Class Note for ME 350 with Professor Wei at UA-Static Machine Components
Class Note for ME 350 with Professor Wei at UA-Static Machine Components
Popular in Course
Popular in Department
This 30 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Alabama - Tuscaloosa taught by a professor in Fall. Since its upload, it has received 37 views.
Reviews for Class Note for ME 350 with Professor Wei at UA-Static Machine Components
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/06/15
Gears kinematics stress analysis 350 Static Machine Components Tm fga aas STRAIGHT SPUR HELIEAL SFUH iHEHRINGBDNE Nomenclature of spmgmgdMgear teeth Capylighk 39rhe mam Hm Campinas Kw Penul39ss A Tnp land producum nr display Face a dum Ad 639 Flank t cimg DaLlcndum Widlhof spam 65 f 9 Clearance Filch 77Q radius Dcdcndum Clearance Gift It circle Nomenclature of spurgear teeth Pinion the smaller of two mating gear Gear the larger of two mating gear Meshing the action of interlocking with another object Pitch cycle a theoretical circle upon which all calculations are usually based Tooth thickness Module m the ratio of the pitch diameter to the number of teeth Addendum a the radial distance between the top land and the pitch cycle Whole depth h the sum of the addendum and the dedendum Clearance c the amount by which the dedendum in a given gear exceeds the addendum of its mating gear Nomenclature of spurgear teeth PzNd mzdN p7TdN7Tm pP7T where P d i metral pitch teeth per inch N Number of teeth d pitch diameter in m module mm p 2 Circular pitch Conjugate amml Hnll Companies inc Pen Mating gear teeth acting against each other to produce rotary are similar to cams When the tooth profiles or cams are designed so as to produce a constant angular velocity ratio during meshing these are said to have conjugate action It is possible arbitrarily to select any profile for one tooth and then to nd a pro le for the meshing tooth that will give conjugate action One of these solutions is the involuteprofilc Point of contact c Line of action bc Intersection of ac with 00 pitch pointP Circles drawn through point P from each center Radius of each circle is called the pitch radius To transmit motion at a constant angular velocity ratio the pitch point must remain xed all the lines of action for every instantaneous point of contact Animation for line of action Generation of involute curve Base circle Pitch Cirth Base circle Mathematic stegs of generating involute curve 0r dlipluyJ Copynghl 39fhe mwHillcmnpam m Femmimnlequuadlmmpmd Base circle lnvo 1 Lite 0 Y I Divide the base circle into a number of equal parts and construct radial lines 0A0 0A1 0A2 0A3 Construct perpendiculars AIBI AZBZ A3B3 Along Aan lay o the distance n XAOAI Connecting B1B2B3 Fundamentals When two gears are in mesh their pitch circles roll on one another Without slipping Designate the pitch radii as r and r2 and the angular velocities as co and 602 respectively Then the pitchline velocity is V T1W1l T2w2 The relation between the radii on the angular velocities is Q 002 71 Suppose we specify that an 18tooth pinion is to mesh With a 30tooth gear and that the diametral pitch of the gearset is to be 2 teeth per inch The pitch diameters of the pinion and gear are respectively mpgz wm Circles of a gear layout Camquotgm a m MonwHm Campinas m Pernusxmn mqmm l 1m mme ch M mwplay Dcdendum cu39cle Pitch circle HI Base circle Involute zl Pitch circle Base circle Basepitch cycle radius and pressure angle Pitch circle Pressure line Pressure line Generating line Line of action Base circle Same thing l I Pressure angle usually has values of 20 or 25 Gear teeth geometry The addendum 6 and dedentum b distances to the pitch cycle for standard interchangeable teeth are UP and 125P respectively a1P 191 25P BMW Circular pitch p7rP Circular pitch Tooth thickness l p2 The clearance c ba025P The base pitch is related to the circular pitch by the equation Pb Pc608 Example A gearset consists of a l6tooth pinion driving a 40tooth gear The diametral pitch is 2 and the addendum and dedendum are UP and 125P respectively The gears are cut using a pressure angle of 20 a Compute the circular pitch the center distance and the radii of the base circles b In mounting these gears the center distance was incorrectly made 1A in larger Compute the new values of the pressure angle and the pitchcircle diameters Contact ratio Copynghl 9 The McGrawHil CompaniesV inc Permission required for repmducu39on or display Arc of Arc of approach 1 recess qr K Addendum h l Circle Addendum circle Pitch circle Motion Interference The contact of portions of tooth pro les that are not conjugate is called interference The smallest number of teeth on a spur pinion and gear onetoone gear ratio which can eXist without interference is Np 2k 2 Np 38m2 1 1 3szn gb where kl for fulldepth teeth 08 for stub teeth If the mating gear has more teeth than the pinion that is mGNGNPm is more than one then Np is given as Np m m2 1 2ms7jn2gbgt Tooth systems Standardized by the American Gear Manufacturers Association AGMA Table 131 Yuo39h System Pressure Angle as deg Addendum 0 Dedendum b SIcndord and MI depth 20 I m or Im I 25mm I 25m Commoniy Used Team 135Pu m I 35m Systems for Spur Gears 22 mm Im I39MPm 25m I 35qu Lasm 25 mm lm I25FIQI 25m I 35mm I 35m Sum 20 o BPIaI o Em IRI m m We 34 TooIh SIzes in GenemI Come 2 2 21 3 A s 3 I0 I2 I5 Uses he 70 m 3140 B s130 ch IzoI I50 200 PIEiPHEU L125 15 2 2 5 3 ll 5 O 8 1012161015 3211050 NedCiIaIce I I23 1375 I 7522 275 35 45 55 7 9 II IA Is 22 28 30 45 Straight BevelGear Teeth Tooth Proponions For Working demh m 20P 20 Srraighx BeveJGeor Clearance r 10188P 0002 m Teeth Addendum 0 gear H 0i Gem who Equivo rem 90 who when 1 90 1 Face w drh F 03Ao 01F whlchever is smo ler Pimonllo 15 4 13 memum numbev a lt6th Genvl 17 20 30 Straight BevelGear Teeth Table 13 4 Qua Formula Formula 5 quotd d TD I Addendum E Exrerncn39 gems Propomcns iov Helm Pu 1 d 5 Dedendum 7 Standard Gem s dws cnce D N mm mm dwame av Gear Duiswde dwametsv o 2n Fuses w c a man dmme ev N Pmion nmswc39e dznmarev d 2 P cos w Novmn am my Ihxckness39 Gem run home39sv D 7 2b Pmmn bust aua dmsq Pmmn mm dvameter u 2b meme gears Gem we dwumerer Dam Cemev d skjnce D d Bone he ix eng e Vnn39 Inn y cosdzy hmde dramewer D 7 2a m may D 2L7 mu mmelmons we in WM and angles me in mm 5 is le nmmn hnikhzih Gear trains 773 7712 d 2n2 N3 d3 Where n revolutions or revmin N number of teeth d pitch diameter drive gear The gear that receives energy from a power source such as an electric motor A drive gear transmits power to a meshing driven gear to perform work driven gear The gear that receives motion from the drive gear on a machine Driven gears often turn tools or components idler gear A gear that is used to keep the direction of motion consistent between a drive gear and a driven gear Gear trains Copynghl The McGrawHIH Companies Inc Penmsslon required for reproducljm or display Gears 2 3 and 5 are drivers While 3 4 and 6 are driven members We de ne the train value 6 as product of driving tooth numbers e production of driven tooth numbers Gear trains nL is the speed of the last gear in the train and nF is the speed of the first 71L 2 BYLF As a rough guideline a train value of up to 10 to I can be obtained with one pair of gears Greater ratios can be obtained in less space and with fewer dynamics problems by compounding additional pairs of gears A twostage compound gear train below can obtain a train value of up to 100 to I Example gear box design A gearbox is needed to provide a 301 i1 percent increase in speed while minimizing the overall gearbox size Specify appropriate teeth numbers w mgrm tum Since the ratio is greater than 101 but less than 1001 a twostage compound gear train is needed The portion to be accomplished in each stage is 3005547 7 For this ratio assuming a typical 200 pressure angle the minimum number of teeth to avoid interference is The number of teeth necessary for the mating gears is 16m 8764 i 88 The overall train value is e 88168816 3025 This is within the 1 percent tolerance If a closer tolerance is desired then increase the pinion size to the next integer and try again Compound reverted geartrain It is sometimes desirable for the input shaft and the output shaft of a two stage compound geartrain to be inline as shown This con guration is called a compound reverted geartrain This requires the distance between the shafts to be the same for both stages of the train Which adds to the complexity of the design task The distance constraint is d22 d32 d42 d52 With PNd and assuming constant diametral pitch in both stages Hi i l ego II II I I l I V l rr A N2 N3 N4 N5 5253 39I ll A lf ff Force analysis Notations Gears are numbered as 2 3 4 shafts are designated as a b c Force exerted by gear 2 against gear 3 as F23 The force of gear 2 against shaft a is F 2a Radial and tangential directions are distinguished by superscripts r and t Gczu39 Force analysis FBD of the pinion Hi I Fm Ha 4 F32 I Tng Torque and power transimission With the defined radial and tangential force components We now define Wt 2 F32 as the transimitted load The applied torque and the trans mitted load are seen to be related by the equation d T Wt 2 where we have used T Ta2 and d 2 d2 to obtain a general relation The power H transmitted through a rotating gear can be obtained from the standard relationship of the product of torque T and angular velocity w H 2 Tea 2 Wtd2w Gear data is often tabulated using pitch line velocity which is the linear velocity of a point on the gear at the radius of the pitch circle thus V d2w Converting this to customary units gives V 7Tdn12 where V pitch line velocity ftmin d 2 gear diameter in n 2 gear speed revmin Many gear design prOb39emS Wi SDGCify the power and speed so it iS Convenient to rearrange the torque force relationship as H Wt33000 V where Wt transmitted load lbf H 2 power hp V pitch line velocity ftmin The above equation in SI is H Wt 60000 7min where Wt transmitted load kN H 2 power kW d 2 gear diameter mm n speed revmin Example power transmission Pinion 2 in the Figure runs at 1750 revmin and transmits 25kW to idler gear 3 The teeth are cut on the 20quot full depth system and have a module of m25mm Draw a freebody diagram of gear 3 and show all the forces act upon it Example power transmission FBD
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'