Presentation on theme: "Chapter 6: Mechanical Properties"— Presentation transcript:
1 Chapter 6: Mechanical Properties ISSUES TO ADDRESS...• Stress and strain: What are they and stress-strain curve.• Elastic behavior: When loads are small, how muchdeformation occurs? What materials deform least?• Plastic behavior: At what point does permanent deformation occur? What materials are most resistant to permanent deformation?• Ductility, Toughness, Resilience, Hardness
2 CONCEPTS OF STRESS AND STRAIN If a load is static or changes relatively slowly with time and is applied uniformly over a cross section or surface of a member, the mechanical behavior may be ascertained by a simple stress–strain test;There are four principal ways in which a load may be applied: namely,tension, compression, shear, and torsion
4 Elastic Deformation d F F d 1. Initial 2. Small load 3. Unload bondsstretchreturn toinitialFdLinear-elasticNon-Linear-Elastic means reversibleTime independent!
5 Plastic Deformation (Metals) 1. Initial2. Small load3. UnloadplanesstillshearedFdelastic + plasticbondsstretch& planesshearplasticFdlinearelasticplasticPlastic means permanent!
6 Engineering Stress F t s = A F t s = A m N or in lb • Tensile stress, s:original areabefore loadingArea, AFts=Ao2fmNorinlb• Shear stress, t:Area, AFts=Ao Stress has units: N/m2 or lbf/in2
7 Common States of Stress • Simple tension: cableAo= cross sectionalarea (when unloaded)Fos=FAsSki lift• Torsion (a form of shear): drive shaftMAo2RFscot=MRI
8 OTHER COMMON STRESS STATES (1) • Simple compression:AoBalanced Rock, ArchesNational Park(photo courtesy P.M. Anderson)Canyon Bridge, Los Alamos, NMos=FANote: compressivestructure member(s < 0 here).(photo courtesy P.M. Anderson)
9 OTHER COMMON STRESS STATES (2) • Bi-axial tension:• Hydrostatic compression:Fish under waterPressurized tank(photo courtesyP.M. Anderson)(photo courtesyP.M. Anderson)sz> 0qs < 0h
10 Engineering Strain w e = d L d e = w q q g = Dx/y = tan • Tensile strain:d/2Low• Lateral strain:e=dLodeL=wo = L=L-L0• Shear strain:qxqg = Dx/y = tany90º - qStrain is alwaysdimensionless.90º
11 Stress-Strain Testing • Typical tensile test machine• Typical tensile specimengaugelengthLoad measuredspecimenextensometerStrain measuredWe can obtained many important properties of materials from stress-strain diagram
12 (linear)Elastic Properties of materials • Modulus of Elasticity, E :(also known as Young's modulus)• Hooke's Law:s = E esLinear-elasticEeFsimpletensiontest
13 Poisson's ratio, n e n = - • Poisson's ratio, n: Units: Lmetals: n ~ 0.33 ceramics: n ~ 0.25 polymers: n ~ 0.40Units:E: [GPa] or [psi]n: dimensionless
14 Mechanical Properties Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metalThe magnitude of E is measured of the resistance to separation of adjacent atoms, that is, the interatomic bonding forcesSo E α (dF/dr)
15 T – E diagram Plot of modulus of elasticity versus temperature for tungsten, steel, andaluminum.
16 Other Elastic Properties simpletorsiontestMtGg• Elastic Shearmodulus, G:t = G gstressstrain• Elastic Bulkmodulus, K:pressuretest: Init.vol =Vo.Vol chg.= DVPP = -KDVo• Special relations for isotropic materials:2(1 + n)EG=3(1 - 2n)K
17 Young’s Moduli: Comparison GraphiteCeramicsSemicondMetalsAlloysComposites/fibersPolymers0.280.61Magnesium,AluminumPlatinumSilver, GoldTantalumZinc, TiSteel, NiMolybdenumGraphiteSi crystalGlass-sodaConcreteSi nitrideAl oxidePCWood( grain)AFRE( fibers)*CFREGFRE*Glass fibers onlyCarbonfibers onlyAramid fibers onlyEpoxy only0.40.824610001200TinCu alloysTungsten<100><111>Si carbideDiamondPTFEHDPLDPEPPPolyesterPSPETCFRE( fibers)FRE( fibers)*FRE(|| fibers)*E(GPa)Composite data based onreinforced epoxy with 60 vol%of alignedcarbon (CFRE),aramid (AFRE), orglass (GFRE)fibers.109 Pa
18 Useful Linear Elastic Relationships • Simple tension:• Simple torsion:a=2MLopr4GM = moment= angle of twist2roLod=FLoEAdL=-nFwoEAFAod/2LLow• Material, geometric, and loading parameters allcontribute to deflection.• Larger elastic moduli minimize elastic deflection.
21 Plastic (Permanent) Deformation (at lower temperatures, i.e. T < Tmelt/3)• Simple tension test:Elastic+Plasticengineering stress, sat larger stressYielding stressyElasticinitiallypermanent (plastic)after load is removedepengineering strain, eplastic strain
22 Plastic properties of materials Yield Strength, syTensile Strength, TSDuctilityToughnessResilience, Ur
23 Yield Strength, sy y = yield strength sy • Stress at which noticeable plastic deformation hasoccurred.when ep = 0.002tensile stress,sengineering strain,esyp= 0.002y = yield strengthNote: for 2 inch sample = = z/z z = in
24 Yield Strength : Comparison Graphite/Ceramics/SemicondMetals/AlloysComposites/fibersPolymersYield strength,sy(MPa)PVCHard to measure,since in tension, fracture usually occurs before yield.Nylon 6,6LDPE7020406050100103020034567Tin (pure)Al(6061)aagCu(71500)hrTa(pure)TiSteel(1020)cd(4140)qt(5Al-2.5Sn)WMo (pure)cwHard to measure,in ceramic matrix and epoxy matrix composites, sincein tension, fracture usually occurs before yield.HDPEPPhumiddryPCPETRoom T valuesa = annealedhr = hot rolledag = agedcd = cold drawncw = cold workedqt = quenched & tempered
25 Tensile Strength, TS TS engineering stress strain engineering strain • Maximum stress on engineering stress-strain curve.ystrainTypical response of a metalF = fracture orultimatestrengthNeck – acts as stress concentratorengineeringTSstressengineering strain• Metals: occurs when noticeable necking starts.• Polymers: occurs when polymer backbone chains arealigned and about to break.
27 Ductility Ductility can be expressed quantita • Plastic tensile strain at failure:x 100LEL%of-=Percent elongationEngineering tensile strain,eEngineeringtensilestress,ssmaller %ELlarger %ELbrittleLfAoAfLoductile• Another ductility measure:100xARA%of-=Percent reduction of area
28 DuctilityBrittle materials are approximately considered to be those having a fracture strain of less than about 5%.Brittle materials have little or no plastic deformation upon fracture
29 Toughness • Energy to break a unit volume of material or it is a measure of the ability of a material to absorbenergy up to fracture( or impact resistance)• Approximate by the area under the stress-straincurve.very small toughness(unreinforced polymers)Engineering tensile strain,eEngineeringtensilestress,ssmall toughness (ceramics)large toughness (metals)Brittle fracture: elastic energy( no apparent plastic deformation takes place fracture) Ductile fracture: elastic + plastic energy(extensive plastic deformation takes place before fracture)
30 Fracture toughness is a property indicative of a material’s resistance to fracture when a crack is presentFor a material to be tough, it must display both strength and ductility; and often, ductile materials are tougher than brittle ones
31 Resilience, UrAbility of a material to store energy when it is deformed elastically, and then, upon unloading, to have this energy recoveredEnergy stored best in elastic regionIf we assume a linear stress-strain curve this simplifies toyr21Ues@Area under - curve taken to yielding(the shaded area)
32 Modulus of resilience( is the area under curve)Thus, resilient materials are those having high yield strengths and low module of elasticity; such alloys would be used in spring applications.
34 Hardness • Resistance to permanently indenting the surface, or is a measure of a materials resistance to deformation bysurface indentation or by abrasion.• Large hardness means:--resistance to plastic deformation or cracking incompression.--better wear properties.e.g.,10 mm sphereapply known forcemeasure sizeof indent afterremoving loaddDSmaller indentsmean largerhardness.increasing hardnessmostplasticsbrassesAl alloyseasy to machinesteelsfile hardcuttingtoolsnitrideddiamond
35 Hardness: Measurement RockwellNo major sample damageEach scale runs to 130 but only useful in rangeMinor load kgMajor load (A), 100 (B) & 150 (C) kgA = diamond, B = mm ball of steel, C = diamondthe hardness can be taken directly from the machine, so it is a quick testHB = Brinell Hardness- P= 500, 1500, 3000 kgTS (psia) = 500 x HBTS (MPa) = 3.45 x HB(The HB and tensile strength relationship)
37 problemMaterialYielding strength( MPa)Tensile strengthStrain at fractureFracture strength(MPa)Elastic modulus(GPa)A3103400.23265210B1001200.4105150C4155500.15500D7008500.14720EFracture before yielding650350a) Which experience the greatest percent reduction in area? Why?b) Which is the strongest? Why?c) Which is the stiffest? Why?d) Which is the hardest? Why?
38 Material B will experience the greatest percent area reduction since it has the highest strain at fracture, and, therefore is most ductile.Material D is the strongest because it has the highest yield and tensile strengths.Material E is the stiffest because it has the highest elastic modulus.stiffness=E=σ/ε, the higher E the material more stiffestMaterial D is the hardest because it has the highest tensile strength.
41 ( ) Hardening s e • An increase in sy due to plastic deformation. large hardeningsmall hardeningy1• Curve fit to the stress-strain response:The region of the true stress-strain curve from the onset of plastic deformation to the point at which necking begins may be approximated bysT=Ke()n“true” stress (F/Ai )“true” strain: ln(L/Lo)hardening exponent:n =0.15 (some steels)to n =0.5 (some coppers)(n and K constants)
42 Design or Safety Factors • Design uncertainties mean we do not push the limit.• Factor of safety, NOften N isbetween1.2 and 4Safe stress, orThe choice of an appropriate value of N is necessary. If N is too large, then component overdesign will result, that is, either too much material or a material having a higher-than-necessary strength will be used. Values normally range between1.2 and 4.0.Selection of N will depend on a number of factors, including economics, previous experience, the accuracy with which mechanical forces and material properties may be determined, and, most important, the consequences of failure in terms of loss of life and/or property damage.
43 Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5.1045 plaincarbon steel:sy= 310 MPaTS = 565 MPaF = 220,000NdLo5d = m = 6.7 cm
46 Summary • Stress and strain: These are size-independent measures of load and displacement, respectively.• Elastic behavior: This reversible behavior oftenshows a linear relation between stress and strain.To minimize deformation, select a material with alarge elastic modulus (E or G).• Plastic behavior: This permanent deformationbehavior occurs when the tensile (or compressive)uniaxial stress reaches sy.• Toughness: The energy needed to break a unitvolume of material.• Ductility: The amount of plastic strain that has occurredat fracture.
47 problemA brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa ( psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.
48 We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 1.9 mm (0.075 in.). It is first necessary to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, ifε(test) < ε(yield)deformation is elastic, and the load may be computed using Equations 6.1 and 6.5. However, ifε(test) > ε(yield)computation of the load is not possible inasmuch as deformation is plastic and we have neither a stress-strain plot nor a mathematical expression relating plastic stress and strain. We compute these two strain values as
49 Therefore, computation of the load is not possible since ε(test) > ε(yield).
50 problemA cylindrical metal specimen having an originaldiameter of 12.8 mm(0.505 in.) and gauge length of mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm (0.320 in.), and the fractured gauge length is mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.
51 This problem calls for the computation of ductility in both percent reduction in area and percent elongation. Percent reduction in area is computed using Equation 6.12 asin which d0 and df are, respectively, the original and fracture cross-sectional areas. Thus,While, for percent elongation, we use Equation 6.11 as