Solutions Manualto accompany
MACHINE ELEMENTS INMECHANICAL DESIGN
Robert L. Mott
Fourth Edition
PEARSON---1'11'111 jl'PHall
10 9 8 7 6 5 4 3 2 1
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Instructors of classes using Mott, Machine Elements in Mechanical Design, Fourth Edition, may reproduce materialfrom the solutions manual for classroom use.
ISBN 0-13-140873-9
CONTENTS
1
iii
Solutions Manual for
Machine Elements in Mechanical Design, 4th ed.By: Robert L. Mott
Table of Contents...... ii
MDESIGN software Included on the CD in the book , iii
Spreadsheets included on the CD in the Solutions Manual , x
Solutions to end-of-chapter problems
Chapter 1: The Nature of Mechanical Design 1
Chapter 2: Materials in Mechanical Design...... 3
Chapter 3: Stress and Deformation Analysis... ... ... ... . .. ... ... ... ... ... ... ... .. 9
Chapter 4: Combined Stresses and Mohr's Circle 31
Chapter 5: Design for Different Types of Loading , 42
Chapter 6: Columns , 64
Chapter 7: Belt Drives and Chain Drives , 77
Chapter 8: Kinematics of Gears .,. ... ... ... .. . ... . .. .. . .. . .. . 87
Chapter 9: Spur Gear Design _. 108
Chapter 10: Helical Gears, Bevel Gears, and Wormgearing 151
Chapter 11: Keys, Couplings, and Seals 180
Chapter 12: Shaft Design .. . .. .. .. .. ... .. .. .. .. 184
Chapter 13: Tolerances and Fits , 206
Chapter 14: Rolling Contact Bearings 211
Chapter 16: Plain Surface Bearings 217
Chapter 17: Linear Motion Elements 224
Chapter 18: Fasteners, 226
Chapter 19: Springs 228
Chapter 20: Machine Frames, Bolted Connections, and Welded Joints 243
Chapter 21: Electric Motors and Controls 254
Chapter 22: Motion Control: Clutches and Brakes , 257
cted bye m ang, or
naterial
MACHINE ELEMENTS IN MECHANICAL DESIGN, 4TH EDITION
By: Robert L. Mott
Published by Prentice-Hall, Inc.
MDES/GN - Mechanical Design Software
Comments On Its Use For Problem Solutions
IntroductionTh.e.designof machine elements jnherently .involves extensive procedures, complex calculations,
and many design decisions. Data must be found from numerous charts and tables. Furthermore, design
is typically iterative, requiring the designer to try several options for any given element, leading to the
repetition ·ofdesign calculations with new data or new design decisions. This is especially true·for
complete mechanical devices containing several components as the interfaces between components are
considered. Changes to one component often require changes to mating elements. Use of computer
aided mechanical design software can facilitate the design process by perfonning many of the tasks while
leaving the major design decisions to the creativity and judgment of the designer or engineer.
We emphasize that users of computer software must have a solid
understanding of the principles of design and stress analysis toensure that design decisions are based on reliable foundations. We
advlsetbat·thesoftware be used only after mastering a given designmethodology by careful study and practicing manual techniques.
Then, data with known results can be applied to the software as a
check on the understanding of the program's Input and output.
Included in this book is the MDES/GN mechanical design software created by the TEDATA
Company. Derived from the very successful MDESIGN mec software produced for the European market,
the U.S. version of MDESIGN employs standards and design methods that are in typical use in North
America. Many of the textual aids and design procedures come directly from this book, Machine Elements
in Mechanical Design.
Topics for which the MDESIGN software can be used as a supplement to this book include:
Beam stress analysis Beam deflections Mohr's circle Columns
Belt drives Chain drives Spur gears Helical gears
Shafts Keys Power screws Springs
Rolling contact bearings Plain surface bearings Bolted connections Fasteners
Clutches BrakesSpecial icons are placed in the book at places where use of one of the 28 modules in the
software is pertinent.
v
Basic Features of MOESIGNMOESIGN soflware is a useful tool for problem solving and design. Users must first ensure that
the software is appropriatetothe nature of the problemlo be solved. While the 28 modules cover a wide
range of applications within the field of machine design, not every problem can be solved. The soflware is
very user friendly and each module contains several textual and graphic aids that explain the technical
"bases on which the module is constructed .and.the data that must be entered by the user. Input screens
prompt the user to define the problem and to make basic design decisions. Several modules contain
extensive data bases that allow the user to consider multiple optional designs and compare them to
select the more optimum choice. Some modules also provide a parametric analysis feature.
After all necessary data are entered, the user applies the Calculate feature to cause the module
to perform the required analyses and produce the output in an easy to read format. The Calculate feature
is represented on the tooioaras an image of a blue calculator with a yellow gear. Altematively, the user
can use the function key, F10. Input and output can be reported in several different units selected by the
user. Conversions, if needed, are automatically accomplished by the software.
The output always contains lists of pertinent input values and computed results. Many modules
augment the basic output with graphical displays of the results and pictorial views of the element being
designed. Some modules include cautionary comments when results do not meet design requirements,
advising the user to redesign the element.
DeKfiptions of the Modules
The following sections list the 28 modules organized into 13 groups and provide brief discussions
about each module. Included are the chapter where each module can be used, a basic description of the
type of problems that can be solved, the type of input data required, and the type of output results that are
produced. NOTE: The computed results from MOESIGN may not match exactly those reported in the
book for example problems or end of chapter problems. Values for some parameters may be selected by
a user from charts and graphs In the book, whereas the program may compute those giving slightly
different results. For some modules, particularly Ball and Roller Bearings, databases in the program aredifferent from those in the book.
Columns Group: Pertinent to Chapter 6 in the book.
Column Analysis Module: Sections 6-6, 6.7, 6·11 and 6.12
Centrally loaded straight or crooked columns and eccentrically loaded straight columns are
analyzed.lnput data required are: column length, end fIXity, material, cross section shape and dimensions,amount of crookedness if any and amo t f . . .. ,un 0 eccentncity of the hne of action of the load if any. Output forstraight, ce~tratty loaded columns includes critical buckling load and allowable load. Output for crooked
and eccentncally loaded columns includes maximum stresses. For eccentrically loaded columns, thernaxmum lateral deflection of the column is also computed.
-u_-------------------Column Design Module: Section 6-10
Given the required load on a column of a given length with specified end fixity, the program
computes the minimum acceptable diameter of'acolumn with a solid circular cross section.
Seams Group:Pertinent to Chapters 3 and 5 in the book.
Statically Detenninate Beams Module:This is a very extensive beam analysis program that allows the application of loads in virtually any
direction including concentrated forces, distributed loads, concentrated moments, and torques. It offers
visualization of the beam, its supports, and its loading pattem. Careful data entry is advised.
Statically Indeterminate Beams Module:This program analyzes beams on more than two supports with a large variety of loading types.
Combined Stress and Mohr's Circle: Pertinent to Chapters 4 and 5Given the stress condition on an element, the program completes Mohr's circle, presents
maximum principal stresses, maximum shear stress, and the orientation of those elements. The Mohr's
circle is drawn automatically.
V-Selt Drives: Pertinent to Chapter 7, Section 7-4This tool for designing V-belt drives is based on an extensive data base of ratings for three sizes
of V.belts. The input data screen allows the specification of power to be transmitted, speeds, approximate
center distance, and service factor. The program presents a list of suitable belt types, sizes, and standard
lengths, and sheave sizes from which the user selects one combination. The performance of that design
is then displayed. Numerous subsequent altemative designs can be easily created.
Chain Drives: Pertinent to Chapter 7, Section 7-6An aid for designing roller chain drives, this program includes an extensive data base of the latest
power ratings for roller chain drives from the American Chain Association. The input data screen allows
the specification of power to be transmitted, speeds, approximate center distance, number of chains, and
service factor. The program presents a list of suitable chain sizes and sprockets from which the user can
select one combination. Single or multiple strands of chain can be used.
Gearing Group: Pertinent to Chapters 8, 9, and 10Separate modules are provided for Spur Gearing, Helical Gearing, Bevel Gearing, and
wormgearing. Each module basically follows the procedures described in the book. Input data include
the diametral pitch, number of pinion teeth, input speed, output speed, certain material factors, overload
factor, reliability factor, and others depending on the module. The program provides guidance for the
quality number and the specification of the number of teeth in the gear and its face width. The geometry
factors for bending and pitting resistance are computed by the program for some modules, while others
require the user to input those using tables and charts from the book or from charts reproduced in the
vii
t d sizes of pertinent geometric features, forces on gearhel screens. Outputs include actual outpu spee , '.
teeth, and tooth stresses. The acceptability of stresses is evaluated by the ~rogram with suggestions for
the types of materials to be specified. Multiple designs can be tried very quickly to work toward anoptimum final result.
Pertinent to Chapter 11Keys Group:
separate modules are provided for Parallel Keys and Woodruff Keys, using design pro~dures
similar to those presented in the book. Input can be either the torque to be transmitted or the .combination
of power and shaft speed. The shaft diameter isspecified,likely based on prior stress analysts, Matenals
for the key, the shaft, and the hUbof the power transmitting element are selected from a list taken from
Appendix 3 in the book.The yield strengths are then automatically inserted. The usespeclnss a design
factor and the length of the hub. The program determines the appropriate size forthe 'key according to the
shaft size and the stress analyses using procedures similar to those outlined in the book.
Bearings Group: Pertinent to Chapters 14 (Rolling contact) and 16 (Plain surface)
Ball and Roller Bearings: Chapter 14
This program aids in the specification of it commercially available bali or roller bearing fromextensive databases for t6 different types of bearings from two widely known manufacturers, FAG and
SKF. The user selects the preferred type, and provides detafor redial and thrust loads, speeds, desired
hours of 'life, and size limitations. New factorsihat are not discussed in the book are also specified:
Viscosity Grade for the lubricant to be applied (typical choice is 66); Operating Temperature (typical
choice is 40"C); and Contamination Factor (tYPical choice is 1.0). The program then produces a list of
possible'bearings, organized in oroer of the bore size. The User selects oneandtl1e program completes
the calculation of performance, giVing the projected life of the bearing in hours that is compared with the
desired life. SUbsequent trials may be done qulte quickly to work toward an optimum design.
NOTE: The basicdynarnic /()~ data .repodedfr1r any [Jivenbearing wiN not match those given in
the book because the bearings come from different Sources and because an additional tector, a23. calledthe Material and Lubrication Factor, is applied. The value of this factor is prOprietary to a givenmanufacturer andfypically yields a higher value for basic dynamic load, C, than is reported in the book.
Plain Surface Bearings: .Chapter 16, Section 16-6
This program assists in the deSign of boundary lubricated plain Surface bearings using the
methOd presented in Section 16-5 of the book. The user enters data for radial/oad, speed, minimum shaft
diameter, and the desired ratio of bearing length to diameter. The program then computes the required
pVvalue and recommends a material from Table 16-1 in the book that has an acceptable rating for pv,
NOTE: Data for the wear factor, K, and the coefficient of dynamic friction are not available in thebook version of the software and are repOrted in the output W!'th I
-va ues of 0.000.
Vl1l
Power Screws: Pertinent to Chapter 17 Linear Motion Elements; Section 17-2
The procedures described in 'Section 17-2 of the book are implemented by this program to design
a power screw with Acme threads having a 14 1/2' thread angle. The user enters data for the load to be
moved, the distance traveled, and the time to move the load. The material for the screw isseleeled from
lists of steels, aluminums, cast irons, copper alloys, bronzes, and zinc alloys. The strength data are
automatically entered by the program. The design value for the coefficient of frielion is entered. The
output includes the dimensions of the standard Acme screw threads from Table 17-1 in the book. The
program checks the tensile stress in the screw and the shear stress 'in the threads. The minimum length
of MQagement'of the threads with a nut is reported. The program also computes the torque required \0
raise and lower the load, the efficiency, the linear speed of the nut, the rotational speed of the screw, and
the power required to drive the screw. !fthe screw is loaded in compression,it should be analyzed for
column buckling using the Column Analysis module.
Joints Group: Pertinent to Chapters 18
Fasteners: Chapter 18, section 18-4
This module completes the procedure outlined in Section 18-4 in the book for designing and
analyzing bolted joints that provide clamping loads. Input data include the total load applied to the joint
and the number of bolts. A Demand Factor, k, is specified that gives the allowable percent of the proof
load of the screw material, often taken to be 75%. The factor, k1, is based on the lubrication present and
is setteo.tsfor typical conditions urness the threads are thoroughly cleaned. After starting the Calculate
process.awmoowasks fortheSAEmaterial grade forthe bolt as listed ln Table 18-1 in the book. The
output includes the required size of bolt, taken from the list in Table 16-4 in the book, and the tightening
torque.
Springs Group: Chapter 19
This group contains three modules for Helical Compression Springs, Helicat Extension
Springs, and Helical Torsion Springs. Each module implements the design procedures developed in
the book and uses data from the book for spring wire sizes and material properties.
Helical Compression Springs: Section 19..0
Method 1 for the design of helical compression springs illustrated in Example Problem 19-2 is
used in this module. The user supplies values for forces and lengths, end type, wire type, type of service
{Iight,average,severe),and an inilialestimateofthe mean diameter of the spring and the design shear
stress (typicaltyinthe range from 60 to 140ksi). The program determines an appropriate wire diameter,
computes the actual stresses, and outputs the geometry of the spring.
Helical Extension Springs: Section 19-7
The method for designing helical extension springs illustrated in Example Problem 19-4 is used in
this module. The user supplies values for forces and lengths, end type, wire type, type of service (light,
average, severe), and an initial estimate of the mean diameter of the spring and the design shear stress
.. f 80 t 140 kSI·) The program determines an appropriate wire diameter,(tYPically In the range rom a .
computes the actual stresses, and outputs the geometry of the spring.
Helical Torsion Springs: Sect jon 19-8
The method for designing helical torsion springs illustrated in Example Problem 19-5 is used in
this module. The user supplies values for moments and angles of rotation, end type, wire type, type of
service (light, average, severe), and an initial estimate of the mean diameter of the spring and the design
bending stress (typically in the range from 11010 190ksi). The program determines an appropriate wire
diameter, computes the actual stresses, and outputs the geometry of the spring.
Joints Group:
Bolted Connections: Chapter 20, Section 20-3
This module determines the minimum required diameter of the bolts in a connection subjected to
direct shear and shear due to a moment applied to the member similar to that shown in Figure 20-5 in the
book. The bolted connection may be comprised of any array of bolls for which the distance from the
centroid of the array 10any.individual bolt is th.e same. Examples area rectangular alTay .of four.bolts or a
circular array of any number of bolts. The analysis considers only a single force to be applied to the
connection. If more than one force is applied, the resultant of all applied forces must be determined by the
user for input to the program. Both the magnitude and the Orientation of the line or action .of the force
must be known. The program assumes that the bolls are subjected to single shear. If they are in doubleshear, the applied force should be divided by 2.0.
Inputs required are the shear lead, the ·number of bolts, the perpendicular distance from the line
of action oflhe load to the centroid of the boll pattem, the radial distance from the centroid to any boll, the
x and y distances from the centroid to the bolt, and the angle of inclination of the applied load. The angle,Cl, is zero for a vertically downward load. Graphic and textual helps are provided.
The output consists of the forces on Ihe bott, the required diameter and the nearest standard bolldiameter as Shown in Table 18-4 in the book.
Welded Joints: Chapter 20, Section 20-4
The method Of Section 20-4 of the :book is used in this module to compute the required size ofweld to carry .a specified force using a weld geometry selected from those shown in Figure 20-8 in thebook. This figure is included on help screens in the program.
Inputs required are the type of jOint, key dimensions of the joint, the location of the load, and theallowable force per inch of weld length taken from Table 2Q,3 that 'ISshe hi. th
wn ona e p screen In eprogram. The term, bending console length, ab, is the distance, a, shown under the Bending column ofFigure 20-8. Similarly, Ihe term, twistirrg console length at, is the d'istan·ce a f th T. I n
., . , ,rom e J orslon co um .OUtputs inclUde geometry factors ·of the weld pattern, bending momenl, twisting moment, forces
on the weld, and Ihe required weld leg size.
x
Clutches and Brakes Group: Chapter 22This group contains five modules for Cone or Plate-Type Clutches or Brakes and Short Shoe,
Long Shoe, and Band Brakes. Each module implements the design procedures developed in the book.
Plate-Type Clutch or Brake: Section 22-11The procedure illustrated in Example Problem 22-6 is used in this module for the analysis of a
plate-type clutch or brake. The user inputs the desired friction torque, the nonmalactuation force
available, the coefficient of friction, the rotational speed, and a reasonable value for the ratio of the
outside radius to the inside radius (say 1.50). The program computes the required mean radius, the
outside radius, the inside radius, the frictional power absorbed and the wear ratio (hplin\
Cone Clutch or Brake Section 22-13
The procedure illustrated in Example Problem 22-7 is used in this module for the analysis of a
cone clutch or brake. The user inputs the desired friction torque, the mean radius of the drum, the cone
angle, and the coefficient offriction. The program computes the required axial actuation force.
Short Shoe Drum Brake: Section 22-14
The procedure illustrated in Example Problem 22-8 is used in this module for the analysis of short
shoe drum brakes. The user inputs the friction torque desired and proposes the drum diameter,
coefficient of friction, and basic dimensions of the actuation system. The program computes the required
actuation force.
Long Shoe Drum Brake: Section 22-14
The procedure illustrated in Example Problem 22-9 is used in this module for the analysis of long
shoe drum brakes. The user inputs the drum speed, friction torque desired, drum radius, coefficient of
friction, design value for maximum pressure, and the basic dimensions of the actuation system. The
program computes the required width of the brake pad, the actuation force, the frictional power, and the
wear ratio (hp/in\
Band Brake Section 22-15
The procedure illustrated in Example Problem 22-10 is used in this module for the analysis of
band brakes. The user inputs the drum speed, braking torque desired, design value for maximum
pressure, drum radius and width, coefficient of friction, the angle of wrap of the band on the drum, and the
basic dimensions of the actuation system. The program computes the friction torque, the actuation force,
the friction power, and the average wear ratio (hp/in2). Subsequent iterations are quickly done to work
toward an optimum design.
MACHINE ELEMENTS IN MECHANICAL DESIGNFourth Edition
Robert L. Mott
Prentice-Hall Publishing Company
Description of Spreadsheets Included on CD in Solutions Manual
Introduction
The Solutions Manual for this book inciudes a computer disk that contains 26 computational aids that arekeyed to the book. The files are written as Microsoft Excei spreadsheets uSing Version 2002 onWindowsXP.
Many of the spreadsheets appear in the text. Others were prepared to Produ~ solutions for the.Solutions Manual. The given spreadsheets include data and results from certain figures In the text. fromcertain example problems, or for certain problems from the end of chapters containing the analysts anddesign procedures featured in the programs.
The following sections give brief descriptions of each spreadsheet. Many are discussed in the text inmore extensive detail. It is expected that you will verify all of the elements of each spreadsheet beforeusing them for solutions to specific problems.
Using the Spreadsheets: It is recommended that you maintain the givenspreadsheets as they initially appear on the disk, considering them to be master copies. To use aprogram for SOlving other problems, call it up in Excel and use the ·Save as. command to nameit something different. For instance, the Original program cal/ed Column Analysis should beconsidered the master. Use "Save as" and call it, for example, Column Analysis _ Working. Thenuse that version for general problem Solving.
You should study the concepts and the solution techniques for each type of problem before using thespreadsheets. You should work sample problems by hand first. Then enter the appropriate data into thespreadsheet to verify the solution. In most spreadsheets, the data that need to be entered are identifiedby gray-shaded areas and by italic type.
Descriptions of Spreadsheets
The descriptions are given here in the order that the Subjects for the spreadsheets are covered in thetext. The files are listed on the disk in alphabetical order.
Co~umnAnalysis: Chapter 6. Analyzes straight columns of uniform cross section to detemnine thecntlca~ buckling load and the allowable load. The spreadsheet shows results for Example Problem 6-1 asgiven In Figure 6-9 on page 242. U.S. Customary units are used. A description is given in Section 6-8.The process IS esse~tlally the same as that shown in the flow chart of Figure 6-4. Note that a shortmacro program In Visual BaSICISused to decide w~ether the column is iong (EUler) or short (J. B.Johnson) and to complete the calculation of the cntlcal bUCkling load Be th t Excel programenables macros. . sure a your
Column Analysis SI: .Chapter 6. Same as COlumn AnalYSis: exce t SI't d The solution toExample Problem 6-21S shown as given in Figure 6-10 on page 243. p UOis are use .
CirCUlar Column Analysis: Chapter 6. Special version of Col A I '. t .cproperties of a COlumnwith a solid cirCUlar cross Section are c umn na ysls In which th~ g.eome nspreadsheet can be used as an iterative des' t I omputed When the diameter IS input. Thea circular cross section to carry a given load Ig~ oOFto determine the required diameter of a column with
. ee Igure 6-14 on page 249.
xii
;
Crooked Column Analysis: Chapter 6. Section 6-11. Analyzes the allowable load on a column ofconstant cross section with a given amount of crookedness. Data from Example Problem 6-4 are used
as shown in Figure 6-16 on page 252.Eccentric Column Analysis: Chapter 6. Section 6-12. Computes the required yield strength of thematerial and the resulting maximum deflection of the middle of a column that is loaded eccentrically.Data from Example Problem 6-6 are used as shown in Figure 6-18 on page 256.
Chain Drive Design: Chapter 7. Design of roller chain drives as described in Section 7-6. User mustobtain rated power data from Tables 7-5, 7-6, or 7-7 to specify a suitable chain number and number of
teeth in the smaller sprocket.Contact Ratio-Spur GeatS: Chapter 8. Computes the contact ratio for spur gears using the procedure
shown on page 317 in Section 8-4.Bevel Gear Geometly: Chapter 8. Computes the geometric features of straight bevel gears using theformulas listed in Table 8-7 in Section 8-8 and illustrated in Example Problem 8-6 on page 337. Twoidentical programs are shown side-by-side. One shows the results of Example Problem 8-6 and the other
can be used to solve any given problem.Wormgearing Geometry, CD, VR: Chapter 8. Computes essential geometric features of a worm andwormgear, the center distance between their shafts, and the velocity ratio. Uses procedure from Section8-10 as illustrated in Example Problem 8-7. The spreadsheet was used to complete Problems 52-57 at
the .end of the chapter.Gear Geometry: Chapter 8. Computes the geometric features of spur and helical gears using therelationships in Sections 8-4 and 8-7. Can be used for Problems 1-9 and 41-44.
GeatS VR Design: Chapter 8. Aids in the specification of the number of teeth in a pinion and gear toproduce a specified velocity ratio. Uses a procedure similar to that shown in Section 8-13 on pages 350-357 and illustrated in Table 8-9. An integer is entered for the number of teeth in the pinion. Theprogram computes the required approximate number of teeth in the gear to produce the given velocityratio. The user then enters an integer for the actual number of gear teeth. The program identifies thecombination of numbers of teeth that produces the minimum differential between the desired ratio andthe actual ratio. The spreadsheet was used to complete Problems 62-65 at the end of the chapter.
Spur Gear Forces: Chapter 9. Computes the tangential, radial, and normal forces on spur gear teethof a given design transmitting a given power at a given pinion speed. It uses the method of Section 9-3.The spreadsheet was used to complete Problems 1-6 at the end of Chapter 9. The results for Problems 1
and 2 are shown in the master.Spur GeatS-Design-U.S.: Chapter 9. Performs a complete design analysis for a pair of spur gears,including the essential geometry, tangential force, required bending stress number, and required contactstress number. All modifying factors for stress calculations as described in Sections 9-8 to 9-12 areincluded. The data from Example Problem 9-5 are shown in the given spread sheet as illustrated inFigure 9-28 on pages 416-417. An extensive discussion of the spreadsheet is given in Section 9-14 onpages 415-419. An added feature of the spreadsheet on the CD is the computation of the requiredhardness (HB) for through-hardened Grade 1 steel using the equations in Figures 9-10 and 9-11. Theuser can then specify suitable materials and list them at the bottom of the spreadsheet.
Geometry Factor-I-Pitting: Chapter 9. Computes the value of the geometry factor, I, used in thecalculation of contact stress for spur gears in Equation 9-25 on page 401 of the text. Program uses the
algorithm from Appendix A20.Spur Gears-Design-U.S .•With I: Chapter 9. Same as Spur Gears-Design except the geometry factor, I,is computed within the program instead of being input by the user. The program Geometry Factor·l-Pitting is integrated within $pur GeatS-Design. One additional input value is needed for the pressure
angle.Spur GeatS-Design-SI: Chapter 9. Similar to Spur GeatS-Deslgn: except SI metric data are used asdescribed in Section 9-13 and illustrated in Example Problem 9-6. Data from Example Problem 9-6 are
used in the given spreadsheet.
XIII
Spur Gears-Capacity-U.S.: Chapter 9. Section 9-16. Determi.nes the ~ower transmitting capaclty of agiven set of spur gears considering both bending strength and pitting resistance. The user must Input theallowable bending stress and allowable contact stress based on the material specified for the pinion andthe gear using Figures 9-10 to 9-15 and Tables 9-3 and 9-4. The spreadsheet includes the computationof the required bending stress number, Sat, and contact stress number, sac, based on user-enteredhardness (HB) for through-hardened Grade 1 steel using the equations in Figures 9-10 and 9-11. Theuser must transcribe these values into the spreadsheet if, in fact, this kind of material is specified.
Plastic Gears· Design: Chapter 9. Completes the design of plastic gears using the procedure frompages 440-1. Data are shown for Example Problem 9-6.
Helical Gears-Design: Chapter 10. Computes the forces on helical gear teeth as described in Section10-2 and illustrated in Example Problem 10-1. Completes the design analysis for a pair of helical gearsas described in Sections 10-3 to 10-5 and illustrated in Example Problem 10-2. Used for the solutions toProblems 1-11 at the end of Chapter 10.
Helical Gears-capacity: Chapter 10. Similar to Spur Gears-Capacity: with modifications for thespecial geometry of helical gear teeth. Used for the solutions to Problems 12 and 13 at the end ofChapter 10. The user must input the allowable bending stress and allowable contact stress based on thematerial specified for the pinion and the gear using Figures 9-10 to 9-15 and Tables 9-3 and 9-4. Thespreadsheet includes the computation of the required bending stress number, sa" and contact stressnum~r, s~,based on user-entered hardness (HB) for through-hardened Grade 1 steel using theequations In Figures 9-10 and 9-11. The user must transcribe these values into the spreadsheet if infact, this kind of material is specified. '
Bevel Gears - Design: Chapter 10. Computes forces and stresses on bevel gears.
Wormgearing - Design: Computes some geometry values, forces, and stresses for wormgearing.
Shaft Design: Chapter 12. Computes the minimum acceptable diameter for shafts using Equation 12-2~~n ':th bending and torsion are present and Equation 12-16 when only vertical shearing stress is~odifyi~9 f:~~so~~~~~~r~~s :or torques, forces, bending moments, pertinent material strengths,at several selected sections of t~en~~'ftand .~tressconcentration factor. The program is typically appliedgiven spreadsheet uses data from one I as ;. ustrated in Design Ex~mple 12-1 on pages 548-552. TheFigure 12-19 on page 561. A discussio ocafth
lonon the shaft .In.Deslgn Example 12-1 as illustrated inno e spreadsheet IS In Section 12.10.
Foree Fits: Chapter 13, Section 13-8 Str f .between mating members assembled ~ith e~es or Force Fits. Computes the pressure at the interfaceresulting stresses and deformations for the ~ I~terference fit (See Section 13-6.) Also computes thefrom Example Problem 13-2 are Shown in th a Ing members usmq the procedure on pages 587-8. Datae example.Spring Design-Method 1: Chapter 19 Section .method from Example Problem 19-2 to d' 19-6. The given spreadsheet uses data and theto fit given geometrical limitations. See ~~J~rn~~f~ helical compression spring for a given loading andSpring Design-Method 2' Ch . e on page 753 and the accompanying discussion.
. . . apter 19. Similar to S rI DdeSigning to a set of geometricallimitalions S Ex P ng eslgn·Method 1 without the restriction ofand the aCCOmpanying diSCUSSion. . ee ample Problem 19-3, Figure 19-17 on page 753,
xiv
CHAPTER 1THE NATURE OF MECHANICAL DESIGN
Problems 1 - 14 require the specification of functions and designrequirements for design projects and have no unique solution.
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CHAPTER 2MATERIALS IN MECHANICAL DESIGN
1. Ultimate tensile strength is the apparent stress at thepeak of the stress-strain curve.
2. Yield point is the value of the apparent stress from thestress-strain curve at which there is a large increase instrain with no increase in stress. It is the point wherethe stress-strain curve exhibits a horizontal slope.
3. Yield strength is the apparent stress from the stress-strain curve at which there is a large increase in strainwith little increase in stress for materials that do notexhibit a yield point. The offset method is used bydrawing a line parallel to the straight part of the stress-strain curve through a value of 0.2\ on the strain axis.
4. Many low alloy steels exhibit a yield point.5. The proportional limit is the apparent stress on the stress-
strain curve at which the curve deviates from a straightline. At this value, the material is usually stillelastic. The elastic limit is the apparent stress at whichthe material is deformed plastically and will not return toits original size and shape.
6. Hooke's law applies to that portion of the stress-straincurve that is a straight line for which stress isproportional to strain.
7. The modulus of elasticity is a measure of the stiffness ofa material.
8.. The percent elongation is a measure of the ductility of amaterial.
9. The material is not ductile. Materials having a percentelongation greater than 5\ are considered to be ductile.
10. poisson's ratio is the ratio of the lateral strain in amaterial to the axial strain when subjected to a tensileload.
11. From ~Q. 2-5 G = E/[2(1+Vl] = (114 GPal/[2(l+O.33l]G = 42.9 GPa
12. Hardness - 52.8 HRC (Approximate; Appendix J~)
13. Tensile strength ~ 235 ksi (Approximate; Appendix 19)
14.-17. Errors in given stat_ents:
14. A hardness of HB 750 is extremely hard, characteristic ofthe hardest steels in the as-quenched or surface hardenedcondition. Appendix 3 shows annealed steels to havehardness values in the approximate range of HB 120 to 230.
15. Hardness on the HRB scale is normally limited to HRB 100.16. Hardness on the HRC scale is normally no lower than HRC 20.17. The relationship between hardness and tensile strength is
only valid for steels.16. Charpy and Izod tests measure impact strength.19. Iron and carbon. Other elements are often present.20. In addition to iron and carbon, AISI 4340 steel contains
nickel, chromium, and molybdenum. (Table 2-3)21. Approximately 0.40% carbon in AISI 4340 steel.22. Low-carbon:
Medium-carbon:High-carbon:
Less than 0.3%0.30% to 0.50%0.50% to 0.95%
23. Typically a bearing steel contains 1.0% carbon.Lead is added to AISI 12L13 steel to improve machinability.Shafts are often made from AISI 1045 4140, 4640, 5150,6150, and 8650 steels. (Table 2-1/),Gears are often made f5150, 6150, and 8650 rom AISI 1045, 4140, 4340, 4640,steels. (Table 2-lf)The blades of a post hole diresistance, high strength gger should have good wearsteel is a reasonable ch ~ and good ductility. AISI 1060
Ol.ce.
AISI 5160 OQT 1000 is hicontaining approXimate~ gh-carbon, chromium steel,It was heat treated by ~ O'fO% carbon and 0.60% chromium.temperature, quenched ine~i ng above its upper criticaldegrees Fahrenheit It h 1, and then tempered at 1000ksi or 1040 MPa) a~d as fairly high strength (Sy ~ 151
good ductility (14% elongation) .
24.25.
26.
27.
26.
4
31. Induction hardening is a heat treating process in which thearea to be hardened is subjected to a high-frequencyelectric current created by a coil, inducing current flownear the surface of the part and causing local heating.After sufficient time to bring the surface to a temperatureabove the upper critical temperature of the material, thepart is quenched to harden the surface.
32. Some carburizing grades of steels are AISI 1015, 1020,1022, 1117, l11S, 411S, 4320, 4620, 4820, 8620 and 9310.The carbon content ranges from 0.10% to 0.20%.
29. In general, a high hardness with good ductility aredesirable for machine parts and tools subjected to impactloads as seen by a shovel. A hardness of HRC 40corresponds to approximately HE 375 and is consideredmoderately hard. While this is a good level, even a highervalue up to HRC 50 (HB 475) would be better, providedductility is fairly high, say about 15% elongation.Appendix 3 shows some forms of oil-quenched AISI 1040 andnone listed have sufficiently high hardness. Appendix 4-1shows the same material quenched in water and tempered.AISI 1040 WQT 700 has a hardness of HB 401 (HRC 43) withapproximately 20% elongation and a yield point of 92 ksi.
30. Through hardening involves heating the entire part followedby quenching to achieve the hardened condition. Except forsome variation in thick sections, the part is hardenedthroughout. But no chemical composition changes occur. Incarburizing, the chemical composition of the surface ischanged by the infusion of carbon. Thus, carburizingresults in a hard surface while the core is softer.
33. The AISI 200 and 300,series of stainless steels arenonmagnetic.
34. Chromium gives stainless steels good corrosion resistance.35. ASTM f\qq:tstructural steel is used for most wide-flange
beams.36. HSLA structural steels are high-strength, low-alloy steels
having yield strengths in the range of 42 - 100 ksi (290 -700 MPa.
37. Three types of cast iron are gray iron, ductile iron, andmalleable iron.
3S. ASTM A4S-S3, Grade 30 is a gray iron with a tensilestrength of 30 ksi (207 MPa); no yield strength; less than1% elongation (brittle); modulus of elasticity (stiffness)of 15x106 psi (103 GPa).
Problem 38. (continued)ASTM A536-84, Grade 100-70-03 is a ductile,iron with atensile strength of 100 ksi (689 MFa); a y~eld strength of70 ksi (483 MFa); 3% elongation (brittle); modulus o~elasticity (stiffness)of 22xl06 psi (152 GPa) .ASTM A47-84, Grade 35018 is a malleable iron with a tensilestrength of 53 ksi (365 MFa); a yield strength of 35 ksi(241 MFa); 18% elongation (ductile); modulus of elasticity(stiffness) of 25xl06 psi (172 GPa) .ASTM A220, Grade 70003 is a malleable iron with a tensilestrength of 85 ksi (586 MFa); a yield strength of 70 ksi(483 MFa); 3% elongation (brittle); modulus of elasticity(stiffness) of 26xl06 psi (179 GPa) .
39. Powdered metals are preformed in a die under high pressureand sintered at a high temperature to fuse the particles.Re-pressing after sintering is sometimes used.Parts made from Zamak 3 zinc casting alloy typically havegood dimensional accuracy and smooth surfaces, a tensilestrength of approximately 41 Ksi (283 MPa), a yieldstrength of 32 Ksi (221 MFa), 10% elongation, and a modulusof elasticity of 12.4xlO' psi (85 GPa). (Appendix 10)
40.
41. Type D tool steels are typically used for stamping dies,punches, and gages. (Table 2-5)
42. The suffix 0 on aluminum 6061-0 indicatescondition. the annealed43. The suffix H on aluminum 3003-H14 indicates h 'strain hardened. t at ~t was44. The suffix T on aluminum 6061-T6 'treated. ~ndicates that it was heat45. Aluminum 7001-T6 has the h' h~ 98 ksi (676 MFa)' y' ld ~g est strength; tensile strength, ~e strength = 91 ksi (627 MFa) .
Aluminum all 606oy 1 is one of the most versatile.Three typical uses of t't 'structures, chemical pr~can~~m alloys are aerospacehardware, ess~ng equipment, and marineBronze is an alloy ofphosphorus, nickel z,Copper with tin, aluminum lead,, ~nc, mangan 'ese, or silicon.
46.47.
48.
49. Bronze C86200 is a manganese bronze casting alloy with atensile strength of 95 ksi (655MPa); yield strength of 48ksi (331 MFa); 20% elongation (ductile); modulus ofelasticity of 15x106 psi (103 GPa) .
50. Bronze is used for gears and bearings.51. Thermosetting plastics undergo a chemical change during
forming resulting in a structure of cross-linked molecules.The process cannot be reversed or repeated. Thermoplasticmaterials can be formed repeatedly by reheating because themolecular structure is essentially unchanged duringprocessing.
PEr;52. a) Gears: Nylon, polycarbonate, acetal,~polyurethane
elastomer, phelolic. b) Helmets: ABS and polycarbonates.c) Transparent shield: Acrylic. d) Structural housing: per,ABS, polycarbonate, acrylic, PVC, phenolic, polyester/glasscomposite. e) Pipe: ABS, PVC. f) Wheels: Polyurethaneelastomer. g) Switch parts: polyimide, phenolic, PIT.
53. Designers of parts to be made from composite materials cancontrol 1) base resin, 2) reinforcing fibers, 3) amount offibers, 4) orientation of fibers, 5) number of layers, 6)overall thickness, 7) orientation of layers, 8)combinations of types of materials.
54. Composite materials are comprised of two or more differentmaterials, typically a resin reinforced by fibers.
55. Resins used for composites include polyesters, epoxies,polyimides.
56. Reinforcing fibers used for composites are glass, boron,aramid, and carbon/graphite.
57. Sporting equipment is made from glass/epoxy, boron/epoxy,and graphite/epoxy composites.
58. Aerospace structures are made from glass/epoxy,boron/epoxy, graphite/epoxy, and aramid/epoxy composites.
59. Sheet molding compound is typically a glass/polyestercomposite.
60. SMC's are used for auto and truck body panels and largehousings.
61. Reinforcing fibers are produced as continuous filaments,chopped fibers, roving, fabric, yarn, and mats.
1
.,...-------------------62.
63.
64.
65.
66.
67.
68.
69.
Wet processing of composites involves,the layup of fabricreinforcing sheets on a form, saturat~on of the sheetswith the resin, and curing under heat and pressure.preimpregnated composite mat~rials are ~roduced with theresin already on the fibers ~n a conven~ent form, calleda prepreg. The prepreg is layered onto the form andcured.SMC's are preimpregnated fabric sheets formed in a moldand cured simultaneously under heat and pressure.Pultrusion is a process of coating the fiberreinforcement as it is pulled through a heated die toproduce a continuous form such as tubing, structuralshapes, rod, and hat sections used to stiffen aircraftstructures.In the filament winding process, continuous filaments areplaced around a mandrel in a controlled pattern and thencured. The process is used for pipe, pressure vessels,rocket motor cases, containers and enclosures.Specific strength is the ratio of the strength of amaterial to its specific weight.speci~ic stiffness is the ratio of the modulus ofelast~city of a material to its specific weight.ManY,c~osites have significantly higher values ofspec~f~c strength and specific stiffness than metals.
70 • 73 refer to P1gure 2-lZ and Table 2-~.General conclusions from Questions 70 - 73: The specificstrengths,of the metals listed range from 0.l94xl06 to1.00xI06 ~n approximat 1 f 'stiffnesses'are very e y a actor of 5.0. The specif~capproximat 1 nearly equal for all metals listed,e y 1.0xI0S in The 'f ' hcomposites I' d . spec~ ~c strengths of t e~sterange 1 87 t ' ,than any of the metal . 0 4.86xI06 ~n, much h~gherstiffness about 2/3 t~t G~ass/epoxy has a specificcomposites listed ran 0 the metals. The otherthe metals. ge from 2.2 to 8. .3 times as stiff as
See Section 2-18 for answers to Questions 74 to 78.
CHAPTER 3STRESS AND DEFORMATION ANALYSIS
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CHAPTER 4COMBINED STRESSES AND MOHR'S CIRCLE
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8. z.t1,:: T_-' "Y,o) =.$1.14; tP,.' ~7°e'"24- 10'- 2pl,.=- '11.M"; fIJr= ".30 ac...., .<r,'_Y~ d"",,=YoM5.0-,.
9. zP.,..o rO:(Gtl/s-j .. sa.O',j Pr=2.,°CI;JZp,.. 9,·-Z.(J,c .3~0.;¢ra"· CCI.J
11"'~"4.tT&.a
- W;;~;...;lfo......j_*iil.l,.,..-II.-..J......!t:r~ .
-I~HII:-o__ -I-~~~--.--h~~~.l.-
/0. 2¢r= To.:! ''''h6S):' ,3,J·7°J ~ ~ /L8So CW
2167"'" 90'_ 2fJ,." $"6•.3'./ ~r.Z9.IS· eGW.rye. -80"'~ ~ c-IIJ.311A. "-""....I!S""If~
"'tZlDI'l"""__ -i~.";;"~:""_""""'_"*-'~~~-~
/I. ~¢tr:/1Jo".... (""ltS)=:J1.4'J ¢cr=lS".a"CW2.;"." q,'-2?,. =- sa.Vo; ire Z.9.z.·C'c-w~-8eM'" di .....,I.".,,... dA,' -1Si'l4
I. CT,(=i $/NlilPo.
23
17';= /1~foS3.t = 1'8.2.""t'.. jl1i.,/If"-SJ.Z ·61./J",f4,.Z (Jrr= ra:! (otO/)S) = Ve-S',' fJrr=Z¢"CW
cry: eUMt'... 0&.6/.,,,,11. 'i" 0 p,It'.'" D,,~LIMTO PA~
-r:.Altz f1VJI1"--/I'r '1$' ro tI;
1tN"4)
13.
9-''';' ('"/")=JI." TZ(,a-a ItO'-9 .. IVc9.V';cj,.=7!tt' Ca-J
11i= IJ~+ '.3 al'1l.JM 1'", j "i .1I$"-74.J =JI.1l'1fo.cr,.= ISol1"- dJ rO ~'Qf"IC_
TO 'AI'''''1'""", -'1$:7 ",,, ...A.,. 'IS" n iiiIf..., "i
If.
/5. Z¢,.= 90'" ~r= liS' ec.w<ry= ... -9'I'I#\"
1-r-t-----~iI-""".Jr...,.o
J't
z tilT" rAN -I (ID'YJsO) = z r;.7;'t/J"..- /'1.9 'cW q; ·SJ+-'IDJ~)ls:! Mt'...Pr =)0.,- CCr,J Oi. - 5" -Yll3 • -:35JMP ....
--+
. /00
OA., ;: '1";1 C;:&DN.. / 'iS3nll
//
2fJ,,'" 180'-"1'_" f'YLS): IUo; tilT'" (,I'eN2~$ '10°-fit" "(lyzr):= )z.'; ~ .. " °cw
tri.2-sz..1.
2 tfltr:=o: ; tllT-o·2 ¢r- .. fldo. f1.r- ¥,sO c cw~
OJ=1M' = ZlO""A...in... 8'"t= -1l.OI"!P ...
"'1',=0GIVEN ~ PH/Nellt'll.ST!lEa eLEM/!"Nr
~r ..Sl)M'"
N Ax //"1'''' !:116f411..J'rREU ELEME",r
0".. "-+---tr--.l...--ir-~O"y=
o
.---OJ " d"r .. ¥DJt.s1O""z. ;: t1'j ;: OI<.S;
..z'tr=dj ;tr"b';2~'" 10' ~ ~r=~· ccwor= ..
---@--;~--_--':f-*~+~-""'Z.lai1';1;:· t7Av =2IJICS;
GiVEN'~ PIJNCJM~ JlflflClMUM S/frAR.ST"L6~ IFLFMENr ,srRIi'SS FLEMeKr
r
zz: :z tPtr = 'lb' ; ¢rr = 'l.$"CtI'I;l¢r=Oo ; 4-0·
a;=oy=o;=;=,
q;= ~= 'Io;a/az." -1'Q'= -'IoksJ
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,=:t:!:;.~y=-i,/(5ia:...
o;;v8 ~~.TN.' ('f'htJ);63.V·; rJoJ'31.,.ce...,,;14,.. 90'1-lAr= /S3.y': ~"';7&.7.t:e1NOJ .. ss:« '1'1.7 = 7'1.U:s.;Ol = lS-'I'/.7= -9.7J:si
"'y
36
l.zs' ';UJ'tT:' 711#-' ('% )::31. 7l; ¢tr:' IS;' 0 cw
2f).r-= '1o'-2.'tr=S8.l7' ;¢r=Z"," ccwa; = ~ +12..9=2.3.9,a/ J &..= 1/-12.'/= -I. '1ks-:
u-"'''o---.,;.
--i:::~=LUS:.~~=6.8~1
O&.·-I.,~
U. dl. ¢o-'" J&'- 711H" fe'3,.stJ) =1'I1.~;¢~ 7/.6·CW
Z.¢r '"lPtr "'9D'",S:J.2S-; /JrsU•.'· CfII ~ ""'-0;. -SOO+~'c!b = 'f180 fS/1T"a. = -Soo - '1480= -sBC I'si
~·Jl.SD'sl.
trr=
~1"'"".. 19
11j(:.
'1'Yy
27. 0; ...z ....+-1l.9- J.2a 1'1""-G"c.'" 200 -/28 = 7ZM'o-<1'", ., 07"..,...... ot/z =16 11./"11'....2~lT= TAN-t F'YJoo) =36.b(','¢tr"I?Jocwa; =/oo/'IA..
37
p---------------a;» Zoo +' 'I ,. 2. 6'/ ""PII- t'0i = 2llO-611' = 1361'11'0.-d}:' a'/Jt1/JJf:: 0"":/, :a/,j2.MI'",2~tr= 1JIH-< (lrSfJ) =18.,,'; ~tT:aI').Jcw
rry: ISfI/'fP., Oi.~/J'I'1?!Jio.!:...t-----_tt:;:;___:;:oF~_ttr::.l...::.".cT,·
0; '"0 ; t1) := - S8[j-l8tJ= -8hBhl' ...
02- -S81!J+ZBo",,-30fJJ.J1.
1"fVfM· Oih:= 'I3YU..:If)o-= /80 - TitN-f {'zr2S2.) =/.$f.SO~·17.J·ccw
CTy=~a:
0;"o
eJ/=O ; 03= -18U-ISD=-J31.SJIA..11'"z.:: -IS?S 1-ISlJ= -37.S.JlI' ...
ilrlll'l' = q-~ = 16B.BA./ ...
:J.¢tr = /80 - r,w /'%J1.S):: 1.s6. 'I •fJtr :: 78.212
CC IN' OJ ..-331$-+F:::::::::==-=~4=-_--r~-!I--
_rq
Oi
38
tr.=o
us/!" D::: D.stlo III 1""6((:SHAFT ABC. SrlZ~ GtEII1I;NToN SoTT/JM.
FRtJM FIG>. 3-11: l'1iJ = 2.S2. LINJI. ; riI·2tJ1) LINN.
~ = 17'/)~.J.1:: 1rfO.sIJI_;//Zl ~ €J.tJl2Z7IJJ'"8 = .Me = 2SZ. LINN. = Zo5.J8 ~ =Gif
i! ",t)IU 71N iZ!, • 1l'D~~:..2r. O.OZ.V.sI//II~-ra = -r., = 2..0 La·/JI· = /1.2L.S~'
~e IJ.Dl'lSY/N'
R= r;,,~"//0 Z'9z+IZZU'l..: IS"'" '03: -r0; = /IU.'" r /.r96'" 2." z rs- hiOi .. /026' -IS"'''~ = -56" ,~.
2 ¢. .. 1AN-I UZZa")= '19.97 ..~ LlOU,f •
¢rr - 2.'1-.98" ccw24- =Z~(T +-9II'-I.1'1.'n°fJr-= 69,98"
Oi.
(JSE o-/.,stJO/!J. F=~S".HAPrA8c• .rrtewa IrLEIOfDllr tJN4.rr.".,.
rtr"", SOLclnlW Foil. PR,tJUl'f J-3D" Fl6.J 73' :He='l:s7l.U./6;~="~DOlQ./il.jZ .. ffDYi~ =- /J.J.1I.1/IJ:!; 4" 1'11l/A: = 'Isn.Uo/N/a.:J.J/J/"lc a s(){) I'J/ ""dj(
r, effO~/' ='22-0.66Z7111.:J/'h= 741~,= 6J'OO/O.6t.2.7 -96.SCJl's/,f..T;.,AIf = Vt5,Dfj ....,..f'-SS s:= 118 70 l"SiOJ - 6'100 T 11870 .. 18 770 I'SIdL ~ 6"100-// ell)::: -'/'I70l's;2 rJo- = 1M'" t:",;:): .5Y,/{6 0; tfJo-=2'1.l"
•2¢,..= 2PtT~"tJ·"/W."'·;4=7l..Z.
"'it~'"10 U71s-1;,,,,-7$,,,,~
+
ITi
Use 0- Z. ZS'm. rOlf: SIIAtt:rACt:.. : .sreess eLCAfENr ON I1Dm,."
rlOH S'/UvnMI ~ ffoUG"1'I 2-3/, FIt; .1-.J~/'18~86'1oa.,w.j rc'l~IJ~C .: lrD:&J. - /.118 IN:' OJ - "'Ie-= 8l~(U.lNN//.llaIN] :: 71U ;Os/ rt!llS/I"J _
tp."'D'#' ::;lZ:: ';'.J.J 7/1/; 74" -r;/b = ~ tlJ'III/z.2Jh~=171~;/i'cr;.".l/- VI111+-JI631> ... Je&,po<· rCj =.38'3 +-.11&7· 77JO'Sftrz. "1~61-3361 .. -'1,s;Z~IT''"rA-.J-' fl%ea) «z.cs:~ = j,jjO C lI>J
24- .. '1o'-Z;r =87$'4- • '1.1.1.1 0 CCw
386J
us« 0:> r".o.- ...nut SHtPr 1+6: SrRt:u cur.NeHr ON8.TTIM.
Ro,., .J"'CJTlW 1'01 Pte4/JU", 3-3l..;FI6< J~8: /'f;4 "·I,'-S'~b."",·7,f .o,J7~1.J;,...t = /Tl)Y.h ..a Z1Z._!; .!-%::.2SbN.....J' /O~ :-Sl.97 ",r'a. ~0I11'-'?6tl'1Q#'t .11'1/1, -li - 'p. J /Z27l.",.,,,,,.., ,
,. - 2: 5lfY,..,... J' 7; .. .L= 37S',;{._ • /0 __ ~/s:.Z8;ff1:..-;:-:::--=--:r==;::::==:::;:::-- __ ~i,. 2i'.5Ylf........ ~ ,.../i''''1,:iAY:: VIS:l.B -, 24'181. ~~S1"""tJj = -U.,/8.,. 34.s1 • +'.01",PI..:T"L· ~2.6.Y8-se.s» - -.s7.as",~..G( = Tao. (IS.Z%,.\\5 )"10.0 •2 ¢tr = /80 "-ri :: ISO'\2l~:: -ZS-" c c ..,2al,..:: 900rl( = flo'Ill,...:: 60° CIA)
'10
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4i
D '" '1.00 IN j A" 7f'0n•IZ.s7 h.I'1.; i:p:: tr%,.12..$'71Nez(J= Z-_ -7SaIlU.I = -S"9h8's:
A - IZ.s?IN"
_L:. 200CfJt.'·IN = 1S9Z.PS;,- e,l /1. .S711J J
1 iT"I= -S'l68'~1*'
1;;&/$9,0Oi -s'/~8 di~+:r~---::;~....M..=...."-tr- zrJtr:: ta.."t-::;~
2t!a=Zlfj.08'¢ltr '" IV. DY· cf/J
2¢,., ..,9D·-ZPtr= 61.'IL·
¢r = 30/" •ccu
D=1.o_ ' A" ~~ 3/,/__ ".::..-==.:.;=~ I 'I
(1;- J:.. :1kJoiJN ... IIY. bill'.. 11FAI.[I,#I• A 'IN_''
l!,.:..!L!!.. ~ IS1/_""S/l, J
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r "l= OAv,". /I'(.t,1'o: S7.J ~+- o,K ~£ t---tr-- 1?"1'"",..:: {n.3 ~ z.ac.~~ 1irJ- ~4I:: 2'1Z.Zf1 1'4-
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CHAPTER 5DESIGN FOR DIFFERENT TYPES OF LOADING
Stress Ratio
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trH;4J< =- 3 S"ttJ AI/ L = '11,1. 11IA./1",sYM/Of
(7M'" =- S£JSAI//1:::' 6.31 hI',,"
r"" '" OS-a> rS-{)())j2 ~2"poAi
0-....::2t'oo~ '" Z$,,s n I'D..
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STIf:ESJ
tJ 1---------:7)"1",,:':"1::-
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0-a.- "" t7)14J;. -0-... ::66..7 -Jo.. ;:;¥(,7~PA.R :: 0-""10/ _ -;, 7/ 'I
/trI'1Ar- ·/~6.7=-O, tJ
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tr, = -ILo L8/ - _ 7."-A '11,,., /fJ./'IN .....- ...,IV'S,
r_ ::(abO -/~}i= 37tJ LS
er_ c::. 370l4/A:::. 23/3 PSt'
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trmlN= /SlJUh,H.YIN~= IJSBI'S/F..., :: (LeOo+ISO)/l ==97r-LlJCT :: 97S"L8/A::. tl8z8Ps;tr '"&-M4,or- 0-.....=/I.Z'7-,tn.9:: 7'17ops;k''' crhl/~ _ IJS8/ _
tJ"'1t4.t.- //G ZJ/l- O.OB)
o h'----~_.,f_--T;:;;;
0'".... I-+--.....Jr--+---'---\--
------------------tr::: F'il ' 14.= ffO_';, IT0.0,..-.): 7. ...
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Fm = (f80~3'l»/z =: .so» Na":..... :::: f7o,vM =: .IJo.6 1'11"'4-
tr4- = Ilb.l - 80,' =- 2'1,? /'f1"A,R = SO. 'Y//I.J ::: d. '162W 8~1't"': cr = .a: : 5 = 1.5'I/,IJ:J FaI(J' y
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L"Ulc.~sr O-ev.£r. ~ rl'lWl./J·,N ~ qJSII'~:MAr /"SYj;r.)
a;",,, = 38'/0""" s: 2'19YI'S; 18•• ;----,1.Sl/ IAI J If
0-... =('1ZSI t-L,/9' ): S'Zz..,s,·2.
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M
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0;. 1-+__ +__+-"_+_
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o-:~ :5=3.0'IIN3F'LSY~1.7
nh4t< =(500l~)(&J1"')= 30Q:l'Lts·'A/O="',
1'1",,# = (sOOI.l)OI/N)= S~LI!.·/t'I· nl
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trhAJf =: 31NDtJL,.'N = '1S{,SI'5,'$. Of/N'J.
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tr_" /2.sD~oY" '11/l I".sl
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[1J I3EAn!
c.
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c;:IL-' 1 ~{D~Sll<l----'V .
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8fiI;V I .s.~;l6<lJ't ..V
01b b
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cr=. 11 &NiJlilG/'1: /,/S;o#·_':=. S, - 1'1'LI""".........·0 D!
.s,tenu; IS A SuPP.tlTED Cfhvr'Lflrt<1l..- CJ4J~(0) Itf'P AIIf-3.
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t:: L07'f!0" A//_,.. z.SoLye AJI{ P:
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p:: '1(".'18 JlIYa M"
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SUM'" JI(lf
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0',.. =un., rro,v)!;!., S'I.lI1/.... 51'/(£55
U-.. = tr,.,."-tr,,, : 1II7,I-Sf.I: '18./,111'~
R: o-~~ .....= IO,~I7.' :0,"47
I> f---if------~f------i: .R..
Endurance Strength
/t' f/ 10'10CD; SA: lJol<~i ) 'AO'" 15~S" (F/<f. s-v /for /l.OLLO?Pl</~nt¥~,,_)0. 7S-/lvON! - t! s=. 0.90 lfd:;. 50 q} : t:_ =;'0, C~r: /.o .err. 8ENru",~R= '19 Yo / CI!= O,~I
5_ I", .r,.. Cr c.: tsr c~ .:(Js,KS,)to.9/)f/-o)(/,oJ[ P.K/) = /8.t ~~,.
ItIS/S)60 Ol/r/30{)j S....=79J-M ......;S ...:3acnpc. /'1AcHINElJ
l.o -- PIA. ~ t!~= II. "Ie· ~ C .. , I. 0 •C s r =. I. O' I?:::. "'IPI--e.Je=O.1?1• I J
$",' - (~oo)(".'fO)(J")(/.O){O.K1) =U'f ""A..
ItIS/ '1/30 W (Jr /30D ; .5-0(= 6761'11"'.... " 5.....=:. zeo HPJ:<.. I1/tCHIIJa>; 1>e=.ZB.D""'"
20"6of/ecr--Cc=".tJ7; C,..-/.o, Csc""/-a.; R"99:to-CJt"'(J,1?1
S 1'1 i ~ ~ L O.!J7 -)( j,o) ( 1.0) ( tJ. fI) =:. /83 11t7A.-
At5/ 301 Sm"C/Las ~:n.: S-'<=/SlJJ<.s/; S'",= Sz..~;/ M'+CH/A/t;.7J
O.60/#Il/A - (!s e (1.93; C .... =. /.0 / t!sr -:'O.so @YllttJ/ Ii? ="'1..,,r-C'1l. ,t). ,}S'
S ....' ; (s;;l)(o.93){ 1.0)((JJfO)(~.?') = ~9.o K(/
Itsrn 1f2-'1Z. "1M..: S..u =- 6J "'5// 5...e-Ji/.oK.s; I'7ACHINE2:J/).)J;r"" -1:.=1-0 J ~-" /.0; Ccr-=1.11;~=.9q % -t!L.(J,EI
S_ (= f;l t(o){I· oJ( /'01./. d)(".rl)::19.VKsi
Design and Analysis
Problems 15 - 18 are open-ended design problems for which there is no unique answer. The GeneralDesign Procedure from Section 5-10 should be used. The loading and support conditions should becompared with the cases described in Section 5-9 to determine the appropriate design stress. A designfactor should be specified using the guidelines in Section 5-i When needed. the endurance streng1hshould be computed from Equation 5-4 in Section 5-'I.
15. The link is subjected to a fluctuating normal stress. Use Case G from Section 5-9. See also thesolution for Problem 1.
16. The rod is subjected to a fluctuating normal stress. Use Case G from Section 5-9. See also lhesolution for Problem 4.
17. The strut is SUbjectedto a fluctuating normal stress. Use Case G from Section 5-9. See also thesolution for Problem 2.
18. The latch part is subjected to a fluctuating normal stress. Use Case G from Section 5-9. See alsothe solution for Problem 5.
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70
ECCENTRIC COLUMN ANALYSIS Data from:RI5.~ii;;;,l;Solves Equation 6-13 for design stress and Equation 6-14 for maximum deflection
.-~'dBt!iifQi\.'V4i:!l!QI!i!l":!i1'~~"Qi..:::_-.:i~_$!' Use consistent U.S. Customary units.
FINAL RESULTS
Computed Values:
Eq. Length, L. = KL = 126.0 in
Column const., C c = 113.5
Argument of sec = 0.769 for strengthValue of secant = 1.3913
Argument of sec = 0.444 for deflectionValue of secant = 1.1073
Slender. ratio, KLJr= 90.6
Column is: short
Req'd yield strength = 46,929 psiMust be less than actual yield strength:
s = 46,000 psiMax Deflection, y.... = 0.322 In
'It
c
ECCENTRIC COLUMN ANALYSIS Data from: F~~'-Solves Equation 6-13 for design stress and Equation 6-14 for maximum deflection
, .--d.T~jL]jjll~lnr·,_.~~!, Use consistent U.S. Customary units
Com utlld Values:NOTE: This solution considers the eccentricload with bending about the horizontal axis.
Eq. Length, L. = KL = 40.0 In
Column const., C c = 70.2
Argument of sec = 0.855 for strengthValue of secant = 1.5236
Argument of sec = 0.494 for deflectionValue of secant = 1.1355
Slender. ratio, KUr = 92.4
Column is: longFINAL RESULTS
Req'd yield strength = 39,964 psiMust be less than actual yield strvngtl/:
5 = 40,000 psiMax Deflection, y.... = 0.237 In
See also Solution 38B for buckling aboutthe thinner vertical axis.
COLUMN ANALYSIS PROGRAM
Eq. Length, L~ = KL = 40.0 in
Refer to Figure 6-4 for analysis logic\":j!c!lQ!!iEfQl1t'!!m,,:JlJi'$:m' It$', Use consistent U.S. Customary units.
Computed Values:
Column const., Cc = 70.2
NOTE: Cross section properties taken withrespect to the vertical axis because the load iscentral to that axis. But buckling is expectedabout the axis through the thin (0.40 in) section.
Slender. ratio, KLJr= 347.8
Column is: long
Critical Buckling Load = 4I91b
Allowable Load = 183 IbThis value governs the design, not soIutlon 3BA
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73
-------------iJ!I
ECCENTRIC COLUMN ANALYSIS Data from: Pfti6renf.~SolvesEquation6-13 for designstress and Equation 6-14 for maximum deflection
,. i'it8r,d~iI!!I?Silil""'''~''!''r .. , ...,'''''Xes", I Use consistent 51M . .. ". '~'" ".... • ".".•"J... ..'tV. ...A.. etneunit[
Com uted Values:
Eq. Length, L. = KL = 750.0 mm
Column const., C c = 63.9
Argument of sec =Value of secant = 0.811 for strenglb
1.4512
Argument of sec =Value of secant =
0.468 fordefledj~1.1205
Slender. ratio, KUr = 102.9
Column is; longFINAL RESULTS
Req'd yield strength .. 389 MPIMust be less than actual yield strength:
.. 9" MPIMax Deflection, y.... .. 2.41 mm
Piston rod Is safe for p•.. 6200 N.
Eq. Length, L. = KL = 156.0 In
COLUMN ANALYSIS PROGRAMRefer to Figure 6-4 for analysis logic.l!i!4.~:,",i!!¥I!,r!fI*~!ii~inj·_-':~";:~.$'~ Use consistent U.S. Customary units.
Com u18d Values:NOTE: Analysis of straight pipe. See alsoSolution 40B for crooked pipe.
Column const., C c = 128.3
Slender. ratio, KUr = 198.2
Column is: long
Critical Buckling Load = 8,101 IbStraight Pipe
Allowable Load = 2,700 IbSee also Solution 40B for crooked pi
75"
Solves Equation6-11 for AllowableLoadCROOKED COLUMN ANALYSIS
Use consistent u.s. Customaryun~:
Com uted Values:
Eq. Length, L. = KL = 156.0 In
Column const., C c = 128.3
Euler buckling load = 8101 Ib
C, in Eqn. 6-11 = ·22074C 2 in Eqn. 6-11 = 3.483E+07
Slender. ratio. KUr = 198.2
Column is: 1000gStnJlght Column
Critical Bucklin Load = 8,101 Ib
This value governs the use of the I •
Crooked ColumnAllowable Load. 1,711 Ib
See solution for slrel ht pipe; Problem 4OA.
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CHAPTER 7BELT DRIVES AND CHAIN DRIVES
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Problems 38-42 are design problems for chain drives for which there are no
unique solutions. The general procedure is illustrated above for one possible
solution for Problem 38. This and the other design problems are shown on the
following pages using the spreadsheet from the CD-ROM that is included with
this Solutions Manual. Data for design power from Tables 7-5,7~, or 7-7must
be used to ensure that the selected chain has sufficient capacity,
8/
�_r------------
CHAIN DRIVE DESIGNInitial'n ut Data:
Actual chain length:Computed actual center distance:
Actual center distance:Angle of wrap-Driver sprocket:Angle of wrap-Driven sprocket:
102.00 in39.938 pitches39.938 in _ ~173.8 degrees rSIl.@Id,be' ifl'lt~teNlfaifj20!m~.188.4 degrees
CHAIN DRIVE DESIGNInitial In ut Data:
Actual output speed:Pitch diameter-Driver sprocket:Pitch diameter-Driven s rocket:
Com
------------ .. 00
Initial In ut Data:CHAIN DRIVE DESIGN
from vendor
11.95 hp rating at 2200 rpm
Actual output speed:Pitch diameter-Dnver sprocket:Pilch diameter-Driven s rocket:
8S
CHAIN DRIVE DESIGNInitial In ut Data:
from vendor
CHAPTER 8KINEMATICS OF GEARS
Gear Geometry
[] N~'/'/./ P,,:: /2.. 4. D-Al/P.r='I'I/n.-;;.'6?/II'
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~ /1/ = l2 d' P", so j.1S'
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sr.. b: }·U'/J.1~= ().?/ '(3/11
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e. a.. ,/,,,-D.US-. IN.
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a... /)= '''''/eo = Z.ZS"OIN.
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~ N- 28 j Pd "'/8
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J. '--;I.e. GL • "" .. O· ""-
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a.b.
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; _ 1,,l.r(l) ~ /';1.S'",., .....
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89
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N- 28 j /1It co f),e
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Veloc:ity Ratio
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Housing Dimensions
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X" 10.700 1/oJ. P" Pol f.l
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Gear Trains - Analysis
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f),.. AI.../~,. 20//6 = ;'UO/N/ 1'11".,..'"' ""',.... J7sp = IflY tf(.M ~IJ D • N,/'If!_ 34/'/1 = 2.371" rv 1.2,/39
IJIf' ~ /II~/IJJ~ 18/12 .. =: I.Sla
ry- of. II- ~. ~ =.£t. /.9 ra 09~A Ale 81J' -;;-.. ,
hI_ ... h1/III/TV '"' 17~A.D 'I =' If! 'Iyy ~t*1 Cw
Helical GearsEJ HEl...lcAL ~E"Afl. f;, ...8, ¢c" /'Ih·, 1V::.~.rre-1?I, F=2.00/N#ELlJ{ AN61-C"" 1/1a 30 o.Clt?CPI.Att.. ~/TUt «er o/~.: 77"/0''= O.39Z7JIJ.NaRltfltL. C,lGIJLAIl 1717r.1+·P#v .. fJ' CN rr-~.J9Z7)CIJ(?IJV· O.:J'IO/N.
NDKMlfL. DIIf""e:rHrt- 1"1 n:H- ~aI '" !?J/GDS1'- 'Ycosf!o'): ·9.Z.1ll
A;rIJ1l,..PI1Z.H:: ~= I'd:".., » B~.')= o.6BO/N
P/Tt.H DIAMw1"F7t.. = D,.= )I//~.. Y.5"/a a 5. 6L1;"/1V.
IV o.e1'1I'tL fill G~SV.t." /+)1/~ I.t;' # ¢... = r""-I [tzJ,... Pot c.,.o...,]¢...= ,-.-1 [t--.I17',.r) c.ro.tJo~}:=12.'2·
FII"", = Z.o(Jn~/O,I.&J'Ii= 2.9'/ -",.;rIAL I'ITCH5J IAI FAeFWIO";'
Ht:LlCAL Gel'll JV:: fB J ~",,J'" /:1.) ¢...alD~ F=I.S01N J JI'= ~.fJ'. 71'/1"11- Stir p.! .. P_ CQJ y= 1.2.·c"sf.',r)=8.'I8S
p.= tT/".'I8S# O..3i"i11'l : ;'... = p- e-1'= %D&' ffdz=,o.Z.6/~/i'.
F',.,.. ~~. ,~_ 1).3'O'N _ 0..110'/01 .. 0. '" N/6 ~ f<s/~ /t-.T b-Y - .~ /1,,- 1&'I4rc,1';6S""1"f
tP~· r--I p;.""J-"""~'(C-1.D~]= 21.20
- e- Y LG=I/.r J .FIIS-= /.SOO/Ii/0.3'bIN = ¥.f)$"AK//h.. fll'rutlti IN FA-c~ WID7»
HELIcAL GIFItIi. 1I::.1~,!?J" &.J (A -llf1.,·, If·r~ F =/.OO/N
.1::-# lT/~.. %...tJ..5236IJJ. ; ~ = p c.o."If. f·e-.!Ir):= 0,.170 //01.
f",,,:: PtI/CAsy• r~{/S"a,6.,/,fS'; f'x= J:t;M.'/'= ,':r-if = O.SZ;'.IN .
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I"+'! 1IB-ICAL Ger/tfL )1'72.; P...d =2'f J' tP.. = 11/'1.' j r'O.z.$/NJ f= 'loS·.1":# "";'4.1 Blrr!!t. : e..d CI4 1'...z« we 'If" --/6.91
r- ~1.97 =- O.18$'1 IJII. j 1"'", = p.C<rS"'. p./,gs';/J/·!D.I,/r"(},/Jp,?/M
~ ...'/t<- t...o.IMf/I:;-¥,s'='tJ./t9S7J1lI.; Dr;,:%» ~/.9'J= '/.2n.lp.
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Set?'" I'.lZO(!,lt:rI I.i9 tJ'" N£«r PII6E Poll. roaM/lt4-S A-NO Sjl~64U
BEVEL GEAR GEOMETRY
'~~fJ~:~f.f.~WgllfM~ili~f@~:~!~jMtnf:~I~;fJH~:;GIVEN DATAN.oD'llileui)i' .fllt41mfh'Iii'Diiinrell:al itee'mt§JiL~~·COMPUTED VALUESGear ratioPitch diameter: PinionPitch diameter: GearPitch cone angle: PinionP~ch cone angle: GearOuter cone distance
COMPUTED VALUESGear ratioPitch diameter: PinionPitch diameter: GearPitch cone angle: PinionPitch cone angle: GearOuter cone distance
3.0002.500 in7.500 in18.435 degrees71.565 degrees3.953 in
2.0002.500 in5.000 in
26.565 degrees63.435 degrees2.795 in
Nominal face width 1.186 inMaximum face width (a) 1.318 inMaximum face width (b) 1.667 inf!ff1!l'!IE~~9~t~~:~[il{1JFfW:W!]~~~!!\'!!I'i]:I\WiijJ!!il1!f
Nominal face widthMaximum face width (a)Maximum face width (b)!l1Ir?OfFace widtlJ
0.839 in0.932 in1.000 in
" O.llOQ i~~ ~Mean cone distanceRatio Am/AoMean circular pitchmean wor1ling depthClearanceMean whole depthmean addendum factorGear mean addendumPinion mean addendumGear mean dedendumPinion mean dedendumGear dedendum anglePinion dedendum angleGear outer addendumPinion outer addendumGear outside diameterPinion outside diameter
3.328 in0.8420.441 in0.281 in0.035 in0.316 in0.2420.068 in0.213 in0.248 in0.103 in4.257 degrees1.774 degrees0.087 in0.259 in7.555 in2.992 in
Mean cone distanceRatio Am/AoMean circular pitchmean wor1ling depthClearanceMean whole depthmean addendum factorGear mean addendumPinion mean addendumGear mean dedendumPinion mean dedendumGeardedendum anglePinion dedendum angleGear outer addendumPinion outer addendumGear outside diameterPinion outside diameter
2.345 in0.8390.264 in0.168 in0.021 in0.189 in0.2830.047 in0.120 in0.141 in0.068 in3.450 degreeS1.670 degrees0.061 in0.148 in5.054 in2.764 in
9'1
Given: N» = 18; NG = 72; Pd = 12; 20° pressure angle.Computed values:
Gear ratio mG = NdNp = 72/18 = 4.000
Pitch diameter: Pinion d = Np/Pd = 18/12 = 1.500 inPitch diameter: Gear D = Nc/Pd = 72/12 = 6.000 in
Pitch cone angle: Pinion r= tan-1(NplNc) = tan" (18/72) = 14.03°Pitch cone angle: Gear F = tan-1(NdNp) = tan" (72/18) = 75.96°
Outer cone distance Ao = 0.5D/sin(I) = 0.5(6.00 in)/sin(75.96°) = 3.092 in
Face width must be specified: F = 0.800 in Based on the following guidelines:
Nominal face width: Fnom = 0.30 Ao = 0.30(3.092 in) = 0.928 inMaximum face width: Fmax = A.l3 = (3.092 in)/3 = 1.031 in
or F_ = 10/Pd = 10/12 = 0.833in
Mean cone distance Am = Am(]= Ao - 0.5F = 3.092 in - 0.5(0.80 in) = 2.692 in
Ratio (A",IA,J = (2.692/3.092) = 0.871 [This ratio occurs in several following calculations]
Mean circular pitch pm = (wPd)(A",IA,J = (w12)(0.871) = 0.228 in
Mean working depth h = (2.00/P,J(A",IA,J = (2.00/12)(0.871) = 0.145 in
Clearance c=0.125h=0.125(0.145 in)=0.018 in
Gear outside diameter
hm = h + c = 0.145 in + 0.018 in = 0.163 inCJ = 0.210 + 0.290/(mc)2 = 0.210 + 0.290/(4.00)2 = 0.228
ao = CJ h = (0.228)(0.145 in) = 0.033 inap=h-aG =0.145 in -0.033in=0.1l2in
bG=hm-aG =0.163 in-0.033 in=0.130inb» = h« -ap = 0.163 in - 0.112 in = 0.051 in
&:;= tan-1(bdAmaJ = tan-1(0.130/2.692) = 2.76°8p = tan-1(bP/AmaJ= tan-1(0.05112.692) = 1.09°
aoG= ao + 0.5FtanopaoG = (0.033 in) + (0.5)(0.80 in)tan(1.09°) = 0.0406 in
aoP= ap + O.5F tan&:;aop = (0.112 in) + (0.5)(0.80 in)tan(2.76°) = 0.1313 in
Do = D + 2aoG cos TDo = 6.000 in + 2(0.0406 in)cos(75.96°) = 6.020 in
do=d+2aoPcosrdo = 1.500 in + 2(0.1313 in)cos(14.04°) = 1.755 in
9~
Mean whole depth
Mean addendum factor
Gear mean addendumPinion mean addendum
Gear mean dedendumPinion mean dedendum
Gear dedendum anglePinion dedendum angle
Gear outer addendum
Pinion outer addendum
Pinion outside diameter
u-'--- -------rBEVEL GEAR GEOMETRY
-~
COMPUTED VALUESGear ratiopnch diameter. Pinionpnch diameter. Gearpnch cone angle: PinionPitch cone angle: GearOuter cone distance
COMPUTED VALUESGear ratiopnch diameter: PinionPitch diameter. Gearpnch cone angle: Pinionpnch cone angle: GearOuter cone distance
4.0001.500 in6.000 in14.036 degrees75.964 degrees3.092 in
4.0000.500 in2.000 in
14.036 degrees75.964 degrees1.031 in
Nominal face width 0.928 inMaximum face width (a) 1.031 inMaximum face width (b) 0.833 in
1191i,g;r;~~~~~~:jj;4l~~g,!~~f1:'II!!mr'H;,~t,illfiii
Nominal face widthMaximum face width (a)Maximum face width (b)!!\!~ce width ti'
0.309 in0.344 in0.313 inl!l~jn
Mean cone distanceRatio Am/AoMean circular pitchmean working depthClearanceMean whole depthmean addendum factorGear mean addendumPinion mean addendumGear mean dedendumPinion mean dedendumGear dedendum anglePinion dedendum angleGear outer addendumPinion outer addendumGear outside diameterPinion outside diameter
2.692 in0.8710.228 in0.145 in0.018 in0.163 in0.2280.033 in0.112 in0.130 in0.051 in2.767 degrees1.090 degrees0.041 in0.131 in6.020 in1.755 in
Mean cone distanceRatio Am/AoMean circular pitchmean working depthClearanceMean whole depthmean addendum factorGear mean addendumPinion mean addendumGear mean dedendumPinion mean dedendumGear dedendum anglePinion dedendum angleGear outer addendumPinion outer addendumGear outside diameterPinion outside diameter
0.881 in0.8540.084 in0.053 in0.007 in0.060 in0.2280.012 in0.041 in0.048 in0.019 in3.113 degrees1.227 degrees0.015 in0.049 in2.007 in0.596 in
BEVEL GEAR GEOMETRY
'!l~;d~'~~f0~~~g,E"J"l!ilP~ftf!{'t'6;~i7},t,*!F;2GIVEN DATAfiif
&SlI~:~'COMPUTED VALUESGear ratioPitch diameter. PinionPitch diameter: GearPitch cone angle: PinionPitch cone angle: GearOuter cone distance
COMPUTED VALUESGear ratioPitch diameter: PinionPitch diameter: GearPitch cone angle: PinionPitch cone angle: GearOuter cone distance
3.0002.000 in6.000 in
18.435 degrees71.565 degrees
3.162 in
3.0000.250 in0.750 in
18.435 degrees71.565 degrees
0.395 in
Nominal face width 0.949 inMaximum face wid1h (a) 1.054 inMaximum face width (b) 1.250 in!B~lflT1!ll~iilW!i:!!h~)!filrl~W,~j];r~:l1';Q@,~fiir,~r;11l1::111)1:
Nominal face wid1h 0.119 inMaximum face width (a) 0.132 inMaximum face width (b) 0.208 in!fl!!f!~iiffi'@~IM~!li:1l!r,'i!1!1:liii19f;~~~lmiil!1111:111',i;
2.662 in0.8420.331 in0.210 in0.026 in0.237 in0.2420.051 in0.159 in0.186 in0.077 in3.992 degrees1.663 degrees0.065 in0.194 in6.041 in2.369 in
Mean cone distanceRatio Am/AoMean circular pitchmean working depthClearanceMean whole depthmean addendum factorGear mean addendumPinion mean addendumGear mean dedendumPinion mean dedendumGear dedendum anglePinion dedendum angleGear outer addendumPinion outer addendumGear outside diameterPinion outside diameter
0.333 in0.8420.055 in0.035 in0.004 in0.039 in0.2420.008 in0.027 in0.031 in0.013 in5.316 degrees2.217 degrees0.011 in0.032 in0.757 in0.311 in
Mean cone distanceRatio Am/AoMean circular pitchmean working depthClearanceMean whole depthmean addendum factorGear mean addendumPinion mean addendumGear mean dedendumPinion mean dedendumGear dedendum anglePinion dedendum angleGear outer addendumPinion outer addendumGear outside diameterPinion outside diameter
'17
Wormgearing
WOlMGFlffi/Jl6: Dw=/.lSo'N,:IAlw"/, !?J.::/p.i¢ ......./y,s,Nt; .. '10; F- o, 62S/K. -CS/N6LE" 'T1f1?G/fD
LEAl? =ANItl. I'lrc.H~ OeCdJAl.P/nJ+ - J/pcJ = %.:: tJ.YY2hv'.~/1NGtC= A= rs: k~)= r....L,"~~)-_~_.r_7·,(IDPe-NO"", -a..= Yt&. Z&=O.lO{)/N.j lJFDENO""". ~~S1:a/iS"1/N
WtJbr DllrJ'/OE" fhA •... D,w= DIN~la..:: /.l.rtJ rJ.U./I.J" /. '1SD/N.WtJil"! ROfJr OIA,'" D"w" J)w-l6 '"1.l.m-Z{P,IIS7) 6I.tJ/8'/N,
C;~1h2P/f"CHO/A, • DG.. H6k .. YOllo.::-Y.t¥Jd/N.
CeHf7T/t. OU rtf/KeF = C .: ~(f +b",,)h = f!.,Od rczs» )/2..=2.6ZS'/JI.VeuJc.lTY RA-T7o=V,= 1Y4"w1u"" -¥Ofi.:: ~
NOTE: On the following two pages are the results of Problems 52-57 giving pertinentgeometric properties of worms and wonngears and their velocity ratios. The detailedcalculations follow the pattern illustrated above for Problem 52. The equations come fromSection 8-10, Equations 8-33 10 8-38.
Compare the results to discern how variations in geometry such as diametral pitch and thenumber of threads in the worm affect the overall results. This is especially pertinenllaProblem 53 in Whichthree different designs for wOrm!wonngear sets provide the samevelocity ratio. The single threaded worm produces the smallest center distance and overall~jze of the reducer. But note, also, that it has the smallest lead angle. The lead angleIncreases as the number of threads is increased. On the positive side, the small lead anglemakes ~e redue:erself-lacking. On the negative side, the small lead angle results in lowermechanical effiCiencyas will be shown in Chapter 10, Section 10-11. The designer muslbalance these advantages and disadvantages for each application.
'18
WORMGEARING PROBLEM: 52 WORMGEARING PROBLEM: 53AINPUTDATA INPUT DATA
Worm pitch diameter = 1.250 in Worm pitch diameter = 1.000 inDiametral pitch = 10 Diametrat pitch = 12
No. of worm threads = 1 No. of worm threads = 1No. of gear teeth = 40 No. of gear teeth = 20Face width of gear = 0.625 in Face width of gear = 0.500 in
COMPUTED RESULTS COMPUTED RESULTSCircular pitch of gear = 0.3142 in Circular pitch of gear = 0.2618 inAxial pitch of worm = 0.3142 in Axial pitch of worm = 0.2618 inLead of the worm = 0.3142 in Lead of the worm = 0.2618 in
Lead angle = 4.574 deg Lead angle = 4.764 degAddendum = 0.100 in Addendum = 0.083 inDedendum = 0.116 in Dedendum = 0.096 in
Worm outside diameter = 1.450 in Worm outside diameter = 1.167 inWorm root diameter = 1.019 in Worm root diameter = 0.807 inGear pitch diameter = 4.000 in Gear pitch diameter = 1.667 in
Center distance = 2.625 in Center distance = 1.333 inVelocity ratio = 40.00 Velocity ratio = 20.00
WORMGEARING PROBLEM: 53B WORMGEARING PROBLEM: 53CINPUTDATA INPUT DATA
Worm pitch diameter = 1.000 in Worm pitch diameter = 1.000 inDlametral pitch = 12 Diametral pitch = 12
No. of worm threads = 2 No. of worm threads = 4No. of gear teeth = 40 No. of gear teeth = 80Face width of gear = 0.500 in Face width of gear = 0.500 in
COMPUTED RESULTS COMPUTED RESULTSCircular pitch of gear = 0.2618 in Circular pitch of gear = 0.2618 inAxial pitch of worm = 0.2618 in Axial pitch of worm = 0.2618 inLead of the worm = 0.5236 in Lead of the worm = 1.0472 in
Lead angle = 9.462 deg Lead angle = 18.435 degAddendum = 0.083 in Addendum = 0.083 inDedendum= 0.096 in Dedendum = 0.096 in
Worm outside diameter = 1.167 in Worm outside diameter = 1.167 inWorm root diameter = 0.807 in Worm root diameter = 0.807 inc:oearpitch diameter = 3.333 in Gear pitch diameter = 6.667 inCenter distance = 2.167 in Center distance = 3.833 inVelocity ratio = 20.00 Velocity ratio = 20.00
/00
Gear Trains - Analysis
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Gear Trains - Kinematic Design
VELOCITYRATIOFORGEARS PROBLEM 62DESIREDVR = 3.1416;: tr
NP NG NGAct VR-Act DIFF =Des VR - VR Act
16 50.27 50 3.1250 0.0165917 53.41 53 3.1176 0.0239518 56.55 57 3.1667 0.0250719 59.69 60 3.1579 0.0163020 62.83 63 3.1500 0.00841
xxi 211 65.97 166 I 3.1429 0.00126 XX22 69.12 69 3.1364 0.0052323 72.26 72 3.1304 0.0111624 75.40 75 3.1250 0.01659
Min dill = 0.00126
VELOCITYRATIOFORGEARS PROBLEM 64DESIREDVR= 6.1644 em
NP NG NGAct VR·Act DIFF =Des VR - VR Act
16 98.63 99 6.1875 0.0230917 104.80 105 6.1765 0.01206
XX 1181 110.96 1111 I 6.1667 0.00225 XX19 117.12 117 6.1579 0.0065220 123.29 123 6.1500 0.0144121 129.45 129 6.1429 0.0215622 135.62 136 6.1818 0.0174023 141.78 142 6.1739 0.00950XX 1241 147.95 1148 I 6.1667 0.00225 XX.,-",D l!(Wit\.
SDL<ln,",s Min dlff = 0.00225
VELOCITY RATIO FOR GEARS PROBLEItI63DESIRED VR = 1.7321 " v'j
NP NG NG VR DIFF=Actual Actual DesVR·VRk1
16 27.71 28 1.7500 0.0179517 29.44 29 1.7059 0.0261718 31.18 31 1.7222 0.0098319 32.91 33 1.7368 0.0047920 34.64 35 1.7500 0.0179521 36.37 3R 1 714~ 1).01m
XXI221 38.11 138 I 1.7273 0.00478 XX23 39.84 40 1.7391 0.0070824 41.57 42 1.7500 0.01795
Min dlff = 0.00478
VELOCITY RATIO FOR GEARS PROBLEItI65DESIREDVR = 7.42
NP NG NG VR DIFF=Actual Actual Des VR •VRAct
16 118.72 119 7.4375 0.0175017 126.14 126 7.4118 0.0082418 133.56 134 7.4444 O.02AU
XX 1191140.98 1141 I 7.4211 0.00105 XX
20 148.40 148 7.4000 0.0200021 155.82 156 7.4286 0.0085722 163.24 163 7.4091 0.0109123 170.66 171 7.4348 0.0147824 178.08 178 7.4167 0.00333
Min dlff" 0.00105 -
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ID3
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17ql DESIGN: mlAl='i2001¥1"I, l3,o</f1Do~/LrliSt! c.",SINI8J /lIEU'. !-JUTltIvD~1"It::EW'Z- ,...> IN f'tU&F1"I 17,LET vlt-z ~ til>. WOt?M 6~ DLln=-, Ale == /, /110" SO
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/07
CHAPTER 9SPUR GEAR DESIGN
Forces on Spur Gear Teeth
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J) C ~ '/vP f-/VG = 20 r7Z =- 3. 83.s NJz I'J 2C/Z-)
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r& =- 6300,/}>;>.. L.~c>ool7.r) = 'Ill US-//IIAlt; ¥ ~6-/
,.) INt - rp .: 270 U,/,N :. 3.2Y LISD,/Z /.6.t7"N/7
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A S/M/tAtt. nC"r'HoO /s UJcO FUR ?tI('I!(..~J 2-6.
5/'tJ2c/t05..11€Er 5OLu17IM/J AR.E SHO",uJ ON :ntE' P/JU4W1N0
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loa
Ij----'Ff';;o:;:rc:::e::s:-:o::n:-:S;:p::u~r:-:G:;:e::a:':r~T~e~et::h:-----
RESULTS:a Gear speed =b VR=mG=c pinion PO =
gear PO =d center distance = C =e pilch line speed =f torque on pinion shaft =
torque on gear shaft =g tangential force =h radial force =I normal force =
Problem 2Chapter 9
486.13.6001.6676.0003.833764270972324118345
rpm
inininfIIminIb inIb inIbIbIb
RESULTS:a Gear speed = 304.4 rpmb VR=mG= 3.778c pinion PO = 3.600 in
gear PO = 13.600 ind center distance = C = 8.600 ine pitch line speed = 1084 fIIminf torque on pinion shaft = 2739 Ib in
torque on gear shaft = 10348 Ib ing tangential force = 1522 Ibh radial force = 5541bI nonnal force = 1620lb
/0'1
Forces on Spur Gear Teeth
Problem 3 "Chapter 9
RESULTS:a Gear speed = 752.7 rpm
b VR=mG= 4.583
c pinion PO = 1.000 ingear PO = 4.583 in
d center distance = C = 2.792 ine pitch line speed = 903 ftIminf torque on pinion shaft = 13.70 Ib in
torque on gear shaft = 62.77 Ibing tangential force = 27.40 Ibh radial force = 9.97 IbI normal force = 29.16 Ib
Problem 4Chapter 9
RESULTS:a Gear speed = 486.1 rpmb VR=mG= 3.600c pinion PO = 1.667 in
gear PO = 6.000 ind center distance = C = 3.833 ine pitch line speed = 764 ftImin
torque on pinion shaft = 270lbintorque on gear shaft = 972 Ib in
g tangential force = 3241bh radial force = 151 IbI normal force = 3581b
//0
Forces on Spur Gear Teeth
Problem 5. Chapler9.
RESULTS:a Gear speed = 304.4 rpm
b VR= mG= 3.778
c pinion PO = 3.600 ingear PO = 13.600 in
d center distance = C = 8.600 ine pitch line speed = 1084 ftIminf torque on pinion shaft = 2739 Ibin
torque on gear shaft = 10348 Ibing tangential force = 1522 Ib
h radial force = 710 IbI normal force = 1680 Ib
RESULTS:a ·Gearspeed=b VR=mG=c pinion PO =
gear PO =d center distance = C =e pitch line speed =
torque on pinion shaft =torque on gear shaft =
g tangential force :h radial force =I normal force =
752.7 'rpm4.5831.000 in4.583 in2.792 in903 ftImin
13.70 Ib in62.77 Ib in27.401b12.781b30.241b
ILl
Gear Manufacture and Quality
7. See Section 9-4. Fonn milling, shaping, and hobbing.
For Problems 8 - 16 refer to Table 9-2 for recommended quality numbers. Some judgment is requiredbased on more detailed knowledge of the application. The total composite tolerance (fCT) is estimatedfrom Table 9-1. Rough interpolation has been used. If more precise data are required, refer to AGMAStandard 2000-88 Gear Classification and Inspection Handbook.
8. Grain harvester. Use Q = 5. For Pd = 8, Np = 40, No = 100: TCT p = 0.0130 in; TCT 0 = 0.0150 in.9. Printing press: Q = 10. For Pd = 20, Np = 40, No = 100: TCT p = 0.0015 in; TCT G = 0.0017 in.10. Auto transmission: Q = 10. For Pd = 8, Np= 40, No = 100: TCTp = 0.0047 in; TCT 0 = 0.0055 in.11. Gyroscope: Q = 14. For Pd = 32, Np = 40, No = 100: TCT p = 0.00031 in; TCT 0 = 0.00035 in.
12. Using Q = 10 as compared with Q = 5, with all other data being equal, requires tooth accuracyapproximately three times greater.
13. Quality number is increasing because of the application. The size of the teeth is also decreasingbecause of the increasing numerical value of the diametral pitch. This results in a dramatic increaseof the accuracy required in the tooth fonn as represented by the TCT values. For the pinion the TeTmoves from 0.0130 in to 0.0047 in to 0.00031 in. Similar reductions for the gear.
For Problems 14, 15, and 16, use the lower part of Table 9-2. For precision machine tool, the qualitynumber choice is related to the pitch line speed of the gears, with a higher quality number for fasterspeeds because the dynamic effects become more pronounced. See solutions for Problems 1,2, and3for pitch line speeds.
14. Pitch line speed = 764 fIImin (From Prob. 1) Use Q = 8. For Pd = 12, Np = 20, No = 72: TCTp =0.0037 in; TCT0 = 0.0044 in. (Rough interpolation)
15. Pitch line speed = 1064 fIImin (From Prob. 2) Use Q = 10. For Pd = 5, Np = 18, No = 68: TCTp =0.0037 in; TCT G = 0.0039 in. (Rough interpolation)
16. Pitch line speed = 764 fIImin (From Prob. 3) Use Q = 10. For Pd = 24, Np = 24, No = 110: TeT p=0.0013 In; TCT 0 = 0.0016 in. (Rough interpolation)
Gear Materials
Answers for Problems 17 - 25 are found in Section 9-7. Only brief statements are given here.
17. Bending stre~s are created by the tang~ntial force on the gear teeth acting in a manner similar toth.at on a.cantilever. The maximum bending stress occurs in the root of the tooth where it blendswith the mvolute tooth fonn. High levels of contact stress (sometimes called Hertz stress) occur inthe face of the t~eth near the pitch line as forces are exerted between the pinion and the gear teeth.The probable failure mode is pitting of the tooth surface.
18. ~e~:t:;t~h=~~drdanrdeSssglVo'fethallOWatbl~bending stress numbers and allowable contact stress numberse ma enal of the teeth.
19. Gear steels are typically medium carbon pi . IIhardening using a quenching d t . am or a oy steels that are heat treated by through-3140 4140 4340 6150 t an empenng process. Examples are given in Section 9-7. 1040,
I I • • e c.
20. The AGMA recommends hardnesses from HB 180 to HB 400.21. Grade 1 steel is typical commercial qualit G d' .
quality controls on the alloy content and y. ra es 2 and 3 require progressively more stnngenthigher grades. See also AGMA Sta d rdcle2aOnlinessof the materials. Costs also increase for the
n a 04-B89 Gear Materials and Heat Treatment.
II~
22. Grades 2 and 3 may be specified for high-speed aerospace applications, turbine engine drivensystems, ship propulsion drives, and high-capacity industrial gears such as those in steel rolling mills.
23. Flame or induction hardening, carburizing, and nitriding.
24. AGMA Standard 2001-C95 (or latest revision).
25. Gray cast iron, ductile (nodular) iron, and bronze. See Table 9-4.26. Allowable bending stress numbers are obtained from Figure 9-10 or by computing values from the
equations given in the figure. Use Figure 9-/1 for allowable contact stress number.
a) Grade 1, 200 HB: Sat = 28 200 psi; Sao=93 500 psi
b) Grade 1, 300 HB: Sat= 26 000 psi; Sac=125 700 psi
c} Grade 1, 400 HB: Sat = 23 700 psi; Sac=157 900 psi
d) Grade 1, 450 HB: Using HB > 400 in not recommended.
e) Grade 2, 200 HB: Sat = 36 800 psi; Sac=104 100 psi
f) Grade 2,300 HB: Sat = 47 000 psi; Sac=139 000 psi
g) Grade 2, 400 HB: Sat= 57 200 psi; Sac=173 900 psi
27. Grade 1, 300 HB; Grade 2,192 HB from Figure 9-/0.
28. 55-64 HRC. See Table 9-3.29. From Appendix 5: AIS11020, AISI 4118, AISI 8620 as examples.
30. 50-54 HRC with materials having good hardenability. See Tabte 9-3.
31. AIS14140. AISI 4340, AISI 6150. All must have good hardenability.
32. 83.5 on the HR15N hardness scale used for thin-case materials.
33. Data are found in Tables 9-3, 9-4, and 9-11,a} Sat= 45 000 psi; sao=170 000 psi
b} Sat= 45 000 psi; Sao=175 000 psi
c} Sat = 55 000 psi; Sac=180 000 psi
d} Sat= 65 000 psi; Sac=225 000 psi
e) Sat = 55 000 psi; Sao= 180000 psi
f) Sat= 39 000 psi; sao=155 000 psi
g} Sat = 51 500 psi; Sac=168000 psi
h} Sat = 63 500 psi; Sac=216 000 psi
i} Sat = 5 000 psi; sao=50 000 psi
j} Sat = 13 000 psi; Sao=75 000 psi
k} Sat = 27 000 psi; sao=92 000 psi
I} Sat = 5 700 psi; sao=30 000 psi
m} Sat = 23 600 psi; sao=65 000 psi
n} Sat = 12000 psi; Notlisted
0) Sal= 9 000 psi; Not listed
34. Depth = 0.027 in. Figure 9-'1:1..35. Depth = 0.90 mm. Figure 9- JJ...
1/3
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