A machine shaft, supported on two identical taper roller bearings A and B, is shown in Fig.15.13. It is subjected to a radial force of 30 kN and a thrust force of 10 kN. The thrust is taken by Bearing A alone. The shaft rotates at 300 rpm. The machine is intermittently used and the expected life

Question

A machine shaft, supported on two identical taper roller bearings A and B, is shown in Fig.15.13. It is subjected to a radial force of 30 kN and a thrust force of 10 kN. The thrust is taken by Bearing A alone. The shaft rotates at 300 rpm. The machine is intermittently used and the expected life [latex] L_{10 h} [/latex] of the bearings is 4000 h. The minimum acceptable diameter of the shaft, where the bearings are mounted, is 60 mm. Select suitable taper roller bearings for the shaft.

Accepted Answer

[latex] ext { Given } K_{a}=10 kN \quad n=300 rpm \quad d=60 mm [/latex].

[latex] L_{10 h }=4000 h [/latex].

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,

[latex] F_{r A} imes 300=30000 imes 100 \quad ext { or } \quad F_{r A}=10000 N [/latex].

Taking moments about the bearing A,

[latex] F_{r B} imes 300=30000 imes 200 ext { or } F_{r B}=20000 N [/latex].

Also,

[latex] K_{a}=10000 N [/latex].

Step II Tentative selection of bearing

[latex] F_{r A}0 [/latex].

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

[latex] F_{a A}=F_{a B}+K_{a} [/latex] (a).

[latex] F_{a B}=\frac{0.5 F_{r B}}{Y} [/latex] (b).

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

Y	e	Designation	C	B	D	d
1.4	0.43	32012 X	76 500	23	95	60
1.8	0.33	33012	85 800	27	95
1.5	0.40	33112	110 000	30	100	60
1.5	0.40	30212	91 300	23.75	110
1.5	0.40	32212	119 000	29.75	110
1.5	0.40	33212	157 000	38	110
1.1	0.54	T5ED060	157 000	39	115
1.8	0.33	T2EE060	183 000	40	115
0.72	0.83	T7FC060	145 000	37	125	60
1.7	0.35	30312	161 000	33.5	130
0.72	0.83	31312	134 000	33.5	130
1.7	0.35	32312	216 000	48.5	130
1.1	0.54	32312 B	205 000	48.5	130

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,

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

[latex] F_{a A}=F_{a B}+K_{a}=7692.31+10000=17692.31 N [/latex].

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

[latex] F_{r A}=10000 N \quad F_{a A}=17692.31 N [/latex].

[latex] K_{a}=10000 N [/latex].

Therefore,

[latex] \frac{F_{a A}}{F_{r A}}=\frac{17692.31}{10000}=1.769 [/latex].

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

[latex] \frac{F_{a A}}{F_{r A}}>e [/latex].

From Eq. (15.12),

[latex] P=0.4 F_{r}+Y F_{a} ext { when }\left(F_{d} / F_{r} ight)>e [/latex] (15.12).

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

= 27 000 N
From Eq. (15.9),

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

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

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

[latex] C=P\left(L_{10} ight)^{0.3}=27000(72)^{0.3}=97401.02 N [/latex].

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.

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

[latex] F_{a A}=F_{a B}+K_{a}=6666.67+10000 [/latex].

= 16 666.67 N
Since,

[latex] \frac{F_{a A}}{F_{r A}}=\frac{16666.67}{10000}=1.67 \quad herefore \frac{F_{a A}}{F_{r A}}>e [/latex].

[latex] herefore \quad P=0.4 F_{r}+Y F_{a} [/latex].

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

[latex] C=P\left(L_{10} ight)^{0.3}=29000(72)^{0.3} [/latex].

= 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,

[latex] F_{r B}=20000 N [/latex].

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

Therefore,

[latex] \frac{F_{a B}}{F_{r B}}=\frac{6666.67}{20000}=0.33 \quad ext { and } \quad e=0.4 [/latex].

[latex] \frac{F_{a B}}{F_{r B}}

From Eq. (15.11),

[latex] P=F_{r} ext { when }\left(F_{a}/ F_{r} ight) \leq e [/latex] (15.11).

[latex] P=F_{r B}=20000 N [/latex].

[latex] C=P\left(L_{10} ight)^{0.3}=20000(72)^{0.3} [/latex].

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

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

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