1.2Stresses in the Members of a Structure9When SI metric units are used, Pis expressed in newtons (N) and Ain square meters (m"), so the stress will be expressed in N/m.This unitiscalledapascal (Pa).However,thepascalisanexceedinglysmallquantityand often multiples of this unitmust be used:the kilopascal (kPa),themegapascal (MPa), and thegigapascal (GPa):1 kPa = 10° Pa = 10°N/m21 MPa=10°Pa=10°N/m21GPa=10°Pa=10°N/m2When U.S.customaryunits areused, force P is usually expressed inpounds (lb) or kilopounds (kip), and the cross-sectional area A is given insquare inches (in"). The stress then is expressed in pounds per squareinch (psi) or kilopounds per square inch (ksi).(a)(b)Fig. 1.13An exampleof simpleeccentricloadingConcept Application1.1Considering the structure of Fig.1.1 on page 5, assume that rod BC ismade of a steel with a maximum allowable stress al=165 MPa. Canrod BC safely support the load to which it will be subjected? The mag-nitude of theforce FBc in therod was 5o kN.Recalling that the diam-eter of the rod is 20 mm, use Eq. (1.5) to determine the stress createdin the rod by the given loading.P=FBc=+50kN=+50×103N(20mm）A=mP=T=㎡(10×10-3m)=314×10-6m22+50X10°NP.=+159×10°Pa=+159MPaA=314×10-6mSince is smallerthan all of the allowable stressinthe steel used,rodBC can safely support theload.To be complete, our analysis ofthe given structure should also includethecompressivestressinboomAB,aswellasthestressesproducedinthepins and their bearings. This will be discussed later in this chapter. Youshould also determine whether the deformations produced by the givenloadingareacceptable.Thestudyof deformationsunderaxial loadswillbethesubject of Chap.2.Formembers in compression, the stability of themember (ie.,its abilityto supportagiven load without experiencing a sud-den change in configuration) will be discussed in Chap. 10.tThe principal SI and U.S. Customary units used in mechanics are listed in tables insidethe front cover of this book. From the table on the right-hand side, I psi is approximatelyequal to 7kPa and 1ksi approximately equal to 7 MPa
1.2 Stresses in the Members of a Structure 9 To be complete, our analysis of the given structure should also include the compressive stress in boom AB, as well as the stresses produced in the pins and their bearings. This will be discussed later in this chapter. You should also determine whether the deformations produced by the given loading are acceptable. The study of deformations under axial loads will be the subject of Chap. 2. For members in compression, the stability of the member (i.e., its ability to support a given load without experiencing a sudden change in configuration) will be discussed in Chap. 10. Fig. 1.13 An example of simple eccentric loading. C M d d (a) (b) P' P' P P † The principal SI and U.S. Customary units used in mechanics are listed in tables inside the front cover of this book. From the table on the right-hand side, 1 psi is approximately equal to 7 kPa and 1 ksi approximately equal to 7 MPa. Concept Application 1.1 Considering the structure of Fig. 1.1 on page 5, assume that rod BC is made of a steel with a maximum allowable stress sall 5 165 MPa. Can rod BC safely support the load to which it will be subjected? The magnitude of the force FBC in the rod was 50 kN. Recalling that the diameter of the rod is 20 mm, use Eq. (1.5) to determine the stress created in the rod by the given loading. P 5 FBC 5 150 kN 5 150 3 103 N A 5 pr 2 5 pa 20 mm 2 b 2 5 p110 3 1023 m2 2 5 314 3 1026 m2 s 5 P A 5 150 3 103 N 314 3 1026 m2 5 1159 3 106 Pa 5 1159 MPa Since s is smaller than sall of the allowable stress in the steel used, rod BC can safely support the load. When SI metric units are used, P is expressed in newtons (N) and A in square meters (m2 ), so the stress s will be expressed in N/m2 . This unit is called a pascal (Pa). How ever, the pascal is an exceedingly small quantity and often multiples of this unit must be used: the kilopascal (kPa), the megapascal (MPa), and the gigapascal (GPa): 1 kPa 5 103 Pa 5 103 N/m2 1 MPa 5 106 Pa 5 106 N/m2 1 GPa 5 109 Pa 5 109 N/m2 When U.S. customary units are used, force P is usually expressed in pounds (lb) or kilopounds (kip), and the cross-sectional area A is given in square inches (in2 ). The stress s then is expressed in pounds per square inch (psi) or kilopounds per square inch (ksi).† bee98233_ch01_002-053.indd 9 11/15/13 9:42 AM
10Introduction-ConceptofStressThe engineer's role is not limited tothe analysis of existing struc-tures andmachinessubjectedtogiven loadingconditions.Ofevengreaterimportanceisthedesignofnewstructuresandmachines,thatistheselectionofappropriatecomponentstoperformagiventask.ConceptApplication1.2As an example of design,let us return to the structure of Fig.1. onpage 5 and assume that aluminum with an allowable stress all100 MPa is tobe used.Since the force in rod BCis still P=FBc=50kNunderthegiven loading,fromEq.(1.5),wehavePP50X10°N500 × 10-m2A=Call=A100×10°PaCalland sinceA=A500×10-6m12.62×103m=12.62mmTTTd=2r=25.2mmTherefore, an aluminum rod 26 mm or morein diameter will beadequate.1.2BShearing StressThe internal forces and the corresponding stresses discussed in Sec. 1.2Awere normal to the section considered. A very different type of stress isobtained when transverse forces P andp'are applied to a memberAB(Fig.1.14).Passing a section at C between the points of application of thetwoforces(Fig.1.15a),you obtainthediagramofportionACshowninPCA(a)C(b)Fig.1.14 OpposingFig. 1.15transverse loads creatingThisshowstheresultinginternal shearshear on memberAB.forceonasectionbetweentransverseforces
10 Introduction—Concept of Stress 1.2B Shearing Stress The internal forces and the corresponding stresses discussed in Sec. 1.2A were normal to the section considered. A very different type of stress is obtained when transverse forces P and P9 are applied to a member AB (Fig. 1.14). Passing a section at C between the points of application of the two forces (Fig. 1.15a), you obtain the diagram of portion AC shown in Concept Application 1.2 As an example of design, let us return to the structure of Fig. 1.1 on page 5 and assume that aluminum with an allowable stress sall 5 100 MPa is to be used. Since the force in rod BC is still P 5 FBC 5 50 kN under the given loading, from Eq. (1.5), we have sall 5 P A A 5 P sall 5 50 3 103 N 100 3 106 Pa 5 500 3 1026 m2 and since A 5 pr 2 , r 5 B A p 5 B 500 3 1026 m2 p 5 12.62 3 1023 m 5 12.62 mm d 5 2r 5 25.2 mm Therefore, an aluminum rod 26 mm or more in diameter will be adequate. Fig. 1.14 Opposing transverse loads creating shear on member AB. A B P' P Fig. 1.15 This shows the resulting internal shear force on a section between transverse forces. A C A C B (a) (b) P P P P' The engineer’s role is not limited to the analysis of existing structures and machines subjected to given loading conditions. Of even greater importance is the design of new structures and machines, that is the selection of appropriate components to perform a given task. bee98233_ch01_002-053.indd 10 11/15/13 9:42 AM
1.2StressesintheMembersofaStructure11Fig.1.15b. Internal forces must exist in the plane of the section, and theirresultant is equaltoP.These elementaryinternal forces arecalled shearingforces, and the magnitude P of their resultant is the shear in the section.Dividing the shear P by the area A of the cross section, you obtain theaverage shearing stress in the section.Denoting the shearing stress by theGreek letter (tau), writeP(1.8)TaveAThe value obtained is an average value of the shearing stress overthe entire section. Contrary to what was said earlier for normal stresses,the distribution of shearing stresses across the section cannotbe assumedto be uniform.As you will see in Chap.6, theactual value ofthe shearingstressvariesfromzeroatthe surfaceofthememberto amaximumvalueTmax that may be much larger than the average value Tave-Photo 1.3 Cutaway view of a connection with a bolt in shear.Shearing stresses are commonlyfound in bolts,pins, and rivets usedFEAto connect various structural members and machine componentsEB(Photo1.3).ConsiderthetwoplatesAandB,whichareconnectedbyabolt CD (Fig.1.16). If the plates are subjected to tension forces of magni-Dtude F, stresses will develop in the section of bolt correspondingtotheplane EE'.Drawing the diagrams of the bolt and of the portion locatedFig.1.16Bolt subject to single shear.above theplaneEE'(Fig.1.17), the shear P in the section is equal toF.The average shearing stress in the section is obtained using Eq. (1.8) bydividing the shear P =F by the area A of the cross section:P_E(1.9)Tave=AACCE1P福LD(a)(b)Fig.1.17(a) Diagram of bolt in single shear;(b)section E-E'of thebolt
1.2 Stresses in the Members of a Structure 11 Fig. 1.15b. Internal forces must exist in the plane of the section, and their resultant is equal to P. These elementary internal forces are called shearing forces, and the magnitude P of their resultant is the shear in the section. Dividing the shear P by the area A of the cross section, you obtain the average shearing stress in the section. Denoting the shearing stress by the Greek letter t (tau), write tave 5 P A (1.8) The value obtained is an average value of the shearing stress over the entire section. Contrary to what was said earlier for normal stresses, the distribution of shearing stresses across the section cannot be assumed to be uniform. As you will see in Chap. 6, the actual value t of the shearing stress varies from zero at the surface of the member to a maximum value tmax that may be much larger than the average value tave. Photo 1.3 Cutaway view of a connection with a bolt in shear. Fig. 1.16 Bolt subject to single shear. C D A F B E' E F' Fig. 1.17 (a) Diagram of bolt in single shear; (b) section E-E’ of the bolt. C C D F E E P (a) (b) F F' Shearing stresses are commonly found in bolts, pins, and rivets used to connect various structural members and machine components (Photo 1.3). Consider the two plates A and B, which are connected by a bolt CD (Fig. 1.16). If the plates are subjected to tension forces of magnitude F, stresses will develop in the section of bolt corresponding to the plane EE9. Drawing the diagrams of the bolt and of the portion located above the plane EE9 (Fig. 1.17), the shear P in the section is equal to F. The average shearing stress in the section is obtained using Eq. (1.8) by dividing the shear P 5 F by the area A of the cross section: tave 5 P A 5 F A (1.9) bee98233_ch01_002-053.indd 11 11/15/13 9:42 AM
Introduction-Concept of Stress12EHThe previous bolt is said to be in single shear.Different loading situ-ationsmayarise,however.For example,if spliceplates CandD areusedCKto connectplates Aand B (Fig.1.18), shearwill takeplacein bolt HJ inFAeach of the two planes KK'and LL'(and similarly in bolt EG).The boltsLLDare said to be in double shear.To determine the average shearing stress ineach plane, draw free-body diagrams of bolt HJ and of the portion of theboltlocated betweenthetwoplanes (Fig.1.19).Observing thatthe shearGPineach ofthe sections isP=F/2,theaverageshearingstressisFig.1.18Bolts subject to double shear.PF/2F(1.10)TaveAA2AHFcKKLFr(b)(a)Fig.1.19(a) Diagram of bolt indouble shear;(b) section K-K'and L-L'of thebolt.1.2CBearingStressinConnectionsBolts, pins, and rivets create stresses in the members they connectalongthe bearing surfaceor surface of contact.Forexample,consideragain the two plates A and B connected by a bolt CD that were dis-cussed in the preceding section (Fig. 1.16). The bolt exerts on plate A aforceP equal and oppositetotheforceFexerted bytheplateonthebolt(Fig.1.20).TheforcePrepresentstheresultantofelementaryforcesdistributed on the inside surface of a half-cylinder of diameter d and ofFig.1.20Equalandoppositeforcesbetweenlength t equal to the thickness of the plate. Since the distribution ofplate and bolt, exerted over bearing surfaces.theseforces-and ofthecorrespondingstresses-isquite complicated,inpractice one uses anaverage nominal value b of the stress, calledthe bearing stress, which is obtained by dividing the load Pby the areaof therectanglerepresentingtheprojectionoftheboltontheplatesec-tion (Fig.1.21). Since this area is equal to td, where t is the plate thickness and d the diameter of the bolt, we havePP(1.11)btdAFig.1.21Dimensions for calculatingbearing stress area.1.2DApplication to the Analysis andDesign of Simple StructuresWearenowinapositiontodeterminethestressesinthemembersandconnections of various simple two-dimensional structures and to designsuch structures. This is illustrated through the following ConceptApplication
12 Introduction—Concept of Stress The previous bolt is said to be in single shear. Different loading situations may arise, however. For example, if splice plates C and D are used to connect plates A and B (Fig. 1.18), shear will take place in bolt HJ in each of the two planes KK9 and LL9 (and similarly in bolt EG). The bolts are said to be in double shear. To determine the average shearing stress in each plane, draw free-body diagrams of bolt HJ and of the portion of the bolt located between the two planes (Fig. 1.19). Observing that the shear P in each of the sections is P 5 Fy2, the average shearing stress is tave 5 P A 5 Fy2 A 5 F 2A (1.10) Fig. 1.20 Equal and opposite forces between plate and bolt, exerted over bearing surfaces. A C D d t F P F' Fig. 1.21 Dimensions for calculating bearing stress area. A d t Fig. 1.19 (a) Diagram of bolt in double shear; (b) section K-K’ and L-L’ of the bolt. K L H J K' L' F FC FD F P P (a) (b) Fig. 1.18 Bolts subject to double shear. K B A L E H G J C D K' L' F' F 1.2C Bearing Stress in Connections Bolts, pins, and rivets create stresses in the members they connect along the bearing surface or surface of contact. For example, consider again the two plates A and B connected by a bolt CD that were discussed in the preceding section (Fig. 1.16). The bolt exerts on plate A a force P equal and opposite to the force F exerted by the plate on the bolt (Fig. 1.20). The force P represents the resultant of elementary forces distributed on the inside surface of a half- cylinder of diameter d and of length t equal to the thickness of the plate. Since the distribution of these forces—and of the corresponding stresses—is quite complicated, in practice one uses an average nominal value sb of the stress, called the bearing stress, which is obtained by dividing the load P by the area of the rectangle representing the projection of the bolt on the plate section (Fig. 1.21). Since this area is equal to td, where t is the plate thickness and d the diameter of the bolt, we have sb 5 P A 5 P td (1.11) 1.2D Application to the Analysis and Design of Simple Structures We are now in a position to determine the stresses in the members and connections of various simple two-dimensional structures and to design such structures. This is illustrated through the following Concept Application. bee98233_ch01_002-053.indd 12 11/15/13 9:42 AM
1.2StressesintheMembersofaStructure13ConceptApplication1.3Returning to the structure of Fig.1.l, we will determine the normalstresses,shearingstressesandbearing stresses.Asshown inFig.1.22the20-mm-diameterrodBChasflatendsof20X40-mmrectangularcross section, while boomABhas a 30× 50-mm rectangular crosssection andisfitted withaclevisatendB.Bothmembersarecon-nectedatBbyapinfromwhichthe30-kNloadissuspendedbymeansofaU-shapedbracket.BoomABissupported atAbyapinfitted intoa double bracket,while rod BC is connected at C to a single bracket.Allpinsare25mmindiameter.yd=25mmC人20mmFlatendTOPVIEWOFRODBC40mmd=20mmTd=20mmyd=25mmY1600mmFRONTVIEWFlat end50mmF800mmQ=30kNQ=30kNENDVIEW125mm20mm30mmH20mm125mmBTOPVIEWOFBOOMABd=25mmFig.1.22Components of boomusedtosupport30kN load.Normal Stress inBoomABandRodBC.As found in Sec.l.1A,theforce inrod BCis Fac=50kN (tension)and the areaof its circular crosssectionisA=314×10-m?.Thecorrespondingaveragenormal stressisBc=+159MPa.However,theflatparts of the rod arealsoundertensionandatthenarrowestsection.Wheretheholeislocated, wehaveA=(20mm)40mm-25mm)=300×10-6m(continued)
1.2 Stresses in the Members of a Structure 13 Normal Stress in Boom AB and Rod BC. As found in Sec. 1.1A, the force in rod BC is FBC 5 50 kN (tension) and the area of its circular cross section is A 5 314 3 1026 m2 . The corresponding average normal stress is sBC 5 1159 MPa. However, the flat parts of the rod are also under tension and at the narrowest section. Where the hole is located, we have A 5 120 mm2140 mm 2 25 mm2 5 300 3 1026 m2 800 mm 50 mm Q 30 kN Q 30 kN 20 mm 20 mm 25 mm 30 mm 25 mm d 25 mm d 25 mm d 20 mm d 20 mm d 25 mm 40 mm 20 mm A A B B B C C B FRONT VIEW TOP VIEW OF BOOM AB END VIEW TOP VIEW OF ROD BC Flat end Flat end 600 mm Fig. 1.22 Components of boom used to support 30 kN load. (continued) Concept Application 1.3 Returning to the structure of Fig. 1.1, we will determine the normal stresses, shearing stresses and bearing stresses. As shown in Fig. 1.22, the 20-mm-diameter rod BC has flat ends of 20 3 40-mm rectangular cross section, while boom AB has a 30 3 50-mm rectangular cross section and is fitted with a clevis at end B. Both members are connected at B by a pin from which the 30-kN load is suspended by means of a U-shaped bracket. Boom AB is supported at A by a pin fitted into a double bracket, while rod BC is connected at C to a single bracket. All pins are 25 mm in diameter. bee98233_ch01_002-053.indd 13 11/15/13 9:42 AM