EE "BSUIR"
Department of Engineering Graphics
“DETERMINATION OF DISPLACEMENTS BY THE MOR METHOD. VERESHCHAGIN'S RULE"
MINSK, 2008
Let's now consider general method displacement determination, 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 select 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 auxiliary state on movements caused by the action of forces of the load state.
Job external forces will be equal to the product of the unit force and the desired displacement ya
and the work of internal forces on absolute value 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 along 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
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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 parabolic diagrams are 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
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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.
Positive value deflection shows that the point of application of the load P moves downwards (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
the 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 below 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 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 diagram of moments 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 solid. 1988
Stepin P.A. Strength of materials. 1990
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In the general case (rod of variable cross-section, complex system loads) the Mohr integral is determined by numerical integration. In many practically important cases, when the stiffness of the section is constant along the length of the rod, the Mohr integral can be calculated using Vereshchagin's rule. Let us consider the definition of the Mohr integral in the section from a to 6 (Fig. 9.18).
Rice. 9.18. Vereshchagin's rule for calculating the Mohr integral
Diagrams of the moment from a single force factor consist of straight segments. Without loss of generality, we assume that within the area
where A and B are the parameters of the line:
The Mohr integral on the section of constant cross section under consideration has the form
where F is the area under the curve (the area of the diagram of bending moments from external forces in section z).
where is the abscissa of the center of gravity of the area.
Equality (109) is valid when the sign does not change within the area and can be considered as an element of the area of the diagram. Now from relations (107) -(109) we obtain
Moment from a unit load in a section
An auxiliary table for using Vereshchagin's rule is given in Fig. 9.19.
Notes. 1. If the diagram from the action of external forces on a section is linear (for example, under the action of concentrated forces and moments), then the rule can be applied in reverse form: multiply the area of the diagram from a single force factor by the ordinate of the diagram corresponding to the center of gravity of the area. This follows from the above proof.
2. Vereshchagin’s rule can be extended to the Mohr integral in general view(Equation (103)).
Rice. 9.19. Areas and positions of centers of gravity of moment diagrams
Rice. 9.20. Examples of determining deflection and rotation angles using Vereshchagin’s rule
The main requirement is the following: within the site, the internal force factors from a unit load must be linear functions along the axis of the rod (linear diagrams!).
Examples. 1. Determine the deflection at point A of the cantilever rod under the action of a concentrated moment M (Fig. 9.20, a).
The deflection at point A is determined by the formula (for brevity, the index is omitted)
The minus sign is due to the fact that they have different signs.
2. Determine the deflection at point A in the cantilever rod under the action of a distributed load.
The deflection is determined by the formula
Diagrams of the bending moment M and shearing force Q from the external load are shown in Fig. 9.20, b, below in this figure are diagrams under the action of a unit force. Next we find
3. Determine the deflection at point A and the angle of rotation at point B for a two-support beam loaded with a concentrated moment (Fig. 9.20.).
The deflection is determined by the formula (we neglect the shear deformation)
Since the diagram of the moment from a unit force is not depicted by one line; then we divide the integral into two sections:
The angle of rotation at point B is equal to
Comment. From the above examples it is clear that Vereshchagin’s method in simple cases allows you to quickly determine deflections and angles of rotation. It is only important to apply a single rule of signs for If you agree, when bending a rod, to build diagrams of bending moments on the “stretched fiber” (see Fig. 9.20), then it is immediately easy to see the positive and negative values moments.
A special advantage of Vereshchagin's rule is that it can be used not only for rods, but also for frames (Section 17).
Restrictions on the application of Vereshchagin's rule.
These restrictions follow from the derivation of formula (110), but let us pay attention to them again.
1. The diagram of the bending moment from a unit load should be in the form of one straight line. In Fig. 9.21, and shows the case when this condition is not met. The Mohr integral must be calculated separately for sections I and II.
2. The bending moment from an external load within the section must have the same sign. In Fig. 9.21, b shows the case when Vereshchagin’s rule should be applied for each section separately. This limitation does not apply to the moment from a single load.
Rice. 9.21. Restrictions when using Vereshchagin's rule: a - the diagram has a break; b - the diagram has different signs; c - the rod has different sections
3. The stiffness of the rod within a section must be constant, otherwise the integration should be extended separately to sections with constant stiffness. Limitations on constant stiffness can be avoided by plotting diagrams.
Determining displacements in systems consisting of rectilinear elements of constant rigidity can be significantly simplified by using a special technique for calculating an integral of the form. Due to the fact that the integrand includes the product of efforts that 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 multiplied diagrams is, for example, rectilinear; in this case (Fig. The second diagram can have any shape (straight, broken or curvilinear).
Let's substitute the value into the expression
where is the differential area of the diagram (Fig. 17.11).
The integral represents the static moment of the area of the diagram relative to the axis (Fig. 17.11).
This static moment can be expressed differently:
where is the abscissa of the center of gravity of the diagram area
But since (see Fig. 17.11)
(26.11)
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 (rectilinear) diagram, taken under the center of gravity of the area of the first diagram.
A method for multiplying diagrams was proposed in 1925 by a student at the Moscow Institute of Engineers railway transport A. N. Vereshchagin, and therefore it is called Vereshchagin’s rule (or method).
Note that the left side of expression (26.11) differs from the Mohr integral in the absence of section rigidity in it. Consequently, the result of the multiplication of diagrams performed according to Vereshchagin’s rule to determine the desired displacement must be divided by the value of rigidity.
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 and (Fig. 18.11, a), then it does not matter what to take: the product of the area of the diagram by the ordinate under its center of gravity from the diagram or the product Qkyt of the area Q of the diagram by the ordinate under (or above) its center gravity from the 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. 11.18.b, we get
(27.11)
In parentheses of this formula, the product of the left ordinates of both diagrams and the product of the right ordinates are taken with a coefficient equal to two, and the products of ordinates located on different sides - with a coefficient equal to one.
Using formula (27.11), you can multiply diagrams that look like “twisted” trapezoids; in this case, the products of ordinates that have the same signs are taken with a plus sign, and different ones - with a minus sign. In the case, for example, shown in Fig. 18.11, b, the result of multiplying diagrams in the form of a “twisted” and an ordinary trapezoid is equal to , and in the case shown in Fig. 18.11, g, equal
Formula (27.11) is also applicable when one or both diagrams being multiplied have 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. 18.11, d, equal
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. 18.11, e.
When one of the diagrams (Fig. 19.11) is outlined according to square parabola(from a uniformly distributed load q), then for multiplication with another diagram it is considered as the sum (in the case shown in Fig. 19.11, a) or the difference (in the case shown in Fig. 19.11, b) trapezoidal and parabolic diagrams
The result of multiplying the diagrams shown in Fig. 19.11, a, is equal after substitution into it we get
The result of multiplying the diagrams shown in Fig. 19.11, b, is equal after substitution into it - and we get
In both expressions obtained, the sums in parentheses are the sums of the products of the extreme ordinates of both diagrams with the quadruple product of the middle ordinates.
There are cases when none of the multiplied diagrams is straight, but one of them (or both) is limited by broken straight lines. In these cases, to multiply diagrams, they are first divided into sections within each of which at least one diagram is straight. So, for example, when multiplying the diagrams shown in Fig. 20.11, a, b, you can divide them into two sections and present the result of multiplication as a sum. You can, by multiplying these same diagrams, divide them into three sections, as shown in Fig. 20.11, c, d; in this case, the result of multiplying the diagrams is equal to
When using Vereshchagin's rule, one has to calculate the areas of different geometric shapes and determine the positions of their centers of gravity. In this regard, in Table. Figure 1.11 shows the area values and coordinates of the centers of gravity of the most common geometric figures.
As an example, consider the use of Vereshchagin’s method to determine the deflection of point C (under force ) of the beam shown in Fig. 16.11, a; At the same time, we take into account the action of bending moments and transverse forces.
The single state of the beam, as well as diagrams of the internal forces in it caused by the load and the unit force are shown in Fig. 16.11, b, b, d, e, f.
According to formula (24.11), using Vereshchagin’s method when multiplying diagrams, we find
This result coincides with the result obtained by integration.
Let us now determine the horizontal displacement of point C of the frame shown in Fig. 21.11, a. Moments of inertia cross sections frame racks and crossbars are shown in the figure; .
The actual state of the frame is shown in Fig. 21.11, a. The diagram of bending moments for this condition (load diagram) is shown in Fig. 21.11, b.
In a single state, a force equal to one is applied to point C of the frame in the direction of the desired displacement (i.e., horizontal).
Table 1.11
(see scan)
The diagram of bending moments M for this state (unit diagram) is shown in Fig. 21.11, at.
The signs of bending moments on the diagrams may not be indicated, since it is known that the ordinates of the diagrams are plotted on the side of the compressed fibers of each element.
By multiplying the load diagram with the unit diagram according to Vereshchagin’s method (Fig. 21.11, b, c) and taking into account the different values of the moments of inertia of the cross sections of the racks and the frame crossbar, we find the required displacement of point C:
The minus sign when multiplying diagrams is taken because diagrams and M are located on different sides of the frame elements, and, therefore, bending moments and M have different signs.
The negative value of the resulting displacement of point C means that this point does not shift in the direction of the unit force (Fig. 21.11, c), but in the opposite direction, i.e. to the right.
Let us now give some practical instructions on the application of the Mohr integral to various cases of calculating displacements.
It is advisable to determine displacements in beams whose section stiffness is constant along the entire length or within individual sections by calculating the Mohr integral using Vereshchagin’s rule. The same applies to frames made of straight rods of constant or step-variable stiffness.
When the stiffness of the sections of a structural element changes continuously along its length, the displacements must be determined by direct (analytical) calculation of the Mohr integral. Such a structure can be calculated approximately by replacing it with a system with elements of step-variable stiffness, after which Vereshchagin’s method can be used to determine the displacements.
Vereshchagin's method can be used not only in determining displacements, but also in determining potential energy.
Determination of movements. O. Mohr's method in combination with Simpson's method (formula)
To determine any movement (linear or angular) in Mohr's method beam is being considered in two states: real and auxiliary. Auxiliary state it turns out as follows: first, the entire specified load must be removed, then a “unit force factor” must be applied in the place where the displacement is required to be determined, and in the direction of this desired displacement. Moreover, when we determine linear movement (beam deflection), then as a “single force factor” is taken concentrated force, and if you need to find rotation angle, then you should attach concentrated couple.
Next, in the same arbitrary section of both states (that is, both real and auxiliary), analytical expressions for the bending moment are compiled, which are substituted into the formula called "Mohr integral":
where: sign Σ applies to all areas beams,
A EI – bending rigidity on the site.
In many cases Mohr integration can be avoided And apply the method "multiplying" diagrams. One such way is Simpson's way over which the value of the Mohr integral over a section of length ℓ calculated using the following formula:
Here it is indicated: a, b And With – respectively, the extreme and average ordinates of the diagram of bending moments of the actual state M,
– extreme and middle ordinates of the diagram of bending moments, but only an auxiliary state.
Sign rule: if both “multiplied” ordinates in two diagrams are located on one side of the diagram axis (that is, they are of the same sign), then before their product we must put a sign "plus: what if they on opposite sides from the axis of the diagram, then in front of the product we put a sign "minus".
It should be borne in mind that methods of “multiplying” diagrams (in addition to the Simpson method are also known Vereshchagin's method) only applicable if available two conditions:
- The bending rigidity of the beam in the area under consideration must be constant (EI= Const),
- One of the two diagrams of moments in this section should be necessarily linear. In this case, both diagrams should not have fracture
If there are several areas on a beam satisfying the specified two conditions, the formula for determining displacements takes the form:
If the result calculations turns out positive, then, therefore, the direction of the desired movement coincides with the direction of the “unit force factor”(), and if the result is negative, then the desired movement occurs in the direction opposite to this factor.
Simpson's formula, written in moments, looks like this: the displacements (deflection or rotation angle) are equal
Where li – section length;
EIi – beam stiffness on the site;
M F – values of bending moments from the load diagram, respectively, the site;
– values of bending moments from a single diagram, respectively at the beginning, in the middle and at the end plot.
When multiplying diagrams it will be useful to determine ordinate diagrams of bending moments:
, Where 
Task
Determine the angle of rotation of the section on the left support φ A
1) Find actual state support reactions .
2) We build diagram of the moments of the actual stateM.
3) Selecting an auxiliary state to determine the angle of rotation φ A.
4) Finding the support reactions of the auxiliary state
We “react” to the minus sign.
5) We construct a diagram of the moments of the auxiliary state:
6) “Multiplying” the diagrams
Since one of them (namely) is linear over the entire span and does not have a fracture, and the diagram M also without a fracture, then in the Simpson formula there will be only one section, and then
The plus sign indicates that the section A turns towards the “single moment”
prosopromat.ru
Simpson's formula for determining displacements
To determine displacement using Simpson's formula, you need to:
- Build load diagram moments (diagram of moments from the action of all external loads).
- Build single diagram moments. To do this, in the section where it is necessary to determine the linear displacement (deflection), apply a unit force, and to determine the angular displacement, apply a unit moment, and from this unit factor, construct a diagram of bending moments.
- Multiply the diagrams (load and unit) using a formula called Simpson’s formula:
Where l i– section length;
EI i– stiffness of the beam in the area;
cargo diagrams, respectively
– values of bending moments with single diagrams, respectively
If the ordinates of the diagrams are located on one side of the beam axis, then when multiplying, the “+” sign is taken into account; if on different sides, then the “-” sign is taken into account.
prosopromat.ru
2.8 Basic options for multiplying diagrams
It is obvious that the variety of applied
loads and geometric patterns
designs leads to different, with
geometry point of view, multiplyable
diagrams To implement Vereshchagin's rule
need to know the areas of geometric
figures and coordinates of their centers of gravity.
Figure 29 shows some basic
options arising in practical
calculations.
To multiply diagrams complex shape
they need to be broken down into simpler ones.
For example, to multiply two diagrams,
having the shape of a trapezoid, you need one of them
split into a triangle and a rectangle,
multiply the area of each of them by
ordinate of the second diagram located
under the appropriate center of gravity,
and add up the results. Likewise
are also used to multiply a curvilinear
trapezoids onto any linear diagram.
If you do the above steps
in general form, we obtain for such
complex cases formulas convenient for
use in practical calculations
(Fig. 30). So, the result of multiplication
two trapezoids (Fig. 30, a):
Rice. 29
Using formula (2.21), you can multiply and
diagrams that look like “twisted”
trapezium (Fig. 30, b), but at the same time the product
ordinates located on different sides
from the axes of the diagrams, taken into account with the sign
minus.
If one of the plots being multiplied is outlined
along a square parabola (which corresponds to
uniformly distributed load
load), then to multiply with
second (necessarily linear) diagram
it is considered as a sum (Fig. 30, c) or
difference (Fig. 30, d) trapezoidal and
parabolic diagrams. Result
multiplication in both cases is determined
formula:
but the value of f is determined
in different ways (Fig. 30, c, d).
Rice. 30
There may be cases where none of the
diagrams are not multiplyable
straightforward, but at least one of them
limited by broken straight lines.
To multiply such diagrams by them
pre-divided into sections,
within each of which at least
At least one diagram is straight.
Consider using the rule
Vereshchagin using specific examples.
Example 15. Determine the deflection in
mid-span and left turn angle
supporting section of a beam loaded
evenly distributed load
(Fig. 31, a), using Vereshchagin’s method.
Sequence of calculation method
Vereshchagina - the same as in the method
Mora, so let's consider three states
beams: cargo – in action
distributed load q; to him
corresponds to the diagram M q (Fig. 31, b),
and two single states - during action
strength 
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
previously by Mohr's method (see example 13). Should
pay attention to the fact that
multiplication of diagrams was performed for
half the beam, and then, due to symmetry,
the result doubled. If the area
of the entire diagram M q multiplied by
located under its center of gravity
ordinate of the diagram 
(
on
Fig. 31, c), then the amount of movement will be
completely different and incorrect because
diagram 
limited by a broken line. On
the inadmissibility of such an approach is already
stated 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 that located under its center
gravity ordinate diagram 
(
,
Fig. 31, d), since the diagram 
limited by a straight line:
This result also coincides with
the result obtained by the previous method
Mora (see example 13).
Rice. 31
Example 16. Define horizontal
and vertical movement of point A in
frame (Fig. 32, a).
As in the previous example, to solve
three problems need to be considered
frame states: cargo and two single.
Diagram of moments M F corresponding
first state, presented on
Fig. 32, b. To calculate horizontal
movements are applied at point A along
direction of the desired movement (i.e.
horizontal) force 
,
and to calculate the vertical
moving 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 the AB section there is a trapezoid (diagram M F)
divided into a triangle and a rectangle,
after which the triangle from the diagram 
"multiplied"
for each of these figures. At the aircraft site
curved trapezoid divided into
curvilinear triangle and rectangle,
and for multiplying diagrams in the SD section
formula (2.21) was used.
The “–“ sign obtained during the calculation 
,
means that point A moves along
horizontal not to the left (in this direction
force applied 
),
and to the right.
Here the sign “-” means that the point
And it moves down, not up.
Note that single diagrams of moments,
built from strength 
,
have the dimension of length, and unit
diagrams of moments built from the moment 
,
are dimensionless.
Example 17. Define vertical
moving point A flat-spatial
systems (Fig. 33, a).
Fig.23
As is known (see Chapter 1), in transverse
sections of plane-spatial rods
systems arise three internal
force factors: shear force Q y,
bending moment M x and torque
moment M cr. Since the influence
shear force per displacement
insignificantly (see example 14,
Fig.27), then when calculating the displacement
Mohr and Vereshchagin's method of six
only two terms remain.
To solve the problem, we will construct diagrams
bending moments M x,q and torque
moments M cr,q from external load
(Fig. 33, b), and then at point A we apply force 
in the direction of the desired movement,
those. vertical (Fig. 33, c), and build
single diagrams of bending moments 
and torques 
(Fig. 33, d).
Arrows on torque diagrams
twist directions shown
relevant areas
flat-spatial system.
Vertical movement of point A:
When multiplying torque diagrams
the work is taken with a “+” sign,
if the arrows indicating the direction
torsion, codirectional, and with the sign "
– ” – otherwise.
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Multiplying diagrams using the Vereshchagin method
To calculate, you need to perform the following operations:
1. Construct diagrams of bending moments Mr And Mk depending on the given and unit loads of the beam, respectively. With complex loading of the beam (Fig. 19, A) follows: either diagram Mr break down into the simplest parts for which the area and the position of the center of gravity are known (Fig. 19, b), or (preferably) construct a diagram Mr in stratified form (Fig. 19, c).
If the beam has a stepwise variable section, the diagram Mr must, in addition, be divided into sections within which the section rigidity is constant.
2. At each section, multiply the area ω of one of the diagrams (for example, diagram Mr) per ordinate Ms another diagram (for example, diagram Mk) under the center of gravity of the first diagram and divide the resulting product by the step coefficient j.
In this case, the ordinate Ms should be taken on a diagram, which in the area under consideration changes according to a linear law (without a break). If the diagram is broken, it should be divided into sections within which it will be linear.
3. Calculate the sum of the terms specified in paragraph 2.
Formula for determining movement using the method under consideration
where the summation is carried out over all sections of the beam
The areas and coordinates of the centers of gravity of some diagrams are given in Table. 11. The results of multiplying frequently occurring load and unit diagrams are given in table. 12.
Example. Determine the angle of rotation values IN stepped beam (see Fig. 19, a).
Having determined the support reactions A and B , let's build a diagram Mr in fig. 19, b And V non-stratified and stratified diagrams are shown Mr. Applying a unit moment to point B of the beam freed from load, we construct a unit diagram M1(Fig. 19. g).
Using the layered diagram Mp, according to formula 36 and table. 12 we determine the desired angle of rotation of section B:
Fig. 20
Example. Determine the deflection at point K of a beam of constant cross-section (Fig. 20, a).
By applying a unit force to point K, freed from the given load of the beam, we will construct a unit diagram of bending moments Mk (Fig. 20, b).
Having determined the support reactions from a given load
Let's cut off the console and replace it with force qa and moment (Fig. 20, c).
Let's construct a stratified diagram M (for each type of load separately), approaching the break point of a single diagram Mk on both sides (Fig. 20, i ).
According to formula (36) using table. 12 determine the required displacement
Order a solution Payment method
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Determination of displacements in a beam using Simpson's formula
For the beam, determine the linear and angular displacements at points A, B, C, having previously selected the I-beam section from the strength conditions.
Given:a=2 m,b=4 m, s=3 m,F=20 kN, M=18 kNm,q=6 kN/m, σadm=160 MPa, E=210 5 MPa
1) Draw a diagram of the beam and determine the support reactions. In a hard seal it occurs 3 reactions - vertical and horizontal, as well as supporting moment. Since there are no horizontal loads, the corresponding reaction is zero. In order to find the reactions at point E, we compose equilibrium equations.
∑F y = 0 q7-F+R E =0
R E =-q7+F=-67+20=-22kN(the sign indicates that
We'll find support moment in rigid embedment, for which we solve the equation of moments relative to any selected point.
∑M C: -M E -R E 9-F6-q77/2-M=0
M E =-18-229+649/2=-18-198+147=-69kNm(the sign indicates that the reaction is directed towards reverse side, we show this in the diagram)
2) We build a load diagram M F - a diagram of moments from a given load.
To construct diagrams of moments, we find moments at characteristic points. IN point B we determine the moments from both right and left forces, since a moment is applied at this point.
To construct a moment diagram on the line of action of a distributed load (sections AB and BC) we need additional points to plot a curve. Let's define the moments in the middle these areas. These are the moments in the middle of sections AB and BC 15.34 kNm and 23.25 kNm. We are building cargo diagram.
3) To determine linear and angular displacements at a point, it is necessary to apply at this point, in the first case, unit force (F=1) and build a diagram of the moments, in the second case, single moment (M=1) and construct a moment diagram. We build diagrams of unit loads for each point - A, B and C.
4) To find displacements we use Simpson's formula.
Where l i – length of the section;
EI i– beam stiffness in the area;
M F– values of bending moments from the load diagram, respectively at the beginning, in the middle and at the end of the section;
– values of bending moments from a single diagram, respectively at the beginning, in the middle and at the end of the section.
If the ordinates of the diagrams are located on one side of the beam axis, then the “+” sign is taken into account when multiplying; if they are on different sides, then the “-” sign is taken into account.
If the result is with a “-” sign, then the desired displacement in direction does not coincide with the direction of the corresponding unit force factor.
Let's consider application of Simpson's formula using the example of determining displacements at point A.
Let's define deflection, multiplying the load diagram by the unit force diagram.
The deflection turned out with a "-" sign means the desired displacement the direction does not coincide with the direction of the unit force (directed upward).
Let's define rotation angle, multiplying the load diagram by the diagram from a single moment.
The rotation angle turned out to be with a "-" sign This means that the desired displacement in direction does not coincide with the direction of the corresponding unit moment (directed counterclockwise).
5) To determine specific displacement values, it is necessary to select a section. Let's select the cross-section of the I-beam
Where Mmax- This maximum moment on the load moment diagram
We select by assortment I-beam No. 30 with W x = 472 cm 3 and I x = 7080 cm 4
6) Determine the displacements at the points revealing section stiffness: E – modulus of longitudinal elasticity of the material or Young’s modulus (2 10 5 MPa),J x – axial moment of inertia of the section
Deflection at point A (up)
Rotation angle (counterclockwise)
If you need to build curved beam axis, then the beam is drawn without load, and deflections in the corresponding directions are laid at the points - a smooth curve is constructed - the curved axis of the beam.
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Multiplying diagrams according to the rule, method or Mohr-Vereshchagin method
Hello! In this article we will learn to determine the movements of cross sections during bending: deflections and angles of rotation, according to Vereshchagin’s method (method, rule). Moreover, this rule is widely used not only in determining displacements, but also in revealing the static indetermination of systems using the method of forces. I will tell you about the essence of this method, how diagrams of varying complexity are multiplied and when it is beneficial to use this method.
What do you need to know to successfully master the materials in this lesson?
It is imperative to know how the diagram of bending moments is constructed, because In this article we will work with this diagram.
Vereshchagin and his method, rule or method
A.K. Vereshchagin in 1925 proposed a simpler method for solving (formulas) of the Mohr integral. He proposed, instead of integrating two functions, to multiply diagrams: multiply the area of one diagram by the ordinate of the second diagram under the center of gravity of the first. This method can be used when one of the diagrams is straight, the second can be any. In addition, the ordinate is taken from a straight-line diagram. When both diagrams are rectilinear, then it does not matter at all whose area to take and whose ordinate. Thus, diagrams according to Vereshchagin are multiplied according to the following formula:
\(( V=( M )_( F ) )\cdot \overline ( M ) =( \omega )_( C )\cdot ( \overline ( M ) )_( C ) \)
The multiplication of diagrams according to Vereshchagin is illustrated: C is the center of gravity of the first diagram, ωс is the area of the first diagram, Mc is the ordinate of the second diagram under the center of gravity of the first.
Area and center of gravity of diagrams
When using the Vereshchagin method, the entire area of the diagram is not taken at once, but in parts, within sections. The diagram of bending moments is stratified into simple figures.
Any diagram can be divided into only three figures: a rectangle, right triangle and parabolic segment.
Multiplication of diagrams according to Vereshchagin
In this block of the article I will show special cases of multiplying diagrams according to Vereshchagin.
Rectangle to rectangle
\(( V=( M )_( F ) )\cdot \overline ( M ) =( b\cdot h\cdot c ) \)
Rectangle to Triangle
\(( V=( M )_( F ) )\cdot \overline ( M ) =( b\cdot h\cdot \frac ( 1 )( 2 ) \cdot c ) \)
Triangle to rectangle
\(( V=( M )_( F ) )\cdot \overline ( M ) =( \frac ( 1 )( 2 ) \cdot b\cdot h\cdot c ) \)
Segment to rectangle
\(( V=( M )_( F ) )\cdot \overline ( M ) =( \frac ( q\cdot ( l )^( 3 ) )( 12 ) \cdot c ) \)
Segment per triangle
\(( V=( M )_( F ) )\cdot \overline ( M ) =( \frac ( q\cdot ( l )^( 3 ) )( 12 ) \cdot \frac ( 1 )( 2 ) \cdot c ) \)
Special cases of stratification of diagrams into simple figures
In this block of the article I will show special cases of stratification of diagrams into simple figures, in order to be able to multiply them according to Vereshchagin.
Rectangle and triangle
Two triangles
Two triangles and a segment
Triangle, rectangle and segment
An example of determining displacements: deflections and rotation angles according to Vereshchagin
Now I propose to consider a specific example with the calculation of the movements of cross sections: their deflections and angles of rotation. Let's take a steel beam that is loaded with all kinds of loads and determine the deflection of section C, as well as the angle of rotation of section A.
Constructing a diagram of bending moments
First of all, we calculate and construct a diagram of bending moments:
Construction of single moment diagrams
Now, for each desired displacement, it is necessary to apply a unit load (a dimensionless value equal to one) and construct unit diagrams:
- For deflections, unit forces are applied.
- For turning angles, single moments are applied.
Moreover, the direction of these loads is not important! The calculation will show the correct direction of movement.
For example, after calculation, the deflection value turned out to be positive, which means that the direction of movement of the section coincides with the direction of the previously applied force. The same goes for turning angles.
Multiplication of diagram sections according to Vereshchagin
After all preparatory work: constructing a diagram of bending moments, dividing it into elementary figures and constructing single diagrams from loads applied in the places and direction of the desired displacements, you can proceed directly to multiplying the corresponding diagrams.
As already written above, linear diagrams You can multiply in any order, that is, take the area of any diagram: main or unit, and multiply by the ordinate of another. But usually, in order not to get confused in the calculations, the areas are taken main diagram of bending moments, in this lesson we will adhere to the same rule.
Determination of section deflection C
We multiply the corresponding diagrams from left to right and calculate the deflection of section C using the Mohr-Vereshchagin method:
\[ ( V )_( C )=\frac ( 1 )( E( I )_( x ) ) (\frac ( 1 )( 2 ) \cdot 6\cdot 3\cdot \frac ( 2 )( 3 ) \cdot 2+\frac ( 1 )( 2 ) \cdot 6\cdot 2\cdot \frac ( 2 )( 3 ) \cdot 2)=\frac ( 20kN( m )^( 3 ) )( E( I ) _( x ) ) \]
Let's imagine that the beam being calculated has a cross section in the form of an I-beam No. 24 according to GOST 8239-89, then the deflection of the beam will be equal to:
\[ ( V )_( C )=\frac ( 20kN( m )^( 3 ) )( E( I )_( x ) ) =\frac ( 20\cdot ( 10 )^( 9 )N\cdot ( cm )^( 3 ) )( 2\cdot ( 10 )^( 7 )\frac ( H )( ( cm )^( 2 ) ) \cdot 3460( cm )^( 4 ) ) =0.289 cm \]
Determining the angle of rotation of section C
We multiply the corresponding diagrams from left to right and calculate the angle of rotation of the section C using the Mohr-Vereshchagin rule:
\[ ( \theta )_( C )=\frac ( 1 )( E( I )_( x ) ) (-\frac ( 1 )( 2 ) \cdot 6\cdot 3\cdot \frac ( 1 )( 3 ) \cdot 1)=-\frac ( 3kN( m )^( 2 ) )( E( I )_( x ) ) \]
\[ ( ( \theta ) )_( C )=-\frac ( 3kN( m )^( 2 ) )( E( I )_( x ) ) =-\frac ( 3\cdot ( 10 )^( 7 )Н\cdot ( cm )^( 3 ) )( 2\cdot ( 10 )^( 7 )\frac ( Н )( ( cm )^( 2 ) ) \cdot 3460( cm )^( 4 ) ) =- 0.0004rad\]
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Trapezoid and Simpson formulas
Let's take advantage
Vereshchagin's rule for multiplication
two straight-line diagrams having the form
trapezoid. Let us divide both trapezoids into
triangles whose areas and
the positions of the centers of gravity are easy
are determined.
Diagram
M F
ω 1
C 1 C 2
ω 2
Diagram
We
received formula
trapezoid,
according to
which the products of the corresponding
left and right ordinates of diagrams are necessary
double, and the products of the cross
take the ordinates as single ones, and the resulting
multiply the amount by one sixth of the length
diagram.
Let's consider
the case when the load diagram is presented
a square parabola, and the unit diagram
– trapezoid.
ω P.S.
Along with
with the extreme ordinates we also indicate the average ones.
Let's break it down
curved diagram on a trapezoid and
parabolic segment.
We will produce
multiplication of corresponding figures.
Expression
I T
we have. We'll find
.
Square
parabolic segment:
Ordinate
single plot under the center of gravity
parabolic segment:
After
substitution we get formula
Simpson:
Work
two diagrams is equal to the sum of the products
extreme ordinates and quadruple
product of average ordinates multiplied
by one sixth of the length of the diagrams.
§7. Force calculation of statically indeterminate rod systems (SNS).
Statically
indeterminate systems (INS) have
advantages and disadvantages compared
with statically definable systems
(SOS).
Advantages:
SNA
have greater survivability
operation under load than SOS. IN
SOS almost all elements
are equally tense, and therefore they have
strength reserves only within the limits
safety factor k
=1,5
– 2. If at least one element goes
to the limit state, the entire structure
will receive unacceptable from the point of view
standards for calculating deformation or collapse.
SNS is an unequally stressed structure
and during the transition of the most intense
element to its limit state,
there is a redistribution of efforts
from the increased load on less stressful
elements.
SNS,
due to the presence of unnecessary connections and redundant
rigidity of individual elements, less
more deformative than SOS, i.e. they have less
linear angular movements.
Flaws:
SNA
are more complex to calculate than SOS, which
explained by the presence of excess
(extra) connections. Calculation complexity
SNA is proportional to the third power
number of extra connections, i.e.
.
For example, if for two systems n 1
=1,
n 2
=4
,
That
t 1
=
α
,
t 2
=64α,
those. calculation time increases by 64 times.
IN
SNA distribution of forces in elements
depends on their geometric dimensions,
the definition of which, in turn,
is the main task of resistance
materials. Thus, there arises
necessity of a priori assignment
bending stiffnesses and transverse
sections of individual rods: (EY)
k =α
k (EY),
which leads to ambiguity
constructive solutions.
More
successful assignment of stiffnesses, depending
from understanding the essence of resistance tasks
materials will lead to the creation of more
optimal designs.
IN
SNS may appear difficult
predictable in size
stress-strain state,
caused by temperature changes
and independent settlement of supports. Change
the temperature of one of the elements causes
appearance of temperature stresses
in all SNA rods. Same as inaccuracy
manufacturing one of the rods or
the displacement of one bond causes the appearance
installation stresses in all rods.
In SOS such tensions do not arise.
Let's consider
basic methods for calculating SNA when
static impact of loads.
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.
Rice. 28
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)
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 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).
Rice. 30
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. 31a), 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).
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 states. 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.
