716 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears ©The McGra-Hfl Mechanical Engineering Elements Companies,2008 Design,Eighth Edition 718 Mechanical Engineering Design Table 14-2 Number of Number of Values of the Lewis Form Teeth Y Teeth Y Factor Y(These Values 12 0.245 28 0.353 Are for a Normal 13 0.261 30 0.359 Pressure Angle of20°， 14 0.277 34 0.371 Full-Depth Teeth,and a 15 0.290 38 0.384 Diametral Pitch of Unity 16 0.296 43 0.397 in the Plane of Rotation) 17 0.303 50 0.409 18 0.309 60 0.422 19 0.314 75 0.435 20 0.322 100 0.447 21 0.328 150 0.460 22 0.331 300 0.472 24 0.337 400 0.480 26 0.346 Rack 0.485 The use of Eq.(14-3)also implies that the teeth do not share the load and that the greatest force is exerted at the tip of the tooth.But we have already learned that the con- tact ratio should be somewhat greater than unity,say about 1.5,to achieve a quality gearset.If,in fact,the gears are cut with sufficient accuracy,the tip-load condition is not the worst,because another pair of teeth will be in contact when this condition occurs.Examination of run-in teeth will show that the heaviest loads occur near the middle of the tooth.Therefore the maximum stress probably occurs while a single pair of teeth is carrying the full load,at a point where another pair of teeth is just on the verge of coming into contact. Dynamic Effects When a pair of gears is driven at moderate or high speed and noise is generated,it is certain that dynamic effects are present.One of the earliest efforts to account for an increase in the load due to velocity employed a number of gears of the same size,mate- rial,and strength.Several of these gears were tested to destruction by meshing and loading them at zero velocity.The remaining gears were tested to destruction at various pitch-line velocities.For example,if a pair of gears failed at 500 Ibf tangential load at zero velocity and at 250 Ibf at velocity Vi,then a velocity factor,designated K.of 2 was specified for the gears at velocity Vi.Then another,identical,pair of gears running at a pitch-line velocity Vi could be assumed to have a load equal to twice the tangen- tial or transmitted load. Note that the definition of dynamic factor K has been altered.AGMA standards ANSI/AGMA 2001-D04 and 2101-D04 contain this caution: Dynamic factor K has been redefined as the reciprocal of that used in previous AGMA standards.It is now greater than 1.0.In earlier AGMA standards it was less than 1.0. Care must be taken in referring to work done prior to this change in the standards
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears 716 © The McGraw−Hill Companies, 2008 718 Mechanical Engineering Design Number of Number of Teeth Y Teeth Y 12 0.245 28 0.353 13 0.261 30 0.359 14 0.277 34 0.371 15 0.290 38 0.384 16 0.296 43 0.397 17 0.303 50 0.409 18 0.309 60 0.422 19 0.314 75 0.435 20 0.322 100 0.447 21 0.328 150 0.460 22 0.331 300 0.472 24 0.337 400 0.480 26 0.346 Rack 0.485 Table 14–2 Values of the Lewis Form Factor Y (These Values Are for a Normal Pressure Angle of 20°, Full-Depth Teeth, and a Diametral Pitch of Unity in the Plane of Rotation) The use of Eq. (14–3) also implies that the teeth do not share the load and that the greatest force is exerted at the tip of the tooth. But we have already learned that the contact ratio should be somewhat greater than unity, say about 1.5, to achieve a quality gearset. If, in fact, the gears are cut with sufficient accuracy, the tip-load condition is not the worst, because another pair of teeth will be in contact when this condition occurs. Examination of run-in teeth will show that the heaviest loads occur near the middle of the tooth. Therefore the maximum stress probably occurs while a single pair of teeth is carrying the full load, at a point where another pair of teeth is just on the verge of coming into contact. Dynamic Effects When a pair of gears is driven at moderate or high speed and noise is generated, it is certain that dynamic effects are present. One of the earliest efforts to account for an increase in the load due to velocity employed a number of gears of the same size, material, and strength. Several of these gears were tested to destruction by meshing and loading them at zero velocity. The remaining gears were tested to destruction at various pitch-line velocities. For example, if a pair of gears failed at 500 lbf tangential load at zero velocity and at 250 lbf at velocity V1, then a velocity factor, designated Kv , of 2 was specified for the gears at velocity V1. Then another, identical, pair of gears running at a pitch-line velocity V1 could be assumed to have a load equal to twice the tangential or transmitted load. Note that the definition of dynamic factor Kv has been altered. AGMA standards ANSI/AGMA 2001-D04 and 2101-D04 contain this caution: Dynamic factor Kv has been redefined as the reciprocal of that used in previous AGMA standards. It is now greater than 1.0. In earlier AGMA standards it was less than 1.0. Care must be taken in referring to work done prior to this change in the standards
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hill m Mechanical Engineering Elements Companies,2008 Design,Eighth Edition Spur and Helical Gears 719 In the nineteenth century,Carl G.Barth first expressed the velocity factor,and in terms of the current AGMA standards,they are represented as 600+V K= (cast iron,cast profile) (14-4a 600 1200+V K= (cut or milled profile) (14-46) 1200 where V is the pitch-line velocity in feet per minute.It is also quite probable,because of the date that the tests were made,that the tests were conducted on teeth having a cycloidal profile instead of an involute profile.Cycloidal teeth were in general use in the nineteenth century because they were easier to cast than involute teeth.Equation(14-4a) is called the Barth equation.The Barth equation is often modified into Eq.(14-4b),for cut or milled teeth.Later AGMA added 50+√F K= (hobbed or shaped profile) (14-5a 50 Ku= 78+√7 (shaved or ground profile) (14-56) 78 In SI units,Eqs.(14-4a)through (14-5b)become Ky= 3.05+V (cast iron,cast profile) (14-6al 3.05 6.1+V K= (cut or milled profile) (14-66) 6.1 Ky= 3.56+F (hobbed or shaped profile) (14-6c 3.56 K= 5.56+√厅 (shaved or ground profile) (14-6d 5.56 where Vis in meters per second (m/s). Introducing the velocity factor into Eg.(14-2)gives 0= KW'P FY (14-7刀 The metric version of this equation is KWr 0= (14-8) FmY where the face width F and the module m are both in millimeters (mm).Expressing the tangential component of load W in newtons (N)then results in stress units of megapascals (MPa). As a general rule,spur gears should have a face width F from 3 to 5 times the circular pitch p. Equations(14-7)and(14-8)are important because they form the basis for the AGMA approach to the bending strength of gear teeth.They are in general use for
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears © The McGraw−Hill 717 Companies, 2008 Spur and Helical Gears 719 In the nineteenth century, Carl G. Barth first expressed the velocity factor, and in terms of the current AGMA standards, they are represented as Kv = 600 + V 600 (cast iron, cast profile) (14–4a) Kv = 1200 + V 1200 (cut or milled profile) (14–4b) where V is the pitch-line velocity in feet per minute. It is also quite probable, because of the date that the tests were made, that the tests were conducted on teeth having a cycloidal profile instead of an involute profile. Cycloidal teeth were in general use in the nineteenth century because they were easier to cast than involute teeth. Equation (14–4a) is called the Barth equation. The Barth equation is often modified into Eq. (14–4b), for cut or milled teeth. Later AGMA added Kv = 50 + √V 50 (hobbed or shaped profile) (14–5a) Kv = 78 + √V 78 (shaved or ground profile) (14–5b) In SI units, Eqs. (14–4a) through (14–5b) become Kv = 3.05 + V 3.05 (cast iron, cast profile) (14–6a) Kv = 6.1 + V 6.1 (cut or milled profile) (14–6b) Kv = 3.56 + √V 3.56 (hobbed or shaped profile) (14–6c) Kv = 5.56 + √V 5.56 (shaved or ground profile) (14–6d) where V is in meters per second (m/s). Introducing the velocity factor into Eq. (14–2) gives σ = KvWt P FY (14–7) The metric version of this equation is σ = KvWt FmY (14–8) where the face width F and the module m are both in millimeters (mm). Expressing the tangential component of load Wt in newtons (N) then results in stress units of megapascals (MPa). As a general rule, spur gears should have a face width F from 3 to 5 times the circular pitch p. Equations (14–7) and (14–8) are important because they form the basis for the AGMA approach to the bending strength of gear teeth. They are in general use for
18 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hill Mechanical Engineering Elements Companies,2008 Design,Eighth Edition 720 Mechanical Engineering Design estimating the capacity of gear drives when life and reliability are not important con- siderations.The equations can be useful in obtaining a preliminary estimate of gear sizes needed for various applications. EXAMPLE 14-1 A stock spur gear is available having a diametral pitch of 8 teeth/in,a 1-in face,16 teeth,and a pressure angle of 20 with full-depth teeth.The material is AISI 1020 steel in as-rolled condition.Use a design factor of nd=3 to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications. Solution The term moderate applications seems to imply that the gear can be rated by using the yield strength as a criterion of failure.From Table A-20,we find S=55 kpsi and Sy=30 kpsi.A design factor of 3 means that the allowable bending stress is 30/3= 10 kpsi.The pitch diameter is N/P=16/8=2 in,so the pitch-line velocity is V=Td=π(21200 12= =628 ft/min 12 The velocity factor from Eq.(14-4b)is found to be K,=1200+V=1200+628 =1.52 1200 1200 Table 14-2 gives the form factor as Y=0.296 for 16 teeth.We now arrange and sub- stitute in Eq.(14-7)as follows: W= FYoa_1.5(0.296)10000 =3651bf KuP 1.52(8) The horsepower that can be transmitted is W1V365(628) Answer hp=33000=33000 =6.95hp It is important to emphasize that this is a rough estimate,and that this approach must not be used for important applications.The example is intended to help you under- stand some of the fundamentals that will be involved in the AGMA approach. EXAMPLE 14-2 Estimate the horsepower rating of the gear in the previous example based on obtaining an infinite life in bending. Solution The rotating-beam endurance limit is estimated from Eq.(6-8) S%=0.5Sr=0.5(55)=27.5kpsi To obtain the surface finish Marin factor k we refer to Table 6-3 for machined surface. finding a =2.70 and b=-0.265.Then Eq.(6-19)gives the surface finish Marin factor ka as ka=a50=2.70(55)-0265=0.934
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears 718 © The McGraw−Hill Companies, 2008 720 Mechanical Engineering Design EXAMPLE 14–1 A stock spur gear is available having a diametral pitch of 8 teeth/in, a 11 2 -in face, 16 teeth, and a pressure angle of 20◦ with full-depth teeth. The material is AISI 1020 steel in as-rolled condition. Use a design factor of nd = 3 to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications. Solution The term moderate applications seems to imply that the gear can be rated by using the yield strength as a criterion of failure. From Table A–20, we find Sut = 55 kpsi and Sy = 30 kpsi. A design factor of 3 means that the allowable bending stress is 30/3 = 10 kpsi. The pitch diameter is N/P = 16/8 = 2 in, so the pitch-line velocity is V = πdn 12 = π(2)1200 12 = 628 ft/min The velocity factor from Eq. (14–4b) is found to be Kv = 1200 + V 1200 = 1200 + 628 1200 = 1.52 Table 14–2 gives the form factor as Y = 0.296 for 16 teeth. We now arrange and substitute in Eq. (14–7) as follows: Wt = FYσall Kv P = 1.5(0.296)10 000 1.52(8) = 365 lbf The horsepower that can be transmitted is Answer hp = Wt V 33 000 = 365(628) 33 000 = 6.95 hp It is important to emphasize that this is a rough estimate, and that this approach must not be used for important applications. The example is intended to help you understand some of the fundamentals that will be involved in the AGMA approach. estimating the capacity of gear drives when life and reliability are not important considerations. The equations can be useful in obtaining a preliminary estimate of gear sizes needed for various applications. EXAMPLE 14–2 Estimate the horsepower rating of the gear in the previous example based on obtaining an infinite life in bending. Solution The rotating-beam endurance limit is estimated from Eq. (6–8) S e = 0.5Sut = 0.5(55) = 27.5 kpsi To obtain the surface finish Marin factor ka we refer to Table 6–3 for machined surface, finding a = 2.70 and b = −0.265. Then Eq. (6–19) gives the surface finish Marin factor ka as ka = aSb ut = 2.70(55) −0.265 = 0.934
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hill 719 Mechanical Engineering Elements Companies,2008 Design,Eighth Edition Spur and Helical Gears 721 The next step is to estimate the size factor k.From Table 13-1,the sum of the adden- dum and dedendum is 11.2511.25 1=p+P=8+8 =0.281in The tooth thickness t in Fig.14-1b is given in Sec.14-1 [Eq.(b)]ast=(4lx) when x =3Y/(2P)from Eq.(14-3).Therefore,since from Ex.14-1 Y=0.296 and P=8, 3Y3(0.296) x=2P=28) =0.0555in then t=(4lx)1/2=[4(0.281)0.0555/2=0.250in We have recognized the tooth as a cantilever beam of rectangular cross section,so the equivalent rotating-beam diameter must be obtained from Eq.(6-25): d=0.808hb)1/2=0.808(Ft)1/p=0.8081.50.250j/2=0.495in Then,Eq.(6-20)gives k as -0.107 de 0.495 、-0.107 kb=0.30/ =0.948 0.30 The load factor ke from Eq.(6-26)is unity.With no information given concerning temperature and reliability we will set k=ke=1. Two effects are used to evaluate the miscellaneous-effects Marin factor kf.The first of these is the effect of one-way bending.In general,a gear tooth is subjected only to one-way bending.Exceptions include idler gears and gears used in reversing mechanisms. For one-way bending the steady and alternating stress components are oa =om= o/2 where o is the largest repeatedly applied bending stress as given in Eq.(14-7).If a material exhibited a Goodman failure locus, + Sm=1 Since Sa and Sm are equal for one-way bending,we substitute Sa for Sm and solve the preceding equation for S,giving Sa= SeSu S+Sut Now replace S with/2,and in the denominator replace Swith 0.5S to obtain 0= =1.33S 0.5Sm+Sm0.5+1 Now k=a/S=1.33S/S?=1.33.However,a Gerber fatigue locus gives mean values of +（）=1
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears © The McGraw−Hill 719 Companies, 2008 Spur and Helical Gears 721 The next step is to estimate the size factor kb. From Table 13–1, the sum of the addendum and dedendum is l = 1 P + 1.25 P = 1 8 + 1.25 8 = 0.281 in The tooth thickness t in Fig. 14–1b is given in Sec. 14–1 [Eq. (b)] as t = (4lx)1/2 when x = 3Y/(2P) from Eq. (14–3). Therefore, since from Ex. 14–1 Y = 0.296 and P = 8, x = 3Y 2P = 3(0.296) 2(8) = 0.0555 in then t = (4lx) 1/2 = [4(0.281)0.0555]1/2 = 0.250 in We have recognized the tooth as a cantilever beam of rectangular cross section, so the equivalent rotating-beam diameter must be obtained from Eq. (6–25): de = 0.808(hb) 1/2 = 0.808(Ft) 1/2 = 0.808[1.5(0.250)] 1/2 = 0.495 in Then, Eq. (6–20) gives kb as kb = de 0.30−0.107 = 0.495 0.30 −0.107 = 0.948 The load factor kc from Eq. (6–26) is unity. With no information given concerning temperature and reliability we will set kd = ke = 1. Two effects are used to evaluate the miscellaneous-effects Marin factor kf . The first of these is the effect of one-way bending. In general, a gear tooth is subjected only to one-way bending. Exceptions include idler gears and gears used in reversing mechanisms. For one-way bending the steady and alternating stress components are σa = σm = σ/2 where σ is the largest repeatedly applied bending stress as given in Eq. (14–7). If a material exhibited a Goodman failure locus, Sa S e + Sm Sut = 1 Since Sa and Sm are equal for one-way bending, we substitute Sa for Sm and solve the preceding equation for Sa , giving Sa = S eSut S e + Sut Now replace Sa with σ/2, and in the denominator replace S e with 0.5Sut to obtain σ = 2S eSut 0.5Sut + Sut = 2S e 0.5 + 1 = 1.33S e Now kf = σ/S e = 1.33S e/S e = 1.33. However, a Gerber fatigue locus gives mean values of Sa S e + Sm Sut 2 = 1
720 Budynas-Nisbett:Shigley's IIL Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hill Mechanical Engineering Elements Companies,2008 Design,Eighth Edition 722 Mechanical Engineering Design Setting Sa=Sm and solving the quadratic in Sa gives 4S2 -1+1+ Setting Sa =a/2,Sur =S2/0.5 gives -忌[-1+1+405 =1.66S% and ky =o/S=1.66.Since a Gerber locus runs in and among fatigue data and Goodman does not,we will use k =1.66. The second effect to be accounted for in using the miscellaneous-effects Marin factor k is stress concentration,for which we will use our fundamentals from Chap.6. For a 20 full-depth tooth the radius of the root fillet is denoted rf,where 0.3000.300 rf= P =0.0375in 8 From Fig.A-15-6 1=1_0.0375 a=7=0.250 =0.15 Since D/d=oo,we approximate with D/d =3,giving K,=1.68.From Fig.6-20, q=0.62.From Eq..(6-32) Kr=1+(0.62)(1.68-1)=1.42 The miscellaneous-effects Marin factor for stress concentration can be expressed as 1 1 =142=0.704 k好= The final value of k is the product of the twok factors,that is,1.66(0.704)=1.17.The Marin equation for the fully corrected endurance strength is Se kakpkekakekf Se =0.934(0.948)1)(1)(1)1.17(27.5)=28.5kpsi For a design factor of n=3,as used in Ex.14-1,applied to the load or strength,the allowable bending stress is Se28.5 0all=兰 3 =9.5 kpsi nd The transmitted load W is wr=FYou=1-50.296950 =3471bf KuP 1.52(8) and the power is,with V=628 ft/min from Ex.14-1, WV347(628) hp=33000=33000 =6.6hp Again,it should be emphasized that these results should be accepted only as prelimi- nary estimates to alert you to the nature of bending in gear teeth
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears 720 © The McGraw−Hill Companies, 2008 722 Mechanical Engineering Design Setting Sa = Sm and solving the quadratic in Sa gives Sa = S2 ut 2S e −1 + 1 + 4S2 e S2 ut Setting Sa = σ/2, Sut = S e/0.5 gives σ = S e 0.52 −1 + 1 + 4(0.5)2 = 1.66S e and kf = σ/S e = 1.66. Since a Gerber locus runs in and among fatigue data and Goodman does not, we will use kf = 1.66. The second effect to be accounted for in using the miscellaneous-effects Marin factor kf is stress concentration, for which we will use our fundamentals from Chap. 6. For a 20◦ full-depth tooth the radius of the root fillet is denoted rf , where rf = 0.300 P = 0.300 8 = 0.0375 in From Fig. A–15–6 r d = rf t = 0.0375 0.250 = 0.15 Since D/d = ∞, we approximate with D/d = 3, giving Kt = 1.68. From Fig. 6–20, q = 0.62. From Eq. (6–32) Kf = 1 + (0.62)(1.68 − 1) = 1.42 The miscellaneous-effects Marin factor for stress concentration can be expressed as kf = 1 Kf = 1 1.42 = 0.704 The final value of kf is the product of the two kf factors, that is, 1.66(0.704) = 1.17. The Marin equation for the fully corrected endurance strength is Se = kakbkckd kekf S e = 0.934(0.948)(1)(1)(1)1.17(27.5) = 28.5 kpsi For a design factor of nd = 3, as used in Ex. 14–1, applied to the load or strength, the allowable bending stress is σall = Se nd = 28.5 3 = 9.5 kpsi The transmitted load Wt is Wt = FYσall Kv P = 1.5(0.296)9 500 1.52(8) = 347 lbf and the power is, with V = 628 ft/min from Ex. 14–1, hp = Wt V 33 000 = 347(628) 33 000 = 6.6 hp Again, it should be emphasized that these results should be accepted only as preliminary estimates to alert you to the nature of bending in gear teeth