Question 15.14: A ball bearing is subjected to a radial force which varies i...

\text { Given } P_{\max }=1500 N \quad n=720 rpm .

L_{10 h }=8000 h .

Step I Equivalent load for complete work cycle
As derived in the previous example, the equation for force P at angle of rotation θ is given by,

P=\frac{1}{2} P_{\max }(1-\cos \theta) .

and applying Eq. (15.15),

P_{e}=\left[\frac{1}{N} \int P^{3} d N\right]^{1 / 3}             (15.15).

P_{e}=\left[\frac{1}{N} \int P^{3} d N\right]^{1 / 3} .

=\left[\frac{1}{2 \pi} \int \frac{P_{\max }^{3}}{8}(1-\cos \theta)^{3} d \theta\right]^{1 / 3} .

=\frac{P_{\max .}}{2}\left[\frac{1}{2 \pi} \int(1-\cos \theta)^{3} d \theta\right]^{1 / 3}                 (a).

\text { Also, } \int(1-\cos \theta)^{3} d \theta=\int\left(1-3 \cos \theta+3 \cos ^{2} \theta-\cos ^{3} \theta\right) d \theta .

=\int\left[1-3 \cos \theta+\frac{3(1+\cos 2 \theta)}{2}-\cos \theta\left(1-\sin ^{2} \theta\right)\right] d \theta .

=\int\left(2.5-4 \cos \theta+1.5 \cos 2 \theta+\cos \theta \sin ^{2} \theta\right) d \theta .

=\left[2.5 \theta-4 \sin \theta+0.75 \sin 2 \theta+\frac{\sin ^{3} \theta}{3}\right]               (b).

The following formulae are used in above derivation:

\int \cos \theta d \theta=\sin \theta \quad \int \cos 2 \theta d \theta=\frac{\sin 2 \theta}{2} .

\int \cos \theta \sin ^{2} \theta=\frac{\sin ^{3} \theta}{3} .

Since

+\int_{\pi}^{2 \pi}(1-\cos \theta)^{3} d \theta .

+\left[2.5 \theta-4 \sin \theta+0.75 \sin 2 \theta+\frac{\sin ^{3} \theta}{3}\right]_{\pi}^{2 \pi} .

=[2.5 \pi-0]+[2.5(2 \pi-\pi)]=5 \pi .

\therefore \quad \int_{0}^{2 \pi}(1-\cos \theta)^{3} d \theta=5 \pi             ( c).

from (a) and (c),

P_{e}=\frac{P_{\max }(2.5)^{1 / 3}}{2}=\frac{1500(2.5)^{1 / 3}}{2}=1017.9 N .

Step II Dynamic load carrying capacity of bearing

L_{10}=\frac{60 n L_{10 h }}{10^{6}}=\frac{60(720)(8000)}{10^{6}} .

From Eq. (15.7),

C=P_{e}\left(L_{10}\right)^{1 / 3}                   (15.7).

C=P_{e}\left(L_{10}\right)^{1 / 3}=1017.9(345.6)^{1 / 3} .

= 7143.26 N.