Lecture 9 - Spur gear design - · PDF file9.2 Problem 2 Spur gear . 9.1 PROBLEM 1 – SPUR...

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Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram Indian Institute of Technology Madras Module 2- GEARS Lecture 9 - SPUR GEAR DESIGN Contents 9.1 Problem 1 Analysis 9.2 Problem 2 Spur gear 9.1 PROBLEM 1 – SPUR GEAR DESIGN In a conveyor system a step-down gear drive is used. The input pinion is made of 18 teeth, 2.5 mm module, 20 o full depth teeth of hardness 330Bhn and runs at 1720 rpm. The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face width of wheels is 35 mm. The gears are supported on less rigid mountings, less accurate gears and contact across full face may be assumed. The ultimate tensile strength of pinion and gear materials is 420 and 385MPa respectively. The gears are made by hobbing process. Find the tooth bending strength of both wheels and the maximum power that can be transmitted by the drive with a factor of safety 1.5. The layout diagram is shown in the Fig 9.1. Fig 9.1 Conveyor drive layout

Transcript of Lecture 9 - Spur gear design - · PDF file9.2 Problem 2 Spur gear . 9.1 PROBLEM 1 – SPUR...

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Module 2- GEARS

Lecture 9 - SPUR GEAR DESIGN

Contents

9.1 Problem 1 Analysis

9.2 Problem 2 Spur gear

9.1 PROBLEM 1 – SPUR GEAR DESIGN

In a conveyor system a step-down gear drive is used. The input pinion is made of 18

teeth, 2.5 mm module, 20o full depth teeth of hardness 330Bhn and runs at 1720 rpm.

The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face

width of wheels is 35 mm. The gears are supported on less rigid mountings, less

accurate gears and contact across full face may be assumed. The ultimate tensile

strength of pinion and gear materials is 420 and 385MPa respectively. The gears are

made by hobbing process. Find the tooth bending strength of both wheels and the

maximum power that can be transmitted by the drive with a factor of safety 1.5. The

layout diagram is shown in the Fig 9.1.

Fig 9.1 Conveyor drive layout

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Solution:

The bending fatigue stress is found from AGMA equation as,

tv o mK K

FK

b m J (9.1)

We know that, Z2= Z1 x (N1/N2)

Substituting values from table 1,

Z2= 18 X (1720/860) = 36

Table 9.1 Data given for gear and pinion

N Z m d b

Pinion 1720rpm 18 2.5 mm 45 mm 35 mm

Gear 860 rpm 36 2.5 mm 90 mm 35 mm

Using the values from Table 9.1,

V = π dn/60000 = π x 45 x 1720/60000

= 4.051m/s

We know that 0.5

v

50 (200V)K

50

(9.2)

Table 9.2 J values for pinion and gear

Z J (sharing) Kv Ko Km

Pinion 18 0.338 1.569 1.25 1.6

Gear 36 0.385 1.569 1.25 1.6

The J value is obtained from Fig. 9.2 for sharing teeth as in practice. Ko and Km values

are obtained from Tables 9.3 and 9.4 for the given conditions.

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Fig.9.2 - Geometric Factor J

SPUR GEAR –TOOTH BENDING STRESS (AGMA)

Table 9.3 - Overload factor Ko

Driven Machinery

Source of power Uniform Moderate Shock Heavy Shock

Uniform 1.00 1.25 1.75

Light shock 1.25 1.50 2.00

Medium shock 1.50 1.75 2.25

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Table 9.4 - Load distribution factor Km

Face width ( mm)

Characteristics of Support 0 - 50 150 225 400 up

Accurate mountings, small bearing

clearances, minimum deflection, precision

gears

1.3

1.4

1.5

1.8

Less rigid mountings, less accurate gears,

contact across the full face

1.6

1.7

1.8

2.2

Accuracy and mounting such that less than

full-face contact exists

Over

2.2

Over

2.2

Over

2.2

Over

2.2

For pinion:

(9. 3)

tv o m

t

t

K K

F1.569x1.25x1.6

35x 2.5x0.338

= 0.1061 F

FK

b m J

x

And for Gear:

(9.4)

tv o m

t

t

σ = K K

F1.569x1.25x1.6

35= x

x 2.5x0.385

= 0.0932 F

FK

b m J

Fatigue strength of the material is given by,

σe = σe’ kL kv ks kr kT kf km (9.5)

Table 9.5 Properties of pinion and gear

Prop. σut MPa σ’e=0.5σut MPa kL Kv ks

Pinion 420 210 1 1 0.8

Gear 385 187.5 1 1 0.8

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

SPUR GEAR – PERMISSIBLE TOOTH BENDING STRESS (AGMA)

Endurance limit of the material is given by:

σe = σe’ kL kv ks kr kT kf km (9.6)

Where, σe’ is the endurance limit of rotating-beam specimen

From table 9.5,

kL = load factor = 1.0 for bending loads

kv = size factor = 1.0 for m < 5 mm and

= 0.85 for m > 5 mm

ks = surface factor, is taken from Fig.9.3 based on the ultimate tensile strength of the

material

for cut, shaved, and ground gears.

kr = reliability factor given in Table 9.5.

kT = temperature factor = 1 for T≤ 350oC

= 0.5 for 350 < T ≤ 500oC

Fig.9.3 Surface factor Ks

Reliability of 90%, working temperature <150o C and reversible is assumed.

kf = 1.0 since it is taken in J factor.

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

km = 1.0 for reverse bending assumed here

Table 9.6 K terms of pinion and gear

Prop. kr kT kf km

Pinion 0.897 1.0 1.0 1.0

Gear 0.897 1.0 1.0 1.0

Table 9.7 Reliability factor R

Reliability factor R 0.50 0.90 0.95 0.99 0.999 0.9999

Factor Kr 1.000 0.897 0.868 0.814 0.753 0.702

Permissible bending stress

(9.7) e[ ]

Hence the design equation from bending consideration is,

(n

σ ≤ [ σ ] (9.8)

Factor of safety required = 1.5

```````````````````````Table 9.8 Strength values of pinion and gear

Prop. σe MPa [σ]= σe / s MPa σ MPa FT N

Pinion 150.7 100.5 0.1061 Ft 947

Gear 134.6 89.7 0.0932 Ft 962

Table 9.8 shows that the pinion is weaker than gear. And maximum tangential force that

can be transmitted is: Ft= 947 N

So, the maximum power that can be transmitted is:

W = Ft v / 1000 = 947 x 4.051 /1000

= 3.84 kW

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Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

9.2 PROBLEM 2 – SPUR GEAR DESIGN

In a conveyor system a step-down gear drive is used. The input pinion is made of 18

teeth, 2.5 mm module, 20o full depth teeth of hardness 340Bhn and runs at 1720rpm.

The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face

width of wheels is 35mm. The gears are supported on less rigid mountings, less

accurate gears and contact across full face may be assumed. The ultimate tensile

strength of pinion and gear materials is 420 and 385MPa respectively. The gears are

made by hobbing process. From surface durability consideration, find the maximum

power that can be transmitted by the drive with a factor of safety 1.2 for a life of 108

cycles. Drive layout is shown in the Fig 9.4.

Fig. 9.4 Conveyor drive Layout diagram

Data given:

i = n1/n2 = 1720/860 = 2

Z2= Z1 x i = 18 X 2 = 36

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Table 9.9 Data given for pinion and gear

n Z m d = mZ b

Pinion 1720rpm 18 2.5 mm 45 mm 35 mm

Gear 860 rpm 36 2.5 mm 90 mm 35 mm

Table 9.10 Properties of gear and pinion

Bhn Ø Reliability Life Temp

Pinion 340 20o 99 % 108 <120oC

Gear 280 20o 99 % - <120oC

Solution:

The induced dynamic contact stress is given by equation below,

tH p V o m

1

FC K K K

bd I (9.9)

When both pinion and gear material are made up of steel, from Table 9.11,

Cp = 191 (9.10) MPa

SPUR GEAR – CONTACT STRESS

Table 9.11 Elastic coefficient Cp for spur gears in MPa

Gear material Pinion Material(μ=0.3 in all cases)

Steel Cast iron Al Bronze

Tin Bronze

Steel, E=207Gpa 191 166 162 158

Cast iron, E=131Gpa 166 149 149 145

Al Bronze, E=121Gpa 162 149 145 141

Tin Bronze, E=110Gpa 158 145 141 137

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

(9.11) sin cos iI

2 i 1

Substituting the values from table 10, o osin 20 cos20 2

I 02 2 1

.1071

SPUR GEAR – SURFACE DURABILITY

From table 3 and 4,

V = π dn/60000 = π x 45 x 1720/60000

= 4.051m/s

For hobbed gear,

(9.12) 0.5)

K v

50 (200V

50

Table 9.14 K Values of pinion and gear

Z Kv Ko Km

Pinion 18 1.569 1.25 1.6

Gear 36 1.569 1.25 1.6

Substituting values from Table 14, we have,

t

H p V o m1

t

t

FC K K K

bd I

F191 1.569x1.25x1.6

35x45x0.1071

26.051 F MPa

Surface fatigue strength of the material is given by,

σsf = σsf‘ KL Kr KT (9.13)

From table 10, for steel life is 107 cycles & reliability 99% and from Table 9.15,

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

σsf’ = 28(Bhn) – 69 = 2.8x340 – 69 = 954MPa

KL = 0.9 for 108 cycles from Fig.9.2

KR = 1.0. for 99% reliability from Table 9.10

SPUR GEAR – SURFACE FATIGUE STRENGTH

Table 9.15 Surafce fatigue strength σsf for metallic spur gears (107 cycle life 99%

reliability and temperature <120 0 C)

Material σsf’(MPa)

Steel 2.8 (Bhn)-69MPa

Nodular iron 0.95 (2.8(Bhn)-69MPa)

Cast iron, grade 20 379

Cast iron, grade 30 482

Cast iron, grade 40 551

Tin Bronze, AGMA 2C (11% Sn) 207

Aluminium Bronze (ASTM 148 – 52)

(Alloy 9C – H.T.)

448

Fig. 9.5 Life factor Kl

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

SPUR GEAR – ENDURANCE LIMIT

Table 9.16 Reliability factor KR

Reliability (%) KR

50 1.25

99 1.00

99.9 0.80

SPUR GEAR – ALLOWABLE SURFACE FATIGUE STRESS (AGMA)

We know that,

[ σH ] = σSf / fs = 954/1.2 = 795MPa

For factor of safety fs = 1.2

Design equation is, σH ≤ [ σH ]

26.051 √ Ft = 795 Ft = 931 N

Maximum Power that can be transmitted is,

W = Ft V/1000 = 931x4.051/1000 = 3.51kW

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