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.

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Accepted Answer

The radial loads on the bearings are

[latex]F_{rA} = 6800(4 in/10 in)=2720[/latex] lb
[latex]F_{rB}= 6800(6 \ in/10 \ in)[/latex]
[latex]T_A = 2500[/latex] lb

To use Equation (14-8) [latex]P_{A}=0.4F_{rA}+0.5\frac{Y_{A}}{Y_{B}} F_{rB}+Y_{A}T_{A}[/latex], 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 = 0.40(2720)+0.5\frac{1.75}{1.75} 4080 + 1.75(2500) = 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

TABLE 14-4 Recommended design life for bearings
Application	Design life [latex]L_{10}[/latex], h
Domestic appliances	1000-2000
Aircraft engines	1000-4000
Automotive	1500-5000
Agricultural equipment	3000-6000
Elevators, industrial fans, multipurpose gearing	8000-15 000
Electric motors, industrial blowers, general industrial machines	20 000-30 000
Pumps and compressors	40 000-60 000
Critical equipment in continuous, 24-h operation	100 000-200 000
Source: Eugene A. Avallone and Theodore Baumeister III, eds., Marks' Standard Handbook for Mechanical Engineers. 9th ed. New York: McGraw-Hill. 1986.

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

The required basic dynamic load rating can now be calculated from Equation (14-3), using k = 3.33:

[latex]C_A = P_A(L_d/10^6)^{1/k}[/latex]
[latex]C_A = 7503(8.4 ×10^7/10^6)^{0.30}= 28 400[/latex] lb

Similarly,

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

From Table 14-7, we can choose the following bearings.

TABLE 14-7 Tapered roller bearing data
Bore	Outside diameter	Width	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	12 800
1.7500	4.0000	1.2500	0.970	1.50	21 400
2.0000	4.3750	1.5000	0.975	2.02	26 200
2.5000	5.0000	1.4375	1.100	1.65	29 300
3.0000	6.0000	1.6250	1.320	1.47	39 700
3.5000	6.3750	1.8750	1.430	1.76	47 700
Note: Dimensions are in inches. Load C is in pounds for an [latex]L_{10}[/latex] life of 1 million rev.

Bearing A

d = 2.5000 in D = 5.0000 in
C = 29 300 lb [latex] Y_A[/latex] = 1.65

Bearing B

d = 1.7500 in D = 4.0000 in
C = 21400 lb [latex]Y_B[/latex] = 1.50

We can now recompute the equivalent loads:

[latex]P_A = 0.40(2720)+0.5\frac{1.65}{1.50} 4080 + 1.65(2500) =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.

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

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