Budynas ch12

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

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Unformatted text preview: budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 304 Chapter 12 12-1 Given dmax = 1.000 in and bmin = 1.0015 in, the minimum radial clearance is cmin = bmin − dmax 1.0015 − 1.000 = = 0.000 75 in 2 2 l/d = 1 r = 1.000/2 = 0.500 ˙ r/c = 0.500/0.000 75 = 667 N = 1100/60 = 18.33 rev/s P = W/(ld) = 250/[(1)(1)] = 250 psi Also S = (6672 ) Eq. (12-7): 8(10−6 )(18.33) = 0.261 250 h 0 /c = 0.595 Fig. 12-16: Q/(rcNl) = 3.98 f r/c = 5.8 Q s /Q = 0.5 h 0 = 0.595(0.000 75) = 0.000 466 in Ans. Fig. 12-19: Fig. 12-18: Fig. 12-20: f = 5.8 5.8 = = 0.0087 r/c 667 The power loss in Btu/s is H= 2π f W r N 2π(0.0087)(250)(0.5)(18.33) = 778(12) 778(12) = 0.0134 Btu/s Ans. Q = 3.98rcNl = 3.98(0.5)(0.000 75)(18.33)(1) = 0.0274 in3 /s Q s = 0.5(0.0274) = 0.0137 in3 /s Ans. 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) l/d = 2.5/1.25 = 2 S= (6252 )(10)(10−6 )(19.167) = 0.585 128 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 305 305 Chapter 12 The interpolation formula of Eq. (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 h o /c = 0.54, P/ pmax = 0.42, Q = 4.4 rcNl 1 l/d = , 4 h o /c = 0.31, P/ pmax = 0.28, Q = 5.25 rcNl Equation (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∞ y1 y1/2 y1/4 yl/d 2 2 2 0.96 0.84 3.09 0.77 0.52 3.60 0.54 0.42 4.40 0.31 0.28 5.25 0.88 0.64 3.28 h o = 0.88(0.001) = 0.000 88 in Ans. pmax = 128 = 200 psi Ans. 0.64 Q = 3.28(0.625)(0.001)(19.167)(2.5) = 0.098 in3 /s Ans. 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, µ = 1.75 µreyn S = (6002 ) 1.75(10−6 )(10) = 0.0354 177.78 Figs. 12-16 and 12-21: h o /c = 0.11, P/ pmax = 0.21 h o = 0.11(0.0025) = 0.000 275 in Ans. pmax = 177.78/0.21 = 847 psi Ans. budynas_SM_ch12.qxd 306 12/04/2006 15:24 FIRST PAGES Page 306 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 12-12: SAE 40, µ = 4.5 µreyn S = 0.0354 h o /c = 0.19, 4.5 1.75 = 0.0910 P/ pmax = 0.275 h o = 0.19(0.0025) = 0.000 475 in Ans. pmax = 177.78/0.275 = 646 psi Ans. 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, µ = 2.1 µreyn 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, µ = 5.05 µreyn S = (5002 ) 5.05(10−6 )(12.5) = 0.237 66.7 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 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 307 Chapter 12 12-5 bmin − dmax 2.0024 − 2 = = 0.0012 in 2 2 2 . d r = = = 1 in, l/d = 1/2 = 0.50 2 2 r/c = 1/0.0012 = 833 N = 800/60 = 13.33 rev/s 600 = 300 psi P= 2(1) cmin = Fig. 12-12: SAE 20, µ = 3.75 µreyn S = (8332 ) 3.75(10−6 )(13.3) = 0.115 300 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 2π f W r N 2π(0.004 56)(600)(1)(13.33) H= = 778(12) 778(12) = 0.0245 Btu/s Ans. Q = 5.3rcNl = 5.3(1)(0.0012)(13.33)(1) = 0.0848 in3 /s 12-6 Ans. bmin − dmax 25.04 − 25 = = 0.02 mm 2 2 r = d/2 = 25/2 = 12.5 mm, l/d = 1 ˙ r/c = 12.5/0.02 = 625 N = 1200/60 = 20 rev/s 1250 = 2 MPa P= 252 50(10−3 )(20) = 0.195 S = (6252 ) For µ = 50 mPa · s, 2(106 ) cmin = From Figs. 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 307 budynas_SM_ch12.qxd 308 12/04/2006 15:24 FIRST PAGES Page 308 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The power loss due to friction is H = 2π T N = 2π(0.1125)(20) = 14.14 W Ans. Q s = 0.57Q 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= 1120 = 18.67 rev/s 60 P= 2750 = 1.833 MPa 30(50) S = (6002 ) 60(10−3 )(18.67) = 0.22 1.833(106 ) l 50 = = 1.67 d 30 This l/d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16). From Fig. 12-16, the h o /c values are: y1/4 = 0.18, y1/2 = 0.34, y1 = 0.54, y∞ = 0.89 ho = 0.659 c Substituting into Eq. (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 fr = 4.59 c From Fig. 12-19, the Q/(rcNl) values are: y1/4 = 5.65, Substituting into Eq. (12-16), y1/2 = 5.05, y1 = 4.05, Q = 3.605 rcN l y∞ = 2.95 h o = 0.659(0.025) = 0.0165 mm Ans. f = 4.59/600 = 0.007 65 Ans. Q = 3.605(15)(0.025)(18.67)(50) = 1263 mm3 /s Ans. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 309 309 Chapter 12 12-8 bmin − dmax 75.10 − 75 = = 0.05 mm 2 2 l/d = 36/75 = 0.5 (close enough) ˙ cmin = r = d/2 = 75/2 = 37.5 mm r/c = 37.5/0.05 = 750 N = 720/60 = 12 rev/s 2000 P= = 0.741 MPa 75(36) Fig. 12-13: SAE 20, µ = 18.5 mPa · s S = (7502 ) 18.5(10−3 )(12) = 0.169 0.741(106 ) 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 = 2π T N = 2π(0.51)(12) = 38.5 W Ans. pmax = 0.741/0.315 = 2.35 MPa Ans. Fig. 12-13: SAE 40, µ = 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 = 2π T N = 2π(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 cmin = r/c = 25/0.025 = 1000 l/d = 25/50 = 0.5, N = 840/60 = 14 rev/s 2000 P= = 1.6 MPa 25(50) budynas_SM_ch12.qxd 310 12/04/2006 15:24 FIRST PAGES Page 310 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 12-13: SAE 30, µ = 34 mPa · s 34(10−3 )(14) = 0.2975 1.6(106 ) S = (10002 ) From Figures 12-16, 12-18, 12-19 and 12-20: h o /c = 0.40, f r/c = 7.8, Q s /Q = 0.74, Q/(rcNl) = 4.9 h o = 0.40(0.025) = 0.010 mm Ans. f = 7.8/1000 = 0.0078 T = f W r = 0.0078(2)(25) = 0.39 N · m H = 2π T N = 2π(0.39)(14) = 34.3 W Ans. Q = 4.9rcNl = 4.9(25)(0.025)(14)(25) = 1072 mm2 /s Q s = 0.74(1072) = 793 mm3 /s Ans. 12-10 Consider the bearings as specified by minimum f: + d−t0 , d +t b−0b maximum W: + d−td0 , +t b−0b and differing only in d and d . Preliminaries: l/d = 1 P = 700/(1.252 ) = 448 psi N = 3600/60 = 60 rev/s Fig. 12-16: S = 0.08 ˙ S = 0.20 ˙ minimum f : maximum W: Fig. 12-12: µ = 1.38(10−6 ) reyn µN /P = 1.38(10−6 )(60/448) = 0.185(10−6 ) Eq. (12-7): r = c S µN/P For minimum f: r = c 0.08 = 658 0.185(10−6 ) . c = 0.625/658 = 0.000 950 = 0.001 in budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 311 Chapter 12 311 If this is cmin , b − d = 2(0.001) = 0.002 in The median clearance is c = cmin + ¯ td + tb td + tb = 0.001 + 2 2 and the clearance range for this bearing is c= td + tb 2 which is a function only of the tolerances. For maximum W: r = c 0.2 = 1040 0.185(10−6 ) . c = 0.625/1040 = 0.000 600 = 0.0005 in If this is cmin b − d = 2cmin = 2(0.0005) = 0.001 in c = cmin + ¯ c= td + tb td + tb = 0.0005 + 2 2 td + tb 2 The difference (mean) in clearance between the two clearance ranges, crange , is td + tb td + tb − 0.0005 + 2 2 = 0.0005 in crange = 0.001 + For the minimum f bearing b − d = 0.002 in or d = b − 0.002 in For the maximum W bearing d = b − 0.001 in For the same b, tb and td , we need to change the journal diameter by 0.001 in. d − d = b − 0.001 − (b − 0.002) = 0.001 in Increasing d of the minimum friction bearing by 0.001 in, defines d of the maximum load bearing. Thus, the clearance range provides for bearing dimensions which are attainable in manufacturing. Ans. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 312 312 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 12-11 Given: SAE 30, N = 8 rev/s, Ts = 60°C, l/d = 1, W = 3000 N cmin = d = 80 mm, b = 80.08 mm, bmin − dmax 80.08 − 80 = = 0.04 mm 2 2 r = d/2 = 80/2 = 40 mm 40 r = = 1000 c 0.04 P= 3000 = 0.469 MPa 80(80) Trial #1: From Figure 12-13 for T = 81°C, µ = 12 mPa · s T = 2(81°C − 60°C) = 42°C S = (10002 ) 12(10−3 )(8) = 0.2047 0.469(106 ) From Fig. 12-24, 0.120 T = 0.349 + 6.009(0.2047) + 0.0475(0.2047) 2 = 1.58 P 0.469 0.120 T = 1.58 = 6.2°C Discrepancy = 42°C − 6.2°C = 35.8°C Trial #2: From Figure 12-13 for T = 68°C, µ = 20 mPa · s, T = 2(68°C − 60°C) = 16°C S = 0.2047 20 12 = 0.341 From Fig. 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 = 9.4°C Discrepancy = 16°C − 9.4°C = 6.6°C Trial #3: µ = 21 mPa · s, T = 65°C T = 2(65°C − 60°C) = 10°C S = 0.2047 21 12 = 0.358 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 313 313 Chapter 12 From Fig. 12-24, 0.120 T = 0.349 + 6.009(0.358) + 0.0475(0.358) 2 = 2.5 P 0.469 0.120 T = 2.5 = 9.8°C Discrepancy = 10°C − 9.8°C = 0.2°C O.K. Tav = 65°C Ans. T1 = Tav − T /2 = 65°C − (10°C/2) = 60°C T2 = Tav + T /2 = 65°C + (10°C/2) = 70°C S = 0.358 From Figures 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 h o = 0.68(0.04) = 0.0272 mm Ans. f = 7.5 = 0.0075 1000 T = f W r = 0.0075(3)(40) = 0.9 N · m H = 2π T N = 2π(0.9)(8) = 45.2 W Ans. Q = 3.8(40)(0.04)(8)(80) = 3891 mm3 /s Q s = 0.44(3891) = 1712 mm3 /s Ans. 12-12 Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110°F, N = 1120 rev/min, and l = 2.5 in. For a trial film temperature T f = 150°F Tf µ S 150 2.421 0.0921 Tav = Ts + T (From Fig. 12-24) 18.5 T 18.5°F = 110°F + = 119.3°F 2 2 T f − Tav = 150°F − 119.3°F which is not 0.1 or less, therefore try averaging (T f ) new = 150°F + 119.3°F = 134.6°F 2 budynas_SM_ch12.qxd 314 12/04/2006 15:24 FIRST PAGES Page 314 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Proceed with additional trials Trial Tf µ S T Tav New Tf 150.0 134.6 128.1 125.5 124.5 124.1 124.0 2.421 3.453 4.070 4.255 4.471 4.515 4.532 0.0921 0.1310 0.1550 0.1650 0.1700 0.1710 0.1720 18.5 23.1 25.8 27.0 27.5 27.7 27.8 119.3 121.5 122.9 123.5 123.8 123.9 123.7 134.6 128.1 125.5 124.5 124.1 124.0 123.9 Note that the convergence begins rapidly. There are ways to speed this, but at this point they would only add complexity. Depending where you stop, you can enter the analysis. (a) µ = 4.541(10−6 ) reyn, From Fig. 12-16: S = 0.1724 ho = 0.482, c h o = 0.482(0.002) = 0.000 964 in From Fig. 12-17: φ = 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= 2π T N 2π(9.84)(1120/60) = = 0.124 Btu/s Ans. 778(12) 778(12) (e) From Fig. 12-19: Q = 4.16, rcNl Q = 4.16(1.25)(0.002) 1120 (2.5) 60 = 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) φ pmax = 16° Ans. (g) φ p0 = 82° Ans. (h) T f = 123.9°F Ans. (i) Ts + T = 110°F + 27.8°F = 137.8°F Ans. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 315 315 Chapter 12 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 = 120°F. Below is a partial tabular summary for comparison purposes. cmin 0.001 in Tf T Tmax µ S fr c Q rcN j l Qs Q ho c f Q Qs ho c 0.002 in cmax 0.003 in 125.8 11.5 131.5 3.014 0.053 7 0.775 0 124.0 7.96 128.0 3.150 0.024 9 0.873 4.317 1.881 1.243 4.129 4.572 4.691 0.582 0.824 0.903 0.501 0.225 0.127 0.006 9 0.094 1 0.054 8 0.000 501 0.006 0.208 0.172 0.000 495 0.005 9 0.321 0.290 0.000 382 132.2 24.3 144.3 2.587 0.184 0.499 Note the variations on each line. There is not a bearing, but an ensemble of many bearings, due to the random assembly of toleranced bushings and journals. Fortunately the distribution is bounded; the extreme cases, cmin and cmax , coupled with c provide the charactistic description for the designer. All assemblies must be satisfactory. The designer does not specify a journal-bushing bearing, but an ensemble of bearings. 12-14 Computer programs will vary—Fortran based, MATLAB, spreadsheet, etc. 12-15 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. The task is to iteratively find the average film temperature, T f , which makes Hgen and Hloss equal. The steps for determining cmin are provided within Trial #1 through Trial #3 on the following page. budynas_SM_ch12.qxd 316 12/04/2006 15:24 FIRST PAGES Page 316 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Trial #1: • Choose a value of T f . • Find the corresponding viscosity. • Find the Sommerfeld number. • Find f r/c , then Hgen = 2545 WN c 1050 fr c • Find Q/(rcNl) and Q s /Q . From Eq. (12-15) T = Hloss = 0.103P( f r/c) (1 − 0.5Q s /Q)[Q/(rcN j l)] h CR A(T f − T∞ ) ¯ 1+α • Display T f , S, Hgen , Hloss Trial #2: Choose another T f , repeating above drill. Trial #3: Plot the results of the first two trials. H Hgen Hloss, linear with Tf (Tf )1 (Tf )3 (Tf )2 Tf Choose (T f ) 3 from plot. Repeat the drill. Plot the results of Trial #3 on the above graph. If you are not within 0.1°F, iterate again. Otherwise, stop, and find all the properties of the bearing for the first clearance, cmin . See if Trumpler conditions are satisfied, and if so, ¯ analyze c and cmax . The bearing ensemble in the current problem statement meets Trumpler’s criteria (for n d = 2). This adequacy assessment protocol can be used as a design tool by giving the students additional possible bushing sizes. b (in) tb (in) 2.254 2.004 1.753 0.004 0.004 0.003 Otherwise, the design option includes reducing l/d to save on the cost of journal machining and vender-supplied bushings. budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 317 317 Chapter 12 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 2 Tav ϭ Tf 1 (Tf )2 (Tf )3 (Tf )1 Tf Trial #3: Pick the third T f from the plot and repeat the procedure. If (T f ) 3 and (Tav ) 3 differ by more than 0.1°F, plot the results for Trials #2 and #3 and try again. If they are within 0.1°F, determine the bearing parameters, check the Trumpler criteria, and compare Hloss with the lubricant’s cooling capacity. Repeat the entire procedure for c = cmax to assess the cooling capacity for the maxi¯ mum radial clearance. Finally, examine c = c to characterize the ensemble of bearings. 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) budynas_SM_ch12.qxd 318 12/04/2006 15:24 FIRST PAGES Page 318 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Trial #1: Choose (T f ) 1 = 79°C. From Fig. 12-13, µ = 13 mPa · s. 13(10−3 )(48) = 0.055 4000(103 ) S = (5952 ) fr = 2.3, = 0.85. c 978(106 ) ( f r/c)SW 2 T = 1 + 1.5 2 ps r 4 From Figs. 12-18 and 12-16: From Eq. (12-25), 2.3(0.055)(102 ) 978(106 ) 1 + 1.5(0.85) 2 200(25) 4 = = 76.0°C Tav = Ts + T /2 = 55°C + (76°C/2) = 93°C Trial #2: Choose (T f ) 2 = 100°C. From Fig. 12-13, µ = 7 mPa · s. 7 = 0.0296 S = 0.055 13 fr = 1.6, c From Figs. 12-18 and 12-16: T = = 0.90 1.6(0.0296)(102 ) 978(106 ) = 26.8°C 1 + 1.5(0.9) 2 200(25) 4 Tav = 55°C + 26.8°C = 68.4°C 2 Tav 100 90 80 Tav ϭ Tf (100ЊC, 100ЊC) (79ЊC, 93ЊC) (79ЊC, 79ЊC) 70 (100ЊC, 68.4ЊC) 85ЊC 60 70 80 90 100 Tf Trial #3: Thus, the plot gives (T f ) 3 = 85°C. From Fig. 12-13, µ = 10.8 mPa · s. S = 0.055 10.8 13 = 0.0457 fr = 2.2, = 0.875 c 2.2(0.0457)(102 ) 978(106 ) = 58.6°C T = 1 + 1.5(0.8752 ) 200(25) 4 From Figs. 12-18 and 12-16: Tav = 55°C + Result is close. Choose 58.6°C = 84.3°C 2 85°C + 84.3°C ¯ = 84.7°C Tf = 2 budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 319 319 Chapter 12 µ = 10.8 MPa · s Fig. 12-13: 10.8 13 S = 0.055 fr = 2.23, c T = = 0.0457 = 0.874, ho = 0.13 c 2.23(0.0457)(102 ) 978(106 ) = 59.5°C 1 + 1.5(0.8742 ) 200(254 ) Tav = 55°C + 59.5°C = 84.7°C 2 O.K. From Eq. (12-22) Q s = (1 + 1.5 2 ) π ps rc3 3µl = [1 + 1.5(0.8742 )] π(200)(0.0423 )(25) 3(10)(10−6 )(25) = 3334 mm3 /s h o = 0.13(0.042) = 0.005 46 mm or 0.000 215 in Trumpler: h o = 0.0002 + 0.000 04(50/25.4) = 0.000 279 in Tmax = Ts + Not O.K. T = 55°C + 63.7°C = 118.7°C or 245.7°F O.K. Pst = 4000 kPa or 581 psi n = 1, as done Not O.K. Not O.K. There is no point in proceeding further. 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 dmin = ⇒ d≥ W 600(l /d) 900 = 1.73 in 2(300)(0.5) budynas_SM_ch12.qxd 320 12/04/2006 15:24 FIRST PAGES Page 320 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design In this problem we will take journal diameter as the nominal value and the bushing bore as a variable. In the next problem, we will take the bushing bore as nominal and the journal diameter as free. To determine where the constraints are, we will set tb = td = 0, and thereby shrink the design window to a point. d = 2.000 in We set b = d + 2cmin = d + 2c nd = 2 (This makes Trumpler’s n d ≤ 2 tight) and construct a table. c b d ¯ T f* Tmax ho 0.0010 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019 0.0020 2.0020 2.0022 2.0024 2.0026 2.0028 2.0030 2.0032 2.0034 2.0036 2.0038 2.0040 2 2 2 2 2 2 2 2 2 2 2 215.50 206.75 198.50 191.40 185.23 179.80 175.00 171.13 166.92 163.50 160.40 312.0 293.0 277.0 262.8 250.4 239.6 230.1 220.3 213.9 206.9 200.6 × × × × × × × × Pst Tmax n × fom −5.74 −6.06 −6.37 −6.66 −6.94 −7.20 −7.45 −7.65 −7.91 −8.12 −8.32 *Sample calculation for the first entry of this column. ¯ T f = 215.5°F Iteration yields: ¯ With T f = 215.5°F, from Table 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 0.817(10−6 )(50) 1 = 0.182 S= 0.001 225 From Figs. 12-16 and 12-18: Eq. (12–24): = 0.7, f r/c = 5.5 0.0123(5.5)(0.182)(9002 ) = 191.6°F [1 + 1.5(0.72 )](30)(14 ) 191.6°F . Tav = 120°F + = 215.8°F = 215.5°F 2 For the nominal 2-in bearing, the various clearances sho...
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