Kinematics, Dynamics, and Design of Machinery 2nd Edition by Waldron Kinzel Chapter 8
Short Description
Solutions to Kinematics, Dynamics, and Design of Machinery 2nd Edition by Waldron Kinzel Chapter 8. Profile Cam Design...
Description
Solutions to Chapter 8 Exercise Problems Problem 8.1 A cam that is designed for cycloidal motion drives a flat-faced follower. During the rise, the follower displaces 1 in for 180˚ of cam rotation. If the cam angular velocity is constant at 100 rpm, determine the displacement, velocity, and acceleration of the follower at a cam angle of 60˚. Solution: The equation for cycloidal motion is: y = L 1 sin 2
2 For L = 1, and = 180˚= , then y = L 1 sin 2 = 1 1 sin 2 = 1 sin2 2 2 2
(
) (
)
y˙ = L 1 cos 2 = (1 cos2) 2
2 y˙˙ = 2L sin 2 = 2 sin2
()
The angular velocity is ˙ = 100 rpm = 100 2 = 10.472 rad / s 60 When = 60˚= , 3 y=
(3 21 sin2( 3)) = (13 21 sin(2 3)) = 0.195 in
y˙ = (1 cos2 ) = 10.472 (1 cos(2 3)) = 5.000 in. sec.
()
2
(
2
)
y˙˙ = 2 sin2 = 2 10.472 sin(2 3) = 60.46 in.2 sec Problem 8.2 A constant-velocity cam is designed for simple harmonic motion. If the flat-faced follower displaces 2 in for 180˚ of cam rotation and the cam angular velocity is 100 rpm, determine the displacement, velocity, and acceleration when the cam angle is 45˚. Solution:
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The equation for simple harmonic motion is: y = L 1 cos
2 For L = 2, and = 180˚= , then y = L 1 cos = 2 1 cos = (1 cos ) 2 2
(
)
y˙ =
dy d ( = 1 cos ) = ˙ sin dt dt
y˙˙ =
d 2y d ˙ = ( sin ) = ˙2 cos dt 2 dt
The angular velocity is ˙ = 100 rpm = 100 2 = 10.472 rad / s 60 When = 45˚, y = (1 cos ) = (1 cos45˚) = 1 0.707 = 0.292 in , y˙ = ˙ sin = 10.472 sin45˚= 10.472 (0.707) = 7.405 in s y˙˙ = ˙2 cos = 10.4722 (0.707) = 77.531 in2 s Problem 8.3 A cam drives a radial, knife-edged follower through a 1.5-in rise in 180˚ of cycloidal motion. Give the displacement at 60˚ and 100˚. If this cam is rotating at 200 rpm, what are the velocity (ds/dt) and the acceleration (d2 s/dt2 ) at = 60˚? Solution: The equation for cycloidal motion is: y = L 1 sin 2
2 For L = 1.5, and = 180˚= , then y = L 1 sin 2 = 1.5 1 sin 2 = 1.5 1 sin2 2 2 2
(
) (
y˙ = L 1 cos 2 = 1.5 (1 cos2 )
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)
2
2 2 y˙˙ = 2L sin 2 = 2(1.5) sin2 = 3 sin2
()
()
The angular velocity is ˙ = 200 rpm = 200 2 = 20.944 rad / s 60 When = 60˚= , 3
3 1 sin 2 = 0.293 in y = 1.5 1 sin2 = 1.5 2 2 3
) (
(
y˙ = 1.5 (1 cos2 ) = 2
()
)
(
)
1.5(20.944) 1 cos 2 = 15.00 in s 3
(
)
2
y˙˙ = 3 sin2 = 3 20.944 sin 2 = 362.76 in2 s 3 When = 100˚= 100 = 5 , 180 9 5 9 1 sin 10 = 0.915 in y = 1.5 1 sin2 = 1.5 2 2 9
(
) (
)
Problem 8.4 Draw the displacement schedule for a follower that rises through a total displacement of 1.5 inches with constant acceleration for 1/4th revolution, constant velocity for 1/8th revolution, and constant deceleration for 1/4th revolution of the cam. The cam then dwells for 1/8th revolution, and returns with simple harmonic motion in 1/4th revolution of the cam. Solution: The displacement profile can be easily computed using the equations in Chapter 8 using Matlab. The curves are matched at the endpoints of each segment. The profile equations are: For 0 2 y1 = a0 + a1 + a2 2 The boundary conditions at = 0 are y1 = 0 and y'1 = 0 . Therefore, a0 = a1 = 0 So, y1 = a2 2 and y'1 = 2a2 where a2 is yet to be determined.
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For 3 2 4 y2 = b0 + b1
()
The boundary conditions at = are y1 = a2 2 2 2 b0 + b1 = a2 2 2
2
and y'1 = 2a2 2 = a2 . Then
()
and 0 = a2
2
( ) + b + b 2 2
0
1
Also y'2 = b1 = a2 or 0 = a2 b1 For 3 5 4 4 y3 = c0 + c1 + c2 2 y'3 = c1 + 2c2 y"3 = 2c2
3 2 The boundary conditions at = 3 are y2 = a2 + = a2 + = a2 and y'2 = a2 . 4 4 4 2 4 Also, at = 5 , y3 = 1.5 and y'3 = 0 . Then, matching the conditions, 4
(
2 3 3 2 = c0 + c1 + c2 2 4 4 3 y'3 = c1 + 2c2 = a2 4 5 5 2 1.5 = c 0 + c1 + c2 4 4 5 y'3 = c1 + 2c2 =0 4 The boundary condition equations can be written as:
( )
y3 = a2
( )
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)
(
)
0 = a2
2
(2 ) + b + b 2 0
1
0 = a2 b1
2 3 3 2 + c0 + c1 + c2 2 4 4 0 = a2 + c1 + 2c2 3 4 5 5 2 1.5 = c 0 + c1 + c2 4 4 0 = c1 + 2c2 5 4 In matrix form, 2 1 0 0 0 2 2 0 0 1 0 0 0 a2 0 2 3 3 2 b0 0 0 1 4 4 b1 0 2 3 0 = 0 0 0 1 c 2 0 5 c1 0 0 0 0 0 1 1.5 2 c2 5 5 2 0 0 0 1 4 4
( )
0 a2
( )
()
( ) ( )
Solving for the constraints using Matlab, a2 0.2026 b0 -0.5000 b1 0.6366 c0 = -1.6250 c1 1.5915 c2 -0.2026 The equations are then given in the following: For 0 2 y1 = a2 2 = 0.2026 2 For 3 2 4 y2 = b0 + b1 = - 0.5000 + 0.6366
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For 3 5 4 4 y3 = c0 + c1 + c2 2 = -1.6250 + 1.5915 0.2026 2
For 5 3 4 2 y4 = 1.5 For the return, 3 2 , and 2 y5 =
3 L 1 + cos = (1+ cos2 ) 4 2
The displacement diagram is plotted in the following:
Follower Displacement, Velocity, and Acceleration Diagrams
1
Position, max value is: 1.5
Normalized Follower Displacement, Velocity, and Acceleration
Velocity, max value is: 1.5 ( ) Acceleration, max value is: 3 ( 2 )
0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8
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200 Cam Angle
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250
300
350
Problem 8.5 Draw the displacement schedule for a follower that rises through a total displacement of 20 mm with constant acceleration for 1/8th revolution, constant velocity for 1/4th revolution, and constant deceleration for 1/8th revolution of the cam. The cam then dwells for 1/4th revolution, and returns with simple harmonic motion in 1/4th revolution of the cam. Solution: The displacement profile can be easily computed using the equations in Chapter 8 using Matlab. The curves are matched at the endpoints of each segment. The profile equations are:
For 0 4 y1 = a0 + a1 + a2 2 The boundary conditions at = 0 are y1 = 0 and y'1 = 0 . Therefore, a0 = a1 = 0 So, y1 = a2 2 and y'1 = 2a2 where a2 is yet to be determined. For 3 4 8 y2 = b0 + b1
()
The bounary conditions at = are y1 = a2 4 4 2 b0 + b1 = a2 4 4 and 2 0 = a2 + b0 + b1 4 4 Also y'2 = b1 = a2 2 or 0 = a2 b1 2
2
and y'1 = 2a2 = a2 . Then, 4 2
()
()
For 3 8
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y3 = c0 + c1 + c2 2 y'3 = c1 + 2c2 y"3 = 2c2 3 3 are y2 = b0 + b1 and y'2 = b1 . Also, at = , y3 = 20 , and 8 8 y'3 = 0 . Then matching the conditions,
The boundary conditions at =
3 3 3 = c0 + c1 + c2 8 8 8 3 y'3 = c1 + 2c2 = b1 8
( )
y3 = b0 + b1
2
20 = c0 + c1 + c2 2 y'3 = c1 + 2c2 = 0 The boundary condition equations can be written as: 2 0 = a2 + b0 + b1 4 4 0 = a2 b1 2 3 3 3 0 b0 b1 + c0 + c1 + c2 8 8 8 0 = b1 + c1 + c2 3 4 20 = c0 + c1 + c2 2 0 = c1 + 2c2
()
( )
2
In matrix form, 0 0 0 0 = 0 20
() 4 2 0 0 0 0
2
1 0
4 1
1 3 8 0 1 0 0 0 0
0
0
0
0
1 3 8 0 1 0 1 1
0 a2 0 b0 3 2 b1 8 c0 3 c 1 4 2 c2 2
( )
Solving for the constraints using Matlab,
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a2 7.2051 b0 -4.4444 b1 11.3177 c0 = -8.4444 c1 18.1083 c2 -2.8820 The equations are then given in the following:
For 0 4 y1 = a2 2 = 7.2051 2 For 3 4 8 y2 = b0 + b1 = - 4.4444 + 11.3177 For 3 8 y3 = c0 + c1 + c2 2 = -8.4444 + 18.1083 -2.8820 2 3 For 2 y4 = 20 mm For the return, 3 2 , and 2 y5 =
L 1 + cos = 10(1+ cos 2 ) 2
The displacement diagram is plotted in the following:
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Follower Displacement, Velocity, and Acceleration Diagrams
Normalized Follower Displacement, Velocity, and Acceleration
1
Position, max value is: 20 Velocity, max value is: 20 ( ) 2 Acceleration, max value is: 40 ( )
0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8
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250
300
350
Cam Angle
Problem 8.6 Draw the displacement schedule for a follower that rises through a total displacement of 30 mm with constant acceleration for 90˚ of rotation and constant deceleration for 45˚ of cam rotation. The follower returns 15 mm with simple harmonic motion in 90˚ of cam rotation and dwells for 45˚ of cam rotation. It then returns the remaining 15 mm with simple harmonic motion during the remaining 90˚ of cam rotation. Solution: The displacement profile can be easily computed using the equations in Chapter 8 using Matlab. The curves are matched at the endpoints of each segment. The profile equations are:
For 0 2 y1 = a0 + a1 + a2 2 The boundary conditions at = 0 are y1 = 0 and y'1 = 0 . Therefore, a0 = a1 = 0 So, y1 = a2 2 and y'1 = 2a2 - 337 -
where a2 is yet to be determined. For 3 2 4 y2 = b0 + b1 + b2 2
()
The boundary conditions at = and y1 = a2 2 2 2 2 = a2 b0 + b1 + b2 2 2 2 and 2 2 0 = a2 + b0 + b1 + b2 2 2 2
()
2
and y'1 = 2a2 = a2 . Then, 2
()
()
()
Also y'2 = b1 + b2 = a2 or 0 = a2 + b1 + b2 The boundary conditions at = 3 are y2 = 30 and y'2 = 0 . Then, 4 2 3 3 = 30 b0 + b1 + b2 4 4
( )
and 0 = b1 + 2b2
( 34 ) = b + b (32 ) 1
2
The four boundary condition equations can be summarized as: 2 0 = a2 + b0 + b1 + b2 2 2 2 0 = a2 + b1 + b2
()
()
3 3 + b2 4 4 0 = b1 + b2 3 2 In matrix form, 30 = b0 + b1
( )
2
2
( )
2 0 2 0 = 30 0 0 0
()
2 0 1 1 3 4 0 1 1
2 2 a2 b0 3 2 b 1 4 3 b2 2 7.5
() ( )
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Solving for the constraints using Matlab, a2 8.1057 b0 -60.0000 = b1 76.3944 b2 -16.2114 The equations are then given in the following:
For 0 2 y1 = a2 2 = 8.1057 2 For 3 2 4 y2 = b0 + b1 + b2 2 = -60.0000 + 76.3944 + -16.2114 2 For 3 5 4 4 = 2 and 15 L y3 = 1+ cos = (1+ cos2 ) 2 2 This equation assumes that the curve is 15 mm high and that the curve falls to zero. However, the actual curve begins at a height of 30 mm and returns to only 15 mm. Because of this, we need to add 15 mm to the value for y3 . Then, y3 = 15+ 7.5(1+ cos2 ) For 5 6 4 4 y4 = 15 For 6 2 4 = 2 and 15 L y5 = 1 + cos = (1 + cos2 ) 2 2 The displacement diagram is plotted in the following:
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Follower Displacement, Velocity, and Acceleration Diagrams
Normalized Follower Displacement, Velocity, and Acceleration
1
Position, max value is: 30 Velocity, max value is: 25.4648 ( ) 2 Acceleration, max value is: 32.4228 ( )
0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8
TextEnd
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Cam Angle
Problem 8.7 Draw the displacement schedule for a follower that rises through a total displacement of 3 inches with cycloidal motion in 120 degrees of cam rotation. The follower then dwells for 90˚ and returns to zero with simple harmonic motion in 90˚ of cam rotation. The follower then dwells for 60˚ before repeating the cycle. Solution: The displacement profile can be easily computed using the equations in Chapter 8 using Matlab. The curves are matched at the endpoints of each segment. The profile equations are: 2 For 0 3
=
2 3
and y1 = L
1 2 3 1 sin sin3 =3 2 2 2
(
)
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1 1 2 3 3 cos cos 3 =3 2 2 2 2 9 y"1 = L 2 sin sin3 =3 2
(
y'1 = L
(
)
)
For 2 7 3 6 y2 = 3 For the return, 7 5 _ 6 3
=
2
and y3 =
L 1+ cos = 1.5(1+ cos2 ) 2
For the remainder of the cycle, 5 2 3 and y4 = 0 The displacement diagram is plotted in the following:
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Follower Displacement, Velocity, and Acceleration Diagrams
Normalized Follower Displacement, Velocity, and Acceleration
1
Position, max value is: 3 Velocity, max value is: 3 ( ) 2 Acceleration, max value is: 6 ( )
0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8
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Cam Angle
Problem 8.8
Follower Travel
A cam returns from a full lift of 1.2 in during its initial 60˚ rotation. The first 0.4 in of the return is half-cycloidal. This is followed by a half-harmonic return. Determine 1 and 2 so that the motion has continuous first and second derivatives. Draw a freehand sketch of y', y'', and y''' indicating any possible mismatch in the third derivative. Not to Scale
0.4 in 1.2 in
0˚
60˚ β1
β2
Solution: The first part of the return is made up of a cycloidal curve and the second part is made up of a harmonic curve. This is shown schematically in the figure below.
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2β 2
Follower Travel
β1 β 2
Harmonic Curve
0.4" 1.2"
Cycloidal Curve
0.4" 0˚
60˚ β2
β1
2 β1 The range for the cycloidal curve is given by 0 1 21 and the range for the harmonic curve is given by
1 2 2 1 + 2 Also,
2 = 1 (1 2) The general form for the cycloidal equation for a return is given in Section 7.8 as y = L 1 + 1 sin 2
2 Half of the cycloidal return is 0.4 so return is 0.8. The range for 1 is 21 . As indicated in the figure above, the cycloidal curve is offset from the horizontal axis by 0.4". Therefore, this much must be added to y. The cycloidal equation for the return is y1 = L1 1 1 + 1 sin 21 + 0.4 = 0.8 1 1 + 1 sin 1 + 0.4 21 1 21 2 21 2 = 0.8 1.5 1 + 1 sin 1
21 2 1
(1)
The harmonic curve is given by Eq. (812). Half of the harmonic return is (1.2"-0.4") = 0.8 so that the whole return is 1.6". The range for 2 is 22 . Therefore, the equation for the harmonic part of the return is: y2 = L2 1+ cos 2 = 0.81+ cos 2 2 2 2 22
(2)
We also know that 2 = 1 (1 2) and 2 = 1 3 - 343 -
Eqs. (1) and (2) can be reduced so that the only unknown is 1 . To solve for the unknown, we can equate the slopes at 1 = 1 . For the cycloidal equation, y'1 = 0.8 1 cos 1
21 1 and at 1 = 1 y'1 = 0.8 1 cos 1 = 0.8 21 1 1
(3)
For the harmonic equation y' 2 = 0.4 sin 2
2 22
(4)
At 1 = 1, 2 = 2 . Therefore, y' 2 = 0.4 sin 2 = 0.4 sin 2 = 0.4 2 22 2 22 2
(4)
Equation Eqs. (3) and (4) give the following equation 0.8 = 0.4 1 2 or 2 = = 1 2 / 3 1 This equation can be easily solved for 1 . The result is: 1 =
2 = 0.4073436 radians. 3( + 2)
We can now write y, y', and y" for each part of the curve. For 1 0.4073436 y = 0.8 1.5 1 + 1 sin 1
2 21 1 y' = 0.8 1 cos 1
21 1 y"= 0.82 sin 1
21 1 and for 0.4073436 / 3 y = 0.81+ cos 2 22
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y' = 0.4 sin 2
2 2 2 2 y"= 0.22 cos 2
2 2 2
where
2 = 1 (1 2) and 2 = 1 3 The results are plotted in the following
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Problem 8.9 Assume that s is the cam-follower displacement and is the cam rotation. The rise is 1.0 cm after 1.0 radian of rotation, and the rise begins and ends at a dwell. The displacement equation for the follower during the rise period is
n
s = hCi i =0
i
If the position, velocity, and acceleration are continuous at = 0, and the position and velocity are continuous at = 1.0 rad, determine the value of n required in the equation, and find the coefficients Ci if ˙ = 2 rad/s. Note: Use the minimum possible number of terms. S Β
1.0
Dwell
Dwell Α
θ, rad
1.0
Solution: First determine the number of terms required. There are a total of five conditions to match; therefore, the number of terms is 5 making n = 4. The conditions to match are: 2 At = 0 , s = ds = d s2 = 0 d d
At = , s = h = 1.0
Now,
ds = 0 d i
2
3
n s = Ci = C0 + C1 + C2 + C3 + C4 i=0 2
ds = C1 + 2C2 + 3C3 + 4C4 d
3
and d2s = 2C2 + 6C3 + 12C4 d 2 2 2 2
2
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4
Applying the conditions at = 0 , s = C0 = 0 ds = C1 C = 0 1 d d2s = 2C2 = 0 C = 0 2 d 2 2 Applying the conditions at = , 3
4
s = C3 + C4 = C3 + C4 = 1 2
3
ds = 3C3 + 4C4 = 3C3 + 4C4 = 0 d or 3C3 + 4C4 = 0 Solving for the constants, C3 = 4 and C4 = 3 Therefore, 3
s = 4 3
4
Problem 8.10 Resolve Problem 8.9 if = 0.8 rad and ˙ = 200 rad/s. Solution: First determine the number of terms required. There are a total of five conditions to match; therefore, the number of terms is 5 making n = 4. The conditions to match are: 2 At = 0 , s = ds = d s2 = 0 d d
At = , s = h = 1.0
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Now,
ds =0 d n
s = Ci i= 0
i
2
3
= C0 + C1 + C3 + C4 + C2 2
ds C1 2C2 3C3 4C = + + 4 + d
4
3
and d 2 s 2C2 6C3 12C4 = + 2 + 2 d 2 2
2
Applying the conditions at = 0 , s = C0 = 0 ds C1 = C1 = 0 d d 2 s 2C2 = = 0 C2 = 0 d 2 2 Applying the conditions at = , 3
4
s = C3 + C4 = C3 + C4 = 1 2
3
ds 3C3 4C 3C 4C = + 4 = 3 + 4 =0 d or 3C3 + 4C4 = 0 Solving for the constants, C3 = 4 and C4 = 3 Therefore, 3
s =4
3
4
Notice that this solution is EXACTLY the same as that for 8.9. The results are independent of both and ˙ . This is one of the reasons for normalizing the problem with respect to .
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Problem 8.11 For the cam displacement schedule given, h is the rise, is the angle through which the rise takes place, and s is the displacement at any given angle . The displacement equation for the follower during the rise period is i
5
s = h ai i =0
Determine the required values for a0 ... a5 such that the displacement, velocity, and acceleration functions are continuous at the end points of the rise portion. S Β
h
Dwell
Dwell Α
β
θ, rad
Solution: There are a total of six conditions to match; therefore, the number of terms is 6 making n = 5. The conditions to match are: 2 At = 0 , s = ds = d s2 = 0 d d
At = , s = h and
Now,
ds = d2s = 0 d d 2 i 2 3 4 5 s = hCi + C2 = h C0 + C1 + C3 + C4 + C5 i=0 n
2 3 4 ds = h C1 + 2C2 + 3C3 + 4C4 + 5C5 d
and 2 3 d2s = h 2C2 + 6C3 + 12C4 + 20C5 2 d 2 2
2
2
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Applying the conditions at = 0 , s = C0 = 0 ds = h C1 C = 0 1 d d2s = h 2C2 = 0 C = 0 2 d 2 2 Applying the conditions at = , 4 5 3 s = h C3 + C4 + C5 = h[C3 + C4 + C5] = h
or C3 + C4 + C5 = 1 2 3 ds = h 3C3 + 4C4 = h 3C3 + 4C4 + 5C5 = 0
d
or 3C3 + 4C4 + 5C5 = 0 2 3 d2s = h 6C3 + 12C4 + 20C5 = h 6C3 + 12C4 + 20C5 = 0 2
d 2 2
2 2 2 2
or 6C3 + 12C4 + 20C5 = 0 Solving for the constants, C3 = 10; C4 = 15; C5 = 6 Therefore, 4 5 3
s = h 10 15 + 6
Problem 8.12 Resolve Problem 8.11 if h = 20 mm and = 120˚. Solution: There are a total of six conditions to match; therefore, the number of terms is 6 making n = 5. The conditions to match are: ds d 2 s At = 0 , s = = =0 d d 2 - 350 -
At = , s = h and
Now,
ds d 2 s = =0 d d 2 n
s = h Ci i= 0
and
i 2 3 4 5
= h C0 + C1 + C3 + C4 + C5 + C2
C 2C 3C3 2 4C4 3 5C5 4 ds = h 1 + 2 + + +
d 2C2 6C3 12C4 2 20C5 3 d 2s = h 2 + 2 + 2 + d 2 2
Applying the conditions at = 0 , s = C0 = 0 ds C = h 1 C1 = 0 d d 2s 2C = h 22 = 0 C2 = 0 d 2 Applying the conditions at = _
or
4 5 3 s = h C3 = h[C3 + C4 + C5 ] = h + C4 + C5
C3 + C4 + C5 = 1
or
3C 2 4C4 3 ds 3C3 4C4 5C5 = h 3 + = h + + = 0 d
3C3 + 4C4 + 5C5 = 0
or
6C3 12C4 2 20C5 3 d 2s 6C3 12C4 20C5 = h 2 + 2 + 2 = h 2 + 2 + 2 = 0 2 d 6C3 + 12C4 + 20C5 = 0
Solving for the constants,
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C3 = 10; C4 = 15; C5 = 6 Therefore, 4 5 3
s = h 10 15 +6
If h=20 mm and = 120˚, then, s = 2010 120
( )
3
4
5 120
( ) ( )
15
120
+6
Here, is assumed to be given in degrees. Note that the values for h and do not enter the problem until the last step. Problem 8.13 Resolve Problem 8.11 if h = 2 in and = 90˚. Solution: There are a total of six conditions to match; therefore, the number of terms is 6 making n = 5. The conditions to match are: ds d 2 s At = 0 , s = = =0 d d 2 At = , s = h and
Now,
ds d 2 s = =0 d d 2 n
s = h Ci i= 0
and
i 2 3 4 5
= h C0 + C1 + C3 + C4 + C5 + C2
C1 2C2 3C3 2 4C4 3 5C5 4 ds = h + + + +
d 2C2 6C3 12C4 2 20C5 3 d 2s = h 2 + 2 + 2 + 2 d 2
Applying the conditions at = 0 , s = C0 = 0 - 352 -
ds C = h 1 C1 = 0 d d 2s 2C = h 22 = 0 C2 = 0 2 d Applying the conditions at = _
or
4 5 3 s = h C3 = h[C3 + C4 + C5 ] = h + C4 + C5
C3 + C4 + C5 = 1
or
3C 2 4C4 3 ds 3C 4C 5C = h 3 + 4 + 5 = 0 = h 3 +
d 3C3 + 4C4 + 5C5 = 0
or
6C3 12C4 2 20C5 3 d 2s 6C 12C 20C = h 23 + 2 4 + 2 5 = 0 = h 2 + 2 + 2 2 d
6C3 + 12C4 + 20C5 = 0
Solving for the constants, C3 = 10; C4 = 15; C5 = 6 Therefore, 4 5 3
s = h 10 15 +6
If h=2 in and = 90˚, then, s = 210 90
( )
3
15
4
5 90
( ) ( ) 90
+6
Here, is assumed to be given in degrees. Note that the values for h and do not enter the problem until the last step.
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Problem 8.14 Assume that s is the cam-follower displacement and is the cam rotation. The rise is h after degrees of rotation, and the rise begins at a dwell and ends with a constant velocity segment. The displacement equation for the follower during the rise period is
n s = hCi i =0
i
If the position, velocity, and acceleration are continuous at = 0 and the position and velocity are continuous at = , determine the n required in the equation, and find the coefficients Ci that will satisfy the requirements if s = h = 1.0.
S 45˚
h
Dwell
θ
β
Solution: First determine the number of terms required. There are a total of five conditions to match; therefore, the number of terms is 5 making n = 4. The conditions to match are: 2 At = 0 , s = ds = d s2 = 0 d d
At = , s = h = 1.0
Now,
ds = tan 45˚= 1.0 d i
2
3
n s = Ci = C0 + C1 + C2 + C3 + C4 i=0
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4
2
ds = C1 + 2C2 + 3C3 + 4C4 d
3
and d2s = 2C2 + 6C3 + 12C4 d 2 2 2 2
2
Applying the conditions at = 0 , s = C0 = 0 ds = C1 C = 0 1 d d2s = 2C2 = 0 C = 0 2 d 2 2 Applying the conditions at = , 3
4
s = C3 + C4 = C3 + C4 = 1 2
3
ds = 3C3 + 4C4 = 3C3 + 4C4 = 1 d or 3C3 + 4C4 = Solving for the constants, C3 = 4 and C4 = 3 Therefore, 3
s = (4 ) + ( 3)
4
Problem 8.15 A follower moves with simple harmonic motion a distance of 20 mm in 45˚ of cam rotation. The follower then moves 20 mm more with cycloidal motion to complete its rise. The follower then dwells and returns 25 mm with cycloidal motion and then moves the remaining 15 mm with harmonic motion in 45˚. Find the intervals of cam rotation for the cycloidal motions and dwell by matching velocities and accelerations, then determine the equations for the displacement (S) as a function of for the entire motion cycle. Solution: - 355 -
This is a curve matching problem. To begin the problem, consider the equations for harmonic and cycloidal motions: Harmonic: y=
L 1 cos 2
y'=
L sin 2
y"=
2L cos 2 2
Cycloidal: 1 2 y=L sin 2 y'= L
1 1 2 cos
2 2 y"= L 2 sin
Follower Travel
For both the harmonic and cycloidal motions, we must determine L and . There are a total of four curves, so we need to determine four L’s and four ’s The geometry is shown in the following figure.
1
3
Cycloidal Curve
20 Harmonic Curve
Cycloidal Curve
25
20
15
Harmonic Curve
45˚
45˚ 2 4
We can treat the rise and return separately, and then determine the dwell to ensure that there is a full cycle of motion. - 356 -
For the rise section, assume that the harmonic curve is half-harmonic, and the cycloidal curve is half cycloidal. This will allow us to match the curves at their inflections points and will ensure curvature continuity. For the harmonic curve,
2 =
2
and L2 = 40 Therefore, the harmonic curve is L 40 y = 1 cos = (1 cos2 ) = 20(1 cos2 ) 2 2 The slope equation is L y'= sin = L( sin2 ) 2 Or the cycloidal curve, L1 = 40 and 1 21 1 21 y = L1 1 sin sin = 40 1 1 1 1 2 1 2 The slope equation is 1 1 21 y'= L1 cos 1 1 1 To find 1 , equate the slopes at the midpoint, then, 1 1 2 1 y'= L2 sin2 2 = L1 cos 2 1 1 1 2 or 40( sin 2 ) = 40
1 1 cos 1 1
or
(sin 2 ) = 1 1 cos 1
Then,
1
2 =1 1 or
1 = 2 The rise part of the curve is given by, For < 4
- 357 -
y = 20(1 cos2 ) The cycloidal curve starts at = ( 2 1 ) / 2 = 2 / 2 = 0.2146 2 Therefore,
( )
1 = + 0.2146 1 Then for 4 < < 4 + 2 1 21 [ + 0.2146] 1 sin [ + 0.2146] y = 40 1 sin = 40 1 2 1 2 2 The angular distance to the dwell is 2 = + 1 = + = 1+ 4 2 4 2 4
(
)
For the return, use the same procedure. The harmonic part of the return is given by L y4 = 4 1+ cos 4 2 4 Where L4 = 2(15) = 30 and
4 =
2
Therefore, y4 = 15(1+ cos24 ) For the cycloidal curve, 1 23 y3 = L3 L3 3 sin 3 3 2 Where L3 = 2(25) = 50 To determine 3 , equate the slopes for the two curves at their midpoints. Then, y'4 =
L4 4 1 1 2 3 sin cos = y'3 = L3 2 4 4 2 3 3 3 2
or
( 2 ) = 50 1 1 cos
30 sin
3
3
Solving for 3 , 2 3 = 3 5 Or - 358 -
10 3 The cycloidal part of the curve will start at
3 =
= 2 ( 3 + 4 ) / 2 Therefore, the dwell period will be + 1 < < 2 ( + ) / 2 3 4 4 2 And the dwell distance will b, y = 40 The cycloidal part of the curve occurs when 2 ( 3 + 4 ) / 2 < < 2 3 / 2 And y3 = L3 L3
3 1 23 sin 3 3 2
The curve will be shifted because L3 is 50. Therefore, y = y3 10 = L3 L3
3 1 2 3 sin 10 3 3 2
The second harmonic part of the curve occurs when 2 3 / 2 < < 2 And y4 =
L4 1+ cos 2 4
The displacement diagram is shown in the following:
- 359 -
Follower Displacement, Velocity, and Acceleration Diagrams
Normalized Follower Displacement, Velocity, and Acceleration
1
Position, max value is: 40 Velocity, max value is: 40 ( ) Acceleration, max value is: 80 ( 2 )
0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6
TextEnd
-0.8
0
50
100
150
200
250
300
350
Cam Angle
Problem 8.16
Displacement, cm
Construct the part of the profile of a disk cam that follows the displacement diagram shown below. The cam has a 5-cm-diameter pitch circle and is rotating counterclockwise. The follower is a knifeedged, radial, translating follower. Use 10-degree increments for the construction. 1.5 1.0 0.5 0˚
45˚ Cam Rotation Angle
90˚
Solution: Start by subdividing the follower diagram into 10˚ increments of cam rotation. Next draw the cam pitch circle and lay off radial lines in the clockwise direction. The follower displacements can then be taken directly from the displacement diagram. Use a smooth curve to draw the cam profile.
- 360 -
Displacement, cm
1.5 1.0 0.5 90˚ 0
1
2
3
4
5
6
7
8
10
9
Cam Rotation Angle
Pitch Circle 0 1 2 3 4 5 6 7 10
8
9
Problem 8.17
Follower Travel, mm
Construct the profile of a disk cam that follows the displacement diagram shown below. The follower is a radial roller and has a diameter of 10 mm. The base circle diameter of the cam is to be 40 mm and the cam rotates clockwise. 30
15
0 0˚
60˚
120˚
180˚
Cam Rotation
Solution:
- 361 -
240˚
300˚
360˚
Use the follower diagram subdivisions of 20˚. Next draw the cam pitch circle and lay off radial lines in the counterclockwise direction. The follower displacements can then be taken directly from the displacement diagram. Draw the pitch curve. Draw the cam follower circles on the pitch curve. Use a smooth curve to draw the cam profile tangent to the follower circles. 5 6
4
3
7 Prime Curve
2 8 1 Prime Circle 0
9 Base Circle 17 Cam 10
16 15 11
14 12
13
Problem 8.18
Total Follow Travel
Accurately sketch one half of the cam profile (stations 0-6) for the cam follower, base circle, and displacement diagram given below. The base circle diameter is 1.2 in.
1.0
0 Base Circle
- 362 -
1 2 11 10
3
4 9
5
Station Point Numbers
6 8
7
Solution: First draw the displacement schedule to scale, and divide the cam into 12 equal sections. The cam is rotating in the clockwise direction so that the cam is layed out in the counterclockwise direction. Then draw the locations of the follower. The cam profile is then drawn such that the cam is tangent to the follower in the different positions. The cam is drawn as follows:
4
3
2
5
1
6 7
8 9 10
12 11
Problem 8.19 Lay out a cam profile using a harmonic follower displacement (both rise and return). Assume that the cam is to dwell at zero lift for the first 100˚ of the motion cycle and to dwell at a 1 in lift for cam angles from 160˚ to 210˚. The cam is to have a translating, radial, roller follower with a 1-in roller diameter, and the base circle radius is to be 1.5 in. The cam will rotate clockwise. Lay out the cam profile using 20˚ plotting intervals. Solution: The displacement profile can be easily computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are: For 0 100˚ y=0 For 100˚ 160˚ y = L 1 cos
2 where L = 1, = 100˚ , and = 160˚100˚= 60˚. For 160˚ 210˚ y=1 - 363 -
For 210˚ 360˚ y = L 1+ cos 2 where L = 1, = 210˚, and = 360˚210˚= 150˚ The displacement diagram is given below followed by a table of values for y at 20˚ increments of .
Theta
Displacement
0.000 20.000 40.000 60.000 80.000 100.000 120.000 140.000 160.000 180.000 200.000 220.000 240.000 260.000 280.000 300.000 320.000 340.000 360.000
0.000 0.000 0.000 0.000 0.000 0.000 0.250 0.750 1.000 1.000 1.000 0.989 0.905 0.750 0.552 0.345 0.165 0.043 0.000
To lay out the cam, first draw the prime circle which has a radius of 1.5" + 0.5" = 2.0". Then lay off radial lines at 20˚ increments and label the lines in the counterclockwise direction. Draw circle arcs corresponding to the two dwells and lay off the distances for the other displacements along the other radial lines. Draw 1" diameter circles through the endponts of the distances layed off along the radial lines, and fit a smooth curve which is tangent to the circles corresponding to the roller follower.
- 364 -
1 inch
100˚ 140˚
120˚ Prime circle
160˚
0˚
340˚ 200˚
320˚ 220˚ 300˚ 240˚ 260˚
280˚
Problem 8.20 Lay out a cam profile using a cycloidal follower displacement (both rise and return) if the cam is to dwell at zero lift for the first 80˚ of the motion cycle and to dwell at 2-in lift for cam angles from 120˚ to 190˚. The cam is to have a translating, radial, roller follower with a roller diameter of 0.8 in. The cam will rotate counterclockwise, and the base circle diameter is 2 in. Lay out the cam profile using 20˚ plotting intervals. Solution: The displacement profile can be easily computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are: For 0 80˚ y=0 For 80˚ 120˚ y = L 1 sin 2
2 where L = 2, = 80˚, and = 120˚80˚= 40˚. For 120˚ 190˚ y=2 For 190˚ 360˚ - 365 -
y = L 1 + 1 sin 2
2 where L = 2, = 190˚ , and = 360˚190˚= 170˚ The displacement diagram is given below followed by a table of values for y at 20˚ increments of .
Theta 0.000 20.000 40.000 60.000 80.000 100.000 120.000 140.000 160.000 180.000 200.000 220.000 240.000 260.000 280.000 300.000 320.000 340.000 360.000
Displacement 0.000 0.000 0.000 0.000 0.000 1.000 2.000 2.000 2.000 2.000 1.997 1.932 1.718 1.344 0.883 0.452 0.154 0.021 0.000
To lay out the cam, first draw the prime circle which has a radius of 1.0" + 0.4" = 1.4". Then lay off radial lines at 20˚ increments and label the lines in the clockwise direction. Draw circle arcs corresponding to the two dwells and lay off the distances for the other displacements along the other radial lines. Draw 0.8" diameter circles through the endponts of the distances layed off along the radial lines, and fit a smooth curve which is tangent to the circles corresponding to the roller follower. Notice the poor pressure angles in the range 80˚ 120˚. This would probably make the cam unacceptable.
- 366 -
260˚ 240˚ 280˚
220˚
1 inch 300˚
320˚
200˚
340˚ Prime circle 180˚
0˚
80˚
100˚
120˚
Problem 8.21 Lay out a cam profile assuming that an oscillating, roller follower starts from a dwell for 0˚ to 140˚ of cam rotation, and the cam rotates clockwise. The rise occurs with parabolic motion during the cam rotation from 140˚ to 220˚. The follower then dwells for 40˚ of cam rotation, and the return occurs with parabolic motion for the cam rotation from 260˚ to 360˚. The amplitude of the follower rotation is 35˚, and the follower radius is 1 in. The base circle radius is 2 in, and the distance between the cam axis and follower rotation axis is 4 in. Lay out the cam profile using 20˚ plotting intervals such that the pressure angle is 0 when the follower is in the bottom dwell position. Solution: The displacement profile can be easily computed using the equations in Chapter 8 using a spreadsheet or MATLAB program. Remember that the parabolic motion is represented by two curves in each rise and return region. The curves are matched at the midpoints of the rise and return. The profile equations are: For 0 140˚ =0 For the first part of the rise, 140˚ 180˚ , and = 2L
2
where L = 35˚, = 140˚ , and = 220˚140˚= 80˚. For the second part of the rise, 180˚ 220˚ , and
- 367 -
2 = L 1 2 1
where L = 35˚, = 140˚ , and = 220˚140˚= 80˚. For 220˚ 260˚
= 35˚ For the first part of the return, 260˚ 310˚, and 2 = L1 2
where L = 35˚, = 260˚, and = 360˚260˚= 100˚ For the second part of the return, 310˚ 360˚, and
2 = 2L1 where L = 35˚, = 260˚, and = 360˚260˚= 100˚ The displacement diagram is given below followed by a table of values for at 20˚ increments of .
Theta 0.000 20.000 40.000 60.000 80.000 100.000 120.000 140.000
Follower
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
Angle 160.000 180.000 200.000 220.000 240.000 260.000 280.000 300.000
- 368 -
4.375 17.500 30.625 35.000 35.000 35.000 32.200 23.800
320.000 340.000
11.200 2.800
360.000
0.000
To lay out the cam, first draw the prime circle which has a radius of 2.0" + 1.0" = 3.0". Next draw the pivot circle for the follower pivot. The radius of the pivot circle is 4". Draw the follower in the initial position ( = 0˚ ) to determine the follower length (r3 ) and the position on the pivot circle corresponding to = 0˚ . As indicated in Example 8.5, the length r3 is given by r3 = r12 (rb + r0) = 42 (2 + 1)2 = 2.646" Identify the point on the pivot circle corresponding to = 0˚ , lay off the radial lines at 20˚ increments from this point, and label the lines in the counterclockwise direction. Draw lines from the intersections of the radal lines with the pivot circle tangent to the prime circle. Then lay off the angular displacements from these tangent lines. Locate the center of the follower by the distance r3 from the pivot circle along these lines. Draw 1" radius circles through the endponts of the distances layed off along these lines, and fit a smooth curve which is tangent to the circles corresponding to the roller follower. The cam profile is shown in the following figure.
- 369 -
140˚ 160˚
120˚
180˚
100˚ 1 inch
200˚
80˚
220˚
60˚
240˚
40˚
260˚
20˚
280˚
0˚ 300˚
340˚ 320˚
Problem 8.22 Lay out the rise portion of the cam profile if a flat-faced, translating, radial follower's motion is uniform. The total rise is 1.5 in, and the rise occurs over 100˚ of can rotation. The follower dwells for 90˚ of cam rotation prior to the beginning of the rise, and dwells for 80˚ of cam rotation at the end of the rise. The cam will rotate counterclockwise, and the base circle radius is 3 in. Solution: The displacement profile can be easily computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are: For 0 90˚ s=0
- 370 -
For 90˚ 190˚ , s = L where L = 1.5, = 90˚, and = 190˚90˚= 100˚. For 190˚ 270˚ , s = 1.5 For 270˚ 360˚ s = L1 where L = 1.5, = 270˚, and = 360˚270˚= 90˚ The displacement diagram is given below followed by a table of values for at 20˚ increments of .
Theta 0.000 20.000 40.000 60.000 80.000 100.000 120.000 140.000 160.000 180.000 200.000 220.000 240.000 260.000 280.000 300.000 320.000 340.000 360.000
Follower
0.000 0.000 0.000 0.000 0.000 0.150 0.450 0.750 1.050 1.350 1.500 1.500 1.500 1.500 1.333 1.000 0.667 0.333 0.000
- 371 -
Angle
We will lay out only the rise portion of the cam. To lay out the cam, first draw the base circle which has a radius of 3". Then lay off radial lines at 20˚ increments and label the lines in the clockwise direction. Draw circle arcs corresponding to the two dwells and lay off the distances for the other displacements along the other radial lines. Then lay off the distances from the base circle along the radial lines. Draw a line perpendicular to the offset lines at the indicated distance from the base circle. Finally fit a smooth curve tangent to the positions of the follower face indicated by the perpendicular lines. The dwells and rise part of the resulting cam are shown below.
260˚ 240˚
280˚
1 inch
220˚
200˚
180˚
0˚
20˚
160˚
40˚
140˚ 120˚
60˚
100˚ 80˚
- 328 -
Problem 8.23 In the sketch shown, the disk cam is used to position the radial flat-faced follower in a computing mechanism. The cam profile is to be designed to give a follower displacement S for a counterclockwise cam rotation according to the function S = k2 starting from dwell. For 60˚ of cam rotation from the starting position, the lift of the follower is 1.0 cm. By analytical methods, determine the distances R and L when the cam has been turned 45˚ from the starting position. Also calculate whether cusps in the cam profile would occur in the total rotation of 60˚.
Point of Contact
Spring
Starting Position
S
R θ
ω
L
Dwell 2.5 cm radius
Solution: Find k first given the values of S and at the given point. Note that must be converted to radians to ensure consistent units. S = k2 Therefore, k = S2 = 1 2 = 0.9119 [ / 3] Now, R = rb + S = rb + k 2 = 2.5+ 0.9119 2 Where rb is the radius of the base circle. Also, L = dR = 2k = 2(0.9119) = 1.8238 d At = , 4
and
2
( ) = 3.063 cm
R = 2.5+ 0.9119 2 = 2.5 + 0.9119 4
- 329 -
()
L = 1.8238 = 1.8238 = 1.432 cm 4 Cusps will not occur if
or
2 rb + S + d S 0 d 2
rb + S + 2k 0 Because, (2.5+ S + 1.18238 = 3.6824 + S) 0 and S is positive for all , there are no cusps for any value of . Problem 8.24 Determine the cam profile assuming that the translating cylindrical-faced follower starts from a dwell from 0˚ to 80˚, and the cam rotates clockwise. The rise occurs with cycloidal motion during the cam rotation from 80˚ to 180˚. The follower then dwells for 60˚ of cam rotation, and the return occurs with simple harmonic motion for the cam rotation from 240˚ to 360˚. The amplitude of the follower translation is 3 cm, and the follower radius is 0.75 cm. The base circle radius is 5 cm, and the offset is 0.5 cm. Solution: The displacement profile can be computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations (displacement and first and second derivatives) are: For 0 80˚ y=0 For 80˚ 180˚ y = L 1 sin 2
2 y' = L 1 cos 2
y"= 2L2 sin 2 where L = 3, = 80˚, and = 180˚80˚= 100˚. For 180˚ 240˚ y=3 For 240˚ 360˚
- 330 -
y = L 1+ cos 2 y' = L sin 2 2
y"= L cos 2 where L = 3, = 240˚, and = 360˚240˚= 120˚ The cam which will generate this follower displacement can be determined using the equations given in Table 8.8. These equations have been programmed in the MATLAB program rf_cam.m. The program calls an m-file called follower.m which contains the equations for the follower displacement. This file is given as follows: function [f]= follower(tt,rise) % % This function determines the follower displacement and derivatives % for a full rotation cam. The routine is set up for the displacement % schedule in Problem 8.24. % The input values are: %theta %rise % % % % %
= cam angle (deg) = maximum follower displacement
The results are returned in the variable f where f(1) is displacement, f(2) is the derivative of the displacement respect to theta, and f(3) is the second derivative with to theta. The derivatives are not used in this problem, required by the program rf.cam.m.
the with respect but they are
% find the correct interval. fact=pi/180; theta=tt*fact; if tt < 80 f(1)=0; f(2)=0; f(3)=0; end if tt >= 80 & tt < 180 beta=100*fact; theta=(tt-80)*fact; f(1)=rise*((theta/beta)-(1/(2*pi))*sin(2*pi*theta/beta)); f(2)=(rise/beta)*(1-cos(2*pi*theta/beta)); f(3)=(2*rise*pi/beta^2)*sin(2*pi*theta/beta); end if tt>=180 & tt< 240 f(1)=rise; f(2)=0; f(3)=0; end if tt >= 240 & tt 2L 4L2
(7.24.1)
For the second part of the rise region 2 4L + rb > L 1 2 1 2
(7.24.2)
and for the return region, 2
rb > L 1+ cos + L cos 2 2
(7.24.3)
To find the minimum value of the base circle for the rise and return, we must compute the value of rb for all in the rise and return ranges. This is most easily done with a program such as MATLAB. When this is done, the minimum value in the rise region is 5.3755 cm and the minimum value in the return region is 0.375 cm. Therefore, the minimum base radius to be used is 5.3755 cm. The cam which will generate this follower displacement can be determined using the equations given in Table 8.9, and these equations have also been programmed in the MATLAB program ff_cam.m. The program calls an m-file called follower.m which contains the equations for the follower displacement. This file is given as follows: function [f]= follower(tt,rise) % This function determines the follower displacement and derivatives % for a full rotation cam. The routine is set up for the displacement % schedule in Problem 8.27. % The input values are: %theta %rise % % % %
= cam angle (deg) = maximum follower displacement
The results are returned in the variable f where f(1) is the displacement, f(2) is the derivative of the displacement with respect to theta, and f(3) is the second derivative with respect to theta.
% find the correct interval. fact=pi/180; theta=tt*fact; if tt < 80 f(1)=0; f(2)=0; f(3)=0; end if tt >= 80 & tt < 130 beta=100*fact; theta=(tt-80)*fact; f(1)=rise*2*(theta/beta)^2; f(2)=(4*rise/beta)*(theta/beta); f(3)=4*rise/beta;
- 338 -
end if tt >= 130 & tt < 180 beta=100*fact; theta=(tt-80)*fact; f(1)=rise*(1-2*(1-theta/beta)^2); f(2)=(4*rise/beta)*(1-theta/beta); f(3)=-4*rise/beta; end if tt>=180 & tt< 240 f(1)=rise; f(2)=0; f(3)=0; end if tt >=240 & tt =100 & tt =190 & tt= 270 beta=90*fact; theta=(tt-270)*fact; ff=pi*theta/beta; f(1)=(rise/2)*(1+cos(ff)); f(2)=-pi*rise/(2*beta)*sin(ff); f(3)=-(rise/2)*(pi*beta)^2*cos(ff); end
- 351 -
The basic program input is specified in the following. Default values are used for noncritical values. Cam Synthesis for Oscillating Flat-Faced Follower Enter 1 for file input and 2 for interactive input [1]: 2 Enter input file name (off_camio.dat): prob8p32.dat Enter base circle radius [2]: 5 Enter distance between fixed pivots [6]: 8 Length of follower face [1.5*r1]: Enter follower offset [0.5]: 0.5 Enter follower rise (deg) [15]: 20 Enter cam rotation direction (CW(-), CCW(+)) [-]: + Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 352 -
Problem 8.33 Solve Problem 8.32 if the cam rotates clockwise. Solution: Everything is the same except for the direction of motion of the cam. The basic program input is specified in the following. Default values are used for noncritical values. Cam Synthesis for Oscillating Flat-Faced Follower Enter 1 for file input and 2 for interactive input [1]: 2 Enter input file name (off_camio.dat): prob8p33.dat Enter base circle radius [2]: 5 Enter distance between fixed pivots [6]: 8 Length of follower face [1.5*r1]: Enter follower offset [0.5]: 0.5 Enter follower rise (deg) [15]: 20 Enter cam rotation direction (CW(-), CCW(+)) [-]: Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
- 353 -
Problem 8.34 Design the cam system assuming that an oscillating, flat-faced follower starts from a dwell for 0˚ to 100˚ of cam rotation and the cam rotates counterclockwise. The rise occurs with uniform motion during the cam rotation from 100˚ to 200˚. The follower then dwells for 40˚ of cam rotation, and the return occurs with parabolic motion for the cam rotation from 240˚ to 360˚. The oscillation angle is 20˚. Solution: First establish the equations for the displacement profile. The displacement profile can be computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are: For 0 100˚ y=0 For 100˚ 200˚
- 354 -
y = L y' = L y"= 0 where L = 20˚, = 100˚ , and = 200˚100˚= 100˚. For 200˚ 240˚ y=20˚ For the first part of the return, 240˚ 300˚, and 2 y = L 1 2
y' = 4L y"= 4L2 where L = 20˚, = 240˚, and = 360˚240˚= 120˚ For the second part of the return, 300˚ 360˚, and y = 2L1
2
y' = 4L 1 y"= 4L2 where L = 20˚, = 240˚, and = 360˚240˚= 120˚ The cam which will generate this follower displacement can be determined using the equations given in Tables 8.12 and 8.13 once the cam base radius, distance between cam and follower pivots, the offset distance, and the length of the follower are known.. The program calls the m-file called o_follower3.m which contains the equations for the follower displacement. This file is given as follows: function [f]= o_follower3(tt,rise) % This function determines the follower displacement and derivatives % for a full rotation cam. The routine is set up for the displacement % schedule in Problem 8.34. The input values are:
- 355 -
%theta = cam angle (deg) %rise = maximum follower rotation % The results are returned in the variable f where f(1) is the % displacement, f(2) is the derivative of the displacement with % respect to theta, and f(3) is the second derivative with respect % to theta. % find the correct interval. fact=pi/180; theta=tt*fact; if tt =100 & tt =200 & tt=240 & tt=300 beta=120*fact; theta=(tt-240)*fact; ff=theta/beta; f(1)=2*rise*(1-ff)^2; f(2)=-(4*rise/beta)*(1-ff); f(3)=(4*rise/beta); end
Before running the program, we need to establish values for the base radius and the distance between pivots. The offset distance is arbitrarily taken as 0. One option is to experiment with the program orf_cam and determine the types of values which will work. It will be found that the base circle must be fairly large to avoid cusps. Similarly, the distance between pivots must be fairly large to avoid interference with the cam and follower pivot. One set of values which will work is a base radius of 3 inches and a distance between pivots of 5 inches. These values and the others given in the problem can be input into the program to determine the cam geometry. The basic program input is as follows. Default values are used for noncritical values. Cam Synthesis for Oscillating Flat-Faced Follower Enter 1 for file input and 2 for interactive input [1]: 2 Enter input file name (off_camio.dat): prob8p34.dat Enter base circle radius [2]: 3 Enter distance between fixed pivots [6]: 5 Length of follower face [1.5*r1]:
- 356 -
Enter Enter Enter Enter
follower offset [0.5]: 0 follower rise (deg) [15]: 20 cam rotation direction (CW(-), CCW(+)) [-]: + angle increment for design (deg) [10]: 0.5
The graphical results from the program are given in the following three plots. The displacement diagram does not show the acceleration "spikes" at the beginning and end of the dwell; however, the discontinuities in the radius of curvature are indicated. This cam would perform poorly in a high speed application.
- 357 -
Problem 8.35 Design the cam system assuming that an oscillating, flat-faced follower starts from a dwell for 0˚ to 50˚ of cam rotation and the cam rotates clockwise. The rise occurs with cycloidal motion during the cam rotation from 50˚ to 200˚. The follower then dwells for 90˚ of cam rotation, and the return occurs with harmonic motion for the cam rotation from 290˚ to 360˚. The oscillation angle is 25˚. Solution: First establish the equations for the displacement profile. The displacement profile (along with the first and second derivatives) can be computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are: For 0 50˚ y=0 For 50˚ 200˚ y = L 1 sin 2
2 y' = L 1 cos 2
y"= 2L2 sin 2 where L = 25, = 50˚, and = 200˚50˚= 150˚. For 200˚ 290˚ y=25 For 290˚ 360˚
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y = L 1+ cos 2 y' = L sin 2 2
y"= L cos 2 where L = 25, = 290˚, and = 360˚290˚= 70˚ The cam which will generate this follower displacement can be determined using the equations given in Tables 8.12 and 8.13 once the cam base radius, distance between cam and follower pivots, and the offset distance. The length of the follower must be specified, but its value is not critical as long as it is long enough for the follower to remain tangent to the cam. The default value in the program can be used. The program calls an m-file called o_follower4.m which contains the equations for the follower displacement. This file is given as follows: function [f]= o_follower(tt,rise) % This function determines the follower displacement and derivatives % for a full rotation cam. The routine is set up for the displacement % schedule in Problem 8.35. The input values are: %theta = cam angle (deg) %rise = maximum follower rotation % The results are returned in the variable f where f(1) is the % displacement, f(2) is the derivative of the displacement with % respect to theta, and f(3) is the second derivative with respect % to theta. % find the correct interval. fact=pi/180; theta=tt*fact; if tt < 50 f(1)=0; f(2)=0; f(3)=0; end if tt >= 50 & tt =200 & tt< 290 f(1)=rise; f(2)=0; f(3)=0; end if tt >= 290 beta=70*fact; theta=(tt-290)*fact; f(1)=(rise/2)*(1+cos(pi*theta/beta)); f(2)=-(pi*rise/(2*beta))*sin(pi*theta/beta); f(3)=-(rise/2)*(pi/beta)^2*cos(pi*theta/beta); end
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Before running the program, we need to establish values for the base radius, the distance between pivots, and the offset distance. One option is to experiment with the program off_cam and determine the types of values which will work. It will be found that the base circle must be fairly large to avoid cusps. Similarly, the distance between pivots must be fairly large to avoid interference with the cam and follower pivot. One set of values which will work is a base radius of 2 inches and a distance between pivots of 4 inches. The offset distance is arbitrarily taken as 0. These values and the others given in the problem can be input into the program to determine the cam geometry. The program input is as follows. Default values are used for noncritical values. Cam Synthesis for Oscillating Flat-Faced Follower Enter 1 for file input and 2 for interactive input [1]: 2 Enter input file name (off_camio.dat): prob8p35.dat Enter base circle radius [2]: 2 Enter distance between fixed pivots [6]: 4 Length of follower face [1.5*r1]: Enter follower offset [0.5]: 0 Enter follower rise (deg) [15]: 25 Enter cam rotation direction (CW(-), CCW(+)) [-]: Enter angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
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Problem 8.36 Determine the cam profile assuming that the translating knife-edged follower starts from a dwell from 0˚ to 80˚ and rotates clockwise. The rise occurs with cycloidal motion during the cam rotation from 80˚ to 180˚. The follower then dwells for 60˚ of cam rotation, and the return occurs with simple harmonic motion for the cam rotation from 240˚ to 360˚. The amplitude of the follower translation is 4 cm. The base circle radius is 5 cm, and the offset is 0.5 cm. Solution: We can treat a knife-edged radial follower as a roller follower with a roller radius of zero. However, we must first establish the equations for the displacement profile. The displacement profile (along with the first and second derivatives) can be computed using the equations in Chapter 8 in a spreadsheet or MATLAB program. The profile equations are: For 0 80˚
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y=0 For 80˚ 180˚ 1 2
y = L sin 2 y' =
L 2 1 cos
y"=
2L 2 sin 2
where L = 4, = 80˚, and = 180˚80˚= 100˚. For 180˚ 240˚ y=5 For 240˚ 360˚ y=
L 1 + cos 2
y' =
L sin 2 2
L y"= cos 2 where L = 5, = 240˚, and = 360˚240˚= 120˚ The cam which will generate this follower displacement can be determined using the equations given in Table 8.8 if we set r0 = 0. The equations are coded in the program called rf_cam. The program calls an m-file called follower.m which contains the equations for the follower displacement. This file is given as follows: function [f]= follower(tt,rise) % This function determines the follower displacement and derivatives % for a full rotation cam. The routine is set up for the displacement % schedule in Problem 8.36. The input values are: %theta = cam angle (deg) %rise = maximum follower rotation % The results are returned in the variable f where f(1) is the % displacement, f(2) is the derivative of the displacement with % respect to theta, and f(3) is the second derivative with respect % to theta. % find the correct interval. fact=pi/180;
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theta=tt*fact; if tt < 80 f(1)=0; f(2)=0; f(3)=0; end if tt >= 80 & tt =180 & tt< 240 f(1)=rise; f(2)=0; f(3)=0; end if tt >= 240 beta=120*fact; theta=(tt-240)*fact; f(1)=(rise/2)*(1+cos(pi*theta/beta)); f(2)=-(pi*rise/(2*beta))*sin(pi*theta/beta); f(3)=-(rise/2)*(pi/beta)^2*cos(pi*theta/beta); end
Before running the program, we need to establish values for the base radius, the distance between pivots, and the offset distance. One option is to experiment with the program orf_follower and determine the types of values which will work. It will be found that the base circle must be fairly large to avoid cusps. Similarly, the follower pivot must be fairly large to avoid interference with the cam and follower pivot. One set of values which will work is a base radius of 2 inches and a distance between pivots of 4 inches. The offset distance is arbitrarily taken as 0. These values and the others given in the problem can be input into the program to determine the cam geometry. The program input is as follows: Cam Synthesis for Axial Roller Follower Enter Enter Enter Enter Enter Enter Enter Enter
1 for file input and 2 for interactive input [1]: 2 input file name (rf_camio.dat): prob8p36.dat base circle radius [2]: 5 radius of cylindrical or roller follower [0.5]: 0 follower offset [1]: 0 follower rise [2]: 5 cam rotation direction (CW(-), CCW(+)) [-]: cam angle increment for design (deg) [10]: 1
The graphical results from the program are given in the following three plots.
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Problem 8.37 A radial flat-faced follower is to move through a total displacement of 20 mm with harmonic motion with the cam rotates through 30˚. Find the minimum radius of the base circle that is necessary to avoid cusps. Solution: The displacement profile can be computed using the equations in Chapter 8. The profile equation for the rise is_ For 0 30˚ y=
L 1 cos 2
y'=
L sin 2 2
L y"= cos 2 where L = 20 mm and = 30˚ . Cusps do not occur when rb > f ( ) f" ( ) The minimum base circle radius is 2
rb = L 1 cos L cos 2 2 2 L L = + 1 cos
2 2 2 = L + L (35) cos6 = 18L cos6 = L + L 1
cos 2 2 /6 /6 2 2
( )
The value on the right is a maximum when = 30˚ . Then, the minimum base circle radius is rb = 18L = 18(20) = 360 mm
Problem 8.38 A radial flat-faced follower is to move through a total displacement of 3 in with cycloidal motion with the cam rotates through 90˚. Find the minimum radius of the base circle that is necessary to Solution: - 365 -
The displacement profile can be computed using the equations in Chapter 8. The profile equation for the rise is: For 0 90˚ 1 2 y=L sin 2 y'=
L 2 1 cos
y"=
2L 2 sin 2
where L = 3 in and = 90˚ = / 2 . Cusps do not occur when rb > f ( ) f" ( ) The minimum base circle radius is 1 2 2L 2 sin 2 sin 2 L L 2 2L 2 = + sin 2 sin 2 2L L 8L 2 15 = + sin4 sin4 = L sin4 2 2
rb = L
[
]
The value on the right is a maximum when
drb = 0 . Then, d
drb 2 60 L =L cos4 = [2 30cos 4 ] = 0 d 2
[
]
or cos 4 = 2 / 30 = 1/15. or 1 = acos(1/15) = 21.54˚, 66.54˚ = 0.376 rad, 1.161 rad 4 Substituting in the two values: 15 sin86.16 = 6.42 [ 2 152 sin 4 ] = 3 2(0.376) 2 2 15 2(1.161) 15 = L[ sin4 ] = 3 sin 266.16 = 9.363 in 2 2
rb1 = L rb 2
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Therefore, the minimum base circle radius required is rb = 9.363 in
Problem 8.39 A radial roller follower is to move through a total displacement of L=19 mm with harmonic motion while the cam rotates 60˚. The roller radius is 5 mm. Use the program supplied with the book and find the minimum radius necessary to avoid cusps during the interval indicated. Solution: The displacement profile can be computed using the equations in Chapter 8. When 0 60˚ y=
L 1 cos 2
y'=
L sin 2
y"=
L cos 2
2
Now, L = 19 mm and
=
3
To determine if there are cusps, run the program CAM2 under MAINMENU. Substitute in a base circle radius of 0, the minimum base circle radius and run the analysis. From this it will be apparent that there are no cusps for any base circle radius.
Problem 8.40 A radial roller follower is to move through a total displacement of L=45 mm with cycloidal motion. The roller radius is 5 mm, and the cam rotates 90 degrees during the rise. Use the program supplied with the book and find the minimum radius necessary to avoid cusps during the interval. Solution:
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The displacement profile can be computed using the equations in Chapter 8. When 0 90˚ 1 2 y=L sin 2 y'=
L 2 1 cos
y"=
2L 2 sin 2
where L = 45 mm and = 90˚ = / 2 _ To determine if there are cusps, run the program CAM2 under MAINMENU. Substitute in a base circle radius of 0, the minimum base circle radius and run the analysis. From this it will be apparent that there are no cusps for any base circle radius. Problem 8.41 Assume that a flat-faced translating follower is used with the displacement schedule in Problem 8.10. Determine if a cusp is present at = 60˚. Solution: When = 60˚ =
_ 3
1 1 3 1 y = sin 2( 3) = sin(2 3) = 0.195 in 3 2 2
(
y˙ =
)
10.472 (1 cos2 ) = (1 cos(2 3)) = 5.000 in. sec .
()
y˙˙ = 2
2
sin2 = 2
(
10.472
2
) sin(2 3) = 60.46 secin.
2
To avoid a cusp, rb > f ( ) f" ( ) Now f ( ) = 0.195 and
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f "( ) = 2
1
2
( ) sin2 = (2 ) sin(2 3) = 0.551
Therefore, rb > 0.195 0.551 or rb > 0.746 Therefore, any base radius will work and there will be no cusp. Problem 8.42 Assume that a flat-faced translating follower is used with the displacement schedule in Problem 8.12. Determine if a cusp is present at = 90˚. Solution: When, = 90˚ = _ 2 f ( ) = L
y'=
1 1 1 2 2 1 sin sin 2( 2) = sin = 0.5 in =1 2 2 2 2
(
2 L L 1 cos = (1 cos2 )
f "( ) =
2L 2 sin2 = sin = 0
To avoid a cusp, rb > f ( ) f" ( ) Now f ( ) = 0.5 and f "( ) = 0 Therefore, rb > 0.5 0 or
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)
rb > 0.5 Therefore, any base radius will work and there will be no cusp.
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