Download Lecture 28

LECTURE #28
MOTION CONTROL
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
Machine systems can operate with intermittent motion. Starting and stopping operations can cycle
frequently.
Motion control elements permit machine systems to achieve intermittent motion.
Motion Control elements:
•
•
Clutches: Devices used to transmit power on an intermittent basis by connecting and/or disconnecting
a driven component to and/or from the prime mover.
– Motor operates efficiently at continuous speeds.
– Avoids accelerating and/or de-accelerating the rotor of the motor each time a driven component
of a machine needs to be cycled.
Brakes: Device that absorbs the kinetic energy of a system and thus controls the motion of the system
by slowing down the system and/or bringing the system to rest.
Functions of a clutch:
•
Connect a rapidly turning shaft to one that is initially stationary.
•
Cause two shafts to turn at the same speed and to do so in a manner that shock is not produced.
•
Limit torque that is transmitted or to prevent torque from being transmitted in a reverse direction.
Types of clutches:
•
Positive contact
– Square Jaw
– Spiral Jaw
•
Friction clutches
– Plate
– Cone
Positive Contact Clutches:
•
Transmits power from the driving shaft to the driven shaft by means of jaws or teeth.
Advantages
a) No slip operation
b) Develop very little heat since they do not depend on friction
c) Generally lighter and less costly than a friction clutch of similar torque capacity
Disadvantages
a) Cannot be engaged at high speeds
b) Shock accompanies engagement at any speed
c) When both driven and drive shafts are at rest, some relative motion may be required to engage.
Disc clutch:
Disc Clutch
Transmits torque from the input shaft to the output shaft due to the friction force developed
between the two plates or discs.
Input disc is free to move axially along the shaft but is splined, or keyed to shaft so disc will rotate
with shaft.
The friction force that can be transmitted depends on axial force developed between discs.
Axial force can be applied in several ways.
1. Mechanical means
a) Levers
b) Springs
c) linkages
2. Hydraulic or pneumatic
3. Electro-magnetic means
Advantages
1. Very little shock during engagement due to slippage
2. Can be used for high speed engagements.
Disadvantages
1. Slippages not suitable for applications that require positive transmission
2. wear
If the discs or plates used in the clutch are relatively flexible, it is possible to obtain relatively uniform
pressure on the friction surfaces when the actuating forces presses the disc together.
If the plates are relatively rigid, the wearing of the friction surface is approximately uniform after an
initial wearing-in has taken place.
When analyzing the clutch, either the assumption of uniform wear or uniform pressure must be made.
A conservative approach is to design with the assumption of uniform wear.
wear may be assumed to be directly proportional to pressure and velocity at a particular point on the
plate of the clutch. Recall velocity is proportional to the radius.
Accordingly W  kP * r
Uniform wear implies local pressure P times r is constant.
Where
W  Wear
Note: If disc or plates used in clutch are relatively
flexible, it is possible to obtain relatively uniform
k  Proportionality Constant
pressure on the friction surfaces.
P  Local Pressure
Alternatively, if the plates are rigid, the wearing of
the plates is relatively uniform after an initial wearing
radius to arbitrary point on disc
r
of the friction surface.
Under the assumption of uniform wear, the product of pressure and radius is a constant
W
 P*r  k
k
Since W and k both are constants
Where
k  proportionality constant
Since pr=k, as r increases then p decreases
Alternatively as r decreases, p increases.
Accordingly the maximum pressure must occur at the minimum radius.
K=pr = pmax ri => p=pmax ri
r
Solving for P from the above equation
P  Pmax
ri
r
Pressure varies from a max at ri to a minimum at ro where ro Outer Radius
Actuating force = force pressing the
two discs together
Actuating force is normal to the friction
surface. It can be obtained by
multiplying the pressure between the
friction surfaces by the area of the
surfaces – since pressure varies with ri
we must integrate to obtain F.
For the case of uniform wear, the following relationship exists between the activation force and the
pressure P existing at r
Fa 
ro

r
p 2 rdr 
ro

r
p max rro 2 rdr  2 rp max
i
i
ro
 dr
ri
ro
 2 rip max[r  ]
ri
 2 rip max[ro  ri ]
The torque that can be transmitted by a disc type of clutch can also be related to the activating force.
For the case of uniform wear. The frictional torque for the elemental annular area must be integrated
over the area of the dics.
T
ro
ro
2
2
rF
fdr  r[ f ] p 2 rdr  frip max[r0  ri ]


ri
ri
T   frip max[r 0  ri][r 0  ri]
T
Faf [r 0  ri ]
2
f=coefficient of friction
Brake and Clutch Material Properties
Material
Maximum Drum
Temperature, 0F
Coefficients of
friction, f
Maximum
Allowable
Pressure, psi
Metal on metal
500-600
0.25
200-250
Wood on metal
200
0.2-0.3
50-90
Leather on metal
150-200
0.3-0.4
15-40
Molded Blocks
500-600
0.25-0.5
100-150
Asbestos on metal
in oil
500
0.35-0.45
50-150
Sintered metal on
cast iron in oil
450
0.2
400
Example : A single plate disc clutch, with both sides effective, is to be used in an automobile. The friction
material that the clutch is made of has a coefficient of friction of 0.3 and a maximum allowable pressure
15 psi. If the outside radius of the clutch is 10 in., and the inside radius is 8 in., determine the torque that
can be transmitted and the actuating force required of the springs.
Solving for torque on a single side
T1 
Torque transmitted for 2 sides
T2  2T1 
Actuating force required to transmit torque.
Fa 
Wear rating: The previous relationships concerning the torque that can be transmitted by a clutch
involved the mean radius of the disc. The area of the disc did not influence torque.
Several different combinations of ro and ri can produce the same Rm but different areas. Although the
different combinations of ro and ri lead to the same Rm and therefore are capable of transmitting the
same level of torque, the difference in the areas will effect the rate at which the disc wears and/or the
amount of power that can be accommodated with a specified wear rate.
WR 
Pf
A
Where
WR 
Wear rating
A  Area of friction surface
Pf  Power transmitted by frictional surfaces of discs
Note:
WR  .04hp / in 2 for frequent applications, a conservative rating
 .10hp / in 2 for average service
 .40hp / in 2 for infrequently used brakes allowed to cord some what between applications
Pf  T f 
Where
T f  amount of torque that can be transmitted by clutch/ brake
  angular velocity of disc
Pf 
Tf n
63,000
Where
n = speed of rotation in rpm
Two approaches are available for establishing the torque requirement demanded from a clutch
and/or brake.
1)
T 
CPk
n
where
C  Conversion factor for units
P  Power of prime mover
n  Speed of prime mover rpm
k  Service factor based on application
T  Torque level required from clutch
2a)
T  J m
where
J m  Mass moment of inertia of load presented to motor
  angular acceleration rad/sec2
For constant α, α=∆ / ∆t
w= angular speed rads/sec
equation 2 may be rewritten:
2b)
  
T  Jm


t


Equation 2b enables the designer to relate the torque required to produce a change in rotational
speed ∆ in a given amount of time ∆t,
Equation 2b enables the designer to assess the impact of time to start and/or time to stop on the
requirements of the clutch.
Given
Example :
J m  2.473slug  ft 2
ni  0
n f  550rpm
t  2.50 sec
Calculate Torque T required to accelerate inertia with specifications given above.
T=
Energy Absorption/ Heat Dissipation
When using a clutch to accelerate an object or a brake to stop a rotating object, heat is generated due
to friction. Heat is transferred from the clutch and/or brake to the surroundings and equilibrium
temperature is established.
Temperature of components can effect life time.
The energy to be absorbed or dissipated by the unit on operational cycle is equal to the change in
kinetic energy of the components being accelerated or stopped.
E  KE 

1
J m i2   2f
2

where
E  energy absorbed or dissipated
J m  Mass moment of inertia
i  Initial Angular velocity
 f  Final Angular velocity
KE  Change in Kinetic energy
Cone Clutch:
Cone clutches have the advantage of being able to transmit a larger torque than disc clutches with the
same outside diameter and actuating force.
The reason for this capability is the increased frictional area and the wedging action that takes place.
Cone clutches find their widest use in relatively low peripheral speed applications.
Cone clutch analysis
F
X
0
Fa  F ' cos   N sin   0
Fa  F ' cos   N sin 
Fa  fN cos   N sin 
Fa  N (sin   f cos  )
N
Fa
sin   f cos 
Assuming the friction force is isotropic and acts at the mean radius Rm 
r0  ri
2
T f  fNRm
Substituting for N from above expression
Tf 
fFa Rm
sin   f cos 
Compute the axial force required for a cone brake if it is to exert a braking torque of 50 lb-ft. The
mean radius of the cone is 5.0 in. Use f=.25. Try cone angles of 100, 120 and 150
α
100
120
150
Fa