Question

A single row deep groove ball bearing has a dynamic load capacity of 50 kN and expected bearing life ($L_{90}$) is 40 million revolution when operating under the following work cycle:

(i) Radial load of 6 kN at 800 rpm for 70% of the time.

(ii) Radial load of 'X' kN at 500 rpm for 30% of the time.

What will be the value of X?

• A. None of these
• B. 18
• C. 24
• D. 30

Answer 1

The correct option is C 24

Given , $C=50kN,L_{90}=40M.rev$

We know that, $L_{90}=(\frac{C}{P_c}{)}^3$

$40=\left(\frac{50}{P_c}\right)$

$\Rightarrow P_c^3=3125$

We know that,

$P_c=\sqrt[3]{3\frac{(P_1{)}^3\times n_1+(P_2{)}^3\times n_2}{n_1+n_2}}$

$P_c=\sqrt[3]{3\frac{(6{)}^3\times (800\times 0.7)+(x^3)\times (500\times 0.3)}{800\times 0.7+500\times 0.3}}$

$P_r^3=\frac{6\times 560+x^3\times 150}{560+150}$

$\frac{(3125\times 710)-(6^3\times 560)}{150}=x^3$

$x^3=13985.266$

$x=24.093kN$