5. ELEMENTS OF STATICS AND DYNAMICS

5. ELEMENTS OF STATICS AND

DYNAMICS





State Engineering University of Armenia

Department of Mechanics and Machine Science

Chair of Applied Mechanics

Applied Mechanics Syllabus

Assoc. Prof. K.V. Hovhannisyan

5.1. The basic concepts of statics

Force: A quantitative measure of mechanical interaction between material

bodies. Force is vector (Fig.5.1) and for that except direction and magnitude

the point of application is mentioned too. The point of application of force and

its direction define the line of action of force. The unit of force in the

International System of Units (SI) is called the Newton (N).



The group of forces that act on a body or a system of bodies is called the

system of forces and is denoted by F , F2 ,..., Fn  .

1





If the system of forces F , F2 ,..., Fn  acting on the free body is replaced

with another system of forces F1, F2 ,..., Fn  and kinematics (rest or mobile)

1



of body does not changed then the systems of forces are called equivalent

F , F ,..., F  ~ F, F,..., F 

1 2 n 1 2 n





In case when the system of forces F , F2 ,..., Fn  is equivalent to F force,

1

then the force is called adequate to that force system

F , F ,..., F  ~ F

1 2 n

If the kinematics of a body does not changed under the action of a system of

forces F , F2 ,..., Fn  then that system of forces is called balanced or equal to

F1 , F2 ,..., Fn  ~ 0

1

zero

A rigid body is considered to be in equilibrium if it is at rest or moving with a

constant velocity. Since the effect of a force is to change the linear velocity

and the effect of a moment or couple is to change the angular velocity, then the

body will be in equilibrium if the sum of all the forces acting on it is zero (ΣF

= 0), and if the sum of all the moments acting on it is zero (ΣM = 0).

The F force that after adding to system of forces F1 , F2 ,..., Fn  becomes the

system of forces to the balance system is called balanced



F , F ,..., F , F ~ 0

1 2 n

5.2. The moment of a force about a point. The

moment of a point about an axis

The moment of force is characterized from the turning effect of body. This

effect into arbitrary point depends on the value of force (module) and the

distance from that point to the line of action of force (arm).

Moment of a force about a point: Vector product of a radius vector

from the point to the line of action of the force and the force itself.

Moment arm: Shortest distance to the line of action of a force from a

given point.

Moment of force F (Fig .5.2) about the point O is denoted as M0 ( F) and is

equal to

M0 ( F)  F h

If all forces of the system of forces being in the same plane the moments of all

forces about any point in that plane are perpendicular to that plane. In that case

the moments of forces are considered as algebraic values with plus or minus



M0 F   F  h

signs



If the rotation is due to counterclockwise then the moment is positive,

otherwise is negative (Fig.5.3).

The moment of a force about an axis: Component along a given

axis of the moment of a force about any point on the axis. The algebraic

value that is equal to moment of projection of moment on the perpendicular

to that axes any arbitrary surface about the intersecting point with the

surface. That moment is positive, if from positive direction of axis the force

moves counterclockwise, in other case is negative. The moment of force F

about axis z (Fig.5.4) is denoted Mz F and equal



Mz F   FXY  h

where the FXY is the projection of force on any surface XOY perpendicular to

the z axis and h is the arm of FXY from the point O of intersection of z axis

and XOY surface. Therefore in order to find the moment of force F about

Z-axis we pass any surface perpendicular to the axis, project this force on that

surface and find the value of moment of projection about the point of

intersection of z-axis and that surface. Moment of force about axis is equal to

zero if the force is parallel to the surface (the projection being point) and if the

line of action of force intersect the axis (the arm is equal to zero). Therefore, if

the force and axis are in the same surface then the moment of force about an

axis is equal to zero.

Between moments of forces about a point and an axis passed through that

point there is a following dependence that is writing without proof.

The moment of force into given axis is equal to projection of moment of

force on that axis into any point lying on that axis

MX  MOX , MY  MOY , MZ  MOZ

5.3. Couple of forces and their moments



Couple: Pair of parallel forces that are equal in magnitude, but opposite in

sense.

Moment of a couple: Vector sum of the moments about any point in

space of the forces that form a given couple.

Let us consider ( F, F ) couple ( F   F ) of forces (Fig.5.5). This system of

forces can not become to adequate and simultaneously cannot be in

equilibrium. The surface of the couple of forces is called the surface of couple

of forces, and the perpendicular distance between lines of action of forces is

called arm.

In the space the moment of couple of forces is defined by vector that is equal

to the product of absolute value of one of this forces and arm and is

perpendicular to the surface of action of couple of forces and is directed so the

body turning due to counterclockwise. The value of the moment is



MF, F    F  h

5.4. Resultant vector and resultant moment of

force system

In statics the two main problems are considered:

- addition of forces and transfer of force system to the simplicities view,

- determination of equilibrium conditions of force systems.

The main theorem of statics helps us to solve these problems for any

mechanism. In that theorem the following important definitions are used.



Resultant vector: Geometrical sum of all forces of given

force system F , F ,..., F 

F

n

1 2 n

F k



k 1

Resultant moment: Moment equal to the vector sum of the moments of

all the forces of a system about a chosen point



 M F 

n

M0  0 k



k 1

Lemma: The force applied on a solid body does not change its

action if to displace it parallel into any other point on the body

and adding couple of forces, which moment is equal to moment

of force about displacing point.



Let in point A on the body F the force is applied (Fig.5.6). The kinematics

of body does not changed if in the any arbitrary point B balanced force system

F, F ~ 0 will be applied so that F  F  F . The system of forces

 

F, F is couple of forces that can be presented as MF, F  . Therefore





F ~ F, F, F  ~ F, M

where



M  MF, F   BA  F  MB F

5.5. The main theorem of statics

Theorem: Arbitrary force system acting on a solid body is adequate to one

force acting in any point of the body (reduced point) that is equal to the

resultant vector of the force system and one couple of force which moment is

equal to the resultant moment of force system into reduced point.

Let an arbitrary force system F1 , F2 ,..., Fn  is applied on body (Fig.5.7a). Using

the lemma above mentioned to displace all forces into arbitrary point 0. In the

end the forces F, F2,..., Fn  and moments ( 1 , M2 ,..., Mn) of couple of forces are

1

 M

applied in center 0 of the body (Fig.5.7b).

Here F  F

k k Mk  M0 Fk   OAk  Fk  rk  Fk

k  1,2,...n

The F, F2,..., Fn  force system and moments system (M1 , M2 ,..., Mn) of couple

1



of forces can be replace (Fig.5.7c) by resultant F vector and resultant

M0 moment n n

F   Fk    Fk M0   M0 Fk    rk  Fk 

n n





k1 k1

(5.1)



k 1 

k 1

Apply OXYZ coordinate system where its initial point O coincides with

reduced center. Projecting Eq. (5.1) to the axis X, Y and Z according to the

famous theorem of geometry: projection of sum of vector to any axis is equal

to the algebraic sum of vectors projection to the same axis, then we have





n

(5.2)



n

FX  FZ   FkZ

n

FkX FY  FkY

k1

k1 k1





 M F   M F 

n n

M0 X  M X  X k

M0 Z  M Z  Z k



k 1 

k 1



 M F 

n

M0 Y  M Y  Y k



k 1



With the projection of the resultant vector and resultant moment and their

connected cosines can be found the F and M vectors’ magnitude and

direction

F  FX  FY  FZ

2 2 2

M  M2  M2  M2

X Y Z





   

cos F, ^ X  FX F cos F, ^ Y  FY F 

cos F, ^ Z  FZ F 

cos F, Z  F

^

Z F cosM, Y   M

^

Y M 

cos M, ^ Z  MZ M 

5.6. Varignon’s theorem

Theorem: If the resultant of a plane system of forces exists then its

moment about any centre is equal to the algebraic sum of the moments of all

forces, forming that system, about the same centre.

Let the system of forces F1 , F2 ,..., Fn  ~ F act on the body is reduced to an

adequate force (Fig.5.8) and that force F is applied on the point C of body.

The position of that point in OXYZ system find out with the r radius vector.

Define the moment of each force Fi into point O and sum up them.



 M F   r  F  r  F  ...  r  F  r  F  F  ...  F 

n



0 k 1 2 n 1 2 n



k 1





Changing the geometry sum of forces with their adequate force F1  F2  ...  Fn  F

we have



       

n n

or

M0 Fk  r  F M0 F  M0 Fk

k1 k1





The theorem is proved.

5.7. Equilibrium conditions for an arbitrary

system of forces

From the main theorem of statics for equilibrium of the system of forces

F1 , F2 ,..., Fn  it is necessary and sufficient the equilibrium of ( F, M0 )

system of forces may be ensured in case if

M0   M0 Fk   0

n

F   Fk  0

n



(5.3)

k1 k1

Conditions (5.3) are equilibrium sufficient conditions for force system.

Therefore, for equilibrium of any arbitrary force system it is necessary and

sufficient that resultant vector and resultant moment of the force system are

equal to zero.

If the vector Eq. (5.3) is projected onto the axes of OXYZ coordinate

system, where the initial point O coincides with the reduced center, then we

have the conditions of equilibrium of F1 , F2 ,..., Fn  force system by projections,

which are called equilibrium equations



n





n n



F

k1

kX

0

k1

F 0

kY

k1

F 0

kZ

(5.4)





 M F   0  M F   0  M F   0

n n

n



Y k Z k

X k

k1

k1 k1

5.8. Equilibrium conditions for a plane system of

forces

Let F1 , F2 ,..., Fn  is any plane force system and its all forces are in the XOY

plane of any OXYZ coordinate system (Fig.5.9). Therefore, from the six

conditions of equilibrium of force system the three conditions would be

automatically sufficient

 M F   0  M F   0

n n



F

n



kZ 0 X k Y k

k1 k1

k1



The moment of forces into Z-axis will be equal to the scalar sum of moments

into point O. For that system of forces to be equilibrium it is necessary and

sufficient that the three following conditions are occurs



n



F F

n



 M F   0

n



kX

0 0 O k (5.7)

kY

k1 k1 k1

5.9. The constraints and their reactions

Free body: A body, which can move in any direction.

Constraint: 1.Restriction on the motion of a system. 2. Any condition that

reduces the degree of freedom of a system.

For example, for a book lying on the desk the latter is the constraint and for the

block hanged from the rope the rope is the constraint. The force by which the

body acts on the constraint and the reaction by which motion of the body is

restricted are equal in value and opposite in direction. Reactions of constraints

often are called passive forces and the other forces that act on the body are

active. Reactions of constraints are dependent as from active forces that act on

the body and as the motion of the body.

The analysis of the states of motion and equilibrium is based on the following

principle: every body can be considered as a free body if it will be released of

constraints the latter ones being substituted by their reactions.

In Table 5.1 the constraints of kinematic pairs applied on mechanism and

their reactions are given.

 A constraint in the form of perfectly smooth surface (Fig.5.10a,b). The

considered constraint restricts the motion of body only in the direction

of the general normal passed at the point of this contact. Therefore after

releasing the body from the constraint the reaction of that constraint

should be directed along the normal.

 A constraint in the form of no stretchable thread or rope (Fig.5.11). The

thread restricts motion of the body only in its direction; therefore the

reaction of the constraint is directed along the thread out of the body.

 A constraint in the form of a weightless rod hanged by hinges (Fig.5.12).

The reaction of the con-straint is directed along the rod (out of the body

or into it).

 Constraints of beams are used in calculation schemes of beams, bending

rods and suspension girders. In general there are three types of these

constraints.

 A constraint in the form of movable hinge (Fig.5.13a) and A constraint

of b), which provides the body an opportunity to turn about the hinge

and move on the plane of the rollers.

 A constraint in the form of immovable hinge (Fig.5.13a) and B constraint

of b), which differs from the previous one only by the opportunity of

turning the body about the hinge. The reaction force A of the

constraint can have any direction in the plane perpendicular to the

axis of the hinge, that is in the plane XAY, FthereforeFwe denote it as

AX AY

and projections.

 A constraint in the form of rigidly embedded body (Fig.5.19c), which in

F

the point of fixation excludes displacement of the body in any direction.

The reaction of this constraint consists of a force A in the plane of

FAX FAY

unknown direction, which usually is resolved into

MA and

components along the axis and of a couple of forces having

moment.

5.10. The primary concepts of dynamics.

External and internal forces

Dynamics: Branch of theoretical mechanics dealing with the motion and

equilibrium of bodies and mechanical systems under the action of forces.

Here the inertia idea is most important. The same force or the same force

system act on different bodies with different effects. In the same time interval

they go different ways, get different velocities and change their velocity in

different ways.

Inertia: The property of material bodies, which is connected to their

speeds change fast or slowly. As quantities characteristics of the inertia of

material body the mass of body is adopted. In classical mechanics the mass of

bodies is scalar constant quantity, which does not, related to the position and

kinematics of body. The unit of mass (m) in the SI system is kg.

Material point: Material body without size but with some mass. For

example, the body moving rectilinearly can be represented by a material point.

The body the size of which is essentially small in comparison with its trace

also can be represented as material point.

Mechanical system: System in which the main properties are mass,

stiffness and damping.

Mechanical system of material points: A group of material points

where the position and motion of each point is connected with the position

and motion of another points.

The mass of mechanical system is equal to the sum of all material points’



m

n

masses

m k

k1





For mechanical system the mass or inertia center S of mechanical system

is called that point which radius vector rS is determined as



m r

n

rS  k k m

k1





The coordinates of mass center in OXYZ coordinate system are



m X

n



m Y m Z

n n

XS  k k m YS  m ZS  k k m

k k

k1 k1 k1

The forces, which are acting on the system, are external or internal. The

external are called forces that act on points of system from points

irrelevant to this system. The forces of mutual interaction of points of the

same system are called internal forces. The external forces are represented

by e exponent and the internal forces by i, which are equivalent to initial

letters of words external and internal.

The internal forces have the following properties

- their geometrical sum or resultant vector is equal to zero

n



k 1

Fki  0

i

where F is equivalent to applied on k-th point internal forces,

k







-about any point the geometrical sum of moments is equal to zero





  

n

MO Fki  0

k1

5.11. The main rules of dynamics

1. The inertia rule (Galilee, 1638): The released from external

influence material point save its rest or translational motion until the applied

forces compel it to change its state. Here “the released from external influence

material point” is that no forces act on it or applied forces are balancing. This

motion of material point is called the motion inertial.

2. The main rule of dynamics: Acceleration of free material point is

proportional to the applied on it force and has same direction. This rule is

represented with the following equation

m a  F

The proof of this rule is the weight force and gravity g acceleration G  m  g.

3. The rule of action and reaction: Two material points on each

other involve equal force but in opposite direction.

4. The rule of force separate action: If on material point act many

forces, then in the result of that forces is acceleration, which is equal to the

geometric sum of all accelerations which the body will have after acting of



a

each force n

a k

k 1

5.12. Dalamber’s principle

Let us consider the system which contains n points. Let the k-th point with the

mass mk under the action of external forces Fke and internal forces Fki takes

acceleration a k . For this point the equation of motion is

Fke  Fki  m k  a k  0

in which FkI   m k  a k (5.7)

Then we get

Fke  Fki  FkI  0 k=1,2,...,n (5.8)

where vector FkI that is equal to the product of mass of k-th point and its

acceleration and is opposite directed to this acceleration is called inertia force.

For all points of a mechanical system from Eq. (5.8) we get



F F F

n n n



k

e

 k

i

kI 0

or k1 k1 k1





F e  F i  FI  0 (5.9)

where F e , F i and FI are resultant vectors of external and inertia forces

applied on the system.

Dalamber’s principle: If in any instant except internal and external

forces acting on points of the system corresponding inertia forces will be

added conditionally then the system of forces will be in equilibrium.

The property of Dalamber’s rule is applying it in the problems of dynamics

then equations of motion are composed like equilibrium equations of statics.

Besides on points of system only act internal and external forces, but inertial

forces does not influence on the motion of points. They are applied

conditionally that gives us the possibility to solve dynamics problems like

above mentioned statics problems.

If multiply Eq. (5.8) by radius vector rk of k-th point concerning to pole O

and take sum of all products then we get



 r  F    r  F    r  F   0

n n n

e i

k k k k k kI

k1 k1 k1

e

Here the first adding is the moment M 0 , of external forces about point O,

i

second is the moment M0 of internal forces and third is the moment of

inertial forces of system. So we get



M0  M0  MI  0

e i

(5.10)

Therefore, from the Eq. (5.9) and (5.10) we can get this

conclusion: In any time interval of the motion of mechanical

system the sum of resultant vectors of internal and external

forces and conditionally applied inertial forces and the sum of

resultant moments about any point are equal to zero.



Mentioned that the resultant internal force F i and resultant

i

moment M 0 about any point of mechanical system are equal to

zero the Eq. (5.9) and (5.10) are changed like





F  FI  0

e , M0  MI  0

e







As difference with statics equations these equations are called

kinetostatics equations.

5.13. Resultant moment and resultant vector

of inertia forces

From Eq. (5.7) and (5.9) we find out that the inertia force of a solid body is

equal to the inertia force of the center of mass accounting that mass of the

body is centered in that point.

FI   m  aS (5.11)

where a S is acceleration of mass center.

It is clear that axial inertial force in turning motion of a body has the same role

as mass in case of rectilinear motion. It means that the axial inertia moment is

the inertia of body in turning motion. Therefore Eq. (5.11) become



MI   IS  

(5.12)

where IS the inertia moment of body about axis is passed through mass center

and is the angular acceleration of that. Sign (-) in Eq. (5.11) and (5.12) shows

that vectors FI and MI are directed opposite to vectors a S and  .

Let find inertial forces and moments of couple of forces for different cases.

 The body performs translation (Fig.5.14a). On this case   0 and MI  0

and the inertial force is defined from Eq. (5.11).



 The body has a plane of symmetry and rotates about an axis passing

through the centre of mass and which is perpendicular to that plane

(Fig.5.14b). On this case aS  0 and FI  0 and the moment of

inertia couple of forces is defines from Eq. (5.12).



 The body has a plane of symmetry and rotates about an axis not passing

through the centre of mass and which is perpendicular to that plane

(Fig.5.20c) or performs plane parallel motion (Fig.5.14d). On this case

from Eq. (5.11) and (5.12) can be defined direction and magnitude of

inertial couple of forces and moment of inertia of the couple of forces.

We can replace the inertia force FI and the moment of inertia couple of MI

forces by one force applied in any point K (Fig.5.14e). Then take equal to

the moment of inertia couple of forces MI the inertial force FI resolved

into equal FI , FI couple of forces which arm h can be defined from

equation h  MI FI .

5.14. Axial inertia moment of solid body about

a pole and coordinate axes

The inertia moment of body about a given axis is called that

scalar value which is equal to the sum of products of the masses of the

individual particles (elements of mass) of a solid body and the squares of

their distances from a given axis



n

IZ  mk  h2 k (5.13)

k1

From the definition it follows that axial inertial moment always is positive and

the unit in the SI system is kg·m2

From the Hiughans-Shtainer theorem the inertia moment I Z of a body about

any axis can be determined if the mass m of the body and its moment of inertia

IS about parallel axis passed through mass center s are known

(5.14)

IZ  IS  m   2



where  is the distance between the given axis from the s center of mass.