The shaft shown in Figure 14–15 carries a transverse load of 6800 lb and a thrust load of 2500 lb. The thrust is resisted by bearing A. The shaft rotates at 350 rpm and is to be used in a piece of agricultural equipment. Specify suitable tapered roller bearings for the shaft.

Question

The shaft shown in Figure 14–15 carries a transverse load of 6800 lb and a thrust load of 2500 lb. The thrust is resisted by bearing [latex]A[/latex]. The shaft rotates at 350 rpm and is to be used in a piece of agricultural equipment. Specify suitable tapered roller bearings for the shaft.

Accepted Answer

The radial loads on the bearings are
[latex]F_{rA}[/latex] = (6800 lb)(4 in/10 in) = 2720 lb
[latex]F_{rB}[/latex] = (6800 lb)(6 in/10 in) = 4080 lb
[latex]T_A[/latex] = 2500 lb

To use Equation (14–7), we must assume values of [latex]Y_A[/latex] and [latex]Y_B[/latex]. Let’s use [latex]Y_A = Y_B = 1.75[/latex]. Then

[latex]P_{A}=0.4 F_{r A}+0.5 \frac{Y_{A}}{Y_{B}} F_{r B}+Y_{A} T_{A}[/latex] (14-7)

[latex]P_{A}=0.4 (2720 lb)+0.5 \frac{1.75}{1.75} (4080 lb)+1.75(2500 lb) = 7503[/latex] lb

[latex]P_B = F_{rB} = 4080[/latex] lb

Using Table 14–4 as a guide, let’s select 4000 h as a design life. Then the number of revolutions would be

[latex]L_d[/latex] = (4000 h)(350 rpm)(60 min/h) = [latex]8.4 × 10^7[/latex] rev

The required basic dynamic load rating can now be calculated from Equation (14–3) ([latex]C=P_d(L_d/10^6)^{1/k}[/latex]), using [latex]k = 3.33[/latex]. Note that the exponent [latex]1/k[/latex] is 1/3.33, resulting in a value of 0.30.

[latex]C_A=P_d(L_d/10^6)^{1/k}[/latex]

[latex]C_A = (7503 lb)(8.4 × 10^7/10^6)^{0.30} = 28400[/latex] lb

Similarly,
[latex]C_B = (4080 lb)(8.4 × 10^7/10^6)^{0.30} = 15 400[/latex] lb

From Table 14–7, we can choose the following bearings.
Bearing [latex]A[/latex]
[latex]d = 2.5000[/latex] in [latex]D = 5.0000[/latex] in
[latex]C = 29 300[/latex] lb [latex]Y_A = 1.65[/latex]

Bearing [latex]B[/latex]
[latex]d = 1.7500[/latex] in [latex]D = 4.0000[/latex] in
[latex]C = 21 400[/latex] lb [latex]Y_B = 1.50[/latex]
We can now recompute the equivalent loads:

[latex]P_{A}=0.40(2720 lb) + 0.5 \frac{1.65}{1.50}(4080 lb) + 1.65(2500 lb) = 7457[/latex] lb

[latex]P_B = F_{rB} = 4080[/latex] lb

From these, the new values of [latex]C_A = 28 200[/latex] lb and [latex]C_B = 15 400[/latex] lb are still satisfactory for the selected bearings.

TABLE 14–4 Recommended Design Life for Bearings
Application	Design life [latex]L_{10}[/latex], h
Domestic appliances, instruments, medical apparatus	1000–2000
Aircraft engines	1000–4000
Automotive	1500–5000
Agricultural equipment, hoists, construction machines	3000–6000
Elevators, industrial fans, multipurpose gearing, rotary crushers, cranes	8000–15 000
Electric motors, industrial blowers, general industrial machines, conveyors	20 000–30 000
Pumps and compressors, textile machinery, rolling mill drives	40 000–60 000
Critical equipment in continuous, 24-h operation; power plants, ship drives	100 000–200 000

TABLE 14–7 Tapered Roller Bearing Data
Bore, d	Outside diameter, D	Width, T	a	Thrust factor, Y	Basic dynamic load rating, C
1.0000	2.5000	0.8125	0.583	1.71	8370
1.5000	3.0000	0.9375	0.690	1.98	12800
1.7500	4.0000	1.2500	0.970	1.50	21400
2.0000	4.3750	1.5000	0.975	2.02	26200
2.5000	5.0000	1.4375	1.100	1.65	29300
3.0000	6.0000	1.6250	1.320	1.47	39700
3.5000	6.3750	1.8750	1.430	1.76	47700

Question 14.5: The shaft shown in Figure 14–15 carries a transverse load of...

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