Solucionario(Ingles) Diseño en Ingenieria Mecanica_shigley(8ed)(Capitulo 11 en Adelante)

FIRST PAGES

Chapter 11

11-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life xD , in multiplesof rating life, is

xD = 30 000(300)(60)106 = 540 Ans.

The design radial load FD is

FD = 1.2(1.898) = 2.278 kNFrom Eq. (11-6),

C10 = 2.278{ 540

0.02 + 4.439[ln(1/0.9)]1/1.483}1/3

= 18.59 kN Ans.Table 11-2: Choose a 02-30 mm with C10 = 19.5 kN. Ans.Eq. (11-18):

R = exp{

[540(2.278/19.5)3 0.02

4.439

]1.483}

= 0.919 Ans.

11-2 For the Angular-contact 02-series ball bearing as described, the rating life multiple is

xD = 50 000(480)(60)106 = 1440

The design load is radial and equal to

FD = 1.4(610) = 854 lbf = 3.80 kNEq. (11-6):

C10 = 854{

14400.02 + 4.439[ln(1/0.9)]1/1.483

}1/3= 9665 lbf = 43.0 kN

Table 11-2: Select a 02-55 mm with C10 = 46.2 kN. Ans.Using Eq. (11-18),

R = exp{

[

1440(3.8/46.2)3 0.024.439

]1.483}

= 0.927 Ans.

budynas_SM_ch11.qxd 12/04/2006 15:25 Page 289

Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali

FIRST PAGES

290 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

11-3 For the straight-Roller 03-series bearing selection, xD = 1440 rating lives from Prob. 11-2solution.

FD = 1.4(1650) = 2310 lbf = 10.279 kN

C10 = 10.279(

14401

)3/10= 91.1 kN

Table 11-3: Select a 03-55 mm with C10 = 102 kN. Ans.Using Eq. (11-18),

R = exp{

[

1440(10.28/102)10/3 0.024.439

]1.483}= 0.942 Ans.

11-4 We can choose a reliability goal of

0.90 = 0.95 for each bearing. We make the selec-tions, find the existing reliabilities, multiply them together, and observe that the reliabilitygoal is exceeded due to the roundup of capacity upon table entry.

Another possibility is to use the reliability of one bearing, say R1. Then set the relia-bility goal of the second as

R2 = 0.90R1or vice versa. This gives three pairs of selections to compare in terms of cost, geometry im-plications, etc.

11-5 Establish a reliability goal of

0.90 = 0.95 for each bearing. For a 02-series angular con-tact ball bearing,

C10 = 854{

14400.02 + 4.439[ln(1/0.95)]1/1.483

}1/3

= 11 315 lbf = 50.4 kNSelect a 02-60 mm angular-contact bearing with C10 = 55.9 kN.

RA = exp{

[

1440(3.8/55.9)3 0.024.439

]1.483}= 0.969

For a 03-series straight-roller bearing,

C10 = 10.279{

14400.02 + 4.439[ln(1/0.95)]1/1.483

}3/10= 105.2 kN

Select a 03-60 mm straight-roller bearing with C10 = 123 kN.

RB = exp{

[

1440(10.28/123)10/3 0.024.439

]1.483}= 0.977



FIRST PAGES

Chapter 11 291

The overall reliability is R = 0.969(0.977) = 0.947, which exceeds the goal. Note, usingRA from this problem, and RB from Prob. 11-3, R = 0.969(0.942) = 0.913, which stillexceeds the goal. Likewise, using RB from this problem, and RA from Prob. 11-2,R = 0.927(0.977) = 0.906.

The point is that the designer has choices. Discover them before making the selection de-cision. Did the answer to Prob. 11-4 uncover the possibilities?

11-6 Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. ForFr = 8 kN and Fa = 4 kN

xD = 5000(900)(60)106 = 270Eq. (11-5):

C10 = 8{

2700.02 + 4.439[ln(1/0.90)]1/1.483

}1/3= 51.8 kN

Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm withC0 = 37.5 kN.

FaC0

= 437.5

= 0.107Table 11-1:

Fa/(V Fr ) = 0.5 > eX2 = 0.56, Y2 = 1.46

Eq. (11-9):Fe = 0.56(1)(8) + 1.46(4) = 10.32 kN

Eq. (11-6): For R = 0.90,C10 = 10.32

(2701

)1/3= 66.7 kN > 61.8 kN

Trial #2: From Table 11-2 choose a 02-80 mm having C10 = 70.2 and C0 = 45.0.Check:

FaC0

= 445

= 0.089

Table 11-1: X2 = 0.56, Y2 = 1.53Fe = 0.56(8) + 1.53(4) = 10.60 kN

Eq. (11-6):

C10 = 10.60(

2701

)1/3= 68.51 kN < 70.2 kN

Selection stands.Decision: Specify a 02-80 mm deep-groove ball bearing. Ans.



FIRST PAGES


11-7 From Prob. 11-6, xD = 270 and the final value of Fe is 10.60 kN.

C10 = 10.6{

2700.02 + 4.439[ln(1/0.96)]1/1.483

}1/3= 84.47 kN

Table 11-2: Choose a deep-groove ball bearing, based upon C10 load ratings.Trial #1:Tentatively select a 02-90 mm.

C10 = 95.6, C0 = 62 kNFaC0

= 462

= 0.0645

From Table 11-1, interpolate for Y2.

Fa/C0 Y20.056 1.710.0645 Y20.070 1.63

Y2 1.711.63 1.71 =

0.0645 0.0560.070 0.056 = 0.607

Y2 = 1.71 + 0.607(1.63 1.71) = 1.661Fe = 0.56(8) + 1.661(4) = 11.12 kN

C10 = 11.12{

2700.02 + 4.439[ln(1/0.96)]1/1.483

}1/3= 88.61 kN < 95.6 kN

Bearing is OK.Decision: Specify a deep-groove 02-90 mm ball bearing. Ans.

11-8 For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.90 andFr = 12 kN

xD = 4000(750)(60)106 = 180

C10 = 12(

1801

)3/10= 57.0 kN Ans.



FIRST PAGES

Chapter 11 293

11-9

Assume concentrated forces as shown.

Pz = 8(24) = 192 lbfPy = 8(30) = 240 lbfT = 192(2) = 384 lbf in

T x = 384 + 1.5F cos 20 = 0

F = 3841.5(0.940) = 272 lbf

MzO = 5.75Py + 11.5RyA 14.25F sin 20 = 0;

thus 5.75(240) + 11.5RyA 14.25(272)(0.342) = 0RyA = 4.73 lbf

M yO = 5.75Pz 11.5RzA 14.25F cos 20 = 0;

thus 5.75(192) 11.5RzA 14.25(272)(0.940) = 0RzA = 413 lbf; RA = [(413)2 + (4.73)2]1/2 = 413 lbfFz = RzO + Pz + RzA + F cos 20 = 0

RzO + 192 413 + 272(0.940) = 0RzO = 34.7 lbfF y = RyO + Py + RyA F sin 20 = 0

RyO + 240 4.73 272(0.342) = 0RyO = 142 lbfRO = [(34.6)2 + (142)2]1/2 = 146 lbf

So the reaction at A governs.Reliability Goal:

0.92 = 0.96

FD = 1.2(413) = 496 lbf

B

O

z

11 12

"

RzO

RyO

Pz

Py

T

F 20

RyA

RzAA

T

y

2 34

"x



FIRST PAGES


xD = 30 000(300)(60/106) = 540C10 = 496

{ 5400.02 + 4.439[ln(1/0.96)]1/1.483

}1/3= 4980 lbf = 22.16 kN

A 02-35 bearing will do.Decision: Specify an angular-contact 02-35 mm ball bearing for the locations at A and O.Check combined reliability. Ans.

11-10 For a combined reliability goal of 0.90, use

0.90 = 0.95 for the individual bearings.

x0 = 50 000(480)(60)106 = 1440

The resultant of the given forces are RO = [(387)2 + 4672]1/2 = 607 lbfand RB = [3162 + (1615)2]1/2 = 1646 lbf .At O: Fe = 1.4(607) = 850 lbf

Ball: C10 = 850{

14400.02 + 4.439[ln(1/0.95)]1/1.483

}1/3= 11 262 lbf or 50.1 kN

Select a 02-60 mm angular-contact ball bearing with a basic load rating of 55.9 kN. Ans.At B: Fe = 1.4(1646) = 2304 lbf

Roller: C10 = 2304{

14400.02 + 4.439[ln(1/0.95)]1/1.483

}3/10= 23 576 lbf or 104.9 kN

Select a 02-80 mm cylindrical roller or a 03-60 mm cylindrical roller. The 03-series rollerhas the same bore as the 02-series ball. Ans.

z

20

16

10

O

FA

RO

RBB

A

C

y

x

FC20



FIRST PAGES

Chapter 11 295

11-11 The reliability of the individual bearings is R =

0.999 = 0.9995

From statics,RyO = 163.4 N, RzO = 107 N, RO = 195 NRyE = 89.2 N, RzE = 174.4 N, RE = 196 N

xD = 60 000(1200)(60)106 = 4320

C10 = 0.196{

43400.02 + 4.439[ln(1/0.9995)]1/1.483

}1/3= 8.9 kN

A 02-25 mm deep-groove ball bearing has a basic load rating of 14.0 kN which is ample.An extra-light bearing could also be investigated.

11-12 Given:Fr A = 560 lbf or 2.492 kNFr B = 1095 lbf or 4.873 kN

Trial #1: Use K A = K B = 1.5 and from Table 11-6 choose an indirect mounting.0.47Fr A

K A< ? >

0.47Fr BK B

(1)(0)

0.47(2.492)1.5

< ? >0.47(4.873)

1.50.781 < 1.527 Therefore use the upper line of Table 11-6.

Fa A = FaB = 0.47Fr BK B = 1.527 kNPA = 0.4Fr A + K A Fa A = 0.4(2.492) + 1.5(1.527) = 3.29 kNPB = Fr B = 4.873 kN

150

300

400

A

O

F zAF yA

E

RzE

RyE

FC

C

RzORyO

z

x

y



FIRST PAGES


Fig. 11-16: fT = 0.8Fig. 11-17: fV = 1.07Thus, a3l = fT fV = 0.8(1.07) = 0.856Individual reliability: Ri =

0.9 = 0.95

Eq. (11-17):(C10) A = 1.4(3.29)

[40 000(400)(60)

4.48(0.856)(1 0.95)2/3(90)(106)]0.3

= 11.40 kN

(C10)B = 1.4(4.873)[

40 000(400)(60)4.48(0.856)(1 0.95)2/3(90)(106)

]0.3= 16.88 kN

From Fig. 11-15, choose cone 32 305 and cup 32 305 which provide Fr = 17.4 kN andK = 1.95. With K = 1.95 for both bearings, a second trial validates the choice of cone32 305 and cup 32 305. Ans.

11-13R =

0.95 = 0.975

T = 240(12)(cos 20) = 2706 lbf in

F = 27066 cos 25

= 498 lbf

In xy-plane: MO = 82.1(16) 210(30) + 42RyC = 0RyC = 181 lbfRyO = 82 + 210 181 = 111 lbf

In xz-plane: MO = 226(16) 452(30) 42Rzc = 0RzC = 237 lbfRzO = 226 451 + 237 = 12 lbfRO = (1112 + 122)1/2 = 112 lbf Ans.RC = (1812 + 2372)1/2 = 298 lbf Ans.

FeO = 1.2(112) = 134.4 lbfFeC = 1.2(298) = 357.6 lbfxD = 40 000(200)(60)106 = 480

z

14"

16"

12"

RzO

RzC

RyOA

B

C

RyC

O

451210

226

T

T

82.1

x

y



FIRST PAGES

Chapter 11 297

(C10)O = 134.4{

4800.02 + 4.439[ln(1/0.975)]1/1.483

}1/3= 1438 lbf or 6.398 kN

(C10)C = 357.6{

4800.02 + 4.439[ln(1/0.975)]1/1.483

}1/3= 3825 lbf or 17.02 kN

Bearing at O: Choose a deep-groove 02-12 mm. Ans.Bearing at C: Choose a deep-groove 02-30 mm. Ans.There may be an advantage to the identical 02-30 mm bearings in a gear-reduction unit.

11-14 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust.The shaft floats within the endplay of the second (Roller) bearing. Since the thrust forcehere is larger than any radial load, the bearing absorbing the thrust is heavily loaded com-pared to the other bearing. The second bearing is thus oversized and does not contributemeasurably to the chance of failure. This is predictable. The reliability goal is not

0.99,

but 0.99 for the ball bearing. The reliability of the roller is 1. Beginning here saves effort.Bearing at A (Ball)

Fr = (362 + 2122)1/2 = 215 lbf = 0.957 kNFa = 555 lbf = 2.47 kN

Trial #1:Tentatively select a 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN.

FaC0

= 2.4763.0

= 0.0392

xD = 25 000(600)(60)106 = 900

Table 11-1: X2 = 0.56, Y2 = 1.88Fe = 0.56(0.957) + 1.88(2.47) = 5.18 kN

FD = f A Fe = 1.3(5.18) = 6.73 kN

C10 = 6.73{

9000.02 + 4.439[ln(1/0.99)]1/1.483

}1/3= 107.7 kN > 90.4 kN

Trial #2:Tentatively select a 02-95 mm angular-contact ball with C10 = 121 kN and C0 = 85 kN.

FaC0

= 2.4785

= 0.029



FIRST PAGES


Table 11-1: Y2 = 1.98Fe = 0.56(0.957) + 1.98(2.47) = 5.43 kN

FD = 1.3(5.43) = 7.05 kNC10 = 7.05

{900

0.02 + 4.439[ln(1/0.99)]1/1.483}1/3

= 113 kN < 121 kN O.K.Select a 02-95 mm angular-contact ball bearing. Ans.Bearing at B (Roller): Any bearing will do since R = 1. Lets prove it. From Eq. (11-18)when (

af FDC10

)3xD < x0 R = 1

The smallest 02-series roller has a C10 = 16.8 kN for a basic load rating.(0.42716.8

)3(900) < ? > 0.02

0.0148 < 0.02 R = 1 Spotting this early avoided rework from

0.99 = 0.995.

Any 02-series roller bearing will do. Same bore or outside diameter is a common choice.(Why?) Ans.

11-15 Hoover Ball-bearing Division uses the same 2-parameter Weibull model as Timken:b = 1.5, = 4.48. We have some data. Lets estimate parameters b and from it. InFig. 11-5, we will use line AB. In this case, B is to the right of A.

For F = 18 kN, (x)1 = 115(2000)(16)106 = 13.8This establishes point 1 on the R = 0.90 line.

10

10

2

10

181 2

39.6

100

1

10

13.8 72

1100 x

2 log x

F

A B

log F

R 0.90R 0.20



FIRST PAGES

Chapter 11 299

The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameterWeibull distribution, x0 = 0 and points A and B are related by [see Eq. (20-25)]:

xA = [ln(1/0.90)]1/b (1)xB = [ln(1/0.20)]1/b

and xB/xA is in the same ratio as 600/115. Eliminating

b = ln[ln(1/0.20)/ ln(1/0.90)]ln(600/115) = 1.65 Ans.

Solving for in Eq. (1)

= xA[ln(1/RA)]1/1.65 =1

[ln(1/0.90)]1/1.65 = 3.91 Ans.

Therefore, for the data at hand,

R = exp[

(

x

3.91

)1.65]

Check R at point B: xB = (600/115) = 5.217

R = exp[

(5.217

3.91

)1.65 ]= 0.20

Note also, for point 2 on the R = 0.20 line.log(5.217) log(1) = log(xm)2 log(13.8)

(xm)2 = 72

11-16 This problem is rich in useful variations. Here is one.Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of(0.99)1/6 = 0.9983.Shaft a

FrA = (2392 + 1112)1/2 = 264 lbf or 1.175 kNFrB = (5022 + 10752)1/2 = 1186 lbf or 5.28 kN

Thus the bearing at B controls

xD = 10 000(1200)(60)106 = 720

0.02 + 4.439[ln(1/0.9983)]1/1.483 = 0.080 26

C10 = 1.2(5.2)(

7200.080 26

)0.3= 97.2 kN

Select either a 02-80 mm with C10 = 106 kN or a 03-55 mm with C10 = 102 kN. Ans.



FIRST PAGES


Shaft bFrC = (8742 + 22742)1/2 = 2436 lbf or 10.84 kNFrD = (3932 + 6572)1/2 = 766 lbf or 3.41 kN

The bearing at C controls

xD = 10 000(240)(60)106 = 144

C10 = 1.2(10.84)(

1440.0826

)0.3= 122 kN

Select either a 02-90 mm with C10 = 142 kN or a 03-60 mm with C10 = 123 kN. Ans.Shaft c

FrE = (11132 + 23852)1/2 = 2632 lbf or 11.71 kNFrF = (4172 + 8952)1/2 = 987 lbf or 4.39 kN

The bearing at E controlsxD = 10 000(80)(60/106) = 48

C10 = 1.2(11.71)(

480.0826

)0.3= 94.8 kN

Select a 02-80 mm with C10 = 106 kN or a 03-60 mm with C10 = 123 kN. Ans.

11-17 The horizontal separation of the R = 0.90 loci in a log F-log x plot such as Fig. 11-5will be demonstrated. We refer to the solution of Prob. 11-15 to plot point G (F =18 kN, xG = 13.8). We know that (C10)1 = 39.6 kN, x1 = 1. This establishes the unim-proved steel R = 0.90 locus, line AG. For the improved steel

(xm)1 = 360(2000)(60)106 = 43.2We plot point G (F = 18 kN, xG = 43.2), and draw the R = 0.90 locus AmG parallelto AG

10

10

2

10

18G G

39.655.8

100

1

10

13.8

11002

x

log x

F

A

AmImproved steel

log F

Unimproved steel

43.2

R 0.90

R 0.90

13

13



FIRST PAGES

Chapter 11 301

We can calculate (C10)m by similar triangles.log(C10)m log 18

log 43.2 log 1 =log 39.6 log 18log 13.8 log 1

log(C10)m = log 43.2log 13.8 log(

39.618

)+ log 18

(C10)m = 55.8 kNThe usefulness of this plot is evident. The improvement is 43.2/13.8 = 3.13 fold in life.This result is also available by (L10)m/(L10)1 as 360/115 or 3.13 fold, but the plot showsthe improvement is for all loading. Thus, the manufacturers assertion that there is at leasta 3-fold increase in life has been demonstrated by the sample data given. Ans.

11-18 Express Eq. (11-1) asFa1 L1 = Ca10L10 = K

For a ball bearing, a = 3 and for a 02-30 mm angular contact bearing, C10 = 20.3 kN.K = (20.3)3(106) = 8.365(109)

At a load of 18 kN, life L1 is given by:

L1 = KFa1= 8.365(10

9)183

= 1.434(106) revFor a load of 30 kN, life L2 is:

L2 = 8.365(109)

303= 0.310(106) rev

In this case, Eq. (7-57) the Palmgren-Miner cycle ratio summation rule can be ex-pressed as

l1L1

+ l2L2

= 1Substituting,

200 0001.434(106) +

l20.310(106) = 1

l2 = 0.267(106) rev Ans.

11-19 Total life in revolutionsLet:

l = total turnsf1 = fraction of turns at F1f2 = fraction of turns at F2



FIRST PAGES


From the solution of Prob. 11-18, L1 = 1.434(106) rev and L2 = 0.310(106) rev.Palmgren-Miner rule:

l1L1

+ l2L2

= f1lL1

+ f2lL2

= 1from which

l = 1f1/L1 + f2/L2l = 1{0.40/[1.434(106)]} + {0.60/[0.310(106)]}= 451 585 rev Ans.

Total life in loading cycles4 min at 2000 rev/min = 8000 rev

6 min10 min/cycle

at 2000 rev/min = 12 000 rev20 000 rev/cycle

451 585 rev20 000 rev/cycle

= 22.58 cycles Ans.Total life in hours (

10min

cycle

)(22.58 cycles

60 min/h

)= 3.76 h Ans.

11-20 While we made some use of the log F-log x plot in Probs. 11-15 and 11-17, the principaluse of Fig. 11-5 is to understand equations (11-6) and (11-7) in the discovery of the cata-log basic load rating for a case at hand.Point D

FD = 495.6 lbflog FD = log 495.6 = 2.70

xD = 30 000(300)(60)106 = 540

log xD = log 540 = 2.73K D = F3DxD = (495.6)3(540)

= 65.7(109) lbf3 turnslog K D = log[65.7(109)] = 10.82

FD has the following uses: Fdesign, Fdesired, Fe when a thrust load is present. It can includeapplication factor af , or not. It depends on context.



FIRST PAGES

Chapter 11 303

Point BxB = 0.02 + 4.439[ln(1/0.99)]1/1.483

= 0.220 turnslog xB = log 0.220 = 0.658

FB = FD(

xD

xB

)1/3= 495.6

( 5400.220

)1/3= 6685 lbf

Note: Example 11-3 used Eq. (11-7). Whereas, here we basically used Eq. (11-6).log FB = log(6685) = 3.825

K D = 66853(0.220) = 65.7(109) lbf3 turns (as it should)Point A

FA = FB = C10 = 6685 lbflog C10 = log(6685) = 3.825

xA = 1log xA = log(1) = 0

K10 = F3AxA = C310(1) = 66853 = 299(109) lbf3 turnsNote that K D/K10 = 65.7(109)/[299(109)] = 0.220, which is xB . This is worth knowingsince

K10 = K DxB

log K10 = log[299(109)] = 11.48

Now C10 = 6685 lbf = 29.748 kN, which is required for a reliability goal of 0.99. If weselect an angular contact 02-40 mm ball bearing, then C10 = 31.9 kN = 7169 lbf.

0.11

0.658

10

101

102

2

1022

103

495.6

6685

3

1044

103

3

x

log x

F

A

D

B

log F

540



FIRST PAGES

Chapter 12

12-1 Given dmax = 1.000 in and bmin = 1.0015 in, the minimum radial clearance is

cmin = bmin dmax2 =1.0015 1.000

2= 0.000 75 in

Also l/d = 1r = 1.000/2 = 0.500

r/c = 0.500/0.000 75 = 667N = 1100/60 = 18.33 rev/sP = W/(ld) = 250/[(1)(1)] = 250 psi

Eq. (12-7): S = (6672)[

8(106)(18.33)250

]= 0.261

Fig. 12-16: h0/c = 0.595Fig. 12-19: Q/(rcNl) = 3.98Fig. 12-18: f r/c = 5.8Fig. 12-20: Qs/Q = 0.5

h0 = 0.595(0.000 75) = 0.000 466 in Ans.

f = 5.8r/c

= 5.8667

= 0.0087

The power loss in Btu/s is

H = 2 f Wr N778(12) =

2(0.0087)(250)(0.5)(18.33)778(12)

= 0.0134 Btu/s Ans.Q = 3.98rcNl = 3.98(0.5)(0.000 75)(18.33)(1) = 0.0274 in3/s

Qs = 0.5(0.0274) = 0.0137 in3/s Ans.

12-2cmin = bmin dmax2 =

1.252 1.2502

= 0.001 inr

.= 1.25/2 = 0.625 inr/c = 0.625/0.001 = 625

N = 1150/60 = 19.167 rev/s

P = 4001.25(2.5) = 128 psi

l/d = 2.5/1.25 = 2

S = (6252)(10)(106)(19.167)

128= 0.585


Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali

FIRST PAGES

Chapter 12 305

The interpolation formula of Eq. (12-16) will have to be used. From Figs. 12-16, 12-21,and 12-19For l/d = , ho/c = 0.96, P/pmax = 0.84, Q

rcNl= 3.09

l/d = 1, ho/c = 0.77, P/pmax = 0.52, QrcNl

= 3.6

l/d = 12

, ho/c = 0.54, P/pmax = 0.42, QrcNl

= 4.4

l/d = 14

, ho/c = 0.31, P/pmax = 0.28, QrcNl

= 5.25

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

l/d y y1 y1/2 y1/4 yl/dho/c 2 0.96 0.77 0.54 0.31 0.88P/pmax 2 0.84 0.52 0.42 0.28 0.64Q/rcNl 2 3.09 3.60 4.40 5.25 3.28

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

pmax = 1280.64 = 200 psi Ans.

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


3.005 3.0002

= 0.0025 in

r.= 3.000/2 = 1.500 in

l/d = 1.5/3 = 0.5r/c = 1.5/0.0025 = 600

N = 600/60 = 10 rev/s

P = 8001.5(3) = 177.78 psi

Fig. 12-12: SAE 10, = 1.75 reyn

S = (6002)[

1.75(106)(10)177.78

]= 0.0354

Figs. 12-16 and 12-21: ho/c = 0.11, P/pmax = 0.21ho = 0.11(0.0025) = 0.000 275 in Ans.

pmax = 177.78/0.21 = 847 psi Ans.



FIRST PAGES


Fig. 12-12: SAE 40, = 4.5 reyn

S = 0.0354(

4.51.75

)= 0.0910

ho/c = 0.19, P/pmax = 0.275ho = 0.19(0.0025) = 0.000 475 in Ans.

pmax = 177.78/0.275 = 646 psi Ans.


3.006 3.0002

= 0.003

r.= 3.000/2 = 1.5 in

l/d = 1r/c = 1.5/0.003 = 500

N = 750/60 = 12.5 rev/s

P = 6003(3) = 66.7 psi

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

S = (5002)[

2.1(106)(12.5)66.7

]= 0.0984

From Figs. 12-16 and 12-21:ho/c = 0.34, P/pmax = 0.395

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

pmax = 66.70.395 = 169 psi Ans.

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

S = (5002)[5.05(106)(12.5)

66.7

]= 0.237

From Figs. 12-16 and 12-21:ho/c = 0.57, P/pmax = 0.47

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

pmax = 66.70.47 = 142 psi Ans.



FIRST PAGES

Chapter 12 307


2.0024 22

= 0.0012 in

r.= d

2= 2

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

r/c = 1/0.0012 = 833N = 800/60 = 13.33 rev/sP = 600

2(1) = 300 psi

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

S = (8332)[

3.75(106)(13.3)300

]= 0.115

From Figs. 12-16, 12-18 and 12-19:ho/c = 0.23, r f/c = 3.8, Q/(rcNl) = 5.3

ho = 0.23(0.0012) = 0.000 276 in Ans.f = 3.8

833= 0.004 56

The power loss due to friction is

H = 2 f Wr N778(12) =

2(0.004 56)(600)(1)(13.33)778(12)

= 0.0245 Btu/s Ans.Q = 5.3rcNl

= 5.3(1)(0.0012)(13.33)(1)= 0.0848 in3/s Ans.


25.04 252

= 0.02 mmr = d/2 = 25/2 = 12.5 mm, l/d = 1

r/c = 12.5/0.02 = 625N = 1200/60 = 20 rev/sP = 1250

252= 2 MPa

For = 50 mPa s, S = (6252)[50(103)(20)

2(106)]

= 0.195

From Figs. 12-16, 12-18 and 12-20:ho/c = 0.52, f r/c = 4.5, Qs/Q = 0.57

ho = 0.52(0.02) = 0.0104 mm Ans.f = 4.5

625= 0.0072

T = f Wr = 0.0072(1.25)(12.5) = 0.1125 N m



FIRST PAGES


The power loss due to friction isH = 2T N = 2(0.1125)(20) = 14.14 W Ans.

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


30.05 30.002

= 0.025 mm

r = d2

= 302

= 15 mm

r

c= 15

0.025= 600

N = 112060

= 18.67 rev/s

P = 275030(50) = 1.833 MPa

S = (6002)[

60(103)(18.67)1.833(106)

]= 0.22

ld

= 5030

= 1.67

This l/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, y1/2 = 0.34, y1 = 0.54, y = 0.89

Substituting into Eq. (12-16), hoc

= 0.659From Fig. 12-18, the f r/c values are:

y1/4 = 7.4, y1/2 = 6.0, y1 = 5.0, y = 4.0

Substituting into Eq. (12-16), f rc

= 4.59

From Fig. 12-19, the Q/(rcNl) values are:y1/4 = 5.65, y1/2 = 5.05, y1 = 4.05, y = 2.95

Substituting into Eq. (12-16), QrcN l

= 3.605

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.



FIRST PAGES

Chapter 12 309


75.10 752

= 0.05 mml/d = 36/75 = 0.5 (close enough)

r = d/2 = 75/2 = 37.5 mmr/c = 37.5/0.05 = 750

N = 720/60 = 12 rev/sP = 2000

75(36) = 0.741 MPa

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

S = (7502)[

18.5(103)(12)0.741(106)

]= 0.169

From Figures 12-16, 12-18 and 12-21:ho/c = 0.29, f r/c = 5.1, P/pmax = 0.315

ho = 0.29(0.05) = 0.0145 mm Ans.f = 5.1/750 = 0.0068T = f Wr = 0.0068(2)(37.5) = 0.51 N m

The heat loss rate equals the rate of work on the filmHloss = 2T N = 2(0.51)(12) = 38.5 W Ans.pmax = 0.741/0.315 = 2.35 MPa Ans.

Fig. 12-13: SAE 40, = 37 MPa sS = 0.169(37)/18.5 = 0.338

From Figures 12-16, 12-18 and 12-21:ho/c = 0.42, f r/c = 8.5, P/pmax = 0.38

ho = 0.42(0.05) = 0.021 mm Ans.f = 8.5/750 = 0.0113T = f Wr = 0.0113(2)(37.5) = 0.85 N m

Hloss = 2T N = 2(0.85)(12) = 64 W Ans.pmax = 0.741/0.38 = 1.95 MPa Ans.


50.05 502

= 0.025 mmr = d/2 = 50/2 = 25 mm

r/c = 25/0.025 = 1000l/d = 25/50 = 0.5, N = 840/60 = 14 rev/s

P = 200025(50) = 1.6 MPa



FIRST PAGES


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

S = (10002)[

34(103)(14)1.6(106)

]= 0.2975

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

ho/c = 0.40, f r/c = 7.8, Qs/Q = 0.74, Q/(rcNl) = 4.9ho = 0.40(0.025) = 0.010 mm Ans.f = 7.8/1000 = 0.0078T = f Wr = 0.0078(2)(25) = 0.39 N mH = 2T N = 2(0.39)(14) = 34.3 W Ans.Q = 4.9rcNl = 4.9(25)(0.025)(14)(25) = 1072 mm2/s

Qs = 0.74(1072) = 793 mm3/s Ans.

12-10 Consider the bearings as specified by

minimum f : d+0td , b+tb0maximum W: d +0td , b

+tb0

and differing only in d and d .Preliminaries:

l/d = 1P = 700/(1.252) = 448 psiN = 3600/60 = 60 rev/s

Fig. 12-16:minimum f : S = 0.08maximum W: S = 0.20

Fig. 12-12: = 1.38(106) reynN/P = 1.38(106)(60/448) = 0.185(106)

Eq. (12-7):r

c=

S

N/P

For minimum f :r

c=

0.08

0.185(106) = 658

c = 0.625/658 = 0.000 950 .= 0.001 in



FIRST PAGES

Chapter 12 311

If this is cmin,b d = 2(0.001) = 0.002 in

The median clearance is

c = cmin + td + tb2 = 0.001 +td + tb

2and the clearance range for this bearing is

c = td + tb2

which is a function only of the tolerances.For maximum W:

r

c=

0.2

0.185(106) = 1040

c = 0.625/1040 = 0.000 600 .= 0.0005 inIf this is cmin

b d = 2cmin = 2(0.0005) = 0.001 in

c = cmin + td + tb2 = 0.0005 +td + tb

2

c = td + tb2

The difference (mean) in clearance between the two clearance ranges, crange, is

crange = 0.001 + td + tb2 (

0.0005 + td + tb2

)= 0.0005 in

For the minimum f bearingb d = 0.002 in

or

d = b 0.002 inFor the maximum W bearing

d = b 0.001 inFor 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 d of the minimum friction bearing by 0.001 in, defines d of the maximum loadbearing. Thus, the clearance range provides for bearing dimensions which are attainablein manufacturing. Ans.



FIRST PAGES


12-11 Given: SAE 30, N = 8 rev/s, Ts = 60C, l/d = 1, d = 80 mm, b = 80.08 mm,W = 3000 N

cmin = bmin dmax2 =80.08 80

2= 0.04 mm

r = d/2 = 80/2 = 40 mmr

c= 40

0.04= 1000

P = 300080(80) = 0.469 MPa

Trial #1: From Figure 12-13 for T = 81C, = 12 mPa sT = 2(81C 60C) = 42C

S = (10002)[

12(103)(8)0.469(106)

]= 0.2047

From Fig. 12-24,

0.120TP

= 0.349 + 6.009(0.2047) + 0.0475(0.2047)2 = 1.58

T = 1.58(

0.4690.120

)= 6.2C

Discrepancy = 42C 6.2C = 35.8CTrial #2: From Figure 12-13 for T = 68C, = 20 mPa s,

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

S = 0.2047(

2012

)= 0.341

From Fig. 12-24,

0.120TP

= 0.349 + 6.009(0.341) + 0.0475(0.341)2 = 2.4

T = 2.4(

0.4690.120

)= 9.4C

Discrepancy = 16C 9.4C = 6.6CTrial #3: = 21 mPa s, T = 65C

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

S = 0.2047(

2112

)= 0.358



FIRST PAGES

Chapter 12 313

From Fig. 12-24,

0.120TP

= 0.349 + 6.009(0.358) + 0.0475(0.358)2 = 2.5

T = 2.5(

0.4690.120

)= 9.8C

Discrepancy = 10C 9.8C = 0.2C O.K.Tav = 65C Ans.

T1 = Tav T/2 = 65C (10C/2) = 60CT2 = Tav + T/2 = 65C + (10C/2) = 70CS = 0.358

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

hoc

= 0.68, f r/c = 7.5, QrcN l

= 3.8, QsQ = 0.44

ho = 0.68(0.04) = 0.0272 mm Ans.

f = 7.51000

= 0.0075

T = f Wr = 0.0075(3)(40) = 0.9 N mH = 2T N = 2(0.9)(8) = 45.2 W Ans.Q = 3.8(40)(0.04)(8)(80) = 3891 mm3/s

Qs = 0.44(3891) = 1712 mm3/s Ans.

12-12 Given:d = 2.5 in,b = 2.504 in,cmin = 0.002 in, W = 1200 lbf,SAE = 20, Ts = 110F,N = 1120 rev/min, and l = 2.5 in.For a trial film temperature Tf = 150F

Tf S T (From Fig. 12-24)150 2.421 0.0921 18.5

Tav = Ts + T2 = 110F +18.5F

2= 119.3F

Tf Tav = 150F 119.3Fwhich is not 0.1 or less, therefore try averaging

(Tf )new = 150F + 119.3F2 = 134.6F



FIRST PAGES


Proceed with additional trials

Trial NewTf S T Tav Tf

150.0 2.421 0.0921 18.5 119.3 134.6134.6 3.453 0.1310 23.1 121.5 128.1128.1 4.070 0.1550 25.8 122.9 125.5125.5 4.255 0.1650 27.0 123.5 124.5124.5 4.471 0.1700 27.5 123.8 124.1124.1 4.515 0.1710 27.7 123.9 124.0124.0 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 pointthey would only add complexity. Depending where you stop, you can enter the analysis.

(a) = 4.541(106) reyn, S = 0.1724

From Fig. 12-16: hoc

= 0.482, ho = 0.482(0.002) = 0.000 964 in

From Fig. 12-17: = 56 Ans.(b) e = c ho = 0.002 0.000 964 = 0.001 04 in Ans.

(c) From Fig. 12-18: f rc

= 4.10, f = 4.10(0.002/1.25) = 0.006 56 Ans.

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

H = 2T N778(12) =

2(9.84)(1120/60)778(12) = 0.124 Btu/s Ans.

(e) From Fig. 12-19: QrcNl

= 4.16, Q = 4.16(1.25)(0.002)(

112060

)(2.5)

= 0.485 in3/s Ans.

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

3/s Ans.

(f) From Fig. 12-21: Ppmax

= 0.45, pmax = 12002.52(0.45) = 427 psi Ans.

pmax = 16 Ans.

(g) p0 = 82 Ans.(h) Tf = 123.9F Ans.(i) Ts + T = 110F + 27.8F = 137.8F Ans.



FIRST PAGES

Chapter 12 315

12-13 Given: d = 1.250 in, td = 0.001in, b = 1.252 in, tb = 0.003in, l = 1.25 in, W = 250 lbf,N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120F.Below is a partial tabular summary for comparison purposes.

cmin c cmax0.001 in 0.002 in 0.003 in

Tf 132.2 125.8 124.0T 24.3 11.5 7.96Tmax 144.3 131.5 128.0 2.587 3.014 3.150S 0.184 0.053 7 0.024 9 0.499 0.775 0 0.873f rc

4.317 1.881 1.243

QrcNjl

4.129 4.572 4.691

QsQ 0.582 0.824 0.903

hoc

0.501 0.225 0.127

f 0.006 9 0.006 0.005 9Q 0.094 1 0.208 0.321Qs 0.054 8 0.172 0.290ho 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 thedistribution is bounded; the extreme cases, cmin and cmax, coupled with c provide thecharactistic description for the designer. All assemblies must be satisfactory.The designer does not specify a journal-bushing bearing, but an ensemble of bearings.

12-14 Computer programs will varyFortran based, MATLAB, spreadsheet, etc.

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

Given the average film temperature, establish the bearing properties. Given a sump temperature, find the average film temperature, then establish the bearing

properties. Now we acknowledge the environmental temperatures 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 andHloss equal. The steps for determining cmin are provided within Trial #1 through Trial #3on the following page.



FIRST PAGES


Trial #1: Choose a value of Tf . Find the corresponding viscosity. Find the Sommerfeld number. Find f r/c , then

Hgen = 25451050W N c( f r

c

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

T = 0.103P( f r/c)(1 0.5Qs/Q)[Q/(rcNjl)]

Hloss = hCR A(Tf T)1 + Display Tf , S, Hgen, Hloss

Trial #2: Choose another Tf , repeating above drill.Trial #3:Plot the results of the first two trials.

Choose (Tf )3 from plot. Repeat the drill. Plot the results of Trial #3 on the above graph.If you are not within 0.1F, iterate again. Otherwise, stop, and find all the properties ofthe bearing for the first clearance, cmin . See if Trumpler conditions are satisfied, and if so,analyze c and cmax .

The bearing ensemble in the current problem statement meets Trumplers criteria(for nd = 2).

This adequacy assessment protocol can be used as a design tool by giving the studentsadditional possible bushing sizes.

b (in) tb (in)2.254 0.0042.004 0.0041.753 0.003

Otherwise, the design option includes reducing l/d to save on the cost of journal machin-ing and vender-supplied bushings.

H HgenHloss, linear with Tf

(Tf)1 (Tf)3 (Tf)2Tf



FIRST PAGES

Chapter 12 317

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 for c = cminTrial #1:

Choose an initial Tf . Find the viscosity. Find the Sommerfeld number. Find f r/c, ho/c, and . From Eq. (12-24), find T .

Tav = Ts + T2 Display Tf , S, T , and Tav.

Trial #2: Choose another Tf . Repeat the drill, and display the second set of values for Tf ,

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

Trial #3:Pick the third Tf from the plot and repeat the procedure. If (Tf )3 and (Tav)3 differ by morethan 0.1F, plot the results for Trials #2 and #3 and try again. If they are within 0.1F, de-termine the bearing parameters, check the Trumpler criteria, and compare Hloss with thelubricants cooling capacity.

Repeat the entire procedure for c = cmax to assess the cooling capacity for the maxi-mum radial clearance. Finally, examine c = c to characterize the ensemble of bearings.

12-17 An adequacy assessment associated with a design task is required. Trumplers criteriawill do.

d = 50.00+0.000.05 mm, b = 50.084+0.0100.000 mmSAE 30, N = 2880 rev/min or 48 rev/s, W = 10 kN

cmin = bmin dmax2 =50.084 50

2= 0.042 mm

r = d/2 = 50/2 = 25 mmr/c = 25/0.042 = 595

l = 12

(55 5) = 25 mml /d = 25/50 = 0.5

p = W4rl

= 10(106)

4(0.25)(0.25) = 4000 kPa

Tav

2

1

Tf(Tf)1(Tf)2 (Tf)3

Tav Tf



FIRST PAGES


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

S = (5952)[

13(103)(48)4000(103)

]= 0.055

From Figs. 12-18 and 12-16:f rc

= 2.3, = 0.85.

From Eq. (12-25), T = 978(106)

1 + 1.52( f r/c)SW 2

psr4

= 978(106)

1 + 1.5(0.85)2[

2.3(0.055)(102)200(25)4

]

= 76.0CTav = Ts + T/2 = 55C + (76C/2) = 93C

Trial #2: Choose (Tf )2 = 100C. From Fig. 12-13, = 7 mPa s.S = 0.055

(713

)= 0.0296


= 1.6, = 0.90

T = 978(106)

1 + 1.5(0.9)2[

1.6(0.0296)(102)200(25)4

]= 26.8C

Tav = 55C + 26.8C2 = 68.4C

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

S = 0.055(

10.813

)= 0.0457


= 2.2, = 0.875

T = 978(106)

1 + 1.5(0.8752)[

2.2(0.0457)(102)200(25)4

]= 58.6C

Tav = 55C + 58.6C2 = 84.3C

Result is close. Choose Tf = 85C + 84.3C2 = 84.7C

100

Tav

Tf60 70 80 90 100

(79C, 93C)

(79C, 79C)

85C(100C, 68.4C)

(100C, 100C)

90

80

70

Tav Tf



FIRST PAGES

Chapter 12 319

Fig. 12-13: = 10.8 MPa s

S = 0.055(

10.813

)= 0.0457

f rc

= 2.23, = 0.874, hoc

= 0.13

T = 978(106)

1 + 1.5(0.8742)[

2.23(0.0457)(102)200(254)

]= 59.5C

Tav = 55C + 59.5C2 = 84.7C O.K.

From Eq. (12-22)

Qs = (1 + 1.52)psrc3

3l

= [1 + 1.5(0.8742)][(200)(0.0423)(25)

3(10)(106)(25)]

= 3334 mm3/sho = 0.13(0.042) = 0.005 46 mm or 0.000 215 in

Trumpler:ho = 0.0002 + 0.000 04(50/25.4)

= 0.000 279 in Not O.K.Tmax = Ts + T = 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, weve performed elements of the design task. Now lets do it more completely.First, remember our viewpoint.

The values of the unilateral tolerances, tb and td , reflect the routine capabilities of thebushing vendor and the in-house capabilities. While 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 journal which satisfies Trumplers con-straint of Pst 300 psi.

Pst = W2dl 300

W2d2 l /d

300 d

W600(l /d)

dmin =

9002(300)(0.5) = 1.73 in



FIRST PAGES


In this problem we will take journal diameter as the nominal value and the bushing boreas a variable. In the next problem, we will take the bushing bore as nominal and the jour-nal diameter as free.

To determine where the constraints are, we will set tb = td = 0, and thereby shrinkthe design window to a point.We set d = 2.000 in

b = d + 2cmin = d + 2cnd = 2 (This makes Trumplers nd 2 tight)

and construct a table.

c b d Tf* Tmax ho Pst Tmax n fom

0.0010 2.0020 2 215.50 312.0 5.740.0011 2.0022 2 206.75 293.0 6.060.0012 2.0024 2 198.50 277.0 6.370.0013 2.0026 2 191.40 262.8 6.660.0014 2.0028 2 185.23 250.4 6.940.0015 2.0030 2 179.80 239.6 7.200.0016 2.0032 2 175.00 230.1 7.450.0017 2.0034 2 171.13 220.3 7.650.0018 2.0036 2 166.92 213.9 7.910.0019 2.0038 2 163.50 206.9 8.120.0020 2.0040 2 160.40 200.6 8.32*Sample calculation for the first entry of this column.Iteration yields: Tf = 215.5FWith Tf = 215.5F, from Table 12-1

= 0.0136(106) exp[1271.6/(215.5 + 95)] = 0.817(106) reynN = 3000/60 = 50 rev/s, P = 900

4= 225 psi

S =(

10.001

)2[0.817(106)(50)225

]= 0.182

From Figs. 12-16 and 12-18: = 0.7, f r/c = 5.5Eq. (1224):

TF = 0.0123(5.5)(0.182)(9002)

[1 + 1.5(0.72)](30)(14) = 191.6F

Tav = 120F + 191.6F2 = 215.8F.= 215.5F

For the nominal 2-in bearing, the various clearances show that we have been in contactwith the recurving of (ho)min. The figure of merit (the parasitic friction torque plus thepumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we willplace the top of the design window at cmin = 0.002 in, and b = d + 2(0.002) = 2.004 in.At this point, add the b and d unilateral tolerances:

d = 2.000+0.0000.001 in, b = 2.004+0.0030.000 in



FIRST PAGES

Chapter 12 321

Now we can check the performance at cmin , c , and cmax . Of immediate interest is the fomof the median clearance assembly, 9.82, as compared to any other satisfactory bearingensemble.

If a nominal 1.875 in bearing is possible, construct another table with tb = 0 andtd = 0.

c b d Tf Tmax ho Pst Tmax fos fom

0.0020 1.879 1.875 157.2 194.30 7.360.0030 1.881 1.875 138.6 157.10 8.640.0035 1.882 1.875 133.5 147.10 9.050.0040 1.883 1.875 130.0 140.10 9.320.0050 1.885 1.875 125.7 131.45 9.590.0055 1.886 1.875 124.4 128.80 9.630.0060 1.887 1.875 123.4 126.80 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+0.0000.001 in, b = 1.881+0.0030.000 inThe ensemble median assembly has fom = 9.31.

We just had room to fit in a design window based upon the (ho)min constraint. Furtherreduction in nominal diameter will preclude any smaller bearings. A table constructed for ad = 1.750 in journal will prove this.

We choose the nominal 1.875-in bearing ensemble because it has the largest figureof 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 similar to that of Prob. 12-18 and the tables will change slightly. In thetable for a nominal b = 1.875 in, note that at c = 0.003 the constraints are loose. Set

b = 1.875 ind = 1.875 2(0.003) = 1.869 in

For the ensemble

b = 1.875+0.0030.001, d = 1.869+0.0000.001Analyze at cmin = 0.003, c = 0.004 in and cmax = 0.005 inAt cmin = 0.003 in: Tf = 138.4F, = 3.160, S = 0.0297, Hloss = 1035 Btu/h and theTrumpler conditions are met.At c = 0.004 in: Tf = 130F, = 3.872, S = 0.0205, Hloss = 1106 Btu/h, fom =9.246 and the Trumpler conditions are O.K.At cmax = 0.005 in: Tf = 125.68F, = 4.325 reyn, S = 0.014 66, Hloss =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.



FIRST PAGES


12-20 From Table 12-1, 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[1360/(1.8C + 127)] (C in C)(mPa s) = 6.89(0.0141) exp[1360/(1.8C + 127)] (C in C)

For SAE 30 at 79C = 6.89(0.0141) exp{1360/[1.8(79) + 127]}

= 15.2 mPa s Ans.

12-21 Originallyd = 2.000+0.0000.001 in, b = 2.005+0.0030.000 in

Doubled,d = 4.000+0.0000.002 in, b = 4.010+0.0060.000

The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were carriedout. Some of the results are:

TrumplerPart c S Tf f r/c Qs ho/c Hloss 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.Hloss is related by a 9898/1237 or an 8-fold increase. The existing ho is related by a 2-foldincrease. Trumplers (ho)min is related by a 1.286-fold increase

fom = 82.37 for double size fom = 10.297 for original size} an 8-fold increase for double-size

12-22 From Table 12-8: K = 0.6(1010) in3 min/(lbf ft h). P = 500/[(1)(1)] = 500 psi,V = DN/12 = (1)(200)/12 = 52.4 ft/minTables 12-10 and 12-11: f1 = 1.8, f2 = 1Table 12-12: PVmax = 46 700 psi ft/min, Pmax = 3560 psi, Vmax = 100 ft/min

Pmax = 4

FDL

= 4(500)(1)(1) = 637 psi < 3560 psi O.K.

P = FDL

= 500 psi V = 52.4 ft/min

PV = 500(52.4) = 26 200 psi ft/min < 46 700 psi ft/min O.K.



FIRST PAGES

Chapter 12 323

Solving Eq. (12-32) for t

t = DLw4 f1 f2 K V F =

(1)(1)(0.005)4(1.8)(1)(0.6)(1010)(52.4)(500) = 1388 h = 83 270 min

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

12-23 Estimate bushing length with f1 = f2 = 1, and K = 0.6(1010) in3 min/(lbf ft h)

Eq. (12-32): L = 1(1)(0.6)(1010)(2)(100)(400)(1000)

3(0.002) = 0.80 in

From Eq. (12-38), with fs = 0.03 from Table 12-9 applying nd = 2 to Fand hCR = 2.7 Btu/(h ft2 F)

L .= 720(0.03)(2)(100)(400)778(2.7)(300 70) = 3.58 in

0.80 L 3.58 inTrial 1: Let L = 1 in, D = 1 in

Pmax = 4(2)(100)(1)(1) = 255 psi < 3560 psi O.K.

P = 2(100)1(1) = 200 psi

V = (1)(400)12

= 104.7 ft/min > 100 ft/min Not O.K.

Trial 2: Try D = 7/8 in, L = 1 in

Pmax = 4(2)(100)(7/8)(1) = 291 psi < 3560 psi O.K.

P = 2(100)7/8(1) = 229 psi

V = (7/8)(400)12

= 91.6 ft/min < 100 ft/min O.K.

PV = 229(91.6) = 20 976 psi ft/min < 46 700 psi ft/min O.K.

f1 = 1.3 + (1.8 1.3)(

91.6 33100 33

)= 1.74

L = 0.80(1.74) = 1.39 in

V f133 1.391.6 f1

100 1.8



FIRST PAGES


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

Pmax = 4(2)(100)(7/8)(1.5) = 194 psi < 3560 psi O.K.

P = 2(100)7/8(1.5) = 152 psi, V = 91.6 ft/min

PV = 152(91.6) = 13 923 psi ft/min < 46 700 psi ft/min O.K.D = 7/8 in, L = 1.5 in is acceptable Ans.

Suggestion: Try smaller sizes.



FIRST PAGES

Chapter 14

14-1

d = NP

= 226

= 3.667 inTable 14-2: Y = 0.331

V = dn12

= (3.667)(1200)12

= 1152 ft/min

Eq. (14-4b): Kv = 1200 + 11521200 = 1.96

W t = Td/2

= 63 025Hnd/2

= 63 025(15)1200(3.667/2) = 429.7 lbf

Eq. (14-7):

= KvWt P

FY= 1.96(429.7)(6)

2(0.331) = 7633 psi = 7.63 kpsi Ans.

14-2d = 16

12= 1.333 in, Y = 0.296

V = (1.333)(700)12

= 244.3 ft/min

Eq. (14-4b): Kv = 1200 + 244.31200 = 1.204

W t = 63 025Hnd/2

= 63 025(1.5)700(1.333/2) = 202.6 lbf

Eq. (14-7): = KvW

t PFY

= 1.204(202.6)(12)0.75(0.296) = 13 185 psi = 13.2 kpsi Ans.

14-3d = mN = 1.25(18) = 22.5 mm, Y = 0.309

V = (22.5)(103)(1800)

60= 2.121 m/s

Eq. (14-6b): Kv = 6.1 + 2.1216.1 = 1.348

W t = 60Hdn

= 60(0.5)(103)

(22.5)(103)(1800) = 235.8 N

Eq. (14-8): = KvWt

FmY= 1.348(235.8)

12(1.25)(0.309) = 68.6 MPa Ans.


Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali

FIRST PAGES


14-4d = 5(15) = 75 mm, Y = 0.290

V = (75)(103)(200)

60= 0.7854 m/s

Assume steel and apply Eq. (14-6b):

Kv = 6.1 + 0.78546.1 = 1.129

W t = 60Hdn

= 60(5)(103)

(75)(103)(200) = 6366 N

Eq. (14-8): = KvWt

FmY= 1.129(6366)

60(5)(0.290) = 82.6 MPa Ans.

14-5d = 1(16) = 16 mm, Y = 0.296

V = (16)(103)(400)

60= 0.335 m/s

Assume steel and apply Eq. (14-6b):

Kv = 6.1 + 0.3356.1 = 1.055

W t = 60Hdn

= 60(0.15)(103)

(16)(103)(400) = 447.6 N

Eq. (14-8): F = KvWt

mY= 1.055(447.6)

150(1)(0.296) = 10.6 mm

From Table A-17, use F = 11 mm Ans.

14-6d = 1.5(17) = 25.5 mm, Y = 0.303

V = (25.5)(103)(400)

60= 0.534 m/s

Eq. (14-6b): Kv = 6.1 + 0.5346.1 = 1.088

W t = 60Hdn

= 60(0.25)(103)

(25.5)(103)(400) = 468 N

Eq. (14-8): F = KvWt

mY= 1.088(468)

75(1.5)(0.303) = 14.9 mm

Use F = 15 mm Ans.



FIRST PAGES

Chapter 14 351

14-7

d = 245

= 4.8 in, Y = 0.337

V = (4.8)(50)12

= 62.83 ft/min

Eq. (14-4b): Kv = 1200 + 62.831200 = 1.052

W t = 63 025Hnd/2

= 63 025(6)50(4.8/2) = 3151 lbf

Eq. (14-7): F = KvWt P

Y= 1.052(3151)(5)

20(103)(0.337) = 2.46 in

Use F = 2.5 in Ans.

14-8

d = 165

= 3.2 in, Y = 0.296

V = (3.2)(600)12

= 502.7 ft/min

Eq. (14-4b): Kv = 1200 + 502.71200 = 1.419

W t = 63 025(15)600(3.2/2) = 984.8 lbf

Eq. (14-7): F = KvWt P

Y= 1.419(984.8)(5)

10(103)(0.296) = 2.38 in

Use F = 2.5 in Ans.

14-9 Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309.

V = (2.25)(600)12

= 353.4 ft/min

Eq. (14-4b): Kv = 1200 + 353.41200 = 1.295

W t = 63 025(2.5)600(2.25/2) = 233.4 lbf

Eq. (14-7): F = KvWt P

Y= 1.295(233.4)(8)

10(103)(0.309) = 0.783 in



FIRST PAGES

Using coarse integer pitches from Table 13-2, the following table is formed.

P d V Kv W t F

2 9.000 1413.717 2.178 58.356 0.0823 6.000 942.478 1.785 87.535 0.1524 4.500 706.858 1.589 116.713 0.2406 3.000 471.239 1.393 175.069 0.4738 2.250 353.429 1.295 233.426 0.782

10 1.800 282.743 1.236 291.782 1.16712 1.500 235.619 1.196 350.139 1.62716 1.125 176.715 1.147 466.852 2.773

Other considerations may dictate the selection. Good candidates are P = 8 (F = 7/8 in)and P = 10 (F = 1.25 in). Ans.

14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309.

V = (36)(103)(900)

60= 1.696 m/s

Eq. (14-6b): Kv = 6.1 + 1.6966.1 = 1.278

W t = 60(1.5)(103)

(36)(103)(900) = 884 N

Eq. (14-8): F = 1.278(884)75(2)(0.309) = 24.4 mm

Using the preferred module sizes from Table 13-2:

m d V Kv W t F

1.00 18.0 0.848 1.139 1768.388 86.9171.25 22.5 1.060 1.174 1414.711 57.3241.50 27.0 1.272 1.209 1178.926 40.9872.00 36.0 1.696 1.278 884.194 24.3823.00 54.0 2.545 1.417 589.463 12.0154.00 72.0 3.393 1.556 442.097 7.4225.00 90.0 4.241 1.695 353.678 5.1746.00 108.0 5.089 1.834 294.731 3.8888.00 144.0 6.786 2.112 221.049 2.519

10.00 180.0 8.482 2.391 176.839 1.82412.00 216.0 10.179 2.669 147.366 1.41416.00 288.0 13.572 3.225 110.524 0.96120.00 360.0 16.965 3.781 88.419 0.72125.00 450.0 21.206 4.476 70.736 0.54732.00 576.0 27.143 5.450 55.262 0.40640.00 720.0 33.929 6.562 44.210 0.31350.00 900.0 42.412 7.953 35.368 0.243

Other design considerations may dictate the size selection. For the present design,m = 2 mm (F = 25 mm) is a good selection. Ans.




FIRST PAGES

Chapter 14 353

14-11

dP = 226 = 3.667 in, dG =606

= 10 in

V = (3.667)(1200)12

= 1152 ft/min

Eq. (14-4b): Kv = 1200 + 11521200 = 1.96

W t = 63 025(15)1200(3.667/2) = 429.7 lbf

Table 14-8: Cp = 2100

psi [Note: using Eq. (14-13) can result in wide variation inCp due to wide variation in cast iron properties]

Eq. (14-12): r1 = 3.667 sin 202 = 0.627 in, r2 =10 sin 20

2= 1.710 in

Eq. (14-14): C = Cp[

KvW t

F cos

(1r1

+ 1r2

)]1/2

= 2100[

1.96(429.7)2 cos 20

(1

0.627+ 1

1.710

)]1/2= 65.6(103) psi = 65.6 kpsi Ans.

14-12dP = 1612 = 1.333 in, dG =

4812

= 4 in

V = (1.333)(700)12

= 244.3 ft/min

Eq. (14-4b): Kv = 1200 + 244.31200 = 1.204

W t = 63 025(1.5)700(1.333/2) = 202.6 lbf

Table 14-8: Cp = 2100

psi (see note in Prob. 14-11 solution)

Eq. (14-12): r1 = 1.333 sin 202 = 0.228 in, r2 =4 sin 20

2= 0.684 in

Eq. (14-14):

C = 2100[

1.202(202.6)F cos 20

(1

0.228+ 1

0.684

)]1/2= 100(103)

F =(

2100100(103)

)2 [1.202(202.6)cos 20

](1

0.228+ 1

0.684

)= 0.668 in

Use F = 0.75 in Ans.



FIRST PAGES

14-13dP = 245 = 4.8 in, dG =

485

= 9.6 in

Eq. (14-4a):

V = (4.8)(50)12

= 62.83 ft/min

Kv = 600 + 62.83600 = 1.105

W t = 63 025H50(4.8/2) = 525.2H

Table 14-8: Cp = 1960

psi (see note in Prob. 14-11 solution)

Eq. (14-12): r1 = 4.8 sin 20

2= 0.821 in, r2 = 2r1 = 1.642 in

Eq. (14-14): 100(103) = 1960[

1.105(525.2H )2.5 cos 20

(1

0.821+ 1

1.642

)]1/2

H = 5.77 hp Ans.

14-14dP = 4(20) = 80 mm, dG = 4(32) = 128 mm

V = (80)(103)(1000)

60= 4.189 m/s

Kv = 3.05 + 4.1893.05 = 2.373

W t = 60(10)(103)

(80)(103)(1000) = 2387 N

Cp = 163

MPa (see note in Prob. 14-11 solution)

r1 = 80 sin 202 = 13.68 mm, r2 =128 sin 20

2= 21.89 mm

C = 163[

2.373(2387)50 cos 20

(1

13.68+ 1

21.89

)]1/2= 617 MPa Ans.

14-15 The pinion controls the design.Bending YP = 0.303, YG = 0.359

dP = 1712 = 1.417 in, dG =3012

= 2.500 in

V = dPn12

= (1.417)(525)12

= 194.8 ft/min

Eq. (14-4b): Kv = 1200 + 194.81200 = 1.162Eq. (6-8): Se = 0.5(76) = 38 kpsiEq. (6-19): ka = 2.70(76)0.265 = 0.857

Eq. (14-6a):

Table 14-8:

Eq. (14-12):

Eq. (14-14):




FIRST PAGES

Chapter 14 355

l = 2.25Pd

= 2.2512

= 0.1875 in

x = 3YP2P

= 3(0.303)2(12) = 0.0379 in

t =

4(0.1875)(0.0379) = 0.1686 inde = 0.808

0.875(0.1686) = 0.310 in

kb =(

0.3100.30

)0.107= 0.996

kc = kd = ke = 1, k f1 = 1.66 (see Ex. 14-2)r f = 0.30012 = 0.025 in (see Ex. 14-2)r

d= r f

t= 0.025

0.1686= 0.148

Approximate D/d = with D/d = 3; from Fig. A-15-6, Kt = 1.68.From Fig. 6-20, with Sut = 76 kpsi and r = 0.025 in, q = 0.62. From Eq. (6-32)

Kf = 1 + 0.62(1.68 1) = 1.42Miscellaneous-Effects Factor:

k f = k f 1k f 2 = 1.65(

11.323

)= 1.247

Se = 0.857(0.996)(1)(1)(1)(1.247)(38 000)= 40 450 psi

all = 40 7702.25 = 18 120 psi

W t = FYPallKv Pd

= 0.875(0.303)(18 120)1.162(12)

= 345 lbfH = 345(194.8)

33 000= 2.04 hp Ans.

Wear1 = 2 = 0.292, E1 = E2 = 30(106) psi

Eq. (14-13): Cp =

1

2(

1 0.292230(106)

) = 2285

psi

r1 = dP2 sin =1.417

2sin 20 = 0.242 in

r2 = dG2 sin =2.500

2sin 20 = 0.428

1r1

+ 1r2

= 10.242

+ 10.428

= 6.469 in1

Eq. (14-12):

Eq. (7-17):

Eq. (14-3):

Eq. (b), p. 717:Eq. (6-25):

Eq. (6-20):



FIRST PAGES

From Eq. (6-68),(SC )108 = 0.4HB 10 kpsi

= [0.4(149) 10](103) = 49 600 psi

Eq. (14-14):

C,all = (SC )108

n= 49 600

2.25= 33 067 psi

W t =(33 067

2285

)2 [0.875 cos 201.162(6.469)

]= 22.6 lbf

H = 22.6(194.8)33 000

= 0.133 hp Ans.Rating power (pinion controls):

H1 = 2.04 hpH2 = 0.133 hp

Hall = (min 2.04, 0.133) = 0.133 hp Ans.

See Prob. 14-15 solution for equation numbers.Pinion controls: YP = 0.322, YG = 0.447Bending dP = 20/3 = 6.667 in, dG = 33.333 in

V = dPn/12 = (6.667)(870)/12 = 1519 ft/minKv = (1200 + 1519)/1200 = 2.266Se = 0.5(113) = 56.5 kpsika = 2.70(113)0.265 = 0.771

l = 2.25/Pd = 2.25/3 = 0.75 inx = 3(0.322)/[2(3)] = 0.161 int =

4(0.75)(0.161) = 0.695 in

de = 0.808

2.5(0.695) = 1.065 inkb = (1.065/0.30)0.107 = 0.873kc = kd = ke = 1r f = 0.300/3 = 0.100 inr

d= r f

t= 0.100

0.695= 0.144

From Table A-15-6, Kt = 1.75; Fig. 6-20, q = 0.85; Eq. (6-32), K f = 1.64k f 2 = 1/1.597, k f = k f 1k f 2 = 1.66/1.597 = 1.039Se = 0.771(0.873)(1)(1)(1)(1.039)(56 500) = 39 500 psi

all = Se/n = 39 500/1.5 = 26 330 psi

W t = FYPallKv Pd

= 2.5(0.322)(26 330)2.266(3) = 3118 lbf

H = W t V/33 000 = 3118(1519)/33 000 = 144 hp Ans.


14-16



FIRST PAGES

Chapter 14 357

Wear

Cp = 2285

psir1 = (6.667/2) sin 20 = 1.140 inr2 = (33.333/2) sin 20 = 5.700 in

SC = [0.4(262) 10](103) = 94 800 psiC,all = SC/nd = 94 800/

1.5 = 77 404 psi

W t =(

C,allCp

)2 F cos Kv

11/r1 + 1/r2

=(77 404

2300

)2 (2.5 cos 202.266

)(1

1/1.140 + 1/5.700)

= 1115 lbf

H = Wt V

33 000= 1115(1519)

33 000= 51.3 hp Ans.

For 108 cycles (revolutions of the pinion), the power based on wear is 51.3 hp.Rating powerpinion controls

H1 = 144 hpH2 = 51.3 hp

Hrated = min(144, 51.3) = 51.3 hp Ans.

14-17 Given: = 20, n = 1145 rev/min, m = 6 mm, F = 75 mm, NP = 16 milled teeth,NG = 30T, Sut = 900 MPa, HB = 260, nd = 3, YP = 0.296, and YG = 0.359.Pinion bending

dP = mNP = 6(16) = 96 mmdG = 6(30) = 180 mmV = dPn

12= (96)(1145)(10

3)(12)(12)(60) = 5.76 m/s

Eq. (14-6b): Kv = 6.1 + 5.766.1 = 1.944

Se = 0.5(900) = 450 MPaa = 4.45, b = 0.265

ka = 4.51(900)0.265 = 0.744l = 2.25m = 2.25(6) = 13.5 mmx = 3Y m/2 = 3(0.296)6/2 = 2.664 mmt =

4lx =

4(13.5)(2.664) = 12.0 mm

de = 0.808

75(12.0) = 24.23 mm

Eq. (14-13):Eq. (14-12):

Eq. (6-68):



FIRST PAGES

kb =(

24.237.62

)0.107= 0.884

kc = kd = ke = 1r f = 0.300m = 0.300(6) = 1.8 mm

From Fig. A-15-6 for r/d = r f /t = 1.8/12 = 0.15, Kt = 1.68.Figure 6-20, q = 0.86; Eq. (6-32),

K f = 1 + 0.86(1.68 1) = 1.58k f 1 = 1.66 (Gerber failure criterion)k f 2 = 1/K f = 1/1.537 = 0.651k f = k f 1k f 2 = 1.66(0.651) = 1.08Se = 0.744(0.884)(1)(1)(1)(1.08)(450) = 319.6 MPa

all = Send

= 319.61.3

= 245.8 MPa

Eq. (14-8): W t = FY mallKv

= 75(0.296)(6)(245.8)1.944

= 16 840 N

H = T n9.55

= 16 840(96/2)(1145)9.55(106) = 96.9 kW Ans.

Wear : Pinion and gearEq. (14-12): r1 = (96/2) sin 20 = 16.42 mm

r2 = (180/2) sin 20 = 30.78 mm

Eq. (14-13), with E = 207(103) MPa and = 0.292, givesCp =

[1

2(1 0.2922)/(207 103)]

= 190

MPa

Eq. (6-68): SC = 6.89[0.4(260) 10] = 647.7 MPa

C,all = SCn

= 647.71.3

= 568 MPa

Eq. (14-14): W t =(

C,allCp

)2 F cos Kv

11/r1 + 1/r2

=(568

191

)2(75 cos 201.944

)(1

1/16.42 + 1/30.78)

= 3433 N

T = WtdP2

= 3433(96)2

= 164 784 N mm = 164.8 N m

H = T n9.55

= 164.8(1145)9.55

= 19 758.7 W = 19.8 kW Ans.Thus, wear controls the gearset power rating; H = 19.8 kW. Ans.




FIRST PAGES

Chapter 14 359

14-18 Preliminaries: NP = 17, NG = 51

dP = NPd =176

= 2.833 in

dG = 516 = 8.500 inV = dPn/12 = (2.833)(1120)/12 = 830.7 ft/min

Eq. (14-4b): Kv = (1200 + 830.7)/1200 = 1.692

all = Synd

= 90 0002

= 45 000 psi

Table 14-2: YP = 0.303, YG = 0.410

Eq. (14-7): W t = FYPallKv Pd

= 2(0.303)(45 000)1.692(6) = 2686 lbf

H = WtV

33 000= 2686(830.7)

33 000= 67.6 hp

Based on yielding in bending, the power is 67.6 hp. (a) Pinion fatigue

Bending

Eq. (2-17): Sut .= 0.5HB = 0.5(232) = 116 kpsiEq. (6-8): Se = 0.5Sut = 0.5(116) = 58 kpsiEq. (6-19): a = 2.70, b = 0.265, ka = 2.70(116)0.265 = 0.766

Table 13-1: l = 1Pd

+ 1.25Pd

= 2.25Pd

= 2.256

= 0.375 in

Eq. (14-3): x = 3YP2Pd

= 3(0.303)2(6) = 0.0758

Eq. (b), p. 717: t =

4lx =

4(0.375)(0.0758) = 0.337 in

Eq. (6-25): de = 0.808

Ft = 0.808

2(0.337) = 0.663 in

Eq. (6-20): kb =(

0.6630.30

)0.107= 0.919

kc = kd = ke = 1. Assess two components contributing to k f . First, based uponone-way bending and the Gerber failure criterion, k f 1 = 1.66 (see Ex. 14-2). Second,due to stress-concentration,

r f = 0.300Pd =0.300

6= 0.050 in (see Ex. 14-2)

Fig. A-15-6:r

d= r f

t= 0.05

0.338= 0.148



FIRST PAGES

Estimate D/d = by setting D/d = 3, Kt = 1.68. From Fig. 6-20, q = 0.86, andEq. (6-32)

K f = 1 + 0.86(1.68 1) = 1.58

k f 2 = 1K f =1

1.58= 0.633

k f = k f 1k f 2 = 1.66(0.633) = 1.051Se = 0.766(0.919)(1)(1)(1)(1.051)(58) = 42.9 kpsi

all = Send

= 42.92

= 21.5 kpsi

Wt = FYPallKv Pd

= 2(0.303)(21 500)1.692(6) = 1283 lbf

H = Wt V

33 000= 1283(830.7)

33 000= 32.3 hp Ans.

(b) Pinion fatigueWearFrom Table A-5 for steel: = 0.292, E = 30(106) psiEq. (14-13) or Table 14-8:

Cp ={

12[(1 0.2922)/30(106)]

}1/2= 2285

psi

In preparation for Eq. (14-14):

Eq. (14-12): r1 = dP2 sin =2.833

2sin 20 = 0.485 in

r2 = dG2 sin =8.500

2sin 20 = 1.454 in(

1r1

+ 1r2

)= 1

0.485+ 1

1.454= 2.750 in

Eq. (6-68): (SC )108 = 0.4HB 10 kpsiIn terms of gear notation

C = [0.4(232) 10]103 = 82 800 psiWe will introduce the design factor of nd = 2 and because it is a contact stress apply itto the load W t by dividing by

2.

C,all = c2

= 82 8002

= 58 548 psi




FIRST PAGES

Chapter 14 361

Solve Eq. (14-14) for Wt:

Wt =(58 548

2285

)2 [ 2 cos 201.692(2.750)

]= 265 lbf

Hall = 265(830.7)33 000 = 6.67 hp Ans.

For 108 cycles (turns of pinion), the allowable power is 6.67 hp.(c) Gear fatigue due to bending and wear

Bending

Eq. (14-3): x = 3YG2Pd

= 3(0.4103)2(6) = 0.1026 in

Eq. (b), p. 717: t =

4(0.375)(0.1026) = 0.392 inEq. (6-25): de = 0.808

2(0.392) = 0.715 in

Eq. (6-20): kb =(

0.7150.30

)0.107= 0.911

kc = kd = ke = 1r

d= r f

t= 0.050

0.392= 0.128

Approximate D/d = by setting D/d = 3 for Fig. A-15-6; Kt = 1.80. Use K f =1.80.

k f 2 = 11.80 = 0.556, k f = 1.66(0.556) = 0.923

Se = 0.766(0.911)(1)(1)(1)(0.923)(58) = 37.36 kpsi

all = Send

= 37.362

= 18.68 kpsi

Wt = FYGallKv Pd =

2(0.4103)(18 680)1.692(6) = 1510 lbf

Hall = 1510(830.7)33 000 = 38.0 hp Ans.

The gear is thus stronger than the pinion in bending.Wear Since the material of the pinion and the gear are the same, and the contactstresses are the same, the allowable power transmission of both is the same. Thus,Hall = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish SC for108/3 revolutions.



FIRST PAGES

(d) Pinion bending: H1 = 32.3 hpPinion wear: H2 = 6.67 hp

Gear bending: H3 = 38.0 hpGear wear: H4 = 6.67 hp

Power rating of the gear set is thusHrated = min(32.3, 6.67, 38.0, 6.67) = 6.67 hp Ans.

14-19 dP = 16/6 = 2.667 in, dG = 48/6 = 8 in

V = (2.667)(300)12

= 209.4 ft/min

W t = 33 000(5)209.4

= 787.8 lbf

Assuming uniform loading, Ko = 1. From Eq. (14-28),Qv = 6, B = 0.25(12 6)2/3 = 0.8255

A = 50 + 56(1 0.8255) = 59.77Eq. (14-27):

Kv =(

59.77 + 209.459.77

)0.8255= 1.196

From Table 14-2,NP = 16T , YP = 0.296NG = 48T , YG = 0.4056

From Eq. (a), Sec. 14-10 with F = 2 in

(Ks)P = 1.192(

2

0.2966

)0.0535= 1.088

(Ks)G = 1.192(

2

0.40566

)0.0535= 1.097

From Eq. (14-30) with Cmc = 1

Cp f = 210(2.667) 0.0375 + 0.0125(2) = 0.0625Cpm = 1, Cma = 0.093 (Fig. 14-11), Ce = 1Km = 1 + 1[0.0625(1) + 0.093(1)] = 1.156

Assuming constant thickness of the gears K B = 1mG = NG/NP = 48/16 = 3

With N (pinion) = 108 cycles and N (gear) = 108/3, Fig. 14-14 provides the relations:(YN )P = 1.3558(108)0.0178 = 0.977(YN )G = 1.3558(108/3)0.0178 = 0.996




FIRST PAGES

Chapter 14 363

Fig. 14-6: JP = 0.27, JG = 0.38From Table 14-10 for R = 0.9, K R = 0.85

KT = Cf = 1

Eq. (14-23) with mN = 1 I = cos 20 sin 20

2

(3

3 + 1)

= 0.1205

Table 14-8: Cp = 2300

psi

Strength: Grade 1 steel with HB P = HBG = 200Fig. 14-2: (St )P = (St )G = 77.3(200) + 12 800 = 28 260 psiFig. 14-5: (Sc)P = (Sc)G = 322(200) + 29 100 = 93 500 psiFig. 14-15: (Z N )P = 1.4488(108)0.023 = 0.948

(Z N )G = 1.4488(108/3)0.023 = 0.973Fig. 14-12: HB P/HBG = 1 CH = 1Pinion tooth bendingEq. (14-15):

( )P = W t KoKv Ks PdFKm K B

J= 787.8(1)(1.196)(1.088)

(62

)[(1.156)(1)0.27

]= 13 167 psi Ans.

Factor of safety from Eq. (14-41)

(SF )P =[

StYN/(KT K R)

]= 28 260(0.977)/[(1)(0.85)]

13 167= 2.47 Ans.

Gear tooth bending

( )G = 787.8(1)(1.196)(1.097)(

62

)[(1.156)(1)0.38

]= 9433 psi Ans.

(SF )G = 28 260(0.996)/[(1)(0.85)]9433 = 3.51 Ans.Pinion tooth wear

Eq. (14-16): (c)P = Cp(

W t KoKv KsKm

dP FCfI

)1/2P

= 2300[

787.8(1)(1.196)(1.088)(

1.1562.667(2)

)(1

0.1205

)]1/2= 98 760 psi Ans.

Eq. (14-42):

(SH )P =[

Sc Z N/(KT K R)c

]P

={

93 500(0.948)/[(1)(0.85)]98 760

}= 1.06 Ans.



FIRST PAGES

Gear tooth wear

(c)G =[(Ks)G

(Ks)P

]1/2(c)P =

(1.0971.088

)1/2(98 760) = 99 170 psi Ans.

(SH )G = 93 500(0.973)(1)/[(1)(0.85)]99 170 = 1.08 Ans.

The hardness of the pinion and the gear should be increased.

14-20 dP = 2.5(20) = 50 mm, dG = 2.5(36) = 90 mm

V = dPnP60

= (50)(103)(100)

60= 0.2618 m/s

W t = 60(120)(50)(103)(100) = 458.4 N

Eq. (14-28): Ko = 1, Qv = 6, B = 0.25(12 6)2/3 = 0.8255A = 50 + 56(1 0.8255) = 59.77

Eq. (14-27): Kv =[59.77 + 200(0.2618)

59.77

]0.8255= 1.099

Table 14-2: YP = 0.322, YG = 0.3775Similar to Eq. (a) of Sec. 14-10 but for SI units:

Ks = 1kb = 0.8433(m F

Y

)0.0535(Ks)P = 0.8433

[2.5(18)

0.322

]0.0535 = 1.003 use 1(Ks)G = 0.8433

[2.5(18)

0.3775

]0.0535> 1 use 1

Cmc = 1, F = 18/25.4 = 0.709 in, Cp f = 1810(50) 0.025 = 0.011Cpm = 1, Cma = 0.247 + 0.0167(0.709) 0.765(104)(0.7092) = 0.259

Ce = 1K H = 1 + 1[0.011(1) + 0.259(1)] = 1.27

Eq. (14-40): K B = 1, mG = NG/NP = 36/20 = 1.8Fig. 14-14: (YN )P = 1.3558(108)0.0178 = 0.977

(YN )G = 1.3558(108/1.8)0.0178 = 0.987Fig. 14-6: (YJ )P = 0.33, (YJ )G = 0.38Eq. (14-38): YZ = 0.658 0.0759 ln(1 0.95) = 0.885

Y = Z R = 1Sec. 14-15:




FIRST PAGES

Chapter 14 365

Eq. (14-16):

Eq. (14-42):

Eq. (14-23) with mN = 1:Z I = cos 20

sin 20

2

(1.8

1.8 + 1)

= 0.103

Table 14-8: Z E = 191

MPa

Strength Grade 1 steel, given HB P = HBG = 200Fig. 14-2: (F P )P = (F P )G = 0.533(200) + 88.3 = 194.9 MPaFig. 14-5: (H P )P = (H P )G = 2.22(200) + 200 = 644 MPaFig. 14-15: (Z N )P = 1.4488(108)0.023 = 0.948

(Z N )G = 1.4488(108/1.8)0.023 = 0.961Fig. 14-12: HB P/HBG = 1 ZW = 1Pinion tooth bending

( )P =(

W t KoKv Ks1

bmtK H K B

YJ

)P

= 458.4(1)(1.099)(1)[

118(2.5)

][1.27(1)

0.33

]= 43.08 MPa Ans.

(SF )P =(

F P

YNYYZ

)P

= 194.943.08

[0.977

1(0.885)]

= 4.99 Ans.

Gear tooth bending

( )G = 458.4(1)(1.099)(1)[

118(2.5)

][1.27(1)

0.38

]= 37.42 MPa Ans.

(SF )G = 194.937.42[

0.9871(0.885)

]= 5.81 Ans.

Pinion tooth wear

(c)P =(

Z E

W t KoKv Ks

K Hdw1b

Z RZ I

)P

= 191

458.4(1)(1.099)(1)[

1.2750(18)

][1

0.103

]= 501.8 MPa Ans.

(SH )P =(

H P

c

Z N ZWYYZ

)P

= 644501.8

[0.948(1)1(0.885)

]= 1.37 Ans.

Gear tooth wear

(c)G =[(Ks)G

(Ks)P

]1/2(c)P =

(11

)1/2(501.8) = 501.8 MPa Ans.

(SH )G = 644501.8[

0.961(1)1(0.885)

]= 1.39 Ans.

Eq. (14-15):

Eq. (14-41):



FIRST PAGES

14-21Pt = Pn cos = 6 cos 30 = 5.196 teeth/in

dP = 165.196 = 3.079 in, dG =4816

(3.079) = 9.238 in

V = (3.079)(300)12

= 241.8 ft/min

W t = 33 000(5)241.8

= 682.3 lbf, Kv =(

59.77 + 241.859.77

)0.8255= 1.210

From Prob. 14-19:YP = 0.296, YG = 0.4056

(Ks)P = 1.088, (Ks)G = 1.097, K B = 1mG = 3, (YN )P = 0.977, (YN )G = 0.996, K R = 0.85

(St )P = (St )G = 28 260 psi, CH = 1, (Sc)P = (Sc)G = 93 500 psi(Z N )P = 0.948, (Z N )G = 0.973, Cp = 2300

psi

The pressure angle is:

Eq. (13-19): t = tan1(

tan 20cos 30

)= 22.80

(rb)P = 3.0792 cos 22.8 = 1.419 in, (rb)G = 3(rb)P = 4.258 ina = 1/Pn = 1/6 = 0.167 in

Eq. (14-25):

Z =[(

3.0792

+ 0.167)2

1.4192]1/2

+[(

9.2382

+ 0.167)2

4.2582]1/2

(

3.0792

+ 9.2382

)sin 22.8

= 0.9479 + 2.1852 2.3865 = 0.7466 Conditions O.K. for usepN = pn cos n = 6 cos 20 = 0.4920 in

Eq. (14-21): mN = pN0.95Z =0.492

0.95(0.7466) = 0.6937

Eq. (14-23): I =[

sin 22.8 cos 22.82(0.6937)

](3

3 + 1)

= 0.193

Fig. 14-7: J P = 0.45, J G = 0.54




FIRST PAGES

Chapter 14 367

Fig. 14-8: Corrections are 0.94 and 0.98JP = 0.45(0.94) = 0.423, JG = 0.54(0.98) = 0.529

Cmc = 1, Cp f = 210(3.079) 0.0375 + 0.0125(2) = 0.0525Cpm = 1, Cma = 0.093, Ce = 1Km = 1 + (1)[0.0525(1) + 0.093(1)] = 1.146

Pinion tooth bending

( )P = 682.3(1)(1.21)(1.088)(5.196

2

)[1.146(1)

0.423

]= 6323 psi Ans.

(SF )P = 28 260(0.977)/[1(0.85)]6323 = 5.14 Ans.

Gear tooth bending

( )G = 682.3(1)(1.21)(1.097)(5.196

2

)[1.146(1)

0.529

]= 5097 psi Ans.

(SF )G = 28 260(0.996)/[1(0.85)]5097 = 6.50 Ans.

Pinion tooth wear

(c)P = 2300{

682.3(1)(1.21)(1.088)[

1.1463.078(2)

](1

0.193

)}1/2= 67 700 psi Ans.

(SH )P = 93 500(0.948)/[(1)(0.85)]67 700 = 1.54 Ans.

Gear tooth wear

(c)G =[

1.0971.088

]1/2(67 700) = 67 980 psi Ans.

(SH )G = 93 500(0.973)/[(1)(0.85)]67 980 = 1.57 Ans.

14-22 Given: NP = 17T , NG = 51T , R = 0.99 at 108 cycles, HB = 232 through-hardeningGrade 1, core and case, both gears.Table 14-2: YP = 0.303, YG = 0.4103Fig. 14-6: JP = 0.292, JG = 0.396

dP = NP/P = 17/6 = 2.833 in, dG = 51/6 = 8.5 inPinion bendingFrom Fig. 14-2:

0.99(St )107 = 77.3HB + 12 800= 77.3(232) + 12 800 = 30 734 psi



FIRST PAGES

Fig. 14-14: YN = 1.6831(108)0.0323 = 0.928V = dPn/12 = (2.833)(1120/12) = 830.7 ft/min

KT = K R = 1, SF = 2, SH =

2

all = 30 734(0.928)2(1)(1) = 14 261 psi

Qv = 5, B = 0.25(12 5)2/3 = 0.9148A = 50 + 56(1 0.9148) = 54.77

Kv =(

54.77 + 830.754.77

)0.9148= 1.472

Ks = 1.192(

2

0.3036

)0.0535= 1.089 use 1

Km = Cm f = 1 + Cmc(Cp f Cpm + CmaCe)Cmc = 1Cp f = F10d 0.0375 + 0.0125F

= 210(2.833) 0.0375 + 0.0125(2)

= 0.0581Cpm = 1Cma = 0.127 + 0.0158(2) 0.093(104)(22) = 0.1586

Ce = 1Km = 1 + 1[0.0581(1) + 0.1586(1)] = 1.2167

Eq. (14-15):K = 1W t = F JPall

KoKv Ks Pd Km K B

= 2(0.292)(14 261)1(1.472)(1)(6)(1.2167)(1) = 775 lbf

H = Wt V

33 000= 775(830.7)

33 000= 19.5 hp

Pinion wear

Fig. 14-15: Z N = 2.466N0.056 = 2.466(108)0.056 = 0.879MG = 51/17 = 3

I = sin 20 cos 20

2

(3

3 + 1)

= 1.205, CH = 1




FIRST PAGES

Chapter 14 369

Fig. 14-5: 0.99(Sc)107 = 322HB + 29 100= 322(232) + 29 100 = 103 804 psi

c,all = 103 804(0.879)2(1)(1) = 64 519 psi

Eq. (14-16): W t =(

c,all

Cp

)2 FdP IKoKv Ks KmCf

=(

64 5192300

)2 [ 2(2.833)(0.1205)1(1.472)(1)(1.2167)(1)

]

= 300 lbf

H = Wt V

33 000= 300(830.7)

33 000= 7.55 hp

The pinion controls therefore Hrated = 7.55 hp Ans.

14-23l = 2.25/Pd , x = 3Y2Pd

t =

4lx =

4(

2.25Pd

)(3Y2Pd

)= 3.674

Pd

Y

de = 0.808

Ft = 0.808

F(

3.674Pd

)Y = 1.5487

F

Y

Pd

kb =1.5487

F

Y/Pd

0.30

0.107

= 0.8389(

F

YPd

)0.0535

Ks = 1kb = 1.192(

F

YPd

)0.0535Ans.

14-24 YP = 0.331, YG = 0.422, JP = 0.345, JG = 0.410, Ko = 1.25. The service conditionsare adequately described by Ko. Set SF = SH = 1.

dP = 22/4 = 5.500 indG = 60/4 = 15.000 in

V = (5.5)(1145)12

= 1649 ft/minPinion bending

0.99(St )107 = 77.3HB + 12 800 = 77.3(250) + 12 800 = 32 125 psiYN = 1.6831[3(109)]0.0323 = 0.832



FIRST PAGES

Eq. (14-17): (all)P = 32 125(0.832)1(1)(1) = 26 728 psi

B = 0.25(12 6)2/3 = 0.8255A = 50 + 56(1 0.8255) = 59.77

Kv =(

59.77 + 164959.77

)0.8255= 1.534

Ks = 1, Cm = 1

Cmc = F10d 0.0375 + 0.0125F

= 3.2510(5.5) 0.0375 + 0.0125(3.25) = 0.0622

Cma = 0.127 + 0.0158(3.25) 0.093(104)(3.252) = 0.178Ce = 1

Km = Cm f = 1 + (1)[0.0622(1) + 0.178(1)] = 1.240K B = 1, KT = 1

Eq. (14-15): W t1 =26 728(3.25)(0.345)

1.25(1.534)(1)(4)(1.240) = 3151 lbf

H1 = 3151(1649)33 000 = 157.5 hpGear bending By similar reasoning, W t2 = 3861 lbf and H2 = 192.9 hpPinion wear

mG = 60/22 = 2.727

I = cos 20 sin 20

2

(2.727

1 + 2.727)

= 0.1176

0.99(Sc)107 = 322(250) + 29 100 = 109 600 psi(Z N )P = 2.466[3(109)]0.056 = 0.727(Z N )G = 2.466[3(109)/2.727]0.056 = 0.769

(c,all)P = 109 600(0.727)1(1)(1) = 79 679 psi

W t3 =(

c,all

Cp

)2 FdP IKoKv Ks KmCf

=(

79 6792300

)2[ 3.25(5.5)(0.1176)1.25(1.534)(1)(1.24)(1)

]= 1061 lbf

H3 = 1061(1649)33 000 = 53.0 hp




FIRST PAGES

Chapter 14 371

Gear wearSimilarly, W t4 = 1182 lbf, H4 = 59.0 hpRating

Hrated = min(H1, H2, H3, H4)= min(157.5, 192.9, 53, 59) = 53 hp Ans.

Note differing capacities. Can these be equalized?

14-25 From Prob. 14-24:

W t1 = 3151 lbf, W t2 = 3861 lbf,W t3 = 1061 lbf, W t4 = 1182 lbf

W t = 33 000Ko HV

= 33 000(1.25)(40)1649

= 1000 lbf

Pinion bending: The factor of safety, based on load and stress, is

(SF )P = Wt1

1000= 3151

1000= 3.15

Gear bending based on load and stress

(SF )G = Wt2

1000= 3861

1000= 3.86

Pinion wear

based on load: n3 = Wt3

1000= 1061

1000= 1.06

based on stress: (SH )P =

1.06 = 1.03Gear wear

based on load: n4 = Wt4

1000= 1182

1000= 1.18

based on stress: (SH )G =

1.18 = 1.09Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06,1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors

(SF )P , (SF )G , (SH )P , (SH )Gare

3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2

and the threat is again from pinion wear. Depending on the magnitude of the numbers,using SF and SH as defined by AGMA, does not necessarily lead to the same conclusionconcerning threat. Therefore be cautious.



FIRST PAGES

14-26 Solution summary from Prob. 14-24: n = 1145 rev/min, Ko = 1.25, Grade 1 materials,NP = 22T , NG = 60T , mG = 2.727, YP = 0.331, YG = 0.422, JP = 0.345,JG = 0.410, Pd = 4T /in, F = 3.25 in, Qv = 6, (Nc)P = 3(109), R = 0.99Pinion HB : 250 core, 390 caseGear HB : 250 core, 390 caseKm = 1.240, KT = 1, K B = 1, dP = 5.500 in, dG = 15.000 in,V = 1649 ft/min, Kv = 1.534, (Ks)P = (Ks)G = 1, (YN )P = 0.832,(YN )G = 0.859, K R = 1Bending

(all)P = 26 728 psi (St )P = 32 125 psi(all)G = 27 546 psi (St )G = 32 125 psi

W t1 = 3151 lbf, H1 = 157.5 hpW t2 = 3861 lbf, H2 = 192.9 hp

Wear = 20, I = 0.1176, (Z N )P = 0.727,

(Z N )G = 0.769, CP = 2300

psi

(Sc)P = Sc = 322(390) + 29 100 = 154 680 psi

(c,all)P = 154 680(0.727)1(1)(1) = 112 450 psi

(c,all)G = 154 680(0.769)1(1)(1) = 118 950 psi

W t3 =(

112 45079 679

)2(1061) = 2113 lbf, H3 = 2113(1649)33 000 = 105.6 hp

W t4 =(

118 950109 600(0.769)

)2(1182) = 2354 lbf, H4 = 2354(1649)33 000 = 117.6 hp

Rated power

Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp Ans.Prob. 14-24

Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hpThe rated power approximately doubled.

14-27 The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell285 core and Brinell 580600 case. Table 14-3:

0.99(St )107 = 55 000 psi




FIRST PAGES

Chapter 14 373

Modification of St by (YN )P = 0.832 produces(all)P = 45 657 psi,

Similarly for (YN )G = 0.859(all)G = 47 161 psi, and

W t1 = 4569 lbf, H1 = 228 hpW t2

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Solucionario(Ingles) Diseño en Ingenieria Mecanica_shigley(8ed)(Capitulo 11 en Adelante)