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).
1. Dynamics of a particle - revisionDynamics
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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
1. Dynamics of a particle - revisionDynamics
5
dynamics
dynamicskinematics
only motion motion, massesand forces
1. Dynamics of a particle - revisionDynamics
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
1. Dynamics of a particle - revisionDynamics
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. Dynamics of a particle - revisionDynamics
8
the number of DOF decreasesif the motion is restricted by joints
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. Dynamics of a particle - revisionDynamics
9
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. Dynamics of a particle - revision
the number of DOF decreasesif the motion is restricted by joints
Dynamics
10
y direction translation
x direction translation
z axis rotationzx
y
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. Dynamics of a particle - revision
the number of DOF decreasesif the motion is restricted by joints
Dynamics
11
the number of DOF decreasesif the motion is restricted by joints
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. Dynamics of a particle - revision
NO vertical translationindependent horizontal translationindependent rotation2 DOF
Dynamics
12
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. The dynamics of particle - repetition
the number of DOF decreasesif the motion is restricted by joints
NO vertical translationwithout sliding in the touch pointhorizontal translation and rotation relate one to the other1 DOFrolling
Dynamics
13
the number of DOF decreasesif the motion is restricted by joints
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)
3 DOF(3 translations)
6 DOF(3 translations and 3 rotations)
1. Dynamics of a particle - revisionDynamics
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
1. Dynamics of a particle - revision
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
1. Dynamics of a particle - revision
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
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
A) Uniform motion – velocity is constant
these are the relationships valid onlyfor uniform motion
the motion of a particleDynamics
18
1. Dynamics of a particle - revision
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
the motion of a particleDynamics
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 .
1. Dynamics of a particle - revision
B) Uniformly accelerated motion – acceleration is constant
(assuming uniformly accelerated motion)
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
the motion of a particleDynamics
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.)
1. Dynamics of a particle - revision
C) Non-uniform motion
amplitude [m]
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a damping environment
( )tstatesteady e1vv ⋅β−−⋅= _
β= g
v statesteady_
the motion of a particleDynamics
23
vga ⋅β−=
vgdt
dv ⋅β−=
dtvg
dv =⋅β−
( )te1g
v ⋅β−−⋅β
=
v
t
vsteady state T
63% vsteady state 95% vsteady state
( )tstatesteady e1vv ⋅β−−⋅= _
t=2·T t=4·T t=3·T t=5·T t=T
tangent
β= 1
T time constant [s]y, v, a
( )tstatesteady e1vv ⋅β−−⋅= _
β= g
v statesteady_
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a damping environment
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a damping environment
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a gravitational field
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a gravitational field
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a gravitational field
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a gravitational field
the motion of a particleDynamics
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
1. Dynamics of a particle - revision
C) Non-uniform motion
motion within a gravitational field
the motion of a particleDynamics
30
1. Dynamics of a particle - revision
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
1. Dynamics of a particle - revision
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
1. Dynamics of a particle - revision
Newton’s 2nd law – the law of force
the equation of motion
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
1. Dynamics of a particle - revisionDynamics
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
1. Dynamics of a particle - revision
Newton’s 2nd law – the law of force
m a Fi⋅ =∑v r
the equation of motion
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
1. Dynamics of a particle - revisionDynamics
36
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
2. The analytical mechanics
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 =
2. The analytical mechanicsDynamics
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 potential energy
the force F=Gvaries with height
the potential energy - positional
2. The analytical mechanicsDynamics
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.
2. The analytical mechanics
the potential energy
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
2. The analytical mechanics
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
2. The analytical mechanics
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
2. The analytical mechanics
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
2. The analytical mechanics
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 work principle ... in statics
2. The analytical mechanics
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 ⋅=
the virtual work principle ... in statics
2. The analytical mechanics
π=
π=ω
B
v
A
v BA
the kinematical method in statics
Dynamics
51
the virtual work principle ... in statics
2. The analytical mechanics
F - given
determine R = ?
Dynamics
52
9 equations of equilibrium
9 unknown forces
the virtual work principle ... in statics
2. The analytical mechanics
the free-body diagramF - given
Dynamics
53
R
Fth
e vi
rtua
l mot
ion
the virtual work principle ... in statics
2. The analytical mechanics
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
the virtual work principle ... in statics
2. The analytical mechanicsDynamics
55
α⋅δ⋅=δ⋅= cosFFArr F
r
δr
α
the virtual work principle ... in statics
2. The analytical mechanicsDynamics
56
A C
B
the virtual work principle ... in statics
2. The analytical mechanics
F - given
determine R = ?
Dynamics
57
F
RA C
Bthe virtual motion
the virtual work principle ... in statics
2. The analytical mechanicsDynamics
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
the virtual work principle ... in statics
2. The analytical mechanicsDynamics
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
2. The analytical mechanicsDynamics
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
2. The analytical mechanics
– 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
the Lagrange equations of the 2nd kind
2. The analytical mechanicsDynamics
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⋅+=π 22222 e2eT sinsin llllllll
( )( )[ ] ( )[ ] 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
the generalized coordinate
the generalized velocity
the Lagrange equations of the 2nd kind
2. The analytical mechanicsDynamics
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⋅⋅⋅−φ⋅+=π 22222 e2eT sinsin llllllll
( )( )[ ] ( )[ ] ( ) 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
the generalized coordinate
the generalized velocity
the Lagrange equations of the 2nd kind
2. The analytical mechanics
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 ⋅⋅<>
the Lagrange equations of the 2nd kind
2. The analytical mechanicsDynamics
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
the Lagrange equations of the 2nd kind
2. The analytical mechanicsDynamics
66
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
3. Vibrations with 1 degree of freedomDynamics
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
3. Vibrations with 1 degree of freedomDynamics
70
The free vibration– no external force as the reason for vibration.I.
The forced vibration- caused by an external force.
types of vibration
3. Vibrations with 1 degree of freedomDynamics
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
3. Vibrations with 1 degree of freedomDynamics
72
free forced
damped
undampedvibration
types of vibration
3. Vibrations with 1 degree of freedomDynamics
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
3. Vibrations with 1 degree of freedomDynamics
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
the free undamped vibration
20m
k Ω=
( ) ( )[ ] 0tCktm
kCm 0000 =φ+⋅Ω⋅⋅+
φ+⋅Ω⋅⋅−⋅ sinsin
3. Vibrations with 1 degree of freedomDynamics
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 :
3. Vibrations with 1 degree of freedomDynamics
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 free undamped vibration
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 :
3. Vibrations with 1 degree of freedomDynamics
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⟩
3. Vibrations with 1 degree of freedomDynamics
78
r
m
φ
ω,ε
atan
G
ε⋅= rat2
n ra ω⋅= φ=ω &
S
the free undamped vibration
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 free undamped vibration
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.
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
3. Vibrations with 1 degree of freedom
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
the free damped vibration
3. Vibrations with 1 degree of freedom
( )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
the free damped vibration
3. Vibrations with 1 degree of freedom
( )( ) ( )[ ]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
the free damped vibration
3. Vibrations with 1 degree of freedom
( )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
the free damped vibration
3. Vibrations with 1 degree of freedom
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
the free damped vibration
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedomDynamics
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
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
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
the particular solution( )φ−⋅ω⋅= txx apart sin
ωωωω – the circular frequencyof the exciting force
the amplitude xaand phase lead φφφφof the particular solutionwill be discussed later
the particular solutiont
x
xa ωπ⋅= 2
T
the solution is the assembly of a “homogenous solution”and a “particular solution”
Dynamics
96
k
x
bFD=b·v
FS=k·x
v, a
tF Fa
T T
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
the particular solution( )φ−⋅ω⋅= txx apart sin
the solution is the assembly of a “homogenous solution”and a “particular solution”
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
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
( ) ( )( ) ( )( ) ( )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
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
( ) ( )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
ω⋅δ⋅+ω−Ω⋅=
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
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
( ) ( )φ−⋅ω⋅= txx at sin
( ) ( )22220
aa
2
1
m
Fx
ω⋅δ⋅+ω−Ω⋅=
220
2
ω−Ωω⋅δ⋅=φ arctan
( ) ( )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
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
( ) ( )φ−⋅ω⋅= txx at sin
( ) ( )22220
aa
2
1
m
Fx
ω⋅δ⋅+ω−Ω⋅=
220
2
ω−Ωω⋅δ⋅=φ arctan
( ) ( )222
aa
21
1
k
Fx
η⋅ξ⋅+η−⋅=
21
2
η−η⋅ξ⋅=φ arctan
stata x
k
F =
the static deformation
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
( )tFxkxbxm a ⋅ω⋅=⋅+⋅+⋅ sin&&&
m
harmonically changing exciting force
3. Vibrations with 1 degree of freedom
F = Fa·sin(ω·t)
Fa – the amplitude
ωωωω – the circular frequencyof the exciting force
( ) ( )φ−⋅ω⋅= txx at sin
( ) ( )22220
aa
2
1
m
Fx
ω⋅δ⋅+ω−Ω⋅=
220
2
ω−Ωω⋅δ⋅=φ arctan
( ) ( )222
aa
21
1
k
Fx
η⋅ξ⋅+η−⋅=
21
2
η−η⋅ξ⋅=φ arctan
stata x
k
F =
the static deformation
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
3. Vibrations with 1 degree of freedom
the cause the consequence
( ) ( )22220
aa
2
1
m
Fx
ω⋅δ⋅+ω−Ω⋅=
220
2
ω−Ωω⋅δ⋅=φ arctan
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
3. Vibrations with 1 degree of freedomDynamics
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
3. Vibrations with 1 degree of freedom
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
3. Vibrations with 1 degree of freedom
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
106
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
4. The vibration with 2 degree of freedomthe free vibrationDynamics
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
4. The vibration with 2 degree of freedom
0cD =⋅
the free vibrationDynamics
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
4. The vibration with 2 degree of freedomthe free vibration
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
4. The vibration with 2 degree of freedomthe free vibration
−−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
4. The vibration with 2 degree of freedomthe free vibration
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
4. The vibration with 2 degree of freedom
direction 1direction 2
direction 1direction 2
x
the free vibrationDynamics
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( )
4. The vibration with 2 degree of freedom
direction 1direction 2
direction 1direction 2
x
the free vibrationDynamics
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
4. The vibration with 2 degree of freedom
direction 1direction 2
direction 1direction 2
the free vibrationDynamics
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.
4. The vibration with 2 degree of freedomthe free vibrationDynamics
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 =⋅+⋅ &&
4. The vibration with 2 degree of freedomthe free vibrationDynamics
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
,
,
4. The vibration with 2 degree of freedomthe free vibrationDynamics
121
4. The vibration with 2 degree of freedomthe free vibration
( ) 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
F1=Fa1·sin(ω·t) F2=Fa2·sin(ω·t)
( ) 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
F1=Fa1·sin(ω·t) F2=Fa2·sin(ω·t)
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
126
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
5. Translation and rotationthe types of motion
2D
Dynamics
129
rotation
5. Translation and rotationthe types of motion
one line does not change its position2D
Dynamics
130
general plane motion
5. Translation and rotationthe types of motion
2D
Dynamics
131
general plane motion
5. Translation and rotationthe types of motion
2D
Dynamics
132
5. Translation and rotationthe types of motion
translation
no lines change their direction
3D
Dynamics
133
3D
5. Translation and rotationthe types of motion
one point does not change its position
spherical motion
Dynamics
134
posuv
rotace
spiral motion
5. Translation and rotationthe types of motion
the body rotates about a specific axisand translates in the direction of this axis
3D
Dynamics
135
5. Translation and rotationthe types of motion
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
no lines change their direction
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
5. Translation and rotationthe translation - kinematics
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
5. Translation and rotationthe translation - kinematics
no lines change their direction1, 2 or 3 degrees of freedom
the time derivative
Dynamics
139
the direct linear translation
All the points move on the same trajectory,with the same velocity and the same acceleration.
5. Translation and rotationthe translation - kinematics
no lines change their direction
Dynamics
140
R
the circular translation
All the points move on the same trajectory,with the same velocity and the same acceleration.
5. Translation and rotationthe translation - kinematics
no lines change their direction
Dynamics
141
the cycloid translation
All the points move on the same trajectory,with the same velocity and the same acceleration.
5. Translation and rotationthe translation - kinematics
no lines change their direction
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
5. Translation and rotationthe translation - dynamics
the equation of motion
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
5. Translation and rotationthe translation - dynamics
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.
5. Translation and rotationthe translation - dynamics
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
the equation of motion
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
the rotation - dynamics
the equation of motion
5. Translation and rotation
∑=ε⋅ iCRCR MI _
the d’Alembert principle
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)
the rotation - dynamics
the equation of motion
5. Translation and rotation
∑=ε⋅ iCRCR MI _
the d’Alembert principle
T2
Tnn
TTtt
CRD
rmamD
rmamD
IM
⋅ω⋅=⋅=
⋅ε⋅=⋅=ε⋅=
0M
0F
0F
CRi
yi
xi
=
=
=
∑∑∑ the reaction solution
ΣF – including d’Alembert forces
the equation of motion
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
the equation of motion
~∑=⋅ iFamrr ∑=ε⋅ iMI
rotationtranslation
the rotation - dynamics 5. Translation and rotation
analogy
the equation of motion
~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
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.
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
the rotation - dynamics 5. Translation and rotationDynamics
155
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.
0rGI =φ⋅⋅+φ⋅ &&
the calculation of the moment of inertia Ifrom measured vibration periodT :
T
2 π⋅=Ω
2
rGI
Ω⋅=
I
rG ⋅=Ω
the rotation - dynamics 5. Translation and rotationDynamics
156
∫ ⋅=m
2 dmrIdm
r
m
S
r = const
2
m
2
m
2 rmdmrdmrI ⋅=⋅=⋅= ∫∫
the thin ring
the rotation - dynamics 5. Translation and rotation
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
the rotation - dynamics 5. Translation and rotation
the moment of inertia
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⋅⋅=
the rotation - dynamics 5. Translation and rotation
the moment of inertia
Dynamics
159
∫ ⋅=m
2 dmrI
m
h
R
the rotation - dynamics 5. Translation and rotation
the moment of inertia
the cylinder
Dynamics
160
∫ ⋅=m
2 dmrI
m
h
r dr
( ) hdrr2hdSdVdm ⋅⋅⋅π⋅⋅ρ=⋅⋅ρ=⋅ρ=
dr2·π·r
dS
the rotation - dynamics 5. Translation and rotation
the moment of inertia
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 rotation - dynamics 5. Translation and rotation
the moment of inertia
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
the rotation - dynamics 5. Translation and rotation
the moment of inertia
∫ ⋅=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 rotation - dynamics 5. Translation and rotation
the thin rectangular plate
the ball
Dynamics
164
the rotation - dynamics 5. Translation and rotation
the firm publication
Dynamics
165
firm publication
the rotation - dynamics 5. Translation and rotationDynamics
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)
the rotation - dynamics 5. Translation and rotationDynamics
167
6. General plane motionkinematics
translation
general plane motion
rotation
the plane motion :
all points movein planesparallel to one another
Dynamics
168
translation
rotation
translation
1 DOF
2 DOF
3 DOF
translation
rotation
6. General plane motionkinematics
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
6. General plane motionkinematics
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
6. General plane motionthe analytical solutionDynamics
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
rolling without sliding
φφφφ·r
cycloid curve
6. General plane motionthe analytical solutionDynamics
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
6. General plane motionthe pole method
the pole– the instantaneous center of zero velocity
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
6. General plane motionthe pole method
only velocities !
not accelerations !
point A moves along the circular trajectory
the
Bpo
int m
oves
on
the
circ
ular
traj
ecto
ry
the pole– the instantaneous center of zero velocity
Dynamics
178
A
B π(t-∆t)
π(t+∆t)
π(t)
6. General plane motionthe pole method
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
6. General plane motionthe pole method
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
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 fixed pole curve
the moving pole curve
6. General plane motionthe pole method
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
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 moving pole curve
6. General plane motionthe pole method
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
6. General plane motionthe pole method
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)
the fixed pole curve
the moving pole curve
the fixed pole curve
the moving pole curve
Dynamics
183
φ,ω
π
rv ⋅ω=v
r
rolling without sliding
ππππ pole
only velocities !
not accelerations
!
6. General plane motionthe pole method
the fixed pole curve
the moving pole curve
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
6. General plane motionmotion decompositionbasic decomposition
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
6. General plane motionmotion decompositionbasic decomposition
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
6. General plane motionmotion decomposition
rolling without sliding
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
6. General plane motionmotion decomposition
rolling without sliding
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
6. General plane motionmotion decomposition
rolling without sliding
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
the resulting motion = the frame motion + the relative motion
general plane motion = rotation + translation
general plane motion = rotation + rotation
translation = translation + translation
6. General plane motionmotion decompositiongeneral decomposition
basic decomposition
Dynamics
193
= +
vr
relvr
framevr
frameω
relframe vvvrrr +=
relvr
framevr v
r
rv frameframe ⋅ω=
r
the resulting motion = the frame motion + the relative motion
6. General plane motionmotion decompositiongeneral decomposition
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 resulting motion = the frame motion + the relative motion
6. General plane motionmotion decompositiongeneral decomposition
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
the d’Alembert principle
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
6. General plane motiondynamics
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
6. General plane motiondynamics
the centre of gravity ≡ the reference point
the equation of motion
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
6. General plane motiondynamics
the centre of gravity ≡ the reference point
the equations of motion
the reaction solution
Dynamics
202
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”
7. Spherical motionkinematics
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.
7. Spherical motionkinematics
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
7. Spherical motionkinematicsDynamics
208
the motion of precession
the motion of nutation
the motion of primary rotation
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
2. ϑϑϑϑ – the angle of nutation about the ξξξξ1 axis
3. φφφφ – the angle of primary rotation about the ζζζζ axis
7. Spherical motionkinematics
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 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
2. ϑϑϑϑ – the angle of nutation about the ξξξξ1 axis
3. φφφφ – the angle of primary rotation about the ζζζζ axis
7. Spherical motionkinematics
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
7. Spherical motionkinematics
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
7. Spherical motionkinematicsDynamics
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
7. Spherical motionkinematics
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
7. Spherical motionkinematics
in body coordinate system ξηζξηζξηζξηζ
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=
7. Spherical motionkinematics
the circumferential acceleration in fixed coordinate system xyz
Dynamics
215
in body coordinate system ξηζξηζξηζξηζ
( ) 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
7. Spherical motionkinematics
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
the d’Alembert principle
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
7. Spherical motiondynamics
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 primary rotation
the precession
the primary rotation
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
7. Spherical motiondynamics
the precession
the primary rotation
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 two uniform rotations about two concurrent axis of constant angle
7. Spherical motiondynamics
the precession
const=ψ&
const=φ&
the body is axially symmetrical,the symmetry axis is coincident with the axis of primary rotation
the heavy flywheel
the primary rotation
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 two uniform rotations about two concurrent axis of constant angle
7. Spherical motiondynamics
the body is axially symmetrical,the symmetry axis is coincident with the axis of primary rotation
the heavy flywheel
the primary rotation
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 coordinate- one or more independent coordinates, determining the position of the mechanism
8. The mechanismsDynamics
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.
8. The mechanismsDynamics
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
8. The mechanismsDynamics
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 classification
the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion
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
8. The mechanismsDynamics
226
the classification with respect to the number of degrees of freedom
The mechanism classification
the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion
posu
v
rotace
the mechanisms with 1 DOF
8. The mechanismsDynamics
227
the classification with respect to the number of degrees of freedom
The mechanism classification
the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion
the mechanisms with 1 DOF
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
the mechanisms with 2 DOF
φφφφ and ψψψψ – two mechanism coordinates
the car goes through the curve –the wheels rotate with different speed
8. The mechanismsDynamics
228
the classification with respect to the number of degrees of freedom
The mechanism classification
the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion
the mechanisms with 1 DOF
the mechanisms with 2 DOF
the mechanisms with more DOF
the number of the mechanism DOF is not limited the mechanism with 7 DOF
8. The mechanismsDynamics
229
the classification with respect to the number of degrees of freedom
The mechanism classification
the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion
the mechanisms with 1 DOF
the mechanisms with 2 DOF
the mechanisms with more DOF
... 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
8. The mechanismsDynamics
230
the classification with respect to the number of degrees of freedom
The mechanism classification
the planar mechanisms (2D)all members perform the planar motion in reciprocally parallel planesthe space mechanisms (3D)at least one member perform space motion
the mechanisms with 1 DOF
the mechanisms with 2 DOF
the mechanisms with more DOF
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 3 members mechanisms
the 4 members mechanisms
the more members mechanisms
8. The mechanismsDynamics
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.
8. The mechanismsDynamics
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)
8. The mechanismsDynamics
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
8. The mechanismsDynamics
234
fixes 2 translations
allows rotation
The planar (2D), idealized joints are :
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
8. The mechanismsDynamics
235
transmits forces
does not transmit moment
the frame
the body the body
the body
The planar (2D), idealized joints are :
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
8. The mechanismsDynamics
236
The planar (2D), idealized joints are :
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
8. The mechanismsDynamics
237
The planar (2D), idealized joints are :
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
transmits force and moment
does not transmit force
the sliding joint – the kinematic scheme
8. The mechanismsDynamics
238
The planar (2D), idealized joints are :
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 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
8. The mechanismsDynamics
239
ε⋅=ω⋅=φ⋅=
ra
rv
rx
r φ
x
The planar (2D), idealized joints are :
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 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
8. The mechanismsDynamics
240
z
φ
The planar (2D), idealized joints are :
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 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)
8. The mechanismsDynamics
241
z
φ
The planar (2D), idealized joints are :
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 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)
does not transmit force and moment
transmits force
the sliding pin joint – the kinematic scheme
8. The mechanismsDynamics
242
φ
The planar (2D), idealized joints are :
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 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,the general joint fixes 1 translation, allows rotation and translation,
(independent)
fixes 1 translation
allows translation and rotation (independent)
sliding in touch point
8. The mechanismsDynamics
243
φ
fixes 1 translation
allows translation and rotation (independent)
does not transmit force and moment
transmits force
sliding in touch point
The planar (2D), idealized joints are :
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 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,the general joint fixes 1 translation, allows rotation and translation,
(independent)
8. The mechanismsDynamics
244
not exactly the same way for the real joints
The planar (2D), idealized joints are :
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 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,the general joint fixes 1 translation, allows rotation and translation,
(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
8. The mechanismsDynamics
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
8. The mechanismsDynamics
246
The space (3D), idealized joints are :
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
8. The mechanismsDynamics
247
The space (3D), idealized joints are :
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 rotationsfixes 2 translations and 3 rotationsallows 1 translationtransmits 2 forces and 3 momentsdoes not transmit the force
the sliding joint
allows translation
z
8. The mechanismsDynamics
248
The space (3D), idealized joints are :
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 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)
8. The mechanismsDynamics
249
The space (3D), idealized joints are :
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 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
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
8. The mechanismsDynamics
250
The space (3D), idealized joints are :
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 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
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
8. The mechanismsDynamics
251
The space (3D), idealized joints are :
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 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
the 4th class joint the sliding-rotating joint fixes 2 translations and 2 rotationsallows independent translation and rotation
the double-rotating joint fixes 3 translations and 1 rotationallows 2 independent rotationstransmits 3 forces and moment,does not transmit 2 moments
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
8. The mechanismsDynamics
252
The space (3D), idealized joints are :
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 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
the 4th class joint the sliding-rotating joint fixes 2 translations and 2 rotationsallows independent translation and rotation
the double-rotating joint fixes 3 translations and 1 rotationallows 2 independent rotationstransmits 3 forces and moment,does not transmit 2 moments
the 3rd class joint the spherical joint fixes 3 translations,allows 3 independent rotationstransmits 3 forces,does not transmit 3 moments
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
8. The mechanismsDynamics
253
The space (3D), idealized joints are :
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 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
the 4th class joint the sliding-rotating joint fixes 2 translations and 2 rotationsallows independent translation and rotation
the double-rotating joint fixes 3 translations and 1 rotationallows 2 independent rotationstransmits 3 forces and moment,does not transmit 2 moments
the 3rd class joint the spherical joint fixes 3 translations,allows 3 independent rotationstransmits 3 forces,does not transmit 3 moments
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
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
8. The mechanismsDynamics
254
The space (3D), idealized joints are :
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 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
the 4th class joint the sliding-rotating joint fixes 2 translations and 2 rotationsallows independent translation and rotation
the double-rotating joint fixes 3 translations and 1 rotationallows 2 independent rotationstransmits 3 forces and moment,does not transmit 2 moments
the 3rd class joint the spherical joint fixes 3 translations,allows 3 independent rotationstransmits 3 forces,does not transmit 3 moments
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
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
8. The mechanismsDynamics
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
8. The mechanismsDynamics
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 members mechanisms
the four-joints mechanism
The types of mechanisms
transmits rotationin 1 : 1 ratio
translation
8. The mechanismsDynamics
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 types of mechanisms
the four members mechanisms
the centric coulisse mechanism the centric coulisse mechanism
the excentric coulisse mechanism
the right-angled coulisse
8. The mechanismsDynamics
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
The types of mechanisms
the four members mechanisms
8. The mechanismsDynamics
261
3
2
4
1
1
1
6
5
The types of mechanisms
the more-members planar mechanisms
the sewing machine mechanism
8. The mechanismsDynamics
262
3
2
4
1
1
The types of mechanisms
the space mechanisms
the 3D crank mechanism
8. The mechanismsDynamics
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 mechanism drive ratio 9. The kinematics of mechanisms
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 mechanism drive ratio 9. The kinematics of mechanisms
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
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
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
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
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
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
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
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
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
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
273
the variators
konst=ωω
=vstupní
výstupníp
pvstupnívýstupní ⋅ε=ε
qp 2vstupnívstupnívýstupní ⋅ω+⋅ε=ε
0d
dpq =
φ=
pvstupnívýstupní ⋅ω=ω
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
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
the generalized 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
the generalized velocity
The mechanism drive ratio 9. The kinematics of mechanismsDynamics
275
the mechanism s u1
u3
u2
the analytical solution
1. the position problem
( )sfu =
2. the velocity solution
mechanism with 1 DOF - degree of freedom
( )( ) ( ) ( )
dt
ds
ds
du
dt
duv tsts
out ⋅==
( )vst
t vdt
ds=
( )( )s
s pds
du=
( ) vstsvýst vpv ⋅=
the mechanism as „geometric converter“
the travel function
DRthe drive ratio
the generalized velocity
The mechanism drive ratio 9. The kinematics of mechanisms
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
the analytical solution
1. the position problem
( )sfu =
2. the velocity solution
mechanism with 1 DOF - degree of freedom
the mechanism as „geometric converter“
the travel function
The mechanism drive ratio 9. The kinematics of mechanisms
3. the acceleration solution
( ) 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 mechanism drive ratio 9. The kinematics of mechanisms
the mechanism as „geometric converter“
1. the position problem
the travel function
2. the velocity solution
3. the acceleration solution
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
1. the position problem
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 analytical solution
9. The kinematics of mechanisms
mechanism with 1 DOF - degree of freedom
1. the position problem
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
( )sfu =the vector method
the analytical solution
9. The kinematics of mechanisms
mechanism with 1 DOF - degree of freedom
1. the position problem
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 =&&
( )sfu =the vector method
the analytical solution
9. The kinematics of mechanisms
mechanism with 1 DOF - degree of freedom
1. the position problem
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
9. The kinematics of mechanismsDynamics
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
9. The kinematics of mechanismsDynamics
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
9. The kinematics of mechanismsDynamics
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
9. The kinematics of mechanismsDynamics
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 :
the motion decomposition
9. The kinematics of mechanismsDynamics
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
the Coriolis decomposition
given : find :
the motion decomposition
9. The kinematics of mechanismsDynamics
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
the Coriolis decomposition
given : find :
the motion decomposition
9. The kinematics of mechanismsDynamics
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
ABv 2B ⋅ω= ABa 2Bt ⋅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
the Coriolis decomposition
given : find :
the motion decomposition
9. The kinematics of mechanismsDynamics
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
ABv 2B ⋅ω= ABa 2Bt ⋅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
the Coriolis decomposition
given : find :
the motion decomposition
9. The kinematics of mechanismsDynamics
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
9. The kinematics of mechanismsDynamics
292
The crank mechanism
the crank - piston mechanism
9. The kinematics of mechanismsDynamics
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
the crank - piston 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
=
=
9. The kinematics of mechanismsDynamics
294
The crank mechanism
the crank - piston 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
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
9. The kinematics of mechanismsDynamics
295
The crank mechanism
the crank - piston 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
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
9. The kinematics of mechanismsDynamics
296
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
the free body diagram method10. The dynamics of mechanisms
- the kinematic solution
Dynamics
301
φφφφ
F
ω,εω,εω,εω,ε
v,a
MM
R
e·cosφφφφ
εεεε
MM
φφφφR
F
a
e
φ⋅⋅−=ε⋅ coseRMI FRam −=⋅
the free body diagram method10. The dynamics of mechanisms
- the free body diagram - the equations of motionof the single bodies
- the kinematic solution
φ⋅⋅ω−φ⋅⋅ε= sincos eea 2
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
ω,εω,εω,εω,ε
v,a
MM
e
the free body diagram method10. The dynamics of mechanisms
- the mathematic solution
φ⋅⋅ω−φ⋅⋅ε= sincos eea 2
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
ω,εω,εω,εω,ε
v,a
MM
e
the free body diagram method10. The dynamics of mechanisms
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
ω,εω,εω,εω,ε
v,a
MM
e
( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos eFMememI 2222
the free body diagram method10. The dynamics of mechanisms
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 reduction method
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
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to translation
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
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to translation
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
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to translation
meqv
Feqv
the equation of motion
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
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to translation
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
10. The dynamics of mechanisms
the alternative
the reduction method
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
10. The dynamics of mechanisms
the alternative
the reduction method
reality alternative
the reduction to rotation
2
2
121
2
2
123eqv r
rII
r
rrmI
⋅++
⋅⋅=
Dynamics
312
r3
ωωωω1
M G
ωωωω2
ωωωω,εεεεI 1, r1
I 2, r2
m
x,v,a
ω⋅=⋅−ω⋅= eqv1 MvGMP
Meqv
I eqv
the reality
10. The dynamics of mechanisms
the alternative
the reduction method
reality alternative
the reduction to rotation
2
112 r
r⋅ω=ω 32
11 r
r
rv ⋅⋅ω=
2
13eqv r
rrGMM ⋅⋅−=
Dynamics
313
r3
ωωωω1
M G
ωωωω2
ωωωω,εεεεI 1, r1
I 2, r2
m
x,v,aMeqv
I eqv
the reality
10. The dynamics of mechanisms
the alternative
the reduction method
red2red
red Md
dI
2
1I =ω⋅
φ⋅+ε⋅
I eqv = const :
redred MI =ε⋅
0d
dIred =φ
the equation of motion
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⋅
⋅++
⋅⋅
the reduction to rotation
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
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to rotation
φ⋅⋅ω= cosrv
ω⋅=⋅−ω⋅= eqvMvFMP
φ⋅⋅−= cosrFMM eqv
Dynamics
315
ωωωω,εεεε
φ⋅⋅−= cosrFMM eqv
v, a
φ
ω, εω, εω, εω, ε
M
mI
r
x
F
φ⋅⋅+= 22eqv rmII cos
the equation of motion
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
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to rotation
( ) φ⋅⋅−=φ⋅φ⋅φ⋅⋅−φ⋅φ⋅⋅+ coscossincos rFMrmrmI 2222 &&&
Meqv
I eqv
Dynamics
316
ωωωω,εεεεv, a
φ
ω, εω, εω, εω, ε
M
mI
r
x
F
the equation of motion
red2red
red Md
dI
2
1I =ω⋅
φ⋅+ε⋅
( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos rFMrmrmI 2222
the reality
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to rotation
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
ω, εω, εω, εω, ε
M
mI
r
x
F
the equation of motion
red2red
red Md
dI
2
1I =ω⋅
φ⋅+ε⋅
( ) φ⋅⋅−=ω⋅φ⋅φ⋅⋅−ε⋅φ⋅⋅+ coscossincos rFMrmrmI 2222
the reality
10. The dynamics of mechanisms
the alternative
the reduction method
the reduction to rotation
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 =⋅
10. The dynamics of mechanisms
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