12. Spur Gear Design and Selection Standard Proportions Forces On

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12. Spur Gear Design and Standard proportions selection Objectives • American Standard Association (ASA) • Apply principles learned in Chapter 11 to actual design and selection of spur gear systems. • American Gear Manufacturers Association • Calculate forces on teeth of spur gears, including impact forces associated with velocity and clearances. (AGMA) • Determine allowable force on gear teeth, including the factors necessary due to angle of involute of tooth shape and materials selected for gears. • Brown and Sharp • Design actual gear systems, including specifying materials, manufacturing accuracy, and other factors necessary for complete spur gear design. • 14 ½ deg; 20 deg; 25 deg pressure angle • Understand and determine necessary surface hardness of gears to minimize or prevent surface wear. • Understand how lubrication can cushion the impact on gearing systems and cool • Full depth and stub tooth systems them. • Select standard gears available from stocking manufacturers or distributors.

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Specifications for standard gear teeth Forces on spur gear teeth

Item Full depth & pitches Full depth & 14½° full •F= Transmitted force coarser than 20 pitches finer depth t than 20 •Fn = Normal force or separating Pressure 20° 25° 20° 14½° force angle •Fr = Resultant force Addendum 1.0/P 1.0/P 1.0/P 1/P • θ = pressure angle (in.) •Fn = Ft tan θ Dedendum 1.250/P 1.250/P 1.2/P + 0.002 1.157/P F (in.) F = t r cos θ P N RAO 3 P N RAO 4

Forces on spur gear teeth Example Problem 12-1: Forces on Spur Gear Teeth T n 63,000 P • Power, P; P = or T = • 20-tooth, 8 pitch, 1-inch-wide, 20° pinion 63,000 n transmits 5 hp at 1725 rpm to a 60-tooth gear.

• Torque, T = Ft r and r = Dp /2 • Determine driving force, separating force, and • Combining the above we can write maximum force that would act on mounting shafts. 2 T Ft = Dp

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1 Example Problem 12-1: Forces on Spur Gear Teeth Example Problem 12-1: Forces on Spur Gear Teeth (cont’d.) − Find transmitted force: • 20-tooth, 8 pitch, 1-inch-wide, 20° pinion transmits 5 hp at 1725 rpm to a 60-tooth gear. (12-3) 2T • Determine driving force, separating force, and maximum force that would act on F = t D mounting shafts. p (2-6) (2)183 in-lb Tn Ft = = 146 lb P = 2.5 in 63,000 − Find separating force: 63,000P T = (12-1) n

Fn = Ft tan θ (63,000)5 T = = 183 in-lb 1725 Fn = 146 lb tan 20°

Fn = 53 lb

− Find pitch circle: − Find maximum force: (12-2) (11-4) F F = t Np r cos θ Dp = Pd 146 lb 20 teeth Fr = D = = 2.5 in cos 20° p 8 teeth/in diameter

Fr = 155 lb P N RAO 7 P N RAO 8

Example Problem 12-2: Surface Speed Surface Speed

• In previous problem, determine the surface speed:

• Surface speed (Vm) is often referred to as pitch-line speed (12-7) Vm = π D n π D n or p (12-5) •Vm = ft/min π D n 12 V = p m 12

π Dp n ft V = π 2.5 in 1725 rpm •Vm = m/min -- Metric units m 12 in 1000 Vm = 1129 ft/min

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Forces on Gear Tooth Strength of Gear Teeth

• Lewis form factor method

Figure 14.20 Forces acting on individual gear tooth.

2 Lewis equation Table 12.1 Lewis form factors (Y)

Sn Y b Fs = Pd

•Fs = Allowable dynamic bending force (lb) 2 •Sn = Allowable stress (lb/in ). Use endurance limit and account for the fillet as the stress concentration factor • b = Face width (in.) • Y = Lewis form factor (Table 12.1)

•Pd = Diametral pitch

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Example Problem 12-3: Strength of Gear Teeth Example Problem 12-3: Strength of Gear Teeth (cont’d.)

• In Example Problem 12-1, determine the force allowable (F ) on these teeth if the s − Gear: pinion is made from an AISI 4140 annealed steel, the mating gear is made from AISI 1137 hot-rolled steel, and long life is desired. Sn = .5 (88 ksi) = 44 ksi − Pinion: (Table 12-1)

Sn = .5 Su = .5 (95 ksi) = 47.5 ksi Y = .421 (12-9) 44,000 (1) .421 Fs = Sn b Y 8 Fs = Pd

Fs = 2316 lb − Find Lewis form factor (Y) from Table 12-1, assuming full-depth teeth:

Y = .320

47,500 (1) .320 − Use Fs = 1900 lb for design purposes. F = s 8

Fs = 1900 lb

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Classes of Gears Force Transmitted

• Transmitted load depends on the accuracy of the • Transmitted load depends on the accuracy of the gears gears • Gear Manufacture • A dynamic load factor is added to take care of – Casting this. – Machining • Forming •Ft = Transmitted force • Hobbing •Fd = Dynamic force • Shaping and Planing • Commercial 600 + V – Forming F = m F – stamping d 600 t

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3 Classes of Gears Design Methods 1200 + V • Carefully cut F = m F • Strength of gear tooth should be greater d 1200 t than the dynamic force; Fs ≥ Fd 0.5 • Precision 78 + Vm • You should also include the factor of safety, Fd = Ft 78 Nsf F s ≥ F 50 + V0.5 N d • Hobbed or shaved F = m F sf d 50 t

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Table 12.2 Service Factors Face width of Gears

• Relation between the width of gears and the diametral pitch 1 < Nsf < 1.25 Uniform load without shock 1.25 < N < 1.5 Medium shock, frequent starts 8 12.5 sf < b < P P 1.5 < Nsf < 1.75 Moderately heavy shock d d

1.75 < Nsf < 2 Heavy shock

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Example Problem 12-4: Design Methods Example Problem 12-5: Design Methods • If, in Example Problem 12-1, the gears are commercial grade, determine dynamic load and, based on force allowable from Example Problem 12-3, would this be an acceptable • Spur gears from the catalog page shown in Figure 12-3 are made from a .2% carbon design if a factor of safety of 2 were desired? steel with no special heat treatment. • Use surface speed and force transmitted from Example Problems 12-2 and 12-3. • What factor of safety do they appear to use in this catalog? – Dynamic load: • Try a 24-tooth at 1800 rpm gear for example purposes. 600 + V m (12-10) F = F • From Appendix 4, an AISI 1020 hot-rolled steel would have .2% carbon with an Sy = 30 d 600 t ksi and Su = 55 ksi. 600 + 1129 F = (146 lb) − Therefore: d 600 Sn = .5 Su Fd = 421lb

Sn = .5 (55 ksi) = 27.5 ksi – Comparing to force allowable:

− Find Dp : Fs ≥ Fd (11-4) N sf N 1900 lb D = p ≥ 421 p P 2 d 950 lb > 421lb 24 D = = 2 in • This design meets the criteria. p 12 P N RAO 23 P N RAO 24

4 Example Problem 12-5: Design Methods (cont’d) Example Problem 12-5: Design Methods (cont’d)

− Find Vm : − Set Fs = Fd and solve for Ft :

(12-5) ⎛ 600 + Vm ⎞ F = F = F d s ⎜ 600 ⎟ t π D n ⎝ ⎠ V = p m 12 ⎛ 600 + 942 ⎞ 519 lb = ⎜ ⎟ F ⎝ 600 ⎠ t ft Vm = π 2 in (1800 rpm) 12 in Ft = 202 lb

(12-3) Vm = 942 ft/min

⎛ Dp ⎞ T = F ⎜ ⎟ − Find Fs : ⎝ 2 ⎠

(12-9) ⎛ 2 in ⎞ T = 202 lb ⎜ ⎟ S b Y ⎝ 2 ⎠ F = n s P d T = 202 in-lb

(from Table 12-1) (2-6) Tn 202 (1800) P = = = 5.8 hp Y = .302 63,000 63,000 27,500 (¾) .302 F = or s 12 (12-6) F = 519 lb F V 202 (942) s P = t m = = 5.8 hp P N RAO 25 33,000 P 33,000N RAO 26

Example Problem 12-5: Design Methods (cont’d) Example Problem 12-6: Design Methods

− Compared to catalog: • Pair of commercial-grade spur gears is to transmit 2 hp at a speed of 900 rpm of the pinion and 300 rpm for gear. • If class 30 cast iron is to be used, specify a possible design for this problem. • The following variables are unknown: P , D , b, N , N , . hp - calculated d p t sf θ N = • As it is impossible to solve for all simultaneously, try the following: sf hp - catalog – Pd = 12, Nt = 48, θ = 14½°, Nsf = 2 – Solve for face width b. 5.8 N = = 1.4 sf 4.14 − Miscellaneous properties:

(11-4) • Appears to be reasonable value. Np 48 • Manufacturer may also have reduced its rating for wear purposes as these Dp = = = 4 in Pd 12 are not hardened gears.

Ng = Np Vr = 48(3) = 144 teeth

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Example Problem 12-6: Design Methods (cont’d.) Example Problem 12-6: Design Methods (cont’d.) ⎛ 600 + Vm ⎞ (12-10) – Dynamic force Fd = ⎜ ⎟ Ft ⎝ 600 ⎠ − Surface speed: ⎛ 600 + 943 ⎞ Fd = ⎜ ⎟ 70 (12-5) ⎝ 600 ⎠ π D n Fd = 180 lb V = p m 12 – Since width b is the unkown: F 4 in 900 rpm s ≥ F Vm = π d 12 in/ft Nsf and Vm = 943 ft/min S b Y F = n − Finding force on teeth: s Pd (12-8) (12-6) Sn b Y = Fd 33,000 hp Nsf Pd Ft = Vm

33,000 (2) –Class 30 CI; Su = 30 ksi; Sn = .4 Su (.4 is used because cast iron): F = t 943 –Sn = 12 ksi –Y = .344 Ft = 70 lb (Table 12-1) P N RAO 29 P N RAO 30

5 Example Problem 12-6: Design Methods (cont’d.) − Substituting: • To increase the dynamic beam strength of

(12-9) the gear

Sn b Y = Fd Nsf Pd – Increase tooth size by decreasing the

12,000 b .344 diametral pitch = 180 2 (12) – Increase face width upto the pitch diameter b = 1.0 inches of the pinion − Check ratio of width to pitch: – Select material of greater endurance limit (12-14) 8 12.5 < b < – Machine tooth profiles more precisely Pd Pd – Mount gears more precisely 8 12.5 < 1 < 12 12 – Use proper lubricant and reduce .66 < 1 < 1.04 contamination • This is an acceptable design. • Many other designs are also possible. P N RAO 31 P N RAO 32

Buckingham Method of Gear Design Fig. 12.4 Expected error in tooth profiles

• It offers greater flexibility • Expected error is based on different-pitch teeth • More conservative design

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Table 12.3 Values of C for e = 0.001 inch Buckingham Method of Gear Design

Material 14 ½ deg 20 deg

Gray iron and Gray iron 800 830 0.05 Vm (bC + Ft ) Fd = Ft + 0.5 Gray iron and steel 1,100 1,140 0.05 Vm + (bC + Ft ) Steel and steel 1,600 1,660

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6 Example Problem 12-7: Buckingham Method of Gear Fig. 12.5 Recommended maximum error in gear teeth Design and Expected Error

• A pair of steel gears is made from annealed AISI 3140. • Each gear has a surface hardness of BHN = 350. • The pinion is a precision, 16 pitch, 20° involute, with 24 teeth 1 inch wide. • The gear has 42 teeth. • To transmit 3 hp at a speed of 3450 rpm for a safety factor of 1.4, is this a suitable design? (Appendix 4)

Su = 95 ksi

Sn = .5 Su = .5(95 ksi) = 47.5 ksi

(11-4)

Np 24 Dp = = = 1.5 in Pd 16

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Example Problem 12-7: Buckingham Method of Gear Example Problem 12-7: Buckingham Method of Gear Design and Expected Error (cont’d.) Design and Expected Error (cont’d.) − Find torque: − Find surface speed:

(2-6) (12-5)

Tn π Dp n P = Vm = 63,000 12

P (63,000) π 1.5 in 3450 rpm T = V = n m 12 ft/in

3 (63,000) V = 1355 ft/min T = m 3450 − Find force allowable (Fs): T = 55 in-lb (12-9)

− Find force transmitted: Sn b Y Fs = Pd (12-3) 2T (from Table 12-1) Ft = Dp Y = .337

2(55 in-lb) 47,500 (1) .337 Ft = F = 1.5 in s 16

Ft = 73 lb F = 1000 P N RAO 39 s P N RAO 40

Example Problem 12-7: Buckingham Method of Gear Design and Expected Error (cont’d.) Wear strength (Buckingham) • Expected error from Figure 12-4: e = .0005 • The value of C from Table 12-3 for steel and steel and 20° involute angle gears is 1660: C = .0005 (1000) 1660 ⎛ 2 D ⎞ C = 830 F = D b K ⎜ g ⎟ • Solve for dynamic load using Buckingham’s equation: W P g ⎜ ⎟ Dp + Dg .05V (bC + F ) ⎝ ⎠ F = F + m t d t 1/ 2 (12-15) .05 Vm + (bC + Ft ) .05 (1355) (1• 830 + 73) Fw = tooth wear strength (lb) Fd = 73 + .05 (1355) + (1 • 830 + 73)1/ 2 Dp = diametral pitch of pinion (in.) Fd = 699 lb

Fs ≥ Fd Dg = diametral pitch of gear (in.) N sf 1000 b = face width (in.) ≥ 699 1.4 714 > 699 Kg = load stress factor (Appendix 13)

• This meets the criteria. P N RAO 41 P N RAO 42

7 Example Problem 12-8: Wear of Gears Example Problem 12-8: Wear of Gears (cont’d.)

• In prior example problem, verify the surface is suitable for wear considerations. – Substituting into equation 12-16:

For wear use Nss = 1.2 Fw = 1.5 (1) 1.27 (270) − Wear formula: (12-16) Fw = 514 Fw = Dp b Q Kg

• This would not be suitable. Try if surfaces each had a BHN = 450.

− Find Q : K = 470 g (from Appendix 13) (12-17) Fw = 1.5 (1) (1.27) (470) 2 Ng Q = Fw = 895 Ng + Np Fw 2 (42) ≥ Fd Q = 42 + 24 N sf 895 Q = 1.27 > 699 1.2 (from Appendix 13) 746 > 699 Kg = 270 • This would now be acceptable if the gear teeth were hardened to a BHN of 450. P N RAO 43 P N RAO 44