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 stress in 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 select a 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 yield strength of steel St3sp5 - 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 maximum weight, then we must equate the values ​​of the yield strength and stress that will arise in the beam (b = 245 MPa = 245,000 kN*m2).

The design process of modern buildings and structures is regulated a huge amount various building codes and regulations. In most cases, standards require certain characteristics to be ensured, for example, deformation or deflection of floor slab beams under static or dynamic load. For example, SNiP No. 2.09.03-85 determines for supports and overpasses the deflection of the beam is no more than 1/150 of the span length. For attic floors this figure is already 1/200, and for interfloor beams and even less - 1/250. Therefore, one of the mandatory design stages is to perform a beam deflection calculation.

Ways to perform deflection calculations and tests

The reason why SNiPs establish such draconian restrictions is simple and obvious. The smaller the deformation, the greater the margin of strength and flexibility of the structure. For a deflection of less than 0.5%, the load-bearing element, beam or slab still retains elastic properties, which guarantees normal redistribution of forces and maintaining the integrity of the entire structure. As the deflection increases, the building frame bends, resists, but stands; when the permissible value is exceeded, the bonds break, and the structure loses its rigidity and load-bearing capacity like an avalanche.

• Use an online software calculator, which is “hardwired” standard conditions, and nothing more;
• Use ready-made reference data for various types and types of beams, for various support load patterns. It is only necessary to correctly identify the type and size of the beam and determine the desired deflection;
• Calculate permissible deflection hands and their heads, most designers do this, while controlling architectural and construction inspectors prefer the second method of calculation.

For your information! To really understand why it is so important to know the magnitude of the deviation from the initial position, it is worth understanding that measuring the amount of deflection is the only accessible and reliable way to determine the condition of the beam in practice.

By measuring how much the ceiling beam has sagged, you can determine with 99% certainty whether the structure is in disrepair or not.

Method of performing deflection calculations

Before starting the calculation, you will need to remember some dependencies from the theory of strength of materials and draw up a calculation diagram. Depending on how correctly the diagram is executed and the loading conditions are taken into account, the accuracy and correctness of the calculation will depend.

We use the simplest model loaded beam shown in the diagram. The simplest analogy of a beam can be a wooden ruler, photo.

In our case, the beam:

1. It has a rectangular cross-section S=b*h, the length of the supporting part is L;
2. The ruler is loaded by a force Q passing through the center of gravity of the bent plane, as a result of which the ends rotate by small angleθ, with deflection relative to the initial horizontal position , equal to f ;
3. The ends of the beam rest hingedly and freely on fixed supports, accordingly, there is no horizontal component of the reaction, and the ends of the ruler can move in any direction.

To determine the deformation of a body under load, use the formula of the elastic modulus, which is determined by the ratio E = R/Δ, where E is a reference value, R is force, Δ is the amount of deformation of the body.

Calculate moments of inertia and forces

For our case, the dependence will look like this: Δ = Q/(S E) . For a load q distributed along the beam, the formula will look like this: Δ = q h/(S E) .

What follows is the most important point. The above Young diagram shows the deflection of a beam or the deformation of a ruler as if it were crushed under a powerful press. In our case, the beam is bent, which means that at the ends of the ruler, relative to the center of gravity, two bending moments are applied with different sign. The loading diagram for such a beam is given below.

To transform Young's dependence for the bending moment, it is necessary to multiply both sides of the equality by the shoulder L. We obtain Δ*L = Q·L/(b·h·E) .

If we imagine that one of the supports is rigidly fixed, and an equivalent balancing moment of forces M max = q*L*2/8 will be applied to the second, respectively, the magnitude of the beam deformation will be expressed by the dependence Δх = M x/((h/3) b (h/2) E). The quantity b h 2 /6 is called the moment of inertia and is designated W. The result is Δx = M x / (W E) the fundamental formula for calculating a beam for bending W = M / E through the moment of inertia and bending moment.

To accurately calculate the deflection, you will need to know the bending moment and moment of inertia. The value of the first can be calculated, but specific formula for calculating a beam for deflection will depend on the conditions of contact with the supports on which the beam is located and the method of loading, respectively, for a distributed or concentrated load. The bending moment from a distributed load is calculated using the formula Mmax = q*L 2 /8. The given formulas are valid only for a distributed load. For the case when the pressure on the beam is concentrated at a certain point and often does not coincide with the axis of symmetry, the formula for calculating the deflection must be derived using integral calculus.

The moment of inertia can be thought of as the equivalent of a beam's resistance to bending load. The magnitude of the moment of inertia for a simple rectangular beam can be calculated using the simple formula W=b*h 3 /12, where b and h are the cross-sectional dimensions of the beam.

From the formula it is clear that the same ruler or board rectangular section may have a completely different moment of inertia and amount of deflection if you put it on supports traditional way or put it on edge. No wonder almost all elements rafter system roofs are made not from 100x150 timber, but from 50x150 boards.

Real sections building structures may have the most different profiles, from square, circle to complex I-beam or channel shapes. At the same time, determining the moment of inertia and the amount of deflection manually, “on paper”, for such cases becomes a non-trivial task for a non-professional builder.

Formulas for practical use

In practice, most often the opposite task is faced - to determine the safety margin of floors or walls for a specific case based on a known deflection value. In the construction business, it is very difficult to assess the safety factor using other, non-destructive methods. Often, based on the magnitude of the deflection, it is necessary to perform a calculation, evaluate the safety factor of the building and the general condition load-bearing structures. Moreover, based on the measurements taken, it is determined whether the deformation is acceptable, according to the calculation, or whether the building is in emergency condition.

Advice! In the matter of calculating the limit state of a beam based on the amount of deflection, the requirements of SNiP provide an invaluable service. By setting the deflection limit in a relative value, for example, 1/250, building codes greatly facilitate the determination emergency condition beams or slabs.

For example, if you intend to buy a finished building that has stood for quite a long time on problematic soil, it would be useful to check the condition of the ceiling based on the existing deflection. Knowing everything permissible norm deflection and the length of the beam, one can assess without any calculation how critical the condition of the structure is.

Construction inspection during deflection assessment and assessment bearing capacity overlap goes a more complicated way:

• Initially, the geometry of the slab or beam is measured and the deflection value is recorded;
• Based on the measured parameters, the assortment of the beam is determined, then the formula for the moment of inertia is selected using the reference book;
• The moment of force is determined by the deflection and moment of inertia, after which, knowing the material, you can calculate the actual stresses in a metal, concrete or wooden beam.

The question is why is it so difficult if the deflection can be obtained using the formula for calculation for a simple beam on hinged supports f=5/24*R*L 2 /(E*h) under a distributed force. It is enough to know the span length L, profile height, design resistance R and elastic modulus E for a specific floor material.

Advice! Use in your calculations existing departmental collections of various design organizations, in which all the necessary formulas to determine and calculate the maximum loaded state.

Conclusion

Most developers and designers of serious buildings act in a similar way. The program is good, it helps to very quickly calculate the deflection and basic loading parameters of the floor, but it is also important to provide the customer with documentary evidence of the results obtained in the form of specific sequential calculations on paper.

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 tensile zone they are compressed.

Assumptions for deriving formulas. Normal voltages

1) The hypothesis of plane sections is fulfilled.

2) Longitudinal fibers do not press on each other and, therefore, under the influence of normal stresses, linear tension or compression operates.

3) Deformations of fibers do not depend on their position along the cross-sectional width. Consequently, normal stresses, changing along the height of the section, remain the same along the width.

4) The beam has at least one plane of symmetry, and all external forces lie in this plane.

5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same.

6) The relationships between the dimensions of the beam are such that it works under conditions flat bend no warping or curling.

At pure bend beams on platforms in its section act 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, beam strength condition according to normal voltages 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.

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 to 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 we'll install the consoles additional point E, since in this section 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 embedment. 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, dangerous is 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=9.03 cm.

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.

Let's consider beam 1st span

Let's define fictitious reactions for the first span beam using tabular formulas (see table “Fictitious support reactions....»)

Beam 2nd span

Beam 3rd span

5. Compose 3 x moments equation for two points– intermediate supports – support 1 and support 2. This is what they will be two missing equations to solve the problem.

The 3-moment equation in general form:

For point (support) 1 (n=1):

For point (support) 2 (n=2):

We substitute all known quantities, taking into account that the moment at the zero support and at the third support are equal to zero, M 0 =0; M 3 =0

Then we get:

Let's divide the first equation by factor 4 for M 2

Divide the second equation by the factor 20 at M 2

Let's solve this system of equations:

We subtract the second from the first equation and get:

We substitute this value into any of the equations and find M 2

Straight bend. Plane transverse bending Constructing diagrams of internal force factors for beams Constructing diagrams of Q and M using equations Constructing diagrams of Q and M using characteristic sections (points) Strength calculations for direct bending of beams Principal stresses during bending. A complete check of the strength of beams. The concept of the center of bending. Determination of displacements in beams during bending. Concepts of deformation of beams and conditions for their rigidity Differential equation of the curved axis of a beam Method of direct integration Examples of determining displacements in beams by the method of direct integration Physical meaning of integration constants Method of initial parameters (universal equation of the curved axis of a beam). 1.3, b). Rice. 1.3 When calculating the bending moment in a given section, the moments of external forces lying to the left of the section are considered positive if they are directed clockwise. For the right side of the beam - vice versa. It is convenient to determine the sign of the bending moment by the nature of the deformation of the beam. The bending moment is considered positive if, in the section under consideration, the cut-off part of the beam bends convexly downward, i.e., the lower fibers are stretched. In the opposite case, the bending moment in the section is negative. There are differential relationships between the bending moment M, shear force Q and load intensity q. 1. The first derivative of the shear force along the abscissa of the section is equal to the intensity of the distributed load, i.e. Positive ordinates of the M diagram are laid down, and negative ordinates are laid upward, i.e., the M diagram is constructed from the side of the stretched fibers. The construction of Q and M diagrams for beams should begin with determining the support reactions. For a beam with one clamped end and the other free end, the construction of diagrams Q and M can be started from the free end, without determining the reactions in the embedment. 1.2. The construction of Q and M diagrams using the Beam equations is divided into sections within which the functions for the bending moment and shear force remain constant (do not have discontinuities). The boundaries of the sections are the points of application of concentrated forces, pairs of forces and places of change in the intensity of the distributed load. At each section, an arbitrary section is taken at a distance x from the origin of coordinates, and for this section equations for Q and M are drawn up. Using these equations, diagrams of Q and M are constructed. Example 1.1 Construct diagrams of transverse forces Q and bending moments M for a given beam (Fig. 1.4,a). Solution: 1. Determination of support reactions. We compose equilibrium equations: from which we obtain The reactions of the supports are determined correctly. The beam has four sections Fig. 1.4 loads: CA, AD, DB, BE. 2. Construction of diagram Q. Section CA. In section CA 1, we draw an arbitrary section 1-1 at a distance x1 from the left end of the beam. We define Q as the algebraic sum of all external forces acting to the left of section 1-1: The minus sign is taken because the force acting to the left of the section is directed downward. The expression for Q does not depend on the variable x1. Diagram Q in this section will be depicted as a straight line parallel to the abscissa axis. Section AD. On the section we draw an arbitrary section 2-2 at a distance x2 from the left end of the beam. We define Q2 as the algebraic sum of all external forces acting to the left of section 2-2: 8 The value of Q is constant in the section (does not depend on the variable x2). The Q plot on the section is a straight line parallel to the abscissa axis. Plot DB. On the site we draw an arbitrary section 3-3 at a distance x3 from the right end of the beam. We define Q3 as the algebraic sum of all external forces acting to the right of section 3-3: The resulting expression is the equation of an inclined straight line. Section BE. On the site we draw a section 4-4 at a distance x4 from the right end of the beam. We define Q as the algebraic sum of all external forces acting to the right of section 4-4: 4 Here the plus sign is taken because the resultant load to the right of section 4-4 is directed downward. Based on the obtained values, we construct Q diagrams (Fig. 1.4, b). 3. Construction of diagram M. Plot m1. We define the bending moment in section 1-1 as the algebraic sum of the moments of forces acting to the left of section 1-1. Using this method, the values ​​of Q and M are calculated in characteristic sections. The characteristic sections are the boundary sections of sections, as well as sections where a given internal force factor has an extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of the differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct diagrams Q and M for the beam shown in Fig. 1.6, a. Rice. 1.6. Solution: We start constructing the Q and M diagrams from the free end of the beam, while the reactions in the embedment do not need to be determined. The beam has three loading sections: AB, BC, CD. There is no distributed load in sections AB and BC. Shear forces are constant. The Q diagram is limited to straight lines parallel to the x-axis. Bending moments vary linearly. Diagram M is limited by straight lines inclined to the abscissa axis. There is a uniformly distributed load on section CD. Transverse forces vary according to a linear law, and bending moments - according to the law of a square parabola with convexity in the direction of the distributed load. At the boundary of sections AB and BC, the transverse force changes abruptly. At the boundary of sections BC and CD, the bending moment changes abruptly. 1. Construction of diagram Q. We calculate the values ​​of transverse forces Q in the boundary sections of sections: Based on the calculation results, we construct diagram Q for the beam (Fig. 1, b). From diagram Q it follows that the transverse force on section CD is equal to zero in the section located at a distance qa a q from the beginning of this section. In this section, the bending moment has a maximum value. 2. Constructing diagram M. We calculate the values ​​of bending moments in the boundary sections of sections: At the maximum moment in the section Based on the calculation results, we construct diagram M (Fig. 5.6, c). Example 1.4 Using a given diagram of bending moments (Fig. 1.7, a) for a beam (Fig. 1.7, b), determine the acting loads and construct diagram Q. The circle indicates the vertex of a square parabola. Solution: Let's determine the loads acting on the beam. Section AC is loaded with a uniformly distributed load, since the diagram M in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting clockwise, since in diagram M we have a jump upward by the magnitude of the moment. In the NE section, the beam is not loaded, since the M diagram in this section is limited by an inclined straight line. The reaction of support B is determined from the condition that the bending moment in section C is equal to zero, i.e. To determine the intensity of the distributed load, we create an expression for the bending moment in section A as the sum of the moments of forces on the right and equate it to zero. Now we determine the reaction of support A. To do this, we will compose an expression for bending moments in the section as the sum of the moments of forces on the left. The design diagram of the beam with a load is shown in Fig. 1.7, c. Starting from the left end of the beam, we calculate the values ​​of transverse forces in the boundary sections of the sections: Diagram Q is shown in Fig. 1.7, d. The considered problem can be solved by drawing up functional dependencies for M, Q in each section. Let's choose the origin of coordinates at the left end of the beam. In the AC section, the diagram M is expressed by a square parabola, the equation of which has the form Constants a, b, c are found from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the equation of the parabola, we obtain: The expression for the bending moment will be Differentiating the function M1 , we obtain the dependence for the transverse force. After differentiating the function Q, we obtain an expression for the intensity of the distributed load. In the section NE, the expression for the bending moment is presented in the form of a linear function. To determine the constants a and b, we use the conditions that this straight line passes through two points, the coordinates of which are known. We obtain two equations: ,b from which we have a 20. The equation for the bending moment in the section NE will be After double differentiation of M2, we will find. Using the found values ​​of M and Q, we construct diagrams of bending moments and shear forces for the beam. In addition to the distributed load, concentrated forces are applied to the beam in three sections, where there are jumps on diagram Q and concentrated moments in the section where there is a shock on diagram M. Example 1.5 For a beam (Fig. 1.8, a), determine the rational position of the hinge C, at which the largest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Construct diagrams of Q and M. Solution Determination of support reactions. Despite the fact that the total number of support links is four, the beam is statically determinate. The bending moment in the hinge C is zero, which allows us to create an additional equation: the sum of the moments about the hinge of all external forces acting on one side of this hinge is equal to zero. Let us compile the sum of the moments of all forces to the right of the hinge C. The diagram Q for the beam is limited by an inclined straight line, since q = const. We determine the values ​​of transverse forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation from which the diagram M for the beam is limited by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are written respectively as follows: From the condition of equality of moments, we obtain a quadratic equation for the desired parameter x: Real value x2x 1.029 m. We determine the numerical values ​​of transverse forces and bending moments in characteristic sections of the beam. Figure 1.8, b shows the diagram Q, and in Fig. 1.8, c – diagram M. The problem considered could be solved by dividing the hinged beam into its constituent elements, as shown in Fig. 1.8, d. At the beginning, the reactions of the supports VC and VB are determined. Diagrams of Q and M are constructed for the suspended beam SV from the action of the load applied to it. Then they move to the main beam AC, loading it with an additional force VC, which is the pressure force of the beam CB on the beam AC. After that, diagrams Q and M are built for beam AC. 1.4. Strength calculations for direct bending of beams Strength calculations based on normal and shear stresses. When a beam bends directly in its cross sections, normal and tangential stresses arise (Fig. 1.9). 11) For beams made of brittle materials with sections that are asymmetrical with respect to the neutral axis, if the diagram M is unambiguous (Fig. 1.12), it is necessary to write down two strength conditions - the distance from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; P – permissible stresses for tension and compression, respectively. Fig.1.12. Considering the left side of the beam, we obtain The diagram of transverse forces is shown in Fig. 1.14, c. The diagram of bending moments is shown in Fig. 5.14, g. 2. Geometric characteristics of cross section 3. The highest normal stresses in section C, where Mmax acts (modulo): MPa. The maximum normal stresses in the beam are almost equal to the permissible ones. 4. The highest tangential stresses in section C (or A), where max Q acts (modulo): Here is the static moment of the half-section area relative to the neutral axis; b2 cm – section width at the level of the neutral axis. 5. Tangential stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 cm3 is the static moment of the area of ​​the section located above the line passing through point K1; b2 cm – wall thickness at the level of point K1. Diagrams  and  for section C of the beam are shown in Fig. 1.15. Example 1.7 For the beam shown in Fig. 1.16, a, required: 1. Construct diagrams of transverse forces and bending moments along characteristic sections (points). 2. Determine the dimensions of the cross section in the form of a circle, rectangle and I-beam from the condition of strength under normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of beam sections according to tangential stress. Given: Solution: 1. Determine the reactions of the beam supports. Check: 2. Construction of diagrams Q and M. Values ​​of transverse forces in characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, load intensity q = const. Consequently, in these areas the Q diagram is limited to straight lines inclined to the axis. In section DB, the intensity of the distributed load is q = 0, therefore, in this section, the diagram Q is limited to a straight line parallel to the x axis. The Q diagram for the beam is shown in Fig. 1.16, b. Values ​​of bending moments in characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section in which Q = 0: Maximum moment in the second section Diagram M for the beam is shown in Fig. 1.16, c. 2. We create a strength condition based on normal stresses, from which we determine the required axial moment of resistance of the section from the expression determined by the required diameter d of a beam of a circular section. Area of ​​a circular section. For a beam of a rectangular section. Required height of the section. Area of ​​a rectangular section. Determine the required number of the I-beam. Using the tables of GOST 8239-89, we find the nearest higher value of the axial moment of resistance 597 cm3, which corresponds to I-beam No. 33 with the characteristics: A z 9840 cm4. Tolerance check: (underload by 1% of the permissible 5%) the nearest I-beam No. 30 (W 2 cm3) leads to significant overload (more than 5%). We finally accept I-beam No. 33. We compare the areas of the round and rectangular sections with the smallest area A of the I-beam: Of the three sections considered, the most economical is the I-beam section. 3. We calculate the highest normal stresses in the dangerous section 27 of the I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-beam section The diagram of normal stresses in the dangerous section of the beam is shown in Fig. 1.17, b. 5. Determine the highest shear stresses for the selected sections of the beam. a) rectangular section of the beam: b) round section of the beam: c) I-beam section: Tangential stresses in the wall near the flange of the I-beam in dangerous section A (right) (at point 2): The diagram of tangential stresses in dangerous sections of the I-beam is shown in Fig. 1.17, c. The maximum tangential stresses in the beam do not exceed the permissible stresses Example 1.8 Determine the permissible load on the beam (Fig. 1.18, a), if 60 MPa, the cross-sectional dimensions are given (Fig. 1.19, a). Construct a diagram of normal stresses in a dangerous section of a beam at an allowable load.