The reliability that meets the uses.
Answer to Problem 5P
The reliabilities that meets the uses are 0.953 for R03−series and 0.94 for R02−series.
Explanation of Solution
Write the expected reliability goal for each bearing.
R02−series=R03−series=√overall reliability (I)
Here, reliability of 02 series is R02−series and the reliability of 03 series is R03−series.
Write the expression for multiple of rating life for bearing.
xD=LDLR (II)
Here, multiple of rating life for design is xD, desired life is LD, and the rating life is LR.
Write the regression equation for bearing load life.
FBxB1a=FDxD1aFB=FD(xDxB)1a (III)
Here, the bearing load for design is FD, the basic bearing load is FD, bearing life in revolutions is L, arbitrary constant is a and multiple of rating life for basic is xB.
Write the equation for bearing life.
L=60ln (IV)
Here, the rating life in hour is l and rating speed in revolutions per minute is n.
Write the equation for reliability along a constant load line.
RD=exp [−(xB−x0θ−x0)b]xB=x0+(θ−x0)(ln1RD)1b (V)
Here, the characteristic parameter is θ, and Weibull parameters are x0 and b.
Write the expression for catalog rating in terms of application factor.
C10=afFB (VI)
Here, catalog rating is C10 and application factor is af.
Write the expression for reliability for 02 series angular ball bearing.
R02−series=exp[−(xD(afFDC10)a−xoθ−xo)] (VII)
Write the expression for reliability for 03 series angular ball bearing.
R03−series=exp[−(xD(afFDC10)a−xoθ−xo)] (VIII)
Write the expression for overall reliability.
R=R02−series×R03−series (IX)
Here, overall reliability is R.
Conclusion:
Substitute 0.90 for overall reliability in Equation (I).
R02−series=R03−series=√0.90=0.95
Substitute 40 kh for l and 520 rpm for n in Equation (II).
LD=(60 min/h)(40 kh)(520 rev/min)=(60 min/h)(40 kh)(1000 h1 kh)(520 rev/min)=(2400000 min)(520 rev/min)=1248×106 rev
Substitute 106 rev for LR and 1248×106 rev for LD in Equation (II).
xD=1248×106 rev106 revxD=1248
Thus, the multiple of rating life of bearing is 1248.
Substitute 0.95 for RD, 0.02 for x0, 4.459 for θ and 1.483 for b in Equation (V).
xB=0.02+(4.459−0.02)(ln10.95)11.483=0.02+(4.457)(0.051293)0.674=0.02+(4.457)(0.135076)=0.622
For ball bearing, the value of constant a is 3.
Substitute 0.622 for xB, 1248 for xD, 3 for a and 725 lbf for FD in Equation (III).
FB=(725 lbf)(12480.622)13=(725 lbf)(2006.4308)13=(725 lbf)(12.612)=9144.2075 lbf
Substitute 1.4 for af and 9144.2075 lbf for FB in Equation (VI).
C10=1.4×9144.2075 lbf=(12801.8905 lbf)(4.45 N1 lbf)=(56968.41273 N)(1 kN1000 N)≈57 kN
Refer table 11-2 “Dimensions and Load Rating of Ball Bearing” to obtain the ball bearing at catalog rating of 57 kN as 65 mm with catalog life 63.7 kN.
Substitute 0.02 for xo, 1248 for xD, 3 for a, 1.4 for af 4.439 for θ−xo and 725 lbf for FD in Equation (II).
R02−series=exp[−(1248(1.4(725 lbf)63.7 kN)a−0.024.439)1.483]=exp[−(1248(1.4((4.44822 N1 lbf)×725 lbf)(63.7 kN)(1000 N1 kN))a−0.024.439)1.483]=exp(−(0.42444.439)1.483)=0.97
For 03 series cylindrical roller bearing:
Substitute 0.90 for overall reliability in Equation (I).
R02−series=R03−series=√0.90=0.95
Substitute 40 kh for l and 520 rpm for n in Equation (II).
LD=(60 min/h)(40 kh)(520 rev/min)=(60 min/h)(40 kh)(1000 h1 kh)(520 rev/min)=(2400000 min)(520 rev/min)=1248×106 rev
Substitute 106 rev for LR and 1248×106 rev for LD in Equation (II).
xD=1248×106 rev106 revxD=1248
Substitute 0.95 for RD, 0.02 for x0, 4.439 for θ and 1.483 for b in Equation (IV).
xB=0.02+(4.439−0.02)(ln10.95)11.483=0.02+(4.419)(0.05129)0.674=0.02+(4.419)(0.135248)=0.617
For roller bearing, the value of constant a is 10/3.
Substitute 0.617 for xB, 1248 for xD, 10/3 for a and 2235 lbf for FD in Equation (II).
FB=(2235 lbf)(12480.617)310=(2235 lbf)(2022.6914)0.3=(2235 lbf)(9.81248)=21930.8928 lbf
Substitute 1.4 for af and 21930.8928 lbf for FB in Equation (V).
C10=1.4×21930.8928 lbfC10=(30703.24992 lbf)(4.45 N1 lbf)C10=(136629.4621 N)(1 kN1000 N)C10=136.6 kN
Refer to table 11-3 “Dimensions and basic Load Rating for cylindrical roller bearing:03 series” to obtain the cylindrical bearing at catalog rating of 138 kN as 65 mm.
Substitute 0.02 for xo, 1248 for xD, 3 for a, 1.4 for af 4.439 for θ−xo, 136.6 kN for C10 and 2235 lbf for FD in Equation (II).
R02−series=exp[−(1248(1.4(2235 lbf)136.6 kN)a−0.024.439)1.483]=exp[−(1248(1.4((4.44822 N1 lbf)×2235 lbf)(136.6 kN)(1000 N1 kN))a−0.024.439)1.483]=0.953
Substitute 0.953 for R03−series and 0.97 for R02−series in Equation (VIII).
R=0.97×0.953=0.924
Here, 0.924>, so the overall reliability exceeds the reliability goal. Therefore, this combination has no use.
Take 0.94 for R02−series and 0.917 for R03−series.
Substitute 0.917 for R03−series and 0.94 for R02−series in Equation (VIII).
R=0.94×0.917=0.861
Here, 0.861<0.90, so the overall reliability is less than reliability goal . Therefore, this combination can be used.
Take 0.94 for R02−series and 0.953 for R03−series.
Substitute 0.953 for R03−series and 0.94 for R02−series in Equation (VIII).
R=0.94×0.953=0.89582
Here, 0.89582<0.90, so the overall reliability is less than the reliability goal. Therefore, this combination can be used.
Thus, the reliabilities that meets the uses are 0.953 for R03−series and 0.94 for R02−series.
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Chapter 11 Solutions
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
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