Part III Design of Power Transmission Systems

Chapter 10 Design of Gears Key Terms: spur gears helical gears bevel gears pinion gear driving gear driven gear tangential force radial force axial force normal force module face width

10.1 Introduction Advantages and Disadvantages of Gears Advantages: high efficiency, High reliability, Constant transmission ratio, compactness, etc. Disadvantages: Cost much, can not be used in the case of large center distance, etc.

2. Types of Gears 1)Classification by working condition encased gearing open gearing semi-open gearing 2) Classification by the case hardness of gear tooth soft tooth surface (case hardness  350HBS) hard tooth surface (case hardness > 350HBS)

10.2 Gear failures and Design Criteria 1) breakage

2) pitting

3) gluing

4) abrasive wear

5) ridging

2. Design Criteria 1)Cased gear: Contact Strength Bending Strength 2)Opened gear:

3. Gear Materials (Table 10.1) 1）forged steel carbon steel alloy steel quench、 temper、 (Q&T)、 normalize casehardening carburizing nitriding Heat-treating

2) Cast Steel 3) Cast Iron gray cast iron nodular iron 4) Non-metal

10.8 Gear Blank Design 1) Gear shafts (e < 2mt)

2) Gears with solid hub (da<200mm)

3) Gears with thinned web (da>200~500mm)

4) Spoked gears (da > 400~1000mm)

Styles Gear shaft Solid hub

Styles spoke Gear shaft

Styles Thinned web

10.9 Efficiency and Lubrication in Gear Sets mesh efficiency, stirring efficiency, bearing efficiency

2. Lubrication 1) bath lubrication (v<12m/s)

2) splash lubrication 3) atomized lubrication (v> 12m/s)

10.4 Design of Spur Gears

Nomenclature of Spur Gear

1. Forces on Spur Gear Teeth Normal Force Fn Radial Force Fr Tangential Force Ft Pinion Gear Direction:

Magnitude: T1——nominal torque d1——reference diameter of pinion ——pressure angle

Example 1 Mating 2-module, 200 gears are shown. Gear 1 is driving at 960 rpm and transmitting 25 kW. Gear 2 is driving a conveyor.Draw the radial and tangential forces on gears. Compute their magnitude. Solutions

2. Calculated Load Where Ftc—— calculated load F —— nominal load K—— load factor

K=KAKVKK Where KA—— application factor KV——dynamic factor K——load partition factor K——load distribution factor \

KA KA is an application factor considers that the load variation, vibration, shock, speed changes which may result in peak loads greater than nominal loads. Table 10.2

KV Kv is a dynamic factor which accounts for the higher actual load caused by the deformation of gear teeth, the base pitch error, profile error, backlash, etc. for spur gears: Kv=1.05~1.4 for helical gears: Kv=1.02~1.2

Modified Tooth Face base pitch pb1>pb2

K K is a load partition factor which reflects the non-uniformity of tooth loading among two or more teeth due to manufacture error and teeth deformation. for spur gears: K=1~1.2 for helical gears: K=1~1.4

K K is a load distribution factor which reflects the non-uniformity of tooth loading over the face width of the teeth, arising from gear and mounting inaccuracy, elastic deformation of shafts and bearings, etc. for soft surface: K=1~1.2 for hard surface: K=1.1~1.35

Fig.10.9 Uneven distribution load on gear teeth

Crowning gear teeth

3.Contact Fatigue Strength of Spur Gear Teeth Design Criterion: H[H] Where H——contact stress [H]——allowable contact stress

Contact Stress Fn: normal force L: contact length of cylinders 1, 2: Poisson’s ratio E1，E2：elastic modulus 1, 2: radius of curvature

let Z——coefficient of contact ratio ZE——elastic coefficient 弹 table 10.3

Mating Teeth

let ZH——zone factor for standard spur gear ZH=2.5

Contact Strength Check Formula: Design Formula: d——face width factor

Notes If H1> H3 then gear 3 is stronger than gear 1 • reference diameter or center distance is related to contact strength • Where [H]=min{[H1], [H2]}

4.Bending Fatigue Strength of Spur Gear Teeth Design Criterion: F[F] Where F——bending stress [F]——allowable bending stress

Gear teeth Analyzed as a cantilever beam

YFa——tooth form factor, table 10

Ysa——stress concentration factor , table 10.4 Y——coefficient of contact ration

Bending Strength Check Formula: Design Formula:

Notes If F1> F3 then gear 3 is stronger than gear 1 module m is related to bending strength Where

Example 2 A spur gear set, its parameters are shown below by table. Try to compare which gear is easier to pitting, which one is easier to breakage? gear m z b YFa YSa [F] [H] 1 3 17 60 2.97 1.52 390 500 2 45 50 2.35 1.68 370 470

5.Allowable Stresses and Design Parameters Allowable Contact Stress: Hlim——endurance limit for contact strength KHN——life factor for contact strength SH——factor of safety for contact strength

(2) Allowable Bending Stress: Flim——endurance limit for bending strength KFN——life factor for bending strength SF——factor of safety for bending strength YST——stress correction factor ， YST=2.0

(3) Endurance limits Hlim, Flim Fig.10.14 and Fig.10.15 (4) Life factors KHN, KFN Fig.10.16 and Fig.10.17 N=60njLh N——expected number of load cycles n ——rotational speed of the gear j——number of load application Lh——design life

(5) Factor of safety SH, SF Table 10.5 (6) Design parameters 1)Pressure angle 

2) Number of teeth z, module m for encased gears : z1=20~40; for open gears: z1=17~20; z2=uz1; m: table 10.6 3) Face width factor table 10.7 b2=b, b1=b+(5~10)mm

4) Quality Classes of Gears P200, Table 10.8

10.6 Design of helical Gears Right hand helix Key Terms: virtual spur gears virtual number of teeth Left hand helix

1. Forces on Helical Gear Teeth Pinion Normal Forces Fn : Radial Force Fr Tangential Force Ft Axial Force Fa (Right or Left Hand Law)

Right or Left Hand Law Ft1= -Ft2 Fr1= -Fr2 Fa1= -Fa2

Magnitude: Where: b——base helix angle n——normal pressure angle , n=200 t——transverse pressure angle

Example 3 The pinion of a pair of helical gears (shown below) has a normal module of 2, a normal pressure angle of 200, 32 teeth, and a helix angle of 150. If the pinion is rotating at 750 rpm while transmitting 20kW, draw and compute the tangential force, the axial force, and the radial force.

Solution

2. Contact Fatigue Strength of Helical Gear Teeth Check Formula: Z——helix angle factor, ZH——zone factor, Fig.10.12 Z——coefficient of contact ratio, Z=0.75~0.88

Design Formula:

Y——contact ratio factor, 3. Bending Fatigue Strength of Helical Gear Teeth Check Formula: Y——helix angle factor, Y=0.85~0.92 YFa——tooth form factor, table 10.4, YSa——stress concentration factor, table 10.4 Y——contact ratio factor,

Design Formula:

10.7 Design of Straight Bevel Gears large end

1. Straight Bevel Gear Geometry Ov2 d1——reference diameter dm——mean reference diameter dv——reference diameter of virtual spur gears

2. Forces on Straight Bevel Gear Teeth Normal Force Fn: Radial Force Fr Tangential Force Ft Axial Force Fa (point to the large end) Ft1= -Ft2 Fr1= -Fa2 Fa1= -Fr2

Example 3 A bevel-helical gear-type speed reducer is shown below, in order to counteract the axial forces on the shaft II, draw tangential forces, radial forces, axial forces on bevel gears and helical gears. determine left hand helix or right hand helix on helical gears.

3. Contact Fatigue Strength on Straight Bevel Gear Teeth

ZE——elastic coefficient, table 10.3 ZH——zone factor, ZH=2.5 K=KA Kv KK KA——application factor, table 10.2 Kv——dynamic factor, Kv=1.1~1.4 K—— load partition factor, K=1; K——load distribution factor, K=1.1~1

4. Bending Fatigue Strength on Straight Bevel Gear Teeth

YFa, Ysa—according to , obtained from table10.4

Conclusion 1)Failure modes of gears; 2) Force analysis of gear tooth; 3) Design criteria of Gears; 4) Design of contact fatigue strength & bending fatigue strength.

Fig.10.14

Fig.10.15

Fig.10.12

The end Thanks!