Calculation of the Mohr integral using Vereshchagin's rule. Simpson's formula for multiplying diagrams - Determining displacements. O. Mohr's method in combination with Simpson's method (formula) Multiplying a diagram by itself

It is obvious that the variety of applied loads and geometric designs of structures leads to different, from the point of view of geometry, multiplied diagrams. To implement Vereshchagin's rule, you need to know the areas of geometric figures and the coordinates of their centers of gravity. Figure 29 shows some of the main options that arise in practical calculations.

To multiply diagrams of complex shapes, they must be broken down into simple ones. For example, to multiply two diagrams that look like a trapezoid, you need to divide one of them into a triangle and a rectangle, multiply the area of ​​each of them by the ordinate of the second diagram, located under the corresponding center of gravity, and add the results. The same applies to multiplying a curved trapezoid by any linear diagram.

If the above steps are carried out in general form, we will obtain formulas for such complex cases that are convenient for use in practical calculations (Fig. 30). Thus, the result of multiplying two trapezoids (Fig. 30, a):

Rice. 29

Using formula (2.21), you can also multiply diagrams that have the form of “twisted” trapezoids (Fig. 30, b), but in this case the product of ordinates located on opposite sides of the diagram axes is taken into account with a minus sign.

If one of the multiplied diagrams is outlined along a square parabola (which corresponds to loading with a uniformly distributed load), then for multiplication with the second (necessarily linear) diagram it is considered as the sum (Fig. 30, c) or the difference (Fig. 30, d) of trapezoidal and parabolic diagrams. The result of multiplication in both cases is determined by the formula:

(2.22)

but the value of f is determined differently (Fig. 30, c, d).

Rice. thirty

There may be cases when none of the multiplied diagrams is rectilinear, but at least one of them is limited by broken straight lines. To multiply such diagrams, they are first divided into sections, within each of which at least one diagram is rectilinear.

Let's consider the use of Vereshchagin's rule using specific examples.

Example 15. Determine the deflection in the middle of the span and the angle of rotation of the left supporting section of the beam loaded with a uniformly distributed load (Fig. 31, a) using the Vereshchagin method.

The sequence of calculations using Vereshchagin’s method is the same as in Mohr’s method, so we will consider three states of the beam: cargo - under the action of a distributed load q; it corresponds to the diagram M q (Fig. 31, b), and two individual states - under the action of force
applied at point C (diagram
, Fig. 31, c), and moment
, applied at point B (diagram
, Fig. 31, d).

Beam deflection in the middle of the span:

A similar result was obtained earlier by Mohr's method (see example 13). Attention should be paid to the fact that the multiplication of diagrams was performed for half of the beam, and then, due to symmetry, the result was doubled. If the area of ​​the entire diagram M q is multiplied by the ordinate of the diagram located under its center of gravity
(
in Fig. 31, c), then the amount of displacement will be completely different and incorrect since the diagram
limited by a broken line. The inadmissibility of such an approach has already been indicated above.

And when calculating the angle of rotation of the section at point B, you can multiply the area of ​​the diagram M q by the ordinate of the diagram located under its center of gravity
(
, Fig. 31, d), since the diagram
limited by a straight line:

This result also coincides with the result obtained previously by Mohr's method (see example 13).

Rice. 31

Example 16. Determine the horizontal and vertical movements of point A in the frame (Fig. 32, a).

As in the previous example, to solve the problem it is necessary to consider three states of the frame: cargo and two single. The diagram of the moments M F corresponding to the first state is presented in Fig. 32, b. To calculate the horizontal displacement, we apply a force at point A in the direction of the desired displacement (i.e. horizontally)
, and to calculate the vertical displacement force
apply vertically (Fig. 32, c, d). Corresponding diagrams
And
are shown in Fig. 32, d, f.

Horizontal movement of point A:

When calculating
in section AB, the trapezium (diagram M F) is divided into a triangle and a rectangle, after which the triangle from the diagram
"multiplied" by each of these figures. In the BC section, the curvilinear trapezoid is divided into a curvilinear triangle and a rectangle, and formula (2.21) is used to multiply diagrams in the SD section.

The "-" sign obtained during the calculation
, means that point A does not move horizontally to the left (force is applied in this direction
), and to the right.

Here the "-" sign means that point A is moving down, not up.

Note that single moment diagrams constructed from the force
, have the dimension of length, and unit diagrams of moments constructed from the moment
, are dimensionless.

Example 17. Determine the vertical displacement of point A of the plane-spatial system (Fig. 33, a).

Fig.23

As is known (see Chapter 1), three internal force factors arise in the cross sections of the rods of a plane-spatial system: transverse force Q y, bending moment M x and torque M cr. Since the influence of the transverse force on the magnitude of the displacement is insignificant (see example 14, Fig. 27), when calculating the displacement by the Mohr and Vereshchagin method, only two of the six terms remain.

To solve the problem, we will construct diagrams of bending moments M x, q and torques M cr, q from an external load (Fig. 33, b), and then at point A we will apply a force
in the direction of the desired movement, i.e. vertical (Fig. 33, c), and construct single diagrams of bending moments
and torques
(Fig. 33, d). The arrows on the torque diagrams show the directions of twisting of the corresponding sections of the plane-space system.

Vertical movement of point A:

When multiplying torque diagrams, the product is taken with a “+” sign if the arrows indicating the direction of torsion are co-directional, and with a “-” sign otherwise.

EE "BSUIR"

Department of Engineering Graphics

“DETERMINATION OF DISPLACEMENTS BY THE MOR METHOD. VERESHCHAGIN'S RULE"

MINSK, 2008

Let us now consider a general method for determining displacements, suitable for any linearly deformable system under any load. This method was proposed by the outstanding German scientist O. Mohr.

Let, for example, you want to determine the vertical displacement of point A of the beam shown in Fig. 7.13, a. We denote the given (load) state by the letter k. Let us choose an auxiliary state of the same beam with unit

force acting at point A and in the direction of the desired displacement. We denote the auxiliary state by the letter i (Fig. 7.13,6).

Let us calculate the work of external and internal forces of the auxiliary state on the displacements caused by the action of the forces of the load state.

The work of external forces will be equal to the product of a unit force and the desired displacement ya

and the work of internal forces in absolute value is equal to the integral

(1)

Formula (7.33) is Mohr’s formula (Mohr’s integral), which makes it possible to determine the displacement at any point of a linearly deformable system.

In this formula, the integrand of MiMk is positive if both bending moments have the same sign, and negative if Mi and Mk have different signs.

If we were to determine the angular displacement at point A, then in state i we should apply a moment equal to one (without dimension) at point A.

Denoting by the letter Δ any movement (linear or angular), we write Mohr’s formula (integral) in the form

(2)

In the general case, the analytical expression Mi and Mk can be different in different sections of a beam or an elastic system in general. Therefore, instead of formula (2), one should use the more general formula

(3)

If the rods of the system work not in bending, but in tension (compression), as, for example, in trusses, then Mohr’s formula has the form

(4)

In this formula, the product NiNK is positive if both forces are tensile or both are compressive. If the rods simultaneously work in bending and tension (compression), then in ordinary cases, as comparative calculations show, displacements can be determined taking into account only bending moments, since the influence of longitudinal forces is very small.

For the same reasons, as noted earlier, in ordinary cases the influence of shear forces can be ignored.

Instead of directly calculating the Mohr integral, you can use the grapho-analytical technique “method of multiplying diagrams,” or Vereshchagin’s rule.

Let's consider two diagrams of bending moments, one of which Mk has an arbitrary outline, and the other Mi is rectilinear (Fig. 7.14, a and b).

(5)

The value MKdz represents the elementary area dωk of the diagram Mk (shaded in the figure). Thus,

(6)

hence,

(8)

But represents the static moment of the area of ​​the diagram Mk relative to some axis y passing through the point O, equal to ωkzc, where ωk is the area of ​​the moment diagram; zc is the distance from the y-axis to the center of gravity of the Mk diagram. From the drawing it is clear that

where Msi is the ordinate of the diagram Mi, located under the center of gravity of the diagram Mk (under point C). Hence,

(10)

i.e., the required integral is equal to the product of the area of ​​the diagram Mk (any shape) by the ordinate of the rectilinear diagram Msi located under its center of gravity. The value of ωкМсi is considered positive if both diagrams are located on the same side of the rod, and negative if they are located on different sides. A positive result of multiplying diagrams means that the direction of movement coincides with the direction of a unit force (or moment).

It must be remembered that the ordinate Msi must be taken in a straight-line diagram. In the particular case when both diagrams are rectilinear, you can multiply the area of ​​any of them by the corresponding ordinate of the other.

For bars of variable cross-section, Vereshchagin’s rule for multiplying diagrams is not applicable, since in this case it is no longer possible to remove the value EJ from under the integral sign. In this case, EJ should be expressed as a function of the abscissa of the section and then the Mohr integral (1) should be calculated.

When changing the rigidity of a rod stepwise, integration (or multiplication of diagrams) is carried out for each section separately (with its own EJ value) and then the results are summed up.

In table 1 shows the areas of some simple diagrams and the coordinates of their center of gravity.

Table 1

Type of diagram	Area of ​​the diagram	Distance to center of gravity

To speed up calculations, you can use ready-made diagram multiplication tables (Table 2).

In this table, in the cells at the intersection of the corresponding elementary diagrams, the results of multiplying these diagrams are given.

When breaking down a complex diagram into elementary ones, presented in table. 1 and 7.2, it should be borne in mind that the parabolic diagrams were obtained from the action of only one distributed load.

In cases where in a complex diagram, curved sections are obtained from the simultaneous action of concentrated moments, forces and a uniformly distributed load, in order to avoid errors, the complex diagram should first be “layered”, i.e., divided into a number of independent diagrams: from the action of concentrated moments, forces and from the action of a uniformly distributed load.

You can also use another technique that does not require stratification of the diagrams, but only requires the selection of the curvilinear part of the diagram along the chord connecting its extreme points.

We will demonstrate both methods with a specific example.

Let, for example, you want to determine the vertical displacement of the left end of the beam (Fig. 7.15).

The total diagram of the load is presented in Fig. 7.15, a.

Table 7.2

The diagram of the action of a unit force at point A is shown in Fig. 7.15, city

To determine the vertical displacement at point A, it is necessary to multiply the load diagram by the unit force diagram. However, we note that in the section BC of the total diagram, the curvilinear diagram is obtained not only from the action of a uniformly distributed load, but also from the action of a concentrated force P. As a result, in the section BC there will no longer be an elementary parabolic diagram given in Tables 7.1 and 7.2, but according to essentially a complex diagram for which the data in these tables is invalid.

Therefore, it is necessary to stratify the complex diagram according to Fig. 7.15, and to the elementary diagrams presented in Fig. 7.15, b and 7.15, c.

Diagram according to Fig. 7.15, b was obtained only from concentrated force, diagram according to Fig. 7.15, c - only from the action of a uniformly distributed load.

Now you can multiply the diagrams using the table. 1 or 2.

To do this, you need to multiply the triangular diagram according to Fig. 7.15, b to the triangular diagram according to Fig. 7.15, d and add to this the result of multiplying the parabolic diagram in Fig. 7.15, in the trapezoidal diagram of the BC section according to Fig. 7.15, d, since in section AB the ordinates of the diagram according to Fig. 7.15, in are equal to zero.

Let us now show the second method of multiplying diagrams. Let's look again at the diagram in Fig. 7.15, a. Let us take the origin of reference in section B. We show that within the limits of the curve LMN, bending moments can be obtained as the algebraic sum of the bending moments corresponding to the straight line LN, and the bending moments of the parabolic diagram LNML, the same as for a simple beam of length a, loaded with a uniformly distributed load q:

The largest ordinate in the middle will be equal to .

To prove this, let’s write the actual expression for the bending moment in the section at a distance z from point B

(A)

Let us now write the expression for the bending moment in the same section, obtained as the algebraic sum of the ordinates of the straight line LN and the parabola LNML.

Equation of line LN

where k is the tangent of the angle of inclination of this line

Consequently, the equation of bending moments obtained as the algebraic sum of the equation of the straight line LN and the parabola LNMN has the form

which coincides with expression (A).

When multiplying diagrams according to Vereshchagin’s rule, you should multiply the trapezoid BLNC by the trapezoid from the unit diagram in the section BC (see Fig. 7.15, d) and subtract the result of multiplying the parabolic diagram LNML (area ) by the same trapezoid from the unit diagram. This method of layering diagrams is especially beneficial when the curved section of the diagram is located in one of the middle sections of the beam.

Example 7.7. Determine the vertical and angular displacements of the cantilever beam at the point where the load is applied (Fig. 7.16).

Solution. We construct a diagram of bending moments for the load state (Fig. 7.16, a).

To determine the vertical displacement, we select the auxiliary state of the beam with a unit force at the point of application of the load.

We construct a diagram of bending moments from this force (Fig. 7.16, b). Determining vertical displacement using Mohr's method

Bending moment value due to load

The value of the bending moment from a unit force

We substitute these values ​​of МР and Mi under the integral sign and integrate

The same result was previously obtained by a different method.

A positive deflection value indicates that the point of application of the load P is moving downward (in the direction of the unit force). If we directed a unit force from bottom to top, we would have Mi = 1z and, as a result of integration, we would get a deflection with a minus sign. The minus sign would indicate that the movement is not up, but down, as it is in reality.

Let us now calculate the Mohr integral by multiplying the diagrams according to Vereshchagin’s rule.

Since both diagrams are rectilinear, it does not matter which diagram to take the area from and which to take the ordinate from.

The area of ​​the load diagram is equal to

The center of gravity of this diagram is located at a distance of 1/3l from the embedment. We determine the ordinate of the diagram of moments from a unit force, located under

center of gravity of the load diagram. It is easy to verify that it is equal to 1/3l.

Hence.

The same result is obtained from the table of integrals. The result of multiplying diagrams is positive, since both diagrams are located at the bottom of the rod. Consequently, the point of application of the load shifts downward, i.e., along the accepted direction of the unit force.

To determine the angular displacement (angle of rotation), we select an auxiliary state of the beam in which a concentrated moment equal to unity acts at the end of the beam.

We construct a diagram of bending moments for this case (Fig. 7.16, c). We determine the angular displacement by multiplying the diagrams. Load diagram area

The ordinates of the diagram from a single moment are equal to unity everywhere. Therefore, the desired angle of rotation of the section is equal to

Since both diagrams are located below, the result of multiplying the diagrams is positive. Thus, the end section of the beam rotates clockwise (in the direction of the unit moment).

Example: Using the Mohr-Vereshchagin method, determine the deflection at point D for the beam shown in Fig. 7.17..

Solution. We build a layered diagram of moments from the load, i.e. we build separate diagrams from the action of each load. In this case, for the convenience of multiplying diagrams, it is advisable to construct stratified (elementary) diagrams relative to the section, the deflection of which is determined in this case relative to section D.

In Fig. 7.17, a shows a diagram of bending moments from reaction A (section AD) and from load P = 4 T (section DC). Diagrams are built on compressed fiber.

In Fig. 7.17, b shows diagrams of moments from reaction B (section BD), from the left uniformly distributed load (section AD) and from a uniformly distributed load acting on section BC. This diagram is shown in Fig. 7.17, b on the DC section from below.

Next, we select the auxiliary state of the beam, for which we apply a unit force at point D, where the deflection is determined (Fig. 7.17, c). The diagram of moments from a unit force is shown in Fig. 7.17, d. Now let’s multiply diagrams 1 to 7 by diagrams 8 and 9, using diagram multiplication tables, taking into account the signs.

In this case, diagrams located on one side of the beam are multiplied with a plus sign, and diagrams located on opposite sides of the beam are multiplied with a minus sign.

When multiplying diagram 1 and diagram 8 we get

Multiplying plot 5 by plot 8, we get

Multiplying plots 2 and 9 gives

Multiply plots 4 and 9

Multiply diagrams 6 and 9

Summing up the results of multiplying diagrams, we get

The minus sign shows that point D does not move downwards, as the unit force is directed, but upwards.

The same result was obtained earlier using the universal equation.

Of course, in this example, it was possible to stratify the diagram only in section AD, since in section DB the total diagram is rectilinear and there is no need to stratify it. In the BC section, delamination is not required, since from a unit force in this section the diagram is equal to zero. The stratification of the diagram in the section BC is necessary to determine the deflection at point C.

Example. Determine the vertical, horizontal and angular displacements of section A of the broken rod shown in Fig. 7.18, a. The cross-sectional rigidity of the vertical section of the rod is EJ1; the cross-sectional rigidity of the horizontal section is EJ2.

Solution. We construct a diagram of bending moments due to load. It is shown in Fig. 7.18, b (see example 6.9). To determine the vertical displacement of section A, we select the auxiliary state of the system shown in Fig. 7.18, c. At point A, a unit vertical force is applied, directed downward.

The diagram of bending moments for this state is shown in Fig. 7.18, c.

We determine vertical displacement using Mohr's method, using the method of multiplying diagrams. Since there is no diagram M1 on the vertical rod in the auxiliary state, we multiply only diagrams related to the horizontal rod. We take the area of ​​the diagram from the load state, and the ordinate from the auxiliary state. The vertical displacement is

Since both diagrams are located below, we take the result of the multiplication with a plus sign. Consequently, point A moves downward, i.e., in the direction of the unit vertical force.

To determine the horizontal movement of point A, we select an auxiliary state with a horizontal unit force directed to the left (Fig. 7.18, d). The moment diagram for this case is presented there.

We multiply diagrams MP and M2 and get

The result of multiplying diagrams is positive, since the multiplied diagrams are located on the same side of the rods.

To determine the angular displacement, we select the auxiliary state of the system according to Fig. 7.18.5 and construct a diagram of bending moments for this state (in the same figure). We multiply diagrams MP and M3:

The result of multiplication is positive, since the multiplied diagrams are located on one side.

Therefore, section A rotates clockwise

The same results would be obtained using tables
multiplying diagrams.

The view of the deformed rod is shown in Fig. 7.18, e, while the displacements are greatly increased.

LITERATURE

Feodosiev V.I. Strength of materials. 1986

Belyaev N.M. Strength of materials. 1976

Kraskovsky E.Ya., Druzhinin Yu.A., Filatova E.M. Calculation and design of instrument mechanisms and computer systems. 1991

Rabotnov Yu.N. Mechanics of deformable solids. 1988

Stepin P.A. Strength of materials. 1990

And his handwritten notes ended up in the hands of the clerk of the Ambassadorial Prikaz, from whom they were received. Other biographical information is extracted only from the text of the “Walk” itself. Why did Afanasy Nikitin call his work “Walking across Three Seas”? The author himself gives us the answer to this question: “Behold, I wrote my sinful “Walking across the Three Seas”, the 1st Derbensky (Caspian) Sea, Doria...

Notes that an indispensable condition for the implementation of any communicative act must be “mutual knowledge of the realities of the speaker and the listener, which is the basis of linguistic communication,” they are called “background knowledge” in linguistics. According to her correct remark, “the meaning of the word used in a given native language to designate such completely different ones from the point of view of Central European culture...

Determining displacements in systems consisting of rectilinear elements of constant rigidity can be significantly simplified by using a special calculation technique

integral of the form

Due to the fact that the integrand includes the product of efforts Mm and Mn, which are the ordinates of diagrams constructed for a single and real state, this technique is called the method of multiplying diagrams. It can be used in the case when one of the diagrams being multiplied, for example Mt, is rectilinear; in this case (Fig. 5.17)

Mm = (x + a) tan a.

The second diagram M p can have any shape (straight, broken

or curvilinear).

Let's substitute the value of M m into the expression

where M n dx= dΩ n is the differential area Ω n of the diagram M n (Fig. 5.17),

Integral represents the static moment of the area Ω n of the diagram M p relative to the 0-0 axis (Fig. 5.17). This static moment can be expressed differently:

where xc is the abscissa of the center of gravity of the diagram area Mn. Then

But since (see Fig. 5.17)

(5.26)

Thus, the result of multiplying two diagrams is equal to the product of the area of ​​one of them by the ordinate of the other (straight-line) diagram, taken under the center of gravity of the area of ​​the first diagram.

The method of multiplying diagrams was proposed in 1925 by A.K. Vereshchagin, a student at the Moscow Institute of Railway Engineers, and therefore it is called Vereshchagin’s rule (or method),

Note that the left side of expression (5.26) differs from the Mohr integral in the absence of rigidity of the section EJ in it. Consequently, the result of multiplying diagrams according to Vereshchagin’s rule to determine the desired displacement must be divided by the stiffness.

It is very important to note that the ordinate must be taken from a straight-line diagram. If both diagrams are straight, then the ordinate can be taken from any diagram. So, if you need to multiply rectilinear diagrams Mi and Mk (Fig. 518, a), then it does not matter what to take: the product yk of the area of ​​the diagram Mi by the ordinate yk under its center of gravity from the diagram Mk or the product Ω_k yi of the area of ​​the diagram M k by ordinate уi under (or above) its center of gravity from the Mg diagram.

When two diagrams in the form of a trapezoid are multiplied, there is no need to find the position of the center of gravity of the area of ​​one of them. You should divide one of the diagrams into two triangles and multiply the area of ​​each of them by the ordinate under its center of gravity from the other diagram. For example, in the case shown in Fig. 518, b, we get

In parentheses of this formula, the product ac of the left ordinates of both diagrams and the product bd of the right ordinates are taken with a coefficient equal to two, and the products ad and bc of ordinates located on different sides - with a coefficient equal to one.

Using formula (5.27), you can multiply diagrams that look like “twisted” trapezoids; in this case, the products of ordinates with the same signs are taken with a plus sign, and different ones - with a minus sign. In the case, for example, shown in Fig. 5.18c, the result of multiplying diagrams in the form of a “twisted” and an ordinary trapezoid is equal to (l/6) (2ac-2bd+ad-bc), and in the case shown in Fig. 5.18, g, is equal to (l/6) (-2ac-2bd+ad+bc).

Formula (5.27) is also applicable when one or both diagrams being multiplied are in the form of a triangle. In these cases, the triangle is treated as a trapezoid with one extreme ordinate equal to zero. The result, for example, of multiplying the diagrams shown in Fig. 5.18, d, equal to (l/6) (2ac+ad).

Multiplying a diagram in the form of a “twisted” trapezoid by any other diagram can be done by dividing the “twisted” trapezoid into two triangles, as shown in Fig. 5.18, e.

Lecture No. 6. Calculation of statically indeterminate flat rod systems: beams, frames, trusses.

Lecture outline:

1. Method of forces.

1.1. Main system. Major unknowns.

1.2. System of canonical equations of the force method for calculating the action of an external load.

1.3. Calculation of statically indeterminate systems by the method of forces.

2. Method of movement.

2.1. Selecting unknowns and determining their number.

2.2. Determining the number of unknowns

2.3. Main system

2.4. Canonical equations

3. Fundamentals of calculation of systems using the finite element method.

Lecture 13 (continued). Examples of solutions for calculating displacements using the Mohr-Vereshchagin method and problems for independent solution

Defining displacements in beams

Example 1.

Determine the movement of a point TO beams (see figure) using the Mohr integral.

Solution.

1) We compose an equation for the bending moment from an external force M F .

2) Apply at the point TO unit force F = 1.

3) We write down the equation of bending moment from a unit force.

4) Determine movements

Example 2.

Determine the movement of a point TO beams according to Vereshchagin's method.

Solution.

1) We are building a cargo diagram.

2) We apply a unit force at point K.

3) We build a single diagram.

4) Determine the deflection

Example 3.

Determine the angles of rotation on the supports A And IN

Solution.

We construct diagrams from a given load and from individual moments applied in sections A And IN(see picture). We determine the required displacements using Mohr integrals

,

, which we calculate using Vereshchagin’s rule.

Finding the plot parameters

C 1 = 2/3, C 2 = 1/3,

and then the rotation angles on the supports A And IN

Example 4.

Determine the angle of rotation of the section WITH for a given beam (see figure).

Solution.

Determining support reactions R A =R B ,

, , R A = R B = qa.

We construct diagrams of the bending moment from a given load and from a single moment applied in the section WITH, where the rotation angle is sought. We calculate the Mohr integral using Vereshchagin's rule. Finding the plot parameters

C 2 = -C 1 = -1/4,

and along them the desired movement

Example 5.

Determine the deflection in the section WITH for a given beam (see figure).

Solution.

Diagram M F(Fig. b)

Support reactions:

BE: , ,

, R B + R E = F, R E = 0;

AB: , R A = R IN = F; , .

We calculate moments at characteristic points, M B = 0, M C = Fa and build a diagram of the bending moment from a given load.

Diagram(Fig. c).

In cross section WITH, where the deflection is sought, we apply a unit force and construct a bending moment diagram from it, first calculating the support reactions BE - , , = 2/3; , , = 1/3, and then moments at characteristic points , , .

2. Determination of the desired deflection. Let's use Vereshchagin's rule and first calculate the parameters of the diagrams and:

,

Section deflection WITH

Example 6.

Determine the deflection in the section WITH for a given beam (see figure).

Solution.

WITH. Using Vereshchagin’s rule, we calculate the parameters of the diagrams ,

and find the desired deflection

Example 7.

Determine the deflection in the section WITH for a given beam (see figure).

Solution.

1. Constructing diagrams of bending moments.

Support reactions:

, , R A = 2qa,

, R A + R D = 3qa, R D = qa.

We construct diagrams of bending moments from a given load and from a unit force applied at a point WITH.

2. Determination of movements. To calculate the Mohr integral, we use Simpson’s formula, sequentially applying it to each of the three sections into which the beam is divided.

PlotAB :

PlotSun :

PlotWITH D :

Required movement

Example 8.

Determine the section deflection A and section rotation angle E for a given beam (Fig. A).

Solution.

1. Constructing diagrams of bending moments.

Diagram M F(rice. V). Having determined the support reactions

, , R B = 19qa/8,

, R D = 13qa/8, we build diagrams of transverse force Q and bending moment M F from a given load.

Diagram(Fig. d). In cross section A, where the deflection is sought, we apply a unit force and construct a diagram of the bending moment from it.

Diagram(Fig. e). This diagram is constructed from a single moment applied in the section E, where the rotation angle is sought.

2. Determination of movements. Section deflection A we find using Vereshchagin's rule. Epure M F at the sites Sun And CD We break it down into simple parts (Fig. d). We present the necessary calculations in the form of a table.

	-qa 3 /6	2qa 3 /3	-qa 3 /2				-qa 3 /2
C i
		-qa 4 /2	5qa 4 /12	-qa 4 /6	-qa 4 /12			-qa 4 /24

We get.

The minus sign in the result means that the point A does not move downwards, as the unit force was directed, but upwards.

Section rotation angle E we find in two ways: by Vereshchagin’s rule and by Simpson’s formula.

According to Vereshchagin’s rule, multiplying diagrams M F and, by analogy with the previous one, we obtain

,

To find the angle of rotation using Simpson's formula, we calculate the preliminary bending moments in the middle of the sections:

The required displacement, increased by EI x once,

Example 9.

Determine at what value of the coefficient k section deflection WITH will be equal to zero. When the value is found k construct a diagram of the bending moment and depict an approximate view of the elastic line of the beam (see figure).

Solution.

We construct diagrams of bending moments from a given load and from a unit force applied in the section WITH, where the deflection is sought.

According to the conditions of the problem V C= 0. On the other hand, . Integral on the plot AB we calculate using Simpson’s formula, and in the section Sun– according to Vereshchagin’s rule.

We find in advance

Moving a section WITH ,

From here , .

When the value is found k determine the value of the support reaction at the point A: , , , from which we find the position of the extremum point on the diagram M according to the condition .

Based on the moment values ​​at characteristic points

We build a diagram of the bending moment (Fig. d).

Example 10.

IN cantilever beam shown in the figure.

Solution.

M from the action of an external concentrated force F: M IN = 0, M A = –F 2l(linear plot).

According to the conditions of the problem, it is necessary to determine the vertical displacement at IN points IN cantilever beam, therefore we build a unit diagram of the action of a vertical unit force F i = 1 applied at the point IN.

Considering that the cantilever beam consists of two sections with different bending rigidities, diagrams and M We multiply using Vereshchagin’s rule by sections separately. Diagrams M and multiply the first section using the formula , and the diagrams of the second section - as the area of ​​the diagram M second section Fl 2 / 2 to ordinate 2 l/3 diagrams of the second section under the center of gravity of the triangular diagram M the same area.

In this case the formula gives:

Example 11.

Determine the vertical movement of a point IN single-span beam shown in the figure. The beam has a constant bending rigidity along its entire length. EI.

Solution.

We build a diagram of bending moments M from the action of an external distributed load: M A = 0; M D = 0;

Apply at the point IN unit vertical force F i = 1 and build a diagram (see figure):

where R a = 2/3;

Where R d = 1/3, so M a = 0; M d = 0; .

Let's divide the beam in question into 3 sections. Multiplying diagrams of the 1st and 3rd sections does not cause difficulties, since we multiply triangular diagrams. In order to apply Vereshchagin’s rule to the 2nd section, let’s split the diagram M 2nd section into two components of the diagram: rectangular and parabolic with area (see table).

Center of gravity of the parabolic part of the diagram M lies in the middle of the 2nd section.

So the formula using Vereshchagin's rule gives:

Example 12.

Determine the maximum deflection in a two-support beam loaded with a uniformly distributed load of intensity q(see picture).

Solution.

Finding bending moments:

From a given load

From a unit force applied at a point WITH where the deflection is sought.

We calculate the required maximum deflection that occurs in the middle section of the beam

Example 13.

Determine the deflection at a point IN beam shown in the figure.

Solution.

We construct diagrams of bending moments from a given load and a unit force applied at a point IN. To multiply these diagrams, the beam must be divided into three sections, since a single diagram is limited to three different straight lines.

The operation of multiplying diagrams in the second and third sections is carried out simply. Difficulties arise when calculating the area and coordinates of the center of gravity of the main diagram in the first section. In such cases, constructing layered diagrams greatly simplifies the solution of the problem. In this case, it is convenient to take one of the sections conditionally as stationary and construct diagrams for each of the loads, approaching this section from the right and left. It is advisable to take the section at the fracture site as a stationary one in the diagram of unit loads.

A layered diagram in which the section is taken as the stationary one IN, is shown in the figure. Having calculated the areas of the component parts of the layered diagram and the corresponding ordinates of the unit diagram, we obtain

Example 14.

Determine the displacements at points 1 and 2 of the beam (Fig. a).

Solution.

Here are the diagrams M And Q for beams at A=2 m; q=10 kN/m; WITH=1,5A; M=0,5qa 2 ; R=0,8qa; M 0 =M; =200 MPa (Fig. b And V).

Let us determine the vertical displacement of the center of the section where the concentrated moment is applied. To do this, consider a beam in a state under the influence of only a concentrated force applied at point 1 perpendicular to the axis of the beam (in the direction of the desired displacement) (Fig. d).

Let's calculate the support reactions by composing three equilibrium equations

Examination

The reactions were found correctly.

To construct a diagram, consider three sections (Fig. d).

1 plot

2nd section

Section 3

Using these data, we construct a diagram (Fig. e) from the side of the stretched fibers.

Let us determine by Mohr's formula using Vereshchagin's rule. In this case, a curved diagram in the area between the supports can be represented as the addition of three diagrams. Arrow

The minus sign means that point 1 moves up (in the opposite direction).

Let us determine the vertical displacement of point 2, where the concentrated force is applied. To do this, consider a beam in a state under the influence of only a concentrated force applied at point 2 perpendicular to the axis of the beam (in the direction of the desired displacement) (Fig. e).

The diagram is constructed similarly to the previous one.

Point 2 moves up.

Let us determine the angle of rotation of the section where the concentrated moment is applied.

The disadvantage of Mohr's method is the need to obtain the values ​​of the internal force factors included in the integrand expressions of formulas (2.18) and (2.19), in general form, as functions of z, which becomes quite labor-intensive even with two or three partition sections in beams and especially in frames

It turns out that this drawback can be avoided if direct integration in Mohr’s formulas is replaced by the so-called multiplying diagrams. Such a replacement is possible in cases where at least one of the multiplied diagrams is rectilinear. All systems consisting of straight rods meet this condition. Indeed, in such systems, the diagram constructed from a generalized unit force will always be rectilinear.

The method of calculating the Mohr integral by replacing direct integration by multiplying the corresponding diagrams is called Vereshchagin's method (or rule) and is as follows: in order to multiply two diagrams, at least one of which is rectilinear, you need to multiply the area of ​​one diagram (if there is a curved diagram, then its area must be) by the ordinate of the other diagram, located under the center of gravity of the first.

Let us prove the validity of this rule. Let's look at two diagrams (Fig. 28). Let one of them (Mn) be a load one and have a curved outline, and the second one corresponds to a unit load and is linear.

From Fig. 28 it follows that Let us substitute the values ​​into the expression

where is the differential area of ​​the diagram Mn.

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The integral represents the static moment of area relative to the O – O1 axis, while:

where zc is the abscissa of the center of gravity of the area, then:

Considering that we get:
(2.20)
Expression (2.20) determines the result of multiplying two diagrams, and not moving. To obtain the displacement, this result must be divided by the stiffness corresponding to the internal force factors under the integral sign.

Basic options for multiplying diagrams

It is obvious that the variety of applied loads and geometric designs of structures leads to different, from the point of view of geometry, multiplied diagrams. For implementation Vereshchagin's rules you need to know the areas of geometric figures and the coordinates of their centers of gravity. Figure 29 shows some of the main options that arise in practical calculations.

For multiplying diagrams complex forms, they must be broken down into simple ones. For example, to multiply two diagrams that look like a trapezoid, you need to divide one of them into a triangle and a rectangle, multiply the area of ​​each of them by the ordinate of the second diagram, located under the corresponding center of gravity, and add the results. The same applies to multiplying a curved trapezoid by any linear diagram.

If the above steps are carried out in general form, we will obtain formulas for such complex cases that are convenient for use in practical calculations (Fig. 30). Thus, the result of multiplying two trapezoids (Fig. 30, a):

(2.21)

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Using formula (2.21), you can also multiply diagrams that have the form of “twisted” trapezoids (Fig. 30, b), but in this case the product of ordinates located on opposite sides of the diagram axes is taken into account with a minus sign.

If one of multiplyable diagrams is outlined along a square parabola (which corresponds to loading with a uniformly distributed load), then for multiplication with the second (necessarily linear) diagram it is considered as the sum (Fig. 30, c) or the difference (Fig. 30, d) of trapezoidal and parabolic diagrams. The result of multiplication in both cases is determined by the formula:
(2.22)

but the value of f is determined differently (Fig. 30, c, d).

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There may be cases when none of the multiplied diagrams is rectilinear, but at least one of them is limited by broken straight lines. To multiply such diagrams, they are first divided into sections, within each of which at least one diagram is rectilinear.
Consider using Vereshchagin's rules on specific examples.

Example 15. Determine the deflection in the middle of the span and the angle of rotation of the left supporting section of the beam loaded with a uniformly distributed load (Fig. 31,a), Vereshchagin's method.

Calculation sequence Vereshchagin's method– the same as in Mohr’s method, therefore we will consider three states of the beam: load – under the action of a distributed load q; it corresponds to the diagram Mq (Fig. 31, b), and two single states - under the action of a force applied at point C (diagram, Fig. 31, c), and a moment applied at point B (diagram, Fig. 31, d) .

Beam deflection in the middle of the span:

A similar result was obtained earlier by Mohr's method (see example 13). Attention should be paid to the fact that the multiplication of diagrams was performed for half of the beam, and then, due to symmetry, the result was doubled. If the area of ​​the entire diagram Mq is multiplied by the ordinate of the diagram located under its center of gravity (in Fig. 31, c), then the amount of displacement will be completely different and incorrect since the diagram is limited by a broken line. The inadmissibility of such an approach has already been indicated above.

And when calculating the angle of rotation of the section at point B, you can multiply the area of ​​the diagram Mq by the ordinate of the diagram located under its center of gravity (Fig. 31, d), since the diagram is limited by a straight line:

This result also coincides with the result obtained previously by Mohr's method (see example 13).

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Example 16. Determine the horizontal and vertical movements of point A in the frame (Fig. 32, a).

As in the previous example, to solve the problem it is necessary to consider three states of the frame: cargo and two single. The diagram of the moments MF corresponding to the first state is presented in Fig. 32, b. To calculate the horizontal movement, we apply force at point A in the direction of the desired movement (i.e., horizontally), and to calculate the vertical movement, we apply the force vertically (Fig. 32, c, e). The corresponding diagrams are shown in Fig. 32, d, f.

Horizontal movement of point A:

When calculating in section AB, the trapezium (diagram MF) is divided into a triangle and a rectangle, after which the triangle from the diagram is “multiplied” by each of these figures. In the BC section, the curvilinear trapezoid is divided into a curvilinear triangle and a rectangle, and formula (2.21) is used to multiply diagrams in the SD section.

The "-" sign obtained during the calculation means that point A moves horizontally not to the left (the force is applied in this direction), but to the right.