# A bearing supports a radial load of 7000 N and a thrust load of 2100 N. The desired life of the ball bearing is 160 × 106 revolutions at 300 rpm. If t

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A bearing supports a radial load of 7000 N and a thrust load of 2100 N. The desired life of the ball bearing is 160 × 106 revolutions at 300 rpm. If the load is uniform and steady, service factor is 1, radial factor is 0.65, thrust factor is 3.5, k = 3 and rotational factor is 1, the basic dynamic load rating of a bearing will be nearly
1. 65 kN
2. 75 kN
3. 85 kN
4. 95 kN

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Correct Answer - Option 1 : 65 kN

Concept:

Equivalent bearing load (W) is given by:

W = XVFr + YFa

Life of a bearing is given by:

${L_{10}} = {\left( {\frac{C}{W}} \right)^n}$

where

L10 = basic life rating in million revolutions

n = constant = 3 for ball bearing and  $\frac{{10}}{3}$ for roller bearing

Calculation:

Given:

Fr = 7000 N, Fa = 2100 N, X = 0.65, V = 1, Y = 3.5, L10 = 160 million revolutions

Equivalent bearing load (W) is given by:

W = XVFr + YFa

W = (0.65 × 1 × 7000) + (3.5 × 2100) = 11900 N

Now,

Life of a bearing is:

${L_{10}} = {\left( {\frac{C}{W}} \right)^n}$

$C = W \times {\left( {{L_{10}}} \right)^{\frac{1}{n}}}$

C = 11900 × (160)1/3 = 64603.14 N = 64.6 kN ≈ 65 kN