Wash away that jealousy of the spacious system of forces. Theoretical mechanics. Statics. Food for self-verification

As a result, the system of forces is changing from equal, її head vector and head moment to zero:

Number of vector equalities is produced up to the next six scalar equalities:

yakі are called the minds of the jealous expanse enough system forces.

The first three minds show the equalness of the head vector to zero, the next three - the zero equalness of the head moment of the system of forces.

In the minds of a jealous person, everything is to be blamed fiery forces- as active (set), and reactions of calls. Rest far behind the unknown, and the minds of the jealous become the equals of the appointment of the unknown - the equals of the jealous.

If the maximum number of equals is more than six, then the task of equalizing the body under a sufficiently large system of forces can be assigned to six unknown reactions. For a large number of unknown leaders, they become statically insignificant.

And one more respect. If the head vector and head moment if the center O reaches zero, then the stench reaches zero unless there is some other center. Tse directly exclaims the material about changing the center of the attraction (to bring it independently). Otzhe, as you know the equal bodies are fixed in one coordinate system, the stench is fixed in any other non-robust coordinate system. Otherwise, it seems that the choice of coordinate axes pіd hour ordering rіvnyan іvnovagi as a whole is more than enough.

Rectangular slab (Fig. 51 a) is flattened in a horizontal position by a spherical hinge O, a bearing A and a cable BE, and the points are on the same vertical. At point D to the slab, a force was applied perpendicular to the side OD and heeled up to the slab area under the cut of 45 °. Calculate the tension of the cable and the reaction of the supports at the points of Vin A, as well as i.

To complete the task, we look at the equal plate. Before the active forces P, G, we add the reaction of the ties - the warehouse reaction of the spherical hinge, the reaction of the bearing, the reaction of the cable. The coordinate axes Oxyz are entered at once (Fig. 51 b). It is obvious that the collection of forces has been taken away, establishing a fairly spacious system, which forces do not have.

For the signing of the uninhabited, we become jealous of the jealousy.

We start from equal projections of forces on the whole:

It is understandable that the projection of the calculation is calculated in two steps; the projection of the force T on the plane, distance, projecting on the axis (more on the axis, parallel), is known (div. Fig. 51,b):

In this way of underwire design, it is manually corristuvatsya, if the line of force and that all do not change. We add up:

Rivnyannya momentіv forces shdo osі maє vglyad:

Moments of forces at equal days, shards of force or force change all x (), or parallel ї th. In both vibrations, the moment of force should be equal to zero (div. p. 41).

Calculation of the moment of force is often easier, as the force is laid out in the same order of warehouses and speeded up by the Varignon theorem. At to this particular type tse manually zrobiti for strength. Laying out її on horizontal and vertical warehouses, we can write.

A fairly spacious system of forces, like a plane, can be brought to the center Pro and replace with one resultant force and a couple with a moment. Rozmirkovuyuchi so that for equal tsієї system of forces it is necessary that sufficient, R= 0 і M o \u003d 0. Ale vectors can only turn to zero if all projections on the coordinate axes return to zero, then if R x= R y= R z = 0 i M x= M y= M z \u003d 0 or, if the fiery forces satisfy the minds:

Σ X i = 0; Σ Mx(Pi) = 0;

Σ Y i = 0; Σ M y(Pi) = 0;

Σ Z i = 0; Σ Mz(Pi) = 0.

In this way, for the equalization of the space system of forces, it is necessary and sufficient that the sum of the projections of the forces of the system on the skin from the coordinate axes, as well as the sum of the moments of all the forces of the system along the skin axis, should be equal to zero.

For the acquisition of more simple systems, it is recommended that the axis be carried out in such a way that the stench permeated more than the unknown forces, or they were perpendicular to them (as it does not complicate the calculation of projections and moments of other forces).

A new element in the folded equalities is the calculation of the moments of forces along the coordinate axes.

In the case of inversions, if it is important to point out from the armchair, why the most important moment of the given force is like a axis, it is recommended to depict on the additional armchair a projection of the body, which is viewed (at once by force) on a plane perpendicular to the axis of the axis.

In quiet slopes, if, when calculating the moment, it is difficult to blame the assigned projection of the force on the vertical plane or the shoulders of the projection, it is recommended to spread the force on two mutually perpendicular warehouses (of which one is parallel to the coordinate Variation axes), rather theoremist.

Example 5. Frame AB(Fig. 45) is trimmed in a rіvnovazі by a hinge BUT and haircut ND. On the edge of the ramie there is vagage vaga R. Significantly, the reaction of the hinge and susilla in the shear.

Fig.45

We look at the jealousy of the rami at once from the vantage.

I will be a rozrahunkov’s scheme, having depicted the frame with a free body and showing all the forces that can be exerted on it: the reactions of the svyazkіv and vaga vantage R. These forces establish the system of forces, as if they were quite scattered on the flat.

Fold the bagan so evenly, so that the skin bulges by one unknown force.

Our boss has a point BUT, de dodano nevidomі ta ; mottled W, de twitch lines of unknown forces that; mottled D- The point of the cross line of the line dії forces і. We store equal projections of forces on the whole at(for all X projecting is not possible, because won perpendicular to a straight line AC).

I, first of all, fold equal, more one more respect. Although on the rozrahunkovy scheme there is a force, it is drawn so that the shoulder її is not easy to know, then when the moment is determined, it is recommended to spread the force vector forward on two, more straightened. For this task, we place the force on two: i (Fig. 37) such that their modules

We store equal:

We know from another level:

3rd third

І from the first

Bo wiyshlo S<0, то стержень ND there will be squeezing.

Example 6. Straightforward police wagon R utrimuetsya in a horizontal position with two shears РЄі CD, attach to the wall at the point E. Haircuts of the same length, AB = 2 a, EO = a. Significantly zusilla in shears and loop reactions BUTі At.

Fig.46

Let's look at the equal slab. We will be using the Rozrahunkov scheme (Fig. 46). The reactions of the loops are usually shown by two forces perpendicular to the axis of the loop: .

Forces establish a system of forces, enough spread out in space. We can put together 6 rubles. Nevidomikh - tezh sixth.

Yakі rivnyannya fold - you need to think. It’s so bad that the stinks were simpler and that they had less unknowing.

Let's put it like this:

Z equal (1) is taken: S1 = S2. Topics (4): .

З (3): Y A = Y B i, (5), . Mean Z equal (6), because S 1 \u003d S 2 slid Z A \u003d Z B. Then (2) Z A =Z B =P/4.

From trikutnika, de, slid,

To that Y A \u003d Y B \u003d 0.25 P, ZA \u003d Z B 0.25 P.

To reconcile the solution, you can add one more equal and wonder what you are satisfied with when you know the meanings of the reactions:

The task is written correctly.

Pretty spacious system of forces a system of forces is called, the lines of which do not lie in one plane.

Sounds are screaming umova rіvnovagi dovіlnoї prostorovoї system of forces.

In a geometric form: for a fairly spacious system of forces, it is necessary and sufficient that the head vector and head moment of the system equal to zero

R = 0, M o = 0 .

In an analytical form: for a fairly spacious system of forces to be equal, it is necessary and sufficient that the sum of the projections of the forces on three coordinate axes and the sum of the moments of the forces of the forces along these axes is equal to zero

ΣF kx = 0 , ΣFky = 0 , ΣF kz = 0 ,

M x (Fk) = 0 , M y (Fk) = 0 , Mz(Fk) = 0 .

The center of the vaga. Methods for assigning the center of gravity. Coordinates of the center of gravity of a flat body and folded overhangs.

Vaga Center

The center of gravity of the body is the point of reporting the force of gravity (equally gravitational forces).

The center of gravity is to divide between two people with vantages for a healthy marriage of their wt.

Appointed to the center of importance

Appointed to the center of heaviness of a prevalent body by the path of the last folding of forces, which is to be born on the edge of the third part - the head of the plant; won't be easier than tіl porіvnjano simple form.

Let the body add up with two weights m 1 and m 2, joined by a shear (Fig. 126). If the weight of the shear is small, if it is matched with the masses m 1 and m 2, then it can be used to fight. On the skin s mass diє gravitational force:

P 1 = m 1 g, P 2 = m 2 g;

offensive stench straightened vertically to the bottom, that is parallel to one to one. As we already know, equal to two parallel forces are applied at the point O, as it stands for the mind

Otzhe, the center of gravity to divide between the two masses of the masses of the salutary mass. Yakshcho tіlo podvіsity at point O, it will be lost at rіvnovazi.

Assignment of coordinates to the center of importance

Methods for assigning coordinates to the center of gravity Vykhodyachi z otrimanih earlier zagalnyh formulas, you can show ways to assign the coordinates of the centers in the importance of solid bodies: 1. Analytical (path integration). 2 Method of symmetry. As if the body is flat, the whole is the center of symmetry, the center of gravity lies equally near the plane of symmetry, the axis of symmetry is the center of symmetry. 3 Experimental (method of lifting the body). 4 Rozbity. The body breaks down on the last number of parts, the skin is the center of gravity C that area S vіdomі. For example, the projection of the body on the plane xOy(Figure 1.8) you can pay at the sight of two flat figures with areas S1і S2 (S = S 1 + S 2). Centers of gravity of these figures are changed at the points C 1 (x 1 , y 1)і C 2 (x 2 , y 2). Todi coordinate the center of your body Figure 1.8 Okremy vipadok to the way of breaking. Vіn zastosovuєtsya to tіl, scho mayut virіzi, like the center of the vag of the body without virіzu and virіzanoї parts of the house. For example, it is necessary to know the coordinates of the center of gravity of a flat figure (Figure 1.9): Baby 1.9

Center of gravity of homogeneous flat bodies

(flat figures)

It is even more often necessary to designate the center of gravity of various flat bodies and geometric flat figures of a folding form. For flat bodies it is possible to write down: V = Ah, de A – figure area, h – height.

The same after the substitution in the record, another formula is taken:

; ; ,

de Ak - the area of ​​the part of the cut; hk, yk - coordinates of the CG of the parts of the cut.

Viraz is called the static moment of the area (Sy.).

The coordinates of the center of gravity of the cut can be seen through a static moment:

Axes that pass through the center of gravity are called central axes. The static moment about the central axis is equal to zero.

Assignment of coordinates to the center of gravity of flat figures

Note. The center of gravity of a symmetrical figure is located on the axis of symmetry.

The center of gravity of the shear is in the middle of the height. The position of the center of heaviness of simple geometric figures can be secured according to the given formulas (Fig. 8.3: a) - colo; b) - square, rectangle; c) - trikutnik; d) - pivkolo).

As it was stated in § 4.4, it is possible to write down three equal projections (4.16) and three moments (4.17):

, , . (7.14)

If the body is completely fixed, then the forces that are on the new one, are in balance and equal (7.13) and (7.14) serve to designate supporting reactions. Obviously, there can be fluctuations, if these equals are not enough for the purpose of supporting reactions; such a statically non-visible system is not perceptible.

For an expanse system of parallel forces, the equalization of the parallel forces looks like (§ 4.4[‡]):

, , . (7.15)

We can now look at the fluctuations, if the body is fixed too often, then. ties put on the body do not guarantee equal body. You can say chotiri okremi vipadki.

1. Solid body maє one unbreakable point. Otherwise, apparently, it is attached to an indestructible point behind the help of a perfect spherical hinge.

Let's place the cob of a non-destructive coordinate system near the qiu point. Diya zvyazku to the point BUT replace the reaction; shards are not at home behind the module and behind the direct one, then we її can be represented by looking at three independent warehouses , , ,

Rivnyannia rіvnovagi (7.13) and (7.14) at the same time write down at the sight:

1) ,

2) ,

3) ,

4) ,

5) ,

Remain three equal times not to avenge warehouse reactions, so that the line of force could pass through the point BUT. From the same time, the central equalization is established by the fallows between the active forces necessary for the equalization of the body, moreover, the three first equalities can be victorious in the designation of the warehouse reactions.

in such a manner, mental equalness of a solid body, which has one non-violent point, equalness to zero of the skin is the sum of the algebra of moments of all active forces of the system, although there are three axes .

2. Tіlo maє two unbroken chickens. Tse, for example, mother of the place, as if it were attached to two unruly points behind the auxiliary hinges.

We select the cob of coordinates at the point BUT and direct all uzdovzh lines to pass through the points BUTі At. Replacing the links with reactions, directing warehouse reactions with coordinate axes. Significantly stand between points BUTі At through a; then equal equal equal (7.13) and (7.14) will be written to the attacker:

1) ,

2) ,

3) ,

4) ,

5) ,

Remain equal not to avenge the forces of reaction and establish a link between the active forces, necessary for the equalization of the body. Otzhe, mental equalness of a solid body, that there are two non-violent points, equalness to zero of the algebraic sum of the moments of all active forces applied to the body, how the axle can pass through non-violent points . First p'yat equal to serve as a sign of nevіdomih warehouse reactions , , , , , .

Respectfully, warehouses cannot be appointed. On the third day, less than the sum + i, therefore, the task of skin care is not significant for a solid body, it is statically non-significant. However, even to the point At there is not a spherical, but a cylindrical hinge (that is, a bearing), which does not cross over to the late forging of the body of the axis of the wrapping, then the factory becomes statically original.

The body can be indestructible all the wrapping, vzdovzh that it can be forged without rubbing. Tse means what in points BUTі At Cylindrical hinges (bearings) are known, moreover, the warehouse reaction of the axis wrapping is equal to zero. Otzhe, the jealousy of the jealousy will wake up looking:

1) ,

2) ,

4) ,

5) ,

Two equals (7.18), zokrema, third and sixth, superimpose an exchange system of active forces, other equals serve as a designated reaction.

The body spirals at three points onto a smooth surface, and the support points do not lie on one straight line. Significantly qi points through BUT, Atі W that sumisny with a flat ABC coordinate plane ahu. Having replaced the connection with vertical reactions, we can write down the equation (7.14) in this way:

3) ,

4) ,

5) ,

The third - p'yate equal can be signified by unknown reactions, and first, friend and superior equal, think about what the active forces of that necessary equal body will show. Obviously, for the equal of the body, it is necessary to win the minds, , , the shards at the points of support can be blamed more than the reaction of the accepted directive.

If the body spirals onto a horizontal plane more and more at three points, then the target becomes statically invisible, so that with every reaction there will be styles, more points, and only three.

Task 7.3. To know the head vector and the head moment of the system of forces shown in fig. Forces are applied to the tops of the cube and straightening the ribs, moreover , . Dovzhina ribs of the cube are old a.

The projections of the head vector are known from formulas (4.4):

, , .

Yogo module is old. Direct cosines will be

, ;

, ;

, .

The head image vector in fig.

,

and the module of the head moment according to the formula (4.8)

Now the direct cosine of the head moment is significant:

, ;

, .

The head moment is shown in Fig. Kut mizh vectors i are calculated according to the formula (4.11) that

We know the inter-regions from the minds:

,

.

Zvіdsi know

,

.

On fig. required area, prompted at , shaded. With all the surface of the plate, you will be safe.

Establish three types of equal equal equal to the flat system of forces. The first, main view blazes without intermediary minds of jealousy:

;

and write like this:

;
;
.

Two others see the jealousy of the jealousy can also be taken from the minds of the jealousy:

;
;
,

de straight AB not perpendicular to axis x;

;
;
.

Krapki A, B і C do not lie on one straight line.

On the view of a flat system of forces with the minds of a fairly spacious system of forces є two vector equalities:

.

If you want to project on a right-angle coordinate system, then you need to take into account the equalness of the space system of forces:

Task 1. Determination of the reaction of the supports of the stacked structure (Double heating system)

The design is made up of two laman haircuts. ABCі CDE, from the point C with a non-destructive cylindrical hinge and attach to a non-destructive surface. xOy or for the help of unruly cylindrical hinges (НШ ), or a ruhomy cylindrical hinge (ПШ) and a hard mortgage (ЖЗ). The area of ​​the stiffened cylindrical hinge becomes a cut   from the sky Ox. Point coordinates A,B,C,D і E, as well as the method of fixing the structure is induced in Table. 1. The design is vanquished equally by different variations of intensity q, perpendicular to the distance of її zastosuvannya, a pair of forces with a moment M that two are energized by forces і . Rіvnomіrno razpodіlene navantazhennya applied in such a way that yоgo іvnоdіyuche pragna turn the structure around the point O against the course of the Godinnikov arrow. Dіlyanki programs qі M, as well as program points і , their direct-direction modules are shown in Table. 2. Units of values ​​that are set: q- kilonewton per meter (kN/m); M- kilonewton meter (kNm); і – kilonewton (kN); Kuti,i following the positive direction of the axis Ox against the course of the year's arrow, as if the stench is positive, and under the hour of the year's arrow - as if negative.

Vyznachte reactions of external and internal links of the construction.

Vkazіvki to vikonannya zavdannya

On the coordinate plane xOy it is necessary to induce points A,B, C,D,E; draw lamani shear ABC,CDE; show ways of fixing tsikh tіl mіzh with oneself and with an indestructible area xOy. Potim, taking the data from the table. 2, take advantage of the structure of the two with moderate forces і , evenly rozpodіlenim vantazhennyam intensity q and a pair of forces with an algebraic moment M. So, as the manager had to keep a steady warehouse body, they gave the need to encourage one more little one, having depicted a new body of the body ABCі CDE. Zovnishni (points A,E) that inner (dot W) the link on both babies should be replaced by a different reaction, and evenly a difference in the preference - by a equal one
(l- Dovzhina of the plot of the addition of the advancement), straightened at the bik of the advancement and applied to the middle of the plot. The oskіlki konstruktsіya, scho razglyadєєєєєєєє z dvoh tіl, then for znahodzhennya reactions zv'yazkіv need to fold six rivnyan rivnovagi. There are three options for the completion of this task:

a) fold three equals for the folded body and three for the body ABC;

b) fold three equals for the folded body and three for the body CDE;

c) add up three equal equals for tіl ABCі CDE.

butt

Given:A (0;0,2);At (0,3:0,2);W (0,3:0,3);D (0,7:0,4);E (0,7:0);
kN/m,
kN, β = - 45˚, that
kN, γ = - 60˚,
kNm.

Significance reactions of external and internal links of the structure.

Solution. Rozіb'єmo design (Fig. 7, a) at the point W in warehouse ABCі CDE(Mal. 7, b,in). Replace hinges Aі B similar reactions, warehouses of which are significant in fig. 7. At the point C imaginably folders
- forces of interaction between the parts of the structure, moreover .

Table 1

Job options 1

Table 2

Dani before the date 1

Incremental differentiation of intensity intensity q instead of equal , kN:

Vector satisfy the positive direct axis y kut φ, which is not easy to know for the coordinates of the point C і D (div. Fig. 7, a):

For the completion of the task, we speed up the first kind of equalization of equalities, writing them down for the left and right parts of the structure. When folding, equal moments are selected as moment points of a point A- for the left E- for the right part of the structure, to allow the difference between the two equals and the difference between the two
і .

Rivnyannya rіvnovagi for the body ABC:

We can see the strength like a sum of warehouses:
de. Todi rіvnyannya rіvnovag for the body CDE can be recorded at a glance

.

Virіshimo svіlno іvnyannya іvnіnіv, in advance pіdstavshi vіdomі znachennya in them.

Vrakhovuchi, what is behind the axiom about the equivalence of the forces of both and the opposite
, we know from the taken system, kN:

Todі z rivnyan rivnovagi, scho lost, tіl ABC і CDE it is not easy to calculate the reactions of internal and external connections, kN:

The results are presented in a table: