Lecture 1 - 2@Page

8/6/2019 Lecture 1 - 2@Page

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Machine Design Lecture 1: Kinematics and equilibrium of rigid body

Page 1 2011 Politecnico di Torino

Machine Design

Unit 4 Lecture 1Kinematics and equilibrium of rigid body

2

Kinematics and equilibrium of rigid body

Plane kinematicsEquilibrium between external load and reactionInternal forces and moments diagrams


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Kinematics and equilibrium of rigid body

Plane kinematics

4

Plane kinematics

Rigid-body rotationInfinitesimal rigid-body rotationConstraints

Plane structure kinematics


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Plane kinematics

Rigid-body rotation

6

In plane rigid-body rotation (1/ 8)

A rigid-body displacement consists of a simultaneous translationand rotation of the body without changing its shape and size. Therigid body motion is characterized by the requirement that any twopoints in the body remains equidistant.The continuum rigid-body moves, in the plane, from originalposition to final position , thus following a rigid-body

displacement.

B

A

A

B


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In plane rigid-body rotat ion (2/ 8)

Segment AB moves to ABVectors AA and BB are the displacements ofpoints A and B respectively.

A

B

B

A

8

Axes a and b are perpendicular to segmentsAA and BB respectively, being HA and HB theirmiddle points. Axes a and b intersect at C.

b a

A

B

B

AHA

C

HB

In plane r igid-body rotat ion (3/ 8)


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11


Lets consider a new segment AD rotated of anangle about A

a

A

AHA

C

D

D

B

B

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Angles:CB=CBBD=BD

CD=CD


a

A

AHA

C

D

D

B

B

since CA=CA and AD= AD, therefore trianglesCAD and CAD are equal, because two sides are equalin length and one angle is of the same measure.


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13

Point C lies both on the axis of DD and on the axisof AA, then it is center of rotation of the segment

AD.The process can be repeated for every segmentswithin the body.


but then: CD=CDa

A

AHA

C

D

D

HD

d

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In plane rigid-body rotation: closure 1

All the segments within the rigid body rotate aboutthe same point C called center of rotation (rigid,finite).

On the other way: the motion of a rigid body aftera finiterotation and translation can be alwaysdescribed by a rigid rotation about the center ofrotation C.


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It follows thatAD=AD+AA=AD+DD AA=DDthat is, the angles opposite to segments(displacements) AA e BB are equals.

In plane rigid-body rotation: closure 2

a

A

AHA

C

D

D

HD

dAsAD=AD,that is, theangles oppositeto sides AD and

AD are equal,

Plane kinematics

Infinitesimal rigid rotation


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17

For the limit case when AB approaches to AB,that is, the displacements are infinitesimal:

we define the instantaneouskinematics around the initialposition of the segment AB.

Instantaneous kinematics (1/ 5)

AA

B

B

C

HA

B

A

AA=A ; BB=B

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As for finiterotation, the angleopposite to A and

to B is of the samemeasure, namely. This is true forthe displacement ofany point P withinthe body P.


AA

BB

C

HA

B

A


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19

The displacements are:

and then for any point P the

first-order approximationgives:


ddA 2CA sin

2

d2 CA CA d

2

= =

=

AA

BB

C

HA

B

AdP = CPddP = CPd

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And, moreover, forinfinitesimal displacementswhen and

that is:

and C reaches the finalposition for the giveninstantaneous motion.

0


A

B

CA, B 0

CAA ,CBB 90 ' '

B

A


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A

B

C

This is the limit case: it is notpossible to show thedisplacement, but it is possibleto show the velocity

obviously with a different scale

from that used fordisplacements.Velocity shows theinfinitesimaldisplacement

P=A, B, others

dP CPd=

dt

dCP

dt

dP

t

Plim

0t

==

22

Determination of kinematics (1/ 11)

The infinitesimal rotation of an angle d about Cdisplaces point P of segment dP:

dP i d CP= 123

i = operator rotating vector CP of90 in the same direction andsense of rotation vector d. P

C

d

dP


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Then:any infinitesimal instantaneous motion of a body canbe seen as an infinitesimal rotation of an angle dabout a centre ofinstantaneous rotation C.The centre of instantaneous rotation change duringtime, unless C is a fixed point (for example when thebody is constrained by a hinge).

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If point C is at infinite the motion is a rigidtranslation. This is the case when the endpoints ofa segment within the body undergo the samedisplacements.

A

HA HB

A

B

B

( )

C

( )


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The center of instantaneous rotation alwaysexists, then any instantaneous motion is ainfinitesimal rigid rotation about a center ofinstantaneous rotation C.It may be convenient to describe the motionwithout using the center of instantaneous rotationC; for example the motion of point B can be

described by using the motion of a generic pointA: lets see how.

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dB = dA + i d AB


C exists then:

It follows that:

dB = i d CBdB = i d (CA + AB) = i d CA + i d AB

dA


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dB = dA + i d AB

Determination of k inematics (6/ 11)

That is, the motion of point B can be describedsuperimposing an infinitesimally small translationdA and an infinitesimally small rotation about Aof the segment AB.

A

B

A

B

d

the same

rotation for allthe segmentswithin thebody

dA

dA

i d AB

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The scalar product dBZ gives, as idAB isorthogonal toAB:

Displacements dA and dB are not independent ofeach other. Defining Z as the axis parallel todirection AB:


Z =AB

AB

=0

dB Z = dA Z + (i d AB) Z dA Z


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This is because wedeal with a rigidbody, wheresegments AB and

AB does not

change theirlength.

Then the projected components of dB and dAon axis Z has the same value.


A

B

A

B

d dA

dAdB

Z

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With a given dA the only allowable dBs are thosesatisfying the condition stated before; thegraphical representation of the equation follows:


=

=

Allowable dBs

A

B

dA


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The properties shown for infinitesimal rigidrotation hold, with a little approximation, also forvery small finite rigid rotation. Very small finiterigid rotation means that the displacements theyproduce are negligible if compared with thelength of the segments that undergone therotation.

Determination of kinematics (10/11)

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From a mathematical standpoint small rotationsare from 0 to 8.

A B

C

AB CB cos=

Determination of kinematics (11/11)

21cos

2Series expansion gives

K99,0cos =and if: < 8

length CB differs from AB less than 1%


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Plane kinematics

Constraints

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We consider punctual constraints that restrainlinear or angular displacements at a point.Constraints are classified according to the degrees

of freedom, linear or angular displacements, thatthey restrain.

Types of constraints (1/ 5)


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Car:restrains the displacement indirection

B

Symbols


Av

A Bvh

Description

v v(A) = 0

(External) Hinge:restrains displacements indirections

The hinge restrains any rigidtranslations of the body in theplane.

v and h ;

v(A) = 0 ; h(A) = 0

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v(A) = 0 ; h(A) = 0 ; (A) = 0


A

v

h

Clamp:restraints any translation

and rotation

B

BA

B

or

A

Prismatic joint:restrains displacement indirection and rotation ;

(equivalent constraint calleddouble pendulum)

Symbols Description

h(A) = 0 ; (A) = 0

h


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These were external constraints, that is, they restrain thedegrees of freedom of the element of the structure withrespect to a fixed reference frame external to the structure(the external world)


There are also internal constraints, that is, they restrainthe degrees of freedom of one element with respect to thedegrees of freedom of another element within the samestructure.

DB

A

External hingeInternal hinge

Car

C

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Internal hinge: All of the body rotations

are allowed; only one body can bedisplaced freely in theplane: the others mustfollow it.

F A

Symbols Description

D

B

E

C

v

h

vh

vh

vh


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The number and type of constraints define thekinematic behavior of the structure.

Constraints and structures (1/ 2)

Body #1

Body #2 Body #3

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Constraints and structures (2/ 2)

Body #1Body #2 Body #3

Segments drawn within the bodies do not alwayssymbolize bars or beams, but in more generalsense they are segments linking restrained points

(internal or external) of bodies whatever theirshape may be.


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Plane kinematics

Kinematics of plane structures

42

A kinematically determinate structure or mechanismcan be defined as a structure where, if it is possibleto find point displacements compatible with the

constraints, those point displacements are unique.The structure has no possible point displacementscompatible with zero member extensions, at least toa first-order approximation.

Kinematics determination

body #1body #2 body #3


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43

Lets give a vertical displacement to point A: dvA

Example A (1/ 7)

A simple example:(Remark: the given displacements and rotations areinfinitesimal

DBA

C

Adv dv

dhd

44

Example A (2/ 7)

verticalprojection

DBA C

Adv

verticalprojection

AC A ACdB dA i d AB dv i d AB= + = +

dv

dhd

ABddv ACA

ACddvACdidAdC ACAAC +=


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Example A (3/ 7)

verticalprojection

Constraints:

DBA

C

Adv

ACd

dv

dhd

CDddvCDdidCdD CDCCD +=

ABdv

d0dB AAC ==

46

v

Example A (4/ 7)

From vertical projection it follows that

=

==

ABBC

dvABAC

1dvACABdv

dvdv AAA

AC

DBA

C

Adv

ACd Cdv

P

=

==

ABBP

dvABAP

1dvAPABdv

dvdv AAA

AP

and that for any generic point P


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Example A (5/ 7)

Adv

A BP C

DBA

C

Adv

ACd Cdv

ZAP

Cdv

=

AB

ABAPAP Z

ZZdvdv

Segment AC:displacement inA is given;

displacement inB is null, then:

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Example A (6/ 7)

0CDddvdv CDCD ==

DBA

C

Adv CDd

CdvC

Cdv dD 0=

DPThe displacement at point C isknown, then:

dv

dhd

CDdv

d CCD =

Note carefully: positive displacementsand rotations are according to theconvention on the right


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Example A (7/ 7)

Adv

AB C

DBA

C

Adv

ACd Cdv

Cdv dD 0=

D

The diagram of the vertical displacements is:

50

Example B (1/ 3)

AB C

=

=

C

Allowabledisplacements dD,compatible withrigid segment CD

D

B

D

CA

E F

Cdv

Cdv

Adv

Adv


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Example B (2/ 3)

=

C

D

E

FD

B

D

CA

E F

Only the verticaldisplacement iscompatible with thekinematics (D,E,F).

Adv

Cdv Fdv

Ddv

52

Example B (3/ 3)

B

D

CA

EF

B

D

CA

E F

The diagram of the vertical displacements is:

Adv


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Example C (1/ 3)

B

D

C

AE

If the constraints are not sufficient, the motion ofthe structure is not univocally determined once thedisplacement of one degree of freedom is given.The structure is then kinematically indeterminate.

=

=

F

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Example C (2/ 3)

B

D

CA

E

Point D can move to any direction, because point Ecan translate horizontally and body (D,E,F) canrotate about E.

=

=

F


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Example C (3/ 3)

B

D

C

AE

The kinematics at (D,E,F) is indeterminate.

F

B

D

CA

E F

56

Example D (1/ 2)

When the number of the constraints does notallow any displacement, the structure iskinematically over-determinate.

Constraints limit displacements when theexternal loads act on the structure (obviouslyuntil the collapse of the structure).


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There is alsoa horizontalcomponent

there is only

a verticalcomponent

Example D (2/ 2)

B

D

CA

Example

The displacements at C are not compatible: no motion allowed for the structure

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Lecture 1 - 2@Page