67
PART II
DESIGN OF MECHANISMS
69
Chapter 5
Cam Design 5.1
The reciprocating radial roller follower of a plate cam is to rise 50 mm with simple harmonic motion in 180D of cam rotation and return with simple harmonic motion in the remaining 180° . If the roller radius is 10 mm and the prime-circle radius is 50 mm, construct the displacement diagram, the pitch curve, and the cam profile for clockwise cam rotation.
5.2
A plate cam with a reciprocating flat-face follower has the same motion as in Problem 5.1. The prime-circle radius is 50 mm, and the cam rotates counterclockwise. Construct the displacement diagram and the cam profile, offsetting the follower stem by 20 mm in the direction that reduces the bending stress in the follower during rise.
70
5.3
Construct the displacement diagram and the cam profile for a plate cam with an oscillating radial flat-face follower that rises through 30° with cycloidal motion in 150° of counterclockwise cam rotation, then dwells for 30D , returns with cycloidal motion in 120° , and dwells for 60° . Determine the necessary length for the follower face, allowing 5 mm clearance at each end. The prime-circle radius is 30 mm, and the follower pivot is 120 mm to the right.
Notice that, with the prime circle radius given, the cam is undercut and the follower will not reach positions 7 and 8. The follower face length shown is 200 mm but can be made as short as 195 mm (position 9)from the follower pivot. Ans.
71
5.4
A plate cam with an oscillating roller follower is to produce the same motion as in Problem 5.3. The prime-circle radius is 60 mm, the roller radius is 10 mm, the length of the follower is 100 mm, and it is pivoted at 125 mm to the left of the cam rotation axis. The cam rotation is clockwise. Determine the maximum pressure angle.
From a graphical analysis, ϕ max = 39° 5.5
Ans.
For a full-rise simple harmonic motion, write the equations for the velocity and the jerk at the midpoint of the motion. Also, determine the acceleration at the beginning and the end of the motion. Using Eqs. (5.18) we find ⎛θ 1⎞ πL π πL y′ ⎜ = ⎟ = sin = 2 2β ⎝ β 2 ⎠ 2β
Ans.
⎛θ 1⎞ π 3L π π 3L y′′′ ⎜ = ⎟ = − sin = − 2 2β 3 2β 3 ⎝ β 2⎠
Ans.
⎛θ ⎞ π 2L π 2L y′′ ⎜ = 0 ⎟ = cos 0 = 2β 2 ⎝β ⎠ 2β 2
Ans.
⎛θ ⎞ π 2L π 2L y′′ ⎜ = 1⎟ = cos π = − 2β 2 ⎝β ⎠ 2β 2
Ans.
72
5.6
For a full-rise cycloidal motion, determine the values of θ for which the acceleration is maximum and minimum. What is the formula for the acceleration at these points? Find the equations for the velocity and the jerk at the midpoint of the motion. Using Eqs. (5.19), we know that acceleration is an extremum when jerk is zero. This happens when cos 2πθ β = 0 or when θ β = 1 4 or θ β = 3 4 .
5.7
⎛ θ 1 ⎞ 2π L π 2π L ′′ y′′ ⎜ = ⎟ = sin = = ymax 2 β2 ⎝ β 4⎠ β2
Ans.
⎛ θ 3 ⎞ 2π L 3π 2π L ′′ =− = ymin sin y′′ ⎜ = ⎟ = 2 2 β 4 2 β β ⎝ ⎠
Ans.
⎛θ 1⎞ L 2L y′ ⎜ = ⎟ = (1 − cos π ) = β ⎝ β 2⎠ β
Ans.
⎛ θ 1 ⎞ 4π 2 L 4π 2 L ′′′ y ⎜ = ⎟= cos π = − β3 β3 ⎝ β 2⎠
Ans.
A plate cam with a reciprocating follower is to rotate clockwise at 400 rev/min. The follower is to dwell for 60° of cam rotation, after which it is to rise to a lift of 62.5 mm. During 25 mm of its return stroke, it must have a constant velocity of 1000 mm/s. Recommend standard cam motions from Sec. 5.7 to be used for high speed operation and determine the corresponding lifts and cam rotation angles for each segment of the cam. The curves shown are initially only sketches and not drawn to scale. They suggest the standard curve types which might be chosen. The actual choices are shown in the table below.
ω=
( 400 rev/min )( 2π rad/rev ) ( 60 s/min )
= 41.888 rad/s cw To match the given velocity condition in Seg. DE we must have y 4 = y4′ω 1000 mm/s = y4′ ( 41.888 rad/s )
y4′ = − L4 β 4 = −23.87 mm/rad − ( 25 mm ) β 4 = −23.87 mm/rad
−2 L5 β 5 = y4′ = −0.954 930 in/rad
β 4 = 1.047339 rad = 60° Matching the first derivatives at D and E we find L5 = 11.9366β 5 (1)
73
−π L3 ( 2 β 3 ) = y4′ = −0.954 930 in/rad
L3 = 15.1981β 3
(2)
Matching the second derivatives at C we find −131.707 ( 62.5 ) β 22 = −π 2 L3 4β 32
β 22 = 133.4476β32 L3
(3)
For geometric continuity, we have L1 + L2 = L3 + L4 + L5 or L3 + L5 = 37.5 mm (4) β1 + β 2 + β 3 + β 4 + β 5 = 2π or β 2 + β 3 + β 5 = 4.188 790 rad (5) Equations (1) to (5) are now solved simultaneously for β 2 , L3 , β 3 , L5 , and β 5 . The results are summarized in the following table: Seg. Type Eq. L, mm β, rad β, deg AB dwell --0 1.047 198 60.000 BC 8th order poly. (5.20) 62.5 1.083 747 62.094 CD half harmonic (5.26) 2.03 0.133 763 7.664 DE constant velocity --25 1.047 198 60.000 EA half cycloidal (5.31) 35.47 2.971 280 170.242 5.8
Repeat Problem 5.7 except with a dwell that is to be for 20° of cam rotation. The procedure is the same as for Problem 5.7. The results are: Seg. Type Eq. L, mm β, rad AB dwell --0 0.349 066 th BC 8 order poly. (5.20) 62.5 1.852 072 CD half harmonic (5.26) 5.93 0.390 658 DE constant velocity --25 1.047 198 EA half cycloidal (5.31) 31.56 2.644 191
5.9
β, deg 20.000 106.116 22.383 60.000 151.501
If the cam of Problem 5.7 is driven at constant speed, determine the time of the dwell and the maximum and minimum velocity and acceleration of the follower for the cam cycle. The time duration of the dwell is Δt = β1 ω = 1.047 198 rad 41.888 rad/s = 0.025 s Ans. Working from the equations listed, the maximum and minimum values of the derivatives in each segment of the cam are as follows: Seg. Eq. ′ ′ ′′ ′′ ymax ymin ymax ymin AB --0 0 0 BC (5.20) 102.48 0 280.346 CD (5.26) 0 -23.86 0 DE ---23.87 -23.87 0 EA (5.31) 0 -23.873 12.62 ′ ω = (102.48 mm/rad )( 41.888 rad/s ) = 4.292 m/s y max = ymax
0 -280.346 -280.346 0 0 Ans.
′ ω = ( −23.87 mm/rad )( 41.888 rad/s ) = 1 m/s y min = ymin
Ans.
′′ ω 2 = ( 280.346 mm/rad ) ( 41.888 rad/s ) = 0.4919 m/s 2 ymax = ymax
Ans.
′′ ω 2 = ( −280.346 mm/rad ) ( 41.888 rad/s ) = −0.4919 m/s 2 ymin = ymin
Ans.
2
2
2
2
74
5.10
A plate cam with an oscillating follower is to rise through 20° in 60D of cam rotation, dwell for 45° , then rise through an additional 20° , return, and dwell for 60° of cam rotation. Assuming high-speed operation, recommend standard cam motions from Section 5.7 to be used, and determine the lifts and cam-rotation angles for each segment of the cam. From the sketches shown (not drawn to scale), the curves types identified in the table below were chosen. Next, equating the second derivatives at D, the remaining entries in the table were found. L L −5.26830 32 = −5.26830 42
β3
β4
β L = 4 = 2.000 β L3 2 4 2 3
β 4 = 2β3
(
)
β 3 + β 4 = 1 + 2 β 3 = 195° β 3 = 80.772° β 4 = 114.228° Seg. AB BC CD DE EA 5.11
Type cycloidal dwell 8th order poly. 8th order poly. dwell
Eq. (5.19) --(5.20) (5.23) ---
L, deg 20.000 0 20.000 40.000 0
β, rad 1.047 198 0.785 399 1.409 731 1.993 661 1.047 198
β, deg 60.000 45.000 80.772 114.228 60.000
Determine the maximum velocity and acceleration of the follower for Problem 5.10, assuming that the cam is driven at a constant speed of 600 rev/min.
ω = ( 600 rev/min )( 2π rad/rev ) ( 60 s/min ) = 62.832 rad/s Working from the equations listed, the maximum and minimum values of the derivatives in each segment of the cam are as follows: Seg. Eq. ′ ′ ′′ ′′ ymax ymin ymax ymin AB (5.19) 0.666 667 0 2.000 000 BC --0 0 0 CD (5.20) 0.440 004 0 0.925 344 DE (5.23) 0 -0.622 264 0.925 402 EA --0 0 0 ′ ω = ( 0.666 667 rad/rad )( 62.832 rad/s ) = 41.89 rad/s y max = ymax ′′ ω 2 = ( 2.000 rad/rad )( 62.832 rad/s ) = 7 896 rad/s 2 ymax = ymax 2
-2.000 000 0 -0.924 344 -0.925 402 0 Ans. Ans.
75
5.12
The boundary conditions for a polynomial cam motion are as follows: for θ = 0 , y = 0 , and y ′ = 0 ; for θ = β , y = L, and y′ = 0 . Determine the appropriate displacement equation and the first three derivatives of this equation with respect to the cam rotation angle. Sketch the corresponding diagrams.
Since there are four boundary conditions, we choose a cubic polynomial
( β) ( β) ( β) 3C C 2C θ θ y′ = β+ β ( β )+ β( β) y = C0 + C1 θ
+ C2 θ
2
+ C3 θ
3
2
1
2
3
Then from the boundary conditions:
( β = 0) = C = 0 C y′ ( θ = 0 ) = β β =0 y ( θ = 1.0 ) = C + C = L β 3C 2C y′ ( θ = 1.0 ) = β β+ β =0 y θ
0
1
2
3
3
2
C0 = 0 C1 = 0 C2 = 3 L C3 = −2 L
Therefore the equation and its three derivatives are: 2 3 2 3 ⎡ ⎤ y θ = 3L θ − 2L θ = L ⎢3 θ −2 θ ⎥ β β β β β ⎣ ⎦ 2 2 ⎡ ⎤ = 6L θ − 6L θ = 6L ⎢ θ − θ y′ θ β β β β β β⎣ β β ⎥⎦ 6L 6L ⎡ θ ⎤ 12 L 2 θ y′′ θ β = β2 − β = β 2 ⎢⎣1 − 2 β ⎥⎦ β
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) y′′′ ( θ ) = − 12L β β 3
5.13
Ans. Ans. Ans.
Ans.
Determine the minimum face width using 2.5 mm allowances at each end, and determine the minimum radius of curvature for the cam of Problem 5.2. Referring to Problem 5.2 for the data and figure, L = 50 mm β = 180° = π rad R0 = 50 mm From Eqs. (5.18) and (5.21) for simple harmonic motion, ′ = π L ( 2 β ) = 25 mm ′ = −π L ( 2 β ) = −25 mm ymax ymax From Eq. (5.35) ′ + allowances Face width=y′max − ymin
Face width= ( 25 mm ) − ( −25 mm ) + 2 ( 2.5 m ) = 55 mm
From Eq. (5.33)
Ans.
76
ρ = R0 + y + y′′ L⎛ L πθ ⎞ ⎛ L πθ ⎞ ⎜ 1 − cos ⎟ + ⎜ cos ⎟ = R0 + (constant) β ⎠ ⎝2 β ⎠ 2⎝ 2 ρ = ( 50 mm ) + ( 50 mm ) 2 = 75 mm = R0 +
5.14
Ans.
Determine the maximum pressure angle and the minimum radius of curvature for the cam of Problem 5.1. Referring to Problem 5.1 for the data and figure, L = 50 mm β = 180° = π rad R0 = 50 mm Rr = 10 mm For simple harmonic motion, Eq. (5.18) can be substituted into Eq. (5.43) to give sin θ . This can be differentiated and dφ dθ set to zero to find the angle tan φ = 3 − cos θ θ = 70.53° at which φmax = 19.47° . However, it is much simpler to use the nomogram of Fig. 5.28 we find φmax = 20° directly. For the accuracy needed, the nomogram is considered sufficient. Ans. From Fig. 5.30a, using R0 L = 1.0 , we get ( ρ min + Rr ) R0 = 1.43 . This gives
ρ min = 1.43R0 − Rr = 1.43 ( 50 mm ) − (10 mm ) = 61.5 mm 5.15
Ans.
A radial reciprocating flat-face follower is to have the motion described in Problem 5.7. Determine the minimum prime-circle radius if the radius of curvature of the cam is not to be less than 12.5 mm. Using this prime-circle radius, what is the minimum length of the follower face using allowances of 3.75 mm on each side? From
Problem
P5.9,
′ = 102.48 mm/rad , ymax
′ = −23.87 mm/rad , ymin
′′ = −280.346 mm/rad 2 ymin Therefore, from Eq. (5.34), ′′ − y = (12.5 mm ) − ( −280.346 mm ) − ( 62.5 mm ) = 230.34 mm R0 > ρ min − ymin
Ans.
Also, from Eq. (5.35), ′ − ymin ′ + allowances = (102.48 mm ) − ( −23.87 mm ) + 2 ( 3.75 mm ) = 133.85 mm Face width = ymax Ans. 5.16
Graphically construct the cam profile of Problem 5.15 for clockwise cam rotation.
77
5.17
A radial reciprocating roller follower is to have the motion described in Problem 5.7. Using a prime-circle radius of 500 mm, determine the maximum pressure angle and the maximum roller radius that can be used without producing undercutting. We will use the nomogram of Fig. 5.28 to find the maximum pressure angle in each segment of the cam. Calculations are shown in the following table. Asterisks are used to signify values used with the nomogram to adjust half-return curves to equivalent full return curves, and to adjust baseline. Seg. φmax , deg R0* , mm L* , mm R0* L* β * , deg BC 500 62.5 CD 558.435 4.065 EA 500 70.935 For the total cam, φmax = 12 deg
8.0 137.4 7.0
62.1 15.3 340.5
12 1 3 Ans.
Also we use Figs. 5.32 and 5.33 to check for undercutting. Again, asterisks are used to denote values that are adjusted for use with the charts. Note that doubling as was done for use of the nomogram is not necessary since we do have figures for half-harmonic and half-cycloidal cam segments. Note also that segment EA need not be checked since undercutting only occurs in segments with negative acceleration. β , deg L, mm Seg. R0* , mm R0* L Rrmax , ( ρmin + Rr ) R0* mm BC 500 62.5 8.0 62.1 0.725 362.5 CD 560.4675 2.0325 275.8 7.7 0.680 380 To avoid undercutting for the entire cam, Rr < 362.5 mm . 5.18
Ans.
Graphically construct the cam profile of Problem 5.17 using a roller radius of 20 mm. The cam rotation is to be clockwise.
78
5.19
A plate cam rotates at 300 rev/min and drives a reciprocating radial roller follower through a full rise of 75 mm in of cam rotation. Find the minimum radius of the primecircle if simple harmonic motion is used and the pressure angle is not to exceed 25D . Find the maximum acceleration of the follower. Using
φmax = 25°
β = 180° , Fig. 5.28 gives
and
R0 = 0.75 L = 0.75 ( 75 mm ) = 56 mm
ω=
( 300 rev/min )( 2π
′′ = ymax
rad/rev )
60 s/min π 2 ( 0.075 m )
R0 L = 0.75 .
Ans.
= 31.416 rad/s
π 2L = = 0.037 5 m/rad 2 2 2 2β 2 (π rad )
′′ ω 2 = ( 0.037 5 m/rad 2 ) ( 31.416 rad/s ) = 37.0 m/s 2 ymax = ymax
Ans.
Repeat Problem 5.19 except that the motion is cycloidal. Fig. 5.28 gives R0 L = 0.95 . Therefore R0 = 0.95L = 0.95 ( 75 mm ) = 71 mm
Ans.
2
5.20
′′ = ymax
2π L
β
2
=
2π ( 0.075 m )
(π
rad )
2
= 0.047 7 m/rad 2
′′ ω 2 = ( 0.047 7 m/rad 2 ) ( 31.416 rad/s ) = 47.1 m/s 2 ymax = ymax
Ans.
Repeat Problem 5.19 except that the motion is eighth-order polynomial. Fig. 5.28 gives R0 L = 0.95 . Therefore R0 = 0.95L = 0.95 ( 75 mm ) = 71 mm
Ans.
2
5.21
′′ = ymax
5.2683L
β
2
=
5.2683 ( 0.075 m )
(π
rad )
2
= 0.040 0 m/rad 2
′′ ω 2 = ( 0.040 0 m/rad 2 ) ( 31.416 rad/s ) = 39.5 m/s 2 ymax = ymax 2
5.22
Therefore
Ans.
Using a roller diameter of 20 mm, 180° determine whether the cam of Problem 5.19 will be undercut. Using R0 L = 0.75 and β = 180° , Fig. 5.30a gives ( ρ min + Rr ) R0 = 1.55 .
ρ min = 1.55 ( 56 mm ) − ( 20 mm ) = 66.8 mm > 0 ; this cam is not undercut.
Ans.
79
5.23
Equations (5.36) and (5.37) describe the profile of a plate cam with a reciprocating flatface follower. If such a cam is to be cut on a milling machine with cutter radius Rc , determine similar equations for the center of the cutter.
In complex polar notation, using Eq. (5.32) and using u and v to denote the local rectangular part coordinates of the cam shape, the loop closure equation is ue jθ + jve jθ = jR0 + jy + y′ + jRc
Dividing this by e jθ u + jv = j ( R0 + Rc + y ) e − jθ + y′e − jθ
Now separating this into real and imaginary parts we find u = ( R0 + Rc + y ) sin θ + y′ cos θ v = ( R0 + Rc + y ) cos θ − y′ sin θ Ans.
