An SKF 6210 angular-contact ball bearing has an axial load Fa of 400 lbf and a radial load Fr of 500 lbf applied with the outer ring stationary. The basic static load rating C0 is 4450 lbf and the basic load rating C10 is 7900 lbf. Estimate the 10 life at a speed of 720 rev/min.

Question

An SKF 6210 angular-contact ball bearing has an axial load [latex]F_a[/latex] of 400 lbf and a radial load [latex]F_r[/latex] of 500 lbf applied with the outer ring stationary. The basic static load rating [latex]C_0[/latex] is 4450 lbf and the basic load rating [latex]C_{10}[/latex] is 7900 bf. Estimate the [latex]\mathscr{L}_{10}[/latex] life at a speed of 720 rev/min.

Accepted Answer

[latex]V = 1 \ and \ {F_a}/{C_0} = {400}/{4450} = 0.090[/latex]. Interpolate for e in Table 11–1:

Table 11–1			[latex]{F_a}/{\left(VF_r\right)} \le e[/latex]		[latex]{F_a}/{\left(VF_r\right)} \gt e[/latex]
Equivalent Radial Load Factors for Ball Bearings	[latex]{F_a}/{C_0}[/latex]	e	[latex]X_1[/latex]	[latex]Y_1[/latex]	[latex]X_2[/latex]	[latex]Y_2[/latex]
	0.014*	0.19	1.00	0	0.56	2.30
	0.021	0.21	1.00	0	0.56	2.15
	0.028	0.22	1.00	0	0.56	1.99
	0.042	0.24	1.00	0	0.56	1.85
	0.056	0.26	1.00	0	0.56	1.71
	0.070	0.27	1.00	0	0.56	1.63
	0.084	0.28	1.00	0	0.56	1.55
	0.110	0.30	1.00	0	0.56	1.45
	0.17	0.34	1.00	0	0.56	1.31
	0.28	0.38	1.00	0	0.56	1.15
	0.42	0.42	1.00	0	0.56	1.04
	0.56	0.44	1.00	0	0.56	1.00
	∗Use 0.014 if [latex]{F_a}/{C_0} \lt 0.014. [/latex]

[latex] {F_a}/{C_0} [/latex]	e
0.084	0.28
0.090	e	from which [latex] e =0.285 [/latex]
0.110	0.30

[latex]{F_a}/{\left(V F_r\right )} = {400}/{\left[\left(1\right)500\right]} = 0.8 \gt 0.285[/latex]. Thus, interpolate for [latex]Y_2[/latex]:

[latex] {F_a}/{C_0} [/latex]	[latex] Y_2 [/latex]
0.084	1.55
0.09	[latex] Y_2 [/latex]	from which [latex] Y_2 = 1.527 [/latex]
0.110	1.45

From Eq. (11–9),

[latex]F_e=X_iVF_r+Y_i F_a[/latex]

[latex]F_e=X_2 V F_r+Y_2 F_a=0.56\left(1\right)500+1.527\left(400\right)=890.8[/latex] lbf

With [latex]\mathscr{L}_D=\mathscr{L}_{10} \ and \ F_D=F_e [/latex] , solving Eq. (11–3)
[latex]C_{10}=F_R=F_D\left(\frac{L_D}{L_R} \right) ^{{1}/{a}}=F_D\left(\frac{\mathscr{L}_D n_D 60}{\mathscr{L}_R n_R 60} \right) ^{{1}/{a}}[/latex]

for [latex]\mathscr{L}_{10}[/latex] gives

[latex]\mathscr{L}_{10}=\frac{60 \mathscr{L}_{R} n_R}{60 n_D}\left(\frac{C_{10}}{F_e}\right)^a=\frac{10^6}{60\left(720\right) }\left(\frac{7900}{890.8} \right)^3 =16 150 [/latex] h.

Question 3.11.4: An SKF 6210 angular-contact ball bearing has an axial load F...

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