Budynas ch12

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Unformatted text preview: Chapter 12 12-1 Given dmax = 1.000 in and bmin = 1.0015 in, the minimum radial clearance is cmin = Also 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 S = (6672 ) 8(10-6 )(18.33) = 0.261 250 h 0 /c = 0.595 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. f = 5.8 5.8 = = 0.0087 r/c 667 Eq. (12-7): Fig. 12-16: Fig. 12-19: Fig. 12-18: Fig. 12-20: The power loss in Btu/s is H= 2 f W r N 2(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. 12-2 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) (6252 )(10)(10-6 )(19.167) = 0.585 128 l/d = 2.5/1.25 = 2 S= Chapter 12 305 The interpolation formula of Eq. (12-16) will have to be used. From Figs. 12-16, 12-21, and 12-19 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 1 l/d = , 4 h o /c = 0.54, h o /c = 0.31, P/ pmax = 0.42, P/ pmax = 0.28, Q = 4.4 rcNl Q = 5.25 rcNl Equation (12-16) is easily programmed by code or by using a spreadsheet. The results are: l/d h o /c P/ pmax Q/rcNl y 0.96 0.84 3.09 y1 0.77 0.52 3.60 y1/2 0.54 0.42 4.40 y1/4 0.31 0.28 5.25 yl/d 0.88 0.64 3.28 2 2 2 h o = 0.88(0.001) = 0.000 88 in Ans. 128 = 200 psi Ans. 0.64 pmax = Q = 3.28(0.625)(0.001)(19.167)(2.5) = 0.098 in3 /s Ans. 12-3 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. 12-12: SAE 10, = 1.75 reyn S = (6002 ) 1.75(10-6 )(10) = 0.0354 177.78 P/ pmax = 0.21 Figs. 12-16 and 12-21: h o /c = 0.11, h o = 0.11(0.0025) = 0.000 275 in Ans. pmax = 177.78/0.21 = 847 psi Ans. 306 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design Fig. 12-12: SAE 40, = 4.5 reyn 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. 12-4 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. 12-14: SAE 10W, = 2.1 reyn 2.1(10-6 )(12.5) = 0.0984 S = (500 ) 66.7 2 From Figs. 12-16 and 12-21: 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. 12-14: SAE 20W-40, = 5.05 reyn S = (5002 ) From Figs. 12-16 and 12-21: 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 5.05(10-6 )(12.5) = 0.237 66.7 Chapter 12 307 12-5 cmin = 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) 3.75(10-6 )(13.3) = 0.115 300 Fig. 12-12: SAE 20, = 3.75 reyn S = (8332 ) From Figs. 12-16, 12-18 and 12-19: 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 2 f W r N 2(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 12-6 cmin = 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 = 50 mPa s, 2(106 ) From Figs. 12-16, 12-18 and 12-20: 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 308 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design The power loss due to friction is H = 2 T N = 2(0.1125)(20) = 14.14 W Ans. Q s = 0.57Q 12-7 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= P= 1120 = 18.67 rev/s 60 2750 = 1.833 MPa 30(50) 60(10-3 )(18.67) = 0.22 1.833(106 ) S = (6002 ) l 50 = = 1.67 d 30 This l/d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16). From Fig. 12-16, the h o /c values are: y1/4 = 0.18, Substituting into Eq. (12-16), From Fig. 12-18, the f r/c values are: y1/4 = 7.4, Substituting into Eq. (12-16), y1/2 = 6.0, y1 = 5.0, y = 4.0 y1/2 = 0.34, y1 = 0.54, y = 0.89 ho = 0.659 c fr = 4.59 c y1/2 = 5.05, y1 = 4.05, Q = 3.605 rcN l y = 2.95 From Fig. 12-19, the Q/(rcNl) values are: y1/4 = 5.65, Substituting into Eq. (12-16), 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. Chapter 12 309 12-8 cmin = bmin - dmax 75.10 - 75 = = 0.05 mm 2 2 l/d = 36/75 = 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 2000 P= = 0.741 MPa 75(36) Fig. 12-13: SAE 20, = 18.5 mPa s S = (7502 ) From Figures 12-16, 12-18 and 12-21: 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 = 2 T N = 2(0.51)(12) = 38.5 W Ans. pmax = 0.741/0.315 = 2.35 MPa Ans. Fig. 12-13: SAE 40, = 37 MPa s S = 0.169(37)/18.5 = 0.338 From Figures 12-16, 12-18 and 12-21: 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 = 2 T N = 2(0.85)(12) = 64 W Ans. pmax = 0.741/0.38 = 1.95 MPa Ans. 12-9 bmin - dmax 50.05 - 50 = = 0.025 mm 2 2 r = d/2 = 50/2 = 25 mm 18.5(10-3 )(12) = 0.169 0.741(106 ) 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) 310 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design Fig. 12-13: SAE 30, = 34 mPa s S = (10002 ) 34(10-3 )(14) = 0.2975 1.6(106 ) Q s /Q = 0.74, Q/(rcNl) = 4.9 From Figures 12-16, 12-18, 12-19 and 12-20: h o /c = 0.40, f r/c = 7.8, 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 = 2 T N = 2(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. 12-10 Consider the bearings as specified by minimum f: maximum W: and differing only in d and d . Preliminaries: l/d = 1 P = 700/(1.252 ) = 448 psi N = 3600/60 = 60 rev/s Fig. 12-16: minimum f : maximum W: Fig. 12-12: S = 0.08 S = 0.20 = 1.38(10-6 ) reyn N /P = 1.38(10-6 )(60/448) = 0.185(10-6 ) Eq. (12-7): r = c For minimum f: r = c 0.08 = 658 0.185(10-6 ) S N/P + d-t0 , d + d-td0 , +t b-0b +t b-0b . c = 0.625/658 = 0.000 950 = 0.001 in 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 td + tb 2 and the clearance range for this bearing is c= 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 2 td + tb td + tb = 0.0005 + 2 2 The difference (mean) in clearance between the two clearance ranges, crange , is crange = 0.001 + td + tb td + tb - 0.0005 + 2 2 = 0.0005 in 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. For the minimum f bearing 312 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design 12-11 Given: SAE 30, N = 8 rev/s, Ts = 60C, 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 12-13 for T = 81C, = 12 mPa s T = 2(81C - 60C) = 42C S = (10002 ) From Fig. 12-24, 0.120 T = 0.349 + 6.009(0.2047) + 0.0475(0.2047) 2 = 1.58 P T = 1.58 0.469 0.120 = 6.2C 12(10-3 )(8) = 0.2047 0.469(106 ) Discrepancy = 42C - 6.2C = 35.8C Trial #2: From Figure 12-13 for T = 68C, = 20 mPa s, T = 2(68C - 60C) = 16C S = 0.2047 From Fig. 12-24, 0.120 T = 0.349 + 6.009(0.341) + 0.0475(0.341) 2 = 2.4 P T = 2.4 0.469 0.120 = 9.4C 20 12 = 0.341 Discrepancy = 16C - 9.4C = 6.6C Trial #3: = 21 mPa s, T = 65C T = 2(65C - 60C) = 10C S = 0.2047 21 12 = 0.358 Chapter 12 313 From Fig. 12-24, 0.120 T = 0.349 + 6.009(0.358) + 0.0475(0.358) 2 = 2.5 P T = 2.5 0.469 0.120 = 9.8C O.K. Discrepancy = 10C - 9.8C = 0.2C Tav = 65C Ans. T1 = Tav - T2 = Tav + S = 0.358 From Figures 12-16, 12-18, 12-19 and 12-20: ho = 0.68, c f r/c = 7.5, Q = 3.8, rcN l Qs = 0.44 Q T /2 = 65C - (10C/2) = 60C T /2 = 65C + (10C/2) = 70C 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 = 2 T N = 2(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. 12-12 Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110F, N = 1120 rev/min, and l = 2.5 in. For a trial film temperature T f = 150F Tf 150 2.421 Tav = Ts + S 0.0921 T (From Fig. 12-24) 18.5 T 18.5F = 110F + = 119.3F 2 2 T f - Tav = 150F - 119.3F which is not 0.1 or less, therefore try averaging (T f ) new = 150F + 119.3F = 134.6F 2 314 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design Proceed with additional trials Trial Tf 150.0 134.6 128.1 125.5 124.5 124.1 124.0 New Tf 134.6 128.1 125.5 124.5 124.1 124.0 123.9 2.421 3.453 4.070 4.255 4.471 4.515 4.532 S 0.0921 0.1310 0.1550 0.1650 0.1700 0.1710 0.1720 T 18.5 23.1 25.8 27.0 27.5 27.7 27.8 Tav 119.3 121.5 122.9 123.5 123.8 123.9 123.7 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) = 4.541(10-6 ) reyn, From Fig. 12-16: S = 0.1724 h o = 0.482(0.002) = 0.000 964 in ho = 0.482, c From Fig. 12-17: = 56 Ans. (b) e = c - h o = 0.002 - 0.000 964 = 0.001 04 in Ans. (c) From Fig. 12-18: 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= 2 T N 2(9.84)(1120/60) = = 0.124 Btu/s Ans. 778(12) 778(12) Q = 4.16, rcNl Q = 4.16(1.25)(0.002) = 0.485 in3 /s Ans. From Fig. 12-20: (f) From Fig. 12-21: 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) pmax = 16 Ans. (g) p0 = 82 Ans. (h) T f = 123.9F Ans. (i) Ts + T = 110F + 27.8F = 137.8F Ans. 1120 (2.5) 60 (e) From Fig. 12-19: Chapter 12 315 12-13 Given: d = 1.250 in, td = 0.001in, b = 1.252 in, tb = 0.003in, l = 1.25 in, W = 250 lbf, N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120F. Below is a partial tabular summary for comparison purposes. cmin 0.001 in Tf T Tmax S fr c Q rcN j l Qs Q ho c f Q Qs ho 132.2 24.3 144.3 2.587 0.184 0.499 4.317 4.129 0.582 0.501 0.006 9 0.094 1 0.054 8 0.000 501 c 0.002 in 125.8 11.5 131.5 3.014 0.053 7 0.775 0 1.881 4.572 0.824 0.225 0.006 0.208 0.172 0.000 495 cmax 0.003 in 124.0 7.96 128.0 3.150 0.024 9 0.873 1.243 4.691 0.903 0.127 0.005 9 0.321 0.290 0.000 382 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 journal-bushing bearing, but an ensemble of bearings. 12-14 12-15 Computer programs will vary--Fortran based, MATLAB, spreadsheet, etc. 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. 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. 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. (12-15) T = Hloss = 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 0.103P( f r/c) (1 - 0.5Q s /Q)[Q/(rcN j l)] h CR A(T f - T ) 1+ (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 0.1F, 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) 2.254 2.004 1.753 tb (in) 0.004 0.004 0.003 Otherwise, the design option includes reducing l/d to save on the cost of journal machining and vender-supplied bushings. Chapter 12 317 12-16 Continue to build a skill with pressure-fed 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. (12-24), 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 Tav Tf 2 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 0.1F, plot the results for Trials #2 and #3 and try again. If they are within 0.1F, 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. 12-17 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) 318 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design Trial #1: Choose (T f ) 1 = 79C. From Fig. 12-13, = 13 mPa s. S = (5952 ) 13(10-3 )(48) = 0.055 4000(103 ) From Figs. 12-18 and 12-16: From Eq. (12-25), fr = 2.3, = 0.85. c 978(106 ) ( f r/c)SW 2 T = 1 + 1.5 2 ps r 4 = 2.3(0.055)(102 ) 978(106 ) 1 + 1.5(0.85) 2 200(25) 4 T /2 = 55C + (76C/2) = 93C = 76.0C Tav = Ts + Trial #2: Choose (T f ) 2 = 100C. From Fig. 12-13, = 7 mPa s. 7 = 0.0296 S = 0.055 13 From Figs. 12-18 and 12-16: T = fr = 1.6, c = 0.90 1.6(0.0296)(102 ) 978(106 ) = 26.8C 1 + 1.5(0.9) 2 200(25) 4 26.8C = 68.4C 2 Tav (79 C, 93 C) Tf (100 C, 100 C) Tav = 55C + Tav 100 90 80 70 (79 C, 79 C) (100 C, 68.4 C) 85 C 60 70 80 90 100 Tf Trial #3: Thus, the plot gives (T f ) 3 = 85C. From Fig. 12-13, = 10.8 mPa s. S = 0.055 From Figs. 12-18 and 12-16: 10.8 13 = 0.0457 fr = 2.2, = 0.875 c 2.2(0.0457)(102 ) 978(106 ) = 58.6C T = 1 + 1.5(0.8752 ) 200(25) 4 58.6C = 84.3C 2 Tav = 55C + Result is close. Choose 85C + 84.3C = 84.7C Tf = 2 Chapter 12 319 Fig. 12-13: = 10.8 MPa s S = 0.055 fr = 2.23, c T = 10.8 13 = 0.0457 ho = 0.13 c = 0.874, 2.23(0.0457)(102 ) 978(106 ) = 59.5C 1 + 1.5(0.8742 ) 200(254 ) 59.5C = 84.7C 2 O.K. Tav = 55C + From Eq. (12-22) Q s = (1 + 1.5 2 ) ps rc3 3l (200)(0.0423 )(25) 3(10)(10-6 )(25) = [1 + 1.5(0.8742 )] = 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 = 55C + 63.7C = 118.7C or 245.7F 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. 12-18 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 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 Pst 300 psi. W Pst = 300 2dl W 2d 2 l /d 300 d W 600(l /d) dmin = 900 = 1.73 in 2(300)(0.5) 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. We set d = 2.000 in b = d + 2cmin = d + 2c nd = 2 and construct a table. c 0.0010 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019 0.0020 b 2.0020 2.0022 2.0024 2.0026 2.0028 2.0030 2.0032 2.0034 2.0036 2.0038 2.0040 d 2 2 2 2 2 2 2 2 2 2 2 T f* 215.50 206.75 198.50 191.40 185.23 179.80 175.00 171.13 166.92 163.50 160.40 Tmax 312.0 293.0 277.0 262.8 250.4 239.6 230.1 220.3 213.9 206.9 200.6 ho 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 (This makes Trumpler's n d 2 tight) *Sample calculation for the first entry of this column. T f = 215.5F Iteration yields: With T f = 215.5F, from Table 12-1 = 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. 12-16 and 12-18: Eq. (1224): TF = = 0.7, f r/c = 5.5 0.0123(5.5)(0.182)(9002 ) = 191.6F [1 + 1.5(0.72 )](30)(14 ) 191.6F . Tav = 120F + = 215.8F = 215.5F 2 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 cmin = 0.002 in, and b = d + 2(0.002) = 2.004 in. At this point, add the b and d unilateral tolerances: d = 2.000+0.000 in, -0.001 b = 2.004+0.003 in -0.000 Chapter 12 321 Now we can check the performance at cmin , c, and cmax . Of immediate interest is the fom of the median clearance assembly, -9.82, as compared to any other satisfactory bearing ensemble. If a nominal 1.875 in bearing is possible, construct another table with tb = 0 and td = 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 Tmax 194.30 157.10 147.10 140.10 131.45 128.80 126.80 ho Pst Tmax fos fom -7.36 -8.64 -9.05 -9.32 -9.59 -9.63 -9.64 The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our design window. d = 1.875+0.000 in, b = 1.881+0.003 in -0.001 -0.000 The ensemble median assembly has fom = -9.31. We just had room to fit in a design window based upon the (h o ) 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. 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 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 d = 1.869+0.000 -0.001 Analyze at cmin = 0.003, c = 0.004 in and cmax = 0.005 in At cmin = 0.003 in: T f = 138.4F, = 3.160, S = 0.0297, Hloss = 1035 Btu/h and the Trumpler conditions are met. At c = 0.004 in: T f = 130F, = 3.872, S = 0.0205, Hloss = 1106 Btu/h, fom = -9.246 and the Trumpler conditions are O.K. At cmax = 0.005 in: T f = 125.68F, = 4.325 reyn, S = 0.014 66, Hloss = 1129 Btu/h and the Trumpler conditions are O.K. The ensemble figure of merit is slightly better; this bearing is slightly smaller. The lubricant cooler has sufficient capacity. 322 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design 12-20 From Table 12-1, Seireg and Dandage, 0 = 0.0141(106 ) reyn and b = 1360.0 (reyn) = 0.0141 exp[1360/(T + 95)] = 0.0141 exp[1360/(1.8C + 127)] For SAE 30 at 79C = 6.89(0.0141) exp{1360/[1.8(79) + 127]} = 15.2 mPa s Ans. (T in F) (C in C) (C in C) (mPa s) = 6.89(0.0141) exp[1360/(1.8C + 127)] 12-21 Originally Doubled, d = 2.000+0.000 in, -0.001 d = 4.000+0.000 in, -0.002 b = 2.005+0.003 in -0.000 b = 4.010+0.006 -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 S Tf f r/c Qs h o /c Hloss ho Trumpler ho f 0.007 3.416 0.0310 135.1 0.1612 6.56 0.1032 0.897 9898 0.000 722 0.000 360 0.005 67 0.0035 3.416 0.0310 135.1 0.1612 0.870 0.1032 0.897 1237 0.000 361 0.000 280 0.005 67 The side flow Q s differs because there is a c3 term and consequently an 8-fold increase. Hloss is related by a 9898/1237 or an 8-fold increase. The existing h o is related by a 2-fold increase. Trumpler's (h o ) min is related by a 1.286-fold increase fom = -82.37 fom = -10.297 12-22 for double size for original size } an 8-fold increase for double-size From Table 12-8: K = 0.6(10-10 ) in3 min/(lbf ft h). P = 500/[(1)(1)] = 500 psi, V = D N /12 = (1)(200)/12 = 52.4 ft/min Tables 12-10 and 12-11: Table 12-12: f 1 = 1.8, f2 = 1 Pmax = 3560 psi, Vmax = 100 ft/min P Vmax = 46 700 psi ft/min, Pmax = P= 4 F 4(500) = = 637 psi < 3560 psi O.K. DL (1)(1) F = 500 psi DL V = 52.4 ft/min P V = 500(52.4) = 26 200 psi ft/min < 46 700 psi ft/min O.K. Chapter 12 323 Solving Eq. (12-32) for t t= (1)(1)(0.005) DLw = = 1388 h = 83 270 min 4 f1 f2 K V F 4(1.8)(1)(0.6)(10-10 )(52.4)(500) Ans. Cycles = N t = 200(83 270) = 16.7 rev 12-23 Estimate bushing length with f 1 = f 2 = 1, and K = 0.6(10-10 ) in3 min/(lbf ft h) Eq. (12-32): L= 1(1)(0.6)(10-10 )(2)(100)(400)(1000) = 0.80 in 3(0.002) From Eq. (12-38), with f s = 0.03 from Table 12-9 applying n d = 2 to F and h CR = 2.7 Btu/(h ft2 F) . 720(0.03)(2)(100)(400) L= = 3.58 in 778(2.7)(300 - 70) 0.80 L 3.58 in Trial 1: Let L = 1 in, D = 1 in Pmax = P= V = 4(2)(100) = 255 psi < 3560 psi O.K. (1)(1) 2(100) = 200 psi 1(1) (1)(400) = 104.7 ft/min > 100 ft/min 12 Not O.K. Trial 2: Try D = 7/8 in, L = 1 in Pmax = P= V = 4(2)(100) = 291 psi < 3560 psi (7/8)(1) 2(100) = 229 psi 7/8(1) (7/8)(400) = 91.6 ft/min < 100 ft/min 12 O.K. O.K. O.K. P V = 229(91.6) = 20 976 psi ft/min < 46 700 psi ft/min V 33 91.6 100 f1 1.3 f1 1.8 f 1 = 1.3 + (1.8 - 1.3) 91.6 - 33 100 - 33 = 1.74 L = 0.80(1.74) = 1.39 in 324 Solutions Manual Instructor's Solution Manual to Accompany Mechanical Engineering Design Trial 3: Try D = 7/8 in, L = 1.5 in Pmax = 4(2)(100) = 194 psi < 3560 psi O.K. (7/8)(1.5) 2(100) P= = 152 psi, V = 91.6 ft/min 7/8(1.5) P V = 152(91.6) = 13 923 psi ft/min < 46 700 psi ft/min D = 7/8 in, L = 1.5 in is acceptable Ans. O.K. Suggestion: Try smaller sizes. ...
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