EntranceBringing a flat arbitrary system of forces to the center. Cases of bringing a plane system of forces to a given point
A flat system of arbitrarily located forces.
Conditions for equilibrium of pairs of forces.
If a rigid body is acted upon by several pairs of forces, arbitrarily located in space, then by successively applying the parallelogram rule to each two moments of the pairs of forces, any number of pairs of forces can be replaced by one equivalent pair of forces, the moment of which equal to the sum moments of given pairs of forces.
Theorem. For equilibrium of pairs of forces applied to a solid body, it is necessary and sufficient that the algebraic sum of the projections of the moments of pairs of forces onto each of the three coordinate axes is equal to zero.
Let us consider the case of force transfer to an arbitrary point that does not lie on the line of action of the force.
Let's take a force F applied at point C. We need to transfer this force parallel to itself to a certain point O. Let us apply at point O two forces F" and F", oppositely directed, equal in value and parallel to the given force F, i.e. F" = F" = F. The application of these forces at point O does not change the state of the body, since they are mutually balanced. The resulting system of three forces can be considered as consisting of a force F" applied at point O, and a pair of forces FF" with a moment M = Fa. This pair of forces is called annexed, and its arm a is equal to the arm of force F relative to point O.
Thus, when a force F is brought to a point not lying on the line of action of the force, an equivalent system is obtained, consisting of a force the same in magnitude and direction as the force F, and an attached pair of forces, the moment of which is equal to the moment of the given force relative to the point casts:
As an example of force reduction, consider the action of force F on the end C of a clamped rod (Fig. 28, b). After bringing the force F to point O of the clamped section, we find in it a force F1 equal and parallel to the given one, and an attached moment M equal to the moment of the given force F relative to the reduction point O,
1.4.2 Bringing a plane system of forces to a given point
The described method of bringing one force to a given point can be applied to any number of forces. Let us assume that forces F1, F2, F3, F4 are applied at points of the body A, B, C and D (Fig. 30).
It is required to bring these forces to point O of the plane. Let us first present the force F1 applied at point A. Let us apply at point O two forces F1" and F1"", parallel to it and directed towards opposite sides. As a result of bringing force F1, we obtain force F1" applied at point O, and a pair of forces F1" F1"" with shoulder a1. By doing the same with force F2 applied at point B, we obtain force F2" applied at point O, and a pair of forces with shoulder a2, etc.
We replaced the flat system of forces applied at points A, B, C and D with converging forces F1, F2, F3, F4 applied at point O, and pairs of forces with moments equal to the moments of the given forces relative to point O:
The forces converging at a point can be replaced by one force F"hl, equal to the geometric sum of the components,
This force, equal to the geometric sum of given forces, is called the main vector of the force system and denote F"ch.
Based on the rule for adding pairs of forces, they can be replaced by the resulting pair, the moment of which is equal to the algebraic sum of the moments of the given forces relative to point O and is called main point relative to the reference point
Consequently, in the general case, a flat system of forces as a result of reduction to a given point O is replaced by an equivalent system consisting of one force (the main vector) and one pair (the main moment).
It is necessary to understand that the main vector F"ch is the resultant of a given system of forces, since this system is not equivalent to one force F"ch. Only in the special case when main point becomes zero, the main vector will be the resultant of this system of forces. Since the main vector is equal to the geometric sum of the forces of a given system, neither its magnitude nor its direction depend on the choice of the center of reduction. The value and sign of the main moment Mgl depend on the position of the center of reduction, since the arms of the component pairs depend on the relative position of the forces and the point (center) relative to which the moments are taken.
The following cases of bringing a system of forces may occur:
1. - general case; the system is reduced to the main vector and to the main moment.
2. ; the system is reduced to one resultant equal to the main vector of the system.
3. ; the system is reduced to a pair of forces whose moment is equal to the main moment.
4. ; the system is in equilibrium, that is, for the equilibrium of a plane system of forces it is necessary and sufficient that its main vector and main moment be simultaneously equal to zero.
It can be proven that in the general case, when, there is always a point about which the main moment of forces is equal to zero.
Let us consider a plane system of forces that is brought to point O, that is, replaced by the main vector applied at point O and the main moment. For definiteness, we assume that the main moment is directed clockwise, i.e. . Let us represent this main moment as a pair of forces FF", the modulus of which will be chosen equal to the modulus of the main vector, i.e. We will apply one of the forces that make up the pair at the center of reduction O, the other force at point C, the position of which will be determined from the condition: Therefore .
Let's arrange a pair of forces so that the force F"" is directed in the direction opposite to the main vector F"ch. At point O we have two equal mutually opposite forces F"ch and F"", directed along the same straight line; they can be discarded (according to the third axiom). Consequently, relative to point C, the main moment of the system of forces under consideration is equal to zero, and the system is reduced to the resultant.
We will replace the flat system of forces applied at points A, B, C, D:
1) forces F 1 ’, F 2 ’, F 3 ’, F 4 ’, applied at point O;
2) in pairs of forces:
F 1 F 1 ': M 1 = M o (F 1) = F 1 a 1
F 2 F 2 ': M 2 = M o (F 2) = F 2 a 2
F 3 F 3 ': M 3 = M o (F 3) = F 3 a 3
F 4 F 4 ': M 4 = M o (F 4) = F 4 a 4
The forces F 1 ’, F 2 ’, F 3 ’, F 4 ’ converging at point O can be replaced by one force (resultant) F ch:
F gl = F 1 ' + F 2 ' + F 3 ' + F 4 ' = F 1 + F 2 + F 3 + F 4
F ch– the main vector of the force system.
The resulting force pairs can be replaced by the resulting pair, the moment of which M ch:
M gl = M 1 + M 2 + M 3 + M 4 = Σ M i = Σ M o (F i)
M gl - main point regarding the reference point.
The plane system of forces at a given point O is replaced by an equivalent system consisting of one force (the main vector) and one pair (the main moment).
Theorem on the moment of the resultant (Varignon’s theorem)
The moment of the resultant plane system of forces relative to an arbitrary point is equal to the algebraic sum of the moments of the component forces relative to the same point.
M o (F Σ)= Σ M o (F i)
Equilibrium equations for a plane system of forces
F GL = 0;
M gl = ΣM o (F i) = 0.
The module of the main vector can be determined through projections onto coordinate axes all forces of the system.
F GL = (ΣF іх) 2 +(ΣF іу) 2 =0 from this the equilibrium equations follow:
Σ F іх =0
Σ F іу =0
Σ M o (F i)=0
Other forms of equilibrium equations:
Σ M A (F i)=0
Σ М В (F і)=0 (ABCs do not lie on the same
Σ M C (F i)=0 direct)
Σ M A (F i)=0 (x axis is not perpendicular
Σ M B (F і)=0 direct AB)
Σ F іх =0
For a system of parallel forces, choosing one of the projection axes parallel to these forces (y-axis), and the other perpendicular to them (x-axis), we obtain two equilibrium equations:
Σ F іу =0
Σ M o (F i)=0
Σ M A (F i)=0
Σ M V (F i)=0
Algorithm for solving problems
1. Select an object of equilibrium (body or point): we will consider equilibrium relative to...
We show everything in the picture active forces, including bond reactions.
3. Select a coordinate system - it is advisable to direct the coordinate axes parallel or perpendicular to the required forces.
We compose equilibrium equations for the object of study.
Σ F іх =0
Σ F іу =0
Σ M o (F i)=0
From the resulting equations we determine unknown quantities (we determine reactions).
We check the correctness of the solution of the equations.
Σ M p (F i)=0
Σ M e (F i)=0
5. Support devices beam systems
Articulating support
Hinged-fixed form and rigid seal (pinching)
Subject:
"Center of gravity.
Geometric characteristics of flat sections"
Plan
1. Center of parallel forces and its coordinates.
2. Center of gravity of areas. Statistical moments of areas.
3. Solving problems to determine the coordinates of the center of gravity of a flat composite figure.
4. Polar and axial moments of inertia.
5. Axial moments of inertia about parallel axes.
6. Determination of moments of inertia of composite sections using tables of normal assortment.
1. Center of parallel forces and its coordinates
Let a system of parallel forces F 1, F 2, F 3, ..., Fn be given; the coordinates of points C 1, C2, C3, ..., Cn of application of these forces are known (Fig. 42, b). Let us denote the point of application by the resultant letter C, and its coordinates we denote x c, y c.
FΣ = F 1 + F 2 + F 3+…. + Fn = ΣF i . (1)
FΣ xс = F 1 x 1 + F 2 x 2 + F 3 x 3 +… + Fnxn = Σ F i x i ,
x c = F 1 x 1 + F 2 x 2 + F 3 x 3 +… + Fnxn / FΣ = Σ F i x i / FΣ
FΣ = F 1+ F 2+ F 3+…+ Fn= Σ F іх c =
= F 1 x 1 + F 2 x 2 + F 3 x 3 +… + Fnxn / F 1+ F 2+ F 3+…+ Fn= Σ F i x i / F i (2)
Theorem . StrengthF , without changing its action on the body, can be transferred from the point of its application A to any center of adduction O, while attaching a couple of forces with a moment to the bodyM , geometrically equal to the moment M ABOUT (F ) of this force relative to the center of reduction.
Let the force be given F, lying in the horizontal plane OXY parallel to the OX axis (Fig. 1.41).
According to the Poinsot method, instead of force F, applied at point A, a force is obtained F 1, equal in magnitude to force F, but applied at point O and attached couple of forces , whose vector moment M= M ABOUT ( F).
According to the theorem on the equivalence of pairs of forces, the associated pair of forces can be replaced by any other pair of forces with the same vector moment.
1.15. Bringing an arbitrary system of forces to a given center
Theorem . Any arbitrary system of forces acting on a body can be reduced in the general case to a force and a pair of forces.
This process of replacing a system of forces with one force and a pair of forces is called bringing the system of forces to a given center .
P
Consistently applying the Poinsot method to each of the given system of forces, we will bring it to an arbitrary center O. As a result of this, we obtain a system of forces ( F 1 , …, F n), applied at the center O, and an associated pair of forces with a moment M= Σ M ABOUT ( F i). Adding up strength F 1 , …, F n according to the parallelogram rule, we obtain their resultant R* , equal to the geometric sum of the given forces and applied at the center of reduction.
The geometric sum of all the forces of the system is called the main vector of the force system and, in contrast to the resultant R, denote R * .
Vector M= Σ M ABOUT ( F i) are called the main moment of the system of forces relative to the center of reduction.
This result can be formulated as follows: forces arbitrarily located in space can be reduced to one force equal to their main vector and applied at the center of reduction and to a pair of forces with a moment equal to the main moment of all forces relative to the center of reduction.
The choice of the center of reduction does not affect the magnitude and direction of the main vector R*, but affects the modulus and direction of the main moment M. Main vector R* is a free vector and can be applied to any point on the body.
1.16. Analytical equilibrium conditions for a plane arbitrary system of forces
Flat arbitrary force system – a system of forces whose lines of action are arbitrarily located in the same plane.
The lines of action of a plane arbitrary system of forces intersect at various points.
N
Consistently applying the Poinsot method for each of the forces F i, let us carry out a parallel transfer of forces from points A i to the beginning O of the reference system OXYZ. According to this method, the force F i will be equivalent to the force F i applied at point O, and the attached pair of forces with a moment M i = M ABOUT ( F i ) . In this case, M i = ± F i h i , where h i is the force arm F i relative to the center of reduction O. At the end of this work we obtain a converging system of forces ( F i ,…, F n) and a convergent system of vector moments M i = M ABOUT ( F i) attached pairs of forces applied at the center of reduction. Adding the force vectors, we get chapters
new vector R* = Σ F i and the main moment of an equivalent pair of forces M = Σ M ABOUT ( F i).
Thus, flat arbitrary system of forces (F i ,…, F n ) is equivalent to one force R* = Σ F i and a pair of forces with moment M = Σ M ABOUT (F i ).
When solving problems, statics use projections of force on coordinate axes and algebraic moments of forces relative to a point.
In Fig. Figure 1.44 shows a flat arbitrary system of forces reduced to the main vector of forces, the module of which R*= 
and an equivalent pair of forces with algebraic moment M = Σ M O ( F i).
IN
Geometric equilibrium condition of any system of forces is expressed by vector equalities: R* = Σ F i = 0; M= Σ M ABOUT ( F i) = 0.
When solving problems, it is necessary to determine reactions R i E external connections imposed on the mechanical system. At the same time, active forces F i E applied to this system are known. Since active forces F i E and bond reactions R i E belong to the category of external forces, then it is advisable to express the geometric condition for the equilibrium of a system of external forces by vector equalities:
Σ F i E + Σ R i E = 0;
Σ M A ( F i E) + Σ M A ( R i E) = 0.
For the system of external forces to be in equilibrium, it is necessary and sufficient that the geometric sum of the active forces F i E and reactions R i E external connections and the geometric sum of the moments of active forces M A ( F i E ) and reactions of external relations M A ( R i E ) relative to an arbitrary point A were equal to zero.
Projecting these vector equalities onto the coordinate axes of the reference system, we obtain analytical conditions for the equilibrium of a system of external forces . For a flat arbitrary system of forces, these equations take the following form:
Σ 
+ Σ 
= 0;
Σ 
+ Σ 
= 0;
Σ M A ( F i E) + Σ M A ( R i E) = 0,
where Σ 
,
Σ 
– respectively, the sum of projections of active forces onto the coordinate axes OX, OY; Σ 
,
Σ 
– the sum of projections of reactions of external connections on the coordinate axes OX, OY; Σ M A ( F i E) – sum of algebraic moments of active forces F i E relative to point A; Σ M A ( R i E) – sum of algebraic moments of reactions R i E external connections relative to point A.
The set of these formulas is the first (basic) form of the equilibrium equations for a plane arbitrary system of external forces .
Thus , for the equilibrium of a flat arbitrary system of external forces applied to a mechanical system, it is necessary and sufficient that the sum of the projections of active forces and reactions of external connections on two coordinate axes and the sum of the algebraic moments of active forces and reactions of external connections relative to an arbitrary point A are equal to zero.
There are other forms of equilibrium equations for a plane arbitrary system of forces.
Second form is expressed by a set of formulas:
Σ 
+ Σ 
= 0;
Σ M A ( F i E) + Σ M A ( R i E) = 0;
Σ M V ( F i E) + Σ M V ( R i E) = 0.
For the equilibrium of a flat arbitrary system of external forces applied to a body, it is necessary and sufficient that the sum of the projections of the forces on the coordinate axis and the sum of the algebraic moments of forces relative to arbitrary points A and B are equal to zero.
Third form equilibrium equations is expressed by a set of formulas:
Σ M A ( F i E) + Σ M A ( R i E) = 0;
Σ M V ( F i E) + Σ M V ( R i E) = 0;
Σ M С ( F i E) + Σ M С ( R i E) = 0.
For the equilibrium of a flat arbitrary system of external forces applied to a body, it is necessary and sufficient that the sums of the algebraic moments of these forces relative to arbitrary points A, B and C are equal to zero.
When using the third form of equilibrium equations, points A, B and C should not lie on the same straight line.
Lecture 5
Summary: Bringing force to a given center. Bringing a system of forces to a given center. Equilibrium conditions for a spatial system of parallel forces. Equilibrium conditions for a plane system of forces. Three-moment theorem. Statically definable and statically indeterminate problems. Equilibrium of the system of bodies.
BRINGING THE SYSTEM OF FORCES TO A SPECIFIED CENTER. CONDITIONS OF EQUILIBRIUM
Bringing force to a given center.
The resultant of a system of converging forces is directly found using the addition of forces according to the parallelogram rule. Obviously, a similar problem can be solved for an arbitrary system of forces if we find a method for them that allows us to transfer all the forces to one point.
Theorem on parallel force transfer . A force applied to an absolutely rigid body can, without changing the effect it exerts, be transferred from a given point to any other point of the body, adding a couple with a moment equal to the moment of the transferred force relative to the point where the force is transferred.
Let a force be applied at point A. The effect of this force does not change if two balanced forces are applied at point B. The resulting system of three forces is a force equal to, but applied at point B and a pair with a moment. The process of replacing a force with a force and a pair of forces is called bringing the force to a given center B.
Bringing a system of forces to a given center.
Main theorem statics (Poinsot).
Any arbitrary system of forces acting on a rigid body can, in general, be reduced to a force and a pair of forces. This process of replacing a system of forces with one force and one pair of forces is called bringing the system of forces to a given center.
The main vector of the system strength is called a vector equal to the vector sum of these forces.
The main point of the system strength relative to point O of the body, a vector is called equal to the vector sum of the moments of all forces of the system relative to this point.
Formulas for calculating the main vector and main moment
Formulas for calculating the modulus and direction cosines
main vector and main moment
Conditions for the equilibrium of a system of forces.
Vector shape.
For the equilibrium of an arbitrary system of forces applied to a rigid body, it is necessary and sufficient that the main vector of the force system is equal to zero and the main moment of the force system relative to any center of reduction is also equal to zero.
Algebraic form.
For the equilibrium of an arbitrary system of forces applied to a rigid body, it is necessary and sufficient that the three sums of the projections of all forces on the axis Cartesian coordinates were equal to zero and the three sums of the moments of all forces relative to the three coordinate axes were also equal to zero.
Conditions for the equilibrium of a spatial system
parallel forces.
A system of parallel forces acts on the body. Let's place the Oz axis parallel to the forces.
Equations 
For the equilibrium of a spatial system of parallel forces acting on a solid body, it is necessary and sufficient that the sum of the projections of these forces be equal to zero and the sum of the moments of these forces relative to two coordinate axes perpendicular to the forces are also equal to zero.
- projection of force onto the Oz axis.
FLAT FORCE SYSTEM.
Equilibrium conditions for a plane system of forces.
A plane system of forces acts on the body. Let's place the Ox and Oy axes in the plane of action of the forces.
Equations 
For the equilibrium of a plane system of forces acting on a solid body, it is necessary and sufficient that the sums of the projections of these forces onto each of the two rectangular coordinate axes located in the plane of action of the forces are equal to zero and the sum of the moments of these forces relative to any point located in the plane of action the forces were also zero.
Three-moment theorem.
For the equilibrium of a plane system of forces acting on a rigid body, it is necessary and sufficient that the sums of the moments of these forces of the system relative to any three points located in the plane of action of the forces and not lying on the same straight line are equal to zero.
Statically definable and statically indeterminate problems.
For any plane system of forces acting on a rigid body, there are three independent equilibrium conditions. Consequently, for any plane system of forces, no more than three unknowns can be found from equilibrium conditions.
In the case of a spatial system of forces acting on a rigid body, there are six independent equilibrium conditions. Consequently, for any spatial system of forces, no more than six unknowns can be found from equilibrium conditions.
Problems in which the number of unknowns is not more number independent equilibrium conditions for a given system of forces applied to a rigid body are called statically definable.
Otherwise, the problems are statically indeterminate.
Equilibrium of the system of bodies.
Let us consider the equilibrium of forces applied to a system of interacting bodies. The bodies can be connected to each other using hinges or in another way.
The forces acting on the system of bodies under consideration can be divided into external and internal.
External are called the forces with which the bodies of the system under consideration are acted upon by bodies that are not included in this system of forces.
Internal are called the forces of interaction between the bodies of the system under consideration.
When considering the equilibrium of forces applied to a system of bodies, one can mentally divide the system of bodies into individual solid bodies and apply the equilibrium conditions obtained for one body to the forces acting on these bodies. These equilibrium conditions will include both external and internal forces of the system of bodies. Inner forces Based on the axiom of the equality of action and reaction forces at each point of articulation of two bodies, they form an equilibrium system of forces.
Let us demonstrate this using the example of a system of two bodies and a plane system of forces.
If we create equilibrium conditions for each solid system of bodies, then for body I 
.
for body II 
In addition, from the axiom about the equality of action and reaction forces for two interacting bodies we have 
.
The presented equalities are the conditions for the equilibrium of external forces acting on the system.
Sealing reaction.
Let's consider a beam, one end of which AB is embedded in the wall. This type of fastening of the end of the beam AB is called sealing at the point B. Let a plane system of forces act on the beam. Let us determine the forces that must be applied to point B of the beam if part of the beam AB is discarded. Distributed reaction forces are applied to the beam section (B). If these forces are replaced by elementary concentrated forces and then brought to point B, then at point B we obtain a force (the main vector of reaction forces) and a pair of forces with a moment M (the main vector of reaction forces relative to point B). Moment M called the closing moment or directive moment. The reaction force can be replaced by two components and .
The seal, unlike a hinge, creates not only a reaction unknown in magnitude and direction, but also a pair of forces with an unknown moment M in the seal.
The moment of force F relative to a given point O is the product of the magnitude of the force by its arm, that is, by the length of the perpendicular lowered from point O to the line of action of this force.
If the force F tends to rotate the body around a given point O in the direction opposite to the clockwise movement, then we agree to consider the torque of the force F relative to the point O as positive; if the force tends to rotate the body around point O in the direction coinciding with the direction of movement of the clock hand, then the moment of the force relative to this point will be considered negative. Hence,
If the line of action of force F passes through this point Oh, then the moment of force F about this point is equal to zero.
The addition of forces located arbitrarily on a plane can be performed in two ways:
1) sequential addition;
2) bringing a given system of forces to an arbitrarily chosen center.
The first method becomes cumbersome when large number components of forces and is not applicable for a spatial system of forces, the second method is general, simpler and more convenient.
If a system of forces is given, located arbitrarily in one plane, then by transferring all these forces to a point O arbitrarily chosen in this plane, called the center of reduction, we obtain the force applied in this center
and a couple with the moment
The geometric sum of the forces of a given system is called the equal vector of this system of forces.
The algebraic sum of the moments of forces of a plane system relative to some point O on the plane of their action is called the principal moment of this system of forces relative to this point O.
The main moment changes with the change in the center of reduction; the dependence of the main moment on the choice of the center of reduction is expressed by the following formula:
where and are two different centers of reduction.
Since the force R and the couple with the moment resulting from bringing a given plane system of forces to the center O lie in the same plane, they can be reduced to a single force applied at some point. This force is the resultant of this plane system of forces.
Thus, if , then the system of forces is reduced to one resultant that does not pass through the center of reduction O. In this case, the moment of the resultant relative to any point will be equal to the algebraic sum of the moments of all given forces relative to the same point (Varignon’s theorem).
If the origin of coordinates is chosen at the center of reduction and the projections of all forces on the coordinate axes and the coordinates of the points of application of these forces are known, then the moment of the resultant is found by the formula
If, as a result of bringing a system of forces to a given center, it turns out that the main vector of this system is equal to zero, and its main moment is different from zero, then this system is equivalent to a pair of forces, and the main moment of the system is equal to the moment of this pair and does not depend in this case on the choice adduction center. If then the system is reduced to the resultant applied at the reduction center O.
If and , then the system of forces is in equilibrium. All cases encountered when adding the forces of a plane system can be presented in the form of a table. 3.
Table 3
We will consider the equilibrium of a plane system of forces in the next paragraph, and now we will move on to solving problems on the addition of forces of a plane system.
Example 13. A plane system of four forces is given, the projections X and Y of these forces onto the coordinate axes, the x and y coordinates of the points of their application are given in the table. 4.
Table 4
Bring this system to the origin and then find the line of action of the resultant.
Solution. Let us find the projections of the main vector of a given system of forces onto the coordinate axes using formula (14)
We find the main point using formula (15)
Let be the point of the line of action of the desired resultant. Then
On the other hand, by Varignon’s theorem we have:
Hence,
This is the equation of the line of action of the resultant.
Example 14. Find the resultant of four forces acting on the sides of a regular hexagon, the direction of which is indicated in Fig. 30 if .
Solution. Let us choose the center O of the hexagon as the center of reduction and find the main vector R and the main moment of this system of forces relative to the center O. Since , then the main vector R is equal to , and the main moment
In order to find the moment of force relative to point O, we lower the perpendicular CM from point O to the line of action of this force. Since the force tends to rotate the hexagon around point O clockwise, then