Solucionario(Ingles) Diseño en Ingenieria Mecanica_shigley(8ed)(Capitulo 11 en Adelante)
-
Upload
juansantillan -
Category
Documents
-
view
47 -
download
1
Transcript of 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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 290
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 291
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
292 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 292
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 293
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
294 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 294
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 295
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
296 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 296
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 297
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
298 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 298
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 299
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
300 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 300
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 301
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
302 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 302
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch11.qxd 12/04/2006 15:25 Page 303
Philadelphia UniversityMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 304
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.
12-3cmin = bmin dmax2 =
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 305
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
306 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
12-4cmin = bmin dmax2 =
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 306
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
Chapter 12 307
12-5cmin = bmin dmax2 =
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.
12-6cmin = bmin dmax2 =
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 307
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
308 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
12-7cmin = bmin dmax2 =
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 308
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
Chapter 12 309
12-8cmin = bmin dmax2 =
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.
12-9cmin = bmin dmax2 =
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 309
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
310 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 310
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 311
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
312 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 312
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 313
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
314 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 314
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 315
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
316 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 316
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 317
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
318 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
From Figs. 12-18 and 12-16:f rc
= 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
From Figs. 12-18 and 12-16:f rc
= 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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 318
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 319
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
320 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 320
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 321
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
322 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 322
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 323
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
324 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
budynas_SM_ch12.qxd 12/04/2006 15:24 Page 324
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabaliPhiladelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 349
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
FIRST PAGES
350 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
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.
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 350
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 351
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
352 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 352
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 353
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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):
354 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 354
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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):
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 355
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
356 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
14-16
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 356
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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):
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 357
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
358 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 358
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 359
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
360 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 360
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 361
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
362 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 362
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 363
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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:
364 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 364
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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):
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 365
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
366 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 366
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 367
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
368 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 368
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 369
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
370 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 370
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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.
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 371
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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
372 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design
budynas_SM_ch14.qxd 12/05/2006 17:39 Page 372
Philadelphia University_jordanMechanical Engineering Design 8theng.ahmad jabali
-
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