me 352 - machine design i name of student summer …€¦ · the kutzbach mobility criterion can be...

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ME 352 - Machine Design I Name of Student_____________________________

Summer Semester 2014 Lab Section Number__________________________

EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, July 2nd, 2014

Use the blank paper provided for your solutions. Write on one side of the paper only. Where necessary, you can use the figures provided on the exam to show vectors and instant centers. Any work that cannot be followed is assumed to be in error. At the completion of your exam, please staple each problem separately. Staple your crib sheet to the end of Problem 1.

Problem 1 (25 Points). For the planar mechanism in the position shown in Figure 1: (i) Determine the mobility using the Kutzbach criterion. Clearly number each link and label the lower pairs and the higher pairs on the given figure. (ii) Define suitable vectors for a kinematic analysis of the mechanism. Label and show the direction of each vector clearly on the given figure. (iii) Write the vector loop equation(s) that are required for a kinematic analysis of the mechanism. Clearly identify suitable input(s) for the mechanism. List: (a) known quantities; (b) unknown variables; and (c) any constraints. If you identified constraints in part (b) then write the constraint equations.

Figure 1. A planar mechanism.

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Problem 2 (25 Points). For the mechanism in the position shown in Figure 2, gear 3 is rolling without slipping on the ground link 1 at the point of contact C. The radius of gear 3 is 3 75 mm, the radius of

the ground link 1 is 1 125 mm, and the length of link 4 is AB 90 mm. The constant angular velocity

of the input link 2 is 2 25 rad / s counterclockwise.

(i) Write the vector loop equation(s) that are suitable for a kinematic analysis of this mechanism. Draw your vectors clearly on the figure. (ii) Using the vector loop equation(s) determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and 4, and the linear velocity of the slider (link 5). Give the magnitudes and directions of these three vectors.


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Problem 3 (25 Points). For the mechanism in the position shown in Figure 3, the input link 2 is rotating counterclockwise with a constant angular velocity 2 30 rad / s .

(i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the given Kennedy circle, show the locations of all the instant centers on Figure 3. Using the locations of the instant centers, determine: (iii) The first-order kinematic coefficients of links 3, 4, and 5. (iv) The angular velocities of links 3, 4, and 5. Give the magnitudes and directions of each vector.

Figure 3. A planar mechanism. (The figure is drawn full size.)

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Problem 4 (25 Points). For the mechanism in the position shown in Figure 4, link 3 is rolling without slipping on the input link 2 at the point of contact C. The link lengths are 3O A 90 mm, AB 150 mm,

and the radius of link 3 is 3 85 mm. The constant velocity of the input link 2 is 2V 15 i m / s.

(i) Write the vector loop equation(s) that are suitable for a kinematic analysis of the mechanism. Draw your vectors clearly on the given figure. (ii) Using the vector loop equation(s) determine the first-order kinematic coefficients for the mechanism. (iii) Determine the angular velocities of links 3 and 4. Give the magnitudes and directions of each vector. (iv) Determine the velocity of pin B connecting link 4 to the slider (link 5). Give the magnitude and direction of this vector.


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Solution to Problem 1. (i) 6 Points. The links, the lower pairs, and the higher pairs of the planar mechanism are as shown in Figure 1a.

Figure 1a. The links and kinematic pairs of the planar mechanism.

The Kutzbach mobility criterion can be written as

1 2M 3(n 1) 2 j 1j (1)

The number of links, the number of lower pairs (or 1j joints), and the number of higher pairs (or 2j

joints), respectively, are

1 2n 6, j 6, and j 2 (2)

Substituting Equation (2) into Equation (1), the mobility of the mechanism can be written as

M 3(6 1) 2 x 6 1x 2 (3a)

Therefore, the mobility of the mechanism is

M 15 12 2 1 (3b)

This is the correct answer for the mobility of the mechanism in this position, that is, for a single input there is a unique output.

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(ii) 5 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Figure 1b.

Figure 1b. The vectors for the mechanism.

(iii) 14 Points. Three independent vector loop equations are required for a kinematic analysis of the mechanism. The three vector loop equations can be written as

Loop 1: ? ?

2 3 4 1 0I

R R R R

(4)

Loop 2: 1 ? ?

22 7 6 11 0C

R R R R

(5)

Loop 3: 2? ?

45 5 111 0C

R R R

(6)

The input link is chosen to be link 2 (see Figure 1b) and the input variable is the angle θ2. Other

suitable inputs are link 4 and link 5 since they are pinned to the ground. Link 6 would not be the best option for the input despite being pinned to ground. The reason is that a

clockwise rotation of link 6 could cause joint separation, so the remaining links would not be driven by link 6 for all positions of the mechanism.

Link 3 is not a practical input because it does not have a connection to the ground link.

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(a) The seven known quantities of the mechanism are

2 2 3 4 4 5 5 22 2 7 6 6E, AE, , , , , and RR O R R O A R O B R O C R CD O D (7)

Additionally, the magnitudes and the directions of the ground vectors 1R

, 11R

, and 111R

are known.

(b) The six unknown variables are

3 4 5 6 7 45, , , , , and R (8)

(c) There are two constraints. The two constraint equations are identified as follows.

Constraint 1: The angle between the vectors 2R and 22R is constant, that is

22 2 (9)

Constraint 2: The directions for vectors 4R and 45R are always equal, that is

45 4 (10)

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Solution to Problem 2. (i) 6 Points. A suitable set of vectors for a kinematic analysis of the mechanism is shown in Figure 2.

Figure 2. The vectors for a kinematic analysis of the mechanism.

The vector loop equation (VLE) for the mechanism can be written as

? ?

2 4 5 11 1 0I

R R R R R

(1)

The input is the angular position of link 2 (connecting O2 to A) which is the arm of the gear train (links 1, 2, and 3). The length of the arm is R2 = 200 mm, that is, the sum of the radii of the ground link 1 and link 3. Also, in the given position θ2 = 0°, θ4 = 90°, θ5 = 30°, θ11 = 0°, and θ1 = 90°. (ii) 13 Points. The X and Y components of Equation (1) are

2 2 4 4 5 5 11 11 1 1cos cos cos cos cos 0R R R R R (2a)

and

2 2 4 4 5 5 11 11 1 1sin sin sin sin sin 0R R R R R (2b)

Differentiating Equations (2) with respect to the input position θ2 gives

' '2 2 4 4 4 5 5sin sin cos 0R R R (3a)

and ' '

2 2 4 4 4 5 5cos cos sin 0R R R (3b)

Writing Equations (3) in matrix form gives

4 4 5 4 2 2

4 4 5 5 2 2

R sin cos R sin

R cos sin R R cos

(4)

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The determinant of the coefficient matrix in Equation (4) is

4 4 5 4 4 5sin sin cos cosDET R R (5a)

which simplifies by the sum-difference trigonometric identity to

4 4 5cos( )DET R (5b)

Substituting θ4 = 90° and θ5 = 30° into Equation (5b) gives

0 0 04 4cos (90 30 ) cos 60DET R R (6a)

Therefore, the determinant of the coefficient matrix is

(90 mm) cos 60 45 mmDET (6b)

Note that the determinant, see Equation (5b), is zero when θ4 = 120° or 300°, that is, when link 4 is perpendicular to the line of sliding of link 5.

Substituting the known data into Equation (4) gives

4

5

90sin(90 ) mm cos(30 ) 200sin(0 ) mm

R90cos(90 ) mm sin(30 ) 200cos(0 ) mm

(7a)

which simplifies to '4'5

0 mm90 mm 3 / 2

200 mm0 mm 1/ 2 R

(7b)

Using Cramer’s rule, the first-order kinematic coefficient for link 4 can be written as

'4

0 mm 3 / 2 3200 mm200 mm 1/ 2 2

45 mm 45 mm

(8a)

Therefore, the first-order kinematic coefficient for link 4 is

'4

20 3 mm mm3.849

9 mm mm (8b)

The positive sign indicates that link 4 is rotating counterclockwise for a counterclockwise rotation of the input link.

The first-order kinematic coefficient for link 5 is

2'5

90 mm 0 mm

0 mm 200 mm 18000 mm400 mm

45 mm 45 mmR

(9)

The positive sign indicates that link 5 is moving up the inclined plane for a counterclockwise rotation of the input link.

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The rotation of gear 3 is constrained by the rolling contact equation between gear 3 and the ground link. The rolling contact equation can be written as

3 1 2

1 3 2

(10a)

Differentiating Equation (10a) with respect to the input position θ2 gives

' '3 1 2

' '1 3 2

(10b)

The correct sign here is negative because there is external rolling between links 3 and 1. Since the arm (link 2) is the input then, by definition, the first-order kinematic coefficient '

2 1. Also, since link 1

denotes the ground link then the angle θ1 does not change. Therefore, by definition, the first-order kinematic coefficient '

1 0. Substituting the known kinematic coefficients into Equation (10b) gives

3'

1 3 1

0 1

(11a)

Rearranging Equation (11a), the first-order kinematic coefficient for gear 3 can be written as

13

3

' 1

(11b)

Substituting the known data into Equation (11b), the first-order kinematic coefficient for gear 3 is

3' 8 mm mm

1 2.6673 mm mm

125 mm75 mm

(12)

The positive sign indicates that link 3 is rotating counterclockwise for a counterclockwise rotation of the input link 2. (iii) 6 Points. The angular velocity of gear 3 can be written from the chain rule as

'3 23

(13a)

Substituting Eq. (12) into Equation (13a), the angular velocity of gear 3 is

3 2.667 x 25 rad/s 66.667 rad/sk k

(13b)

The angular velocity of link 4 can be written from the chain rule as

'4 24

(14a)

Substituting Equation (8) into Equation (14a), the angular velocity of link 4 is

420 3 mm 500 3

25 rad/s rad/s9 mm 9

k k

(14b)

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Therefore, the angular velocity of link 4 is

4 96.225 rad/sk

(14c)

The velocity of link 5 can be written from the chain rule as

'5 5 2R R

(15a)

Substituting Equation (9) into Equation (15a), the velocity of link 5 is

5 400 mm x 25 rad/s 10,000 mm/sR

(15b)

The velocity of link 5 is in the direction of the line of sliding, up the inclined plane, that is

5 10,000 30 mm/sV

(16)

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Solution to Problem 3. (i) 5 Points. The number of links in the mechanism is five, therefore the total number of instant centers for this mechanism is

( 1) 5 410

2 2

n nN

(1)

There are six primary instant centers, namely: I12, I23, I34, I45, I15, and I13. The instant centers I12, I23, I34, I45, and I15 are primary because they are the given pin joints. The instant center I13 is also regarded as a primary instant center because this instant center must lie somewhere on the line that passes through the pin of link 3 and be perpendicular to the ground slot. From Kennedy’s theorem, the primary instant center I13 must also lie on the line connecting the instant centers connecting I12I23. Note that these two lines are parallel in this position, therefore, I13 is at infinity.

There are four secondary instant centers, namely: I14, I35, I24, and I25.

(ii) 12 Points. The locations of the ten instant centers are as shown in Figure 3.

Figure 3. The locations of the ten instant centers.

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The procedure to find the four secondary instant centers is outlined by the following steps: (i) The point of intersection of the line through I34I13 and the line through I15I45 is the instant center I14. (ii) The point of intersection of the line through I13I15 and the line through I34I45 is the instant center I35. (iii) The point of intersection of the line through I23I34 and the line through I12I14 is the instant center I24. (iv) The point of intersection of the line through I23I35 and the line through I12I15 is the instant center I25.

(iii) 5 Points. The general notation of the first-order kinematic coefficient of link j can be written as

1

1

' ij i

ij j

I I

jI I

R

R (2)

where the correct sign is determined from the location of the relative instant center relative to the locations to the two absolute instant centers. (Refer to Sections 3.17 and 3.18 of the text book for more details).

The first-order kinematic coefficient of link 3 can be written as

23 12

23 13

'3

I I

I I

R

R (3)

From the scaled drawing, see Figure 3, the distance I13I23 is infinite and the distance

I12I23 = 0.771 in = 19.582 mm (4)

Substituting these values into Equation (3), the first-order kinematic coefficient of link 3 is

'3

0.771in0 in/in

in

(5)

Therefore, the motion of link 3 is pure translation (in this position). The first-order kinematic coefficient of link 4 can be written as

24 12

24 14

'4

I I

I I

R

R (6)

The correct sign for the first-order kinematic coefficient of link 4 is negative because the relative instant center I24 lies between the two absolute instant centers I12 and I14. Check: By inspection, since the input link is rotating counterclockwise then link 4 must rotate clockwise. From the scaled drawing, the distances are measured as

I12I24 = 1.652 in = 41.958 mm and I14I24 = 3.856 in = 97.943 mm (7)

Substituting Equation (7) into Equation (6), the first-order kinematic coefficient of link 4 is

'4

1.652 in0.428 in/ in

3.856 in (8)

The negative sign indicates that link 4 is rotating clockwise for a counterclockwise rotation of the input link.

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The first-order kinematic coefficient of link 5 can be written as

25 12

25 15

'5

I I

I I

R

R (9)

The correct sign for the first-order kinematic coefficient of link 5 is negative because the relative instant center I25 lies between the two absolute instant centers I12 and I15. Check: By inspection, since the input link is rotating counterclockwise then link 5 must rotate clockwise. From the scale drawing, the distances are measured as

I12I25 = 2.149 in = 54.580 mm and I15I25 = 2.965 in = 75.315 mm (10)

Substituting Equation (10) into Equation (9), the first-order kinematic coefficient of link 5 is

'5

2.149 in0.725 in/in

2.965 in (11)

The negative sign indicates that link 5 is rotating clockwise for a counterclockwise rotation of the input link. (iv) 3 Points. The angular velocity of link 3 can be written as

'3 3 2

(12a)

Substituting Equation (5) and the input angular velocity into Equation (12a), the angular velocity of link 3 is

3 (0 in/in)(30 rad/s) 0k (12b)

The angular velocity of link 4 can be written as

'4 4 2

(13a)


4 ( 0.428 in/in)(30 rad/s) 12.84 rad/sk k (13b)

The angular velocity of link 5 can be written as

'5 5 2

(14a)


5 ( 0.725 in/in)(30 rad/s) 21.75 rad/sk k (14b)

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Solution to Problem 4. (i) 3 Points. A suitable set of vectors for a kinematic analysis of the mechanism is shown in Figure 4.

Figure 4. A suitable set of vectors for a kinematic analysis of the mechanism.

The vector loop equation (VLE) for this mechanism can be written as

? ?

2 7 3 4 5 0I RC

R R R R R

(1)

where θ2 = 180°, θ7 = 90°, θ3 = 90°, and θ5 = 150° in this position.

The input is link 2 (the rack), and 2R

is the vector from point E on the ground to point C (that is, the

point of contact between link 2 and link 3).

The arm (link 7), that is, the vector 7R

, is the vector from the point of contact C to O3 and is always

perpendicular to the rack.

The vector 3R

is the vector from O3 to A, 4R

is the vector from B to A, and 5R

is the vector from point

E to pin B along the line of sliding of link 5. (ii) 15 Points. The angle θ4 can be obtained from the triangle O3AB. From the sine rule, the angle

1 3

4

sinR

R

(2)

Then the supplement to the angle α is

1 34

4

180 sinR

R

(3a)

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Substituting the known dimensions into Equation (3a) gives

14

90 mm180 sin 143.13

150 mm

(3b)

The rolling contact constraint between the rack (link 2) and the pinion (gear 3) can be written as

2 3 3 7( )R (4a)

The correct sign here is negative because a decrease in the length of the vector R2 (when the rack moves to the right) causes a counterclockwise rotation of gear 3. Differentiating Equation (4a) with respect to the input position, the first-order kinematic coefficient of link 2 can be written as

' ' '2 3 3 7( )R (4b)

Since link 2 is the input then by definition the kinematic coefficient of the input link is

'2 1R (5a)

Also, the arm (link 7) is always perpendicular to the rack, that is, the angle θ7 does not change. Therefore, the kinematic coefficient of the arm is

'7 0 (5b)

Substituting Equations (5) into Equation (4b), and rearranging, the first-order kinematic coefficient of link 3 can be written as

'3

3

1

(6a)

Substituting the radius of gear 3 into Equation (6a), the first-order kinematic coefficient of link 3 is

'3

1 10.0118

85 mm mm (6b)

The negative sign indicates that link 3 is rotating counterclockwise for the given velocity of the input link (that is, the input vector is becoming shorter as the input link moves to the right).

The VLE in Equation (1) can be written in terms of the X and Y components as

2 2 7 7 3 3 4 4 5 5cos cos cos cos cos 0R R R R R (7a)

and

2 2 7 7 3 3 4 4 5 5sin sin sin sin sin 0R R R R R (7b)

Differentiating Equations (7) with respect to the input position gives

' ' ' '2 2 3 3 3 4 4 4 5 5cos sin sin cos 0R R R R (8a)

and ' ' ' '2 2 3 3 3 4 4 4 5 5sin cos cos sin 0R R R R (8b)

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Writing Equations (8) in matrix form and substituting in Equation (6a) gives

'2 2 3 3'

34 4 5 4'

4 4 5 5 '2 2 3 3

3

1cos sin

sin cos

cos sin 1sin cos

R RR

R RR R

(9)

The determinant of the coefficient matrix in Equation (9) is

4 4 5 4 4 5sin sin cos cosDET R R (10a)

which can be written as

4 4 5cos( )DET R (10b)

The determinant is zero when θ4 = 60° or 240°, that is, when link 4 is perpendicular to the line of sliding for link 5. Substituting the known data into Equation (10b), the determinant of the coefficient matrix is

150 cos(143.13 150 ) mm 148.923 mmDET (11)

Substituting the known information into Equation (9) gives

'4'5

1 mm1cos(180 ) 90 sin(90 )

150sin(143.13 ) mm cos(150 ) 85 mm

150cos(143.13 ) mm sin(150 ) 1 mm1sin(180 ) 90 cos(90 )

85 mm

R

(12a)

Simplifying this equation gives

'4

'5

90 mm3 190 mm85 mm2mm1 0120 mmmm2

R

(12b)

Using Cramer’s Rule, the first-order kinematic coefficient of link 4 can be written as

'4

90 mm 31

85 mm 2mm 1

0mm 2148.923 mm

(13a)

Therefore, the first-order kinematic coefficient of link 4 is

' 44

1 90 mm12 170 mm 1.975 10

148.923 mm mm

(13b)

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The negative sign indicates that link 4 is rotating counterclockwise as the rack is moving to the right, that is, as the vector R2 is decreasing in length.

Using Cramer’s Rule, the first-order kinematic coefficient of the slider link 5 is

'5

90 mm90 mm 1

85 mmmm

120 mm 0mm

148.923 mmR

(14a)

Therefore, the first-order kinematic coefficient of the slider link 5 is

2

'5

10800 mm120 mm85 mm 0.0474

148.923 mm mmR

(14b)

The negative sign here indicates that link 5 is sliding upward as the rack is moving to the right, that is, as the vector R2 is decreasing in length. (iii) 5 Points. The angular velocity of link 3 can be written as

'3 3 2R

(15)

where 2R

has equal magnitude and opposite direction of the input velocity V2, that is, the vector R2

decreases in length if the input velocity V2 is to the right (and the vector R2 increases in length if the input velocity V2 is to the left).

Substituting Equation (5b) into Equation (15), the angular velocity of the pinion link 3 is

13

1 3000mm ( 15000 mm/s) rad/s 176.471 rad/s

85 17

(16a)

The positive sign indicates that link 3 is rotating counterclockwise, that is

3 176.471 rad/sk (16b)

The angular velocity of link 4 can be written as '

4 4 2R

(17)

Substituting Equation (13) into Equation (17), the angular velocity of link 4 is

4 14 ( 1.975 10 mm )( 15000 mm/s) 2.962 rad/s (18a)

The positive sign indicates that link 4 is rotating counterclockwise, that is

4 2.962 rad/sk (18b)

(iv) 2 Points. The velocity of pin B is equal to the velocity of link 5 because the slider is in pure translation. The velocity of pin B and the velocity of link 5 can be written as

19

'5 5 5 2BV V R R R

(19)

Substituting Equation (14) into Equation (19), the velocity of pin B and link 5 are

5 5 ( 0.0474 mm/mm)( 15000 mm/s) 711 mm/sBV V R

(20a)

The positive sign indicates that the slider is moving upward along the line of sliding, that is

5 711 150 mm/sBV V

(20b)

me 352 - machine design i name of student summer …€¦ · the kutzbach mobility criterion can be...

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