me 352 - machine design i name summer semester … me 352 - machine design i name_____ summer...

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ME 352 - Machine Design I Name_____________________________ Summer Semester 2008 Lab. Div.__________________________

FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Wednesday, July 30th, 2008 Please show all your work for your solutions on the blank white paper. Write on one side of the paper only. Credit will be given for clear presentation. Problem 1 (25 Points). For the mechanism in the position shown in Figure 1, the kinematic coefficients are 1

3 2 ,−′ = − mθ 23 6.928 ,−′′ = − mθ 4 1.732 ,′ = −R ND and 1

4 8 .−′′ = −R m The velocity and the

acceleration of the input link 2 are 2V 5 j m / s= − and 22A 20 j m / s= − and the force acting on link 2

is F 200 j N.= − The length of link 3 is AB 1 m= and the distance 2 3 0.5 m.=G G A linear spring is attached between points O and A with a free length L 0.5 m= and a spring constant K 2500 N/m.= A viscous damper with a damping coefficient C 45 N.s/m= is connected between the ground and link 4. The masses and mass moments of inertia of the links are 2 0.75 kg,m = 3 2.0 kg,m = 4 1.5 kg,m =

22 0.25 N.m.s ,=GI 2

3 1.0 N.m.s=GI and 24 0.35 N.m.s .=GI Assume that gravity acts in the negative

Y-direction (as shown in the figure) and the effects of friction in the mechanism can be neglected. (i) Determine the first-order kinematic coefficients of the linear spring and the viscous damper. (ii) Determine the equivalent mass of the mechanism. (iii) Determine the magnitude and direction of the horizontal force P that is acting on link 4.

Figure 1. A Planar Mechanism.

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Problem 2 (25 Points). Part A. A rotating steel shaft is simply supported by two rolling element bearings at A and B as shown in Figure 2. Gears 1, 2, and 3 are rigidly attached to the shaft at locations C, D, and E, respectively. The gear 1 at location C weighs 50 N, the gear 2 at location D weighs 20 N, the gear 3 at location E weighs 30 N, and the weight of the shaft can be neglected. The first critical speeds of the shaft with each gear attached separately are 11 250 rad/s,ω = 22 650 /s,radω = and 33 200 /s,= radω respectively. (i) Determine the first critical speed of the shaft with all three gears using the Dunkerley approximation. (ii) Determine the influence coefficients 11 22a , a and 33a . (iii) If gear 1 is now placed at location D and gear 2 is placed at location C, then determine the first critical speed of the new system using the Dunkerley approximation.

Figure 2. A Rotating Shaft with Three Gears. Part B. The weights of two flywheels rigidly attached to a rotating shaft are 1W 55 N= and 2W 95 N.= From the exact equation, the first and second critical speeds of the shaft with the two flywheels are

1 450 rad/s=ω and 2 725 rad/s.=ω The known influence coefficients of the shaft are 5

11a 9.60 x 10 cm/N−= and 521a 9.25 x 10 cm/N.−= Determine the influence coefficient 22a .

Recall that the exact equation for the first and second critical speeds of a rotating shaft with two mass disks was derived in lecture and 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

+ ± + − −=

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Problem 3 (25 Points). Part A. The angular speed of three mass particles rigidly attached to the shaft shown in Figure 3(a) is a constant 350 rev/min. The masses of the three particles are 1m 3 kg,= 2m 2 kg,= and 3m 1.5 kg.= The radial distances of the three mass particles, from the shaft axis, are 1R 45 mm,= 2R 65 mm,= and

3R 50 mm.= (i) Determine the magnitudes of the bearing reaction forces at locations A and B. (ii) Determine the magnitudes of the masses that must be removed in the correcting planes (1) and (2). The radial distances of the masses to be removed in the correcting planes, from the shaft axis, are

C1 C2R R 70 mm.= = (iii) Determine the angular orientations, relative to the X-axis, of the masses to be removed in the correcting planes (1) and (2).

Figure 3(a). A Rotating Shaft with Three Mass Particles.

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Problem 3 (Continued). Part B. The three mass particles in Figure 3(a) are now replaced by a continuous mass system (denote as 2) as shown in Figure 3(b). The angular speed of the rotating shaft is a constant 180 rev/min. The magnitudes of the bearing reaction forces at locations A and B are specified as 21 A(F ) 75 N= at 75− ° and B21(F ) 25 N= at 150° , respectively. The radial distances of the two correcting masses in the correcting planes (1) and (2) are C1 C2R R 30 mm.= =

Determine the two correcting masses and the angular orientations of these two masses that must be added in the correcting planes (1) and (2).

Figure 3(b). A Rotating Shaft with a Continuous Mass System.

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ME 352 - Machine Design I Name_____________________________ Summer Semester 2008 Lab. Div.__________________________ Problem 4 (25 Points). The two cranks of the two-cylinder engine shown in Figure 4 are perpendicular. The effective mass of each piston is 1m m 10 kg= = and 2m 2m 20 kg,= = the length of each connecting rod is L 75 cm,= and the length of the throw of each crank is R 15 cm.= The angular velocity of the crankshaft is a constant ω θ 50 rad/s= = counterclockwise. (i) Determine the X and Y components of the resultant of the first harmonic forces (i.e., the primary shaking forces). (ii) Determine the magnitudes and the locations (i.e., the angles) of the two correcting forces created by the two correcting masses. (iii) Draw your answers on the figure to the right below where the reference line is shown at the crank angle 0 .= oθ

Figure 4. A Two Cylinder Engine.

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Solution to Problem 1. (i) 8 points. The vectors for the linear spring are shown in Figure 2.1.

Fig. 2.1. The Vectors for the Linear Spring. The vector loop for the linear spring can be written as

?

2 0V IV

SR R− = (1a) which can be written as the scalar equation

2SR R= (1b)

Differentiating Equation (1b) with respect to the input position 2R , the first-order kinematic coefficient of the spring is

1′ = +SR ND (2) Note that the sign is positive because for the negative input, the length of the linear spring is decreasing. Also, note that the first-order kinematic coefficient of the mass center of the input link 2 is

2 1′ ′= = +G SY R ND (3) The vectors for the viscous damper are shown in Figure 2.2.

Fig. 2.2. The Vectors for the Viscous Damper. The vector loop for the damper can be written as

?

49 0VV V VV

CR R R− − = (4)

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The X and Y components of Equation (4) are

9 9 4 4cos cos cos 0C CR R Rθ θ θ− − = (5a) and

9 9 4 4sin sin sin 0C CR R Rθ θ θ− − = (5b)

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

4 4cos 0′ ′− − =C CR cos Rθ θ (6a) and

4 4sin sin 0′ ′− − =C CR Rθ θ (6b)

Substituting 00=Cθ and 04 0θ = into Equation (6a), the first-order kinematic coefficient of the viscous

damper is 4 1.732′ ′= − = +CR R ND (7)

The sign agrees with our intuition, for a negative input the length of the viscous damper is decreasing. (ii) 8 Points. The equivalent mass of the mechanism can be written as

4

2=

=∑EQ jj

m A (8)

For Link 2: ( )2 2 22 2 G2 G2 G2 2A m X Y I′ ′ ′= + + θ (9)

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

??

2 2

IV

GR R= (10) The X and Y components of Equation (10) are

2 0=GX (11a) and

2 2=GY R (11b)


2 0′ =GX ND (12a) and

2 2R 1GY ND′ ′= = + (12b)


12 0 0GX m−′′ = = (13a)

and 1

2 2 0 −′′ ′′= =GY R m (13b)

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Substituting Equations (12) and (13) into Equation (9) gives

( ) ( )22 22 0.75 0 1 0.25 0 0.75= + + = +A kg (14)

For Link 3: ( ) 233

23

2333 θ ′+′+′= GGG IYXmA (15)

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

?? ?

3 2 3

IV V

G GR R R= + (16) The X and Y components of Equation (16) are

3 2 2 3 3cos cos 0.25G GX R R mθ θ= + = (17a) and

3 2 2 3 3sin sin 0.433G GY R R mθ θ= + = (17b)


3 2 3 3 3cos sin 0.866G GX R NDθ θ θ′ ′= − = − (18a) and

3 2 3 3 3sin cos 0.5G GY R NDθ θ θ′ ′= + = + (18b)

Then differentiating Equations (18) with respect to the input position 2R gives

2 13 3 3 3 3 3 3sin cos 4.0G G GX R R mθ θ θ θ −′′ ′′ ′= − − = − (19a)

and 2 1

3 3 3 3 3 3 3cos sin 0G G GY R R mθ θ θ θ −′′ ′′ ′= − = (19b) Substituting the known data and Equations (18) and (19) into Equation (15) gives

32 2 2A 2.0 ( 0.866) 0.5 1.0 ( 2) 6 kg⎛ ⎞⎜ ⎟

⎝ ⎠= − + + − = + (20)

For Link 4: ( ) 244

24

2444 θ ′+′+′= GGG IYXmA (21a)

Note that G4 4X R 1.732 m,′ ′= = − therefore, Equation (21a) can be written as

( ) ( )2 224 1.5 [ 1.732 0 ] 0.35 0 4.5= − + + = +A kg (21b)

Therefore, the equivalent mass of the mechanism from Equation (8) is

0.75 6 4.5 11.25EQm kg= + + + = + (22) (iii) 9 points. The power equation for the mechanism can be written as

2 4F.V P.V+ = + + fdWdT dUdt dt dt

(23a)

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Substituting the time rate of change of energy terms into the right-hand side of Equation (23a) gives

( )4 4 4

3 2 22 4 2 2 2 2 0 2

2 2 2 2F.V P.V

= = =

′ ′ ′+ = + + + − +∑ ∑ ∑j j j G j s S S S Cj j j

A R R B R m g Y R K R R R R C R R (24b)

The linear velocity of link 2 and the force acting on link 2 are both in the same direction (i.e., both downward). Assuming that the force P is in the same direction as the velocity of link 4, that is, to the right, then Eq. (24b) can be written as

( )4 4 4

3 2 22 4 2 2 2 2 0 2

2 2 2 2F V P V

= = =

′ ′ ′+ = + + + − +∑ ∑ ∑j j j G j s S S S Cj j j

A R R B R m g Y R K R R R R C R R (24a)

The velocity of the input link 2 is 2 2V R= + and the velocity of link 4 is ,4 4V R= + therefore, Equation (24a) can be written as

( )4 4 4

3 2 22 4 2 2 2 2 0 2

2 2 2 2FR PR

= = =

′ ′ ′+ + = + + + − +∑ ∑ ∑j j j G j s S S S Cj j j

A R R B R m g Y R K R R R R C R R (24b)

Canceling the input velocity 2R in Equation (24b) gives the equation of motion for the mechanism, that is

( )4 4 4

2 24 2 2 0 2

2 2 2

F PR= = =

′ ′ ′ ′+ + = + + + − +∑ ∑ ∑j j j G j s S S S Cj j j

A R B R m g Y K R R R C R R (25)

where the first-order coefficient of link 4 is given as 4 1.732 NDR′ = − . The sum of the B terms can be written as

4 4

2 2 2

12

jj

j j

dAB

dθ= =

=∑ ∑ (26)

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′ ′′ ′ ′′ ′ ′′= + + (27a) which can be written as

( ) ( )2 0.75 0 0 0.25 0 0B = + + = (27b)

For Link 3: ( )3 3 3 3 3 3 3 3 3G G G G GB m X X Y Y I θ θ′ ′′ ′ ′′ ′ ′′= + + (28a) which can be written as

13 2 [( .866)( 4) (0.5)(0)] 1( 2)( 6.928) 20.78B kg m−= − − + + − − = + ⋅ (28b)

For Link 4: ( ) 444444444 θθ ′′′+′′′+′′′= GGGGG IYYXXmB (29a) which can be written as

( )( ) 1

4B 1.5[( 1.732)( 8) (0)(0)] 0.35 0 0 20.78 kg m−= − − + + = + ⋅ (29b)

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Adding Equations (27b), (28b), and (29b) gives

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1 2 32

0 20.78 20.78 41.56jj

B B B B kg m−

=

= + + = + + = + ⋅∑ (30)

The change in potential energy due to gravity is

4

2j G j

jm g Y

=

′∑ (31)

For link 2: 2 2Gm g Y ′ (32a)

Which can be written as 2 (0.75)(9.81) 7.36= = +m g N (32b)

For link 3: 3 3Gm g Y ′ (33a)

Which can be written as 3 3 (2)(9.81)( 2) 39.24′ = − = −Gm g Y N (33b)

For Link 4: 4 4Gm g Y ′ (34a)

Which can be written as 4 4 (1.5)(9.81)(0) 0Gm g Y N′ = = (34b) Substituting Equations (32b), (33b) and (34b) into Equation (31) gives

4

2

7.36 39.24 0 31.80=

′ = + − + = −∑ j G jj

m g Y N (35)

Then substituting Equations (5), (10), (22), (30) and (35) into Equation (25) gives

( )2 24 2 2 0 2F PR 11.25R 41.56R 31.80 ( 1) ( 1.732) R′+ + = + + − + − + + +S S SK R R C (36a)

Rearranging this equation, the force acting on link 4 can be written as

( )2 22 2 0 2

4

1 F 11.25 R 41.56 R 31.80 ( 1.732) RR

⎡ ⎤= − + + − + − + +⎣ ⎦′ S S SP K R R C (36b)

The input velocity 2R 5 m/s= − , the input acceleration 22R 20 m/s= − and the force F 200 N.= −

Substituting these values and the known data into this equation, the force acting on link 4 can be written as

( )2 21 200 11.25( 20) 41.56( 5) 31.80 2500 0.866 0.5 45 ( 1.732) ( 5)1.732

⎡ ⎤= + + − + − − + − + + −⎣ ⎦−P (37a)

or as

[ ]1 200 225 1039 31.80 915.0 675.01.732

= + − + − + −−

P (37b)

or as

11

1222.21.732

+=

−P (37c)

Therefore, the force acting on link 4 is

705.66= −P N (38) The negative sign indicates that the force P (acting on link 4) is acting to the left; that is, in the opposite direction to the velocity of the output link 4. Recall that the force P was assumed to be acting to the right.

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Solution to Problem 2. Part A. (i) 5 Points. From the Dunkereley approximation, the first critical speed of the shaft can be written as

2 2 2 21 11 22 33

1 1 1 1= + +

ω ω ω ω (1) where the double numbered subscripts correspond to the critical speeds of the shaft with each gear attached searately.

Substituting the known critical speeds corresponding to each gear into Eq. (1), the first critical speed of the shaft is

52 2 2 2

1

1 1 1 1 2.114 10250 650 200ω

−= + + = × (2) which can be written as

1 5

1 /2.114 10−=

×rad sω (3a)

Therefore, the first critical speed of the shaft is

1 217.47 /= rad sω (3b) (ii) 5 Points. The influence coefficients for each location can be determined using the mass at that location and the corresponding critical speed, that is

1 11 211

1m aω

= (4) Rearranging Equation (4) and solving for the influence coefficient gives

611 2 2

1 11

1 1 3.136 10 /(50 / 9.8)(250)

a m Nmω

−= = = × (5a) Similarly, the influence coefficient

622 2 2

2 22

1 1 1.160 10 /(20 / 9.8)(650)

a m Nm ω

−= = = × (5b) Similarly, the influence coefficient

633 2 2

3 33

1 1 8.167 10 /(30 / 9.8)(200)

a m Nm ω

−= = = × (5c) (iii) 5 Points. When gears 1 and 2 are interchanged, the first critical speed of the new system is

11 2 22 12 21 33

1 1a m a mω ω

= + + (6)

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Equation (6) can be written as

6 6 62 2

1

1 1(3.136 10 )(20 / 9.8) (1.160 10 )(50 / 9.8) 37.318 10200ω

− − −= × + × + = × (7a) or as

1 6

1 /37.318 10−=

×rad sω (7b)

Therefore, the first critical speed of the news ystem is

1 163.70 /= rad sω (8) Part B. (i) 5 Points. The exact solution for the first and second critical speeds of the shaft can be written as

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

2 21 2

( ) ( ) 4 ( )1 1,2

a m a m a m a m m m a a a aω ω

+ ± + − −= (1)

The sum of the first and second critical speeds of the shaft can be written as

11 1 22 22 21 2

1 1+ = +a m a m

ω ω (2)

Substituting the known values into Equation (2) gives

5222 2

1 1 55 959.60 x 10450 725 980 980

a−+ = + (3)

Rearranging this equation gives 5

22 2 2

95 1 1 559.60 x 10980 450 725 980

a −= + − (4)

which can be written as

5 5 5220.97 0.494 10 0.190 10 0.539 x 10− − −= × + × −a

(5) Therefore, the influence coefficient is

622 1.49 10 /a cm N−= × (6)

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Solution to Problem 3(a). (i) 5 Points. The constant angular velocity of the rotating shaft is

(2 rad/rev)(350 rev/min) 36.65 rad/s(60 s/min)

πω = = (1)

The inertial forces of the three rotating mass particles are

2 21 1 1 (3 kg)(0.045 m)(36.65 rad/s) 181.34 NF m Rω= = = (2a)

2 2

2 2 2 (2 kg)(0.065 m)(36.65 rad/s) 174.62 NF m R ω= = = (2b) and

2 23 3 3 (1.5 kg)(0.050 m)(36.65 rad/s) 100.74 NF m R ω= = = (2c)

Therefore, the inertial forces of the three rotating mass particles can be written as

1 181.34 N 60 90.67 157.05 NF i j= ∠ ° = + (3a)

2 174.62 N 150 151.23 87.31 NF i j= ∠ ° = − + (3b) and

3 100.74 N 300 50.37 87.24 NF i j= ∠ ° = − (3c) The sum of the inertial forces of the three rotating mass particles can be written as

1 2 3 10.19 157.12 N 157.45 N 93.71F F F F i j= + + = − + = ∠ °∑ (4) Therefore, the reaction forces at the two bearings A and B can be written as

( )1 2 3 10.19 157.12 NA BF F F F F i j+ = − + + = − (5) The sum of the two correcting forces can be written as

( )1 2 1 2 3 10.19 157.12 NC CF F F F F i j+ = − + + = − (6) The sum of the moments about the correcting plane (2) is

(2) 0∑ =M (7a) which can be written as

3 2 1 10.165 0.215 0.065 0.305 0Ck F k F k F k F× + × + × + × = (7b) or as

1 1 47.84 60.03C Y C XF i F j i j N− = − − (7c) Therefore, the inertial force in the correcting plane (1) can be written as

1 60.03 47.84 76.76 321.5CF i j N= − = ∠ ° (8)

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

11 2

1

= CC

C

FmRω

(9a)


1 2

76.76 0.816(0.070 )(36.65 / )C

Nm kgm rad s

= = (9b)

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

1 321.5Cθ = ° (10a) Since the mass is to be subtracted then the angle must be rotated 180°. The orientation of the mass, from the X-axis, that is to be removed in correcting plane (1) is

2 321.5 180 141.5Cθ = °+ ° = ° (10b) Substituting Equation (8) into Equation (6a), the force in the second correcting plane is

2 10.19 157.12 60.03 47.84 49.84 109.28CF i j i j i j N= − − + = − − (11a) which can be written as

2 120.11 245.5CF N= ∠ ° (11b) The correcting mass in correcting plane (2) can be written as

22 2

2

= CC

C

FmRω

(12a)


2 2

120.11 1.277(0.070 )(36.65 / )C

Nm kgm rad s

= = (12b)

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

2 245.5Cθ = ° (13a) Since the mass is to be subtracted then the angle must be rotated 180°. The orientation of the mass, from the X-axis, that is to be removed in correcting plane (2) is

2 245.5 180 65.5Cθ = °+ ° = ° (13b) (iii) 5 Points. The sum of the moments about bearing A can be written as

0∑ =AM (14a) which can be written as

3 2 10 0.140 0.090 0.240 0.375A BM k F k F k F k F∑ = = × + × + × + × (14b)

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Therefore, the reaction force at bearing B is

88.90 40.54BY BXF i F j i j− = − + (15a) which can be written as

40.54 88.89 97.70 245.5BF i j N N= − − = ∠ ° (15b) Substituting Equation (15b) into Equation (5), the reaction force at bearing A is

50.73 68.23 85.02 306.6AF i j N N= − = ∠ ° (16) Solution to Problem 3(b). (i) 5 Points. The sum of the bearing reaction forces must be equal to the sum of the correcting forces; i.e.,

0=∑F (1a) which can be written as

1 2 0+ − − =A B C CF F F F (1b) or as

1 2A B C CF F F F+ = + (1c) The sum of the moments about the correcting plane (2) is

(2) 0∑ =M (2a) which can be written as

10.255 0.125 0.025 0A C Bk F k F k F× + × − × = (2b) Substituting the known data into Equation (2b), the first correcting force can be written as

10 0.1 ( 21.65 12.50 ) 0.4 0.5 (19.41 72.44 )Ck i j k F k i j= − × − + + × + × − (3a) which can be written as

1 1 93.68 29.68C Y C XF i F j i j− = + (3b) Therefore, the correcting force in the correcting plane (1) is

1 29.68 93.68 98.27 98.4CF i j N N= − + = ∠ ° (3c)

Substituting andA BF F and Equation (3c) into Equation (1b) gives

2( 21.65 12.50 ) (19.41 72.44 ) ( 29.68 93.68 ) 0− + + − − − + − =Ci j i j i j F (4) Therefore, the correcting mass in the correcting plane (2) is

2 27.44 153.62 156.05 280.1CF i j N N= + − = ∠ ° (5) Therefore, the correcting mass in the correcting plane (1) is

11 2

1

= CC

C

FmR ω

(6a)

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Substituting the known date into Eq. (6a), the correction mass is

1 2

98.27 9.22(0.030 )(18.85 / )

= =CNm kg

m rad s (6b)

The angular location of the correcting mass from the X-axis is

1 98.4Cθ = ° (7) The correcting mass in the correcting plane (2) is

22 2

2

= CC

C

FmR ω

(8a)


2 2

156.05 14.64(0.030 )(18.85 / )

= =CNm kg

m rad s (8b)

The angular location of the correcting mass from the X-axis is

2 280.1Cθ = ° (9)

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Solution to Problem 4. (i) 5 Points. The X and Y components of the first harmonic 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 first harmonic 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 1, the angles 1 120 ,ψ = + ° 2 60 ,ψ = − ° and 2 270 90 .φ = ° = − ° Substituting these angles and 2

1P P m Rω= = and 22 2 2P P m Rω= = into Equations (1a) and (2a), the X-component of the resultant

of the first harmonic forces can be written as

1 2 cos( 60 )cos60 2 cos( 60 90 )cos ( 60 )X X XS S S P Pθ θ= + = − ° °+ + °− ° − ° (3a) Therefore, the X-component of the resultant of the first harmonic forces is

1.116 cos 0.933 sinXS P Pθ θ= + + (3b) Also, substituting the angles into Equations (1b) and (2b), the Y-component of the resultant of the first harmonic forces can be written as

1 2 cos( 60 )sin 60 2 cos( 60 90 )sin ( 60 )Y Y YS S S P Pθ θ= + = − ° °+ + °− ° − ° (4a) Therefore, the Y-component of the resultant of the first harmonic forces is

1.067 cos 0.116 sinYS P Pθ θ= − − (4b) The magnitude of the resultant of the first harmonic force is

2 2 20.884 1.5cos 2.33cos sinP X YS S S P θ θ θ= + = + + (5a) and the direction of the resultant of the first harmonic force is

1 1 1.067cos 0.116sintan tan1.116cos 0.933sin

Y

X

SS

θ θθ θ

− − − −=

+ + (5b)

The primarry shaking force (or the first harmonic force) is shown in Figure 2.

Recall that the X and Y components of the resultant of the first harmonic forces 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

cos sinYS C Dθ θ= + (6b)

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The first harmonic forces can always be balanced by a pair of rotating masses, or in some special cases a single mass, as shown in 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 first harmonic forces (or 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

1.116 , 0.933 , 1.067 , and 0.116A P B P C P D P= + = + = − = − (12)

20

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

( ) ( )2 21

1 1.116 0.116 0.933 1.067 2.2362

F P P P P P= + − + − − = (13a)

and

( ) ( )2 22

1 1.116 0.116 0.933 1.067 1.2392

F P P P P P= − − + + − = (13b) The magnitude of P is

2 250 0.1 25 3,125= = × × =P m R Nω (14) Therefore, the two correcting forces are

21 1 1 2.236 6,987.5F m R P Nω= = = and 2

2 2 2 1.239 3,871.9F m R P Nω= = = (15) Substituting Equation (12) into Equation (11a), the location angle for the first correcting mass is

1[( 0.933 ) ( 1.067 )]tan 2.000[( 1.116 ) ( 0.116 )]

P PP P

γ + − −= = −− + + −

(16a)

Since the denominator (or the cosine of the angle) is negative then the location angle is

1 296.6oγ = (16b) Substituting Equation (12) into Equation (11b), the location angle for the second correcting mass is

2[( 0.933 ) ( 1.067 )]tan 0.109[( 1.116 ) ( 0.116 )]

P PP P

γ + + −= = +− + − −

(17a)

Since the denominator (or the cosine of the angle) and the numerator (or the sine of the angle) are negative then the location angle is

2 186.2oγ = (17b)

For the crank position 0 ,oθ = the answers are shown on Figure 2.

21

Figure 2. The Magnitudes and the Locations of the Two Correcting Masses. For the position 0 .oθ =

me 352 - machine design i name summer semester … me 352 - machine design i name_____ summer...

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