Question

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

Answer 1

Correct Answer - Option 1 : 65 kN

Concept:

Equivalent bearing load (W) is given by:

W = XVFr + YFa

where X = radial factor, V = service factor, F_r = radial load, Y = thrust factor, F_a =thrust load

Life of a bearing is given by:

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

where

L₁₀ = basic life rating in million revolutions

C = dynamic load-carrying capacity

W = equivalent dynamic load

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

Calculation:

Given:

F_r = 7000 N, F_a = 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