Consider a constrained housing as depicted in Fig. 11–19 with two direct-mount tapered roller bearings resisting an external thrust Fae of 8000 N. The shaft speed is 950 rev/min,the desired life is 10 000 h, the expected shaft diameter is approximately 1 in. The reliability goal is 0.95 The applica

Question

Consider a constrained housing as depicted in Fig. 11–19 with two direct-mount tapered roller bearings resisting an external thrust [latex]F_{ae}[/latex] of 8000 N. The shaft speed is 950 rev/min,the desired life is 10 000 h, the expected shaft diameter is approximately 1 in. The reliability goal is 0.95. The application factor is appropriately [latex]a_f = 1[/latex].
(a) Choose a suitable tapered roller bearing for [latex]A[/latex].
(b) Choose a suitable tapered roller bearing for [latex]B[/latex].
(c) Find the reliabilities [latex]R_A \ , \ R_B \ , \ and \ R[/latex].

Accepted Answer

(a) By inspection, note that the left bearing carries the axial load and is properly labeled as bearing [latex]A[/latex].

The bearing reactions at A are

[latex]F_{rA} = F_{rB} = 0 \ F_{aA} = F_{ae} = 8000[/latex] N.

Since bearing [latex]B[/latex] is unloaded, we will start with [latex]R = R_A = 0.95[/latex] .
With no radial loads, there are no induced thrust loads. Eq. (11–16)

[latex]if \ F_{iA} \le \left(F_{iB}+F_{ae} ight\}\begin{cases} F_{eA}=0.4 F_{rA}+K_A\left(F_{iB}+F_{ae} ight)\ F_{eB}=F_{rB}\end{cases} [/latex]

is applicable.
[latex]F_{eA}=0.4 F_{rA}+K_A\left(F_{iB}+F_{ae} ight)= K_A F_{ae}[/latex]

If we set [latex]K_A = 1[/latex] , we can find [latex]C_{10}[/latex] in the thrust column and avoid iteration:

[latex]F_{eA} = \left(1 ight)8000 = 8000 \ N \ F_{eB} = F_{rB} = 0[/latex]

The multiple of rating life is

[latex]x_D =\frac{L_D}{L_R}=\frac{\mathscr{L}_D n D60}{L_R}=\frac{\left(10 000 ight)\left(950 ight)\left(60 ight)}{90\left(10^6 ight)}= 6.333[/latex]

Then, from Eq. (11–7),

[latex]C_{10}\doteq a_fF_D \left[\frac{x_D}{x_0+\left( heta -x_0 ight)\left(1-R_D ight)^{{1}/{b}} } ight]^{{1}/{a}} \ \ R \ge 0.90[/latex]

for bearing [latex]A[/latex]

[latex]C_{10} = a_f F_{eA} \left[\frac{x_D}{4.48 \left(1 − R_D ight)^{{2}/{3}}} ight]^{{3}/{10}} \ = \left(1 ight)8000 \left[\frac{6.33}{4.48 \left(1 − 0.95 ight)^{{2}/{3}}} ight]^{{3}/{10}}= 16 159[/latex] N

Figure 11–15 presents one possibility in the 1-in bore (25.4-mm) size: cone,HM88630, cup HM88610 with a thrust rating

[latex] \left(C_{10} ight) _a = 17 200[/latex] N.

(b) Bearing B experiences no load, and the cheapest bearing of this bore size will do,including a ball or roller bearing.
(c) The actual reliability of bearing A, from Eq. (11–21), is

[latex]R_A\doteq 1-\left\{\frac{x_D}{4.48\left[{C_{10}}/{\left(a_f F_D ight) } ight]^{{10}/{3}} } ight\} ^{{3}/{2}} \ \doteq 1-\left\{\frac{6.333}{4.48\left[{17200}/{\left(1 imes 8000 ight) } ight]^{{10}/{3}} } ight\} ^{{3}/{2}}=0.963[/latex]

which is greater than 0.95, as one would expect. For bearing [latex]B[/latex],

[latex]F_D = F_{eB} = 0 \ R_B \doteq 1-\left[\frac{6.333}{0.85\left({17200}/{0} ight)^{{10}/{3}}} ight]^{{3}/{2}} =1-0=1[/latex]

as one would expect. The combined reliability of bearings [latex]A \ and \ B[/latex] as a pair is

[latex]R = R_A R_B = 0.963 \left(1 ight) = 0.963[/latex]

which is greater than the reliability goal of 0.95, as one would expect.

Question 3.11.11: Consider a constrained housing as depicted in Fig. 11–19 wit...

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