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《固体力学基础》课程教学资源（文献资料）Beer-2015-Mechanics of Materials-Seventh Edition

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Introduction-Conceptof Stress4IntroductionIntroductionThe study of mechanics of materials provides future engineers with themeans of analyzing and designing various machines and load-bearing1.1REVIEW OFTHEstructures involving thedetermination of stresses and deformations.ThisMETHODSOF STATICSfirstchapterisdevotedtotheconceptof stress.1.2STRESSES IN THESection 1.1 is a short review of the basic methods of statics and theirMEMBERSOFAapplication to determine the forces in the members of a simple structureSTRUCTUREconsistingofpin-connectedmembers.Theconceptofstressinamember1.2AAxial Stressofastructureandhowthatstresscanbedeterminedfromtheforceinthe1.2BShearingStressmemberwill bediscussed in Sec.1.2.Youwill considerthenormal stresses1.2C Bearing Stress in Connectionsin a member under axial loading,the shearing stresses caused by theappli-1.2D Application to the Analysis andDesign of Simple Structurescationof equal and oppositetransverseforces,and thebearingstressescreated by bolts and pins in the members they connect.1.2EMethodofProblemSolutionSection1.2endswithadescriptionofthemethodyoushould use1.3STRESS ONANOBLIQUEin the solution of an assignedproblemand adiscussionof thenumericalPLANE UNDER AXIALLOADINGaccuracy.These concepts will be applied in the analysis of the members ofthe simple structure considered earlier.1.4STRESS UNDER GENERALAgain,a two-forcemember under axialloadingis observedinLOADING CONDITIONS;Sec.1.3wherethestressesonanobliqueplaneincludebothnormal andCOMPONENTSOFSTRESSshearing stresses,while Sec.1.4 discussesthatsix components are required1.5DESIGNto describe the state of stress at a point in a body under the most generalCONSIDERATIONSloading conditions.1.5ADeterminationoftheUltimateFinally, Sec. 1.5 is devoted to the determination of the ultimateStrengthofaMaterialstrength from test specimens and the use of a factor of safetyto compute1.5BAllowableLoadandAllowableStress:Factorof Safetythe allowable loadfor a structural component made of that material.1.5CFactorofSafetySelection1.5DLoadandResistanceFactorREVIEW OFTHEMETHODS1.1DesignOF STATICSConsider the structure shown in Fig. 1.1, which was designed to supporta30-kNload.ItconsistsofaboomABwitha30×50-mmrectangularcross section and a rod BCwith a 20-mm-diameter circular cross section.TheseareconnectedbyapinatBandaresupported bypinsand bracketsatA and C,respectively.Firstdrawa free-bodydiagram ofthe structurebydetaching itfromitssupports atAand Cand showing thereactionsthatthese supports exert on thestructure (Fig.1.2).Note that the sketchof thestructure has been simplified by omitting all unnecessary details.Many ofyoumayhaverecognized atthispointthatABandBCaretwo-forcemem-bers.Forthose of you whohavenot,wewill pursue ouranalysis,ignoringthat factand assuming thatthe directions of thereactions at Aand C areunknown.Each of thesereactions are represented by two components:Aand Ay at A, and Cx and Cy at C.The equilibrium equations are.+Mc=0:A(0.6m)-(30kN)(0.8m)=0o(1.1)A,= +40kNF= 0:A+ C=0(1.2)C,=-40kNC =-A+ F, = 0:Ay + Cy - 30 kN = 0Photo1.11Cranebooms usedto loadandunloadA, + C, = +30kN(1.3)ships

Introduction The study of mechanics of materials provides future engineers with the means of analyzing and designing various machines and load-bearing structures involving the determination of stresses and deformations. This first chapter is devoted to the concept of stress. Section 1.1 is a short review of the basic methods of statics and their application to determine the forces in the members of a simple structure consisting of pin-connected members. The concept of stress in a member of a structure and how that stress can be determined from the force in the member will be discussed in Sec. 1.2. You will consider the normal stresses in a member under axial loading, the shearing stresses caused by the application of equal and opposite transverse forces, and the bearing stresses created by bolts and pins in the members they connect. Section 1.2 ends with a description of the method you should use in the solution of an assigned problem and a discussion of the numerical accuracy. These concepts will be applied in the analysis of the members of the simple structure considered earlier. Again, a two-force member under axial loading is observed in Sec. 1.3 where the stresses on an oblique plane include both normal and shearing stresses, while Sec. 1.4 discusses that six components are required to describe the state of stress at a point in a body under the most general loading conditions. Finally, Sec. 1.5 is devoted to the determination of the ultimate strength from test specimens and the use of a factor of safety to compute the allowable load for a structural component made of that material. 1.1 REVIEW OF THE METHODS OF STATICS Consider the structure shown in Fig. 1.1, which was designed to support a 30-kN load. It consists of a boom AB with a 30 3 50-mm rectangular cross section and a rod BC with a 20-mm-diameter circular cross section. These are connected by a pin at B and are supported by pins and brackets at A and C, respectively. First draw a free-body diagram of the structure by detaching it from its supports at A and C and showing the reactions that these supports exert on the structure (Fig. 1.2). Note that the sketch of the structure has been simplified by omitting all unnecessary details. Many of you may have recognized at this point that AB and BC are two-force members. For those of you who have not, we will pursue our analysis, ignoring that fact and assuming that the directions of the reactions at A and C are unknown. Each of these reactions are represented by two components: Ax and Ay at A, and Cx and Cy at C. The equilibrium equations are. 1l o MC 5 0: Ax10.6 m2 2 130 kN210.8 m2 5 0 Ax 5 140 kN (1.1) 1 y o Fx 5 0: Ax 1 Cx 5 0 Cx 5 2Ax Cx 5 240 kN (1.2) 1x o Fy 5 0: Ay 1 Cy 2 30 kN 5 0 Ay 1 Cy 5 130 kN (1.3) Introduction 1.1 REVIEW OF THE METHODS OF STATICS 1.2 STRESSES IN THE MEMBERS OF A STRUCTURE 1.2A Axial Stress 1.2B Shearing Stress 1.2C Bearing Stress in Connections 1.2D Application to the Analysis and Design of Simple Structures 1.2E Method of Problem Solution 1.3 STRESS ON AN OBLIQUE PLANE UNDER AXIAL LOADING 1.4 STRESS UNDER GENERAL LOADING CONDITIONS; COMPONENTS OF STRESS 1.5 DESIGN CONSIDERATIONS 1.5A Determination of the Ultimate Strength of a Material 1.5B Allowable Load and Allowable Stress: Factor of Safety 1.5C Factor of Safety Selection 1.5D Load and Resistance Factor Design 4 Introduction—Concept of Stress Photo 1.1 Crane booms used to load and unload ships. bee98233_ch01_002-053.indd 4 11/15/13 9:42 AM

6IntroductionConceptofStressNotethat thereactionatA is directed alongtheaxis of theboomABandcauses compression in that member. Observe that the components Cxand C,of thereaction at Care,respectively,proportional to thehorizontaland vertical components of the distancefromB to C and thattheFBCreaction atC is equal to50kN,is directed along the axisof the rod BC,FBCand causes tension in that member.30kNThese results could have been anticipated by recognizing that ABandBCaretwo-forcemembers,ie.,membersthataresubjectedtoforcesBFABatonlytwopoints,thesepointsbeingAandBformemberAB,andBandFABCformemberBC.Indeed,foratwo-forcememberthelinesof actionoftheresultants oftheforcesacting at each of thetwopoints are equal and30kNopposite and pass through both points.Usingthis property,we could have(a)(b)obtained a simpler solution by considering the free-body diagram of pin B.The forces on pin B, Fas and Fso, are exerted, respectively, by membersFig.1.4Free-bodydiagramofboom's jointBandassociated forcetriangle.AB and BC and the30-kNload (Fig.1.4a).PinB is showntobein equi-librium by drawing the corresponding force triangle (Fig. 1.4b).SinceforceFsc is directed along member BC, its slope isthe sameas that of BC, namely, 3/4.We can, therefore,write the proportionFAB =FBC30kN453from whichFAB=40kNFBc=50kNForcesF'aBandFscexertedbypinBonboomABandrodBCareequaland opposite to FaB and Fc (Fig. 1.5),FBCDFRFDRBFABBF'ABAFig.1.5Free-body diagrams of two-forceFig.1.6Free-body diagrams of sections of rod BCmembersABand BCKnowingtheforcesatthe endsof eachmember, wecan now deter-mine the internal forces in these members.Passing a section at some arbi-trary point D of rod BC, we obtain two portions BD and CD (Fig.1.6). Since50-kNforcesmustbeappliedatDtobothportionsoftherodtokeeptheminequilibrium,aninternalforceof50kNisproducedinrodBCwhena30-kN load is applied at B.From the directions of theforces FBc and Fscin Fig.1.6 we see that the rod is in tension.A similar procedure enablesustodeterminethattheinternal forceinboomABis 4okNandisincompression

6 Introduction—Concept of Stress Note that the reaction at A is directed along the axis of the boom AB and causes compression in that member. Observe that the components Cx and Cy of the reaction at C are, respectively, proportional to the horizontal and vertical components of the distance from B to C and that the reaction at C is equal to 50 kN, is directed along the axis of the rod BC, and causes tension in that member. These results could have been anticipated by recognizing that AB and BC are two-force members, i.e., members that are subjected to forces at only two points, these points being A and B for member AB, and B and C for member BC. Indeed, for a two-force member the lines of action of the resultants of the forces acting at each of the two points are equal and opposite and pass through both points. Using this property, we could have obtained a simpler solution by considering the free-body diagram of pin B. The forces on pin B, FAB and FBC, are exerted, respectively, by members AB and BC and the 30-kN load (Fig. 1.4a). Pin B is shown to be in equilibrium by drawing the corresponding force triangle (Fig. 1.4b). Since force FBC is directed along member BC, its slope is the same as that of BC, namely, 3/4. We can, therefore, write the proportion FAB 4 5 FBC 5 5 30 kN 3 from which FAB 5 40 kN FBC 5 50 kN Forces F9 AB and F9 BC exerted by pin B on boom AB and rod BC are equal and opposite to FAB and FBC (Fig. 1.5). Fig. 1.4 Free-body diagram of boom’s joint B and associated force triangle. (a) (b) FBC FBC FAB FAB 30 kN 30 kN 3 5 4 B Fig. 1.5 Free-body diagrams of two-force members AB and BC. FAB F' AB FBC F' B BC A B C C D B D FBC FBC F' BC F' BC Fig. 1.6 Free-body diagrams of sections of rod BC. Knowing the forces at the ends of each member, we can now determine the internal forces in these members. Passing a section at some arbitrary point D of rod BC, we obtain two portions BD and CD (Fig. 1.6). Since 50-kN forces must be applied at D to both portions of the rod to keep them in equilibrium, an internal force of 50 kN is produced in rod BC when a 30-kN load is applied at B. From the directions of the forces FBC and F9 BC in Fig. 1.6 we see that the rod is in tension. A similar procedure enables us to determine that the internal force in boom AB is 40 kN and is in compression. bee98233_ch01_002-053.indd 6 11/15/13 9:42 AM

1.2Stresses in the Members of a Structure71.2STRESSESIN THEMEMBERSOF A STRUCTURE1.2AAxial StressNNAIn the preceding section, we found forces in individual members.This isthe first and necessary step in the analysis of a structure.However it doesnot tell us whether the given load can be safely supported.Rod BC of theexample considered in the preceding section is a two-force member and,therefore, the forces Fec and Fsc acting on its ends B and C (Fig. 1.5) aredirected along the axis of the rod.Whether rod BC will break or not underPhoto 1.2Thisbridgetruss consists of two-forcethis loading depends upon the value found for the internal force Fso themembersthatmaybeintensionorincompressioncross-sectional area of the rod, and the material of which the rod is made.Actually,theinternalforceFecrepresentstheresultantofelementaryforcesdistributed over the entire area A of the cross section (Fig. 1.7). The averageFBCFig.1.7 Axial forcerepresents the resultantofdistributedelementaryforces.intensity of these distributed forces is equal to the force per unit area,Fec/A, on the section. Whether or not the rod will break under the givenloading depends upon the ability of the material to withstand the corresponding valueFsc/Aof the intensityof the distributed internal forces.Let us look at the uniformly distributed force using Fig.1.8.Theforce per unit area, or intensity of the forces distributed over a given sec-(a)(b)tion, is called the stress and is denoted by the Greek letter (sigma).TheFig.1.8(a) Member with an axial load.stress inamemberof cross-sectional area A subjectedtoan axial loadp(b) Idealized uniform stress distribution at anis obtained by dividing the magnitude P of the load by the area A:arbitrarysection.P(1.5)AAFApositive sign indicates a tensile stress (member in tension),and a nega-AAtive sign indicatesa compressive stress (member in compression)As shown in Fig. 1.8, the section through the rod to determine theinternal force in the rod and the corresponding stress is perpendicular to theaxis of the rod. The corresponding stress is described as a normal stress.Thus, Eq. (1.5) gives the normal stress in a member under axial loading:Note that in Eq. (1.5), represents the aerage value of the stress overthe cross section, ratherthanthestress ata specific point of the cross section.To define the stress at a given point Q of the cross section, consider a smallarea A (Fig. 1.9). Dividing the magnitude of F by A, you obtain the averagevalueofthe stress overA.LettingAapproachzero,thestress atpointQisFig.1.9SmallareaAA,atanarbitrarycrossAFJ(1.6)T=section point carries/axial AF in this axial member

1.2 Stresses in the Members of a Structure 7 1.2 STRESSES IN THE MEMBERS OF A STRUCTURE 1.2A Axial Stress In the preceding section, we found forces in individual members. This is the first and necessary step in the analysis of a structure. However it does not tell us whether the given load can be safely supported. Rod BC of the example considered in the preceding section is a two-force member and, therefore, the forces FBC and F9 BC acting on its ends B and C (Fig. 1.5) are directed along the axis of the rod. Whether rod BC will break or not under this loading depends upon the value found for the internal force FBC, the cross-sectional area of the rod, and the material of which the rod is made. Actually, the internal force FBC represents the resultant of elementary forces distributed over the entire area A of the cross section (Fig. 1.7). The average Fig. 1.7 Axial force represents the resultant of distributed elementary forces. A F FBC BC A Fig. 1.8 (a) Member with an axial load. (b) Idealized uniform stress distribution at an arbitrary section. (a) (b) A P A P' P' P Fig. 1.9 Small area DA, at an arbitrary cross section point carries/axial DF in this axial member. P' Q A F intensity of these distributed forces is equal to the force per unit area, FBCyA, on the section. Whether or not the rod will break under the given loading depends upon the ability of the material to withstand the corresponding value FBCyA of the intensity of the distributed internal forces. Let us look at the uniformly distributed force using Fig. 1.8. The force per unit area, or intensity of the forces distributed over a given section, is called the stress and is denoted by the Greek letter s (sigma). The stress in a member of cross-sectional area A subjected to an axial load P is obtained by dividing the magnitude P of the load by the area A: s 5 P A (1.5) A positive sign indicates a tensile stress (member in tension), and a negative sign indicates a compressive stress (member in compression). As shown in Fig. 1.8, the section through the rod to determine the internal force in the rod and the corresponding stress is perpendicular to the axis of the rod. The corresponding stress is described as a normal stress. Thus, Eq. (1.5) gives the normal stress in a member under axial loading: Note that in Eq. (1.5), s represents the average value of the stress over the cross section, rather than the stress at a specific point of the cross section. To define the stress at a given point Q of the cross section, consider a small area DA (Fig. 1.9). Dividing the magnitude of DF by DA, you obtain the average value of the stress over DA. Letting DA approach zero, the stress at point Q is s 5 lim ¢Ay0 ¢F ¢A (1.6) Photo 1.2 This bridge truss consists of two-force members that may be in tension or in compression. bee98233_ch01_002-053.indd 7 11/15/13 9:42 AM

IntroductionConcept of Stress8PIn general, the valuefor the stress g at a given point Q of the sectionis different from that for the average stress given by Eq. (1.5),and o is1found to vary across the section.In a slender rod subjected to equal andopposite concentrated loads P and p' (Fig. 1.10a), this variation is smallinasectionawayfromthepointsofapplicationoftheconcentrated loads(Fig.1.1oc), but it is quite noticeable in theneighborhood of these points(Fig. 1.10b and d).It follows from Eq.(1.6)that themagnitude of the resultant of thedistributed internal forcesisdAButthe conditions of eguilibriumofeach of theportions of rod shown inFig. 1.10 require that this magnitude be equal to the magnitude P of theconcentrated loads.Therefore,(b)(c)(d)(a)gdA(1.7)DFig. 1.10Stress distributions at different sectionsalongaxiallyloadedmember.whichmeans that the volume under each of the stress surfaces in Fig.1.10must be equal tothe magnitude P of theloads.However,this is the onlyinformation derived from staticsregardingthe distribution of normalstresses in the various sections of the rod.The actual distribution ofstresses in any given section is statically indeterminate.To learn moreabout this distribution, it is necessary to consider the deformations result-ingfromtheparticularmodeofapplicationoftheloadsattheendsof therod.Thiswill bediscussed further in Chap.2.In practice,itis assumed that thedistribution of normal stresses inFig.1.11Idealizeduniformstressdistributionan axially loaded member is uniform, except in the immediate vicinity ofimpliestheresultantforcepassesthroughthecrossthe points of application of the loads.The value of the stress is then equalsection'scenter.to ave and can be obtained from Eq. (1.5).However, realize that when weassumeauniformdistributionof stressesinthesection,itfollowsfromelementary staticstthat the resultant P of the internal forces must beapplied at the centroid C of the section (Fig. 1.11).This means that a uni-form distribution of stress is possible only if the line ofaction of the concentratedloadsPandP'passesthroughthecentroidofthesectionconsidered(Fig.1.12).Thistypeofloadingiscalled centricloadingand willtakeplacein all straighttwo-force members found in trusses and pin-connectedstructures, such as the one considered in Fig. 1.1. However, if a two-forcemember isloaded axially,but eccentrically,as shown in Fig.1.13a,the con-C.ditions of equilibrium of the portion of member in Fig.1.13b show that theinternal forces in a given section must be equivalent to a force P appliedat thecentroid of the section and a coupleM of moment M=Pd. Thisdistributionof forces-the correspondingdistribution of stresses-cannotbe uniform.Nor can thedistribution of stresses be symmetric.This pointwill be discussed in detail in Chap. 4.tSee Ferdinand P. Beer and E.Russell Johnston, Jr., Mechanics for Engineers, 5th ed.Fig.1.12CentricloadinghavingresultantforcesMcGraw-Hill, New York, 2008, or Vector Mechanics for Engineers, 10th ed., McGraw-Hill,New York, 2013, Secs.5.2and 5.3.passingthroughthecentroidofthesection

8 Introduction—Concept of Stress In general, the value for the stress s at a given point Q of the section is different from that for the average stress given by Eq. (1.5), and s is found to vary across the section. In a slender rod subjected to equal and opposite concentrated loads P and P9 (Fig. 1.10a), this variation is small in a section away from the points of application of the concentrated loads (Fig. 1.10c), but it is quite noticeable in the neighborhood of these points (Fig. 1.10b and d). It follows from Eq. (1.6) that the magnitude of the resultant of the distributed internal forces is #dF 5 # A sdA But the conditions of equilibrium of each of the portions of rod shown in Fig. 1.10 require that this magnitude be equal to the magnitude P of the concentrated loads. Therefore, P 5 #dF 5 # A s dA (1.7) which means that the volume under each of the stress surfaces in Fig. 1.10 must be equal to the magnitude P of the loads. However, this is the only information derived from statics regarding the distribution of normal stresses in the various sections of the rod. The actual distribution of stresses in any given section is statically indeterminate. To learn more about this distribution, it is necessary to consider the deformations resulting from the particular mode of application of the loads at the ends of the rod. This will be discussed further in Chap. 2. In practice, it is assumed that the distribution of normal stresses in an axially loaded member is uniform, except in the immediate vicinity of the points of application of the loads. The value s of the stress is then equal to save and can be obtained from Eq. (1.5). However, realize that when we assume a uniform distribution of stresses in the section, it follows from elementary statics† that the resultant P of the internal forces must be applied at the centroid C of the section (Fig. 1.11). This means that a uniform distribution of stress is possible only if the line of action of the concentrated loads P and P9 passes through the centroid of the section considered (Fig. 1.12). This type of loading is called centric loading and will take place in all straight two-force members found in trusses and pin-connected structures, such as the one considered in Fig. 1.1. However, if a two-force member is loaded axially, but eccentrically, as shown in Fig. 1.13a, the conditions of equilibrium of the portion of member in Fig. 1.13b show that the internal forces in a given section must be equivalent to a force P applied at the centroid of the section and a couple M of moment M 5 Pd. This distribution of forces—the corresponding distribution of stresses—cannot be uniform. Nor can the distribution of stresses be symmetric. This point will be discussed in detail in Chap. 4. (a) (b) (c) (d) P' P' P' P' P Fig. 1.10 Stress distributions at different sections along axially loaded member. Fig. 1.11 Idealized uniform stress distribution implies the resultant force passes through the cross section’s center. C P Fig. 1.12 Centric loading having resultant forces passing through the centroid of the section. C P P' † See Ferdinand P. Beer and E. Russell Johnston, Jr., Mechanics for Engineers, 5th ed., McGraw-Hill, New York, 2008, or Vector Mechanics for Engineers, 10th ed., McGraw-Hill, New York, 2013, Secs. 5.2 and 5.3. bee98233_ch01_002-053.indd 8 11/15/13 9:42 AM

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