Question

Figure P14.28 shows two bearings supporting a 1000-rpm shaft and gear. The bearing on the left carries a 5-kN radial load and
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Answer #1

Given data 1

N= 1000 rpm

F, = 5 kN

F_{a}=1\textup{ kN}

\textup{From Table 14.3, for light-moderate impact}\left ( K_{a} \right )=1.5

L_{h}=5000\textup{ h}

\textup{reability}=1-0.02=0.98

\textup{From Table 14.3, with 98 percent reliability }\left ( K_{r} \right )=0.33

Solution →

\textup{Life in million revolutions are}

L=L_{h}*N*60

L=5000*1000*60=3*10^{8}\textup{ rev}

\textup{The life corresponding to rated capacity}\left ( L_{R} \right )=90*10^{6}

\frac{F_{a}}{F_{r}}=\frac{1}{5}=0.2< 0.35

\textup{For a radial bearing, Eq. (14.3) gives}

F_{e}=F_{r}=5\textup{ kN}

\textup{For a radial bearing, from Eq. (14.5b), the required value of rated capacity for the application}C_{req}=F_{e}K_{a}*\left ( \frac{L}{K_{r}L_{R}} \right )^{0.3}

C_{req}=5*1.5*\left ( \frac{3*10^{8}}{0.33*90*10^{6}} \right )^{0.3}

C_{req}=15.00\textup{ kN}

\textup{From Table 14.2, a 200 radial bearing with a rated capacity of 16 kN would be sufficient for left bearing with bore 65 mm}

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