1)

Now it remains to calculate these movements (deflections).

Loading main system given load(rice .G) and we'll build load diagramM F (rice. d ).

IN T. IN Let's apply and build an ep. (rice. hedgehog ).

Using Simpson's formula we determine deflection due to active load.

Now let's define deflection from the action of “extra” reaction R B , for this we load the main system R B (rice. h ) and build a diagram of the moments from its action M R (rice. And ).

We compose and solve equation (1):

Let's build ep. Q And M (rice. k, l ).

Building a diagram Q.

Let's build a diagram M method characteristic points. We place points on the beam - these are the points of the beginning and end of the beam ( D,A ), concentrated moment ( B ), and also mark the middle of a uniformly distributed load as a characteristic point ( K ) is an additional point for constructing a parabolic curve.

We determine bending moments at points. Rule of signs cm. - .

The moment in IN we will define it as follows. First let's define:

Full stop TO let's take in middle area with a uniformly distributed load.

Building a diagram M . Plot AB – parabolic curve(umbrella rule), area ВD – straight slanted line.

For a beam, determine the support reactions and construct diagrams of bending moments ( M) and shear forces ( Q).

1. We designate supports letters A And IN and direct support reactions R A And R B .

Compiling equilibrium equations.

Examination

Write down the values R A And R B on design scheme.

2. Constructing a diagram shear forces method sections. We arrange the sections on characteristic areas(between changes). According to the dimensional thread - 4 sections, 4 sections.

sec. 1-1 move left.

The section passes through the area with evenly distributed load, mark the size z 1 to the left of the section before the start of the section. The length of the section is 2 m. Rule of signs For Q - cm.

We build according to the found value diagramQ.

sec. 2-2 move on the right.

The section again passes through the area with a uniformly distributed load, mark the size z 2 to the right from the section to the beginning of the section. The length of the section is 6 m.

Building a diagram Q.

sec. 3-3 move on the right.

sec. 4-4 move on the right.

We are building diagramQ.

3. Construction diagrams M method characteristic points.

Feature point- a point that is somewhat noticeable on the beam. These are the points A, IN, WITH, D , and also a point TO , wherein Q=0 And bending moment has an extremum. also in middle console we will put an additional point E, since in this area under a uniformly distributed load the diagram M described crooked line, and it is built at least according to 3 points.

So, the points are placed, let's start determining the values ​​​​in them bending moments. Rule of signs - see.

Sites NA, AD – parabolic curve(the “umbrella” rule for mechanical specialties or the “sail rule” for construction specialties), sections DC, SV – straight slanted lines.

Moment at a point D should be determined both left and right from point D . The very moment in these expressions Excluded. At the point D we get two values ​​with difference by the amount m – leap by its size.

Now we need to determine the moment at the point TO (Q=0). However, first we define point position TO , designating the distance from it to the beginning of the section as unknown X .

T. TO belongs second characteristic area, its equation for shear force(see above)

But the shear force incl. TO equal to 0 , A z 2 equals unknown X .

We get the equation:

Now knowing X, let's determine the moment at the point TO on the right side.

Building a diagram M . The construction can be carried out for mechanical specialties, putting aside positive values up from the zero line and using the “umbrella” rule.

For a given design of a cantilever beam, it is necessary to construct diagrams of the transverse force Q and the bending moment M, and perform a design calculation by selecting a circular section.

Material - wood, design resistance of the material R=10MPa, M=14kN m, q=8kN/m

There are two ways to construct diagrams in a cantilever beam with a rigid embedment - the usual way, having previously determined the support reactions, and without determining the support reactions, if you consider the sections, going from the free end of the beam and discarding the left part with the embedding. Let's build diagrams ordinary way.

1. Let's define support reactions.

Evenly distributed load q replace with conditional force Q= q·0.84=6.72 kN

In a rigid embedment there are three support reactions - vertical, horizontal and moment; in our case, the horizontal reaction is 0.

We'll find vertical ground reaction R A And supporting moment M A from equilibrium equations.

In the first two sections on the right there is no shear force. At the beginning of a section with a uniformly distributed load (right) Q=0, in the background - the magnitude of the reaction R A.
3. To construct, we will compose expressions for their determination in sections. Let's construct a diagram of moments on fibers, i.e. down.

(the diagram of individual moments has already been constructed earlier)

We solve equation (1), reduce by EI

Static indetermination revealed, the value of the “extra” reaction has been found. You can start constructing diagrams of Q and M for a statically indeterminate beam... We sketch the given diagram of the beam and indicate the magnitude of the reaction Rb. In this beam, reactions in the embedment can not be determined if you move from the right.

Construction Q plots for a statically indeterminate beam

Let's plot Q.

Construction of diagram M

Let us define M at the extremum point - at the point TO. First, let's determine its position. Let us denote the distance to it as unknown " X" Then

We are building a diagram of M.

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 of the cross section located on one side of the layer in which the tangential stresses are determined, I x is the moment of inertia of the entire cross section, b is the width of the section in the place where 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:

Design and verification calculations. For a beam with constructed diagrams of internal forces, select a section in the form of two channels from the condition of strength under normal stresses. Check the strength of the beam using the shear stress strength condition and the energy strength criterion. Given:

Let's show a beam with constructed diagrams Q and M

According to the diagram of bending moments, it is dangerous section C, in which M C = M max = 48.3 kNm.

Normal stress strength condition for this beam has the form σ max =M C /W X ≤σ adm . It is necessary to select a section from two channels.

Let's determine the required calculated value axial moment of resistance of the section:

For a section in the form of two channels, we accept according to two channels No. 20a, moment of inertia of each channel I x =1670cm 4, Then axial moment of resistance of the entire section:

Overvoltage (undervoltage) at dangerous points we calculate using the formula: Then we get undervoltage:

Now let's check the strength of the beam based on strength conditions for tangential stresses. According to shear force diagram dangerous are sections on section BC and section D. As can be seen from the diagram, Q max =48.9 kN.

Strength condition for tangential stresses has the form:

For channel No. 20 a: static moment of area S x 1 = 95.9 cm 3, moment of inertia of the section I x 1 = 1670 cm 4, wall thickness d 1 = 5.2 mm, average flange thickness t 1 = 9.7 mm , channel height h 1 =20 cm, shelf width b 1 =8 cm.

For transverse sections of two channels:

S x = 2S x 1 =2 95.9 = 191.8 cm 3,

I x =2I x 1 =2·1670=3340 cm 4,

b=2d 1 =2·0.52=1.04 cm.

Determining the value maximum shear stress:

τ max =48.9 10 3 191.8 10 −6 /3340 10 −8 1.04 10 −2 =27 MPa.

As seen, τ max<τ adm (27MPa<75МПа).

Hence, the strength condition is satisfied.

We check the strength of the beam according to the energy criterion.

From consideration diagrams Q and M follows that section C is dangerous, in which they operate M C =M max =48.3 kNm and Q C =Q max =48.9 kN.

Let's carry out analysis of the stress state at the points of section C

Let's define normal and shear stresses at several levels (marked on the section diagram)

Level 1-1: y 1-1 =h 1 /2=20/2=10cm.

Normal and tangent voltage:

Main voltage:

Level 2−2: y 2-2 =h 1 /2−t 1 =20/2−0.97=9.03 cm.

Main stresses:

Level 3−3: y 3-3 =h 1 /2−t 1 =20/2−0.97=9.03cm.

Normal and shear stresses:

Main stresses:

Extreme shear stress:

Level 4−4: y 4-4 =0.

(in the middle the normal stresses are zero, the tangential stresses are maximum, they were found in the strength test using tangential stresses)

Main stresses:

Extreme shear stress:

Level 5−5:

Normal and shear stresses:

Main stresses:

Extreme shear stress:

Level 6−6:

Normal and shear stresses:

Main stresses:

Extreme shear stress:

Level 7−7:

Normal and shear stresses:

Main stresses:

Extreme shear stress:

In accordance with the calculations performed stress diagrams σ, τ, σ 1, σ 3, τ max and τ min are presented in Fig.

Analysis these diagram shows, which is in the section of the beam dangerous points are at level 3-3 (or 5-5), in which:

Using energy criterion of strength, we get

From a comparison of equivalent and permissible stresses it follows that the strength condition is also satisfied

(135.3 MPa<150 МПа).

The continuous beam is loaded in all spans. Construct diagrams Q and M for a continuous beam.

1. Define degree of static indetermination beams according to the formula:

n= Sop -3= 5-3 =2, Where Sop – number of unknown reactions, 3 – number of static equations. To solve this beam it is required two additional equations.

2. Let us denote numbers supports from zero in order ( 0,1,2,3 )

3. Let us denote span numbers from the first in order ( ι 1, ι 2, ι 3)

4. We consider each span as simple beam and build diagrams for each simple beam Q and M. What pertains to simple beam, we will denote with index "0", that which relates to continuous beam, we will denote without this index. Thus, is the shear force and bending moment for a simple beam.

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 is a special case of 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 a pair of external forces causing 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 bending plane (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 in absolute value is greater 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.

Bending deformation consists in curvature of the axis of a straight rod or in a change in the initial curvature of a straight rod (Fig. 6.1). Let's get acquainted with the basic concepts that are used when considering bending deformation.

Rods that bend are called beams.

Clean called bending, in which the bending moment is the only internal force factor arising in the cross section of the beam.

More often, in the cross section of the rod, along with the bending moment, a transverse force also arises. This bending is called transverse.

Flat (straight) called bending when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section.

At oblique bend the plane of action of the bending moment intersects the cross section of the beam along a line that does not coincide with any of the main central axes of the cross section.

We begin our study of bending deformation with the case of pure plane bending.

Normal stresses and strains during pure bending.

As already mentioned, with pure plane bending in the cross section, of the six internal force factors, only the bending moment is nonzero (Fig. 6.1, c):

Experiments carried out on elastic models show that if a grid of lines is applied to the surface of the model (Fig. 6.1, a), then with pure bending it deforms as follows (Fig. 6.1, b):

a) longitudinal lines are curved along the circumference;

b) the contours of the cross sections remain flat;

c) the contour lines of the sections intersect everywhere with the longitudinal fibers at right angles.

Based on this, it can be assumed that in pure bending, the cross sections of the beam remain flat and rotate so that they remain normal to the curved axis of the beam (flat sections in bending hypothesis).

Rice. 6.1

By measuring the length of the longitudinal lines (Fig. 6.1, b), you can find that the upper fibers lengthen when the beam bends, and the lower ones shorten. Obviously, it is possible to find fibers whose length remains unchanged. A set of fibers that do not change their length when a beam is bent is called neutral layer (n.s.). The neutral layer intersects the cross section of the beam in a straight line, which is called neutral line (n.l.) section.

To derive a formula that determines the magnitude of normal stresses arising in the cross section, consider a section of the beam in a deformed and undeformed state (Fig. 6.2).

Rice. 6.2

Using two infinitesimal cross sections, we select an element of length
. Before deformation, sections bounding the element
, were parallel to each other (Fig. 6.2, a), and after deformation they bent slightly, forming an angle
. The length of the fibers lying in the neutral layer does not change when bending
. Let us denote the radius of curvature of the trace of the neutral layer on the drawing plane by the letter . Let us determine the linear deformation of an arbitrary fiber
, located at a distance from the neutral layer.

The length of this fiber after deformation (arc length
) is equal to
. Considering that before deformation all fibers had the same length
, we find that the absolute elongation of the fiber under consideration

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.

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