AGMA 93FTM5-1993 Optimal Gear Design for Equal Strength Teeth Using Addendum Modification Coefficients《利用齿顶修正系数进行等强度轮齿的齿轮优化设计》.pdf

《AGMA 93FTM5-1993 Optimal Gear Design for Equal Strength Teeth Using Addendum Modification Coefficients《利用齿顶修正系数进行等强度轮齿的齿轮优化设计》.pdf》由会员分享，可在线阅读，更多相关《AGMA 93FTM5-1993 Optimal Gear Design for Equal Strength Teeth Using Addendum Modification Coefficients《利用齿顶修正系数进行等强度轮齿的齿轮优化设计》.pdf（14页珍藏版）》请在麦多课文档分享上搜索。

1、93FTM5Optimal Gear Design for EqualStrength Teeth Using AddendumModification Coefficientsby: C. H. SuhUniversity of ColoradoAmerican Gear Manufacturers AssociationTECHNICAL PAPEROptimal Gear Design for Equal Strength Teeth UsingAddendum Modification CoefficientsC. H. Suh, University of ColoradoThe s

2、tatementsandopinionscontainedhereinare those ofthe authorand shouldnotbe construedasanofficial action oropinion of the American Gear ManufacturersAssociation.ABSTRACT:The addendummodificationcoefficient or shift factor is first defined and explainedin detail in terms of gear designterminology. Twoty

3、pes of helical gear design equationsare derived, one with the truegenerating shift factor, and theother with the conventional shiftor addendum modification factor usedin the United States. Nakadas equation and acantilever beam equation usedto design equalbending strength teeth are reviewed and discu

4、ssed. Then a new designmethod to synthesizeequal strengthteethbetweenmatingpinionand gear,whichmayhave differentmaterialproperties,is presented. Thismethod uses the shiftfactorsas designvariablesand an effectivenonlinearequation solver tofind thesolution. Variousnumericalexamples are given to illusl

5、rate this new method.Copyright 1993American Gear Manufacturers Association1500King Street, Suite 201Alexandria, VLrginia,22314October, 1993ISBN: 1-55589-598-0OPTIMAL GEAR DESIGN FOR EQUAL STRENGTH TEETHUSING ADDENDUM MODIFICATION COEFFICIENTSC. H. Suh, ProfessorUniversity of Colorado, Boulder Colora

6、do 80309first published an equation containing x_Introduction : and x2, the addendum modificationThere must be many advantages in the coefficients of the pinion and gear,involute tooth form that has enabled it which provides for equal toothto practically eradicate all other tooth thicknesses of the

7、pinion and gear on theforms that have existed during the last base circle, and stated this tooth formseveral decades. One of the major to approximate the Maag gear. Mabie et.advantages is as stated by Buckingham i al. 4 used cantilever beam stressthat when the involute tooth form theory as a model f

8、or predicting gearprovides conjugate action it becomes the stresses and an interpolation process toonly tooth form that yields a constant balance the stresses in mating teeth. Inpressure anqle from point to point. It is this paper, three methods to balance thewell kno_.mn, however, that the most ben

9、ding fatigue safety factors based onimportant attribute is themanufacturability of involute gears: the the AGMA bending stress formulae, and theinvolute gears can be cut by a straight- solution of nonlinear equations, aredescribed and illustrated with variousedge cutter. A “straight-line cutter“ can

10、not only be easily manufactured but also numerical examples.easily checked, and is thus easilymaintained for the high precisionproduction of gears. This paper is List of S_mbols :focused on yet another importantadvantage of the involute tooth form - Cr operating center distancethe addendum modificat

11、ion coefficient or Dz operating pitch diameter ofinvolute shift factor for gears with pinionvariable or non-standard center distance, D2 operating pitch diameter ofwhich can be used to balance the teeth gearstrength, increase the contact ratio and Dbz base circle diameter of pinionDb2 base circle di

12、ameter of gearminimize the weight for mating gears. Doz outside diameter of pinionIt is generally known that Maag 2 Do2 outside diameter of gearwas one of the first to introduce gears ee sum of hob offsetswith addendum modification coefficients. (ee = ez + e2)Maag used the coefficient to balance the

13、 ez hob offset for pinionbending strengths of the teeth for mating (ez = xzgm_)pinion and gear, but the method as well e2 hob offset for gearas most criteria used in the design of (e2 = x2gm_)his Maag gear were proprietary and F effective face widthremained undisclosed to gear design h modified toot

14、h whole depthengineers for several decades. Nakada 3 ha hob addendumJ AGMA J geometry factor Definitions of Addendum Modificationk top clearance factor Coefficient :Ka AGMA application factor In many European countries and JapanKB AGMA rim thickness factor the term addendum modificationKL AGMA life

15、factor coefficient or rack shift factor, x, isK_ AGMA load-distribution factor meant to include both shifting of theKR AGMA reliability factor rack for addendum modification andK_ AGMA temperature factor shifting of the rack for thinning of theKs AGMA size factor teeth to obtain backlash. While Amer

16、icanKv AGMA dynamic factor gear designers, following the AGMAm module for spur gear standards 5, refer to two separatem_ gear ratio shift coefficients - the generating rackm. normal module shift coefficient, xg applies to them_ transverse modulecompletely finished tooth, and includesnI pinion operat

17、ing speed shifting of the rack for addendumn2 gear operating speed modification plus shifting of the rackNI pinion tooth numberfor backlash, x, the rack shiftN2 gear tooth number coefficient for addendum modification,p, normal base pitchQv AGMA Quality number only includes the shifting of the rackne

18、cessary for the specified addendumR_ base circle radius of pinion modification.R_z base circle radius of gearSF b desired safety factor forbending The relationship between the shiftcoefficients is summarized as follows:T transmitted torquexx sum of addendum modificationcoefficients xg (real shift fa

19、ctor)xI pinion addendum modification = x (in Europe)coefficient or pinion shiftfactor A snx2 gear addendum modification = x (in USA)coefficient or gear shift 2mntan_nfactorxgI pinion generating (real) shiftfactor General Equations of Profile Shiftedxg2 gear generating (real) shift Helical Gears :fac

20、tor The following equations are theY, Yc coefficient of increase in general equations for profile shiftedcenter distance helical gears, with derivations given inWn normal load Appendix A. There are two forms of theseWt tangential load equations because of the above mentionedalternative definitions f

21、or the shiftangle between Wt and Wno tool protuberance coefficient of the rack.Ac n total backlash on base circleACnl amount pinion tooth thinned Desiqn Equations of Helical Gears withfor backlash on base circle x_, (the qeneratinq shift factor)Acn2 amount gear tooth thinned forbacklash on base circ

22、le inv_ z = 2 tan_n Ixg1+xgz+YclAs n total backlash on pitch circle N1+-_z ) (i)As._ amount pinion tooth thinnedfor backlash on pitch circle + in_Ash2 amount gear tooth thinned forbacklash on pitch circlePao tool tip radius Cr - m= + ym nstandard transverse pressure 2 cos (2)angle_= standard normal

23、pressure angle ( NI )_r operating pressure angle Dol = + 2bending stress cos_ mn (3)UalI AGMA allowable bending stress + 2 ( y - xg2 ) m=Cat bending strength_t AGMA bending stressstandard helix angle I N2 )_b base helix angle Do2 = cos$ + 2 In= (4)+ 2 (y-xg I ) m=h=(2+k)m n(5)- (xgl +x_ -y) m=As an

24、example of designing equalwhere tooth thickness gears consider thetan_ - tan_n following problem.cos_ Example1NI+N_ (cos_ Given: NI = 14, N2 = 15, _n = 200,Y - 2cos_ _cos_r i/ x_1 = 0.1812Find the gear shift factor, xg2, whichmakes the gears tooth thickness equal tothe pinions tooth thickness on the

25、 base(Acn circle. Using Nakada s equation1 theYc = 2sin_l_sin2_cos2_ n k mn solution is readily found.Solution:inv_n - 0.1607xg2 = xgI 2tan_nDesiqn Equations of Helical Gears with x,the US Conventional Shift Factor_ Although, the tooth thicknesses are theinv_ r = 2 tan_n x1+xz, + inv_ (6) same for t

26、hese mating gears, it does not_NI+N2 follow that the bending stresses underoperation will be the same.N_ +N zCr - 2 cos_ mn + Yma (7) Designing for Equal Bending Strength :NI 1 Method i: In case of same materialDol = + 2 ma properties for the pinion and qear._cos ) (8) Mabie et. al. 4 assume that th

27、e+ 2 ( y -x 2 ) mn + 2k 2 gear tooth is acting as a cantilever beamand that the load is applied at the tip( n2 ) of the tooth and carried by a single pair= + 2 ma of teeth. Using traditional cantileverD2 cos_ (9) beam theory the bending stress is+ 2 ( y - xI ) ma + 2k I calculated, and then the adde

28、ndummodification needed to balance thebending strengths are found by anh = (2+k)m n - (xl+x2-y) ma iterative process. The first method(i0) presented here extends this work by+ (k1+k z) making use of the AGMA formulae forcalculating the bending stress whichconsiders both the tangential and radialACn

29、= Aca_ + Acn2 (ii) components of the loading, as well as thestress concentration factor, and bysolving for the required addendumwhere modification directly from a describingnonlinear equation.Ac_ Acn2kl = 2ninOn k2 - 2ninOn The AGMA formula for bending stress(without the appropriate stress factors)i

30、s given by the equation:Nakadas Equation : u - Wt (12)Fm nJNakada 3 first published thecriteria which establishes equal tooththickness for mating gears on the base Rewriting this stress equation into thecircle. The equation is stated as: form introduced in reference 4 so thatthe load and face width

31、need not bexgz - xg_ = inv_ n considered, results inN2 - NI 2tall_a ,F _ cosp (13)Wn maJDerivation of this equation is given inAppendix B. The J-factor is calculated using themethod described in AGMA paper p139.036 . The method used here for solving theproblem is similar to Mabie 4, but Four numeric

32、al examples were given withinstead of a systematic trial and error the paper of Mabie et el. 4, and theseapproach to finding the correct addendum were all resolved here by the new methodmodification, a describing nonlinear described above for comparison purposes.equation is solved directly for the N

33、otice that the results are close whensolution, considering the differences between thetwo methods.Alqorithm (i)i) User specifies: NI, N_,m_, _n, and Cr Method 2: In case of different materialproperties with known tooth numbers.2) Calculate the operating pressure This algorithm describes a methodangl

34、e, _=: for distributing the addendummodification coefficients between the_r = cos-l_n (NI+ N2) cos_n pinion and gear which are composed ofdifferent materials. The method is based2Cr on full AGMA bending stress formulae asgiven in AGMA Standard 908-B89 5, andnonlinear equation solution techniques.In

35、addition, backlash may be specified.3) Calculate the sum of the hob offset,ee: The problem is statedas follows:(NI+N 2) (inv_r-in _) mnee = eI + e2 = 2tenOn Given the following specified designparameters for a set of helical gears,find the addendum modificationcoefficients which equalize the designe

36、l can now be written in terms of e2 assafety factors for bending fatiguee_= ee - e2. failure.4) Use reference 6 to now solve for theJ-factors of the pinion and gear. Required Input:5) Finally the single unknown e2 can be _at_, _at2, NI, N2. mn, T, n I , Cr, F, _n, _,solved using the following equal

37、bending hao, Pao, _o, Asnl, ASn2, Qv, Ka, KL, KR, _,stress equation, where the angle _ is Kscalculated using the method of Mabie 4:For the following algorithm 6atl and oat2(cos_ cos_ may either be the same or different.(14)i -!Example 2NI = 20, N2 = 30, m_ = 1/5 in., 9n = 20, Cr = 5.25 inAddendum Mo

38、dification Stress Factore = x_m=, in (o%_/Wn)Cantilever AGM_A Cantilever AGMABeam Method Method Beam Method Methodpinion 0.15787 0.13176 9.952 11.353gear 0.13286 0.15890 10.18 11.353Example 3NI : 35, N2 = 44, m_ = I/i0 in., #n = 20o, Cr = 3.90 inAddendum Modification Stress Factor (_F/W n)e = x_m=,

39、inCantilever AGMA Cantilever AGMABeam Method Beam MethodMethod Methodpinion -0.017098 -0.01896 37.47 35.047gear -0.030397 -0.02853 35.53 35.047Example 4Nl = 20, N2 = 51, mn = 1/12 in., _n = 20o, Cr = 3.033333 inAddendum Modification Stress Factor (Ol_IW=)e = x_mn, in.Cantilever AGMA Cantilever AGMAB

40、eam Method Beam MethodMethod Methodpinion Not given 0.03506 28.12 31.807gear Not given 0.04655 26.95 31.807Example 5NI : 41, N= = 51, mn = 1/12 in., _n = 20o, Cr = 3.90033 inAddendum Modification Stress Factor (_F/Wn)e = x0mn, in.Cantilever AGMA Cantilever AGMABeam Method Beam MethodMethod Methodpin

41、ion Not given 0.03439 27.13 32.998gear Not given 0.03678 27.09 32.998equal safety factors. The sum of theAlqorithm (2) : addendum modification coefficients iscalculated from the Eq. (6) as:The gear ratio m_, and gear speed n2 are (NI+N2) (inv_z_inv_)calculated as: xx = xI + xz = 2ta/l_ nN2m_- NInI w

42、heren2 - (rpm)MG_r = COS-Iran (NI+/V2)2CzCOS*The transmitted power is calculated fromthe torque T and pinion speed nI as:P- Tnl (kZ9 Subsequently, xI = xx - x2. Therefore, the9549.3 problem becomes one dimensional for x2.The operating pitch diameters, DI and D2, The outside diameters, Dol and Dos, a

43、reare then calculated as: calculated from Eqs. (8) and (9) as2C r (ram) Do_ = Na + 2 - 2x 2 mn + tan_ nD_- m_+ i cos_9o2=(N2 As_Finally, the pitch-line velocity v_ and Other AGMA stress factors can now betangential load Wt are determined as: calculated.=niD1 (res) Dynamic Factor:vt- 60000W_ -1000Pvt

44、 (BO Kv = A + _BA = 50 + 56 (I - B)The problem now becomes how to assign (12-Qv)2/3addendum modification coefficients, xI B -and xa, to the gears so that they have 45LALoad distribution factor:(“alll =(“111 (15)Km = 1.25 + 0.00110 F _ Ut _i_on _ Ut g_ror as input by user whereThe I and J geometry fa

45、ctors are aall - aa_KLcalculated using the analytical methods KTK agiven in AGMA 908-B89 5 and AGMA paperp139.03 6 which are easily programmed at = WtKa I KsKmKBwith a computer. An initial guess is _ Fm-_s jprovided for the one unknown in thisproblem, x2, so that all other values Example 6will be in

46、 terms of this one temporaryvalue. The solution is now obtained by The following is an example tosolving iteratively using the Newton- demonstrate the new method and to provideRaphson method for the one unknown, x2, a numerical sample so that futurewhich satisfies the equation investigation of the m

47、ethod as well ascomputer algorithm is possible. Noticethat the safety factors are equal for thepinion and gear.*SHIFTED GEAR DESIGN USING AGMA FORMULAE* DESIGN CONDITION *MATERIALPINION: 240 BhnGEAR : 180 BhnHARDNESS (HB) sat (MPa) sac (MPa) E (MPa) uPINION 240.0 210.0 720.0 200000.0 0_3000GEAR 180.

48、0 170.0 590.0 200000.0 0.3000POWER TRANSMITTED P = 13.05143 (Kw)TORQUE TRANSMITTED T = 173.10000 (Nm)REVOLUTIONS OF PINION nl = 720.00000 (rpm)REVOLUTIONS OF GEAR n2 = 486.00000 (rpm)MODULE (NORM) m = 2.42477 (mm)TOOTH NUMBER - PINION zl = 42.0TOOTH NUMBER - GEAR z2 = 62.0CENTER DISTANCE C = 146.600

49、00 (mm)FACE WIDTH F = 36.30000 (mm)HELIX ANGLE: psi = 30.00000 (deg.)* TOOL PARAMETERS *STANDARD NORMAL TOOL PRESSURE ANGLE (degree):20.0TOOL ADDENDUM FOR UNITY MODULE (hao/Mn) :1.355TOOL TIP RADIUS FOR UNITY MODULE (rao/Mn) :0.4276AMOUNT OF PROTUBERANCE FOR UNITY MODULE (del/Mn):0.0AMOUNT PINION TOOTH THINNED FOR BACKLASH (snl/Mn) :0.024AMOUNT GEAR TOOTH THINNED FOR BACKLASH (sn2/Mn) :0.024* AGMA STRESS FORMULA FACTORS *AGMA ACCUR