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ME 352 - Machine Design I Name___________________________________ Fall Semester 2009 Lab. Div.________________________________

FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Friday, December 18th, 2009 Write your solutions on the blank paper that is provided. Write on one side of the paper only. Where necessary, you can use the figures that are provided to show vectors. Staple each problem separately, and attach your crib sheet to the end of your solution to Problem 1. Problem 1 (25 Points). For the mechanism in the position shown in Figure 1, the known kinematic coefficients are 4R 0.26 m / rad,′ = − 34R 0.37 m / rad′ = + , 2

4R 0.45 m / rad ,′′ = − 234R 0.37 m / rad′′ = −

(where 2 44R G G= and 4 334R G G= ), 3

0.35 m/rad,GX ′ = − 3

0.35 m/rad,GY ′ = + 2

30.35 m/rad ,GX ′′ = −

and 2

30.35 m/rad .GY ′′ = − The geometry, the free length and spring rate of the linear spring, the damping

constant of the viscous damper, and the masses and second moments of mass about the mass centers of links 2, 3, and 4 are:

2G A 2 4G G AB OSR K C 2m 3m 4m

2GI 3GI

4GI

m m m m N/m Ns/m kg kg kg 2kg-m 2kg-m 2kg-m0.5 0.25 0.5 0.6 50 15 0.5 5 4 3 7 6

If gravity is acting vertically downward and friction can be neglected then determine:

(i) The kinetic energy of the mechanism. (ii) The first-order kinematic coefficient of the linear spring. (iii) The first-order kinematic coefficient of the viscous damper. (iv) The magnitude and direction of torque 12T acting on input link 2 from the equation of motion.

Figure 1. A planar mechanism.

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ME 352 - Machine Design I Name_____________________________ Fall Semester 2009 Lab. Div.__________________________

Problem 2 (25 Points).

Part A. The weights of masses 1m and 2m which are rigidly attached to the rotating shaft shown in Figure 2 are 140 N and 60 N, respectively. The shaft is rotating counterclockwise with a constant angular velocity ω 100 rad/s= . From a deflection analysis, the influence coefficients for the shaft are

116a 2 10 cm/N,−= × 22

6a 12 10 cm/N,−= × and 12 216a a 4 10 cm/N.−= = × Assuming that the mass of the

shaft can be neglected then determine: (i) The first and second critical speeds of the shaft using the exact equation. Is the operating speed of the shaft acceptable? (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. (iv) The first critical speed of the shaft if the mass 1m is moved to location 2 and the mass 2m is moved to location 1.

Figure 2. A rotating shaft with two mass disks. Part B. The first and second critical speeds of a rotating shaft with two flywheels rigidly attached are

1 375 rad/sω = and 2 615 rad/s.ω = The weights of the two flywheels are 1W 65 N= and 2W 80 N= and the known influence coefficients of the shaft are 5

11a 5.90 x 10 cm/N−= and 521a 2.74 x 10 cm/N.−=

Determine the influence coefficient 22a .

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ME 352 - Machine Design I Name_____________________________ Fall Semester 2009 Lab. Div.__________________________

Problem 3 (25 Points). Part A. The shaft shown in Figure 3(a) is rotating with a constant angular velocity ω 80 rad/s.= The masses of the two particles are 1m 4 kg= and 2m 2 kg= and the radial distances are 1R 30 mm= and

2R 50 mm.= 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 65 mm,= = for dynamic balance. Show the angular orientations of the correcting masses on the right hand side of Figure 3(a).

Figure 3(a). A rotating shaft with two mass particles.

Part B. The system shown in Figure 3(b), rotating with a constant angular velocity ω 50 rad/s,= is dynamically balanced by adding correction masses 1m 3 kg= and 2m 2 kg= at radii 1R 5 cm= and

2R 15 cm.= A decision has been made to use planes 1 and 3 instead of planes 1 and 2. Determine the magnitudes and locations of the new correcting masses 1 new(m ) and 3 new(m ) at the same radii

1R 5 cm= and 3R 15 cm.= Show the locations of these masses on the left hand side of Figure 3(b).

Figure 3(b). A continuous mass system on a rotating shaft.

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ME 352 - Machine Design I Name_____________________________ Fall Semester 2009 Lab. Div.__________________________ Problem 4 (25 Points). The effective mass of each piston in the two-cylinder engine shown in Figure 4 is 1 2m m m 5 kg,= = = the length of each connecting rod is L 55 cm,= and the length of the throw of each crank is R 10 cm.= The crankshaft is rotating counterclockwise with a constant angular velocity ω θ 45 rad/s.= = Part A. Determine the magnitude and direction of the primary shaking force in terms of crank angle θ. Part B. If correcting masses are required to balance the primary shaking force then determine: (i) The magnitudes and directions of the inertial forces created by these correcting masses. (ii) The magnitudes and angular locations of the correcting masses. Given C C1 2

R R 20 cm.= =

Part C. Show your answers to Parts A and B on the right-hand figure below. Note that for this part of the problem, the reference line is specified at the crank angle θ 0 .o=

Figure 4. A two-cylinder engine.

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Solution to Problem 1. The vector loop for this mechanism is shown in Figure 1.

Figure 1. The vector loop for kinematic analysis of mechanism. The vector loop equation (VLE) can be written as

2 34 4 0R R R− − = (1)

where 2 44R G G= and 4 334R G G= . The X and Y components of Equation (1) can be written as

2 2 34 34 4 4cos cos cos 0R R Rθ θ θ− − = (2a) and

2 2 34 34 4 4sin sin sin 0R R Rθ θ θ− − = (2b) Differentiating Equations (2) with respect to the input position θ2 gives

2 2 34 34 4 4sin cos cos 0R R Rθ θ θ′ ′− − − = (3a) and

2 2 34 34 4 4cos sin sin 0R R Rθ θ θ′ ′+ − − = (3b) Equations (3) can be written in matrix form as

34 4 34 2 2

34 4 4 2 2

cos cos sinsin sin cos

R RR R

θ θ θθ θ θ

′− − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′− − −⎣ ⎦ ⎣ ⎦⎣ ⎦

(4)

Substituting 2 45 ,θ = ° 4 0 ,θ = ° 34 105 ,θ = ° and 2 0.5m,R = into Equation (4) gives

34

4

0.2588 1 0.35360.9659 0 0.3536

RR′+ − +⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′− −⎣ ⎦ ⎣ ⎦⎣ ⎦ (5)

Using Cramers rule, the first-order kinematic coefficients are

34R 0.37 m / rad′ = + and 4R 0.26 m / rad′ = − (6) Differentiating Equations (3) with respect to the input position θ2 gives

2 2 34 34 4 4cos cos cos 0R R Rθ θ θ′′ ′′− − − = (7a) and

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2 2 34 34 4 4sin sin sin 0R R Rθ θ θ′′ ′′− − − = (7b) Equations (7) can be written in matrix form as

34 4 34 2 2

34 4 4 2 2

cos cos cossin sin sin

R RR R

θ θ θθ θ θ

′′− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′′− − ⎣ ⎦ ⎣ ⎦⎣ ⎦

(8)

Substituting the known data into Equation (8) gives

34

4

0.2588 1 0.35360.9659 0 0.3536

RR′+ − +⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′− +⎣ ⎦ ⎣ ⎦⎣ ⎦ (9)

Using Cramers rule, the second-order kinematic coefficients are

234R 0.37 m / rad′′ = − and 2

4R 0.45 m / rad′′ = − (10) (i) 8 Points. The kinetic energy of the mechanism can be written as

22

1T2 EQI θ= (1a)

where the equivalent mass moment of inertia of the mechanism can be written as

2 3 4A A AEQI = + + (1b)

For Link 2: ( )2 2 22 2 G2 G2 G2 2A m X Y I′ ′ ′= + + θ (2)

The X and Y components of the center of mass of the input link 2 are

2 0GX = and 2 0GY = (3a)

Differentiating Equations (3a) with respect to the input position 2θ gives

2 0GX ′ = and 2 0GY ′ = (3b) Substituting Equations (3b) and the known data into Equation (2) gives

( ) ( )22 2 22 0.5 0 0 3 1 3 kg-mA = + + = + (4)

For Link 3: ( ) 233

23

2333 θ ′+′+′= GGG IYXmA (5)

The vector loop for the center of mass of link 3 can be written as

?? ?

3 2

V

GR R= (6)

The X and Y components of Equation (6), with 2 0.5 mR = and 2 45θ = ° , are

3 2 2cos 0.354GX R mθ= = and 3 2 2sin 0.354GY R mθ= = (7)

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Differentiating Equations (7) with respect to the input position 2θ gives

3 2 2sin 0.354 m/radGX R θ′ = − = − (9a) and

3 2 2cos 0.354 m/radGY R θ′ = + = + (9b)

Also, differentiating Equations (9) with respect to the input position 2θ gives

23 2 2cos 0.35 m/radGX R θ′′ = − = − (10a)

and 2

3 2 2sin 0.35 m/radGY R θ′′ = − = − (10b) Substituting Equations (9) and the known data into Equation (6) gives

2]32 2 2A 5[( 0.35) ( 0.35) 7 (0) 1.225 kg m= − + + + = + − (11)

For Link 4: ( ) 244

24

2444 θ ′+′+′= GGG IYXmA (12a)

Since XG 44

R 0.26 m / rad′ ′= = − and YG40′ = then Equation (12a) can be written as

( ) ( )2 22 24 4 [ 0.26 0 ] 6 0 0.27 kg mA = − + + = + − (12b)

Substituting Equations (5), (11), and (12b) into Equation (1b), the equivalent mass moment of inertia is

23 1.225 0.27 4.50 kg mEQI = + + + = − (13) Then substituting Equation (13) and the input angular velocity into Equation (1a), the kinetic energy is

21T (4.50) ( 5) 56.3 Nm2

= − = (14) (ii) 7 points. The vector loop for the linear spring can be written as

1 2

??0S

VV VVR R R− − = (15)

where 1R is the vector from pin O2 to pin B. The X and Y components of Equation (15) are

1 1 2 2cos cos cos 0S SR R Rθ θ θ− − = (16a) and

1 1 2 2sin sin sin 0S SR R Rθ θ θ+ − = (16b)

Differentiating Equations (16) with respect to the input position 2θ gives

2 2sin sin 0S S S S SR cos R Rθ θ θ θ′ ′− + + = (17a) and

2 2sin cos cos 0S S S S SR R Rθ θ θ θ′ ′− − − = (17b) Then writing Equations (17) in matrix form and using Cramer’s rule gives

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2 2

2 2

sin sincos cos

sin coscos sin

S S

S SS

S S S

S S S

R RDET

R RR

RDET

R

θ θθ θθ θθ θ

+ −⎡ ⎤⎢ ⎥− +⎣ ⎦′ =

+ −⎡ ⎤⎢ ⎥− −⎣ ⎦

(18a)

Therefore, the first-order kinematic coefficient of the spring can be written as

2 22 2

sin( ) sin( )S SS S

S

R RR RRθ θ θ θ− −′ = = + −

− (18b)

Since 0Sθ = ° then Equation (18b) can be written as

2 2 3sinS GR R Yθ′ = + = (18c)

Therefore, the first-order kinematic coefficient of the spring is

30.5sin 45 0.354 m/radS GR Y′ = = + ° = + (18d)

The positive sign indicates that for the negative input, the length of the linear spring is decreasing. Check: From a study of the velocity of point A, the first-order kinematic coefficient of the spring is

3( 0.35) 0.35 m/radS GR X′ ′= − = − − = + (18e)

(iii) 3 points. The vector for the viscous damper can be written as

?

4

V VV

CR R= (19)

Differentiating Equation (19) with respect to the input position 2θ gives

4 0.26 m/radCR R′ ′= = − (20a) The negative sign indicates that the length of the damper is increasing for the negative input. Note that the velocity of G4 is not the same as the velocity of G3 , i.e., the first-order kinematic coefficient of the damper is

4 30.35 m/radC G GR X X′ ′ ′= ≠ ≠ − (20b)

(iv) 7 points. The power equation can be written as

12 2T . F .V fB B

dWdT dUdt dt dt

ω + = + + (21a)

or as

( )4 4 4

3 2 212 2 2 2 2 2 0 2

2 2 2 2T . F .VB B j j j G j s S S S C

j j j

A B m g Y K R R R C Rω α ω ω ω ω ω= = =

′ ′ ′+ = + + + − +∑ ∑ ∑ (21b)

or as

( )4 4 4

3 2 212 2 4 2 2 2 2 0 2

2 2 2 2T F RB j j j G j s S S S C

j j jA B m g Y K R R R C Rω α ω ω ω ω ω

= = =

′ ′ ′+ + = + + + − +∑ ∑ ∑ (21c)

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Note in Equation (21c) that the unknown direction of the torque is assumed to be acting in the same direction as the given angular velocity of the input link 2, that is, in the clockwise direction. Canceling the input angular velocity, the equation of motion is

( )4 4 4

2 212 4 2 2 0 2

2 2 2T FB j j j G j s S S S C

j j jR A B m g Y K R R R C Rα ω ω

= = =

′ ′ ′ ′+ + = + + + − +∑ ∑ ∑ (22)

The sum of the B terms can be written as

4

2( )

j j j j jj j G G G G G j jj

B m X X Y Y I θ θ=

′ ′′ ′ ′′ ′ ′′= + +∑ (23)

For Link 2: 2 2 2 2 2 2 2 2 2( )G G G G GB m X X Y Y I θ θ′ ′′ ′ ′′ ′ ′′= + + (24)

Differentiating Equations (5) with respect to the input position 2θ gives

2 2 0GX R′′ ′′= = and 2 0GY ′′ = (25) Substituting Equations (25) and the given data into Equations (24) gives

2 0B = (26)

For Link 3: ( )3 3 3 3 3 3 3 3 3G G G G GB m X X Y Y I θ θ′ ′′ ′ ′′ ′ ′′= + + (27a) Therefore, Equation (27a) can be written as

3 5 [( 0.35)( 0.35) (0.35)( 0.35)] 7(0)(0) 0B = − − + − + = + (27b)

For Link 4: ( ) 444444444 θθ ′′′+′′′+′′′= GGGGG IYYXXmB (28a)

Since XG 44R 0.45 m / rad′′ ′′= = − and YG4

0,′′ = Equation (28a) can be written as

( ) ( ) 2

4B 4 0.26 0.45 0.468 kg m= − − = + − (29) Substituting Equations (26), (27), and (29) into Equation (14) gives

42

2 3 42

0 0 0.468 0.468 kg-mjj

B B B B=

= + + = + + = +∑ (30)

The change in the potential energy due to gravity. For link 2:

22 (0.5)(9.81)(0) 0Gm g Y ′ = = (31a)

For link 3: 3 3

(5)(9.81)( 0.35) 17.17 NmGm g Y ′ = + = (31b) For link 4:

4 4(4)(9.81)(0) 0Gm g Y ′ = = (31c)

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Adding Equations (31) gives

4

20 17.17 0 17.17 Nmj G j

jm g Y

=

′ = + + = +∑ (32)

Substituting 2 5 rad / sω = − , 22 3 rad / sα = + , Equations (13a), (18), (20), (30), and (32) and the known

data into Equation (33), the torque acting on link 2 is

( )2 212T 250( 0.26) 4.50( 3) 0.468( 5) 17.17 50 0.5 0.6 ( 0.35) 15 ( 0.26) ( 5) Nm+ − = + + − + + − + + − − (33)

The torque can be written as

12T 65 13.50 11.7 17.17 1.75 5.07 Nm− = + + + − −

(34a) or as

12 65 35.6 NmT − = (34b) or as

12 100.6 NmT = + (34c) Recall that the torque was assumed to be acting in the same direction as the angular velocity of the input link, that is, the clockwise direction. Therefore, the positive sign in Equation (34c) indicates that this assumption was wrong, that is, the torque T12 is, in fact, acting in the counterclockwise direction.

The torque acting on link 2 is

12 100.6 NmT k= − (35)

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Solution to Problem 2. Part A. (i) 6 Points. The exact equation for the first and second critical speeds can be written as

211 1 22 2 11 1 22 2 11 22 12 21 1 2

2 21 2

( a m a m ) ( a m a m ) 4 ( a a a a ) m m1 1,ω ω 2

+ ± + − −=

(1) Substituting the influence coefficients and the masses into Equation (1) gives

8 8 8 8 2 8 8 8 8

2 21 2

(2 10 )(140) (12 10 )(60) [(2 10 )(140) (12 10 ) (60)] 4[ (2 10 )(12 10 ) (4 10 )(4 10 )](140)(60)1 1,ω ω 2x9.81

− − − − − − − −× + × ± × + × − × × − × ×= (2a)

which can be written as

8 8 2 16 16

2 21 2

1000 10 (1000 10 ) 4[ (24 10 16 10 )8400]1 1,ω ω 2x9.81

− − − −× ± × − × − ×= (2b)

or as 8 6 16 16

2 21 2

1000 10 10 10 (32 10 )84001 1,ω ω 2x9.81

− − −× ± × − ×= (2c)

Simplifying this equation gives

8 16

2 21 2

1 1 (1000 10 ) 731200 10,ω ω 2x9.81

− −× ± ×= (3a)

which can be written as 8 2 8 2

2 21 2

1 1, 94.55 10 , 7.385 10ω ω

s s− −= × × (3b)

Therefore, the first critical speed can be written is

2 2 21 8

1ω /94.55 10

rad s−=× (4a)

that is 1ω 1028.42 /rad s= (4b)

The second critical speed can be written as

2 2 22 8

1ω /7.385 10

rad s−=× (5a)

that is 2ω 3679.72 /rad s= (5b)

The operating speed of the shaft is much less than the first critical speed of the shaft. Therefore, the operating speed of the shaft is acceptable. (ii) 5 Points. The Rayleigh-Ritz equation can be written as

2 1 1 2 21 2 2

1 1 2 2

W x W xω gW x W x⎡ ⎤+

= ⎢ ⎥+⎣ ⎦ (6)

where the deflections are

12

6 6 81 11 1 12 2x a W a W 2 10 140 4 10 60 cm 520 x 10 m− − −= + = × × + × × = (7a)

and 6 6 8

2 21 1 22 2x a W a W 4 10 140 12 10 60 cm 1280x 10 m− − −= + = × × + × × = (7b) Substituting Equations (7) into Equation (6), the Rayleigh-Ritz equation can be written as

8 82 2 21 8 2 8 2

140 (520 x 10 ) 60 (1280 x 10 )ω 9.81 rad /s140 (520 x 10 ) 60 x(1280 x 10 )

− −

− −

⎡ ⎤+= ⎢ ⎥+⎣ ⎦

(8a)

which can be written as

2 2 21

7728 768ω 9.81 x 10 rad /s37856 98304⎡ ⎤+

= ⎢ ⎥+⎣ ⎦ (8b)

or as 2 2 21

79.81 x 1496 x 10ω rad /s136160

= (8b) Therefore, the first critical speed of the shaft is

1ω 1038.19 rad / sec= (9) Note that the Rayleigh-Ritz equation to the first critical speed of the shaft is greater than the exact answer, see Eq. (4b). This is consistent with the fact that the Rayleigh-Ritz equation is an upper bound. (iii) 4 Points. The Dunkerley approximation to the first critical speed of the shaft can be written as

11 1 22 221

1 a m a mω

= + (10)

Substituting the numerical values into Equation (10), the Dunkerley approximation to the first critical speed of the shaft is

8 8 221

1 140 602 10 12 10 secω 9.81 9.81

− −= × × + × × × (11a)

which can be written as

8 8 8 221

1 28.5423 10 73.3945 10 101.9368 10 secω

− − −= × + × = × (11b)

Therefore, the Dunkerley approximation to the first critical speed of the shaft is

1ω 990.45 rad / sec= (12) Note that the the Dunkerley approximation to the first critical speed of the shaft is less than the exact answer, see Eq. (4b). This is consistent with the fact that the Dunkerley approximation is a lower bound to the first critical speed. (iv) 5 Points. When the two masses are interchanged then Equation (1) can be written as

211 2 22 1 11 2 22 1 11 22 12 21 1 2

2 21 2

( a m a m ) ( a m a m ) 4 ( a a a a ) m m1 1,ω ω 2

+ ± + − −= (13)

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Substituting the influence coefficients and the masses into Equation (13), the first and second critical speeds of the new system can be written as

8 8 8 8 2 8 8 8 8

2 21 2

( (2 10 ) (60) (12 10 )(140)) ((2 10 ) (60) (12 10 ) (140)) 4( (2 10 )(12 10 ) (4 10 )(4 10 ) ) (140)(60)1 1,ω ω 2x9.81

− − − − − − − −× + × ± × + × − × × − × ×= (14)

Simplifying this equation gives

8 8 2 16 16

2 21 2

(1800 10 ) (1800 10 ) 4( (24 10 16 10 )(8400)1 1,ω ω 2x9.81

− − − −× ± × − × − ×= (15a)

which can be written as

8 6 16 16

2 21 2

(1800 10 ) 3.24 10 10 (32 10 ) (8400)1 1,ω ω 2x9.81

− − −× ± × × − ×= (15b)

Therefore, the first and second critical speeds can be written as

8 16

2 21 2

1 1 (1800 10 ) 2971200 10,ω ω 2x9.81

− −× ± ×= (16a)

that is 8 2 8 2

2 21 2

1 1, 179.60 10 , 3.89 10ω ω

s s− −= × × (16b)

The first critical speed can be written as

2 2 21 8

1ω rad /s179.60 10−

=× (17a)

Therefore, the first critical speed is

1ω 746.19rad/s= (17b) Note that this answer is less than the answer in Part (i). Part B. 6 Points. The sum of the roots (i.e., the first and second critical speeds squared) can be written from Equation (1) as

11 1 22 22 21 2

1 1+ = +a m a m

ω ω (1)

Substituting the known values into Equation (1) gives

5222 2

1 1 65 80(5.9 x 10 )375 615 981 981

a−+ = + (2)

Rearranging this equation gives 5

22 2 2

80 1 1 65(5.9 x 10 )981 375 615 981

a −= + − (3)

which can be written as

14

6 6 6220.08155 7.111 10 2.644 10 3.909 x 10a − − −= × + × − (4a)

or as 6

220.08155 5.846 10a −= × (4b)

Therefore, the influence coefficient is

622 71.69 10 cm/Na −= × (5a)

or 5

22 7.17 10 cm/Na −= × (5b)

Check: Substitute Equation (5b) and the given influence coefficient 511a 5.90 x 10 cm/N−= and

521a 2.74 x 10 cm/N−= into the exact equation, that is

2

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

( a m a m ) ( a m a m ) 4 ( a a a a ) m m1 1,ω ω 2

+ ± + − −= (6)

This gives the same answers for the first and second critical speeds, that is

1 375 rad/sω = and 2 615 rad/sω =

15

Solution to Problem 3. Part A. 11 Points. The inertial forces of the two rotating mass particles are

2 21 1 1 (4 kg)(0.03 m)(80 rad/s) 768 NF m Rω= = = (1a)

and 2 2

2 2 2 (2 kg)(0.05 m)(80 rad/s) 640 NF m R ω= = = (1b) Therefore, the inertial forces of the two rotating mass particles can be written as

1 768 N 180 768 0 NF i j= ∠ ° = − + (2a) and

2 640 N 120 320 554.26 NF i j= ∠ ° = − + (2b) The sum of the inertial forces of the two rotating mass particles can be written as

1 2 1088 554.26 N 1221.04 N 153.00F F F i j= + = − + = ∠ °∑ (3) Therefore, the reaction forces at bearings A and B can be written as

( )1 2 1088 554.26 NA BF F F F i j+ = − + = + − (4)

The sum of the moments about bearing A can be written as

0.25 0.4 ( 768 0 ) 0.6 ( 320 554.26 ) 0Bk F k i j k i j− × − × − + − × − + = (5) The X and Y components of Equation (5) can be written as

0.25 0.4 ( 768) 0.6( 320 )BXF = − − − − (6a) and

0.25 0.4( 0) 0.6(554.26 )BYF = − − (6b) Therefore, the X and Y components of the force at bearing B are

1996.8 NBXF = and 1330.22 NBYF = − (7a)

Therefore, the force at bearing B can be written as

1996.8 1330.22 NBF i j= + − (7b) Substituting Equation (7b) into Equation (4) gives

1996.8 1330.22 1088 554.26 NAF i j N i j+ − = + − (8a) Therefore, the force at bearing A can be written as

908.8 775.96 NAF i j= − + (8b) The sum of the two correcting forces can be written from Equation (4) as

( )1 2 1 2 1088 554.26 NC CF F F F i j+ = − + = + − (9)

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The sum of the moments about correcting plane (2) can be written as

1 10.2 0.2 0Ck F k F× + × = (10a) or as

1 1 0 768 NC Y C XF i F j i j− = − (10b) Therefore, the inertial force in correcting plane (1) can be written as

1 768 0 768 N 0CF i j= + = ∠ ° (11) The correcting mass in correcting plane (1) can be written as

11 2

1

= CC

C

FmRω

(12a)

which can be written as

1 2

768 N 1.846 kg(80 rad/s) (0.065 m)Cm = = (12b)

The orientation of the correcting mass in correcting plane (1) is

1 0Cθ = ° (13a) The orientation of the mass from the X-axis, that is to be removed in correcting plane (1), is

2 0 180 180Cθ = °+ ° = ° (13b) Substituting Equation (11) into Equation (9), the force in the second correcting plane is

2 1088 554.26 768 0 320 554.26 NCF i j i j i j= − − + = − (14a) which can be written as

2 640 N 300CF = ∠ ° (14b) The correcting mass in correcting plane (2) can be written as

22 2

2

= CC

C

FmRω

(15a)

which can be written as

2 2

640 N 1.538 kg(80 rad/s) (0.065 m)Cm = = (15b)

The orientation of the correcting mass in the correcting plane (2) is

2 300Cθ = ° (16a) The orientation of the mass from the X-axis, that is to be removed in correcting plane (2), is

2 300 180 120Cθ = °− ° = ° (16b) These answers are shown in Figure 3.1.

17

Figure 3.1. The locations of the correction masses.

Part B. 14 Points. The inertial forces due to the two original correcting masses are

2 2 21 1 1 (3 kg)(0.05 m) 0.15 NF m Rω ω ω= = = (1a)

and 2 2 2

2 2 2 (2 kg)(0.15 m) 0.3 NF m R ω ω ω= = = (1b) Therefore, the inertial forces due to the two original correcting masses can be written as

2 21 0.15 N 90 0 0.15 NF i jω ω= ∠ ° = + (2a)

and 2 2 2

2 0.3 N 300 0.15 0.260 NF i jω ω ω= ∠ ° = + − (2b) The sum of the inertial forces due to the two original correcting masses can be written as

2 2 21 2 0.15 0.11 N 0.186 N 323.75F F F i jω ω ω= + = + − = ∠ °∑ (3)

Therefore, the reaction forces at the two bearings A and B can be written as

( ) 2 21 2 0.15 0.11 NA BF F F F i jω ω+ = − + = − + (4)

Two Procedures: (i) For dynamic balance, the sum of the moments about the correcting planes (1) and (3) due to the new correcting masses must be the same as the sum of the moments about the old correcting planes (1) and (2) due to the original correcting masses. (ii) Also for dynamic balance, the sum of the moments about any point in the shaft must be zero. Therefore use the sum of the moments to determine the reaction forces at A and B for the unbalanced system. Then use these answers to find the correcting masses in planes (1) and (3).

For example, use procedure (i). Consider the sum of the moments about the correcting plane (3).

1 1 20.18 ( ) 0.18 0.08newk F k F k F× = × + × (5) Substituting Equations (2) into Equation (5) and performing the cross-products, the X and Y components of Equation (6a) can be written as

18

2 21 1( ) ( ) 0.0344 0.0666Y new X newF i F j i j Nω ω− + = − + (6)

Rearranging Equation (6b), the inertial force in the correcting plane (1) can be written as

2 2 21( ) 0.0666 0.0344 0.075 N 27.3newF i jω ω ω= + + = ∠ ° (7)

The correcting mass in the correcting plane (1) can be written as

11 new 2

1

( ) FmRω

= (8a)

which can be written as 2

1 new 2

0.075 N( ) 1.5 kg(0.05 m)

m ωω

= = (8b)

The orientation of the correcting mass in correcting plane (1) is

1 27.3Cθ = + ° (9)

The sum of the moments about the correcting plane (1) due to the original correcting masses is equal to the sum of the moments about the same correcting plane due to the new correcting masses, that is

3 20.18 ( ) 0.1newk F k F− × = − × (10a) or as

2 23 3( ) ( ) 0.1444 0.0833 NY new X newF i F j i jω ω− + = + + (10b)

Rearranging Equation (10b), the inertial force in correcting plane (3) can be written as

2 2 23( ) 0.0833 0.1444 0.167 N 300newF i jω ω ω= − = ∠ ° (11)

The correcting mass in correcting plane (3) can be written as

33 new 2

3

( )( ) newFmRω

= (12a)

which can be written as 2

3 new 2

0.167 N( ) 1.113 kg(0.15 m)

m ωω

= = (12b)

The orientation of the correcting mass in the correcting plane (3) is

3 300Cθ = ° (13) The correcting masses for the continuous mass system are shown in Figure 3.2. Check: Use Procedure (ii). For dynamic balance, the sum of the two new inertial forces must be the same as the sum of the two original inertial forces, that is

1 3 1 2( ) ( )new newF F F F+ = + (14) Substituting Equations (7) and (9) into the left hand side of Equation (14) gives

2 2 2 21 3( ) ( ) 0.0666 0.0334 0.0833 0.1444 Nnew newF F i j i jω ω ω ω+ = + + + − (15a)

19

which can be written as

2 21 3( ) ( ) 0.15 0.11 Nnew newF F i jω ω+ = + − (15b)

Note that this agrees with the answer given by Equation (3).

Figure 3.2. The correcting masses for the continuous mass system.

20

Solution to Problem 4. (i) 5 Points. The X and Y components of the primary shaking force for cylinder 1 can be written as

1 1 1 1cos( )cosXS P θ ψ ψ= − (1a) and

1 1 1 1cos( )sinYS P θ ψ ψ= − (1b) The X and Y components of the primary shaking force for cylinder 2 can be written as

2 2 2 2 2cos( )cosXS P θ ψ φ ψ= − + (2a) and

2 2 2 2 2cos( )sinYS P θ ψ φ ψ= − + (2b)

From Figure 4.1, the angles 1 45 ,ψ = + ° 2 135 ,ψ = ° and 2 180 .φ = ° Substituting these angles and 2

1 2P P P m Rω= = = into Equations (1aa) and (2a), the X-component of the resultant of the primary shaking force can be written as

1 2S S SX X X= + (3a) or as

cos( 45 )cos 45 cos( 135 180 )cos135XS P Pθ θ= − ° °+ − °+ ° ° (3b) Therefore, the X-component of the resultant of the primary shaking force is

0 cos 1 sinXS P Pθ θ= + (3c) Also, substituting the angles into Equations (1b) and (2b), the Y-component of the resultant of the primary shaking force can be written as

1 2S S SY Y Y= + (4a) or as

cos( 45 )sin 45 cos( 135 180 )sin135YS P Pθ θ= − ° ° + − °+ ° ° (4b) Therefore, the Y-component of the resultant of the primary shaking force is

1 cos 0 sinYS P Pθ θ= + (4c) The magnitude of the resultant of the primary shaking force is

2 2P X YS S S P= + = (5a)

and the direction of the resultant of the primary shaking force is

1 1 costan tan 90sin

Y

X

SS

θτ θθ

− −= = = − (5b)

Recall that the X and Y components of the resultant of the primary shaking force for any multi-

cylinder reciprocating engine (see page 651 in the Uicker, et al., text book) can be written in the form

cos sinXS A Bθ θ= + (6a) and

21

cos sinYS C Dθ θ= + (6b)

The primary shaking force can be balanced by a pair of rotating masses, or in some special cases a single mass, see Figure 19.28, page 651, in the Uicker, et al., text book. The correcting mass m1 creates a correcting force F1 at a location angle of 1( )θ γ+ from the X-axis and the correcting mass m2 creates a correcting force F2 at a location angle of 2( )θ γ− + from the X-axis. For balance, these two correcting forces plus the resultant of the primary shaking force must be equal to zero; i.e.,

1 1 2 2cos sin cos ( ) cos [ ( ) ] 0A B F Fθ θ θ γ θ γ+ + + + − + = (7a) and

1 1 2 2cos sin sin ( ) sin [ ( ) ] 0C D F Fθ θ θ γ θ γ+ + + + − + = (7b)

Expanding these two equations, in terms of the angles of 1 2, , andθ γ γ , and rearranging, gives

1 1 2 2 1 1 2 2( cos cos ) cos ( sin sin ) sin cos sinF F F F A Bγ γ θ γ γ θ θ θ+ − + = − − (8a) and

1 1 2 2 1 1 2 2( sin sin ) cos ( cos cos ) sin cos sinF F F F C Dγ γ θ γ γ θ θ θ− + − = − − (8b) To satisfy Equations (8), for all values of the crank angle θ , the necessary conditions are

1 1 2 2cos cosF F Aγ γ+ = − (9a)

1 1 2 2sin sin+ = +F F Bγ γ (9b)

1 1 2 2sin sinF F Cγ γ− = − (9c) and

1 1 2 2cos cosF F Dγ γ− = − (9d) Solving Equations (9), the correcting forces are

( ) ( )2 21

12

F A D B C= + + − (10a)

and

( ) ( )2 22

12

F A D B C= − + + (10b) Also, from Equations (9), the location angles of the correcting forces are

1tan( )−

=− +

B CA D

γ (11a) and

2tan( )+

=− −

B CA D

γ (11b)

Comparing Equation (3b) with Equation (6a) and comparing Equation (4b) with Equation (6b), the

coefficients are

0, 1 , 1 , and 0A B P C P D= = = = (12)

22

Substituting Equation (12) into Equations (10), the two correcting forces are

( ) ( )2 21

1 0 0 1 1 02

F P P P P= + + − = (13a)

and

( ) ( )2 22

1 0 0 1 1 12

F P P P P P= − + + = (13b)

The magnitude is 2 25 0.1 45 1,012.5 NP m Rω= = × × = (14)

Therefore, the two correcting forces are

21 1 1 0C CF m R ω= = and 2

2 2 2 1 1,012.5 NC CF m R Pω= = = (15a)

Given C C1 2R R 20 cm,= = the non-zero correcting force is

22 2 2 2

2

1,012.5 N 2.5 kg0.2 x 45 m/sC

C

FmR ω

= = = (15b)

Substituting Equation (12) into Equation (11a), the location angle for the first correcting mass is

1[1 1 ]tan

[0 0]P P undefinedγ −

= =− +

(16a)

This is consistent with the mass being zero, that is, the first correcting mass is not needed.

Substituting Equation (12) into Equation (11b), the location angle for the second correcting mass is

2[1 1 ]tan

[0 0]P Pγ +

= = ∞− −

(17a)

Since the numerator is positive, the angle for the second correcting mass is

2 90oγ θ= − (17b)

For the crank position 0 ,oθ = the magnitude of the resultant of the primary shaking force from Equation (6a) is

1012.5 NPS P= = (18a) The direction of the resultant of the primary shaking force from Equation (6b) is

1 11tan tan0

τ − −= = ∞−

(18b)

that is 90oτ = (18b)

which implies that the resultant of the primary shaking force is vertically upward.

The answers are shown on Figure 4.2.

23

Figure 4.2. The Magnitudes and the Locations of the Two Correcting Masses. For the position 0 .oθ =