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
bartleby

Videos

Textbook Question
Chapter 11, Problem 23P

11–22 to

11–26 An 02-series single-row deep-groove ball bearing is to be selected from Table 11–2 for the application conditions specified in the table. Assume Table 11–1 is applicable if needed. Specify the smallest bore size from Table 11–2 that can satisfy these conditions.

Chapter 11, Problem 23P, 1122 to 1126 An 02-series single-row deep-groove ball bearing is to be selected from Table 112 for

Expert Solution & Answer
Check Mark
To determine

The bore size of single-row deep-grove ball bearing.

Answer to Problem 23P

The bore size of single-row deep-grove ball bearing is 02-90mm.

Explanation of Solution

Write the expression for Design life.

    (xD)=60LDNDL10                                                                      (I)

Here, desired life is LD, desires speed is ND and rating life (106) is L10.

Write the expression for catalog rating life.

    (C10)=afFr[xDx0+(θx0)(ln(1R))1b]1a                                (II)

Here, application factor is af, reliability is R, radial load is Fr, design life is xD.and a,b and x0 are constant.

Write the expression for ration of axial load and Co.

    k1=FaCo                                                                                 (III)

Here, axial force is Fa and Co is a constant.

Write the expression for load factor.

    k2=Fa(VFr)                                                                            (IV)

Here, axial force is Fa, radial load is Fr and rotation factor is V.

Write the expression for equivalent radial load.

    Fe=X2VFr+Y2Fa                                                                   (V)

Here, axial force is Fa, radial load is Fr,rotation factor is V and X2, Y2 are constants.

Conclusion:

Substitute 10×103h for LD, 400rev/min for ND and 106 for L10 in Equation (I).

    (xD)=60×10×103h×400rev/min106=240×106106=240

Refer to table of “Weibull parameters Rating lives for manufacturer 2” to obtain x0=0.02, (θx0)=4.439 and b=1.483.

Substitute 1 for af, 8kN for Fe, 240 for xD, 0.02 for x0, 4.439 for (θx0), 1.483 for b, 0.99 for R and 3 for a in Equation (II).

    (C10)=(1)(8kN)[2400.02+(4.439)(ln(10.99))11.483]13=1×8kN[2400.2196]13=82.4kN

Refer to Table 112 “Dimensions and Load Ratings for Single-Row 02- Series Deep-Groove and Angular-Contact Ball Bearings” for 02-80mm deep groove to obtain C10=70.2kN and Co=45.0kN.

Substitute 2kN for Fa and 45kN for Co in Equation (III).

    FaCo=2kN45kN=0.044

Refer to Table 111 “Equivalent Radial Load Factor for Ball Bearings”, to obtain e=0.243 at at FaCo=0.088.

Substitute 2kN for Fa, 1 for V and 8kN for Fr in Equation (IV).

    Fa(VFr)=2kN1(8kN)=0.25

Refer to Table 111 “Equivalent Radial Load Factor for Ball Bearings”, at FaCo=0.044 and FaCo>e to obtain X2=0.56 and Y2=1.85.

Substitute 2kN for Fa, 1 for V, 8kN for Fr, 0.56 for X2 and 1.85 for Y2 in Equation (V).

    Fe=(0.56×1×8kN)+1.85(2kN)=4.48+3.7=8.18kN

Substitute 1 for af, 8.18kN for Fe, 240 for xD, 0.02 for x0, 4.439 for (θx0), 1.483 for b, 0.99 for R and 3 for a in Equation (II).

    (C10)=(1)(8.18kN)[2400.02+(4.439)(ln(10.99))11.483]13=(1)(8.18kN)[2400.2196]13=84.26kN

Since C10=84.26kN, that is greater than 70.2kN, so this bearing cannot be expected to carry load.

Refer to Table 112 “Dimensions and Load Ratings for Single-Row 02 Series Deep-Groove and Angular-Contact Ball Bearings” for 02-85mm deep groove to obtain C10=83.2kN and.Co=53.0kN.

Substitute 2kN for Fa and 53kN for Co in Equation (III).

    FaCo=2kN53kN=0.038

Refer to Table 111 “Equivalent Radial Load Factor for Ball Bearings”, to obtain e=0.234 at FaCo=0.088.

Substitute 2kN for Fa, 1 for V and 8kN for Fr in Equation (IV).

    Fa(VFr)=2kN1(8kN)=0.25

Refer to Table 111 “Equivalent Radial Load Factor for Ball Bearings”, at FaCo=0.038 and FaCo>e to obtain X2=0.56 and Y2=1.89.

Substitute 2kN for Fa, 1 for V, 8kN for Fr, 0.56 for X2 and 1.89 for Y2 in Equation (V).

    Fe=(0.56×1×8kN)+1.89(2kN)=4.48+3.78=8.26kN

Substitute 1 for af, 8.26kN for Fe, 240 for xD, 0.02 for x0, 4.439 for (θx0), 1.483 for b, 0.99 for R and 3 for a in Equation (II).

    (C10)=(1)(8.26kN)[2400.02+(4.439)(ln(10.99))11.483]13=(1)(8.26kN)[2400.2196]13=85.08kN

Since C10=85.08kN, that is greater than 83.2kN, so this bearing cannot be expected to carry load.

Refer to Table 112 “Dimensions and Load Ratings for Single-Row 02 Series Deep-Groove and Angular-Contact Ball Bearings” for 0290mm deep groove to obtain C10=95.26kN and.Co=62kN.

Substitute 2kN for Fa and 62kN for Co in Equation (III).

    FaCo=2kN62kN=0.032

Refer to Table 111 “Equivalent Radial Load Factor for Ball Bearings”, to obtain e=0.226 at FaCo=0.032.

Substitute 2kN for Fa, 1 for V and 8kN for Fr in Equation (IV).

    Fa(VFr)=2kN1(8kN)=0.25

Refer to Table 111 “Equivalent Radial Load Factor for Ball Bearings”, at FaCo=0.032 and FaCo>e to obtain X2=0.56 and Y2=1.95.

Substitute 2kN for Fa, 1 for V, 8kN for Fr, 0.56 for X2 and 1.95 for Y2 in Equation (V).

    Fe=(0.56×1×8kN)+1.95(2kN)=4.48kN+3.9kN=8.38kN

Substitute 1 for af, 8.38kN for Fe, 240 for xD, 0.02 for x0, 4.439 for (θx0), 1.483 for b, 0.99 for R and 3 for a in Equation (II).

    (C10)=(1)(8.38kN)[2400.02+(4.439)(ln(10.99))11.483]13=(1)(8.38kN)[2400.2196]13=86.314kN

Since C10=86.3kN<95.6kN, so this bearing can expected to carry load.

Thus, the bore size of single-row deep-grove ball bearing is 02-90mm.

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
A Self-aligning medium series bearing for bores of 85 mm is operating at 300 r.p.m for an average life of 9 years at 10 hours per day, is acted by a 15 kN radial load and 10 kN thrust load. The bearing is subjected to a light shock load and the inner ring is sationary. Determine the suitable self-aligning bearing. Mention the advantage of using self align bearing.
A rolling contact bearing is subjected to the following work cycle: (a) Radial load of 6000 N at 150 r.p.m. for 25% of the time; (b) Radial load of 7500 N at 600 r.p.m. for 20% of the time; and (c) Radial load of 2000 N at 300 r.p.m. for 55% of the time. The inner ring rotates and loads are steady. Select a bearing for an expected average life of 2500 hours.
Consider the belt drive having a diameter of 150 mm that is mounted on the overhung end of ashaft, 30 mm from the nearest bearing. The stepped shaft is made of steel having Su = 965 MPa(140 ksi). The shaft diameter, d, is 30 mm. The bearing bore, D, is 36 mm. The fillet with amachined surface has a radius, r, of 3 mm. What is the safety factor with respect to eventualfatigue failure of the shaft?

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
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
    • SEE MORE QUESTIONS
    Recommended textbooks for you
  • Elements Of Electromagnetics
    Mechanical Engineering
    ISBN:9780190698614
    Author:Sadiku, Matthew N. O.
    Publisher:Oxford University Press
    Mechanics of Materials (10th Edition)
    Mechanical Engineering
    ISBN:9780134319650
    Author:Russell C. Hibbeler
    Publisher:PEARSON
    Thermodynamics: An Engineering Approach
    Mechanical Engineering
    ISBN:9781259822674
    Author:Yunus A. Cengel Dr., Michael A. Boles
    Publisher:McGraw-Hill Education
  • Control Systems Engineering
    Mechanical Engineering
    ISBN:9781118170519
    Author:Norman S. Nise
    Publisher:WILEY
    Mechanics of Materials (MindTap Course List)
    Mechanical Engineering
    ISBN:9781337093347
    Author:Barry J. Goodno, James M. Gere
    Publisher:Cengage Learning
    Engineering Mechanics: Statics
    Mechanical Engineering
    ISBN:9781118807330
    Author:James L. Meriam, L. G. Kraige, J. N. Bolton
    Publisher:WILEY
  • Elements Of Electromagnetics
    Mechanical Engineering
    ISBN:9780190698614
    Author:Sadiku, Matthew N. O.
    Publisher:Oxford University Press
    Mechanics of Materials (10th Edition)
    Mechanical Engineering
    ISBN:9780134319650
    Author:Russell C. Hibbeler
    Publisher:PEARSON
    Thermodynamics: An Engineering Approach
    Mechanical Engineering
    ISBN:9781259822674
    Author:Yunus A. Cengel Dr., Michael A. Boles
    Publisher:McGraw-Hill Education
    Control Systems Engineering
    Mechanical Engineering
    ISBN:9781118170519
    Author:Norman S. Nise
    Publisher:WILEY
    Mechanics of Materials (MindTap Course List)
    Mechanical Engineering
    ISBN:9781337093347
    Author:Barry J. Goodno, James M. Gere
    Publisher:Cengage Learning
    Engineering Mechanics: Statics
    Mechanical Engineering
    ISBN:9781118807330
    Author:James L. Meriam, L. G. Kraige, J. N. Bolton
    Publisher:WILEY
    BEARINGS BASICS and Bearing Life for Mechanical Design in 10 Minutes!; Author: Less Boring Lectures;https://www.youtube.com/watch?v=aU4CVZo3wgk;License: Standard Youtube License