Mechanical Properties of Metals 6.1 Elastic and Plastic Deformation

Chapter 6Mechanical Properties of MetalsMechanical Properties refers to the behavior of material whenexternal forces are appliedStress and strain ⇒ fractureFor engineering point of view: allows to predict the ability of a componentor a structure to withstand the forces applied to itFor science point of view: what makes materials strong → helps us todesign a better new oneLearn basic concepts for metals, which have the simplest behaviorReturn to it later when we study ceramics, polymers, composite materials,nanotubesChapter 616.1 Elastic and Plastic Deformation• Metal piece is subjected to a uniaxial force ⇒ deformation occurs• When force is removed:- metal returns to its original dimensions ⇒ elastic deformation (atoms return to theiroriginal position)- metal deformed to an extent that it cannot fully recover its original dimensions ⇒plastic deformation (shape of the material changes, atoms are permanentlydisplaced from their positions)FL 0A 0 AL=L 0+∆ LChapter 6F21

6.2 Concept of Stress and StrainLoad can be applied to the material by applying axial forces:NotdeformedTensionFCompressionL 0A 0AL=AFL=L 0+∆ LL 0+∆ LFF∆L can be measured as a function of the applied force; area A 0 changes in responseChapter 63Stress (σ) and Strain (ε)Block ofmetalAFFL=L 0+∆ LStress (σ)• defining F is not enough ( F and A can vary)• Stress σ stays constantFσ =A•UnitsForce / area = N / m 2 = Pausually in MPa or GPaStrain (ε) – result of stress• For tension and compression: change in length of asample divided by the original length of sample∆Lε =LChapter 642

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Elastic Properties of Materials• Most materials will get narrow when stretched and thicken when compressed• This behaviour is qualified by Poisson’s ratio, which is defined as the ratio of lateraland axial strainεPoisson's _ Ratio : ν = −εxz= −εεyz• the minus sign is there because usually if ε z > 0, and ε x + ε y < 0 ⇒ ν > 0• It can be proven that we must have ν ≤ ½; ν = ½ is the case when there is novolume change( l + ∆l)( l + ∆l)( l + ∆l) = l × l × lxxyyzzxyzChapter 67Poisson’s Ratio, ν•For isotropic materials (i.e. material composed of many randomly - oriented grains) ν= 0.25• For most metals:0.25 < ν < 0.35•If ν = 0 :means that the width of the material doesn’t change when it is stretched orcompressed• Can be:ν < 0(i.e. the material gets thicker when stretched)Chapter 684

6.3 Modulus of elasticity, or Young’s Modulus• Stress and strain are properties that don’t depend on the dimensions of thematerial (for small ε), just type of the material• E – Young’s Modulus, Paσ ( stress)E = ε ( strain)• Comes from the linear range in the stress-strain diagram• many exceptions…Behavior is related to atomic bonding between the atomsMaterialMetalsPolypropeleneRubberHydrogels and live cellsYoung’s Modulus [GPa]20-1001.5-20.01

Tensile Test1. Modulus of elasticity2. Yield strength at 0.2% offset3. Ultimate tensile strength4. Percent elongation at fracture5. Percent reduction in area at fractureChapter 611Other tensile test characteristics:• Yield strength (at 0.2% offset)• Ultimate Tensile Strength (UTS): themaximum strength reached in the stressstraincurveF ∆Lσ == E ×A(original)L• Percent elongation at fracture (measure ofductility of the metal)• Percent reduction in area at fractureAinitial− Afinal% _ reduction _ in _ area =× 100%AinitialChapter 6126

True and Engineering Stressσengineeringσ=trueFAinitial=FA∆L= E ×LinstantChapter 6136.4 HardnessHardness: a measure of the resistance of a material to plastic(permanent) deformationMeasured by indentation• indenter material (ball, pyramid, cone) is harder than thematerial being tested (i.e.: tungsten carbide, diamond)• indenter is pressed at 90 o• hardness is based on the depth of the impression or its crosssectionalareaSeveral common hardness tests: hardness numbers can becalculatedMaterial strength and hardness are relatedHardness test is nondestructive ⇒ often usedChapter 6147

Hardness tests: hardness numbersChapter 6156.5 Plastic deformations of single crystal metalsA rod of a single crystal Zn (hcp) stressedbeyond its elastic limit:• slipbands: slip of metal atoms on specificcrystallographic planes (slip planes)•slip is predominately along the basal planesChapter 6A rod of a single crystal Cu (fcc) duringplastic deformation:• slip lines: 50-500 atoms apart• slipbands: separated by ~>10,000atomic planes168

Other mechanical characteristics• Ductility: amount of plastic deformation that occurs before fracture- if ductility is high, the material can be deformed by applying stresses.Ex.: gold- if it is low, material breaks first, without significant deformation (material is brittle)- depend on T: at low T many metals become brittle and can break as a glass• Resilience: ability to have high yield strength and low E.Ex.: good springs• Toughness: ability to absorb energy up to a fractureChapter 617Mechanism of Slip deformationthe group of atoms do NOT slide over eachother during plastic shear deformation ⇒the process requires too much energyThe process takes less energy!!!Chapter 6189

Motion of DislocationsIn the metal slip mechanism,dislocations move through the metalcrystals like wave fronts, allowingmetallic atoms to slide over each otherunder low shear stress ⇒deformation without fractureChapter 619Slip SystemsTypically slip planes are the most densely packed planes (less energy is required tomove from one position to another), which are the farthest separatedCombination of a slip plane and a slip direction: slip systemChapter 62010

Slip systems observed in crystal structuresFor hcp crystals: 3 slip systems, restricts their ductilityChapter 621Schmid’s LawF cosλcosφτr==AoFAoSlip process begins within the crystal when theshear stress on the slip plane in slip directionreaches critical resolved shear stress τ chcp (Zn, Mg): 0.18, 0.77 MPafcc (Cu): 0.48 MPaFrτr=AF r= F cosλslip _ planecos λ cosφ= σ cosλcosφSchmid’s lawAoAslip _ plane=cosφChapter 62211

Q.: A stress of 75 MPa is applied in the [0 01] direction on an fcc single crystal.Calculate(a) the resolved shear stress acting on the (111) [-101] slip system and,(b) (b) the resolved shear stress acting on the (111) [-110] slip system.Chapter 623Mechanical TwinningAnother important plastic deformation mechanism (low T)Schematic diagram of surfaces of a deformedmetal after (a) slip and (b) twinningfrom G. GottsteinChapter 62412

6.6 Plastic Deformations in Polycrystalline Metals• Majority of engineering alloys and metals arepolycrystalline• Grain boundaries – act as diffusion barriers fordislocation movements• In practice: fine grain materials are stronger andharder (but less resistant to creep and corrosion)• Strength and grain size are related by Hall-Pelchequation:kσy= σo+dσ o and k – constantsas grain diameter decreases, theyield strength of the materialincreasesChapter 625Pile-up of dislocationsSchematicObserved in stainlesssteel (TEM)Grain shape changes with plastic deformationDislocation arrangement changesChapter 62613

6.7 Cold Plastic Deformation for Strengthening of Metals• The dislocation density increases with increased cold deformation• New dislocations are created by the deformation and must interact with those alreadyexisting• As the dislocation density increases with deformation, it becomes more and moredifficult for the dislocations to move through the existing dislocations⇒ Thus the metal work or strain hardens with cold deformationPercent cold work versus tensile strength andelongation for unalloyed oxygen-free copperCold work is expressed as a percent reduction incross-sectional area of the metal being reduced.Chapter 6276.8 Superplasticity in MetalsSuperplasticity: the ability of some metals to deform plastically by 1000-2000% at hightemperature and low loading ratesEx.: Ti alloy (6Al – 4V) 12% @ RT, typical tensile test load rates~1000% @ 840 o C, lower loading ratesRequirements:1. The material must possess very fine grain size (5-10mm) and be highly strain-ratedensity2. A high loading T (>0.5 T m ) is required3. A low and controlled strain rate in the range of 0.01-0.00001 s - 1 is requiredChapter 62814

Nanocrystalline Metals• Nanocrystalline metals: d < 10-50nmσy= σo+• Consider Cu: σ o = 25 MPa, k = 0.11 MPa m 0.5 (from Table 6.5)σσIs this possible?10µmkd0.11MPa25MPa+−8=100.11MPa25MPa+−51010 nm=?Different dislocation mechanism: grain boundary sliding, diffusion, etcChapter 629Summary•Introduced stress, strain and modulus of elasticityFσ =A• Plastic deformations of single crystal metals- In the single crystal metal - slip mechanism: dislocations move through the metalcrystals like wave fronts, allowing metallic atoms to slide over each other under low shearstress- Slip process begins within the crystal when the shear stress on the slip plane in slipdirection reaches critical resolved shear stress τ c-Schmid’slaw:F cosλcosφFτr== cos λ cosφ= σ cosλcosφA A• Plastic deformations in polycrystalline metals- Strength and grain size are related by Hall-Pelch equation:• Nanocrystalline materialso∆Lε =LoChapter 6σ ( stress)E = ε ( strain)σy= σo+kd3015