Answer:
F[tex]_D[/tex] for A > F[tex]_D[/tex] 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;
[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]
where F[tex]_R[/tex] is the catalog rating( 2.12 kN)
L[tex]_R[/tex] is the rating life ( 3000 hours )
n[tex]_R[/tex] is the rating speed ( 500 rev/min )
F[tex]_D[/tex] is the desired load
L[tex]_D[/tex] is the desired life ( L₀ )
n[tex]_D[/tex] is the the desired speed ( n₀ )
Now as we know, a = 3 for ball bearings
so we substitute
[tex]2.12( 3000 * 500 * 60 )^{1/3[/tex] = [tex]F_D( L_0n_060)^{1/3[/tex]
950.0578 = [tex]F_D( L_0n_0)^{1/3} 3.914867[/tex]
950.0578 / 3.914867 = [tex]F_D( L_0n_0)^{1/3}[/tex]
242.6794 = [tex]F_D( L_0n_0)^{1/3}[/tex]
F[tex]_D[/tex] for A = (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN
Therefore the load that bearing A can carry is (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN
Next is Bearing B
[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]
F[tex]_R[/tex] = 7.5 kN, [tex](L_Rn_R60) = 10^6[/tex]
Also, for ball bearings, a = 3
so we substitute
[tex]7.5(10^6)^{1/3[/tex] = [tex]F_D(L_0n_060)^{1/3}[/tex]
750 = [tex]F_D(L_0n_0)^{1/3} 3.914867[/tex]
750 / 3.914867 = [tex]F_D(L_0n_0)^{1/3}[/tex]
191.5773 = [tex]F_D(L_0n_0)^{1/3}[/tex]
F[tex]_D[/tex] for B = ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN
Therefore, the load that bearing B can carry is ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN
Now, comparing the Two results above,
we can say;
F[tex]_D[/tex] for A > F[tex]_D[/tex] for B
Hence, Bearing A can carry the larger load
