20872249 spur gear design by iit madras

8/6/2019 20872249 Spur Gear Design by IIT Madras

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Lecture 9 - Spur gear design

CONTENTS

1. Problem 1 Analysis2. Problem 2 Analysis3. Problem 3 Spur gear design

PROBLEM 1 SPUR GEAR DESIGN

In a conveyor system a step-down gear drive is used. The input pinion is made of 18 teeth, 2.5mm module, 20

ofull depth teeth of hardness 330 Bhn and runs at 1720 rpm. The driven gear is

of hardness 280 Bhn 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 facemay be assumed. The ultimate tensile strength of pinion and gear materials are 420 and 385 MPa

respectively. The gears are made by hobbing process. Find the tooth bending strength of bothwheels 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 figure.

Conveyor drive Layout diagram


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Solution:The bending fatigue stress is found from AGMA equation:

Z2= Z1 x (N1/N2) = 18 X (1720/860) = 36

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

V = dn/60000 = x 45 x 1720/60000

= 4.051m/s

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 Graph 2 for sharing teeth as in practice. Ko and Km values are

obtained from Tables 3 and 4 respectively for the given conditions.

tv o m K K

FK (1)

b m J=

0.5

v

50 (200V)K (2)

50

+=


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Graph 2 - Geometric Factor J:

SPUR GEAR TOOTH BENDING STRESS (AGMA)

Table 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


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SPUR GEAR PERMISSIBLE TOOTH BENDING STRESS (AGMA)

Endurance limit of the material is given by:

e = e kL kv ks krkT kfkm (27)

Where, e endurance limit of rotating-beam specimenkL = 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 Graph 4 based on the ultimate tensile strength of the material

for cut, shaved, and ground gears.kr= reliability factor given in Table 5.

kT = temperature factor, = 1 for T 350oC

= 0.5 for 350 < T 500oC

Graph 4 Surface factor Ks

Reliability of 90%, working temperature


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Fatigue strength of the material is:

e = e kL kv ks krkT kfkm (6)

Permissible bending stress

Hence the design equation from bending consideration is :

[ ] (8)

Factor of safety required is : s = 1.5

Prop. e MPa []= e / s MPa

MPa FT N

Pinion 150.7 100.5 0.1061 Ft 947

Gear 134.6 89.7 0.0932 Ft 962

The table shows that the pinion is weaker than gear.

Maximum tangential force that can be transmitted is: Ft= 947 N

Maximum power that can be transmitted is :

W = Ft v / 1000 = 947 x 4.051 /1000= 3.84 kW

e[] (7)

n

=


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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 340 Bhn and runs at 1720 rpm. The driven gear is

of hardness 280 Bhn 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 are 420 and 385 MPa

respectively. The gears are made by hobbing process. From surface durability consideration, findthe maximum power that can transmitted by the drive with a factor of safety 1.2 for a life of 10

8

cycles. Drive layout is shown in the figure.

Conveyor drive Layout diagram

Given Data: i = n1/n2 = 1720/860 = 2Z2= Z1 x i = 18 X 2 = 36

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


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Bhn Reliability

Life

Temp

Pinion 340 20o

99 % 108


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SPUR GEAR SURFACE DURABILITY


Driven Machinery


Moderate Shock

Heavy Shock

Uniform 1.00 1.25 1.75

Light shock 1.25 1.50 2.00


Table 3. Load distribution factor Km

Face width b ( mm)

Characteristics of Support 0 - 50 150 225 400 up

Accurate mountings, small bearingclearances, 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

V = dn/60000 = x 45 x 1720/60000= 4.051m/s

For hobbed gear

0.5

v 50 (200V)K (12)50

+=


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Z Kv Ko Km

Pinion 18 1.569 1.25 1.6

Gear 36 1.569 1.25 1.6


Surface fatigue strength of the material is

sf= sf KL KrKT (13)

For steel 107

cycles life & 99% reliability from Table 4sf = 28(Bhn) - 69 = 2.8x340 69

= 954 MPa

KL = 0.9 for 108cycles from Fig.1

KR= 1.0. for 99% reliability from Table 5

SPUR GEAR SURFACE FATIGUE STRENGTH

tH p V o m

1

t

t

F C K K K

bd I

F191 1.569x1.25x1.6

35x45x0.1071

26.051 F MPa

===


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SPUR GEAR ENDURANCE LIMIT

Table 5. Reliability factor KR

Reliability (%) KR

50 1.25

99 1.00

99.9 0.80

SPUR GEAR ALLOWABLE SURFACE FATIGUE STRESS (AGMA)

[ H ] = Sf / fs = 954/1.2 = 795 MPaFor factor of safety fs = 1.2

Design equation is : H [ H ]

26.051

Ft = 795 Ft = 931 NMaximum Power that can be transmitted

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


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Problem 3 - Design of Spur gear

A pair of gears is to be designed to transmit 30 kW power from a pinion running at 960 rpm to a

gear running at 320 rpm. Design the gears so that they can last for 108

cycles. Assume 20o

full

depth involute spur gear for the system. Motor shaft diameter is 30 mm.

Data: W = 30 kW ; n1= 960 rpm; n2 = 320 rpm; Life = 108

cycles ; 20o

full depth involute spur

gear .

Solution: i = n1 / n2 = 960 / 320 = 3

In order to keep the size small and meet the centre distance Z1 = 17 chosenZ2 = i Z1 = 3 x 17 = 51

Torque:

From Lewis equation we have for pinion

Steel C50 WQT with 223 Bhn hardness of and tensile strength of 660 MPa and steel C45 OQT

with hardness 205 Bhn and tensile strength of 600 MPa are assumed for the pinion and the gear.For C50 form the data book is permissible static bending stress []p = 542 MPa and for C45 []g= 487 MPa and face width b= 10m is chosen for both wheels.

12 n 2x x960

100.48rad/ s60 60

= ==

1

w 30x1000T 298.57Nm

100.48= = =

t 1p p2

1

F 2T [] (1)

b Y m bYZ m= =


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Problem 3 - Design of Spur gear- Buckingham approach

Equation (1) we have

For the pinion Y = 0.25808 for Z1= 17, from the Table 1

For the gear, Y = 0.39872, for Z2 = 51 from the Table 1;For gear, Y[]g = 0.39872x 487 = 194.17For pinion, Y[]p = 0.25808 x 542 = 139.87

1p p3

1

T [] (2 )

5YZ m=


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SPUR GEAR - TABLE 1 FOR MODIFIED LEWIS FORM FACTOR

For the same face width hence pinion will be weaker and considered for the design.

p 3 3 3

1

T 298.57 x 1000 13610 (3 )

5YZ m 5x0.25808x 17m m

542MPa

= = =


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m = 2.93 mm, . Since motor shaft diameter is 30 mm, to get sufficiently large pinionm = 4 mm is taken

Wheel Z m b=10m d V =wrv Material

Hardness

Pinion 17 4mm 40 mm 68mm 3.42 m/s

C 50 223

Gear 51 4mm 40 mm 204mm

3.42 m/s

C 45 205

We will now use Buckingham dynamic load approach for the design

Ft = T1/r1 = 298.57/0.034 = 8781 NBuckingham dynamic load:

For V=3.42 m/s permissible error is e= 0.088 mm from graph6. From table 7, if we choose I

class commercial cut gears, expected error is 0.050 for m=4mm. In order to keep the dynamicload low precision cut gears are chosen. e = 0.0125

SPUR GEARS PERMISSIBLE ERROR GRAPH 6

40

ti

t

9.84V(Cb+F )F ( )

9.84V + .4696 Cb+F

=


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SPUR GEARS EXPECTED ERROR IN TOOTH PROFILE TABLE 7

SPUR GEARS VALUE OF C TABLE 6


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C = 11860e = 11860 x 0.0125 = 148.25Substituting the values Ft = 8781 N ,

C = 148.25, V=3.42 m/s, b= 40mm in eqn (4)

Buckingham dynamic load:

Fd = Ft + Fi = 8781 + 5464 = 14245 N

Beam strength of the pinion Ftp = bYm []p= 40x0.25808 x4x542 = 22381 N

Since Ftp(22381)> Fd(14245) the design is safe from tooth bending failure consideration.

Wear strength of the pinion is:

Cp = 191 MPa0.5

from the table for steel vs steel

Substituting i =3, =200 we get I= 0.1205

SPUR GEAR BUCKINGHAM CONTACT STRESS EQUATION

87815464 5

0 8781i

9.84x3.42(148.25x40+ )F N ( )

9.84x3.42+ .4696 148.25x40+

= =

2

Hts 1

p

[ ]F bd I (6)

C

=

o osincos i sin20 cos20 3I 0.1205

2 i 1 2 3 1= = =+ +


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Surface fatigue strength of the pinion material is sf= sf KL KRKTsf = 2.8(Bhn) 69 MPa

= 2.8 x 223-69 = 555.4 MPa

KL= 0.9 for 108

cycles life from graph1

KR= 1.0 taken for 99 reliabilityKT = 1.0 for operating temperature


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Table 5 Reliability factor KR

Reliability (%) KR

50 1.25

99 1.00

99.9 0.80

Surface fatigue strength of the pinion material is sf= sf KL KRKT= 555.4x0.9x1x1 = 500 MPa

Assuming a factor of safety s = 1.1

[H] = sf/s = 500/1.1 = 455 MPa

Wear strength of the pinion is:

Since Fts (1860)


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= 2.8 x 475-69 = 1261 MPa

KL= 0.9 for 108




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Since Fts (12439) < Fd (14245) still Not safe. Hence increase the module to 5mm.


Hardness


C 50 475


4.27 m/s

C 45 450

With new dimensions Fd = 16098 N

Fts 19436 N. Since Fts > Fd , the revised design is safe from surface fatigue (pitting)considerations.

Problem 3 - Design of Spur gear-AGMA Approach

AGMA equation for tooth bending stress

Ft = 8781 N, b= 10m, Kv =1.15 assumed

SPUR GEAR TOOTH BENDING STRESS (AGMA)

Ko = 1.25 is taken assuming uniform power source and moderate shock load from the table3Km = 1.3 taken assuming accurate mounting and precision cut gears for face width of about

50mm.J = 0.34404 for pinion Z1= 17 mating with gear Z2=51For gear J = 0.40808

2 2

Hts 1

p

[ ] 1032F =bd I =52x68x0.1205 =12439N

C 191

tv o m K K

FK

b m J=

0.50.5 0.5

v

78 (200V) 78 (200x3.42)K 1.15

78 78

+= = =


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Driven Machinery


Moderate Shock

Heavy Shock

Uniform 1.00 1.25 1.75

Light shock 1.25 1.50 2.00



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Face width ( mm)


Accurate mountings, small bearing

clearances, minimum deflection, precisiongears

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




e = e kL kv ks krkT kfkmThe pinion is of steel C50 OQT with 223Bhn hardness and tensile strength of 660 MPa and the

gear is of C45 OQT 30 with hardness 210 Bhn and tensile strength of 465 MPa.

For pinion e = 0.5 ut = 0.5 x 660 = 330 MPakL = 1 for bending, kV = 1 assumed expecting m to be


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Substituting the value in the equation

m = 3.55 mm take m=4 mm.

From this module the dimensions calculated are given in Table form


Hardness


C 50 223


3.42 m/s

C 45 205

Ft = T1 / 2 = 29854/34 = 8781N

The tooth has to be checked from surface durability considerations now.

The contact stress equation of AGMA is given below:

1v o m2 2

1

298540 x 1.15 x 1.25 x 1.3 K K

10m x 17xm x 0.34404

2TK []

b Z m J= =

3

9539213 6.

m= =

tH p V o m

1

F C K K K

bd I=


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Cp = 191 MPa0.5

from the table for steel vs steelSubstituting i =3, =200 we get I= 0.1205



Driven Machinery


Moderate Shock

Heavy Shock

Uniform 1.00 1.25 1.75

Light shock 1.25 1.50 2.00



Face width b ( mm)


Accurate mountings, small bearingclearances, 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




o osincos i sin20 c os20 3I 0.1205

2 i 1 2 3 1

= = =+ +

0.50.5 0.5

v

78 (200V) 78 (200x3.42)K 1.15

78 78

+= = =


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Ko = 1.25 and Km = 1.3 assumed as in the case of bending stress calculation

H = 1209 MPa


= 2.8 x 223-69 = 555.4 MPa

KL= 0.9 for 108


KR= 1.0 taken for 99 reliabilityKT = 1.0 for operating temperature [H ] (455)The design is not safe and surface fatigue failure will occur.Solutions:

Increase the surface hardness of the material to 475 Bhn and also increase the b to 13m = 13 x 4= 52 mm

tH p V o m

1

F 8781x1.15x1.25x1.3 C K K K 191

bd I 40x68x0.1205= =


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= 2.8 x 475-69 = 1261 MPa

KL= 0.9 for 108




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Surface fatigue strength of the pinion material is sf= sf KL KRKT= 1261x0.9x1x1 = 1135 MPa

Assuming a factor of safety s = 1.1

[H] = sf/s = 1135 /1.1 = 1032 MPa

As H (1185) >[H ] (1032) the design is not safe from surface durability considerations.Hence increase the module to 5mm and take b=13m

H =948 MPa < [H ] (1032 MPa) . Hence the design is safe now from surface durabilityconsiderations.

Final specification of the pinion and gear are given in the table

Wheel Z m b=13m d

Pinion 17 5mm 65 mm 85mm


Wheel Material Steel Hardness

Manufacturing quality

Pinion C50 OQT 475 Bhn Precision cut

Gear C45 OQT 450 Bhn Precision cut

tH p V o m

1

F 8781x1.15x1.25x1.3 C K K K 191

bd I 52x68x0.1205= =

tH p V o m

1

F 8781x1.15x1.25x1.3 C K K K 191

bd I 65x85x0.1205= =

20872249 spur gear design by iit madras

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