Exam Problems for Machine Design | ME 3180, Study notes of Machine Design

Download Exam Problems for Machine Design | ME 3180 and more Study notes Machine Design in PDF only on Docsity! ME. 21S © Some suggestions for the exam: Read the entire exam (4 problems) first. When answering the questions, be as complete as possible. Clearly state all your assumptions. Ple: And remember, I cannot give credit for empty pages, so attempt all questions. Problem 1 On the following pages, a description is given of a so-called overdrive unit for an automobile. As I have told you in class, these overdrive units are planetary (or epicyclical) gear systems which are mounted behind a regular transmission in order to provide additional gearing. An exploded view of the overdrive unit is shown in Figure 6.13. In Figure 6.12, a sectional view of the overdrive unit is shown. In Figures 6.10 and 6.11, the working principle of the overdrive is shown and explained in the accompanying text. Please note that the planet carrier is the arm and the annulus is the ring gear. In a nutshell the working of the overdrive is as follows. In Fig. 6.10, the transfer of the input to output motion is shown when the overdrive is not in use. If the overdrive is not in use, there is a 1 to 1 “direct drive” ratio between the rotation of the input shaft (the so- called third motion shaft) and the output shaft which transfers the rotation and power to the differential of the car. The rotation from the input shaft is transferred to the output shaft by means of a unidirectional clutch (item 5, Fig. 6.10) which completely bypasses the planetary gear system. In Figure 6.11, the overdrive unit is shown in the engaged state. If the overdrive is in use, the sun wheel remains stationary with respect to the housing by means of a brake ring (item 2, Fig. Fig 6.11). The planet carrier is directly connected to the input shaft and starts rotating with the same rotational speed as the input shaft. The transfer of the rotation through the planet wheels causes the annulus (ring gear) to overrun the uni- directional clutch and give an increased speed to the output shaft, i.e., “overdrive”. Problem 2 —- Your next task is to determine the bearings required to support the output shaft of the overdrive unit from Problem 1. In the overdrive unit of Problem 1, a number of ball bearings can be identified supporting the output shaft, that is, items 77 and 81 in Figure 6.13 (see also Figure 6.12). One of the bearings supporting the output shaft is a SKF decp groove ball bearing. The axial and radial load on this single ball bearing is as follows: F, = 200 Ibs F, = 300 Ibs The overdrive unit is to last for at least five years while operating 10 hours a week with an average speed of 2000 revolutions per minute. Furthermore, the bore of the bearings should be 50 mm, normal clearances are used, and we would like to minimize the weight of the bearing. Question: Which SKF bearing designation would you recommend to be used for this bearing? (15 points) Note: Use the SKF bearing data given in the pages following this problem. Do not use bles in your text! !  Deep groove ball bearings Equivalent dynamic bearing load For single bearings and bearing pairs arranged in tandem P =F, P= XF, + YFa when F,/F, < @ when F,/F, > € Tha X and Y factors required for the cal- culation of the equivalent bearing toad of deep groove ball bearings are dependent ‘on the ratio of the axial load F, to the basic static load rating Co. They are also influenced by the magnitude of the radial interna! clearance; increased clearance enables heavier axial loads to be carried. If the bearings are mounted with the usual fits (tolerance [5 to n6 depending on shaft diameter, and J7 for the housing) the values of e, X and Y given in the upper table opposite can be used to calculate the equivalent load. For bearing pairs in tandem, the vaiues given under “C3 clearance” should be used. Ifa greater clearance than Normal is chosen because a reduction in clearance will be obtained in operation, for example as a result of strong heating of the inner ting, then the values of the factors under Normal clearance should be used. For bearing pairs in tandem, F, and F, are the forces acting on the bearing pair. For bearing pairs arranged back-to-back or face-to-face P=F,+YiFa when F,/F, § & P= 0,75 F, + Yor, whenF,/F, > e F, and F, are the forces acting on the bearing pair. The values for factors e, v1 and Y; for different values of F,/Cy are given in the lower table opposite. Equivalent static bearing load For single bearings and bearing pairs in tandem P, = 0,6 F, + 0,5 Fy When Py < Fy, Po = F; should be used. For paired bearings, F, and F, are the forces acting on the bearing pair. For bearing pairs arranged back-to-back or face-to-face PHF, +47 Fe F, and F, are the forces acting on the bearing pair. Axlal load carrying capacity if deep groove ball bearings are subjected to a purely axial load, this axial joad should generally nat exceed the value of 0,5 Cg. Small bearings and light series bearings (Diameter Series 8, 9, 0 and 1} should not be subjected to a load greater than 0,25 Cy. Excessive axial toads can lead to an appreciable reduction in bear- ing life. Gh. { Caloulation factors for eingie row deep groove ball bearings totace i Rip ey Ye Moga"? age Ta) “aa O10 en 04 cia EB fe PD, “iden tran innnn pra baste hes 18: GABA WS Lt Pe emmy gob pes {ea Oty. ee 4 Ser Bearing pairs arranged back-to-back or faoet Me is hee CA clearance . v ea x .¥ " 175° 04 , 044 | 142 1e2 0a O44 | 1,98 44a O44 1 044 | 127 13 048 044 | 1,168 $14 Oa. 044 | 1:05 1 O58 044 1 single row Deep groove ball bearings . d 35-55 mm With full outer With recessed outer fing shoulders sing shoulders } Principat Basle toad ratings ——Fatiguo~=—‘Speed ratings Mass ——_Dosignation Dimensions Abutment and titet cimenatons dyramic static, toad Ldrication dimensions grease se ee & OR ee mom N N remin kg = mm am 3547: OT 4780 3200 168 12000 16000 © 0,080 38987 495 - = O87 a 55 100 9580-6 200 250 11900 14000 © 0,080 “16 48 - 08 39 51 O86 @ 9 12400 8 160 375 jo000 18000 0.11 440 a3 a 6 14 15900 10 440 10000 13000 «= 0.16 } 47 936 557 4 40. 87 1 72 «1? «© 28800 15300 «= eas. sooo 611000 0,20 489° 806 627 14 0 41S 655 1 80 21 «= $9200 19009 «a5 B50 «10000048 495 661 692 15 40 72) 18 I 100 25 «© $5300 31000 «= 1280) 7000 a 500 095 74 BOG 18 43245 | 0 52 7 4940 3.450 198 11900 14099 9,034 4 497 «485 - 08 az si @ 12 13800 9300 425 too00 13000 ©0112 7 552 os 44 SB | @ 9 3 9150 440 sso iz002 0a 44 57 os a2 wD | 68 615 «= 16800 11600 «490 9500 12000 0.39 ! 4920 591 611 1 4% 63) #0 18 80700 «19.000 «=a 8500 10000 © 0.37 626 679 08 11 89465 735 1 | 90 28 41000 24.000 71020 7500 9000 063 561 747 777 45 e245 | 110 27 «= 63700-36500 «1530 «= 8700. 8 000 125 ozs a8 4a tor 45 6050 ©4300 228 9500 12000 © 0,040 4 487 845 - 08 a? 8G | 44000 9 800 485, 9000 14000 Ota : 529 608 - Of 4 64 96 | Bie 2 Ge Ie fa, # Bo 33200 21600 BTS 7500 © 9-000 oat { s76 729 752 41 06157851 | 52700 31500 134 6709 8900 @21 87 867 18 2 48 76100 45000 = 1900 © 6000 7 000, 155 i 69 988 2 Bott 50 6240 4750 250 9000 11000 0,052 ‘ Bo 547 BOS - 09 82 6 03 14600 10400 S00 asco 10000014 588 659 - 06 S4 68 08 16900 11400 «580 8500 10000 «Os. 80 704 - 06 54 7 08 21600 16000710 #500 10600 O28 soy 706 72a +t 5B 751 Bs 100 29200 B80 7000 |B 500 046 ge mi By 4 Bab aS Br ioo 52000. 2200 500 8900 74 108 Bt) Bt tte ear) 6320 © 200 228 e5co 10000 8 602 67 - 08 §7 70 09 & 4 jes t40m © aos, Pup Boe & @ = ts & & 6 80 . ~ ¥ 808 80 18 = 28100-24200 = 800. 7500 © 9000 ‘ 63 701 818 ti 6t5 335 1 joo 21 43600 28000 1250 «© 8300 7 500 9 8 884 18 gs oe 15 120 bo 74500 48.000 1800 5600 6 700 i 753° 10t 104 2 ein 44093°=«= 98.500 g2000 2600 «= 5000» 6 000 es 618 8s sto t