Chapter 2 Graphical Analysis of Mechanisms with Revolute Joints or Fixed Slides

1
Chapter 2 Graphical Analysis of Mechanisms with
Revolute Joints or Fixed Slides

• EML 4930 Kinematics
• Fall 2005

2
2.1 Introduction

• why does the chapter title mention only
  mechanisms with revolute joints or fixed slides?

fixed slide
moving slide
3
2.1 Introduction (cont)

• graphical vs. analytical position, velocity, and
  acceleration analyses
• advantages of graphical technique
• position, velocity, and acceleration analyses can
  be accomplished quickly for a single position
• good alternate technique to check analytical
  program
• provides better understanding of problem
• advantages of analytical technique
• can perform analysis quickly for many positions

4
2.2 Graphical Position Analysis

• suppose we had a four bar mechanism with AB 4
  cm BC 12 cm CD 10 cm AD 8 cm f 60
• graphically determine theposition of all the
  links

f
5
2.2 Graphical Position Analysis (cont)

• suppose we had a four bar mechanism with AB 4
  cm, BC 12 cm, CD 10 cm, AD 8 cm f
  60
• step 1 draw the frame AD and locate the link
  AB

f
6
2.2 Graphical Position Analysis (cont)

• suppose we had a four bar mechanism with AB 4
  cm, BC 12 cm, CD 10 cm, AD 8 cm f
  60
• step 2 draw circle at B (rad BC) and
  circle at D (rad CD)

f
7
2.2 Graphical Position Analysis (cont)

• suppose we had a four bar mechanism with AB 4
  cm, BC 12 cm, CD 10 cm, AD 8 cm f
  60
• step 3 intersections of circles give possible
  locations for pt. C

f
8
2.2 Graphical Position Analysis (cont)

• suppose we had a four bar mechanism with AB 4
  cm, BC 12 cm, CD 10 cm, AD 8 cm f
  60
• step 4 identify possible configurations selec
  t desired

f
9
2.3 Planar Velocity Polygons

• suppose you have two points in a rigid body that
  is moving relative to ground
• from the figure it is seen that rB rA
  rB/Awhere all vectors are expressed inthe
  ground coordinate system
• differentiating this equation gives
• rA and rB are vectors whose tails originate at
  the origin of the fixed frame thus they
  represent the coordinates of points A and B

10
2.3 Planar Velocity Polygons (cont)

• the time rate of change of the coordinates of
  points A and B are the velocities of these points
• thus
• rB/A is a vector whose tail originatesat the
  moving point A and whosehead ends at the
  moving point B
• this vector will not change in length (magnitude)
• it will change in direction

11
2.3 Planar Velocity Polygons (cont)

• lets change the direction by an angle d? in a
  small time increment dt
• the change in the vector rB/A is dr
• the magnitude of the change is dr rB/A d? (a)
• as dt ? 0, d? ? 0 and the anglebetween dr and
  rB/A approaches 90
• if ? is the angular velocity of the moving
  body, d? ? dt (b)
• substituting (b) into (a) and rearranging gives

Note that bold items are vectors and non-bold
items are the magnitudes of those vectors.
12
2.3 Planar Velocity Polygons (cont)

• in the limit as dt?0
• we have the magnitude of thechange in the vector
  rB/A and weknow the direction of the changein
  rB/A is perpendicular to rB/A.
• it turns out that the change in rB/A can be
  written as

this will give the correct magnitude
anddirection for the change in rB/A
Note that bold items are vectors and non-bold
items are the magnitudes of those vectors.
13
2.3 Planar Velocity Polygons (cont)

• thus
• all vectors are expressed in the fixed coordinate
  system
• knowing the velocity of a point in the moving
  body and the angular velocity of the body means
  that the velocity of any point in the moving body
  can be determined

14
2.4 Graphical Acceleration Analysis

• taking the derivative of the position equation
  gave the velocity equation for this case
• now take the derivative of thevelocity equation

15
2.4 Graphical Acceleration Analysis (cont)
16
2.4 Graphical Acceleration Analysis

• summary points A and B are in the same rigid
  body

tangential acceleration
radial acceleration
all vectors written in fixed coordinate system
17
2.5 Graphical Analysis of a Four-Bar Mechanism

• given
• constant mechanism parameters AB2.5, BC6,
  CD4, AD4
• input position, vel , and acc f120 ?210
  rad/sec a2 0
• find
• angular vel and acc of links 3 and 4
• linear vel and acc of any point in any of the
  bodies (point E)

18
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
0

• vB2 vA2 ?2 ? rB2/A2 solve for vB2
• vC3 vB3 ?3 ? rC3/B3
• vC4 vD4 ?4 ? rC4/D4
• equate vC4 vC3
• vB3 ?3 ? rC3/B3 ?4 ? rC4/D4

D M
D M
D
D M
D M
0
D
D M
D
D
D M
D M
D M
19
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
vE3 29.3/sec
20
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
0
0
solve for aB2
D M
D M
D M
D M
D M
D M
D M
D M
D
0
D M
D M
D M
D M
D
equate aC4 aC3
21
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
0
0
solve for aB2
D M
D M
D M
D M
D M
D M
D M
D M
D
'
22
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
0
D M
D M
D M
D M
D
c'3, c4
Note that a4 is negative because the direction of
the vector is opposite to k?rC4/D4.
23
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
c'3, c4
Note that a4 is negative because the direction of
the vector is opposite to k?rC4/D4.
24
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
D M
D M
D M
D M
D M
D M
c'3, c4
'
e3
25
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)

• New Problem

Determine ?3, a3, vC4, aC4
26
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
0

• vB2 vA2 ?2 ? rB2/A2 solve for vB2
• vC3 vB3 ?3 ? rC3/B3
• direction of vC4 and thusvC3 is known
• calculate ?3

D M
D M
D
D M
D M
direction of vC4
Note that the angular velocity ?3 is negative.
direction of ?3 ? rC3/B3
27
2.5 Graphical Analysis of a Four-Bar Mechanism
(cont)
0
0
solve for aB2
D M
D M
D M
direction of aC4 and thus aC3is known
D M
D M
D M
D M
D M
D
D
direction of aC4
direction of a3? rC3/B3
28
2.7 The Velocity Image Theorem

• knowing the linear velocity of one point on a
  moving body plus the angular velocity of the body
  means that the linear velocity of any other point
  can be calculated
• vB vA ? ? rB/A
• once the velocity analysis is complete, knowing
  the linear acceleration of one point and the
  angular acceleration of the body means that the
  linear acceleration of any other point can be
  calculated

29
2.7 The Velocity Image Theorem (cont)

• Points B, C, and E are in the same body
• The velocity image theorem states
• triangle bce in the velocity diagram will be
  similar to triangle BEC
• triangle bce will be rotated by 90 in the
  positive ? direction
• triangle bce will be scaled by ?

30
2.7 The Velocity Image Theorem (cont)

• proof
• the vector rR/P has been rotated 90 in direction
  of ?
• ? ? rR/P ? rR/P

? ? rR/P
31
2.7 The Velocity Image Theorem (cont)
The velocity image theorem provides an easy way
to find the velocities of other points in the
body once the linear velocity of two points is
known.
?
32
2.7 The Velocity Image Theorem (cont)
The velocity image theorem provides an easy way
to find the velocities of other points in the
body once the linear velocity of two points is
known.
?
?
?
33
2.7 The Velocity Image Theorem (cont)
The velocity image theorem provides an easy way
to find the velocities of other points in the
body once the linear velocity of two points is
known.
?
?
?
?
34
2.7 The Velocity Image Theorem (cont)
The velocity image theorem provides an easy way
to find the velocities of other points in the
body once the linear velocity of two points is
known.
?
?
?
?
35
2.7 The Velocity Image Theorem (cont)
The velocity image theorem provides an easy way
to find the velocities of other points in the
body once the linear velocity of two points is
known.
?
?
?
?
?
?
36
2.7 The Velocity Image Theorem (cont)
The velocity image theorem provides an easy way
to find the velocities of other points in the
body once the linear velocity of two points is
known.
?
?
?
?
?
?
37
2.8 The Acceleration Image Theorem

• provides an easy way to find the acceleration of
  additional points in the body once the linear
  acceleration of two points is known
• triangle pqr will be similar to triangle PQR
• triangle pqr is rotated by an angle
• ? p tan-1 (a/?2)
• the scale factor is

38
2.9 Solution by Inversion
39
2.9 Solution by Inversion (cont)