Question

Two ball bearings from different manufacturers are being considered for a certain application. Bearing A has a catalog rating of 2.12 kN based on a catalog rating system of 3000 hours at 500 rev/min. Bearing B has a catalog rating of 7.5 kN based on a catalog that rates at 106 cycles. For a given application, determine which bearing can carry the larger load.

Answer 1

Answer:

F$_D$ for A > F$_D$ for B

Hence, Bearing A can carry the larger load

Explanation:

Given the data in the question,

First lets consider an application which requires desired speed of n₀ and a desired life of L₀.

Lets start with Bearing A

so we write the relation between desired load and life catalog load and life;

$F_R(L_Rn_R60)^{1/a}$ = $F_D(L_Dn_D60)^{1/a}$

where F$_R$ is the catalog rating( 2.12 kN)

L$_R$ is the rating life ( 3000 hours )

n$_R$ is the rating speed ( 500 rev/min )

F$_D$ is the desired load

L$_D$ is the desired life ( L₀ )

n$_D$  is the the desired speed ( n₀ )

Now as we know, a = 3 for ball bearings

so we substitute

$2.12( 3000 * 500 * 60 )^{1/3$  =  $F_D( L_0n_060)^{1/3$    

950.0578 = $F_D( L_0n_0)^{1/3} 3.914867$    

950.0578 / 3.914867 = $F_D( L_0n_0)^{1/3}$

242.6794 =   $F_D( L_0n_0)^{1/3}$

F$_D$ for A =  (242.6794 / $( L_0n_0)^{1/3}$ ) kN

Therefore the load that bearing A can carry is  (242.6794 / $( L_0n_0)^{1/3}$ ) kN

Next is Bearing B

$F_R(L_Rn_R60)^{1/a}$ = $F_D(L_Dn_D60)^{1/a}$

F$_R$ = 7.5 kN, $(L_Rn_R60) = 10^6$

Also, for ball bearings, a = 3

so we substitute

$7.5(10^6)^{1/3$ = $F_D(L_0n_060)^{1/3}$

750 =  $F_D(L_0n_0)^{1/3} 3.914867$

750 / 3.914867  =  $F_D(L_0n_0)^{1/3}$

191.5773 = $F_D(L_0n_0)^{1/3}$

F$_D$ for B = ( 191.5773 / $(L_0n_0)^{1/3}$ ) kN

Therefore, the load that bearing B can carry is  ( 191.5773 / $(L_0n_0)^{1/3}$ ) kN

Now, comparing the Two results above,

we can say;

F$_D$ for A > F$_D$ for B

Hence, Bearing A can carry the larger load