Question 15.9: A machine shaft, supported on two identical taper roller bea...

\text { Given } K_{a}=10 kN \quad n=300 rpm \quad d=60 mm .

L_{10 h }=4000 h .

Step I Radial and axial forces on bearings
Refer to forces acting on the shaft in the vertical plane as shown in Fig. 15.13. Taking moments about the bearing B,

F_{r A} \times 300=30000 \times 100 \quad \text { or } \quad F_{r A}=10000 N .

Taking moments about the bearing A,

F_{r B} \times 300=30000 \times 200 \text { or } F_{r B}=20000 N .

Also,

K_{a}=10000 N .

Step II Tentative selection of bearing

F_{r A}<F_{r B} \text { and } K_{a}>0 .

The above conditions are similar to Case 2(a) illustrated in Fig. 15.11. For this case,

F_{a A}=F_{a B}+K_{a}             (a).

F_{a B}=\frac{0.5 F_{r B}}{Y}               (b).

Tentatively, we will consider the following taper roller bearings from Table 15.6, which are available for the shaft of 60 mm diameter.

It is also observed that the values of Y vary from 0.72 to 1.8. Taking an average value of 1.3 as the first trial value,

F_{a B}=\frac{0.5 F_{r B}}{Y}=\frac{0.5(20000)}{1.3}=7692.31 N .

F_{a A}=F_{a B}+K_{a}=7692.31+10000=17692.31 N .

Bearing A is more critical because it is subjected to maximum load. For the bearing A,

F_{r A}=10000 N \quad F_{a A}=17692.31 N .

K_{a}=10000 N .

Therefore,

\frac{F_{a A}}{F_{r A}}=\frac{17692.31}{10000}=1.769 .

It is observed from the above table that for all values of e,

\frac{F_{a A}}{F_{r A}}>e .

From Eq. (15.12),

P=0.4 F_{r}+Y F_{a} \text { when }\left(F_{d} / F_{r}\right)>e           (15.12).

P=0.4 F_{r}+Y F_{a}=0.4(10000)+1.3(17692.31) .

= 27 000 N
From Eq. (15.9),

L_{10}=\frac{60 n L_{10 h }}{10^{6}}               (15.9).

L_{10}=\frac{60 n L_{10 h }}{10^{6}}=\frac{60(300)(4000)}{10^{6}} .

= 72 million rev.
The dynamic load carrying capacity of the bearing A is given by,

C=P\left(L_{10}\right)^{0.3}=27000(72)^{0.3}=97401.02 N .

It is observed that Bearing No. 33112 (C = 111 000 N) may be suitable.
Step III Final check for selection
For Bearing No. 33112,

e = 0.4        Y = 1.5           C = 111 000 N.

F_{a B}=\frac{0.5 F_{r B}}{Y}=\frac{0.5(20000)}{1.5} 6666.67 N .

F_{a A}=F_{a B}+K_{a}=6666.67+10000 .

= 16 666.67 N
Since,

\frac{F_{a A}}{F_{r A}}=\frac{16666.67}{10000}=1.67 \quad \therefore \frac{F_{a A}}{F_{r A}}>e .

\therefore \quad P=0.4 F_{r}+Y F_{a} .

= 0.4(10 000) + 1.5 (16 666.67)
= 29 000 N.

C=P\left(L_{10}\right)^{0.3}=29000(72)^{0.3} .

= 104 615.91 N < 110 000 N
Therefore, Bearing No. 33112 (C = 111 000 N) is suitable for the application. We will check whether Bearing No. 33112 is also suitable at B. For the bearing B,

F_{r B}=20000 N .

F_{a B}=\frac{0.5 F_{r B}}{Y}=\frac{0.5(20000)}{1.5}=6666.67 N .

Therefore,

\frac{F_{a B}}{F_{r B}}=\frac{6666.67}{20000}=0.33 \quad \text { and } \quad e=0.4 .

\frac{F_{a B}}{F_{r B}}<e .

From Eq. (15.11),

P=F_{r} \text { when }\left(F_{a}/ F_{r}\right) \leq e               (15.11).

P=F_{r B}=20000 N .

C=P\left(L_{10}\right)^{0.3}=20000(72)^{0.3} .

= 72 148.91 N < 110 000 N.
Therefore, Bearing No. 33112 is selected at A as well as at B.