mechanics statics dynamics - vŠb-tuo dynamics mechanics statics dynamics the forces acting on rigid...

1

Dynamics

mechanics

dynamicsstatics

The forces acting on rigid bodies.

The forces acting on moving bodies.The relationshipbetween forces and motion

The mechanics of external forces (the mechanics of rigid bodies).The mechanics of internal forces (the mechanics of flexible bodies).

1. Dynamics of a particle - revision

Jiří PodešvaFaculty of Mechanical Engineering, VŠB – Technical University of OstravaOstrava, Czech Republic

2

Newton’s 1st law – law of inertia

A body stays at rest or at a constant velocity if no force acts upon it.

Newton’s 2nd law – law of force

A force acting upon a body leads to a change of velocity thatis directly proportional to the acting force.

The coefficient of proportionality is the body mass.

Famrr =⋅

mass · acceleration = force

Newton’s 3rd law – law of action and reaction

The actions of two bodies upon each other are always equalin magnitude and opposite in direction.

Isaac Newton (1642-1727)„Philosophiae Naturalis Principia Mathematica” (1687).

1. Dynamics of a particle - revisionDynamics

3

Any two objects exert a gravitational force of attraction upon each other. Themagnitude of the force is proportional to the product of the gravitational massesof the objects, and inversely proportional to the square of the distance between them.

m1 m2

r

Gr

Gr

221

r

mmG

⋅⋅κ=

κκκκ = 6,67·10-11 kg-1·m3·s-2 - gravitational constant,m1 - the mass,m2 - the mass,r - the distance between bodies.

221 sm 819

rm

g −⋅=⋅κ= ,

gmG ⋅=

m1 = 5,98·1024 kg - the Earth’s mass,r = 6 378 km - the Earth’s radius.

On the Earth’s surface then :

The gravitational force is then :

where g is the gravitational acceleration :

Newton’s law of gravitation

Isaac Newton (1642-1727)„Philosophiae Naturalis Principia Mathematica” (1687).


4

The particle - has no dimension, but has a certain mass

The rigid body - has certain dimensions, is rigid, undeformable

The chain of bodies - mechanism - the relative position of one body changes with respect to another


5

dynamics

dynamicskinematics

only motion motion, massesand forces


6

z

y

x

zyx ,,

x

y

222 Ryx =+22 xRy −±=

φ⋅=φ⋅=

cos

sin

Ry

Rx φ

x

The degree of freedom (DOF)the possible, independent motion.

φ

independent coordinate


7

particle body

on a track(1D)

1 DOF

in a plane(2D)

2 DOF(2 translations)

3 DOF(2 translations and 1 rotation)

in space(3D)


6 DOF(3 translations and 3 rotations)


8

the number of DOF decreasesif the motion is restricted by joints

particle body

on a track(1D)

1 DOF

in a plane(2D)



in space(3D)




9

particle body

on a track(1D)

1 DOF

in a plane(2D)



in space(3D)





Dynamics

10

y direction translation

x direction translation

z axis rotationzx

y

particle body

on a track(1D)

1 DOF

in a plane(2D)



in space(3D)





Dynamics

11


particle body

on a track(1D)

1 DOF

in a plane(2D)



in space(3D)




NO vertical translationindependent horizontal translationindependent rotation2 DOF

Dynamics

12

particle body

on a track(1D)

1 DOF

in a plane(2D)



in space(3D)



1. The dynamics of particle - repetition


NO vertical translationwithout sliding in the touch pointhorizontal translation and rotation relate one to the other1 DOFrolling

Dynamics

13


particle body

on a track(1D)

1 DOF

in a plane(2D)



in space(3D)




14

time tunit [s]units [min, hr, ...]

track, route, coordinate s, x, y, ...unit [m]units [cm, km, ...]

velocity vunit [m/s, m·s-1]units [km/hr]

acceleration aunit [m/s2, m·s-2]

1. Dynamics of a particle - revisionthe motion of a particleDynamics

15

sdt

dsv &==

vdt

dva &==

sdt

sda

2

2

&&==

ds

dvva ⋅=

( )ds

vd

2

1a

2

⋅=

the velocity is the first derivative of a track with respect to time

these are the generally validrelationshipsbetween time, track, velocity and acceleration


acceleration is the first derivative of a velocity with respect to time

acceleration is the second derivative of a track with respect to time

acceleration is the first derivative of a velocity with respect to a track, multiplied by velocity

acceleration is one half of the first derivative of a square velocity with respect to a track

the motion of a particleDynamics

16

C) Non-uniform motion – everything changes

sdt

dsv &==

vdt

dva &==

sdt

sda

2

2

&&==

ds

dvva ⋅=

( )ds

vd

2

1a

2

⋅=

A) Uniform motion – velocity is constant

B) Uniformly accelerated motion- acceleration is constant


these are the generally validrelationshipsbetween time, track, velocity and acceleration

With respect to the behaviorof a track, velocity and acceleration over time we can distinguish


17

0dt

dva ==

s

t

s0

t

sv

∆∆= 0sss −=∆

0ttt −=∆tvs ∆⋅=∆

( )00 ttvss −⋅=−

0stvs +⋅=

the velocity is constant, the acceleration is zero

s - the instant tracks0 - the initial track – the initial conditiont - the instant timet0 - the initial time, usually t0=0


A) Uniform motion – velocity is constant

these are the relationships valid onlyfor uniform motion


18


0vtav +⋅=

( ) 200 vssa2v +−⋅⋅=

002

21 stvtas +⋅+⋅⋅=

s

t

s0

v

t

v0

v

s

v0

a

vvt 0−=

B) Uniformly accelerated motion – acceleration is constant

these are the relationships valid onlyfor the uniformly accelerated motion

the initial conditions:s0 - the initial trackv0 - the initial velocity


19

tav ⋅=

The sports car accelerates from zero to v = 100 km/hr (27.8 m/s)in time t = 5 s .

acceleration is then a = 5.6 m/s2.

221 tas ⋅⋅= the track is then s = 70 m .


B) Uniformly accelerated motion – acceleration is constant

(assuming uniformly accelerated motion)


20

φ

v

ry

harmonic motion – the track changes harmonically (same velocity and accel.)

( )0try φ+⋅ω⋅= sin

π⋅ω=

2f

ωπ⋅== 2

f

1T

r

v=ω

r amplitude [m]

frequency [Hz]

circular frequency [s-1]

period [s]

φ phase lead [-]

r

tT

T

y

φ0

ω

number of cycles per second

time of one cycle


C) Non-uniform motion


21

φ

v

ry

( )0try φ+⋅ω⋅= sin

( )0tryv φ+⋅ω⋅ω⋅== cos&

( )ya

trva2

02

⋅ω−=

φ+⋅ω⋅ω⋅−== sin&

r

max. velocity [m/s]ω⋅r2r ω⋅ max. acceleration [m/s2]

t

y

φ0

ω

the oscillation of a particle mass on a flexible link

r

T

T

harmonic motion – the track changes harmonically (same velocity and accel.)



amplitude [m]


22

vga ⋅β−=

vgdt

dv ⋅β−=

dtvg

dv =⋅β−

∫∫ =⋅β−

t

0

v

0

dtvg

dv

( )[ ] t

0

v0 tvg

1 =⋅β−⋅β−

ln

The solution with zero initial conditions :

( ) ( )[ ] tgvg1 =−⋅β−⋅β−

lnln

( )te1g

v ⋅β−−⋅β

=

tvg

11 =

⋅β−⋅

β−ln

tg

vg1 =⋅β−⋅β−

ln

y, v, a



motion within a damping environment

( )tstatesteady e1vv ⋅β−−⋅= _

β= g

v statesteady_


23

vga ⋅β−=

vgdt

dv ⋅β−=

dtvg

dv =⋅β−

( )te1g

v ⋅β−−⋅β

=

v

t

vsteady state T

63% vsteady state 95% vsteady state


t=2·T t=4·T t=3·T t=5·T t=T

tangent

β= 1

T time constant [s]y, v, a


β= g

v statesteady_





24

( )tstatesteady e1v

dt

dyv ⋅β−−⋅== _

( ) dte1vdy tstatesteady ⋅−⋅= ⋅β−

_

( ) ( )∫∫∫ ⋅−⋅=⋅−⋅= ⋅β−⋅β−t

0

tstatesteady

t

0

tstatesteady

y

0

dte1vdte1vdy __

t

0

tstatesteady e

1tvy

⋅β−

−⋅= ⋅β−_

β−+⋅

β−−⋅= ⋅β− 1

e1

tvy tstatesteady_

( )

−⋅β

−⋅= ⋅β− tstatesteady e1

1tvy _

tvy statesteady ⋅= _y

t

y, v, a





25

( ) ( )2

2

22 hR

Rgm

hR

mM

r

mMG

+⋅⋅=

+⋅⋅κ=⋅⋅κ=

κκκκ = 6.67·10-11 kg-1·m3·s-2 - the gravitational constant,M = 5.98·1024 kg - the Earth’s mass,R = 6 378 km - the Earth’s radius.

Gm

Earth R

h

on the Earth’s surface (y=0) :

gmR

mMG

2⋅=⋅⋅κ=

2RgM ⋅=⋅κ

22 81.9

smg

R

M ==⋅κ



motion within a gravitational field


26

Gm

Země R

y

v, a

( )2

2

yhR

Rg

dy

dvva

−+⋅=⋅=

( )2

2

yhR

RgmG

−+⋅⋅=

h

free fall from a height of h

( )∫∫ ⋅−+

⋅=⋅y

02

2v

0

dyyhR

Rgdvv

y

0

2221

yhR

1Rgv

−+⋅⋅=⋅

+−

−+⋅⋅=⋅

hR

1

yhR

1Rgv 22

21





27

Gm

Země R

y

v, a

( )2

2

yhR

Rg

dy

dvva

−+⋅=⋅=

( ) ( ) ( )hRyhR

Ryg2v

2

y +⋅−+⋅⋅⋅=

( ) hg2v hy ⋅⋅≅=Rh <<

( )2

2

yhR

RgmG

−+⋅⋅=

( ) hR

Rhg2v hy +

⋅⋅⋅==

h

the drop velocity :

free fall from a height of h





28

v, a

G

m

Země R

y

( )2

2

yR

Rga

+⋅−=

( )2

2

yR

RgmG

+⋅⋅=

v0

( )2

2

yR

Rg

dy

dvv

+⋅−=⋅

( ) ( )∫∫∫ ⋅+⋅⋅−=⋅+⋅−=⋅ −

y

0

22y

02

2v

0v

dyyRRgdyyR

Rgdvv

[ ] ( ) y

0

2

y

0

12v

0v2

21

yR

1Rg

1

yRRgv

+⋅⋅=

−+⋅⋅−=⋅

−

a vertical throw upward





29

v, aG

m

Země R

y

v0 ( )yR

yRg

R

1

yR

1Rgvv 22

02

21

+−⋅⋅=

−

+⋅⋅=−⋅

yR

yRg2vv 2

0 +⋅⋅⋅−=

20

20

vRg2

Rvh

−⋅⋅⋅=

skm 11Rg2v0 /≅⋅⋅< skm 11Rg2v0 /≅⋅⋅>

( ) Rg2vvv 20y

ystatesteady ⋅⋅−==

∞→lim_( ) 0v hy ==

the particle stops at the height of h the particle draws apart for ever

( )2

2

yR

RgmG

+⋅⋅=

a vertical throw upward





30


the curved trajectorythe motion of a particle

( ) ( ) vvv ttt ∆+=∆+rr

vdt

dv

t

va

0t

&rr==

∆∆=

→∆lim

( )tvr

( )ttv ∆+

r

velvr

∆

•

•

( )tvr

( )ttv ∆+

r

smvr

∆

r r ra a at n= +

dt

dvat =

R

va

2

n =

A(t)

trajectory( )ttr ∆+r

( )trr

rr

∆

O

A(t+∆t)( )tvr ( )tvv ∆+

rta

r

t

nna

r

R – the radius of curvature

Dynamics

31

mF

m = 2 kg

a = 1.5 m/s2

F = 3 N

a

m·a = F


Newton’s 2nd law – the law of force

m – mass [kg]

m a Fi⋅ =∑v r

a – acceleration [m/s2]

F – force [N]

the equation of motion

The relationship between mass, force and motion.

Dynamics

32

m a Fi⋅ =∑v r

α

fm

G

F

N

Ty

xa

TNFGFam i

rrrrrv +++==⋅ ∑TFGFam xix −α⋅−α⋅==⋅ ∑ cossin

fNFGam ⋅−α⋅−α⋅=⋅ cossin

0FGNFam yiy =α⋅−α⋅−==⋅ ∑ sincosay = 0

ax = a

α⋅+α⋅= sincos FGN

( )α⋅+α⋅⋅−α⋅−α⋅=⋅ sincoscossin FGfFGam

( ) ( )α⋅+α⋅−α⋅−α⋅=⋅ sincoscossin fFfGam




Dynamics

33

Jean Le Rond d’Alembert (1717-1783)

d’Alembert principle

amDvr

⋅−=

0DFi

rrr=+∑amD ⋅=

the equationsof equilibrium)

1.

2.

a

mF D

F - D = 0D = m·a

m·a = F


34

Jean Le Rond d’Alembert (1717-1783)

d’Alembert principle

amDvr

⋅−=

0DFi

rrr=+∑amD ⋅=

the equationsof equilibrium)

1.

2.

a

mF D

F - D = 0D = m·a



m a Fi⋅ =∑v r


mF

a

m·a = F

DO NOT MIX !

Dynamics

35

( ) ( )α⋅+α⋅−α⋅−α⋅=⋅ sincoscossin fFfGam

the kinetostatic task the dynamic task

a required motion is given,for example the acceleration a,determine the force F=? needed to reach the required motion

( )α⋅+α

⋅−α⋅−α⋅=sincos

cossin

f

amfGF

the force F is given,determine the motion,the acceleration a=?

( ) ( )m

fFfGa

α⋅+α⋅−α⋅−α⋅= sincoscossin

amD ⋅=the equations of equilibrium - algebraic

sa &&=the differential equations

0Fi =∑

α

m

G

Ff

Na

y

x

T


37

m a F⋅ =r r

Fdt

vdm

rr

=⋅

( )d m v

dtF

⋅=

rr

( )d m v F dt⋅ = ⋅r r

( )d m v m v m v F dtm v

m v t

⋅ = ⋅ − ⋅ = ⋅⋅

⋅

∫ ∫r r r r

r

r

0

1

1 00

r rp m v= ⋅

( )r rI F dtt

t

= ⋅∫0

Ippprrrr =−=∆ 01

the momentum

the force impulse

[kg·m·s-1]

[N·s ≈ kg·m·s-1]0p

r

pr

∆ppp 01

rrr∆+=

the law of the momentum change

if the force is constant : tFI ⋅=rr

01 ppprrr

−=∆the momentum change –the change of amount,the change of direction

p0 - the momentum at the beginning,p1 - the momentum at the end of the event.

The physical quantitiesderived from the equation of motion.

I.The laws of change.

2. The analytical mechanicsDynamics

38

m a F⋅ =r r

( )F

ds

vd

2

1m

2

=⋅⋅

( )F

ds

vmd 221

=⋅⋅

( ) dsFvmd 221 ⋅=⋅⋅

( ) ∫∫ ⋅=⋅⋅−⋅⋅=⋅⋅⋅⋅

⋅⋅ s

202

1212

1

vm

vm

221 dsFvmvmvmd

212

1

202

1

221

K vmE ⋅⋅=

∫ ⋅=s

dsFAr

the kinetic energy

the work

[J ≈ kg·m2·s-2]

[N·m ≈ kg·m2·s-2]

AEEE 0K1KK =−=∆

the law of kinetic energy change

EK0 – the kinetic energy at the beginning,EK1 – the kinetic energy at the end of the event.

if the force is constant(both amount and direction) : sFA

rr⋅=

2. The analytical mechanicsThe physical quantitiesderived from the equation of motion.

II.The laws of change.

Dynamics

39

∫ ⋅=s

dsFAr

δ⋅⋅=⋅= cossFsFArr

Fr

δ sr

NFr

WFr

Fr

δ sr

δ⋅= cosFFW δ⋅= sinFFN

working component

sFsFA P ⋅δ⋅=⋅= cos

°>δ 90

°<δ 90

positive work – work done

negative work – work consumed

0=δ

°=δ 90

10 =cos

090 =°cos zero work – not-done

( ) 090 <°>δcos

°=δ 180 1180 −=°cos

the workthe work is a scalar product of force and track,the angle between them must be taken into account :

0sFA >⋅=→0sFA >δ⋅⋅=→ cos

0A =→0sFA <δ⋅⋅=→ cos

sFA ⋅−=→

the scalar product

not-working component

2. The analytical mechanicsThe physical quantitiesderived from the equation of motion.

II.

Dynamics

40

PdA

dt

F ds

dtF v= = ⋅ = ⋅

r rr r

the power

[N·m·s-1 ≈ W]

∫ ⋅=s

dsFAr

[N·m ≈ kg·m2·s-2 ≈ J]

δ

δ

WFr

NFr

Fr

Fr

vr

vr

δ⋅⋅=⋅= cosvFvFPrr

δ⋅= cosFFW δ⋅= sinFFN

vFvFP W ⋅δ⋅=⋅= cos

The physical quantities,derived from the equation of motion.

II.

2. The analytical mechanics

the work

the velocity

Dynamics

41

the potential energyAsdFEs

P =⋅= ∫rr

hgmdygmdygmdyFAh

0

h

0

h

0

⋅⋅=⋅⋅=⋅⋅=⋅= ∫∫∫

gmGF ⋅==y 2 31

0EP =the zero potential energy level – the choice

hgmEP ⋅⋅=

G

F=Gm

G

F=Gm

G

F=Gm


the potential energy - positional

Dynamics

42

G

F=Gm

Earth R

y

AsdFEs

P =⋅= ∫rr

the potential energy

κκκκ = 6,67·10-11 kg-1·m3·s-2 - gravitational constant,M = 5,98·1024 kg - the Earth’s mass,R = 6 378 km - the Earth’s radius,r - the total distance from

the Earth’s centre,y - the height above

the Earth’s surface.

the force F=Gvaries with height

( ) ( )2

2

22 yR

Rgm

yR

mM

r

mMG

+⋅⋅=

+⋅⋅κ=⋅⋅κ=

on the Earth’s surface :

22

RgM gmR

mMG ⋅=⋅κ⇒⋅=⋅⋅κ=

( )∫ ⋅=h

0

y dyFA

g

0EP =


43

R

y

( ) ( )∫∫ ⋅+

⋅⋅⋅=⋅=h

02

2h

0

y dyyR

1RgmdyFA

( ) hR

Rhgm

hRR

hRgmA 2

+⋅⋅⋅=

+⋅⋅⋅⋅=

0EP =

AsdFEs

P =⋅= ∫rr

+

−⋅⋅⋅=

+−⋅⋅⋅=

hR

1

R

1Rgm

yR

1RgmA 2

h

0

2

E m g hR

R hP = ⋅ ⋅ ⋅+

h«R 1hR

R ≅+

hgmEP ⋅⋅≅

AEP =the potential energy is equal to work :

for a small height above the Earth’s surface is approx. :

G

F=Gm

Earth


the force F=Gvaries with height

the potential energy - positional


44

AsdFEs

P =⋅= ∫rr

y

F

F = k·y k – the stiffness

JE3

Fy

3

⋅⋅⋅= l llll – the beam length,

E – the Young modulusJ – the quadratic

moment of inertia

3

JE3k

l

⋅⋅=

The force F acts on the fixed beam, the beam deformation is y.



yFykdyykdyFA 212

21

y

0

y

0

⋅⋅=⋅⋅=⋅⋅=⋅= ∫∫

yFykE 212

21

P ⋅⋅=⋅⋅=AEP = the potential energy - deformation

the potential energy is equal to work :

Dynamics

45

The law of the total mechanical energy conservation

constant=+= PKT EEE

m

h

v0 = 0

EK0 = 0

EP0 = m·g·h

EP1 = 0

EK1 = ½·m·v12

1P1K0P0K EEEE +=+

0vmhgm0 212

1 +⋅⋅=⋅⋅+

hg2v1 ⋅⋅=

v1 ≠ 0

0EP =

The total mechanical energy conserves.

the zero potential energy level


constant=+= PKT EEE

the conservative system

Dynamics

46(the forces which do not create potential energy)

α

v h

s

m

G

F

T N

konst≠+= PKT EEEv=?

The law of the total mechanical energy change


the non-conservative system

EP1 = m·g·hEK1 = ½·m·v1

2

EP0 = 0

EK0 = ½·m·v02

AEE 0T1T +=

sTsFvm0vmhgm 202

1212

1 ⋅−⋅α⋅+⋅⋅+=⋅⋅+⋅⋅ cos

hgmsTsFvm0vm 202

1212

1 ⋅⋅−⋅−⋅α⋅+⋅⋅+=⋅⋅ cos

m

hgmsTsFvmv

21

202

1

1 ⋅⋅⋅−⋅−⋅α⋅+⋅⋅= cos

ET1 ET0 A

α⋅= sinsh

The change of the total mechanical energy is equal to the work of non-conservative forces.

α⋅+α⋅= sincos FGN

NfT ⋅=

Dynamics

47

α

v h

s

m

G

F

T N

m

h


The law of the total mechanical energy conservation / change

the way of the dynamic solution,

based on the total mechanical energy analysis,

is called “energy balance solution”

Dynamics

48

the virtual work principle ... in statics

F - given

determineR = ?

F

R

A

B

A

B

0M

0F

0F

i

iy

ix

=

=

=

∑∑∑

_

_

NA

NB

the

free

-bod

y di

agra

m


the virtual work principle is the alternative to the equations of equilibrium

Dynamics

49

F

R

A

B

δA

δB

BA RFA δ⋅−δ⋅= the virtual work

AEK =∆ the law of the kinetic energy change

0EK =∆

⇓0RFA BA =δ⋅−δ⋅=

Pt

A =∆

AA vt

=∆δ

BB vt

=∆δ

the power

0vRvFP BA =⋅−⋅=

the motion status does not change



the virtual motion

the virtual work principle

(the virtual power principle)

Dynamics

50

F

R

A

B

0vRvF BA =⋅−⋅

vA

vB

π

ω

π⋅ω⋅=π⋅ω⋅ BRAF

π⋅=π⋅ BRAF

ππ⋅=

B

AFR F

R

A

B

0M i =∑ π_NA

NB

π

B

A

v

vFR ⋅=



π=

π=ω

B

v

A

v BA

the kinematical method in statics

Dynamics

51



F - given

determine R = ?

Dynamics

52

9 equations of equilibrium

9 unknown forces



the free-body diagramF - given

Dynamics

53

R

Fth

e vi

rtua

l mot

ion



δF

δR

Dynamics

54

R

x

y

vx

vy

llll

222 yx llll=+0yy2xx2 =⋅⋅+⋅⋅ &&

0vyvx yx =⋅+⋅

0vFvR yx =⋅−⋅−

y

x

v

v

x

y −=

x

y

v

vFR ⋅−=

F

φ=⋅=

tan

F

y

xFR

φ



55

α⋅δ⋅=δ⋅= cosFFArr F

r

δr

α



56

A C

B



F - given

determine R = ?

Dynamics

57

F

RA C

Bthe virtual motion



58

F

RA C

BvB

vC

π

ωF

RvB

vCα

α

0vRvF CB =⋅+⋅ rrrr

0180vRvF CB =°⋅⋅+α⋅⋅ coscos

0vRvF CB =⋅−α⋅⋅ cos

π=

π=ω

C

v

B

v CB

ππ⋅=

B

Cvv BC

ππ⋅=α⋅⋅

B

CvRvF BB cos

ππ⋅α⋅=

C

BFR cos

ABAC

B −α

=πcos

α⋅=π tanACC



59

“the work” of the d’Alembert forces will be taken into account

Gy, v, a

φ, ω, ε

D

MD

S≡T

r

amD ⋅=ε⋅= SD IM

r

v=ωr

a=ε

0MvDvG D =ω⋅−⋅−⋅

0r

vMvDvG D =⋅−⋅−⋅

0r

1MDG D =⋅−−

Gr

1

r

aIam S =⋅⋅+⋅

Gar

Im

2S =⋅

+

the virtual work principle ... in dynamics


60

ii

P

i

K

i

K Qq

E

q

E

v

E

dt

d =∂∂+

∂∂−

∂∂

qi ... the generalized coordinate

a system with n degrees of freedom

vi ... the generalized velocity ii qv &=

i = 1...n

t ... time

EK ... the kinetic energy

EP ... the potential energy

Qi ... the generalized force ∑ ∂∂

⋅=j i

jji q

rFQ

Fj – the real forces

j = 1...m

rj – the radius vectorof the point of applicationof the force Fj

the Lagrange equations of the 2nd kind


– an independent coordinate determining the position of the system

Dynamics

61

G

F

y

φ

e

llll φ

π

ωω, ε

vT

φ=ω &

the generalized coordinate

φ⋅= sinlllly

the generalized velocity

2T2

12T2

1K IvmE ω⋅⋅+⋅⋅=

π⋅ω= TvT

T( )φ−°⋅⋅⋅−+=π 90ye2yeT 222 cos

φ⋅⋅⋅−φ⋅+=π 22222 e2eT sinsin llllllll

( )( )[ ] 2T

2221

K Ie2emE ω⋅+φ⋅⋅−⋅+⋅⋅= sinllllllll ( )( )[ ] ω⋅+φ⋅⋅−⋅+⋅=ω T

22K Ie2emd

dEsinllllllll

( )( )[ ]( )[ ] ω⋅φ⋅φ⋅φ⋅⋅−⋅⋅⋅+

+ε⋅+φ⋅⋅−⋅+⋅=

ω

&cossin

sin

e22m

Ie2emd

dE

dt

dT

22K

llllllll

llllllll

( )( )[ ]( )[ ] 2

T22K

e22m

Ie2emd

dE

dt

d

ω⋅φ⋅φ⋅⋅−⋅⋅⋅+

+ε⋅+φ⋅⋅−⋅+⋅=

ω

cossin

sin

llllllll

llllllll

( ) 2K e2md

dE ω⋅φ⋅φ⋅⋅−⋅⋅=φ

cossinllllllll

φ⋅⋅⋅= sinegmEP φ⋅⋅⋅=φ

cosegmd

dEP

φ⋅⋅=φ

⋅= cosllllFd

dyFQ



62

G

F

y

φ

e

llll φ

π

ωω, ε

vT

φ=ω &

φ⋅= sinlllly π⋅ω= TvT

T( )φ−°⋅⋅⋅−+=π 90ye2yeT 222 cos


( )( )[ ] ( )[ ] φ⋅⋅=φ⋅⋅⋅+ω⋅φ⋅φ⋅⋅−⋅⋅+ε⋅+φ⋅⋅−⋅+⋅ coscoscossinsin llllllllllllllllllll Fegme2mIe2em 2T

22

( )( )[ ] ( )[ ] ( ) φ⋅⋅⋅−⋅=ω⋅φ⋅φ⋅⋅−⋅⋅+ε⋅+φ⋅⋅−⋅+⋅ coscossinsin egmFe2mIe2em 2T

22llllllllllllllllllll

φ−

ω d

dE

d

dE

dt

d KK

φd

dEP

φ⋅=d

dyFQ





63

G

F

y

φ

e

llll φ

π

ωω, ε

vT

φ=ω &

φ⋅= sinlllly π⋅ω= TvT

T( )φ−°⋅⋅⋅−+=π 90ye2yeT 222 cos


( )( )[ ] ( )[ ] ( ) φ⋅⋅⋅−⋅=ω⋅φ⋅φ⋅⋅−⋅⋅+ε⋅+φ⋅⋅−⋅+⋅ coscossinsin egmFe2mIe2em 2T

22llllllllllllllllllll

( )( )[ ] ( )[ ] ( ) φ⋅⋅⋅−⋅=

φ⋅φ⋅φ⋅⋅−⋅⋅+φ⋅+φ⋅⋅−⋅+⋅ coscossinsin egmFdt

de2m

dt

dIe2em

2

2

2

T22

llllllllllllllllllll

2

2

dt

d φ=ε





φω⋅ω=ε

d

d

( )( )[ ] ( )[ ] ( ) φ⋅⋅⋅−⋅=ω⋅φ⋅φ⋅⋅−⋅⋅+φω⋅ω⋅+φ⋅⋅−⋅+⋅ coscossinsin egmFe2m

d

dIe2em 2

T22

llllllllllllllllllll

Dynamics

64

G

F

y

φ

llll

ωω, ε T

the prismatic rod

llll⋅= 21e 0e2 =⋅−llll

( )[ ] ( ) φ⋅⋅⋅⋅−⋅=ε⋅+⋅⋅ cosllllllllllll 21

T2

21 gmFIm

2121

T mI llll⋅⋅=

φ⋅

−⋅=ε cos2

g

m

F3

llll

φ⋅α=ε cos

φ⋅α=φ cos&&φ⋅α=φω⋅ω cos

d

d

∫∫φω

φ⋅φ⋅α=ω⋅ω00

dd cos

φ⋅α=ω⋅ sin221

e=llll/2

α

( ) φ⋅α⋅=ω φ sin202

g

m

F3 <>

−⋅=αllll

gmF 21 ⋅⋅<>



65

z, v

φ, ω

r

v=ω

G

αααα

r

S ≡≡≡≡ T F

b

x = b-z·cosα

EP = 0

22

2212

21

K vr

Im

2

1IvmE ⋅

+⋅=ω⋅⋅+⋅⋅=

vr

Im

dv

dE2

K ⋅

+=

ar

Imv

r

Im

dv

dE

dt

d22

K ⋅

+=⋅

+=

&

0dz

dEK =( )α⋅−⋅⋅= sinzhgmEP

h

α⋅⋅−= singmdz

dEP

α⋅−=⋅= cosFdz

dxFQi

i

P

i

K

i

K Qq

E

q

E

v

E

dt

d =∂∂+

∂∂−

∂∂

α⋅−α⋅⋅=⋅

+ cossin Fgmar

Im

2

rolling without sliding

h = b·tgα+ r·tg2α+ r·cosα



67

F

F

∆llll

FdG

nD84

3

⋅⋅

⋅⋅=∆llll

llll∆⋅⋅⋅

⋅=nD8

dGF

3

4φd

φD

here : G – the shear modulus [Pa, MPa],(the material property),

d – the wire diameter [m, mm],D – the spring diameter [m, mm],n – the number of spring screws [-].

the stiffness

llll∆⋅⋅⋅

⋅=nD8

dGF

3

4

k – the stiffness[N/m, N/mm]

nD8

dGk

3

4

⋅⋅⋅=

k

F=∆llll llll∆⋅= kF

k–

the

sprin

g st

iffne

ssk

3. Vibrations with 1 degree of freedomDynamics

68

k

F

F

FSllll∆⋅= kFS

The spring force– the reaction,the response of the spring to the deformation.

∆llll

k–

the

sprin

g st

iffne

ss

the stiffness

llll∆⋅= kF


69

k

F

F

FSy

F(y)=k·y

potential energy (deformation energy)

ykF ⋅=

( )2

21

00

y kdyykdyFA llll

llllllll

∆⋅⋅=⋅⋅=⋅= ∫∫∆∆

llllllll ∆⋅⋅=∆⋅⋅== FkAE 212

21

P

llllllll ∆⋅⋅=∆⋅⋅= FkA 212

21

l∆⋅= kF


70

The free vibration– no external force as the reason for vibration.I.

The forced vibration- caused by an external force.

types of vibration


71

Undamped vibrationno physical phenomenon, which will decrease the vibration, is present.The vibration will last forever.

II. Damped vibrationdue to some physical phenomenon,the vibration will decreaseuntil it vanishes

a v

isco

us

liqu

id

types of vibration


72

free forced

damped

undampedvibration

types of vibration


73

v, ax

m

FS=k·x

S

i

Fam

Fam

−=⋅

=⋅ ∑

20m

k Ω=

k

0xkxm =⋅+⋅ &&m

the free undamped vibration

0xx 20 =⋅Ω+&&

( )

( )

( )t2

t

tt

tt

ex

ex

ex

⋅λ

⋅λ

⋅λ

⋅λ=

⋅λ=

=

&&

&

002

0

20

2

t20

t2

i1

0

0ee

Ω⋅=Ω⋅−=Ω−=λ

=Ω+λ

=⋅Ω+⋅λ ⋅λ⋅λ

( ) ( )00ti

t tCeCx 0 φ+⋅Ω⋅=⋅= ⋅Ω⋅ sin


74

v, ax

m

FS=k·x

S

i

Fam

Fam

−=⋅

=⋅ ∑

( )( )

( )002

0

000

00

tCxa

tCxv

tCx

φ+⋅Ω⋅Ω⋅−==

φ+⋅Ω⋅Ω⋅==φ+⋅Ω⋅=

sin

cos

sin

&&

&

k

0xkxm =⋅+⋅ &&m

( )[ ] ( )[ ] 0tCktCm 00002

0 =φ+⋅Ω⋅⋅+φ+⋅Ω⋅Ω⋅−⋅ sinsin


20m

k Ω=

( ) ( )[ ] 0tCktm

kCm 0000 =φ+⋅Ω⋅⋅+

φ+⋅Ω⋅⋅−⋅ sinsin


75

x v, a

x

t

T T

T C

C∆t = φ0/Ω0

( )00 tCx φ+⋅Ω⋅= sin

m

k0 =Ω

π⋅Ω=2

f 0

0

2

f

1T

Ωπ⋅==

the natural circular frequency [s-1]

the natural frequency [Hz]the number of cyclesper second

the period [s]the time of one cycle

The integration constants :C the amplitude [m]φφφφ0 the phase lead [rad]

result from the initial conditions :t=0 ... x=x0 – the initial displacement,

v=v0 – the initial velocity.

( )( )000

00

Cv

Cx

φ⋅Ω⋅=φ⋅=cos

sin

20

202

0

vxC

Ω+=

0

000 v

x Ω⋅=φ arctan

Parameters, resulting from substitution :k

( )( )000

00

tCxv

tCx

φ+⋅Ω⋅Ω⋅==φ+⋅Ω⋅=

cos

sin

&

mthe free undamped vibration

Two groups of parameters :


76

x v, a( )00 tCx φ+⋅Ω⋅= sin

m

k0 =Ω

π⋅Ω=2

f 0

0

2

f

1T

Ωπ⋅==

the natural circular frequency [s-1]

the natural frequency [Hz]the number of cyclesper second

the period [s]the time of one cycle

The integration constants :C the amplitude [m]φφφφ0 the phase lead [rad]

result from the initial conditions :t=0 ... x=x0 – the initial displacement,

v=v0 – the initial velocity.

( )( )000

00

Cv

Cx

φ⋅Ω⋅=φ⋅=cos

sin

20

202

0

vxC

Ω+=

0

000 v

x Ω⋅=φ arctan

Parameters, resulting from substitution :k m


the integration constants are :

the alternative mathematic description :( ) ( ) ( )tBtAtCx 0000 ⋅Ω⋅+⋅Ω⋅=φ+⋅Ω⋅= sincossin

where : 0CA φ⋅= sin 0CB φ⋅= cos

( ) ( )tBtAvx 0000 ⋅Ω⋅Ω⋅+⋅Ω⋅Ω⋅−== cossin&

Ax0 = 00 Bv Ω⋅=

( )( ) ( ) ( ) ( )tCtC

tC

0000

00

⋅Ω⋅φ⋅+⋅Ω⋅φ⋅==φ+⋅Ω⋅

sincoscossin

sin

t=0 ... x=x0 – the initial displacement,v=v0 – the initial velocity.

0xA =0

0vB

Ω=

and finally :22 BAC +=

B

A0 arctan=φ

Two groups of parameters :


77

x v, a

x

t

T T

T C

C∆t = φ0/Ω0

( )00 tCx φ+⋅Ω⋅= sin

A note about the arctanfunction :The arctanfunction always has 2 roots.

Ex. : arctan(0,5) = 26,6ºbut also : arctan(0,5) = 206,6º

Or : arctan(-1) = -45ºbut also : arctan(-1) = 135º

( ) ( ) ( )tBtAtCx 0000 ⋅Ω⋅+⋅Ω⋅=φ+⋅Ω⋅= sincossin

0CA φ⋅= sin 0CB φ⋅= cos

0xA =0

0vB

Ω=

22 BAC +=B

A0 arctan=φ

k mthe free undamped vibration

the alternative mathematic description :

where :

and finally :A>0

B<0

A<0

B>0φ0 ∈⟨90º,180º⟩

φ0A

B

C

⟨0,90º⟩

⟨180º,270º⟩ ⟨270º,360º⟩


78

r

m

φ

ω,ε

atan

G

ε⋅= rat2

n ra ω⋅= φ=ω &

S


the mathematical pendulum :

3. Vibrations with 1 degree of freedom

The mathematical pendulum(the idealization of a real pendulum).The particle of the mass mon the weightless wire of the length r.

φ⋅−==⋅ ∑ sin_ GFam itt φ⋅−==⋅ ∑ cos_ GSFam inn

φ⋅⋅−=ε⋅⋅ singmrm

0gr

0gr

=φ⋅+φ⋅

=φ⋅+ε⋅

sin

sin&&

2rmGS ω⋅⋅+φ⋅= cos

φ=ε &&

φφφφ [º]

1º5º

15º

φφφφ [rad]

0,0174530,0872660,261799

sin φφφφ

0,0174520,0871560,258819

error

0,005 %0,13 %1,15 %

0gr =φ⋅+φ⋅ &&

r·sinφφ≅φ)

sin

Dynamics

79

r

m

φ

ω,ε

atan

G

S


the mathematical pendulum :


The mathematical pendulum(the idealization of a real pendulum).The particle of the mass mon the weightless wire of the length r.

0gr =φ⋅+φ⋅ &&

0xkxm =⋅+⋅ && 0gr =φ⋅+φ⋅ &&

the analogical solution :

( )0tCx γ+⋅Ω⋅= sin ( )0tC γ+⋅Ω⋅=φ sin

m

k=Ω

2

202

0

vxC

Ω+=

0

00 v

x Ω⋅=γ arctan

the initial conditions :t = 0 ...φφφφ = φφφφ0 the initial angleωωωω = ωωωω0 the initial

angular velocity

r

g=Ω

2

202

0CΩω+φ=

0

00 ω

Ω⋅φ=γ arctan

the circular frequency

the amplitude

the phase lead

Dynamics

80

k

x v, a

bFD=b·v

FS=k·x

DS

i

FFam

Fam

−−=⋅

=⋅ ∑

20m

k Ω= δ=⋅m2

b

0xkxbxm =⋅+⋅+⋅ &&&

Substitution :

m

the free damped vibration


0xx2x 20 =⋅Ω+⋅δ⋅+ &&&

( )

( )

( )t2

t

tt

tt

ex

ex

ex

⋅λ

⋅λ

⋅λ

⋅λ=

⋅λ=

=

&&

&

220

20

2

20

2

t20

tt2

i

02

0ee2e

δ−Ω⋅±δ−=Ω−δ±δ−=λ

=Ω+λ⋅δ⋅+λ

=⋅Ω+⋅λ⋅δ⋅+⋅λ ⋅λ⋅λ⋅λ

( )( ) ( )0

ttit teCeCx φ+⋅Ω⋅⋅=⋅= ⋅δ−⋅δ−Ω⋅ sin

220 δ−Ω=Ω

k – stiffnessb – damping coefficient

Dynamics

81

k

x v, a

bFD=b·v

FS=k·x

DS

i

FFam

Fam

−−=⋅

=⋅ ∑

20m

k Ω= δ=⋅m2

b

0xkxbxm =⋅+⋅+⋅ &&&

Substitution :

m



( )0t teCx φ+⋅Ω⋅⋅= ⋅δ− sin

m

k0 =Ω the natural circular frequency of undamped vibration [s-1]

(not present directly in the solution)

m2

b

⋅=δ the decay constant[s-1]

220 δ−Ω=Ω

π⋅Ω=

2f the natural frequency[Hz] number of cycles per second

Ωπ⋅== 2

f

1T the period[s] time of one cycle

the natural circular frequency of damped vibration[s-1]

0

0

0

Ω>δΩ<δΩ=δ critical damping

sub-critical damping

super-critical damping

Dynamics

82

k

x v, a

bFD=b·v

FS=k·x

DS

i

FFam

Fam

−−=⋅

=⋅ ∑

20m

k Ω= δ=⋅m2

b

0xkxbxm =⋅+⋅+⋅ &&&

Substitution :

m



( )( ) ( )[ ]00

t

0t

tteCxv

teCx

φ+⋅Ω⋅δ−φ+⋅Ω⋅Ω⋅⋅==

φ+⋅Ω⋅⋅=⋅δ−

⋅δ−

sincos

sin

&

Finally C and φφφφ0 are integration constants, determined from initial conditions:

( ) ( )[ ]( ) ( ) ( ) ( )[ ]tABtABexv

tBtAext

t

⋅Ω⋅Ω⋅+δ⋅−⋅Ω⋅δ⋅−Ω⋅⋅==⋅Ω⋅+⋅Ω⋅⋅=

⋅δ−

⋅δ−

sincos

sincos

&

t=0 ... x=x0 – the initial displacement, v=v0 – the initial velocity.

0xA =Ω

δ⋅+= 00 xvB 22 BAC +=

B

A0 arctan=φ

Ax0 =Or alternatively :

( )00 Cx φ⋅= sin ( ) ( )[ ]000 Cv φ⋅δ−φ⋅Ω⋅= sincos

δ⋅−Ω⋅= ABv0

Dynamics

83

k

x v, a

bFD=b·v

FS=k·x

DS

i

FFam

Fam

−−=⋅

=⋅ ∑

20m

k Ω= δ=⋅m2

b

0xkxbxm =⋅+⋅+⋅ &&&

Substitution :

m



( )0t teCx φ+⋅Ω⋅⋅= ⋅δ− sin

t

T

∆t = φ0/Ω

teC ⋅δ−⋅x the period

Dynamics

84

k

x v, a

bFD=b·v

FS=k·x

DS

i

FFam

Fam

−−=⋅

=⋅ ∑

20m

k Ω= δ=⋅m2

b

0xkxbxm =⋅+⋅+⋅ &&&

Substitution :

m



teCx ⋅δ−⋅=

t

xteC ⋅δ−⋅

δ=τ 1

C

τ - the time constant

tangent

t=τx=37% C

t=τ C370eCeCeC 1 ⋅=⋅=⋅=⋅ −δδ−τ⋅δ− ,

t=3·τ

t=5·τ

C050eCeCeC 33

3 ⋅=⋅=⋅=⋅ −δδ−τ⋅δ− ,

C0070eCeCeC 55

5 ⋅=⋅=⋅=⋅ −δδ−τ⋅δ− ,

t=3·τ t=5·τx=0,7% Cx=5% C

Dynamics

85

k

x v, a

bFD=b·v

FS=k·x

DS

i

FFam

Fam

−−=⋅

=⋅ ∑

20m

k Ω= δ=⋅m2

b

0xkxbxm =⋅+⋅+⋅ &&&

Substitution :

m



teCx ⋅δ−⋅=

t

xteC ⋅δ−⋅

δ=τ 1

C

τ - the time constant

tangent

t=τx=37% C

t=3·τ t=5·τx=0,7% Cx=5% C

05 10 15 20

1

t0

δ=0,8; τ=1,25

δ=0,1; τ=10 small damping – slow decay,large damping – quick decay.

C370eCeCeC 1 ⋅=⋅=⋅=⋅ −δδ−τ⋅δ− ,

C050eCeCeC 33

3 ⋅=⋅=⋅=⋅ −δδ−τ⋅δ− ,

C0070eCeCeC 55

5 ⋅=⋅=⋅=⋅ −δδ−τ⋅δ− ,

Dynamics

86

∆llll

k1 k2

F

F

FS1 FS2

llll∆⋅= 11S kF

llll∆⋅= 22S kF

spring assembly


the parallel assembly

2S1S FFF +=

llllllll ∆⋅+∆⋅= 21 kkF

( ) llll∆⋅+= 21 kkF

llll∆⋅= TkF

21T kkk +=

F

kT

llll∆⋅= TkF

The total stiffness is the sum of partial stiffnesses.

the parallel assemblythe deformation ∆llll is common for both springs, the spring forces FS1 and FS2 are summed

Dynamics

87

k1

k2

F

FS1

FS2

F

llll0+∆llll1

llll0+∆llll2

∆llllT=∆llll1+∆llll2FS2

spring assembly


the serial assembly

llll∆⋅= TkF

F

kT

111S kF llll∆⋅= 222S kF llll∆⋅=

1

1S1 k

F=∆llll2

2S2 k

F=∆llll

FFF 2S1S ==

2

2S

1

1S

T21T k

F

k

F

k

F +==∆+∆=∆ llllllllllll

21T k

1

k

1

k

1 +=

The reciprocal total stiffness is the sum of reciprocal partial stiffnesses.

the serial assemblythe deformations ∆llll1 and ∆llll2 are summed, the spring forces FS1 a FS2 are equal

21

21

21

T kk

kk

k1

k1

1k

+⋅=

+=

Dynamics

88

k1

k2

F ∆llll

F

FS1

FS2

llll∆⋅= 11S kF

llll∆⋅= 22S kF

2S1S FFF +=

spring assembly


the parallel assembly !

llllllll ∆⋅+∆⋅= 21 kkF

( ) llll∆⋅+= 21 kkF

llll∆⋅= TkF

21T kkk +=

F

kT

llll∆⋅= TkF

The total stiffness is the sum of partial stiffnesses.

the parallel assemblydeformation ∆llll is common for both springs, the spring forces FS1 and FS2 are summed

Dynamics

89

JE3

Fy

3

⋅⋅⋅= l

m m

kbending

R

F

ykRF bending⋅==

3bending

JE3k

l

⋅⋅=

bending vibration


E – the Young modulus[Pa, MPa] (material)

J – the square moment of inertia[m4, mm4] (profile)

llll – the beam length [m, mm]

kbending – the bending stiffness[N/m, N/mm]

Dynamics

90

F

R

m

JE48

Fy

3

⋅⋅⋅= l

y

m

ykRF bending⋅==

3bending

JE48k

l

⋅⋅=

bending vibration


91

k

x

bFD=b·v

FS=k·xF

v, a

constant external force : F = konst,harmonically changing external force : F = Fa·sin(ωωωω·t)

tF

∑=⋅ iFam

Fxkxbxm =⋅+⋅+⋅ &&&

Fvbxkam +⋅−⋅−=⋅FFFam BD +−−=⋅

the forced vibration


m

Dynamics

92

k

G

G

FS dyny

llll0ll ll

0–

the

free

leng

th

∆llllstatFS stat

m

m

the static deformationthe equilibrium position

∑=⋅ iFam

totalSFGam _−=⋅

totalkGam llll∆⋅−=⋅( )ykGam stat +∆⋅−=⋅ llll

statkGykym llll∆⋅−=⋅+⋅ &&

0ykym =⋅+⋅ &&

ykkGam stat ⋅−∆⋅−=⋅ llll

FS dynFS stat

= 0

statstatD kFG llll∆⋅== _

∆llllstat

the vibration under the constant force


k

Gstat =∆llll

statstatS kFG llll∆⋅== _

dynSstatStotalS FFF ___ +=

( )tstattotalS ykkF ⋅+∆⋅= llll_

( )ttotalS ykGF ⋅+=_

Dynamics

93

k

x

b

v, a

F = Fa·sin(ω·t)

tF Fa

T T

Fa – the amplitude [N]

ωωωω – the circular frequencyof the exciting force [s-1]

f – the frequency [Hz]

π⋅ω=

2f

T – the period [s]

ωπ⋅== 2

f

1T

ΩΩΩΩ0 – the natural circularfrequency [s-1]

f – the natural frequency [Hz]

π⋅Ω=2

f 0

T - the period [s]

0

2

f

1T

Ωπ⋅==

m

k0 =Ω

number of changesof the exciting forcefrom positive to negativeand back per second

the number of cycles per second

theseparameters

mustnotbe

confused

the timeof one cycle

the timeof one change

m

harmonically changing exciting force


the parametersof the exciting force

the parametersof natural (free) vibration

Dynamics

94

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T

( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&

m



F = Fa·sin(ω·t)

Fa – the amplitude

ωωωω – the circular frequencyof the exciting force

the homogenous solution ( )0

thom teCx φ+⋅Ω⋅⋅= ⋅δ− sin

0xkxbxm =⋅+⋅+⋅ &&&

220

0

m2

bm

k

δ−Ω=Ω

⋅=δ

=Ω

the integration constants C a φφφφ0 results from the initial conditions

the homogenous solution

t

x

Ωπ⋅= 2

T

ΩΩΩΩ – the natural circular frequency

the solution is the assembly of a “homogenous solution”and a “particular solution”

Dynamics

95

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)



the particular solution( )φ−⋅ω⋅= txx apart sin


the amplitude xaand phase lead φφφφof the particular solutionwill be discussed later

the particular solutiont

x

xa ωπ⋅= 2

T


Dynamics

96

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)



the particular solution( )φ−⋅ω⋅= txx apart sin


the homogenous solution ( )0

thom teCx φ+⋅Ω⋅⋅= ⋅δ− sin

0xkxbxm =⋅+⋅+⋅ &&&

( ) ( ) ( )φ−⋅ω⋅+φ+⋅Ω⋅⋅=+= ⋅δ− txteCxxx a0t

parthomt sinsin

the steady state

the homogenous solutionthe total solution

the particular solutiont

x

the transient process

Dynamics

97

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)



( ) ( )( ) ( )( ) ( )φ−⋅ω⋅ω⋅−=

φ−⋅ω⋅ω⋅=

φ−⋅ω⋅=

txx

txx

txx

2at

at

at

sin

cos

sin

&&

&

( ) ( )22220

aa

2

1

m

Fx

ω⋅δ⋅+ω−Ω⋅=

220

2

ω−Ωω⋅δ⋅=φ arctan π=°∈φ ,, 01800

= ?= ?

Dynamics

98

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)



( ) ( )φ−⋅ω⋅= txx at sin

xa

t

F(t) x(t)ωφ=∆t

ωπ⋅= 2

T

F, x

Fa ωφ=∆t

[s]

[rad]

the exciting force

the response

( ) ( )22220

aa

2

1

m

Fx


220

2

ω−Ωω⋅δ⋅=φ arctan

the amplitudexa – maximum displacement,phase delayφφφφ - represents the time delay of the response x(t) with respect to the exciting force F

Dynamics

99

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)




( ) ( )22220

aa

2

1

m

Fx


220

2


( ) ( )222

aa

21

1

k

Fx

η⋅ξ⋅+η−⋅=

21

2

η−η⋅ξ⋅=φ arctan

stata x

k

F =

the static deformation

0Ωδ=ξ

0Ωω=η the tuning

the damping ratio0Ω⋅η=ω 0Ω⋅ξ=δ

m

k20 =Ω

Dynamics

100

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)




( ) ( )22220

aa

2

1

m

Fx


220

2


( ) ( )222

aa

21

1

k

Fx

η⋅ξ⋅+η−⋅=

21

2


stata x

k

F =


2a

220

aa

1

1

k

F1

m

Fx

η−⋅=

ω−Ω⋅=

°=π=φ 180

0=φ 0Ω<ω

0Ω>ω

1<η

1>η

if

ifnebo

the displacement is in the same phasewith the exciting force

The solution for undamped vibration -δδδδ≅0, ξξξξ≅0 (more precisely small damped δδδδ<<Ω0, ξξξξ<<1).

the displacement is in the opposite phaseto the exciting force

Dynamics

101

k

x

bFD=b·v

FS=k·x

v, a

tF Fa

T T


m



F = Fa·sin(ω·t)




( ) ( )22220

aa

2

1

m

Fx


220

2


( ) ( )222

aa

21

1

k

Fx

η⋅ξ⋅+η−⋅=

21

2


stata x

k

F =


The solution for undamped vibration -δδδδ≅0, ξξξξ≅0 (more precisely small damped δδδδ<<Ω0, ξξξξ<<1).

2a

220

aa 1

1

k

F1

m

Fx

η−⋅=

ω−Ω⋅=

0=φthe amplitude xa is positive ifωωωω<Ω0, ηηηη<1, the same phase,the amplitude xa is negative ifωωωω>Ω0, ηηηη>1, the opposite phase

Dynamics

102

( ) ( )φ−⋅ω⋅= txx at sin( ) ( )tFF at ⋅ω⋅= sin

Fa, ω xa, φ

amplitude and phase characteristics


the cause the consequence

( ) ( )22220

aa

2

1

m

Fx


220

2


Dynamics

103

2

xa

xst

1 η

δ>0

δ=0

0

( ) ( ) ( )22220

aa

2

1

m

Fx

ω⋅δ⋅+ω−Ω⋅=ω

( ) ( ) ( )222

aa

21

1

k

Fx

η⋅ξ⋅+η−⋅=η

0Ωω=η

0Ω⋅η=ω

1. ωωωω=0, ηηηη=0. ( ) stata

0a xk

Fx ===η

2. ωωωω≅ΩΩΩΩ0, ηηηη≅1. The resonance.

3. ωωωω>>ΩΩΩΩ0, ηηηη>>1, xa(ηηηη→∞) → 0

3.

1. 2. the resonance

variable

variable

amplitude characteristic


104

2

xa

xst

1 η

δ>0

δ=0

0

( ) ( ) ( )22220

aa

2

1

m

Fx

ω⋅δ⋅+ω−Ω⋅=ω

( ) ( ) ( )222

aa

21

1

k

Fx

η⋅ξ⋅+η−⋅=η

0Ωω=η

0Ω⋅η=ω2. the resonance

variable

variable

amplitude characteristic


2. ωωωω≅ΩΩΩΩ0, ηηηη≅1. The resonanceappears if the exciting frequencyis near to the natural frequency.The amplitude reaches extremely high value.!The total maximum of the characteristicis at the tuning of :

η ξres = − ⋅1 2 2

This means slightly ωωωω<Ω0, ηηηη<1.

2

stresa

12

xx

ξ−⋅ξ⋅=_

The amplitude in resonance is :

Dynamics

105

phase characteristic


0Ωω=η

0Ω⋅η=ω

( ) 220

2

ω−Ωω⋅δ⋅=φ ω arctan

( ) 21

2

η−η⋅ξ⋅=φ η arctan

21 η0

δ>0

δ=0

90º

0

180º

φ

Dynamics

107

m

kL kR

m

kL1 =Ω

m

kR2 =Ω

m

?=Ω

direction 1direction 2

90º

4. The vibration with 2 degree of freedomthe free vibrationDynamics

108

m

kL kR

x

y

FSL FSR

ax

ay

α⋅−α⋅−=⋅α⋅+α⋅−=⋅

cossin

sincos

SRSLy

SRSLx

FFam

FFam

α

αRRSR

LLSL

kF

kF

l

l

∆⋅=∆⋅=

xy

α⋅=∆ cos' xLl

α⋅=∆ sin'' yLl

α⋅+α⋅−=∆α⋅+α⋅=∆

cossin

sincos

yx

yx

R

L

l

l

x

yα⋅−=∆ sin' xRl

α⋅=∆ cos'' yRl


109

m

kL kR

x

y

FSL FSR

ax

ay

α

α

( ) ( )( ) ( ) α⋅α⋅+α⋅−⋅−α⋅α⋅+α⋅⋅−=⋅

α⋅α⋅+α⋅−⋅+α⋅α⋅+α⋅⋅−=⋅coscossinsinsincos

sincossincossincos

yxkyxkam

yxkyxkam

RLy

RLx

( ) ( )( ) ( ) 0ykkxkkam

0ykkxkkam2

R2

LRLy

RL2

R2

Lx

=⋅α⋅+α⋅+⋅α⋅α⋅−+⋅

=⋅α⋅α⋅−+⋅α⋅+α⋅+⋅

cossincossin

cossinsincos

k11 k12

k21 k22

0ykxkym

0ykxkxm

2221

1211

=⋅+⋅+⋅=⋅+⋅+⋅

&&

&&

=

⋅

+

⋅

0

0

y

x

kk

kk

y

x

m0

0m

2221

1211

&&

&&

0uKuM =⋅+⋅ &&the mass matrixthe stiffness matrixthe column matrix (vector) of displacementsthe column matrix (vector) of accelerationsu&&

MKu

k12= k21k11>0k22>0

4. The vibration with 2 degree of freedom

α⋅−α⋅−=⋅α⋅+α⋅−=⋅

cossin

sincos

SRSLy

SRSLx

FFam

FFam

RRSR

LLSL

kF

kF

l

l

∆⋅=∆⋅=

α⋅+α⋅−=∆α⋅+α⋅=∆

cossin

sincos

yx

yx

R

L

l

l

the free vibrationDynamics

110

0ykxkym

0ykxkxm

2221

1211

=⋅+⋅+⋅=⋅+⋅+⋅

&&

&&0uKuM =⋅+⋅ &&

( )( )tCy

tCx

y

x

⋅Ω⋅=⋅Ω⋅=

sin

sin

( )( )tCy

tCx2

y

2x

⋅Ω⋅Ω⋅−=

⋅Ω⋅Ω⋅−=

sin

sin

&&

&&

( ) ( ) ( )( ) ( ) ( ) 0tCktCktCm

0tCktCktCm

y22x212

y

y12x112

x

=⋅Ω⋅⋅+⋅Ω⋅⋅+⋅Ω⋅Ω⋅⋅−

=⋅Ω⋅⋅+⋅Ω⋅⋅+⋅Ω⋅Ω⋅⋅−

sinsinsin

sinsinsin

( )( ) 0CmkCk

0CkCmk

y2

22x21

y12x2

11

=⋅Ω⋅−+⋅

=⋅+⋅Ω⋅−

( )t⋅Ω⋅= sincu

=y

x

C

Cc

the column matrix (vector)of amplitudes

( )t2 ⋅Ω⋅Ω⋅−= sincu&&

( ) ( ) 0cKcM =⋅Ω⋅Ω⋅⋅+⋅Ω⋅Ω⋅⋅− tt 22 sinsin

( ) 0cMK =⋅⋅Ω− 2

0CdCd

0CdCd

y22x21

y12x11

=⋅+⋅

=⋅+⋅

Ω⋅−Ω⋅−

=

=

22221

122

11

2221

1211

mkk

kmk

dd

ddD


0cD =⋅


111

0CdCd

0CdCd

y22x21

y12x11

=⋅+⋅

=⋅+⋅

the trivial solution0C

0C

y

x

==

the non-trivial solution

0CdCd

0CdCd

y22x21

y12x11

=⋅+⋅

=⋅+⋅ the linearly dependent equations(the 2. eq. is the multiple of the 1. eq.)

0CdCd y12x11 =⋅+⋅

x12

11y C

d

dC ⋅−=

0Cd

ddCd x

12

1122x21 =⋅⋅−⋅

0dddd 21122211 =⋅−⋅

0mkk

kmk

dd

dd2

2221

122

11

2221

1211 =Ω⋅−

Ω⋅−= the frequency determinant

the infinite number of solutions for Cx and Cy exists,the only we can calculate is their ratio

4. The vibration with 2 degree of freedomthe free vibration

11

12

y

x

d

d

C

C −=

Dynamics

112

0mkk

kmk2

2221

122

11 =Ω⋅−

Ω⋅−

( ) ( ) 0kkmkmk 21122

222

11 =⋅−Ω⋅−⋅Ω⋅−

( ) 0kkkkmkkm 211222112

221142 =⋅−⋅+Ω⋅⋅+−Ω⋅

( ) ( ) ( )2

21122211222

22112211

m2

kkkkm4mkkmkk

⋅⋅−⋅⋅⋅−⋅+±⋅+

=λ

the bi-quadratic equation

( ) ( ) ( )m2

kkkk4kkkk 211222112

22112211

⋅⋅−⋅⋅−+±+

=λ

m

kL1 =λ

m

kR2 =λ

m

kL1 =Ω

m

kR2 =Ω

( )α⋅+α⋅=

α⋅α⋅−==α⋅+α⋅=

2R

2L22

RL2112

2R

2L11

kkk

kkkk

kkk

cossin

cossin

sincos

the frequency determinant

( ) 0kkkkmkkm 21122211221122 =⋅−⋅+λ⋅⋅+−λ⋅ where λ = Ω2


2121 ,, λ=Ω Ω1 – smaller, Ω2 – greater

Dynamics

113

direction 1direction 2 m

kL kP

m

kL1 =Ω

m

kR2 =Ω

x12

11y C

d

dC ⋅−=

x

y

12

211

12

11

x

y

k

mk

d

d

C

C Ω⋅−−=−=

( ) α⋅α⋅−==α⋅+α⋅=cossin

sincos

RL2112

2R

2L11

kkkk

kkk

α=Ω⋅−−= tan12

2111

x

y

k

mk

C

C

α

α

α−=Ω⋅−−=

tan

1

k

mk

C

C

12

2211

x

y

m direction 1direction 2

direction 1

mode shape 1

direction 2

mode shape 2


−−Ω⋅−Ω⋅−=

ΩΩ

ΩΩ

1212

2211

2111

2x1x

2y1y

kk

mkmkCC

CC

__

__ the modal matrixx

y

Ω1 Ω2

Dynamics

114

direction 1direction 2 m

kL kP

m

kL1 =Ω

m

kR2 =Ω

x12

11y C

d

dC ⋅−=

x

y

12

211

12

11

x

y

k

mk

d

d

C

C Ω⋅−−=−=

α

α

m direction 1direction 2


the normalization (scaling) of mode shapes

the normalization to unit

1

1thanless

C

C

x

y __=

the normalization with respect to the mass matrix

1C

CCC

x

yxy =

⋅⋅M

Dynamics

115

m

kL kP

y

α

α

( ) ( ) ( )( ) ( ) ( )222y2111y1t

222x2111x1t

tCtCy

tCtCx

φ+⋅Ω⋅⋅κ+φ+⋅Ω⋅⋅κ=

φ+⋅Ω⋅⋅κ+φ+⋅Ω⋅⋅κ=

sinsin

sinsin

1y

1x

1y1

1x1

C

C

C

C =⋅κ⋅κ

2y

2x

2y2

2x2

C

C

C

C =⋅κ⋅κ

the integration constants κκκκ1, κκκκ2, φφφφ1and φφφφ2results from the initial conditions

t=0 ... x(t=0) = x0, y(t=0) = y0,vx(t=0) = vx0, vy(t=0) = vy0

m




x


116

m

kL kP

y

α

α

( ) ( ) ( )( ) ( ) ( )222y2111y1t

222x2111x1t

tCtCy

tCtCx

φ+⋅Ω⋅⋅κ+φ+⋅Ω⋅⋅κ=

φ+⋅Ω⋅⋅κ+φ+⋅Ω⋅⋅κ=

sinsin

sinsin

t=0 ... x(t=0) = x0 = 10 mmy(t=0) = y0 = 5 mmvx(t=0) = vx0 = -0,05 m/svy(t=0) = vy0 = 0,1 m/s

m

10 mm -11 mm

6 mm 16 mm

0.025 0 0.025

0.025

0.0250.025

0.025−

y t( )

0.0250.025− x t( )




x


117

m

kL kP

x

y

α

α

( ) ( ) ( )( ) ( ) ( )222y2111y1t

222x2111x1t

tCtCy

tCtCx

φ+⋅Ω⋅⋅κ+φ+⋅Ω⋅⋅κ=

φ+⋅Ω⋅⋅κ+φ+⋅Ω⋅⋅κ=

sinsin

sinsin

t=0 ... x(t=0) = x0 = 10 mmy(t=0) = y0 = 6,5 mmvx(t=0) = vx0 = 0,1 m/svy(t=0) = vy0 = 0,065 m/s

m

24 mm 0 mm

15 mm 0 mm





118

the summary :

during the vibration with n degrees of freedom n natural frequenciesappears,

these are arranged in order of their magnitude f1 < f2 < f3 ...

(one natural frequency is a simple number)

to every one natural frequency one mode shapecorresponds,

giving the ratio of the single degrees of freedom amplitudes.

(one mode shape is the column matrix with n rows,

equal to the number of degrees of freedom)

the natural frequencies and mode shapes are the modal characteristics,

the resulting vibration is the linear combination of all mode shapes,

the coefficients of the linear combination are determined

in dependence on the initial conditions.


119

m1 m2ka kckb

x2x1

FDa FDb FDb FDc

v1, a1 v2, a2

DbDa11 FFam +−=⋅ DcDb22 FFam −−=⋅

1aaaDa xkkF ⋅=∆⋅= l ( )12bbbDb xxkkF −⋅=∆⋅= l 2cccDc xkkF ⋅=∆⋅= l

( )12b1a11 xxkxkam −⋅+⋅−=⋅ ( ) 2c12b22 xkxxkam ⋅−−⋅−=⋅

( )( ) 0xkkxkam

0xkxkkam

2cb1b22

2b1ba11

=⋅++⋅−⋅=⋅−⋅++⋅

=

⋅

+−−+

+

⋅

0

0

x

x

kkk

kkk

x

x

m0

0m

2

1

cbb

bba

2

1

2

1

&&

&&

0xKxM =⋅+⋅ &&


120

m1 m2ka kckb

x2x1

FDa FDb FDb FDc

v1, a1 v2, a2

0mkkk

kmkk

22

cbb

b12

ba =⋅Ω−+−

−⋅Ω−+frekvenční determinant

( ) ( ) 0kmkkmkk 2b2

2cb1

2ba =−⋅Ω−+⋅⋅Ω−+

( ) ( )[ ] ( ) ( ) 0kkkkkmkkmkkmm 2bcbba

21cb2ba

421 =−+⋅++Ω⋅⋅++⋅+−Ω⋅⋅a cb

0cba 24 =+Ω⋅−Ω⋅

a2

ca4bb 22

21 ⋅⋅⋅−=Ω m

,

⋅Ω−+=

=1

21ba

b

12

111

mkk

k

C

Cc

,

,

⋅Ω−+=

=1

22ba

b

22

212

mkk

k

C

Cc

,

,


121


( ) 0uMK =⋅⋅Ω− 2the generalized eigenvalue problem

the natural circular frequenciesthe eigenfrequencies

the mode shapesthe eigenvectors

the number of natural frequencies Ω1,2, ...and the number of mode shapes u⟨1,2, ...⟩

is equal to the number of DOF

( ) 0u1A =⋅⋅Ω− 2the special eigenvalue problem

KMA ⋅= −1

=

nn2n1n

n22212

n12111

uuu

uuu

uuu

,,,

,,,

,,,

K

MOMM

K

K

Uthe modal matrix

Ω1 ΩnΩ2

x1x2

xn

Dynamics

122

m1 m2

ka kckb

x2x1 v1, a1 v2, a2

( ) ( )( ) ( )tFFxkkxkam

tFFxkxkkam

2a22cb1b22

1a12b1ba11

⋅ω⋅==⋅++⋅−⋅⋅ω⋅==⋅−⋅++⋅

sin

sin

( )( )

⋅ω⋅⋅ω⋅

=

⋅

+−−+

+

⋅

tF

tF

x

x

kkk

kkk

x

x

m0

0m

2a

1a

2

1

cbb

bba

2

1

2

1

sin

sin

&&

&&

fxKxM =⋅+⋅ &&

F1=Fa1·sin(ω·t) F2=Fa2·sin(ω·t)

( )( )txx

txx

2a2

1a1

⋅ω⋅=⋅ω⋅=

sin

sin

( )( ) 2

222a2

122

1a1

xtxx

xtxx

⋅ω−=⋅ω⋅ω⋅−=

⋅ω−=⋅ω⋅ω⋅−=

sin

sin

&&

&&

4. The vibration with 2 degree of freedomthe forced vibrationDynamics

123

m1 m2

ka kckb

x2x1 v1, a1 v2, a2


( ) aa2 fxMK =⋅⋅ω−

( )( )

( )( )

( )( )

⋅ω⋅⋅ω⋅

=

⋅ω⋅⋅ω⋅

⋅

+−−+

+

⋅ω⋅⋅ω−⋅ω⋅⋅ω−

⋅

tF

tF

tx

tx

kkk

kkk

tx

tx

m0

0m

2a

1a

2a

1a

cbb

bba

2a2

1a2

2

1

sin

sin

sin

sin

sin

sin

=

⋅

⋅ω−

+−−+

2a

1a

2a

1a

2

12

cbb

bba

F

F

x

x

m0

0m

kkk

kkk

aa fxD =⋅D – the dynamic stiffness matrix

( ) 0cMK =⋅⋅Ω− 2

4. The vibration with 2 degree of freedomthe forced vibrationDynamics

124

m1 m2

ka kckb

x2x1 v1, a1 v2, a2


the solution of the steady state forced vibration amplitudes – the system of algebraic equationsthe operation shape of vibration

( ) 0cMK =⋅⋅Ω− 2

( ) aa2 fxMK =⋅⋅ω−

ω ≅ Ω – the resonance

the natural frequencies and mode shapes

( )( ) 2a2a2

2cb1ab

1a2ab1a12

ba

Fxmkkxk

Fxkxmkk

=⋅⋅ω−+⋅−=⋅−⋅⋅ω−+

4. The vibration with 2 degree of freedomthe forced vibration

1. the exciting forces have the different phase angle,2. including damping.The solution in complex numbers

Dynamics

125

4. The vibration with 2 degree of freedomthe forced vibration

m2

ka

kb y1

F1(t) = F1a·sin(ω·t)

m1

y2

0 20 40 60 80 100 120 140

1

2

y1a

y2a y1a

y2a y1a

y2a

ω =

Ω0_

1

ω ω =

Ω0_

2

ωan

ti

anti-resonance

the zero vibration amplitude in y1

Dynamics

127

translation

general plane motion

rotation

the plane motion :

all points movein the planesparallel one to the other

5. Translation and rotationthe types of motion

spiral motion

spherical motion

general space motion

the space motiontranslation

Dynamics

128

no lines change their directiontranslation


2D

Dynamics

129

rotation


one line does not change its position2D

Dynamics

130



2D

Dynamics

131



2D

Dynamics

132


translation

no lines change their direction

3D

Dynamics

133

3D


one point does not change its position

spherical motion

Dynamics

134

posuv

rotace

spiral motion


the body rotates about a specific axisand translates in the direction of this axis

3D

Dynamics

135


general space motion

3D

Dynamics

136

η

ζ ξ

x

z

y

A

O Ω

1, 2 or 3 degrees of freedom

x,y,z – the fixed coordinate system(not moving), its origin is O

ξ,η,ζ – the body coordinate system–fixed to the body and moving with it,its origin is Ω

ξ//x, η//y, ζ//z

A – the common point of the body

5. Translation and rotationthe translation - kinematics


Dynamics

137

η

ζ ξ

x

z

y

A

Ωrr

Arr

Ω ΩAr

r

P

rA – the radius vectorof the point A

with respect to xyz

rΩ – the radius vectorof the point Ω

with respect to xyz,the position of the body

in space

rAΩ – the radius vectorof the point A

with respect to ξηζ,the position of A inside the body

ΩΩ += AA rrr rrr


no lines change their direction1, 2 or 3 degrees of freedom

Dynamics

138

η

ζ ξ

x

z

y

A

PΩrr

Arr

ΩΩAr

r

ΩΩ === avva AAr

&r

&rr

ΩΩ +== AAA rrrv &r

&r

&rr

0rAr

&r =Ω

Ω= vvArr

Ω= aaArr

All the points move on the same trajectory,with the same velocity and the same acceleration.

the time derivativeΩΩ += AA rrr rrr


no lines change their direction1, 2 or 3 degrees of freedom

the time derivative

Dynamics

139

the direct linear translation




Dynamics

140

R

the circular translation




Dynamics

141

the cycloid translation




Dynamics

142

amD rr⋅−=

0DFi

rrr=+∑

dm

dmdm

dm

a

aa

adD

D

dDdD

dD

CG

dm

dmdm

dm

dG

G

CGdG

dG

dG

5. Translation and rotationthe translation - dynamics

∑=⋅ iFamrr

the equation of motion the d’Alembert principle

the point of applicationof the dÁlembert forceis in the centre of gravity

Dynamics

143

∑=⋅ iFamrr

φ

r

G

BA

m

at

φ

T

b

r

G

b

r

CD

BA

mT

φ⋅=⋅ cosGam t

( ) ( )02

0 r

g2 φ−φ⋅⋅+ω=ω φ sinsin

φ⋅⋅=ε⋅⋅ cosgmrm

φ⋅=ε cosr

g

φ⋅=φω⋅ω cos

r

g

d

d

( ) ( ) ( )022

0 gr2rrv φ−φ⋅⋅⋅+⋅ω=⋅ω= φφ sinsin

φ⋅φ⋅=ω⋅ω dr

gd cos

∫∫φ

φ

ω

ω

φ⋅φ⋅=ω⋅ω00

dr

gd cos

[ ] [ ]φφ

ω

ω φ⋅=ω⋅ 002

21

r

gsin



Dynamics

144

amD rr⋅−=

0DFi

rrr=+∑

G

T

Dt

Dn

SCSD

CD

BA

y

x

b

r

G

b

r

CD

BA

mT

( )

φ−φ⋅⋅+ω⋅⋅=

=ω⋅⋅=⋅=

φ⋅⋅=⋅=

02

0

2nn

tt

rg

2rm

rmamD

gmamD

sinsin

cos


the d’Alembert principle

0Fxi =∑ 0Fyi =∑ 0M i =∑

K=CS K=DSφ⋅=ε cosrg

the three equations of equilibrium to solve :1) the equation of motion,2) the reaction forces.

the d’Alembert forcein the centre of gravity (tangentialand normal component)

Dynamics

145

∑=⋅ iFamrr amD rr

⋅−=

0DFi

rrr=+∑

b

r

G

b

r

CD

BA

mT

Assembly of equations of motion – centralize all mass into the mass particle.


To solve forces (usually reaction forces) – use the d’Alembert principle –to locate the d’Alembert force within the centre of gravity,to assemble the equations of equilibrium and solve forces.From equations of equilibrium the equation of motion can be derived.

Dynamics

146

ω, ε φ

o

5. Translation and rotationthe rotation - kinematics

one line does not change its position 1 DOFevery point runs on the circular trajectory of the radius R

φ

φ=φ=ω &

dt

d

( )φ

ω⋅=φω⋅ω=φ=φ=ω=ω=ε

d

d

2

1

d

d

dt

d

dt

d 2

2

2&&&

the rotation angle

the angular velocity

the angular acceleration

[°, rad, revolute]

[rad/sec, rev/min]

[rad/sec2]

Dynamics

147

5. Translation and rotationthe rotation - kinematics

one line does not change its position 1 DOFevery point runs on the circular trajectory of the radius R

φ

φ=φ=ω &

dt

d

ω=ω=ε &dt

d

the rotation angle

the angular velocity

the angular acceleration

[°, rad, revolute]

[rad/sec, rev/min]

[rad/sec2]

R

S

φ, ω, ε

tar

vr

nar

Rv ⋅ω=

Rat ⋅ε=

Ra 2n ⋅ω=

Rs ⋅φ=

rv rrr ×ω=

ratrrr ×ε=

vanrrr ×ω=

v the circumferential velocity

at the tangential acceleration

an the normal acceleration

r the radius vector

ω, ε φ

o

ωr

rr vrR

Dynamics

148

∑=⋅ iFamrr

ω, ε

m

CR

the rotation - dynamics


5. Translation and rotation

∑=ε⋅ iCRCR MI _

m – the body mass (not presentin the equation of motion)

ICR - the moment of inertia with respectto the centre of rotationCR[kg·m2]

ε - the angular acceleration[rad/s2]

M – the force moment[N·m]

Dynamics

149






T2

Tnn

TTtt

CRD

rmamD

rmamD

IM

⋅ω⋅=⋅=

⋅ε⋅=⋅=ε⋅=

ω, ε

aCGn

aCGt

CR CG

Dt

Dn MD

m, ICR rT

the d’Alembert forces (and moment)

∑=⋅ iFamrr

Dynamics

150

CR

Dt

Dn MD x

y

Ry

Rx

ω, ε

the primary, external forces (actions)

the reaction forces

the d’Alembert forces (and moment)






T2

Tnn

TTtt

CRD

rmamD

rmamD

IM

⋅ω⋅=⋅=

⋅ε⋅=⋅=ε⋅=

0M

0F

0F

CRi

yi

xi

=

=

=

∑∑∑ the reaction solution

ΣF – including d’Alembert forces


2CR2

1K IE ω⋅⋅= the kinetic energy

Dynamics

151

rotationtranslation

the track [m, mm]s, x, ... ~ the angle [rad, °]φ

the velocity [m/s] ~ the angular velocity [rad/s]sv &= φ=ω &

the acceleration [m/s2] ~ the angular acceleration [rad/s2]

dsdv

vsva ⋅=== &&&φω⋅ω=φ=ω=ε

dd

&&&

example – uniformly accelerated motion

002

21

0

stvtas

vtav

+⋅+⋅⋅=

+⋅=

002

21

0

tt

t

φ+⋅ω+⋅ε⋅=φ

ω+⋅ε=ω~

~

the rotation - dynamics 5. Translation and rotation

analogy

Dynamics

152

the force [N]F, G, ... ~ the force moment [N·m]M

the mass [kg]m ~ the moment of inertia

[kg·m2]

I


~∑=⋅ iFamrr ∑=ε⋅ iMI

rotationtranslation


analogy


~the kinetic energy

221

K vmE ⋅⋅= 221

K IE ω⋅⋅=

~the work ∫ ⋅= sdFA rrthe work ∫ φ⋅= dMA[N·m]

[J][J]

[N·m]

~the power vFP rr⋅= the power[W] ω⋅= MP [W]

the kinetic energy change law AEEE 0K1KK =−=∆ [J ~ N·m]

the kinetic energy

Dynamics

153

r

m, I

φ

ω,ε

G

CG

the physical pendulum : The physical pendulumThe body of the mass m and moment of inertia I.The distance between the centre of rotationand the centre of gravity is r.

r·sinφ

0rGI

0rGI

rGI

=φ⋅⋅+φ⋅

=φ⋅⋅+ε⋅φ⋅⋅−=ε⋅

sin

sin

sin

&&

linearizationsinφφφφ ≅ φφφφ

0rGI =φ⋅⋅+φ⋅ &&

the rotation - dynamics 5. Translation and rotationDynamics

154

r

m, I

φ

ω,ε

G

CG


0rGI =φ⋅⋅+φ⋅ &&

0xkxm =⋅+⋅ && 0rGI =φ⋅⋅+φ⋅ &&

( )0tCx γ+⋅Ω⋅= sin ( )0tC γ+⋅Ω⋅=φ sin

the analogical solution :

m

k=Ω

2

202

0

vxC

Ω+=

0

00 v

x Ω⋅=γ arctan

the initial conditions :t = 0 ...φφφφ = φφφφ0 the initial angleωωωω = ωωωω0 the initial

angular velocity2

202

0CΩω+φ=

0

00 ω

Ω⋅φ=γ arctan

the circular frequency

the amplitude

the phase lead

I

rG ⋅=Ω


155

r

m, I

φ

ω,ε

G

CG


0rGI =φ⋅⋅+φ⋅ &&

the calculation of the moment of inertia Ifrom measured vibration periodT :

T

2 π⋅=Ω

2

rGI

Ω⋅=

I

rG ⋅=Ω


156

∫ ⋅=m

2 dmrIdm

r

m

S

r = const

2

m

2

m

2 rmdmrdmrI ⋅=⋅=⋅= ∫∫

the thin ring


the moment of inertia

Dynamics

157

∫ ⋅=m

2 dmrIdm

r

m

S

x dx

dm

m

l

∫ ⋅=m

2 dmxI dxm

dmdx

mdm ⋅=⇒=

ll

∫∫ ⋅⋅=⋅⋅=ll

ll 0

2

0

2 dxxm

dxm

xI

3m

3xm

I3

0

3l

ll

l

⋅=

⋅=

2m31

I l⋅⋅=

the thin prismatic rod



Dynamics

158

∫ ⋅=m

2 dmrIdm

r

m

S

∫ ⋅=m

2 dmxI dxm

dmdx

mdm ⋅=⇒=

ll

∫∫−−

⋅⋅=⋅⋅=2

2

22

2

2 dxxm

dxm

xI/

/

/

/

l

l

l

lll

x dx

dm

m

l

the thin prismatic rod

43

1m

3

xmI

32

2

3l

ll

l

l

⋅⋅=

⋅=

−

/

/

2m121

I l⋅⋅=



Dynamics

159

∫ ⋅=m

2 dmrI

m

h

R



the cylinder

Dynamics

160

∫ ⋅=m

2 dmrI

m

h

r dr

( ) hdrr2hdSdVdm ⋅⋅⋅π⋅⋅ρ=⋅⋅ρ=⋅ρ=

dr2·π·r

dS



the cylinder

Dynamics

161

∫ ⋅=m

2 dmrI

2Rm21

I ⋅⋅=

m

R

h

( ) hdrr2hdSdVdm ⋅⋅⋅π⋅⋅ρ=⋅⋅ρ=⋅ρ=

hRm

hSm

Vm

2 ⋅⋅π=

⋅==ρ

drrRm

2hdrr2hR

mdm 22 ⋅⋅⋅=⋅⋅⋅π⋅⋅

⋅⋅π=

4R

Rm

24r

Rm

2drrRm

2drrRm

2rI4

2

R

0

4

2

R

0

32

R

02

2 ⋅⋅=

⋅⋅=⋅⋅⋅=⋅⋅⋅⋅= ∫∫



the cylinder

Dynamics

162

2CG emII ⋅+=

m

e

CG

the Steiner theorem

ICG- the moment of inertiato the axis crossing the centre of gravity,

I - the moment of inertiato the parallel shifted axis.

I CGI



∫ ⋅=m

2 dmrI

to the parallel shifted axis

Dynamics

163

r

m

the thin circular plate

241

T rmI ⋅⋅=

a

m

b

2121

xT bmI ⋅⋅=_x

z y ( )22121

zT bamI +⋅⋅=_2

121

yT amI ⋅⋅=_

rm

a

( )2312

41

T armI ⋅+⋅⋅=

the cylinder

r

m

2103

T rmI ⋅⋅=

the cone the pyramid

a

m

b

( )22201

T bamI +⋅⋅=

rm

252

T rmI ⋅⋅=


the thin rectangular plate

the ball

Dynamics

164


the firm publication

Dynamics

165

firm publication


166

3D CAD modeling

PRINT MASS PROPERTIES ASSOCIATED WITH THE CURRENTLY SELECTED VOLUMESTOTAL NUMBER OF VOLUMES SELECTED = 1 (OUT OF 1 DEFINED)***********************************************SUMMATION OF ALL SELECTED VOLUMES

TOTAL VOLUME = 0.11537E+08TOTAL MASS = 0.92296E-01CENTER OF MASS: XC=-0.14674E-03 YC= 0.0000 ZC= 0.0000

*** MOMENTS OF INERTIA ***ABOUT ORIGIN ABOUT CENTER OF MASS PRINCIPAL

IXX = 1752.3 1752.3 1752.3 IYY = 1752.3 1752.3 1752.3 IZZ = 3392.2 3392.2 3392.2 IXY = 0.55354E-03 0.55354E-03IYZ = 0.46905E-04 0.46905E-04IZX = -0.62350E-04 -0.62350E-04PRINCIPAL ORIENTATION VECTORS (X,Y,Z):

0.993 -0.116 0.000 0.116 0.993 0.000 0.000 0.000 1.000(THXY= -6.635 THYZ= 0.000 THZX= 0.000)


167

6. General plane motionkinematics

translation


rotation

the plane motion :

all points movein planesparallel to one another

Dynamics

168

translation

rotation

translation

1 DOF

2 DOF

3 DOF

translation

rotation


1, 2 or 3 degrees of freedom

Dynamics

169

1 DOF

2 DOF

one independent motion

rolling without slipping

sliding in the touch-point

x, v, a

φ, ω, ε

rolling without slipping

translation

rotationr

sliding in the touch-point

independent translation and rotationx, v, a

φ, ω, ε


two independent motions

Dynamics

170

- the analytical solution

- the pole method

- motion decomposition

6. General plane motionkinematicsDynamics

171

the velocity solution

the geometry solution

xA

yB

A

B

lBB a ,vrr

AA a ,vrr

sv &=

ds

dvvsva ⋅=== &&&

A2A

2

AA

A

BBB v

x

x

dt

dx

dx

dy

dt

dyv ⋅

−

−=⋅==l

22B

2A yx l=+

2A

2B xy −= l

Av ( )xAp

A2

AA2

AA

B apvqapvdx

dpa ⋅+⋅=⋅+⋅=

AAA

AB apv

dt

dx

dx

dpa ⋅+⋅⋅=

dt

dvpv

dt

dp

dt

dva A

AB

B ⋅+⋅==

( )xAq

Av

the basic scheme

the acceleration

solution

vA, aA given,determine vB, aB

6. General plane motionthe analytical solutionDynamics

172

sv &=

ds

dvvsva ⋅=== &&&

φ⋅= coslAx φ⋅= sinlBy

dt

d

d

dx

dt

dxv AA

A

φ⋅φ

==

φ⋅φ⋅−= &l sinAv

φ⋅ω⋅−= sinlAv

dt

d

d

dy

dt

dyv BB

B

φ⋅φ

==

φ⋅φ⋅= &l cosBv

φ⋅ω⋅= coslBv

φ⋅φ⋅ω⋅−φ⋅ω⋅−== &l&l cossindt

dva A

A

φ⋅ω⋅−φ⋅ε⋅−= cossin 2Aa ll

φ⋅φ⋅ω⋅−φ⋅ω⋅== &l&l sincosdt

dva B

B

φ⋅ω⋅−φ⋅ε⋅= sincos 2Ba ll

xA

yB

A

B

lBB a ,vrr

AA a ,vrr

φ

ω,ε


173

the geometry solution

( )sAB fs = sA, sB – the generalized coordinates – longitudinal or angular

dt

ds

ds

ds

dt

dsv A

A

BBB ⋅==

( ) AsAB vpv ⋅= vA, vB – the generalized velocities – longitudinal or angular

( )dt

dvpv

dt

dp

dt

vpd

dt

dva A

AAB

B ⋅+⋅=⋅==

AAA

AB apv

dt

ds

ds

dpa ⋅+⋅⋅=

( ) ( ) AsA2

AsAB apvqa ⋅+⋅=aA, aB – the generalized accelerations – longitudinal or angular

( )A

BsA ds

dsp =

( ) 2A

B2

AsA

ds

sd

ds

dpq ==

AA v

dt

ds =

AA a

dt

dv =

6. General plane motionthe analytical solution

the acceleration solution

the velocity solution

Dynamics

174

rx ⋅φ=

x v,a

φ,ω,ε φ = 360º = 2·π ≅ 6,28 rad

x = 2·π·r

r

rv ⋅ω=ra ⋅ε=

φ


φφφφ·r

cycloid curve


175

B

AAB y

xvv ⋅=

xA

yB

A

B

lBvr

vAr

nB

nA

π

6. General plane motionthe pole method

B

AA

y

v

A

v =π

=ω

B

AAAB y

xvxBv ⋅=⋅ω=π⋅ω=

A

B

l nA

nB

π

ω

Bvr

Avr

the pole– the instantaneous center of zero velocity

Dynamics

176

only velocities !

not accelerations

!A

B

C

π

nC

nB

nA

ω

Cvr

Avr



Dynamics

177

R

va

2

n =

A

B

lBvr

vAr

nB

nA

π

A

B

l nA

nB

π

ω

Bvr

Avr

aBn=0

aAn=0

point A moves along the line trajectory

the

Bpo

int m

oves

on

the

line

traj

ecto

ry

Bna

Ana


only velocities !

not accelerations !

point A moves along the circular trajectory

the

Bpo

int m

oves

on

the

circ

ular

traj

ecto

ry


Dynamics

178

A

B π(t-∆t)

π(t+∆t)

π(t)


the pole is at different points at different momentsa set of points representing a pole in the past, present and future is the pole curve

Dynamics

179

A

B π(t)

the moving pole curve

A

B π(t-∆t)

π(t+∆t)

π(t)

the fixed pole curve

a set of points of a polein a fixed space,in a fixed coordinate system,the fixed pole curve



a set of points of a polein a moving space,in a body coordinate system,the moving pole curve

Dynamics

180

A

B π(t-∆t)

π(t+∆t)

π(t)





the pole curves touch one anotherat the point where the pole is at present

Dynamics

181

A

B

π(t-∆t)

π(t+∆t)

π(t)C

D

Ethe fixed pole curve



the general plane motion can be interpretedas the rolling of a moving pole curve on a fixed pole curve

ω

(no matter if it corresponds to the real technical realization)

Dynamics

182

B

the A point runs on the line trajectory

the

Bpo

int r

uns

on th

e lin

e tr

ajec

tory

rolling

AA

B

π(t-∆t)

π(t+∆t)

π(t)C

D

E


the general plane motion can be interpretedas the rolling of a moving pole curve on a fixed pole curve

(no matter if it corresponds to the real technical realization)





Dynamics

183

φ,ω

π

rv ⋅ω=v

r


ππππ pole

only velocities !

not accelerations

!




Dynamics

184

translation

A

B

A

B vtrans

vA A

B

rotationvrot

vB

vA

+

the superposition of translation and rotation

given : vA, aA – the velocity and acceleration of the point A,determine : vB, aB – the velocity and acceleration of the point B.

6. General plane motionmotion decompositionbasic decomposition

Dynamics

185

translation

A

B

rotation

BAAB vvvrrr +=

rotBtransBB vvv __

rrr +=

BAAB aaarrr

+=

vtrans

vrot

vB

vA

= vBA

A – the reference point

= vA

A

B

vB

vA

6. General plane motionmotion decomposition

the superposition of translation and rotation

basic decomposition

Dynamics

186

Avr

φ=

tanA

B

vv

A

B

φφφφ

BAvr+ Bv

r=vB

vA

vBA ⊥⊥⊥⊥ ABφφφφ

φ=

sinA

BA

vv

φ⋅==ω

sinllllllll

ABA vv

llll

ωωωω

vB

vA


Dynamics

187

Aar BBAA aaa

rrr=+

A

B

φφφφ

BAnar

+ Bar

=aB

aA

aBAt ⊥⊥⊥⊥ ABφφφφ

φφ+φ⋅+

φ=

tan

cossin

tan BAnA

B aa

a

llll

ωωωωBAta

r+ aBAn |||||||| AB

φφφφ

llllllll

⋅ω== 22

BABAn

va

0aaa BAtBAnA =φ⋅−φ⋅+ sincos

BBAtBAn aaa =φ⋅+φ⋅ cossin

φ+

φ=

tansinBAnA

BAt

aaa

llll

BAta=ε

, ε, ε, ε, ε

aB

aAhorizo

ntal

vertical


Dynamics

188

rvA ⋅ω=

ω

vA,aA

BA

raA ⋅ε= BAAB vvvrrr +=

translation + rotation

φ

BAvr

Avr

Bvr

⊥ ABC

(⊥ BC)

φx

y

xBAAxB vvv __ +=

ψ

yBAyB vv __ =

bvBA ⋅ω=

r

φ⋅⋅ω= sin_ bv yB

φ⋅+⋅= cos_ r

b1vv AxB φ⋅⋅= sin_ r

bvv AyB

( )φ⋅+⋅ω= cos_ brv xB

2yB

2xBB vvv __ +=

xB

yB

v

v

_

_arctan=ψr

vA=ω

Bvr



A –

the

refe

renc

e po

int

b

Dynamics

189

vA,aA

BA φ

⊥ AB

|| AB

C

φx

y

xnBAxtBAAxB aaaa _____ −+=

ynBAytBAyB aaa _____ +=

ba tBA ⋅ε=_

r

( ) φ⋅⋅ω−φ⋅+⋅ε= sincos_ bbra 2xB

2yB

2xBB aaa __ +=

xB

yB

a

a

_

_arctan=γ

ba 2nBA ⋅ω=_

BAAB aaarrr

+=nBAtBAAB aaaa __

rrrr++=

nBAa _

r

tBAa _

r

Aar

Bar

φ

φ⋅⋅ω+φ⋅⋅ε= cossin_ bba 2yB

rvA ⋅ω=

ω,ε

raA ⋅ε=

r

aA=εr

vA=ω

γ

translation + rotation



A –

the

refe

renc

e po

int

b

Dynamics

190

vA,aA

BA φ

C

⊥ BC

x

y

ψ ba tBA ⋅ε=_

rb

2yB

2xBB aaa __ +=

xB

yB

a

a

_

_arctan=γ

ba 2nBA ⋅ω=_

BAAB aaarrr

+=nBAtBAAB aaaa __

rrrr++=

Bar

|| BC

tBa _

r

nBa _

r

γ

ψ

( )ψ−γ⋅= cos_ BtB aa

( )ψ−γ⋅= sin_ BnB aa

rvA ⋅ω=

ω,ε

raA ⋅ε=translation + rotation



A –

the

refe

renc

e po

int

Dynamics

191

the resulting motion = the frame motion + the relative motion

shore

the

fram

e m

otio

n

the

rela

tive

circ

ling

6. General plane motionmotion decompositiongeneral decomposition

Dynamics

192

general plane motion = translation + rotation

unášivá rotace

relativní posuv

unášivá rotace

relativní rotace

unášivý pohyb - posuvný kruhový

relativní pohyb - posuvný přímočarý

translation

rotation


general plane motion = rotation + translation

general plane motion = rotation + rotation

translation = translation + translation


basic decomposition

Dynamics

193

= +

vr

relvr

framevr

frameω

relframe vvvrrr +=

relvr

framevr v

r

rv frameframe ⋅ω=

r



Dynamics

194

= +

ar

relvr

frame_tar

frameω

frameε

frame_nar

relar

relframe vvvrrr +=

Correlframe aaaarrrr

++=

relframeCor v2arrr

×ω⋅=

relar

tframea _

r

Correltframenframe aaaaarrrrr

+++= __

nframea _

r

ar

Corar

rr

va 2

frame

2frame

nframe ⋅ω==_

r

ra frametframe ⋅ε=_

the Coriolis acceleration



the general decomposition

the Coriolis decomposition

Dynamics

195

the frame motion

the relative motion

φd

dsd ⋅φ

dsφd

0dt

dsd →⋅φ

221 dtadsd ⋅⋅=⋅φ

frameω relv

relframeCor v2arrr

×ω⋅=

relframeCor v2a ⋅ω⋅=

Coraframeω

relv

dt

ds

dt

d2

dt

dsd2a

2⋅φ⋅=⋅φ⋅=

6. General plane motionthe Coriolis accelerationDynamics

196

frameωr

relωr

relframeresult ω+ω=ωrrr

relframeResal ω×ω=εrrr

Resalrelframeresult ε+ε+ε=εrrrr

6. General plane motionthe Resal angular accelerationDynamics

197

A

ε⋅= AD IM

MD

Dt

Dn

Dtr

0M =∑0Fx =∑ 0Fy =∑

the equations of equilibrium

the reactionsthe primary forces

ππππ

translationB

the translation and rotation superposition

Atr amD ⋅=GGtt rmamD ⋅ε⋅=⋅=

G2

Gnn rmamD ⋅ω⋅=⋅=

r G–

the

dist

ance

bet

wee

nce

ntre

of g

ravi

ty a

nd r

efer

ence

poi

nt

G


rotation

the d’Alembert forces

basic decomposition + the d’Alembert principle

6. General plane motiondynamics

A – the reference point

0M =∑ πthe equation of motion

Dynamics

198

A

Bthe kinetic energy

2G2

12G2

1k IvmE ω⋅⋅+⋅⋅=

rotationkntranslatiokk EEE __ +=

the reference point ≡≡≡≡ the centre of gravity

G ωωωω

vG

basic decomposition


2A2

12A2

1 Ivm ω⋅⋅+⋅⋅

centre of gravity

Dynamics

199

0M

0F

0F

i

iy

ix

=

=

=

∑∑∑

_

_

rolling without sliding – the general plane motion with 1 degree of freedom

x, v, a

φ, ω, ε

ra

rv

rx

⋅ε=⋅ω=⋅φ=

N

L

MD

0rmD

0rmD

IM

amD

G2

n

Gt

CD

tr

=⋅ω⋅=

=⋅ε⋅=ε⋅=

⋅=

the centre of gravity ≡ the reference point

ππππ

0M i =∑ π_the equation of motion

0rGrDM pD =α⋅⋅−⋅+ sin

G

αααα

α⋅⋅=⋅⋅+ε⋅ sinrGramIS

α⋅⋅=⋅⋅+⋅ sinrGramr

aIS

α⋅=⋅

+ sinGamr

I2S

0F

0F

ix

iy

=

=

∑∑

_

_ α⋅= cosGN

trDGL −α⋅= sin

2C

C

rmI

IGL

⋅+⋅α⋅= sin

fNL ⋅≤ frmI

I2

C

C ≤⋅+

⋅αtan

the not-sliding condition

the reaction solution

Dtr

rC ≡≡≡≡ G

252

C rmI ⋅⋅=the ball

α⋅=⋅⋅ sin, Gam41

6. General plane motiondynamicsDynamics

200

0rmD

0rmD

IM

amD

G2

n

Gt

CD

tr

≠⋅ω⋅=

≠⋅ε⋅=ε⋅=

⋅=x, v, a

φ, ω, ε

ra

rv

rx

⋅ε=⋅ω=⋅φ=

N

L

Dtr

MD 0M

0F

0F

i

iy

ix

=

=

=

∑∑∑

_

_

ππππ G

αααα

G

Dt

Dn

C

( )2fa ωφ= ,

the differential equation of II. order, non-linear

non-uniform motion !

rG

0 0.2 0.4 0.6 0.8

50

100

150

0

ω

t

rolling without sliding – the general plane motion with 1 degree of freedom




Dynamics

201

x, v, a

φ, ω, ε

ra

rv

rx

⋅ε=⋅ω=⋅φ=

N

F = N·f

MD

0rmD

0rmD

IM

amD

G2

n

Gt

CD

tr

=⋅ω⋅=

=⋅ε⋅=ε⋅=

⋅=

0M

0F

0F

i

iy

ix

=

=

=

∑∑∑

_

_

0M iC =∑ _

0rFM D =⋅−

G

αααα

0GFD tr =α⋅−+ sin

0F iy =∑ _ α⋅= cosGN

Dtr

rC ≡≡≡≡ G

rfGIC ⋅⋅α⋅=ε⋅ cos

0F ix =∑ _

( )α⋅−α⋅=⋅ cossin fGam

- 2 degrees of freedom- independent translation and rotation- 2 independent equations of motion

with sliding – the general plane motion with 2 degrees of freedom



the equations of motion

the reaction solution

Dynamics

203

the point is called “the center of the spherical motion”

střed sférického pohybu

S

k1o1

k2o2rám

o3

konst

3 DOF

7. Spherical motionkinematics

one point does not change its position (stays motionless)

2 DOF

Dynamics

204

the point is called “the center of the spherical motion”


one point does not change its position (stays motionless)

1 DOF

Dynamics

205

the basic kinematic quantities are angular velocity ωωωωand angular acceleration εεεε.

both value and direction of these change during spherical motion

the vector of the instantaneous angular velocity ωωωωdetermines the instantaneous axis of zero velocity(every moment going through the center of spherical motion).

the elementary motion can be interpreted asthe instantaneous rotation about the instantaneous axis of zero velocity

7. Spherical motionkinematicsDynamics

206

the set of lines, being instantaneous axisof zero velocity in past, present and future,is the pole cone

in fixed space – fixed pole conein moving space – moving pole cone

the pole cones touch one the otherin the present instantaneous axisof zero velocity

The spherical motion can be interpreted asthe rolling of the moving pole cone on the fixed pole cone

the basic kinematic quantities are angular velocity ωωωωand angular acceleration εεεε.


the instantaneous axisof zero velocity

the fixed pole conethe moving pole cone

ωωωω - the instantaneous angular velocitythe center of the spherical motion

Dynamics

207

the Euler’s angles fixed coordinate system xyz , body coordinate system ξηζξηζξηζξηζboth CS with common origin in the center of spherical motion.

1. ψψψψ – the angle of precession about the z axis

x≡ξ0 ξ1

ψ

ψ y≡η0

z≡ζ0≡ζ1

η1

2. ϑϑϑϑ – the angle of nutation about the ξξξξ1 axis

x≡ξ0

ϑ

y≡η0

z≡ζ0≡ζ1

η1

ϑη2

ξ1≡ξ2

ζ2

uzlová přímka

3. φφφφ – the angle of primary rotation about the ζζζζ axis

x≡ξ0

φ y≡η0

z≡ζ0≡ζ1

η1

φη2

ξ1≡ξ2

ξ

η

ζ2≡ζ


208

the motion of precession

the motion of nutation

the motion of primary rotation






the precession the primary rotation

the nutationthe motion of precessionthe frame rotation about the fixed axis (z)the motion of nutationthe relative rotation about the axis, doing precessionthe motion of primary rotationthe relative rotation about axis, doing precession and nutation

Dynamics

209

the motion of precession

the motion of nutation

the motion of primary rotation






the precession

the primary rotation

the motion of precessionthe frame rotation about the fixed axis (z)the motion of nutationthe relative rotation about the axis, doing precessionthe motion of primary rotationthe relative rotation about axis, doing precession and nutation

Dynamics

210

the direction angles of the vector :

ψ+ϑ⋅φ=ω

ψ⋅ϑ+ψ⋅ϑ⋅φ−=ω

ψ⋅ϑ+ψ⋅ϑ⋅φ=ω

&&

&&

&&

cos

sincossin

cossinsin

z

y

x

φ+ϑ⋅ψ=ω

φ⋅ϑ−φ⋅ϑ⋅ψ=ω

φ⋅ϑ+φ⋅ϑ⋅ψ=ω

ζ

η

ξ

&&

&&

&&

cos

sincossin

cossinsin

ϑ⋅φ⋅ψ⋅+φ+ϑ+ψ==ω+ω+ω=ω+ω+ω=ω ζηξ cos&&&&&Kr

22222222z

2y

2x

ωω=α xcos

ωω

=β ycosωω=γ zcos

with respect to the x axis

the Euler’s kinematic equations the angular velocity ωωωω


with respect to the z axiswith respect to the y axis

ωr x

y

z

ωxωz

ωy

γ

βα

ωr

Dynamics

211

ψ+ϑ⋅φ=ω

ψ⋅ϑ+ψ⋅ϑ⋅φ−=ω

ψ⋅ϑ+ψ⋅ϑ⋅φ=ω

&&

&&

&&

cos

sincossin

cossinsin

z

y

x

φ+ϑ⋅ψ=ω

φ⋅ϑ−φ⋅ϑ⋅ψ=ω

φ⋅ϑ+φ⋅ϑ⋅ψ=ω

ζ

η

ξ

&&

&&

&&

cos

sincossin

cossinsin

ϑ⋅φ⋅ψ⋅+φ+ϑ+ψ==ω+ω+ω=ω+ω+ω=ω ζηξ cos&&&&&Kr

22222222z

2y

2x

( ) ( ) ( ) kxyjzxiyz

zyx

kji

rv yxxzzyzyx

rrr

rrr

rrr ⋅⋅ω−⋅ω+⋅⋅ω−⋅ω+⋅⋅ω−⋅ω=ωωω=×ω=rvrrr ×ω=

yzv zyx ⋅ω−⋅ω=zxv xzy ⋅ω−⋅ω=

xyv yxz ⋅ω−⋅ω=

( ) ( ) ( ) ttt

ttt

kji

kji

rvrrr

rrr

rrr ⋅ξ⋅ω−η⋅ω+⋅ζ⋅ω−ξ⋅ω+⋅η⋅ω−ζ⋅ω=ζηξ

ωωω=×ω= ηξξζζηζηξ

η⋅ω−ζ⋅ω= ζηξv

ζ⋅ω−ξ⋅ω= ξζηv

ξ⋅ω−η⋅ω= ηξζv

the circumferential velocity

x

yz

the cyclic change

ξ

ηζ

the Euler’s kinematic equations the angular velocity ωωωω


212

( ) kjikji zyxzyx

r&

r&

r&

rrr&rr

⋅ω+⋅ω+⋅ω=⋅ω+⋅ω+⋅ω=ω=ε•

in fixed coordinate system xyz

kji zyx

rrrr⋅ε+⋅ε+⋅ε=ε xx ω=ε & zz ω=ε &yy ω=ε &

ψ⋅ψ⋅ϑ−ψ⋅ϑ+ψ⋅ϑ⋅ψ⋅φ+ψ⋅ϑ⋅ϑ⋅φ+ψ⋅ϑ⋅φ=ε sincoscossinsincossinsin &&&&&&&&&&x

ψ⋅ψ⋅ϑ+ψ⋅ϑ+ψ⋅ϑ⋅ψ⋅φ+ψ⋅ϑ⋅ϑ⋅φ−ψ⋅ϑ⋅φ−=ε cossinsinsincoscoscossin &&&&&&&&&&y

ψ+ϑ⋅ϑ⋅φ−ϑ⋅φ=ε &&&&&& sincosz

( ) ttttttttt kkjjiikji&rr

&&rr

&&rr

&rrr

&rr⋅ω+⋅ω+⋅ω+⋅ω+⋅ω+⋅ω=⋅ω+⋅ω+⋅ω=ω=ε ζζηηξξ

•

ζηξ

tttttt kjikji&r&r&rr

&r

&r

&r

⋅ω+⋅ω+⋅ω+⋅ω+⋅ω+⋅ω=ε ζηξζηξ

tir

the radius vector of the point 1,0,0 tjr

tkr

0,1,0 0,0,1

rvrrrr&r ×ω== tt ivi

rrr&r ×ω== tt jvjrrr&r ×ω== tt kvk

rrr&r ×ω==

( ) ( ) ( )=×ω⋅ω+×ω⋅ω+×ω⋅ω=⋅ω+⋅ω+⋅ω ζηξζηξ tttttt kjikjirrrrrr&r&r&r

( ) 0kji ttt

rrrrrrr=ω×ω=⋅ω+⋅ω+⋅ω×ω= ζηξ

the angular acceleration εεεε


in body coordinate system ξηζξηζξηζξηζ

Dynamics

213

( ) kjikji zyxzyx

r&

r&

r&

rrr&rr

⋅ω+⋅ω+⋅ω=⋅ω+⋅ω+⋅ω=ω=ε•

kji zyx

rrrr⋅ε+⋅ε+⋅ε=ε xx ω=ε & zz ω=ε &yy ω=ε &

ψ⋅ψ⋅ϑ−ψ⋅ϑ+ψ⋅ϑ⋅ψ⋅φ+ψ⋅ϑ⋅ϑ⋅φ+ψ⋅ϑ⋅φ=ε sincoscossinsincossinsin &&&&&&&&&&x

ψ⋅ψ⋅ϑ+ψ⋅ϑ+ψ⋅ϑ⋅ψ⋅φ+ψ⋅ϑ⋅ϑ⋅φ−ψ⋅ϑ⋅φ−=ε cossinsinsincoscoscossin &&&&&&&&&&y

ψ+ϑ⋅ϑ⋅φ−ϑ⋅φ=ε &&&&&& sincosz

( ) ttttttttt kkjjiikji&rr

&&rr

&&rr

&rrr

&rr⋅ω+⋅ω+⋅ω+⋅ω+⋅ω+⋅ω=⋅ω+⋅ω+⋅ω=ω=ε ζζηηξξ

•

ζηξ

ttt kjirrrr

⋅ε+⋅ε+⋅ε=ε ζηξ ξξ ω=ε & ζζ ω=ε &ηη ω=ε &

φ⋅φ⋅ϑ−φ⋅ϑ+φ⋅ϑ⋅φ⋅ψ+φ⋅ϑ⋅ϑ⋅ψ+φ⋅ϑ⋅ψ=εξ sincoscossinsincossinsin &&&&&&&&&&

φ⋅φ⋅ϑ−φ⋅ϑ−φ⋅ϑ⋅φ⋅ψ−φ⋅ϑ⋅ϑ⋅ψ+φ⋅ϑ⋅ψ=εη cossinsinsincoscoscossin &&&&&&&&&&

φ+ϑ⋅ϑ⋅ψ−ϑ⋅ψ=εζ&&&&&& sincos

in fixed coordinate system xyzthe angular acceleration εεεε



Dynamics

214

( ) vrrrrvarrrr&rrr&rrr&rr

×ω+×ε=×ω+×ω=×ω== •

( ) ( ) ( ) kxyjzxiyz

zyx

kji

r yxxzzyzyx

rrr

rrr

rr⋅⋅ε−⋅ε+⋅⋅ε−⋅ε+⋅⋅ε−⋅ε=εεε=×ε

yza zyx ⋅ε−⋅ε= xya yxz ⋅ε−⋅ε=zxa xzy ⋅ε−⋅ε=

( ) ( ) ( ) kvvjvvivv

vvv

kji

v xyyxzxxzyzzy

zyx

zyx

rrr

rrr

rr⋅⋅ω−⋅ω+⋅⋅ω−⋅ω+⋅⋅ω−⋅ω=ωωω=×ω

yzzyx vva ⋅ω−⋅ω=zxxzy vva ⋅ω−⋅ω=

xyyxz vva ⋅ω−⋅ω=

kajaiaa zyx

rrrr⋅+⋅+⋅= yzzyzyx vvyza ⋅ω−⋅ω+⋅ε−⋅ε=

xyyxyxz vvxya ⋅ω−⋅ω+⋅ε−⋅ε=

zxxzxzy vvzxa ⋅ω−⋅ω+⋅ε−⋅ε=


the circumferential acceleration in fixed coordinate system xyz

Dynamics

215


( ) vrrrrvarrrr&rrr&rrr&rr

×ω+×ε=×ω+×ω=×ω== •

( ) ( ) ( ) ttt

ttt

kji

kji

rrrr

rrr

rr⋅ξ⋅ε−η⋅ε+⋅ζ⋅ε−ξ⋅ε+⋅η⋅ε−ζ⋅ε=

ζηξεεε=×ε ηξξζζηζηξ

( ) ( ) ( ) ttt

ttt

kvvjvvivv

vvv

kji

vrrr

rrr

rr⋅⋅ω−⋅ω+⋅⋅ω−⋅ω+⋅⋅ω−⋅ω=ωωω=×ω ξηηξζξξζηζζη

ζηξ

ζηξ

η⋅ε−ζ⋅ε= ζηξa ξ⋅ε−η⋅ε= ηξζaζ⋅ε−ξ⋅ε= ξζηa

ηζζηξ ⋅ω−⋅ω= vva

ζξξζη ⋅ω−⋅ω= vvaξηηξζ ⋅ω−⋅ω= vva

ttt kajaiaarrrr

⋅+⋅+⋅= ζηξ ηζζηζηξ ⋅ω−⋅ω+η⋅ε−ζ⋅ε= vva

ξηηξηξζ ⋅ω−⋅ω+ξ⋅ε−η⋅ε= vva

ζξξζξζη ⋅ω−⋅ω+ζ⋅ε−ξ⋅ε= vva


the circumferential acceleration

Dynamics

216

the Euler’s equations of motion

( )( )( ) ∑

∑∑

ζξηηξζζ

ηζξξζηη

ξηζζηξξ

=−⋅ω⋅ω+ε⋅

=−⋅ω⋅ω+ε⋅

=−⋅ω⋅ω+ε⋅

i

i

i

MIII

MIII

MIII

_

_

_

the kinetic energy

ζηηζζξξζ

ηξξηζζ

ηηξξ

ω⋅ω⋅−ω⋅ω⋅−

−ω⋅ω⋅−ω⋅⋅+

+ω⋅⋅+ω⋅⋅=

DD

DI

IIE2

21

2212

21

K

ξξ ⋅−= TamD

( )[ ]( )[ ]( )[ ]ξηηξζζζ

ζξξζηηη

ηζζηξξξ

−⋅ω⋅ω+ε⋅−=

−⋅ω⋅ω+ε⋅−=

−⋅ω⋅ω+ε⋅−=

IIIM

IIIM

IIIM

D

D

D

_

_

_

ηη ⋅−= TamD

ζζ ⋅−= TamD


0MM

0MM

0MM

Di

Di

Di

=+

=+

=+

ζζ

ηη

ξξ

∑∑∑

__

__

__

0DF

0DF

0DF

i

i

i

=+

=+

=+

ζζ

ηη

ξξ

∑∑∑

_

_

_

?

?

?

===

⇒

ζ

η

ξ

R

R

R

7. Spherical motiondynamics

the d’Alembert force ... and moment

the

Eul

er’s

equa

tions

of m

otio

n

Dynamics

217


the special cases

the heavy flywheel (the Lagrange flywheel)

the body is axially symmetrical,the symmetry axis is coincidentwith the axis of primary rotation

supported in the center of spherical motion

G

the zero force flywheel (the Euler flywheel)

the body is axially symmetrical,the symmetry axis is coincidentwith the axis of primary rotation

supported in the center of spherical motionwhich is coincident with the center of gravity

the precession


the precession


only the gravitational force G acts

G

Dynamics

218

the two uniform rotations about two concurrent axis of constant angle

the body is axially symmetrical,the symmetry axis is coincident with the axis of primary rotation

konst

const=ψ&

const=φ&

0=ϑ

=ϑ&

const


the precession


zero nutation

the heavy flywheel

Dynamics

219

konst

ϑ=const

ϑ⋅⋅ω⋅=⋅= ψ sinG2

Gnn rmamD

( ) ψφφ

ψ ω×ω⋅

ϑ⋅

ωω

⋅−+=rrr

cosIIIM ssG

ψωr

φωr

G

Dn

MG

I s

rG

I = I G + m·rG2

ψφ ω×ω⋅=rrr

sG IM a) I s = I

b) ϑ = 90º

c) ωψ << ωφ

ϑ⋅ω⋅ω=ω×ω ψφψφ sinrr

( ) ϑ⋅ω⋅ω⋅

ϑ⋅

ωω

⋅−+= ψφφ

ψ sincosIIIM ssG

ϑ⋅ω⋅ω⋅= ψφ sinsG IM

I G

aGn

rG·sinϑ

the kinetic energy

( )[ ]2s

22K II

2

1E φψψ ω+ϑ⋅ω⋅+ϑ⋅ω⋅⋅= cossin



the precession

const=ψ&

const=φ&


the heavy flywheel


Dynamics

220

( ) ϑ⋅ω⋅ω⋅

ϑ⋅

ωω

⋅−+=

=ϑ⋅⋅

ψφφ

ψ sincos

sin

III

rG

ss

G

GG MrG =ϑ⋅⋅ sin

( ) ψφφ

ψ ω⋅ω⋅

ϑ⋅

ωω

⋅−+=⋅ cosIIIrG ssG

konst

ϑ=const

ψωr

φωr

GaGn

MG

I s

rG

I GG

rG·sinϑ

I = I G + m·rG2

the precession

const=ψ&

const=φ&

ϑ⋅⋅ω⋅=⋅= ψ sinG2

Gnn rmamD

( ) ψφφ

ψ ω×ω⋅

ϑ⋅

ωω

⋅−+=rrr

cosIIIM ssG




the heavy flywheel


Dynamics

221

The mechanism serve the purpose to transmit the force and to transform the motion.

tran

slat

ion

rotation

The mechanism is the system (chain) of rigid bodies, linked one to the other by joints.

8. The mechanismsDynamics

222

The basic termsthe member of mechanism

- the body, do not change the shape

the frame – the member fixed to the ground (the Earth)

the kinematic pair – the pair of members, linked together by joint

the joint coordinate- the coordinate, determining the relative position of the members

one to the other

the driving member, the powered member - the member on the driving end of the chain- the member performing the function for which the mechanism was designed

the input member, the output member- the member on the beginning and on the end of the chain

the number of degrees of freedom (DOF)- the number of independent motions the mechanism is able to perform

the kinematic schema- the geometric sketch of the mechanism, simplified as more as possible

the mechanism coordinateone or more independent coordinates, determining the position of the mechanism


223

The kinematic schemeThe geometric sketch of the mechanism, simplified as more as possible ....- to retainthe dimensions necessaryfor the kinematicfunction of the mechanism,- to suppressthe dimensions not importantfor the kinematic function of the mechanism.

The crank mechanism ... and its kinematic scheme

the frame

the crank

the connecting rod

the piston

the frame

the connecting rod length

the crank length

The crankshaft designed as the eccentric pin.

the crank

the pin

ojnice

the eccentricity - - the function length of the crank

The kinematic scheme can be very different from the real mechanism design.


224

The mechanism classification

the planar mechanism the space mechanism

the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion

3

2

4

1

1


225

the classification with respect to the number of degrees of freedom

The instant position of the mechanism is unambiguously determined by the coordinateswhich number is equal to the number of the mechanism degrees of freedom (DOF)

The number of mechanism degrees of freedom is equalto the number of independent coordinates, determining the mechanism position



the mechanism coordinateone or more independent coordinates, determining the position of the mechanism;the number of the mechanism coordinates is equal to the number of the mechanism DOF


226




posu

v

rotace

the mechanisms with 1 DOF


227





the differential satellites

the driving shaft

the wheels

the differential cage

the differential gear

5

1 1

4 3

2

φ ψ

ω5 ω2

the mechanism with 2 DOF


φφφφ and ψψψψ – two mechanism coordinates

the car goes through the curve –the wheels rotate with different speed


228






the mechanisms with more DOF

the number of the mechanism DOF is not limited the mechanism with 7 DOF


229







... with respect to the drive ratio

the mechanisms with constant drive ratio

ω1

ω2

konst=ωω=

1

2DR

the mechanisms with changing drive ratiokonst≠

ω= v

DR

v

ω


230







the mechanisms with constant drive ratio

the mechanisms with changing drive ratio

1 2

1

2

3 2

1 1

1

2

2

1 1

4

3

... with respect to the drive ratio

... with respect to the number of members

the 2 members mechanisms



the more members mechanisms


231

y

ψ

φ

A

C

B

3

2

1

1

The only one coordinate is independent, the other two can be calculated.One degree of freedom.

yBCAB

BCAB

=ψ⋅+φ⋅ψ⋅=φ⋅

coscos

sinsin

The number of the mechanism DOF we can determine intuitively or calculate.

( ) 21 c2c11n3i ⋅−⋅−−⋅=

( ) ∑=

⋅−−⋅=5

1jjcj1n6i

( ) 12211133i =⋅−⋅−−⋅=

The number of DOF formulai – the number of DOF,n – the number of members

(including the frame),c1 – the number of the 1st class joints

(fixing 1 DOF),c2 – the number of the 2nd class joints

(fixing 2 DOF).

The space mechanismcj – the number of the j class joints

(fixing j DOF).

The number of the mechanism degrees of freedom (DOF)

The number of mechanism DOFis equal to the number of the independent mechanism coordinates.


232

The joints are either planar (2D) or space (3D)

The joints

The planar joints (2D)– links two mechanism members,performing planar motion in reciprocally parallel planes

The space joints (3D)– links two mechanism members,from which at least one perform space motion

The joints are either idealized or real.

The idealized joints– the friction is negligible

The real joints– the friction is taken into account

We are interested in :

The joint mechanical properties with respect to statics

the transmition of forces

The joint mechanical properties with respect to kinematics

the relative motion of one member with respect to the other the joint allows or fixes

The joint is of the j class if it fixes j relative motions (fixes j degrees of freedom)


233

The planar (2D), idealized joints are :

The joints

the 3rd class joint the perfect fixing

fixes both 2 translations and rotation transmits 2 forces and moment

the frame

the body

the body

the body

the perfect fixing – the kinematic scheme

fixes 3 motions, does not allow neither one


234

fixes 2 translations

allows rotation


The joints

the 3rd class joint the perfect fixingthe 2nd class joints the pin joint

fixes 3 motions, does not allow neither onefixes 2 translations, allows rotation


235

transmits forces

does not transmit moment

the frame

the body the body

the body


The joints

the 3rd class joint the perfect fixingthe 2nd class joints the pin joint

fixes 2 translations

allows rotation

the pin joint – the kinematic scheme

fixes 3 motions, does not allow neither onefixes 2 translations, allows rotation


236


The joints

the 3rd class joint the perfect fixing fixes 3 motions, does not allow neither onethe 2nd class joints the pin joint fixes 2 translations, allows rotation

fixes 1 translation and rotation, allows translationthe sliding joint

fixes 1 translation and rotation

allows translation


237


The joints


fixes 1 translation and rotation, allows translationthe sliding joint

fixes 1 translation and rotation

allows translation

transmits force and moment

does not transmit force

the sliding joint – the kinematic scheme


238


The joints


fixes 1 translation and rotation, allows translationthe sliding jointfixes 1 translation, allows rotation and translation,

related one to the otherthe rolling joint

r φ

x

ε⋅=ω⋅=φ⋅=

ra

rv

rx

fixes 1 translation, allows rotation and translation,related one to the other

no sliding in touch point


239

ε⋅=ω⋅=φ⋅=

ra

rv

rx

r φ

x


The joints


fixes 1 translation and rotation, allows translationthe sliding jointfixes 1 translation, allows rotation and translation,

related one to the otherthe rolling joint

transmits forces,does not transmit moment

fixes 1 translation, allows rotation and translation,related one to the other

no sliding in touch point


240

z

φ


The joints


fixes 1 translation and rotation, allows translationthe sliding jointfixes 1 translation, allows rotation and translation,the rolling joint

the 1st class joints the sliding pin joint fixes 1 translation, allows rotation and translation,(independent)

fixes 1 translation

allows translation and rotation (independent)


241

z

φ


The joints



the 1st class joints the sliding pin joint fixes 1 translation, allows rotation and translation,(independent)

fixes 1 translation


does not transmit force and moment

transmits force

the sliding pin joint – the kinematic scheme


242

φ


The joints



the 1st class joints the sliding pin joint fixes 1 translation, allows rotation and translation,the general joint fixes 1 translation, allows rotation and translation,

(independent)

fixes 1 translation


sliding in touch point


243

φ

fixes 1 translation


does not transmit force and moment

transmits force

sliding in touch point


The joints




(independent)


244

not exactly the same way for the real joints


The joints




(independent)

the joint transmits such force or moment,in what direction it fixes the translation or rotat ion

if the joint fixes translation in certain direction, it transmits appropriate forceif the joint fixes rotation about certain axis, it transmits appropriate moment

if the joint allows translation in certain direction, it does not transmit appropriate forceif the joint allows rotation about certain axis, it does not transmit appropriate moment


245

The space (3D), idealized joints are :

The joints

the 6th class joint the perfect fixing fixes 3 translations and 3 rotationstransmits 3 forces and 3 moments


246


The joints

the 6th class joint the perfect fixing fixes 3 translations and 3 rotations

the 5th class joint the rotating joint fixes 3 translations and 2 rotationsallows 1 rotationtransmits 3 forces and 2 momentsdoes not transmit the momentto the axis of rotation

allows rotation φ


247


The joints


the 5th class joint the rotating joint fixes 3 translations and 2 rotationsfixes 2 translations and 3 rotationsallows 1 translationtransmits 2 forces and 3 momentsdoes not transmit the force

the sliding joint

allows translation

z


248


The joints


the 5th class joint the rotating joint fixes 3 translations and 2 rotationsfixes 2 translations and 3 rotationsthe sliding joint

the spiral joint fixes 2 translations and 2 rotationsallows translation and rotation relates one to the other

z φ

ε⋅π⋅

=

ω⋅π⋅

=

φ⋅π⋅

=

2

sa

2

sv

2

sz

s – the spiral lead- the translation corresponding to the 360º rotation (2·π ≅ 6,28 rad)


249


The joints




the 4th class joint the sliding-rotating joint

allows independent translation and rotation

z φ

fixes 2 translations and 2 rotationsallows independent translation and rotationtransmits 2 forces and 2 momentsdoes not transmit force and moment


250


The joints




the 4th class joint the sliding-rotating joint fixes 2 translations and 2 rotationsallows independent translation and rotation

allows two rotations

φ

ψ

the double-rotating joint fixes 3 translations and 1 rotationallows 2 independent rotationstransmits 3 forces and moment,does not transmit 2 moments


251


The joints






allows three rotations

ψ, ϑ, φ

the 3rd class joint the spherical joint fixes 3 translations,allows 3 independent rotationstransmits 3 forces,does not transmit 3 moments


252


The joints







the 2nd class joint the sliding spherical joint fixes 2 translations, allows translation and 3 independent rotationstransmits 2 forces, does not transmit the force and 3 moments

z ψ, ϑ, φ

allows translation and three rotations


253


The joints








the 1st class joint the general kinematic pair fixes 1 translations, allows 2 translations and 3 rotationstransmits the force, does not transmit the 2 forces and 3 moments

allows all motions except one translation normal to the common tangential plane


254


The joints








the 1st class joint the general kinematic pair fixes 1 translations, allows 2 translations and 3 rotationstransmits the force, does not transmit the 2 forces and 3 moments

not exactly the same way for the real joints

the joint transmits such force or moment,in what direction it fixes the translation or rotat ion

if the joint fixes translation in certain direction, it transmits appropriate forceif the joint fixes rotation about certain axis, it transmits appropriate moment

if the joint allows translation in certain direction, it does not transmit appropriate forceif the joint allows rotation about certain axis, it does not transmit appropriate moment


255

the three members mechanisms

the cam mechanism

3

2

1

1

the theoretical outline 3

2

1

1

3

2

1

1

3 2

1

1

3 2

1

1

3

2

1

3

1

2

The types of mechanisms 8. The mechanismsDynamics

256

3

2

1

1

the three members mechanisms

the mechanism with the general kinematic pair

The types of mechanisms 8. The mechanismsDynamics

257

3 2

1 1

4

3 2

1

1 4

excentricity

1

1

2 4 3

the four members mechanisms

the crank mechanism

The types of mechanisms

the centric crank mechanism the excentric crank mechanism

the excentric crank mechanism


258

1

2 4

3

1

1 1

2 4

3

the paralelogram

1 1

2 4 3

D

C B

A

1 1

2

4

3

D

C B

A


the four-joints mechanism


transmits rotationin 1 : 1 ratio

translation


259

1

4

3 2

excentricity

1

3

4

2

1

4

3 2

1

4

3

2

1

3

4

2

1

1

3

4

2

1

the coulisse mechanism



the centric coulisse mechanism the centric coulisse mechanism

the excentric coulisse mechanism

the right-angled coulisse


260

3

2 4

1

3 2

1

4

nesouosost

transmits rotation in 1:1 ratio with certain misalignment of the axis

the Oldham clutch




261

3

2

4

1

1

1

6

5


the more-members planar mechanisms

the sewing machine mechanism


262

3

2

4

1

1


the space mechanisms

the 3D crank mechanism


263

The mechanism drive ratio

the analytical solution of the orthogonal coulisse mechanism

y φ ω, ε

r

a ,vrr

φ⋅= sinry

φ⋅φ⋅−φ⋅φ⋅=

φ⋅φ⋅=

sincos

cos2rry

ry&&&&&

&&

ε=φ

ω=φ&&

&

ay

vy

==

&&

&

φ⋅⋅ω−φ⋅⋅ε=φ⋅⋅ω=

sincos

cos

rra

rv2

DR - the drive ratio

the mechanism with variable drive ratio ( )φ=ω

= fv

p

the DR derivative

9. The kinematics of mechanismsDynamics

264

t

t

ε=φ

ω=φ&&

&

kolečko

talíř φ

ψ

R r

ωt,εt

ωk, εk

v = ωt·R = ωk·r

Rs ⋅φ=

r

s=ψ

r

R⋅φ=ψ

pr

Rttk ⋅ω=⋅ω=ω

pr

Rttk ⋅ε=⋅ε=ε

rs ⋅ψ=

r

R⋅φ=ψ

r

R⋅φ=ψ &&

k

k

ε=ψω=ψ

&&

& r

Rtk ⋅ω=ω &&

The mechanism drive ratio 9. The kinematics of mechanisms

the analytical solution of the chain mechanism

the mechanism with constant drive ratio

DRthe drive ratio

const==φψ=

r

R

d

dp

Dynamics

265

pr

Rttk ⋅ω=⋅ω=ω

kolečko

talíř φ

ψ

R r

ωt,εt

ωk, εk

v = ωt·R = ωk·r

Rs ⋅φ=

const==φψ=

r

R

d

dp

pr

Rttk ⋅ε=⋅ε=ε

r

R⋅φ=ψ

r

R⋅φ=ψ &&

r

Rtk ⋅ω=ω &&

qp 2ttk ⋅ω+⋅ε=ε0

d

dpq =

φ=

The DR – drive ratio is constant,does not vary

The acceleration of the driving and driven mechanism memberare in the same ratio (DR) as velocities


the analytical solution of the chain mechanism

Dynamics

266

R

r

ω1, ε1

ω2, ε2

2 1

v = ω1·r = ω2·R

R

rp =

p

p

12

12

⋅ε=ε⋅ω=ω

the driving wheel (small)

the driven wheel (large)

p < 1

2

1

r

Ri

ωω== i > 1

R

r ω1, ε1

1

2 ω2, ε2

γ δ

s

δγ=

δ⋅γ⋅==

sin

sin

sin

sin

s

s

R

rp

p

p

12

12

⋅ε=ε⋅ω=ω

the deceleration drive

The mechanism drive ratio 9. The kinematics of mechanismsDynamics

267

r1

1 2 3

R2

r2R3

ω1

ω2 ω3

2

112 R

r⋅ω=ω

3

223 R

r⋅ω=ω

3

2

2

113 R

r

R

r ⋅⋅ω=ω

p

p

13

13

⋅ε=ε⋅ω=ω

3

2

2

12312 R

r

R

rppp ⋅=⋅=

2

112 R

rp =

3

223 R

rp =

the drive ratios are multiplied


the combination of drive ratios

Dynamics

268

1 2

ω1 ω2

R

r

ω1, ε1

ω2, ε2

2 1

v = ω1·r = ω2·R the non sliding rolling with friction transmision

the teeth – the mechanical restraint against sliding

1 2

ω1 ω2

the pitch circle

1 2

ω1 ω2

R r

the pitch circle


269

the gear

1 2

ω1 ω2

ω1 ω2

the pushing surface

the pushed surface

the mechanism with varying DR

p=ω2/ω1

φ1p=ω2/ω1

φ1

If the kinematic pair ofpushing and pushed surfaceshould represent the constantdrive ratio mechanismthe both surfaces must havethe specific shapeof epicycloids


270

the double gear

r1

1 23

R2

r2R3

ω1

ω2

ω3

1

2

3

předloha

p

p

13

13

⋅ε=ε⋅ω=ω

3

2

2

12312 R

r

R

rppp ⋅=⋅=

ω1 1

3 2

ω3

r1 R

3

2

1 ω1 ω3

r2

r3

the satellite gear

π

A

SSv

r

Avr

SA v2v ⋅=

1

A1 r

v=ω

3

S3 r

v=ω

3

1

A

1

3

S

1

3

r2

r

v

r

r

vp

⋅=⋅=

ωω=

213 rrr +=

21 r2rR ⋅+=

1

1

21

1

rR

r

rr

r

2

1p

+=

+⋅=

the outer wheel

the satellite

the carrier

the central wheel


271

ω1 1

3 2

ω3

r1 R

3

2

1 ω1 ω3

r2

r3

the satellite gear

π

A

SSv

r

Avr

SA v2v ⋅=

1

A1 r

v=ω

3

S3 r

v=ω

3

1

A

1

3

S

1

3

r2

r

v

r

r

vp

⋅=⋅=

ωω=

the outer wheel

the satellite

the carrier

the central wheel


272

the pulley block

v,a

vl=2·v

v

vl=2·v

π

v2v ⋅=l

a2a ⋅=l

v,a

vl=4·v

2·v 2·v

v

2·v

π

4·v

r r/3

v4v ⋅=l

a4a ⋅=l


273

the variators

konst=ωω

=vstupní

výstupníp

pvstupnívýstupní ⋅ε=ε

qp 2vstupnívstupnívýstupní ⋅ω+⋅ε=ε

0d

dpq =

φ=

pvstupnívýstupní ⋅ω=ω


274

the mechanism s u1

u3

u2

the analytical solution

s – the driving member coordinate- the input coordinate- the coordinate of the mechanism

the number of the coordinates of mechanism is equal to the number of the mechanism DOF

u – the driven member coordinate- the output coordinate


length or angle

the number of the output coordinates is arbitrary

1. the position problem

( )sfu =

2. the velocity solution

mechanism with 1 DOF - degree of freedom

( )( ) ( ) ( )

dt

ds

ds

du

dt

duv tsts

out ⋅==

( )in

t vdt

ds=

( )( )s

s pds

du=

( ) insout vpv ⋅=

the mechanism as „geometric converter“

the travel function

DRthe drive ratio



275

the mechanism s u1

u3

u2



( )sfu =



( )( ) ( ) ( )

dt

ds

ds

du

dt

duv tsts

out ⋅==

( )vst

t vdt

ds=

( )( )s

s pds

du=

( ) vstsvýst vpv ⋅=


the travel function

DRthe drive ratio



3. the acceleration solution

( )( ) ( )( ) dt

dvpv

dt

dp

dt

vpd

dt

dva in

sinsinsout

out ⋅+⋅=⋅

==

( )( ) dt

dvpv

dt

ds

ds

dpa in

sins

out ⋅+⋅⋅=

invdt

ds = inin a

dt

dv =

( )( )s

s qds

dp=

( ) ( )2

insinsout vqapa ⋅+⋅=

the DR derivative

the generalized acceleration

Dynamics

276

the mechanism s u1

u3

u2



( )sfu =




the travel function



( ) insout vpv ⋅=

( ) ( )2

insinsout vqapa ⋅+⋅=

( )( )s

s pds

du=

( )sfu =

( )( )s

s qds

dp=

the conversion functions

the travel function

DR - the drive ratio

the DR derivative

Dynamics

277

the mechanism r u1

u3

u2 s ( )srfu ,=

( ) ( ) dss

udr

r

udu srsr ⋅

∂∂

+⋅∂

∂= ,,

dt

ds

s

u

dt

dr

r

u

dt

duvvýst ⋅

∂∂+⋅

∂∂==

( ) ( )sr2vstsr1vstvýst 2pv1pvv ,, ⋅+⋅=

( )( )

r

u1p sr

sr ∂∂

= ,, ( )

( )

s

u2p sr

sr ∂∂

= ,,

the total differentiate

12qvv2

22qv11qv

2pa1paa

2vst1vst

22vst

21vst

2vst1vstvýst

⋅⋅⋅++⋅+⋅+

+⋅+⋅=

( )( )

r

1p11q sr

sr ∂∂

= ,, ( )

( )

s

2p22q sr

sr ∂∂

= ,,

( )( ) ( )

r

2p

s

1p12q srsr

sr ∂∂

=∂

∂= ,,

,

the analytical solution mechanism with 2 DOF - degrees of freedom




the travel function



Dynamics

278

( )sfu =the trigonometry method

The trigonometry method meansthe intuitive use of several geometriclaws, rules and solutions,applied to the mechanism geometry

φ b r

x

4

3 2

1

ω32, ε32

a ,vrr

φ⋅⋅⋅−+= cosbr2brx 222

( ) φ⋅⋅⋅−+=φ cosbr2brx 22

the analytical solution mechanism with 1 DOF - degree of freedom

9. The kinematics of mechanisms


the travel function

Dynamics

279

( )sfu =the vector method

the kinematic scheme - the chain of members

the closed vector loop

φ1

φ2

φi

φn

1rr

2rr

irr

nrr

0rrrrrn

1iini21

rrrK

rK

rr==+++++ ∑

= 0r

0r

n

1iii

n

1iii

=φ⋅

=φ⋅

∑

∑

=

=

sin

cos





the travel functionThe vector method meansthe replacement of the kinematic schemeby the vector chain – the closed loop.The equations of the closed vector chainrepresents the position problem solution

Dynamics

280

3

2

1

C

B

A

ψ

4

φ

ω2,ε2

ω3, ε3

r

z H

ω4, ε4

3434 a ,vrr

ψ

φ

Hr

zr

rr

0Hzrrrrr

=++

0Hzr

0zr

=+ψ⋅−φ⋅=ψ⋅−φ⋅

coscos

sinsin

φ⋅+φ⋅=ψ

φ⋅+φ⋅=ψ

cos

sinarctan

cos

sintan

rH

r

rH

r

2224

24

qp

p

ω⋅+ε⋅=ε

ω⋅=ω

( )( )

( )( ) ( )

2

2

d

d

d

dpq

d

dp

φψ

=φ

=

φψ

=

φφφ

φφ






the travel function

Dynamics

281

3

2

1

C

B

A

ψ

4

φ

ω2,ε2

ω3, ε3

r

z H

ω4, ε4

3434 a ,vrr

ψ

φ

Hr

zr

rr

0Hzrrrrr

=++

0Hzr

0zr

=+ψ⋅−φ⋅=ψ⋅−φ⋅

coscos

sinsin

0zzr

0zzr

=ψ⋅ψ⋅+ψ⋅−φ⋅φ⋅−

=ψ⋅ψ⋅−ψ⋅−φ⋅φ⋅

sincossin

cossincos

&&&

&&&

2ω=φ& 4ω=ψ& 34vz =&

0zzz2

zrr

0zzz2

zrr

2

2

2

2

=ψ⋅ψ⋅+ψ⋅ψ⋅+ψ⋅ψ⋅⋅++ψ⋅−φ⋅φ⋅−φ⋅φ⋅−

=ψ⋅ψ⋅+ψ⋅ψ⋅−ψ⋅ψ⋅⋅−

−ψ⋅−φ⋅φ⋅−φ⋅φ⋅

cossinsin

coscossin

sincoscos

sinsincos

&&&&&

&&&&&

&&&&&

&&&&&

2ε=φ&& 4ε=ψ&& 34az =&&






the travel function

Dynamics

282

π

4

3

D

CB

A

2

1

1

nB nC

ω4=?

ω2

Bvr

Cvr

given : ω2

find : ω4

π

3

CB

ω3

Bvr

Cvr

ABv 2B ⋅ω=π

=ωB

vB3

ππ⋅=π⋅ω=

B

CvCv B3CCD

C

B

AB

CD

v2

C4

π⋅π

⋅ω==ω

the pole solution


283

4

3

D

CB

A

2

1

γ

ψ

φ

π

C B ψ-γ

φ+γ

180º-(φ+ψ)

s

v

ADCDBCAB

ADCDBCAB

−=ψ⋅−γ⋅−φ⋅=ψ⋅+γ⋅+φ⋅

sinsinsin

coscoscos

( )[ ] ( )

( ) ( )γ+φπ=

γ−ψπ=

=ψ+φ

=ψ+φ−°

sinsin

sinsin

CB

BC

180

BC

( )( )ψ+φ

γ−ψ⋅=πsin

sinBCB

( )( )ψ+φ

γ+φ⋅=πsin

sinBCC

ADV

ADS

the pole solution


284

A

π

B nA

nB

Bvr

Avr

3

2

1

C

B

A

4

ω2

π

nB

nC

D

nD •

Bvr

Dvr

the pole solution


285

the basic decomposition

A

C

B 4 3

2

1

ω2,ε2 CC a ,vrr

CBBC vvvrrr +=

CBnCBtBnBtCnCt aaaaaarrrrrr

+++=+

A

C B

A

C B C B

+ =

the resulting motion the general plane motion

the carrying motion translation

the relative motion rotation + =

CBBC aaarrr

+=

Bvr

CBvr

Cvr

Car

Bnar

CBnar

CBtar Bta

rABAB

BC

AB

BCBC

BC

va

2CB

CBn =AB

va

2B

Bn =

given : find :

the motion decomposition


286

A

C

B 4 3

2

1

ω2,ε2 CC a ,vrr

CBBC vvvrrr +=

CBnCBtBnBtCnCt aaaaaarrrrrr

+++=+CBBC aaa

rrr+=

A

C

B

φ ψ

φ ψ φ

Bvr

CBvr

Cvr

ψ φ

φ

ψ

Btar

Bnar

CBtar

CBnar

Car

ψ⋅−φ⋅=ψ⋅−φ⋅=

coscos

sinsin

CBB

CBBC

vv0

vvv

ψ⋅−ψ⋅−φ⋅−φ⋅=ψ⋅+ψ⋅−φ⋅+φ⋅=

sincossincos

cossincossin

CBnCBtBnBt

CBnCBtBnBtC

aaaa0

aaaaa

BC

va

2CB

CBn =AB

va

2B

Bn =

the basic decomposition

given : find :



287

C

B

D A

ω2,ε2

4 3

2

1

A BtB a ,v

rrCtC a ,v

rr

B

AC

D

B

CD

=

+

B

A

C D

ω2,ε2

aBn

δ

ψ

φ ω41,ε41

BtB a ,vrr

3434 a ,vrr

3441B vvvrrr +=

ψ⋅+δ⋅=φ⋅ψ⋅+δ⋅−=φ⋅

sincoscos

cossinsin

3441B

3441B

vvv

vvv

δ ψ

φ

34vr

Bvr

41vr

B

4 : 1

3 : 4

BD

v4141 =ω

ABv 2B ⋅ω=

AB

BC

BD

B

C

D


given : find :



288

δ

ψ

φ

a41t

φ

δ 41na

r

34ar

Corar

Btar

Bnar

B

B

AC

D

B

C

D

B

CD

=

+Cor3441B aaaa

rrrr++=

BDBD

va 2

41

241

n41 ⋅ω==

3441Cor v2a ⋅ω⋅=

C

B

D A

ω2,ε2

4 3

2

1

A BtB a ,v

rrCtC a ,v

rr

B

A

C D

ω2,ε2

aBn

δ

ψ

φ ω41,ε41

BtB a ,vrr

3434 a ,vrr

Cor34t41n41BtBn aaaaaarrrrrr

+++=+

4 : 1

3 : 4

BD

v4141 =ω

BD

a t4141 =ε

ABv 2B ⋅ω= ABa 2Bt ⋅ε=

ABa 22Bn ⋅ω=

BC

AB

ABBD

BD BCψ⋅+ψ⋅+δ⋅+δ⋅=

=φ⋅+φ⋅ψ⋅+ψ⋅+δ⋅−δ⋅=

=φ⋅−φ⋅

cossincossin

cossin

sincossincos

cossin

Cor34t41n41

BtBn

Cor34t41n41

BnBt

aaaa

aa

aaaa

aa


given : find :



289

B

AC

D

B

CD

=

+

BDBD

va 2

41

241

n41 ⋅ω==

3441Cor v2a ⋅ω⋅=

C

B

D A

ω2,ε2

4 3

2

1

A BtB a ,v

rrCtC a ,v

rr

4 : 1

3 : 4

BD

v4141 =ω

BD

a t4141 =ε


ABa 22Bn ⋅ω=

3441C vvvrrr +=

CDv 4141 ⋅ω=

C

D

ω41,ε41

γ

ψ 3434 a ,vrr

41t41 a ,vrr

41nar

Corar

γ

ψ

ν Cxvr

CyvrCv

r41v

r

34vr

C

BC

CD

ψ⋅−γ⋅=γ⋅−ψ⋅=

sincos

sincos

3441Cy

4134Cx

vvv

vvv

B

C

D


given : find :



290

B

A

C D

B

CD

=

+

BDBD

va 2

41

241

n41 ⋅ω==

3441Cor v2a ⋅ω⋅=

C

B

D A

ω2,ε2

4 3

2

1

A BtB a ,v

rrCtC a ,v

rr

4 : 1

3 : 4

BD

v4141 =ω

BD

a t4141 =ε


ABa 22Bn ⋅ω=

C

D

ω41,ε41

γ

ψ 3434 a ,vrr

41t41 a ,vrr

41nar

Corar

CD

CD

Cor3441C aaaarrrr

++=Cor34t41n41C aaaaa

rrrrr+++=

CDa 41t41 ⋅ε=CDa 2

41n41 ⋅ω=

γ

ψ

µ γ

ψ

Cxar

Cyar

Corar

31ar

34ar

41tar

41nar

C

ψ⋅+ψ⋅−γ⋅−γ⋅=ψ⋅+γ⋅−γ⋅−ψ⋅=

cossinsincos

sinsincoscos

Cor34n41t41Cy

Cort41n4134Cx

aaaaa

aaaaa

BC

BC

B

C

D


given : find :



291

3 2

1

1 4

excentricity

The crank mechanism

the crank - piston mechanism

3 2

1 1

4

the centric crank mechanism the excentric crank mechanism


292

The crank mechanism



293

0

0,5

1

1,5

2

2,5

3

3,5

4

4,5

5

0 30 60 90 120 150 180 210 240 270 300 330 360

The crank mechanism


A C

B

φψ

ωA, εA

ωC, εC

x vC, aC

r

b φ⋅−ψ⋅=ψ⋅=φ⋅coscos

sinsin

rbx

br

φ⋅−φ⋅−= cossin rrbx 222

x

φ

b - r·cosφ

( )dt

d

d

rrbd

dt

dxv

222

C

φ⋅φ

φ⋅−φ⋅−==

cossin

( ) AC pv ω⋅= φ ( )( )

φ= φ

φ d

dxp

dt

dp

dt

d

d

dpa

dt

dp

dt

dp

dt

dva

AAC

AA

CC

ω⋅+ω⋅φ⋅φ

=

ω⋅+ω⋅==

( ) ( ) A2

AC pqa ε⋅+ω⋅= φφ ( )( ) ( )

2

2

d

xd

d

dpq

φ=

φ= φφ

φ

AA

A

dt

ddt

d

ε=ω

ω=φ

AC

C

adt

dv

vdt

dx

=

=


294

The crank mechanism


A C

B

φψ

ωA, εA

ωC, εC

x vC, aC

r


sinsin

rbx

br


AA

A

dt

ddt

d

ε=ω

ω=φ

AC

C

adt

dv

vdt

dx

=

=

φ⋅φ⋅+ψ⋅ψ⋅−=

ψ⋅ψ⋅=φ⋅φ⋅&&&

&&

sinsin

coscos

rbx

br

CC

C

dt

ddt

d

ε=ω

ω=ψ

φ⋅−ψ⋅=ψ⋅=φ⋅coscos

sinsin

rbx

br

ACC

CA

rbv

br

ω⋅φ⋅+ω⋅ψ⋅−=ω⋅ψ⋅=ω⋅φ⋅sinsin

coscos

( )φ⋅ψ−φ⋅⋅ω=ψ⋅φ⋅⋅ω=ω

costansin

cos

cos

rv

b

r

AC

AC

CAC

CA

brv

br

ω⋅ψ⋅−ω⋅φ⋅=ω⋅ψ⋅=ω⋅φ⋅

sinsin

coscos

( ) ( )( ) ( )CCAAC

CCAA

brv

br

ω⋅ψ+ω⋅ψ⋅ψ⋅−ω⋅φ+ω⋅φ⋅φ⋅=

ω⋅ψ⋅ψ−ω⋅ψ⋅=ω⋅φ⋅φ−ω⋅φ⋅&&&&&

&&&&

sincossincos

sincossincos

( ) ( )( ) ( )C

2CA

2AC

2CC

2AA

bra

br

ε⋅ψ+ω⋅ψ⋅−ε⋅φ+ω⋅φ⋅=

ω⋅ψ−ε⋅ψ⋅=ω⋅φ−ε⋅φ⋅

sincossincos

sincossincos

( )ψ⋅

ω⋅ψ⋅+ω⋅φ−ε⋅φ⋅=εcos

sinsincos

b

br 2C

2AA

C


295

The crank mechanism


A C

B

φψ

ωA, εA

ωC, εC

x vC, aC

r


sinsin

rbx

br


AA

A

dt

ddt

d

ε=ω

ω=φ

AC

C

adt

dv

vdt

dx

=

=

φ⋅φ⋅+ψ⋅ψ⋅−=

ψ⋅ψ⋅=φ⋅φ⋅&&&

&&

sinsin

coscos

rbx

br

CC

C

dt

ddt

d

ε=ω

ω=ψ

φ⋅−ψ⋅=ψ⋅=φ⋅coscos

sinsin

rbx

br

ACC

CA

rbv

br

ω⋅φ⋅+ω⋅ψ⋅−=ω⋅ψ⋅=ω⋅φ⋅sinsin

coscos

( )φ⋅ψ−φ⋅⋅ω=ψ⋅φ⋅⋅ω=ω

costansin

cos

cos

rv

b

r

AC

AC

CAC

CA

brv

br

ω⋅ψ⋅−ω⋅φ⋅=ω⋅ψ⋅=ω⋅φ⋅

sinsin

coscos

( ) ( )( ) ( )CCAAC

CCAA

brv

br

ω⋅ψ+ω⋅ψ⋅ψ⋅−ω⋅φ+ω⋅φ⋅φ⋅=

ω⋅ψ⋅ψ−ω⋅ψ⋅=ω⋅φ⋅φ−ω⋅φ⋅&&&&&

&&&&

sincossincos

sincossincos

( ) ( )( ) ( )C

2CA

2AC

2CC

2AA

bra

br

ε⋅ψ+ω⋅ψ⋅−ε⋅φ+ω⋅φ⋅=

ω⋅ψ−ε⋅ψ⋅=ω⋅φ−ε⋅φ⋅

sincossincos

sincossincos

( )ψ⋅

ω⋅ψ⋅+ω⋅φ−ε⋅φ⋅=εcos

sinsincos

b

br 2C

2AA

C-1

-0,8

-0,6

-0,4

-0,2

0

0,2

0,4

0,6

0,8

1

0 30 60 90 120 150 180 210 240 270 300 330 360

-3

-2

-1

0

1

2

3vCωC vC

-1,5

-1

-0,5

0

0,5

1

1,5

0 30 60 90 120 150 180 210 240 270 300 330 360

-4

-3

-2

-1

0

1

2

3

4aCεC aC

εC


297

the free body diagram method

the reduction method

G2G1

10. The dynamics of mechanismsDynamics

298

G2G1

a

a = ?m2

m1

αααα

f

I

the free body diagram method10. The dynamics of mechanismsDynamics

299

G2

G1

a

a

εεεε

S1

S1

S2

S2αααα

r

N

I

m2

m1

r

a=εFf

the free body diagram method10. The dynamics of mechanisms

- the free body diagram

- the equations of motionof the single bodies

222 SGam −=⋅

rSrSr

aI 12 ⋅−⋅=⋅

TGSam 111 −α⋅−=⋅ sin

α⋅= cos1GN

( )α⋅+α⋅−=⋅ cossin fGSam 111

fGfNF 1f ⋅α⋅=⋅= cos

- the kinematic solution

- the mathematic solution

( )α⋅+α⋅+⋅= cossin fGamS 111

amGS 222 ⋅−=

( )α⋅+α⋅−=⋅

++ cossin fGGar

Imm 12221the equation of motion

Dynamics

300

φ⋅+= sinery

φ⋅⋅ω=φ⋅φ⋅== coscos eeyv && ω=φ&

φ⋅φ⋅ω⋅−φ⋅ω⋅== &&& sincos eeva

φ⋅⋅ω−φ⋅⋅ε= sincos eea 2

ε=ω&

F

ω,εω,εω,εω,ε

v,a

MM

φφφφ

e·si

nφ

y =

r+

e·si

nφe

r



Dynamics

301

φφφφ

F


v,a

MM

R

e·cosφφφφ

εεεε

MM

φφφφR

F

a

e

φ⋅⋅−=ε⋅ coseRMI FRam −=⋅


- the free body diagram - the equations of motionof the single bodies



Dynamics

302

( )φ⋅⋅−=φ⋅⋅⋅+ε⋅

φ⋅⋅+⋅−=ε⋅φ⋅⋅−=ε⋅

coscos

cos

cos

eFMeamI

eFamMI

eRMI

FRam −=⋅ FamR +⋅=

( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos eFMememI 2222

φφφφ

F


v,a

MM

e


- the mathematic solution


Dynamics

303

given : φφφφ, ωωωω, εεεε, F.find : M. ( ) 2222 ememIeFM ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅++φ⋅⋅= cossincoscos

0 9 0 1 8 0 2 7 0 3 6 0 4 5 0 5 4 0 6 3 0 7 2 0

- 1 0 0

1 0 0

2 0 0 R [ N ]

M [ N · m ]

φφφφ [ º ]

0

ωωωω = const, εεεε = 0, F = const

φφφφ

F


v,a

MM

e


the kinetostatic task ( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos eFMememI 2222

Dynamics

304

( ) φ⋅⋅−=φ⋅φ⋅φ⋅⋅−φ⋅φ⋅⋅+ coscossincos eFMememI 2222 &&&

t φφφφ ωωωω εεεε v a R

the numerical solution

φφφφ

F


v,a

MM

e

( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos eFMememI 2222


the dynamic taskgiven : F, M.find : φ=φ(t), ω=ω(t), ε=ε(t).

Dynamics

305

r3

ωωωω1

M G

meqv

Feqv

ωωωω2

I 1, r1

I 2, r2

m

x,v,a x,v,a

the reduction to translation

the reality

10. The dynamics of mechanisms

the alternative


the three links

Dynamics

306

r3

ωωωω1

M G

ωωωω2

I 1, r1

I 2, r2

m

x,v,a x,v,a

2eqv2

12212

22212

1121

k vmvmIIE ⋅⋅=⋅⋅+ω⋅⋅+ω⋅⋅=

reality alternative

the kinematic relations

32 r

v=ω1

2

31 r

r

r

v ⋅=ω

the kinetic energy

the reality


the alternative



meqv

Feqv

2

32

2

31

21eqv r

1I

r

1

r

rImm

⋅+

⋅⋅+=

Dynamics

307

r3

ωωωω1

M G

ωωωω2

I 1, r1

I 2, r2

m

x,v,a x,v,a

vFvGMP eqv1 ⋅=⋅−ω⋅=the power

Gr

1

r

rMF

31

2eqv −⋅⋅=

the reality


the alternative



meqv

Feqv

the kinematic relations

32 r

v=ω1

2

31 r

r

r

v ⋅=ωreality alternative

Dynamics

308

r3

ωωωω1

M G

ωωωω2

I 1, r1

I 2, r2

m

x,v,a x,v,a

eqv2eqv

eqv Fvdx

dm

2

1am =⋅⋅+⋅

the reality


the alternative



meqv

Feqv


2

32

2

31

21eqv r

1I

r

1

r

rImm

⋅+

⋅⋅+=

Gr

1

r

rMF

31

2eqv −⋅⋅=

0dx

dmeqv =const=eqvm

eqveqv Fam =⋅

Gr

1

r

rMa

r

1I

r

1

r

rIm

31

2

2

32

2

31

21 −⋅⋅=⋅

⋅+

⋅⋅+

Dynamics

309

AEK =∆the kinetic energy change law the work

Pt

A

t

EK =∆

=∆

∆ the power

Pdt

dEK =

2eqv2

1K vmE ⋅⋅=

3eqv21

eqv2eqv

21

eqveqv2eqv

21K v

dx

dmavmv

dt

dx

dx

dmavm

dt

dvv2mv

dt

dm

dt

dE ⋅⋅+⋅⋅=⋅⋅⋅+⋅⋅=

⋅⋅⋅+⋅⋅=

adt

dv = vdt

dx =

the reality


the alternative



vFP eqv ⋅=

vFvvdx

dmamv

dx

dmavm eqv

2eqv21

eqv3eqv

21

eqv ⋅=⋅

⋅⋅+⋅=⋅⋅+⋅⋅

eqv2eqv

21

eqv Fvdx

dmam =⋅⋅+⋅

Dynamics

310

r3

ωωωω1

M G

ωωωω2

ωωωω,εεεεMeqv

I eqv

I 1, r1

I 2, r2

m

x,v,a

the reality


the alternative


the reduction to rotation

Dynamics

311

r3

ωωωω1

M G

ωωωω2

ωωωω,εεεεI 1, r1

I 2, r2

m

x,v,a

2eqv2

12212

22212

1121

k IvmIIE ω⋅⋅=⋅⋅+ω⋅⋅+ω⋅⋅=2

112 r

r⋅ω=ω 32

11 r

r

rv ⋅⋅ω=

Meqv

I eqv

the reality


the alternative


reality alternative


2

2

121

2

2

123eqv r

rII

r

rrmI

⋅++

⋅⋅=

Dynamics

312

r3

ωωωω1

M G

ωωωω2


I 2, r2

m

x,v,a

ω⋅=⋅−ω⋅= eqv1 MvGMP

Meqv

I eqv

the reality


the alternative


reality alternative


2

112 r

r⋅ω=ω 32

11 r

r

rv ⋅⋅ω=

2

13eqv r

rrGMM ⋅⋅−=

Dynamics

313

r3

ωωωω1

M G

ωωωω2


I 2, r2

m

x,v,aMeqv

I eqv

the reality


the alternative


red2red

red Md

dI

2

1I =ω⋅

φ⋅+ε⋅

I eqv = const :

redred MI =ε⋅

0d

dIred =φ


2

13eqv r

rrGMM ⋅⋅−=

2

2

121

2

2

123eqv r

rII

r

rrmI

⋅++

⋅⋅=

2

13

2

2

121

2

2

123 r

rrGM

r

rII

r

rrm ⋅⋅−=ε⋅

⋅++

⋅⋅


Dynamics

314

ωωωω,εεεεMeqv

I eqv

2eqv2

12212

21

k IvmIE ω⋅⋅=⋅⋅+ω⋅⋅=

φ⋅⋅ω= cosrv

φ⋅⋅+= 22eqv rmII cos

φ⋅⋅ω=φ⋅φ⋅=

φ⋅=

cos

cos

sin

rv

rx

rx&&

F

v, a

φ

ω, εω, εω, εω, ε

M

mI

x

r

the reality


the alternative



φ⋅⋅ω= cosrv

ω⋅=⋅−ω⋅= eqvMvFMP

φ⋅⋅−= cosrFMM eqv

Dynamics

315

ωωωω,εεεε

φ⋅⋅−= cosrFMM eqv

v, a

φ


M

mI

r

x

F

φ⋅⋅+= 22eqv rmII cos


red2red

red Md

dI

2

1I =ω⋅

φ⋅+ε⋅

φ⋅φ⋅⋅⋅−=φ

sincos2red rm2d

dI

( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos rFMrmrmI 2222

the reality


the alternative



( ) φ⋅⋅−=φ⋅φ⋅φ⋅⋅−φ⋅φ⋅⋅+ coscossincos rFMrmrmI 2222 &&&

Meqv

I eqv

Dynamics

316

ωωωω,εεεεv, a

φ


M

mI

r

x

F


red2red

red Md

dI

2

1I =ω⋅

φ⋅+ε⋅


the reality


the alternative



Meqv

I eqv

The 1. type task - kinetostatic

given : φ(t), ω(t), ε(t),

find : M = M(φ)

( ) φ⋅⋅+ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+= coscossincos rFrmrmIM 2222

Dynamics

317

ωωωω,εεεεv, a

φ


M

mI

r

x

F


red2red

red Md

dI

2

1I =ω⋅

φ⋅+ε⋅


the reality


the alternative



Meqv

I eqv

The 2. type task - dynamic

given : F, M

find : the motionφ = f(t), ω, ε

( ) φ⋅⋅−=φ⋅φ⋅φ⋅⋅−φ⋅φ⋅⋅+ coscossincos rFMrmrmI 2222 &&&

t φφφφ ωωωω εεεε

5 10 15 20

10

0

t [s] 0

ω [s-1]

Dynamics

318

- more work, longer solution(the system of equations of motion)

- the solution includesthe joint forces and moments

- shorter, less complicated(one equation of motion)

eqv2eqv

eqv Fvdx

dm

2

1am =⋅⋅+⋅

eqveqv Fam =⋅


the reduction methodthe free body diagram method

the comparison

- the solution does not includethe joint forces and moments

- the solution allows to includethe friction forces and moments(through the joint forces / moments)

- the solution does not allowto include the friction forcesand moments

- there is no essential differencein application to the mechanismswith constant and varying drive ratio

- there is differencein application to the mechanismswith constant and varying drive ratio

advantages and disadvantages

Dynamics

mechanics statics dynamics - vŠb-tuo dynamics mechanics statics dynamics the forces acting on rigid...

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