made into a “standard” shape and size. It has a constant circular cross section with enlarged ends, so that failure will not occur at the grips. d0 ! 0.5 in. To perform a tension or compression test apunch specimen material Before testing, two small marksof arethe placed alongisthe specimen’s uniform length. Measurements are taken of both the specimen’s initial made into a “standard” shape and size. It has a constant circular cross L0 ! 2 in. A0, and L0 between the area, gauge-length section with enlargedcross-sectional ends, so that failure willthenot occur at distance the grips. punchpunch marks.marks For example, when along a metalthe specimen is used in a tension Fig. 3–1 testing, two small Before are placed specimen’s test it generally has an initial diameter of d0 = 0.5 in. (13 mm) and a uniform length. Measurements are taken of both the specimen’s initial L0 ! 2 in. Bölüm 3. Malzemelerin Mekanik gauge length of L0 = 2 in. (50 mm), Fig. 3–1. In order to apply an axial Özellikleri A , L0 ends cross-sectional area, and the gauge-length distance between the seated into ofATthe the are usually 8 2 C HAPTER 3 M E C H A N I Cload A 0L Pwith R O P Eno R T bending IES OF M E R Ispecimen, ALS punch marks. For example, when a metal specimen is used in a tension Fig. 3–1 3 Çekme ve Basınç ball-and-socket joints. A testing machine like the one shown in Fig. 3–2 is Testleri 82 C H Atest P T E it R 3 M E C H Ahas N I C Aan L Pinitial RO P E R to Tdiameter I Estretch S O F Mthe AT R I0A L d =S 0.5atin.a very generally ofEspecimen (13 mm) and a rate until it then used slow, constant d0 ! 0.5 in. Tomachine perform a tension or compression testaxial a to specimen the material is 2 in.The gauge length of L0 = fails. (50 mm), Fig. 3–1. Intoorder to load apply an is designed read the required maintainof this made into a “standard” shape and seated size. It into has a constant circular cross uniform stretching. load with no bending of the specimen, the ends are usually d0 ! 0.5 in. To perform a tension orthe compression test a specimen of the material is frequent intervals during recorded of not the applied section with enlarged so data that will occur at the grips. ball-and-socket joints. AAt testing machine like theends, onetest, shown inisfailure Fig. 3–2 made into a the “standard” shape and size. It hasa is a constant circular cross readout. load P, as read on dial of the machine or taken from digital Before testing, two small punch marks are placed along the specimen’s then used to stretch thesection specimen a very slow, constant until withatenlarged ends, so that rate failure willitmarks not occur at the grips. Also,uniform the elongation d = L L between the punch on the Typical steel specimen with attached 0 length. Measurements are taken ofthis both the specimen’s initial fails. totesting, read the load required to maintain L0 !The 2 in. machine is designed strain gauge. Before two small punch marks are placed along the specimen may be measured using either a caliper or a mechanical or specimen’s ! cross-sectional area, A0, and the gauge-length distance L0 between the uniform stretching. optical devicelength. called an extensometer. This of dof(delta) then used uniform Measurements arevalue taken both isthe specimen’s initial L0Fig. ! in. punch the marks. For example, whenof a the metal specimen is used in a tension At23–1 frequent intervals during test, data is ,recorded the applied to calculate the average normal strain in specimen. Sometimes, A L cross-sectional area, and the gauge-length distance between the Test sonrası (deformasyon yapmış durum) noktalar arası 0uzaklık L kumpas ile ölçülerek 0 in.read test it machine generally has an initial of dpossible (13 mm) and a readout. load P, as read on the however, dialpunch of the orexample, taken from a diameter digital 0 = 0.5 to this measurement is not taken, since it is also marks. For when a metal specimen is used in a tension Fig. 3–1 L =an 2electrical-resistance in. gauge length (50 mm), Fig.on3–1. Ingauge, orderwhich to apply an axial Also, the elongation = Lit-directly L0 hesaplanır. between the punch marks strain byofusing strain steel specimen with attachedUzama/kısalma bulunur. miktarıthe ! d test Bu işlem yerine strain-gage (birim dthe generally has0 an initial diameter of (13 mm) and a 0 = 0.5 in. with no bending specimen, the ends aregauge usually looksload likeusing the one shown in of Fig.the 3–3. The operation of this is seated into uge. specimen may be measured either a caliper or a mechanical or 3 L0electrical = 2 in. resistance gauge length of in (50 mm),of Fig. 3–1.thin In order to apply an axial based on the change a very piece in Fig. 3–2 is ball-and-socket joints. Aoftesting machine like thewire oneorshown device called an extensometer. This value d (delta) is then used deformasyon optical ölçer) kullanılabilir. load with no bending of the specimen, the isends are usually into of metal foil under strain. Essentially the Sometimes, gauge cemented to the seated 3 then used to stretch thespecimen. specimen at a very slow, constant rate until it to calculate the average normal strain in the ball-and-socket joints. A testing machine like the one shown in Fig. 3–2 is specimen along its length.isIf designed the cementtoisread very strong in comparisonto tomaintain this fails. The machine the however, this measurement isused not taken, sincethe it is also possible to load read required specimen at a very thethen gauge, thento thestretch gauge is in effect an integral part slow, of theconstant specimen,rate until it uniform stretching. the strain directly by so using electrical-resistance strain gauge, which that when the specimen is strained the direction ofrequired the gauge, fails.an The machine is designed toinread the load tothe maintain this At frequent intervals during the test, data ismeasuring recordedthe of the applied looks like the one shown in Fig. 3–3. The operation of this gauge wire and specimen will experience the same strain. Byis uniform stretching. load resistance P, asresistance readof onthe the dial of the machine or taken from a digital readout. based on the change in electrical of a very thin wire or piece electrical wire, the gauge may be calibrated to read At frequent intervals during the test, data is recorded of the applied ! Typical steel specimen Also,normal the elongation d = L cemented - L0 between punch marks on the attachedstrain. values strain directly. of metal with foil under gauge totaken thethefrom loadofEssentially P, as read onthe the dial of is the machine or a digital readout. strain gauge. specimen along its length.specimen may is bevery measured using either a caliper or a mechanical or If the cement strong in comparison to 3.2 T HE STRESS –S TRAIN D IAGRAM 83 Also, the elongation d = L L between the punch marks on the Typical steel specimen with attached 0 movable optical devicean called an extensometer. This value of d (delta) is then used the gauge, then the gauge is in effect partusing of the specimen, strain gauge. upper specimen may beintegral measured either a caliper or a mechanical or to iscalculate the average normal inthe the specimen. Sometimes, so that when the specimen strained in the direction of thestrain gauge, optical device called an extensometer. This value of d (delta) is then used load however, this measurement is not taken, since it is also possible to read wire and specimen will experience the same strain. By measuring the to calculate the average normal strain in the specimen. Sometimes, dial The Stress–Strain Diagram the strain directly by using an electrical-resistance strain gauge, which electrical resistance of the wire, the gauge may be calibrated to read tension however, this measurement is not taken, since it is also possible to read specimen looks like the one shown in Fig. 3–3. The operation of this gauge is values of normal strain directly. the strain directly by using an electrical-resistance strain gauge, which L0, change in electrical resistance of a very thin wire or piece t is not feasible to prepare a test specimen to match the motor size, A0 and based on the andlooks load like the one shown in Fig. 3–3. The operation of this gauge is f each structural member. Rather, the test results mustcontrols beofreported so under strain. Essentially the gauge is cemented to the metal foil based on the change in electrical resistance of a very thin wire or piece hey apply to a member of any size. To achieve this, the load and specimen along its length. If the cement is very strong in comparison to of metal foil under strain. Essentially the gauge is cemented to the orresponding deformation data are used to calculate various values then of the gauge is in effect an integral part of the specimen, the gauge, specimen along its length. If the cement is very strong in comparison to load he stress and corresponding straindial in the specimen. A plot so of that the results when the specimen is strained in the direction of the gauge, the the Tgauge, then the gauge is in effect an part of the specimen, HE STRESS –STRAIN Dwill IAGRAM 83integral Electrical–resistance roduces a curve called the stress–strain diagram. There3.2 are two ways in wire and specimen experience the same strain. By measuring the so that when strain the specimen is strained in the direction of the gauge, the gauge ! described. electrical resistance ofexperience the wire, the beBy calibrated to read which it is normally wire and specimen will thegauge same may strain. motor 3measuring the Fig. 3–3 Fig. 3–2 values of normal strain directly. and load electrical resistance of the wire, the gauge may be calibrated to read controls he Stress–Strain Diagram Conventional Stress–Strain Diagram. We can values determine the strain directly. of normal C H A P T E R 3 movable M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S ominal or engineering stress by dividing the applied load P by Gerilme-Birim Deformasyon Diyagramıthe upper ble to prepare a test specimen to match size,calculation A0 and L0,assumes that pecimen’s original cross-sectional area Athe movable crosshead 0. This s reported so upper true fracture stress tural member. Rather, the the test cross resultssection must be he stress is constant over and throughoutload the gauge s¿f crosshead o a member load and dial ength. We haveof any size. To achieve this, theElectrical–resistance load tension strain gauge g deformation data are used to calculate various values ofdial specimen ultimate tensionin the specimen. A plot of the results motor corresponding strain Fig. 3–3 su Fig. 3–2 stress fracture P specimen rve called the stress–strain diagram. in and load s = There are two ways stress proportional limit(3–1) motor controls s A0 f rmally described. elastic limit crosshead 3.2 and load yield controls stress sY spl 3 nal Stress–Strain can determine the Likewise, the nominal Diagram. or engineeringWe strain is found directly from the engineering stress byordividing the the applied load the train gauge reading, by dividing change in P thebyspecimen’s gauge Electrical–resistance iginal cross-sectional area A . This calculation assumes that 0 ength, d, by the specimen’s original gauge length L0. Here the strain is strain gauge constant over the crossthroughout section andthe throughout the gauge ssumed to be constant region between the gauge points. Electrical–resistance strain gauge strain necking Fig. 3–2 elastic yielding ve Thus, region hardening Fig. 3–2 ! s = P A0 ! P = d L0 ! elastic behavior plastic behavior P Fig. 3–3 Fig. 3–3 (3–1) and true (3–2) Conventional stress-strain diagrams for ductile material (steel) (not to scale) Fig. 3–4 nominal or engineering strain is and found directly from the the vertical If the corresponding values of s P are plotted so that reading, or by dividing the change in the specimen’s gauge xis is the stress and the horizontal axis is the strain, the resulting curve is the original gauge length the strain is L0. Here Elastic Behavior. Elastic behavior of the material occurs when alledspecimen’s a conventional stress–strain diagram. Realize, however, that two etress–strain constant throughout the region between the gauge points. thematerial strains inwill thebespecimen are within diagrams for a particular quite similar, but the light orange region shown in 3–4. Here the curve is actually will never be exactly the same. This isFig. because the results actually depend a straight line throughout most of this region, so that the stress is proportional to the strain. The material n variables such as the material’s composition, microscopic in this region is said to be linear elastic. The upper stress limit to this d mperfections, the way it is manufactured, the rate of loading, and the P = (3–2) linear relationship is called the proportional limit, spl. If the stress emperature during the time L0 of the test. slightly exceeds the proportional limit, the curve tends to bend and !30 A0 is a decreasing load, since constant when calculating Whensupports this diagram is plotted it has a form shown by the light-blue 3.3 . However, STRESS–STRAIN BEHAVIOR OF DUCTILE AND BRITTLE MATERIALS 87 s = the P>A s – P from the true diagram, s – P curveengineering in Fig. 3–4. stress, Note that conventional and true diagrams 0 the actual area A3.3 within the–S necking is always decreasing until are practically coincident when the strainregion is small. The differences STRESS TRAIN BEHAVIOR OF DUCTILE AND BRITTLE MATERIALS 87 3.3 Sin TRESS TRAIN BEHAVIOR OF Drange, UCTILEstress, AND BRITTLE MATERIALS 87 fracture, s¿f, and begin so thetomaterial actually sustains increasing between the diagrams appear the–Sstrain-hardening Stress–Strain of more Ductile since s = P>A. Behavior where the magnitude of strain becomes significant. In particular, there a large divergence within the necking region. Here it can be andisBrittle Materials Stress–Strain Behavior of s –Ductile P diagram seenStress–Strain from the conventional the specimen actually Behavior of that Ductile and Brittle Materials A supports a decreasing load, since is constant when on calculating 0 brittle, depending Materials canand be classified as either being ductile or Brittle Materials s = P>A . s – P engineering stress, However, from the true diagram, This steel specimen clearly shows the necking 0 their stress–strain characteristics. 3.3 STRESS –STRAIN BEHAVIOR OF DUCTILE AND BRITTLE MATERIALS 8 7 that occurred just before the specimen failed. ls can be classified as either ductile or brittle, depending on decreasing until the actual area Abeing within the necking region is always Materials can besclassified either being ductile or brittle, depending on stress, This resulted in the formation of a “cup-cone” ess–strain characteristics. ¿f, and fracture, soasmaterial the material actually sustains increasing Ductile Materials. Any that can be subjected to large shape at the fracture location, which is their stress–strain characteristics. since s = P>A. is called a ductile material. Mild steel, as strains before it fractures characteristic of ductile materials. Stress–Strain Behavior of Ductile ediscussed Materials. Anyismaterial that can be subjected to choose large Failure previously, a typical example. Engineers often ductile of a Necking 3 and Materials Ductile Materials. Any material that can besteel, subjected to large before itBrittle fractures called ductile material. ductile material materials for design becausea these materials areMild capable ofas absorbing ! itis fractures ! strains before is called a ductile material. Mild steel, as (b) dsshock previously, is aastypical example. Engineers often they choose (a) 3 energy, and if being they become overloaded, usually exhibit can be or classified either ductile brittle, depending onwillductile discussed previously, is amaterials typicalorexample. Engineers often choose ductile s for design because these are capable of absorbing 3 This steel specimen clearly shows the necking ess–strain characteristics. large deformation before failing. Fig. 3–5capable of absorbing materials for design because these they materials are that occurred just before the specimen failed. energy, and if they become overloaded, will usually exhibit One way to specify the ductility of a material is to report its percent This resulted in the formation of a “cup-cone” shock before or energy, and if(Düktil) they overloaded, they will usuallyGerilme-Birim exhibit Sünek Gevrek Deformasyon İlişkisi eformation Materials. Any material that become can ve be subjected toMalzemelerin large failing. elongation or percent reduction in area Mild at the time of fracture. The shape at the fracture location, which is before itspecify fractures is ductility called a ductile material. steel, its as percent large deformation before failing. way to the of a material is to report elongation is the specimen’s fracture strain expressed as a characteristic of ductile materials. Sünek malzeme: Kırılma öncesi büyük birimitsdeformasyon yapabilen dpercent previously, isway a typical example. Engineers often ductile One to specify the ductility of time achoose material is to report percent 3 malzeme on or percent reduction in area at the of fracture. The percent. Thus, if the specimen’s original gauge length is L and its length Failure of a Necking are capable of absorbing s for design because these materials 0 elongation or percent reduction in strain area atexpressed the timeductile of afracture. The elongation isLthe specimen’s fracture as material energy, and ifisthey overloaded, they usually exhibit Yumuşak (mild) çelikwill için düktilitenin belirlenmesi: at fracture then f,become percent elongation is the specimen’s fracture strain expressed as a (b) (a) Thus, if the specimen’s original gauge length is L and its length formation before failing. 0 Thus, if the specimen’s original gauge length is L and its length way the ductility of a material is to report its percent 0 re istopercent. , then Lspecify - L3–5 Lf Fig. f 0 on oratpercent reduction in area at the time The fracture is Percent Lf, then (3–3) 1100%2 elongation = of fracture. L0 elongation is the ! specimen’s fracture as a L0 expressed Lf -strain Thus, if the specimen’s original gauge L0 L and itsL length (3–3) Percent elongation = length is1100%2 f 0 L0 this re is seen Lf, then (3–3) 1100%2 Percent = value would As in Fig. 3–6, be 38% for a mild = 0.380, Pf elongation Lf: since Kırılma/kopma anındaki uzunluk Lo:Başlangıç uzunluğu L0 steel specimen. - L0 Lf this in Fig. since would be 38% fordeğer) a mild = 0.380, The3–6, percent reduction areavalue is another to specify ductility. It is (3–3) 1100%2 Percent elongation = in (yumuşak !Pf 3–6, çelikway için tipik As seen in Fig. since P L 0f = 0.380, this value would be 38% for a mild ecimen. defined within the region of necking as follows: steel specimen. ercent reduction in area is another way to specify ductility. It is kesit Af ve başlangıçtaki kesit Ao a göre de Düktilite çekmewould deneyinde kopma anındaki in Fig. 3–6, since 38% forway a mild 0.380, this value Pf = reduction The percent infollows: area is be another to specify ductility. It is A A within the region of necking as 0 f cimen. (3–4) Percent area =as follows: 1100%2 defined within thereduction region of of necking ercent reduction in incelenebilir. area is another way to specify ductility. A0 It is A A f within the region of necking as follows: 0 (3–4) Percent reduction of area = 1100%2 A0 - Af A Here A0 is the specimen’s original area and Af is the area 0 (3–4) Percent reduction of area = 1100%2 A0 - Across-sectional f A(3–4) 0 of 60%. Percent = steel has1100%2 of the neckreduction at fracture. Mild a typical value ! of area A0 is the specimen’s original cross-sectional areamolybdenum, and Af is the area Besides steel, other metals such as brass, and zinc may s (ksi) Here is the specimen’s original cross-sectional area and the area Awhereby A 60 eck at fracture. Mild steel has a typical value of 60%. 0 f issonuç Yukarıdaki formülün yumuşak çelik için tipik değeri %60 tır. also ductile stress–strain characteristics similar is the exhibit specimen’s original cross-sectional area and Af is the area to steel, s (ksi) sYS ! 51 the neck at fracture. Mild steel has a typical value of 60%. esthey steel, other metals such as brass, molybdenum, and zinc may eck atof fracture. Mild steel has a typical value of 60%. undergo elastic stress–strain behavior, yielding at sconstant stress, 50 s (ksi) (ksi) and 60 Besides steel, other metals such similar asuntil brass, molybdenum, zinc may es steel, other metals such asfinally brass, molybdenum, and to zinc may whereby ibit ductile stress–strain characteristics steel, strain hardening, and necking fracture. In60 most metals, sYS ! 5160 ibit ductile stress–strain characteristics similar to steel, whereby also exhibit ductile stress–strain characteristics similar to steel, whereby dergo elastic stress–strain behavior, yielding at constant stress, 40 sYS !50 51One however, constant yielding will not occur beyond the elastic range. sYS ! 51 dergothey elastic stress–strain behavior, yielding behavior, at constant yielding stress, 50 undergo elastic stress–strain at constant stress, ardening, and finally necking until fracture. In most metals, 50 metal for which this is the case is aluminum. Actually, this metal often ardening, and finally necking until fracture. In most metals, strain hardening, and occur finally necking fracture. In 40 metals,30 r, constant yielding will theuntil elastic range. notyielding have a will well-defined yieldbeyond point, and consequently it ismost standard 40 One r,does constant not not occur beyond the elastic range. One however, constant yielding will not occur beyond the elastic range. One 40 which this is the is strength aluminum. Actually, thisoften metal oftencalled rrpractice which this the case is aluminum. Actually, metal toisdefine acase yield using athis graphical procedure 20 30 the 30 metal for which this is the is aluminum. Actually, thisand metal often t have have well-defined yield and consequently it)isisstandard aawell-defined yield point, and case consequently it isin.>in. standard offset method. Normally apoint, 0.2% strain (0.002 chosen, from 30 to define anot strength using a graphical called thecalled does have well-defined yieldprocedure point, and consequently standard10 20 theit is to define ayield yield strength a graphical procedure 20 this point on the ausing line parallel to the initial straight-line portion Pa axis, ethod. Normally 0.2% strain (0.002 ) using is chosen, and fromand in.>in.in.>in. practice toadefine adiagram yield strength called ethod. Normally a 0.2% strain (0.002 )aThe isgraphical chosen, from this the is drawn. point procedure where linethe 20 10 P (in./in.) ntof on offset the P stress–strain axis, a line parallel to the initial straight-line portion 10 method. Normally a 0.2% strain ( ) is chosen, and from 0.002 in.>in. nt on the axis, a line parallel to the initial straight-line portion P 0.005 0.010 intersects the curve the point yieldwhere strength. An example of the stress–strain diagram is defines drawn. The this line 0.002 P (in./in.) 10 point on determining the axis, astrength. line parallel to the initial straight-line portion stress–strain diagram is drawn. The point where line alloy 0.005 construction for the yield for aluminum is (0.2%0.010 ts thethis curve defines theP yield Anstrength example of an thethis P (in./in.) offset) Yield strength 0.002 for an aluminum alloy 0.005 0.010 of the stress–strain diagram is drawn. The point where this line tstion the curve defines the yield strength. An example of the for determining theFrom yield strength for anthe aluminum alloy is (0.2% P (in./in.) offset) shown in Fig. 3–7. the graph, yield 8strength isC s 51 ksi 8 H AYS P T E= R 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S 0.002 Yield strength for an aluminum alloy 0.005 0.010 Aluminyum alaşım diyagram: ! example curve defines the için yield strength. of offset) the n(352 Fig.intersects 3–7. From the the graph, the yield strength sYS = 51 ksi An ction for determining the yield strength forisan aluminum alloy is (0.2% 3–7 MPa). Yield0.002 strength for Fig. an aluminum alloy Fig.is3–7 (0.2% offset) construction the yield strength aluminum alloy na).Fig. 3–7. From for thedetermining graph, the yield strength is sfor = 51 ksi YS an Yield strength for an aluminum alloy s (ksi) strength is s = 51 ksi Fig. 3–7 Pa). shown in Fig. 3–7. From the graph, the yield YS 2.0 Fig. 3–7 (352 MPa). 3.3 3.3 1.5 s (ksi) 1.0 3 sf " 22 20 B !0.06 !0.05 !0.04 !0.03 !0.02 !0.01 A 0.5 0.01 !20 Doğal lastik için diyagram: ! 2 4 6 8 s–P diagram for natural rubber 10 P (in./in.) !40 !60 Fig. 3–8 !80 !31 C !100 !120 s–P diagram for gray cast iron L PROPERTIES M AT E R I A L S OF Gevrek Malzemeler Gevrek malzemeler kırılma öncesi çok az akar (yield) veya hiç akmaz. Gri dökme demir örneği: s (ksi) sf " 22 20 B !0.06 !0.05 !0.04 !0.03 !0.02 !0.01 A 0.01 P (in./in.) !20 10 P (in./in.) !40 !60 !80 !100 !120 C s–P diagram for gray cast iron ! Fig. 3–9 3.3 STRESS –STRAIN BEHAVIOR OF STRESS–STRAIN BEHAVIOR OF DUCTILE AND3.3 BRITTLE MATERIALS 8 9DUCTILE A Bir gevrek malzemenin çekme altındaki kırılması/kopması s (ksi) (st)max " 0.4 2 !0.0030 !0.0025!0.0020!0.0015!0.0010!0.0005 !0.0030 !0.0025!0.0020!0.0015! P (in./in.) Realize that the yield strength is not a physical property of the 0 0.0005 material, since it is a stress that causes a specified permanent strain in the !2 material. In this text, however, we will assume that the yield strength, yield point, elastic limit, and proportional limit all coincide unless ! An exception would be natural rubber, Basınç altında otherwise stated. which in fact dışarı doğru şişme:! !4 Compression causes Compression causes Tension failure of Tension failure of (sc)max " 5 material to bulgeare outa brittle does not even have a proportional limit, since stress and strain not material material to bulge out a brittle material 3 (b) which 89 (a)UCTILE linearly Instead, as shown in AND Fig. B3–8, material, is (a) (b) 3.3 Srelated. TRESS–STRAIN BEHAVIOR OF D RITTLEthis MATERIALS !6 known as a polymer, exhibits nonlinear elastic behavior. Betonarme karışım için Fig.tipik 3–10 gerilme-birim deformasyon diyagramı Fig. 3–10 Wood is a material that is often moderately ductile, and as a result it is s!P diagram for typical concrete mix s!P diagra s (ksi) usually designed to respond only to elastic loadings. The strength Fig. 3–11 F characteristics of wood vary greatly from one species to 2 another, and for (st)max " 0.4 each species they depend on the moisture content, age, and the size and Materials exhibit little or no before that exhibit little or no yielding before !0.0025!0.0020!0.0015!0.0010!0.0005 Brittle Materials. arrangement !0.0030 ofBrittle knots inMaterials. the wood. Since wood isthat a fibrous material, itsyieldingMaterials P (in./in.) failure are referred to as brittle materials. Gray cast iron is an example, 0 0.0005 referred to as brittle materials. Gray cast iron is an example, tensile or compressive characteristics will differ greatly failure when itare is loaded a stress–strain tensionhaving as shown by portion AB of thein tension as shown by portion AB of the a stress–strain diagram either parallel having or perpendicular to diagram its grain.inSpecifically, wood splits !2 curve in Fig. 3–9. Here fracture at (152 MPa) took place at sf = 22 ksi (152 MPa) took place s = 22 ksi curve Fig. and 3–9. Here fracture easily when it is loaded in tension perpendicularf to its in grain, initially at an imperfection or microscopic crack and then spread rapidly initially an imperfection or microscopic crack and then spread rapidly consequently tensile loads are almost always intended to beatapplied acrossofthe specimen, causing complete fracture. Since the appearance of !4 the across specimen, causing complete fracture. Since the appearance of parallel to the grain wood members. Compression causes brittle materials do not have (s ) " 5 material to bulge out initial cracks in a specimen is quite random, c max in a specimen initial cracks is quite random, brittle materials do not have 3 a well-defined tensile fracture stress. Instead the average fracture stressstress. Instead the average fracture stress (b) a well-defined tensile fracture !6 from a set of observed tests is generally reported. A typical failed from a set of observed tests is generally reported. A typical failed specimen is shown in Fig. 3–10a. specimen is shown in Fig. 3–10a. Steel rapidly loses its strength when Steel s!P diagram for typical concrete mix ! Compared with their behavior in tension, heated. For this reason engineers such often as brittle as in heate Compared materials, with their such behavior tension, brittle materials, require main structural members to be Fig. 3–11 gray cast iron, exhibit a much higher resistance toiron, axialexhibit compression, requi gray az castise a gelir. much as higher resistance to axial compression, as Çelikteki karbon miktarı fazla ise gevrek, sünek hale insulated in case of fire. evidenced by portion AC of the curve in evidenced Fig. 3–9. For this case any cracks by portion AC of the curve in Fig. 3–9. For this case any cracks insula or imperfections in the specimen tend to close up, and the load sıcaklıklarda at exhibit little orDüşük no yielding before sıcaklıklarda malzemeler dahaorsert ve gevrek olurlar. Yüksek ise up, daha imperfections inasthe specimen tend to close and as the load increases the material will generally bulge or become barrel shaped as s (ksi) erials. Gray cast iron is an example, increases the material will generally bulge or become barrel shaped as s (ksi) the strains Fig. 3–10b. the strains become larger, Fig. 3–10b. 40# F vebecome sünek olurlar. nsion as shown byyumuşak portion AB of the larger, 9 Like took gray place cast iron, concrete is classified as a brittle material, and it sf = 22 ksi (152 MPa) Like gray cast iron, concrete is classified as a brittle material, and it 9 also has a low strength capacity in tension. Thea characteristics of its 8in tension. The characteristics of its 8 scopic crack and then spread rapidly also has low strength capacity stress–strain diagram depend primarily on the mix of concrete (water, 7 te fracture. Since the appearance of stress–strain diagram depend primarily on the mix of concrete (water, 7 sand, gravel, and cement) and the time and temperature of curing. A !32 andom, brittle materials do not have 110# F sand, gravel, and cement) and the6 time and temperature of curing. A 6 typical example of a “complete” stress–strain diagram for concrete is . Instead the average fracture stress typical example of a “complete” 5stress–strain diagram for concrete is given in Fig. 3–11. By inspection, its maximum compressive strength enerally reported. A typical failed given in Fig. 3–11. By inspection, its maximum compressive strength 5 is almost 12.5 times greater than its tensile strength, 1sc2max = 5 ksi 4 s = EP (3–5) s = EP (3–5) the previous section, the stress–strain diagrams for most Here E exhibit represents the constant of proportionality, materials a linear relationship between stress which and is called the modulus of elasticity Young’s modulus, after Thomas Young, he elastic region. Consequently, an of increase innamed stress causes Here E represents theorconstant proportionality, which is called the who published an account of it in 1807. ate increaseofinelasticity strain. This fact wasmodulus, discovered by Robert modulus or Young’s named after Thomas Young, Equation 3–5and actually represents the equation of the initial straight76who using springs is known published an account of itasin Hooke’s 1807. law. It may be lined portion of the stress–strain diagram up to the proportional limit. thematically Hook, Equationas 3–5Hooke actually Kanunu: represents (Robert the equation of the1676) initial straightFurthermore, the modulus of elasticity represents the slope of this line. lined portion of the stress–strain diagram up to the proportional limit. Since strain is dimensionless, Eq. 3–5, E will have the same units as = EP offrom (3–5) Furthermore, thesmodulus elasticity represents the slope of this line. stress, such as ! psi, ksi, or pascals. As an example of its calculation, Since strain is dimensionless, from Eq. 3–5, E will have the same units as consider constant the stress–strain diagram for steelis shown in Fig. 3–6. Here esents proportionality, which calledModülü the stress,the such as E: psi,of ksi, or pascals. As an example of its calculation, Elastisite Modülü veya Young s = 35 ksi and P = 0.0012 in.>in., so that pl pl lasticity orthe Young’s modulus,diagram named after Thomas Young, consider stress–strain for steel shown in Fig. 3–6. Here ds an itGerilme-birim in = 35 ksiofand Ppl1807. =s0.0012 in.>in., deformasyon ilişkisindeki orantılılık (proportionality) sınırlarını kullanarak: pl account pl 35 ksiso that 3 E = the equation = = 29110 2 ksi –5 actually represents of the initial straightPpl 0.0012 in.>in. s 35 the ksi proportional3 limit. of the stress–strain diagram up to E = = = 29110 2 ksi theAs modulus the slope of this Ppl represents in.>in. shownof in elasticity Fig. 3–13, the0.0012 proportional limit for aline. particular type of ! dimensionless, from Eq. 3–5, E will have the same units asgrades of steel, steel alloy depends on its carbon content; however, most As ksi, shown in Fig. 3–13, the example proportional limit for a particular type of s from psi, or pascals. As an of its calculation, için farklı karbon içerikleri elastisite the softestÇelik rolled steel to the hardest tool steel, have modülünde about the önemli değişikliğe sebep olmaz: steel alloy depends onfor its carbon content; stress–strain diagram steel shown in however, Fig. 3–6. most Heregrades of steel, s (ksi)sosteel from softestin.>in., rolled and Ppl the = 0.0012 that to the hardest tool steel, have about the E = spl Ppl = s (ksi) 35 ksi 180 0.0012 in.>in. 180 160 = 2911032 spring ksi steel (1% carbon) spring steel (1% carbon) a particular n Fig. 3–13, the proportional limit for type of 160 140 pends on its carbon content; however, most grades of steel, est rolled steel to140 the hardest toolhard steel, steelhave about the 120 s (ksi) 180 160 140 120 100 80 60 40 20 (0.6% carbon) heat treated hard steel (0.6% carbon) heatsteel treated machine 100 80 spring steel (0.6% carbon) (1% carbon) machine steel 80 60 structural steel (0.6% carbon) (0.2% carbon) 60 40 structural soft steel steel (0.2% (0.1% carbon) carbon) 40 20hard steel soft steel (0.6% carbon) (0.1% carbon) 20heat treated 0.002 0.004 0.006 0.008 0.01 ! machine steel Fig. 3–13 (0.6% carbon) 0.002 0.004 0.006 0.008 0.01 120 100 structural steel (0.2% carbon) soft steel (0.1% carbon) 0.002 0.004 0.006 0.008 0.01 P (in./in.) P (in./in.) Fig. 3–13 P (in./in.) Fig. 3–13 !33 3.5 Strain Energy As a material is deformed by an external loading, it tends to store energy internally throughout its volume. Since this energy is related to the strains in the material, it is referred to as strain energy. To obtain this strain energy consider a volume element of material from a tension Birim Deformasyon Enerjisi test specimen. It is subjected to uniaxial stress as shown in Fig. 3–15. s This stress develops a force ¢F = s ¢A = s1¢x ¢y2 on the top and bottom faces of the element after the element of length ¢z undergoes a 3 vertical displacement P ¢z. By definition, work is determined by the product of the force and displacement in the direction of the force. Since the force is increased uniformly from zero to its final magnitude ¢F !z when the displacement P ¢z is attained, the work done on the element by the force is equal to the average force magnitude 1¢F>22 times the 92 C H A P T E R 3 M E C H A N I C A L P R O P E Rdisplacement T I E S O F M AT E R L S This “external work” on the element is equivalent to PI A¢z. !x !y the “internal work” or strain energy stored in the element—assuming ARPOTPEERR 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S T I E S O F M AT E R I A L S L P R O P E R T I E S O F M AT E R I A L S 92energy is lost C H Ain P Tthe ER 3 Mof E C heat. H A N I CConsequently, A L P R O P E R T I E Sthe O Fstrain M AT Eenergy RIALS that no form s TF E RM3AT! EM EC ATis E R¢U I A L S = 11 ¢F2 P ¢z = 11 s ¢x ¢y2 P ¢z. Since the volume of the O P E RC T IHEA S PO RIA L SH A N I C A L P R O P E R T I E S O F M ¢U 2 2 Strain Energy 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S 1 Fig. 3–15 element is ¢V =H A¢x ¢y ¢z, then 9 2 C H A P T E R 3 M E C N I C A L P R O P Ebütün R T I¢U E S O= F 2 MsP AT E¢V. Rboyunca IALS Strain Energy Strain Energy Bir cisim dış yük altında deformasyon yaptığında hacmi bir iç enerji O F M AT E R I A L S M E C H A N I C A LStrain P E R T I E S O F M AT E R I A L S it is sometimes convenient specify the Energy strain HAPTER 3 MPE RCO H A N I C AEnergy L P R O P E R T I E S O F M As AT Ea R Imaterial A L SFor applications, is deformed by an external loading, it tends totostore energy Strain Strain Energy energy per unit volume of material. This is called the strain-energy Strain Energy depolar. Bu enerji içindeki birim deformasyonlara bağlı olduğundan As aan material iscisim deformed external loading, it tends toHstore energy internally throughout Since energy is related tobuna the birim As a material is deformed by external loading, it by tends to storeits energy 9an 2 and C Hvolume. APTER 3 M E Cthis A N I C A L P R O P E R T I E S O F M AT E R I A L S Strain Energy density, itstore canit expressed As a material is deformed by an external loading, tends toSince energy throughout its volume. this energytoisasasrelated to As the strains init the isthe referred strain energy. To obtain is this internally throughout itsinternally volume. Since this energy is material, related tobe Strain Energy a material deformed by an external load Strain Energy aSince material is deformed byto an external loading,To it tends tothis store energy deformasyon enerjisi denmektedir. Eksenel bir çubuk için birim Energy As ainternally material deformed by external loading, itenergy tends to store energy strains inAs the itenergy is referred as strain energy. obtain throughout its volume. thisEnergy is related the çekmeye consider a to volume element ofmaruz material from a tension Strain strains in theis material, itStrain is an referred tomaterial, as strain strain energy. To obtain this internally throughout its volume. Since thi As aits material is deformed by an aexternal loading, itthe tends to this store¢U energya1 istension internally throughout its volume. Since energy related to the nternally throughout volume. Since energy is related to energy consider volume element of material from strains in the material, itvolume is referred tothis as energy. To this strain energy consider astrain element ofstrain material from aobtain tension test specimen. ItYukarıdaki is subjected to As uniaxial stressisas shown inby Fig. a= material deformed an3–15. external loading, it tends to to as stors strains in the material, italt is referred deformasyon enerjisini belirleyelim. şekilde, normal gerilme elemanın üst Energy ve u = sP (3–6) rial is deformed by an external loading, it tends to store energy strains in the material, it is referred to as strain energy. To obtain this Strain internally throughout its volume. Since this energy is related to the As ais material is by ansubjected loading, tends to as store energy trains in the material, its is to asis strain energy. Toexternal obtain this test specimen. It is to uniaxial stress shown Fig. 3–15. As adeformed material deformed by an loading, it tends to store teststrain specimen. Itconsider subjected to uniaxial stress as shown in Fig. 3–15. This stress develops ait ¢F = s ¢A =in s1¢x ¢y2 on the top and energy areferred volume element ofexternal material from aforce tension ¢V 2 energy internally throughout its volume. Since this energy is related strain energy consider a volume element o strain energy consider athe volume element oftomaterial from a tension throughout its volume. Since this energy tothis strains in thestress material, it of isisreferred to as strain energy. To¢y2 obtain this top internally throughout its Since energy isafter related the train energy consider aforce volume element material from as tension s stress This develops arelated force ¢F =the ¢A = s1¢x the internally throughout its volume. Since energy ison to and the This develops ¢F to = uniaxial s ¢Avolume. = stress s1¢x ¢y2 on top this and strains inrelated the material, itundergoes is referred as strain to energy. To ob bottom faces of the element the element of length ¢z a istosubjected test specimen. It isa subjected as shown in Fig. 3–15. yüzeyinde ! kuvvetini oluşturmaktadır. Burada ! kenarı test specimen. Itdeformed uniaxial st test specimen. It is subjected to uniaxial stress as shown in Fig. 3–15. isuzayacaktır As a material by an external load the material, itIt isisstrain referred tomaterial, asconsider strain To obtain this a referred volume element ofelement material from a ¢z tension 3 specimen. strains in energy the it isthe toafter as¢z strain energy. obtain this est to uniaxial stress aslength shown in Fig. 3–15. bottom faces ofenergy. element the ofTo length undergoes a a volume strains in the material, itIf is referred to as strain energy. To obtain this bottom faces of thesubjected element after the element of undergoes a strain energy consider element of material from a vertical displacement P ¢z. By definition, work is determined by the This stress develops a force ¢F = s ¢A = s1¢x ¢y2 on the top and the material behavior is linear elastic, then Hooke’s law applies, s This stress develops a force ¢F = s ¢A = s1¢x ¢y2 on the top and internally throughout its volume. Since thi s This stress develops a force ¢F = s ¢A = testaenergy specimen. It is material subjected uniaxial as shown Fig. 3–15.a It rgy consider astrain volume element ofs from adetermined tension consider a consider volume element ofstress material from adetermined tension his stress develops force ¢F = ¢A =product s1¢x ¢y2 onforce the topby and vertical displacement Pto¢z. By definition, work isingöre byisthe strain energy a volume element of material from tension test specimen. subjected to uniaxial stress as shown in F vertical displacement P ¢z. By definition, work is the of the and displacement in the direction of the force. Since ve yapacağı deplasman ! olacaktır. Tanıma iş, kuvvet ile kuvvetin etkime bottom faces ofThis the after theaaselement ofin ¢z=therefore undergoes aon s = EP, and we can express the elastic strain-energy density bottom faces of the after the element of length ¢zstrains undergoes a of in the material, itinis referred as s bottom faces the element after the to elemen stress develops force ¢F =length selement ¢A s1¢x ¢y2 the top and men. isofsubjected toelement uniaxial stress Fig. 3–15. specimen. Itspecimen. is subjected to uniaxial stress as shown in Fig. 3–15. ottomItfaces oftest the element after the element of length ¢z undergoes adirection product of the and displacement in the of the force. test Itshown isdirection subjected to uniaxial stress as shown in Fig. 3–15. This stress develops amagnitude force ¢F ¢F = s ¢A = s1¢x ¢y2 on the product the force and displacement inforce the the force. Since the force issof increased uniformly from zero to its Since final 3 terms of the uniaxial stress as vertical displacement P ¢z. By definition, work is determined by the vertical displacement P ¢z. By definition, work is determined by the strain energy consider a volume element o bottom faces of element the element length ¢z undergoes aoftop vertical displacement definition, sertical develops a increased force ¢F =the s ¢A =the s1¢x ¢y2 on the top stress develops a deplasmanın force ¢F = s ¢A =and onattained, the topthe and force is increased from zero to its final magnitude ¢F P uniformly ¢z. By definition, work isforce determined by the s displacement This stress develops auniformly ¢F =s1¢x sof ¢A =elde ¢y2 on the and çarpımı ile edilir. Kuvvet sıfırdan başlayıp ! P ¢z.ofBy !This z doğrultusundaki bottom faces the element after the element length ¢z und the force isof from zero toafter its final magnitude ¢F when the displacement P¢y2 ¢z iss1¢x work done onspecimen. the element product of the force and displacement in the direction of the force. Since product the force and displacement in the direction of the force. Since test It is subjected to uniaxial st 3 vertical displacement Pthe ¢z. By definition, is determined by the ces of the element after the element of ¢zthe awork product thedefinition, force and displacement in the bottom ofisthe element after element of element length ¢z a element displacement Pundergoes ¢z attained, the work done on the roduct of the force andfaces displacement in length the direction ofisafter the force. Since bottom faces of the element the of undergoes length ¢z undergoes aP ¢z.of vertical displacement By work is determined when the displacement Pwhen ¢z attained, done on the element by the force is equal to thefrom average force magnitude 1¢F>22 times the 2 the force iswork increased uniformly zero to its final magnitude ¢F sthe This stress develops a uniformly force ¢F = s ¢A = the force is¢z. increased uniformly from zero to its final magnitude ¢F stimes 1üniform product of the and displacement in the direction of force. Since splacement P By definition, work is determined by the deplasmanına karşılık gelen ve son şiddeti olan ! değerine bir artış gösterdiğinden, vertical displacement P ¢z. By definition, work is determined by the by the force is equal to the average force magnitude 1¢F>22 the he force is increased uniformly from zero to its final magnitude ¢F the force is increased from zero vertical displacement P ¢z. By definition, work is determined by the product of the force and displacement in the direction of the for by!zthe force is equal to the average force magnitude 1¢F>22 times the the (3–7) after the elemen u on = the displacement P ¢z. This “external work” element is element equivalent to when the displacement P ¢z is attained, work done on the bottom faces of the element ! x when the displacement P ¢z is attained, the work done on the element ! z the force is increased uniformly from zero to its final magnitude ¢F displacement Pdisplacement ¢z. This “external work” on the element isthe equivalent to uniformly the force and product displacement inforce the work” direction of the Since of“external the and in on the direction the force. when the displacement ¢z is attained, the work done the element 2isforce. E product of the force and displacement theof direction ofSince Since when the displacement is attained, the y P the force from zeroP ¢z to its final magnit displacement onthe the element is the equivalent to 3force. !xs P ¢z.!This “internal work” orin strain energy stored inincreased the byto the force is equal to average force magnitude 1¢F>22 times the vertical displacement ¢z. By definition, !y by the force iseleman equal to the average force 1¢F>22 times the when the displacement Pmagnitude ¢z is attained, work done on the element üzerinde gerçekleştirilen iş the ortalama kuvvet şiddeti ! element—assuming ile deplasman ! P the the “internal work” or strain energy stored in the element—assuming !magnitude zin the force is increased from zero to its final magnitude ¢F y the force isuniformly equal to average force 1¢F>22 times the is increased from zero its final magnitude ¢F the force isuniformly increased uniformly from zero to its final magnitude ¢Fthe when the displacement P ¢z is attained, the work done on the by force is equal to average force m the “internal work” or the strain energy stored the element—assuming that no energy is“external lost in thework” form ofon heat. Consequently, the strain energy displacement Pthe ¢z. This the element is equivalent to force and displacement in the product of the s that no energy is lost in form of heat. Consequently, the strain energy by force is equal to the average force magnitude 1¢F>22 times the ! x the displacement P ¢z. This “external work” on the element is equivalent to when displacement P ¢z is attained, the work done on the element isplacement P ¢z. This “external work” on the element is equivalent to displacement P ¢z is attained, the work done on the element when the displacement P ¢z attained, the work done on element 1 1 by the force is equal to the average force magnitude 1¢F>22 t displacement P ¢z. This “external work” on that no energy is lost in the form of heat. Consequently, the strain energy y ! Modulus of In the particular, when thekaybı stress s reaches s ¢UPiswork” ¢U= =1or ¢F2 P Resilience. ¢z =¢z. 12 stored s ¢x biçiminde ¢y2 Pvolume ¢z.element—assuming Since the volume of the 1 eşittir. 112 strain the=This “internal energy the force is increased uniformly from zero !the xin değerinin çarpımına Sistemde ısıya dönüşme bir enerji olmadığını 1 work” 1P ¢U is ¢U 1 ¢F2 ¢z s ¢x ¢y2 P Since of the displacement ¢z. “external work” on the element is equivalent to by the force is equal to the average force magnitude 1¢F>22 times the he “internal work” or strain energy stored in the element—assuming y ! by the force is equal to the average force magnitude 1¢F>22 times the “internal or strain energy stored in the element—assuming ce is equal to the average force magnitude 1¢F>22 times the s displacement P ¢z. This “external work” on the element is equiv 1 2 2 pl ¢Uthe is ¢U = 1 ¢F2 P ¢z = 1 s ¢x ¢y2 P ¢z. Since the volume of the !x work” or3–6 strain stored the proportional the strain-energy density, as“internal calculated by Eq. !¢y z limit, ! xform 2 2 1¢z, Fig. 3–15 element ¢x then ¢U = 2 sP ¢V. that no=energy is!¢z, lost in = the of heat. Consequently, the the strain energy when the displacement P ¢z isenergy attained, the yis ¢V Fig. 3–15 element ¢x ¢y then ¢U =1 the sPelement—assuming ¢V. s isdisplacement the work” or stored in Pthen ¢z. This “external work” on the element is equivalent to resilience, hat no energy in“internal the form ofisheat. Consequently, the strain energy displacement P 1strain ¢z. This “external work” on the element is equivalent to i.e., ent P ¢z. work” on the element isenergy equivalent to noisThis energy is lost in the form of¢V heat. Consequently, the strain energy the “internal work” or strain energy stored inform the of element—a 2to 1 element ¢V =lost ¢x ¢y ¢z, ¢U = sP ¢V. ! x!ythat !x1“external or 3–7, is referred as the modulus of that no energy is lost in the heat. Con varsayarsak, eleman üzerindeki dış iş, iç işe veya eleman içinde depolanmış olan birim 2 ¢U is ¢U = 1 ¢F2 P ¢z = 1 s ¢x ¢y2 P ¢z. Since the volume of the For applications, it is sometimes convenient to specify the strain by the force is equal to the average force m 1 1that 1is 2 of 2 s the For it sometimes convenient to specify theis strain the work” oris strain energy stored inthe element—assuming ¢U iswork” ¢U =or12= ¢F2 Penergy =sometimes 1stored s1applications, ¢x ¢y2 Pthe ¢z. Since the volume ofstrain the no in form heat. Consequently, the “internal work” or strain energy stored inthe the element—assuming 1 heat. Consequently, 1 nalFor strain in the element—assuming that no energy energy lost in is the¢U form of the strai 1strain ¢U is ¢U Pis energy ¢z slost ¢x ¢y2 P¢V ¢z. Since the volume of the 2=“internal applications, it¢z convenient to specify 2 element ¢U = 1 ¢F2 P ¢z = 1 s ¢x ¢y2 P ¢ s Fig. 3–15 12 ¢F2 is = ¢x ¢y ¢z, then ¢U = sP ¢V. displacement P ¢z. This “external work” on energy per unit volume of material. This is called the strain-energy 1 1 1 2 2 !the 1=¢V. energy per unit volume of form material. This isthe called that no energy lost the of heat. Consequently, the strain energy lement isper ¢V =that ¢x ¢y ¢z, then ¢U =the sP energy isConsequently, lost in form heat. Consequently, strain ¢U ¢U =material. 1then ¢F2 PThis 1in s ¢x ¢y2 PSonuç ¢z. Since volume ofstrain-energy the ¢U isxenergy ¢U = 112 ¢F2 P ¢z = 112 s 2¢x ¢y2 P ¢z. Since the volum ergy iss lost in the form heat. the strain energy 1 2¢z 2enerjisine 2of !ythe element isunit ¢V =no ¢x ¢z, ¢U =is sP ¢V. energy volume is called the strain-energy deformasyon eşit olacaktır. olarak:! uisof r ¢yof For applications, it is sometimes convenient to specify the strain 2 Fig. 3–15 the “internal work” or strain energy stored element is ¢V = ¢x ¢y ¢z, then ¢U = s 1 1 1 1 density, and it can be expressed as 1 1 2 density, and it can be expressed as2¢z. ¢U isPas ¢U = 1convenient Pvolume ¢z =3–15 1specify ¢x ¢y2 Pthe ¢z.volume Since the volume of2 the applications, isit¢x sometimes to specify theSince strain ¢U is =is 1¢V Pconvenient ¢z 12 s ¢x ¢y2 of is the = ¢x ¢y ¢z, then ¢U = sP ¢V. Fig. element ¢Vthe ¢y ¢z, then ¢U = 2 sP ¢V. =For 112For ¢F2 P it ¢zcan =element 1it12 ¢U s ¢y2 ¢z. Since of the 2=¢F2 2Ps 2 ¢F2 density, and be expressed applications, is sometimes to the pl energy perthe unit volume of material. is called strain-energy 1= s¢x that noapplications, energy is lost itin is thesometimes form of heat. Con 1 strainThis 1 1 For conven 1 ¢x Fig. 3–15 element is ¢V = ¢x ¢y ¢z, then ¢U = sP ¢V. s nergy per unit volume of material. This is called the strain-energy element is ¢V = ¢y ¢z, then ¢U = sP ¢V. For applications, it is sometimes convenient to specify the strain it is sometimes convenient ur olacaktır. =For sapplications, (3–8) 1 to specify th 2 ¢V = ¢xper ¢y unit ¢z, then ¢U = 2material. sP ¢V. This 2the ! strain-energy plPpl = Hacim ! of olduğuna energy volume called density, and itis can begöre expressed as ¢U is ¢U =unit 112 ¢F2 P ¢z of = 1material. ¢y2 P ¢ 2specify Eenergy permaterial. volume 2 s ¢x ¢U 1strain For it is the sometimes to the2 strain ensity, anditit is cansometimes be expressed as applications, For applications, it is to sometimes convenient the ¢U is convenient 1to specify energy perconvenient unit volume of material. This called the strain-energy energy per unit volume of This is called the This strai plications, specify strain u =3–15 = sP (3–6) ¢U 1 density, and it can be expressed as Fig. element is ¢V = ¢x ¢y ¢z, then ¢U = 12 s u = = sP (3–6) density, and it can be expressed as energy per unit volume of material. This is called the strain-energy energy per unit volume of material. This is called the strain-energy ¢V 2 u = = sP (3–6) density, and it can be expressed as Uygulamalarda birim hacimdeki birim deformasyon enerjisinin belirlenmesi kimi zaman density, and it can be expressed as P ¢V 2 r unit volume of Pmaterial. This¢V is called strain-energy For applications, it is sometimes conven pl 2 the expressed and as u = ¢U = 1 sP density,as anddensity, it can be expressed ¢U 1it can beas (3–6) d it can be expressed energy per unit volume This From the elastic region of the stress–strain diagram, Fig. 3–16a, notice u = = sP (3–6) Modulus of resilience u ¢V 2 daha uygun olmaktadır. birim deformasyon enerjisi yoğunluğu denmektedir veof material. ¢U 1 If Buna r ¢U 1 ¢U 1expressed the material behavior is linear elastic, then Hooke’s law applies, ¢U 1 ¢V 2= behavior If the material is linear elastic, then Hooke’s law applies, u = sP (3–6) density, and it can be as that u is equivalent to the shaded triangular area under the diagram. r u = = sP u = = sP If the material behavior is linear elastic, then Hooke’s law applies, u = = sP (3–6) (a) ¢U 1 ¢V 2 1express s = ¢U EP, and therefore weelastic canresilience express therepresents elastic strain-energy density in 2 to s = 1EP, and therefore we can the strain-energy density in ¢V ¢V 2 ¢V 2 a material’s ¢U Physically the ability of the material u = = sP (3–6) u = = sP (3–6) s = EP, and therefore we can express the elastic strain-energy density in Ifuniaxial the material behavior linear elastic, then Hooke’s law applies, u = Fig. = sPthe (3–6) ¢V isstress 2 as any of stress as terms of the ¢V 2uniaxial If theofmaterial behavior is elastic, then Hooke’s law applies, 3–16 absorb energy damage todensity ¢U 1 ¢Vterms terms the uniaxial stress as 2linear s =şeklinde EP, and therefore we canwithout express thepermanent elasticlineer strain-energy in ! ifade edilmektedir. Malzeme elastik the ise, material. Hooke kanunu, u =Hooke’s = law sP If the is linear elastic, then Hooke’s law = EP, and material therefore we express theof elastic strain-energy in If the material behavior is linear elastic, then Ifbehavior thecan material behavior linear elastic, thenapplies, Hooke’s law applies, If the material behavior is linear elastic, ¢V 2 terms theisuniaxial stress density as 2 2 and the material behavior isthe linear elastic, then Hooke’s law applies,we can express the elastic strain-energy d erms the uniaxial stress as If EP, s of = EP, and therefore we can express the elastic strain-energy density in s 1s material behavior iscan linear then Hooke’s law applies, 1= s =linear EP, therefore express strain-energy density intherefore 2we s = EP, and therefore we can express the ela material behaviorIf elastic, then Hooke’s lawelastic, applies, ! sisthe ,sand geçerli ve elastik birim deformasyon enerjisi yoğunluğu eksenel gerilme stherefore 1olur (3–7) u =elastic u strain-energy = density = EP, and we can express the elastic density in terms of the uniaxial stress as (3–7) suniaxial =terms EP, and therefore we can express the elastic strain-energy in terms of the stress as (3–7) u = 2 E of the uniaxial stress as 2 2 E terms of the uniaxial stress asis linear elastic, If the material behavior nd therefore we can express the elastic strain-energy density in 1s 2 2E of the uniaxial s terms of theterms uniaxial stress as stress as (3–7) u = s 1 s = EP, and therefore we can express the ela he uniaxial stress as (3–7)2 E u = 2 2In particular, when the stress s reaches 2 s 1 2 E terms of the uniaxial stress as Modulus of Resilience. s Resilience. 1of 1 solacaktır. Modulus In particular, when the stress s reaches 2 cinsinden: ! Inuparticular, u = 1 s2 2 stress s 1s Modulus of Resilience. when the reaches (3–7) u = (3–7) = s 1 2 E the proportional limit, the strain-energy density, as calculated by Eq. 3–6 u = spl 2 (3–7)s reaches = the density, calculated by Eq. 3–6 2 Eu limit, 2 density, E theof 1s (3–7)theasstress uproportional =Resilience. s Instrain-energy particular, the proportional limit,u the calculated by Eq. 3–6 2 E2 2 resilience, E 3–7,InisModulus referred towhen as as the modulus of i.e.,of when (3–7) =orstrain-energy 2referred Estress 1s Modulus of Resilience. particular, the s reaches s or 3–7, is to as the modulus resilience, i.e., proportional limit, the strain-energy density, as calculated by Eq. 3–6 or 3–7, is referred to as the2modulus resilience, i.e., E theof uthe = stress s Modulus of Resilience. In particular, when he proportional limit, the strain-energy density, as calculated by Eq. 3–6 2 E or 3–7, is referred assthe modulus ofstress resilience, i.e., of Resilience. In to particular, when the s reaches Modulus Modulus oftouModulus Resilience. Inof when the stress s reaches sparticular, Modulus of Resilience. In when particular, when stress s reaches plresilience, Modulus of Resilience. In particular thesthe proportional limit, the strain-energy density, as calculated by r 3–7, is referred as the modulus i.e., of Resilience. In particular, the stress reaches r 2 the proportional limit, the strain-energy density, calculated by Eq. 3–6 theResilience. proportional limit, the density, assstrain-energy calculated byasdensity, Eq. 2 by Eq. s of In particular, when the reaches s thestrain-energy proportional limit, the calculated 3–6 pl as 1pl 1 s3–6 orby 3–7, is13–6 referred to asproportional the modulus limit, of resilience, i.e., 2stress the the strain-energy dens s theor proportional limit, the strain-energy density, as calculated Eq. pl ur 1 s urby = s3–6 (3–8) 3–7, isthe referred to as the modulus ofEq. resilience, i.e., Modulus of Resilience. In particular pl as the 1 is referred 1 to plPplof=resilience, 3–7, modulus or 3–7, is referred modulus of resilience, i.e., tional limit, the density, as ur E = spli.e., Ppls2= (3–8) to as the modulus of resili 2 or 3–7, is referred uor s Pplthe = calculated (3–8) orstrain-energy 3–7,toisas referred toplas modulus of resilience, i.e., s2 r = pl pl 2 E 1 1 the proportional limit, the strain-energy dens 2 2 2 2 E eferred to as the modulus of resilience, i.e., ur = splPpl = (3–8) 1 1 spl ur P or 3–7, is referred to as the 2 modulus of resili 2 2 E ur ur = splPpl =P (3–8) 1 1 spl Ppl 2 2 the elastic2 region E 21 of the stress–strain s2pl ur s 1 u = s P = 2 From diagram, Fig. 3–16a, notice r pl pl 1s 1 pl P s of Presilience ur 1 2 uFrom su (3–8) 2 E plP 1=spl the 1s elastic region of diagram, Fig. 3–16a, 2notice ur the stress–strain r1diagram, plP pl pl the elastic == (3–8) 1 1s From region of uthe 3–16a, Modulus of resilience u stress–strain pl =notice equivalent triangular area under the diagram. (3–8) 2plE s2plthe Pplr shaded =Fig. (3–8) r u=rr1isss plplPplur= = to 2 2 E 1 that u = s P = (a) r pl pl From the elastic region of the stress–strain diagram, Fig. 3–16a, notice that u is equivalent to the shaded triangular area under the diagram. 2 2 E 2 2 E 1 1s that ur isofequivalent to area under the diagram. Modulus resilience r P ability of the material to ur = ur(a) splthe Ppl shaded = (3–8) Physically atriangular material’s represents the Ppl resilience ur = 2 splPpl = 2 rom the elastic region of the stress–strain diagram, Fig. 3–16a, notice P 2 2 E that u is equivalent to the shaded triangular area under the diagram. P Physically a material’s resilience represents the ability of the material to r Physically a material’s resilience represents the ability of the material to to the material. 2 2 Fig. 3–16 P(a) absorb energy without anyunder permanent damage From the elastic region of thetostress–strain diagram, Fig. 3–16 hat ur is equivalent to3–16 the shaded triangular area the udiagram. Modulus of resilience Physically athe material’s resilience the ability the material r any Fig. absorb energy without permanent damage toPofthe material. absorb energy without any permanent damage to the material. Ppl represents From the elastic region of stress–strain diagram, Fig. 3–16a, notice From the elastic region of the stress–strain diagram, Fig. 3–16a, notice ence u that unotice is the equivalent to the shaded triangular area under the d lus of resilience urFrom the r P rto hysically a material’s resilience represents the ability of the to 3–16 region ofenergy stress–strain diagram, Fig. 3–16a, absorb without anymaterial permanent damage material. (a) Parea the elastic region of the stress–strain diagram, Fig. 3–16a, notice e ur FromFig. pl that ur elastic isthat equivalent to the shaded triangular under theunder diagram. From the elastic region of stress–strain ur is equivalent to the shaded triangular area the diagram. Physically a material’s resilience represents thethe ability of the ma Modulus of resilience u (a) energy r elastic region of the stress–strain diagram, Fig. 3–16a, notice bsorb without any permanent damage to the material. thatPhysically ur is to equivalent to triangular the shaded triangular area under the diagram. From the elastic region of the stress–strain that ur is equivalent the shaded area under the diagram. 3 ! 4 Modulus ofof resilience uof a material’s resilience represents the ability the material Physically a material’s resilience represents the ability the to material toany r energy that u is equivalent to the shaded triangul Fig. 3–16 absorb without permanent damage to the material. r equivalent to the shaded triangular area under the diagram. (a) a material’s resilience represents the material to that ur is equivalent to the shaded triangul Physically a Physically material’s resilience represents the ability of the the ability material to Fig. 3–16 absorb energy any permanent damage to theofmaterial. absorbwithout energy without any permanent damage to the material. (a) Physically a material’s resilience represents t a material’s resilience represents the ability of the material toto the material. absorb energy anydamage permanent damage Physically a material’s resilience represents t absorb energy without any without permanent to the material. Fig. 3–16 absorb energy without any permanent dama ergy without any permanent damage to the material. Fig. 3–16 absorb energy without any permanent dama 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3 MECHANICAL PROPERTIES OF M AT E R I A L S T I E S O F M AT E R I A L S A N I C A L P R O P E R T I E S O F M AT E R I A L S OHE PAC EPR E M IA LI ESASL OPFR O R SI C 3O C N E RETRI EI A S LO AAPNTIECRA L3 P RC M HTTAEI N AF LM P ERAT OH PEA ERR TI C MPAT SF A L P R O P E R T I E S O F M AT E R I A L S CHAPTER 3 96 CHAPTER 3 MECHANICAL PROPERTIES OF M AT E R I A L S M AT E R I A L S HAPTER 3 M E C H A N I C A L P R O P E R T I E SEXAMPLE O F M AT E R I A L S M E C9H6A N I C A L P RC O P E R T I E S O F M AT E R I A L S 3.3 An aluminum rod shown in Fig. 3–20a has a circular cross section and is XAMPLE 3.3 3.3 An aluminum rod shown in Fig. 3 subjected to an axial load of 10 kN. If a portion of the stress–strain s (MPa) minum rod in Fig. 3–20a hasinain circular cross section andcross is approximate subjected to an axial load of 10 Anshown aluminum rod shown Fig. 3–20a hasdetermine a circular section and is diagram is shown Fig. 3–20b, the elongation EXAMPLE 3.3 An aluminum rod in 3–20a has a circular An aluminum rod shown in Fig.shown 3–20a hasFig. a circular cross sectioncross and issection and is ! oftothe LE 3.3 An aluminum shown in Fig. 3–20a has a circular cross section and is ed sto(MPa) ansubjected axial load ofanrod 10 kN. If a portion of the stress–strain diagram is shown in Fig. 3–20b, axial 10 kN. a portion stress–strain rod when theof load is applied. Take Eof =the 70 GPa. An aluminum rod shown inload Fig. 3–20a has aIf circular cross section and is If al subjected to an axial load of 10 kN. portion of theCstress–strain subjected to an10 axial load of 10 kN. If a stress–strain portion of athe stress–strain subjected to an axial load of kN. If a portion of the m isFshown in Fig. 3–20b, determine the approximate elongation 96 H A P T E R 3 E Crod Hsection AN I C Aand Lthe P is Rload O P Eis RT IES 60 ofM the when applie An aluminum rod shown in Fig. 3–20a has a circular cross diagram in Fig.of3–20b, determine the approximate elongation 56.6 subjected to is anshown axial load 10 kN. If aFig. portion of the stress–strain s diagram (MPa) Dairesel kesitli bir aluminyum çubuk 103–20b, kNhas eksenel yüke maruzdur. için şekilde F Aluminyum isshown shown in Fig. determine approximate elongation is shown in 3–20b, determine the elongation An rod in70 Fig. 3–20a aapproximate circular cross section and is a portion of 50axial the diagram shown in Fig.aluminum 3–20b, determine the approximate elongation od whenof the load iswhen applied. Take E =diagram 70 GPa. al 20 mm subjected to an load of 10 kN. If the stress–strain the rod the load is applied. Take E = GPa. 15 diagram is shown in of Fig. 3–20b, determine the approximate elongation oftothe rod when is applied. Take Ealof=the 70 stress–strain GPa. the rodiswhen the load isalload applied. Take Emm 70 GPa. sYportion ! 40 56.6 60 al = subjected anTake axial ofload 10 kN. If 2 of the rodverilen when load E =the GPa. F the al diagram is aCshown ingöre, Fig. 3–20b, determine thealtında approximate elongation A applied. B70 gerilme-birim deformasyon ilişkisi verildiğine verilen yükleme meydana 50of theFrod when the load is applied. Take E = 70 GPa. al M EPC H!A N I C A L P R O P E R T I20 E Smm Odiagram F M AT E Ris I A15 Lshown S mm in Fig. 3–20b, determine the30approximate elongation A 0.0450 20 mm 60 of the rod when the load is applied. Take E = 70 GPa. 15 mm Y ! 40BC 10 kN 56.6 al 20 mm 20 mm 20 10 kN 15 mm EXAMPLE 3.3 15 mm F PBC ! 0.0450 20 mm A 10 kN 15 mm Take 50 ofB the rod when the is applied. 70 GPa. değerini kullanınız. C 30 A yaklaşık B load 20 mm CB 3 gelecek uzama belirleyiniz. ! Eal =10 B 15 A mm A miktarını C F C s ! 40 10 kN 400 B .0450 600 mm 20 mm 04 20 0.06 P10 C mm 15 mm 0.0450 10 kN O ! 0.0450AY PA BC ! BCkN 10BkN 10 kN 10 kN 10 kN C 0.02 0.04 0.06 30 0.045010 20 mm 10 kN 10 kNA15 mm s (MPa) B (a) C 10 kN 10 kN 600 mm 20 O 0.04 600400 mmmm 400 mm 0.0450 600kN mm 400 mm (b) mm 0.02 0.04 0.06 02 0.06 APBC !600 B400 mm 10 C 3 10 600 mm in Fig. Fig. 400 mm An aluminum shown 3–20a has a circular cross section and is 6 (a) 600rod P(b) (a) 400 mm3–20 (a) mm BC ! 0.0450 (a) 10 kN 10 kN (b) subjectedO to an 0.02 axial load of 10 kN. If a portion of the stress–strain 600 mm 400 mm 0.04 0.06 (a) 10 kN (a) 600 mm An aluminu subjected to diagram isFigs of the rod w Fig. 56.6 60 (a) Fig.3–20b, 3–20 determine Fig.400 3–20 diagram is 3–20 shown in(b) Fig. the approximate 600Fig. mm3–20 mmelongation 0.06 F SOLUTION 50 SOLUTION Fig. 3–20 of the rod when theFig. load is applied. Take Eal = 70 GPa. 3–20 (a) For the analysis we will neglect the localized deformations at the point (b) F sY ! 40Fig. 3–20 For the analysis we will neglect t ION SOLUTION SOLUTION of loadSOLUTION application and where20the rod’s cross-sectional area suddenly 30 mm 15 mm analysis we the will analysis neglect the localized deformations atdeformations thethe point For the analysis we will neglect deformations Forwill the neglect analysis we will neglect localized deformations at the point at the point of load application and where t SOLUTION Fig. 3–20 For we the localized atthe thelocalized point SOLUTION changes. (These effects will be discussed in Sections 4.1 and 4.7.)20 ( These effects will be Bdeformations SOLUTION ofthe load application and where cross-sectional area suddenly changes. application and where rod’s cross-sectional area suddenly of and where the rod’s cross-sectional area suddenly PBC ! 0.0450 Cthe Forload the application analysis weload willAapplication neglect the localized at therod’s point of and where rod’s cross-sectional suddenly 10 kN of the analysis we the will neglect the localized deformations atarea the point 3 Throughout the midsection of each segment the normal stress and Throughout thepoint midsection ! 0.0450 PBCFor 10 10 kNbe kN in changes. ( These effects will bein discussed Sections 4.1 and 4.7.) For the analysis we10will neglect localized deformations at the changes. ( These effects willinand be discussed Sections 4.1 and the 4.7.) s. (These discussed in Sections 4.1 4.7.) ofeffects load application and where the rod’s cross-sectional area suddenly changes. (will These effects will be discussed Sections 4.1 and 4.7.) of load application and where the rod’s cross-sectional area suddenly SOLUTION deformation are uniform. deformation are uniform. Throughout the midsection ofstress each segment theOnormal stress and the midsection each segment the4.7.) normal stress and of load application and where the rod’s area suddenly houtchanges. theThroughout midsection of Throughout each segment the normal stress and changes. (These effects will be discussed inof Sections 4.1 and the midsection of each the normal and 0.02cross-sectional 0.04 In order 0.06 ( These will beanalysis discussed in Sections 4.1we and 4.7.) For the wesegment will neglect the localized deformations at the point 600 mm 400 mm 0.06 to the elongatio Ineffects order to find the elongation of the rod, must first obtain the ! ! deformation are uniform. are uniform. (stress These and effects will be discussed in Sections 4.1 andfind 4.7.) ationThroughout aredeformation uniform. Throughout thedeformation midsection of each segment thechanges. normal are uniform. the midsection of application each segment the normal stress and of load and where the rod’s cross-sectional area suddenly strain. This is done by calculating (b) strain. This isorder done byfind calculating the stress, then using the stress–strain (a) In order to find the elongation of the rod, we must first obtain the In to the elongation of the rod, we must first obtain the Throughout the the midsection of each segment the normal stress and rder deformation to find elongation of the rod, we must first obtain the first obtain Inthe order touniform. finduniform. the elongation of the rod, we must deformation are are changes. (These effects will be discussed in Sections 4.1 and 4.7.) diagram. The normal stress within Çözüm: Çubuk farklı iki kesit bölgesinde oluşmaktadır. Herbir bölge için önce gerilmeleri ve diagram.The normal stress within each segment is strain. This is done by calculating the stress, then using the stress–strain strain. This is done by calculating the stress, then using the stress–strain deformation are uniform. This is done calculating the stress, then using the3–20 stress–strain This istodone by calculating the then the stress–strain Inbyorder find the elongation of stress, the rod, weusing must first obtain the Fig. Instrain. order to find the elongation of the rod, we must first obtain the Throughout the midsection of each segment the normal stress and The normal stress withinto each is diagram. The diagram. normal stress within each is segment Insegment order the elongation of the rod, we must first obtain the m.The normal stress within each segment is each diagram. stress within segment is using strain. This isnormal done by calculating the stress, the stress–strain buna bağlı olarak ve diyagramı kullanarak birimfind deformasyonları hesaplamalıyız. Buradan strain. This isThe done by calculating the stress, then3then using the stress–strain deformation are uniform. 101 strain. This is done by calculating the stress, then using the stress–strain P 2 Nis 10110 P each diagram.The normal within segment SOLUTION diagram. The normal stressstress within each segment is sABSOLUTION = = InsAB order to = find the elongation of the rod, we must first obtain the 3normal 3 31.83 = = MPa diagram. The stress within each segment is 2Pdeformations N 10110 2 N 10110 deplasmana A 2 3 p10. 3 p10.01 analysis we This will neglect the localized at the point A 2strain. NPgeçebiliriz. 10110 2=N P 10110 PFor the sm2 = 31.83 = isABdone by=calculating the using theMPa stress–strain s =stress, 31.83then MPa AB = 2 2 For the ana sAB = of =loadsapplication = 31.83 MPa A = rod’s 31.83 MPa m2 suddenly 3 p10.01 m2 p10.01area the cross-sectional AB = 2 = and 3 where 2 N2A 10110 diagram.The normal stress 2N P 10110 A A 3 p10.01Pm2 p10.01 m2 10 3 within each segment is P changes. ( These be discussed in Sections 4.1 and P4.7.) 10110 2 N 10110 2 N == effects = 31.83 MPa = Pwill sAB s=AB = 31.83 MPa sBCof = load = app 3 2 3 2 s = = 31.83 MPa = A 10110 2 N p10.01 m2 10110 2 N s = = = 56.59 MPa AB A P p10.01 m2 A P 2 BC p10. Throughout midsection segment =the =normal stress and p10.01 changes. (T =A56.59 MPa m2 10110 2of N each 101103the 2N sABC 3= = sBCm2=2 56.59 MPa p10.0075 P P 3 2 2 2 N 10110 A A p10.0075 m2 P p10.0075 m2 = = 56.59 MPa sBC = deformation = sBC =are uniform. = 56.59 MPa 3 Throughou 3 10110 2m2 N=2 of the 2 NsAB = 31.83 MPa the 1011032 N = rod, we must A P 210110 A In p10.0075 p10.0075 Pto find m2 2 order the elongation first obtain P From the stress–strain diagr A s = = = 56.59 MPa p10.01 m2 sBC =BC = = 56.59 MPa deformatio 2 m22 sstress–strain = is = 56.59 MPa From stress–strain the material inthe segment BC = AB A A strain. Thisthe is done byp10.0075 calculating the stress, then using p10.0075 m2diagram, 2 AB is strained elastically since s From the stress–strain diagram, the material in segment From the stress–strain diagram, the material in segment AB is A AB 6 p10.0075 m2 ! strainedThe In order elastically since law, Using 6 each sY =segment 40 MPa. s 3 Using Hooke’s normal stress within is AB Fromdiagram. thediagram, stress–strain diagram, in segment AB is m the stress–strain the elastically material inthe segment AB is strained elastically since Hooke’s law, 6 s = 40 MPa. s 10110 2 N strained since Using Hooke’s law, 6 s = 40 MPa. smaterial P AB Y AB Y strain. This sBC = in = AB 56.59 MPa strained Using Hooke’s 6 Using sthe 40= MPa. sMPa. d elastically since Hooke’s law, 6 sYsince = 40 AB Y = 2 law, AB stress–strain From diagram, the material inFrom segment iskullanarak: From the elastically stress–strain diagram, material segment AB is !sthe olduğuna göre, Hooke kanununu sAB A p10.0075 m2 the stress–strain diagram, the material in segment AB is 31.8311 6 3 2 Pa 6 6 law, PAB = diagram.Th = 2Using NsAB 10110 sAB strained elastically Using Hooke’s 6 =s = 40 MPa. strained elastically sincesince Hooke’s law, sY 40 MPa. sAB s6 P 31.83110 2 Pa 31.83110 2 Pa AB Y s31.83110 strained elastically since sAB 6 sY = 40 MPa. Using Hooke’s law,Eal AB 701109 = = 0.0004547 mm>mm = = 31.83 MPa = 6 P = = = 0.0004547 mm>mm P = = = 0.0004547 mm>mm AB 6 PABs= AB 9 AB A 2 Pa 2 9 9 31.83110 s 31.83110 2 Pa E sAB 70110 Pa Ediagram, p10.01270110 m2 AB al Pa al2 Pa 70110 al the E stress–strain the 2material in segment AB is =From = 0.0004547 mm>mm PAB = = PAB = 9 = 0.0004547 6 9 6 mm>mm 6 31.83110 2 Pa EPa 31.83110 2 Pa sAB 2s 70110 2 Pa Eal al 70110 31.83110law, 2 Pa AB The material within segme s strained elastically since Using Hooke’s 6 s = 40 MPa. s 3 AB AB Y PAB! P=AB = = = 0.0004547 2 N mm>mm = 0.0004547 mm>mm P 9 = 10110 P = = = 0.0004547 mm>mm 9 AB material within segment BCE is strained 9plastically, since sBC 7 sY = 40 MPa. From the Eal material material segment is MPa strained plastically, since The within segment BC is= BC strained plastically, since EThe sBC =2 PaThe 56.59 70110 2=within Pa al 70110 70110 2 Pa al 2 m2 From the graph, for PsMPa, sA 7p10.0075 splastically, = graph, 40 MPa. = Pdeğeri 56.59 From the graph, sisYsegment = 40 MPa. =MPa, 56.59 From the for swithin =7 40 MPa. sBCfor = !s 56.59 BC Y BC L BC BC since BC L MPa, The material BC is strained BC 7 s Y BC BC 6plastically, material within strained since ! ssegment olduğuna göre, grafikten için! PBC L ! 0.045 mm>mm. The approxima 31.83110 2 Pa s AB 0.045 mm>mm . The approximate elongation of the rod is therefore 0.045 mm>mm . The approximate elongation of the rod is therefore 0.045 . The approximate elongation ofMPa, the rod isLtherefore From graph, for PBC 7 From smaterial = within 40mm>mm MPa. sBCPplastically, = 56.59 the graph, for sY = The 40sMPa. s = BC 56.59 Lplastically, BC Y Pthe = since 0.0004547 mm>mm material segment BC is strained BC BC9The AB The within segment is=MPa, strained since material within segment BC is strained plastically, since E 70110 2 rod Pa isinPbölgedeki Çubuktaki uzama miktarı herbir uzama miktarlarının toplamı olarak d = ©PL = 0.0004547 al is From theFrom stress–strain diagram, the material segment AB is . The approximate elongation of56.59 the therefore mm>mm The elongation of the rod therefore the graph, for sBC. 0.045 7BCsapproximate =solacaktır. 40 MPa. s = MPa, L From the graph, for P s 7mm>mm = 40 MPa. s = 56.59 MPa, L Y BC BC From the graph, PBC L s 7 s = 40 MPa. Y BC= 0.00045471600 BC BCmm2 Y =Y ©PL mm2 + mm2 0.04501400 mm2 for sBC = 56.59 MPa, dsince ==©PL =6d 0.00045471600 + Hooke’s 0.04501400 strained elastically Using law, s = 40 MPa. s AB d = ©PL 0.00045471600 mm2 + 0.04501400 mm2 = 18.3 mm 0.0450.045 mm>mm . The approximate elongation of the rod is therefore mm>mm . The approximate elongation of the rod is therefore 0.045 mm>mm . The approximate elongation of the rod is therefore aşağıdaki gibi hesaplanır. =mm2 18.3 mm Ans. = 18.3 mm d = ©PL ==mm2 0.00045471600 mm2 + 0.04501400 mm2 The material within segment BC is strained plastically, since d = ©PL = 0.00045471600 0.04501400 18.3 + mm Ans. Ans. 6 31.83110 Pa From the graph, sBC 40 MPa. sAns. = 0.00045471600 56.59 MPa, PBCmm2 L + 0.04501400 mm2 s7ABsY = == 18.3 mm d = d©PL = 0.00045471600 mm2 + 20.04501400 mm2 BC = From the d for = ©PL ©PL mm2 + 0.04501400 mm2 = 18.3 mm Ans. PAB=0.045 =0.00045471600 = = 0.0004547 mm>mm 9 mm>mm . The approximate elongation of the rod is therefore E 70110 2 Pa = 18.3 mm Ans. strained ela al = 18.3 mm Ans. = 18.3 mm Ans. 0.02 0.04 ! d = segment ©PL = 0.00045471600 mm2 + 0.04501400 The material within BC is strained plastically, sincemm2 = 18.3the mm graph, for sBC = 56.59 MPa, PBC L sBC 7 sY = 40 MPa. From 0.045 mm>mm. The approximate elongation of the rod is therefore d = ©PL = 0.00045471600 mm2 + 0.04501400 mm2 = 18.3 mm Ans. PA The mat sBC 7 sY 0.045 mm>m Ans. d = = !35 amount d, and its radius contracts by an amount d¿. Strains in the 3 lateral or radial direction are, longitudinal or axial direction and in the respectively, 102 C H A P T E R 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S d d¿ and Plat = Plong = r L Consider a bar having an original radius r and length to the tensile force P in Fig. 3–21. This force elongate amount d, and its radius contracts by an amount d¿ longitudinal or axial direction and in the lateral or rad respectively, 3.6that within Poisson’s Ratio In the early 1800s, the French scientist S. D. Poisson realized the Poisson Oranı Plong = d L and Plat = d¿ r elastic range the ratio of these strains is a constant, since the deformations In the earlyto1800s, the French S. D. Poisson realize a deformable an axial tensile scientist force, not only as Poisson’s ratio, body is subjected d and d¿ are proportional. This constant is referred toWhen elastic range the ratio of these strains is a constant, since t it elongate but it also contracts laterally. For example, if a rubber particular material n (nu), and it has a numerical value that is unique for adoes and are proportional. This constant is referred to as d d¿ band is stretched, it can be noted that both the thickness and width of the that is both homogeneous and isotropic. Stated mathematically it is (nu), and it has a numerical value that is unique for a pa n band are decreased. Likewise, a compressive force acting on a body causes that is both homogeneous and isotropic. Stated mathema it to contract in the direction of the force and yet its sides expand laterally. Plat (3–9) n = Consider a bar having an original radius r and length L and subjected Plong Plat to the tensile force P in Fig. 3–21. This force elongates the n =bar - by an 3 amount d,(positive and its radius contracts by an amount d¿. Strains Pinlongthe The negative sign is included here since longitudinal elongation longitudinal or axial direction and in the lateral or radial direction are, strain) causes lateral contraction (negative strain), and vice versa. Notice The negative sign is included here since longitudinal elon respectively, that these strains are caused only by the axial or longitudinal force P; i.e., strain) causes lateral contraction (negative strain), and v no force or stress acts in a lateral direction in order to strain the material d d¿ and strains Plat =are caused only by the axial or longitud Plong =that these in this direction.! ! r in a lateral direction in order to st noLforce or stress acts Poisson’s ratio is a dimensionless quantity, and for most nonporous in this direction. 1 altında 1 thebir In early 1800s, the French uzama/kısalmanın scientist S. D. Poisson realized thatenine within the Eksenel veya basınç cisimde eksenel yanında When 4 the block is compressed solids it has a value that isçekme generally between and rubber values of 3 . Typical Poisson’s ratio is asince dimensionless quantity, and for elastic range the ratio of these strains is a constant, the deformations (negative strain) its sides will expand 1 1 n for common engineering materials are listed on the inside back cover. solids it has a value that is generally between 4 and 3 . T doğrultuda daralma/genişleme meydana gelecektir. (positive strain). The ratio of these strains and are proportional. This constant is referred to as Poisson’s ratio, d d¿ For an “ideal material” having no lateral deformation when it is stretched for common engineering materials are listed on the in remains constant. and it has ainnumerical nvalue that is unique for a particular material nit(nu), or compressed Poisson’s ratio will be 0. Furthermore, will be shown For an “ideal material” having no lateral Orijinal yarıçapı r ve uzunluğu Lthat olan ve eksenel çekme etkisine maruz birmathematically çubuk gözönüne isATboth isotropic. Stated it is deformation wh ERST IO E SFCO M FA AT MTEthat AT E3 LRSIthe A LM 102 CS HEmaximum A R IC 3 A L MP ERCOHPAENRITCIAE LS PORFO M Pfor E AT R T IPoisson’s OLFS M E R I A Lhomogeneous Sis 0.5. C H A P Tand 102 ER 3 MECHAN I C A L P R O P E R T I E S O F M AT E R I A L S Sec. 10.6 possible value ratio H P ERRI A CPHTAEN EERSI A or compressed Poisson’s ratio will be 0. Furthermore, it Therefore 0 … nalalım. … 0.5. Sec. 10.6 Pthat the maximum possible value for Poisso lat O F M AT E R I A L S (3–9) n = - 0 … n … 0.5. Therefore d/2 Plong Poisson’s Poisson’s Ratio Ratio 6 3.6 Poisson’s 3.6 Poisson’s Ratio Ratio 3.6 Poisson’s Ratio d/2 P Ltensile The negative included here since longitudinal (positive ndeformable a deformable body body is subjected is subjected toWhen an to axial an aaxial tensile force, notnot only d/2force, When a deformable bodysign is subjected to tensile an axial tensile force, not elongation onlydeformable deformable body isonly subjected toisan axial force, not only oisson’s Ratio When body is subjected P avice strain) causes lateral contraction (negative strain), and versa. Notice it longate elongate butbut it also it also contracts contracts laterally. laterally. For example, example, ifit aalso ifrubber a but rubber does itbut elongate it also contracts For example, ifdoes a rubber L does itFor elongate contracts laterally. laterally. For example, if a rubber it elongate but it also d/2 contracts that these strains arethat caused by the axial longitudinal force P; i.e., band isitand stretched, it be noted bothonly the thickness andofor width of the tretched, is stretched, it can it can besubjected noted be noted that that both thethe thickness thickness and width width of the ofcan the band istensile stretched, can be noted that both the thickness and width the Original Shape ormable body is to anboth axial force, not only Final Shape band is stretched, it can be noted that no force or stress aacts in a lateral direction to strain the material band decreased. compressive force onin aorder body causes are decreased. decreased. Likewise, a compressive a compressive force force acting acting onare body arubber body causes causes band are decreased. aLikewise, compressive force acting onacting a body causes gate but it Likewise, also contracts laterally. For example, ifaon aLikewise, Original Shape band are decreased. Likewise, a compr Final inof this direction. toexpand contract in the direction ofand the force yetexpand its sides laterally. expand laterally. ontract ract initthe in the direction the of the force force and and yet yet itsrsides itsitand sides expand laterally. laterally. it to contract in the direction of the force yet itsand sides ched, can bedirection notedofthat both the thickness width the it to contract in the direction of the for When the rubber block is compressed PPoisson’s ratio is a dimensionless quantity, and for most nonporous Tension Consider a bar having anradius original radius r andLlength L and subjected d¿subjected aexpand bar having an original r and length and subjected der nsider a bar a Likewise, bar having having ana original an original radius radius rConsider and r and length length and L and subjected creased. aLbody causes r an original r ! compressive (negative strain) itsforce sidesacting will on 1 Consider 1 a bar having P solids itby has value that is generally between andby3.an Typical values of to the tensile force Pan ina Fig. 3–21. This force elongates to the tensile force Pbar inbar Fig. This force elongates the bar the by 4bar an etnsile tensile force force P (positive in Pof Fig. inthe Fig. 3–21. 3–21. This This force force elongates the the by an3–21. in the direction force and yet itsofelongates sides expand laterally. Tension strain). The ratio these strains d¿ T 3 to the tensileback force P in Fig. 3–21. n for common engineering materials are listed on the inside cover. amount , and its radius contracts by an amount Strains in the d d¿. amount , and its radius contracts by an amount Strains in the d d¿. nt , and , and its its radius radius contracts contracts by by an an amount amount Strains Strains in the in the d d d¿. d¿. Fig. 3–21 remains constant. a bar having an original radius r and lengthaltında L and subjected Çubuk, eksenel kuvvet ! kadar kadar kısalacaktır. Buna göre 3 uzayacak ve yarıçapı ! amount its radius contracts d, and longitudinal or an axial direction in theorlateral ordirection radial direction are, For “ideal material” having no lateral deformation when it is stretched orthe axial and in theand lateral radial are, nal udinal or axial orPaxial direction and and in the in longitudinal the lateral lateral or radial or radial direction are, are, le force indirection Fig. 3–21. This force elongates bardirection by an Fig. 3–21direction and in longitudinal or axial respectively, boyuna ve birim d¿. deformasyonlar ctively, ely, its radius contracts and by enine an respectively, amount Strains or in compressed the sırasıylaPoisson’s ratio will be 0. Furthermore, it will be shown in respectively, Sec.are, 10.6 that the dmaximum possible value for Poisson’s ratio is 0.5. or axial direction andd indthe lateral or direction d¿ d d¿ d¿ radial d¿ and Plat = and = Plong = 0Plong Therefore … n= …Plat 0.5. = = andandPlat P=lat = PlongPlong d r L r L r olacaktır. r L L ve ! and Plong = ! L d/2 d d¿ In thethat early 1800s, the FrenchS.scientist S. D.realized Poisson that realized thatthe within the early 1800s, thethat French scientist D. Poisson within rly early 1800s, 1800s, the the French French scientist S.Plat D.S.In Poisson D.the Poisson realized realized within within the the = scientist and = Plong 1800 lü yılların Fransız bilimadamı S.D. Poisson, elastik bölgede bu1800s, birim r başlarında L elastic range the ratio of these strains is a constant, since the deformations In the early the French scientist S elastic range the ratio of these strains is a constant, since the deformations cnge range thethe ratio ratio of these of these strains strains is aisconstant, a constant, since since thethe deformations deformations P L and are proportional. This constant is referred to as Poisson’s ratio, d d¿ elastic range the ratio of these strains i and are proportional. This constant is referred to as Poisson’s ratio, d d¿ d/2 are are proportional. proportional. This This constant constant is referred is referred to as to Poisson’s as Poisson’s ratio, ratio, d¿ 1800s, the French scientist S. D. Poisson realized that within the deformasyonlar oranının sabit olduklarının farkına varmıştır. Çünkü boyuna ve enine (nu), and it has a numerical value that is unique for a particular material n (nu), and it has a numerical value that is unique for a particular material n and are proportional. This consta d d¿ nd ),the and it has it has a numerical a numerical value value that that is unique is unique for for a particular a particular material material ratio of these strains is a constant, since the deformations that is both ithomogeneous andOriginal isotropic. Stated mathematically it(nu), is and it has a numerical value tha that is both homogeneous and isotropic. Stated mathematically it is n Shape oth s proportional. both homogeneous homogeneous and and isotropic. isotropic. Stated Stated mathematically mathematically is it is This constant is referred to as Poisson’s ratio, Final Shape deformasyonlar orantılıdır. Bu sabit Poisson oranı olarak adlandırılmış ve ! sembolü ile that is both homogeneous and isotrop t has a numerical value that is unique for a particular material Plat r Plat PHomojen ve izotropik malzeme buPlatoran: lat mathematically homogeneous andgösterilmiştir. isotropic. it is bir n = (3–9) (3–9) n = - için (3–9) (3–9) n =n -=Stated P Plong P Tension long d¿ PlongPlong n = Plat (3–9) n = The negative sign ishere included here since longitudinal elongation (positive The negative sign is included (positive Fig. 3–21 Psince ative egative signsign is included is included here here longitudinal longitudinal elongation elongation (positive (positivesince longitudinal elongation longsince strain) causes lateral contraction (negative strain), and vice versa. Notice strain) causes lateral contraction (negative strain), and vice versa. Notice ! The negative sign is included here sin auses ) causes lateral lateral contraction contraction (negative (negative strain), strain), andand vicevice versa. versa. Notice Notice that these strains are caused only by or thelongitudinal axial or longitudinal force P; i.e., that these strains are caused only by the axial force P; i.e., e strains signstrains is included here since longitudinal elongation (positive hese areare caused caused only only by the by the axial axial or longitudinal or longitudinal force force P; i.e., P; i.e., strain) causes lateral contraction (neg no force or stress acts indirection a lateral direction in strain order to strain the material no force or stress acts in a lateral in order to the material Buradaki eksi işareti, boyuna ve enine deformasyonların birbirlerinin aksi biçimde, biri uzama es contraction (negative strain), vice rce orlateral stress or stress acts acts in ainlateral a lateral direction direction in order inand order to strain toversa. strain theNotice the material material that these strains are caused only by th in this direction. in this direction. rains are caused byblock the axial or longitudinal force P; i.e., she rection. direction. no force or stress acts in a lateral direc When the only rubber is compressed rubber block is (+)compressed iken diğerinin kısalma (-)is olmasından kaynaklanmaktadır. deformasyon Poisson’s ratio is a dimensionless andBurada for mostenine nonporous Poisson’s ratio a dimensionless quantity, quantity, and for most nonporous stress acts in a lateral direction in order toand strain the material (negative strain) its sides will expand n’s sson’s ratio ratio is a is dimensionless a dimensionless quantity, quantity, and for for most most nonporous nonporous e strain) its sides will expand 1 1 1 1 in this direction. a value that is generally between and 3. values Typical of values of solids that is generally between and 3.isTypical 1it has 1 solids 1a value 1 it has 4 compressed (positive strain). ratio of these strains When the rubber 4 block strain). ratio of these strains tion. has it has a value aThe value that that isoluşurken, generally isThe generally between between Typical Typical values of of Poisson’s ratio is a dimensionless enine bir values kuvvetin etkimediğine dikkati çekmemiz gereklidir. 4 and 4 and 3 .common n3.for engineering materials listed on theback inside back cover. Poisson n fordoğrultuda common engineering materials are listedare on the inside cover. remains constant. constant. (negative strain) its sides will expand ratio engineering is aengineering dimensionless quantity, and nonporous mmon common materials materials areare listed listed onfor the onmost the inside inside back back cover. cover. solids it has a value that is generally For an “ideal material” having no lateral deformation when it is stretched For1 an “ideal material” having(positive no lateral deformation it isstrains stretched strain). The 1/4 ratio when of these oranı çoğu boşluksuz için ile 1/3 arasında bir değerdir. a“ideal value that ishaving generally between andve13. Typical values of malzeme ndeal material” material” having no boyutsuzdur lateral no lateral deformation deformation when when it is it stretched is stretched 4 n for common engineering materials or compressed Poisson’s ratio will be 0. Furthermore, it will be shown in or compressed Poisson’s ratioremains will beconstant. 0. Furthermore, it will be shown in on engineering materials are listed on the inside back cover. ressed mpressed Poisson’s Poisson’s ratio ratio willwill be 0. be Furthermore, 0. Furthermore, it will it will be shown be shown in For isan0.5. “ideal material” having no later Sec. 10.6 that the in maximum possible for Poisson’s Sec. 10.6 that the maximum possible value forvalue Poisson’s ratio is ratio 0.5. l that material” having no lateral deformation when it is stretched 10.6 that thethe maximum maximum possible possible value value for for Poisson’s Poisson’s 0.5. or compressed Poisson’s ratio will be 0ratio … isn 0.5. …is 0.5. Therefore 0Therefore … n … ratio 0.5. eed fore 0 Poisson’s …0 n……n 0.5. …ratio 0.5. will be 0. Furthermore, it will be shown in Sec. 10.6 that the maximum possibl d/2 d/2 at the maximum possible value for Poisson’s ratio is 0.5. Therefore 0 … n … 0.5. d/2 d/2 … n … 0.5. P P d/2 L L !36 P P d/2 d/2 d/2 L P Original Original Shape Shape L L d/2 d/2 d/2 Original Shape Original Shape FinalFinal Shape Shape Final Shape Final Shape P 3.6 NICAL PROPERTIES OF EXAMPLE 3.4 Kitabınızın arka kapağında yaygın mühendislik malzemeleri için bu oranın tipik değerleri aa 3.6 es of Typical Engineering Materialsa Ratio Poisson’s I Units) POISSON’S RA M AT E R I A L S AverageMechanical MechanicalProperties PropertiesofofTypical TypicalEngineering EngineeringMaterials Materials Average verilmiştir. (SIUnits) Units) shown in Fig. 3–22. If an A bar made of A-36 steel has the (SI dimensions POISSON’S RATIO When a deformable body is subjected to an force axial tensile only to the bar, determine the3.6 axial of P = force, 80 kNnot is applied change Strength does (MPa) it elongate Ultimate Strength (MPa) Coef. of Therm. Modulus of Modulus of Yield Strength (MPa) Ultimate Strength (MPa) Modulus of Modulus of Yield Strength (MPa) Ultimate Strength (MPa) but it also contracts laterally. For and example, if a rubber 3.6 Pafter OISSON’S RATIO inPoisson’s its length the change in the dimensions of its cross section sY su % Elongation in Density Expansion ss ss Materials %%Elongation Materials Densityrr Elasticity ElasticityEEa Rigidity RigidityGG Elongationinin YY uu band is stretched, itb can be noted that both the3 thickness and width of the -6 b bb bb 3 applying the load. The material behaves elastically. (10 )>°C Comp. Shear Tens. Comp. Shear 50 mm specimen (Mg>m (GPa) (GPa) Tens. Mg>mn) ) (GPa) (GPa) Tens. Comp. Comp. Shear Shear Tens. Tens. Comp. Comp. Shear Shear 5050mm mmspecimen specimen band are decreased. Likewise, a compressive force acting on a body causes EXAMPLE 3.4(Ratio Metallic Metallic of the force it to contract in the direction sides expand laterally. EXAMPLE 3.4and yetPits ! 80 kN Consider a bar having an original radius r and length L and subjected A bar made of A-36 steel has the shown Fig. 3–22. an 414 172 469 469 290 2014-T6 0.35 23 dimensions 2.79 73.1 2727 414 414 172 469 290 1010 2014-T6 10 2.79 73.1 414 in 414 172 If469 469 469 290 Aluminum Aluminum to the tensile force P in Fig. 3–21. This force elongates the bar by an axial force ofA-36 P = steel 80 is the applied to the determine the131 change Wrought Alloys A bar made has shown If an290 255 131 290 Wrought 290 Alloys 186 12 of 0.35 24dimensions 6061-T6 2.71 68.9 2626 bar, 255 255 290 186 1212 6061-T6 2.71kN 68.9 255 in Fig. 255 3–22. 131 290 290 186 amount d, and its radius contracts anchange amount Strains in determine the d¿. dimensions axial of and P by =the 80 kN is applied to the bar, change in itsforce length in the of its crossthe section after – – 179 Cast – ASTM 0.28 12 Gray 7.19 67.0 2727 –– –– 179 669 –– 0.6 Gray ASTM20 200.6 in the lateral 7.19 67.0 179 669 0.6 Iron Cast669 Iron direction longitudinal or axial and or radial direction y– –after in its length and the The change in the dimensions ofare, its cross–section applying the load. material behaves elastically. Alloys Alloys – – 276 572 – 5 A-197 0.28 12 Malleable ASTM 7.28 172 6868 –– –– 276 572 –– 55 Malleable ASTM A-197 7.28 172 – 276 572 respectively, applying the load. The material behaves elastically. 703 924 0.35 0.35 0.28 0.28 0.28 0.28 70.0 70.0 –– 241 241x 241 241 –– 3535 0.35 0.35 345 345 –– 655 655 655 655 –– 2020 0.34 0.34 Magnesium Magnesium – early 276 1800s, 276the French 152 1 0.30 26 [Am 1.83 44.7 1818 152 [Am1004-T61] 1004-T61] 1.83 44.7 152 In the scientist S. D. Poisson realized that within the Alloy Alloy 152 152 –– 276 276 152 152 11 0.30 0.30 – 207 0.35 0.35 345 345 – 345 250 Poisson’s Poisson’s Ratio Ration n 70.0 1.5 m 70.0 70.0 152 50 mm 10 1 0 3C 241 Copper 241 Copper Alloys 655 Alloys 655 Red Red–Brass BrassC83400 C83400 dP35 ! 80 kN 0.35 8.74 8.74 =P !2080and P – long Bronze C86100 Bronze C86100 kN Plat = 0.34 8.83 8.83 L d¿ r 18 101 101 3737 17 103 103 3838 elastic range the ratio of these strains is a constant, since the deformations y – d¿ 400 400 Structural – 30 0.32 12 y 250 A36 7.85 200 7575 ratio, 250 Structural A36 constant 7.85 200 250 250 are proportional. This is referred to as Poisson’s d and Steel Steel 50 mm – 517 itAlloys – 40 that is unique 0.27 17 Stainless 304 7.86 7575 207 Stainless 304value 7.86 193 2073–22207 and has517a numerical for a193 particular material n (nu), Fig. Alloys x207 50 mm 1.5 200 m 800 homogeneous 800 Tool –L2 22 0.32 12 8.16 75 703 Tool L2 and isotropic. 8.16 mathematically 200 75 is 703 x 703 703 that– is both Stated it 1.5 m – 1,000 Titanium Titanium 1,000 Alloy Alloy – [Ti-6Al-4V] [Ti-6Al-4V] 16 n = - Plat Plong 0.36 4.43 4.43 9.4 120 120 4444 924 924 100 mm 100 mm SOLUTION 100 mm The normal stress in the bar is The negative sign is included here since longitudinal elongation (positive z 400 400 400 400 –– 3030 0.32 0.32 –– 517 517 517 517 –– 4040 0.27 0.27 –– 800 800 800 800 –– 2222 0.32 0.32 1,000 1,000 1,000 1,000 –– 1616 0.36 0.36 P924 ! 80 kN– – 924 z 276 276 –– (3–9) P ! 80 kN Nonmetallic Nonmetallic P ! 80 kN z Fig. 3–22 –– 12 – – – Strength 0.15 11 3–22 Low 2.38 22.1 –– –– 1212 –– –– –– –– 0.15 Low Strength – 2.38 22.1 0.15 Fig. Concrete Concrete strain) causes lateral contraction (negative strain), and vice versa. Notice 3 – 38 – – – Strength 0.15 11 High 2.38 29.0 – – – 38 – – – – 0.15 High Strength – 2.38 29.0 – – – 38 – – – – 0.15 80110 2 N P that these strains are caused only by the axial or longitudinal sz = force = P; i.e., = 16.011062 Pa – – 717 483acts 20.3 0.34 – Athe–material Plastic 49 131 –– –m2 –– 717 483 20.3 2.8 0.34 Plastic Kevlar 49 2.8 1.45order to 131 – 10.1 m210.05 – 717 483 20.3 2.8 0.34 no force or stress inKevlar a lateral direction 1.45 in strain – – 90 131 – – 0.34 – Reinforced 30% Glass 1.45 72.4 – – – – 90 131 – – 0.34 Reinforced 30% Glass 1.45 72.4 – – – – 90 131 – – 0.34 in this direction. SOLUTION essed From the table the inside back cover for A-36 steel Est = 200 GPa, Poisson’sc ratio is aSOLUTION dimensionless quantity, and foron most nonporous Wood Wood d e c c dd dd e e pand– The normal stress in0.47 the is 1 bar 1 – in the z––direction 26d 6.2 –Fir 0.29 – 2.1 2626 6.2 –– 0.29 Douglas Fir 13.1 – – is –– –– 2.1 6.2 0.29 Douglas 0.47 13.1 2.1 and so the strain The normal stress in the bar is solids it has a value that is generally between and Typical values of . Select Structural SelectdStructurald 3 e 4 c c dd dd e e rains– – 2.5c 36 6.7 – 0.31 – White Spruce 3.60 9.65 – – – – 2.5 36 6.7 – 0.31 White Spruce 3.60 9.65 – – – – 2.5 36 6.7 – 0.31 Grade Grade n for common engineering materials are listed on the inside back cover. 3 32 N ! ! PP ! 80110 80110 6 2 sN 16.0110 2 Pa For an “ideal material” having no lateral when itz is stretched 6 62 Pa = == 16.0110 szsz=deformation = =16.0110 2 Pa P = = =mechanical 80110-6working 2working mm>mm aa z A 10.1 m210.05 m2 9 tion, mechanical working of the specimen, or heat treatment. For a more exact value Specific values may vary for a particular material due to alloy or mineral composition, ofofthe ororheat or compressed Poisson’s ratio will be 0. Furthermore, it will be shown in Specific values may vary for a particular material due to alloy or mineral composition, mechanical thespecimen, specimen, heattreatment. treatment.For Fora amore moreexa ex A 10.1 m210.05 m2 Poisson oranının mümkün olan maksimum değeri 0.5’tir (Bkz. hacim modülü hesabı). Böylece E 200110 2 Pa st reference books for the material should be consulted. reference books for the material should be consulted. 3.6 POISSON’S RATIO 103 Sec. 10.6 thatb the maximum possible value for Poisson’s ratio is 0.5. bThe yield and ultimate strengths for ductile materials can be assumed equal for both tension and compression. oth tension and compression. The yield and ultimate strengths for ductile materials can be assumed equalsteel for both and compression. From thetable tableonon the inside back cover A-36 steel 200 GPa, From the the inside back cover forfor A-36 EstEtension = =200 GPa, st Therefore ! 0 … n … 0.5. olmaktadır. cc The axial elongation of the bar is therefore Measured perpendicular to the Measuredand perpendicular tostrain thegrain. grain. and so the in the z direction is so the strain in the z direction is dd Measured grain. Measuredparallel parallelto tothe thed/2 grain. 3.6 POISSON’S RATIO 103 -6 EXAMPLE 3.4 perpendiculartotothe Ans. = load Pload = [80110 2]11.5 m2 = 120 mm zLiszisapplied 6dz6the grain when along the grain. the grain. Deformationmeasured measuredperpendicular the grain when the applied along the grain. ! eeDeformation szsz 16.0110 16.01102 Pa 2 Pa P -6 -6 PzPz= L = =80110 2 mm>mm = == 80110 2 mm>mm 9 d/2 9 E Using Eq. 3–9, where from the imal insideedilmiş back bir n A bar made of A-36 steel has the dimensions shown in Fig. 3–22.as Iffound an E 200110 2 Pa 200110 2 Pa st st st A36 çeliğinden (Tablodan ! =! ve ! = 0.32 değerleri alınır) – EXAMPLE 3.4 the to lateral contraction strains in both the x and y directions are axial force of P = 80 kN iscover, applied the bar, determine the change Original Shape Final Shape The axial elongation of the bar is is therefore in its length and the change in the dimensions of its cross section after çubuk şekilde gösterildiği gibi 80 kN eksenel çekme kuvvetine maruzdur. Yük uygulandıktan The axial elongation of the bar therefore A bar made of A-36 steel has the dimensions shown in Fig. 3–22. If an -6 Px =elastically. Py = -nstPz = -0.32[80110 2] = -25.6 mm>m applying load.isThe material behaves r the -6-6 the change axial force P = the 80 boyundaki kN applied to bar, determine Ans. dzdve = P L = [80110 2]11.5 m2m2= =120 mm sonraof çubuk çubuk kesiti değişimleri hesaplayınız. Malzeme z z Ans. = P L = [80110 2]11.5 120 mm P boyutlarındaki z z z Tension d¿ in its length and the change in the dimensions of its cross section after of the cross section are Thus the changes in the dimensions ! 80 material kN applying the load.Pelastiktir. The elastically. Using 3–9, the nstn = =0.32 davranışı UsingEq. Eq.behaves 3–9,where where foundfrom from theinside insideback back 0.32asasfound st -6 Fig. 3–21 cover, the lateral contraction strains in both the x and y directions Ans. d = P L = -[25.6110 2]10.1 m2 = are -2.56 x in x both the x and y directions cover, the lateral contractionxstrains are mm P ! 80 kN -6 d = PyLy = -6 -[25.6110 2]10.05 m2 = -1.28 mm PxP = =PyP = =- n-n 2] mm>m -6 = - 25.6 stPzP = =y- 0.32[80110 -0.32[80110 2] = -25.6 mm>m y x st z y 3 mmcross section are Thus the changes in the dimensions of50the Thus the changes in the dimensions of the cross section are x y 1.5 m -6 dx = PxLx = - [25.6110 2]10.1 m2 = - 2.56 mm -6 50 mm dx = PxLx = -[25.6110 2]10.1 m2 = -2.56 mm x ! 80 kN -6 1.5 m dy = PyLy = - [25.6110 -6 2]10.05 Pm2 = - 1.28 mm dy = PyLy = -[25.6110 2]10.05 m2 = -1.28 mm P 100 ! 80mm kN Fig. 3–22 100 mm ! Ans. Ans. Ans. Ans. z z Fig. 3–22 SOLUTION The normal stress in the bar is SOLUTION The normal stress in the bar isP 8011032 N sz = = = 16.011062 Pa A 310.1 m210.05 m2 3 Ans. !37 PP!!80 80kN kN 100 mm 100 mm 100 mm zz z 100 100mm mm zz Fig.Fig. 3–223–22 Fig. 3–22 Fig. Fig.3–22 3–22 SOLUTION SOLUTION SOLUTIONÇözüm: SOLUTION The normal stress in the bar is SOLUTION The Thenormal normalstress stressininthe thebar barisis The in the bar is Thenormal normalstress stress in the bar is Eksenel çekme gerilmesi: 3 3 32 N 80110 N P=PP = 80110 80110 33 2 2 6 6 62 Pa s = 16.0110 22NN N = =16.0110 szsz=PzP =A= 80110 PaPa 80110 = 16.0110 66 2 2 10.1 m210.05 m2 ssz == A=A == 16.0110 10.1 m210.05 m2 10.1 m210.05 m2 = 10.1 16.011022Pa Pa z A m210.05 m2 A 10.1 m210.05 m2 ! table From the on the inside back cover for A-36 steel = 200 GPa, From the table on the inside back cover for steel Est stE=st=200 GPa, From the table on the inside back cover forA-36 A-36 steel 200 GPa, From the table on the inside back cover for A-36 steel EEst E == 200 GPa, and so the strain in the z direction is From the table on the inside back cover for A-36 steel 200 GPa, and ininthe z zdirection isisdeformasyonu: st andso sothe thestrain strain the direction Eksenel uzama birim and andso sothe thestrain strainin inthe thezzdirection directionisis 6 sz 16.0110 16.0110 2 Pa 6 s=s PaPa -6 6 262 16.0110 z z 16.0110 6 P = = 80110 2 mm>mm -6 s 2 Pa z z 9 s 16.0110 2 Pa P = = =80110 2 2mm>mm -6 -6 P = = =80110 80110 mm>mm z Est 200110 9 9 2=Pa -62 mm>mm PPz z==z EE = = = 80110 2 mm>mm 200110 2 Pa 9 st 200110 2 Pa ! z EEstst st 200110 200110922Pa Pa The axial elongation of the bar is therefore The isistherefore Eksenelofthe uzama deformasyonu: Theaxial axialelongation elongation ofthe thebar bar therefore The Theaxial axialelongation elongationof of thebar barisistherefore therefore -6 Ans. dz P=LPzL = [80110 2]11.5 m2 120 = 120 mm z [80110 -6-6-6 Ans. 2]11.5 mm Ans. = [80110 2]11.5 m2 = 120 mm zP z z=[80110 -62]11.5 Ans. ddzdz=d=z=PP= = m2m2 = =120 mm zL zL z Ans. zLz = [80110 2]11.5 m2 = 120 mm ! z Using Eq. 3–9, where nst = 0.32 as found from the inside back Using Eq. where asasfound from back Using Eq.3–9, 3–9, where found fromthe theinside inside back nst=0.32 =0.32 0.32 Using Eq. where as from back nnstnst= Using Eq. 3–9, where as found found from the inside backenine 0.32 cover, the3–9, lateral contraction strains in both the xthe andinside y directions are st = hesaplayarak Poisson etkisini çubuk enkesitindeki cover, the lateral contraction strains in the x xand y ydirections are cover, the lateral contraction strains inboth both the and directions are cover, the lateral contraction strains in both the x and y directions are cover, the lateral contraction strains in both the x and y directions are -6 Px = Py = -nstPz = -0.32[80110 2] = -25.6 mm>m -6 -6-6 -0.32[80110 -25.6 mm>m = P=y=-=n-n -0.32[80110 -25.6 mm>m PPxPx=P=x! =PP PstPzPst= 0.32[80110 =2] mm>m -62] 2] zP z=-= yPy= st-n n = 0.32[80110 2] ==-= -25.6 25.6 mm>m x y st z Thus the changes in the dimensions of the cross section are Thus the changes inthe the dimensions of the cross section are Thus the changes in the dimensions of the cross section are Thus the changes dimensions the cross section are Çubuk enkesitindeki deformasyonlar: Thus the changes inin the dimensions ofof the cross section are -6 Ans. dx = PxLx = -[25.6110 -6-6-6 2]10.1 m2 = -2.56 mm ddxdx=d=x=PP LxPxLxL =x=-=[25.6110 m2 ===-= 2.56 mm -62]10.1 Ans. = -[25.6110 2]10.1 m2 -2.56 mm Ans. Ans. -[25.6110 2]10.1 m2 -2.56 mm xP Ans. m2 2.56 mm x xL x x= - [25.6110 2]10.1 -6 d L= P=yL-y [25.6110 Ans. = -[25.6110 -6-6-6 2]10.05 m2 = -1.28 mm ddydy=d=y=PPy= m2 ===-= mm -62]10.05 Ans. =-[25.6110 -[25.6110 2]10.05 m2 -1.28 mm Ans. Ans. 2]10.05 m2 -1.28 mm yPL yL yL yP y=y= Ans. [25.6110 2]10.05 m2 -1.28 1.28 mm y ! y birim deformasyonları belirleyelim: y Çubuğun x ekseni doğrultusundaki kesit boyutu 100 mm iken deformasyon sonrası (100-0.00256) mm olacaktır. Çubuğun y ekseni doğrultusundaki kesit boyutu 50 mm iken deformasyon sonrası (50-0.00128) mm olacaktır. !38 homogeneous and isot M pure AT E R I A LS to shear, equilibrium requires that equal shear stresses y element uniformly, Fig. eveloped on four faces of the element. These stresses txy must 3 gxy gxy measures the angul d toward or away from diagonally opposite corners of the 2 originally along the x an as shown inStress–Strain Fig. 3–23a. Furthermore, if the material is e Shear Diagram The behavior of a ma ous and isotropic, then this shear stress will distort the laboratory using specim g xy niformly, Fig. 3–23b. As mentioned in Sec. 2.2, shear strain was shown that when a small element of the material is them to a torsional loa 2 res the angular distortion of the that element relative to the sides pure shear, equilibrium requires equal shear stresses x Kayma Gerilme - Birim Deformasyon Diyagramı torque and the resultin p g along the x and y axes. oped on four faces of the element. These stresses txy must ! xy explained in Chapter 5 2 1can 0 4 be H Ain P TaE R bulunması 3 M E C H A N durumu) ICAL PROPE R T I E S O F M ATkayma E R I A L S gerilmeleri avior ofora away material subjected pure shear studied Basit kaymato(sadece kayma gerilmelerinin durumunda oward from diagonally opposite corners of Cthe stress and shear strain, (b) using specimens in theFurthermore, shape of thiniftubes and subjecting shown in Fig. 3–23a. the material is example of such a diag aşağıda şekilde torsional loading. If measurements are made of they applied and isotropic, then thisgösterildiği shear stress will yönlenmişlerdir. distort the Fig. 3–23 Like the tension test, th d the Fig. resulting of twist, inthen the methods to be The Shear Stress–Strain Diaga ormly, 3–23b.angle As mentioned Sec.by 2.2, shear strain linear-elastic behavior Malzeme homojen ve the izotropik ise kayma gerilmesi, elemanda üniform biçimde aşağıda in Chapter 5, the dataofcan used to determine sheartxy the angular distortion the be element relative to thethe sides Also, strain hardening and a shear stress–strain diagram plotted. sebep An In Sec. 1.5basit it was showntesti thatverileri when a And smallfinally, elemen ngshear the xstrain, and y axes. gösterildiği gibi bir biçim değişimine olacaktır. Laboratuvarda kayma reached. th f such a diagram for a ductile material is shown in Fig. subjected to pure shear, equilibrium requires that equw or of a material subjected to pure shear can be studied in 3–24. a until it reaches a point kullanılarak birsubjecting kayma gerilme-birim deformasyonu edilebilir. ension test, thisinmaterial when subjected toand shear will exhibit must be elde developed on four faces of themost element. These ing specimens the shape ofaşağıdaki thin tubesgibi For engineering x tic behavior andIf itmeasurements will have a defined proportional limit tpl. be directed toward or away from diagonally opposit sional loading. are made of the applied t behavior is linear, and s (a) in hardening will occur untilthen an by ultimate shear stress tu is element, as shown in Fig. 3–23a. Furthermore, if he resulting angle of twist, the methods to be 104 C H Ato P Tto E R lose 3 M H A N I Cstrength A L P R O P E R T I E S O F M AT E R I A L S And finally, the data material begin itsE Cshear homogeneous and isotropic, then this shear stress Chapter 5, the can will be used determine the shear y tu element uniformly, Fig. 3–23b. As mentioned in Sec. 2.2 chesstrain, a point where it fractures, tf. ear and a shear stress–strain diagram plotted. An 3 tf gxyelastic gxy measures the angular distortion of the element rela t engineering likematerial the one just described, the uch a diagram materials, for a ductile is shown in Fig. 3–24. y 2 originally along the x andDiagram y axes. son linear, Hooke’s law subjected for shear can be written as The tShear Stress–Strain test, and this so material when to shear will exhibit pl Here Gtoispure called thecas The behavior of a material subjected shear txy behavior and it will have a defined proportional limit tpl. rigidity. Its value repre laboratory using specimens in the of thin that when a small element of shape material is tube gxy 1.5 it was shown hardening will occur until an ultimate shear stress tu isIn Sec. (3–10) 2 t = Gg is, Gstresses Typi = tpl>g pl. mad to a torsional loading. If that measurements are to pureG shear,them equilibrium requires that equal shear finally, the material will begin to lose its shear strengthsubjected x listed on the inside back torque theelement. resulting angle of twist, then by the be developed on four facesand of the These txy must p must s a point where it fractures, tf. g stresses x ! gxy gpl explained gin gr 5, the data u Chapter Gcorners will be the as can be used to dete 2 be directed toward or away from diagonally opposite of thesame ! one ! elastic of ! ngineering like the just described, s called thematerials, shear modulus of elasticity or thethe modulus radians, a dimensionless and shear strain, and a shear stress–strain diag (b) (a)can be written as element, as shown in stress Fig. 3–23a. near, andrepresents so Hooke’sthe law for shear Fig. 3–24Furthermore, if the material is s value of the line on the t–g diagram, It will be shown Se example of such diagram for will a ductile material is sho Lineerslope elastik malzemede kayma gerilmeleri için Hooke kanunu şu ashear şekilde yazılabilir: homogeneous and isotropic, then this stress distort the in engineering materials are3–23 = tpl>gpl. Typical values for common Fig. y G are actually related Like the test,inthis material toby s uniformly, Fig. 3–23b. Astension mentioned Sec. 2.2, the when shear subjected strain he inside back cover. that the units of measurementelement for 3 t =Notice g behavior and it will have a defined propo xy (3–10)gxy measures the angularlinear-elastic Gg distortion of the element relative to the sides the same as those in ! for t (Pa 2or psi), since g is measuredoriginally strain hardening will occur until an ultimate s along the x andAlso, y axes. dimensionless quantity. reached. And finally, the material begin The behavior of a material subjected to pure shear can diğer bewill studied in to a lose it Burada G three ye elastik kayma modülü veya Kayma modülünü elled shown Sec. 10.6 that the material constants, E, n, and rijitlik modülü denmektedir. the in shear modulus of elasticity or the modulus of untilin it reaches a point where it fractures, tf. laboratory using specimens the shape of thin tubes and subjecting g ally related by the alue represents theequation slope of the line on2xythe t–g diagram,them to a torsional loading. ForIfmost materials, one just desc measurements are made oflike thethe applied malzeme sabitleri cinsinden aşağıdaki gibi yazabiliriz (bkz. Bölüm 10.6engineering ) Typical values for common engineering materials are >g . x t behavior is linear, and so Hooke’s law for shear be w pl pl torque and the resulting angle of twist, then by the Provided methods E to and be can p g G are xy nside back cover. Notice that the of measurement forexplained in Chapter 5, the data can be used to determine !units the shear E 2 from this equation rat same as those forGt=(Pa or psi), (b) since g is measured(3–11) instress and shear strain, and a shear stress–strain diagram plotted. An in the tu 211 + n2 For example, t = Gg ensionless quantity. tf example of such a diagram for a ductile material is shown in Fig. 3–24. ! the three material Gst = 11.011032 ksi, so t Fig. 3–23 own in Sec. 10.6 that constants, E, n, andLike the tension test, this material when subjected to shear will exhibit y related by the equation tG pl ve E bilinirse linear-elastic behavior and it will have a defined proportional tpl. Buradan görülüyor ki, Poisson oranı formülden hesaplanabilir ve deneysel Here G is called the shear modulus limit of elasticity or E and G are known, the value of n can then be determined Also, strain hardening will occur until an ultimate shear stress t rigidity. Its value represents the slope of the line on t u is equation rather than through experimental measurement. olarak ölçülmesine gerek kalmaz. reached. And finally, thethat material begin to lose its shear strengthengineer is, G =will values for common tpl>g mple, in the case of EA-36 steel, Est =G 2911032 ksi and pl. Typical 3 until it reaches a point where it fractures, . t (3–11) G from = listed on the insidef back cover. Notice that the units of m 3–11, nst = 0.32. 110 2 ksi, so that, g 211Eq. + n2 gu most engineering gpl gr For materials, one described, theor elastic G will be like the the same asjust those for t (Pa psi), since g t behavior is linear, and soradians, Hooke’sa law for shear can be written as dimensionless quantity. Fig. 3–24 It will be shown in Sec. 10.6 that the three material co nd G are known, the value of n can then be determined G are actually related by the equation uation rather thantuthrough experimental measurement. (3–10) t = Gg 3 t , in the case of f A-36 steel, Est = 29110 2 ksi and 3 2 ksi, so that, from Eq. 3–11, nst = 0.32. E tpl G the = modulus of Here G is called the shear modulus of elasticity or 211 + n2 rigidity. Its value represents the slope of the line on the t–g diagram, G that is, G = tpl>gpl. Typical values for common engineering materials are listed on the inside back Provided cover. Notice thatGthe of measurement E and areunits known, the value of for n can then g gu gpl gr G will be the same as those for (Pa or psi), since measured in t g from this equation rather thanisthrough experimenta radians, a dimensionless For quantity. example, in the case of A-36 steel, Est = Fig. 3–24 It will be shown in Sec.G10.6= that the 3three constants, E, n,nand so that, from Eq. 3–11, 11.0110 2 ksi, material st st = 0.32. G are actually related by the equation 3.7 3.7 G = E 211 + n2 (3–11) Provided E and G are known, the value of n can then be determined from this equation rather than through experimental measurement. For example, in the case of A-36 steel, Est = 29110!3392 ksi and Gst = 11.011032 ksi, so that, from Eq. 3–11, nst = 0.32. CHA1 P T0E6 R 3 M E CCHHAANPITCEARL 3P R OMP EE CR THIAE N S IO F LMPAT 106 CA R OEPREI A R LT SI E S C H A P T E R 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S 1 0 6 C H A P T E R 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S OF M AT E R I A L S 106 A TAE C HEA R IEARLISA L S CN H IACNAILC APLR O P PR EORPTEI RE TS I EOSF OMF AT MEAT 106 E106 E RC I AHLEC SRPH ESR TO I EFS M OAT F M AT IA LPSRT E3R 3M EM M E C H A NM I C A L P R O P E R T I EEXAMPLE S O F M AT E R I A L S EXAMPLE 3.6 PTER 3 E C H A N I C A L P R O P E R T I E S O F M 3.6 AT E R I A L S M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S 3.6 T I E S O F M AT E R I A L S EXAMPLE EXAMPLE C H A P T E R 3! M3.6 E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S An aluminum specimen shownspecimen in Fig. 3–26 diameter of a d An aluminum shownhas in aFig. 3–26 has An aluminum specimen shown in Fig. 3–26 has a diameter of and a gauge length of If a force d = 25 mm L = 250 mm. and a gauge length of If a force of 165 kN d = 25 mm L = 250 mm. An aluminyum aluminum specimen shown 0 0 Şekilde verilen numune için in ! 0 Fig. 3–26 has ve ! a0 diameter ofolduğu and a gauge length of If a force of 165 kN d = 25 mm L = 250 mm. 0 0 elongates the gauge length 1.20 mm, determine elongates the gauge length 1.20 mm, determine AMPLE 3.6 and a of gauge length force of 165 daluminum =diameter mmspecimen LFig. 250 minum specimen shown in Fig. 3–263–26 has a0 aluminum of aluminum specimen shown in Fig. has a25 diameter An shown inofinFig. 3–26 hashas aIf aadiameter of ofkN the modulus of the m 0 = 165 165 kN kN An specimen shown 3–26mm. diameter elongates the gauge length 1.20 mm, determine the modulus of causes th An aluminum specimen shown in Fig. 3–26 has a diameter of elasticity. Also, determine by how much force bilinmektedir. 165 kN kuvvet ölçüm uzunluğunu 1.20 mm uzattığına görethe diameter elasticity. Also, by much the force the gauge length 1.20 determine theof modulus ofcauses and and a gauge length of Lof If force of 165 kN =specimen aAn gauge length If force of 165 kN =mm L250 250specimen and a3–26 gauge length of Ifofahow 165 kN dmm. 25 mm L0 L =0mm, 250 mm. An shown in Fig. 3–26 has adetermine diameter 6525 kNmm and aofgauge length of Ifforce a force of 165 kN d=a0elongates =aa25 mm =ofby 250 mm. 0aluminum 0 = mm. aluminum shown in Fig. has aforce diameter 0 has uminum specimen shown in Fig. 3–26 diameter elasticity. Also, determine how much the force causes the diameter and a gauge length of If a of 165 kN d = 25 mm L = 250 mm. An aluminum specimen shown in Fig. 3–26 has a diameter of 165 kN of the specimen to contract. Take and sY = G = 26 GPa 0 0 elasticity. Also, determine by1.20 how much the force causes the diameter the specimen to contract. Take andalsY = 440 MPa. Galmodulus = 26 GPa s the the gauge length determine the modulus of gates gauge length 1.20 mm, determine the modulus ofIf elongates the gauge 1.20 mm, determine the modulus of and a gauge length ofofgauge If akuvvet force of 165 kN d0of =mm, L 250 mm. elongates the length mm, determine the of and a=mm gauge length of aAyrıca force of 165 kN delongates 251.20 mm L0mm, =of 250 mm. 0 =length 0 =length If a force 165 kN 5 mm and a gauge L25 250 mm. the specimen to contract. Take and G = 26 GPa = 440 MPa. s elastisite modülünü belirleyiniz. etkisi ile numunenin çapındaki the length 1.20 determine the modulus of 0gauge al Y and a gauge length of If a force of 165 kN d = 25 mm L = 250 mm. 0 0 Take Gal = 26 GPa and sY = 440 MPa. of the specimen to contract. y.ticity. Also,Also, determine by how much thethe force causes the diameter determine byelongates how much the force causes the diameter elasticity. Also, determine by how much the force causes thethe diameter gauge length 1.20 mm, the modulus ofcauses elasticity. Also, determine by how much force diameter the gauge length mm, the modulus ofthe tes the gaugeelongates length 1.20 mm, determine the modulus of determine elongates the gauge length 1.20 mm, determine the modulus of elasticity. Also, determine by 1.20 how much the force causes the diameter SOLUTION pecimen to contract. Take and G = 26 GPa = 440 MPa. s he specimen toelasticity. contract. Take and G = 26 GPa = 440 MPa. s al Y of the specimen to contract. Take and G = 26 GPa = 440 MPa. s elasticity. Also, determine by how much the force causes the diameter al Y SOLUTION of the specimen to contract. Take and G = 26 GPa = 440 MPa. olarak s Also, determine by how much the force causes the diameter al Y al Y ty. Also, determine by specimen how much causes diameter büzülme ! elasticity. Also, determine by how the diameter of the tothe contract. Take the and much G 26 GPa 440force MPa.causes ve sY =! the SOLUTION 3forcemiktarını al = hesaplayınız. 3 specimen SOLUTION Modulus of Elasticity. The average normal stress isin the of the specimen to contract. Take and G = 26 GPa = 440 MPa. s of the to contract. Take and G = 26 GPa = 440 MPa. s al Y al Y specimen to contract. Take and G = 26 GPa = 440 MPa. s 3 Elasticity. The normal stress in the specimen of the specimen Take andaverage = 26 GPaThe 440 MPa. sY =average al Y to contract.Modulus al of 3 Modulus ofGElasticity. normal stress in the specimen is ON verilmektedir. UTION Modulus of Elasticity. The average normal stress in the specimen is SOLUTION SOLUTION 16511032 N 3 P SOLUTION 2 N= 3 P 165110165110 SOLUTION = 336.1 MPa s = s3 of Elasticity. The The average normal stress in the specimen is 2 N SOLUTION dulus of Elasticity. average normal stress in the specimen is 3 P TION SOLUTION of of Elasticity. The average normal in in thethe specimen Modulus Elasticity. The average specimen is MPa =stress =isMPa 336.1 2=Nis stress Çözüm: Modulus A Modulus of Elasticity. The average normal stress specimen m22 = =m2 336.1 = s normal 2 1p>4210.025 L0 P in thes165110 A 2 1p>4210.025 L d = = 336.1 MPa s = Modulus of Elasticity. The average normal stress in the specimen is A 0 3 Modulus Elasticity. The average stress in specimen isspecimen 3 of 1p>4210.025 m2 us of Elasticity. average normal stress in0 the specimen isnormal Modulus of Elasticity. The average normal is L0 2in the 3 the 3 stress 165110 165110 P P The 1p>4210.025 d0 2 N d0 L02 N 3 2N 165110 2and N m2 P PA 165110 165110 2 N the average normal strain is = 336.1 MPa s = s == d0 = = 336.1 MPa P 3 = average = 336.1 MPa s = =MPa s=32=the the32average normal isMPa 2N AP A1p>4210.025 336.1 sN 2=m2 16511032m2 P 2 = 165110 165110 Nand 165110 Nnormal strain is336.1 2 2 =strain 1p>4210.025 P2and P A A 1p>4210.025 m2 1p>4210.025 m2 L L and the average normal strain is A = = 336.1 MPa s = 0 0 1p>4210.025 =s 336.1 s = d0 d=0 = 336.1 MPa = s =m22 = = 336.1 MPa = MPa 2 d 1.20 mm ! m2A 1p>4210.025 m2A A L0 1p>4210.025 1p>4210.025 m22 A 1p>4210.025 m22 L0dnormal d mm 1.20 mm P = = = 0.00480 mm>mm average strain is the average normal strain is d 1.20 0 and the average normal strain is and the average normal strain is L 250mm>mm mm P = = = 0.00480 d 1.20 mm P = = = 0.00480 mm>mm and the average normal strain is P = = = 0.00480 mm>mm L 250 mm L 250 mm Boyuna birim deformasyon: and the average normal strain is and the average normal e average normal strain and is the average normal strain is strain isL 250 mm d d1.20 1.20 mm mm d d 1.20 mm 1.20 mm Since s 6 sY = 440 MPa, the material behaves elast P = P == = = 0.00480 d mm>mm 1.20 mm = 0.00480 mm>mm P = = =Y 0.00480 mm>mm P = = ==MPa, 0.00480 mm>mm the material behaves elastically. The The s mm 6 ssY 6 = s 440 dSince 1.20 Since the material behaves elastically. 440 MPa, L d L250 mm = 0.00480 mm>mm 250 mm P = d =1.20 dmm 1.20 1.20 mm modulus of elasticity is therefore mm L 250 mm Since the material behaves elastically. The s 6 s = 440 MPa, L 250 mm P = = = 0.00480 mm>mm L 250 mm P = = = 0.00480 mm>mm P = = = 0.00480 mm>mm Y = 0.00480 P = = mm>mm modulus of elasticity is therefore modulus of elasticity is therefore L 250 mm L 250modulus mm L 250 mm elasticity L 250ofmm ! the the material behaves elastically. The The is therefore e6 ssY6 =sY440 material behaves elastically. = MPa, 440 MPa, 336.111062 Pa s The 6 elastically. Since thethematerial behaves elastically. 6 MPa, Since material behaves The ss 6Ys=Y 440 =behaves 440 MPa, 6= Since s 6 sY = 440 MPa, thes material elastically. The 336.1110 2 Pa E = = 70.0 GPa s6 336.1110 2 Pa isSince therefore al The dulus of is therefore Since the elastically. material behaves s the 6 s = 440 MPa, s elastically. material behaves elastically. The sof6elasticity sYelasticity = 440 MPa, material behaves sthe 6ofselasticity MPa, Y Y = 336.1110 2=Pa P =GPa 0.00480 s Ans. Ans. Eolarak = The = 70.0 of elasticity is therefore Since the material behaves elastically. The s 440 6 sis =modulus 440 MPa, modulus of elasticity is therefore modulus therefore al E = = 70.0 GPa Y ! yani malzeme elastik davranmaktadır al kN Eal = modulus of elasticity165 is therefore P Ans. = 70.0 GPa us of elasticity modulus is therefore of elasticity therefore P=0.00480 0.00480 6 6 is 165 kN 165 kN is therefore modulus of P 0.00480 336.1110 2 Pa 336.1110 2 elasticity Pa s s165 6 Contraction 6 6 kN of Diameter. First we will determine Pois 336.1110 Pa 6 336.1110 2 Pa s s s 336.1110 Ans. Eal = = 70.0 GPa Eal == 336.1110 = = 70.0 GPa 2Fig. 6 336.1110 Pa 2 Pa 3–26 s=EAns. 2EPa Contraction of2 Diameter. First we will determine Poisson’s ratio ratio 336.111062 Pa 6= s P s P 0.00480 Ans. = = 70.0 GPa 0.00480 Ans. E = = 70.0 GPa Ans. = = = 70.0 GPa Contraction of Diameter. First we will determine Poisson’s al al for the material using Eq. 3–11. al Fig. = 3–26 Ans. =Diameter. =Ans. 70.0 GPa 2 Pa Contraction of70.0 First we will determine Poisson’s ratio s E336.1110 al = = Ans. Eal = Fig. = 3–26 =Fig. GPa P70.0 E =3–26 GPa 0.00480 Pmaterial 0.00480 P 0.00480 al for the using Eq. 3–11. P 0.00480 Eal for =0.00480 = 70.0 GPa using Eq.Ans. P 165 kN kN 0.00480 for Eq. the material 3–11. Hooke Pkanunu ile,=! material 165 kN165 using 3–11. PthePoisson’s 0.00480 tion of Diameter. FirstFirst we will determine Poisson’s ratioratio traction of Diameter. we will determine E 65 kN Contraction of Diameter. First we of will determine Poisson’s ratio Contraction Diameter. First wewe will determine Poisson’s ratio Contraction of Diameter. First will determine Poisson’s ratioG = E ratio Contraction of Firstbu weoran will determine Poisson’s he material using Eq. 3–11. material using Eq. 3–11. action of Diameter. First weDiameter. will determine Poisson’s ratio Fig. 3–26 Contraction of First weDiameter. will determine Poisson’s ratio Şimdi de Poisson oranını hesaplayalım, çünkü ile büzülme miktarı hesaplanabilir. Fig. 3–26 E Fig. 3–26 211 + n2 G = E for the material using Eq. 3–11. for thethe material using Eq. 3–11. Contraction of Diameter. First we will determine ratio 211 for material using Eq. 3–11. G Poisson’s the material using Eq. 3–11. G =+ n2 = material Eq.the 3–11. material usingfor Eq. 3–11. g. 3–26 usingfor 211 + n2 211 + n2 for the material E E using Eq. 3–11. 70.0 GPa E G =G = E 26 GPa = E E E GPa = 70.0 GPa E G = 211 + n2 211 + n2 70.0 GPa 26 211 + n2 G = 70.0 G = G = G = G = 211 + n2 E 211 =+ n2 26 GPa +211 n2 26211 GPa =211 + n2 + n2 211 + n2 211 + n2 G = 211 + n2 n = 0.347 211 + n2 70.0 70.0 GPaGPa 70.0 GPa211 + n2 70.0 GPa n = 0.347 26 GPa = GPa 26 GPa = 70.0 70.0 GPa 70.0 GPa = GPa + n2+ n226 GPa =70.0 GPa 26 GPa n = 0.347 n = 0.347 26 26 GPa 26 GPa =211 211 211= += n2211 211 + n270.0 GPa Since Pn2 long = 0.00480 mm>mm, then by Eq. 3–9, 211 + n2 + 211 + n2 26 GPa =26211 Since then by Eq. 3–9, P = 0.00480 mm>mm, + n2 GPa = long = 0.347 n = n0.347 ! Since by Eq. 3–9, then by Eq. 3–9, Plat Plong = 0.00480 211Since +mm>mm, n2= Plong = then 0.00480 mm>mm, n 0.347 n = 0.347 n =n 0.347 n = 0.347 = 0.347 Plat n = 0.347 n = -P n = -P long engP= then by 3–9, Eq. 3–9, = 0.00480 mm>mm, then by deformasyonu Eq. mm>mm, P3–9, Plat n = 0.347 long0.00480 Enine birim hesaplayalım: lat Since then by Eq. P = 0.00480 mm>mm, long Since then by Eq. 3–9, P = 0.00480 mm>mm, long nthen =then -by longthen by Eq. 3–9, Plong = 0.00480Since mm>mm, Since Eq.Eq. 3–9, Plong = 0.00480 mm>mm, Since by 3–9,n = - Plong Plong = 0.00480 mm>mm, P Plat then by Eq. 3–9, Plong = 0.00480 mm>mm, long Plat Plat Plat 0.347 = Plat Plat then by Eq. 3–9, = P-latPPlong = 0.00480 mm>mm, n = n-Since 0.347 = 0.00480 mm>mm Plat PlatPlat n = -P Plong long Plat n = -Plat 0.00480 mm>mm n = long n0.347 = - P=- n = n = - PPlong 0.347 = Plong P P 0.00480 mm>mm lat longlong P = - 0.00166 mm>mm long 0.00480 mm>mm Plat Plat Plat n = -Plat Plat = - 0.00166 mm>mmlat 0.3470.347 = -= P Plat 0.347 = 0.347 = long P P Plat lat latmm>mm 0.00480 mm>mm P = 0.00166 0.00480 mm>mm 0.347 = 0.00480 Plat = -0.00166 mm>mmis therefore 0.347 = lat= - - mm>mm The contraction of the diameter 0.347 0.347 = - 0.00480 mm>mm 0.00480 mm>mm Plat The contraction of0.00480 the diameter is therefore 0.00480 mm>mm 0.00480 mm>mm mm>mm 0.347 = P = -0.00166 mm>mm P! lat =lat-0.00166 mm>mm The contractionPlat = -diameter 0.00166 mm>mm of The the is therefore d¿ = 10.001662125 mm2 P = -0.00166 mm>mm 0.00480 mm>mm contraction ofd¿the is therefore Plat = - 0.00166 mm>mm = diameter 10.001662125 mm2 Platlat= -0.00166 mm>mm PlatPlat = = -0.00166 mm>mm -0.00166 mm>mm = 0.0416 mm = 10.001662125 mm2 mm contraction of diameter the diameter is therefore traction of the is therefore The contraction of= the diameter isd¿ therefore = 0.0416 Ans. -0.00166 mm>mm Çap boyunca enine deformasyonu (büzülmeyi) hesaplayalım: d¿ = 10.001662125 mm2 The contraction of the diameter isPlat therefore ntraction of the diameter is therefore The contraction of the diameter iscontraction therefore The contraction of the diameter is therefore The of the diameter is therefore = 0.0416 mm Ans. = 10.001662125 d¿ =d¿10.001662125 mm2mm2 d¿ = 10.001662125 mm2 = 0.0416 mm Ans. d¿ mm2 d¿ = 10.001662125 mm2 The contraction of =the10.001662125 diameter ismm2 therefore = 10.001662125 mm2 d¿ =mm 10.001662125 mm2 = d¿ 0.0416 Ans. = 0.0416 Ans. = 0.0416 mm mm d¿ = 10.001662125Ans. = 0.0416 mm Ans. Ans. d¿ mm = 10.001662125 mm2 = 0.0416 Ans. ! = 0.0416 mm = 0.0416 mm Ans. = 0.0416 mm Ans. EXAMPLE EXAMPLE3.6 3.6 165 kN 3.6 165 kN 165 kN 165 kN = 0.0416 mm Ans. !40

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