MER312 Winter 13 Exam III - Solution - Union Collegerbb.union.edu/courses/mer312/Exams/MER312 2013... · MER312:Adv. Dyn. & Kinematics Exam III Union College Winter 2013 Mechanical

MER312:Adv. Dyn. & Kinematics Exam III Union College Winter 2013 Mechanical Engineering

‐1-

As a student at Union College, I am part of a community that values intellectual effort, curiosity and discovery. I understand that in order to truly claim my educational and academic achievements, I am obligated to act with academic integrity. Therefore, I affirm that I carried out the work on this exam with full academic honesty, and I rely on my fellow students to do the same.

For this exam I understand that:

1. I must work alone in writing out the answers to this exam. 2. I cannot copy solutions to these problems from any person or resource. 3. I cannot use any electronic resources, other than the program I wrote as part of this class,

to assist me in the solution to the questions on this exam. 4. I can use one formula sheet (both sides) that I prepared as a reference for this exam, I

must staple this sheet to the back of my exam, and this sheet cannot contain any solutions to problems.

Signature:

Print Name: Solution


‐2-

PROBLEM 1 (30 pts): A box sits on top of the base shown in Position 1 and needs to be

moved to Position 2, and then to Position 3. (Guess 2=-50, 3=-100, 2=-50, and 3=-80)

1a. Using the program that you developed in class, perform an analytical synthesis to design a linkage that will move the box from positions 1 to 2 to 3, and has ground pivots on the base. Show all work need to calculate the parameters used in your computer model below. Staple a copy of the computer solution directly after the next page of this exam.


‐3-

From the figure inputs to the numerical analytical synthesis spreadsheet need to be calculated.

First and Second Dyad

1 1

2 2

3 3

2 2

21

2 2

31

12

13

2

3

, 0 , 0

, 421 , 963

, 184 ,1400

421 0 963 0 1051

184 0 1400 0 1412

963tan 66.4

421

1400tan 82.5

184

27

88

x y

x y

x y

P P mm mm

P P mm mm

P P mm mm

p mm mm mm mm mm

p mm mm mm mm mm

mm

mm

mm

mm

Free Choices

2

3

2

3

50

100

50

80


‐4-

GIVEN: CHOSEN: FIND: x-coord y-coord.P12 1051.00 2 -50.00 w 864.570 O2 -306.819 -402.654P13 1412.00 -100.00 155.216 A1 -1091.757 -40.228 66.40 z 1092.498 A2 -533.733 431.606 82.50 2.110 A3 186.405 307.4242 27.00 W1x -784.938 P1 0.000 0.000

88.00 W1y 362.426 P2 420.767 963.097Z1x 1091.757 P3 184.303 1399.920Z1y 40.228

-0.3572 0.7660 -0.1090 -0.4540 W1x 420.7668 -0.7660 -0.3572 0.4540 -0.1090 W1y 963.0972 -1.1736 0.9848 -0.9651 -0.9994 Z1x 184.3030 -0.9848 -1.1736 0.9994 -0.9651 Z1y 1399.9201

FIRST DYAD

GIVEN: CHOSEN: FIND: x-coord y-coord.P12 1051.00 2 -50.00 u 966.678 O4 122.789 -364.194P13 1412.00 -80.00 163.084 B1 -802.064 -82.925 66.40 s 806.340 B2 -256.231 525.081 82.50 5.903 B3 239.186 595.4502 27.00 U1x -924.853 P1 0.000 0.000 88.00 U1y 281.269 P2 420.767 963.097

S1x 802.064 P3 184.303 1,399.920S1y 82.925

-0.3572 0.7660 -0.1090 -0.4540 U1x 420.7668 -0.7660 -0.3572 0.4540 -0.1090 U1y 963.0972 -0.8264 0.9848 -0.9651 -0.9994 S1x 184.3030 -0.9848 -0.8264 0.9994 -0.9651 S1y 1399.9201

SECOND DYAD


‐5-

1b. Draw the final synthesized mechanism to scale in Position 1.

O4

O2

A1

B1

123

307

403

364

83

40

802

1092


‐6-

PROBLEM 2 (30 pts): A container of liquid sit on top of a base (A1B1) and is required to move to the position on the lower step ( A3B3).

Above is an illustration of the graphical two positions synthesis. This configuration 1) insures that the orientation of the container remains the same throughout the motion from the top of the base to the top of the lower platform and is the minimum configuration for the container clearing the corner of the base. The black lines represent the construction lines and perpendicular bisectors of those lines; it is along the perpendicular bisectors that the rotopoles or fixed pivots have to be located.

O2

O4

P1

P3

97.5

118

204


‐7-

2a. Using the program that you developed in class, perform an analytical synthesis to design a linkage that will move the box from the top positions (A1B1) to the bottom position ( A3B3), and has ground pivots on the base. Show all work need to calculate the parameters used in your computer model below. Staple a copy of the computer solution directly after this page.

This particular configuration lends itself to the Analytical Two-Position Synthesis – Approach B. The GIVEN parameters that can be calculated from the above figure are found below.

FIRST DYAD

1 1

2 2

1 1

2 2

2 2

21

12

2

, 0 , 0

, 86 184 86 , 130 184 , 130

, 43 , 0

, 43 184 , 130 227 , 130

184 0 130 0 225

130tan 35.2 324.8

184

0 (The orienta

x y

x y

x y

x y

P P mm mm

P P mm mm mm mm mm mm

A A mm mm

A A mm mm mm mm mm

p mm mm mm mm mm

mm

mm

22

1 1 1 1 2

1 11 12

1 1

2

tion of the container does not change)

43 (Dependent on the chosen location of P )

0tan tan 180 (Dependent on chosen location of P )

43

120 (M

x x y y

y y

x x

z P A P A mm

P A mm

P A mm

easured from the figure, not a unique value.)

GIVEN: CHOSEN: FIND: x-coord y-coordP12 225.3 z 43 w 130.077 O2 97.561 -118.081

324.8 180 114.800 A1 43.000 0.000

2 0 2 -120 W1x -54.561 A2 227.103 -129.870W1y 118.081 P1 0.000 0.000

x-coord y-coord P2 184.103 -129.870W1 -54.561 118.081W2 129.542 -11.789Z1 -43.000 0.000Z2 -43.000 0.000

-1.5 0.866025404 W1x 184.1027 -0.5 -0.28868

-0.866025404 -1.5 W1y -129.87 0.288675 -0.5

APPROACH B FIRST DYAD

inverse


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SECOND DYAD

1 1

2 2

1 1

2 2

22 2 2

21 2 1 2 1

2 112

2 1

, 0 , 0

, 86 184 86 , 130 184 , 130

, 43 , 86

, 43 184 , 130 86 227 , 216

184 0 130 0 225

tan ta

x y

x y

x y

x y

x x y y

y y

x x

P P mm mm

P P mm mm mm mm mm mm

B B mm mm

B B mm mm mm mm mm mm

p P P P P mm mm mm mm mm

P P

P P

1

2

22 2 2

1 1 1 1 2

1 11

1 1

130n 35.2 324.8

184

0 (The orientation of the container does not change)

43 86 96.2 (Dependent on the chosen location of P )

tan tan

x x y y

y y

x x

mm

mm

s P B P B mm mm mm

P B

P B

12

2

86116.6 (Dependent on chosen location of P )

43

120 (Measured from the figure, not a unique value.)

mm

mm

GIVEN: CHOSEN: FIND: x-coord y-coordP12 225.3 s 96.2 u 130.077 O4 97.635 -204.099

324.8 116.6 114.800 B1 43.074 -86.018

2 0 -120 U1x -54.561 B2 227.177 -215.888U1y 118.081 P1 0.000 0.000

x-coord y-coord P2 184.103 -129.870U1 -54.5611 118.0810U2 129.5417 -11.7892S1 -43.0744 86.0176S2 -43.0744 86.0176

-1.5 0.866025 U1x 184.1027 -0.5 -0.28868

-0.86603 -1.5 U1y -129.87 0.288675 -0.5

inverse

SECOND DYADAPPROACH B


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2b. Does your linkage design require the base to be modified in any way? Explain. Can it be designed so that the base does not have to be modified? How?

The configuration that was synthesized above minimally satisfies the constraint of not having to modify the base. If the fix pivots are positioned any lower on the bisectors of the moving pivots, the corner of the container will run into the base. If this were allowed to happen, the base will have to be modified. If the fixed pivots are positioned higher than the minimal point on the bisectors of the moving pivots, the gap between the container and the corner of the base will increase in distance. The limit to how far above the base the container can be carried is limited by knowing the fixed pivots must be positioned on the bisector, but below the line between the moving pivots.

Because the synthesized mechanism is designed to be a parallelogram, all points on the container will sweep out a circular path of the same diameter. This means that the line between B and the left corner of the container will remain horizontal during the motion of the container and that the location of the center of the corners path is horizontal and to the left of the center of point B’s path, at a distance that is equal to the distance from point B to the corner. This geometry enabled the minimal configuration to be identified as shown in the figure.

2c. Will the linkage design enable the container to hold a liquid without spilling it as it travels from the upper base position to the lower platform position? Explain. Can the linkage be designed to hold the container in the same position throughout its motion from the upper to lower base? How?

The linkage designed above will always stay in the same orientation throughout the motion of the container from the top of the base to the lower platform. This is assured, as explained above, by the parallelogram nature of the mechanism design. Because the top of the container will remain horizontal, the mechanism is designed to maximize the ability of the container to hold a liquid without spilling it during the motion of the mechanism.


‐10-

PROBLEM 3 (40 pts): Design a single-dwell cam to move a follower from 0 to 2” in 60, fall

2” in 90, and dwell for the remainder. The total cycle must take 2 seconds.

Using Cycloid Functions:

Region 1: 1

11 1 13 3 30 60 , 0 , 60 , 0 1, 0 60 , 0

1 1 1 1 11

1 1 1 1

1 11

1 1 1

21

31 1 1sin 2 2 sin 2 2 sin 6

2 2 2

6 61 cos 2 1 cos 2 1 cos 6

2sin 2

in inrad rad

s h in in

hv

ha

2 2

3 3

1 11

1 1

21 1

131 1 1

36 36sin 2 sin 6

4 216 216cos 2 cos 2 cos 6

in inrad rad

in inrad rad

hj

Region 2: 2

2

52 2 23 6 2 260 150 , , 90 , 0 1, 0 90 , 0

2 2 2 2 22

2 2 2 2

2 22

2 2 2

21 1 1sin 2 2 2 sin 2 2 2 sin 4

2 2 2

4 41 cos 2 1 cos 2 1 cos 4in in

rad rad

s h h in in in in

hv

a

2 2

3 3

2 222

2 2 2

22 2

232 2 2

2 16 16sin 2 sin 2 sin 4

4 64 64cos 2 cos 2 cos 4

in inrad rad

in inrad rad

h

hj

Region 3: 3

3

5 7 73 3 36 6 6150 360 , 2 , 210 , 0 1, 0 210 , 0

0

0

0

0

s

v

a

j


‐11-

0.0000

1.0000

2.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

Displacement

‐4.0000

‐2.0000

0.0000

2.0000

4.0000

6.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

Velocity

‐15.0000

‐10.0000

‐5.0000

0.0000

5.0000

10.0000

15.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

Acceleration

‐100.0000

‐50.0000

0.0000

50.0000

100.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

Jerk


‐12-

Using 3-4-5 Polynomial Functions, starting with a function of the form

2 3 4 5

0 1 2 3 4 5( )s C C C C C C

Region 1: 1

11 1 13 3 30 60 , 0 , 60 , 0 1, 0 60 , 0

1 1 1 12

1 1 1 1

1 1 12

1 1 1

1 1 1

1 1

1

1 3

3

3 4

0, 0 ( 0) 0 , ( 0) 0 , ( 0) 0

(The above BC are used to determine that .)

60 ( 1) 2 , ( 1) 0 , ( 1) 0

( 1) 2

in inrad rad

in inrad rad

s in v a

s in v a

s in C C

0 1 2C = C = C = 0

1

1 1

1 1 1 1

1 1 1 1

1 1 1 12 2 2

1 1 1 1

4 5

5 3 4 5

2 3 41

3 4 5 3 4 5

2 3 41

3 4 5 3 4 5

2

( 1) 0 3 4 5 0 3 4 5

( 1) 0 6 12 20 0 6 12 20

in inrad rad

in inrad rad

C in C C C

v C C C C C C

a C C C C C C

Solving the three equations simultaneously results in C3=20in, C4=-30in, and C5=12in. For this region,

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 12

1 1 11

3 4 5 3 4 5

3 4 5

2 3 4 2 3 431

3 4 5

21

3 4 5

( ) 20 30 12

( ) 3 4 5 60 120 60

( ) 6 12 20

inrad

s C C C in in in

v C C C

a C C C

1 1 12

1 1 1 1

1 1 1 1 13 3

1 1 1 1 11

3 2 39

2 2271

3 4 5

120 360 240

( ) 6 36 60 120 1080 720

inrad

inradj C C C

Region 2: 2

2

52 2 23 6 2 260 150 , , 90 , 0 1, 0 90 , 0

2 2 2 22

2 2 2 2

2 2 22

2 2 2

2

2

2

2 2

3

0, 0 ( 0) 2 , ( 0) 0 , ( 0) 0

(The above BC are used to determine .)

90 ( 1) 2 , ( 1) 0 , ( 1) 0

( 1) 0 2

in inrad rad

in inrad rad

s in v a

s in v a

s in in C

0 1 2C = 2in, C = 0, and C = 0

2 2 2

2 2 2

2 2 2 2

2 2 2 2

2 2 2 22 2 2

2 2 2 2

3 4 5

4 5 3 4 5

2 3 41

3 4 5 3 4 5

2 3 41

3 4 5 3 4

2

( 1) 0 3 4 5 0 3 4 5

( 1) 0 6 12 20 0 6 12 20

in inrad rad

in inrad rad

C C in C C C

v C C C C C C

a C C C C C C

5

Solving the three equations simultaneously results in C3=-20in, C4=30in, and C5=-12in. For this region,

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

2 2 22

2 2 22

3 4 5 3 4 5

3 4 5

2 3 4 2 3 41 2

3 4 5

21

3 4

( ) 2 2 20 30 12

( ) 3 4 5 60 120 60

( ) 6 12

inrad

s in C C C in in in in

v C C C

a C C

2 2 2 22

2 2 2 2

2 2 2 2 23 3

2 2 2 2 22

3 2 34

5

2 281

3 4 5

20 120 360 240

( ) 6 36 80 120 1080 720

inrad

inrad

C

j C C C


‐13-

Region 3: 3

3

5 7 73 3 36 6 6150 360 , 2 , 210 , 0 1, 0 210 , 0

0, 0, 0, 0s v a j

0.0000

1.0000

2.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

3‐4‐5 Polynomial Displacement

‐4.0000

‐2.0000

0.0000

2.0000

4.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

3‐4‐5 Polynomial Velocity

‐15.0000

‐10.0000

‐5.0000

0.0000

5.0000

10.0000

15.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

3‐4‐5 Polynomial Acceleration

‐100.0000

‐50.0000

0.0000

50.0000

100.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

3‐4‐5 Polynomial Jerk


‐14-

Using 4-5-6-7 Polynomial Functions, starting with a function of the form

2 3 4 5 6 7

0 1 2 3 4 5 6 7( )s C C C C C C C C

Region 1: 1

11 1 13 3 30 60 , 0 , 60 , 0 1, 0 60 , 0

1 1 1 1 12 3

1 1 1 1 1

1 1 1 12

1 1 1

1

1 3

0, 0 ( 0) 0 , ( 0) 0 , ( 0) 0 , j( 0) 0

(The above BC are used in determining that .)

60 ( 1) 2 , ( 1) 0 , ( 1) 0 , (

in in inrad rad rad

in inrad rad

s in v a

s in v a j

0 1 2 3C = C = C = C = 0

31

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 12 2

1

4 5 6 7

4 5 6 7 4 5 6 7

3 4 5 61

4 5 6 7 4 5 6 7

14

1) 0

( 1) 2 2

( 1) 0 4 5 6 7 0 4 5 6 7

( 1) 0 12

inrad

inrad

inrad

s in C C C C in C C C C

v C C C C in C C C C

a C

1 1 1

1 1 1 1

1 1 1 1 12 3

1 1 1 1 1

2 3 4 5

5 6 7 4 5 6 7

2 3 41

4 5 6 7 4 5 6 7

20 30 42 0 12 20 30 42

( 1) 0 24 60 120 210 0 24 60 120 210inrad

C C C in C C C C

j C C C C in C C C C

Solving the three equations simultaneously results in C4=70in, C5=-168in, C6=140in, and C7=-40in. For this region,

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

4 5 6 7 4 5 6 7

4 5 6 7

3 4 5 6 3 4 5 631

4 5 4 5

( ) 70 168 140 40

( ) 4 5 6 7 280 840 840 280inrad

s C C C C in in in in

v C C C C

1 1 1 1 1 1 1 1 12 2

1 1 1 1 1 1 1 1 11

1 1 1 1 13 3

1 1 1 1 11

2 3 4 5 2 3 4 591

4 5 4 5

2 3 4271

4 5 4 5

( ) 12 20 30 42 840 3360 4200 1680

( ) 36 60 120 210

inrad

in

a C C C C

j C C C C

1 1 1 1

1 1 1 1

2 3 41680 10080 16800 8400rad


‐15-

Region 2: 2

2

52 2 23 6 2 260 150 , , 90 , 0 1, 0 90 , 0

2 2 2 2 22 3

2 2 2 2 2

2 2 22

2 2 2

2

2 3

0, 0 ( 0) 2 , ( 0) 0 , ( 0) 0 , j( 0) 0

(The above BC are used in determining that .)

60 ( 1) 0 , ( 1) 0 , ( 1) 0

in in inrad rad rad

in inrad rad

s in v a

s in v a

0 1 2 3C = 2in, C = C = C = 0

23

2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2 2

22

2

4 5 6 7

4 5 6 7 4 5 6 7

3 4 5 61

4 5 6 7 4 5 6 7

1

, ( 1) 0

( 1) 0 2 2

( 1) 0 4 5 6 7 0 4 5 6 7

( 1) 0

inrad

inrad

inrad

j

s in in C C C C in C C C C

v C C C C in C C C C

a

2 2 2 22

2 2 2 22

2 2 2 2 22 3

2 2 2 2 22

2 3 4 5

4 5 6 7 4 5 6 7

2 3 41

4 5 6 7 4 5 6 7

12 20 30 42 0 12 20 30 42

( 1) 0 24 60 120 210 0 24 60 120 210inrad

C C C C in C C C C

j C C C C in C C C C

Solving the three equations simultaneously results in C4=-70in, C5=168in, C6=-140in, and C7=40in. For this region,

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

4 5 6 7 4 5 6 7

4 5 6 7

3 4 5 6 3 4 51 2

4 5 4 5

( ) 2 2 70 168 140 40

( ) 4 5 6 7 280 840 840 280inrad

s in C C C C in in in in in

v C C C C

1

1

1 1 1 1 1 1 1 1 12 2

1 1 1 1 1 1 1 1 11

1 1 1 1 13

1 1 1 1 11

6

2 3 4 5 2 3 4 51 4

4 5 4 5

2 3 41

4 5 4 5

( ) 12 20 30 42 840 3360 4200 1680

( ) 36 60 120 210

inrada C C C C

j C C C C

1 1 1 13

1 1 1 1

2 3 48 1680 10080 16800 8400in

rad

Region 3: 3

3

5 7 73 3 36 6 6150 360 , 2 , 210 , 0 1, 0 210 , 0

0, 0, 0, 0s v a j


‐16-

0.0000

2.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

4‐5‐6‐7 Polynomial Displacement

‐5.0000

0.0000

5.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

4‐5‐6‐7 Polynomial Velocity

‐20.0000

‐10.0000

0.0000

10.0000

20.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

4‐5‐6‐7 Polynomial Acceleration

‐100.0000

‐50.0000

0.0000

50.0000

100.0000

0 30 60 90 120 150 180 210 240 270 300 330 360

4‐5‐6‐7 Polynomial Jerk


‐1-

As a student at Union College, I am part of a community that values intellectual effort, curiosity and discovery. I understand that in order to truly claim my educational and academic achievements, I am obligated to act with academic integrity. Therefore, I affirm that I carried out the work on this exam with full academic honesty, and I rely on my fellow students to do the same.

For this exam I understand that:

1. I must work alone in writing out the answers to this exam. 2. I cannot copy solutions to these problems from any person or resource. 3. I cannot use any electronic resources, other than the program I wrote as part of this class,

to assist me in the solution to the questions on this exam. 4. I can use one formula sheet (both sides) that I prepared as a reference for this exam, I

must staple this sheet to the back of my exam, and this sheet cannot contain any solutions to problems.

Signature:

Print Name:

Assignment:

Due Date:


‐2-

PROBLEM 1 (30 pts): A box sits on top of the base shown in Position 1 and needs to be moved

to Position 2, and then to Position 3. (Guess 2=-50, 3=-100, 2=-50, and 3=-80)


‐3-

1a. Using the program that you developed in class, perform an analytical synthesis to design a linkage that will move the box from positions 1 to 2 to 3, and has ground pivots on the base. Show all work need to calculate the parameters used in your computer model below. Staple a copy of the computer solution directly after this page.


‐4-

1b. Draw the final synthesized mechanism to scale in Position 1.


‐5-

PROBLEM 2 (30 pts): A container of liquid sit on top of a base (A1B1) and is required to move to the position on the lower step ( A3B3).


‐6-

2a. Using the program that you developed in class, perform an analytical synthesis to design a linkage that will move the box from the top positions (A1B1) to the bottom position ( A3B3), and has ground pivots on the base. Show all work need to calculate the parameters used in your computer model below. Staple a copy of the computer solution directly after this page.


‐7-

2b. Does your linkage design require the base to be modified in any way? Explain. Can it be designed so that the base does not have to be modified? How?

2c. Will the linkage design enable the container to hold a liquid without spilling it as it travels from the upper base position to the lower platform position? Explain. Can the linkage be designed to hold the container in the same position throughout its motion from the upper to lower base? How?


‐8-

PROBLEM 3 (40 pts): Design a single-dwell cam to move a follower from 0 to 2” in 60, fall

2” in 90, and dwell for the remainder. The total cycle must take 2 seconds.


‐9-


‐10-

Documents

MER312 Winter 13 Exam III - Solution - Union Collegerbb.union.edu/courses/mer312/Exams/MER312 2013... · MER312:Adv. Dyn. & Kinematics Exam III Union College Winter 2013 Mechanical