218 39 Solutions Instructor Manual Chapter 5 Cam Design

8/10/2019 218 39 Solutions Instructor Manual Chapter 5 Cam Design

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PART II

DESIGN OF MECHANISMS


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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 180 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.


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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 30 , 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.


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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, max 39 = Ans.

5.5 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

1sin

2 2 2 2

L Ly

= = =

Ans.

3 3

3 3

1

sin2 22 2

L L

y

= = =

Ans.2 2

2 20 cos 0

2 2

L Ly

= = =

Ans.

2 2

2 21 cos

2 2

L Ly

= = =

Ans.


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5.6 For a full-rise cycloidal motion, determine the values of 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 2 0 = or when 1 4 = or 3 4 = .

max2 2

1 2 2sin

4 2

L Ly y

= = = =

Ans.

min2 2

3 2 3 2sin

4 2

L Ly y

= = = =

Ans.

( )1 2

1 cos2

L Ly

= = =

Ans.

2 2

3 3

1 4 4cos

2

L Ly

= = =

Ans.

5.7 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 thetable below.

( ) ( )

( )

400 rev/min 2 rad/rev

60 s/min

41.888 rad/s cw

=

=

To match the given velocity

condition in Seg.DEwe must have

4 4y y=

( )41000 mm/s 41.888 rad/sy=

4 4 4 23.87 mm/rady L = =

( ) 425 mm 23.87 mm/rad =

4 1.047339 rad 60 = = Matching the first derivatives at D

andEwe find

5 5 42 0.954 930 in/rad L y = = 5 511.9366L = (1)


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( )3 3 42 0.954 930 in/rad L y = = 3 315.1981L = (2)

Matching the second derivatives at Cwe find

( ) 2 2 22 3 3131.707 62.5 4L = 2 2

2 3 3133.4476 L = (3)

For geometric continuity, we have

1 2 3 4 5L L L L L+ = + + or 3 5 37.5 mmL L+ = (4)1 2 3 4 5 2 + + + + = or 2 3 5 4.188 790 rad + + = (5)

Equations (1) to (5) are now solved simultaneously for 2 , 3L , 3 , 5L , and 5 . The

results are summarized in the following table:

Seg. Type Eq. L, mm , rad , degAB 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 , rad , deg

AB dwell --- 0 0.349 066 20.000

BC 8thorder poly. (5.20) 62.5 1.852 072 106.116

CD half harmonic (5.26) 5.93 0.390 658 22.383

DE constant velocity --- 25 1.047 198 60.000

EA half cycloidal (5.31) 31.56 2.644 191 151.501

5.9

If the cam of Problem 5.7 is driven at constant speed, determine the time of the dwell andthe maximum and minimum velocity and acceleration of the follower for the cam cycle.

The time duration of the dwell is1

1.047 198 rad 41.888 rad/s 0.025 st = = = 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.maxy miny maxy miny

AB --- 0 0 0 0

BC (5.20) 102.48 0 280.346 -280.346

CD (5.26) 0 -23.86 0 -280.346

DE --- -23.87 -23.87 0 0EA (5.31) 0 -23.873 12.62 0

( ) ( )max max 102.48 mm/rad 41.888 rad/s 4.292 m/sy y = = = Ans.

( )( )min min 23.87 mm/rad 41.888 rad/s 1 m/sy y = = = Ans.

( ) ( )2 22 2

max max 280.346 mm/rad 41.888 rad/s 0.4919 m/sy y = = = Ans.

( ) ( )2 22 2

min min 280.346 mm/rad 41.888 rad/s 0.4919 m/sy y = = = Ans.


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5.10 A plate cam with an oscillating follower is to rise through 20 in 60 of cam rotation,dwell for 45 , then rise through an additional 20 , return, and dwell for 60 of camrotation. 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.

3 4

2 2

3 4

5.26830 5.26830L L

=

2

4 4

2

3 3

2.000L

L

= =

4 32 =

( )3 4 31 2 195 + = + = 3 80.772 =

4 114.228 =

Seg. Type Eq. L, deg , rad , degAB cycloidal (5.19) 20.000 1.047 198 60.000

BC dwell --- 0 0.785 399 45.000

CD 8th

order poly. (5.20) 20.000 1.409 731 80.772

DE 8th

order poly. (5.23) 40.000 1.993 661 114.228

EA dwell --- 0 1.047 198 60.000

5.11 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.

( ) ( ) ( )600 rev/min 2 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.maxy miny maxy miny

AB(5.19) 0.666 667 0 2.000 000 -2.000 000BC --- 0 0 0 0

CD (5.20) 0.440 004 0 0.925 344 -0.924 344

DE (5.23) 0 -0.622 264 0.925 402 -0.925 402

EA --- 0 0 0 0

( )( )max max 0.666 667 rad/rad 62.832 rad/s 41.89 rad/sy y = = = Ans.

( ) ( )22 2

max max 2.000 rad/rad 62.832 rad/s 7 896 rad/sy y = = = Ans.


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5.12 The boundary conditions for a polynomial cam motion are as follows: for 0= , 0=y ,and 0=y ; for , , and 0y L y = = = . Determine the appropriate displacementequation 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

( ) ( ) ( )2 3

0 1 2 3y C C C C

= + + +

( ) ( )2

31 232 CC C

y

= + +

Then from the boundary conditions:

( ) 00 0y C= = = 0 0C =

( )10 0Cy = = = 1 0C =

( ) 2 31.0y C C L= = + = 2 3C L=

( ) 32 321.0 0CCy = = + = 3 2C L= Therefore the equation and its three derivatives are:

( ) ( ) ( ) ( ) ( )2 3 2 3

3 2 3 2y L L L

= =

Ans.

( ) ( ) ( ) ( ) ( )2 2

6 6 6L L Ly

= = Ans.

( ) ( ) ( )2 2 26 612 1 2L LLy = =

Ans.

( ) 312Ly = Ans.

5.13 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,

50 mmL= 180 rad = = 0 50 mmR = From Eqs. (5.18) and (5.21) for simple harmonic motion,

( )max 2 25 mmy L = = ( )max 2 25 mmy L = =

From Eq. (5.35)

max minFace width=y allowancesy +

( ) ( ) ( )Face width= 25 mm 25 mm 2 2.5 m 55 mm + = Ans.

From Eq. (5.33)


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0

0 01 cos cos (constant)2 2 2

R y y

L L LR R

= + +

= + + = +

( ) ( )50 mm 50 mm 2 75 mm= + = Ans.

5.14 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,

50 mmL= 180 rad = = 0 50 mmR = 10 mmrR =

For simple harmonic motion, Eq. (5.18) can be substituted into Eq. (5.43) to give

sintan

3 cos

=

. This can be differentiated andd d set to zero to find the angle

70.53= at which max 19.47 = . However, it is much simpler to use the nomogram of

Fig. 5.28 we findmax

20 = directly. For the accuracy needed, the nomogram is

considered sufficient. Ans.

From Fig. 5.30a, using0

1.0R L= , we get ( )min 0 1.43rR R + = . This gives

( ) ( )min 01.43 1.43 50 mm 10 mm 61.5 mmrR R = = = Ans.

5.15 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, max 102.48 mm/rady = , min 23.87 mm/rady = ,2

min 280.346 mm/rady = Therefore, from Eq. (5.34),

( ) ( ) ( )0 min min 12.5 mm 280.346 mm 62.5 mm 230.34 mmR y y > = = Ans.

Also, from Eq. (5.35),

( ) ( ) ( )max minFace width allowances 102.48 mm 23.87 mm 2 3.75 mm 133.85 mmy y = + = + =

Ans.

5.16 Graphically construct the cam profile of Problem 5.15 for clockwise cam rotation.


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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. *0R , mm

*L , mm * *0R L

* , deg max , deg

BC 500 62.5 8.0 62.1 12

CD 558.435 4.065 137.4 15.3 1

EA 500 70.935 7.0 340.5 3

For the total cam, max 12 deg = Ans.

Also we use Figs. 5.32 and 5.33 to check for undercutting. Again, asterisks are used todenote 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.

Seg. *0R , mm L, mm

*

0R L , deg ( )*

min 0rR R + max

rR ,

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, 362.5 mmrR < . Ans.

5.18 Graphically construct the cam profile of Problem 5.17 using a roller radius of 20 mm.

The cam rotation is to be clockwise.


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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 prime-

circle if simple harmonic motion is used and the pressure angle is not to exceed 25 .

Find the maximum acceleration of the follower.

Using max 25 = and 180= , Fig. 5.28 gives 0 0.75R L= . Therefore

( )0 0.75 0.75 75 mm 56 mmR L= = = Ans.

( )( )300 rev/min 2 rad/rev31.416 rad/s

60 s/min

= =

( )

( )

222

max 22

0.075 m0.037 5 m/rad

2 2 rad

Ly

= = =

( )( )22 2 2

max max 0.037 5 m/rad 31.416 rad/s 37.0 m/sy y = = = Ans.

5.20 Repeat Problem 5.19 except that the motion is cycloidal.

Fig. 5.28 gives 0 0.95R L= . Therefore ( )0 0.95 0.95 75 mm 71 mmR L= = = Ans.

( )

( )2

max 22

2 0.075 m20.047 7 m/rad

rad

Ly

= = =

( )( )22 2 2

max max 0.047 7 m/rad 31.416 rad/s 47.1 m/sy y = = = Ans.

5.21 Repeat Problem 5.19 except that the motion is eighth-order polynomial.

Fig. 5.28 gives 0 0.95R L= . Therefore ( )0 0.95 0.95 75 mm 71 mmR L= = = Ans.

( )

( )2

max 22

5.2683 0.075 m5.26830.040 0 m/rad

rad

Ly

= = =

( ) ( )22 2 2max max 0.040 0 m/rad 31.416 rad/s 39.5 m/sy y = = = Ans.

5.22 Using a roller diameter of 20 mm, 180 determine whether the cam of Problem 5.19 willbe undercut.

Using 0 0.75R L= and 180= , Fig. 5.30agives ( )min 0 1.55rR R + = .

( ) ( )min 1.55 56 mm 20 mm 66.8 mm 0 = = > ; this cam is not undercut. Ans.


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5.23 Equations (5.36) and (5.37) describe the profile of a plate cam with a reciprocating flat-

face follower. If such a cam is to be cut on a milling machine with cutter radius cR ,

determine similar equations for the center of the cutter.

In complex polar notation, usingEq. (5.32) and using u and v to

denote the local rectangular part

coordinates of the cam shape, the

loop closure equation is

0

j j

cue jve jR jy y jR + = + + +

Dividing this by je

( )0j j

cu jv j R R y e y e

+ = + + +

Now separating this into real andimaginary parts we find

( )0 sin coscu R R y y = + + +

( )0 cos sincv R R y y = + + Ans.

218 39 Solutions Instructor Manual Chapter 5 Cam Design

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