Chapter N = 1100/60 = rev/s. P = W/ (ld) = 1200/ [12.5(25)] = 3.84 N/mm 2 = 3.84 MPa ( ) ( )

Transcription

1 Chapter Given: d max = 5 mm, b min = 5.0 mm, l/d = 1/, W = 1. kn, µ = 55 mpa s, and N = 1100 rev/min. bmin dmax min = = = mm r 5/ = 1.5 mm r/ = 1.5/0.015 = 8. N = 1100/60 = 18. rev/s P = W/ (ld) = 100/ [1.5(5)] =.84 N/mm =.84 MPa Eq. (1-7): 6 r µ N S = = = P ig. 1-16: h 0 / = 0. h 0 = 0.(0.015) = mm ig. 1-18: f r/ = 5.4 f = 5.4/8. = T =f Wr = (100)1.5(10 ) = N m H loss = π TN = π (0.097)18. = 11. W ig. 1-19: Q/(rNl) = 5.1 Q = 5.1(1.5)0.015(18.)1.5 = 19 mm /s ig. 1-0: Q s /Q = 0.81 Q s = 0.81(19) = 177 mm /s _ 1- Given: d max = mm, b min =.05 mm, l = 64 mm, W = 1.75 kn, µ = 55 mpa s, and N = 900 rev/min. bmin dmax.05 min = = = 0.05 mm r / = 16 mm r/ = 16/0.05 = 640 N = 900/60 = 15 rev/s P = W/ (ld) = 1750/ [(64)] = MPa Shigley s MED, 10 th edition Chapter 1 Solutions, Page 1/6

2 l/d = 64/ = Eq. (1-7): r µ N S = = 640 = P Eq. (1-16), igs. 1-16, 1-19, and 1-1 l/d y y 1 y 1/ y 1/4 y l/d h 0 / P/p max Q/rNl h 0 = 0.9 = 0.9(0.05) = 0.0 mm p max = P / 0.65 = 0.854/0.65 = 1.1 MPa Q =.0 rnl =.0(16)0.05(15)64 = 1. (10 ) mm /s _ 1- Given: d max =.000 in, b min =.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and SAE 10 and SAE 40 at 150. bmin dmax min = = = in r.000 / = in l / d = 1.5 / = 0.5 r / = 1.5 / = 600 N = 600 / 60 = 10 rev/s W 800 P = = = psi ld 1.5() ig. 1-1: SAE 10 at 150, µ = 1.75 µ reyn S 6 r µ N 1.75(10 )(10) = = 600 = P igs and 1-1: h 0 / = 0.11 and P/p max = 0.1 = 0.11(0.005) = in p = / 0.1 = 847 psi max ig. 1-1: SAE 40 at 150, µ = 4.5 µ reyn Shigley s MED, 10 th edition Chapter 1 Solutions, Page /6

3 4.5 S = = / = 0.19, P / pmax = 0.75 = 0.19(0.005) = in pmax = / 0.75 = 646 psi _ 1-4 Given: d max =.50 in, b min =.56 in, l = in, W = 800 lbf, and N = 1000 rev/min. bmin dmax min = = = 0.00 r.50 / = 1.65 in l / d = /.50 = 0.9 r / = 1.65 / 0.00 = 54 N = 1000 / 60 = rev/s W 800 P = = = 8.05 psi ld (.5) ig. 1-1: SAE 0W at 150, µ =.40 µ reyn Note to instrutors: Some students may obtain a higher value of visosity (.85) from ig The value from ig. 1-1 is used here as the preferred value sine this figure is speifially for single-visosity oils. S 6 r µ N.40(10 )(16.67) = = 54 = 0.14 P 8.05 rom Eq. (1-16), and igs and 1-1: l/d y y 1 y 1/ y 1/4 y l/d h 0 / P/p max = 0.4 = 0.4(0.00) = in P 8.05 pmax = = = 195 psi ig. 1-14: SAE 0W-40 at 150, µ = 4.4 µ reyn S 6 4.4(10 )(16.67) = 54 = rom Eq. (1-16), and igs and 1-1: Shigley s MED, 10 th edition Chapter 1 Solutions, Page /6

4 l/d y y 1 y 1/ y 1/4 y l/d h 0 / P/p max = 0.58 = 0.58(0.00) = in P 8.05 pmax = = = 178 psi _ 1-5 Given: d max =.000 in, b min =.004 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE 0 at 10. bmin dmax.004 min = = = in d r = = 1 in, l / d = 1 / = 0.50 r / = 1 / = 8 N = 800 / 60 = 1. rev/s W 600 P = = = 00 psi ld (1) ig. 1-1: SAE 0 at 10, µ =.75 µ reyn S 6 r µ N.75(10 )(1.) = = 8 = P 00 rom igs. 1-16, 1-18 and 1-19: The power loss due to frition is / = 0., r f / =.8, Q / ( rnl) = 5. = 0.(0.001) = in f.8 = = π f WrN π( )(600)(1)(1.) H = = 778(1) 778(1) = Btu/s Q = 5.rNl = 5.(1)(0.001)(1.)(1) = in / s _ Shigley s MED, 10 th edition Chapter 1 Solutions, Page 4/6

5 1-6 Given: d max = 5 mm, b min = 5.04 mm, l/d = 1, W = 1.5 kn, µ = 50 mpa s, and N = 100 rev/min. bmin dmax min = = = 0.0 mm r d / = 5 / = 1.5 mm, l / d = 1 r / = 1.5 / 0.0 = 65 N = 100 / 60 = 0 rev/s W 150 P = = = MPa ld 5 r µ N 50(10 )(0) or µ = 50 mpa s, S = = 65 = P (10 ) rom igs. 1-16, 1-18 and 1-0: / = 0.5, f r / = 4.5, Qs / Q = 0.57 = 0.5(0.0) = mm f 4.5 = = T = f Wr = 0.007(1.5)(1.5) = N m The power loss due to frition is H = πt N = π (0.115)(0) = W Q s = 0.57Q The side flow is 57% of Q _ 1-7 Given: d max = 1.5 in, b min = 1.5 in, l = in, W = 60 lbf, µ = 8.5 µ reyn, and N = 110 rev/min. bmin dmax min = = = in r = d / = 1.5 / = 0.65 in r / = 0.65 / = 65 N = 110 / 60 = rev/s W 60 P = = = 48 psi ld 1.5() 6 µ N r 8.5(10 )(18.67) S = = P 48 = l / d = / 1.5 = 1.6 rom Eq. (1-16), and igs. 1-16, 1-18, and 1-19 Shigley s MED, 10 th edition Chapter 1 Solutions, Page 5/6

6 l/d y y 1 y 1/ y 1/4 y l/d h 0 / fr/ Q/rNl h 0 = 0.69 = 0.69(0.001) = in f = 4.9/(r/) = 4.9/65 = Q = 1.6 rnl = 1.6(0.65) 0.001(18.57) = 0.08 in /s _ 1-8 Given: d max = mm, b min = mm, l = 6 mm, W = kn, N = 70 rev/min, and SAE 0 and SAE 40 at 60 C. bmin dmax min = = = 0.05 mm l / d = 6 / 75 = (lose enough) r = d / = 75 / = 7.5 mm r / = 7.5 / 0.05 = 750 N = 70 / 60 = 1 rev/s W 000 P = = = MPa ld 75(6) ig. 1-1: SAE 0 at 60 C, µ = 18.5 mpa s S µ N 6 P 0.741(10 ) r 18.5(10 )(1) = = 750 = rom igures 1-16, 1-18 and 1-1: / = 0.9, f r / = 5.1, P / pmax = 0.15 = 0.9(0.05) = mm f = 5.1 / 750 = T = f Wr = ()(7.5) = 0.51 N m The heat loss rate equals the rate of work on the film H loss = πt N = π(0.51)(1) = 8.5 W p max = 0.741/0.15 =.5 MPa ig. 1-1: SAE 40 at 60 C, µ = 7 mpa s Shigley s MED, 10 th edition Chapter 1 Solutions, Page 6/6

7 S = 0.169(7)/18.5 = 0.8 rom igures 1-16, 1-18 and 1-1: / = 0.4, f r / = 8.5, P / pmax = 0.8 = 0.4(0.05) = 0.01 mm f = 8.5 / 750 = T = f Wr = 0.011()(7.5) = 0.85 N m Hloss = π TN = π (0.85)(1) = 64 W pmax = / 0.8 = 1.95 MPa 1-9 Given: d max = mm, b min = mm, l = 8 mm, W =.4 kn, N = 900 rev/min, and SAE 40 at 65 C. bmin dmax min = = = 0.05 mm r = d / = 56 / = 8 mm r / = 8 / 0.05 = 110 l / d = 8 / 56 = 0.5, N = 900 / 60 = 15 rev/s 400 P = = 1.5 MPa 8(56) ig. 1-1: SAE 40 at 65 C, µ = 0 mpa s µ N 6 P 1.5(10 ) r 0(10 )(15) S = = 110 = 0.69 rom igures 1-16, 1-18, 1-19 and 1-0: / = 0.44, f r / = 8.5, Qs / Q = 0.71, Q / ( rnl) = 4.85 = 0.44(0.05) = mm f = 8.5 / 1000 = T = f Wr = (.4)(8) = 0.51 N m H = π TN = π (0.51)(15) = 48.1 W Q = 4.85rNl = 4.85(8)(0.05)(15)(8) = 146 mm /s Qs = 0.71(146) = 101 mm /s 1-10 Consider the bearings as speified by minimum f : maximum W: d b tb t, d 0 0 tb + t, d 0 d b + Shigley s MED, 10 th edition Chapter 1 Solutions, Page 7/6

8 and differing only in d and d. Preliminaries: l / d = 1 P = W ld = = N = 600 / 60 = 60 rev/s / 700 / (1.5 ) 448 psi ig. 1-16: minimum f : S 0.08 maximum W: S 0.0 ig. 1-1: Eq. (1-7): µ = 1.8(10 6 ) reyn µn/p = 1.8(10 6 )(60/448) = 0.185(10 6 ) r S = µn / P or minimum f : r 0.08 = = (10 ) = 0.65 / 658 = in If this is min, b d = (0.001) = 0.00 in The median learane is t + t t + t = min + = d b d b and the learane range for this bearing is t = whih is a funtion only of the toleranes. d + t b or maximum W: r 0. = = (10 ) = 0.65 / 1040 = in If this is min Shigley s MED, 10 th edition Chapter 1 Solutions, Page 8/6

9 b d = min = (0.0005) = in td + tb td + tb = min + = td + tb = The differene (mean) in learane between the two learane ranges, range, is t + t t + t range = = in d b d b or the minimum f bearing or or the maximum W bearing b d = 0.00 in d = b 0.00 in d = b in or the same b, t b and t d, we need to hange the journal diameter by in. d d = b ( b 0.00) = in Inreasing d of the minimum frition bearing by in, defines d of the maximum load bearing. Thus, the learane range provides for bearing dimensions whih are attainable in manufaturing Given: SAE 40, N = 10 rev/s, T s = 140, l/d = 1, d =.000 in, b =.00 in, W = 675 lbf. bmin dmax.00 min = = = in r = d / = / = 1.5 in r / = 1.5 / = 1000 W 675 P = = = 75 psi ld () Trial #1: rom igure 1-1 for T = 160, µ =.5 µ reyn, T = ( ) = 40 6 µ N r.5(10 )(10) S = = 1000 = P 75 Shigley s MED, 10 th edition Chapter 1 Solutions, Page 9/6

10 rom ig. 1-4, 9.70 T = + + = P P 75 T =.16 =.16 = Disrepany = = 15.6 Trial #: T = 150, µ = 4.5 µ reyn, (0.4667) (0.4667).16 T = ( ) = S = = 75 rom ig. 1-4, T = + + = P P 75 T =.97 =.97 = Disrepany = = 10.7 Trial #: T = 154, µ = 4 µ reyn, (0.6) (0.6).97 T = ( ) = S = = 75 rom ig. 1-4, T = + + = P P 75 T =.57 =.57 = (0.5) (0.5).57 Disrepany = = 0.4 O.K. T av = / = 154 T1 = Tav T / = 154 (8 / ) = 140 T = Tav + T / = (8 / ) = 168 S = 0.4 rom igures 1-16, 1-18, to 1-0: Shigley s MED, 10 th edition Chapter 1 Solutions, Page 10/6

11 f r Q Q s = 0.75, = 11, rn l =.6, Q = 0. = 0.75(0.0015) = in f 11 = = T = f Wr = ()(40) = 0.9 N m π f WrN π Hloss = = = Btu/s Q =.6rN l =. 6(1.5)0.0015(10) = 0.4 in /s Qs = 0.(0.4) = in /s 1-1 Given: d =.5 in, b =.504 in, min = 0.00 in, W = 100 lbf, SAE = 0, T s = 110, N = 110 rev/min, and l =.5 in. P = W/(ld) = 100/(.5) = 19 psi, N = 110/60 = rev/s or a trial film temperature, let T f = 150 Table 1-1: Eq. (1-7): ig. 1-4: µ = exp[171.6/( )] =.441 µ reyn 6 r µ N.5 / S = = = P ( ) ( T = ) = 17.9 T T av f T 17.9 = Ts + = = T = = 1.0 av whih is not 0.1 or less, therefore try averaging for the new trial film temperature, let ( T f ) new = = 14.5 Proeed with additional trials using a spreadsheet (table also shows the first trial) Shigley s MED, 10 th edition Chapter 1 Solutions, Page 11/6

12 Trial T f µ' S T T av T f T av New T f Note that the onvergene begins rapidly. There are ways to speed this, but at this point they would only add omplexity. (a) µ = S = (10 ), rom ig. 1-16: 0.48, 0.48(0.00) in = = = rom ig. 1-17: φ = 56 (b) e = h 0 = = in f r () rom ig. 1-18: = 4.10, f = 4.10(0.00 /1.5) = (d) T = f Wr = (100)(1.5) = 9.84 lbf in H π T N π (9.84)(110 / 60) = = = 778(1) 778(1) 0.14 Btu/s = (e) rom ig. 1-19: Q rnl Q = 4.16(1.5)(0.00) (.5) = in /s 60 Qs rom ig. 1-0: 0.6, Qs 0.6(0.485) 0.91 in /s Q = = = (f) rom ig. 1-1: P p max rom ig. 1-: φ = 16 pmax ( ld ) W / 100 /.5 = 0.45, pmax = = = 47 psi Shigley s MED, 10 th edition Chapter 1 Solutions, Page 1/6

13 (g) rom ig. 1-: φ = 8 p0 (h) rom the trial table, T f = 1.8 (i) With T = 7.5 from the trial table, T s + T = = Given: d = 1.50 in, t d = in, b = 1.5 in, t b = 0.00 in, l = 1.5 in, W = 50 lbf, N = 1750 rev/min, SAE 10 lubriant, sump temperature T s = 10. P = W/(ld) = 50/1.5 = 160 psi, N = 1750/60 = 9.17 rev/s or the learane, = 0.00 ± in. Thus, min = in, median = 0.00 in, and max = 0.00 in. or min = in, start with a trial film temperature of T f = 15 Table 1-1: Eq. (1-7): ig. 1-4: µ = exp[1157.5/( )] =.4 µ reyn 6 r µ N 1.5 / S = = = P ( ) ( T = ) =.9 T T av f T.9 = Ts + = 10 + = 11.4 T = =.6 av whih is not 0.1 or less, therefore try averaging for the new trial film temperature, let ( T f ) new = = 1. Proeed with additional trials using a spreadsheet (table also shows the first trial) Trial T f µ' S T T av T f T av New T f Shigley s MED, 10 th edition Chapter 1 Solutions, Page 1/6

14 With T f = 1.1, T = 4.0, µ =.58 µ reyn, S = , T max = T s + T = = ig. 1-16: h 0 / = 0.50, h 0 = 0.50(0.001) = in = 1 h 0 / = = 0.05 in ig. 1-18: r f / = 4.5, f = 4.5/(0.65/0.001) = ig. 1-19: Q/(rNl) = 4.1, Q = 4.1(0.65)0.001(9.17)1.5 = in /s ig. 1-0: Q s /Q = 0.58, Q s = 0.58(0.0941) = in /s The above an be repeated for median = 0.00 in, and max = 0.00 in. The results are shown below. min in median 0.00 in max 0.00 in T f µ S Τ T max h 0 / h fr/ f Q/(rNl) Q Q s /Q Q s Computer programs will vary In a step-by-step fashion, we are building a skill for natural irulation bearings. Given the average film temperature, establish the bearing properties. Given a sump temperature, find the average film temperature, then, establish the Shigley s MED, 10 th edition Chapter 1 Solutions, Page 14/6

15 bearing properties. Now we aknowledge the environmental temperature s role in establishing the sump temperature. Se. 1-9 and Ex. 1-5 address this problem. Given: d max =.500 in, b min =.504 in, l/d = 1, N = 110 rev/min, SAE 0 lubriant, W = 00 lbf, A = 60 in, T = 70, and α = lbf load with minimal learane: We will start by using W = 600 lbf (n d = ). The task is to iteratively find the average film temperature, T f, whih makes H gen and H loss equal. bmin dmax min = = = 0.00 in N = 110/60 = rev/s P W 600 = = = ld.5 96 psi S 6 r µ N 1.5 µ = = = P µ Table 1-1: µ = exp[171.6/(t f + 95)] H gen 545 f r 545 = WN = f r = 54. f r ( Tf ) ( f 70 ) h A.7 60 / 144 = ( f ) = α CR Hloss T T T = Start with trial values of T f of 0 and 40. Trial T f µ S f r/ H gen H loss 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 = T f +.1. Setting this equal to H loss and solving for T f gives T f = 7. Trial T f µ S f r/ H gen H loss Shigley s MED, 10 th edition Chapter 1 Solutions, Page 15/6

16 whih is satisfatory. Table 1-16: h 0 / = 0.1, h 0 = 0.1 (0.00) = in ig. 1-4: ( ) ( T = ) = 6.1 T 1 = T s = T f T = 7 6.1/ =.8 T max = T 1 + T = = 40.1 Trumpler s design riteria: d = (.5) = in < h 0 O.K. T max = 40.1 < 50 O.K. W st 00 ld =.5 = < 48 psi 00 psi O. K. n d = (assessed at W = 600 lbf) O.K. We see that the design passes Trumpler s riteria and is deemed aeptable. or an operating load of W = 00 lbf, it an be shown that T f = 19., µ = 0.78, S = 0.118, f r/ =.09, H gen = H loss = 84 Btu/h, h 0 =, T = 10.5, T 1 = 4.6, and T max = Given: d = in, b = in, SAE 0, T s = 10, p s = 50 psi, N = 000/60 =. rev/s, W = 4600 lbf, bearing length = in, groove width = 0.50 in, and H loss 5000 Btu/hr. bmin dmax min = = = in r = d/ =.500/ = in r / = 1.750/0.005 = 700 l = ( 0.5)/ = in l / d = 0.875/.500 = 0.5 W 4600 P = = = 4rl psi Shigley s MED, 10 th edition Chapter 1 Solutions, Page 16/6

17 Trial #1: Choose (T f ) 1 = 150. rom Table 1-1, µ = exp[160.0/( )] =.6 µ reyn 6 r µ N.6(10 )(.) S = = 700 = P 751 rom igs and 1-18: = 0.9, f r/ =.6 rom Eq. (1-4), 0.01( f r / ) SW T = ò p r s = = (0.9) T av = T s + T / = / = Trial #: Choose (T f ) = 160. rom Table 1-1 µ = exp[160.0/( )] =.9 µ reyn.9 S = = rom igs and 1-18: = 0.915, f r/ = T = = T av = / = 14.5 Trial #: Thus, the plot gives (T f ) = rom Table 1-1 µ = exp[160.0/( )] =.4 µ reyn Shigley s MED, 10 th edition Chapter 1 Solutions, Page 17/6

18 .4 S = = rom igs and 1-18: = 0.905, f r/ = T = = T av = / = Result is lose. Choose T f = = 15.1 Try 15 Table 1-1: µ = exp[160.0/( )] =.47 µ reyn.47 S = = f r =.4, ò = 0.90, = (.4) ( 4600 ) T = = ( 0.90 ) + 50( ) Tav = / = 15.1 O.K. h 0 = 0.098(0.005) = in T max = T s + T = = Eq. (1-): Q s ( 50) 1.750( ) π psr π = ( ò ) = µ l (.47) 10 ( 0.875) = in /s Trumpler s design riteria: H loss = ρ C p Q s T = 0.011(0.4)1.047(64.1) = Btu/s = 0.877(60 ) = 160 Btu/h O.K (.5) = in > Not O.K. T max = < 50 O.K. P st = 751 psi > 00 psi Not O.K. n = 1, as done Not O.K Given: d = mm, b = mm, SAE 0, T s = 55 C, p s = 00 kpa, N = 880/60 = 48 rev/s, W = 10 kn, bearing length = 55 mm, groove width = 5 mm, and Shigley s MED, 10 th edition Chapter 1 Solutions, Page 18/6

19 H loss 00 W. b d r = d/ = 50/ = 5 mm min max min = = = 0.04 mm r / = 5/0.04 = 595 l = (55 5)/ = 5 mm l / d = 5/50 = 0.5 W P = = = 4 MPa 4rl 4( 5) 5 Trial #1: Choose (T f ) 1 = 79 C. rom ig. 1-1, µ = 1 mpa s. µ N 6 r 1(10 )(48) S = = 595 = P 4(10 ) rom igs and 1-18: = 0.85, f r/ =. rom Eq. (1-5), 978(10 ) ( f r / ) SW T = ò p r 6 4 s 6 978(10 ).(0.055)(10 ) = 76. C 1 1.5(0.85) = 00(5) 4 + T av = T s + T / = / = 9. C Trial #: Choose (T f ) = 100 C. rom ig. 1-1, µ = 7 mpa s. 7 S = = rom igs and 1-18: = 0.90, f r/ = (10 ) 1.6(0.097)(10 ) T = = 6.9 C (0.9) 00(5) T av = / = 68.5 C Shigley s MED, 10 th edition Chapter 1 Solutions, Page 19/6

20 Trial #: Thus, the plot gives (T f ) = 85.5 C. rom ig. 1-1, µ = 10.5 mpa s S = = rom igs and 1-18: = 0.87, f r/ =. T = = 58.9 C 1 1.5(0.87 ) 00(5) 6 978(10 ).(0.0457)(10 ) 4 + T av = / = 84.5 C Result is lose. Choose T f = = 85 C ig. 1-1: µ = 10.5 mpa s 10.5 S = = f r ò = 0.87, =., = (10 ).(0.0457)(10 ) T = = 58.9 C or (0.87 ) 00(5 ) T = / = 84.5 C O.K. av rom Eq. (1-) h 0 = 0.1(0.04) = mm or in T max = T s + T = = 11.9 C or 7 Q s ( 00) 5( 0.04 ) π psr π = ( ò ) = ( 0.87 ) 6 µl ( 10.5) 10 ( 5) = 156 mm /s = = 0.19 in /s H loss = ρ C p Q s T = 0.011(0.4)0.19(18) = 0.48 Btu/s = 1.05(0.48) = 0.65 kw = 65 W not O.K. Trumpler s design riteria: (50/5.4) = in > h 0 Not O.K. T max = 7 O.K. P st = 4000 kpa or 581 psi > 00 psi Not O.K. n = 1, as done Not O.K So far, we have performed elements of the design task. Now let s do it more ompletely. The values of the unilateral toleranes, t b and t d, reflet the routine apabilities of the Shigley s MED, 10 th edition Chapter 1 Solutions, Page 0/6

21 bushing vendor and the in-house apabilities. While the designer has to live with these, his approah should not depend on them. They an be inorporated later. irst we shall find the minimum size of the journal whih satisfies Trumpler s onstraint of P st 00 psi. W Pst = 00 dl W W 00 d d l / d 600( l / d) d min 900 = = 1.7 in (00)(0.5) 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 onstraints are, we will set t b = t d = 0, and thereby shrink the design window to a point. We set d =.000 in b = d + min = d + n d = (This makes Trumpler s n d tight) and onstrut a table. b d T * f T max h 0 P st T max n fom *Sample alulation for the first entry of this olumn. Iteration yields: T = 15.5 With T = 15.5, from Table 1-1 f f Shigley s MED, 10 th edition Chapter 1 Solutions, Page 1/6

22 µ N S = + = 900 = 000 / 60 = 50 rev/s, P = = 5 psi (10 )(50) = = (10 ) exp[171.6 / ( )] 0.817(10 ) reyn rom igs and 1-18: = 0.7, f r/ = 5.5 Eq. (1 4): av 0.01(5.5)(0.18)(900 ) 4 [ (0.7 )](0)(1 ) = = T T = 10 + = or the nominal -in bearing, the various learanes show that we have been in ontat with the reurving of (h 0 ) min. The figure of merit (the parasiti frition torque plus the pumping torque negated) is best at = in. or the nominal -in bearing, we will plae the top of the design window at min = 0.00 in, and b = d + (0.00) =.004 in. At this point, add the b and d unilateral toleranes: d =.000 in, b =.004 in Now we an hek the performane at min,, and max. Of immediate interest is the fom of the median learane assembly, 9.8, as ompared to any other satisfatory bearing ensemble. If a nominal in bearing is possible, onstrut another table with t b = 0 and t d = 0. b d T f T max h 0 P st T max n fom The range of learane is < < in. That is enough room to fit in our design window. d = in, b = in The ensemble median assembly has a fom = 9.1. We just had room to fit in a design window based upon the (h 0 ) min onstraint. urther Shigley s MED, 10 th edition Chapter 1 Solutions, Page /6

23 redution in nominal diameter will prelude any smaller bearings. A table onstruted for a d = in journal will prove this. We hoose the nominal in bearing ensemble beause it has the largest figure of merit This is the same as Prob but uses design variables of nominal bushing bore b and radial learane. The approah is similar to that of Prob and the tables will hange slightly. In the table for a nominal b = in, note that at = 0.00 in the onstraints are loose. Set or the ensemble b = in d = (0.00) = in b = in, d = in Analyze at min = 0.00 in, = in and max = in = 0.00 in: T = 18.4, µ =.160 µ reyn, S = 0.097, H = 105 Btu/h At min and the Trumpler onditions are met. f loss At = in: T f = 10, µ =.87 µ reyn, S = 0.005, H loss = 1106 Btu/h, fom = 9.46 and the Trumpler onditions are O.K. At max = in: T f = 15.68, µ = 4.5µ reyn, S = , H loss = 119 Btu/h and the Trumpler onditions are O.K. The ensemble figure of merit is slightly better; this bearing is slightly smaller. The lubriant ooler has suffiient apaity. 1-0 Table 1-1: µ (µ reyn) = µ 0 (10 6 ) exp [b / (T + 95)] b and T in The onversion from µ reyn to mpa s is given on p. 61. or a temperature of C degrees Celsius, T = 1.8 C +. Substituting into the above equation gives µ (mpa s) = 6.89 µ 0 (10 6 ) exp [b / (1.8 C )] = 6.89 µ 0 (10 6 ) exp [b / (1.8 C + 17)] or SAE 50 oil at 70 C, from Table 1-1, µ 0 = (10 6 ) reyn, and b = rom the equation, µ = 6.89(0.0170) 10 6 (10 6 ) exp {1509.6/[1.8(70) + 17]} Shigley s MED, 10 th edition Chapter 1 Solutions, Page /6

24 = 45.7 mpa s rom ig. 1-1, µ = 9 mpa s The figure gives a value of about 15 % lower than the equation. 1-1 Originally Doubled, d d =.000 in, b =.005 in = in, b = in The radial load quadrupled to 600 lbf when the analyses for parts (a) and (b) were arried out. Some of the results are: Part µ S T f r/ Q f s h 0 / H loss h 0 Trumpler (a) (b) The side flow Q s differs beause there is a term and onsequently an 8-fold inrease. H loss is related by a 9898/17 or an 8-fold inrease. The existing h 0 is related by a -fold inrease. Trumpler s (h 0 ) min is related by a 1.86-fold inrease. 1- Given: Oiles SP 500 alloy brass bushing, L = 0.75 in, D = 0.75 in, T = 70, = 400 lbf, N = 50 rev/min, and w = in. h 0 f Table 1-8: K = 0.6(10 10 ) in min/(lbf ft h) P = / (DL) = 400/ [0.75(0.75)] = 711 psi V = πdn/ 1 = π (0.75)50/1 = 49.1 ft/min rom Table 1-10, interpolation gives V f f 1 = > f 1 = Table 1-11: f = 1.0 Shigley s MED, 10 th edition Chapter 1 Solutions, Page 4/6

25 Table 1-1: PV max = psi ft/min, P max = 560 psi, V max = 100 ft/min P = psi 560 psi O. K. π DL = π 0.75 = < max PV = 711 (49.1) = psi ft/min < psi ft/min O.K. Eq. (1-) an be written as w = 4 f1 fk Vt π DL Solving for t, t 1 π ( 0.75) 0.75( 0.004) 10 π DLw = = 4 f f KV = 1056 h = = min Cyles = Nt = 50 (6 400) = 15.9 (10 6 ) yles 1- Given: Oiles SP 500 alloy brass bushing, w max = 0.00 in for 1000 h, N = 00 rev/min, = 100 lbf, h CR =.7 Btu/(h ft o ), T max = 00 o, f s = 0.0, and n d =. Using Eq. (1-8) with n d for, f s = 0.0 from Table 1-9, and h CR =.7 Btu/(h ft ο ), gives L 70 fsnd N 70(0.0)(100)00 = = 1.79 in Jh 778(.7)(00 70) CR T T ( f ) rom Table 1-1, the smallest available bushing has an ID = 1 in, OD = 1 in, and L = 8 in. With L/D = /1 =, this is inside of the reommendations of Eq. (1-). Thus, for the first trial, try the bushing with ID = 1 in, OD = 1 in, and L = in. Thus, 8 Eq. (1-1): P max = 4 nd 4 (100) 17.psi 560 psi (OK) π DL = π 1() = < nd (100) P = = = 100 psi DL 1() Shigley s MED, 10 th edition Chapter 1 Solutions, Page 5/6

26 Eq. (1-9): V π DN π (1) 00 = = = 5.4 ft/min < 100 ft/min (OK) 1 1 PV = 100(5.4) = 540 psi ft/min < psi ft/min (OK) rom Table 1-10, interpolation gives V f f 1 = > f 1 = Eq. (1-), with Tables 1-8 and 1-10: w 11 f f Kn N t 1.445(1) 6 10 (100) 00(1000) L () 1 d = = = < in in (OK) Answer Selet ID = 1 in, OD = 1 in, and L = in. 8 Shigley s MED, 10 th edition Chapter 1 Solutions, Page 6/6