Lecture 29

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LECTURE #29
BRAKES
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
Brake:
A device used to bring a moving system to rest, to slow its speed or to control its speed to a certain
value under varying conditions.
From previous lecture, the energy absorbed or dissipated by the unit per cycle is equal to the change
in Kinetic energy of the components being accelerated or stopped.
E 
1
2

J m i   f
2
2

where
J m  mass moment of inertia of system being de-accelerated
 i  initial angular velocity of system being de-accelerated
 f  final angular velocity of system being de-accelerated
E  change in energy associated with de-accelerated system
If 
f
 0 then
E 
1
2
J m i
2
The change in energy associated with de-accelerating a body must be absorbed by the material adjacent
to the brake lining. (usually a metal drum)
The temperature increase of the brake is related to the energy that must be absorbed by the brake.
E   C  T
where
E 
frictional energy the brake must absorb
  weight density of the drum material lb/ in3
 
volume of drum material absorbing the energy
C  Specific heat of brake drum material (lbf -ft/lbm 0F)
T  increase in temperature of brake drum
Specific heats of a few materials:
1) Cast iron =101
2) Steel= 93
3) Aluminum = 195
The following table list maximum drum temperatures for some commonly used brake and clutch materials
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
Most brake lining manufacturers include the effect of the rate of energy/ dissipation by giving limiting
values of Pv for given materials
where
P= pressure lb/in2
v= velocity = ft/ min
Pv= ft-lb/in2m or hp/in2
To develop an appreciation for the numbers involved, the following Pv values are typical.
1) Less than 28,000 in applications involving continuous operations and inadequate heat dissipation.
2) Less than 55,000 for intermittent operation and poor heat dissipation but with long periods of
rest.
3) Less than 83,000 for continuous applications with good heat dissipation.
Band Brakes:
A flexible material, usually made of steel, is faced with a friction material that can conform to the
curvature of the drum. The application of a force to the lever establishes tension in the band and
forces the frictional material against the drum creating a torque that slows down the drum
The relationship between the tensions of the brake band, designated as P1 and P2 in figure 1 and the
torque applied to the brake drum are the same as the relationships used with pulleys.
T f  ( P1  P2 ) r
where
T F  braking torque applied to brake down by brake band (ft-lbs)
P1  Maximum band tension (lbs)
P2 
r 
Minimum band tensions (lbs)
radius of brake drum
Since the brake band is not rotating, the terminology used to describe the brake is different from the
terminology used to describe the belt and pulley.
tight side tension -> Maximum band tension
slack side tension -> Minimum band tension
As was the case before:
P1
f
e
P2
  angle of wrap (coverage) of the band in radians
Where
f 
coefficient of friction of brake band
Similar the disc clutches and/or brakes, the maximum pressure that can be applied to the brake band is
determined by the material. The maximum pressure that is associated with the band and drum contact
occurs at the end of the band where the maximum in band tension exists.
P1  Pmax rw
The actuating force w1 required to operate the brake can be related to the tension in the brake band
and the geometry of the lever. For the simple brake band shown in Figure 1(a) sum moments about the
pivot point of the lever to relate w to P2

M
Pivot
 P2 a  wL  0   w 
P2 a
L
For the geometry illustrated in figure 1(b), sum the moments about the pivot point of the lever to
relate w to P1 and P2.
M
w
Pivot
 P2 a  P1 e  wL  0
 P2 a  P1 e 
L
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