### ME321 - Kinematics and Dynamics of Machines Design Process

```ME321
Kinematics and Dynamics of
Machines
Steve Lambert
Mechanical Engineering,
U of Waterloo
7/17/2015
Kinematics and Dynamics




Position Analysis
Velocity Analysis
Acceleration Analysis
Force Analysis


F  ma
We will concentrate on four-bar linkages
7/17/2015

What type of motion is possible?
q
l
s
p
7/17/2015
Grashof’s Criteria

Used to determine whether or not at least one of the links
can rotate 360o
 the sum of the shortest and longest links of a planar
four-bar mechanism cannot be greater than the sum of
the remaining two links if there is to be continuous
relative rotation between the two links.
s + l< p + q
q
l
s
p
7/17/2015
Grashof’s Criteria
l
q
l
s
s
p
q
p
7/17/2015
s+l<p+q
s+l>p+q
Grashof Mechanism
Non-Grashof Mechanism
Grashof Mechanisms (s+l < p+q)
l
l
q
q
s
s
p
Crank-Rocker
p
Rocker-Crank
Shortest link pinned to ground and rotates 360o
7/17/2015
Grashof Mechanisms (s+l < p+q)
s
q
l
p
q
p
s
l
7/17/2015
Double-Rocker
- Both input and output
- Coupler rotates 360o
Change-Point Mechanism
S+l = p+q
l
s
q
p
7/17/2015
Non-Grashof Mechanisms
•Four possible triplerockers
•Coupler does not
rotate 360o
s
q
p
l
7/17/2015
Transmission Angle
• One objective of position
analysis is to determine the
transmission angle, 
• Desire transmission angle
to be in the range:
45o <  < 135o
7/17/2015
coupler

input
output
Position Analysis


Given the length of all links, and the input angle,in, what
is the position of all other links?
Use vector position analysis or analytical geometry
coupler
output
input
in
7/17/2015
Vector Position Analysis
• ‘Close the loop’ of vectors
to get a vector equation with
two unknowns
• Three possible solution
techniques:

RB / A

RA s
O2
• Graphical Solution
• Vector Components
• Complex Arithmetic
7/17/2015

RB

RO2 / O4
O4




RA  RB / A  RO4 / O2  RB
Graphical Solution
• Draw ground and
input links to scale, and
at correct angle
• Draw arcs (circles)
corresponding to length
of coupler and output
•Intersection points
represent possible
solutions
7/17/2015
Vector Component Solution
y, i
R

3
x, j
4
2
O4
O2
‘Close the loop’ to get a vector equation:





R2 cos 2iˆ  sin 2 ˆj  R3 cos3iˆ  sin3 ˆj  R1iˆ  R4 cos 4iˆ  sin 4 ˆj
7/17/2015

Vector Component Solution (con’t)
Rewrite in terms of i and j component equations:
R2 cos 2  R3 cos 3  R1  R4 cos 4
R2 sin  2  R3 sin  3  R4 sin  4
• These represent two simultaneous transcendental
equations in two unknowns:  3 and 4
•Must use non-linear (iterative) solver
7/17/2015
Complex Arithmetic
• Represent (planar) vectors as
complex numbers

R  Rei  Rcos  i sin  
iy
R

x
• Write loop equations in terms of real and imaginary
components and solve as before
7/17/2015
Analytical Geometry
• Examine each mechanism
as a special case, and apply
analytical geometry rules
• For four-bar mechanisms,
draw a diagonal to form two
triangles
• Apply cosine law as
required to determine length
of diagonal, and remaining
angles
7/17/2015
B
3
A


2
O2
4

O4
l 2  a 2  b 2  2ab cos
Limiting Positions for Linkages
• What is the range of output motion for a crack-rocker
mechanism?
7/17/2015
```