Question

The ball bearings are to be selected for an application in which the radial load is 2000 N during 90 per cent of the time and 8000 N during the remaining 10 per cent. The shaft is to rotate at 150 r.p.m. Determine the minimum value of the basic dynamic load rating for 5000 hours of operation with not more than 10 per cent failures.

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

13.8KN (Minimum basic dynamic load rating.)

Explanation:

Given  W₁=2000N    n₁=0.9

           W₂=8000N    n₂=0.1

           N=150 rpm

Step1:

Equivalence constant load(W) for ball bearings (k=3)

   

          W = $\frac{n_{1} (W_{1})^{3} + n_{2} (W_{2})^{3} }{n_{1}+ n_{2}}$

          W= $\frac{0.9 (2000)^{3} + 0.1 (8000)^{3} }{0.1+ 0.9}$

        On solving for W,

          W = 3879.75 N

Step2:

Relation for life in revolutions (L) and life in working hours (L_H) is:

                               $L=60N.L_{H} revolutions$

Given $L_{H}$=5000 hours

$L =60*150*5000\\ L =45000000 =45*10^{6} revolutions$

Step3:

Basic dynamic load Rating is

$c=W(\frac{L}{10^{6} })^{\frac{1}{k} }$

K=3 for ball bearings.

$c=3879.75*(\frac{45*10^{6} }{10^{6} } )^{\frac{1}{3} }$

$c = 3879.75 * 3.556\\c= 13799.85N$

c= 13.8KN (ANS)