Equilibrium of an arbitrary spatial system of forces. Vector and analytical conditions for the equilibrium of an arbitrary spatial system of forces. Equilibrium conditions for an arbitrary spatial system of forces

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 - the 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.

Above (6.5, case 6) it was found that

Given that , , we project formulas (6.18) onto the Cartesian coordinate axes. We have the analytical form of the equations of equilibrium of an arbitrary spatial system of forces:

(6.19)

The last three equations take place due to the fact that the projection of the moment of force about a point on the axis that passes through this point is equal to the moment of force about the axis (formula (6.9)).

Conclusion arbitrary spatial system of forces, which is applied to a rigid body, we must compose six equilibrium equations(6.19), therefore, with the help of these equations, we have the opportunity to determine six unknowns.

Consider the case spatial system of parallel forces. We choose the coordinate system so that the axis Oz was parallel to the lines of action of forces (Fig. 6.11).

So there are three equations left:

Conclusion. When solving balance problems parallel spatial system of forces, which is applied to a rigid body, we must compose three equilibrium equations and we have the possibility with the help of these equations determine three unknown quantities.

In the first lecture on the "Static" section, we found out that there are six varieties of force systems, which may occur in your practice of engineering calculations. In addition, there are two possibilities for the location of pairs of forces: in space and in a plane. Let us reduce all the equilibrium equations for forces and for pairs of forces into one table (Table 6.2), in which in the last column we note the number of unknown quantities that the system of equilibrium equations will allow us to determine.

Table 6.2 - Balance equations for different systems of forces

Type of force system	Equilibrium equations	Number of defined unknowns

Converging plane
Parallel plane ( axis 0 at)	v. A 0xy
Arbitrary flat (in the 0xy plane)	v. A- arbitrary, belonging to the plane 0xy

Table 6.2 continued

Questions for self-control on topic 6

1. How to find the moment of force about the axis?

2. What relationship exists between the moment of force about a point and the moment of the same force about an axis that passes through this point?

3. In what cases is the moment of force relative to the axis equal to zero? When is it the greatest?

4. In what cases is the system of forces reduced to the resultant?

5. In what case is the spatial system of forces given:

- to a pair of forces;

– to the dynamic screw?

6. What is called a static invariant? What do you know about static invariants?

7. Write down the equilibrium equations for an arbitrary spatial system of forces.

8. Formulate the necessary and sufficient condition for the equilibrium of a parallel spatial system of forces.

9. Will the main vector of the system of forces change when the center of reduction changes? And the main point?

Topic 7. FARMS. DEFINITION OF EFFORT

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 load weighing 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.

The case of such an equilibrium of forces corresponds to two equilibrium conditions

M= Mo= 0, R* = 0.

Main moment modules Mo and main vector R* of the system under consideration are determined by the formulas

Mo= (M x 2 + M y 2 + +M z 2) 1/2 ; R*= (X 2 + Y 2 + Z 2) 1/2.

They wound zero only under the following conditions:

M x = 0, M y =0, M z = 0, X=0, Y=0, Z=0,

which correspond to six basic equations of balance of forces, arbitrarily located in space

=0; =0;

=0; (5-17)

=0 ; =0.

The three equations of the system (5-17) on the left are called equations of moments of forces relative to the axes of coordinates, and three on the right - equations of projections of forces on the axis.

Using these formulas, the equation of moments can be represented as

å (y i Z i - z i Y i)=0; å(z i Х i - x i Z i)=0 ; å(x i Y i - y i X i)=0 .(5-18)

Where x i , y i , z i- coordinates of the points of application of the force P; Y i , Z i , X i - projections of this force on the coordinate axes, which can have any direction.

There are also other systems of six equations for the balance of forces, arbitrarily located in space.

Bringing the system of forces to the resultant force.

If the main vector of the system of forces R* is not equal to zero, but the main moment Mo or equal to zero, or directed perpendicular to the main vector, then the given system of forces is reduced to the resultant force.

2 cases are possible.

1st case.

Let R*¹ 0; Mo = 0 . In this case, the forces lead to a resultant, the line of action of which passes through the center of reduction O, and the force R* replaces the given system of forces, i.e. is its resultant.

2nd case.

R*¹0; Mo¹ 0 and Mo ⊥ R*. (fig.5.15).

After bringing the system of forces to the center O, the force is obtained R* , applied in this center and equal to the main vector of forces, and a pair of forces, the moment of which M equal to the main moment Mo all forces relative to the center of reference, and Mo ⊥ R*.

Let's choose the strengths of this pair R' And R equal in absolute value to the main vector R* , i.e. R= R' = R *. Then the leverage of this pair should be taken equal to OK = = M O/R * .Let's draw through the point O the plane I, perpendicular to the moment of the pair of forces M . Power couple R' , R should be in this plane. Let us arrange this pair so that one of the forces of the pair R' was applied at point O and directed opposite to the force R * . Let us restore in plane I at point O the perpendicular to the line of action of the force R * , and at point K at a distance OK = M O/R * from point O we apply the second force of the pair R .

We put the segment OK in such a direction from the point O, so that, looking towards the moment vector M, we see a pair tending to rotate its plane counterclockwise. Then the forces R* And R' , applied at point O, will be balanced, and the force R pair applied at point K will replace the given system of forces, i.e. will be its resultant. The straight line coinciding with the line of action of this force is the line of action of the resultant force. Rice. 5.15 shows the difference between the resultant force R and force R* obtained by bringing forces to the center O.

Resultant R system of forces applied at point K, which has a certain line of action, is equivalent to a given system of forces, i.e. replaces this system.

The strength is R* at point O replaces a given system of forces only in conjunction with a pair of forces with a moment M= Mo .

Strength R* can be applied at any point of the body to which the forces are brought. Only the modulus and direction of the main moment depend on the position of the point Mo .

Varignon's theorem. The moment of the resultant about any point is equal to the geometric sum of the moments of the constituent forces about this point, and the moment of the resultant force about any axis is equal to the algebraic sum of the moments that make up the forces about this axis.

Methods for solving equilibrium problems with an arbitrary spatial system of forces are considered. An example of solving the problem of equilibrium of a plate supported by rods in three-dimensional space is given. It is shown how due to the choice of axes when compiling the equilibrium equations, it is possible to simplify the solution of the problem.

Content

The procedure for solving equilibrium problems with an arbitrary spatial system of forces

To solve the problem of equilibrium of a rigid body with an arbitrary spatial system of forces, it is necessary to choose a rectangular coordinate system and, relative to it, compose the equilibrium equations.

The equilibrium equations for an arbitrary system of forces distributed in three-dimensional space are two vector equations:
the vector sum of the forces acting on the body is zero
(1) ;
the vector sum of the moments of forces, relative to the origin, is equal to zero
(2) .

Let Oxyz be our chosen coordinate system. Projecting equations (1) and (2) onto the axis of this system, we obtain six equations:
the sums of projections of forces on the xyz axis are equal to zero
(1.x) ;
(1.y) ;
(1.z) ;
the sums of the moments of forces about the coordinate axes are equal to zero
(2.x) ;
(2.y) ;
(2.z) .
Here we consider that n forces act on the body, including the reaction forces of the supports.

Let an arbitrary force , with components , be applied to the body at a point . Then the moments of this force relative to the coordinate axes are determined by the formulas:
(3.x) ;
(3.y) ;
(3.z) .

Thus, the procedure for solving the problem, for equilibrium with an arbitrary spatial system of forces, is as follows.

We discard the supports and replace them with reaction forces. If the support is a rod or thread, then the reaction force is directed along the rod or thread.
We choose a rectangular coordinate system Oxyz .
We find the projections of force vectors on the coordinate axes, , and their application points, . The force application point can be moved along a straight line drawn through the force vector. From such a displacement, the values of the moments will not change. Therefore, we choose the most convenient points for the calculation of the application of forces.
We compose three equilibrium equations for forces (1.x,y,z).
For each force, according to the formulas (3.x,y,z), we find the projections of the moments of force on the coordinate axes.
We compose three equilibrium equations for the moments of forces (2.x, y, z).
If the number of variables is greater than the number of equations, then the problem is statically indeterminate. It cannot be solved by static methods. It is necessary to use methods of resistance of materials.
We solve the resulting equations.

Simplification of calculations

In some cases, it is possible to simplify calculations by using an equivalent equilibrium condition instead of Eq. (2).
The sum of the moments of forces about an arbitrary axis AA′ is equal to zero:
(4) .

That is, you can select several additional axes that do not coincide with the coordinate axes. And concerning these axes to make equations (4).

An example of solving the problem of equilibrium of an arbitrary spatial system of forces

The equilibrium of the plate, in three-dimensional space, is maintained by a system of rods.

Find the reactions of the rods supporting a thin uniform horizontal slab in three dimensions. The rod attachment system is shown in the figure. The plate is affected by: gravity G; and force P applied at point A, directed along side AB.

Given:
G= 28 kN; P= 35 kN; a = 7.5 m; b= 6.0 m; c= 3.5 m.

The solution of the problem

First, we will solve this problem in a standard way applicable to an arbitrary spatial system of forces. And then we get a simpler solution, based on the specific geometry of the system, due to the choice of axes when compiling the equilibrium equations.

Solving the problem in a standard way

Although this method will lead us to rather cumbersome calculations, it is applicable to an arbitrary spatial system of forces, and can be used in computer calculations.

Let us discard the bonds and replace them with reaction forces. The connections here are rods 1-6. Instead of them, we introduce forces directed along the rods. The directions of forces are chosen at random. If we do not guess with the direction of any force, we will get a negative value for it.

Draw a coordinate system Oxyz with origin at point O .

We find the projections of forces on the coordinate axes.

For strength we have:
.
Here α 1 - angle between LQ and BQ. From right triangle LQB:
m;
;
.

Forces , and are parallel to the z axis. Their components:
;
;
.

For strength we find:
.
Here α 3 - angle between QT and DT . From right triangle QTD :
m;
;
.

For strength:
.
Here α 5 - angle between LO and LA . From right triangle LOA :
m;
;
.

The force is directed along the diagonal of a rectangular parallelepiped. It has the following projections on the coordinate axes:
.
Here are the direction cosines of the diagonal AQ:
m;
;
;
.

We select the points of application of forces. Let's take advantage of the fact that they can be moved along the lines drawn through the force vectors. So, as a point of application of force, you can take any point on the line TD. Let's take a point T, because for it x and z - coordinates are equal to zero:
.
In a similar way, we select the points of application of the remaining forces.

As a result, we obtain the following values of the force components and points of their application:
; (point B );
; (point Q );
; (point T );
; (point O );
; (point A );
; (point A );
; (point A );
; (point K ).

We compose the equilibrium equations for forces. The sums of projections of forces on the coordinate axes are equal to zero.

;

;

.

We find the projections of the moments of forces on the coordinate axes.
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;

We compose the equilibrium equations for the moments of forces. The sums of the moments of forces about the coordinate axes are equal to zero.

;

;

;

So, we got the following system of equations:
(P1) ;
(P2) ;
(P3) ;
(P4) ;
(P5) ;
(P6) .

This system has six equations and six unknowns. Further, here you can substitute numerical values and get the solution of the system using a mathematical program for calculating a system of linear equations.

But, for this problem, you can get a solution without using computer technology.

An efficient way to solve a problem

We will take advantage of the fact that equilibrium equations can be written in more than one way. You can arbitrarily choose the coordinate system and axes with respect to which the moments are calculated. Sometimes, due to the choice of axes, one can obtain equations that are solved more simply.

Let's take advantage of the fact that, in balance, the sum of the moments of forces about any axis is zero. Let's take the AD axis. The sum of the moments of forces about this axis is zero:
(P7) .
Further, note that all forces except cross this axis. Therefore, their moments are equal to zero. Only one force does not cross the AD axis. It is also not parallel to this axis. Therefore, in order for equation (A7) to hold, the force N 1 should be zero:
N 1 = 0 .

Now let's take the AQ axis. The sum of the moments of forces relative to it is equal to zero:
(P8) .
This axis is crossed by all forces except . Since the force is not parallel to this axis, then for equation (A8) to be satisfied, it is necessary that
N 3 = 0 .

Now let's take the AB axis. The sum of the moments of forces relative to it is equal to zero:
(P9) .
This axis is crossed by all forces except , and . But N 3 = 0 . That's why
.
The moment from the force about the axis is equal to the product of the arm of the force and the projection of the force on the plane perpendicular to the axis. The shoulder is equal to the minimum distance between the axis and the straight line drawn through the force vector. If the twist is in the positive direction, then the torque is positive. If it's negative, then it's negative. Then
.
From here
kN.

The remaining forces can be found from equations (P1), (P2) and (P3). From equation (P2):
N 6 = 0 .
From equations (P1) and (P3):
kN;
kN

Thus, solving the problem in the second way, we used the following equilibrium equations:
;
;
;
;
;
.
As a result, we avoided cumbersome calculations associated with calculating the moments of forces relative to the coordinate axes and obtained a linear system of equations with a diagonal matrix of coefficients, which was immediately resolved.

N 1 = 0 ; N 2 = 14.0 kN; N 3 = 0 ; N 4 = -2.3 kN; N 5 = 38.6 kN; N 6 = 0 ;

The minus sign indicates that the force N 4 directed in the direction opposite to that shown in the figure.