lect position analysis

ME 3507: Theory of Machines

Position Analysis

Dr. Faraz Junejo

Introduction A principal goal of kinematic analysis is to determine

the accelerations of all the moving parts in the

assembly. Why ?The design engineer must ensure that the proposed mechanism or machine will not fail under its operating conditions. Thus the stresses in the materials must be kept well below allowable levels.

From Newton's second law, F = ma, one typically needs to know the accelerations (a) in order to compute the dynamic forces (F) due to the motion of the system's mass (m).

Introduction (contd.)

Kinematic Analysis• We determine relative motion characteristic of

a given mechanism.

• Can be classified into:- Position analysis

- Velocity analysis

- Acceleration analysis

• For all these three type of problems, we can use either:

- Graphical Method or- Analytical Method

Position Analysis

• Given the kinematic dimensions and

position or movement of the input

link determine the position or

movement of all other links

Objective

• Determine the positions of links

and points on mechanisms.

Graphical ApproachIn the graphical method, the kinematic

diagram of the mechanism is drawn to a suitable scale, and

The desired unknown quantities are determined through suitable geometrical constructions and calculations.

Graphical approach

• We will have to do an independent graphical

solution for each of the positions of interest

• None of the information obtained graphically

for the first position will be applicable to the

second position.

• It is useful for checking the analytical results.

Analytical approach

• Derive the general equations of motion

– Solve analytical expressions

– Once the analytical solution is derived for a

particular mechanism, it can be quickly solved

(with a computer) for all positions.

• Graphical Position Analysis – Is more simple then the algebraic approach• Graphical Velocity and Acceleration analysis – Becomes quite complex and difficult then the

algebraic approach

• Graphical analysis is a tedious exercise and

was the only practical method available in the

day B.C.(Before Computer) , not so long ago.

Graphical vs. Analytical approach

Graphical vs. Analytical approach (contd.)

Coordinate System• Global or Absolute: Master frame reference

fixed in space.

• Local: Typically attached to a link at some point of interest.

- This might be a pin joint, a center of gravity, or a line of centers of a link.

- These local coordinate system may be either rotating or non-rotating as we desire.

Position & Displacement (Point motion)

• The position of a point in the plane can be defined by the use of a position vector.

• Polar coordinate / Cartesian coordinate

• A position vector can be expressed in:

– Polar form : a magnitude and angle of vector

– Cartesian form : X and Y components of the vector

Position Vector in Cartesian and Polar Form

Coordinate Transformation

• The system’s origins are coincident and the

required transformation is a rotation.

Coordinate Transformation

Displacement of a point

Is the change in its position and can be

defined as the straight line between the initial

and final position of a point which has moved in

the reference frame.

• Note that displacement is not necessarily the same as the path length which the point may have traveled to get from its initial to final position.

Displacement (contd.)

• Figure a shows a point in two positions, A and B. The

curved line depicts the path along which the point

traveled.

The position vector RBA defines the displacement ofthe point B with respect to point A .

• Figure b defines this situation with respect to a global reference frame XY.

Displacement (contd.)

The vectors RA and RB define, respectively, the absolute positions of points A and B with respect to this global XY reference frame.

Displacement (contd.)

The vector RBA denotes the difference in position, or the displacement, between A and B. This can be expressed as the position difference eq:

RBA= RB – RA or RBA=RBO-RAO

The position of B with respect to A is equal to the (absolute) position of B minis the (absolute) position of A, where absolute means with respect to the origin ofthe global reference frame.

Case 1 – One body in two successive position• position difference

Case 2 – Two bodies simultaneous in separate position• relative or apparent position

Displacement (contd.)

Summary• Cartesian (Rx, Ry)

• Polar (RA, q)• Converting between the two

• Position Difference, Relative position– Difference (one point, two times)– relative (two points, same time)

RBA=RB-RA

xy

yxA

RR

RRR

arctan

22

q qq

sincos

Ay

Ax

RRRR

X

Y

RB

RA

ABRBA

Translation

All points on the body have the same displacement, as

No change in angular orientation

Can be curvilinear or rectilinear translation

Rotation

• Different points in the body

undergo different displacements

and thus there is a displacement

difference between any two

points chosen

• The link now changes its angular

orientation in the reference frame

Complex MotionThe sum of the translation and rotation components.

total complex displacement =translation component + rotation component The total complex displacement of point B can be defined as:

Whereas, the new absolute position of point B w.r.t origin at A is:

Theorems

Euler’s theorem The general displacement of a rigid body with

one point fixed is a rotation about some axis. This applies to pure rotation as mentioned earlier.

Chasles’ theorem describes complex motion Any displacement of a rigid body is equivalent

to the sum of a translation of any one point on that body and a rotation of the body about an axis through that point.

Summary: Translation, Rotation, and Complex motion

• Translation: keeps the same angle

• Rotation: one point does not move, such as A

in preceding examples

• Complex motion: a combination of rotation

and translation

Example: 1

• The path of a moving point is defined by the

equation y = 2x2 – 28. Find the position

difference from point P to point Q, when

3 and 4 xQ

xP RR

Example: 1 (contd.)• The y-components of two vectors can be written as

• Therefore, the two vectors can be written as

• Thus, position difference from point P to Q is

102832 and 428-42 22 yQ

yP RR

j10 ˆ3 and j4 ˆ4 iRiR QP

4.24318043.637

14tan

and 65.15)14((-7)As,

243.415.65j14 7

1

22

q

iRRR PQQP

Remember:Angles will always be measured ccw from +ve x-axis.

Example: 2

2link w.r.t 3link of , 2/3 ntdisplacemeRWhere P

Example: 2 (contd.)

Ans 7.118421.4ˆ879.3ˆ121.2

)1()2(

.ˆ6ˆ90sin6ˆ90cos6)2(

906)22()2(

.ˆ121.2ˆ121.2ˆ45sin3ˆ45cos3)1(

453)21()2(

3

333

3

3

3

3

42

2

41

242

jiR

RRR

jjiR

eR

jijiR

eetreR

P

PPP

P

j

P

P

jtjj

P

q

Example: 2 (contd.)

Ans ˆ3)1()2(

ˆ6ˆ)22()2(

ˆ3ˆ)21()1(

ˆ)2(0)2(

0)2(

2

222

222

222

02

2/32/32/3

2/3

2/3

2/3

2/3

iRRR

iiR

iiR

ittR

etreR

PPP

P

P

P

jjP

qq

Objective of Position Analysis• The main task in position analysis is the find the

output variables knowing: – The input variable – The length of all the links

Objective of Position Analysis (contd.)• As discussed earlier, there are 2 ways of doing this: – Graphical method (use drawing tools) – Analytical method (use equations)• Reminder: All angles are measured counter clockwise

from the positive x-axis, as shown below

Graphical Position Analysis• For any one-DOF linkage, such a four-bar, only one

parameter is needed to completely define the positions of all the links. The parameter usually chosen is the angle of the input link; i.e.

Construction of the graphical solution

1. The ground link (1) and the input link (2) are drawn to a convenient scale such that they intersect at the origin O2 of the global XY coordinate system with link 2 placed at the input angle θ2.

2. Link 1 is drawn along the X axis for convenience.

Construction of the graphical solution (contd.)

3. The compass is set to the scaled length of link 3 (i.e. length b), and an arc of that radius swung about the end of link 2 (point A) i.e. draw an arc centered at end of Link 2 (point A)

Construction of the graphical solution (contd.)

4. Set the compass to the scaled length of link 4 (i.e. length c), and draw another arc centered at end of Link 1 (point O4). Label the intersection of both arcs B and B’

Note that intersection of both arcs B and B’ define the two solution to the position problem for a four-bar linkage which can be assembled in two configurations, called circuits, labeled open and crossed.

O2-A-B-O4 is first config.O2-A-B’-O4 is second config.

First Config. (Open Config.)

• Measure θ3 and θ4 with protractor• Called ‘Open’ configuration because both links

adjacent to the shortest link (Links 1 and 3) do NOT cross each other

Second Config. (Cross Config.)

• Measure θ3’ and θ4’ with protractor (CCW from positive x-axis)

• Called ‘Cross’ configuration because both links adjacent to the shortest link (Links 1 and 3) cross each other

Summary: Example 1Given the length of the links (a,b,c,d), the ground position, and q2. Find q3 and q4

a

d

b

cq3

q4

A

B

O2 O4

bc

q2

Example 1: Graphical Linkage Analysis

• Draw an arc of radius b, centered at A

• Draw an arc of radius c, centered at O4

• The intersections are the two possible positions for the linkage, open and crossed

adq2

b

cq3

q4

A

O2 O4

B1

B2

Summary: Graphical Position Analysis

Shaping machine• A photographic view of general configuration of shaping

machine is shown in Figure. The main functions of shaping machines are to produce flat surfaces.

Example: 2• Model of Slotted quick return mechanism used in

Shaping machines

Cutting toolBull gear rotated at constant speed

Tool holder moves in a slot, in the frame of machine

Block Hinged to the bull gear, and moves up & down in the slotted lever

Slotted lever hinged to frame

Link connecting slotted lever with tool holder

• So, we have a six link mechanism, where continuous uniform rotation of the bull gear is converted into to and fro motion of the cutting tool.

• It can be seen that cutting tool is doing useful work during forward motion/stroke, so we have to maintain a proper cutting speed. However, during return stroke it is not doing any useful work, so we would like to make return stroke faster, hence it is referred as quick return mechanism.

Example: 2 (contd.)

Link 2, O2A Bull gear

Link 3, block that is hinged to bull

gear and goes up & down in the

slotted lever, which is link 4

Link 5 connects slotted lever with

tool holder, which is represented

by link 6. So, we have a 6 link

mechanism.

Here, we have got 5 revolute pairs at 02, O4, A, B and D respectively.

There are two prismatic pair, one between link 1 & 6 in the horizontal direction, second one is between link 3 & 4 along the slotted lever.

Example 2: Statement • Determination of quick return ratio (ratio of the durations

of the forward stroke and the return stroke) & stroke length

of a slotted lever mechanism used in shapers, with constant

angular speed ω2 of input link 2 i.e. bull gear

toolcutting theofmotion return during 2 ofrotation

toolcutting theofmotion forward during 2 ofrotation ;

..

link

linkwhere

rrq

r

f

r

f

q

q

qq

Example 2: Solution

1. Note that Point A moves along the circle drawn whose centre is at O2 with radius O2A. Therefore, this circle represent path of point A i.e. KA.

2. To determine the extreme positions of the link 4 (i.e. slotted lever), we draw two tangents to the circle (representing path of point A) from point O4.

3. Consequently, tangent drawn on right hand side (R.H.S) represent right most position of slotted lever (i.e. link 4), indicated by AR, whereas tangent on L.H.S. represent right most position of slotted lever (i.e. link 4), indicated by AL.

Example 2: Procedure

4. Since the distance O4B does not change, so we can also locate rightmost position of revolute pair at B (indicated by BR), by drawing a circular arc with O4 as centre and radius O4B. In similar manner, on L.H.S. we can locate BL.

5. It should be noted that the distance BD does not changes, as D (i.e. tool holder) moves horizontally. Hence, BR location can be used to locate rightmost position of tool holder (indicated by DR) by drawing a circular arc with BR as centre and radius BD. In similar manner, DL i.e. leftmost position of tool holder can be obtained.

6. Distance between DR and DL represent the stoke length of the cutting tool as per scale of the figure.

Determination of Q.R.R It can be seen that input link O2A rotates from O2AR to O2AL for

forward motion (i.e. right to left), hence the angle between O2AR and O2AL represent qf i.e. rotation of input link (i.e. link 2) during forward motion.

Similarly, it can be seen that for return motion (i.e. left to right) input link travels from O2AL to O2AR now indicating this angle with qr i.e. rotation of input link (i.e. link 2) during return motion.

It can be seen that qf is larger then qr resulting in quick return motion of tool holder.

Example 2: Procedure (contd.)

Example 2: Discussion• It should be noted that if stroke length needs to be

decreased, we need to:

decrease the length of input link O2A, because as a result, tangent from O4 to

circle KA i.e. AR and AL points representing rightmost and leftmost position of

slotted lever will move up, resulting in qf qr . (i.e. qf approaches qr) implying a

decrease in quick return ratio.

It can be concluded, that this mechanism is OK for

producing quick return effect so long the stroke length

is sufficiently large, and quick return effect decreases as

stroke length decreases.

Slotting machine• Slotting machines can simply be considered as vertical shaping

machine where the tool reciprocates vertically.• Unlike shaping machines, slotting machines are generally used to

machine internal surfaces, implying smaller stroke length.

Example: 3

• Determination of quick return ratio of

Whitworth quick return mechanism used in

slotting machines.

• Here, the quick return ratio is independent of

the stroke-length.

• Model of Whitworth quick return mechanism used in Slotting machines

Cutting tool

Kinematic diagram

It is a 6 link mechanism, with five revolute pairs at O2, O4, A, C and D, and two prismatic pairs between link 3 & 4, and link 6 & 1 respectively.

• Link 2 (O2A) is input link that rotates at constant angular speed, and is hinged to fixed link at O2.

• Link 3 is the block that moves along link 4 via a prismatic pair.

• Link 4 is hinged to fixed link at O4.

• Link 4 & Link 5 are connected by a revolute pair at point C.

• Link 5 & Link 6 are connected by a revolute pair as well.

• Link 6 has prismatic pair with fixed link 1 for horizontal motion.

Kinematic diagram description

Example 3: Solution

Example: 4• For a six link mechanism shown below, determine

stroke-length of the output link i.e. the slider 6. Also determine the quick return ratio assuming constant angular speed of link 2.

Example 4: Solution

Exercise: 1

Figure shows a kinematic diagram of a mechanism that is driven by moving link 2. Graphically reposition the links of the mechanism as link 2 is displaced 30° counterclockwise. Determine the resulting angular displacement of link 4 and the linear displacement of point E . Use suitable scale.

dq4= 26o, CCWR E = 0.95 in.

• Link 2 is graphically rotated 30° counterclockwise, locating the position of point B’.

• Being rigid, the shape of link 3 cannot change, and the distance between points B and C remains constant. Because point B has been moved to B’, an arc can be drawn of length rBC , centered at B’. This arc represents the feasible path of point C’. The intersection of this arc with the constrained path of C (obtained by Drawing an arc of radius 3.3 in, centered at D) yields the position of C’.

Exercise: 1 (sol)

• This same logic can be used to locate the position of point E’. The shape of link 5 cannot change, and the distance between points C and E i.e. rCE remains constant.

• Because point C has been moved to C’, an arc can be drawn of length rCE, centered at C’. This arc represents the feasible path of point E’. The intersection of this arc with the constrained path of E yields the position of E’.

• Finally, with the position of C’ and E’ determined, links 3 through 6 can be drawn. The displacement of link 4 is the angular distance between the new (C’D) and original position (CD), approx.: 26 Degrees counterclockwise

• The displacement of point E is the linear distance between the new (E’) and original position (E)of point . approx.: 0.95 in

Exercise: 2

• A point Q moves from A to B along link 3while

link 2 rotates from . Find the

absolute displacement of Q.

120 to30 '22 qq