A ball bearing is subjected to a radial force which varies in sinusoidal way as shown in Fig.15.14 and discussed in Ex. 15.13. The direction of force remains fixed. The amplitude of the force is 1500 N and the speed of rotation is 720 rpm. Determine the dynamic load capacity of the bearing for

Question

A ball bearing is subjected to a radial force which varies in sinusoidal way as shown in Fig.15.14 and discussed in Ex. 15.13. The direction of force remains fixed. The amplitude of the force is 1500 N and the speed of rotation is 720 rpm. Determine the dynamic load capacity of the bearing for the expected life of 8000 h.

Accepted Answer

[latex] ext { Given } P_{\max }=1500 N \quad n=720 rpm [/latex].

[latex] L_{10 h }=8000 h [/latex].

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,

[latex] P=\frac{1}{2} P_{\max }(1-\cos heta) [/latex].

[latex] ext { Considering the work cycle from } heta=0 ext { to } heta=2 \pi [/latex]

and applying Eq. (15.15),

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

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

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

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

[latex] ext { Also, } \int(1-\cos heta)^{3} d heta=\int\left(1-3 \cos heta+3 \cos ^{2} heta-\cos ^{3} heta ight) d heta [/latex].

[latex] =\int\left[1-3 \cos heta+\frac{3(1+\cos 2 heta)}{2}-\cos heta\left(1-\sin ^{2} heta ight) ight] d heta [/latex].

[latex] =\int\left(2.5-4 \cos heta+1.5 \cos 2 heta+\cos heta \sin ^{2} heta ight) d heta [/latex].

[latex] =\left[2.5 heta-4 \sin heta+0.75 \sin 2 heta+\frac{\sin ^{3} heta}{3} ight] [/latex] (b).

The following formulae are used in above derivation:

[latex] \int \cos heta d heta=\sin heta \quad \int \cos 2 heta d heta=\frac{\sin 2 heta}{2} [/latex].

[latex] \int \cos heta \sin ^{2} heta=\frac{\sin ^{3} heta}{3} [/latex].

Since

[latex] \int_{0}^{2 \pi}(1-\cos heta)^{3} d heta=\int_{0}^{\pi}(1-\cos heta)^{3} d heta [/latex]

[latex] +\int_{\pi}^{2 \pi}(1-\cos heta)^{3} d heta [/latex].

[latex] =\left[2.5 heta-4 \sin heta+0.75 \sin 2 heta+\frac{\sin ^{3} heta}{3} ight]_{0}^{\pi} [/latex]

[latex] +\left[2.5 heta-4 \sin heta+0.75 \sin 2 heta+\frac{\sin ^{3} heta}{3} ight]_{\pi}^{2 \pi} [/latex].

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

[latex] herefore \quad \int_{0}^{2 \pi}(1-\cos heta)^{3} d heta=5 \pi [/latex] ( c).

from (a) and (c),

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

Step II Dynamic load carrying capacity of bearing

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

[latex] =345.6 ext { million rev. } [/latex]

From Eq. (15.7),

[latex] C=P_{e}\left(L_{10} ight)^{1 / 3} [/latex] (15.7).

[latex] C=P_{e}\left(L_{10} ight)^{1 / 3}=1017.9(345.6)^{1 / 3} [/latex].

= 7143.26 N.

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

This Question has been Answered.

Already have an account?

Login

Related Answered Questions