_EMPTY_

^o^

_EMPTY_

1'

MECHANICSOF MATERIALSByGLENN MURPHY, C.E., PH.D.Professor of Theoretical and Applied MechanicsIowa State CollegeCQ/10

IRWIN-FARNHAM PUBLISHING COMPANYChicago, Illinois1948

1A4C5MCopyright 1948byIRWIN-FARXHAM PUBLISHING COMPANYPRINTED IN THE UNITED STATES OF AMERICAAll rights reserved. This book or any part thereofmay not be reproduced in any form withoutwritten permission of the publishers.

'(t'//-PREFACE,The purpose of a textbook in mechanics of materials is to assist thestudent in his development of an understanding of the behavior under loadof structural members and machine parts constructed from the commonengineering materials. To be of tangible assistance, such a textbook shoulddirect his attention to those principles which have been found useful inexplaining observed phenomena, it should acquaint him with standardprocedures of analysis in order that he may readily understand the bulk of^ the engineering literature on the subject, it should provide him with a setof problems by means of which he may test his understanding of the subject,and it should serve to make him aware that our knowledge of materials ise> not final and absolute but is constantly growingthat what we know aboutmechanics of materials is a tool which, when intelligently used, is of in-estimable value in the daily tasks of engineering.With these several objectives, this book has been designed as a text-booknot as an encyclopedia. The principles of statics, the characteristicsof the geometry of the loaded member, and the effects of the properties ofthe material have been emphasized in the consideration of each type ofstress situation. The statics, the geometry, and the properties of the ma-terial have been used as the starting point in the development of eachphase of the work, for it is the author's conviction that only through anunderstanding of the relative importance of each does the student obtainperspective of the problems of stress analysis.Discussion is included of those topics which normally comprise the firstcourse in mechanics of materials, strength of materials, or resistance ofmaterials, and the customary formulas have been developed. It is to behoped that a minimum of emphasis will be placed on the formulas as such.Those which the author considers important are printed in boldface type.At appropriate intervals some elementary aspects of design applicationshave been included to help the student in his evaluation of the analysis inits relationship to practical engineering problems.A rather generous set of drill problems has been included. Some of theproblems are very simple, many require more than a superficial under-standing of the subject, and the others require a reasonable degree ofmastery of the text material. The problems are grouped at the ends of theirrespective chapters. Some students may find this practice inconvenient,especially if they are accustomed to books in which the problems aredistributed throughout the text in such a way that any one may be solvedby plugging some numbers into the formula in the preceding paragraph.However, it is the author's observation that those practical engineering

vi PREFACEproblems which he is asked to solve are seldom immediately preceded bythe appropriate formula in italics or boldface type.Answers are given to practically all of the even-numbered problems. Itis hoped that this device will be useful not only to those instructors whowish the student to have answers available as he works the problems, butalso to those instructors who feel that the student should be encouraged todevelop confidence in his own results as is essential in engineering practice.Many individuals have contributed directly or indirectly to the manu-script, and to them the author expresses his appreciation. Particular creditis due to Professor E. H. Ohlsen of Iowa State College for his helpfulcriticism and suggestions, and to Professor H. J. Gilkey, Head of the De-partment of Theoretical and Applied Mechanics at Iowa State College, formany stimulating discussions concerning the general field of Mechanics ofMaterials. Several of the author's colleagues have contributed helpful sug-gestions, particularly in the form of problems.The author is deeply indebted to Frances Murphy for her care in thepreparation of the finished drawings for the figures. Credit is due to SusanBarker for her assistance in the preparation of the manuscript.GLENN MURPHYAMES, IOWAFebruary 1948

TABLE OF CONTENTSCHAPTER 1. STRESS, STRAIN, AND AXIAL LOADINGARTICLE1. Introduction .......2. Objectives of Stress Analysis ....3. Methods of Stress Analysis ....4. Classification of Load-carrying Members .5. Definition of Stress......6. Uniform Stress Distribution and Axial Loading .7. Stress Concentration under Axial Loading .8. Stresses on Inclined Planes of Axially Loaded Members9. Strain Due to Axial Loading ....10. Shearing Strain ......11. Strain Due to Temperature Changes12. Stress-Strain Relationships ....13. Elastic Action ......14. Inelastic Action and Failure ....15. Allowable Working Stress and Factor of Safety .16. Statically Indeterminate Axially Loaded Members17. Thin-walled Pressure Vessels ....18. Dynamic and Repeated LoadingPAGE11235691114161717222326262830CHAPTER 2. JOINTS AND CONNECTIONS FOR AXIALLY LOADED MEMBERS19. General Considerations .20. Types of Welds21. Allowable Stresses in Welds22. Design Considerations23. Types of Riveted Joints .24. Stresses in Lap Joints25. Stresses in Butt Joints .26. Efficiency of a Riveted Joint27. Other Design Procedures28. Design Considerations29. End Fittings .4949515155566467687071CHAPTER 3. TORSION30. General Considerations . . . .31. Geometry of a Circular Torsional Member32. Shearing Stresses in a Circular Torsional Member33. Power Transmission by Torsional Members34. Stresses on Inclined Planes35. Shearing Stresses in Noncircular Sections36. Stress Concentration37. Couplings and Riveted or Bolted Fittings38. Thin-walled Members in Torsion39. Localized Compression or Buckling .40. Stresses beyond the Proportional Limit41. Helical Springs ....42. Statically Indeterminate Composite Torsion Members43. Design Considerations ......8181838588899295959699100CHAPTER 4. STRESSES IN FLEXURAL MEMBERS44. Types of Flexural Members ......44. Types of Flexural45. Flexural Stresses113115

viii TABLE OF CONTENTSARTICLE PAGI,46. Limitations of the Flexure Formula ........ 12147. Shear Diagrams ........... 12348. Moment Diagrams. . . . . . . . . . .12749. Shearing Stress 13150. Stresses beyond the Proportional Limit ....... 13651. Stress Concentration .......... 13752. Beams of Two Materials. ......... 13953. Design Considerations .......... 141CHAPTER 5. DEFLECTION OP FLEXUBAL MEMBERS54. Introduction ............ 16355. Fundamental Geometrical Relationships in a Bent Flexural Member . . 163Double Integration56. General Integration Procedure. . . . . . . . . 165Area Moments57. General Relationships .......... 16958. Application of Theorems. . . . . . . . . .17359. Deflections by Superposition . . . . . . . . .17660. Deflections Caused by Shear . . . . . . . . .17861. Design Considerations .......... 179CHAPTER 6. STATICALLY INDETERMINATE BEAMS^-62. Introduction . 18763. Solution of Statically Indeterminate Beams ...... 18764. Double Integration . . . . . . . . . .18765. Area Moments ........... 19266. Superposition of Structures . . . . . . . . .19667. Theorem of Three Moments 199CHAPTER 7. COLUMNS68. Introduction 20969. Euler Column Theory 21070. Alternate Solution of Differential Equation 21371. Intermediate Columns . . . . . . . . . .21472. Effect of End Conditions 21973. Eccentric Loads on Compression Blocks ....... 22274. Eccentric Loads on Columns ......... 22575. The Secant Formula 22776. Double-Modulus Formulas 23077. Design Considerations .......... 231CHAPTER 8. COMBINED LOADINGS78. Introduction 24379. Principal Stresses 24380. Maximum Shearing Stress ......... 24681. The Mohr Circle 25082. Diagonal Tension 25183. Principal Strains 25384. Theories of Elastic Failure 255CHAPTER 9. DYNAMIC AND REPEATED LOADS85. Importance of Nonstatic Loads ........ 269Dynamic Loads86. Definition 26987. Basis of Evaluation of Equivalent Static Loads...... 270

TABLE OF CONTENTSIXARTICLE88. Axial Dynamic Loads ....89. Torsional Impact Loads ....90. Flexural Impact Loads ....91. Design Considerations for Dynamic LoadsRepeated Loads92. Definition93. Effect of Repeated Loading94. Criterion of Resistance .95. Mechanism of Failure96. Fluctuating Stresses97. Stress ConcentrationPAGE. 271. 273. 274275277278278280281282APPENDIX A. PROPERTIES OF SECTIONSTABLEI. Properties of l-BearasII. Properties of Channels .III. Properties of Equal AnglesIV. Properties of Unequal Angles .295296297298APPENDIX B. ANSWERS TO PROBLEMSAnswers to Problems299INDEXIndex307

_EMPTY_

MECHANICS OF MATERIALS

_EMPTY_

CHAPTER 1STRESS, STRAIN, AND AXIAL LOADING1. Introduction.One important phase of engineering activity is design,and it is one of the most involved, for the development of a good design for astructure or machine requires the integration of information from a varietyof sources. Careful consideration must be given to the function of the com-pleted product; the optimum arrangement of the members must be de-termined; and the most suitable material selected, considering factors suchas availability, workability, cost, durability, and appearance. From anestimate of the loads to which the structure will be subjected, the mosteffective shape and size of the individual members must be determined,and appropriate means of connecting them must be devised. Considerationshould be given to possible manufacturing processes in order that effectiveand economical fabrication processes may be specified, and usually the ap-pearance of the finished product is important.The success of the design is, in the last analysis, largely dependent uponthe judgment and experience of the engineer responsible for the details, butmost of the actual calculations involved in the design are based upon well-established analytical and experimental procedures. These procedures areset up to assist the engineer in important determinations such as ascertain-ing the optimum size for the individual members of the structure or themachine. The field of knowledge comprising the techniques of establishingrelationships among load, size, and shape of members is known as stressanalysis.2. Objectives of Stress Analysis.The objectives of stress analysis areto determine the ways in which individual structural members resist loadand to provide the methods by which the engineer responsible for theirdesign is enabled to answer with confidence the two critical questions:(1) Is the proposed member strong enough but not too strong?(2) Is the proposed member stiff enough without being too stiff?These questions may be rephrased in the form of certain type problems:(1) Given a member of specified dimensions and specified material, findthe maximum load that it will carry before it fails by breaking or by de-forming too much.(2) Given a member of specified dimensions and shape which is to carry aspecified load, find the material which will be best adapted to meet the1

MECHANICS OF MATERIALSChap. 1given requirements and to avoid failure by breaking or by excessivedeformation.(3) Given limiting dimensions of a member which is to carry a specifiedload, find the material which will be best adapted to meet the given re-quirements and to avoid failure by breaking or by excessive deformation.The first of these problems is a problem of investigation, and the othertwo are design problems. In each, the relative arrangement of the memberswithin the structure is assumed to be established by other considerations.3. Methods of Stress Analysis.Both analytical and experimentalmethods of stress analysis are used. The analytical procedures are based on(1) the laws of motion (or equilibrium),(2) verified observations of the characteristics of materials, and

Fig. 1(3) assumptions regarding the behavior of members under load, as justi-fied on the basis of results obtained from their use.After the loads which the structure must carry have been estimated, theforce which each member must carry or resist is evaluated with the aid offree-body diagrams and the laws of motion, as formulated in analyticalmechanics (Statics and Dynamics). Analytical mechanics, together withinformation obtained from observations and measurements, is used toestablish the relationships among loads, dimensions, and material for theindividual members of the structure. In the design of a truss, for example,analytical mechanics is used to determine the forces which the loads developin each member, while stress analysis is used to determine the distribution ofthe forces within individual members. A number of experimental proce-

Art. 4STRESS, STRAIN, AND AXIAL LOADINGdures have been developed for assistance in the analysis of stresses.** Theyconsist essentially in measurements of strain.4. Classification of Load-carrying Members.It is obvious that themaximum load which a given member of a given material will carry is de-pendent upon the orientation of the load with respect to the member andupon the way in which the member is supported. For example, a block ofwood 2 in. by 4 in. in cross section and 18 in. long will carry much more loadif the load is directed along the axis of the member than it will if the memberis supported at each end and the load is applied normal to the longitudinalaxis.Fortunately, any loading (which constitutes a force system) may be re-solved into a series of relatively simple component loadings, each of which

(a)

(CJFig. 2may be considered separately. Their effects may be added in most cases.For example, the resultant force 72 acting at any cross section in a member(such as at section A A of the airplane landing gear unit in Fig. 1) may be re-solved into six componentsthe three forces Fx, Fv, and Fx passing throughthe centroid of the cross section and directed along three orthogonal axes,and three couples or moments Mx, Mv, and M, which lie in the threeorthogonal planes. For convenience, the yz-plane is taken as the plane of thecross section. Similarly, the resultant force acting at another cross sectionBB (near AA} may be resolved into a set of three forces F'x, F'v, and F',which are parallel to the three forces at section A A, and three moments* Murphy, Glenn, "Strain-Measuring Equipment," Advanced Mechanics of Materials(McGraw-Hill Book Co., Inc., 1946), p. 49.t Gilkey, Murphy, and Bergman, "Experimental Aids in Stress Analysis," MaterialsTesting (McGraw-Hill Book Co., Inc., 1941), Chap. 12.

4 MECHANICS OF MATERIALS Chap. 1M'x, M'v, and M'x which lie in planes parallel to the planes of the threemoments of section A A.The two forces Fx and F'x (which are colinear) may be considered toconstitute one component force system acting upon the portion A B of theoriginal member. It is evident that this force system, shown in Fig. 2(a),produces a state of compression on planes parallel to A A and BB in theportion of the member between A A and BB, i.e., the forces tend to shortenthe distance AB and squeeze together adjacent parallel normal planes.The pair of forces Fv and F'v, Fig. 2(6), as well as the pair Fx and F'x,tends to produce what is known as a state of shear between A and B. Thatis, the forces tend to make adjacent parallel planes slide across one another.The moments Mx and M'x, which lie in the planes of the cut sections,tend to twist the portion of the member between AA and BB, Fig. 2(c),while the pair of moments Mv and M'v tends to bend the member, Fig. 2(d).Similarly, the pair of moments Mx and M'x tends to produce bending. Thus,the six possible components tend to develop four possible types of distortionbetween sections AA and BB, which are typical of adjacent cross sectionsin any member. The four component types of loading producing the fourtypes of distortion are:(1) Axial Loading.This is produced by a pair of colinear forces actingalong the longitudinal axis of the section. If the forces are directed towardeach other, a state of compression is developed within the member, while ifthe forces are directed away from each other, a state of tension is developed.Axial loading occurs in members such as struts, tie rods, connecting rods ofengines, bridge trusses, building trusses, and airplane trusses.It is evident that if the loads are not axial, even though they are colinear,bending will be developed in addition to direct compression or tension. Ifthe member is long and slender and loaded in compression, bending orbuckling will result.(2) Shear.A state of shear is developed by forces which lie in the planesof the cross sections. Forces of this type are developed in rivets, bolts, andpins, and in most beams. Shear is usually accompanied by bending.(3) Torsion.A member which is twisted or tends to be twisted by theaction of a pair of couples lying in planes perpendicular to the axis is saidto be subjected to torsion. This action is typical of the loading conditiondeveloped in shafts and many similar machine parts. The resistance ofcircular members to torsion is discussed in Chap. 3.(4) Flexure.A state of flexure or bending is developed when the mem-ber is acted upon by a pair of couples (called bending moment) which lie inplanes perpendicular to the cross section. The bending moment producestension on one side of the member and compression on the other side. Thistype of loading is discussed in Chap. 4. While some bending is present inpractically all load-carrying members, it is the predominant factor in beamsand long slender compression members.

Art. 5 STRESS, STRAIN, AND AXIAL LOADING 5Only rarely does one of these four types of loading exist alone. Practicallyall structural members or machine parts are loaded and supported in such away that they are subjected throughout at least a part of their length totwo or more of the component types of loading. In most cases, however,each of the components may be treated independently of the others andthen- effects added. Exceptions will be considered later.5. Definition of Stress. The effect of an external load on a membermay be indicated as tension, compression, shear, bending, torsion, or combi-nations of these as described in Art. 4. However, in addition to this qualita-tive description, a quantitative measure of the intensity of the effect producedby the loading is essential for design purposes. It is obvious that an axialload of 1000 Ib, for example, may have a different effect upon a 1/2-in.diam rod than upon a 2-in. diam rod of the same material.A numerical index of the intensity of effect at any point on any plane isgiven by the unit stress,* denned as the magnitude of the force per unitarea developed at the point on the plane. It is evaluated asin which S is the unit stress, anddF is the differential force developed on the differential area dA.Differential force and differential area are indicated since the unit stressusually varies over the plane under consideration. From the definition it isapparent that unit stress has the dimensions of force per unit area. It iscommonly expressed in pounds per square inch, abbreviated as psi., or maybe expressed in kips (thousands of pounds) per square inch, abbreviated asksi.Qualitatively, the unit stress may be normal stress (tension or compres-sion) or shearing stress, depending upon whether the differential force act-ing on the given differential area is normal to the area or in the plane of thearea. In general, on planes perpendicular to the axis of the member normalstresses are produced by axial loading and by flexure, while shearing stressesare produced by shear and torsion.Maximum allowable values of unit stress have been established by manyagencies for most engineering materials under various conditions of usage,and a few of them are listed in Table 1.The evaluation of the unit stress which a given load causes at a givenpoint on a given plane in a structural member involves first the determina-* The expression "unit stress" is frequently abbreviated in technical literature to"stress." However, the term "stress" is also used in the literature (particularly in CivilEngineering practice) to denote the total axial force which a member carries. Hence,some confusion may arise unless units are specified. In this text, the terms "stress" and"unit stress" will be used interchangeably to denote intensity of loading expressed asforce per unit area.

MECHANICS OF MATERIALSChap. 1TABLE 1ALLOWABLE WORKING STRESSES FOR A FEW ENGINEERING MATERIALS*ALLOWABLESTHESS (PSI.)TensionShearCompressionFlexureWrought iron12,00080001500012 000Structural steel18,00010,0001400018 000Gray cast iron15,0003,00020,00015000Aluminum alloy 24S-T18,000100001400018 000Concrete f0607501350{White oak180016711331866Douglas fir180012013861800* Values are approximate and are based on average conditions for members without structural defects,t All values established as a fixed proportion of the ultimate compressive strength. Values listed are forstandard 3000-psi. concrete,t Reinforced concrete.tion of the total force which the load causes to be transmitted across thegiven plane; and second, the distribution of force (or the variation in stress)throughout the plane. The first step is usually accomplished by the use ofstatics or dynamics, utilizing a free-body diagram, while the determinationof the distribution of the stress is a problem in stress analysis. Stressdistribution under conditions of axial loading will be discussed in the follow-ing articles, and stress distribution for other types of loading are consideredin subsequent chapters.6. Uniform Stress Distribution and Axial Loading.If a vertical axialload W is applied to the top of a short vertical compressive member asindicated in Fig. 3(a), it will produce stress throughout the member. The

/////////(a)(e)Fig. 3magnitude of the average unit stress on any horizontal plane such as A Amay be investigated by first constructing a free-body diagram of one por-tion of the block bounded by the plane A A. One of the two possible por-tions is shown in Fig. 3(6) and the corresponding free-body diagram is indi-cated in Fig. 3(c). It is evident that a force F must be developed at the cutsection if the upper portion of the block is to remain in equilibrium andthat the magnitude of F must be equal to W plus the weight of the portionof the block above the section A A. If the weight of the block is negligible

Art. 6 STRESS, STRAIN, AND AXIAL LOADING 7in comparison with the applied load, F will equal W and the force F mustbe colinear with W.However, the force developed at the section A A is not concentrated at apoint as indicated in Fig. 3(c) but is distributed over the plane A A. Undercertain conditions the unit stress will be distributed approximately uni-formly and under other conditions it may vary greatly from one point toanother. It is unlikely that the stress is ever actually uniformly distributedin a structural member or machine part; but, if the material is homo-geneous, the stress may be distributed more nearly uniformly than if it isnonhomogeneous. For example, the stress would probably be distributedmore nearly uniformly in a block of fine-grained steel than in a materiallike wood which contains inequalities due to grain, variations in growth,and natural defects such as pitch pockets or knots.The loading conditions necessary for a uniform distribution of stressmay be established by assuming the unit stress to be distributed uniformly.The differential force acting on each differential area in the cross sectionmay be evaluated by Eq. (1). Since these differential forces form a parallelforce system, their resultant, the total force developed on the cross section,may be determined by direct addition of the differential forces, that is, byintegrating Eq. (1).r rAI dF = I SdA. (2)Jo JoSince the unit stress is assumed to be uniformly distributed over thecross section, the term S may be taken outside of the integral sign. ThenEq. (2) integrates toF = SA. (2a)The location of the resultant force developed at the cut section may bedetermined by the application of the Principle of Moments.* The momentequation written with reference to any convenient y-axia isLS x dA. (3)0Since the unit stress S is assumed to be constant, it may again be takenoutside of the integral sign, and F may be replaced by SA, from Eq. (2a),givingrKI = S I x dA.JoSAx1 = S I xdA. (3a)* The moment of the resultant force developed on the cross section is equal to the sumof the moments of the individual differential forces with respect to the same axis.

MECHANICS OF MATERIALS Chap. 1However, by definition of the centroid of an area,1xdA = xA. (4)0Hence,SAxi = SxA, (3b)andxl = x. (3c)That is, the line of action of the resultant must have the same x-coordi-nate as the centroid of the area. A similar moment equation, written withrespect to a convenient x-axis, will show that the ^-coordinate of the line ofaction of the resultant must be equal to the ^-coordinate of the centroidof the area. Therefore, the line of action of the resultant force developedon the cross section must pass through the centroid of the cross section.Since the applied load is colinear with the force F, it also must lie on theline passing through the centroid of the cross section and must, therefore,be an axial load. Thus, a necessary condition for the development of a uni-form stress distribution on any cross section is that the applied load actalong the line forming the loci of the centroids of the cross sections, i.e.,the applied load must be axial.* While this is a necessary condition for thedevelopment of a uniform stress distribution, it is not a sufficient condition.A nonuniform stress distribution symmetrical with respect to the longi-tudinal axis (such as that shown in Fig. 4) will satisfy Eq. (3b). This con-dition may result from nonuniformities in the material or it may be broughtabout by an abrupt change in the cross section of the member.If the line of action of the resultant does not pass through the centroidof the cross section, the resultant may be resolved into a force through thecentroid plus a couple. The force will develop axial loading, whereas thecouple will produce bending, and the stress will not be uniformly distributedacross the cross section.A procedure similar to that used in developing Eqs. (2a) and (3b) may befollowed for shearing stresses with the result that, if the shearing stress is tobe uniformly distributed over the area, the line of action of the resultantshearing force must pass through the centroid of .the area. The magnitude ofthe average shearing unit stress isS. = f, (2b)in whichQ is the magnitude of the shearing force.* This assumes that the member is straight. If the member is curved, the stress at agiven cross section cannot be uniformly distributed unless the line of action of the ap-plied load passes through the centroid of that cross section.

Art. 7STRESS, STRAIN, AND AXIAL LOADINGIn most situations, as will be shown later, shearing stress is not uniformlydistributed over the area resisting the shearing force. Hence Eq. (2b) isconsidered to give the average and not the maximum unit shearing stress.7. Stress Concentration tinder Axial Loading.If an axially loadedmember contains a discontinuity such as a hole or a pair of symmetricalnotches as indicated in Fig. 4, the stress will not be distributed uniformlyo

(a) U>) tc) Id)Fig. 4. Nonuniform stress distribution in axially loaded members.across the minimum cross section or adjacent cross sections but will begreater near the edges of the discontinuity. This phenomenon is known asstress concentration. A quantitative measure of the extent of the stress con-centration is given by the stress concentration factor which, for axialloading, is denned as the ratio of the maximum stress developed by thediscontinuity to the average stress in the gross cross section.*K(5)Some typical values of stress concentration factors are indicated in Fig. 5.It will be noted that the magnitude of the stress concentration factor is de-pendent upon the abruptness or sharpness of the discontinuity. For somediscontinuities, the stress concentration factor may be evaluated mathe-matically, while, for other shapes, experimental techniques are used todetermine the distribution of stress near the discontinuity.Stress concentration also occurs in beams, shafts, and other nonaxiallyloaded members; and values of the concentration factor for those conditionsare indicated in later chapters. The importance of the stress concentrationfactor depends upon the type of loading and the characteristics of the ma-* The stress concentration factor may also be denned as the ratio of the maximumstress to the average stress on the net cross section. In using values of the stress concen-tration factor, it is important that the engineer know which definition of stress concentra-tion factor was used before he employs values of the factor.

10Chap. 1MECHANICS OF MATERIALS/o"?I'IrI_rLower Curves- Jo/id LinesBroken .ines 05

Dotted Lines 77"O O./ O.2 0.3 O.4 O.S 0.6 O.7 O.8Fig. 5. Stress concentration factors for axially loaded members.terial. The concentration factors as given are for static loads. Other typesof loading (impact and repeated loads) may result in different values of thefactors.

Art. 811STRESS, STRAIN, AND AXIAL LOADINGIn general, the phenomenon of stress concentration may be ignored in thedesign of members of ductile materials subjected to static loads, but it ishighly significant in members made of brittle materials, or members sub-jected to impact and repeated loads. Reasons for this will be pointed outlater.8. Stresses on Inclined Planes of Axially Loaded Members.Usuallystresses will be developed on all planes in a loaded member. The techniqueof evaluating the stress on a plane perpendicular to the axis of an axiallyloaded member is indicated in Fig. 3 and Art. 6. The method for evaluatingthe stresses on an inclined plane follows the same general procedure. Ifthe stress is to be evaluated on an inclined plane such as A A in Fig. 6(a),

la)ic)Fig. 6the first step is to construct a free-body diagram of one portion of the mem-ber bounded by the plane on which the stress is desired. One suitable free-body diagram is indicated in Fig. 6(6). The weight of the material is as-sumed to be negligible. The load P on the top of the member is balancedby an equal and opposite force F on the inclined plane. However, this forceF is not normal to the inclined plane nor is it parallel to the inclined plane.Hence, to classify its effect qualitatively as tension, compression, or shear,it must be resolved into two components, N and Q, which are perpendicularand parallel respectively to the inclined plane as indicated in Fig. 6(c). Ifthe plane is assumed to make an angle 6 with the horizontal, the com-ponents N and Q may be evaluated from a force triangle or from the free-body diagram asN = F cos 6, (4a)andQ = F sin(4b)If the resultant of N and Q passes through the centroid of the area whichthe inclined plane delineates on the block and if no discontinuities arepresent, the normal and shearing stresses may be assumed to be uniformlydistributed and may be evaluated as

12 MECHANICS OF MATERIALS Chap. 1Sn = ^, (4c)andS. = -> (4d)respectively. In each equation A is the area of the inclined plane which mayreadily be determined from the original cross-sectional area of the specimenand the angle 6.The forces N and Q are given by Eqs. (4a) and (4b), in which the term Fis equal to the applied load P if the block is in equilibrium.Illustrative ProblemA block of concrete 8 in. high and 4 in. by 4 in. in cross section is subjected to anaxial compressive load of 16,000 Ib. Determine the stress on a plane which is per-pendicular to one pair of faces and which makes an angle of 60 with the horizontal.Solution: The free-body diagram of Fig. 6(c) may be used and the forces N and Qdetermined as:N = 16,000 (0.500) (a)= 8,000 Ib compression,andQ = 16,000 (0.866) (b)= 13,860 Ib shear.The area of the inclined plane is= 32 sq in.The normal stress is, therefore,9 8,000 S. = -32" (d)= 250 psi. compression,and the shearing stress is13,860S- = -32- (6)= 433 psi. shear.The direction of the shearing stress and the fact that the normal stress is compres-sion may be determined from the free-body diagram.It may be shown that in an axially loaded member the maximum normalstress at a point occurs on the plane perpendicular to the direction of theload and that the maximum shearing stress at a point occurs on the planewhich makes an angle of 45 with the direction of the applied load. Themagnitude of the maximum shearing stress is equal to one half of the magni-tude of the maximum normal stress.The fact that shearing stress is developed on inclined planes in axially^"^"^ members is important because many materials are relatively weak in

Art. 813STRESS, STRAIN, AND AXIAL LOADINGshear and will fail along an inclined plane at a lower load than would causefailure on the plane at right angles to the direction of the applied load. Ex-amples are shown in Fig. 7.

Fig. 7. Shearing failure of wood loaded in axial compression.It is apparent that the normal and shearing stresses on a plane whichmakes an angle ( 6) with the horizontal in Fig. Q(d) will be equal in magni-tude to the normal and shearing stresses on the plane investigated in Fig.Ss ay ail1 Ss dx dz/ P' ft *tgJ> /jr fifcFig. 8

14Chap. 1MECHANICS OF MATERIALS6(c). It may also be shown that the shearing stress on a plane which makesan angle of 90 with AA will be equal to the shearing stress on AA. Thismay be proved by considering the equilibrium of the differential element inFig. 8. The two horizontal components of shearing stress are equal sincethey are on adjacent parallel planes, and the two vertical components ofshearing stress are equal for the same reason. Hence, the two horizontalcomponents of shearing forces form a couple, as do the two vertical com-ponents of shearing forces. The equation of moments with respect to an axisthrough the centroid of the element givesSs = S'..That is, a shearing stress on one plane is accompanied by an equal shearingstress on a plane at right angles with it.9. Strain Due to Axial Loading.Whenever a structural member ormachine part is loaded, its shape changes. The magnitude of the change inshape depends on the material but the action is qualitatively identical (ifthe loads are not excessive) for all materials. If a member such as the barindicated in Fig. 9 is subjected to an axial tensile load, it will elongate and

HH-*Fig. 9. Strain in a bar subjectedto tensile loading.its cross section will become smaller. If an arbitrary length L, known as thegage length, is laid off on the surface of a member before loading, it willbe found to have increased after the load is applied. The magnitude of theincrease, ei, is called the total strain, and the ratio of the total strain to the

Art. 9 STRESS, STRAIN, AND AXIAL LOADING 15original length L is called the unit strain. The unit strain* along the axis isdesignated by the symbol ;. Hence, ' /*\i = T (6)The value of unit strain given by Eq. (6) is the strain in a 1-in. gage lengthor the average value of unit strain occurring over the gage length L. Forsituations in which the strain is uniformly distributed throughout the gagelength, Eq. (6) gives the value of the uniform strain at any point; but, ifthe unit strain is not uniformly distributed throughout the gage length,Eq. (6) gives only the value of the average, not the maximum, unit strain.The unit strain at a point may be expressed asdei , .ei = dE (6a)by using an infinitesimal gage length.One example of nonuniform distribution of strain is that occurring in aductile material stressed to the breaking point in tension. In this case theunit strain at failure is much greater near the break than it is at points ashort distance from the break.In an axially loaded member such as that indicated in Fig. 9, strain willoccur in directions at right angles to the applied load as well as in thedirection of the applied load, i.e., the diam of the rod will decrease under theinfluence of the axial tension. The unit strain in the transverse directionmay be defined in the same way as the unit strain in the longitudinal direc-tion,, = y. (6b)in whiche< is the change in the transverse dimensions,t is the transverse dimension.For most materials a definite relationship exists between the magnitudeof the unit strain in the transverse direction and the unit strain in thelongitudinal direction. If the strains are not excessive, the ratio of thetransverse unit strain to the longitudinal unit strain is constant. The ratiois known as Poisson's ratio and is designated by the symbol /*.M = 2~ (6c)* The expression "unit strain" is frequently abbreviated to "strain" in the technicalliterature. In this text, the two terms will be used interchangeably to denote the ratioof the total strain to the original length.

16Chap. 1MECHANICS OF MATERIALSFor axially loaded members a tensile strain in the longitudinal direction isaccompanied by a compressive strain in the transverse direction and viceversa. The value of Poisson's ratio is about 0.30 for steel and about 0.33for most of the other engineering metals. Concrete has a Poisson's ratio ofapproximately 0.20, while the Poisson's ratio for some grades of rubber ap-proaches 0.50.10. Shearing Strain.Axial loading produces longitudinal and trans-verse strains in a block, the sides of which are parallel and perpendicular tothe line of action of the applied load. For example, the block DEFG inFig. 10(a) will be deformed to the rectangle indicated by the dotted lines.

te>Fig. 10Each of the sides changes in length, but the angles remain right angles.However, if the block is oriented at 45 with the longitudinal axis of themember, as the block OABC in Fig. 10(a), it is apparent that the longi-tudinal extension and lateral contraction will deform the square to adiamond-shaped element as indicated by the dotted lines. The lengths ofthe sides will not change appreciably but each of the angles will change,those at A and C becoming smaller and those at O and B becoming larger.If the material is homogeneous and isotropic, the change in angle ateach of the corners will be the same; and the change in angle is a measureof the shearing strain. If the block OABC is drawn in its original position,Fig. 10(6), and the deformed block superimposed upon the original blockin such a way that the sides OA of each coincide, the block will form therhomboid OAB'C' as indicated by the dotted lines in Fig. 10(6). The total

Art. 12 STRESS, STRAIN, AND AXIAL LOADING 17shearing strain occurring in the block may be measured by the distancede (CC"), and the unit shearing strain designated by the letter -v is definedas the ratio of thejptaljhearing strain to the[length dl over.which it occurs.Hence\ - tl

18Chap. 1MECHANICS OF MATERIALSunit strain in a given material. The relationship between unit stress andunit strain for a material may be shown graphically by means of a diagramin which values of unit stress are plotted as ordinates against values of unitstrain as abscissae. The resulting line is known as a stress-strain diagram.Examples of stress-strain diagrams for axially loaded test specimensare shown in Fig. 11. It will be noted that each diagram consists of a

Stra/n(a)Stra/n(UFig. 11. Examples of stress-strain diagrams. (a) Typical for mildsteel except that left portion of curve is expanded horizontally.(6) Typical for timber, concrete, and certain other brittle materials.straight-line portion followed by a curve. An equation may easily be de-veloped for the straight-line portion of a stress-strain diagram. Within thestraight-line range of the diagram, the slope is constant and may be desig-nated by the symbol E.dS(9)The slope E is known as the modulusjiLelastici^, or Young!s_module, ofthe material.Eq. (9) may be integrated within the range in which E is constant, givingS - Si = E(e- 0. (9a)If both /Si and i are zero, i.e., if the curve passes through the origin,S = Ee. (9b)The equation for the curved portion of the diagram of a specific materialis not so easily determined, but a number of equations have been suggestedfor the stress-strain relationships for various materials.*The modulus of elasticity of most metals is but little affected by methodof manufacture, heat treatment, or small percentages of alloys. Themodulus for most of the carbon steels varies from 28,000,000 to 31,000,000* Osgood, W. R., "Stress-Strain Formulas," J. Aeronautical Sciences (Jan. 1946),13:45.

)Fig. 14. Evaluation of elastic strength.parallel to the initial straight-line portion of the stress-strain diagram. Thestress at the intersection of this line and the curve is the yield strength.The construction is illustrated in Fig. 14(a). This method of establishingthe elastic strength has the advantage of being sufficiently flexible to allowfor any desired amount of permanent set. On the other hand, the value ofyield strength is meaningless unless the offset (allowable permanent set) isstipulated.(d) Yield Point. A few engineering materials, notably the low-carbonsteels, have the characteristic of yielding abruptly at a stress of 60 per centor more of the ultimate strength. This characteristic leads to a stress-straindiagram with a flat portion or a dip at stresses slightly above the propor-tional limit as indicated in Fig. 14(6). The stress at which this phenomenonoccurs is a convenient measure of elastic strength and is known as the yieldpoint. It may be defined as the unit stress at which the material exhibits anincrease in strain with no increase in stress, with the understanding thatwith further strain the stress will again increase.

Art. 14 STRESS, STRAIN, AND AXIAL LOADING 25The mechanism by which this inelastic action under a gradually increas-ing strain takes place is known a&^slip and may be shown by microscopicexamination to consist in the sliding or slipping of one portion of a crystalalong another portion of the crystal. In order for the magnitude of slip to bemeasurable with the ordinary laboratory equipment, slip must occur in alarge number of crystals. The slip takes place along certain planes known asslip planes which are always oriented in certain directions with respect tothe geometrical axes of the crystal. Slip planes within individual crystalsare normally parallel, but slip planes in adjacent crystals are not usuallyparallel because the crystals are oriented in different directions.A number of other properties have been denned which may be used ascriteria of the elastic strength. Of all the properties, the yield strength isprobably the most generally applicable, although the yield point (for thosematerials which have a yield point) is the easiest to determine in the labora-tory since no stress-strain diagram is required.If a member is subjected to a constant stress for a long period of time,inelastic deformation may occur and continue to increase in magnitude un-til the member separates into two parts. This type of inelastic action isknown as' creep^nd differs from the inelastic action (known as slip) dis-cussed in tKepreceding paragraphs, in that the inelastic deformation in-creases under constant stress. When slip occurs, the inelastic strain does notincrease in magnitude under constant stress, whereas, when creep occurs,the inelastic strain does increase in magnitude under constant stress. Manyof the metals commonly used in engineering (steel, brass, aluminum al-loys) do not creep at ordinary temperatures but will creep at elevatedtemperatures. Metals with low melting points are prone to creep at roomtemperature.The maximum unit stress to which a material may be subjected withouthaving the inelastic strain exceed a specified amount in a specified time at aspecified temperature is known as the creep limit.(3) Fracture.If a specimen is subjected to gradually increasing axialstrain, the material will develop stress until a maximum value is reached,after which fracture or separation of the specimen into two or more partsoccurs. The fracture may be abrupt (as in a brittle material) or it may bepreceded by a relatively large amount of strain at stresses near the maxi-mum (as in a ductile material). The percentage elongation (in an 8-in. or a2-in. gage length), which is the unit strain at fracture in tension expressedas a percentage, is sometimes used as a measure of ductility.If a member in service is not to fail by fracture, it is obvious that themaximum stress must be kept below a limiting value corresponding to themaximum obtained in the test specimen, which is known as the ultimatestrength. For ductile materials the value of the ultimate strength'in tensionis definite, but the ultimate compressive strength may be indefinite due to

26 MECHANICS OF MATERIALS Chap. 1the large increase in the cross-sectional area which may accompany theplastic yielding. For example, a cylindrical specimen of lead 2 in. in diamand 4 in. high may be compressed to a disc a fraction of an inch thick andseveral inches in diameter without reaching maximum resistance. For suchcases the magnitude of the ultimate strength in compression is rathermeaningless, and the maximum stress permitted in compression is usuallybased on the yield strength.15. Allowable Working Stress and Factor of Safety.The allowableworking stress is denned as the maximum computed stress permitted in thematerial. Its value is usually established by the specifications under whichthe design is being prepared.The factor of safety is defined as the ratio of the strength of the ma-terial to the maximum computed stress in the member. The maximum com-puted stress in general will not exceed the allowable working stress and maybe far below it if factors such as rigidity (rather than stress) control thedesign. The maximum allowable working stress is always established at avalue less than the strength of the material in order to prevent failure.There are three reasons for this:(1) The actual maximum stresses are unknown, and an idealizeddistribution of stress is normally assumed. In most cases the actual maxi-mum stress will exceed the maximum computed stress.(2) The properties of the materials in the member are usually not knownbut are assumed to be equal to the properties in standardized test speci-mens. An allowance must be made for the difference in properties of thematerial in the member and the test specimens.(3) The magnitude of the maximum load which the member must resistis unknown at the time the member is designed. It is impossible to predictexactly what loads any structural member or machine part must carry dur-ing its life, so allowances must be made to provide for the possibility ofloads greater than those estimated for normal operating conditions.In some branches of engineering the term margin of safety is used. It isequal to the factor of safety minus 1. A positive margin of safety indicatesa safe member, and a negative margin presumably indicates an unsafemember.A factor of safety of 2 or a margin of safety of 1 does not mean that themember can carry twice as much load as the design load without failure,since the stress distribution under the higher load may be entirely differentfrom the distribution under the design load.16. Statically Indeterminate Axially Loaded Members.The axiallyloaded members which have been considered in the preceding articles wereassumed to be homogeneous. Hence, the stress (if no stress raisers werepresent) was assumed to be uniformly distributed across any normal cross

Art. 1627STRESS, STRAIN, AND AXIAL LOADINGsection of the member. If the member is not homogeneous, the stress, ingeneral, will not be distributed uniformly. If the member consists of two ormore portions which themselves are homogeneous, the stress distributionwithin each of the portions may be uniform, but the stresses will not havethe same magnitude. In order to determine the magnitude of the stresses,it is necessary to know the distribution of the load among the portions ofthe cross section.Since a parallel force system is involved in the free-body diagram of ashort length of the member, only one equation of equilibrium is available,assuming that the member is symmetrical. Therefore, the distribution ofthe load among the portions of the cross section cannot be determined fromthe equations of equilibrium alone. Such a member is known as a staticallyindeterminate member.In order to determine the distribution of the force among the variouscomponents of a statically indeterminate member, information other thanthat coming from the equations of equilibrium is necessary. In general, in-formation is obtained from the geometrical behavior of the member underload. The details of the solution of a specific problem depend upon thecharacteristics of that problem, but the principal steps involved in thesolution of an axially loaded indeterminate member are given in the follow-ing problem.Illustrative ProblemA concrete post 10 in. square is reinforced with four 1/2-in. square symmetricallylocated steel rods as indicated in Fig. 15(a). Determine the maximum load which theII /Oin.II||-M ,riiiiII-i 1IIIIIIIIMil|ii

(a)Fig. 15post will support if the allowable working stress in the concrete is 800 psi. and de-termine the stress in the steel.

28 MECHANICS OF MATERIALS Chap. 1Solution: A free-body diagram of a portion of the post is indicated in Fig. 15(6).The total load on the post is P, the resultant force developed in the steel is R., andthe resultant force in the concrete is R,.Statics. The equation of equilibrium gives

Art. 1729STRESS, STRAIN, AND AXIAL LOADINGhalf section of the cylinder, as indicated in Fig. 16(a). Three forces areindicatedone due to the internal pressure and the other two due to theresistance of the cylinder to splitting lengthwise. If the free-body diagramis constructed to include the fluid within half of the cylinder, it is evidentthat the resultant internal pressure'is equal to 2prl. The resisting tensileforce developed in each half of the wall of the cylinder may be designated asT. The equation of equilibrium written in the direction of the forces gives2T = 2prl.(11)

Fig. 16If the circumferential stress is uniformly distributed throughout the crosssection of the wall,T = StU, (1la)and5, = f (11b)Eq. (l1b) indicates the average value of the circumferential stress. If thestress is not uniformly distributed throughout the thickness of the cylinderwall, the maximum will be greater than the average value given by Eq.(11b). It may be shown that the maximum circumferential stress is alwaysgreater than the average value and that the maximum occurs at the insidesurface. However, if the diameter of the cylinder is equal to ten times thewall thickness, the maximum stress will be only 10 per cent greater than

30 MECHANICS OF MATERIALS Chap. 1the average stress, and, as the diameter increases in proportion to thethickness, the stress distribution will become more nearly uniform.The magnitude of the longitudinal stress may be found from a free-bodydiagram produced by cutting the cylinder with a transverse plane. It is ap-parent that the resultant force developed by the fluid pressure within thetank, indicated by the arrow in Fig. 16(6), is given byF = p*r*. (12)If the stress is uniformly distributed, as would be expected from symmetry,the resultant force developed in the wall isF' = 2irrtSi. (12a)From the force equation of equilibrium written in the longitudinal direc-tion,F = F', (12b)from whichSt = 2f (12c)It may be shown that the longitudinal stress is uniformly distributedthroughout the wall thickness except near the end of the cylinder or nearany cutouts or other irregularities in the section.A comparison of Eqs. (11b) and (12c) indicates that the average cir-cumferential stress is equal to twice the longitudinal stress. Hence, a thin-walled cylinder under internal pressure would be expected to fail along alongitudinal line rather than along a circumferential line. It should benoted that the stress distribution assumed in this article is not a distribu-tion that actually exists, but under the conditions indicated is satisfactoryfor design procedures. The formulas would not be expected to be validfor the condition when the external pressure is greater than the internalpressure, because of the possibility of the cylinder wall failing by buckling.18. Dynamic and Repeated Loading. The problems discussed thus farin this chapter have dealt with a situation in which the load or the strainis increased gradually and in which a condition of equilibrium exists. Thistype of loading is known as steady loading or static loading. Two othertypes of loading merit consideration because of the difference in the effectswhich they produce in a structural member.(1) Dynamic Loading. A dynamic load is a load which is applied to themember with shock or impact rather than by being applied slowly. Ingeneral, its effect is to produce vibration which will gradually be dampedout, resulting in a final deflection or displacement equal to that produced by

Art. 18 STRESS, STRAIN, AND AXIAL LOADING 31a static load of the same magnitude, provided that failure does not occur.However, for an instant during each cycle of the vibration, strains are de-veloped which are accompanied by higher stresses than would be obtainedif a load of the same magnitude had been applied slowly. If the higher stressexceeds the strength of the material, failure will occur.A dynamic load may be considered equivalent to a static load of suf-ficiently greater magnitude to produce the same maximum distortion that isproduced by the dynamic load. The ratio of the magnitude of the equivalentstatic load (static load which would produce the same distortion as thedynamic load) to the dynamic load is called the load factor. It is evidentthat the load factor in a given instance will depend upon the amount ofshock involved in the impact load or will depend on the velocity with whichthe impact load is traveling. Load factors as high as 18 are used in the de-sign of certain airplanes. This means that the wings, for example, must bedesigned to carry eighteen times the weight of the plane.Methods of evaluating dynamic loads in terms of equivalent staticloads are given in Chap. 9.(2) Repeated Loading.If a load which produces a stress less than theproportional limit of a material is applied to a member once or a dozentimes, the member is undamaged; but, if the same stress is applied severalmillion tunes, failure may occur. For most materials there exists a maximumstress which may be applied indefinitely or an arbitrarily large number oftimes, such as 500,000,000, without producing failure. The magnitude ofthis stress is known as the endurance limit of the material and, as usuallyemployed, it involves the application of alternate equal stresses in tensionand compression. A large number of applications results in fracture becauseeventually some portion of the material is stressed above the proportionallimit. When the stress is removed or reversed, strain hardening occurs;and, with the repeated application of stress, the over-stressed portion be-comes brittle and finally fractures. The presence of the resulting small crackcauses stress concentration which increases the susceptibility of the ad-jacent material to failure in the same manner. Thus, a small crack is formedand gradually works across the member, ultimately resulting in failure ofthe member by fracture. This type of fracture is known as progressivefailure or sometimes, fatigue failure. Methods which have been developedfor the evaluation of load factors and the resistance of a material to a combi-nation of steady loading and repeated loading are discussed in Chap. 9.PROBLEMS1. With the aid of appropriate free-body diagrams, determine the type of loadingdeveloped at the sections indicated in each of the members of Fig. P-l andevaluate the component forces (or couples) which the section must resist.

32Chap. 1MECHANICS OF MATERIALSs/00/h per ftI ^OOO /t>\ ff\1fe-A6ftBJ&Zft^ft(a)

Ct)fc)P-l2. Identify the type of loading developed at the indicated sections in each of thestructural elements shown in Fig. P-2. Evaluate the magnitude of each com-/OO/b

(a)fe)ponent force (or forces and couple). Neglect the friction at all surfaces of con-tact, and neglect the weights of the members.3. Construct appropriate free-body diagrams to determine the type of loading andthe magnitude of the forces developed in each of the sections indicated in themembers of Fig. P-3.4. With the aid of suitable free-body diagrams, identify the type of loading andevaluate the force (or forces and couple) at each of the sections indicated inFig. P-4. Neglect the weights of the members.

Probs. 2-4STRESS, STRAIN, AND AXIAL LOADING/eOteperftHill^-^20/6 per fr4ft ffftSft P/ate

s\!\ \) ff1I >K1 i[1!\CP-3

P-4

34Chap. 1MECHANICS OF MATERIALS5. Determine the magnitude of the stress developed in the 1/8-in. diam controlcable shown in Fig. P-5 due to an applied force of 100 Ib.

P-56. Fig. P-6 shows part of the landing-gear assembly for a glider. Determine theaverage compressive stress developed on section A A if the reaction on thewheel is 2360 Ib.

p-67. Determine the minimum diam of structural steel bolts required for each of theconnections at A, B, and C in the landing-gear assembly of Fig. P-6 for amaximum wheel load of 2360 Ib. Each bolt is in double shear.8. Determine the maximum compressive stress developed on section A A of theconnecting rod shown in Fig. P-8 if it is used to transmit the force from a3-1/8-in. diam piston under a pressure of 300 psi.9. Determine the maximum normal stress developed near the midlength of mem-ber [72L3 of the pin-connected truss shown in Fig. P-4(6) if it is an aluminumalloy tube with an outside diam of 1-1/2 in. and a wall thickness of 0.064 in.

Probs. 5-1235STRESS, STRAIN, AND AXIAL LOADING

10. From the rolled sections given in Tables I to IV (Appendix A), select suitablecross sections for each of the four structural steel truss members marked in Fig.P-2(c) if the loading shown is critical.11. Determine the minimum required diam for the structural steel pins at jointsL0, I/i, and L2 in the truss of section P-2(c) if each pin is in double shear.12. Determine the required diam for the pin in the upper pulley of Fig. P-12 if theaverage shearing stress is not to exceed 12,000 psi.

tortP-12

MECHANICS OF MATERIALSChap. 113. Determine the diam required for the cable of a mine shaft car which is to carrya load of 2400 Ib with a load factor of 1.8. Establish the value of working stresswhich shall have a factor of safety of 2 with respect to failure by slip.14. Determine the minimum dimensions for each of the indicated portions of thetruss in Fig. P-14. The material is Douglas fir. Assume each joint to be pin-connected./4,OOO /b

P-1415. Determine the maximum stress developed in the member of Fig. P-15.IV c> in. d/am.

P-1516. What maximum load may be applied to the member of Fig. P-16 if the maxi-mum stress is not to exceed 24,000 psi.?

P-1617. A plate 1/4 in. thick is to be reduced from a 1-in. width to a 1/2-in. width.Determine the required radius of fillet if the maximum stress is not to exceed24,000 psi. for an applied load of 2000 Ib.18. Determine the maximum permissible axial load for the member in Fig. P-18if the stress is not to exceed 30,000 psi.

P-18

Probs. 13-2537STRESS, STRAIN, AND AXIAL LOADING19. Prove that the maximum shearing stress in an axially loaded member occurs ona plane which makes an angle of 45 with the axis of the plate and show that themagnitude of the stress is one half of the magnitude of the normal stress on atransverse cross section.20. Determine the maximum shearing stress developed near the midlength ofmember EC of Fig. P-6 for the wheel load of 2360 Ib.21. Determine the required cross-sectional area of the member USL4 indicated inFig. P-4(6) if the maximum shearing stress is not to exceed 12,000 psi. and themaximum tensile stress is not to exceed 20,000 psi.22. Determine the maximum allowable load on the 1-5/8-in. by 1-5/8-in. timbermember of Fig. P-22 if the shearing stress parallel to the grain is not to exceed100 psi. and if the maximum normal stress parallel to the grain is not to exceed1000 psi.

P-2223. Determine the maximum shearing stress which the welded joint of Fig. P-23must withstand., //'n

60,

P-2324. A concrete cylinder 3 in. in diam and 6 in. high failed along a plane making anangle of 60 with the horizontal when subjected to an axial vertical load of18,000 Ib. Determine the shearing stress and the normal stress on the plane.25. Determine the stresses on the plane A A in Fig. P-25.3OO psi.

P-25

38Chap. 1MECHANICS OF MATERIALS26. The normal stress on the inclined plane in Fig. P-26 is 3000 psi. tension. De-termine the magnitude of S,.1 I I 8000 />s/:

P-2627. In a certain structural member Sx is always equal to 1/2Sv. Determine themagnitude of the stress on a plane which makes an angle of 30 with Sx.28. The shearing stress on plane A A in Fig. P-28 is 600 psi. Determine the normalstress on plane BB.

P-2829. The control cable for the rudder of a certain airplane is made of 1/8-in. diam.stranded steel wire and has a total length of 34 ft 6 in. Determine the elongationof the wire for a pull of 400 Ib. The modulus of elasticity of the wire may be as-sumed to be 12,000,000 psi.30. A 1-in. diam stranded steel cable is used in a mine hoist. Determine the totalelongation in a length of 800 ft when the cable is subjected to a total tensileforce of 2-1/2 tons. The modulus of elasticity may be assumed to be 12,000,000psi.31. An aluminum tube 6 ft long having an internal diam of 1/2 in. and an externaldiam of 3/4 in. is welded to the end of an aluminum tube having an internaldiam of 3/4 in. and an external diam of 1 in. The length of the second tube is 4ft. Determine the total elongation if an axial load of 2400 Ib is applied to theassembly.32. A 1/4-in. diam steel rod 4 ft long is attached to the end of a brass tube with aninternal diam of 1/4 in. and wall thickness of 1/16 in. What load will be re-quired to stretch the assembly 1/100 in. if the brass tube is 8 ft long?33. A steel rod 1/2 in. in diam and 6 ft long is to be subjected to an axial tensile loadof 1600 Ib. The rod is to be turned down to a diam of 1/4 in. throughout part ofits length, so that the load will cause a total elongation of 1/16 in. Determinethe length of the turned-down portion.

Prohs. 26-4180STRESS, STRAIN, AND AXIAL LOADING34. Determine the change of length of the steel member BC in Fig. P-34 as a re-sult of the 2000-Ib load.'Jin. d/'am

P-3435. The 200-Ib weight shown in Fig. P-35 is supported by 1/8-in. diam copperwires. If the load causes the horizontal member to stretch 0.0025 in., determinethe axial force to which the inclined wire is subjected.6ft

P-3536. If the member L3U? in Fig. P-4(6) is composed of two 2-in. by 2-in. by 1/8-in.steel angles, determine its total elongation due to the indicated loads.37. A 1/2-in. diam steel test specimen stretched 0.0024 in. in a length of 8 in. De-termine the probable load.38. A load of 3000 Ib applied to a 1/2-in. diam aluminum test specimen caused anelongation of 0.0030 in. in a 2-in. gage length. Determine the apparent modulusof elasticity of the material.39. A 3/4-in. diam brass rod was subjected to a total axial load of 20,000 Ib. De-termine the elongation in an 8-in. gage length and the change in diam of the rod.40. Determine the probable change in diam of a 3-1/8-in. diam aluminum pistonsubjected to a cylinder pressure of 300 psi.41. Determine the deflection of point 0 in Fig. P-41 due to the load of 6 kips. Thealuminum alloy tube has an outside diam of 1 in. and a wall thickness of 0.049in.

40Chap 1.MECHANICS OF MATERIALSA/ a//oytube& /a. diamstee/

42. If the outside diam of the alloy tube in Fig. P-41 is 0.75 in. and the maximumpermissible displacement of point O due to the 6-kip load is 0.15 in., whatminimum wall thickness must the tube have?43. Determine the change in diam of the piston of Prob. 40 if the temperaturechanges from -30F to 140F.44. The B-17 airplane has a wing span of 103 ft 9 in. Determine the change in spanif the plane leaves the ground at a temperature of 110F and climbs to an eleva-tion at which the temperature is 20F.45. The Golden Gate Bridge has a span of 4200 ft. Determine the change in lengthof a steel floor system for the total span (if no expansion joints were provided)due to a temperature change of 40F.46. One of the large cement kilns has a length of 450 ft and a diam of 12 ft. De-termine the change in length and diam of the structural steel shell caused by anincrease in temperature of 200F.47. An aluminum tank 3 ft in diam was strengthened by the addition of externalsteel hoops 1/4 in. thick. So far as resistance to internal pressure is concerned,will the strengthening effect be increased or decreased if the temperature of bothmetals increases?48. A steel micrometer designed for measuring diameters of approximately 4 ft wascalibrated at 68F. What error will be introduced in the measurements if theinstrument is used at a temperature of 110F without correcting for temperature?49. A bimetallic element was made by securely attaching a strip of nickel to a stripof copper. Determine the shearing stress developed along the junction betweenthe two metals if the unit is subjected to an increase in temperature of 40F andrestrained from bending.

Probs. 42-5941STRESS, STRAIN, AND AXIAL LOADING50. Determine the load required to produce a deflection of 0.01 in. in the member ofFig. P-41 if the tube has an external diam of 1 in. and a wall thickness of 0.065in.51. Determine the length of the steel member in Fig. P-51 if point A is to deflect1/4 in. as a result of a temperature change of 450F.ffross -j

f/xed distanceP-5152. Determine the required cross-sectional area of member LiL2 of Fig. P-2(c) ifthe member is to be made of structural steel and is to have a factor of safety of2.00 with respect to failure by slip under a 6-kip load at L2.53. Determine the factor of safety with respect to failure by slip of the steel rivetin Fig. P-3(c).54. Determine the required area of member J73L4 in Fig. P^(6) if the member is tobe constructed of aluminum alloy 17S-T with a factor of safety of 2.0 withrespect to failure by slip.55. Determine the factor of safety of the member shown in Fig. P-15 with respectto failure by (a) slip and (6) fracture if the member is structural steel.56. Determine the factor of safety of the steel rod in Fig. P-34 with respect tofailure by fracture.57. Determine the factor of safety with respect to failure by both slip and fractureof the structural steel rod and the aluminum alloy tube shown in Fig. P-41 ifthe tube has an external diam of 1 in. and a wall thickness of 0.065 in.58. An aluminum alloy plate 1 in. thick, 4 in. wide, and 24 in. long is bolted to oneside of a vertical Douglas fir post 4 in. square and 24 in. high. The post is thenloaded with vertical compressive force. Determine the magnitude of the forceand the location of its line of action if it develops a uniformly distributedcompressivfc stress of 500 psi. on a horizontal plane in the fir.59. A 1/2-in. diam steel bolt is run longitudinally through a brass tube having aninside diam of 1/2 in., an outside diam of 3/4 in., and a length of 16 in. Washersare placed at the ends of the bolt to provide uniform bearing on the tube. De-termine the increase in the stress in the brass tube if the nut is tightened one-quarter turn. The bolt has 8 threads per in.

42Chap. 1MECHANICS OF MATERIALS60. A pier of average concrete is 12 in. square and has four 1-in. square steel rein-forcing bars imbedded longitudinally in it. Determine the load which will de-velop a maximum compressive stress of 800 psi. in the concrete. What is themagnitude of the compressive stress in the steel?61. A concrete pier 12 in. square and 2 ft high is to carry a maximum axial compres-sive load of 300,000 Ib with a factor of safety of 4 with respect to failure byfracture. Determine whether or not steel reinforcing is required and if so, howmuch. Evaluate the shortening of the pier as a result of the load.62. A Douglas fir post 4 in. square and 3 ft long is to carry an axial compressive loadof 12,000 Ib. How much may the change in length of the post be decreased bysecurely bolting a 4-in. by 3/4-in. wrought iron plate to one side of the post?Where should the line of action of the load lie for the stress to be uniformlydistributed throughout the fir?63. Member A in the frame represented in Fig. P-63 is a steel bar with a cross-sectional area of 0.60 sq in., B is a wrought iron bar with an area of 1.20 sq in.,and C is assumed to be rigid. The 12,000-Ib load is placed on the rigid member Cin such a position that A elongates 0.0012 in. more than B. Determine the axialunit stress in each bar.///////-W

64.cJZ.OOO /t>P-63Bar C represented in Fig. P-64 is brass and has a cross-sectional area of 2 sq in.The bars are fastened between two rigid supports D and E and are connectedsecurely at B. At a temperature of 70F there is no stress in either bar. Determinethe stress in bar S (steel) when the temperature is 30F if the bar has a cross-sectional area of 1/2 sq in.0cJ"0 /o;*.20 //?.P-6465. Bar AB shown in Fig. P-65 is made of aluminum alloy 17S-T with an area of3 sq in. The post CD is a square concrete post 3 in. on a side. The modulus of

Probs. 60-67 STRESS, STRAIN, AND AXIAL LOADING 43elasticity of the concrete is 4 X 106 psi. The bar EC is assumed to be rigid andhas a slope as shown when the load P is zero. Determine the magnitude of theload P which will make BC horizontal.66.

Determine the magnitude of the total load, P, which will produce an axial com-pressive stress of 1000 psi. in member C of Fig. P-66, if the clearance betweenthe rigid member and C is 0.002 in., as shown when P is zero. The temperatureis assumed to remain constant. Bar A is a 1-1/2-in. by 2-in. brass rod, B is a24S-T bar 2 in. by 1 in. in cross section, and C is a medium concrete block 4 in.by 3 in. in cross section.

/////////////P-6667. Two 3-in. square wooden blocks and a 2-in. by 3-in. cast iron block are to sup-port an axial compressive load as shown in Fig. P-67. The cast iron block is0.004 in. shorter than the wooden blocks. Determine the maximum permissiblemagnitude of the load P if the allowable compressive unit stresses are 900 psi.for the wood and 15,000 psi. for the cast iron. The horizontal member is as-sumed to be rigid.

44Chap. 1MECHANICS OF MATERIALS3 in. in. 3 in.h + + 4

4/n. PIfl/OOCtWoodP-6768. A heavy steel member BC in Fig. P-68 is supported by a 1-3/4-in. square brassrod and a 1-in. square steel rod, CD. The line BC is horizontal when the frame isloaded as shown at a temperature of 110F. Determine the tensile unit stressdeveloped in the brass by the 30,000-Ib load if the line BC is also horizontalwhen the frame is not loaded and the temperature is 10F.

30, OOO tiP-6869. The assembly in Fig. P-69 consists of a 1-in. by 1/2-in. steel bar A, a rigidblock C, and a 1-in. by 2-in. Monel Metal bar B, securely fastened together andattached to rigid supports at the ends. Determine the change in the normalstress in each of the bars developed by a temperature drop of 40F, and the ap-plication of the load P of 25,000 Ib.

Probs. 68-71 STRESS, STRAIN, AND AXIAL LOADING45P-6970. The bar marked \1AS5A-Y////////////2y///////P-7071. Three vertical bars of steel, copper, and aluminum alloy 24S-T support a hori-zontal rigid bar carrying a load P as shown in Fig. P-71. Determine the maxi-mum allowable load P if the supporting bar is to remain horizontal and thefollowing stresses are not to be exceeded:Cross-SectionalAreaAllowable UnitTensile StressBarSteel1 sq in.18,000 psi.Conner . .3 sq in.10,000 psi.24S-T2 sa in.15.000 psi.

46Chap. 1MECHANICS OF MATERIALS

//fV/7.P-7172. A longitudinal joint in a 4-ft diam cylindrical boiler transmits a total stress of4000 Ib per in. of length. Determine the intensity of the internal pressure.73. Two 10-ft diam hemispheres are joined to form a spherical pressure vessel.Determine the total force transmitted per inch of length of joint when theinternal pressure is 200 psi.74. Two cylindrical containers 16 in. in diam and 20 in. long are joined by boltingtheir flanges together as indicated in Fig. P-74. Determine the number of 1-in.diam structural steel bolts required to hold the cylinders together when the as-sembly is subjected to an internal pressure of 200 psi. The increase in the tensilestress in the bolts is not to exceed 8000 psi.BO//?.

P-7475. Determine the minimum wall thickness required for the assembly of Fig. P-74if it is constructed of magnesium alloy C74-S with a factor of safety of 2.50 withrespect to failure by slip.76. Determine the minimum diameter of structural steel bolts required for thecylinder of Fig. P-74 if the increase in the tensile stress in the 8 bolts is not toexceed 10,000 psi. for an increase in internal pressure of 240 psi.77. Two flanged half cylinders 20 in. long and 14 in. in diam are joined by boltingthe flanges together using twelve 1/2-in. diam bolts on each side. Determine the

Probs. 72-84 STRESS, STRAIN, AND AXIAL LOADING 47increase in axial tensile stress in the bolts if the internal pressure in the assemblyis increased 200 psi.78. How many 1/2-in. diam bolts would be required for the assembly of Prob. 77 ifan increase in internal pressure of 240 psi. is not to cause an increase in tensilestress in the bolts of more than 8000 psi?79. A steel boiler 3 ft in diam is welded using a spiral seam which makes an angleof 30 with the longitudinal direction. Determine the magnitude of the normalforce and the shearing force transmitted across the seam due to an increase in in-ternal pressure of 160 psi.80. A cylindrical boiler is to be formed by welding steel plate along a spiral joint.From the standpoint of stress to be transmitted across the joint, would a spiralmaking an angle of 60 with the longitudinal axis of the boiler be preferable toone with a 45 angle? Explain.81. A spherical gas holder 40 ft in diam is subjected to an internal pressure of 180psi. Determine the thickness of structural steel plate required for the holder ifthe tensile stress is not to exceed 12,000 psi.82. Determine the thickness of steel plate required for a cylindrical boiler 6 in. indiam and 20 ft long if the tensile stress in the plate is not to exceed 12,000 psi.for an internal pressure of 240 psi.83. The longitudinal riveted joint in a 48-in. diam cylindrical pressure vessel is de-signed for a maximum stress of 10,420 psi. in the walls which are 1/2 in. thick.Determine the maximum allowable internal pressure in the pressure vessel.84. The oleo strut BC in Fig. P-6 consists of an alloy steel tube with an inside diamof 1-7/8 in., filled with oil, and containing a piston at the lower end. Determinethe minimum wall thickness required for the tube if the maximum wheel loadof 2360 Ib is not to develop a tensile stress of more than 24,000 psi. in the tubewall.

_EMPTY_

CHAPTER 2JOINTS AND CONNECTIONS FOR AXIALLY LOADED MEMBERS1Q. General Considerations.In the design of a structure or a machinecomposed of two or more members, it is inevitable that connections must bedevised to transmit load from one member to another. Of the many differentsystems that have been developed for this purpose at least three are ap-propriate for joining axially loaded members. One consists in overlappingthe two members and attaching them by means of a third unit, such as abolt, rivet, nail, or screw as shown in Fig. 17(a). A second method consistsj --la) (b) (0Fig. 17. Types of connections.in attaching the two members by means of a more uniformly distributedbonding agent, such as glue, or by welding as indicated in Fig. 17(6). Fromthe standpoint of analysis, one method of welding, spot welding, produces ajoint somewhat similar to the riveted or bolted connection. The thirdmethod, suitable for attaching rods (or members of circular cross section),consists in fitting the threaded end of one member into a matching fittingin the other member, such as the socket of Fig. 17(c).20. Types of Welds.Within recent years the techniques of weldinghave been developed and improved to the extent that this method of con-necting two members is now regarded as a standard fabrication procedurerather than as simply a device for making emergency repairs. The usualwelding operation is performed by clamping the two parts to be joined andheating them to a sufficiently high temperature to produce localized fusionmelting the parts together so that a rigid connection is formed uponcooling. Usually a small amount of additional material (weld metal) ismelted in place so that it adheres to both parts being connected. Heat issupplied by a gas flame or electrical current.49

50 MECHANICS OF MATERIALS Chap. 2A discussion of the relative advantages and disadvantages of the variouswelding procedures may be found in any of the current textbooks on ma-terials, in technical periodicals, or in the literature of the American Weld-ing Society..Three types of welds are useful for joining axially loaded members. Theyare butt welds, fillet welds, and spot welds, as illustrated in Fig. 18.

lal BvTt We/d (t) Fi//et k/e/J (c) Spot We/dFig. 18. Types of welds.(1) Butt Weld.This type of connection is formed by preparing theends of the member to be joined, aligning the members with a small gapbetween them, and filling the space with molten metal, using a techniquewhich will insure consolidation of the base metal and the added filler ma-terial. The filler material is usually of the same composition as the basemetal to provide the necessary strength and to reduce electrolytic or chemi-cal corrosion.* A sketch of a butt weld is indicated in Fig. 18(a).If the two members which are joined are axially loaded, it is evident thatthe primary stress which the weld must withstand is normal stress on atransverse cross section. The stress may be tension or it may be compres-sion depending on the direction of the applied load.(2) Fillet Weld.In some circumstances a butt weld is impractical andthe connection between the two members is formed by lapping the membersfor a short distance and welding around the edges of the overlap as indi-cated in Fig. 18(6). If an axial load is applied to the members, it is apparentthat the primary stress in the fillet weld will be shear, although under someconditions tension may be developed. However, in the design or analysisof a fillet weld it is customary to assume that the weld resists only shearingstress and to assume that the area subjected to shear is the length of theweld multiplied by the "throat distance," or thickness of the fillet on a 45line from the inside corner. The throat distance is one-half of the distance tin Fig. 18(6).(3) Spot Weld.If the two members being joined are relatively thin,spot welding may prove more useful than butt welding or fillet welding. The* If the strength of the joint need not approach the strength of the members beingjoined, brazing metal or solder may be used as the filler metal. These materials melt atrelatively low temperatures and may be made to adhere to the parts being joined with-out fusing them.

Art. 22 JOINTS AND CONNECTIONS 51usual technique of forming a spot weld consists in lapping the members tobe joined and passing a heavy localized electric current through the twothicknesses of metal. Under the proper circumstances a small portion ofeach member will be melted, forming a more or less circular connection asindicated in Fig. 18(c). If the two members are subjected to axial loading, itis apparent that the primary stress in the weld will be shear in the plane ofcontact of the members although secondary tensile stresses may be de-veloped.21. Allowable Stresses in Welds.In the design or analysis of a weldedjoint, it is customary to assume that the weld is subjected only to tensionif it is a butt weld or subjected only to shear if it is a fillet weld or spot weld.The strength of a welded joint is dependent upon many factors, includingthe quality of the metal used in the weld, the alteration in properties of thebase metal owing to the annealing effect of the welding operation, and theskill of the operator in preventing local overheating and other defects suchas blowholes and inclusions or irregular portions which serve as stressraisers. The fusion code of the American Welding Society permits a designstress of 16,000 psi. (24,000 psi. allowed as an emergency value) in tensionon the section through the throat of the butt weld in steel. The allowablecompressive stress on the butt weld is taken equal to the allowable compres-sive stress, for the base metal. For a fillet weld the allowable shearing stressis 13,600 psi. in steel (15,000 psi. emergency value). These stresses are forstatic loads only; if the joint is subjected to impact load or to repeated load,the allowable stresses must be reduced correspondingly. For an ultimatestrength of the material in tension of about 60,000 psi., the factor of safetyis approximately 4.Spot welding has been widely used for connecting aluminum sheet. Thestrength of individual spot welds varies from about 6000 psi. to over 30,000psi. depending upon the type of equipment used, the thickness of the sheet,and the characteristics of the alloys involved. For aluminum alloys 2S and3S a minimum ultimate shearing strength of approximately 9000 psi. isdeveloped in a weld connecting 0.040-in. sheet while for alloy 52S the cor-responding strength is about 21,000 psi. For spot welds in alloys 24S and75S a strength of 26,000 psi. may be regarded as the lower range of ultimateshearing strength.The welding procedure is usually adjusted to produce a spot weld havinga diameter equal to the sum of the thicknesses of the sheets plus 1/16 in.22. Design Considerations.The purpose of a welded connection is tohold the members in the desired relative position, to provide a means for asafe transfer of load from one member to another, and to provide the neces-sary rigidity. Adequate but not excessive area of weld must be provided. Inaddition, the welded joint should be designed to keep secondary stresses

52 MECHANICS OF MATERIALS Chap. 2(stresses other than those due to axial loading) as low as possible and toprovide the necessary accessibility to the parts. In many applications aminimum of weight is also desirable.The design procedure involves (1) the determination of the minimumamount of weld area necessary to permit the transfer of the axial load on thebasis of the allowable stresses, and (2) the placement of the weld so that itwill be most effective and will not induce bending or other undesirable per-formance. In addition, it must be so located that fabrication will not beunduly difficult.If the members being connected have the same width and a thicknessgreater than about 1/2 in., butt welding will often prove the most satis-factory. If the prevailing conditions do not permit the use of a butt weld, alap joint with one or two fillet welds, as indicated in Fig. 19(a), may be used.(a) (i>) (c)Fig. 19This type of connection introduces some bending at the joint, since the twomembers attempt to align themselves, as indicated in Fig. 19(c), therebyintroducing undesirable secondary stresses. The bending may be eliminatedby using a butt weld with two cover plates, as indicated in Fig. 19(c),in which each cover plate is welded to the main plates with fillet welds.No bending will occur in a joint of this type, but the weight has been in-creased due to the addition of the cover plates.If an unsymmetrical member, such as an angle, is welded to a symmetri-cal member, such as a flat plate, the location of the weld metal must beconsidered carefully to prevent undue bending or twisting of the joint.Illustrative ProblemA 6-in. by 6-in. by 3/8-in. structural steel angle in which the tensile stress is 10,000psi. is to be welded to a flat plate 10 in. wide. Design a suitable welded connection.Solution: One possible connection would consist of a fillet weld across the end ofthe angle as indicated in Fig. 20(a). The angle has a total cross-sectional area of4.36 sq in., so the total load to be transferred is 43,600 Ib. With an allowable stress of13,600 psi. in the throat of the weld, the length of weld required may be determinedas7 _ 43,600 , .*" 13,600 (0.707) (0.375)= 12.1 in.Obviously, the length available at the end of the angle is only 6 in., so the weld mustextend down the sides of the angle. If the weld were placed along only one side of the

Art. 2253JOINTS AND CONNECTIONSangle as indicated in Fig. 20(6), twisting would be induced. This may be shown byconsidering a free-body diagram of the end of the angle. The resultant force (F)in the angle acts at the centroid of the angle as shown in Fig. 20(c), and the resistingshearing force (Q) developed in the weld acts along the center of the weld. Sincethe two forces are not colinear, the resultant couple will induce twisting, which willincrease the stresses on the weld. In order to prevent the undesirable effects whichmay result from the twisting, the line of action of the resultant force developed bythe weld should be at the same distance from the back of the angle as the line ofaction of the applied force in the angle. This may be accomplished by welding alongthe attached leg of the angle as well as along the back of the angle.The design of the connection then involves the determination of the length ofeach of the welds so that their resultant will act along a line 1.64 in. from the back

(a)(c)

let)(e)Fig. 20(t)of the 6-in. by 6-in. angle. The two unknown lengths xi and x2, Fig. 20(d), may bedetermined from the free-body diagram, the equations of equilibrium, and the as-sumption that the unit stress is constant throughout the entire length of each weld.It is apparent thatQi + Q, = 43,6001b, (b)13,600 (0.707) (0.375) Xi + 13,600 (0.707) (0.375) z2 = 43,600, (c)xi + xi = 12.1 in. (d)The moment equation of equilibrium written with respect to an axis through Qigives43,600 (1.64) = 13,600 (0.707) (0.375) (6) x2, (e)from whichz2 = 33.3 in. (f)Then, from (d),xi = 8.8 in. (g)Hence, for a balanced design, the weld should extend for 8.8 in. along the back andfor 3.3 in. along the attached leg.* If it is desirable to include a weld along the* Practically, the length of each weld should be increased 1/4 in. as allowance forstarting the weld.

54Chap. 2MECHANICS OF MATERIALSend of the angle, the force developed by it will also need to be taken into considera-tion.It will be noted that this design eliminates twisting about an axis perpendicularto the plane in which the welds lie; but, it does not eliminate bending with respectto a transverse axis in the plane of the welds, because the resultant force in the angledoes not lie in the plane of the welds. In addition, the solution is based on the as-sumption that the stress is uniformly distributed along the length of the weld. Thisdistribution is not correct, but it is sufficiently accurate for design purposes. In-evitably stress concentration will exist near the weld, and a distribution other thanthat assumed will exist, but the high factor of safety will, in the case of static loads,provide a sufficient margin to give a safe design. Tests show that a transverse weldis about 30 per cent stronger than a longitudinal weld.That the distribution of stress along a longitudinal fillet weld is not uniform mayreadily be shown by considering the deformations produced in the members whichare welded together. For example, assume that a 1-in. by 6-in. plate is lap-welded to a

6QOO0/i>6O,00C/iFig. 211-in. by 8-in. plate as indicated in Fig. 21, and the unit subjected to an axial load of60,000 lb. If the unit stress is assumed to be distributed uniformly along the weld,the distribution of load carried by the 6-in. plate is indicated by the line ABC inFig. 21. From statics it is apparent that the load carried by the 8-in. plate must beas indicated by the line DEF. Thus, at section GG the load carried by the 8-in.plate will be 40,000 lb, and the load carried by the 6-in. plate will be 20,000 lb.However, the average unit stress at section GG in the 8-in. plate will be 5000 psi.,and the corresponding unit strain is5000es (30) 106= 0.000167,while at the same section the unit strain in the 6-in. plate isC6 6 (30) 106= 0.000111.Obviously the two plates could not act as a unit with this var