Introduction-Concept of Stress14The correspondingaveragevalue of the stressisP50X10°N(oBC)end167.0MPaA300×10-6m2Note that this is an average value. Close to the hole the stress will actu-ally reach a much larger value, as you will see in Sec. 2.1l. Under an50kNincreasingload,therod will fail near oneof theholesratherthan in(a)its cylindrical portion; its design could be improved by increasing thewidth orthethickness of theflatendsoftherod.=25mmRecall from Sec. 1.1A that the force in boom AB is FAB = 40 kND(compression).Sincethe area of theboom's rectangularcross section is50kNA=30mm ×50mm=1.5X10-3m,the average valueofthe normalFstress in themain part of the rod between pinsAand B is(b)40×103N-26.7×10°Pa=-26.7MPaCAB1.5×103m250kNNotethatthe sections of minimumarea atAand B arenotunderstress, sincetheboomisincompression,andthereforepushes onthe(c)pins (insteadofpullingon thepins as rodBCdoes).Fig.1.23Diagrams ofthe singleshear pin at C.Shearing Stress in Various Connections. To determine theshearing stressinaconnectionsuchasa bolt,pin,orrivet,youfirstshow the forces exerted bythevarious members it connects.In thecase of pin C (Fig.1.23a),drawFig.1.23b to showthe 50-kN forceexerted bymemberBC onthepin, and theequal and oppositeforce40kNexerted by the bracket. Drawing the diagram of the portion of the pinlocatedbelowtheplaneDD'whereshearingstressesoccur (Fig.1.23c),notice that the shear in that plane is P = 5o kN.Since the cross-sectional area of thepin is(a)25mmA=Tr=m= π(12.5 × 10-3 m)2 = 491 × 10-6m2d = 25mm2the average value of the shearing stress in the pin at C isEDD50×10NP40kN=102.0MPaTaveA491×10-6mEENote that pin A (Fig.1.24) is in double shear. Drawing the free-bodydiagramsofthepinandtheportion ofpinlocatedbetweenthe(b)planes DD' and EE' where shearing stresses occur, we see thatP= 20kN and40kNP20kN=40.7MPaTave491×10-6m2A(c)Pin B (Fig. 1.25a) can be divided into five portions that are actedFig.1.24 Free-bodydiagramsofuponbyforces exerted bytheboom,rod, andbracket.PortionsDEthe double shear pin at A.(Fig.1.25b)andDG(Fig.1.25c)showthattheshearinsectionEisPe=15kN and the shear in section Gis Pc=25kN.Since the loading(continued)
14 Introduction—Concept of Stress The corresponding average value of the stress is 1sBC2end 5 P A 5 50 3 103 N 300 3 1026 m2 5 167.0 MPa Note that this is an average value. Close to the hole the stress will actually reach a much larger value, as you will see in Sec. 2.11. Under an increasing load, the rod will fail near one of the holes rather than in its cylindrical portion; its design could be improved by increasing the width or the thickness of the flat ends of the rod. Recall from Sec. 1.1A that the force in boom AB is FAB 5 40 kN (compression). Since the area of the boom’s rectangular cross section is A 5 30 mm 3 50 mm 5 1.5 3 1023 m2 , the average value of the normal stress in the main part of the rod between pins A and B is sAB 5 2 40 3 103 N 1.5 3 1023 m2 5 226.7 3 106 Pa 5 226.7 MPa Note that the sections of minimum area at A and B are not under stress, since the boom is in compression, and therefore pushes on the pins (instead of pulling on the pins as rod BC does). Shearing Stress in Various Connec tions. To determine the shearing stress in a connection such as a bolt, pin, or rivet, you first show the forces exerted by the various members it connects. In the case of pin C (Fig. 1.23a), draw Fig. 1.23b to show the 50-kN force exerted by member BC on the pin, and the equal and opposite force exerted by the bracket. Drawing the diagram of the portion of the pin located below the plane DD9 where shearing stresses occur (Fig. 1.23c), notice that the shear in that plane is P 5 50 kN. Since the crosssectional area of the pin is A 5 pr 2 5 pa 25 mm 2 b 2 5 p112.5 3 1023 m2 2 5 491 3 1026 m2 the average value of the shearing stress in the pin at C is tave 5 P A 5 50 3 103 N 491 3 1026 m2 5 102.0 MPa Note that pin A (Fig. 1.24) is in double shear. Drawing the freebody diagrams of the pin and the portion of pin located between the planes DD9 and EE9 where shearing stresses occur, we see that P 5 20 kN and tave 5 P A 5 20 kN 491 3 1026 m2 5 40.7 MPa Pin B (Fig. 1.25a) can be divided into five portions that are acted upon by forces exerted by the boom, rod, and bracket. Portions DE (Fig. 1.25b) and DG (Fig. 1.25c) show that the shear in section E is PE 5 15 kN and the shear in section G is PG 5 25 kN. Since the loading Fig. 1.23 Diagrams of the single shear pin at C. 50 kN (a) C 50 kN (b) Fb D' D d 25 mm 50 kN (c) P Fig. 1.24 Free-body diagrams of the double shear pin at A. (a) 40 kN A (c) 40 kN P P (b) 40 kN Fb Fb D' E' D E d 25 mm (continued) bee98233_ch01_002-053.indd 14 11/15/13 9:42 AM
1.2Stresses in the Members of a Structure15of the pin is symmetric, the maximum value of the shear in pin B isIFAB=20kNPc=25kN, and thelargestthe shearing stresses occur in sections GIFAB=20kNand H,where0=15kPinBPG25kN=50.9MPaDTaveFac-50kNA491×10-6m21Q=15kNBearing Stresses.UseEq-(1.11)todeterminethenominal bearing(a)stressatAinmemberAB.FromFig.1.22,t=30mmandd=25mm.PERecallingthatP=FaB=40kN,wehaveP40kN53.3MPaObtd(30mm)(25mm)ToobtainthebearingstressinthebracketatA,uset=2(25mm)1Q=15kN50mmandd=25mm:(b)140kN32.0MPaO=td(50mm)(25mm)FAB=20kNThe bearing stresses at B in member AB, at B and C in memberBC,and inthebracketat C arefound ina similarway.AQ=15kN(c)Fig.1.25Free-body diagrams forvarious sections atpin B.1.2EMethodofProblemSolutionYou should approach a problem in mechanics as you would approach anactual engineering situation.Bydrawingonyour own experienceand intu-ition about physical behavior, you will find it easier to understand and for-mulatetheproblem.Yoursolutionmustbebased onthefundamentalprinciples of statics and on the principles you will learn in this text Everystep you take in the solution must be justified on this basis, leaving noroomfor your intuition or “feeling After you have obtained an answer, youshould check it.Here again,you may call upon your common sense andpersonal experience.If you are not completely satisfied with the result, youshould carefully check yourformulation ofthe problem,thevalidity of themethods used for its solution,and the accuracy ofyour computations.In general, you can usually solve problems in several different ways;there is no one approach that works best for everybody. However, we havefound that students often find it helpful to have a general set of guidelinesto use for framing problems and planning solutions. In the SampleProblems throughout this text, we use a four-step approach for solvingproblems, which we refer to as the SMART methodology: Strategy,Modeling, Analysis, and Reflect & Think:1.Strategy.Thestatement ofa problem shouldbeclear and precise,andshould contain the given data and indicate what information isrequired. The first step in solving the problem is to decide whatconcepts you have learned that apply to the given situation and
1.2 Stresses in the Members of a Structure 15 of the pin is symmetric, the maximum value of the shear in pin B is PG 5 25 kN, and the largest the shearing stresses occur in sections G and H, where tave 5 PG A 5 25 kN 491 3 1026 m2 5 50.9 MPa Bearing Stresses. Use Eq. (1.11) to determine the nominal bearing stress at A in member AB. From Fig. 1.22, t 5 30 mm and d 5 25 mm. Recalling that P 5 FAB 5 40 kN, we have sb 5 P td 5 40 kN 130 mm2125 mm2 5 53.3 MPa To obtain the bearing stress in the bracket at A, use t 5 2(25 mm) 5 50 mm and d 5 25 mm: sb 5 P td 5 40 kN 150 mm2125 mm2 5 32.0 MPa The bearing stresses at B in member AB, at B and C in member BC, and in the bracket at C are found in a similar way. Fig. 1.25 Free-body diagrams for various sections at pin B. (a) 1 2 FAB 20 kN FBC 50 kN 1 2 FAB 20 kN 1 2 Q 15 kN 1 2 Q 15 kN Pin B D E G H J (b) 1 2 Q 15 kN D E PE (c) 1 2 FAB 20 kN 1 2 Q 15 kN D G PG 1.2E Method of Problem Solution You should approach a problem in mechanics as you would approach an actual engineering situation. By drawing on your own experience and intuition about physical behavior, you will find it easier to understand and formulate the problem. Your solution must be based on the fundamental principles of statics and on the principles you will learn in this text. Every step you take in the solution must be justified on this basis, leaving no room for your intuition or “feeling.” After you have obtained an answer, you should check it. Here again, you may call upon your common sense and personal experience. If you are not completely satisfied with the result, you should carefully check your formulation of the problem, the validity of the methods used for its solution, and the accuracy of your computations. In general, you can usually solve problems in several different ways; there is no one approach that works best for everybody. However, we have found that students often find it helpful to have a general set of guidelines to use for framing problems and planning solutions. In the Sample Problems throughout this text, we use a four-step approach for solving problems, which we refer to as the SMART methodology: Strategy, Modeling, Analysis, and Reflect & Think: 1. Strategy. The statement of a problem should be clear and precise, and should contain the given data and indicate what information is required. The first step in solving the problem is to decide what concepts you have learned that apply to the given situation and bee98233_ch01_002-053.indd 15 11/15/13 9:42 AM
Introduction-Concept of Stress16connect the data to the required information. It is often useful to workbackward from the information you are trying to find: ask yourself whatquantities youneedtoknowtoobtain the answer,and if some ofthesequantities are unknown, how can you find them from the given data.2.Modeling.Thesolution ofmostproblemsencounteredwill requirethatyoufirstdetermine thereactions atthe supports and internal forces andcouples.It is important to include one or several free-body diagrams tosupport these determinations.Draw additional sketches as necessaryto guide the remainder of your solution, such as for stress analyses3. Analysis. After you have drawn the appropriate diagrams, use thefundamental principles of mechanics to write equilibrium equa-tions.These equations can be solved for unknown forces and usedto compute the required stresses and deformations.4.Reflect&Think.Afteryouhaveobtained the answer,check itcarefullyDoes it make sense in thecontext of the original problem?You canoften detectmistakes in reasoning by carryingtheunitsthroughyourcomputations and checking the units obtained for the answer.Forexample, in the design oftherod discussed in Concept Application1.2,the required diameter of the rod was expressed in millimeters, whichis the correct unit for a dimension; if you had obtained another unit,you would know that some mistake had been made.You can often detect errors in computation by substituting thenumerical answerintoan equation thatwasnot used inthe solution andverifying that the equation is satisfied. The importance of correct compu-tations in engineering cannot be overemphasized.Numerical Accuracy.Theaccuracyof thesolution ofa problemdependsupontwoitems:(1)theaccuracyofthegivendata and (2)theaccuracyofthecomputationsperformedThesolutioncannotbemoreaccuratethanthelessaccurateofthesetwoitems.Forexample,iftheloadingofabeamisknowntobe75,000lbwith a possible error of 1ooIbeitherway,the relative error that measuresthedegree ofaccuracyof thedata is100 lb=0.0013= 0.13%75,000 1bTo compute the reaction at one of the beam supports, it would be mean-ingless torecord it as 14,322Ib.The accuracy of the solution cannot begreater than 0.13%,nomatter howaccuratethe computations are,and thepossible error in the answer maybe as large as (0.13/100)(14,322 lb)20Ib.The answer should beproperly recorded as 14,320±20 lb.In engineeringproblems,thedataareseldomknownwithanaccu-racy greater than 0.2%. A practical rule is to use four figures to recordnumbers beginning with a"I"and three figures in all other cases. Unlessotherwise indicated, the data given are assumed to be known with a com-parable degree of accuracy.Aforce of 4o lb, for example, should be read40.0 lb, and a force of 15 Ib should be read 15.00 Ib.The speed and accuracyof calculators and computers makes thenumerical computationsinthesolution of manyproblemsmucheasierHowever, students should not record more significantfiguresthan can bejustified merely because they are easily obtained. An accuracy greaterthan 0.2% is seldom necessary or meaningful in the solution of practicalengineering problems
16 Introduction—Concept of Stress connect the data to the required information. It is often useful to work backward from the information you are trying to find: ask yourself what quantities you need to know to obtain the answer, and if some of these quantities are unknown, how can you find them from the given data. 2. Modeling. The solution of most problems encountered will require that you first determine the reactions at the supports and internal forces and couples. It is important to include one or several free-body diagrams to support these determinations. Draw additional sketches as necessary to guide the remainder of your solution, such as for stress analyses. 3. Analysis. After you have drawn the appropriate diagrams, use the fundamental principles of mechanics to write equilibrium equations. These equations can be solved for unknown forces and used to compute the required stresses and deformations. 4. Reflect & Think. After you have obtained the answer, check it carefully. Does it make sense in the context of the original problem? You can often detect mistakes in reasoning by carrying the units through your computations and checking the units obtained for the answer. For example, in the design of the rod discussed in Concept Application 1.2, the required diameter of the rod was expressed in millimeters, which is the correct unit for a dimension; if you had obtained another unit, you would know that some mistake had been made. You can often detect errors in computation by substituting the numerical answer into an equation that was not used in the solution and verifying that the equation is satisfied. The importance of correct computations in engineering cannot be overemphasized. Numerical Accuracy. The accuracy of the solution of a problem depends upon two items: (1) the accuracy of the given data and (2) the accuracy of the computations performed. The solution cannot be more accurate than the less accurate of these two items. For example, if the loading of a beam is known to be 75,000 lb with a possible error of 100 lb either way, the relative error that measures the degree of accuracy of the data is 100 lb 75,000 lb 5 0.0013 5 0.13% To compute the reaction at one of the beam supports, it would be meaningless to record it as 14,322 lb. The accuracy of the solution cannot be greater than 0.13%, no matter how accurate the computations are, and the possible error in the answer may be as large as (0.13y100)(14,322 lb) < 20 lb. The answer should be properly recorded as 14,320 6 20 lb. In engineering problems, the data are seldom known with an accuracy greater than 0.2%. A practical rule is to use four figures to record numbers beginning with a “1” and three figures in all other cases. Unless otherwise indicated, the data given are assumed to be known with a comparable degree of accuracy. A force of 40 lb, for example, should be read 40.0 lb, and a force of 15 lb should be read 15.00 lb. The speed and accuracy of calculators and computers makes the numerical computations in the solution of many problems much easier. However, students should not record more significant figures than can be justified merely because they are easily obtained. An accuracy greater than 0.2% is seldom necessary or meaningful in the solution of practical engineering problems. bee98233_ch01_002-053.indd 16 11/15/13 9:42 AM
1.2Stresses in the Members of a Structure17SampleProblem1.1Inthehanger shown,theupperportionof linkABCisin,thickandthe lower portions are each in. thick. Epoxy resin is used to bondthe upper and lower portions together at B. The pin at A has a -in.diameter,while a I-in.-diameter pin is used at C. Determine (a)theshearing stress inpin A, (b)theshearing stressinpin C, (c)thelarg-est normal stress in linkABC, (d)theaverage shearing stress on the1.25imbonded surfaces at B, and (e) the bearing stress in the link at C.6inSTRATEGY:Considerthefreebodyof thehangerto determinetheinternal forceformemberAB and then proceed todeterminethe1.75imshearing and bearing forces applicable to thepins.These forces canthen be used to determine the stresses.iMODELING:Draw thefree-body diagram of thehangerto deter-mine the support reactions (Fig. 1). Then draw the diagrams of the10in.various components of interest showing the forces needed to deter-500Jbmine the desired stresses (Figs.2-6).5in.ANALYSIS:AFACDFree Body:Entire Hanger.Since the link ABCis a two-force mem-ber (Fig.1),the reaction at Ais vertical; the reaction at DisrepresentedAODOD,byits componentsD,and D,Thus,+EMD=0:(500 lb)(15 in.) FAc(10 in) = 05in.10in.FAc=+750IbFAc= 750 IbtensionELOCa. Shearing Stress in Pin A.Since this -in.-diameter pin is insingle shear (Fig. 2), write500lbFig.1 Free-body diagram ofFAChanger.750lbTA=6790psiTAr(0.375 in.)2A7501b=7501bb.Shearing Stress in PinC.Since this -in.-diameter pin is indouble shear (Fig.3),writeFAC3751b-in. diameterTc=7640psiTC=Am(0.25in.)2Fig.2Pin A.=7501bFAC=3751lyFAC=3751bI -in. diameterFig.3Pin C.(continued)
1.2 Stresses in the Members of a Structure 17 Sample Problem 1.1 In the hanger shown, the upper portion of link ABC is 3 8 in. thick and the lower portions are each 1 4 in. thick. Epoxy resin is used to bond the upper and lower portions together at B. The pin at A has a 3 8-in. diameter, while a 1 4-in.-diameter pin is used at C. Determine (a) the shearing stress in pin A, (b) the shearing stress in pin C, (c) the largest normal stress in link ABC, (d) the average shearing stress on the bonded surfaces at B, and (e) the bearing stress in the link at C. STRATEGY: Consider the free body of the hanger to determine the internal force for member AB and then proceed to determine the shearing and bearing forces applicable to the pins. These forces can then be used to determine the stresses. MODELING: Draw the free-body diagram of the hanger to determine the support reactions (Fig. 1). Then draw the diagrams of the various components of interest showing the forces needed to determine the desired stresses (Figs. 2-6). ANALYSIS: Free Body: Entire Hanger. Since the link ABC is a two-force member (Fig. 1), the reaction at A is vertical; the reaction at D is represented by its components Dx and Dy. Thus, 1l oMD 5 0: 1500 lb2115 in.2 2 FAC110 in.2 5 0 FAC 5 1750 lb FAC 5 750 lb tension a. Shearing Stress in Pin A. Since this 3 8-in.-diameter pin is in single shear (Fig. 2), write tA 5 FAC A 5 750 lb 1 4p10.375 in.2 2 tA 5 6790 psi ◀ b. Shearing Stress in Pin C. Since this 1 4-in.-diameter pin is in double shear (Fig. 3), write tC 5 1 2FAC A 5 375 lb 1 4p10.25 in.2 2 tC 5 7640 psi ◀ 6 in. 7 in. 1.75 in. 5 in. 1.25 in. 10 in. 500 lb A B C D E 5 in. 500 lb 10 in. A D Dx FAC Dy E C Fig. 1 Free-body diagram of hanger. 750 lb FAC 5 750 lb -in. diameter 3 8 A Fig. 2 Pin A. -in. diameter FAC 5 750 lb 1 4 FAC 5 375 lb 1 2 FAC 5 375 lb 1 2 C Fig. 3 Pin C. (continued) bee98233_ch01_002-053.indd 17 11/15/13 9:42 AM
Introduction-ConceptofStress18C.Largest Normal Stress in Link ABC.The largest stress isfoundwheretheareaissmallest;thisoccursatthecrosssectionatA(Fig.4) where the -in.hole is located. We have750lb750lbFACCA=2290psiOA=0.328in2Anet（in.）(1.25in.-0.375in.）d.Average Shearing Stress at B.We note that bonding existsonbothsides oftheupperportion ofthelink (Fig.5)andthatthe shearforceoneachside isF,=(750lb)/2=375lb.Theaverage shearingstressoneachsurfaceisF375lb人T=171.4psiTBA(1.25 in.)(1.75 in.)e.Bearing Stress in Link at C.For each portion of the link(Fig.6),Fi=375 lb, and the nominal bearing area is (0.25 in.)(0.25 in.)=0.0625in.Fi375lb4o=6000psiOb=0.0625 in2AFAc=750lb1.25in.1.25in.T1.75 in.g-in.diameterF,=F2=FAc=375lbFACFig.4LinkABCsectionatA.Fig.5 Element AB.375lbF=375IbI-in.diameterFig.6LinkABCsectionatC.REFLECTandTHINK:Thissampleproblemdemonstratestheneedtodrawfree-bodydiagramsoftheseparatecomponents,carefullycon-sidering the behavior in each one.As an example,based on visualinspectionofthehangeritisapparentthatmemberACshouldbeintensionforthegivenload,andtheanalysisconfirmsthis.Hadacom-pression result been obtained instead, a thorough reexamination oftheanalysiswouldhavebeenrequired
18 Introduction—Concept of Stress c. Largest Normal Stress in Link ABC. The largest stress is found where the area is smallest; this occurs at the cross section at A (Fig. 4) where the 3 8-in. hole is located. We have sA 5 FAC Anet 5 750 lb 1 3 8 in.211.25 in. 2 0.375 in.2 5 750 lb 0.328 in2 sA 5 2290 psi ◀ d. Average Shearing Stress at B. We note that bonding exists on both sides of the upper portion of the link (Fig. 5) and that the shear force on each side is F1 5 (750 lb)/2 5 375 lb. The average shearing stress on each surface is tB 5 F1 A 5 375 lb 11.25 in.211.75 in.2 tB 5 171.4 psi ◀ e. Bearing Stress in Link at C. For each portion of the link (Fig. 6), F1 5 375 lb, and the nominal bearing area is (0.25 in.)(0.25 in.) 5 0.0625 in2 . sb 5 F1 A 5 375 lb 0.0625 in2 sb 5 6000 psi ◀ REFLECT and THINK: This sample problem demonstrates the need to draw free-body diagrams of the separate components, carefully considering the behavior in each one. As an example, based on visual inspection of the hanger it is apparent that member AC should be in tension for the given load, and the analysis confirms this. Had a compression result been obtained instead, a thorough reexamination of the analysis would have been required. 375 lb F1 375 lb -in. diameter 1 4 1 4 in. Fig. 6 Link ABC section at C. -in. diameter 3 8 in. 1.25 in. 3 8 FAC Fig. 4 Link ABC section at A. FAC 750 lb 1.25 in. 1.75 in. F2 F1 A B F1 F2 FAC 375 lb 1 2 Fig. 5 Element AB. bee98233_ch01_002-053.indd 18 11/15/13 9:42 AM