1

Example 1: Following data is given for a steel spur gear transmitting 7.5 kW
power running at 1440 rpm to a machine running at 480 rpm. Approximate
centre distance = 240 mm, Allowable bending stress for pinion and gear are 200
and 160 respectively. Surface hardness is 450 BHN. Tooth system is 20 full
depths involutes. Design a spur gear drive for above application.
Solution:
Given data: Power = 7.5 kW, Centre distance = 240 mm, Pinion speed = 1440 rpm
Gear speed = 480 rpm, Allowable bending stress at pinion = 200 MPa
Allowable bending stress at gear =160 MPa, Surface hardness = 450 BHN
Centre distance, a =

= 240

= i =

= 3 =

= 3

a =

= 240

= 120 mm and

= 360 mm
Assume,

= 18

= m

m = 6.6667 mm
Take, m = 7 mm (From table)

= 126 mm,

= 378 mm
a = 252 mm 240 mm (Given)
Take,

= 20

= 20 * m mm
m = 6 mm

= 120 mm and

= 360 mm
a =

= 240 mm

0.154
.

0.154
.

= 0.1084

3 = 60

0.154
.

0.154
.

0.1388

= 200 0.1084 = 21.68

=160 * 0.1388 = 22.208

<

From above condition Pinion is weaker for beam strength.
Checking For Beam Strength:

bm

Assume, face width of gear b= 10 m mm

= 200 * 60 * 0.1084 * 3.14 * 6 = 24507.07 N
3. DESIGN OF SPUR GEAR

2

Power, P =

7.5 10

2 1440
60

T = 49761.16 Nmm

=
.

829.35N
For 20 full depth steel-steel combination;
C = 11860 *e and e = 0.064 (from PSG design data book)

= 9.0432

+
. . .
. . .

= 829.35+
..
.

= 22552.36 N

>

so design is safe.
Checking For Wear Strength:

Q =

= 1.5

= 2.75BHN - 68.65 = 2.75 * 450 68.65 = 1168.85 MPa

= 2.1 10
5
MPa (for steel)

= Pre. Angle = 20
K=

=
.

= 3.1787

= 120 * 1.5 * 3.178 * 60 = 34330.14 N

>

Design is safe.

Example 2: A pair of carefully cut mating spur gears has 20
o
full depth of 4mm
module. The number of teeth on pinion and gears are 38 and 115, respectively.
The face width is 40 mm. If the pinion and gear are made of steel with f
bStatic
=
233 MPa and surface hardness of 300 BHN. Calculate the safe power that can be
transmitted when the pinion is run at 1200 rpm.
Solution:

0.154
.

0.154
.

= 0.13
Beam Strength:
Lewis equation
F
s
= f
b
b m y
p
= 233 40 4 0.13
= 15225.41 N
Wear Strength:

3

D
p
= Z
p
m = 384 = 152 mm
V =

= 9.55 m/s
f
es
= 2.74 BHN - 68.65 = 2.74 300 - 68.65 = 753.35 MPa

Q =


=


= 1.503
K=

=
.

Assume, E
p
= E
g
= 2.0710
5
MPa
= 1.3396
F
w
= D
p
b Q K = 152401.5031.3396
= 12241.558 N
For safe design
F
s
F
d
and F
w
F
d
Take, dynamic load (F
d
) equal to either beam strength (F
s
) or wear strength (F
w
)
whichever is less.
F
d
= F
w
= 12241.58 N
C = 11860e (from design data book)
= 296.5 (e = 0.025)
F
d
= F
t
+

12241.58 = F
t
+
. .

. .

Above equation solve by trial and error method
F
t
= 2960 N
T = F
t

= 2960

= 224960 Nmm
= 224.96 Nm
Power transmitting capacity
P =

=
.

= 28263.5 W
= 28.26 kW
Example 3: A pair of a carefully cut spur gear with 20
o
full depth involute teeth
consists of 19 pinion meshing with 40 teeth gear. The pinion shaft is directly
coupled to a single cylinder diesel engine developed power 8 kW at 1500 rpm.
The gear shaft is transmitting a power to a two stage reciprocating air
compressor. There service factor and factor of safety are 1.5 and 3 respectively.
The pinion as well as gear are made of plain carbon steel 45C8 (

= 600
N/mm
2
). The module and face width are 3 mm and 50 mm respectively. The
gears are heat treated to a surface hardness of 450 BHN. Calculate the factor of
safety of the above gear drive.
Solution:
Given Data:

4

= 20
o
full depth involute, carefully cut

= 19,

= 40, Power (

) = 8 kW,

= 1.5, FOS = 3

= 600 MPa, m = 3 mm, b = 50 mm, BHN = 375, N = 1500 rpm

2 N T
60

810
3
=
T

T = 50955 Nmm
Pinion Diameter, D
p
= m * Z
p

= 319 = 57 mm
Transmitted Load, f
t
=

D

1787.8
Here material of pinion and gear are same. So product f
b
will be minimum for
pinion.
Thus pinion will be weaker than gear. So we design a pinion.
Y
p
= 0.124
.

= 0.124
.

= 0.106
f
b
=

200 MPa
Beam Strength:
F
s
= F
b
* b * * m * Y
p

= 200 * 50 * * 3 * 0.10
= 9985.2 N
Dynamic load:
V =

= 4.474 m/s
C = 11860 * e (from design data book)
= 296.5 (e = 0.025 for carefully cut gear)
F
d
= F
t
+

= 1787.8+
. . .
. . .

= 8792.10 N
Wear strength:
f
es
= 2.74 * BHN - 68.65
= 2.74 * 450 - 68.65
= 1164.35 N/mm
2
Q =


= 1.356
K =

5

=
.

Assume, E
p
= E
g
= 2.0710
5
MPa
= 3.2
F
w
= D
p
* b * Q * K
= 57 * 50 * 1.356 * 3.2
= 12366.72 N
Now

=
.
.
= 1.136

=
.
.
= 1.406

>

Factor of safety of gear drive is 1.136

Example 4: A spur pinion having 20 teeth to be made up of plain carbon steel
40C8 (f
ut
= 580MPa) is to be mesh with a gear having 85 teeth to be made up of
Grey cast iron FG260. The pinion shaft is connected to 15 kW, 1440 rpm electric
motor. The starting torque of motor is approximately twice the rated torque. The
tooth system is 20
o
full depth involute. Calculate the module and hardness
required for above gear pair.
Solution:
Given: Power, P = 15 kW, N
p
= 1440 rpm, k
s
= 2,
No. of teeth on pinion, Z
p
= 20
No. of teeth on gear, Z
g
= 85
Ultimate tensile strength of pinion, f
ut
= 580 MPa
Ultimate tensile strength of gear, f
ut
= 260 MPa
Assuming Factor of safety = 3, for both pinion and gear
Bending stress in pinion,
33 . 193
3
580
= = =
FOS
f
f
ut
b
MPa
Bending stress in gear,
66 . 86
3
260
= = =
FOS
f
f
ut
b
MPa
Speed of gear,
p p g g
Z N Z N =
20 1440 85 =
g
N
85
20 1440
=
g
N
82 . 338 =
g
N rpm
Now, for 20
o
full depth, the Lewis form factor is,

6

For pinion,
20
912 . 0
154 . 0
912 . 0
154 . 0 = =
p
p
Z
y 1084 . 0 =
For gear,
85
912 . 0
154 . 0
912 . 0
154 . 0 = =
g
g
Z
y 1432 . 0 =
Now,
95 . 20 1084 . 0 33 . 193 = =
p bp
Y f
41 . 12 1432 . 0 66 . 86 = =
g bg
Y f
here,
g bg p bp
Y f Y f >
Hence, the gear is weak in beam strength than pinion. Hence we will design the gear
in beam strength.
Design of gear in beam strength:
Power,

60
2
g g
T N
P

60
82 . 338 14 . 3 2
1000 15
g
T
=

Nm T
g
97 . 422 =
Torque of gear drive is, Nmm T
g
3
10 97 . 422 =
Tangential force,
g
g
t
D
T
F

=
2

m 85
10 97 . 422 2
3

= N
m
23 . 9952
=

also velocity,
60000
g g
m
N D
V

=

m/sec 507 . 1
60000
82 . 338 85 14 . 3
m
m
V
m
=

=

Assume, face width of gear , b = 10 m and k
s
= 2 (Given)
m V
k
m
v
507 . 1 3
3
3
3
+
=
+
=

Now,
g bg
v
s t
y m b f
k
k F
=

( )
1432 . 0 14 . 3 10 667 . 86
3
507 . 1 3 2 23 . 9952
=
+
m m
m
m

( ) 0 507 . 1 3 0256 . 17
3
= + m m

7

865 . 5 m
Take m = 6 mm , and mm m b 60 6 10 10 = = =
g bg s
y m b f F =

1432 . 0 6 14 . 3 60 66 . 86 =

9 . 14027 = N
Also,
N
m
F
t
7 . 1658
6
23 . 9952 23 . 9952
= = =

Velocity,

Here, for grey cast iron material
c = 8150e and for commercial cut, e = 0.064
6 . 521 064 . 0 8150 = = c

now dynamic load,
( ) { }
( ) { }

+ +
+
+ =
t m
t m
t d
F b c V
F b c V
F F
21
21


( ) { }
( ) { }

+ +
+
+ =
7 . 1658 60 6 . 521 042 . 9 21
7 . 1658 60 6 . 521 042 . 9 21
7 . 1658


N 18506 =

As here
d s
F F

Is not fulfilled hence the design is not safe.
Now take, m = 8 mm,
mm m b 80 8 10 10 = = =
g bg s
y m b f F =

1432 . 0 8 14 . 3 80 66 . 86 =

29 . 24921 = N
Also,
N
m
F
t
02 . 1244
8
23 . 9952 23 . 9952
= = =

Velocity,

Here, for grey cast iron material
c = 8150e and for carefully cut, e = 0.038
7 . 309 038 . 0 8150 = = c

m/sec 0426 . 9 6 507 . 1 507 . 1 = = = m Vm
m/sec 06 . 12 8 507 . 1 507 . 1 = = = m Vm

8

( ) { }
( ) { }

+ +
+
+ =
02 . 1244 80 7 . 309 06 . 12 21
02 . 1244 80 7 . 309 06 . 12 21
02 . 1244
d
F


N 71 . 17139 =

Here , F
s
> F
d
is fulfilled, hence the design is safe under beam strength.
Design in wear strength:
Dynamic load:
N F
d
71 . 17139 =

For safe design the criteria is given by,
d w
F F
And also pinion is always weak in wear strength
d p w
F k Q b D F = =
Here,
619 . 1
20 85
85 2
2
=
+

=
+

=
g p
g
Z Z
Z
Q
71 . 17139 619 . 1 100 200 = k
529 . 0 = k
Now,

+ =
g p
es
E E
Sin
f
k
1 1
4 . 1
2

=
5 5
2
10 1 . 2
1
10 1 . 2
1
20
4 . 1
529 . 0 Sin
f
es

98 . 476 =
es
f N/mm
2

Thus,
65 . 68 74 . 2 = BHN f
es

65 . 68 74 . 2 98 . 476 = BHN
133 . 199 BHN

Thus, for safe design in wear strength the BHN should be 200.

Example 5: The following data is given for a steel spur gear pair transmitting 5
kW power from a shaft rotating at 3000 rpm to another parallel shaft rotating at
1500 rpm.
, , .

,

. Calculate the factor of safety of gear drive.
Solution:

9

Here, the pinion and gear are made up of same material having

630, fos 3
Permissible stress

630
3
210 MPa
Now,

5 10

2 3.14 3000

60

15923.5
Tangential force,

2 15923.5

2 15923.5
4 18
442.31
Velocity,

60000

4 18 3000
60000
11.304
Assume,

3
3

3
3 11.304
0.2097
Now,

4 18 72 mm
Y

0.154
.

0.154
.

0.1033

0.154
0.912

0.154
0.912
36
0.1287
Hence, the pinion as weak as gear because

.
Face width 40 is given,
Design for a beam strength,

210 40 4 0.1033
10898
Dynamic load carrying capacity,

21

21

, c 11860e
11860 0.05 593 (e = 0.05, using first class commercial gears)

442.31
21 11.304593 40 442.321
21 11.304 593 40 442.321

15046.67
As,

, the design is unsafe.

442.31
21 11.304296.5 40 442.321
21 11.304 296.5 40 442.321

8826.976 N
Thus factor of safety in beam strength is,

10901.72
8826.97

1.23
Now in wear strength,

Here,

72 mm and 40
Now,

2

2 36
18 36
1.333

sin
1.4

1

Here, BHN is 400.

2.74 68.65 2.74 400 68.65

1027.35

sin20
1.4

1
2.1 10

1
2.1 10

2.4556

72 40 1.333 2.4556
9427.14 N
Thus factor of safety in wear strength,

9427.14
8826.976

1.067
Factor of safety of gear drive is 1.067.

Example 6: Design and draw a 2-stage spur gear reduction gear box pairs for
operating a set of two belt conveyors for data.
Max. Conveyor Speed = 1.5 m/s
Effective diameter of driving pulley = 250 mm
Amount of torque Transmitted by each driving pulley of conveyor = 200 Nm
Speed of the input shaft (N) = 1920 rpm.
Solution:



11


















P & Q: Two Conveyor Pulleys
1-2 & 3-4: Pair of meshing gears
1 is input gear and 4 is output gear
Effective diameter of driving pulley, d = 250 mm
Speed of input shaft, N
1
= 1920 rpm
Torque on each pulley = 200 Nm
Torque on output shaft = 400 Nm
Now, maximum conveyor speed, V
c
= 1.5 m/s

60
4
N d
V
c

=


60
25 . 0 14 . 3
5 . 1
4
N
=
649 . 114
4
= N rpm
Power transmitted is,
8 . 4
60000
400 64 . 114 14 . 3 2
60000
2
=

=

=
T N
P

kW
Considering slip and other losses take service factor as 1.25
So, Power, 6 8 . 4 25 . 1 = = P kW
Considering gears of same material (steel-steel)
Max. Torque is transmitted by pair of gears 3-4; so we would design for that gear pair.
Module,

, 3

.


1


2

3


4


Q
Input Shaft
Output Shaft

P

12

p
Z m d =
3

Thus velocity,
60000 60000
3
3 3
3
N Z m
N d
V
p

=

=

Now, final speed ratio,
16 74 . 16
64 . 114
1920
4
1
= =
N
N

It is achieved in two stages
So, speed ratio in 1 stage 4 16 = =
So, speed of the gear 2 or 3,
480
4
1920
3 2
= = = N N rpm
Then ,
For 20
0
full depth, assume Z
p
= 18
m
m
N Z m
V
p
45216 . 0
60000
480 18 14 . 3
60000
3
3
=

=

=

m/s
Lewis tooth form factor,
1033 . 0
18
912 . 0
154 . 0
912 . 0
154 . 0 = = =
p
Z
Y
Here, k
s
= 1 and
m V
k
v
45216 . 0 3
3
3
3
3
+
=
+
=
Let, face width, b = 10m mm,
For steel, f
b
= 210 MPa
For gear 3
Power,
60000
480 14 . 3 2
60000
2
3 3 3
T T N
P

=

=



60000
480 14 . 3 2
6
3
T
=
42 . 119
480 14 . 3 2
6 60000
3
=

= T Nm
Thus tangential force,
m m d
T
F
p
t
889 . 13268
18
10 42 . 119 2 2
3
3
=

=
Now,

g bg
v
s t
y m b f
k
k F
=

( )
1033 . 0 14 . 3 10 210
3
45216 . 0 3 1 889 . 13268
=

+
m m
m
m
( ) 0 45216 . 0 3 4932 . 6
3
= + m m


13

055 . 3 m
Take, m 3 mm
30 3 10 10 = = = m b mm

210 10 3 3 0.1033 6130.441 N
48 . 4418
3
889 . 13268 889 . 13268
= = =
m
F
t
N
3564 . 1 3 45216 . 0 45216 . 0
3
= = = m V m/s

(here, V
m
= V
3
)
c = 11860e and e = 0.05
593 05 . 0 11860 = = c

( ) { }
( ) { }

+ +
+
+ =
48 . 4418 30 593 3564 . 1 21
48 . 4418 30 593 3564 . 1 21
48 . 4418
d
F

206 . 7982 = N
As, criteria of
d s
F F is not fulfilled the design is not safe.
Hence,
Now take, m = 4 mm
30 3 10 10 = = = m b mm

210 10 4 4 0.1033 10898.56 N
22 . 3317
4
889 . 13268 889 . 13268
= = =
m
F
t
N
8086 . 1 4 45216 . 0 45216 . 0
3
= = = m V m/s

(here, V
m
= V
3
)
c = 11860e and e = 0.05 (commercial cut gears)
593 05 . 0 11860 = = c

( ) { }
( ) { }

+ +
+
+ =
22 . 3317 40 593 8086 . 1 21
22 . 3317 40 593 8086 . 1 21
22 . 3317
d
F 521 8390. = N
As, criteria of
d s
F F is fulfilled the design is safe under beam strength.
Now, For wear strength,
k Q b d F
p w
=
Where,
72 18 4 = = =
p p
Z m d mm
6 . 1
5
8
5
4 2 2
= =

=
+

=
p
p
g p
g
Z
Z
Z Z
Z
Q


14

For safe design in wear the criteria is,
d w
F F

Take, F
w
= F
d

521 . 8390 = k Q b d
p

521 . 8390 6 . 1 40 72 = k
8208 . 1 = k
Now,

+ =
g p
es
E E
Sin
f
k
1 1
4 . 1
2

=
5 5
2
10 1 . 2
1
10 1 . 2
1
20
4 . 1
8208 . 1 Sin
f
es

63 . 884 =
es
f MPa
Thus,

65 . 68 74 . 2 = BHN f
es


65 . 68 74 . 2 63 . 884 = BHN
9 . 347 BHN

Thus, for safe design in wear strength the suggested BHN is 350.
Example 7: Following data is given for a steel gear pair transmits 7.5 KW power
from an electric motor running at 1440 RPM to a machine running at 480 RPM.
Centre distance is 240 mm. Allowable bending stress for pinion and gear 160
MPa. Surface hardness is 300 BHN and 20
0
stub tooth. K
s
= 1.25. Design a gear
pair for above application.
Solution:
GIVEN DATA:-
Power, P = 7.5 KW, Pinion speed, N
p
= 1440 RPM, Gear speed, N
g
= 480 RPM
Center distance, a = 240 mm, Gear type is 20
0
stub tooth,
Bending stress for pinion and gear, f
b
= 160 MPa
k
s
= 1.25
Speed ratio:-

1440
480
3
Center distance:-

240

240 2

120 =
p
D
mm

15

360 3 120 = =
g
D mm
Assume, Z
P
= 18
p p
Z m D =
18 120 = m
67 . 6 = m mm
Take, m = 7 mm
126 18 7 = = =
p p
Z m D mm
mm 378 3 126 = =
g
D
252
2
360
2
126
= + = a mm
Assume, Z
P
= 19
p p
Z m D =
19 120 = m
314 . 6 = m mm
Take, m = 7 mm
133 19 7 = = =
p p
Z m D mm
399 3 133 = =
g
D
266
2
399
2
133
= + = a mm
Assume, Z
P
= 20

p p
Z m D =
20 120 = m
6 = m mm
Take, m = 6 mm
120 20 6 = = =
p p
Z m D mm
360 3 120 = =
g
D
240
2
360
2
120
= + = a mm
This condition is satisfied
So,take Z
p
= 20
Checking for beam strength,
60 6 10 10 = = = m b mm
For 20
0
stub teeth

p
p
Z
y
95 . 0
175 . 0 =
20
95 . 0
175 . 0 = 1275 . 0 =

p b s
y m b f F =

16


1275 . 0 6 14 . 3 60 160 = 16 . 23060 = N
Now,
Power,
2 2 3.14 1440
60000 60000
N T T
P

= =
3
10 761 . 49
1440 14 . 3 2
5 . 7 60000
=

= T N-mm
Tangential force,
93 . 828
120
10 761 . 49 2 2
3
=

=

=
p
t
D
T
F N
For 20
0
stub teeth and considering carefully cut


e c 12300 = and 030 . 0 = e
369 030 . 0 12300 = = c
Velocity,
0477 . 9
60000
1440 120 14 . 3
60000
=

=

=
p p
m
N D
V

m/s
Dynamic load,
( ) { }
( ) { }

+ +
+
+ =
t m
t m
t d
F b c V
F b c V
F F
21
21


( ) { }
( ) { }
12 . 13606
93 . 828 60 369 0477 . 9 21
93 . 828 60 369 0477 . 9 21
93 . 828 =

+ +
+
+ = N
Here, F
s
> F
d

So, design is safe
Checking for wear strength
k Q b D F
p w
=
Where,
5 . 1
4
3 2
3
3 2 2
=

=
+

=
+

=
p p
p
g p
g
Z Z
Z
Z Z
Z
Q
Also,
35 . 753 65 . 68 300 74 . 2 65 . 68 74 . 2 = = = BHN f
es
N/mm
2

Thus,
3 . 1
10 1 . 2
1
10 1 . 2
1
20
4 . 1
35 . 753 1 1
4 . 1
5 5
2 2
=

+ = Sin
E E
Sin
f
k
g p
es

Hence,

k Q b D F
p w
= 14040 3 . 1 5 . 1 60 120 = = N
As seen, F
w
> F
d

hence the design is safe in wear strength.


17

Example 8: A pair of spur gear with 20
0
full depth involute consist of 18 teeth
pinion meshing with 40 teeth gear. The module is 5 mm and face width is 50 mm.
The pinion is made up of alloy steel , While the gear is made up of plain carbon
steel , For which the permissible bending stress are 250 MPa and 200 MPa
respectively. The gears are heat treated to a surface hardness of 400 BHN. The
service factor is 1.75. If pinion rotates at 1440 rpm, determine rated power
transmitting capacity of gear pair.
Solution:
Given: Power, N
p
= 1440 rpm, k
s
= 1.75,
No. of teeth on pinion, Z
p
= 18 No. of teeth on gear, Z
g
= 40
Ultimate tensile strength of pinion, f
ut
= 250MPa
Ultimate tensile strength of gear, f
ut
= 200MPa
Speed of gear,
p p g g
Z N Z N =
18 1440 40 =
g
N
40
18 1440
=
g
N
648 =
g
N

rpm
Now, for 20
o
full depth, the Lewis form factor is,
For pinion,
18
912 . 0
154 . 0
912 . 0
154 . 0 = =
p
p
Z
y 0.1033 =
For gear,
40
912 . 0
154 . 0
912 . 0
154 . 0 = =
g
g
Z
y 0.1312 =
Now,
825 . 25 1033 . 0 250 = =
p bp
Y f
24 . 26 1312 . 0 200 = =
g bg
Y f
here,
p bp g bg
Y f Y f >
Hence, the pinion is weak in beam strength than gear. Hence we will design the
pinion in beam strength.
Now m =
p
p
Z
D

5 =
18
p
D


p
D = 90 mm
Now,
60000
p p
m
N D
V

=

18

s m V
m
/ 79 . 6
60000
1440 90 14 . 3
=

=

By Lewis equation,
s bp p
F f b m y = = 250 50 5 0.1033 = 20282.9 N
Now, k Q b D F
p w
=
Here,
3793 . 1
18 40
40 2
2
=
+

=
+

=
g p
g
Z Z
Z
Q
2
1 1
1.4
es
p g
f
k Sin
E E

= +




2.74 68.65
es
f BHN =
2.74 400 68.65
es
f =
2
1027.35N/mm
es
f =

=
5 5
2
10 1 . 2
1
10 1 . 2
1
20
4 . 1
35 . 1027
Sin k = 2.45
N F
w
48 . 15203 45 . 2 379 . 1 50 90 = =
F
d
= F
s
OR F
w
whichever is less.
Therefore, F
d
= F
w
= 15203.48 N
Now for dynamic load,
( ) { }
( ) { }

+ +
+
+ =
t m
t m
t d
F b c V
F b c V
F F
21
21

c = 11860 e
here e = 0.056 (commercial cut)
c = 664.16
( ) { }
( ) { }

+ +
+
+ =
t
t
t
F
F
F
50 16 . 664 79 . 6 21
50 16 . 664 79 . 6 21
48 . 15203
Solve the above equation by trail and error method.
480 =
t
F N
p
p
t
D
T
F

=
2

90
2
480
p
T
=

21600 =
p
T Nmm


19

1000 60000
21600 1440 14 . 3 2

= = 3.25 kW

Example 9: The P.C.D of spur pinion and gear are 100 mm and 300 mm
respectively, The pinion is made up of plain carbon steel 40C8 (f
ut
= 600 MPa)
While gear is made up of grey cast iron FG300 Pinion receives 5 kW power at
500 rpm through its shaft. The service factor and F.O.S can be taken as 1.5 each.
Design a gear pair.
Solution:
d
p
= 100 mm, d
g
= 300 mm
N
p
= 500 rpm, P = 5 kW
Ultimate tensile strength of pinion, f
ut
= 600MPa
Ultimate tensile strength of gear, f
ut
= 300MPa

300 MPa

150 MPa
Assume min. number of teeth on pinion = 20

20
100 = m20
m = 5 mm
For pinion,
20
912 . 0
154 . 0
912 . 0
154 . 0 = =
p
p
Z
y 1084 . 0 =
For gear,
60
912 . 0
154 . 0
912 . 0
154 . 0 = =
g
g
Z
y 1388 . 0 =
Now,
52 . 32 1084 . 0 300 = =
p bp
Y f
82 . 20 1388 . 0 150 = =
g bg
Y f
here,
p bp g bg
Y f Y f <
Hence, the gear is weak in beam strength than pinion. Hence we will design the gear
in beam strength.
p p g g
D N D N =
g
N 300 = 500 100
g
N = 166.66 rpm
2
60
g g
N T
P

=
T
g
=

.
= 286.49 = 286.4910
3


20

Tangential force,
g
g
t
D
T
F

=
2
286.491000 2
300

= 1909.933N =

also velocity
60000
g g
m
N D
V

=

617 . 2
60000
66 . 166 300 14 . 3
=

= m / s

g bg s
y m b f F =
b = 10m mm
1388 . 0 5 50 150 =
s
F = 16351.98 N
Now dynamic load,
( ) { }
( ) { }

+ +
+
+ =
t m
t m
t d
F b c V
F b c V
F F
21
21

c = 8150 e
Here from table e = 0.056

c = 456.4
( ) { }
( ) { }

+ +
+
+ =
83 . 1909 50 4 . 456 617 . 2 21
83 . 1909 50 4 . 456 617 . 2 21
83 . 1909
d
F

=

1909.83 + 6404.27 = 8314.10 N

1.96 > 1.5

=
+

=
g p
g
Z Z
Z
Q
F
w
= 12471.15 N

w p d
F D b Q k F = =
12471.15 100 50 1.5
k = 1.662

2
1 1
1.4
es
p g
f
k Sin
E E

= +

=
5 5
2
10 1 . 1
1
10 1 . 2
1
20
4 . 1
662 . 1 Sin
f
es

f
es
= 700.78 N/mm
2


2.74 68.65
es
f BHN =
700.78 = 2.74 BHN 68.65
BHN = 280.81
********** ********** ********** **********