# DESIGN AND DEVELOPMENT; BEARING SELECTION AND DESIGN Essay Example

Bearing Selection and Design

Bearing Selection and Design

Bearing Selection and Design

Taking moment about bearing A, then the reactions at the bearing supports will be obtained as follows. Giving = 2.33KN

And summing vertical forces, then — -2.33+7=0

Giving = 4.66KN

In the Horizontal direction each bearing offers an equal amount of reaction giving an aggregate of 3KN

Thus = 1.5KN and =1.5KN

Combined reactions at bearings;

= 2.7736KN A =

= 4.90213KN B=

Equivalent radial loads are obtained as follows; V= rotational factor which is equal to 1 as the inner race is rotating = constant axial load = constant radial load

Equivalent radial load at bearing B= KN

If the expected life for 90% of the bearing is 1500 hours and taking a load factor as 1 at steady state condition, N is obtained from the equation Where V= 33m/s =  = N=1273revs = 114.5767 Basic dynamic load rating, C,= Where; k=3 for roller bearings

C=4.90213 114.57671/3 1= 23.8095KN

From SKF tables we select the bearing code 6210

2. If bearing no. 61810 is selected, the value of Basic dynamic load rating ,C, from the tables is 6.76KN

L= 3 L= 3 L=2.6223 and this is life at 90% reliability

life at 99% reliability will be obtained as follows; Considering life adjustment factors for operating condition and material as 0.9 and 0.85 respectively and their product equal to b are constants whose value is 1.17 =0.1026 =0.1026 2.6223 =0.26905  0.26905 60 1273  =3.5223 hours

ii. Expected life in km

Speed= 120km/h

Life= 120km/h 3.5223

=422.68km

3. Life of bearing B= life of bearing A= 114.5767 C= C= C= 13.47KN

From SKF table the bearing selected is with the code 61910

4. It’s an overestimate, the reason being that the bearing selection and force application is considered at 99% reliability