Shigley's Mechanical Engineering Design 9th Edition Solutions Manual

1. Chapter 12
12-1
Given: d max = 25 mm, b min = 25.03 mm, l/d = 1/2, W = 1.2 kN,  = 55 mPas, and N =
1100 rev/min.
b  d max 25.03  25

 0.015 mm
cmin  min
2
2
r  25/2 = 12.5 mm
r/c = 12.5/0.015 = 833.3
N = 1100/60 = 18.33 rev/s
P = W/ (ld) = 1200/ [12.5(25)] = 3.84 N/mm2 = 3.84 MPa
Fig. 12-16:
Fig. 12-18:
ra
ft
Eq. (12-7):
2
 55 103 18.33 
 r  N
2
  0.182
 833.3 
S  
6
c P
 3.84 10  


h 0 /c = 0.3

h 0 = 0.3(0.015) = 0.0045 mm
f r/c = 5.4

f = 5.4/833.3 = 0.006 48
Ans.
T =f Wr = 0.006 48(1200)12.5(103) = 0.0972 Nm
D
H loss = 2 TN = 2 (0.0972)18.33 = 11.2 W
Fig. 12-19:
Q/(rcNl) = 5.1 
Ans.
Q = 5.1(12.5)0.015(18.33)12.5 = 219 mm3/s
Fig. 12-20:
Q s /Q = 0.81 
Q s = 0.81(219) = 177 mm3/s
Ans.
______________________________________________________________________________
12-2
Given: d max = 32 mm, b min = 32.05 mm, l = 64 mm, W = 1.75 kN,  = 55 mPas, and N =
900 rev/min.
b  d max 32.05  32

 0.025 mm
cmin  min
2
2
r  32/2 = 16 mm
r/c = 16/0.025 = 640
N = 900/60 = 15 rev/s
Chapter 12, Page 1/26

2. P = W/ (ld) = 1750/ [32(64)] = 0.854 MPa
l/d = 64/32 = 2
2
 55 103 15 
 r  N
2
  0.797
S  
 640 
c P
 0.854 


Eq. (12-7):
Eq. (12-16), Figs. 12-16, 12-19, and 12-21
l/d
y 
y 1
y 1/2
y 1/4
y l/d
h 0 /c
2
0.98
0.83
0.61
0.36
0.92
P/p max
Q/rcNl
2
2
0.84
3.1
0.54
3.45
0.45
4.2
0.31
5.08
0.65
3.20
h 0 = 0.92 c = 0.92(0.025) = 0.023 mm
Ans.
Ans.
ra
ft
p max = P / 0.065 = 0.854/0.65 = 1.31 MPa
Ans.
Q = 3.20 rcNl = 3.20(16)0.025(15)64 = 1.23 (103) mm3/s
______________________________________________________________________________
Given: d max = 3.000 in, b min = 3.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and
SAE 10 and SAE 40 at 150F.
b  d max
3.005  3.000
cmin  min

 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
W
800
P

 177.78 psi
ld 1.5(3)
D
12-3
Fig. 12-12: SAE 10 at 150F, µ  1.75 µreyn
2
6
 r  N
2 1.75(10 )(10) 
S  
 600 
  0.0354
c P
 177.78

Figs. 12-16 and 12-21: h 0 /c = 0.11 and P/p max = 0.21
h0  0.11(0.0025)  0.000 275 in Ans.
pmax  177.78 / 0.21  847 psi Ans.
Fig. 12-12: SAE 40 at 150F, µ  4.5 µreyn
Chapter 12, Page 2/26

3.  4.5 
S  0.0354 
  0.0910
 1.75 
h0 / c  0.19, P / pmax  0.275
h0  0.19(0.0025)  0.000 475 in Ans.
pmax  177.78 / 0.275  646 psi Ans.
______________________________________________________________________________
ra
ft
Given: d max = 3.250 in, b min = 3.256 in, l = 3.25 in, W = 800 lbf, and N = 1000 rev/min.
b  d max
3.256  3.250
cmin  min

 0.003
2
2
r  3.250 / 2  1.625 in
l / d  3 / 3.250  0.923
r / c  1.625 / 0.003  542
N  1000 / 60  16.67 rev/s
W
800
P

 82.05 psi
ld
3(3.25)
Fig. 12-14: SAE 20W at 150F,  = 2.85  reyn
2
6
 r  N
2  2.85(10 )(16.67) 
S  
 542 
  0.1701
82.05
c P


From Eq. (12-16), and Figs. 12-16 and 12-21:
h o /c
l/d
y 
y 1
y 1/2
y 1/4
y l/d
0.923
0.85
0.48
0.28
0.15
0.46
0.923
0.83
0.45
0.32
0.22
0.43
D
12-4
P/p max
ho  0.46c  0.46(0.003)  0.001 38 in
P
82.05
pmax 

 191 psi Ans.
0.43
0.43
Ans.
Fig. 12-14: SAE 20W-40 at 150F,  = 4.4  reyn
S  5422
4.4(10 6 )(16.67)
 0.263
82.05
From Eq. (12-16), and Figs. 12-16 and 12-21:
l/d
y 
y 1
y 1/2
y 1/4
y l/d
h o /c
0.923
0.91
0.6
0.38
0.2
0.58
P/p max
0.923
0.83
0.48
0.35
0.24
0.46
Chapter 12, Page 3/26

4. h0  0.58c  0.58(0.003)  0.001 74 in Ans.
8205 82.05
pmax 

 178 psi Ans.
0.46
0.46
______________________________________________________________________________
Given: d max = 2.000 in, b min = 2.0024 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE
20 at 130F.
2.0024  2
b  d max

 0.0012 in
cmin  min
2
2
d
2
r 
  1 in, l / d  1 / 2  0.50
2
2
r / c  1 / 0.0012  833
N  800 / 60  13.33 rev/s
W
600

 300 psi
P 
ld
2(1)
ra
ft
12-5
Fig. 12-12: SAE 20 at 130F, µ  3.75 µreyn
2
6
 r  N
2  3.75(10 )(13.3) 
 833 
S  
  0.115
300
c P


From Figs. 12-16, 12-18 and 12-19:
D
h0 / c  0.23, r f / c  3.8, Q / (rcNl )  5.3
h0  0.23(0.0012)  0.000 276 in Ans.
3.8
f 
 0.004 56
833
The power loss due to friction is
2 f WrN
2 (0.004 56)(600)(1)(13.33)

778(12)
778(12)
 0.0245 Btu/s Ans.
Q  5.3rcNl
 5.3(1)(0.0012)(13.33)(1)
 0.0848 in 3 / s Ans.
______________________________________________________________________________
H 
12-6
Given: d max = 25 mm, b min = 25.04 mm, l/d = 1, W = 1.25 kN,  = 50 mPas, and N =
1200 rev/min.
Chapter 12, Page 4/26

5. bmin  d max
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
W
1250
P 

 2 MPa
ld
252
cmin 
2
 50(10 3 )(20) 
 r  N
S  
 6252 
  0.195
6
c P
 2(10 ) 
From Figs. 12-16, 12-18 and 12-20:
For µ = 50 MPa · s,
ra
ft
h0 / c  0.52, f r / c  4.5, Qs / Q  0.57
h0  0.52(0.02)  0.0104 mm Ans.
4.5
f 
 0.0072
625
T  f Wr  0.0072(1.25)(12.5)  0.1125 N · m
The power loss due to friction is
H = 2πT N = 2π (0.1125)(20) = 14.14 W
Ans.
Q s = 0.57Q The side flow is 57% of Q Ans.
______________________________________________________________________________
Given: d max = 1.25 in, b min = 1.252 in, l = 2 in, W = 620 lbf,  = 8.5  reyn, and N =
1120 rev/min.
b  d max 1.252  1.25
cmin  min

 0.001 in
2
2
r  d / 2  1.25 / 2  0.625 in
r / c  0.625 / 0.001  625
N  1120 / 60  18.67 rev/s
W
620
P 

 248 psi
ld 1.25(2)
D
12-7
2
 8.5(10 6 )(18.67) 
 r  N
S  
 6252 
  0.250
248
c P


l / d  2 / 1.25  1.6
From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19
Chapter 12, Page 5/26

6.
l/d
y 
y 1
y 1/2
y 1/4
y l/d
h 0 /c
fr/c
Q/rcNl
1.6
1.6
1.6
0.9
4.5
3
0.58
5.3
3.98
0.36
6.5
4.97
0.185
8
5.6
0.69
4.92
3.59
h 0 = 0.69 c = 0.69(0.001) =0.000 69 in
Ans.
f = 4.92/(r/c) = 4.92/625 = 0.007 87
Ans.
Ans.
Q = 1.6 rcNl = 1.6(0.625) 0.001(18.57) 2 = 0.0833 in3/s
______________________________________________________________________________
Given: d max = 75.00 mm, b min = 75.10 mm, l = 36 mm, W = 2 kN, N = 720 rev/min, and
SAE 20 and SAE 40 at 60C.
bmin  d max
75.10  75

 0.05 mm
2
2
l / d  36 / 75  0.48  0.5 (close enough)
r  d / 2  75 / 2  37.5 mm
r / c  37.5 / 0.05  750
N  720 / 60  12 rev/s
W
2000
P 

 0.741 MPa
ld
75(36)
ra
ft
cmin 
Fig. 12-13: SAE 20 at 60C, µ = 18.5 MPa · s
D
12-8
2
18.5(10 3 )(12) 
 r  N
S  
 7502 
  0.169
6
c P
 0.741(10 ) 
From Figures 12-16, 12-18 and 12-21:
h0 / c  0.29, f r / c  5.1, P / pmax  0.315
h0  0.29(0.05)  0.0145 mm Ans.
f  5.1 / 750  0.0068
T  f Wr  0.0068(2)(37.5)  0.51 N · m
The heat loss rate equals the rate of work on the film
H loss = 2πT N = 2π(0.51)(12) = 38.5 W
p max = 0.741/0.315 = 2.35 MPa Ans.
Ans.
Fig. 12-13: SAE 40 at 60C, µ = 37 MPa · s
Chapter 12, Page 6/26

7. S = 0.169(37)/18.5 = 0.338
From Figures 12-16, 12-18 and 12-21:
h0 / c  0.42, f r / c  8.5, P / pmax  0.38
h0  0.42(0.05)  0.021 mm Ans.
f  8.5 / 750  0.0113
T  f Wr  0.0113(2)(37.5)  0.85 N · m
H loss  2 TN  2 (0.85)(12)  64 W Ans.
pmax  0.741 / 0.38  1.95 MPa Ans.
_____________________________________________________________________________
Given: d max = 56.00 mm, b min = 56.05 mm, l = 28 mm, W = 2.4 kN, N = 900 rev/min, and
SAE 40 at 65C.
b  d max
56.05  56

 0.025 mm
cmin  min
2
2
r  d / 2  56 / 2  28 mm
r / c  28 / 0.025  1120
l / d  28 / 56  0.5, N  900 / 60  15 rev/s
2400
 1.53 MPa
P 
28(56)
ra
ft
12-9
D
Fig. 12-13: SAE 40 at 65C, µ = 30 MPa · s
2
 30(10 3 )(15) 
 r  N
S  
 11202 
  0.369
6
c P
 1.53(10 ) 
From Figures 12-16, 12-18, 12-19 and 12-20:
h0 / c  0.44, f r / c  8.5, Qs / Q  0.71, Q / (rcNl )  4.85
h0  0.44(0.025)  0.011 mm Ans.
f  8.5 / 1000  0.007 59
T  f Wr  0.007 59(2.4)(28)  0.51 N · m
H  2 TN  2 (0.51)(15)  48.1 W Ans.
Q  4.85 rcNl  4.85(28)(0.025)(15)(28)  1426 mm3 /s
Qs  0.71(1426)  1012 mm3 /s Ans.
_____________________________________________________________________________
12-10 Consider the bearings as specified by
minimum f :


d t0d , b0tb
maximum W:


d t0d , b0tb
Chapter 12, Page 7/26

8. and differing only in d and d .
Preliminaries:
l /d 1
P  W / (ld )  700 / (1.252 )  448 psi
N  3600 / 60  60 rev/s
Fig. 12-16:
minimum f :
maximum W:
S  0.08
S  0.20
Fig. 12-12:
µ = 1.38(106) reyn
µN/P = 1.38(106)(60/448) = 0.185(106)
Eq. (12-7):
S
µN / P
ra
ft
r

c
For minimum f :
r
0.08

 658
0.185(10 6 )
c
c  0.625 / 658  0.000 950  0.001 in
b  d = 2(0.001) = 0.002 in
D
If this is c min ,
The median clearance is
c  cmin 
t d  tb
t  tb
 0.001  d
2
2
and the clearance range for this bearing is
t  tb
c  d
2
which is a function only of the tolerances.
For maximum W:
r
0.2

 1040
c
0.185(10 6 )
c  0.625 / 1040  0.000 600  0.0005 in
If this is c min
Chapter 12, Page 8/26

9. b  d   2cmin  2(0.0005)  0.001 in
t  tb
t  tb
c  cmin  d
 0.0005  d
2
2
t  tb
c  d
2
The difference (mean) in clearance between the two clearance ranges, c range , is
t d  tb 
t  tb 
  0.0005  d

2
2 

 0.0005 in
crange  0.001 
For the minimum f bearing
b  d = 0.002 in
or
d = b  0.002 in
d = b  0.001 in
ra
ft
For the maximum W bearing
For the same b, t b and t d , we need to change the journal diameter by 0.001 in.
d   d  b  0.001  (b  0.002)
 0.001 in
D
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.
_____________________________________________________________________________
12-11 Given: SAE 40, N = 10 rev/s, T s = 140F, l/d = 1, d = 3.000 in, b = 3.003 in, W = 675
lbf.
3.003  3
b  d max
cmin  min

 0.0015 in
2
2
r  d / 2  3 / 2  1.5 in
r / c  1.5 / 0.0015  1000
675
W
P 

 75 psi
3(3)
ld
Trial #1: From Figure 12-12 for T = 160°F, µ = 3.5 µ reyn,
T  2(160  140)  40F
2
 3.5(10 6 )(10) 
 r  N
S  
 10002 
  0.4667
75
c P


From Fig. 12-24,
Chapter 12, Page 9/26

10. 9.70T
 0.349 109  6.009 40(0.4667)  0.047 467(0.4667) 2  3.16
P
P
75
T  3.16
 3.16
 24.4F
9.70
9.70
Discrepancy = 40  24.4 = 15.6°F
Trial #2: T = 150°F, µ = 4.5 µ reyn,
T  2(150  140)  20F
 4.5 10 6 10 
2
S  1000 
  0.6
75




From Fig. 12-24,
ra
ft
9.70T
 0.349 109  6.009 40(0.6)  0.047 467(0.6) 2  3.97
P
P
75
T  3.97
 3.97
 30.7F
9.70
9.70
Discrepancy = 20  30.7 =  10.7°F
Trial #3: T = 154°F, µ = 4 µ reyn,
D
T  2(154  140)  28F
 4 10 6 10 
2
S  1000 
  0.533
75




From Fig. 12-24,
9.70T
 0.349 109  6.009 40(0.533)  0.047 467(0.533)2  3.57
P
P
75
T  3.57
 3.57
 27.6F
9.70
9.70
Discrepancy = 28  27.6 = 0.4°F
O.K.
T av = 140 +28/2 = 154°F Ans.
T1  Tav  T / 2  154  (28 / 2)  140F
T2  Tav  T / 2  154  (28 / 2)  168F
S  0.4
From Figures 12-16, 12-18, to 12-20:
Chapter 12, Page 10/26

11. h0
fr
Q
Qs
 0.75,
 11,
 3.6,
 0.33
c
c
rcN l
Q
h0  0.75(0.0015)  0.001 13 in Ans.
11
f 
 0.011
1000
T  f Wr  0.0075(3)(40)  0.9 N · m
2  0.011 675 1.5 10
2 f WrN
H loss 

 0.075 Btu/s
778 12 
778 12 
Ans.
Q  3.6rcN l  3.6(1.5)0.0015(10)3  0.243 in 3 /s
Ans.
3
Qs  0.33(0.243)  0.0802 in /s Ans.
_____________________________________________________________________________
12-12 Given: d = 2.5 in, b = 2.504 in, c min = 0.002 in, W = 1200 lbf, SAE = 20, T s = 110°F,
N = 1120 rev/min, and l = 2.5 in.
N = 1120/60 = 18.67 rev/s
ra
ft
P = W/(ld) = 1200/(2.5)2 = 192 psi,
For a trial film temperature, let T f = 150°F
Table 12-1:
Eq. (12-7):
Fig. 12-24:
 = 0.0136 exp[1271.6/(150 + 95)] = 2.441  reyn
6
2
2
 r   N  2.5 / 2  2.44110 18.67
S  

 0.927

192
 c  P  0.002 
192
0.349 109  6.009 40  0.0927   0.047 467  0.0927 2  

9.70 
 17.9F
D
T 
T
17.9
 110 
 119.0F
2
2
T f  Tav  150  119.0  31.0F
Tav  Ts 
which is not 0.1 or less, therefore try averaging for the new trial film temperature, let
(T f ) new 
150  119.0
 134.5F
2
Proceed with additional trials using a spreadsheet (table also shows the first trial)
Chapter 12, Page 11/26

12. Trial
Tf
'
S
T
T av
T f T av
New
Tf
150.0
134.5
127.9
125.3
124.3
124.0
123.8
2.441
3.466
4.084
4.369
4.485
4.521
4.545
0.0927
0.1317
0.1551
0.1659
0.1704
0.1717
0.1726
17.9
22.6
25.4
26.7
27.2
27.4
27.5
119.0
121.3
122.7
123.3
123.6
123.7
123.7
31.0
13.2
5.2
2.0
0.7
0.3
0.1
134.5
127.9
125.3
124.3
124.0
123.8
123.8
Note that the convergence begins rapidly. There are ways to speed this, but at this point
they would only add complexity.
µ  4.545(106 ), S  0.1726
ra
ft
(a)
From Fig. 12-16:
h0
 0.482, h0  0.482(0.002)  0.000 964 in
c
From Fig. 12-17:  = 56°
Ans.
D
(b) e = c  h 0 = 0.002  0.000 964 = 0.001 04 in Ans.
f r
(c) From Fig. 12-18:
 4.10, f  4.10(0.002 /1.25)  0.006 56
c
Ans.
(d) T = f Wr = 0.006 56(1200)(1.25) = 9.84 lbf · in
H 
(e) From Fig. 12-19:
2 T N
2 (9.84)(1120 / 60)

 0.124 Btu/s
778(12)
778(12)
Ans.
Q
 4.16
rcNl
 1120 
3
Q  4.16(1.25)(0.002) 
 (2.5)  0.485 in /s
60 

Qs
From Fig. 12-20:
 0.6, Qs  0.6(0.485)  0.291 in 3/s
Q
Ans.
Ans.
W /  ld  1200 / 2.52
P
(f) From Fig. 12-21:
 0.45, pmax 

 427 psi
pmax
0.45
0.45
From Fig. 12-22:  pmax  16 Ans.
Ans.
Chapter 12, Page 12/26

13. (g) From Fig. 12-22:  p0  82 Ans.
(h) From the trial table, T f = 123.8°F Ans.
(i) With T = 27.5°F from the trial table, T s + T = 110 + 27.5 = 137.5°F Ans.
_____________________________________________________________________________
12-13 Given: d = 1.250 in, t d = 0.001 in, b = 1.252 in, t b = 0.003 in, l = 1.25 in, W = 250 lbf,
N = 1750 rev/min, SAE 10 lubricant, sump temperature T s = 120°F.
P = W/(ld) = 250/1.252 = 160 psi,
N = 1750/60 = 29.17 rev/s
For the clearance, c = 0.002  0.001 in. Thus, c min = 0.001 in, c median = 0.002 in, and
c max = 0.003 in.
For c min = 0.001 in, start with a trial film temperature of T f = 135°F
 = 0.0158 exp[1157.5/(135 + 95)] = 2.423  reyn
Eq. (12-7):
6
2
2
 r   N  1.25 / 2  2.423 10  29.17
S  

 0.1725

160
 c  P  0.001 
Fig. 12-24:
ra
ft
Table 12-1:
160
 0.349 109  6.009 40  0.1725   0.047 467  0.17252  

9.70 
 22.9F
T 
T
22.9
 120 
 131.4F
2
2
T f  Tav  135  131.4  3.6F
D
Tav  Ts 
which is not 0.1 or less, therefore try averaging for the new trial film temperature, let
135  131.4
 133.2F
2
Proceed with additional trials using a spreadsheet (table also shows the first trial)
(T f ) new 
Trial
Tf
'
S
T
T av
T f T av
New
Tf
135.0
133.2
132.5
132.2
132.1
2.423
2.521
2.560
2.578
2.583
0.1725
0.1795
0.1823
0.1836
0.1840
22.9
23.6
23.9
24.0
24.0
131.4
131.8
131.9
132.0
132.0
3.6
1.4
0.6
0.2
0.1
133.2
132.5
132.2
132.1
132.1
Chapter 12, Page 13/26

14. With T f = 132.1°F, T = 24.0°F,  = 2.583  reyn, S = 0.1840,
T max = T s + T = 120 + 24.0 = 144.0°F
Fig. 12-16:
h 0 /c = 0.50, h 0 = 0.50(0.001) = 0.000 50 in
 = 1  h 0 /c = 1  0.50 = 0.05 in
Fig. 12-18:
r f /c = 4.25, f = 4.25/(0.625/0.001) = 0.006 8
Fig. 12-19:
Q/(rcNl) = 4.13, Q = 4.13(0.625)0.001(29.17)1.25 = 0.0941 in3/s
Fig. 12-20:
Q s /Q = 0.58, Q s = 0.58(0.0941) = 0.0546 in3/s
The above can be repeated for c median = 0.002 in, and c max = 0.003 in. The results are
shown below.
ra
ft
c min 0.001 c median c max 0.003
in
0.002 in
in
Tf
132.1
2.583
0.184
24.0
125.6
3.002
0.0534
11.1
124.1
3.112
0.0246
8.2
h 0 /c
144.0
0.5
131.1
0.23
128.2
0.125
h0
0.00050
0.00069
0.00038

0.50
4.25
0.0068
4.13
0.0941
0.77
1.8
0.0058
4.55
0.207
0.88
1.22
0.0059
4.7
0.321
0.58
0.82
0.90
0.0546
0.170
0.289
D

S

T max
fr/c
f
Q/(rcNl)
Q
Q s /Q
Qs
_____________________________________________________________________________
12-14 Computer programs will vary.
_____________________________________________________________________________
12-15 Note to the Instructor: In the first printing of the 9th edition, the l/d ratio and the
lubrication constant  were omitted. The values to use are l/d = 1, and  = 1. This will be
updated in the next printing. We apologize for any inconvenience this may have caused.
Chapter 12, Page 14/26

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 temperature’s role in establishing the sump
temperature. Sec. 12-9 and Ex. 12-5 address this problem.
Given: d max = 2.500 in, b min = 2.504 in, l/d = 1, N = 1120 rev/min, SAE 20 lubricant, W =
300 lbf, A = 60 in2, T  = 70F, and  = 1.
600 lbf load with minimal clearance: We will start by using W = 600 lbf (n d = 2). The
task is to iteratively find the average film temperature, T f , which makes H gen and
H loss equal.
b  d max
2.504  2.500
cmin  min

 0.002 in
2
2
N = 1120/60 = 18.67 rev/s
W
600

 96 psi
ld
2.52
ra
ft
P 
6
2
2
 r   N  1.25    10 18.67
S  

 0.0760 

96
c P
 0.002 
 = 0.0136 exp[1271.6/(T f + 95)]
Table 12-1:
2545
fr
 f r  2545
WNc 
 600 18.67  0.002 

1050
c
 c  1050
fr
 54.3
c
D
H gen 
H loss 
2.7 60 / 144
 CR A
T f  T   1  1  T f  70 
 1
 0.5625 T f  70 
Start with trial values of T f of 220 and 240F.
Trial T f
220
240

0.770
0.605
S
0.059
0.046
f r/c
1.9
1.7
H gen
103.2
92.3
H loss
84.4
95.6
As a linear approximation, let H gen = mT f + b. Substituting the two sets of values of
T f and H gen we find that H gen =  0.545 T f +223.1. Setting this equal to H loss and
solving
for T f gives T f = 237F.
Chapter 12, Page 15/26

16. 
Trial T f
237
S
0.048
0.627
f r/c
1.73
H loss
94.0
H gen
93.9
which is satisfactory.
Table 12-16:
h 0 /c = 0.21,
h 0 = 0.21 (0.002) = 000 42 in
Fig. 12-24:
T 
96
 0.349 109  6.009 4  0.048   0.047 467  0.0482  

9.7 
 6.31 F
T 1 = T s = T f  T = 237  6.31/2 = 233.8F
T max = T 1 + T = 233.8 + 6.31 = 240.1F
Trumpler’s design criteria:
ra
ft
0.002 + 0.000 04d = 0.002 + 0.000 04(2.5) = 0.000 30 in < h 0
O.K.
T max = 240.1F < 250F O.K.
Wst
300

 48 psi  300 psi
ld
2.52
O.K .
n d = 2 (assessed at W = 600 lbf)
O.K.
D
We see that the design passes Trumpler’s criteria and is deemed acceptable.
For an operating load of W = 300 lbf, it can be shown that T f = 219.3F,  = 0.78, S =
0.118, f r/c = 3.09, H gen = H loss = 84 Btu/h, h 0 = , T = 10.5F, T 1 = 224.6F, and T max =
235.1F.
_____________________________________________________________________________
12-16 Given: d  3.5000.000 in, b  3.5050.005 in , SAE 30, T s = 120F, p s = 50 psi,
0.001
0.000
N = 2000/60 = 33.33 rev/s, W = 4600 lbf, bearing length = 2 in, groove width = 0.250 in,
and H loss  5000 Btu/hr.
b  d max
3.505  3.500
cmin  min

 0.0025 in
2
2
r = d/ 2 = 3.500/2 = 1.750 in
r / c = 1.750/0.0025 = 700
l = (2  0.25)/2 = 0.875 in
Chapter 12, Page 16/26

17. l / d = 0.875/3.500 = 0.25
W
4600
P 

 751 psi
4rl 4 1.750  0.875
Trial #1: Choose (T f ) 1 = 150°F. From Table 12-1,
 = 0.0141 exp[1360.0/(150 + 95)] = 3.63 µ reyn
2
6
 r  N
2  3.63(10 )(33.33) 
S  
 700 
  0.0789
751
c P


From Figs. 12-16 and 12-18:
 = 0.9, f r/ c = 3.6
From Eq. (12-24),
T 
0.0123  3.6  0.0789  46002 
1  1.5(0.9) 2  50 1.7504 


 71.2F
ra
ft

0.0123( f r / c)SW 2
1  1.5  2  ps r 4
T av = T s + T / 2 = 120 + 71.2/2 = 155.6F
Trial #2: Choose (T f ) 2 = 160°F. From Table 12-1
 = 0.0141 exp[1360.0/(160 + 95)] = 2.92 µ reyn
D
 2.92 
S  0.0789 
  0.0635
 3.63 
From Figs. 12-16 and 12-18:
 = 0.915, f r/ c =3
T 
T av
0.0123  3 0.0635  46002 
1  1.5  0.9152   50 1.7504 


= 120 + 46.9/2 = 143.5F
 46.9F
Chapter 12, Page 17/26

18. Trial #3: Thus, the plot gives (T f ) 3 = 152.5°F. From Table 12-1
 = 0.0141 exp[1360.0/(152.5 + 95)] = 3.43 µ reyn
 3.43 
S  0.0789 
  0.0746
 3.63 
From Figs. 12-16 and 12-18:
 = 0.905, f r/ c =3.4
T 
T av
0.0123  3.4  0.0746  46002 
1  1.5  0.9052   50 1.7504 


= 120 + 63.2/2 = 151.6F
Result is close. Choose T f 
Table 12-1:
 63.2F
152.5  151.6
 152.1F
2
Try 152F
 = 0.0141 exp[1360.0/(152 + 95)] = 3.47 µ reyn
D
ra
ft
 3.47 
S  0.0789 
  0.0754
 3.63 
f r
h0
 3.4,   0.902,
 0.098
c
c
0.0123  3.4  0.0754  46002 
T 
 64.1F
1  1.5  0.9022   50 1.7504 


Tav  120  64.1 / 2  152.1F O.K.
h 0 = 0.098(0.0025) = 0.000 245 in
T max = T s + T = 120 + 64.1 = 184.1F
Eq. (12-22):
  50 1.750  0.00253 
 ps rc3
2
Qs 
1  1.5   3  3.47 10 6  0.875 1  1.5  0.9022 


3 l
 1.047 in 3 /s
H loss =  C p Q s T = 0.0311(0.42)1.047(64.1) = 0.877 Btu/s
= 0.877(602) = 3160 Btu/h O.K.
Trumpler’s design criteria:
0.0002 + 0.000 04(3.5) = 0.000 34 in > 0.000 245
Not O.K.
T max = 184.1°F < 250°F O.K.
P st = 751 psi > 300 psi Not O.K.
n = 1, as done Not O.K.
_____________________________________________________________________________
Chapter 12, Page 18/26

19. 12-17 Given: d  50.000.00 mm, b  50.0840.010 mm , SAE 30, T s = 55C, p s = 200 kPa,
0.05
0.000
N = 2880/60 = 48 rev/s, W = 10 kN, bearing length = 55 mm, groove width = 5 mm, and
H loss  300 W.
b  d max
50.084  50
cmin  min

 0.042 mm
2
2
r = d/ 2 = 50/2 = 25 mm
r / c = 25/0.042 = 595
l = (55  5)/2 = 25 mm
l / d = 25/50 = 0.5
10 103 
W
P 

 4 MPa
4rl 4  25  25
Trial #1: Choose (T f ) 1 = 79°C. From Fig. 12-13, µ = 13 MPa · s.
2
ra
ft
13(103 )(48) 
 r  N
S  
 5952 
  0.0552
6
c P
 4(10 ) 
From Figs. 12-16 and 12-18:
 = 0.85, f r/ c = 2.3
From Eq. (12-25),
D
978(106 ) ( f r / c)SW 2
T 
1  1.5  2
ps r 4
978(106 )  2.3(0.0552)(102 ) 

  76.3C
1  1.5(0.85) 2 
200(25) 4


T av = T s + T / 2 = 55 + 76.3/2 = 93.2C
Trial #2: Choose (T f ) 2 = 100°C. From Fig. 12-13, µ = 7 MPa · s.
7
S  0.0552    0.0297
 13 
From Figs. 12-16 and 12-18:
 = 0.90, f r/ c =1.6
T 
978(106 ) 1.6(0.0297)(102 ) 
  26.9C
200(25)4
1  1.5(0.9) 2 


T av = 55 + 26.9/2 = 68.5C
Chapter 12, Page 19/26

20. Trial #3: Thus, the plot gives (T f ) 3 = 85.5°C. From Fig. 12-13, µ = 10.5 MPa · s.
 10.5 
S  0.0552 
  0.0446
 13 
From Figs. 12-16 and 12-18:
 = 0.87, f r/ c =2.2
T 
978(106 )  2.2(0.0457)(102 ) 
  58.9C
1  1.5(0.87 2 ) 
200(25) 4


ra
ft
T av = 55 + 58.9/2 = 84.5C
Result is close. Choose T f 
µ = 10.5 MPa · s
 10.5 
S  0.0552 
  0.0446
 13 
f r
h0
  0.87,
 2.2,
 0.13
c
c
978(106 )  2.2(0.0457)(102 ) 
T 
  58.9C
1  1.5(0.87 2 ) 
200(254 )


Tav  55  58.9 / 2  84.5C O.K.
D
Fig. 12-13:
85.5  84.5
 85C
2
or
138F
From Eq. (12-22)
h 0 = 0.13(0.042) = 0.005 46 mm or 0.000 215 in
T max = T s + T = 55 + 58.9 = 113.9C
or
237°F
   200  25  0.0423  
 1  1.5  0.87 2   

 3 10.5 10 6  25  
3µl




3
3
3
 3156 mm /s  3156  25.4   0.193 in /s
Qs  (1  1.5  2 )
 ps rc 3
H loss =  C p Q s T = 0.0311(0.42)0.193(138) = 0.348 Btu/s
= 1.05(0.348) = 0.365 kW = 365 W not O.K.
Chapter 12, Page 20/26

21. Trumpler’s design criteria:
Not O.K.
0.0002 + 0.000 04(50/25.4) = 0.000 279 in > h 0
T max = 237°F O.K.
P st = 4000 kPa or 581 psi > 300 psi Not O.K.
n = 1, as done Not O.K.
_____________________________________________________________________________
12-18 So far, we’ve performed elements of the design task. Now let’s do it more completely.
The values of the unilateral tolerances, t b and t d , reflect the routine capabilities of the
bushing 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 Trumpler’s constraint
of P st ≤ 300 psi.
W
 300
2dl 
ra
ft
Pst 
W
 300  d 
2d 2l  / d
d min 
W
600(l  / d )
900
 1.73 in
2(300)(0.5)
D
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 t b = t d = 0, and thereby shrink the
design window to a point.
We set
d = 2.000 in
b = d + 2c min = d + 2c
n d = 2 (This makes Trumpler’s n d ≤ 2 tight)
and construct a table.
Chapter 12, Page 21/26

22. c
b
d
Tf *
T max
h o P st T max n
fom
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












































-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:
ra
ft
With T f  215.5F , from Table 12-1
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
6
 1   0.817(10 )(50) 
S 
 
  0.182
225
 0.001  

e = 0.7,
f r/c = 5.5
D
From Figs. 12-16 and 12-18:
Eq. (12–24):
0.0123(5.5)(0.182)(9002 )
 191.6F
[1  1.5(0.7 2 )](30)(14 )
191.6F
Tav  120F 
 215.8F  215.5F
2
TF 
For the nominal 2-in bearing, the various clearances show that we have been in contact
with the recurving of (h o ) min . The figure of merit (the parasitic friction torque plus the
pumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we will
place the top of the design window at c min = 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0.000 in, b  2.0040.003 in
0.001
0.000
Now we can check the performance at c min , c , and c max . Of immediate interest is the
fom of the median clearance assembly,  9.82, as compared to any other satisfactory
bearing ensemble.
Chapter 12, Page 22/26

23. If a nominal 1.875 in bearing is possible, construct another table with t b = 0 and t d = 0.
c
0.0020
0.0030
0.0035
0.0040
0.0050
0.0055
0.0060
b
1.879
1.881
1.882
1.883
1.885
1.886
1.887
d
1.875
1.875
1.875
1.875
1.875
1.875
1.875
Tf
157.2
138.6
133.5
130.0
125.7
124.4
123.4
T max h o P st T max n fom

  7.36
194.30  

  8.64
157.10  



  9.05
147.10

  9.32
140.10  

  9.59
131.45  



  9.63
128.80

  9.64
126.80  
The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our
design window.
b  1.8810.003 in
0.000
ra
ft
d  1.8750.000 in,
0.001
The ensemble median assembly has a fom =  9.31.
We just had room to fit in a design window based upon the (h 0 ) min constraint. Further
reduction in nominal diameter will preclude any smaller bearings. A table constructed for
a d = 1.750 in journal will prove this.
We choose the nominal 1.875-in bearing ensemble because it has the largest figure of
merit. Ans.
_____________________________________________________________________________
D
12-19 This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b and
radial clearance c.
The approach is similar to that of Prob. 12-18 and the tables will change slightly. In the
table for a nominal b = 1.875 in, note that at c = 0.003 in the constraints are “loose.” Set
b = 1.875 in
d = 1.875  2(0.003) = 1.869 in
For the ensemble
b  1.8750.003 in,
0.001
d  1.869 0.000 in
0.001
Analyze at c min = 0.003, c = 0.004 in and c max = 0.005 in
At cmin  0.003 in: T f  138.4, µ  3.160, S  0.0297, H loss  1035 Btu/h and the
Trumpler conditions are met.
At c  0.004 in: T f  130F,  = 3.872, S = 0.0205, H loss = 1106 Btu/h, fom = 9.246
Chapter 12, Page 23/26

24. and the Trumpler conditions are O.K.
At cmax  0.005 in: T f  125.68F,  = 4.325, S = 0.014 66, H loss = 1129 Btu/h and the
Trumpler conditions are O.K.
The ensemble figure of merit is slightly better; this bearing is slightly smaller. The
lubricant cooler has sufficient capacity.
_____________________________________________________________________________
12-20 Table 12-1:
 ( reyn) =  0 (106) exp [b / (T + 95)]
b and T in F
The conversion from  reyn to mPas is given on p. 620. For a temperature of C degrees
Celsius, T = 1.8 C + 32. Substituting into the above equation gives
 (mPas) = 6.89  0 (106) exp [b / (1.8 C + 32+ 95)]
= 6.89  0 (106) exp [b / (1.8 C + 127)]
Ans.
ra
ft
For SAE 50 oil at 70C, from Table 12-1,  0 = 0.0170 (106) reyn, and b = 1509.6F.
From the equation,
 = 6.89(0.0170) 106(106) exp {1509.6/[1.8(70) + 127]}
= 45.7 mPas
Ans.
From Fig. 12-13,  = 39 mPas
Ans.
D
The figure gives a value of about 15 % lower than the equation.
_____________________________________________________________________________
12-21 Originally
d  2.0000.000 in, b  2.0050.003 in
0.001
0.000
Doubled,
d  4.000 0.000 in, b  4.010 0.006 in
0.002
0.000
The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were
carried out. Some of the results are:
Part
(a)
(b)
c

0.007
0.0035
3.416
3.416
S
0.0310
0.0310
Tf
135.1
135.1
f r/c
0.1612
0.1612
Qs
6.56
0.870
h 0 /c
0.1032
0.1032
e
0.897
0.897
H loss
9898
1237
h0
0.000 722
0.000 361
Trumpler
h0
0.000 360
0.000 280
f
0.005 67
0.005 67
The side flow Q s differs because there is a c3 term and consequently an 8-fold increase.
H loss is related by a 9898/1237 or an 8-fold increase. The existing h 0 is related by a 2-fold
increase. Trumpler’s (h 0 ) min is related by a 1.286-fold increase.
Chapter 12, Page 24/26

25. _____________________________________________________________________________
12-22 Given: Oiles SP 500 alloy brass bushing, L = 0.75 in, D = 0.75 in, T  = 70F, F = 400
lbf, N = 250 rev/min, and w = 0.004 in.
K = 0.6(1010) in3min/(lbffth)
Table 12-8:
P = F/ (DL) = 400/ [0.75(0.75)] = 711 psi
V = DN/ 12 =  (0.75)250/12 = 49.1 ft/min
Tables 12-10 and 12-11:
f 1 = 1.8, f 2 = 1.0
Table 12-12: PV max = 46 700 psift/min, P max = 3560 psi, V max = 100 ft/min
Pmax 
4 F
4 400

 905 psi  3560 psi O.K .
 DL  0.752
ra
ft
PV = 711 (49.1) = 34 910 psift/min < 46 700 psift/min
O.K.
Eq. (12-32) can be written as
w  f1 f 2 K
4 F
Vt
 DL
Solving for t,
 DLw

  0.75  0.75  0.004 
4 1.8 1.0  0.6 1010  49.1 400
D
t 
4 f1 f 2 KVF
 833.1 h  833.1 60   49 900 min
Ans.
Cycles = Nt = 250 (49 900) = 12.5 (106) cycles
_____________________________________________________________________________
12-23 Given: Oiles SP 500 alloy brass bushing, w max = 0.002 in for 1000 h, N = 400 rev/min, F
= 100 lbf,  CR = 2.7 Btu/ (hft2F), T max = 300F, f s = 0.03, and n d = 2.
Estimate bushing length with f 1 = f 2 = 1, and K = 0.6(10-10) in3 · min/(lbf · ft · h)
Using Eq. (12-32) with n d F for F,
f1 f 2 Knd FNt 1(1)(0.6)(1010 )(2)(100)(400)(1000)
L

 0.80 in
3w
3(0.002)
From Eq. (12-38), with f s = 0.03 from Table 12-9 applying n d = 2 to F
Chapter 12, Page 25/26

26. and  CR  2.7 Btu/(h · ft 2 · °F)
L 
720 f s nd FN
720(0.03)(2)(100)(400)

 3.58 in
778(2.7)(300  70)
J  CR T f  T 
0.80  L  3.58 in
Trial 1: Let L = 1 in, D = 1 in
4 nd F
4(2)(100)

 255 psi  3560 psi O.K .
 DL
 (1)(1)
nF
2(100)
P  d 
 200 psi
DL
1(1)
 DN  (1)(400)
V 

 104.7 ft/min  100 ft/min Not O.K .
12
12
Pmax 
ra
ft
Trial 2: Try D = 7/8 in = 0.875 in, L = 1 in
4(2)(100)
 291 psi  3560 psi O.K .
 (0.875)(1)
2(100)
P 
 229 psi
0.875(1)
 (0.875)(400)
V 
 91.6 ft/min  100 ft/min
12
Pmax 
O.K .
D
PV = 229(91.6) = 20 976 psi · ft/min < 46 700 psi · ft/min
f1
V
33
1.3
91.6 f 1
100
1.8
O.K.
 91.6  33 
 f1  1.3  (1.8  1.3) 
  1.74
 100  33 
Lnew  f1Lold  1.74  0.80   1.39 in
Trial 3: Try D = 7/8 in = 0.875 in, L = 1.5 in
4(2)(100)
 194 psi  3560 psi O.K .
 (0.875)(1.5)
2(100)
P 
 152 psi, V  91.6 ft/min
0.875(1.5)
PV  152(91.6)  13 923 psi · ft/min  46 700 psi · ft/min
D  7 / 8 in, L  1.5 in is acceptable Ans.
Pmax 
O.K .
Chapter 12, Page 26/26

27. D
ra
ft
Suggestion: Try smaller sizes.
Chapter 12, Page 27/26