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Lect Position Analysis.pptx
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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
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
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