Mechanical Engineering Design Chapter 12 Solutions

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

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


 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


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.


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


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

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

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


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

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:


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)


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.


(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


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

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    T f  70  11  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



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. _____________________________________________________________________________ 0.005 12-16 Given: d  3.5000.000 0.001 in, b  3.5050.000 in , SAE 30, T s = 120F, p s = 50 psi, 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


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


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 1  1.5  0.9022   Qs  1  1.5    6  3 l 3  3.47 10  0.875    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. _____________________________________________________________________________


0.010 12-17 Given: d  50.000.00 0.05 mm, b  50.084 0.000 mm , SAE 30, T s = 55C, p s = 200 kPa, 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


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     6 3µl  3 10.5 10  25    3156 mm3 /s  3156  25.4 3   0.193 in 3 /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.


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.


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 b  2.0040.003 0.001 in, 0.000 in

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

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 0.000 in

ra ft

d  1.8750.000 0.001 in,

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 0.001 in,

d  1.869 0.000 0.001 in

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

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 b  2.0050.003 0.001 in, 0.000 in

Doubled, d  4.000 0.000 b  4.010 0.006 0.002 in, 0.000 in

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

_____________________________________________________________________________ 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


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 .


D

ra ft

Suggestion: Try smaller sizes.


Mechanical Engineering Design Chapter 12 Solutions

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