Unformatted text preview: budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 304 Chapter 12
121 Given dmax = 1.000 in and bmin = 1.0015 in, the minimum radial clearance is
cmin = bmin − dmax
1.0015 − 1.000
=
= 0.000 75 in
2
2 l/d = 1
r = 1.000/2 = 0.500
˙
r/c = 0.500/0.000 75 = 667
N = 1100/60 = 18.33 rev/s
P = W/(ld) = 250/[(1)(1)] = 250 psi Also S = (6672 ) Eq. (127): 8(10−6 )(18.33)
= 0.261
250
h 0 /c = 0.595 Fig. 1216: Q/(rcNl) = 3.98
f r/c = 5.8
Q s /Q = 0.5
h 0 = 0.595(0.000 75) = 0.000 466 in Ans. Fig. 1219:
Fig. 1218:
Fig. 1220: f = 5.8
5.8
=
= 0.0087
r/c
667 The power loss in Btu/s is
H= 2π f W r N
2π(0.0087)(250)(0.5)(18.33)
=
778(12)
778(12) = 0.0134 Btu/s Ans.
Q = 3.98rcNl = 3.98(0.5)(0.000 75)(18.33)(1) = 0.0274 in3 /s
Q s = 0.5(0.0274) = 0.0137 in3 /s Ans.
122 bmin − dmax
1.252 − 1.250
=
= 0.001 in
2
2
.
r = 1.25/2 = 0.625 in cmin = r/c = 0.625/0.001 = 625
N = 1150/60 = 19.167 rev/s
P= 400
= 128 psi
1.25(2.5) l/d = 2.5/1.25 = 2
S= (6252 )(10)(10−6 )(19.167)
= 0.585
128 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 305 305 Chapter 12 The interpolation formula of Eq. (1216) will have to be used. From Figs. 1216, 1221,
and 1219
Q
= 3.09
l/d = ∞, h o /c = 0.96, P/ pmax = 0.84,
For
rcNl
Q
l/d = 1, h o /c = 0.77, P/ pmax = 0.52,
= 3.6
rcNl
1
l/d = ,
2 h o /c = 0.54, P/ pmax = 0.42, Q
= 4.4
rcNl 1
l/d = ,
4 h o /c = 0.31, P/ pmax = 0.28, Q
= 5.25
rcNl Equation (1216) is easily programmed by code or by using a spreadsheet. The results are:
l/d
h o /c
P/ pmax
Q/rcNl
∴ y∞ y1 y1/2 y1/4 yl/d 2
2
2 0.96
0.84
3.09 0.77
0.52
3.60 0.54
0.42
4.40 0.31
0.28
5.25 0.88
0.64
3.28 h o = 0.88(0.001) = 0.000 88 in Ans. pmax = 128
= 200 psi Ans.
0.64 Q = 3.28(0.625)(0.001)(19.167)(2.5) = 0.098 in3 /s Ans.
123
cmin = bmin − dmax
3.005 − 3.000
=
= 0.0025 in
2
2 .
r = 3.000/2 = 1.500 in
l/d = 1.5/3 = 0.5
r/c = 1.5/0.0025 = 600
N = 600/60 = 10 rev/s
P= 800
= 177.78 psi
1.5(3) Fig. 1212: SAE 10, µ = 1.75 µreyn
S = (6002 ) 1.75(10−6 )(10)
= 0.0354
177.78 Figs. 1216 and 1221: h o /c = 0.11, P/ pmax = 0.21 h o = 0.11(0.0025) = 0.000 275 in Ans.
pmax = 177.78/0.21 = 847 psi Ans. budynas_SM_ch12.qxd 306 12/04/2006 15:24 FIRST PAGES Page 306 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 1212: SAE 40, µ = 4.5 µreyn
S = 0.0354
h o /c = 0.19, 4.5
1.75 = 0.0910 P/ pmax = 0.275 h o = 0.19(0.0025) = 0.000 475 in Ans.
pmax = 177.78/0.275 = 646 psi Ans.
124
cmin = bmin − dmax
3.006 − 3.000
=
= 0.003
2
2 .
r = 3.000/2 = 1.5 in
l/d = 1
r/c = 1.5/0.003 = 500 N = 750/60 = 12.5 rev/s
P= 600
= 66.7 psi
3(3) Fig. 1214: SAE 10W, µ = 2.1 µreyn
2.1(10−6 )(12.5)
= 0.0984
S = (500 )
66.7
2 From Figs. 1216 and 1221:
h o /c = 0.34, P/ pmax = 0.395 h o = 0.34(0.003) = 0.001 020 in Ans.
pmax = 66.7
= 169 psi Ans.
0.395 Fig. 1214: SAE 20W40, µ = 5.05 µreyn
S = (5002 ) 5.05(10−6 )(12.5)
= 0.237
66.7 From Figs. 1216 and 1221:
h o /c = 0.57, P/ pmax = 0.47 h o = 0.57(0.003) = 0.001 71 in Ans.
pmax = 66.7
= 142 psi Ans.
0.47 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 307 Chapter 12 125 bmin − dmax
2.0024 − 2
=
= 0.0012 in
2
2
2
. d
r = = = 1 in, l/d = 1/2 = 0.50
2
2
r/c = 1/0.0012 = 833
N = 800/60 = 13.33 rev/s
600
= 300 psi
P=
2(1) cmin = Fig. 1212: SAE 20, µ = 3.75 µreyn
S = (8332 ) 3.75(10−6 )(13.3)
= 0.115
300 From Figs. 1216, 1218 and 1219:
h o /c = 0.23, r f /c = 3.8, Q/(rcNl) = 5.3
h o = 0.23(0.0012) = 0.000 276 in Ans.
3.8
f =
= 0.004 56
833
The power loss due to friction is
2π f W r N
2π(0.004 56)(600)(1)(13.33)
H=
=
778(12)
778(12)
= 0.0245 Btu/s Ans.
Q = 5.3rcNl
= 5.3(1)(0.0012)(13.33)(1)
= 0.0848 in3 /s
126 Ans. bmin − dmax
25.04 − 25
=
= 0.02 mm
2
2
r = d/2 = 25/2 = 12.5 mm, l/d = 1
˙
r/c = 12.5/0.02 = 625
N = 1200/60 = 20 rev/s
1250
= 2 MPa
P=
252
50(10−3 )(20)
= 0.195
S = (6252 )
For µ = 50 mPa · s,
2(106 )
cmin = From Figs. 1216, 1218 and 1220:
h o /c = 0.52, f r/c = 4.5, Q s /Q = 0.57
h o = 0.52(0.02) = 0.0104 mm Ans.
4.5
= 0.0072
f =
625
T = f W r = 0.0072(1.25)(12.5) = 0.1125 N · m 307 budynas_SM_ch12.qxd 308 12/04/2006 15:24 FIRST PAGES Page 308 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The power loss due to friction is
H = 2π T N = 2π(0.1125)(20) = 14.14 W Ans.
Q s = 0.57Q
127
cmin =
r= The side ﬂow is 57% of Q Ans. bmin − dmax
30.05 − 30.00
=
= 0.025 mm
2
2
30
d
=
= 15 mm
2
2 r
15
=
= 600
c
0.025
N= 1120
= 18.67 rev/s
60 P= 2750
= 1.833 MPa
30(50) S = (6002 ) 60(10−3 )(18.67)
= 0.22
1.833(106 ) l
50
=
= 1.67
d
30
This l/d requires use of the interpolation of Raimondi and Boyd, Eq. (1216).
From Fig. 1216, the h o /c values are:
y1/4 = 0.18, y1/2 = 0.34, y1 = 0.54, y∞ = 0.89 ho
= 0.659
c Substituting into Eq. (1216),
From Fig. 1218, the f r/c values are:
y1/4 = 7.4,
Substituting into Eq. (1216), y1/2 = 6.0, y1 = 5.0, y∞ = 4.0 fr
= 4.59
c From Fig. 1219, the Q/(rcNl) values are:
y1/4 = 5.65,
Substituting into Eq. (1216), y1/2 = 5.05, y1 = 4.05,
Q
= 3.605
rcN l y∞ = 2.95 h o = 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 FIRST PAGES Page 309 309 Chapter 12 128 bmin − dmax
75.10 − 75
=
= 0.05 mm
2
2
l/d = 36/75 = 0.5 (close enough)
˙ cmin = r = d/2 = 75/2 = 37.5 mm
r/c = 37.5/0.05 = 750
N = 720/60 = 12 rev/s
2000
P=
= 0.741 MPa
75(36)
Fig. 1213: SAE 20, µ = 18.5 mPa · s
S = (7502 ) 18.5(10−3 )(12)
= 0.169
0.741(106 ) From Figures 1216, 1218 and 1221:
h o /c = 0.29, f r/c = 5.1, P/ pmax = 0.315 h o = 0.29(0.05) = 0.0145 mm Ans.
f = 5.1/750 = 0.0068
T = f W r = 0.0068(2)(37.5) = 0.51 N · m
The heat loss rate equals the rate of work on the ﬁlm
Hloss = 2π T N = 2π(0.51)(12) = 38.5 W Ans.
pmax = 0.741/0.315 = 2.35 MPa Ans.
Fig. 1213: SAE 40, µ = 37 MPa · s
S = 0.169(37)/18.5 = 0.338
From Figures 1216, 1218 and 1221:
h o /c = 0.42, f r/c = 8.5, P/ pmax = 0.38 h o = 0.42(0.05) = 0.021 mm Ans.
f = 8.5/750 = 0.0113
T = f W r = 0.0113(2)(37.5) = 0.85 N · m
Hloss = 2π T N = 2π(0.85)(12) = 64 W Ans.
pmax = 0.741/0.38 = 1.95 MPa Ans.
129 bmin − dmax
50.05 − 50
=
= 0.025 mm
2
2
r = d/2 = 50/2 = 25 mm cmin = r/c = 25/0.025 = 1000
l/d = 25/50 = 0.5, N = 840/60 = 14 rev/s
2000
P=
= 1.6 MPa
25(50) budynas_SM_ch12.qxd 310 12/04/2006 15:24 FIRST PAGES Page 310 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 1213: SAE 30, µ = 34 mPa · s
34(10−3 )(14)
= 0.2975
1.6(106 ) S = (10002 ) From Figures 1216, 1218, 1219 and 1220:
h o /c = 0.40, f r/c = 7.8, Q s /Q = 0.74, Q/(rcNl) = 4.9 h o = 0.40(0.025) = 0.010 mm Ans.
f = 7.8/1000 = 0.0078
T = f W r = 0.0078(2)(25) = 0.39 N · m
H = 2π T N = 2π(0.39)(14) = 34.3 W Ans.
Q = 4.9rcNl = 4.9(25)(0.025)(14)(25) = 1072 mm2 /s
Q s = 0.74(1072) = 793 mm3 /s Ans.
1210 Consider the bearings as speciﬁed by
minimum f: +
d−t0 ,
d +t
b−0b maximum W: +
d−td0 , +t
b−0b and differing only in d and d .
Preliminaries:
l/d = 1
P = 700/(1.252 ) = 448 psi
N = 3600/60 = 60 rev/s
Fig. 1216:
S = 0.08
˙
S = 0.20
˙ minimum f :
maximum W:
Fig. 1212: µ = 1.38(10−6 ) reyn
µN /P = 1.38(10−6 )(60/448) = 0.185(10−6 ) Eq. (127):
r
=
c S
µN/P For minimum f:
r
=
c 0.08
= 658
0.185(10−6 ) .
c = 0.625/658 = 0.000 950 = 0.001 in budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 311 Chapter 12 311 If this is cmin ,
b − d = 2(0.001) = 0.002 in
The median clearance is
c = cmin +
¯ td + tb
td + tb
= 0.001 +
2
2 and the clearance range for this bearing is
c= td + tb
2 which is a function only of the tolerances.
For maximum W:
r
=
c 0.2
= 1040
0.185(10−6 ) .
c = 0.625/1040 = 0.000 600 = 0.0005 in
If this is cmin
b − d = 2cmin = 2(0.0005) = 0.001 in
c = cmin +
¯
c= td + tb
td + tb
= 0.0005 +
2
2 td + tb
2 The difference (mean) in clearance between the two clearance ranges, crange , is
td + tb
td + tb
− 0.0005 +
2
2
= 0.0005 in crange = 0.001 + For the minimum f bearing
b − d = 0.002 in
or
d = b − 0.002 in
For the maximum W bearing
d = b − 0.001 in
For the same b, tb and td , we need to change the journal diameter by 0.001 in.
d − d = b − 0.001 − (b − 0.002)
= 0.001 in
Increasing d of the minimum friction bearing by 0.001 in, deﬁnes d of the maximum load
bearing. Thus, the clearance range provides for bearing dimensions which are attainable
in manufacturing. Ans. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 312 312 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 1211 Given: SAE 30, N = 8 rev/s, Ts = 60°C, l/d = 1,
W = 3000 N
cmin = d = 80 mm, b = 80.08 mm, bmin − dmax
80.08 − 80
=
= 0.04 mm
2
2 r = d/2 = 80/2 = 40 mm
40
r
=
= 1000
c
0.04
P= 3000
= 0.469 MPa
80(80) Trial #1: From Figure 1213 for T = 81°C, µ = 12 mPa · s
T = 2(81°C − 60°C) = 42°C
S = (10002 ) 12(10−3 )(8)
= 0.2047
0.469(106 ) From Fig. 1224,
0.120 T
= 0.349 + 6.009(0.2047) + 0.0475(0.2047) 2 = 1.58
P
0.469
0.120 T = 1.58 = 6.2°C Discrepancy = 42°C − 6.2°C = 35.8°C
Trial #2: From Figure 1213 for T = 68°C, µ = 20 mPa · s,
T = 2(68°C − 60°C) = 16°C
S = 0.2047 20
12 = 0.341 From Fig. 1224,
0.120 T
= 0.349 + 6.009(0.341) + 0.0475(0.341) 2 = 2.4
P
T = 2.4 0.469
0.120 = 9.4°C Discrepancy = 16°C − 9.4°C = 6.6°C
Trial #3: µ = 21 mPa · s, T = 65°C
T = 2(65°C − 60°C) = 10°C
S = 0.2047 21
12 = 0.358 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 313 313 Chapter 12 From Fig. 1224,
0.120 T
= 0.349 + 6.009(0.358) + 0.0475(0.358) 2 = 2.5
P
0.469
0.120 T = 2.5 = 9.8°C Discrepancy = 10°C − 9.8°C = 0.2°C O.K. Tav = 65°C Ans.
T1 = Tav − T /2 = 65°C − (10°C/2) = 60°C T2 = Tav + T /2 = 65°C + (10°C/2) = 70°C S = 0.358
From Figures 1216, 1218, 1219 and 1220:
ho
= 0.68,
c f r/c = 7.5, Q
= 3.8,
rcN l Qs
= 0.44
Q h o = 0.68(0.04) = 0.0272 mm Ans.
f = 7.5
= 0.0075
1000 T = f W r = 0.0075(3)(40) = 0.9 N · m
H = 2π T N = 2π(0.9)(8) = 45.2 W Ans.
Q = 3.8(40)(0.04)(8)(80) = 3891 mm3 /s
Q s = 0.44(3891) = 1712 mm3 /s Ans.
1212 Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110°F,
N = 1120 rev/min, and l = 2.5 in.
For a trial ﬁlm temperature T f = 150°F
Tf µ S 150 2.421 0.0921 Tav = Ts + T (From Fig. 1224)
18.5 T
18.5°F
= 110°F +
= 119.3°F
2
2 T f − Tav = 150°F − 119.3°F
which is not 0.1 or less, therefore try averaging
(T f ) new = 150°F + 119.3°F
= 134.6°F
2 budynas_SM_ch12.qxd 314 12/04/2006 15:24 FIRST PAGES Page 314 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Proceed with additional trials
Trial
Tf µ S T Tav New
Tf 150.0
134.6
128.1
125.5
124.5
124.1
124.0 2.421
3.453
4.070
4.255
4.471
4.515
4.532 0.0921
0.1310
0.1550
0.1650
0.1700
0.1710
0.1720 18.5
23.1
25.8
27.0
27.5
27.7
27.8 119.3
121.5
122.9
123.5
123.8
123.9
123.7 134.6
128.1
125.5
124.5
124.1
124.0
123.9 Note that the convergence begins rapidly. There are ways to speed this, but at this point
they would only add complexity. Depending where you stop, you can enter the analysis.
(a) µ = 4.541(10−6 ) reyn,
From Fig. 1216: S = 0.1724 ho
= 0.482,
c h o = 0.482(0.002) = 0.000 964 in From Fig. 1217: φ = 56° Ans.
(b) e = c − h o = 0.002 − 0.000 964 = 0.001 04 in Ans.
(c) From Fig. 1218: fr
= 4.10,
c f = 4.10(0.002/1.25) = 0.006 56 Ans. (d) T = f W r = 0.006 56(1200)(1.25) = 9.84 lbf · in
H= 2π T N
2π(9.84)(1120/60)
=
= 0.124 Btu/s Ans.
778(12)
778(12) (e) From Fig. 1219: Q
= 4.16,
rcNl Q = 4.16(1.25)(0.002) 1120
(2.5)
60 = 0.485 in3 /s Ans.
From Fig. 1220:
(f) From Fig. 1221: Qs
= 0.6,
Q
P
pmax = 0.45, Q s = 0.6(0.485) = 0.291 in3 /s Ans.
pmax = 1200
= 427 psi Ans.
2.52 (0.45)
φ pmax = 16° Ans. (g) φ p0 = 82° Ans.
(h) T f = 123.9°F Ans.
(i) Ts + T = 110°F + 27.8°F = 137.8°F Ans. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 315 315 Chapter 12 1213 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 = 120°F.
Below is a partial tabular summary for comparison purposes.
cmin
0.001 in
Tf
T
Tmax
µ
S
fr
c
Q
rcN j l
Qs
Q
ho
c
f
Q
Qs
ho c
0.002 in cmax
0.003 in 125.8
11.5
131.5
3.014
0.053 7
0.775 0 124.0
7.96
128.0
3.150
0.024 9
0.873 4.317 1.881 1.243 4.129 4.572 4.691 0.582 0.824 0.903 0.501 0.225 0.127 0.006 9
0.094 1
0.054 8
0.000 501 0.006
0.208
0.172
0.000 495 0.005 9
0.321
0.290
0.000 382 132.2
24.3
144.3
2.587
0.184
0.499 Note the variations on each line. There is not a bearing, but an ensemble of many bearings, due to the random assembly of toleranced bushings and journals. Fortunately the
distribution is bounded; the extreme cases, cmin and cmax , coupled with c provide the
charactistic description for the designer. All assemblies must be satisfactory.
The designer does not specify a journalbushing bearing, but an ensemble of bearings.
1214 Computer programs will vary—Fortran based, MATLAB, spreadsheet, etc. 1215 In a stepbystep 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 temperature’s role in establishing the sump
temperature. Sec. 129 and Ex. 125 address this problem.
The task is to iteratively ﬁnd the average ﬁlm temperature, T f , which makes Hgen and
Hloss equal. The steps for determining cmin are provided within Trial #1 through Trial #3
on the following page. budynas_SM_ch12.qxd 316 12/04/2006 15:24 FIRST PAGES Page 316 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Trial #1:
• Choose a value of T f .
• Find the corresponding viscosity.
• Find the Sommerfeld number.
• Find f r/c , then
Hgen = 2545
WN c
1050 fr
c • Find Q/(rcNl) and Q s /Q . From Eq. (1215)
T =
Hloss = 0.103P( f r/c)
(1 − 0.5Q s /Q)[Q/(rcN j l)]
h CR A(T f − T∞ )
¯
1+α • Display T f , S, Hgen , Hloss
Trial #2: Choose another T f , repeating above drill.
Trial #3:
Plot the results of the ﬁrst two trials.
H Hgen
Hloss, linear with Tf (Tf )1 (Tf )3 (Tf )2 Tf Choose (T f ) 3 from plot. Repeat the drill. Plot the results of Trial #3 on the above graph.
If you are not within 0.1°F, iterate again. Otherwise, stop, and ﬁnd all the properties of
the 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 Trumpler’s criteria
(for n d = 2).
This adequacy assessment protocol can be used as a design tool by giving the students
additional possible bushing sizes.
b (in) tb (in) 2.254
2.004
1.753 0.004
0.004
0.003 Otherwise, the design option includes reducing l/d to save on the cost of journal machining and vendersupplied bushings. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 317 317 Chapter 12 1216 Continue to build a skill with pressurefed bearings, that of ﬁnding the average temperature of the ﬂuid ﬁlm. First examine the case for c = cmin
Trial #1:
•
•
•
•
• Choose an initial T f .
Find the viscosity.
Find the Sommerfeld number.
Find f r/c, h o /c, and .
From Eq. (1224), ﬁnd T .
Tav = Ts + • Display T f , S, T, and Tav . T
2 Trial #2:
• Choose another T f . Repeat the drill, and display the second set of values for T f ,
S, T, and Tav .
• Plot Tav vs T f :
Tav 2 Tav ϭ Tf 1
(Tf )2 (Tf )3 (Tf )1 Tf Trial #3:
Pick the third T f from the plot and repeat the procedure. If (T f ) 3 and (Tav ) 3 differ by more
than 0.1°F, plot the results for Trials #2 and #3 and try again. If they are within 0.1°F, determine the bearing parameters, check the Trumpler criteria, and compare Hloss with the
lubricant’s 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.
1217 An adequacy assessment associated with a design task is required. Trumpler’s criteria
will do.
d = 50.00+0.00 mm, b = 50.084+0.010 mm
−0.05
−0.000
SAE 30, N = 2880 rev/min or 48 rev/s, W = 10 kN
bmin − dmax
50.084 − 50
=
= 0.042 mm
cmin =
2
2
r = d/2 = 50/2 = 25 mm
r/c = 25/0.042 = 595
1
l = (55 − 5) = 25 mm
2
l /d = 25/50 = 0.5
10(106 )
W
=
p=
= 4000 kPa
4rl
4(0.25)(0.25) budynas_SM_ch12.qxd 318 12/04/2006 15:24 FIRST PAGES Page 318 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Trial #1: Choose (T f ) 1 = 79°C. From Fig. 1213, µ = 13 mPa · s.
13(10−3 )(48)
= 0.055
4000(103 ) S = (5952 ) fr
= 2.3,
= 0.85.
c
978(106 ) ( f r/c)SW 2
T =
1 + 1.5 2
ps r 4 From Figs. 1218 and 1216:
From Eq. (1225), 2.3(0.055)(102 )
978(106 )
1 + 1.5(0.85) 2
200(25) 4 = = 76.0°C
Tav = Ts + T /2 = 55°C + (76°C/2) = 93°C Trial #2: Choose (T f ) 2 = 100°C. From Fig. 1213, µ = 7 mPa · s.
7
= 0.0296
S = 0.055
13
fr
= 1.6,
c From Figs. 1218 and 1216:
T = = 0.90 1.6(0.0296)(102 )
978(106 )
= 26.8°C
1 + 1.5(0.9) 2
200(25) 4 Tav = 55°C + 26.8°C
= 68.4°C
2 Tav
100
90
80 Tav ϭ Tf (100ЊC, 100ЊC) (79ЊC, 93ЊC) (79ЊC, 79ЊC) 70 (100ЊC, 68.4ЊC)
85ЊC
60 70 80 90 100 Tf Trial #3: Thus, the plot gives (T f ) 3 = 85°C. From Fig. 1213, µ = 10.8 mPa · s.
S = 0.055 10.8
13 = 0.0457 fr
= 2.2,
= 0.875
c
2.2(0.0457)(102 )
978(106 )
= 58.6°C
T =
1 + 1.5(0.8752 )
200(25) 4 From Figs. 1218 and 1216: Tav = 55°C +
Result is close. Choose 58.6°C
= 84.3°C
2 85°C + 84.3°C
¯
= 84.7°C
Tf =
2 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 319 319 Chapter 12 µ = 10.8 MPa · s Fig. 1213: 10.8
13 S = 0.055
fr
= 2.23,
c
T = = 0.0457 = 0.874, ho
= 0.13
c 2.23(0.0457)(102 )
978(106 )
= 59.5°C
1 + 1.5(0.8742 )
200(254 ) Tav = 55°C + 59.5°C
= 84.7°C
2 O.K. From Eq. (1222)
Q s = (1 + 1.5 2 ) π ps rc3
3µl = [1 + 1.5(0.8742 )] π(200)(0.0423 )(25)
3(10)(10−6 )(25) = 3334 mm3 /s
h o = 0.13(0.042) = 0.005 46 mm or 0.000 215 in
Trumpler:
h o = 0.0002 + 0.000 04(50/25.4)
= 0.000 279 in
Tmax = Ts + Not O.K. T = 55°C + 63.7°C = 118.7°C or 245.7°F O.K.
Pst = 4000 kPa or 581 psi
n = 1, as done Not O.K. Not O.K. There is no point in proceeding further.
1218 So far, we’ve performed elements of the design task. Now let’s do it more completely.
First, remember our viewpoint.
The values of the unilateral tolerances, tb and td , reﬂect the routine capabilities of the
bushing vendor and the inhouse 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 Trumpler’s constraint of Pst ≤ 300 psi.
W
Pst =
≤ 300
2dl
W
2d 2 l /d ≤ 300 dmin = ⇒ d≥ W
600(l /d) 900
= 1.73 in
2(300)(0.5) budynas_SM_ch12.qxd 320 12/04/2006 15:24 FIRST PAGES Page 320 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design In this problem we will take journal diameter as the nominal value and the bushing bore
as a variable. In the next problem, we will take the bushing bore as nominal and the journal diameter as free.
To determine where the constraints are, we will set tb = td = 0, and thereby shrink
the design window to a point.
d = 2.000 in We set b = d + 2cmin = d + 2c
nd = 2 (This makes Trumpler’s n d ≤ 2 tight) and construct a table.
c b d ¯
T f* Tmax ho 0.0010
0.0011
0.0012
0.0013
0.0014
0.0015
0.0016
0.0017
0.0018
0.0019
0.0020 2.0020
2.0022
2.0024
2.0026
2.0028
2.0030
2.0032
2.0034
2.0036
2.0038
2.0040 2
2
2
2
2
2
2
2
2
2
2 215.50
206.75
198.50
191.40
185.23
179.80
175.00
171.13
166.92
163.50
160.40 312.0
293.0
277.0
262.8
250.4
239.6
230.1
220.3
213.9
206.9
200.6 ×
×
×
×
×
×
×
× Pst Tmax n × fom
−5.74
−6.06
−6.37
−6.66
−6.94
−7.20
−7.45
−7.65
−7.91
−8.12
−8.32 *Sample calculation for the ﬁrst entry of this column.
¯
T f = 215.5°F
Iteration yields:
¯
With T f = 215.5°F, from Table 121
µ = 0.0136(10−6 ) exp[1271.6/(215.5 + 95)] = 0.817(10−6 ) reyn
900
N = 3000/60 = 50 rev/s, P =
= 225 psi
4
2
0.817(10−6 )(50)
1
= 0.182
S=
0.001
225
From Figs. 1216 and 1218:
Eq. (12–24): = 0.7, f r/c = 5.5 0.0123(5.5)(0.182)(9002 )
= 191.6°F
[1 + 1.5(0.72 )](30)(14 )
191.6°F
.
Tav = 120°F +
= 215.8°F = 215.5°F
2
For the nominal 2in bearing, the various clearances sho...
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 Trigraph, Bearing, Accompany Mechanical Engineering

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