4: WORKED OUT PROBLEMS ON INVOLUTE SPUR …nptel.ac.in/courses/112106137/pdf/2_4.pdf · Lecture –...

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

Indian Institute of Technology Madras

Module 2 - GEARS

Lecture – 4: WORKED OUT PROBLEMS ON INVOLUTE SPUR GEARS Content 4.1 Introduction

4.2 Gear Calculation -Worked example 1

4.3 Gear Calculation -Worked example 2

4.1 INTRODUCTION

In the earlier lectures, the definition of gear, gear terminology, interference, methods of

eliminating interference, standard tooth systems for spur gears, profile shifted gears,

involutometry and design of gear blanks were dealt in detail. With the background of the

above knowledge, solving of simple gear problems will be dealt in this lecture.

4.2 GEAR CALCULATION – Worked out example 1 In a drive, a velocity ratio of 2.5 with a centre distance of 70 mm is desired. (a) Determine

the pitch diameter of the gears with 20o full depth involute teeth; (b) Is there any

interference in the system? If so, how will you avoid it? (c) Determine the contact ratio, (d)

Find the dedendum, addendum, root diameters and the tip clearance, (e) If the centre

distance is increased by 1.5%, what will be the new pressure angle?

• Given Data: i = 2.5; C = 70 mm; φ = 20o;

• It is an involute full depth tooth system ;

• To be found: m = ?; Z1 = ? ; d1 = ?; Interference exists or not ?;

Z2 = ?; d2 = ?; CR = ?

We know that,

C = (r1 +r2) = 0.5 m ( Z1 + Z2) = 0.5 m (Z1 + i Z1) = 70 mm

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From the above, i.e., m x Z1 = 40

• Possible solutions for standard modules as can be seen from the table:

Solution I:

• m = 2mm, Z1 = 20, Z2 = i. Z1 = 2.5 x 20 = 50

• d1 = m Z1 = 2 x 20 = 40 mm

• d2 =i d1 = 2.5 x 40 = 100 mm

STANDARD MODULES IN mm:

0.3 0.4 0.5 0.6 0.7 0.8 1.0

1.25 1.5 1.75 2.0 2.25 2.5 3.0

3.5 4.0 4.5 5.0 5.5 6.0 6.5

7 8 9 10 11 12 13

14 15 16 18 20 22 24

26 28 30 33 36 39 42

45 50 Further increase is in terms of 5 mm

• Minimum number of teeth that can engage with the gear of 50 teeth without

interference is given by,

2 21 1 2 2

4k(z + k)z + 2 z z =

sin φ

• For full depth gears, k = 1. Substituting the values Z2 = 50, φ = 20o in the above

equation,

2

1 1 2 o

4 x 1(50 1)z 2 z x50

sin 20

2z + 100 z 1744 0

1 1

• i.e., Z1min = 15 is the minimum permissible number of teeth without interference.

Since from the above it is seen that, Z1 > Z1min no interference will occur.

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Fig 4.1. Path of Contact

Contact ratio, CR:

Referring to fig. 4.1, the path of contact:

La = ua + ur

2 2 2 2 2 2

1 1 2 2 1 2 L ( r a ) r cos ( r a ) r cos ( r r ) sin

a

Substituting the values,

o oL ( ) cos ( ) cos ( )sina

. mm

2 2 2 2 2 220 2 20 20 50 2 50 20 20 50 20

9 798

o

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Now CR can be calculated with the formulae below,

L L . .a aCR .

opcos m cos .x x cos

9 798 9 7981 660

5 9012 20 For full depth tooth, the various values calculated are as shown below:

Addendum: a = 1m = 1x2 = 2mm Dedendum: b = 1.25m = 1.25x2 = 2.5 mm Clearance: c = 0.25 m = 0.25x2 = 0.5 mm Pinion root circle diameter: dr1 = d1- 2b = 40 – 2x2.5 = 35 mm Pinion addendum diameter: da1 = d1+ 2a = 40 + 2x2 = 44 mm Gear root circle diameter: dr2 = d2- 2b = 100 – 2x2.5 = 95 mm Gear addendum diameter: da2 = d2+ 2a = 100 + 2x2 = 104 mm

Solution II:

The centre distance is given by, C = (r1 +r2) = 0.5 m ( Z1 + Z2) = 0.5 m (Z1 + i Z1)

= 70 mm

• Simplifying, m x Z1 = 40

• Another possible solution for standard module from the table is,

m = 2.5 mm, Z1 = 16, Z2 = i x Z1 = 40,

d1 = m Z1 = 2.5x16=40 mm,

d2 = m Z2 = 2.5 x 40 =100 mm

• Minimum number of teeth that can engage the gear of 40 teeth without interference

is given by,

2 21 1 2 2

4 k (z +k)z + 2 z z =

sin φ

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• For full depth gears k = 1. Substituting Z2 = 40, φ = 20o, in the above equation,

4(40 + )2z + z x =1 1 2 osin

12 40

20

2z + 80 z 1402 01 1

• i.e., Z1min = 15 is the minimum permissible without interference. Since Z1 > Z1min, no interference will occur. Contact ratio, CR :

Referring to the fig. 4.1, the path of contact is given by,

La = ua + ur

2 2 2 2 2 2

1 1 2 2 1 2 L ( r a ) r cos ( r a ) r cos ( r r ) sin

a


2 2 2 o 2 2 2 o oL (20 2.5 ) 20 cos 20 (50 2.5 ) 50 cos 20 (20 50 ) sin 20a

11.854 mm

Now CR can be calculated as,

11 854 11 854

1 6075 9012 5 20

L L . .a aCR .opcos m cos .x . x cos

For full depth tooth the values are,

Addendum : a =1m = 1x2.5 =2.5mm

Dedendum : b =1.25m =1.25x2.5 = 3.125 mm

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Clearance: c = 0.25 m = 0.25x2.5 = 0.625 mm

[Pinion root circle diameter: dr1= d1- 2b = 40 – 2x3.125 = 33.75 mm

Pinion addendum diameter: da1= d1+ 2a = 40+2x2.5 = 45 mm

Gear root circle diameter: dr2= d2- 2b = 100 – 2x3.125 = 93.75 mm

Pinion addendum diameter: da2= d2+ 2a = 100 + 2x2.5 = 105 mm

New Pressure angle Ø when the centre distance C is increased by 1.5% is given by:

Ø = cos-1( r1 cos20o / r1 1.015) = 22.21o

4.3 GEAR CALCULATIONS- Worked out example 2

A 21 teeth gear has 20o full depth involute teeth with a module of 12mm. (a) Calculate the

radii of pitch circle, base circle and addendum circle (b) Determine the tooth thickness at

base circle, pitch circle and addendum circle. Comment on the top land thickness.

brrcos

inv tan

t1t 2 r ( inv inv

2 r1

) Fig 4.2 Involutometry The terminologies used in the problem are explained in earlier chapters. Hence the figures

referred are self explanatory.

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Fig. 4.3 Layout for tooth thickness determination at any point on tooth profile

Given Data: Pressure angle, Ø =20o = 0.349 rad., Z1 = 21, m= 12mm.

Solution:

r1 = 0.5.m.Z1 = 0.5 x 12 x 21=126 mm

rb = r1.CosØ = 126 x cos 20o = 118.4 mm

a = 1m = 1 x 12 = 12 mm

ra = r1 + a = 126 + 12 = 138 mm

p1 = π.m = π. 12 = 37.14 mm

t1 = 0.5 p1 = 0.5 x 37.14 = 18.57 mm

inv Ø = tan Ø - Ø = tan 0.349 – 0.349 = 0.015 rad

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At the base circle Øb= 0, Hence tooth thickness tb at the base circle is:

t1t 2 r ( inv inv )

b b b2 r1

18.572x 118.4 x( 0.015 0 ) 21.0 mm

2x126

At the addendum circle the pressure angle is given by,

Øa=cos-1(rb/ra) =cos-1(118.4 /138) =0.54 rad

inv Øa = tan 0.54 – 0.54 = 0.059

Hence tooth thickness ta can be calculated as,

t1t 2 r inv inv

a a 2 r1

a


.t x x . . . ma x

18 57

2 138 0 015 0 059 8 202 126

m

Minimum recommended top land thickness is,

tamin = 0.25 m = 0.25 x 12 = 3 mm

From the above, we know that, ta > tamin, hence the tip thickness is adequate and

permissible.

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4: WORKED OUT PROBLEMS ON INVOLUTE SPUR …nptel.ac.in/courses/112106137/pdf/2_4.pdf · Lecture –...

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