Dynamics and control of mechanical systemshirpa/DMS6021/2018/1.2 Kinematics of rigid bodies.pdf · Bedford and W. Fowler, Engineering Mechanics–Staticsand dynamicsprinciples 3

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Dynamics and control of mechanical systemsDate ContentDay 1(03/05)- 05/05

• Review of the basics of mechanics. • Kinematics of rigid bodies – plane motion of rigid bodies, angular

velocity vector, description of velocity and acceleration in relatively moving frames.

Day 2(07/05)

§ Euler angles, Review of methods of momentum and angular momentum of system of particles, inertia tensor of rigid body.

• Dynamics of rigid bodies - Euler's equation, application to motion of symmetric tops and gyroscopes and problems of system of bodies.

Day 3(09/05)

• Kinetic energy of a rigid body, virtual displacement and classification of constraints.

• D’ Alembert’s principle.

Day 4(11/05)

• Introduction to generalized coordinates, derivation of Lagrange's equation from D’ Alembert’s principle.

• Small oscillations, matrix formulation, Eigen value problem and numerical solutions.

Day 5(14/05)

• Modelling mechanical systems, Introduction to MATLAB®, computer generation and solution of equations of motion.

• Introduction to complex analytic functions, Laplace and Fourier transform.

Day 6(16/05)

• PID controllers, Phase lag and Phase lead compensation. • Analysis of Control systems in state space, pole placement, computer simulation

through MATLAB.DMS6021 - Dynamics and Control of Mechanical Systems

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Content4 Introduction4 Symbols and designations4 Kinematics of rigid bodies in a plane4 Instantaneous center of motion/velocity4 Analysis of rigid body motion (velocity and acceleration)

q Graphical methods to determine velocities and accelerations

q Analytical methods to (briefly):

DMS6021 - Dynamics and Control of Mechanical Systems

References1. F. P. Bear, E.R. Johnston and P.J. Cornwell; Vector Mechanics for Engineers, Dynamics, 10th

Edition2. A. Bedford and W. Fowler, Engineering Mechanics – Statics and dynamics principles3. D.T. Greenwood, Advanced dynamics, Cambridge University Press (2006)4. G. Gray, et al.,-Engineering Mechanics - Dynamics-McGray (2010)

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Introduction4 Study of the Kinematics of rigid bodies (mechanisms) :


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Kinematics of rigid bodies4 Study of the Kinematics of rigid bodies (mechanisms) :


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Symbols & designationsPosition, velocity and acceleration

s = position vector of a point (size and orientation)s = size of a position vector sv = velocity vector of a point (size and orientation)v = speed (size of the velocity vector v)a = aceeleration vector of a point (size and orientation)a = size of an acceleration vector aan = normal component of the acc. vector (normal to path of motion)at = tangential component of acc. vector (tangent to path of motion)j = angular positionw = dj/dt = = angular speed à tme derivativea = dw/dt = = angular acceleration

•

j

•

w

•

j


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Symbols & designationsPosition figure, velocity polygon and acc- polygon

A, B, C, etc. =

Points in the position polygon, i.e. line segment Ato B in a position polygon gives the distancebetween point A and B

A’, B’, C’, etc. =

Points in the velocity polygon, i.e.. Line segment from A’ to B’ n a velocity polygon givesthe relative velocity of point B wrt point A (= vBA), while the absolute velocity of point A (vA= vAO) is given by the line segment from O’ to A’.

A’’, B’’, C’’, etc. =

Points in the acceleration polygon, i.e.. Lijesegment from A’’ to B’’ in the acc. polygon givesthe relative acc. of B wrt A (aBA), while absolute acce. Of point A (aA) is given by the line segment O’’ to A’’.


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Symbols & designationsSome examples on use of indices

rAB = distance from point A to point B

vAB = relative velocity of point A wrt point B

vBA = relative velocity of point B wrt point A (= -vAB)

vA = absolute velocity of point A = relative velocity of Awrt a stationary point (= vAO)

vrel = relative velocity

aBAn = normal comp. Of relative acc. aBA

Oij = instantaneous center for relative motion of part i and part j


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Kinematics of rigid bodies4 Kinematics of rigid bodies:

Relations between time and positions, velocities and accelerations of particles composing the rigid body

4 Classifications of rigid body motion


- direction of any straight line inside the body is constant,

- all particles forming the body move in parallel lines.

- All particles have the same velocity and the same acceleration.

- All particles in thebody have the same angualr velocity and angular acceleration

- Sum of translationand rotation

ABAB aaandvv !!!!==

velocityangiskk

rdtrdv

.!"

!!

!!!

!

qww

w

==

´==

kkkdtd !

""!"

!!!

qwaaw====

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Plane motion of a rigid body4 General plane motion:

A motion where all particles in the body move in a parallell plane

4 A general plane motion is composed of two motions


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Plane motion a rigid body4 Position vector of a

particle in motion

nt VVV D+D=D

4 Graphical representation of velocityand acc. vectors

dtd

trrV =

DD

= lim

dtd

tVVa ==

DDlim

rv xω=

raraaaa nt

xxx

n

t

ωωα

==+=

ogder

rV

tVaa n

nn

2

lim ===DD

!V

tn

tr

rVV

tVder

DDD+

=DD )(


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Instantaneous center of motion

For the car to be able to drive along a curve without sliding, all the axles of the wheels indicate towards the same point, which is an instantaneous center of motion (point O)

è The car (at the moment) is undertaking a rotational motion about point O

Instantaneouscenter of zero velocity


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Instantaneous center of motion

Fig. (a): Part 3 moves wrt part 1 (which is stationary part)§ Point C and D follow circular paths about A and B respectively

§ Point O (at the moment) has no velocity in any directionà It is Instantaneous Center – a point where part 3 (at the given instant) undergoes a rotational motion about the pointà Curve OO’ is the motion path that the ICV follows


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Instantaneous center of motionSummaryICV is:1) A point on part 1 (stationary part)

where part 3 rotates about at the momentExample: for point E on part 3

2) A point on part 3 (stationary part) where part 1 rotates about at the moment

3) A crossing point where both part 1 and part 3 have the same absolute motion, i.e. null relative motion, (valid when both are in motion)

13

13

13

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23

2

3

:normalon,Accelarati

;:tangentialon,Accelerati

:

OEOE

EEn

OEEt

OEE

rxrV

rx

IxSpeed

--

-

-

==

=

=

w

a

w

a

a

V


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Number of instantaneous center of velocitySeveral parts in a mechanism: è several instantaneous centers (ICV)# instantaneous centers: = # different ways of selecting to parts

F. ex. The mechanism shown below has 4 parts joined together at joints A, B, C and D

Each joint has two parts having a relative motion against each other.

à Each joint is an ICV


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Types of Instantaneous CentersType 1: Fixed instantaneous centers

– a fixed point on a body where another part is rotating about

f. ex. O12 and O14

Type 2: Permanent instantaneous centers- A common point of two parts that are in motion with the same speed.

This is valid for joints f. ex. O23 and O34

NB: Type 1 and 2 are known (identified) ICVs

Type 3: Imaginary instantaneous centers - an imaginary point on or outside of a mechanism where the part can be imagined to rotate about at the moment

f. ex. O13 and O24

Three main types of ICVs


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Types of Instantaneous Centers

Pin joint mechanism– Each joint is known instantaneous center

Sliding contact- Two points A and B on a rigid body that is in

sliding motion undertakes translation motion, thus it has known instantaneous center

- Q: where is the instantaneous center OAB?

Rolling contact gives known instantaneous center

- Where is the instantaneous center O12?Assume A pure rolling motion!

Three types of known/identified instantaneous centers


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Kennedy´s Theorem4 Kennedy´s Theorem

Gives the basis for a systematic decisions of the number and location of instantaneous center of velocityThe Theorem is expressed as followsFor relative motion of three randomly selected parts, the three ICVs are located on a straight line segment

NB: The Theorem is valid if only one of the parts is in motion or both are in motion

Example: Kennedy´s theorem for two parts in sliding contact

http://web.mit.edu/linkagedemo/www/linkageanimation.htmlDMS6021 - Dynamics and Control of Mechanical Systems

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Kennedy´s Theorem, an example


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Kennedy’s theorem and graphical solution, example

.

1. Determine number of parts in the mechanism: = n, and # ICV: 2. Draw a circle with a random dimeter size.3. Divide the circumference into n equal sections and number them ( 1, 2, … n). 4. Locate the known ICVs on the circle by drawing a chord between the marked points

1 and 2, 2 and 3, etc. to get O12, O23, O34, etc. Location of ICVs at joints are known ICVs.

5. Search for a chord that creates the last side of two triangles (common side to two triangles).6. Draw this chord in the circle using a dashed line until the location of the ICV is found on

the mechanism.7. According to Kennedy’s theorem, the sides of each triangle correspond to ICVs that are on a

straight line. This means that each triangle represents a straight line.8. The intersection between two lines (corr. to two triangles) determines the searched ICV.9. When the ICV is localized, draw the chord with a solid line.10. Go back and repeat the procedure from step 5 until all ICVs are localized on the

mechanism

2)1(

2-

=÷÷ø

öççè

æ nnn

A systematic procedure of using Kennedy’s theorem to find and locate ICVs:


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Kennedys teorem og grafiske løsninger


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Graphical methods of determining velocities and accelerations

aB

aBn + aBt

= aA + aBAaAn + aAt aBAn + aBAt

sr r sr sr sr r

aAt

Velocity polygon

Acceleration polygon

VA = O’A’: perpendicular to CA

VB = O’B’:perpendicular to DBVBA = A’B’:perpendicular to AB


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§http://www.softintegration.com/chhtml/toolkit/mechanism/fig/fourbar/animation.gif

Illustrative example

How will the motion of point P on part 3 look like, when part 2 rotates?


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§ In this mechanism, the crank AB is rotating with a constant angular speed sothat the speed of point B is VB = 5 m/s. At the given position, use graphicalmethod and determine acceleration of point D, which is a corner point of therigid body 3.

0,5=CV

0,5=CBV

O’C’

B’

0,5=BV

Example exercise


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Example exercise

)()()()()(

rrsrsraaaaaa CBtCBnBCBBC ++=+=

)()()()(

rrsrsaaaaa

aaa

DCtDCnCDtDn

DCCD

++=+Þ+=

( ))()()( rrsrs

aaaaaaaa

DBtDBnBDtDn

DBBD

++=+Þ+=


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Analytical approach


jkxiixk

ijxkkxj

kixjjxi

0kxkjxjixi

=-=

=-=

=-=

===

).(rxrxαa

).(),rx(xVxa

).(,rxV

.exp

.

t

2n

ramagn

ramagn

rvmagn

componentsaccandvelocityofressionVector

t

n

aw

wwww

ww

===

===

==

®!!!!

!!!

!!

Relative motion of point B wrt A: - Velocity: VB = VA + VBA- Acceleration: aB = aA + aBA- Acc. Components: aBn + aBt = aAn+ aAt + aBAn+ aBAt

( )jxiy

jyixxkrxV

A2A2

AA2A

w+w-

+w=w=

000

kirxV 2A

AA yx

jww == ji 22 AA xy ww +-=

( )0

00ki

xrxxa 22

AA

n

yx

jk wwww == ( ) rji 2

2222 www -=+-= AA yx

A 4 bar linkage mechanism

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Summary and questions


Next: Review of methods of momentum and angular momentum of system of particles, Euler angles, inertia tensor of rigid body.

The following are covered in ttis part of the lecture4 Kinematics of rigid bodies – analysis and synthesis4 Classification of rigid body motion

4 Parameters to describe a general plane motion4 Instantaneous center of motion/velocity4 Analysis of rigid body motion (velocity and acceleration)

q Graphical methods to determine velocities and accelerationsq Analytical methods to determine velocities and accelerations:

?

Documents

Dynamics and control of mechanical systemshirpa/DMS6021/2018/1.2 Kinematics of rigid bodies.pdf · Bedford and W. Fowler, Engineering Mechanics–Staticsand dynamicsprinciples 3