If the system of forces is in equilibrium, then its main vector and main moment are equal to zero:

These vector equalities lead to the following six scalar equalities:

which are called the equilibrium conditions of an arbitrary spatial system of forces.

The first three conditions express the equality to zero of the main vector, the next three - the equality to zero of the main moment of the system of forces.

Under these conditions of equilibrium, all acting forces must be taken into account - both active (given) and reactions of constraints. The latter are unknown in advance, and the equilibrium conditions become equations for determining these unknowns - equilibrium equations.

Since the maximum number of equations is six, then in the problem of body equilibrium under the action of an arbitrary spatial system of forces, six unknown reactions can be determined. With more unknowns, the problem becomes statically indeterminate.

And one more note. If the main vector and the main moment relative to some center O are equal to zero, then they will be equal to zero relative to any other center. This directly follows from the material on the change of the center of reduction (to prove it yourself). Therefore, if the equilibrium conditions of the body are satisfied in one coordinate system, then they will be fulfilled in any other fixed coordinate system. In other words, the choice of coordinate axes in compiling the equilibrium equations is completely arbitrary.

A rectangular plate (Fig. 51, a) is held 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, a force is applied to the plate perpendicular to the side OD and inclined to the plane of the plate at an angle of 45°. Determine the tension of the cable and the reactions of the supports at the points He A, if and .

To solve the problem, we consider the equilibrium of the plate. To the active forces P, G we add the reactions of the bonds - the components of the reaction of the spherical hinge, the reaction, the bearing, the reaction of the cable. At the same time, we introduce the coordinate axes Oxyz (Fig. 51, b). It can be seen that the resulting set of forces forms an arbitrary spatial system in which the forces are unknown.

To determine the unknowns, we compose equilibrium equations.

We start with the equation of projections of forces on the axis:

Let us explain the definition of the projection, the calculation is carried out in two steps - first, the projection of the force T on the plane is determined, then, projecting on the x axis (more conveniently on the axis parallel to), we find (see Fig. 51, b):

This method of double projection is convenient to use when the line of action of the force and the axis do not intersect. Next, we compose:

The equation of the moments of forces about the axis has the form:

There are no moments of forces in the equation, since these forces either cross the x () axis, or are parallel to it. In both of these cases, the moment of force about the axis is zero (see p. 41).

The calculation of the moment of a force is often facilitated by decomposing the force in a suitable way into its components and using Varignon's theorem. In this case, it is convenient to do this for the force . Decomposing it into horizontal and vertical components, we can write.

ABOUTR= 0 and M R x= R y= R z = 0 and M x= M y= M

Equilibrium conditions for an arbitrary spatial system of forces.

An arbitrary spatial system of forces, like a flat one, can be brought to some center ABOUT and replace with one resultant force and a pair with a moment. Arguing in such a way that for the equilibrium of this system of forces it is necessary and sufficient that at the same time R= 0 and M o \u003d 0. But vectors can vanish only when all their projections on the coordinate axes are equal to zero, i.e. when R x= R y= R z = 0 and M x= M y= M z = 0 or when the acting forces satisfy the conditions

Thus, for the equilibrium of an arbitrary spatial system of forces, it is necessary and sufficient that the sum of the projections of all forces on each of the three coordinate axes and the sum of their moments about these axes be equal to zero.

Principles of solving problems on the balance of a body under the action of a spatial system of forces.

The principle of solving the problems of this section remains the same as for a plane system of forces. Having established the equilibrium of which body will be considered, they replace the bonds imposed on the body with their reactions and make up the conditions for the equilibrium of this body, considering it as free. The required quantities are determined from the obtained equations.

To obtain simpler systems of equations, it is recommended to draw the axes so that they intersect more unknown forces or be perpendicular to them (unless this unnecessarily complicates the calculation of projections and moments of other forces).

A new element in the formulation of equations is the calculation of the moments of forces about the coordinate axes.

In cases where it is difficult to see from the general drawing what the moment of a given force is relative to some axis, it is recommended to depict on the auxiliary drawing the projection of the body in question (together with the force) onto a plane perpendicular to this axis.

In those cases when, when calculating the moment, there are difficulties in determining the projection of the force on the corresponding plane or the shoulders of this projection, it is recommended to decompose the force into two mutually perpendicular components (of which one is parallel to any coordinate axis), and then use the Varignon theorem.

Example 5

Frame AB(fig.45) is kept in balance by a hinge A and rod sun. At the edge of the frame is a weight R. Let us determine the hinge reactions and the force in the rod.

Fig.45

We consider the balance of the frame together with the load.

We build a calculation scheme, depicting the frame as a free body and showing all the forces acting on it: the reactions of the bonds and the weight of the load R. These forces form a system of forces arbitrarily located on the plane.

It is desirable to compose such equations so that each has one unknown force.

In our problem, this is the point A, where the unknowns and are applied; dot WITH, where the lines of action of unknown forces and intersect; dot D- the point of intersection of the lines of action of forces and. Let's make the equation of projections of forces on an axis at(per axle X it is impossible to design, because it is perpendicular to the line AC).

And, before writing equations, we make one more useful remark. If there is a force on the design scheme, located so that its shoulder is not easy, then when determining the moment, it is recommended to first decompose the vector of this force into two, more conveniently directed. In this problem, we decompose the force into two: and (Fig. 37) such that their modules

We make equations:

From the second equation we find . From the third And from the first

So how did it turn out S<0, то стержень sun will be compressed.

There are three types of equilibrium equations for a plane system of forces. The first, main form follows directly from the equilibrium conditions:

;

and written like this:

;
;
.

Two other kinds of equilibrium equations can also be derived from the equilibrium conditions:

;
;
,

where is the line AB not perpendicular to axis x;

;
;
.

Points A, B And C do not lie on the same line.

In contrast to a flat system of forces, the equilibrium conditions for an arbitrary spatial system of forces are two vector equalities:

.

If these relations are projected onto a rectangular coordinate system, then we obtain the equilibrium equations for the spatial system of forces:

Task 1. Determination of reactions of supports of a composite structure (Two-body system)

The design consists of two broken rods ABC And CDE connected at a point C fixed cylindrical hinge and attached to a fixed plane xOy or with the help of fixed cylindrical hinges (НШ ), or a movable cylindrical hinge (PSh) and a rigid seal (ZhZ). The rolling plane of the movable cylindrical hinge is angle   with axle Ox. Point coordinates A,B,C,D And E, as well as the method of fastening the structure are given in table. 1. The structure is loaded with a uniformly distributed load of intensity q, perpendicular to the site of its application, by a pair of forces with a moment M and two concentrated forces And . A uniformly distributed load is applied in such a way that its resultant tends to rotate the structure around the point O counterclockwise. Application sites q And M, as well as application points And , their modules and directions are indicated in Table. 2. Units of set values: q– kilonewton per meter (kN/m); M- kilonewton meter (kNm); And are kilonewton (kN); Angles,andshould be set aside from the positive direction of the axis Ox counterclockwise if positive, and clockwise if negative.

Determine the reactions of external and internal links of the structure.

Instructions for completing the task

On the coordinate plane xOy in accordance with the condition of the task variant (Table 1), it is necessary to construct points A,B, C,D,E; draw broken rods ABC,CDE; indicate the ways of attaching these bodies to each other and to a fixed plane xOy. Then, taking the data from the table. 2, load the structure with two concentrated forces And , uniformly distributed load intensity q and a pair of forces with an algebraic moment M. Since the task examines the balance of a composite body, then you need to build another drawing, depicting the bodies separately on it ABC And CDE. External (points A,E) and internal (point WITH) the bonds in both figures should be replaced with the corresponding reactions, and the uniformly distributed load should be replaced with the resultant
(l is the length of the load application section) directed towards the load and applied to the middle of the section. Since the structure under consideration consists of two bodies, in order to find the reactions of the bonds, it is necessary to compose six equilibrium equations. There are three options for solving this problem:

a) compose three equilibrium equations for a composite body and three for a body ABC;

b) compose three equilibrium equations for a composite body and three for a body CDE;

c) compose three equilibrium equations for bodies ABC And CDE.

Example

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

Define reactions of external and internal links of the structure.

Solution. Let's split the structure (Fig. 7, A) at the point WITH into constituent parts ABC And CDE(Fig. 7, b,V). Let's replace the hinges A And B corresponding reactions, the components of which we indicate in Fig. 7. At the point C depict the components
- forces of interaction between the parts of the structure, and .

Table 1

Job options 1

A

Mounting method designs

x A

y A

x B

y B

x C

y C

x D

y D

x E

y E

T. E

table 2

Data for task 1

	Force			Force			Moment M
		Meaning		Meaning			Meaning			Meaning

Uniformly distributed load intensity q replace the resultant , kN:

Vector forms with positive axis direction y angle φ, which is easy to find from the coordinates of the points C And D (see fig. 7, A):

To solve the problem, we use the first type of equilibrium equations, writing them separately for the left and right parts of the structure. When compiling the equations of moments, we choose as the moment points the points A- for left and E– for the right parts of the structure, which will allow us to solve these two equations together and determine the unknowns
And .

Equilibrium equations for a body ABC:

Imagine the power as the sum of the components:
, Where. Then the equilibrium equations for the body CDE can be written in the form

.

Let's jointly solve the equations of moments, after substituting the known values ​​into them.

Considering that, according to the axiom about the equality of the forces of action and reaction
, from the resulting system we find, kN:

Then from the remaining equilibrium equations of bodies ABC And CDE it is easy to determine the reactions of internal and external bonds, kN:

We present the results of the calculations in a table:

We combine the origin of coordinates with the point of intersection of the lines of action of the forces of the system. We project all the forces onto the coordinate axes and summarize the corresponding projections (Fig. 7.4). We get the projections of the resultant on the coordinate axes:

The modulus of the resultant system of converging forces is determined by the formula

The direction of the resultant vector is determined by the angles

Arbitrary spatial system of forces

Bringing an arbitrary spatial system of forces to the center O.

A spatial system of forces is given (Fig. 7.5, a). Let's bring it to the center O.

Forces must be moved in parallel, thus forming a system of pairs of forces. The moment of each of these pairs is equal to the product of the modulus of force and the distance to the center of reduction.

A beam of forces arises in the center of reduction, which can be replaced by a total force (main vector) F GL (Fig. 7.5, b).

The moments of pairs of forces can be added to obtain the total moment of the system M ch (main moment).

Thus, an arbitrary spatial system of forces is reduced to the main vector and the main moment.

It is customary to decompose the main vector into three components directed along the coordinate axes (Fig. 7.5, c).

Usually, the total moment is decomposed into components: three moments about the coordinate axes.

The absolute value of the main vector (Fig. 7.5b) is

The absolute value of the main moment is determined by the formula.

Equilibrium Equations of the Spatial System of Forces

At equilibrium F ch = 0; M hl = 0. We get six equilibrium equations:

Six equations of equilibrium of the spatial system of forces correspond to six independent possible displacements of the body in space: three displacements along the coordinate axes and three rotations around these axes.

Examples of problem solving

Example 1 On the body in the form of a cube with an edge A\u003d 10 cm, three forces act (Fig. 7.6). Determine the moments of forces about the coordinate axes coinciding with the edges of the cube.

Solution

1. Moments of forces about the axis Oh:

2. Moments of forces about the axis OU.

Example 2 Two wheels are fixed on a horizontal shaft, r 1 = 0.4 m; g 2 \u003d 0.8 m. The remaining dimensions are in fig. 7.7. Force applied to wheel 1 F1, to wheel 2 - force F2= 12 kN, F3= 4kN.

Determine strength F1 and reactions in the hinges A And IN in a state of equilibrium.

Recall:

1. At equilibrium, six equilibrium equations are satisfied.

Moment equations should be written with respect to supports A and B.

2. Forces F 2 \\O x; F 2 \\Oy;F 3 \\Oy.

The moments of these forces about the corresponding axes are equal to zero.

3. The calculation should be completed by checking using additional equilibrium equations.

Solution

1. Determine strength F\, by compiling the equation of the moments of forces about the axis Oz:

2. Determine the reactions in the support A. Two components of the reaction act on the support ( Y A ; X A ).

We compose the equation of the moments of forces about the axis Oh"(in support IN).

Rotation around an axis Oh" not happening:

The minus sign means that the reaction is in the opposite direction.

Rotation around an axis OU" does not occur, we draw up the equation of the moments of forces about the axis OU"(in support IN):

3. We determine the reactions in support B. Two components of the reaction act on the support ( X B , Y B ). We compose the equation of the moments of forces about the axis Oh(support A):

We compose the equation of moments about the axis OU(support A):

4. Check. We use projection equations:

The calculation was done correctly.

Example 3 Determine the numerical value of the force P1 , at which the shaft sun(Fig. 1.21, A) will be in balance. With the found value of the force R 1 determine support reactions.

Forces acting on the gears R And R 1 directed along the tangents to the initial circles of the wheels; strength T And T 1 - along the radii of the wheels; strength A 1 parallel to the axis of the shaft. T \u003d 0.36P, 7T 1 \u003d P 1; A 1 \u003d 0.12P 1.

Solution

The shaft supports shown in fig. 1.21, a, should be considered as spatial hinged supports that prevent linear movements in the directions of the axes And And v(the selected coordinate system is shown in Fig. 1.21, b).

We free the shaft from bonds and replace their action with reactions V B, H B, V C , H C (Fig. 1.21, b). We have obtained a spatial system of forces, for which we compose the equilibrium equations using the selected coordinate system (Fig. 1.21.6):

Where A 1*1.25D/2 - moment relative to the axis And strength A 1 , attached to the right gear.

Moments about the axis And forces T 1 And A 1(applied to the middle gear), P 1 (applied to the right gear) and P are equal to zero, since the forces P, T 1, P 1 are parallel to the axis And, and force A 1 crosses the axis And.

where V C \u003d 0.37P;

where VB=0.37P.

hence the reactions V B And V C defined correctly;

Where A 1 * 1,25D/2- moment about the axis v strength A 1 , applied to the middle gear.

Moments about the axis v forces T, P 1 (applied to the middle gear), A 1 And T 1(applied to the right gear) are equal to zero, since the forces T, R 1 , T 1 parallel to the axis v, force A 1 crosses the axis v.

whence H C \u003d 0.81 P;

whence H C \u003d 1.274 P

Let's make a test equation:

hence the reactions H B And H S defined correctly.

In conclusion, we note that the support reactions turned out with a plus sign. This indicates that the chosen directions V B , H B, V C And H S coincide with the actual directions of bond reactions.

Example 4 The steam engine connecting rod pressure P = 25 kN is transmitted to the middle of the crankshaft journal at the point D at an angle α \u003d 30 ° to the horizon with a vertical arrangement of the cheeks of the knee (Fig. 1.22). A belt pulley is mounted on the end of the shaft. The tension of the driving branch of the belt is twice that of the driven one, i.e. S 1 \u003d 2S 2. Flywheel gravity G = 10 kN.

Determine the tension of the branches of the belt drive and the reaction of the bearings A And IN, neglecting the mass of the shaft.

Solution

We consider the balance of a horizontal crankshaft with a pulley. We apply the given forces in accordance with the condition of the problem P, S 1 , S 2 And G . We release the shaft from the support fasteners and replace their action with reactions V A , H A, V B And N V. We choose the coordinate axes as shown in Fig. 1.22. in hinges A And IN no reactions occur along the axis w, since the tension of the belt branches and all other forces act in planes perpendicular to this axis.

We compose the equilibrium equations:

In addition, according to the condition of the problem, we have one more equation

So there are six unknown efforts here. S 1, S 2, H A, V A, H B And V B and six equations relating them.

Equation of projections on the axis w in the example under consideration turns into the identity 0 = 0, since all forces lie in planes perpendicular to the axis w.

Substituting S 1 \u003d 2S 2 into the equilibrium equations and solving them, we find:

Reaction value H B turned out with a minus sign. This means that in reality its direction is opposite to that taken in Fig. 1.22.

Control questions and tasks

1. Write down the formulas for calculating the main vector of the spatial system of converging forces.

2. Write down the formula for calculating the main vector of a spatial system of arbitrarily located forces.

3. Write down the formula for calculating the main moment of the spatial system of forces.

4. Write down the system of equilibrium equations for the spatial system of forces.

5. Which of the equilibrium equations should be used to determine the reaction of the rod R 1 (Fig. 7.8)?

6. Determine the main moment of the system of forces (Fig. 7.9). The reference point is the origin of coordinates. The coordinate axes coincide with the edges of the cube, the edge of the cube is 20 cm; F 1 - 20kN; F 2 - 30kN.

7. Determine the reaction Xv (Fig. 7.10). A vertical axle with a pulley is loaded with two horizontal forces. Forces F1 And F2 parallel to the axis Oh. AO = 0.3 m; OV= 0.5 m; F 1 = 2kN; F 2 = 3.5 kN.

Recommendation. Write the equation of moments about the axis OU" at the point A.

8. Answer the test questions.

An analytical record of the equilibrium conditions for an arbitrary spatial system of forces is a system of six equations (5.3).

From a mechanical point of view, the first three equations establish the absence of translational, and the last three - angular displacement of the body. In the case of CCC, the equilibrium conditions will be represented by the system of the first three equations. In the case of a system of parallel forces, the system will also consist of three equations: one equation of the sum of the projections of forces on the axis parallel to which the forces of the system are oriented, and two equations of moments about axes that are not parallel to the lines of action of the forces of the system.

BODY CENTER OF GRAVITY

The center of gravity of a solid body is the point through which the line of action of the resultant forces of gravity of the particles of a given body passes, at any location in space.

The coordinates of the center of gravity, point C (Fig. 6.3) can be determined by the following formulas:

It is clear that the finer the partition, the more accurate will be the calculation by formulas (6.7), (6.8). However, the complexity of the calculations can be quite large. In engineering practice, formulas are used to determine the center of gravity of bodies of regular shape.

KINEMATICS

LECTURE 6

Kinematics is a branch of mechanics that considers the movement of bodies and

Points without taking into account the forces applied to them.

6.1. Methods for specifying the movement of a point

It is possible to consider the motion of bodies or points only with respect to some reference systems - a real or conditional body, relative to which the position and movement of other bodies are determined.

Let's consider the three reference systems most used in solving problems and, corresponding to them, three ways of specifying the movement of a point. Their characteristics are reduced to: a) description of the reference system itself; b) determining the position of a point in space; c) specifying the equations of point motion; d) establishment of formulas by which the kinematic characteristics of the motion of a point can be found.

Vector way

This method is used, as a rule, in the derivation of theorems and other theoretical provisions. Its advantage over other methods is its compactness. The center is used as a reference system in this method. ABOUT with a triple of unit vectors – i, j, k (Fig. 8.1). Position in space of an arbitrary point M determined by radius vector, r. Thus, the equation of motion of a point M will be a single-valued function of the radius-vector of time, t :

Comparing the last two definitions, we can conclude that the trajectory of a point is simultaneously the hodograph of its radius vector.

We introduce the concept average speed, V cf (Fig. 8.1):

And true (instantaneous) speed, V:

Direction V coincides with the tangent to the trajectory of the point (Fig. 8.1).

The acceleration of a point is a vector quantity that characterizes the change in the speed of a point:

natural way

relationship between S and time t , is the equation of motion of a point in the natural way of specifying motion:

Point speed, directed along the axis t , is defined as:

point acceleration, A, is in the plane nt and can be decomposed into components:

The physical meaning of this expansion is as follows: the line of action of the tangent component, a t , coincides with the line of action of the velocity vector, V , and reflects the change in the velocity modulus only; normal component of acceleration, a n , characterizes the change in the direction of the line of action of the velocity vector. Their numerical values ​​can be found using the following formulas:

Where

is the radius of curvature of the trajectory at a given point.

Coordinate method

This method is most often used in problem solving. The reference system is a triple of mutually perpendicular axes x , y , z (Fig. 8.3). Point position M determined by its coordinates x M , y M , z M .

The equations of motion of a point are single-valued functions of these coordinates from

and its module:

The direction of the velocity vector in space can be analytically determined using direction cosines:

point acceleration M can be established by its projections on the coordinate axes:

The direction of the acceleration vector in space is determined by the direction cosines.