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ME 352 - Machine Design I Name of Student_____________________________

Summer Semester 2014 Lab Section Number__________________________

FINAL EXAM. OPEN BOOK AND CLOSED NOTES Thursday, August 7th, 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. Any work that cannot be followed is assumed to be in error. At the completion of your exam, please staple each problem separately and staple your crib sheet to the end of Problem 1.

Problem 1 (25 Points). For the linkage shown in Figure 1, a horizontal force EˆF 800 kNi

is acting

on the coupler link 3 at point E and a torque 12T is acting on the crankshaft at 2O to hold the linkage in

static equilibrium. The length of link 2 is 2O A 1m and the distance AB 2 m. Links 2, 3, and 4 are a

steel alloy with a compressive yield strength of 415 MPaycS and modulus of elasticity 210 GPa .E

The links have square cross-sections with width 50 mm.t Ignoring gravity, determine for link 4: (i) The slenderness ratio at the point of tangency between Euler’s formula and Johnson’s equation. (ii) The factor of safety guarding against buckling if the length of link 4 is 4O B 1.25 m.

(iii) The factor of safety guarding against buckling if the length of link 4 is 4O B 2 m.

Figure 1. A four-bar linkage in static equilibrium.

2



Problem 2 (25 Points). For the mechanism shown in Figure 2, the first and second-order kinematic

coefficients of links 3 and 4 are 13 0.019 mm , 4R 0.87 mm/mm, 4 2

3 1.515 10 mm ,

and 14R 0.019 mm . The first and second-order kinematic coefficients of points P and G4 are

PX 0.43 mm/mm, PY 0, 3 -1

PX 4.55 10 mm , -1

PY 0.011 mm , G4X 0.435 mm/mm,

G4Y 0.753 mm/mm, 3 -1

4GX 9.5 10 mm , and -1

G4Y 0.016 mm . The position and constant

velocity of the input link 2 are OA 25 mm and 2V 15 mm/s.i The dimensions are AP 30 mm

and AB 50 mm. The masses are 2m 15 kg, 3m 25 kg, 4m 20 kg , and the mass moments of

inertia about the mass centers are 2

4 2GI 6.25 10 N ms ,

3

3 2GI 3 10 N ms , and

4

3 2GI 1.04 10 N ms . The force PF 300 N and an unknown force, parallel to the inclined plane, is

acting at point E. The free length of the linear spring is 50 mm, the spring stiffness is 15 N/mm, and the damping constant of the viscous damper is 80 N s/mm. Neglecting friction, determine: (i) The first-order kinematic coefficients of the spring and the damper. (ii) The kinetic energy of the mechanism. (iii) The magnitude and direction of the force acting at point E using the equation of motion.

Figure 2. A planar mechanism.

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Problem 3 (25 Points).

Part A. The weights of two pulleys rigidly attached to a rotating shaft are 1W 350 N and 2W 650 N.

The constant speed of the shaft is 225 rad/sk and the influence coefficients for the shaft are 5

11a 3.85 10 mm/N, 522a 13.25 10 mm/N, and 5

12a 6.75 10 mm/N. Assuming that the mass

of the shaft can be neglected, determine: (i) The first and second critical speeds of the shaft using the exact equation. (ii) The first critical speed of the shaft using the Rayleigh-Ritz equation. (iii) The first critical speed of the shaft using the Dunkerley approximation.

Part B. The shaft shown in Figure 3 is rotating with a constant angular velocity 95 rad/s.k The

masses of the three particles are 1m 5 kg, 2m 10 kg, and 3m 8 kg and the radial distances are

1R 60 mm, 2R 90 mm, and 3R 120 mm. For dynamic balance, determine the magnitudes and

angular locations of the correcting masses to be removed in planes (1) and (2), at the radial distances from the shaft axis C1 C2R R 85 mm. Show the locations of the correcting masses on the left-hand

side of Figure 3.

Figure 3. A rotating shaft with three mass particles.

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Problem 4 (25 Points). The effective mass of each piston in the three-cylinder engine shown in Figure 4 is 1 2 3m m m 8 kg. The length of each connecting rod is L 70 cm, and the length of the throw of

each crank is R 20 cm. The constant angular velocity of the crankshaft is 125 rad/s.k Determine: (i) The X and Y-components of the primary shaking force in terms of the crank angle θ. (ii) The magnitudes and directions of the inertia forces created by the correcting masses in terms of θ. (iii) The correcting masses that must be added given that the radii are C C1 2

R R 30 cm.

(iv) Show the correcting masses on the right-hand figure below where the reference line is specified at the crank angle oθ 30 .

Figure 4. A three-cylinder engine.

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Solution to Problem 1. (i) 5 Points. The slenderness ratio at the point of tangency between the Euler column formula and the Johnson parabolic equation can be written from Equation (13.36), page 576, as

2D

yc

CESr

S (1)

Since link 4 has pinned-pinned end conditions then the end-condition constant C = 1. Substituting the given material properties into Equation (1), the slenderness ratio at the point of tangency is

9

6

2 1 210 1099.9426

415 10DSr

(2)

(ii) 12 Points. The forces acting on link 4 can be obtained from a static force analysis of the linkage. The free-body diagram of link 3 is shown in Figure 1a.

Figure 1a. The free-body diagram of link 3.

The sum of external moments about point A can be written as

0AM (3a)

that is

3 43 33 0X Y Y ER F R F (3b)

Substituting the dimensions of the coupler link and the force 800 kNEF into Equation (3b) gives

3432 m 0.4 m 800 10 N 0YF (3c)

that is 3

43 160 10 NYF (3d)

Therefore, the internal reaction force acting on link 4 at pin B is

334 43 160 10 NY YF F (3e)

Link 4 is a two-force member in compression. The line of action of the forces is vertical, that is

14 34 0Y YF F (4a)

Substituting Equation (3e) into Equation (4), the compressive force in link 4 is

6

314 160 10 0YF (4b)

that is 3

14 160 10 NYF (4c)

The compressive load in link 4 will be written as

3160 10 NAPPP (5)

Check: The free-body diagram of link 4 is shown in Figure 1b.

Figure 1b. The free-body diagram of link 4.

The sum of external forces in the X-direction can be written as

0XF (6a)

that is

14 34 0X XF F (6b)

The sum of external forces in the Y-direction can be written as

0YF (7a)

that is

14 34 0Y YF F (7b)

The sum of external moments about pin O4 can be written as

40OM (8a)

that is

4, 34 0Y XR F (8b)

Since the length of link 4 is not zero then the internal reaction force is

7

34 0XF (8c)

Therefore, the internal reaction force acting on link 3 at point B is

43 34 0X XF F (8d)

Substituting Equation (8c) into Equation (6b) gives

14 0XF (9)

Equations (8c) and (9) confirm that the lines of action of the forces F34 and F14 are purely vertical (that is, link 4 is in compression). Substituting Equation (3e) into Equation (7b) gives

314 160 10 0YF (10a)

that is 3

14 160 10 NYF (10b)

Referring back to Figure 1a, the sum of external forces acting on link 3 in the X-direction is

0XF (11a)

that is

23 43 0X X EF F F (11b)

Substituting Equation (8d) and the given external force into Equation (11b) gives

323 0 800 10 0XF (12a)

Therefore, the internal reaction force on link 3 at pin A is

323 800 10 NXF (12b)

By definition, the internal action force on link 2 at pin A is

332 23 800 10 NX XF F (12c)

The sum of external forces acting on link 3 in the Y-direction can be written as

0YF (13a)

that is

23 43 0Y YF F (13b)

Substituting Equation (3d) into Equation (13b) gives

323 160 10 0YF (14a)

that is 3

23 160 10 NYF (14b)

Therefore, the internal reaction force acting on link 2 at pin A is

332 23 160 10 NY YF F (14c)

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The free-body diagram of link 2 is shown in Figure 1c.

Figure 1c. The free-body diagram of link 2.

The sum of external forces in the X-direction can be written as

0XF (15a)

that is

12 32 0X XF F (15b)


312 800 10 0XF (16a)

Therefore, the internal reaction force at pin O2 is

312 800 10 NXF (16b)

The sum of external forces in the Y-direction can be written as

0YF (17a)

that is

12 32 0Y YF F (17b)


312 160 10 0YF (18a)

Therefore, the internal reaction force at pin O2 is

312 160 10 NYF (18b)

The sum of external moments about the ground pin O2 can be written as

9

20OM (19a)

that is

12 2, 32 0Y XT R F (19b)

Substituting Equation (12c) and the given length 2 1 mO A into Equation (19b), the torque is

312 1 800 10 0T (20a)

Therefore, the external torque to hold the mechanism in static equilibrium is

312 800 10 NmT (20b)

The positive sign indicates that the torque is counterclockwise. The slenderness ratio of link 4 can be written as

r

LS

k (21a)

where the radius of gyration is

Ik

A (21b)

Since link 4 has a square cross-section then the area and the area moment of inertia, respectively are

2A t (22a) and

4112I t (22b)

Substituting Equations (2) into Equation (21a), the slenderness ratio of link 4 can be expressed as

4112

2

12r

L LS

tt

t

(23a)

Substituting the width 50 mm 0.05 mt into Equation (23a), the slenderness ratio can be written as

20 12rS L (23b)

where the length L of link 4 must be in meters. For the length L = 1.25 m, then the slenderness ratio from Equation (23b) is

20 12 1.25 86.6025rS (24)

Comparing Equation (24) with Equation (2) we note that

86.60 < 99.94

Therefore, link 4 is a Johnson column. Using the Johnson parabolic equation, the critical load for link 4 can be written from Equation (13.40), page 576, as

10

21

2yc r

CR yc

S SP A S

CE

(25)

Substituting Equation (24), the given geometry, and the material properties into Equation (25) gives

2

6

2 6 39

415 10 86.602510.05 415 10 647.99 10 N

1x 210 10 2CRP

(26)

The factor of safety guarding against buckling is defined, see page 579, as

CR

APP

PN

P (27)

Substituting Equations (5) and (26) into Equation (27), the factor of safety for link 4 is

3

3

647.99 104.05

160 10N

(28)

Therefore, link 4 is predicted not to buckle under the given loading. (iii) 8 Points. If the length of link 4 is L = 2 m then the slenderness ratio from Equation (13b) is

20 12 2 138.5641rS (29)

Comparing Equation (29) with Equation (2) we observe that

138.56 > 99.94

Therefore, link 4 is an Euler column. Using the Euler column formula, see Equation (13.34), page 574, the critical load is

2

2CRr

AEP C

S

(30)

Substituting Equation (29), the given geometry, and the material properties into Equation (30) gives

2 2 9

32

0.05 210 10(1) 269.87 10 N

138.5641CRP

(31)

Substituting Equations (5) and (31) into Equation (27), the factor of safety for link 4 is

3

3

269.87 101.69

160 10N

(32)

Therefore, link 4 is predicted not to buckle under the given loading.

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Solution to Problem 2. A suitable set of vectors for a complete kinematic analysis is shown in Figure 2.

Figure 2. The vector set for a kinematic analysis of the mechanism.

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

? ?

2 3 4 1 0I

R R R R

(1)

where 2R

is the input vector from the origin O to pin A, 3R

is the vector from pin A to pin B, 4R

is the

vector from the ground point D (D is a point fixed on the ground link 1) to pin B and 1R

is the ground

vector from O to D. The length R3 = AB is a constant 50 mm, the angle θ2 is a constant 0°, the angle θ4 is a constant 60°, and R2 = OA is given as 25 mm in this position. The remaining position variables can be obtained from trigonometry. Position analysis. From the triangle APB, the angle α is given as

1 30cos 53.13

50

(2)

Since the line AP is perpendicular to the X-axis then the orientation of the vector 2R

(that is, the angle

of link 3) can be written as

3 90 36.87 (3)

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The distance from point D to pin B can be written as

4

sin 36.8750 34.64 mm

sin120R

(4)

Velocity analysis. The X and Y components of Equation (1) can be written as

2 2 3 3 4 4 1 1cos cos cos cos 0R R R R (5a)

and

2 2 3 3 4 4 1 1sin sin sin sin 0R R R R (5b)

Differentiating Equations (5) with respect to the input position gives

' '2 3 3 3 4 4cos sin cos 0R R (6a)

and ' '

2 3 3 3 4 4sin cos sin 0R R (6b)

Writing Equations (6) in matrix form gives

'3 3 4 23

'3 3 4 24

sin cos cos

cos sin sin

R

R R

(7)

The determinant of the coefficient matrix in Equation (7) is

3 3 4 3 3 4 3 4 3cos cos sin sin cos( )DET R R R (8a)

Note that the mechanism is in a singular position when θ3 = 150° or 330°, that is, when the line AB is perpendicular to the line of sliding of link 4. Substituting the known values into Equation (8a) gives

(50 mm) cos(60 36.87 ) 45.98 mmDET (8b)

Substituting the known values into Equation (7) gives

3 1 '5 2 3

'3445 2

50 mm 1

050 mm R

(9a)

which simplifies to 1 '2 3

'342

30 mm 1

040 mm R

(9b)

Using Cramer’s rule, the first-order kinematic coefficient of link 3 is

12

3 32' 2

3

1

00.01883 rad/mm 0.019 rad/mm

DET DET

(10)

The positive sign indicates that link 3 is rotating clockwise for a negative input (that is, link 2 is moving to the left). Using Cramer’s rule, the first-order kinematic coefficient of link 4 is

13

'4

30 mm 1

40 mm 0 40 mm0.8699 mm/mm 0.87 mm/mmR

DET DET

(11)

The positive sign indicates that link 4 is sliding down the slope as the input link 2 is moving to the left (that is, the length of the vector R4 is decreasing as the length of the input vector R2 is decreasing). Acceleration analysis. Differentiating Equations (6) with respect to the input position gives

' 2 '' ''3 3 3 3 3 3 4 4cos sin cos 0R R R (12a)

and ' 2 '' ''

3 3 3 3 3 3 4 4sin cos sin 0R R R (12b)

Writing Equations (12) in matrix form gives

' 2''3 3 4 3 3 33

' 2''3 3 4 3 3 34

sin cos cos

cos sin sin

R R

R RR

(13a)

Note that the coefficient matrix of Equation (13a) is identical to the coefficient matrix of Equation (7). Substituting known values into Equation (13a) gives

2 11 4''2 53

'' 23 -1342 5

30 mm 50 0.01883 0 mm

40 mm 50 0.01883 0 mmR

(13b)

Using Cramer’s rule, the second-order kinematic coefficient of link 3 is

2 -1 1

22 -12 -1 3

2'' 4 23

40(0.01883) mm

20 3 15 0.01883 mm30(0.01883) mm1.515 10 rad/mm

DET DET

(14)

and the second-order kinematic coefficient of link 4 is

2 -1

2 2-1

'' -1 -14

30 mm 40 0.01883 mm

40 mm 30 0.01883 mm 2500 0.018830.01928 mm 0.019 mmR

DET DET

(15)

Point P Analysis. The vector equation for point P, see Figure 2, can be written as

??

2 33

I C

PR R R

(16)

where the constraint equation is

33 3 3 53.13o (17)

In the given position θ33 = 90°, and R33 = AP = 30 mm. The first and second-order kinematic coefficients of link 3 are given by Equations (10) and (14), respectively. The X and Y components of Equation (16) are

2 2 33 33cos cosPX R R (18a)

14

and

2 2 33 33sin sinPY R R (18b)

Substituting the known values into Equations (18) gives

25 mm 1 30 mm 0 25 mmPX (19a)

and

25 mm 0 30 mm 1 30 mmPY (19b)


' '2 33 33 33cos sinPX R (20a)

and ' '

2 33 33 33sin cosPY R (20b)


' -11 30 mm 1 0.019 mm 0.43 mm/mmPX (21a)

and

' -10 30 mm 0 0.019 mm 0PY (21b)

The velocity of point P can be written as

' '2

ˆ ˆP P PV X i Y j R

(22)

Substituting Equations (21) and the input velocity into Equation (22), the velocity of point P is

ˆ ˆ ˆ0.43 0 15 mm/s 6.45 mm/sPV i j i

(23a)

The magnitude and direction of the velocity of point P is

6.45 180 mm/sPV

(23b)


'' '' ' 233 33 33 33 33 33sin cosPX R R (24a)

and '' '' ' 2

33 33 33 33 33 33cos sinPY R R (24b)


2'' 4 -2 -1 3 -130 mm 1 1.515 10 mm 30 mm 0 0.019 mm 4.545 10 mmPX (25a)

and

2'' 4 -2 -1 -130 mm 0 1.515 10 mm 30 mm 1 0.019 mm 0.01083 mmPY (25b)

The acceleration of point P can be written as

15

' '' 2 ' '' 22 2 2 2

ˆ ˆP P P P PA X R X R i Y R Y R j (26)

Substituting Equations (21) and (25) into Equation (26), the acceleration of point P can be written as

2 21 1ˆ ˆ0.004545 mm 15 mm/s 0.01083 mm 15 mm/sPA i j

(27a)

Therefore, the acceleration of point P is

2ˆ ˆ1.0226 2.4368 mm/sPA i j

(27b)

Therefore, the magnitude and direction of the acceleration of point P are

22.6426 292.77 mm/sPA

(27c)

Point G4 Analysis. The vector loop equation for the mass center G4, see Figure 2, can be written as

4

??

2 3

I

GR R R

(28)

The X and Y components of Equation (28) are

4 2 2 3 3cos cosGX R R (29a)

and

4 2 2 3 3sin sinGY R R (29b)


4

4525 mm 1 50 mm 65 mmGX (30a)

and

4

3525 mm 0 50 mm 30 mmGY (30b)

Check: The height of the mass center of link 4 is

4AP 30 mmGY (30b)


4

' ' '2 2 3 3 3cos sinGX R R (31a)

and

4

' ' '2 2 3 3 3sin cosGY R R (31b)

Substituting Equation (10) and the known values into Equations (31), the first-order kinematic coefficients of the mass center G4 are

4

' 351 1 50 0.01883 0.435 mm/mmGX (32a)

and

4

' 451 0 50 0.01883 0.753 mm/mmGY (32b)

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Check: Since link 4 is in pure translation then the first-order kinematic coefficients can be written as

4

' '4 4cosGX R (33a)

and

4

' '4 4sinGY R (33b)

Substituting Equation (11) and the given angle of sliding, the first-order kinematic coefficients are

4

' 120.8699 0.435 mm/mmGX (34a)

and

4

' 320.8699 0.753 mm/mmGY (34b)

Equations (34) are in agreement with Equations (32). The velocity of the mass center can be written as

4 4 4

' '2

ˆ ˆG G GV X i Y j R (35)

Substituting Equations (34) and the given input velocity into Equation (35), the velocity is

4

ˆ ˆ ˆ ˆ0.435 0.753 15 6.525 11.295 mm/sGV i j i j

(36a)

or in terms of magnitude and direction

413.044 60 mm/sGV

(36b)

Differentiating Equations (31) with respect to the input position, the second-order kinematic coefficients of the mass center G4 can be written as

4

2'' '' '' '2 2 3 3 3 3 3 3cos sin cosGX R R R (37a)

and

4

2'' '' '' '2 2 3 3 3 3 3 3sin cos sinGY R R R (37b)

Substituting Equations (10) and (14) and the known values into Equations (37), the second-order kinematic coefficients of the mass center are

4

2'' 4 3 -13 40 1 50 1.515 10 50 0.01883 9.5 10 mm5 5GX (38a)

and

4

2'' 4 -1340 0 50 1.515 10 50 0.01883 0.016 mm5 5GY (38b)

Check: Since link 4 is pure translation, the second-order kinematic coefficients are

4

'' ''4 4cosGX R (39a)

and

4

'' ''4 4sinGY R (39b)

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Substituting the known kinematic information into Equations (39), the second-order kinematic coefficients of the mass center G4 are

4

'' -110.019 0.0095 mm2GX (40a)

and

4

'' -130.019 0.016 mm2GY

(40b)

Note that Equations (40) are in good agreement with Equations (38). The acceleration of the mass center G4 can be written as

4 4 4 4 4

' '' 2 ' '' 22 2 2 2

ˆ ˆG G G G GA X R X R i Y R Y R j (41)

Substituting Equations (32) and (38) into Equation (41), the acceleration of the mass center G4 is

4

2 2 2ˆ ˆ ˆ ˆ0.435 0 0.0095 15 0.753 0 0.016 15 2.1375 3.7 mm/sGA i j i j

(42a)

Therefore, the magnitude and direction of the acceleration of the mass center is

4

24.273 60 mm/sGA

(42b)

(i) 5 Points. From observation, the first-order kinematic coefficients of the linear spring are

' '2 1SR R (1a)

and ' 0S (1b)

Check. The vector loop equation for the spring, see Figure 2, can be written as

1?C

2 0SR R

(1c)

This vector equation has only one unknown since the angle of the spring is constrained to the line of sliding of link 2, that is

2 0S (1d)

Therefore, the length of the spring from Equation (1a) is

2 25 mmSR R (2)

Differentiating Equation (2), the first-order kinematic coefficient of the spring is

' '2 1SR R (3)

Note that differentiating Equation (1b) gives

' 0S (4)

From observation, the first-order kinematic coefficients of the viscous damper are

18

0.43 mm/mmPCR X (5a)

and ' 0 rad/mmC (5b)

Check. The vector loop equation for the damper, see Figure 2, can be written as

??

11 0C PR R R

(5c)

The X and Y components of Equation (5) can be written as

11 11cos cos cos 0C C P PR R R (6a)

and

11 11sin sin sin 0C C P PR R R (6b)

The length of the damper is equal to the position of the input link 2, that is

2 25 mmCR R (7a)

and the damper is horizontal for the given position, that is

0C (7b)

Alternatively, Equation (6) can be written in terms of the X and Y coordinates of point P, that is

11 11cos cos 0C C PR R X (8a)

and

11 11sin sin 0C C PR R Y (8b)


' ' 'cos sin 0C C C C C PR R X (9a)

and ' ' 'sin cos 0C C C C C PR R Y (9b)

Substituting Equations (6) and the first-order kinematic coefficients of point P into Equations (9) gives

' 'cos0 25 sin 0 0.43 0C CR (10a)

and

' 'sin 0 25 cos0 0 0C CR (10b)

Therefore, the first-order kinematic coefficients of the viscous damper are

' 0.43 mm/mmCR (10c)

and ' 0 rad/mmC (10d)

(ii) 8 Points. The potential energy of the mechanism can be written as

19

4

24

1

2 S SOGU m g y K R R (11a)

Substituting the given information into Equation (11a), the potential energy of the mechanism is

23 3 6120 x 9.81x 30 x10 x15x10 25 50 x10 10.573 Nm2U (11b)

The kinetic energy of the mechanism can be written as

22

1

2 EQT m R (11c)

where the equivalent mass of the mechanism can be written as

4 4

2 2 2

2 2j j jEQ j j G G G j

j j

m A m X Y I

(11d)

Considering Equation (11b), the coefficient for link 2 can be written as

2 2 2

2 2 22 2 2G G GA m X Y I (12)

Since the input link 2 is in pure translation in the X-direction then the coefficient is

2 2 24 22 15 kg 1 0 6.25 10 Nms 0 15 kgA (13)

Similarly, the coefficient for link 3 can be written as

3 3 3

2 2 23 3 3G G GA m X Y I (14)

Substituting the given kinematic coefficient, mass, and mass moment of inertia, and knowing that the center of mass of link 3 is coincident with pin A, the coefficient is

22 2 3 2 -13 25 kg 1 0 3 10 Nms 19 m 26.083 kgA (15)

Finally, the coefficient for link can be written as

4 4 4

2 2 24 4 4G G GA m X Y I (16)

Substituting the given kinematic coefficients, mass, and mass moment of inertia, the coefficient is

2 2 23 24 20 kg 0.435 0.753 1.04 10 Nms 0 15.125 kgA (17)

Substituting Equations (13), (15), and (17) into Equation (12), the equivalent mass of the mechanism is

15 26.083 15.125 56.208 kgEQm (18)

Substituting Equation (18) and the input velocity into Equation (11a), the kinetic energy is

20

2 31 x 56.208( 0.015) 6.322 10 Nm2T (19)

(iii) 12 Points. The power equation for the mechanism can be written as

fP P E E

dWdT dUF V F V

dt dt dt

(20a)

or as

2 2

4 4 43 2 2

2 2 2 2 2 22 2 2

ˆ ˆ ˆ ˆ

j

P P P E E E

j j j G S SO S Cj j j

F X i Y j R F X i Y j R

A R R B R m gy R K R R R R CR R

(20b)

Cancelling the input velocity gives the equation of motion for the mechanism, that is

4 4 4

2 22 2 2

2 2 2

ˆ ˆ ˆ ˆjP P P E E E j j j G S SO S C

j j j

F X i Y j F X i Y j A R B R m gy K R R R CR R

(21)

Consider the left-hand side of Equation (21). The first dot product involving the force PF

can be written

as

ˆ ˆ ˆ ˆcos sinP PP F P F P PF i F j X i Y j (22a)

Substituting the given values into this equation, and performing the dot product, gives

2 22 2

ˆ ˆ ˆ ˆ300 N 0.43 0 91.217 Ni j i j (22b)

The second dot product involving the force EF

can be written as

ˆ ˆ ˆ ˆcos sinE EE F E F E EF i F j X i Y j (23a)

The force EF

is parallel to the inclined plane and the kinematic coefficients of point E are the same as

the kinematic coefficients of point B because link 4 is pure translation. Therefore, Equation (23a) can be written as

ˆ ˆ ˆ ˆcos60 sin 60E B BF i j X i Y j (23b)

Note that if EF

is positive then the force is acting up and to the right; and if the force EF

is negative

then the force is acting down and to the left. Substituting the kinematic coefficients of point B into Equation (23a) and performing the dot product gives

312 2

ˆ ˆ ˆ ˆ0.435 0.753 0.8696E EF i j i j F (24a)

Check: The dot product can also be written as

4 0.87E EF R F (24b)

Consider the right-hand side of Equation (21). The B coefficients can be written as

21

4 4 4

2 2 22

12 j j j j j

jj j G G G G G j j

j j j

dAB m x x y y I

dR

(25)

The coefficient for the input link 2 can be written as

2 2 2 2 22 2 2 2G G G G GB m x x y y I (26a)

that is

42 15 kg 1 0 0 0 6.25 10 0 0 0B (26b)

Similarly, the coefficient for link 3 can be written as

3 3 3 3 33 3 3 3G G G G GB m x x y y I (27a)

Substituting the known kinematic coefficients and mass properties of link 3, the coefficient is

3 2 -1 -23 25 kg 1 0 0 0 3 10 Nms 19 m 151.5 m 8.6355 kg/mB (27b)

Finally, the coefficient for link 4 can be written as

4 4 4 4 44 4 4 4G G G G GB m x x y y I (28a)

Substituting the known kinematic coefficients and mass properties of link 4, the coefficient is

-1 -1 3 24 20 kg 0.435 9.5 m 0.753 16 m 1.04 10 Nms 0 0 323.61 kg/mB (28b)

Therefore, substituting Equations (23), (25), and (27) into Equation (22) gives

4

2

0 8.6355 323.61 332.2455 kg/mjj

B

(29)

Considering Equation (21), the gravitational term in the equation of motion may be written as

2 3 4

4

2 3 42

jj G G G Gj

m gy g m y m y m y

(30a)

Substituting the given masses and kinematic coefficients into Equation (30), the gravitational term is

4

2

9.81[15x 0 25x 0 20( 0.753)] 147.739 Njj G

j

m gy

(30b)

From Equation (21), the term from the spring is

S SO SK R R R (31a)

Substituting Equation (2b) and (3b) and the given spring properties into Equation (32) gives

15,000 0.025 0.05 1 375 NS SO SK R R R (31b)

22

From Equation (21), the term for the viscous damper is

22CCR R (32a)

Substituting Equation (10c), the given input velocity and the given damping coefficient gives

222 80,000 0.43 0.015 221.88 NCCR R (32b)

Substituting Equations (18), (22), (24), (26), (28), (30), and (32b) into Equation (21) gives

291.217 + 0.8696 56.208 kg 0 332.2455 0.015 147.739 375 221.88EF (33)

Rearranging this equation, the magnitude of the force acting at point E is

621.47 NEF (34a)

Since the negative sign indicates that the force is acting down and to the left then the force is

621.47 240 NEF

(34b)

23

Solution to Problem 3. Part A. (12 Points). (i) 6 Points. The exact equation for the first two critical speeds of the shaft can be written from Equation (15.74), page 736, as

2

11 1 22 2 11 1 22 2 1 2 11 22 12 21

2 21 2

41 1,

2

a m a m a m a m m m a a a a

(1a)

In terms of the weights of the two pulleys, the exact equation can be written as

2

11 1 22 2 11 1 22 2 1 2 11 22 12 21

2 21 2

41 1,

2

a W a W a W a W WW a a a a

g

(1b)

Substituting the given weights (in N) and influence coefficients (in m/N) into Equation (1b) gives

12

8 8

2 21 2

28 8

8 8 8 8

350 3.85 10 650 13.25 101 1,

2 9.81

350 3.85 10 650 13.25 101

2 9.81 4 350 650 3.85 10 13.25 10 6.75 10 6.75 10

(2a)

that is 6 6 2

2 21 2

1 1, 5.07645 10 4.94793 10 s

(2b)

Therefore, the first critical speed of the shaft is

1 315.843 rad/s (3a)

and the second critical speed of the shaft is

2 2789.388 rad/s (3b)

(ii) 3 Points. The Rayleigh-Ritz equation for approximating the first critical speed of a shaft can be written from Equation (15.66), page 734, as

1 1 2 221 2 2

1 1 2 2

g W y W y

W y W y

(4)

where 1y and 2y are the total deflections of the shaft at locations 1 and 2. The total deflection of the

shaft at location 1 can be written as

1 11 1 12 2y a W a W (5a)

and the total deflection of the shaft at location 2 can be written as

2 21 1 22 2y a W a W (5b)

24

Substituting the given influence coefficients and the weights of the pulleys into Equations (5) gives

8 8 51 3.85 10 m 350 N 6.75 10 m 650 N 5.735 10 my (6a)

and

8 8 42 6.75 10 m 350 N 13.25 10 m 650 N 1.0975 10 my (6b)

Substituting Equations (6) and the given pulley weights into Equation (4), the first critical speed of the shaft is

125 4

1 2 25 4

9.81 350 5.735 10 650 1.0975 10315.997 rad/s

350 5.735 10 650 1.0975 10

(7)

Note from Equations (3a) and (7) that the Rayleigh-Ritz equation does, indeed, give an upper bound for the first critical speed of the shaft. (iii) 3 Points. The Dunkerley approximation for the first critical speed of a shaft can be written can be written from Equation (15.78), page 736, as

1

11 1 22 2

1

a m a m

(8a)

or

1 21

11 22

1W Wg ga a

(8b)

Substituting the given influence coefficients and the weights of the pulleys into Equation (8b), the first critical speed of the shaft is

2 2

18 8m m

N Nm ms s

1313.837 rad/s

350 N 650 N3.85 10 13.25 10

9.81 9.81

(9)

Note from Equations (3a) and (9) that the Dunkerley approximation does, indeed, give a lower bound for the first critical speed of a shaft. Part B. 13 Points. The inertia force caused by a rotating mass particle can be written as

2i i iF m R

(1a)

Since dynamic balancing does not depend on the speed of the shaft then it is more convenient to look at the vector quantity proportional to the inertia force, that is

2i

i i

Fm R

(1b)

From the given masses, radial distances, and angles, the proportional vectors of the inertia forces from the three mass particles are

1 1 5 kg 60 60 mm 300 60 kg-mmm R

(2a)

25

2 2 10 kg 90 225 mm 900 225 kg-mmm R

(2b)

and

3 3 8 kg 120 300 mm 960 300 kg-mmm R

(2c)

Dynamic balancing is achieved by summing moments about a point along the shaft and solving for the correcting masses. To simplify the process, the moments caused by the inertia forces are summed about each correcting plane. (Recall that dynamic balancing ensures static balancing). Taking moments about the correcting plane (1) gives

1 2 3 21 1 2 2 3 3 2 2 0Cm m m m C Cz m R z m R z m R z m R

(3)

Note the distance vector along the shaft, that is iz

, is positive if the mass particle im is to the left of

correcting plane 1 and negative if to the right of correcting plane (1). Substituting the given distances (in mm) and Equations (2c) into Equation (3) gives

2 2 2 2

ˆ ˆˆ ˆ ˆ ˆ35k 300cos 60 300sin 60 60 k 900cos 225 900sin 225

ˆ ˆˆ ˆ ˆ ˆ90 k 960cos300 960sin 300 160 k 85 cos 85 sin 0C C C C

i j i j

i j m i m j

(4)

Performing the cross-product and collecting like terms, the two resulting scalar equations are

2 285 cos 475.8360 kg-mmC Cm (5a)

and

2 285 sin 285.8381 kg-mmC Cm (5b)

Dividing Equation (5b) by Equation (5a) gives

22

2

sin 285.8381tan

cos 475.8360C

CC

(6a)

Therefore, the angular location of the added correcting mass in plane (2) is

12

285.8381tan 329.01

475.8360C

(6b)

Squaring and adding Equations (5a) and (5b) gives

2 2 22 22 2 285 cos sin 475.8360 285.8381C C Cm (7a)

Therefore, the correcting mass for plane (2) is

2 212 85 475.8360 285.8381 6.530 kgCm (7b)

Taking moments about correcting plane (2) gives

1 2 3 11 1 2 2 3 3 1 1 0Cm m m m C Cz m R z m R z m R z m R

(8)

26

Substituting the given distances (in mm) and Equations (2) into Equation (8) gives

1 1 1 1

ˆ ˆˆ ˆ ˆ ˆ125k 300cos 60 300sin 60 100 k 900cos 225 900sin 225

ˆ ˆˆ ˆ ˆ ˆ250 k 960cos300 960sin 300 160 k 85 cos 85 sin 0C C C C

i j i j

i j m i m j

(9)

Performing the cross-product, equating the i and j components, and rearranging the resulting scalar equations gives

1 185 cos 469.4399 kg-mmC Cm (10a)

and

1 185 sin 1493.8110 kg-mmC Cm (10b)

Dividing Equation (10b) by Equation (10a) gives

11

1

sin 1493.8110tan

cos 469.4399C

CC

(11a)

Therefore, the angular location of the added correcting mass in plane (1) is

11

1493.4399tan 107.45

469.4399C

(11b)

Squaring and adding Equations (10a) and (10b) gives

2 2 22 21 1 185 cos sin 469.4399 1493.8110C C Cm (12a)

Therefore, the correcting mass for plane (1) is

2 211 85 469.4399 1493.8110 18.422 kgCm (12b)

Note that Equations (6b), (7b), (11b) and (12b) give the magnitudes and angular locations of correcting masses to be added in planes (1) and (2). For removing masses in the correcting planes, the mass magnitudes are given by Equations (7b) and (12b) and the angular locations of the removed correcting masses are given as

Remove Add

180Ci Ci (13)

Substituting Equations (6b) and (11b) into Equation (13) gives

1 Remove107.45 180 287.45C (14a)

and

2 Remove329.01 180 149.01C (14b)

Combining Equations (12b) and (14a), the correcting mass to be removed in plane (1) is

1 Remove18.422 kg at 287.45Cm (15a)

27

Combining Equations (7b) and (14b), the correcting mass to be removed in plane (2) is

2 Remove6.530 kg at 149.01Cm (15b)

The angular orientations of the correcting masses to be removed are shown in Figure 3.

Figure 3. The locations of the correcting masses to be removed.

28

Solution to Problem 4. (i) 10 Points. The X and Y-components of the primary shaking force, see page 806, can be written as

cos sinXPS A B (1a)

and cos sin

YPS C D (1b)

where the coefficients are

3

1

cos cosi i i ii

A P

(2a)

3

1

sin cosi i i ii

B P

(2b)

3

1

cos sini i i ii

C P

(2c)

and

3

1

sin sini i i ii

D P

(2d)

where i is the angle from the X-axis to the line of sliding of piston i, and i is the angle from the

reference line on crank 1 to crank i. From Figure 4, the angles are specified as

1 2 390 , 240 , 300 (3a)

and

1 2 30 , 180 , 300 (3b)

Since the pistons have the same effective mass m and the cranks have the same length R then the inertia force of each piston can be written as

2iP P mR (4)

Substituting Equations (3) into Equations (2), the coefficients are

14cos 90 0 cos90 cos 240 180 cos 240 cos 300 300 cos300A P P (5a)

34sin 90 0 cos90 sin 240 180 cos 240 sin 300 300 cos300B P P (5b)

3 34cos 90 0 sin 90 cos 240 180 sin 240 cos 300 300 sin 300C P P (5c)

and

14sin 90 0 sin 90 sin 240 180 sin 240 sin 300 300 sin 300D P P (5d)

Substituting the effective mass of the pistons 1 2 3 8 kg,m m m m the crank length 0.2 m,R and

the angular velocity 125 rad/s into Equation (4), the inertia force of each piston is

28 kg 0.2 m 125 rad/s 25,000 NP (6)

29

Substituting Equations (5) and (6) into Equations (1), and X and Y-components of the primary shaking force in terms of the crank angle θ are

314 425,000 cos 25,000 sin 6,250cos 6,250 3 sin N

XPS (7a)

and

3 3 14 425,000 cos 25,000 sin 18,750 3 cos 6,250sin N

YPS (7b)

(ii) 10 Points. The magnitudes of the two inertia forces can be written as

1

22 211 2F A D B C (8a)

and

1

22 212 2F A D B C (8b)

Substituting Equations (5) into Equations (8), the magnitudes of the inertia forces, in terms of P, are

1

222 3 3 31 1 1 11 2 4 4 4 4 2F P P P P P

(9a)

and

1

222 3 3 3 31 1 12 2 4 4 4 4 2F P P P P P

(9b)

The direction of the first inertia force can be written as

1 1( )F

(10a)

where

1

11

sintan

cos

B C

A D

(10b)

The direction of the second inertia force can be written as

2 2F

(11a)

where

2

22

sintan

cos

B C

A D

(11b)

Substituting Equations (5) into Equations (10b) and (11b), the angles are

3 3 3

1 4 41 1 1

4 4

tan 120P P

P P

(12a)

and

3 3 3

1 4 42 1 1

4 4

tan 270P P

P P

(12b)

30

Therefore, the inertia forces can be written from Equations (9) and (12) as

11 2 120 NF P

and 32 2 270 NF P

(13a)

Substituting Equation (6) into Equations (13a), the inertia forces can be written as

1 12,500 120 NF

and 2 12,500 3 270 NF

(13b)

(iii) 5 Points. The correcting masses can be expressed in terms of the inertia forces as

2i

i

iC

C

Fm

R (14)

Substituting Equations (14), 1 2

0.3 mC C CR R R and the angular velocity of the crank into Equation

(15), the correcting masses are

1 2

12,500 8kg 2.6667 kg

30.3 125Cm (15a)

and

2 2

12,500 3 8 3kg 4.6188 kg

30.3 125Cm (15b)

The locations of the correcting masses for an arbitrary crank angle are shown in Figure 4a.

Figure 4a. The locations of the two correcting masses for an arbitrary crank angle.

31

(iv) 5 Points. The locations of the correcting masses for the crank angle 30 are shown in Figure 4b.

Figure 4b. The locations of the correcting masses for the crank angle 30 .

32

(v) 5 Points. The magnitude and the direction of the primary shaking force can be written as

2 2

X YP P PS S S (17a)

and

1tan Y

X

P

P

S

S

(17b)

Substituting θ = 30° into Equations (7), the X and Y-components of the primary shaking force are

0 NXPS (18a)

and 25,000 N

YPS (18b)

Then substituting Equations (9) into Equations (8), the magnitude and direction of the primary shaking force are

2 20 25,000 25,000 NPS (19a)

and

1 25,000tan 270

0

(19b)

The primary shaking force caused by the unbalanced three cylinder engine is shown in Figure 4c.

Figure 4c. The primary shaking force for the crank angle 30 .

me 352 - machine design i name of student summer semester ... · 6 3 f14y 160 10 0 (4b) that is 3...

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