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CH#14 Spur and Helical Gears

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The Lewis Form Factor

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The Bending stresses

• Examination of run-in teeth will show that the heaviest loads occur near the middle of the tooth.

• 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.

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• Only spur gear will be analysed• Gear bending stress equation “𝜎”

• Bending factor of safety “SF”

14-3 AGMA Stress equation/Analysis

SI units US customary units

SI units US customary units

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14-3 AGMA Stress equation/Analysis…

• The endurance strength “𝜎 ” equation

US customary units

SI units

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• Gear Contact stress “𝜎 ” equation (Wear)

• Wear factor of safety “SH”

14-3 AGMA Stress equation/Analysis…

SI units US customary units

US customary unitsSI units

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14-3 AGMA Stress equation/Analysis…

• The gear contact endurance strength “𝜎 , ” equation is

US customary units

SI units

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Allowable bending stress FP (St)

• For steel, from figures 14-2, 14-3, 14-4

• For other materials, refer to Table 14-3 & 14-4

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Allowable contact stress HP (Sc)

• For through hardened steel use Fig. 14-5• For all other, use Table 14-6 and 14-7

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14-5 Geometry Factors YJ (J) and ZI (I)

• YJ (I) is the bending-strength geometry factor used in the bending-stress equation

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14-5 Geometry Factors YJ (J) and ZI (I)…

• ZI (I) is the surface-strength geometry factor used in the contact-stress equation

• Can be determined as (for external gear)

𝑍𝐼 𝐼𝑐𝑜𝑠∅ 𝑠𝑖𝑛∅

2𝑀𝑚

𝑚 1• Where

• MN = 1 for spur gear• mG is the gear ratio

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14-6 The Elastic Coefficient ZE (Cp)

• The elastic coefficient is calculated from Table 14-8

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achine Design II• used to account for inaccuracies in the manufacture and meshing of

gear teeth in action.

• Inaccuracies produced in the generation of the tooth profile; these include errors in tooth spacing, profile lead, and runout

• Vibration of the tooth during meshing due to the tooth stiffness

• Magnitude of the pitch-line velocity

• Dynamic unbalance of the rotating members

• Wear and permanent deformation of contacting portions of the teeth

• Gear-shaft misalignment and the linear and angular deflection of the shaft

• Tooth friction

• K𝑣 can also be determined using Figure 14-9 (take care of unit system)

14-7 Dynamic factor Kv

𝐾 , (V in m/s)

𝐵 0.25 12 𝑄

𝐴 50 56 1 𝐵

Qv is related to the transmission accuracy grade number:

3 𝑄 123-7: Commercial quality gears8-12: Precision quality

Note: in 2004, AGMA replaced Qv with Av

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• is intended to make allowance for all externally applied loads in excess of the nominal tangential load Wt in a particular application

• Examples include variations in torque from the mean value due to firing of cylinders in an internal combustion engine or reaction to torque variations in a piston pump drive. There are other similar factors such as application factor or service factor.

• These factors are established after considerable field experience in a particular application

14-8 Overload factor K0

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achine Design II• It depends on

– Surface finish affected by cutting, shaving, lapping, grinding, shot peening.

– Residual stresses– Work hardening– The value should be greater than unity if

desired.– For normal condition use ZR = Cf = 1

14-9 Surface Condition factor ZR (Cf)

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• The size factor reflects non-uniformity of material properties due to size. It depends upon– Tooth size

– Diameter of part

– Ratio of tooth size to diameter of part

– Face width

– Area of stress pattern

– Ratio of case depth to tooth size

– Hardenability and heat treatment

𝐾 0.904 𝑏𝑚 𝑌.

(SI units)

𝐾 1.192.

(US customary units)

• You may set Ks = 1 or use the equations above• If Ks is less than 1, use 1

14-10 Size factor Ks

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achine Design II• The load-distribution factor modifies the stress

equations to reflect non-uniform distribution

• KH = Km depends on different constants like

• where

• and Cpf is (in SI Units)

• Note that 0.05 if 0.05

14-11 Load distribution factor KH (Km)

Crowned teeth

𝐶

𝑏10𝑑

0.025 𝑏 25𝑚𝑚

𝑏10𝑑

0.0375 4.92 10 𝑏 25 𝑏 425𝑚𝑚

𝑏10𝑑

0.1109 8.15 10 𝑏 3.53 10 𝑏 425 𝑏 1000𝑚𝑚

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– Where

14-11 Load distribution factor KH (Km)…

Note: This Table in the Book (Table 14-9) is for English Units

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achine Design II• Zw = CH = 1 for pinion

• It is used only for the gear

• It is used if the hardness of the mating gears is not the same as pinion.

• The value is obtained as;𝑍 𝐶 1.0 𝐴 𝑚 1.0

• Where

• 𝐴 8.98 10 8.29 10 if 1.2 1.7

• 𝐴 0 if 1.2

• 𝐴 0.00689 if 1.7

• When Pinion of 48 Rockwell hardness is run with 180-400 BH gear, then use equation 14-37.

14-12 Hardness Ratio factor Zw (CH)

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14-13 Stress cycle factors• YN = Bending stress cycle factor

– The AGMA strengths as given in Figs. 14–2 through 14–4, in Tables 14–3 and 14–4 for bending fatigue, and in Fig. 14–5 and Tables 14–5 and 14–6 for contact-stress fatigue are based on 107 load cycles applied.

– The purpose of the load cycle factor YN is to modify the gear strength for lives other than 107 cycles.

– YN =1 for 107

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14-13 Stress cycle factors…• ZN = Stress cycle factor for pitting resistance

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14-14 Reliability factor YZ (KR)

• It can be determined as

• Or from table

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14-15 Temperature factor Y (Kt)

• Use Y = Kt = 1 for temperature up to 120oC or 250oF

• Use larger than 1 if temperature is large.M

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14-16 Rim thickness factor KB

• When the rim thickness is not sufficient to provide full support for the tooth root, the location of bending fatigue failure may be through the gear rim rather than at the tooth fillet. In such cases, the use of a stress-modifying factor KB or (tR) is recommended.

• It is a function of the backup ratio mB,

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Example 14-4 (in SI units)

A 17-tooth 20° pressure angle spur pinion rotates at 1800 rev/minand transmits 3 kW to a 52-tooth disk gear. The module is 2.5 mm,the face width is 38 mm, and the quality standard is No. 6. Thegears are straddle-mounted with bearings immediately adjacent.The pinion is a grade 1 steel with a hardness of 240 Brinell toothsurface and through-hardened core. The gear is steel, through-hardened also, grade 1 material, with a Brinell hardness of 200,tooth surface and core. The geometry factor YJ,P = 0.30 and YJ,G =0.40. The loading is smooth because of motor and load. Assume apinion life of 108 cycles and a reliability of 0.90, and useYN=1.3558(N)−0.0178 and ZN=1.4488N-0.023. The tooth profile isuncrowned. This is a commercial enclosed gear unit.

a) Find the factor of safety of the gears in bending.b) Find the factor of safety of the gears in wear. c) By examining the factors of safety, identify the threat

to each gear and to the mesh.

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Solution (in SI units)

(a) Based on bending Stress (Steps)

• FOS is given by; 𝑆 → 1 (i = P or G)

• and 𝜎 → 2

Where;

• 𝑊.

0.75 𝑘𝑁

• Loading factor; 𝑲𝒐 𝑷 𝑲𝒐 𝑮 1 (as the loading is smooth)

• Dynamic factor; 𝐾 → 3

• 𝐵 0.826

• 𝐴 50 56 1 𝐵 59.74

• (3) becomes; 𝑲𝒗 𝒑 𝑲𝒗 𝑮 = 1.37

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Solution…

Steps…

• Size Factor; 𝑲𝒔 𝑷 0.904 𝑚𝑏 𝑌 . 0.904 2.5 38 0.303 . 1.117

𝑲𝒔 𝑮 0.904 𝑚𝑏 𝑌 . 0.904 2.5 38 0.412 . 1.126

• Load distribution factor; 𝐾 1 𝐶 𝐶 𝐶 𝐶 𝐶 → 4

Where

• 𝐶 1 ( as the tooth is uncrowned)

• 𝐶 0.0375 4.92 10 𝑏 0.07 (for b = 38 mm, and, use same value of as

0.05 i.e .

0.089 0.05 )

• 𝐶 1 (as the bearings are immediately installed, 0.175)

• 𝐶 𝐴 𝐵 𝑏 𝐶 𝑏 → 5 (b is the face width)

• 𝐴 0.127,𝐵 0.622 10 , 𝐶 1.69 10 (Commercial unit, from Table)

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Solution…

Steps…

• (5) becomes; Cma = 0.15 (Note: you can also read it from Fig. 14-11)

• Ce = 1 (as no information about assembly/condition/compatibility)

• (4) becomes; 𝑲𝑯 𝒑 𝑲𝑯 𝑮 = 1.22

• Geometry factor; YJ)P = 0.30 and YJ)G = 0.40 (Given)

• Allowable bending Strength; 𝜎 0.533𝐻 88.3

𝜎 0.533 240 88.3 216 MPa

𝜎 0.533 200 88.3 195 MPa

• Stress Cycle factor; 𝑌 1.3558 𝑁 . 1.3558 10 .

𝑌 0.977

𝑌 1.6831.

→ 6)

For Grade 1 steelFigure 14-2

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Solution…

Steps…

• Where mG is the gear ratio and is given by; 𝑚 3.06

• (6) becomes; 𝒀𝑵 𝑮 1.3558.

.0.996

• Temperature factor; Y = 1 (Room temperature)

• Reliability factor; 𝒀𝒁 𝑷 𝒀𝒁 𝑮 0.85 (From Table 14-10)

• Put values in (2) and (1), to get;

𝜎0.75 10

2.5 381 1.37 1.117 1 1.22

0.349.12 MPa

𝑆

216 0.9771 0.85

49.125.1

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Solution…

Steps…

• For Gear

𝜎0.75 10

2.5 381 1.37 1.126 1 1.22

0.437.1 MPa

𝑆

195 0.9961 0.85

37.16.1

• Since Factors of Safety are greater than 1, the gear set is safe

based on Bending Stress.

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Solution…

(b) Based on Wear (Steps)

• FOS is given by; 𝑆 → 1

𝜎 𝐶𝐾 𝐾 𝐾 𝐾 𝑍

𝑍𝑊

𝑑 𝑏→ 2

• Contact Fatigue Strength;𝜎 𝑝 2.22𝐵𝐻 200 MPa 732.8 MPa

𝜎 𝐺 2.22𝐵𝐻 200 MPa 644 MPa

• Pitting resistance stress-cycle factor𝑍 1.4488 10 . 0.948

𝑍 1.4488𝑁

𝑚

.

1.4488103.06

.

0.973

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Solution…

(b) Based on Wear (Steps)• Hardness ratio factor

𝑍 𝑝 1 (For pinion, always equal to 1)𝑍 𝐺 1.005 (after using eqn. 14-36)

• Surface condition factor𝑍 1 (for normal condition use always 1)

• Surface strength geometry factor

𝑍𝑐𝑜𝑠20 𝑠𝑖𝑛20

2 13.06

3.06 10.121

• Elastic coefficient factor

𝐶 191 𝑀𝑃𝑎 From Table 14-8

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Solution…

• (2) becomes

𝝈𝒄 𝒑 511 MPa

• (1) becomes𝑺𝑯 𝒑 1.6

• Adopt similar procedure to calculate 𝑺𝑯 𝑮 1.44

• Since Factors of Safety are greater than 1, the gear set issafe based on Wear.

(c) Threat to the gear set

• Compare SF to (SH)2 for the pinion and gear, i.e.

5.1 > (1.6)2 = 5.1 > 2.56

6.1 > (1.44)2 = 6.1 > 2.07

• So threat to the gear set is due to Wear.

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Example 14-5 (Helical gear set)

• Several parameters in this example are thesame as in Example 14-4, with the exceptionthat we are using helical gears.

• For the same loading conditions, the fos arelarger for Helical gears as compared to spurgears. Why?

(Home task)

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Problems

• 14-19, 14-22, 14-24, 14-27, 14-28