Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
Question
Book Icon
Chapter 11, Problem 5P
To determine

The reliability that meets the uses.

Expert Solution & Answer
Check Mark

Answer to Problem 5P

The reliabilities that meets the uses are 0.953 for R03series and 0.94 for R02series.

Explanation of Solution

Write the expected reliability goal for each bearing.

    R02series=R03series=overallreliability (I)

Here, reliability of 02 series is R02series and the reliability of 03 series is R03series.

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[(xBx0θx0)b]xB=x0+(θx0)(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.

    R02series=exp[(xD(afFDC10)axoθxo)] (VII)

Write the expression for reliability for 03 series angular ball bearing.

    R03series=exp[(xD(afFDC10)axoθxo)] (VIII)

Write the expression for overall reliability.

    R=R02series×R03series (IX)

Here, overall reliability is R.

Conclusion:

Substitute 0.90 for overallreliability in Equation (I).

    R02series=R03series=0.90=0.95

Substitute 40kh for l and 520rpm for n in Equation (II).

    LD=(60min/h)(40kh)(520rev/min)=(60min/h)(40kh)(1000h1kh)(520rev/min)=(2400000min)(520rev/min)=1248×106rev

Substitute 106rev for LR and 1248×106rev for LD in Equation (II).

    xD=1248×106rev106revxD=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+(4.4590.02)(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 725lbf for FD in Equation (III).

    FB=(725lbf)(12480.622)13=(725lbf)(2006.4308)13=(725lbf)(12.612)=9144.2075lbf

Substitute 1.4 for af and 9144.2075lbf for FB in Equation (VI).

    C10=1.4×9144.2075lbf=(12801.8905lbf)(4.45N1lbf)=(56968.41273N)(1kN1000N)57kN

Refer table 11-2 “Dimensions and Load Rating of Ball Bearing” to obtain the ball bearing at catalog rating of 57kN as 65mm with catalog life 63.7kN.

Substitute 0.02 for xo, 1248 for xD, 3 for a, 1.4 for af 4.439 for θxo and 725lbf for FD in Equation (II).

    R02series=exp[(1248(1.4(725lbf)63.7kN)a0.024.439)1.483]=exp[(1248(1.4((4.44822N1lbf)×725lbf)(63.7kN)(1000N1kN))a0.024.439)1.483]=exp((0.42444.439)1.483)=0.97

For 03 series cylindrical roller bearing:

Substitute 0.90 for overallreliability in Equation (I).

    R02series=R03series=0.90=0.95

Substitute 40kh for l and 520rpm for n in Equation (II).

    LD=(60min/h)(40kh)(520rev/min)=(60min/h)(40kh)(1000h1kh)(520rev/min)=(2400000min)(520rev/min)=1248×106rev

Substitute 106rev for LR and 1248×106rev for LD in Equation (II).

    xD=1248×106rev106revxD=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+(4.4390.02)(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 2235lbf for FD in Equation (II).

    FB=(2235lbf)(12480.617)310=(2235lbf)(2022.6914)0.3=(2235lbf)(9.81248)=21930.8928lbf

Substitute 1.4 for af and 21930.8928lbf for FB in Equation (V).

    C10=1.4×21930.8928lbfC10=(30703.24992lbf)(4.45N1lbf)C10=(136629.4621N)(1kN1000N)C10=136.6kN

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 138kN as 65mm.

Substitute 0.02 for xo, 1248 for xD, 3 for a, 1.4 for af 4.439 for θxo, 136.6kN for C10 and 2235lbf for FD in Equation (II).

    R02series=exp[(1248(1.4(2235lbf)136.6kN)a0.024.439)1.483]=exp[(1248(1.4((4.44822N1lbf)×2235lbf)(136.6kN)(1000N1kN))a0.024.439)1.483]=0.953

Substitute 0.953 for R03series and 0.97 for R02series 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 R02series and 0.917 for R03series.

Substitute 0.917 for R03series and 0.94 for R02series 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 R02series and 0.953 for R03series.

Substitute 0.953 for R03series and 0.94 for R02series 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 R03series and 0.94 for R02series.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Continuity equation A y x dx D T معادلة الاستمرارية Ly X Q/Prove that ди хе + ♥+ ㅇ? he me ze ོ༞“༠ ?
Q Derive (continuity equation)? I want to derive clear mathematics.
motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

Chapter 11 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 11 - 11-8 to 11-13 For the bearing application...Ch. 11 - 11-8 to 11-13 For the bearing application...Ch. 11 - 11-8 to 11-13 For the bearing application...Ch. 11 - A countershaft carrying two V-belt pulleys is...Ch. 11 - A countershaft carrying two V-belt pulleys is...Ch. 11 - A countershaft carrying two V-belt pulleys is...Ch. 11 - A countershaft carrying two V-belt pulleys is...Ch. 11 - For the shaft application defined in Prob. 3-77,...Ch. 11 - For the shaft application defined in Prob. 3-79,...Ch. 11 - An 02-series single-row deep-groove ball bearing...Ch. 11 - An 02-series single-row deep-groove ball bearing...Ch. 11 - 11-22 to 11-26 An 02-series single-row deep-groove...Ch. 11 - 1122 to 1126 An 02-series single-row deep-groove...Ch. 11 - 1122 to 1126 An 02-series single-row deep-groove...Ch. 11 - 1122 to 1126 An 02-series single-row deep-groove...Ch. 11 - 1122 to 1126 An 02-series single-row deep-groove...Ch. 11 - The shaft shown in the figure is proposed as a...Ch. 11 - Repeat the requirements of Prob. 11-27 for the...Ch. 11 - The shaft shown in the figure is proposed as a...Ch. 11 - Repeat the requirements of Prob. 11-29 for the...Ch. 11 - Shown in the figure is a gear-driven squeeze roll...Ch. 11 - The figure shown is a geared countershaft with an...Ch. 11 - The figure is a schematic drawing of a...Ch. 11 - A gear-reduction unit uses the countershaft...Ch. 11 - The worm shaft shown in part a of the figure...Ch. 11 - In bearings tested at 2000 rev/min with a steady...Ch. 11 - A 16-tooth pinion drives the double-reduction...Ch. 11 - Estimate the remaining life in revolutions of an...Ch. 11 - The same 02-30 angular-contact ball bearing as in...Ch. 11 - A countershaft is supported by two tapered roller...Ch. 11 - For the shaft application defined in Prob. 3-74,...Ch. 11 - For the shaft application defined in Prob. 3-76,...Ch. 11 - Prob. 43PCh. 11 - The gear-reduction unit shown has a gear that is...
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY