上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch14 spur and helical gears

Budynas-Nisbett Shigleys I Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hil m Mechanical Engineering Elements Companies,2008 Design,Eighth Edition 4 Spur and Helical Gears Chapter Outline 14-1 The Lewis Bending Equation 714 14-2 Surface Durability 723 14-3 AGMA Stress Equations 725 14-4 AGMA Strength Equations 727 14-5 Geometry Factors I and J(Z and Y1 731 14-6 The Elastic Coefficient Cp (Z 736 14-7 Dynamic Factor K,736 14-8 Overload Factor K。738 14-9 Surface Condition Factor Cr (Zel 738 14-10 Size Factor K 739 14-11 Load-Distribution Factor Km(KH 739 14-12 Hardness-Ratio Factor CH 741 14-13 Stress Cycle Life Factors YN and ZN 742 14-14 Reliability Factor Ke(Y 743 14-15 Temperature Factor KT (Ya 744 14-16 Rim-Thickness Factor K8 744 14-17 Safety Factors Se and SH 745 14-18 Analysis 745 14-19 Design of a Gear Mesh 755 713

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears © The McGraw−Hill 711 Companies, 2008 Spur and Helical Gears Chapter Outline 14–1 The Lewis Bending Equation 714 14–2 Surface Durability 723 14–3 AGMA Stress Equations 725 14–4 AGMA Strength Equations 727 14–5 Geometry Factors I and J (ZI and YJ) 731 14–6 The Elastic Coefficient Cp (ZE) 736 14–7 Dynamic Factor Kv 736 14–8 Overload Factor Ko 738 14–9 Surface Condition Factor Cf (ZR) 738 14–10 Size Factor Ks 739 14–11 Load-Distribution Factor Km (KH) 739 14–12 Hardness-Ratio Factor CH 741 14–13 Stress Cycle Life Factors YN and ZN 742 14–14 Reliability Factor KR (YZ) 743 14–15 Temperature Factor KT (Yθ) 744 14–16 Rim-Thickness Factor KB 744 14–17 Safety Factors SF and SH 745 14–18 Analysis 745 14–19 Design of a Gear Mesh 755 14 713

12 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 714 Mechanical Engineering Design This chapter is devoted primarily to analysis and design of spur and helical gears to resist bending failure of the teeth as well as pitting failure of tooth surfaces.Failure by bend- ing will occur when the significant tooth stress equals or exceeds either the yield strength or the bending endurance strength.A surface failure occurs when the significant contact stress equals or exceeds the surface endurance strength.The first two sections present a little of the history of the analyses from which current methodology developed. The American Gear Manufacturers Association'(AGMA)has for many years been the responsible authority for the dissemination of knowledge pertaining to the design and analysis of gearing.The methods this organization presents are in general use in the United States when strength and wear are primary considerations.In view of this fact it is important that the AGMA approach to the subject be presented here. The general AGMA approach requires a great many charts and graphs-too many for a single chapter in this book.We have omitted many of these here by choosing a single pressure angle and by using only full-depth teeth.This simplification reduces the complexity but does not prevent the development of a basic understanding of the approach.Furthermore,the simplification makes possible a better development of the fundamentals and hence should constitute an ideal introduction to the use of the general AGMA method.2 Sections 14-1 and 14-2 are elementary and serve as an examination of the foundations of the AGMA method.Table 14-1 is largely AGMA nomenclature. 14-1 The Lewis Bending Equation Wilfred Lewis introduced an equation for estimating the bending stress in gear teeth in which the tooth form entered into the formulation.The equation,announced in 1892, still remains the basis for most gear design today To derive the basic Lewis equation,refer to Fig.14-1a,which shows a cantilever of cross-sectional dimensions F and t,having a length and a load W,uniformly dis- tributed across the face width F.The section modulus //c is Ft-/6,and therefore the bending stress is M 6W'I (a) Gear designers denote the components of gear-tooth forces as W,Wr,Wa or W,W, Wa interchangeably.The latter notation leaves room for post-subscripts essential to free- body diagrams.For instance,for gears 2 and 3 in mesh,W is the transmitted force of 500 Montgomery Street.Suite 350,Alexandria.VA 22314-1560. 2The standards ANSI/AGMA 2001-D04 (revised AGMA 2001-C95)and ANSI/AGMA 2101-D04 (metric edition of ANSUAGMA 2001-D04).Fundamental Rating Factors and Calculation Methods for Imvolute Spur and Helical Gear Teeth,are used in this chapter.The use of American National Standards is completely voluntary:their existence does not in any respect preclude people,whether they have approved the standards or not,from manufacturing,marketing.purchasing,or using products.processes.or procedures not conforming to the standards. The American National Standards Institute does not develop standards and will in no circumstances give an interpretation of any American National Standard.Requests for interpretation of these standards should be addressed to the American Gear Manufacturers Association.[Tables or other self-supporting sections may be quoted or extracted in their entirety.Credit line should read:"Extracted from ANSI/AGMA Standard 2001-D04 or 2101-D04 Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth"with the permission of the publisher.American Gear Manufacturers Association, 500 Montgomery Street.Suite 350.Alexandria,Virginia 22314-1560.]The foregoing is adapted in part from the ANSI foreword to these standards

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears 712 © The McGraw−Hill Companies, 2008 714 Mechanical Engineering Design 1 500 Montgomery Street, Suite 350, Alexandria, VA 22314-1560. 2 The standards ANSI/AGMA 2001-D04 (revised AGMA 2001-C95) and ANSI/AGMA 2101-D04 (metric edition of ANSI/AGMA 2001-D04), Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth, are used in this chapter. The use of American National Standards is completely voluntary; their existence does not in any respect preclude people, whether they have approved the standards or not, from manufacturing, marketing, purchasing, or using products, processes, or procedures not conforming to the standards. The American National Standards Institute does not develop standards and will in no circumstances give an interpretation of any American National Standard. Requests for interpretation of these standards should be addressed to the American Gear Manufacturers Association. [Tables or other self-supporting sections may be quoted or extracted in their entirety. Credit line should read: “Extracted from ANSI/AGMA Standard 2001-D04 or 2101-D04 Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth” with the permission of the publisher, American Gear Manufacturers Association, 500 Montgomery Street, Suite 350, Alexandria, Virginia 22314-1560.] The foregoing is adapted in part from the ANSI foreword to these standards. This chapter is devoted primarily to analysis and design of spur and helical gears to resist bending failure of the teeth as well as pitting failure of tooth surfaces. Failure by bend￾ing will occur when the significant tooth stress equals or exceeds either the yield strength or the bending endurance strength. A surface failure occurs when the significant contact stress equals or exceeds the surface endurance strength. The first two sections present a little of the history of the analyses from which current methodology developed. The American Gear Manufacturers Association1 (AGMA) has for many years been the responsible authority for the dissemination of knowledge pertaining to the design and analysis of gearing. The methods this organization presents are in general use in the United States when strength and wear are primary considerations. In view of this fact it is important that the AGMA approach to the subject be presented here. The general AGMA approach requires a great many charts and graphs—too many for a single chapter in this book. We have omitted many of these here by choosing a single pressure angle and by using only full-depth teeth. This simplification reduces the complexity but does not prevent the development of a basic understanding of the approach. Furthermore, the simplification makes possible a better development of the fundamentals and hence should constitute an ideal introduction to the use of the general AGMA method.2 Sections 14–1 and 14–2 are elementary and serve as an examination of the foundations of the AGMA method. Table 14–1 is largely AGMA nomenclature. 14–1 The Lewis Bending Equation Wilfred Lewis introduced an equation for estimating the bending stress in gear teeth in which the tooth form entered into the formulation. The equation, announced in 1892, still remains the basis for most gear design today. To derive the basic Lewis equation, refer to Fig. 14–1a, which shows a cantilever of cross-sectional dimensions F and t, having a length l and a load Wt , uniformly dis￾tributed across the face width F. The section modulus I/c is Ft 2/6, and therefore the bending stress is σ = M I/c = 6Wt l Ft 2 (a) Gear designers denote the components of gear-tooth forces as Wt , Wr, Wa or Wt , Wr, Wa interchangeably. The latter notation leaves room for post-subscripts essential to free￾body diagrams. For instance, for gears 2 and 3 in mesh, Wt 23 is the transmitted force of

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears ©The McGraw-Hil 13 Mechanical Engineering Elements Companies,2008 Design,Eighth Edition Spur and Helical Gears 715 Table 14-1 Symbol Name Where Found Symbols,Their Names, b Net width of face of narrowest member Eq.(14-161 and Locations* Ce Mesh alignment correction factor Eq.(14-351 C Surface condition factor Eq.14-16) CH Hardness-ratio factor Eq.(14-18) Cma Mesh alignment factor Eq.14-34 Cme Load correction factor Eq.(14-31) Cmf Face load-distribution factor Eq.(14-30) Cp Elastic coefficient Eq.14-13 Cpf Pinion proportion factor Eq.(14-32 Cpm Pinion proportion modifier Eq.(14-33) d Operating pitch diameter of pinion Ex.(14-1) dp Pitch diameter,pinion Eq.(14-22) Pitch diameter,gear Eq.(14-22) E Modulus of elasticity Eq.14-10 F Net face width of narrowest member Eq.I14-15 Pinion surface finish Fig.14-13 H Power Fig.14-17 Ha Brinell hardness Ex.14-3 HBG Brinell hardness of gear Sec.14-12 Hap Brinell hardness of pinion Sec.14-12 hp Horsepower Ex.14-] hr Gear-tooth whole depth Sec.14-16 Geometry factor of pitting resistance Eq.(14-16) Geometry factor for bending strength Eq.14-15) K Contact load factor for pitting resistance Eq.6-651 K Rim-thickness factor Eq.(14-401 K Fatigue stress-concentration factor Eq.(14-91 Load-distribution factor Eq.(14-301 to Overload factor Eq.14-15) Reliability factor Eq.(14-17) K Size factor Sec.14-10 K Temperature factor Eq.(14-17刀 K Dynamic factor Eq.(14-27 m Metric module Eq.(14-15) mg Backup ratio Eq.(14-391 mG Gear ratio (never less than 1) Eq.14-22) mN Load-sharing ratio Eq.14-21 N Number of stress cycles Fig.14-14 NG Number of teeth on gear Eq.(14-22) Np Number of teeth on pinion Eq.I14-22) n Speed E×.14-1 [Continued)

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears © The McGraw−Hill 713 Companies, 2008 Spur and Helical Gears 715 Symbol Name Where Found b Net width of face of narrowest member Eq. (14–16) Ce Mesh alignment correction factor Eq. (14–35) Cf Surface condition factor Eq. (14–16) CH Hardness-ratio factor Eq. (14–18) Cma Mesh alignment factor Eq. (14–34) Cmc Load correction factor Eq. (14–31) Cmf Face load-distribution factor Eq. (14–30) Cp Elastic coefficient Eq. (14–13) Cpf Pinion proportion factor Eq. (14–32) Cpm Pinion proportion modifier Eq. (14–33) d Operating pitch diameter of pinion Ex. (14–1) dP Pitch diameter, pinion Eq. (14–22) dG Pitch diameter, gear Eq. (14–22) E Modulus of elasticity Eq. (14–10) F Net face width of narrowest member Eq. (14–15) fP Pinion surface finish Fig. 14–13 H Power Fig. 14–17 HB Brinell hardness Ex. 14–3 HBG Brinell hardness of gear Sec. 14–12 HBP Brinell hardness of pinion Sec. 14–12 hp Horsepower Ex. 14–1 ht Gear-tooth whole depth Sec. 14–16 I Geometry factor of pitting resistance Eq. (14–16) J Geometry factor for bending strength Eq. (14–15) K Contact load factor for pitting resistance Eq. (6–65) KB Rim-thickness factor Eq. (14–40) Kf Fatigue stress-concentration factor Eq. (14–9) Km Load-distribution factor Eq. (14–30) Ko Overload factor Eq. (14–15) KR Reliability factor Eq. (14–17) Ks Size factor Sec. 14–10 KT Temperature factor Eq. (14–17) Kv Dynamic factor Eq. (14–27) m Metric module Eq. (14–15) mB Backup ratio Eq. (14–39) mG Gear ratio (never less than 1) Eq. (14–22) mN Load-sharing ratio Eq. (14–21) N Number of stress cycles Fig. 14–14 NG Number of teeth on gear Eq. (14–22) NP Number of teeth on pinion Eq. (14–22) n Speed Ex. 14–1 Table 14–1 Symbols, Their Names, and Locations∗ (Continued)

14 Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears ©The McGraw-Hil Mechanical Engineering Elements Companies,2008 Design,Eighth Edition 716 Mechanical Engineering Design Table 14-1 Symbol Name Where Found Symbols,Their Names, np Pinion speed Ex.14-4 and Locations' P Diametral pitch Eq.14-21 [Continued) Pd Diametral pitch of pinion Eq.(14-151 PN Normal base pitch Eq.(14-24) Pn Normal circular pitch Eq.(14-241 Px Axial pitch Eq.(14-191 Q, Transmission accuracy level number Eq.(14-291 R Reliability Eq.(14-381 R Root-mean-squared roughness Fig.14-13 Tooth fillet radius Fig.14-1 rG Pitch-circle radius,gear In standard p Pitch-circle radius,pinion In standard Tbp Pinion base-circle radius Eq.(14-251 fbG Gear base-circle radius Eq.(14-251 Sc Buckingham surface endurance strength E×.14-3 Se AGMA surface endurance strength Eq.(14-181 S AGMA bending strength Eq.14-17 Bearing span Fig.14-10 S1 Pinion offset from center span Fig.14-10 Se Safety factor-bending Eq.(14-41) SH Safely factor-pitting Eq.(14-42) Wor W! Transmitted load Fig.14-1 YN Stress cycle factor for bending strength Fig.14-14 ZN Stress cycle factor for pitting resistance Fig.14-15 B Exponent Eq.(14-44 0 Bending stress Eq.(14-2) oc Contact stress from Hertzian relationships Eq.(14-14) e Contact stress from AGMA relationships Eq.(14-161 Call Allowable bending stress Eq.14-17 Oc,all Allowable contact stress,AGMA Eq.(14-181 必 Pressure angle Eq.(14-12) Transverse pressure angle Eq.(14-23) Helix angle at standard pitch diameter Ex.14-5 Because ANSI/AGMA 2001-95inuedsignificntmo ofnew nomendatre,nd cnind in ANSI/AGMA 21-04 this summary and references are provided for use until the reader's vocabulary has grown. See preferenie following E.()Se.14-1. body 2 on body 3,and W is the transmitted force of body 3 on body 2.When working with double-or triple-reduction speed reducers,this notation is compact and essential to clear thinking.Since gear-force components rarely take exponents,this causes no com- plication.Pythagorean combinations,if necessary,can be treated with parentheses or avoided by expressing the relations trigonometrically

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears 714 © The McGraw−Hill Companies, 2008 716 Mechanical Engineering Design Symbol Name Where Found nP Pinion speed Ex. 14–4 P Diametral pitch Eq. (14–2) Pd Diametral pitch of pinion Eq. (14–15) pN Normal base pitch Eq. (14–24) pn Normal circular pitch Eq. (14–24) px Axial pitch Eq. (14–19) Qv Transmission accuracy level number Eq. (14–29) R Reliability Eq. (14–38) Ra Root-mean-squared roughness Fig. 14–13 rf Tooth fillet radius Fig. 14–1 rG Pitch-circle radius, gear In standard rP Pitch-circle radius, pinion In standard rbP Pinion base-circle radius Eq. (14–25) rbG Gear base-circle radius Eq. (14–25) SC Buckingham surface endurance strength Ex. 14–3 Sc AGMA surface endurance strength Eq. (14–18) St AGMA bending strength Eq. (14–17) S Bearing span Fig. 14–10 S1 Pinion offset from center span Fig. 14–10 SF Safety factor—bending Eq. (14–41) SH Safety factor—pitting Eq. (14–42) Wt or W† t Transmitted load Fig. 14–1 YN Stress cycle factor for bending strength Fig. 14–14 ZN Stress cycle factor for pitting resistance Fig. 14–15 β Exponent Eq. (14–44) σ Bending stress Eq. (14–2) σC Contact stress from Hertzian relationships Eq. (14–14) σc Contact stress from AGMA relationships Eq. (14–16) σall Allowable bending stress Eq. (14–17) σc,all Allowable contact stress, AGMA Eq. (14–18) φ Pressure angle Eq. (14–12) φt Transverse pressure angle Eq. (14–23) ψ Helix angle at standard pitch diameter Ex. 14–5 ∗Because ANSI/AGMA 2001-C95 introduced a significant amount of new nomenclature, and continued in ANSI/AGMA 2001-D04, this summary and references are provided for use until the reader’s vocabulary has grown. † See preference rationale following Eq. (a), Sec. 14–1. Table 14–1 Symbols, Their Names, and Locations∗ (Continued) body 2 on body 3, and Wt 32 is the transmitted force of body 3 on body 2. When working with double- or triple-reduction speed reducers, this notation is compact and essential to clear thinking. Since gear-force components rarely take exponents, this causes no com￾plication. Pythagorean combinations, if necessary, can be treated with parentheses or avoided by expressing the relations trigonometrically

Budynas-Nisbett:Shigley's Ill.Design of Mechanical 14.Spur and Helical Gears T©The McGraw-Hil 715 Mechanical Engineering Elements Companies,2008 Design,Eighth Edition Spur and Helical Gears 717 I Figure 14-1 (a) Referring now to Fig.14-16,we assume that the maximum stress in a gear tooth occurs at point a.By similar triangles,you can write 业= x 1/2 (6) By rearranging Eq.(a), 6W1w1W11 0= F=下21a=下24吾 (d) If we now substitute the value ofx from Eq.(b)in Eq.(c)and multiply the numerator and denominator by the circular pitch p,we find Wp G=F③xP (d) Letting y =2x/3p,we have W 0= (14-11 Fpy This completes the development of the original Lewis equation.The factor y is called the Lewis form factor,and it may be obtained by a graphical layout of the gear tooth or by digital computation. In using this equation,most engineers prefer to employ the diametral pitch in determining the stresses.This is done by substituting P=z/p and Y=zy in Eq.(14-1).This gives WP 0= FY (14-2 where r=2tp (14-3) 3 The use of this equation for Y means that only the bending of the tooth is considered and that the compression due to the radial component of the force is neglected.Values of Y obtained from this equation are tabulated in Table 14-2

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 14. Spur and Helical Gears © The McGraw−Hill 715 Companies, 2008 Spur and Helical Gears 717 Figure 14–1 l F t W t W t W r l t a rf x W (a) (b) Referring now to Fig. 14–1b, we assume that the maximum stress in a gear tooth occurs at point a. By similar triangles, you can write t/2 x = l t/2 or x = t 2 4l (b) By rearranging Eq. (a), σ = 6Wt l Ft 2 = Wt F 1 t 2/6l = Wt F 1 t 2/4l 1 4 6 (c) If we now substitute the value of x from Eq. (b) in Eq. (c) and multiply the numerator and denominator by the circular pitch p, we find σ = Wt p F 2 3 xp (d) Letting y = 2x/3p, we have σ = Wt Fpy (14–1) This completes the development of the original Lewis equation. The factor y is called the Lewis form factor, and it may be obtained by a graphical layout of the gear tooth or by digital computation. In using this equation, most engineers prefer to employ the diametral pitch in determining the stresses. This is done by substituting P = π/p and Y = πy in Eq. (14–1). This gives σ = Wt P FY (14–2) where Y = 2x P 3 (14–3) The use of this equation for Y means that only the bending of the tooth is considered and that the compression due to the radial component of the force is neglected. Values of Y obtained from this equation are tabulated in Table 14–2.

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 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 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 tangen￾tial 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 sub￾stitute 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 under￾stand 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 con￾siderations. 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 adden￾dum 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 prelimi￾nary estimates to alert you to the nature of bending in gear teeth