mechanics of machines dr. mohammad kilani class 4 velocity analysis
TRANSCRIPT
Mechanics of Machines
Dr. Mohammad Kilani
Class 4Velocity Analysis
DERIVATIVE OF A ROTATING VECTOR
Derivative of a Rotating Unit Vector
A unit vector in the θ direction, uθ
, is a vector of unity magnitude
and an angle θ with the x- axis. It is written as:
If uθ
rotates, it angle θ changes with time, the time derivative of uθ
is found by applying the standard differentiation rules on the
expression of uθ
above
jiuθ sincos
jiu
jiu
θ
θ
cossin
cossin
dt
d
dt
ddt
d
dt
d
dt
d
θ
cosθ i
sinθ j
uθ
Derivative of a Rotating Unit Vector
Noting that
the time derivative of the vector uθ
is
sin2cos
cos2sin
jiu
jiu
θ
θ
cossin
sincos
dt
d
dt
d
2
2sin2cos
cossin
uu
jiu
jiu
θ
θ
θ
dt
d
dt
ddt
d
dt
ddt
d
dt
dθ
cosθ i
sinθ j
uθ
Derivative of a Rotating Unit Vector
The derivative of a unit vector whose angle with the x-
axis is θ, (θ changes in time), is a vector whose angle
with the x-axis is θ+π/2 and whose magnitude is dθ/dt.
If we define a vector ω as a vector in the k direction of
magnitude ω = dθ/dt, then
2
sincos
uu
jiu
θ
θ
dt
d
dt
d
θ
uθ
duθ
/dt
θ + π/2
θθ uωu
u
kω
2
dt
d
dt
ddt
d
Derivative of a Rotating Vector
θθθ
θθθ
θθ
θθ
θ
θθ
rωur
uωur
uur
uu
r
ur
dt
dr
dt
d
rdt
dr
dt
ddt
dr
dt
dr
dt
ddt
dr
dt
dr
dt
d
r
2
θ
uθ
duθ
/dt
θ + π/2
A vector rθ
= r uθ
, is a general vector of magnitude r pointing in the θ direction. The time
derivative of rθ
is found by applying the normal differentiation rules on the expression for rθ
Derivative of a Rotating Vector
θ
rθ
ω x rθ
θ + π/2
(dr/dt) uθ
Given a vector rθ
= ruθ
which rotates relative to the
reference coordinates, the time derivative of rθ
has two
components; a component in the direction of uθ
and a
component normal to uθ
in the direction of uθ+π/2
The magnitude of the component of drθ
/dt in the direction
of uθ
is equal to dr/dt; that is the time derivative of the length
of rθ
.
The magnitude of the component of drθ
/dt in the direction
of direction of uθ+π/2
is equal to ωr
VELOCITY ANALYSIS OF FOUR BAR MECHANISMS
Derivative of the Loop Closure Equation for a Four Bar Kinematic Chain
The loop closure equation of a 4-bar kinematic chain is written as
When all the links in the chain are of constant lengths, the equation above
reduces to
0
0
0
0
44113322
4132
4244121132332222
4132
4132
4132
4132
urururururururur
ururururdt
d
rrrrdt
d
rrrr
rrrr
0
0
44113322
244211233222 4132
rωrωrωrω
urururur
Derivative of the Loop Closure Equation for a Four Bar Mechanism
For a four bar mechanism with link 1 fixed we have dθ1/dt = ω
1 =
0.
The vector equation above contains two scalar equations and can
be solved for two unknowns. Knowledge of dθ2/dt allows the
calculation of dθ3/dt and dθ
4/dt .
244233222 432 ururur
444333222
444333222
coscoscos
sinsinsin
rrr
rrr
Derivative of the Loop Closure Equation for a Four Bar Mechanism
Eliminate dθ3/dt by carrying out a dot product with u
θ3 on both sides of the equation
Alternatively, eliminate dθ4/dt by carrying out a dot product with u
θ4 on both sides of
the equation
244233222 432 ururur
2
344
3222
344
32244
34443222
34443222
sin
sin
sin
sin
sinsin
2cos2cos
r
r
r
r
rr
rr
2
343
4222
433
4222
433
42233
43334222
43334222
sin
sin
sin
sin
sin
sin
sinsin
02cos2cos
r
r
r
r
r
r
rr
rr
Angular Velocity Ratio and Mechanical Advantage
The angular velocity ratio mV
is defined as the output angular velocity divided by the input angular velocity. For a four bar mechanism
with link 2 as the input and link 4 as the output this is expressed as
The efficiency of a four bar linkage is defined as the output power over the input power,
Assuming 100% efficiency, which is normally approached by four bar mechanisms, we have
2
4
in
outVm
inin
outout
in
out
T
T
P
P
VT
out
in
inin
out
mm
T
T 1
Angular Velocity Ratio and Mechanical Advantage
The mechanical advantage is defined as the ratio
between the output force to the input force
out
inT
inin
outout
in
outA r
rm
rT
rT
F
Fm
VELOCITY ANALYSIS OF SLIDER-CRANK MECHANISMS
Velocity Analysis of a Slider-Crank Mechanism
Design parameters: r2
, r3
, r4
, θ1
.
Position analysis parameters:
r1
, θ2
, θ3
Velocity analysis parameter.
Find dr1
/dt, dθ2
/dt , dθ3
/dt
To eliminate ω3
dot product both sides by uθ3
r3
r2
r1
r4
rp
132 1233222
4132
ururur
rrrr
2
13
2321
3113222
3113222
cos
sin
cossin
cos2cos
rr
rr
rr
2
133
1223
13331222
cos
cos
coscos
r
r
rr
To eliminate dr1
/dt dot product both sides by
u(θ1+π/2
)
Velocity Analysis of an Inverted Slider-Crank Mechanism
Given r1
, r2
, θ1
, θ2
, ω2
Position analysis: Find r3
, θ3
Velocity analysis: Find dr3
/dt, dθ3
/dt
To eliminate ω3
and find dr1
/dt dot product both sides by uθ3
To eliminate dr1
/dt and find ω3
dot product both sides by u(θ3+π/2
)
2333222
312
312
332
312
ururur
ururur
rrr
32322
33222
sin
2cos
rr
rr
r2
r1
θ1
θ2
r3
θ3
23
2323
333222
cos
cos
r
r
rr
Velocity Analysis of an Inverted Slider-Crank Mechanism
Given r1
, r2
, r4
, θ1
, θ2
, ω2
Position analysis: Find r3
, θ3
, θ4
Velocity analysis: Find dr3
/dt, dθ3
/dt , dθ4
/dt
r3
r2 r
1
r4
34423333222
34
24423333222
41332
4132
32
432
412
2
urururur
urururur
urururur
rrrr
with
Velocity Analysis of an Inverted Slider-Crank Mechanism
r3
r2 r
1
r4
23223
32243
23
322433222
23
32234
4433222
4433222
3
23
3223
333222
2
34423333222
sincos
cossin
cos
sin
2cos
cos
0cos3
32
rr
rrr
r
rrrr
r
r
rrr
rrr
u
r
r
rr
u
urururur
But
by sides both Dot
by sides both Dot
Example
224411
22441112,13
2244113
2244113
2,11
2,14
222
2,12
44
42424224144124
22
21
2
42421241144124
22
21
2
241
152
412
coscoscos
coscossintan
coscossinsin
coscoscoscos
tan2,02
0sincos
sinsincoscos2cos2cos2
cos2cos2cos2
2413
61532
4132
rrr
rrr
rrrba
rrrba
tCA
ACBBtACBttAC
CBA
rrrrrrrrrba
rrrrrrrrrba
urururuba
usururuaur
ururubaur
(2) rquation closure loop from Also
itself by equation the of sides both Dot
(1) Equation Closure Loop
II Loop
I Loop
a
r2
r1
b
r4
r5
s
a
r2
r1
b
r4
r5
s
2tan
cos2
sin2
cos22
4
2221
24
22
21
142
24241
t
barrrrrC
rrB
rrrrA
Example
224411
22441112,13
2244113
2244113
2,11
2,14
222
2,12
24144124
22
21
2
32226213262
3122122
222
125
61
32226213262
131122116122
222
125
215
coscoscos
coscossintan
coscossinsin
coscoscoscos
tan2,02
cos2cos2
cos2cos2cos2cos2
cos2cos2
2,0
cos2cos2cos2cos2
cos2cos2cos2
32615
rrr
rrr
rrrba
rrrba
tCA
ACBBtACBttAC
rrrrrrrba
arsrassr
arrrarsrr
arsrassr
arrrsrarsrr
uaurusurur
(2) rquation closure loop from Also
Let
itself by equation the of sides both Dot
(2) Equation Closure Loop
a
r2
r1
b
r4
r5
s
a
r2
r1
b
r4
r5
s
2tan
cos2
sin2
cos22
4
2221
24
22
21
142
24241
t
barrrrrC
rrB
rrrrA
HW#3
6-17, 6-18 (b+c), 6-21 (b), 6-44.
METHOD OF INSTANT CENTERS
Relative Velocities Between Two Points on a Rigid Body
Given any two points A and B that lie on a
rigid body, let the line AB be a line passing
through A and B, then the components of
the velocity of A and the velocity of B on
the line AB must be equal.
The velocity of point B relative to A must
be normal to the line AB
A
B
A
B
ABBA vvv
Instant Center Relative to the Ground
If line AA’ is drawn normal to the direction of vA
, then the velocity of
any point on the rigid body that falls on line AC must be perpendicular
to the line AA’
In a similar manner, if line BB’ is drawn perpendicular to the direction
of vB
, then the velocity of any point on the rigid body that falls on this
line must be perpendicular to the line BB’
If point I is the intersection of lines AA’ and BB’, then the velocity of
point I must be perpendicular to both AA’ and BB’. This can only
happen when the velocity of point I is zero.
A
BI
A’
B’
IBB
IAAIAA
IAA
IAIA
rv
rvrv
rkv
rkvv
Instant Center Relative to the Ground
Point I determined as before is called the instant center of
zero velocity of the rigid body with respect to the ground.
The direction of the velocity of point A is normal to the line
IA. The direction of the velocity of line B is normal to the
line IB.
The direction of the velocity of any other point C on the
same body must be normal to the line IC.
The magnitude of the velocity of any point on the link is
proportional to its distance from the point I.
A
BI
A’
B’
C
IC
v
IB
v
IA
v ABA
Velocity of a Point on a Link byInstant Center Method
Consider two points A and B on a rigid link. Let vA
and vB
be the velocities
of points A and B at a given instant. If vA
is known in magnitude and
direction and vB
in direction only, then the magnitude of vB
may be
determined by the instantaneous centre method.
Draw AI and BI perpendiculars to the directions vA
and vB
respectively. Let
these lines intersect at I, which is known as instantaneous centre or virtual
centre of the link. The complete rigid link rotates about the centre I at the
given instant. The relations shown may be used to determine the
magnitude of the velocity of point B.
IBB
IAAIAA
IAA
IAIA
rv
rvrv
rkv
rkvv
Instant Center Between Two Rigid Bodies
In the foregoing discussion, the velocities of points A and B
were assumed to be reference to a coordinate system
attached to the ground. The resulting center is called an
instant center relative to the ground.
The velocities of points A and B could as well be taken
relative to coordinate system attached to another body. The
resulting point in this case will have a zero velocity with
respect to that body. In other words, point I will have the
same velocity for both bodies.
A
BI
A’
B’
Instant Center Between Two Rigid Bodies
A instant center between two bodies is a:
A point on both bodies
A point at which the two bodies have no
relative velocity.
A point about which one body may be
considered to rotate around the other body
at a given instant.
A
BI
A’
B’
Instant Center Between Two Rigid Bodies
When two links are connected to one
another by a revolute joint, the center of
the connecting joint is an instant center for
the two links.
When two links are not connected, an
instant center between the two links will
also exist and can be determined if the
velocities of both links is known.
A
A’
Instant Center Between Two Rigid Bodies
The number of instant centers in a constrained kinematic chain is equal to the
number of possible combinations of two links.
The number of pairs of links or the number of instantaneous centers is the number
of combinations of L links taken two at a time. Mathematically, number of instant
centers is,
2
1
LLN
Types of Instantaneous Centers
The instant centers for a mechanism are of the following three
types :
1. Primary (Permanent) instant centers. They can be fixed
or moving
2. Secondary Instant centers (Not permanent)
2
1
nnC
Instant Centers of a Four Bar Mechanism
Consider a four bar mechanism ABCD as shown. The number of instant centers
(N) in a four bar mechanism is given by
The instant centers I12
and I14
are fixed instant centers as they remain in the
same place for all configurations of the mechanism. The instant centers I23
and
I34
are permanent instant centers as they move when the mechanism moves,
but the joints are of permanent nature. The instantaneous centres I13
and I24
are neither fixed nor permanent as they vary with the configuration of the
mechanism.
2
1
nnC
6
2
12
2
144
N A
B
C
D
1 (Ground)
23
4
Location of Instant Centers
When the two links are connected by a pin joint (or pivot joint), the instant
center lies on the center of the pin as. Such an instant center is of
permanent nature. If one of the links is fixed, the instant center will be of
fixed type.
When the two links have a pure rolling contact (without slipping), the
instantaneous centre lies on their point of contact, as this point will have the
same velocity on both links.
When the two links have a sliding contact, the instant center lies at the
center of curvature of the path of contact. This points lies at the common
normal at the point of contact.
Kennedy (or Three Centers in Line) Theorem
Kennedy’s theorem states that any three bodies in plane
motion will have exactly three instant centers, and the three
centers will lie on the same straight line.
Note that this rule does not require that the three bodies be
connected in any way. We can use this rule, in conjunction
with the linear graph, to find the remaining lCs which are not
obvious from inspection.
Example: Instant Centers of a Four Bar Mechanism
Draw a circle with all links numbered around the circumference
Locate as many ICs as possible by inspection. All pin joints will be permanent
ICs . Connect the links numbered on the circle to create a linear graph and
record those found.
Identify a link combination for which the IC has not been found, and draw a
dotted line connecting those two link numbers. Identify two triangles on the
graph which each contains the dotted line and whose two other sides are solid
lines representing the ICs the already found. Use Kennedy’s theorem to locate
the needed IC.
Example: Instant Centers of a Four Bar Mechanism
Example: Instant Centers of a Slider-Crank Mechanism
Example: Instant Centers of a Slider-Crank Mechanism
Example: Instant Centers of a Cam and Follower: Common Normal Method
Example: Instant Centers of a Cam and Follower: Effective Link Method
Velocity Analysis with Instant Centers
Once the instant centers (ICs) of a linkage with
respect to the ground link have been found, they
can be used for a very rapid graphic analysis for
that link.
Note that some of the ICs may be very far removed
from the links. For example, if links 2 and 4 are
nearly parallel, their extended lines will intersect at
a point far away and not be practically available for
velocity analysis.
Velocity Analysis with Instant Centers
From the definition of the instant center, both links sharing the instant center will
have identical velocity at that point.
Instant center I13
involves the coupler (link 3) which is in general plane motion,
and the ground link which is stationary. All points on the ground link have zero
velocity in the global coordinate system, which is embedded in link 1. Therefore,
I13
must have zero velocity at this instant, and it can be considered to be an
instantaneous "fixed pivot" about which link 3 is in pure rotation with respect to
link 1.
A moment later, I13
will move to a new location and link 3 will be "pivoting" about
a new instant center.
Velocity Analysis with Instant Centers
If ω2
is known for the mechanism shown, the magnitude of the velocity of
point A can be computed as vA
= ω2
O2
A Its direction and sense can be
determined by inspection.
Note that point A is also instant center and it has the same velocity as part
of link 2 and as part of link 3. Since link 3 is effectively pivoting about I13
at
this instant, the angular velocity ω3
can be found by ω3
= vA
/AI13
. Once
ω3
is known, the magnitude of vB
can also be found from vB
= ω3
AI13
Once vB
is known, ω4
can also be found from ω4
= vB
/BO4
= vB
/BO4
.
Finally, vC
(or the velocity of any other point on the coupler) can be found
from vC
= ω3
CI13
133 AI
vA
133BIvB