When bending in cross sections, the beams act. Normal bending stresses

Building a diagram Q.

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

Determine bending moments at points. Rule of signs cm. - .

The moment in AT we will determine in the following way. First let's define:

Point To let's take in middle area with a uniformly distributed load.

Building a diagram M . Plot AB – parabolic curve(rule of "umbrella"), plot BD – straight oblique line.

For a beam, determine the support reactions and plot bending moment diagrams ( M) and transverse forces (Q).

1. We designate supports letters BUT and AT and direct the support reactions R A and R B .

Compiling equilibrium equations.

Examination

Write down the values R A and R B on the calculation scheme.

2. Plotting transverse forces method sections. We place 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 section with uniformly distributed load, note the size z 1 to the left of the section before the beginning of the section. Plot length 2 m. Rule of signs for Q - cm.

We build on the found value diagramQ.

sec. 2-2 move right.

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

Building a diagram Q.

sec. 3-3 move right.

sec. 4-4 move to the right.

We are building diagramQ.

3. Construction diagrams M method characteristic points.

characteristic point- a point, any noticeable on the beam. These are the dots BUT, AT, With, D , as well as the point To , wherein Q=0 and bending moment has an extremum. also in middle console 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, we proceed to determine the values ​​​​in them bending moments. Rule of signs - see..

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

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

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

T. To belongs second characteristic area, shear force equation(see above)

But the transverse force in t. To is equal to 0 , a z 2 equals unknown X .

We get the equation:

Now knowing X, determine the moment at a point To on the right side.

Building a diagram M . The construction is feasible for mechanical specialties, postponing positive values up from the zero line and using the "umbrella" rule.

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

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

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

1. Define support reactions.

Uniformly distributed load q replace the 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.

Let's find vertical support reaction R A and reference moment M A from the equilibrium equations.

In the first two sections on the right, there is no transverse force. At the beginning of a section with a uniformly distributed load (right) Q=0, in the back - the magnitude of the reaction R.A.
3. To build, we will compose expressions for their definition on sections. We plot the moment diagram on the fibers, i.e. down.

(the plot of single moments has already been built earlier)

We solve equation (1), reduce by EI

Static Indeterminacy Revealed, the value of the "extra" reaction is found. You can start plotting Q and M diagrams for a statically indeterminate beam... We sketch the given beam scheme and indicate the reaction value Rb. In this beam, the reactions in the termination can not be determined if you go to the right.

Building plots Q for a statically indeterminate beam

Plot Q.

Plotting M

We define M at the point of extremum - at the point To. First, let's define its position. We denote the distance to it as unknown " X". Then

We plot M.

Determination of shear stresses in an I-section. Consider the section I-beam. S x \u003d 96.9 cm 3; Yx=2030 cm 4; Q=200 kN

To determine the shear stress, it is used formula, where Q is the transverse 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 shear 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

Compute maximum shear stress:

Let us calculate the static moment for top shelf:

Now let's calculate shear stresses:

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 for normal stresses. Check the strength of the beam using the shear strength condition and the energy strength criterion. Given:

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

According to the diagram of bending moments, the dangerous is section C, wherein M C \u003d M max \u003d 48.3 kNm.

Strength condition for normal stresses for this beam has the form σ max \u003d M C / W X ≤σ adm . It is necessary to select a section from two channels.

Determine the required calculated value axial section modulus:

For a section in the form of two channels, according to accept two channels №20a, the 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 according to the formula: Then we get undervoltage:

Now let's check the strength of the beam, based on strength conditions for shear stresses. According to diagram of shear forces dangerous are sections in section BC and section D. As can be seen from the diagram, Q max \u003d 48.9 kN.

Strength condition for shear stresses looks like:

For channel No. 20 a: static moment of the area S x 1 \u003d 95.9 cm 3, moment of inertia of the section I x 1 \u003d 1670 cm 4, wall thickness d 1 \u003d 5.2 mm, average shelf thickness t 1 \u003d 9.7 mm , channel height h 1 \u003d 20 cm, shelf width b 1 \u003d 8 cm.

For transverse sections of two channels:

S x \u003d 2S x 1 \u003d 2 95.9 \u003d 191.8 cm 3,

I x \u003d 2I x 1 \u003d 2 1670 \u003d 3340 cm 4,

b \u003d 2d 1 \u003d 2 0.52 \u003d 1.04 cm.

Determining the value maximum shear stress:

τ max \u003d 48.9 10 3 191.8 10 -6 / 3340 10 -8 1.04 10 -2 \u003d 27 MPa.

As seen, τmax<τ adm (27MPa<75МПа).

Hence, strength condition is met.

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

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

Let's spend 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 \u003d h 1 / 2-t 1 \u003d 20 / 2-0.97 \u003d 9.03 cm.

Main stresses:

Level 3-3: y 3-3 \u003d h 1 / 2-t 1 \u003d 20 / 2-0.97 \u003d 9.03 cm.

Normal and shear stresses:

Main stresses:

Extreme shear stresses:

Level 4-4: y 4-4 =0.

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

Main stresses:

Extreme shear stresses:

Level 5-5:

Normal and shear stresses:

Main stresses:

Extreme shear stresses:

Level 6-6:

Normal and shear stresses:

Main stresses:

Extreme shear stresses:

Level 7-7:

Normal and shear stresses:

Main stresses:

Extreme shear stresses:

According to the performed calculations 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 cross 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 the equivalent and allowable stresses, it follows that the strength condition is also satisfied

(135.3 MPa<150 МПа).

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

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

n= Sop -3= 5-3 =2, where Sop - the number of unknown reactions, 3 - the number of equations of statics. To solve this beam, it is required two additional equations.

2. Denote numbers supports with zero in order ( 0,1,2,3 )

3. Denote span numbers from the first in order ( v 1, v 2, v 3)

4. Each span is considered as simple beam and build diagrams for each simple beam Q and M. What pertains to simple beam, we will denote with index "0", which refers to continuous beam, we will denote without this index. Thus, is the transverse force and the bending moment for a simple beam.

Consider beam of the 1st span

Let's define fictitious reactions for the beam of the first span according to tabular formulas (see table "Fictitious support reactions....»)

Beam 2nd span

Beam 3rd span

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

The equation of 3 moments in general form:

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

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

We substitute all the known values, taking into account that the moment on the zero support and on the third support are equal to zero, M 0 =0; M3=0

Then we get:

Divide the first equation by the factor 4 for M 2

We divide the second equation by the factor 20 for M 2

Let's solve this system of equations:

Subtract the second equation from the first equation, we get:

We substitute this value in any of the equations and find M2

Chapter 1

1.1. Basic dependencies of the theory of beam bending

Beams It is customary to call rods working in bending under the action of a transverse (normal to the axis of the rod) load. Beams are the most common elements of ship structures. The axis of the beam is the locus of the centers of gravity of its cross sections in the undeformed state. A beam is called straight if the axis is a straight line. The geometric location of the centers of gravity of the cross sections of the beam in a bent state is called the elastic line of the beam. The following direction of the coordinate axes is accepted: axis OX aligned with the axis of the beam, and the axis OY and oz- with the main central axes of inertia of the cross section (Fig. 1.1).

The theory of beam bending is based on the following assumptions.

1. The hypothesis of flat sections is accepted, according to which the cross sections of the beam, initially flat and normal to the axis of the beam, remain flat and normal to the elastic line of the beam after its bending. Due to this, the beam bending deformation can be considered regardless of the shear deformation, which causes distortion of the beam cross-sectional planes and their rotation relative to the elastic line (Fig. 1.2, a).

2. Normal stresses in areas parallel to the axis of the beam are neglected due to their smallness (Fig. 1.2, b).

3. Beams are considered sufficiently rigid, i.e. their deflections are small compared to the height of the beams, and the angles of rotation of the sections are small compared to unity (Fig. 1.2, in).

4. Stresses and strains are connected by a linear relationship, i.e. Hooke's law is valid (Fig. 1.2, G).

Rice. 1.2. Beam bending theory assumptions

We will consider the bending moments and shearing forces that appear during the bending of the beam in its section as a result of the action of the part of the beam mentally discarded along the section on the remaining part of it.

The moment of all forces acting in the section relative to one of the main axes is called the bending moment. The bending moment is equal to the sum of the moments of all forces (including support reactions and moments) acting on the rejected part of the beam, relative to the specified axis of the section under consideration.

The projection onto the plane of the section of the main vector of forces acting in the section is called the shear force. It is equal to the sum of the projections onto the sectional plane of all forces (including support reactions) acting on the discarded part of the beam.

We confine ourselves to considering the beam bending occurring in the plane XOZ. Such bending will take place in the case when the transverse load acts in a plane parallel to the plane XOZ, and its resultant in each section passes through a point called the center of the bend of the section. Note that for sections of beams with two axes of symmetry, the center of bending coincides with the center of gravity, and for sections with one axis of symmetry, it lies on the axis of symmetry, but does not coincide with the center of gravity.

The load of the beams included in the ship's hull can be either distributed (most often evenly distributed along the axis of the beam, or changing according to a linear law), or applied in the form of concentrated forces and moments.

Let us denote the intensity of the distributed load (the load per unit length of the beam axis) through q(x), an external concentrated force - as R, and the external bending moment as M. A distributed load and a concentrated force are positive if their directions of action coincide with the positive direction of the axis oz(Fig. 1.3, a,b). The external bending moment is positive if it is directed clockwise (Fig. 1.3, in).

Rice. 1.3. Sign rule for external loads

Let us denote the deflection of a straight beam when it is bent in the plane XOZ through w, and the angle of rotation of the section through θ. We accept the rule of signs for bending elements (Fig. 1.4):

1) the deflection is positive if it coincides with the positive direction of the axis oz(Fig. 1.4, a):

2) the angle of rotation of the section is positive if, as a result of bending, the section rotates clockwise (Fig. 1.4, b);

3) bending moments are positive if the beam under their influence bends with a convexity upwards (Fig. 1.4, in);

4) shear forces are positive if they rotate the selected beam element counterclockwise (Fig. 1.4, G).

Rice. 1.4. Sign rule for bend elements

Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε x, located at z from the neutral axis, will be equal to

ε x= −z/ρ ,(1.1)

where ρ is the beam curvature radius in the considered section.

Rice. 1.5. Beam bending scheme

The neutral axis of the cross section is the locus of points for which the linear deformation during bending is equal to zero. Between curvature and derivatives of w(x) there is a dependence

By virtue of the accepted assumption about the smallness of the angles of rotation for sufficiently rigid beams, the valuesmall compared to unity, so we can assume that

Substituting 1/ ρ from (1.2) to (1.1), we obtain

Normal bending stresses σ x according to Hooke's law will be equal

Since it follows from the definition of beams that there is no longitudinal force directed along the axis of the beam, the main vector of normal stresses must vanish, i.e.

where F is the cross-sectional area of ​​the beam.

From (1.5) we obtain that the static moment of the cross-sectional area of ​​the beam is equal to zero. This means that the neutral axis of the section passes through its center of gravity.

The moment of internal forces acting in the cross section relative to the neutral axis, M y will

If we take into account that the moment of inertia of the cross-sectional area relative to the neutral axis OY is equal to , and substitute this value in (1.6), then we obtain a dependence that expresses the basic differential equation for the beam bending

Moment of internal forces in the section relative to the axis oz will

Since the axes OY and oz by condition are the main central axes of the section, then .

It follows that under the action of a load in a plane parallel to the main bending plane, the elastic line of the beam will be a flat curve. This bend is called flat. Based on dependences (1.4) and (1.7), we obtain

Formula (1.8) shows that the normal bending stresses of beams are proportional to the distance from the neutral axis of the beam. Naturally, this follows from the hypothesis of flat sections. In practical calculations, to determine the highest normal stresses, the section modulus of the beam is often used

where | z| max is the absolute value of the distance of the most distant fiber from the neutral axis.

Further subscripts y omitted for simplicity.

There is a connection between the bending moment, the shearing force and the intensity of the transverse load, which follows from the equilibrium condition of the element mentally isolated from the beam.

Consider a beam element with a length dx (Fig. 1.6). Here it is assumed that the deformations of the element are negligible.

If a moment acts in the left section of the element M and cutting force N, then in its right section the corresponding forces will have increments. Consider only linear increments .

Fig.1.6. Forces acting on the beam element

Equating to zero the projection on the axis oz of all efforts acting on the element, and the moment of all efforts relative to the neutral axis of the right section, we get:

From these equations, up to values ​​of a higher order of smallness, we obtain

From (1.11) and (1.12) it follows that

Relationships (1.11)–(1.13) are known as the Zhuravsky–Schwedler theorem. It follows from these relationships that the shear force and bending moment can be determined by integrating the load q:

where N 0 and M 0 - shear force and bending moment in the section corresponding tox=x 0 , which is taken as the origin; ξ,ξ 1 – integration variables.

Permanent N 0 and M 0 for statically determinate beams can be determined from the conditions of their static equilibrium.

If the beam is statically determinate, the bending moment in any section can be found from (1.14), and the elastic line is determined by integrating the differential equation (1.7) twice. However, statically determinate beams are extremely rare in ship hull structures. Most of the beams that are part of ship structures form repeatedly statically indeterminate systems. In these cases, to determine the elastic line, equation (1.7) is inconvenient, and it is advisable to go over to a fourth-order equation.

1.2. Differential equation for beam bending

Differentiating equation (1.7) for the general case, when the moment of inertia of the section is a function of x, taking into account (1.11) and (1.12), we obtain:

where the dashes denote differentiation with respect to x.

For prismatic beams, i.e. beams of constant section, we obtain the following differential equations of bending:

An ordinary inhomogeneous fourth-order linear differential equation (1.18) can be represented as a set of four first-order differential equations:

We further use equation (1.18) or the system of equations (1.19) to determine the beam deflection (its elastic line) and all unknown bending elements: w(x), θ (x), M(x), N(x).

Integrating (1.18) successively 4 times (assuming that the left end of the beam corresponds to the sectionx= x a ), we get:

It is easy to see that the integration constants N a ,M a ,θ a , w a have a certain physical meaning, namely:

N a- cutting force at the origin, i.e. at x=x a ;

M a- bending moment at the origin;

θ a – angle of rotation at the origin;

w a - deflection in the same section.

To determine these constants, it is always possible to make four boundary conditions - two for each end of a single-span beam. Naturally, the boundary conditions depend on the arrangement of the ends of the beam. The simplest conditions correspond to hinged support on rigid supports or a rigid attachment.

When the end of the beam is hinged on a rigid support (Fig. 1.7, a) beam deflection and bending moment are equal to zero:

With rigid termination on a rigid support (Fig. 1.7, b) deflection and angle of rotation of the section are equal to zero:

If the end of the beam (console) is free (Fig. 1.7, in), then in this section the bending moment and the shearing force are equal to zero:

A situation associated with a sliding or symmetry termination is possible (Fig. 1.7, G). This leads to the following boundary conditions:

Note that the boundary conditions (1.26) concerning deflections and angles of rotation are called kinematic, and conditions (1.27) power.

Rice. 1.7. Types of boundary conditions

In ship structures, one often has to deal with more complex boundary conditions, which correspond to the support of the beam on elastic supports or elastic termination of the ends.

Elastic support (Fig. 1.8, a) is called a support having a drawdown proportional to the reaction acting on the support. We will consider the reaction of the elastic support R positive if it acts on the support in the direction of the positive direction of the axis oz. Then you can write:

w =AR,(1.29)

where A- coefficient of proportionality, called the coefficient of compliance of the elastic support.

This coefficient is equal to the drawdown of the elastic support under the action of the reaction R= 1, i.e. A=w R = 1 .

Elastic supports in ship structures can be beams that reinforce the beam under consideration, or pillars and other structures that work in compression.

To determine the compliance coefficient of an elastic support A it is necessary to load the corresponding structure with a unit force and find the absolute value of the subsidence (deflection) at the place of application of the force. A rigid support is a special case of an elastic support with A= 0.

Elastic seal (Fig. 1.8, b) is such a support structure that prevents the free rotation of the section and in which the angle of rotation θ in this section is proportional to the moment, i.e. there is dependence

θ = Â M.(1.30)

Proportionality multiplier  is called the coefficient of compliance of the elastic seal and can be defined as the angle of rotation of the elastic seal at M= 1, i.e.  = θ M= 1 .

A special case of elastic embedding at  = 0 is a hard termination. In ship structures, elastic embeddings are usually beams normal to the one under consideration and lying in the same plane. For example, beams, etc., can be considered elastically embedded on the frames.

Rice. 1.8. Elastic support ( a) and elastic embedding ( b)

If the ends of the beam are long L supported on elastic supports (Fig. 1.9), then the reactions of the supports in the end sections are equal to the shear forces, and the boundary conditions can be written:

The minus sign in the first condition (1.31) is accepted because the positive shearing force in the left reference section corresponds to the reaction acting on the beam from top to bottom, and on the support from bottom to top.

If the ends of the beam are long Lresiliently embedded(Fig. 1.9), then for the reference sections, taking into account the sign rule for the angles of rotation and bending moments, we can write:

The minus sign in the second condition (1.32) is adopted because, with a positive moment in the right reference section of the beam, the moment acting on the elastic attachment is directed counterclockwise, and the positive angle of rotation in this section is directed clockwise, i.e. the directions of the moment and the angle of rotation do not coincide.

Consideration of the differential equation (1.18) and all boundary conditions shows that they are linear with respect to both the deflections and their derivatives included in them, and the loads acting on the beam. Linearity is a consequence of the assumptions about the validity of Hooke's law and the smallness of beam deflections.

Rice. 1.9. A beam, both ends of which are elastically supported and elastically embedded ( a);

forces in elastic supports and elastic seals corresponding to positive
directions of bending moment and shear force ( b)

When several loads act on a beam, each beam bending element (deflection, angle of rotation, moment and shear force) is the sum of the bending elements from the action of each of the loads separately. This very important provision, called the principle of superposition, or the principle of summation of the action of loads, is widely used in practical calculations and, in particular, to reveal the static indeterminacy of beams.

1.3. Initial Parameters Method

The general integral of the beam bending differential equation can be used to determine the elastic line of a single-span beam when the beam load is a continuous function of the coordinate throughout the span. If concentrated forces, moments, or a distributed load act on parts of the beam length (Fig. 1.10), then expression (1.24) cannot be used directly in the load. In this case, it would be possible, by denoting the elastic lines in sections 1, 2 and 3 through w 1 , w 2 , w 3 , write out for each of them the integral in the form (1.24) and find all arbitrary constants from the boundary conditions at the ends of the beam and the conjugation conditions at the boundaries of the sections. The conjugation conditions in the case under consideration are expressed as follows:

at x=a 1

at x=a 2

at x=a 3

It is easy to see that such a way of solving the problem leads to a large number of arbitrary constants, equal to 4 n, where n- the number of sections along the length of the beam.

Rice. 1.10. Beam, on some sections of which loads of different types are applied

It is much more convenient to represent the elastic line of the beam in the form

where the terms behind the double line are taken into account when x³ a 1, x³ a 2 etc.

Obviously, δ 1 w(x)=w 2 (x)−w 1 (x); δ2 w(x)=w 3 (x)−w 2 (x); etc.

Differential equations for determining the corrections to the elastic line δ iw (x) based on (1.18) and (1.32) can be written as

General integral for any correction δ iw (x) to the elastic line can be written in the form (1.24) for x a = a i . At the same time, the parameters N a ,M a ,θ a , w a the changes (jump) make sense, respectively: in the shear force, bending moment, angle of rotation and deflection arrow at the transition through the section x=a i . This technique is called the method of initial parameters. It can be shown that for the beam shown in Fig. 1.10, the elastic line equation will be

Thus, the method of initial parameters makes it possible, even in the presence of discontinuity in loads, to write the equation of an elastic line in a form containing only four arbitrary constants N 0 , M 0 , θ 0 , w 0 , which are determined from the boundary conditions at the ends of the beam.

Note that for a large number of variants of single-span beams encountered in practice, detailed bending tables have been compiled that make it easy to find deflections, angles of rotation and other bending elements.

1.4. Determination of shear stresses during beam bending

The hypothesis of flat sections accepted in the theory of beam bending leads to the fact that the shear deformation in the beam section turns out to be equal to zero, and we do not have the opportunity, using Hooke's law, to determine the shear stresses. However, since, in the general case, shear forces act in the beam sections, the shear stresses corresponding to them should arise. This contradiction (which is a consequence of the accepted hypothesis of flat sections) can be avoided by considering the equilibrium conditions. We will assume that when a beam composed of thin strips is bent, the shear stresses in the cross section of each of these strips are uniformly distributed over the thickness and directed parallel to the long sides of its contour. This position is practically confirmed by the exact solutions of the theory of elasticity. Consider a beam of an open thin-walled I-beam. On fig. 1.11 shows the positive direction of shear stresses in the belts and the profile wall during bending in the plane of the beam wall. Select the longitudinal section I-I and two cross sections element length dx (Fig. 1.12).

Let us denote the shear stress in the indicated longitudinal section as τ, and the normal forces in the initial cross section as T. Normal forces in the final section will have increments. Consider only linear increments, then .

Rice. 1.12. Longitudinal forces and shear stresses
in beam girdle element

The condition of static equilibrium of the element selected from the beam (equality to zero of the projections of forces on the axis OX) will

where ; f- the area of ​​the part of the profile cut off by the line I-I; δ is the thickness of the profile at the section site.

From (1.36) it follows:

Since the normal stresses σ x are defined by formula (1.8), then

In this case, we assume that the beam has a section that is constant along the length. Static moment of a part of the profile (cut-off line I-I) relative to the neutral axis of the beam section OY is an integral

Then from (1.37) for the absolute value of stresses we obtain:

Naturally, the resulting formula for determining shear stresses is also valid for any longitudinal section, for example II -II(see Fig. 1.11), and the static moment S ots is calculated for the cut-off part of the beam profile area relative to the neutral axis, without taking into account the sign.

Formula (1.38), according to the meaning of the derivation, determines the shear stresses in the longitudinal sections of the beam. From the theorem on the pairing of shear stresses, known from the course of the strength of materials, it follows that the same shear stresses act at the corresponding points of the beam cross section. Naturally, the projection of the main shear stress vector onto the axis oz must be equal to the shear force N in this section of the beam. Since in the girdle beams of this type, as shown in Fig. 1.11, shear stresses are directed along the axis OY, i.e. normal to the plane of action of the load, and are generally balanced, the shear force must be balanced by shear stresses in the beam web. The distribution of shear stresses along the height of the wall follows the law of change in the static moment S cut off part of the area relative to the neutral axis (with a constant wall thickness δ).

Consider a symmetrical section of an I-beam with a girdle area F 1 and wall area ω = hδ (Fig. 1.13).

Rice. 1.13. Section of an I-beam

The static moment of the cut-off part of the area for a point separated by z from the neutral axis, will

As can be seen from dependence (1.39), the static moment changes from z according to the law of a quadratic parabola. Highest value S ots , and consequently, shear stresses τ , will turn out at the neutral axis, where z= 0:

The greatest shear stress in the beam web at the neutral axis

Since the moment of inertia of the section of the considered beam is equal to

then the greatest shear stress will be

Attitude N/ω is nothing but the average shear stress in the wall, calculated assuming a uniform distribution of stresses. Taking, for example, ω = 2 F 1 , by formula (1.41) we obtain

Thus, for the beam under consideration, the largest shear stress in the wall at the neutral axis is only 12.5% exceeds the average value of these stresses. It should be noted that for the majority of beam profiles used in the ship's hull, the excess of the maximum shear stresses over the average is 10–15%.

If we consider the distribution of shear stresses during bending in the cross section of the beam shown in Fig. 1.14, it can be seen that they form a moment relative to the center of gravity of the section. In the general case, the bending of such a beam in the plane XOZ will be accompanied by twisting.

Beam bending is not accompanied by twisting if the load acts in a plane parallel to XOZ passing through a point called the center of the bend. This point is characterized by the fact that the moment of all tangential forces in the beam section relative to it is equal to zero.

Rice. 1.14. Tangential stresses during channel beam bending (point BUT - bend center)

Denoting the distance of the center of the bend BUT from the axis of the beam web through e, we write down the condition of equality to zero of the moment of tangential forces relative to the point BUT:

where Q 2 - tangential force in the wall, equal to the shear force, i.e. Q 2 =N;

Q 1 =Q 3 - force in the girdle, determined on the basis of (1.38) by the dependence

The shear strain (or shear angle) γ varies along the height of the beam web in the same way as the shear stresses τ , reaching its greatest value at the neutral axis.

As shown, for beams with corbels, the change in shear stresses along the wall height is very insignificant. This allows further consideration of some average shear angle in the beam web

Shear deformation leads to the fact that the right angle between the plane of the cross section of the beam and the tangent to the elastic line changes by the value γ cf. A simplified diagram of the shear deformation of a beam element is shown in fig. 1.15.

Rice. 1.15. Beam Element Shear Diagram

Denoting the deflection arrow caused by the shear through w sdv , we can write:

Taking into account the sign rule for the shear force N and find the angle of rotation

Insofar as ,

Integrating (1.47), we obtain

Constant a, included in (1.48), determines the displacement of the beam as a rigid body and can be taken equal to any value, since when determining the total deflection arrow from bending w bend and shear w sdv

the sum of the constants of integration will appear w 0 +a determined from the boundary conditions. Here w 0 - deflection from bending at the origin.

We put in the future a=0. Then the final expression for the elastic line caused by the shear will take the form

The bending and shear components of the elastic line are shown in Figs. 1.16.

Rice. 1.16. Flexural ( a) and shear ( b) components of the elastic line of the beam

In the considered case, the angle of rotation of the sections during shear is equal to zero, therefore, taking into account the shear, the angles of rotation of the sections, bending moments and shear forces are associated only with the derivatives of the elastic line from bending:

The situation is somewhat different in the case of the action of concentrated moments on the beam, which, as will be shown below, do not cause shear deflections, but only lead to an additional rotation of the beam sections.

Consider a beam freely supported on rigid supports, in the left section of which acting moment M. The cutting force in this case will be constant and equal

For the right reference section, respectively, we obtain

.(1.52)

Expressions (1.51) and (1.52) can be rewritten as

The expressions in parentheses characterize the relative addition to the angle of rotation of the section caused by the shear.

If we consider, for example, a freely supported beam loaded in the middle of its span by the force R(Fig. 1.18), then the deflection of the beam under the force will be equal to

Bending deflection can be found from beam bending tables. Shear deflection is determined by formula (1.50), taking into account the fact that .

Rice. 1.18. Scheme of a freely supported beam loaded with a concentrated force

As can be seen from formula (1.55), the relative addition to the beam deflection due to shear has the same structure as the relative addition to the angle of rotation, but with a different numerical coefficient.

We introduce the notation

where β is a numerical coefficient depending on the specific task under consideration, the arrangement of supports and the load of the beam.

Let us analyze the dependence of the coefficient k from various factors.

If we take into account that , we obtain instead of (1.56)

The moment of inertia of the beam section can always be represented as

,(1.58)

where α is a numerical coefficient depending on the shape and characteristics of the cross section. So, for an I-beam, according to formula (1.40) with ω = 2 F 1 find I= ωh 2 /3, i.e. α=1/3.

Note that with an increase in the dimensions of the beam corbels, the coefficient α will increase.

Taking into account (1.58), instead of (1.57) we can write:

Thus, the value of the coefficient k significantly depends on the ratio of the span length of the beam to its height, on the shape of the section (through the coefficient α), the device of the supports and the load of the beam (through the coefficient β). The relatively longer the beam ( h/L small), the smaller the effect of shear deformation. For rolled profile beams related to h/L less than 1/10÷1/8, the shift correction can practically not be taken into account.

However, for beams with wide girths, such as, for example, keels, stringers and floors as part of bottom slabs, the effect of shear and at the indicated h/L may be significant.

It should be noted that shear deformations affect not only the increase in beam deflections, but in some cases also the results of the disclosure of the static indeterminacy of beams and beam systems.

General concepts.

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

Bending rods are called beams.

clean called a bend, in which the bending moment is the only internal force factor that occurs 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 occurs. Such a bend is called transverse.

flat (straight) called a bend 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.

With an 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 the study of bending deformation with the case of pure plane bending.

Normal stresses and strains in pure bending.

As already mentioned, with a pure flat bend in the cross section, of the six internal force factors, only the bending moment is non-zero (Fig. 6.1, c):

; (6.1)

Experiments performed on elastic models show that if a grid of lines is applied to the surface of the model(Fig. 6.1, a) , then under pure bending it is deformed 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 lines of the contours of the sections intersect everywhere with the longitudinal fibers at a right angle.

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 bent axis of the beam (flat section hypothesis in bending).

Rice. .

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

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

Rice. .

By two infinitesimal cross sections, we select an element of length. Before deformation, the sections bounding the element were parallel to each other (Fig. 6.2, a), and after deformation, they tilted somewhat, forming an angle. The length of the fibers lying in the neutral layer does not change during bending. Let us designate the radius of curvature of the trace of the neutral layer on the plane of the drawing by a letter. Let us determine the linear deformation of an arbitrary fiber spaced 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 obtain that the absolute elongation of the considered fiber

Its relative deformation

Obviously, 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.

We introduce the assumption that the longitudinal fibers do not press each other during bending. Under this assumption, each fiber is deformed in isolation, experiencing a simple tension or compression, at which. Taking into account (6.2)

, (6.3)

i.e., normal stresses are directly proportional to the distances of the considered points of the section from the neutral axis.

We substitute dependence (6.3) into the expression for the bending moment in the cross section (6.1)

Recall that the integral is the moment of inertia of the section about the axis

Or

(6.4)

Dependence (6.4) is Hooke's law for bending, since it relates the deformation (curvature of the neutral layer) to the moment acting in the section. The product is called the bending stiffness of the section, N m 2.

Substitute (6.4) into (6.3)

(6.5)

This is the desired formula for determining the normal stresses in pure bending of the beam at any point in its section.

For In order to establish where the neutral line is in the cross section, we substitute the value of normal stresses in the expression for the longitudinal force and the bending moment

Insofar as,

then

(6.6)

(6.7)

Equality (6.6) indicates that the axis - the neutral axis of the section - passes through the center of gravity of the cross section.

Equality (6.7) shows that and are the main central axes of the section.

According to (6.5), the greatest stresses are reached in the fibers furthest from the neutral line

The ratio is the axial section modulus relative to its central axis, which means

The value for the simplest cross sections is as follows:

For rectangular cross section

, (6.8)

where is the section side perpendicular to the axis;

The side of the section is parallel to the axis;

For round cross section

, (6.9)

where is the diameter of the circular cross section.

The strength condition for normal stresses in bending can be written as

(6.10)

All obtained formulas are 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, the practice of calculations shows that in the case of transverse bending of beams and frames, when in addition to the bending moment, a longitudinal force and a transverse force also act in the section, you can use the formulas given for pure bending. In this case, the error turns out to be insignificant.

Determination of transverse forces and bending moments.

As already mentioned, with a flat transverse bending in the cross section of the beam, two internal force factors u arise.

Before determining and determine the reactions of the beam supports (Fig. 6.3, a), compiling the equilibrium equations of statics.

To determine and apply the method of sections. In the place of interest to us, we will make a mental section of the beam, for example, at a distance from the left support. Let's discard one of the parts of the beam, for example, the right one, and consider the balance of the left side (Fig. 6.3, b). We will replace the interaction of the beam parts with internal forces and.

Let us establish the following sign rules for and:

• The transverse force in the section is positive if its vectors tend to rotate the considered section clockwise;
• The bending moment in the section is positive if it causes compression of the upper fibers.

Rice. .

To determine these forces, we use two equilibrium equations:

1. ; ; .

2. ;

Thus,

a) the transverse force in the cross section of the beam is numerically equal to the algebraic sum of the projections onto the transverse axis of the section of all external forces acting on one side of the section;

b) the bending moment in the cross section of the beam is numerically equal to the algebraic sum of the moments (calculated relative to the center of gravity of the section) of external forces acting on one side of the given section.

In practical calculations, they are usually guided by the following:

1. If the external load tends to rotate the beam clockwise relative to the considered section, (Fig. 6.4, b), then in the expression for it gives a positive term.
2. If an external load creates a moment relative to the considered section, causing compression of the upper fibers of the beam (Fig. 6.4, a), then in the expression for in this section it gives a positive term.

Rice. .

Construction of diagrams in beams.

Consider a double beam(Fig. 6.5, a) . A beam is acted upon at a point by a concentrated moment, at a point by a concentrated force, and at a section by a uniformly distributed load of intensity.

We define support reactions and(Fig. 6.5, b) . The resultant distributed load is equal, and its line of action passes through the center of the section. Let us compose the equations of the moments with respect to the points and.

Let's determine the transverse force and the bending moment in an arbitrary section located in a section at a distance from point A(Fig. 6.5, c) .

(Fig. 6.5, d). The distance can vary within ().

The value of the transverse force does not depend on the coordinate of the section, therefore, in all sections of the section, the transverse forces are the same and the diagram looks like a rectangle. Bending moment

The bending moment changes linearly. Let's determine the ordinates of the diagram for the boundaries of the plot.

Let us determine the transverse force and the bending moment in an arbitrary section located in a section at a distance from the point(Fig. 6.5, e). The distance can vary within ().

The transverse force changes linearly. Define for the boundaries of the site.

Bending moment

The diagram of bending moments in this section will be parabolic.

To determine the extreme value of the bending moment, we equate to zero the derivative of the bending moment along the abscissa of the section:

From here

For a section with a coordinate, the value of the bending moment will be

As a result, we obtain diagrams of transverse forces(Fig. 6.5, e) and bending moments (Fig. 6.5, g).

Differential dependencies in bending.

(6.11)

(6.12)

(6.13)

These dependencies allow you to establish some features of the diagrams of bending moments and shear forces:

H in areas where there is no distributed load, the diagrams are limited to straight lines parallel to the zero line of the diagram, and the diagrams in the general case are inclined straight lines.

H in areas where a uniformly distributed load is applied to the beam, the diagram is limited by inclined straight lines, and the diagram is limited by quadratic parabolas with a bulge facing the direction opposite to the direction of the load.

AT sections, where, the tangent to the diagram is parallel to the zero line of the diagram.

H and areas where, the moment increases; in areas where, the moment decreases.

AT sections where concentrated forces are applied to the beam, there will be jumps on the magnitude of the applied forces on the diagram, and fractures on the diagram.

In sections where concentrated moments are applied to the beam, there will be jumps in the diagram by the magnitude of these moments.

The ordinates of the diagram are proportional to the tangent of the slope of the tangent to the diagram.

count beam for bending there are several options:
1. Calculation of the maximum load that it will withstand
2. Selection of the section of this beam
3. Calculation of the maximum allowable stresses (for verification)
let's consider general principle of beam section selection on two supports loaded with a uniformly distributed load or a concentrated force.
To begin with, you will need to find a point (section) at which there will be a maximum moment. It depends on the support of the beam or its termination. Below are diagrams of bending moments for schemes that are most common.

After finding the bending moment, we must find the modulus Wx of this section according to 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 the beam and this stress we must compare 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 if an I-beam No. 10 (St3sp5 steel) 2 meters long rigidly embedded in the wall can withstand you if you hang on it. Let your mass be 90 kg.
First, we need to choose a calculation scheme.

This diagram shows that the maximum moment will be in the termination, and since our I-beam has the same 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

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

It will be equal to 39.7 cm3. Convert to cubic meters and get 0.0000397 m3.
Further, according to the formula, we find the maximum stresses that we have 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 allowable stress equal to the yield strength of St3sp5 steel - 245 MPa.

45.34 MPa - right, so this I-beam can withstand a mass of 90 kg.

2. [i] Since we got 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, can withstand.
If we want to find the maximum mass, then the values ​​of the yield strength and the stress that will occur in the beam, we must equate (b \u003d 245 MPa \u003d 245,000 kN * m2).

Straight bend. Flat transverse bending Plotting diagrams of internal force factors for beams Plotting Q and M diagrams according to equations Plotting Q and M diagrams using characteristic sections (points) Calculations for strength in direct bending of beams Principal stresses in bending. Complete verification of the strength of beams Understanding the center of bending Determination of displacements in beams during bending. Concepts of deformation of beams and conditions of their rigidity Differential equation of the bent axis of the beam Method of direct integration Examples of determining displacements in beams by the method of direct integration Physical meaning of the constants of integration Method of initial parameters (universal equation of the bent axis of the beam). Examples of determining displacements in a beam using the method of initial parameters Determination of displacements using the Mohr method. A.K.'s rule Vereshchagin. Calculation of the Mohr integral according to A.K. Vereshchagin Examples of determination of displacements by means of Mohr's integral Bibliography Direct bending. Flat transverse bend. 1.1. Plotting diagrams of internal force factors for beams Direct bending is a type of deformation in which two internal force factors arise in the cross sections of the bar: a bending moment and a transverse force. In a particular case, the transverse force can be equal to zero, then the bend is called pure. With a flat transverse bending, all forces are located in one of the main planes of inertia of the rod and are perpendicular to its longitudinal axis, the moments are located in the same plane (Fig. 1.1, a, b). Rice. 1.1 The transverse force in an arbitrary cross section of the beam is numerically equal to the algebraic sum of the projections onto the normal to the axis of the beam of all external forces acting on one side of the section under consideration. The transverse force in the m-n section of the beam (Fig. 1.2, a) is considered positive if the resultant of external forces to the left of the section is directed upwards, and to the right - downwards, and negative - in the opposite case (Fig. 1.2, b). Rice. 1.2 When calculating the transverse force in a given section, the external forces lying to the left of the section are taken with a plus sign if they are directed upwards, and with a minus sign if downwards. For the right side of the beam - vice versa. 5 The bending moment in an arbitrary beam cross section is numerically equal to the algebraic sum of the moments about the central axis z of the section of all external forces acting on one side of the section under consideration. The bending moment in the m-n section of the beam (Fig. 1.3, a) is considered positive if the resultant moment of external forces is directed clockwise from the section to the left of the section, and counterclockwise to the right, and negative in the opposite case (Fig. 1.3b). 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 with a convexity downward, i.e., the lower fibers are stretched. Otherwise, the bending moment in the section is negative. Between the bending moment M, the transverse force Q and the intensity of the load q, there are differential dependencies. 1. The first derivative of the transverse force along the abscissa of the section is equal to the intensity of the distributed load, i.e. . (1.1) 2. The first derivative of the bending moment along the abscissa of the section is equal to the transverse force, i.e. . (1.2) 3. The second derivative with respect to the abscissa of the section is equal to the intensity of the distributed load, i.e. . (1.3) We consider the distributed load directed upwards to be positive. A number of important conclusions follow from the differential dependencies between M, Q, q: 1. If on the beam section: a) the transverse force is positive, then the bending moment increases; b) the transverse force is negative, then the bending moment decreases; c) the transverse force is zero, then the bending moment has a constant value (pure bending); 6 d) the transverse force passes through zero, changing sign from plus to minus, max M M, otherwise M Mmin. 2. If there is no distributed load on the beam section, then the transverse force is constant, and the bending moment changes linearly. 3. If there is a uniformly distributed load on the beam section, then the transverse force changes according to a linear law, and the bending moment - according to the law of a square parabola, convex inverted towards the load (in the case of plotting M from the side of tensioned fibers). 4. In the section under the concentrated force, the diagram Q has a jump (by the magnitude of the force), the diagram M has a break in the direction of the force. 5. In the section where a concentrated moment is applied, the diagram M has a jump equal to the value of this moment. This is not reflected in the Q plot. Under complex loading, beams build diagrams of transverse forces Q and bending moments M. Plot Q (M) is a graph showing the law of change in the transverse force (bending moment) along the length of the beam. Based on the analysis of diagrams M and Q, dangerous sections of the beam are established. The positive ordinates of the Q diagram are plotted upwards, and the negative ordinates are plotted downwards from the base line drawn parallel to the longitudinal axis of the beam. The positive ordinates of the diagram M are laid down, and the negative ordinates are plotted upwards, i.e., the diagram M is built from the side of the stretched fibers. The construction of diagrams Q and M for beams should begin with the definition of support reactions. For a beam with one fixed end and the other free end, plotting Q and M can be started from the free end without defining reactions in the embedment. 1.2. The construction of diagrams Q and M according to the Balk equations is divided into sections, within which the functions for the bending moment and the shear force remain constant (have no 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. An arbitrary section is taken at each section at a distance x from the origin, and equations for Q and M are drawn up for this section. Plots Q and M are built using these equations. Example 1.1 Construct plots of shear forces Q and bending moments M for a given beam (Fig. 1.4a). Solution: 1. Determination of reactions of supports. We compose the equilibrium equations: from which we obtain The reactions of the supports are defined correctly. The beam has four sections Fig. 1.4 loadings: CA, AD, DB, BE. 2. Plotting Q. Plot SA. On 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 the 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. Plot Q in this section will be depicted as a straight line parallel to the x-axis. Plot AD. On the site, 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 on the section (does not depend on the variable x2). Plot Q on the plot is a straight line parallel to the x-axis. DB site. 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. Plot B.E. 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 build diagrams Q (Fig. 1.4, b). 3. Plotting 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. is the equation of a straight line. Section A 3 Define the bending moment in section 2-2 as the algebraic sum of the moments of forces acting to the left of section 2-2. is the equation of a straight line. Plot DB 4 We define the bending moment in section 3-3 as the algebraic sum of the moments of forces acting to the right of section 3-3. is the equation of a square parabola. 9 Find three values ​​at the ends of the section and at the point with coordinate xk , where Section BE 1 Define the bending moment in section 4-4 as the algebraic sum of the moments of forces acting to the right of section 4-4. - the equation of a square parabola we find three values ​​of M4: Based on the obtained values, we build a plot M (Fig. 1.4, c). In sections CA and AD, plot Q is limited by straight lines parallel to the abscissa axis, and in sections DB and BE, by oblique straight lines. In the sections C, A and B on the diagram Q there are jumps by the magnitude of the corresponding forces, which serves as a check of the correctness of the construction of the diagram Q. In sections where Q  0, the moments increase from left to right. In sections where Q  0, the moments decrease. Under the concentrated forces there are kinks in the direction of the action of the forces. Under the concentrated moment, there is a jump by the moment value. This indicates the correctness of plotting M. Example 1.2 Construct plots Q and M for a beam on two supports, loaded with a distributed load, the intensity of which varies linearly (Fig. 1.5, a). Solution Determination of support reactions. The resultant of the distributed load is equal to the area of ​​the triangle representing the load diagram and is applied at the center of gravity of this triangle. We make up the sums of the moments of all forces relative to points A and B: Plotting Q. Let's draw an arbitrary section at a distance x from the left support. The ordinate of the load diagram corresponding to the section is determined from the similarity of triangles The resultant of that part of the load that is located to the left of the section zero: Plot Q is shown in fig. 1.5, b. The bending moment in an arbitrary section is equal to The bending moment changes according to the law of a cubic parabola: The maximum value of the bending moment is in the section, where 0, i.e. at. 1.5, c. 1.3. Plotting Q and M diagrams by characteristic sections (points) Using the differential relationships between M, Q, q and the conclusions arising from them, it is advisable to build Q and M diagrams by characteristic sections (without formulating equations). Using this method, the values ​​of Q and M are calculated in characteristic sections. The characteristic sections are the boundary sections of the sections, as well as the sections where the 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 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 plotting Q and M diagrams from the free end of the beam, while the reactions in the embedment can be omitted. The beam has three loading areas: AB, BC, CD. There is no distributed load in sections AB and BC. The transverse forces are constant. Plot Q is limited by straight lines parallel to the x-axis. Bending moments change linearly. Plot M is limited to straight lines inclined to the x-axis. On section CD there is a uniformly distributed load. The transverse forces change linearly, and the bending moments change according to the law of a square parabola with a 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. Plotting Q. We calculate the values ​​of transverse forces Q in the boundary sections of the sections: Based on the results of calculations, we build a diagram Q for the beam (Fig. 1, b). It follows from the diagram Q that the transverse force in the section CD is equal to zero in the section spaced at a distance qa a q from the beginning of this section. In this section, the bending moment has a maximum value. 2. Construction of diagram M. We calculate the values ​​of bending moments in the boundary sections of the sections: Example 1.4 According to the given diagram of bending moments (Fig. 1.7, a) for the beam (Fig. 1.7, b), determine the acting loads and plot Q. The circle indicates the vertex of the square parabola. Solution: 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 in a clockwise direction, since on the diagram M we have an upward jump by the magnitude of the moment. In the NE section, the beam is not loaded, since the diagram M 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 compose an expression for the bending moment in section A as the sum of the moments of forces on the right and equate to zero. Now we determine the reaction of support A. To do this, we compose an expression for bending moments in the section as the sum of the moments of forces on the left. The calculation scheme of a beam with a load is shown in fig. 1.7, c. Starting from the left end of the beam, we calculate the values ​​of the transverse forces in the boundary sections of the sections: Plot Q is shown in fig. 1.7, d. The considered problem can be solved by compiling functional dependencies for M, Q in each section. Let's choose the origin of coordinates at the left end of the beam. On the AC section, the plot M is expressed by a square parabola, the equation of which is of the form Constants a, b, c, we find from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the parabola equation, we get: 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 represented as a linear function. To determine the constants a and b, we use the conditions that this line passes through two points whose coordinates are known. We obtain two equations: ,b of which we have a 20. The equation for the bending moment in the section NE will be After a two-fold differentiation of M2, we will find. Based on the found values ​​of M and Q, we build 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 the Q diagram, and concentrated moments in the section where there is a jump on the M diagram. 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). Build diagrams Q and M. Solution Determination of reactions of supports. Despite the fact that the total number of support links is four, the beam is statically determinate. The bending moment in hinge C is equal to zero, which allows us to make 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. Compose the sum of the moments of all forces to the right of the hinge C. 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 whence Plot M for the beam is limited by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the termination are written respectively as follows: From the condition of equality of the moments, we obtain a quadratic equation with respect to the desired parameter x: The real value is x2x 1.029 m. We determine the numerical values ​​of the transverse forces and bending moments in the characteristic sections of the beam. 1.8, c - plot M. The considered problem 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. Plots Q and M are constructed for the suspension 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 the AC beam. 1.4. Strength calculations for direct bending of beams Strength calculation for normal and shear stresses. With a direct bending of a beam, normal and shear stresses arise in its cross sections (Fig. 1.9). 18 Fig. 1.9 Normal stresses are related to the bending moment, shear stresses are related to the transverse force. In direct pure bending, shear stresses are equal to zero. Normal stresses at an arbitrary point of the beam cross section are determined by the formula (1.4) where M is the bending moment in the given section; Iz is the moment of inertia of the section relative to the neutral axis z; y is the distance from the point where the normal stress is determined to the neutral z axis. Normal stresses along the height of the section change linearly and reach the greatest value at the points most distant from the neutral axis. If the section is symmetrical about the neutral axis (Fig. 1.11), then 1.11 the greatest tensile and compressive stresses are the same and are determined by the formula,  - axial moment of section resistance in bending. For a rectangular section with a width b and a height h: (1.7) For a circular section with a diameter d: (1.8) For an annular section   are the inner and outer diameters of the ring, respectively. For beams made of plastic materials, the most rational are symmetrical 20 section shapes (I-beam, box-shaped, annular). For beams made of brittle materials that do not equally resist tension and compression, sections that are asymmetrical about the neutral axis z (ta-br., U-shaped, asymmetrical I-beam) are rational. For beams of constant section made of plastic materials with symmetrical section shapes, the strength condition is written as follows: (1.10) where Mmax is the maximum bending moment modulo; - allowable stress for the material. For beams of constant section made of plastic materials with asymmetric section shapes, the strength condition is written in the following form: (1. 11) For beams made of brittle materials with sections that are asymmetric about the neutral axis, if the diagram M is unambiguous (Fig. 1.12), two strength conditions must be written - the distance from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; P - allowable stresses, respectively, in tension and compression. Fig.1.12. 21 If the bending moment diagram has sections of different signs (Fig. 1.13), then in addition to checking the section 1-1, where Mmax acts, it is necessary to calculate the maximum tensile stresses for the section 2-2 (with the largest moment of the opposite sign). Rice. 1.13 Along with the basic calculation for normal stresses, in some cases it is necessary to check the beam strength for shear stresses. Shear stresses in beams are calculated by the formula of D. I. Zhuravsky (1.13) where Q is the transverse force in the considered cross section of the beam; Szots is the static moment about the neutral axis of the area of ​​the part of the section located on one side of the straight line drawn through the given point and parallel to the z axis; b is the width of the section at the level of the considered point; Iz is the moment of inertia of the entire section about the neutral axis z. In many cases, the maximum shear stresses occur at the level of the neutral layer of the beam (rectangle, I-beam, circle). In such cases, the strength condition for shear stresses is written as, (1.14) where Qmax is the transverse force with the highest modulus; - allowable shear stress for the material. For a rectangular beam section, the strength condition has the form (1.15) A is the cross-sectional area of ​​the beam. For a circular section, the strength condition is represented as (1.16) For an I-section, the strength condition is written as follows: (1.17) d is the wall thickness of the I-beam. Usually, the dimensions of the cross section of the beam are determined from the condition of strength for normal stresses. Checking the strength of beams for shear stresses is mandatory for short beams and beams of any length, if there are large concentrated forces near the supports, as well as for wooden, riveted and welded beams. Example 1.6 Check the strength of a box-section beam (Fig. 1.14) for normal and shear stresses, if MPa. Build diagrams in the dangerous section of the beam. Rice. 1.14 Decision 23 1. Plot Q and M plots from characteristic sections. Considering the left side of the beam, we obtain. The diagram of the transverse forces is shown in fig. 1.14, c. The plot of bending moments is shown in fig. 5.14, g. 2. Geometric characteristics of the cross section 3. The highest normal stresses in the section C, where Mmax acts (modulo): MPa. The maximum normal stresses in the beam are practically equal to the allowable ones. 4. The highest shear 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 is the width of the section at the level of the neutral axis. Fig. 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 part of the section located above the line passing through the point K1; b2 cm is the wall thickness at the level of point K1. Plots  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, it is 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 for normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of the beam sections for shear stresses. Given: Solution: 1. Determine the reactions of the beam supports Check: 2. Plot Q and M diagrams. Values ​​of transverse forces in characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, the load intensity q = const. Therefore, in these sections, the diagram Q is limited to straight lines inclined to the axis. In the section DB, the intensity of the distributed load q \u003d 0, therefore, in this section, the diagram Q is limited to a straight line parallel to the x axis. Diagram Q for the beam is shown in fig. 1.16b. Values ​​of bending moments in the characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section, in which Q = 0: The maximum moment in the second section Diagram M for the beam is shown in fig. 1.16, c. 2. We compose the strength condition for normal stresses, from which we determine the required axial section modulus from the expression determined the required diameter d of a round beam. Round section area For a rectangular beam. Required section height. Rectangular section area. According to the tables of GOST 8239-89, we find the nearest greater value of the axial moment of resistance 597 cm3, which corresponds to the I-beam No. 33 with the characteristics: A z 9840 cm4. Tolerance check: (underload by 1% of the allowable 5%) the nearest I-beam No. 30 (W 2 cm3) leads to a significant overload (more than 5%). We finally accept the I-beam No. 33. We compare the areas of circular and rectangular sections with the smallest area A of the I-beam: Of the three considered sections, the I-section is the most economical. 3. We calculate the largest 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. 1.17b. 5. We determine the largest shear stresses for the selected sections of the beam. a) rectangular section of the beam: b) circular section of the beam: c) I-section of the beam: Shear stresses in the wall near the flange of the I-beam in the dangerous section A (on the right) (at point 2): The diagram of shear stresses in the dangerous sections of the I-beam is shown in fig. 1.17, in. The maximum shear stresses in the beam do not exceed the allowable stresses Example 1.8 Determine the allowable 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 the dangerous section of the beam under the allowable load. Fig 1.18 1. Determination of the reactions of the beam supports. In view of the symmetry of the system 2. Construction of diagrams Q and M from characteristic sections. Shear forces in the characteristic sections of the beam: Diagram Q for the beam is shown in fig. 5.18b. Bending moments in the characteristic sections of the beam For the second half of the beam, the ordinates M are along the axes of symmetry. Diagram M for the beam is shown in fig. 1.18b. 3. Geometric characteristics of the section (Fig. 1.19). We divide the figure into two simple elements: an I-beam - 1 and a rectangle - 2. Fig. 1.19 According to the assortment for I-beam No. 20, we have For a rectangle: Static moment of the sectional area relative to the z1 axis Distance from the z1 axis to the center of gravity of the section Moment of inertia of the section relative to the main central axis z of the entire section according to the formulas for the transition to parallel axes dangerous point "a" (Fig. 1.19) in the dangerous section I (Fig. 1.18): After substituting numerical data 5. With a permissible load in the dangerous section, the normal stresses at points "a" and "b" will be equal: dangerous section 1-1 is shown in fig. 1.19b.