Solutions Mechanical Engineering Design Shigley 7th Edition (1)

7/31/2019 Solutions Mechanical Engineering Design Shigley 7th Edition (1)

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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


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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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 = 0.0005 + 2

. td +t

bAc =

2

The difference (mean) in clearance between the two clearance ranges, c r a n g e , is

c r a n g e = 0.001 + 2 I 0.0005 + 2 I

= 0.0005 in

For the minimum f bearing

b d = 0.002 in

or

d = b 0.002 in

For the maximum Wbearing

d' = b 0.001 in

For the same b, tb and td, we need to change the journal diameter by 0.001 in .

d' d = b 0.001 (b 0.002)

= 0.001 in

Increasing dof the minimum friction bearing by 0.001 i n, defines d'of the maximum load

bearing. Thus, the clearance range provides for bearing dimensions which are attainable

in manufacturing. Ans.


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12-11 Given: SAE 30, N = 8rev/s, Ts = 60C, l/d = 1, d = 80 mm, b = 80.08 mm,

W = 3000N

bmin - dmax 80.08 80

c m i n = = = 0.04 mm2 2

r = d/2 = 80/2 = 40 mm

r 40

- = = 1000c 0.04

3000P = = 0.469 MPa

80(80)

Trial #1: From Figure 12-13 forT = 81C, \x = 12 MPa s

AT = 2(81C 60C) = 42C

12(10- 3

) (8)

S = (1000

2

)

From Fig. 12-24,

0.120AT

0.469(106) 0.2047

0.349 + 6.009(0.2047) + 0.0475(0.2047)2

= 1.58P

(0.469\I = 6.2C0.120 /Discrepancy = 42C 6.2C = 35.8C

Trial #2: From Figure 12-13 for T = 68C, \ = 20 MPa s,

AT = 2(68C 60C) = 16C

(2 0 \ I = 0.34112From Fig. 12-24,

0.120AT0.349 + 6.009(0.341) + 0.0475(0.341)

2

= 2.4P

(

0.469\I = 9.4C

0.120 /

Discrepancy = 16C 9.4C = 6.6C

Trial #3: \ = 21 MPa s, T = 65C

AT = 2(65C 60C) = 10C

(21 \ I = 0.35812

_


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321

From Fig. 12-24,

0.120 AT0.349 + 6.009(0.358) + 0.0475(0.358)

2

= 2.5P

(

0.469\I = 9.8C

0.120 /

Discrepancy = 10C 9.8C = 0.2C O.K.

Tav = 65C Ans.

T1 = Tav AT/2 = 65C (10C/2) = 60C

T 2 = Tav + AT/2 = 65C + (10C/2) = 70C

S = 0.358

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

ho Q Qs = 0.68, fr/c = 7.5, = 3.8, = 0.44c rcN l Q

ho = 0.68(0.04) = 0.0272 mm Ans.

7.5f = = 0.0075

1000

T = fWr = 0.0075(3)(40) = 0 . 9 N m

H = 2n TN = 2n (0.9)(8) = 4 5 . 2 W Ans.

Q = 3.8(40)(0.04)(8)(80) = 3891 mm 3/s

Qs = 0.44(3891) = 1712mm3

/s Ans.

12-12 Given: d = 2.5 in , b = 2.504 i n , c m i n = 0.002 in, W = 1200 lb f ,SAE = 20 , Ts = 110F,

N = 1120 rev/min, and l = 2 .5 in .

For a trial film temperature Tf = 150F

Tf S A T (From Fig. 12-24)

150 2.421 0.0921 18.5

AT 18.5FTav = Ts + = 110F + = 119.3F2 2

Tf Tav = 150F 119.3F

which is not 0.1 orless, therefore try averaging

150F + 119.3F(Tf ) n e w = = 134.6F


11/21


Proceed with additional trials

Trial Ne w

Tf l-i' S AT Tf

150.0 2.421 0.0921 18.5 119.3 134.6

134.6 3.453 0.1310 23.1 121.5 128.1

128.1 4.070 0.1550 25.8 122.9 125.5

125.5 4.255 0.1650 27.0 123.5 124.5

124.5 4.471 0.1700 27.5 123.8 124.1

124.1 4.515 0.1710 27.7 123.9 124.0

124.0 4.532 0.1720 27.8 123.7 123.9

Note that the convergence begins rapidly. There are ways to speed this, but at this point

they would only add complexity. Depending where you stop, you can enterthe analysis.

(a) \jt! = 4.541(10 6 ) , S = 0.1724

hoFrom Fig. 12-16: = 0.482, ho = 0.482(0.002) = 0.000 964 in

c

From Fig. 12-17: 0 = 56 Ans.

(b) e = c ho = 0.002 0.000 964 = 0.001 04 in Ans.

fr(c) From Fig. 12-18: = 4.10, f = 4.10(0.002/1.25) = 0.006 56 Ans.

c

(d) T = fWr = 0.006 56(1200)(1.25) = 9.84 lbf in

2n TN 2n (9.84)(1120/60)H = = = 0.124 Btu/s Ans.

778(12) 778(12)

Q 1120(e) From Fig. 12-19: = 4.16, Q = 4.16(1.25)(0.002M I (2.5)

rcNl 60

= 0.485 in 3/s Ans.

Qs 3From Fig. 12-20: = 0.6, Qs = 0.6(0.485) = 0.291 in

3/s Ans.Q

P 1200

(f) From Fig. 12-21: = 0.45, pmax =

2 = 427 psi Ans.pmax 2. 5

2(0.45)

0p = 16 Ans.

(g)


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323

12-13 Given: d = 1.250 in , ta = O.OOlin, b = 1.252in, tb = 0.003i n, / = 1.25in, W = 250lbf,

N = 1750 rev/min, SA E 10 lubricant , sump temperature Ts = 120F.

Below is a partial tabular summary for comparison purposes.

c

mi n cc

max0.001 in 0.002 in 0.003 in

Tf 132.2 125.8 124.0

AT 24.3 11.5 7.96

Tmax 144.3 131.5 128.0

n' 2.587 3.014 3.150

S 0.184 0.053 7 0.0249

e 0.499 0.775 0 0.873

fr

c4.317 1.881 1.243

QrcNjl

4.129 4.572 4.691

Qs

Q0.582 0.824 0.903

ho

c0.501 0.225 0.127

f 0.006 9 0.006 0.005 9

Q 0.0941 0.208 0.321

Qs 0.0548 0.172 0.290

ho 0.000 501 0.000 495 0.000 382

Note the variations on each line. There is not a bearing, but an ensemble of many bear-

ings, due to the random assembly of toleranced bushings and journals. Fortunately the

distribution is bounded; the extreme cases, c m i n and c m a x , coupled with c provide the

charactistic description for the designer. A l l assemblies must be satisfactory.

The designer does not specify a journal-bushing bearing, but an ensemble of bearings.

12-14 Computer programs w i l l varyFortran based, M A T L A B , spreadsheet, etc.

12-15 In a step-by-step fashion, we are building a skill for natural circulation bearings.

Giv en the average film temperature, establish the bearing properties.

Giv en a sump temperature, find the average film temperature, then establish the bearing

properties.

No w we acknowledge the envi ronmental temperature's role in establishing the sump

temperature. Sec. 12-9 and Ex. 12-5 address this problem.

The task is to iteratively find the average film temperature, Tf, which makes Hgen and

H l o s s equal. The steps for determining c m i n are provided within Trial #1 through Trial #3

on the following page.


13/21


Trial #1:

Choose a value of Tf.

Find the corresponding viscosity.

Find the Sommerfeld number.

Find fr/c, then

2545Hgen = WNc

1050

Find Q/(rcNl) and Qs/ Q. From Eq. (12-15)

0.103P(fr/c)AT = J '

(1 0.5 Qs /Q )[Q/(rcNjl)]

hcR A( Tf Too)H l o s s = \ '

1 + a

Display Tf, S, H g e n , H l o s s

Trial #2: Choose another Tf, repeating above drill.

Trial #3:

Plot the results of the first tw o trials.

Choose (Tf)3 from plot. Repeat the drill. Plot the results ofTrial #3 on the above graph.

I f you are not within 0.1F, iterate again. Otherwise, stop, and find all the properties of

the bearing for the first clearance, c m i n . See i f Trumpler conditions are satisfied, and i f so,

analyze c and c m a x .

The bearing ensemble in the current prob lem statement meets Trumpler's criteria

(fornd = 2).

This adequacy assessment pro tocol can be used as a design tool by giving the students

additional possible bushing sizes.

b (in) tb (in)

2.254 0.004

2.004 0.004

1.753 0.003

Otherwise, the design option includes reducing l/d to save on the cost of journal machin

ing and vender-supplied bushings.


14/21

325

12-16 Continue to build a skill with pressure-fed bearings, that of finding the average tempera

ture of the fluid film. First examine the case forc = c m i n

Trial #1 :

Choose an initial Tf.

Find the viscosity.

Find the Sommerfeld number.

Find fr/c, ho/c, and e.

From Eq. (12-24), find AT.

ATTav = Ts +

2

Display Tf, S, AT, and Tav.

Trial #2:

Choose anotherTf. Repeat the drill, and display the second set of values for Tf,

S, A T, and Tav. Plot Tav vs Tf :

i

i

1

\if>2 Uf)3 U f)l

Trial #3:

Pickthe third Tf from the plot and repeat the procedure. I f(Tf ) 3 and ( Ta v ) 3 differ by more

than 0 . l F , plot the results forTrials #2 and #3 and try again. I fthey are within 0 . l F ,de -

termine the bearing parameters, check the Trumple r cri teria, and compare / / l o s s with the

lubricant's cooling capacity.

Repeat the entire procedure forc = c m a x to assess the cooling capacity for the maxi

mum radial clearance. Finally, examine c = Cto characterize the ensemble of bearings.

12-17 A n adequacy assessment associated with a design task is required. Trumpler 's criteria

w i l l do.

d = 50.00+0 05 mm, b = 50.084+ ' 0 1 0 mm

SAE 30, N = 2880 rev/min or 48 rev/s, W = 10 k N

bmin - dmax 50.084 - 50c m i n = = = 0.042 mm

2 2

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

r/c = 25/0.042 = 595.

1l - (55 5) = 25 mm

2

l'/d = 25/50 = 0.5

W 10(106

)p = = = 4000 kPa

4r l ' 4(0.25)(0.25)


15/21


Trial #1: Choose (Tf ) 1 = 79C. From Fig. 12-13, JZ = 13 MPa s.

13(10 3 )(48) 'S = (595

2

)


From Eq. (12-25), AT =

0.0554000(10

3

)

fr = 2.3, e = 0.85.

c

978(106

) (f r / c ) SW 2

1 + 1.5e2 psr

4

978(106

) 2.3(0.055)(102

)

200(25)41 + 1 . 5(0.85)2

= 76.0C

Tav = Ts + AT/2 = 55C + (76C/2) = 93C

Trial #2: Choose (Tf ) 2 = 100C. From Fig. 12-13, JI= 7MPa s.

S = 0.055 ( 13(7 \

13 0.0296

frFrom Figs. 12-18 and 12-16: = 1.6, e = 0.90

AT978(10

6

)

1 + 1 .5(0.9)2

1.6(0.0296)(102

)

200(25)4

26.8C

26.8CTav = 55C + z = 68.4C2

100

90

80

70

(79C, 93C)

(79C, 79C)

> 85 C

L V_ | 1 L

(100C, 100C)

60 70 80 90 100 f

(100C, 68.4C)

Trial #3: Thus, the plot gives (Tf ) 3 = 85C. From Fig. 12-13, JX = 10.8MPa s.

(10.8 \I = 0.045713fr

From Figs. 12-18 and 12-16: = 2.2, e = 0.875

AT978(10

6

)

1 + 1 .5(0.8752

)

2.2(0.0457)(102

)

200(25)4

58.6C

58.6CTav = 55C + = 84.3C

2

85C + 84.3CResult is close. Choose Tf = = 84.7C

_

_ _


16/21

327

Fig. 12-13: = 10.8MPa s

(10.8\I =13 / 0.0457fr ho = 2.23, e = 0.874, = 0.13

c c

AT978(10

6

)

1 + 1.5(0.8742

)

59.5CTav = 55C + = 84.7C

2

2.23(0.0457)(102

)

200(254

)

O.K.

59.5C

From Eq. (12-22)

Qs = (1 + 1.5e2

)n psrc

3

3l'

n (200)(0.0423

)(25)

3(10)(106

)(25)

Trumpler:

= [1 + 1.5(0.874 2)]

= 3334 mm3

/s

ho = 0.13(0.042) = 0.005 46 mm or 0.000 215 in

ho 0.0002 + 0.000 04(50/25.4)

= 0.000 279 in Not O.K.

T m a x = Ts + AT = 55C + 63.7C = 118.7C or 245.7F O.K.

Pst = 4000 kPa or 581 psi Not O.K.

n = 1, as done Not O.K.

There is no point in proceeding further.

12-18 So far, we've performed elements of the design task. Now let's do it more completely.

First, rememberour viewpoint.

The values of the unilateral tolerances, tb and td, reflect the routine capabilities of the

bushing vendor and the in-house capabilities. Whi le the designer has to live with these,

his approach should not depend on them. They can be incorporated later.

First we shall find the minimum size of the journa l whi ch satisfies Trumpler's con

straint of Pst < 300psi.

W

2dV< 300

W

2d 2 1 ' /d < 300 == d >

W

600(l '/ d)

d

m i n

900

2(300)(0.5)1.73 in

_ _

_


17/21


In this problem we w i l l take journal diameter as the nomina l value and the bushing bore

as a variable. In the next pro blem, we w i l l take the bushing bore as nominal and the jour

nal diameter as free.

To determine where the constraints are, we w i l l set tb = td = 0, and thereby shrink

the design window to a point.

We set d = 2.000 in

b = d + 2 c m i n = d + 2c

nd = 2 (This makes Trumpler's nd < 2 tight)

and construct a table.

d Tma x ho Pst Tmax n fom

0.0010 2.0020 2 215.50 312.0 XS X

-5 .74

0.0011 2.0022 2 206.75 293.0 X -6 .06

0.0012 2.0024 2 198.50 277.0 X -6 .37

0.0013 2.0026 2 191.40 262.8X

-6 .660.0014 2.0028 2 185.23 250.4 X -6 .94

0.0015 2.0030 2 179.80 239.6 X -7 .20

0.0016 2.0032 2 175.00 230.1 X -7 .45

0.0017 2.0034 2 171.13 220.3 X -7 .65

0.0018 2.0036 2 166.92 213.9

f- 7 . 9 1

0.0019 2.0038 2 163.50 206.9 -8 .12

0.0020 2.0040 2 160.40 200.6

s s s s-8 .32

*Sample cal cula tion for the first entry of this colum n.

Iteration yields: Tf = 215.5F

With Tf = 215.5F, from Table 12-1

X = 0.0136(10 - 6 )exp[12 71.6/(2 15.5 + 95)] = 0 .8 17 (1 0 - 6 ) r eyn

900N = 3000/60 = 50 rev/s, P = = 225 psi

4

S1

0.001

0.817(10 - 6 ) (50)0.182


Eq. (12-24):

ATf =

225

e = 0.7, fr/c =

0.0123(5.5)(0.182)(9002

)

5.5

191 .6F[1 + 1.5(0.7 2 )](30)(1 4)

215.8F = 215.5F191 .6F

Tav = 120F + =2

For the nominal 2-in bearing, the various clearances show that we have been in contact

with the recurving of( ^ 0 ) m i n . The figure of merit (the parasitic friction torque plus the

pumping torque negated) is best at c = 0.0018 in . For the nomina l 2-in bearing, we w i l l

place the top of the design window at c m i n = 0.002 i n, and b = d + 2(0.002) = 2.004 in.

A t this poin t, add the b and dunilateral tolerances:

d 2.000+0 of)0 in, b = 2.004+0'00 in

_

_


18/21

329

Now we can check the performance at c m i n , c, and c m a x . Of immediate interest is the fom

of the median clearance assembly, 9.82, as compared to any other satisfactory bearing

ensemble.

I f a nominal 1.875 in bearing is possible, construct another table with tb = 0 and

td = 0.

c b d Tmax ho Pst Tmax fos fom

0.0020 1.879 1.875 157.2 194.30 X S S S 7.36

0.0030 1.881 1.875 138.6 157.10 S S S S 8.64

0.0035 1.882 1.875 133.5 147.10 S S S S 9.05

0.0040 1.883 1.875 130.0 140.10 S S S S 9.32

0.0050 1.885 1.875 125.7 131.45 S S S S 9.59

0.0055 1.886 1.875 124.4 128.80 S S S S 9.63

0.0060 1.887 1.875 123.4 126.80 X S S S 9.64

The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our de

sign window.

d = 1.875+0 o0i in, b = 1.881+ 0 '[ 00 in

The ensemble median assembly has fom =9.31 .

We just had room to fit in a design window based upon the ( h o ) m i n constraint. Further

reduction in no mina l diameterw i l l preclude any smaller bearings. A table constructed for a

d = 1.750 in journal w i l l prove this.

We choose the nominal 1.875-in bearing ensemble because it has the largest figure

of merit. Ans.

12-19 This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b andradial clearance c.

The approach is similarto that of Prob. 12-18 and the tables w i l l change slightly. In the

table for a nominal b = 1.875 in , note that at c = 0.003 the constraints are "loose." Set

b = 1.875 in

d = 1.875 2(0.003) = 1.869 in

For the ensemble

b = 1.875+00 0? , d = 1.869+o0 00

Analyze at c m i n = 0.003, c = 0.004 in and c m a x = 0.005 in

At c m i n = 0.003 in: Tf = 138.4, \' = 3.160, S = 0.0297, H l o s s = 1035 Btu/h and the

Trumplerconditions are met.

At c = 0.004 in : Tf = 130F, \' = 3.872, S = 0.0205, H l o s s = 1106 Btu/h, fo m =

9.246 and the Trumpler conditions are O.K.

At c m a x = 0.005 in: Tf = 125.68F, \' = 4.325 /zreyn, S = 0.01466, H l o s s =

1129 Btu/h and the Trumpler conditions are O.K.

The ensemble figure of merit is slightly better; this bearing is slightly smaller. The lubri

cant cooler has sufficient capacity.


19/21


12-20 From Table 1 2-1 , Seireg and Dandage, 0 = 0.0141(106)reyn and b = 1360.0

(reyn) = 0.0141 exp[1360/(T + 95)] ( T in F)

= 0.0141 exp[1 360/( 1.8C + 127)] (C in C)

( M P a s) = 6.89(0.0141) exp [13 60/ (1. 8C + 127)] (C in C)

For SA E 30 at 79C

= 6.89(0.0141) exp{1360/[1.8(79) + 127]}

= 15.2MPa s Ans.

12-21 Originally

d = 2.000+0 001 in , b = 2.005+0'00 in

Doubled,

d = 4.000+0 00 0 in, b = 4.010+0'00o

The rad ial load quadrupled to 3000 lbfwhen the analyses for parts (a) and (b) were carried

out. Some of the results are:

Trumpler

Part c S fr/c Qs ho/c H l o s s ho ho f

(a) 0.007 3.416 0.0310 135.1 0.1612 6.56 0.1032 0.897 9898 0.000 722 0.000 360 0.005 67

(b) 0.0035 3.416 0.0310 135.1 0.1612 0.870 0.1032 0.897 1237 0.000 361 0.000 280 0.005 67

The side flow Qs differs because there is a c3 term and consequently an 8-fold increase.

H l o s s is related by a 9898/1237 or an 8-fold increase. The existing ho is related by a 2-fold

increase. Trumpler's ( h o ) m i n is related by a 1.286-fold increase

fom = 82.37 for double size 1} an 8-fold increase for double-size

fo m = 10.297 fororiginal size

12-22 From Table 12-8: K = 0.6(10 ) i n min/(lbf ft h). P = 500/[(1)(1)] = 500 psi,

V = nDN/12 = n(1)(200)/12 = 52.4 ft/min

Tables 12-10 and 12-11: f1 = 1 .8, f2 = 1

Table 12-12: PVmax = 46 700 psi ft/min, Pmax = 3560 psi, Vmax = 100 ft/min

4 F 4(500)P m a x = = = 637 psi < 3560 psi O.K.

n DL n

F

P = = 500 psi V = 52.4 ft/minDL

PV = 500(5 2.4) = 26 200 psi ft/min < 46 700 ps i ft/min O.K.


20/21

331

Solving Eq. (12-32) fort

n DLw n(1)(1)(0.005)t = = = 1388 h = 83 270 mi n

4/1 f2 KVF 4(1.8)(1)(0.6)(101 0)(52.4)(500)

Cycles = Nt = 200(83 270) = 16.7rev Ans.

12-23 Estimate bushing length with f1 = f2 = 1, and K = 0.6(101 0 ) i n 3 min/(lbf ft h)

1(1)(0.6)(10 _ 1 0)(2)(100)(400)(1000)Eq. (12-32): L = = 0.80in

3(0.002)

From Eq. (12-38), with fs = 0.03 from Table 12-9 applying nd = 2 to F

and hcR= 2.7 Btu/(h f t 2 F)720(0.03)(2)(100)(400)

L = = 3.58 in

778(2.7)(300 70)0.80 < L < 3.58 in

Trial 1 : Le t L = 1in , D = 1 in

4(2)(100)

P m a x = = 255 psi < 3560 psi O.K.

2(100)P = = 200 psi

1(1)

n (1)(400)V = = 104.7ft/min > 100 ft/min Not O.K.

12

Trial 2: Tr y D = 7 / 8 i n , L = 1 in

4(2)(100)P m a x = = 291 psi < 3560 psi O.K.

n(7/8)(1)

2(100)P = = 229 psi

7/8(1)

n(7/8)(400)V = = 91.6ft/min < 100 ft/min O.K.

12

PV = 229(91.6) = 20 976 psi ft/min < 46700 psi ft/min O.K.

V / 1

33 1.3

91.6

100 1.8(91.6 3 3 \J = 1.74100 33

L = 0.80(1.74) = 1.39 in


21/21


Trial 3: Try D = 7/8 in, L = 1.5 in

4(2)(100)

P m a x = = 194 psi < 3560 psi O.K.n(7/8)(1.5)2(100)

P = = 152 psi, V = 91.6 ft/min7/8(1 .5)

PV = 152(91.6) = 13 923 psi ft/min < 46700 psi ft/mi n O.K.

D = 7 /8 in , L = 1.5 in is acceptable Ans.

Suggestion: Try smaller sizes.

Solutions Mechanical Engineering Design Shigley 7th Edition (1)

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