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Solutions Mechanical Engineering Design Shigley 7th Edition (1)

Apr 04, 2018

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12-1 Given d m a x = 1.000 in and b m i n = 1.0015 i n , the mini mum radial clearance is

Also

Eq. (12-7):

Fig. 12-16:

Fig. 12-19:

Fig. 12-18:

Fig. 12-20:

cmin

b mi n d n 1 .0015 1 .000

2 2

l/d = 1

r = 1.000/2 = 0.500

r/c = 0.500/0.00075 = 667

N = 1100/60 = 18.33 rev/s

P = W/(ld) = 250 /[( 1)( 1)] = 250 psi

8(10- 6

)(18.33) '

0.000 75 in

S (6672

)250

h00/c

Q/(rcNl)

fr/c

Qs / Q

0. 261

0.595

3.98

5.8

0.5

f 0.0087

h0 = 0.595(0.00075) = 0.000466 in Ans.

5.8 5.8- ^

r/c 667

The power loss in Btu/s is

2:TfWrN 2n (0.0087)(250)(0.5)(18.33)H778(12)778(12)

= 0.0134 Btu/s Ans.

Q = 3.98rcNl = 3.98(0.5)(0.000 75)(18.33)(1) = 0.0274 in 3/s

Qs = 0.5(0.0274) = 0.0137 in3

/s Ans.

_ _ _

_ _

_

_

_ _

bmin - dmax 1.252 - 1.250

c m i n = 2 = 2 = 0.001 in

r =.

1 .25/2 = 0.625 in

r/c = 0.625/0.001 = 625

N = 1150/60 = 19.167 rev/s

400

P = = 128 psi1 .25(2.5)

l/d = 2.5/1.25 = 2

(6252

)(10)(10- 6

)(19.167)S = = 0.585

128

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313

The interpolation formula of Eq. (12-16) w i l l have to be used. From Figs. 12-16, 12-21,

and 12-19

For //d = oo, ho/e = 0.96, PI p m a x = 0.84,

//d = 1, ho/e = 0.77, P /pmax = 0.52,

1

l/d = , ho/e = 0.54, P /p m a x = 0.42,2

1l/d = , ho/e = 0.31, P / p m a x = 0.28,

Q

reNl

Q

reNlQ

reNl

Q

3.09

3.6

4.4

5.254 rcNl

Equation (12-16) is easily programmed by code or by using a spreadsheet. The results are:

l/d yiy

1 / 2 y 1 / 4

ho/e 2 0.96 0.77 0.54 0.31 0.88

pP/pPmax 2 0.84 0.52 0.42 0.28 0.64

Q / reNl 2 3.09 3.60 4.40 5.25 3.28

ho = 0.88(0.001) = 0.000 88 in Ans.

128p

max0.64

200 psi Ans.

Q = 3.28(0.625)(0.001)(19.167)(2.5) = 0.098 in3/s Ans.

12-3

e

min

bmin - dmax 3.005 - 3.000

2 2

r =. 3.000/2 = 1 .500 in

l/d = 1 .5/3 = 0.5

r/e = 1 .5/0.0025 = 600

N = 600/60 = 10 rev/s

800

0.0025 in

P1 . 5(3)

Fig. 12-12: SAE 10, ' = 1.75 /zreyn

S = (6002

)

177.78 psi

1.75(10 - 6 )(10)

177.780.0354

Figs. 12-16 and 12-21: ho/e = 0.11, P /p m a x = 0.21

ho = 0.11(0.0025) = 0.000 275 in Ans.

pmax = 177.78/0.21 = 847 psi Ans.

_ _

_

_ _

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314 Instructor's Solution Manual to Accompany Mechanical Engineering Design

Fig. 12-12: SA E 40, = 4.5 reyn

( 4.5 \) = 0.09101 .75ho/e = 0.19, P /p m a x = 0.275

ho = 0.19(0.0025) = 0.000475 in Ans.

pmax = 177.78/0.275 = 646 psi Ans.

12-4bmin - dmax 3.006 - 3.000

e m i n = 2 = 2 = 0.003

r = . 3.000/2 = 1 . 5 in

l/d = 1

r/e = 1 .5/0.003 = 500

N = 750/60 = 12.5rev/s

600

P = = 66.7 psi3(3)

Fig. 12-14: SAE 10W, = 2.1 reyn

'2 .1(10- 6

) (12.5)S = (500 2)

66.70.0984

From Figs. 12-16 and 12-21:

ho/e = 0.34, P /pm a x

= 0.395

ho = 0.34(0.003) = 0.001 020 in Ans.

66.7

p m a x = = 169 psi Ans.0.395

Fig. 12-14: SAE 20W-40, = 5.05 reyn

' 5.05(10- 6 ) (12.5)

S = (500 2)66.7

0.237

From Figs. 12-16 and 12-21:

ho/e = 0.57, P/pmax = 0.47

ho = 0.57(0.003) = 0.001 71 in Ans.

66.7

pm a x = = 142 psi Ans.

0.47

_

_

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315

1 2 - 5

bmin - dmax 2.0024 - 2c m i n = = = 0.0012 in

2 2

d 2r = = - = 1in , l/d = 1/2 = 0.50

2 2

r/c =1/0.0012 =833

N =800/60 = 13.33 rev/s

600P = = 300 psi

2(1)

Fig. 12-12: SAE 20, =3.75 reyn

' 3.75(10- 6

)(13.3)S = (8332

)300

0.115

From Figs. 12-16, 12-18 and 12-19:

ho/c = 0.23, rf/c = 3.8, Q/(rcNl) =5.3

ho = 0.23(0.0012) = 0.000 276 in Ans.

3.8

f = = 0.00456

833

The powerloss due to friction is

2:TfWrN 2n (0.004 56)(600)( 1) (13.33)

H = =778(12) 778(12)

= 0.0245 Btu/s Ans.

Q = 5.3rcNl

= 5.3(1)(0.0012)(13.33)(1)

= 0.0848 in3

/s Ans.

12-6_ bmin - dmax _ 25.04 - 25

c min _

2 2

r ' d/2 = 25/2 = 12.5 mm,

r/c = 12.5/0.02 = 625

N = 1200/60 = 20 rev/s

= 0.02 mm

l/d = 1

1250P =

l o r = 50 MPa s, S

252

(6252

)

2 M Pa

50(10- 3

) (20)2(10

6

)0.195

From Figs. 12-16, 12-18 and 12-20:

ho/c = 0.52, fr/c = 4.5, Qs/Q = 0.57

ho = 0.52(0.02) = 0.0104 mm Ans.

f=

T =

4. 5

6250.0072

fWr = 0.0072(1.25)(12.5) = 0.1125 N m

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316 Instructor's Solution Manual to Accompany Mechanical Engineering Design

The power loss due to friction is

H = 2nTN = 2n(0.1125)(20) = 14.14 W Ans.

Qs = 0.57 Q The side flow is 57% of Q Ans.

12-7bmin - dmax 30.05 - 30.00

c m i n = = = 0.025 mm

d 30r = 15 mm

2 2

r 15- = = 600c 0.025

1120N = = 18.67 rev/s

60

2750

P = = 1.833 MPa30(50)

2 ' 60(10- 3 )(18.67)

S = (600 2)1 . 833(10

6)0.22

l 50= = 1.67d 30

This //d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16).

From Fig. 12-16, the ho/c values are:

y1/4 = 0.18, y i / 2 = 0.34, y i = 0.54, y ^ = 0.89

hoSubstituting into Eq. (12-16), = 0.659

c

From Fig. 12-18, the fr/c values are:

y 1 / 4 = 7.4, y 1 / 2 = 6.0, y 1 = 5.0, yoo = 4.0

fr

Substituting into Eq. (12-16), = 4.59c

From Fig. 12-19, the Q/(rcNl) values are:

y 1 / 4=

5.65, y 1 / 2=

5.05, y 1=

4.05,=

2.95Q

Substituting into Eq. (12-16), = 3.605

rcNl

ho = 0.659(0.025) = 0.0165 mm Ans.

f = 4.59/600 = 0.007 65 Ans.

Q = 3.605(15)(0.025)(18.67)(50) = 1263 mm3

/s Ans.

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317

1 2 - 8 bmin - dmax 75.10 - 75c m i n = = = 0.05 mm

2 2

l/d = 36/75 = 0.5 (close enough)

r = d/2 = 75/2 = 37.5 mm

r/c = 37.5/0.05 = 750

N = 720/60 = 12 rev/s

2000

P = = 0.741 MPa75(36)

Fig. 12-13: SAE 20, 1 = 18.5MPa s

18.5(10- 3

)(12)S = (750 2)

0.741(106)0.169

From Figures 12-16, 12-18 and 12-21:

ho/c = 0.29, fr/c = 5.1, P /pmax = 0.315

ho = 0.29(0.05) = 0.0145 mm Ans.

f = 5.1/750 = 0.0068

T = fWr = 0.0068(2)(37.5) = 0.51 N m

The heat loss rate equals the rate ofworkon the film

//loss = 2:rTN = 2n (0.51)(12) = 38.5W Ans.

p m a x = 0.741/0.315 = 2.35 MPa Ans.

Fig. 12-13: SAE 40, 1 = 37 MPa s

S = 0.169(37)/18.5 = 0.338

From Figures 12-16, 12-18 and 12-21:

ho/c = 0.42, fr/c = 8.5, P/pmax = 0.38

ho = 0.42(0.05) = 0.021 mm Ans.

f = 8.5/750 = 0.0113

T = fWr = 0.0113(2)(37.5) = 0.85 N m

H l o s s = 2n TN = 2n (0.85)(12) = 64 W Ans.

pmax = 0.741 /0.38 = 1 .95 MPa Ans.

_

bmin - dmax 50.05 - 50

c m i n = = = 0.025 mm2 2

r = d/2 = 50/2 = 25 mm

r/c = 25/0.025 = 1000

l/d = 25/50 = 0.5, N = 840/60 = 14 rev/s

2000

P = = 1.6MPa25(50)

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318 Instructor's Solution Manual to Acc omp an y Mechani cal Engineering Design

Fig. 12-13: SAE 30, X = 34 MPa s

2 r34(10- 3 ) (14)

S = (1000 )1.6(10

6

)

From Figures 12-16, 12-18, 12-19 and 12-20:

0.2975

ho/c = 0.40, fr/c = 7.8, Qs / Q = 0.74, Q/(rcNl) = 4.9

ho = 0.40(0.025) = 0.010 mm Ans.

f= 7.8/1000 = 0.0078

T = fWr = 0.0078(2)(25) = 0.39 N m

H = 2n TN = 2n (0.39)(14) = 34.3W Ans.

Q = 4.9rcNl = 4.9(25 )(0.025 )(14)(2 5) = 1072 mm2

/s

Qs = 0.74(1072) = 793 mm3

/s Ans.

12-10 Consider the bearings as specified by

minimum f:

maximum W:

and differing only in dand d'.

Preliminaries:

b+tb

b

- 0

Fig. 12-16:

minimum f:

maximum W:

Fig. 12-12:

Eq. (12-7):

Forminimum f:

IJLN / P =

l/d = 1

P = 700/(1.252

) = 448 psi

N = 3600/60 = 60 rev/s

S = 0.08

S = 0.20

1.38(10- 6

)reyn

1.38(10- 6

)(60/448) = 0.185(10- 6

)

r

c

S

XN / P

0.08658

0 .185(10- 6

)

c = 0.625/658 = 0.000 950 = 0.001 in

_

r

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319

I f this iS C m i n ,

b d = 2(0.001) = 0.002 in

The median clearance is

, td + tb , td + tb

c = c m i n + 2 = 0.001 + 2

and the clearance range forthis bearing is

. td + tbAc =

2

which is a function only of the tolerances.

Formaximum W:

r / 0.2

- = J ^ = 1040

c \ 0

I f this is c m i n

c y 0.185(106 )

c = 0.625/1040 = 0.000 600 = 0.0005 in

b d' = 2c m i n = 2(0.0005) = 0.001 in

, td + tb n^nr- .t

d +t

b

c = c m i n + 2 = 0.00

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