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
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Chapter 11, Problem 34P

A gear-reduction unit uses the countershaft depicted in the figure. Find the two bearing reactions. The beatings are to be angular-contact ball bearings, having a desired life of 50 kh when used at 300 rev/min. Use 1.2 for the application factor and a reliability goal for the bearing pair of 0.96, assuming distribution data from manufacturer 2 in Table 11-6. Select the bearings from Table 11-2.

Problem 11-34

Dimensions in inches.

Chapter 11, Problem 34P, A gear-reduction unit uses the countershaft depicted in the figure. Find the two bearing reactions.

Expert Solution & Answer
Check Mark
To determine

The reaction load at point O

The reaction load at point C

The bore diameter of angular contact ball bearing at point O.

The bore diameter of angular contact ball bearing at point C.

Answer to Problem 34P

The reaction load at point O is 112lbf.

The reaction load at point C is 297lbf.

The bore diameter of angular contact ball bearing at point O is 17 mm.

The bore diameter of angular contact ball bearing at point C is 35 mm.

Explanation of Solution

Write the expression for net force in y direction at point A.

FAy=F1sinθ (I)

Here, net force in y direction at point A is FAy, applied load on point A is F1 and the angle of applied force with horizontal direction is θ.

Write the expression for net force in z direction at point A.

FAz=F1cosθ (II)

Write the expression for net force in y direction at point B.

FBy=FsinθB (III)

Here, net force in y direction at point B is FBy, applied load on point B is F and the angle of applied force with horizontal direction is θB.

Write the expression for net force in z direction at point B.

FBz=FcosθB (IV)

Here, net force in y direction at point B is FBy.

Write the expression for net torque on shaft.

T=0{FAzr1FBzr2}=0 (V)

Here, radius of gear at point A is r1 and radius of pulley at point B is r2.

Write the expression for moment equilibrium in z direction at O.

MOz=0(FAyrA)(FByrB)+(RCyrC)=0 (VI)

Here, the moment is M, the reaction force at C in y direction is RCy, the distance between A and O is rA, the distance between B and O is rB and the distance between C and O is rC.

Write the expression for moment equilibrium in y direction at O.

MOy=0(FAzrA)(FBzrB)+(RCzrC)=0 (VII)

Here, the reaction force at C in z direction is RCz.

Write the expression for static force equilibrium in y direction.

Fy=0ROyFAyFBy+RCy=0 (VIII)

Here, the force is F and the reaction force at O in y direction is ROy.

Write the expression for static force equilibrium in z direction.

Fz=0ROzFAz+FBz+RCz=0 (IX)

Here, the force is F and the reaction force at O in z direction is ROz.

Write the expression for resultant force at point O.

RO=(ROy)2+(ROz)2 (X)

Here, the resultant force at point O is RO.

Write the expression for resultant force at point C.

RC=(RCy)2+(RCz)2 (XI)

Here, the resultant force at point C is RC.

Write the expression for desired reliability.

RD=R1×R2 (XII)

Here, the desired reliability is RD, the reliability of component 1 is R1 and the reliability of component 2 is R2.

Write the expression for multiple of dating life.

xD=LD×nD×60L10 (XIII)

Here, the multiple of rating life is xD, the desired life is LD, the desired speed is nD and life hours of 90% component is L10.

Write the expression for catalogue load rating at point O.

C10O=afRO[xDx0+(θx0){ln(1RD)}1b]1a (XIV)

Here the catalogue load rating is C10O, the application factor is af, the reaction load is RO, minimum value of Weibull life measure is x0, the characteristic parameter is θ, the shape parameter for skewness is b and the factor of power is a.

Write the expression for catalogue load rating at point C.

C10C=afRC[xDx0+(θx0){ln(1RD)}1b]1a (XV)

Here the catalogue load rating is C10C.

Conclusion:

Substitute 240lbf for F1 and 20° for θ in Equation (I).

FAy=240lbfsin20°=(240×0.342)lbf82.1lbf

Substitute 240lbf for F1 and 20° for θ in Equation (II).

FAz=240lbfcos20°=(240×0.9396)226lbf

Substitute 25° for θB in Equation (III).

FBy=Fsin25°

Substitute 25° for θB in Equation (IV).

FBz=Fcos25°

Substitute 225.52lbf for FAz, 12in for r1, 6in for r2 and Fcos25° for FBz in Equation (V).

{225.52lbf12inFcos25°6in}=0F=225.52lbf×12incos25°×6inF=2706.245.4378F498lbf

Substitute 25° for θB and 498lbf for F in Equation (III).

FBy=498lbfsin25°=(498×0.42262)lbf210lbf

Substitute 25° for θB and 498lbf for F in Equation (IV).

FBz=498lbfcos25°=(498×0.90631)lbf451lbf

Substitute 82.1lbf for FAy, 210lbf for FBy, 16 in for rA, 30 in for rB and 42 in for rC in Equation (VI).

(82.1lbf16in)(210lbf30in)+(RCy42in)=0RCy=(1313.6lbfin)+(6300lbfin)42inRCy181lbf

Substitute 82.1lbf for FAy, 210lbf for FBy and 181lbf for RCy in Equation (VIII)

ROyFAyFBy+RCy=0ROy=82.11lbf+210lbf181lbfROy111.1lbf

Substitute 226lbf for FAz, 451lbf for FBz, 16 in for rA, 30 in for rB and 42 in for rC in Equation (VII).

(226lbf16in)(451lbf30in)(RCz42in)=0RCz=(3616lbfin)(13530lbfin)42inRCz236lbf

Substitute 226lbf for FAz, 451lbf for FBz and 236lbf for RCz in Equation (IX)

ROzFAz+FBz+RCz=0ROz=226lbf-451lbf+236lbfROy11lbf

Substitute 111.1lbf for ROy and 11lbf for ROz in Equation (X).

RO=(111.1lbf)2+(11lbf)2=12464.21112lbf

Thus the reaction load at point O is 112lbf

Substitute 181lbf for RCy and 236lbf for RCz in Equation (XI).

RC=(181lbf)2+(236lbf)2=88457297lbf

Thus the reaction load at point C is 297lbf

Substitute 0.96 for (R1×R2) in Equation (XII).

RD=0.96=0.98

Substitute 50000h for LD, 300rev/min for nD and 106h for L10 in Equation (XIII).

xD=50000h×300rev/min×60106h=9×108106=900

Refer to Table 11-6 “Typical Weibull Parameters for Two Manufacturers” to obtain values of xO as 0.02, θ as 4.459 and b as 1.1483.

Substitute 1.2 for af, 112lbf for RO, 900 for xD, 0.02 for x0, 4.459 for θ, 1.483 for b, 0.98 for RD and 3 for a in Equation (XIV).

C10O=1.2×112lbf[9000.02+(4.4590.02){ln(10.98)}11.483]131860lbf(1N0.2248lbf)(1kN1000N)8.28kN

Refer to Table 11-2 “Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearing” to obtain 17 mm bore size angular contact ball bearing corresponding to 8.28kN of catalogue load rating.

Thus the Bore diameter of angular contact ball bearing at point O is 17 mm.

Substitute 1.2 for af, 297lbf for RC, 900 for xD, 0.02 for x0, 4.459 for θ, 1.483 for b, 0.98 for RD and 3 for a in Equation (XV).

C10C=1.2×297lbf[9000.02+(4.4590.02){ln(10.98)}11.483]134932lbf(1N0.2248lbf)(1kN1000N)21.9kN

Refer to Table 11-2 “Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearing” to obtain 35 mm bore size angular contact ball bearing corresponding to 21.9kN of catalogue load rating.

Thus Bore diameter of angular contact ball bearing at point C is 35 mm.

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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...
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