mech1230 dynamics unit 2 - rigid body kinematics (1)

8/13/2019 MECH1230 Dynamics Unit 2 - Rigid Body Kinematics (1)

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2.1

MECH1230

SOLID MECHANICS

Dynamics

Unit 2

RIGID BODY

KINEMATICS

AND MECHANISMS


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2.2

Contents

2.1 Rigid Body Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3

2.1.1 Plane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5

2.1.2 Relative Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6

2.1.3 Relative Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9

2.2 Instantaneous Centre of Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11

2.3 Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12

2.3.1 Diagrammatic representation of mechanisms . . . . . . . . . . . 2.12

2.3.2 Alternative mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.15

2.3.3 Analysis of Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17

2.3.3.1 A single link . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17

2.3.3.2 Two connected links. . . . . . . . . . . . . . . . . . . . . . . . . 2.18

2.3.4 Case studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.22

2.4 Analysis of Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.24

2.4.1 Slider Crank Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.24

2.4.2 The Four Bar Chain Mechanism . . . . . . . . . . . . . . . . . . . . . 2.41

2.4.3 Quick Return Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 2.43


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2.3

2.1 Rigid Body Kinematics

In order to analyse the kinematics of a particle, all that is required is a complete

description of the location of the particle at all instants of time. For a rigid body a

corresponding description of the motion requires the definition of the location and

orientation of the body. Thus a kinematic analysis of a rigid body involves both linear

and angular quantities.

There are five types of rigid body motion:

1. Translation, which may be one of either: rectilinear (a straight line in 3-D space); curvilinear (along a curve in space but with the body having the same

orientation);

co-planar (within a 2-D plane).2. Rotation about a fixed axis.3. General plane motion.4.

Rotation about a fixed point.

5. General motion.In this part of the course, general plane motion is studied since there exists a vast range ofengineering problems which are amenable to solution through application of the

techniques described in the following sections.

When considering plane motion, every particle in the body remains in a single plane. As

all points along all lines drawn perpendicular to the plane have the same motion, only the

motion in a single plane need be considered. This may be, but is not limited to, the plane

on which the mass centre lies.

As particles cannot move out of the plane of motion, the position of the rigid body in

plane motion is completely determined by providing the location of one point and theorientation of one line in the plane of motion. The orientation of the line may be given

by:

1. the angle the line makes with a fixed direction, see Figure 2.1, or by;2. specifying the location of any two points on the line, see Figure 2.2.

School of Mechanical Engineering

MECH 1230 Solid Mechanics


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2.4

Figure 2.1: Position of a point on a rigid body

Figure 2.2: Position of two points on a rigid body

The angular motion of lines in the plane of motion is the same for everystraight line in a

rigid body. Consider the situation depicted in Figure 2.3 which shows a rigid body on

which two linesABand CDare drawn and which are separated by the angle. The line

ABforms an angle ABto the reference direction and the line CDforms an angle CDtothe reference direction. Clearly,

CD=AB+. (2.1)

As the rigid body rotates the angles ABand CDchange but as the body is rigid, the angle

remains fixed. Differentiation of equation (2.1) with respect to time gives:

, ABABCDCD (2.2)

where is the angular velocity or rate of change of angular position. Thus every line in

the body has the same angular velocity, .

A

rA

O x

y A fixed line across the

body

rA

O x

y A fixed line across the

body

rB

B

A


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2.5

Figure 2.3: Angular motion of a rigid body

Similarly, the angular acceleration of every line within the body is the same:

, ABABABCDCDCD (2.3)

where is the angular acceleration or the rate of change of angular velocity.

2.1.1 Plane Motion

Consider any two pointsAandBon a general plane rigid body. Since the body is rigid it

is important to appreciate that the distanceABwill remain constant regardless of the

motion. We view the body at two instants of time, tand t + dt, as shown.

Figure 2.4: Plane motion of a rigid body

D

O x

y

B

A

C

AB

CD

A

B C

BC

B1

C1

Time t Time t + dt

dl

dl

x

y

O

d

A1


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2.6

Figure 2.4 shows that the motion over the time interval dtmay be thought of as:

a translation fromAtoA1of dland fromBtoBof dl; plus, a rotation daboutA1to bringB into positionB1.

A point C has been included in the diagram representing another point on the body suchthat AC is at an angle ofwith respect to AB. It can be seen that the angle remainsunchanged throughout the processes of translation and rotation. It follows therefore that

in moving from positionABto positionA1B1the body has a velocity of translationgiven

by:

,ldt

dlv (2.4)

and a rotational velocityor angular velocity, , given by:

.dt

d (2.5)

Hence every point on the body is treated as having a velocity of translation, v, and every

line joining any two points the angular velocity, .

2.1.2 Relative Velocity

Once again consider a plane rigid body and any two pointsAandBwithin it having

velocity vAand vbrespectively - see Figure 2.5. For a rigid body vAdoes not have to be

equal to vBas there can be rigid body rotation about an axis located at some arbitraryposition in the plane.

Figure 2.5: Relative velocity of two points on a rigid body

A

B

x

y

O

vB

vA

A

B


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2.7

Using the idea of the triangle law for vector addition write the velocity atAandBincomponent form - one parallel to the direction ofABthe other perpendicular toAB- see

Figure 2.6, where:

.vv,vv

vv,vv

BBPerpBAAPerpA

BBParBAAParA

sinsin

coscos

__

__

Figure 2.6: Velocity components

Since the body is rigid the distance betweenAandBmust remain fixed. Consequently

vA_Par= vB_Parand the magnitude of the velocity ofBrelative toA, vBA, is given by:

vBA= vB_PerpvA_Perp, (2.6)

whose direction is perpendiculartoAB- see Figure 2.7.

A

B

x

y

O

vB

vA

vA_ParvA_Perp

vB_ParvB_Perp

A

B


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2.8

Figure 2.7: Velocity of B relative to A

Thus the motion ofBas seen byAis one of rotation aboutA- it follows therefore that the

angular velocity, , of the body is given by:

.d

v BA (2.7)

The absolute velocity of the pointBis then given by:

vB= vA+ vBA. (2.8)

Clearly, these must be added vectorially- see Figure 2.8. Equation (2.8) states that the

absolute velocity ofBis the resultant of the absolute velocity of A and the velocity ofBrelative toA.

A

B

x

y

O

vBA

d

vBA

vA

vB

Parallel to BA


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2.9

Figure 2.8: Vector addition for relative velocities

2.1.3 Relative Acceleration

Since the velocities at the pointsAandBwill, in general, vary with time they will

accelerate. As in the case of relative velocity, the absolute accelerations aAand aBof the

pointsAandBcan be reduced to component form - see Figure 2.9.

Figure 2.9: Acceleration components

So, the acceleration ofBrelative toAhas two components:

( aB_ParaA_Par) alongAB;

( aB_PerpaA_Perp) normal toAB.

If the angular acceleration of the body, , is denoted by , then for a point moving on a

circular path it has accelerations R 2andR, parallel and perpendicular to the radiusR

from the centre of rotation, respectively. (Recall that an angular velocity gives a

tangential, or perpendicular velocity component with respect to AB, and this angularvelocity generates an acceleration that is radial or parallel to AB).

A

B

x

y

O

aB

aA

aA_Par

aA_Perp

aB_Par

aB_Perp

d


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2.10

Now considering the motion ofBrelative toA, which are separated by a distance d, the

components of acceleration are given by:

( aB_ParaA_Par) = 2d alongAB;

( aB_PerpaA_Perp) = d normal toAB.

These accelerations are those that an observer at Awould see.

The angular velocity, is found from the velocities, vA_Perpand vB_Perpand thegeometry.

The angular acceleration, is found from the accelerations, aA_Perpand aB_Perpandthe geometry.

Now,

aB= aA+ aBA. (2.9)

but, aBAhas twocomponents (unlike relative velocity which has only one). Also,

aA= aB+ aAB.

So, for the pointB, aBAhas components d(normal to AB) and 2d(parallel to AB) as

shown in Figure 2.10.

Figure 2.10: Acceleration components of B relative to A

2.2 Instantaneous Centre of Rotation

aB

aA

aBA_Par = 2d

aBA_Perp = d

aBA

Parallel to BA


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2.11

Consider the rigid body shown in Figure 2.11. The pointsAandBhave absolute velocity

vAand vB, respectively.

If vAvBthe rigid body will have an angular velocity at a particular instant. SoAandB

can be considered to be rotating about some point along a perpendicular to the velocity

vectors vAand vBat these points. These perpendiculars lie alongA AandB B

as shown in

Figure 2.11. We conclude therefore, that at this particular instant the body must be

rotating about the point of intersection ofA AandB B

- that is, about the pointI, the

instantaneous centre of rotation.

Notethat in general vAand vBare not constant, soIis not a fixed point - but a pointregarded as being instantaneously at rest.

If is the angular velocity of the body at that instant, then (from v = r:

.

IB

v

IA

v

BA (2.10)

The concept of an instantaneous centre of rotation is a useful and quick way of finding

the velocity of the moving parts of a mechanism. Mechanisms are examined more fullysubsequently but in the meantime consider the schematic of an up-and-over garage door

shown in Figure 2.12 (not to scale). As the door QSopens the slider Smoves to the left,

as it closes the slider moves to the right. IfPRis the pivot arm for the garage door where

is the instantaneous centre of rotation of the garage door in the position shown?


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2.12

2.3 Mechanisms

There exists a vast range of mechanisms, which find application throughout mechanical

engineering. Mechanisms are made up of combinations of links, slides, gears and cams

which may operate at low or high speeds and may either transmit power or modify

motion. The word mechanism is derived from the Greek word meaning machine and in

fact dictionary definitions of the word mechanism include "the working parts of amachine" and system whose parts work together as in a machine". Mechanisms vary

greatly in their size and complexity ranging from simple engine mechanisms to high-speed textile equipment.

2.3.1 Diagrammatic representation of mechanisms

A convenient way to represent the kinematic behaviour of a mechanism is to use a so-

called stick diagram. A stick diagram accurately conveys the ways in which the variouscomponents can move relative to one another. The basic elements of a stick diagram are

shown in Figure 2.13 together with some examples of commonly encountered

mechanisms in stick form in Figure 2.14 in which the arrows indicate a typical input tothe mechanism.

Q


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2.13


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2.14


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2.15

2.3.2 Alternative mechanisms

Many mechanisms involve motion in three-dimensional space, for example an excavator

shovel, vehicle suspension system (because the spring/damper axis is normally angledand the suspension linkages pivot generating complex 3D motion) or robotic device.

Fortunately there are a significant number whose motion can be described in a 2D plane.Some examples found in everyday life are shown in Figure 2.15 and a number ofengineering applications are provided in Figure 2.16.


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2.16


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2.17

2.3.3 Analysis of Links

2.3.3.1 A single link

The application of velocity and acceleration components in polar coordinates (see unit 1,

page 1.26) may be illustrated through consideration of the crank element of a slider crankmechanism. The crank can be represented as a link, OA, of fixed length, which is pivoted

at one end. Figure 2.17 shows how the generalised velocity and acceleration termssimplify for this case.

(Note in the diagrams above, and )

The restriction of fixed length means that all terms involving r and r must be zero. Thevelocity and acceleration components shown are very important as they form the basis for

the graphical method of mechanism analysis which follows.

Figure 2.17


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2.18

2.3.3.2 Two connected links

Now consider a situation involving a second fixed length link,AB, attached to the free

end of the pivoted link OA. This might, for example, represent the connecting rod of aslider crank mechanism. The motion of the linkABcan involve translation in bothXand

Ydirections and rotation. This is depicted in Figure 2.18.

In comparison toAB, the link OAmay only rotate about Oand is said to be kinematically

constrained. This means that the motion of OAis simply a function of the angle 1andthe link geometry.

Velocity

The direction and magnitude of the velocity of the free end of the linkABcan bedetermined by application of the idea of relative velocity:

The velocity of pointArelative to ground at Ois the absolute velocity ofA, which has a

magnitude vAgiven by:

11 OAOAvA (2.11)

which is perpendicular to the line OAas shown in Figure 2.18.

Calculation of the velocity of pointBrelative to pointAis complicated by the fact that

pointAis itself moving. Relative to an observation position atA, pointBcan only have a

r velocity term since its length is fixed. Thus, the velocity of point Brelative to pointA

has a magnitude equal to:

Figure 2.18


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2.19

2ABvBA (2.12)

where 2is the angular velocity of AB and vBAhas a direction which is perpendicular tothe linkAB, see Figure 2.18. From this follows the important result that allows an

expression for the absolute velocity of pointB(velocity ofBrelative to ground at O) to

be derived as:

vB = vA + vBA. (2.13)

A pictorial representation of this result is shown in the velocity vector diagram of Figure

2.19, which provides a means of calculating the magnitude and direction of the resulting

velocity vB.

EXAMPLE TO TRY YOURSELF

Using the data provided below in conjunction with the figure, calculate the resultant

magnitude and direction of the velocity of pointBat this instant.

vB = vA + vBA

Figure 2.18Figure 2.19


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2.20

Data: OA=0.4 m, AB=0.5 m, 1=20, 2=75, 1211 542 rads.,rads Answers: vA= 0.8 ms

-1, vBA= -2.25 ms

-1, vB= 1.91 ms

-1, at an angle of 5.1to the positive

x-axis.

Acceleration

The same principles of relative motion can be applied to the calculation of the

accelerations of the interconnected links.

The pointAon the link OAcan only have two components, namely1

r and2

1r .

However if the link OAis rotating with constant angular velocity, 1then the 1r term is

zero, leaving a single component2

1r directed towards the point O.

The acceleration ofBrelative toAhas two components:

22AB alongAB;

2AB perpendicular toAB;

which are shown in Figure 2.20.


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2.21

The absolute acceleration ofBrelative toAis given by the vector expression:

aB=aA+aBA (2.14)

from which a diagram showing the acceleration vector, Figure 2.21 (which assumes OA

is rotating at constant angular speed), can be drawn. In order to calculate aB, a value for

2is required which is the instantaneous rotational acceleration of the linkAB.

Note that calculations of this type refer to one position of the links. This can be thought

of as equivalent to analysing a still frame taken from a motion picture.

Figure 2.20


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2.22

When considering the dynamic behaviour of a body, or an assembly of bodies, in a

machine the first step is to reduce the actual system to an idealised model by making

appropriate assumptions. One should remember that it is very important to bear in mind

at all times the assumptions made in setting up the model, particularly when interpretingthe results of any analyses. If the assumptions are invalid, the results will also be

invalid. Once the model has been defined, a free body diagram must be drawn before

proceeding further, for each body in the system, showing the forces which act on them.The appropriate equations of motion can then be applied. Next, any kinematic constraints

must be identified and the resulting equations then solved for the required variables.

Figure 2.21


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2.23

2.3.4 Case studies

In order to demonstrate the techniques used to analyse the kinematic behaviour of plane

mechanisms it is appropriate to concentrate on a small number of mechanisms from thevast range available. What follows will concentrate on the analysis of two mechanisms,

namely the slider crank and the four bar chain. Together these make an excellent vehiclethrough which to introduce the basic concepts and techniques associated with the analysisof linkages and mechanisms. The four bar chain typifies a particularly large class of

mechanisms.

Slider cranks are commonly found in internal combustion engines and a typical layout is

shown in Figure 2.22 along with the equivalent stick diagram in Figure 2.23. Note the

differences and the fact that the stick diagram captures all the essential kinematic

information.


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2.24

The main elements of the slider crank are the piston, crank and connecting rod. These

perform the following types of motion:

Piston - Rectilinear (translational) motion;

Crank - Rotational motion; Connecting rod - Combination of translational and rotational motion.

Rocker arm

Valve

Figure 2.22


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2.25

2.4 Analysis of Mechanisms

In this section we focus our attention on the analysis of the slider crank and four barchain. The former is a convenient vehicle by which to introduce the basic concepts

involved; the latter is fundamental in the design of mechanisms - it is simple, fairly

inexpensive, widely used and takes many different shapes and sizes.

2.4.1 Slider Crank Mechanism.

This mechanism is very common and therefore of great interest.

Cylinder

Valves

Connecting Rod

Crankshaft

Crank axis

Pivot of connecting

rod about the crank

Piston

Figure 2.23


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2.26

Method of Analysis

First reduce the mechanism to a `Stick' Position Diagram(SPD) as shown below:

There are three approaches that can be adopted to analyse this and other mechanisms:

1. Trigonometric.- Write down analytic expression(s) for position and then differentiate twice to

obtain explicit expressions for velocity and acceleration.

This is a lengthy process even for simple mechanisms. Almost impossible for complex ones!

2. Graphical - scale drawings.- Gives velocity and acceleration for one position only and hence this process

must be repeated many times to produce an overall picture of the motion for one

complete cycle.

Its use as an aid to physical understanding is questionable. Long winded and tedious.

3. Computational.- This is vector based and is usually implemented on a computer. It is

straightforward to write down the displacement vectors and then to perform

subsequent velocity and acceleration analysis via numerical methods. Thisapproach is:

quick (if the programme already exists); convenient (use the computer to plot out the results); and flexible (easy to solve for any input, different link lengths, etc).

Clearly the latter approach is the preferred one, particularly if the necessary software isreadily available.


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2.27

Trigonometric Analysis Slider Crank

For simple mechanisms, algebraic expression(s) may be formulated to describe theposition of an output of interest relative to some known input. This expression can then

be differentiated once with respect to time to obtain an explicit expression for the velocity

and again to derive an explicit expression for the acceleration. Note that simplifyingassumptions may be required to effect the differentiation.

Consider the slider crank mechanism shown in the figure below, which is assumed to

have its crank driven at a constant speed of rads1

. If the output of interest is the piston

then an expression for piston displacementxas a function of the crank angle is

required.

The main idea is to findxas a function of ,x = f(), and then differentiate this function

w.r.t. time, t, to obtain xv and xa . Now:

rcoslcosx (2.15)

and clearly (from the sine rule),

sinsin

rl rsin() = lsin() (2.16)

Now from equation (2.16) sinl

rsin and:

21

sin1cos 2 since .1cossin 22

So


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2.28

.sinl

r1cos

2

1

2

2

2

(2.17)

Hence substituting for cos() from equation (2.17) into (2.15) gives:

.sinl

r1lrcosx

2

1

2

2

2

(2.18)

The velocity and acceleration of the piston are found by the successive differentiation ofequation (2.18) with respect to time. In order to simplify this process it is appropriate to

expand the term within the square root sign using a Binomial expansion of the form

....,k

21

1mmmk1kl 2

m

which leads to:

;sin2l

rlrcos

;sin2l

r1lrcosx

22

2

2

2

(2.19)

where higher order terms are ignored based on the assumption that they will contain

terms of diminishing magnitude.

Thus since varies with time, the velocity can be determined using the chain rule as:

,d

dx

dt

d

d

dx

dt

dxx (2.20)

such that:

.2sin2l

rrsin

;cos2sin

2l

rrsin

d

dx

2

2

(2.21)

If the input speed is a constant, say , then:


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2.29

.2sin2

sin

2sin2

sin2

l

rr

;l

rrx

(2.22)

Similarly, for acceleration:

,d

xd

dt

d

d

xdx

(2.23)

where:

.lrr

l

rr

d

xd

2coscos

,2cos22

cos

(2.24)

Combination of equations (2.23) and (2.24) and noting once again that leads tothe desired result:

.2cosl

rcosr

;2cosl

rcosrx

2

(2.25)

Equations (2.18), (2.22) and (2.25) can be expressed as a function of time given that is

constant and the crank angle can be written as:

= t. (2.26)

Substitution of (2.26) into equations (2.18), (2.22) and (2.25) leads to the following

expressions for the position, velocity and acceleration of the piston under the assumption

that is constant:

Postion ;tsinl

r1ltrcosx

21

2

2

2

(2.27)

Velocity ;t2sin2l

rtsinrx

(2.28)

Acceleration .tl

rtrx

2coscos2 (2.29)


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2.30

Graphical AnalysisSlider Crank

This is the traditional approach to solving mechanism problems. There are two types of

graphical analysis. The first uses the concept of an instantaneous centre of rotation.

Method of Instantaneous Centre

The following diagram shows a scaled drawing of the displacements for a slider crank at

some particular angle of input, where P is the piston, O the axis of rotation of the crankand A is the pivot between the connecting rod and the crank.

The following steps are involved in performing a graphical analysis:

(i) Draw the mechanism to scale (on graph paper) at the desired position.

(ii) Locate the instantaneous centre,I, in this particular position. This can be

determined as the direction of motion of each end of the connecting rod is known.

The end at the piston must move with the piston which is modeled as a slider and

hence moves horizontally (in the diagram as drawn). The other end of the

connecting rod is at the end of the crank which is rotating with uniform circular


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2.31

motion about O, hence the velocity at A must be a tangent to OA.

(iii) Calculate vA( the tangential velocity of the pointA = r, where here refers

to the angular speed of the crankshaft).

(iv) The angular velocity of the connecting rod (PA) is then given by:

IA

r

IA

v Arod ,

whereIAis scaled from the position diagram.

(v) The velocity of the piston, vPis given by (againIPis scaled from the position

diagram):

.IA

IPvIPv ArodP

WORKED EXAMPLE

Consider the garage door problem introduced earlier, with the dimensions shown. Given

that PR rotates about P at the constant angular speed indicated, find the velocity of the

door and the slider (at this instant) together with the length d. Note that by defining SF inthe position diagram defines the position of the whole mechanism.


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2.32

Solution

To determine the velocities and distances a Scaled Position Diagram (SPD) must bedrawn:

(i) The distance dis easily measured off the above scaled diagram and is found toequal 0.9m.

(ii) The position of the instantaneous centre,I, is as shown in the above scaledposition diagram and lies at the intersection of perpendiculars to the velocity

vectors vSand vRfrom the points S andR respectively.

(iii) .0.349radsrads360

22020degs

111

The velocity of

1

R 0.314msPQ0.90.349vR,

and hence the angular velocity of thedoor is = ..

.

. 1R rads2180441

3140

IR

v (Where IR is measured from the SPD).

(iv) The velocity of the slider, Sis (where IS is measured from the SPD):

.0.174ms1.44

0.80.314

IR

ISvv

IR

v

IS

v 1RS

RS


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2.33

The second is based on drawing scale velocity and acceleration diagrams derived from a

scale position diagram. Note that the velocity and acceleration relate to one instantaneous

position and so many scale drawings are required to obtain a picture of the kinematicbehaviour of a mechanism over a complete cycle. With the advent of computer based

methods the graphical approach is declining in its use. However, from an educational

standpoint, the graphical method does serve to reinforce the concepts behind vector

mechanics. Below, the slider crank on the previous page and a four-bar chain mechanism

is analysed in this way for purely illustrative purposes.

Graphical Analysis for Kinematics using velocity and acceleration diagrams

Another way of finding the velocity and acceleration (at any point) for one position onlyis as follows:


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2.34

(i) Draw a position (stick) diagram of the mechanism to scale (repeated below).

(ii) Construct the velocity vector diagrams (to scale):

Draw vA= r = OAand therefore the magnitude and the directionof this vector are known.

vPmust be horizontal and therefore only its direction is known. The velocity ofPrelative toAmust be perpendicular to the linePA

and thus again only its direction is known.


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2.36

The intersection of the vectors aandaP defines the magnitudes ofthese vectors, and for aPAand from the scaled diagram their values canbe obtained (note that their direction is already known).

WORKED EXAMPLE

A four bar linkage is shown below in which the crank AB rotates clockwise with a

constant angular velocity of 10 rads1

. Using velocity and acceleration diagrams

calculate the angular velocities and accelerations of the links BC and CD. (Note the

diagram is not to scale and fixed pivots A and D should be on the same horizontal line).

aA

aP

a||

a

aPA

Parallel toPA


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2.38

(ii) The velocity of Crelative toDis of an unknown magnitude but its

direction must be perpendicular to CD, i.e. at 24to the vertical.

(iii) The velocity of Crelative toBis also of unknown magnitude but its

direction must be perpendicular toBC, i.e. at 32to the vertical.

(iv) The intersection of the two vectors with unknown magnitude definethe velocity of C, 172 mm s

-1, and the relative velocity of C with respect to

B, 362 mm s-1

.

The angular velocities of the linksBCandDC can now be calculated, i.e.,

Angular velocity ofBC =75

362= 4.83 rad s

1anticlockwise.

Angular velocity ofDC =35

172= 4.91 rad s

1anticlockwise.

3. The acceleration diagram can be constructed as follows:

(i) The acceleration ofBrelative toAis AB 2= 30 100.

Therefore 3000 mm s2

alongBA. Note that there is noAB component


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2.39

since is considered to be constant. (Note however that the angular

rotation rates of CD and BC cannot be considered to be constant).

(ii) The acceleration of the point Crelative toDhas two components, of

magnitude CD 2

CD along CD = 35 (4.91)2= 843.8 mm s

2

and CD CD perpendicular to CDbut its magnitude is as yet unknown.

(iii) The acceleration of the point C relative toBalso has two components,

of magnitudeBC 2BC

along the link CB = 75 (4.83)2= 1749.7mm s

2

andBC BC is perpendicular toBCbut its magnitude is unknown.

(iv) The intersection of the above components now defines the

acceleration of C and the unknown components for both CDandBC,

i.e., perpendicular component for CD = 560 mm s2

and hence

35

560

CD = 16 rad s2

in an anticlockwise direction

and the component forBC = 1780 mm s2

and hence

75

1780CB = 23.7 rad s

2anticlockwise.


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2.40

Computational Analysis

This is a modern, vector based approach which is quick, accurate and general, i.e., it

gives a full description of position, velocity and acceleration and is therefore now the

method of choice. The computational approach relies on the fact that the position vectorsassociated with any mechanism are straightforward to define. The subsequent velocity

and acceleration analysis is performed through numerical methods.

The method involves an interesting blend of numerical analysis and computing, and

hinges on the use of vector displacement/loop equations. The emphasis here, therefore, is

the derivation of such equations for a variety of mechanisms. Once you have these, it issimply a matter of using readily available computer software to solve them.

Slider-Crank

Variables:1, p2which can be either a length or an angle, and in

the case of the slider crank these are enough tocompletely define the mechanism.

Known: a1and a2,the fixed links

Input: the angle of input is q(t).

Displacement Loop Equations

Form the displacement loops by projecting the displacement onto thexandyaxes to givetwo equations:

x-direction a1cos(q) + a2cos(p1) + p2 = 0;

y-direction a1sin(q) + a2sin(p1) = 0; (2.30)

x

y

Arrows indicatepositive

directions

p1

q

a1a2

C

B

A

p2


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2.41

i.e., two equations for two unknowns which are easily solved. Note that you need to be

careful of the signs of the various terms. By defining the angles with respect to the

positivex-axis and working around the loop in an ordered fashion most of the signs willtake care of themselves. For example, in thexdirection equation above, starting fromA

the distance is a1to get toBat an angle of qwith respect to the positivex-axis. As q(as

drawn) is between 90and 180, a1cos(q)will be negative. The next term is fromBto Cand following the same methodology will be a2cos(p1)which again will be negative asp1

is between 180and 270. The final term is from Cback toA, and is effectivelyp2cos(0),hence +p2.

What about 2121 pandp,p,p ?

These can be found by differentiating the displacement equations (2.30) with respect to

time.

Velocity Loop Equations

The velocity loop equations are therefore:

x-direction 21121 sinsin pppaqqa = 0;

y-direction 1121 ppcosaqqcosa = 0. (2.31)

To find the two acceleration loop equations for the two unknown accelerations,

differentiate equation (2.31) w.r.t. time. This is left as an exercise.

In the computational analysis, the loop equations for the displacement, velocity and

acceleration for the slider-crank are solved using commercially available kinematicssoftware. The same is done in the case of loop equations derived for the other

mechanisms shown earlier.

Let us now consider a different mechanism in order to illustrate the graphical and

computational methods described above and the mathematics behind them.


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2.42

2.4.2 The Four Bar Chain Mechanism

Variables: p1, p2

Known: a1, a2, a3and a4, the fixed links

Input: the angle of input is q.

The variables that need to be determined to analyse the mechanism completely are:

(a) p1andp2for any given input, qand therefore the displacements of the

mechanism are known completely ;

(b) 1p and 2p ;

(c) 1p and 2p .

Displacement Loop Equations

The displacement loop equations for this mechanisms are (working around OBCDO):

x-direction

)180cos(apcosapcosaqcosa 423121 = 0;

423121 apcosapcosaqcosa = 0;

y-direction

x

y

Arrows indicate

positive

directions

q

p1

p2

a1

a2

a3

O

B

C

D

a4


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2.43

180sinapsinapsinaqsina 423121 = 0 23121 psinapsinaqsina = 0. (2.32)

Unlike the displacement loop equations for the slider crank mechanism, these equationsare nonlinear (i.e., one cannot write explicit expressions forp1andp2as the variables are

not separable, in fact these two variables are present as the argument of trigonometricfunctions which are intrinsically non-linear). The two equations have to be solvednumerically which requires initial estimates for the two unknownsp1andp2at a known

value of input crank angle q. An algorithm must be employed to increase the accuracy of

the estimates forp1andp2until the desired precision is obtained. This must be done for

every possible value of the input q. The displacement information of the mechanism isthus completely defined.

Velocity Loop Equations

The next stage involves calculating the velocity and acceleration information so that a

complete kinematic description of the mechanism is available for the purpose of lateranalysis. Differentiation of the displacement loop equations with respect to time leads to

the velocity loop equations in 1p and 2p .

x-direction 2231121 psinpapsinpaqsinqa- = 0;

y-direction 2231121 pcospapcospaqcosqa = 0. (2.33)

Assuming the angles q, p1andp2are known and q has been defined, then solution of the

velocity equations for the angular velocities 1p and 2p is simple, both expressions are

linear simultaneous equations (Cramers rule can be applied for two equations with twounknowns).

Acceleration Loop Equations

Differentiation of the velocity loop equations with respect to time yields two further

equations for the angular accelerations 1p and 2p :

direction-x

;pcospapsinpapcospa

psinpaqcosqaqsinqa-

2

2

232231

2

12

112

2

11

0

direction-y

.psinpapcospapsinpa

pcospaqsinqaqcosqa-

2

2

232231

2

12

112

2

11

0

(2.34)


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2.44

Again, the acceleration loop equations are linear and their solution for 1p and 2p is

straightforward particularly as the anglesp1andp2and velocities 1p and 2p are at this

stage known.

Thus all the kinematic information is contained in the two coordinate variablesp1andp2

and in their derivatives.

As demonstrated by the four bar chain, the displacement loop equations are non-linear,simultaneous equations which it is impossible to solve through the derivation of explicit

expressions. This situation is in fact commonly encountered and so a numerical, iterative

technique is used to solve the displacement loop equations and is built into availablekinematics analysis software. There are several numerical methods available to the

engineer, however they all work by using a basic trial and error process, which can be

explained as follows:

1. Begin the process by guessing a solution forp.2. Using this estimate, calculate an incremental change, p, to attempt to get closer

to the solution.

3. Calculate a new estimatep=p+p4. Go back to 2 and keep going round the stages (2 to 4) until some convergence

condition is met (e.g. |p| < an accuracy requirement) and the estimated solution

is close to the actual one

Possible refinements to the technique involve:

1. making the iterative process as fast and as efficient as possible so that the numberof iterations is minimised,

2. ensuring the algorithm is as robust as possible so that it will deal with a widerange of equation sets (which translates into a wide range of mechanisms), and

3. making it stable so that the algorithm always converges to the correct solutionrather that diverging away from it.

Certain simple mechanisms such the slider crank have displacement loop equations

which are explicit and so can be solved directly without recourse to any iterative solvers.

2.4.3 Quick Return Mechanism

WORKED EXAMPLE

Write appropriate loop equations in order to analyse the quick return mechanism shown

below.


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2.45

Solution.

The displacement loop equations for the above mechanism are:

Considering LoopACD:

a4cos(p3) + p4cos(180) + (a1+p1)cos(270) = 0

a4cos(p3)p4 = 0;

a4sin(p3)+ p4sin(180)+ (a1+p1)sin(270) = 0.

a4sin(p3)a1 p1 = 0.

Considering loopABO:

p2cos(p3) + a2cos(180+q)+ p1cos(270) = 0;

p2cos(p3)a2cos(q) = 0;


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2.46

p2sin(p3) + a2sin(180+q)+ p1sin(270) = 0;

p2sin(p3)a2sin(q)p1 = 0.

The above loop equations are not the only ones that could have been chosen. For exampleloopACDcould be replaced by loop OBCD:

a2cos(q) + (a4p2)cos(p3)p4 = 0;

a2sin(q) + (a4p2)sin(p3)a1 = 0.

After all the object is to produce the same number of independent equations as there are

unknowns in the problem.

A point of special interest is the CofM of the linkAC, denoted by Gon the above

diagram. The coordinates of this point relative to the fixed axis, Oare:

xG = a2cos(q) (p2a3)cos(p3);

yG = a2sin(q)(p2a3)sin(p3).

Deriving the loop equations for velocity and acceleration are left as an exercise.

ADDITIONAL EXAMPLES TO TRY

1) The four-bar linkage shown below has the following lengths:A0A =125mm, AB =

275mm, B0B = 225mmandA0B0= 200mm. For the position when the angleA A0B0=

120ocalculate the angular velocity of the output and the coupler if the input angular

velocity 1= 25 rad s1

clockwise.


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2.47


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2) The figure shown below is that of a variable stroke mechanism which allows the stroke

of a reciprocating piston to be varied by altering the position of the fixed point A. Write

down a set of displacement loop equations for the system.

mech1230 dynamics unit 2 - rigid body kinematics (1)

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