1. Materiali per lIngegneriaM.Ferraris1Mechanical properties ofmaterials Materials choice Design issues Correlation requested properties/class of material Mechanical properties independent to time Mechanical properties dependent to time Mechanical properties dependent to temperature

2. Materiali per lIngegneriaM.Ferraris2 (STRESS) and (STRAIN) : definition (STRESS) = F (force, Newton, N) / A (surface, m2) = F/A (N/m2= Pa) (N/mm2= MPa)DEFORMATION or (STRAIN)(adimensional) = (l l0)/l0 = l/l0sample length = linitial sample length = l0tensile stress, compressive stress, bending stress, shear stress 3. Materiali per lIngegneriaM.Ferraris3Tensile and compressionstress/strainA0FFll0 l0FFlA0Tension CompressionStress, = F/A0Strain, = (l l0)/l0= l/l0 4. Materiali per lIngegneriaM.Ferraris4ShearA0FF (Shear stress), = F/A0(Shear strain), = tan 5. Materiali per lIngegneriaM.Ferraris5TorsionTTTorsion = T/A0Shear strain, = tan Ao 6. Materiali per lIngegneriaM.Ferraris6stress/strain curves standardsgauge length, 2 in.reduced section, 2.25 in.diameter, 0.5 in.diameter, 0.75 in.radius 3/8 in.ASTM (American Society for Testing andMaterials)ISOUNI EN 7. Materiali per lIngegneriaM.Ferraris7For most of materials, low stresses correspond toproportionally low deformations (directproportionality)= E (Hooke law, E=Young modulus orelastic modulus)=F/A=l/loF/A=E l/lo 8. Materiali per lIngegneriaM.Ferraris8Tensile test curve on a metalTensile strengthElastic fieldPlastic field =E 9. Materiali per lIngegneriaM.Ferraris9Elastic behaviour Small stresses corresponds to smalldeformation having a direct proportionalitybetween them. Small deformations are immediately recoveredwhen stress is removed E = / = [N/mm2] = [MPa]Young modulus or Elastic Modulus E(Hooke law) 10. Materiali per lIngegneriaM.Ferraris10Elastic modulus and atomic bond E ceramics > E metals >> E polymers Elastic modulus is directly proportional to theatomic bond (and melting temperature) No fracture of bonds in the elastic field Increasing of the bond length when stress isapplied. Immediate recovering of the bond equilibriumlength after stress removal 11. Materiali per lIngegneriaM.Ferraris11CallisterHow atomicbonds areresponsible ofmaterialspropertiesCondon-Morse curves 12. Materiali per lIngegneriaM.Ferraris12Elastic modulus and bond strengthE is directlyproportional to(dF/dr)roStrong bonds:ceramicsWeak bonds: polymers 13. Materiali per lIngegneriaM.Ferraris13Elastic modulus and meltingtemperatureMaterial Tempmelting [ C]E(GPa)TiC 3160 310Al2O3 2045 370W 3410 393Fe 1536 207Cu 1083 111Al 660 70Pb 327 14Polyetilene 130 1glasses - 60-9070-90 14. Materiali per lIngegneriaM.Ferraris14g Only the elastic field 15. Materiali per lIngegneriaM.Ferraris15Al and alloys 65-80 diamond 1000Mg and alloys 45-50 ceramici 300-600steels 195-210legno 10-150.5-1Ti and alloys 85-130 polimeri 0.1-5CFRP 100-200GFRP 10-40glasses 70-90 W 407Some elastic moduli (GPa) 16. Materiali per lIngegneriaM.Ferraris16How to measure the elasticmodulussteelAlstressstrain 17. 17Esercizio 6.1 pagina 122 e progetto 6.1 p.144Esercizi:6.6 (16 mm)6.7 (a:325mm2*245N/mm2=79625 N)(b: 115,24 mm)6.8 (diametro 7,65 mm) 18. Materiali per lIngegneriaM.Ferraris18How to measure the elasticmodulus Drawbacks of tensile tests:Sample preparationDestructive testUnsuitable for brittle materials Measure of the elastic modulus by Tensile,compression or bending tests Non destructive tests 19. Materiali per lIngegneriaM.Ferraris19Elastic modulus by sonic test E= 0.9465 * (2L4/t2) = costant depending on material shape = sonic wave frequency; L= sample length;= material density; t= sample thickness. 20. 20Elastic modulus vs TWsteelAlElastic modulus, GPa 21. Materiali per lIngegneriaM.Ferraris21Elastic modulus of a steel vs T 22. Materiali per lIngegneriaM.Ferraris22lz/2l0xlx/2l0zzyzx==Es. 6.2 pag.124Poisson coefficient ()Metals: about 0.33Ceramics: 0.17 - 0.27Polymers: 0.33 - 0.5 23. Materiali per lIngegneriaM.Ferraris23Non linear elastic behaviour Some polymers, cast iron, concrete,. E calculated locallyStrainStress121 2Secant modulus at 1 = 1/1Tangent modulus at 2 = d/d 24. Materiali per lIngegneriaM.Ferraris24summaryLinear elasticBehaviour(most materials)Non linear elasticbehaviour(some polymers,cast iron,concrete,)Anelastic behaviourTime is required torecover deformationafter stress removal(some polymers) 25. Materiali per lIngegneriaM.Ferraris25Plastic deformation and chemical bond When applied stresses increases and correspondingdeformations are higher than about 0.002, the Hookelaw doesnt work any more. Bonds start breaking and, if possible, to form othersin different positions If the load is removed, the sample recovers only theelastic deformation, but a plastic, permanentdeformation is evident. The border between elastic and plastic deformation isdefined yield strength or proportional limit 26. Materiali per lIngegneriaM.Ferraris26/ for metalsfractureTensilestrengthUpperlowerystrainstressstrictionElastic field Plastic fieldStrain at fractureMetals follow oneof the two curveswww.matcoinc.com/images/sem1a.jpg 27. Materiali per lIngegneriaM.Ferraris27 28. Materiali per lIngegneriaM.Ferraris28Yield strength at 0.2 %strain (0.002) 29. Materiali per lIngegneriaM.Ferraris29Yield strength for differentmaterials y [MPa]steels 250-1400Ni alloys 200-1600Ti alloys 200-1300Cu alloys 60-950Al alloys 100-650Mg alloys 80-300Epoxy resins 30-100nylon 50-90polystirene 35-70abs 50polyetilene 6-30 30. Materiali per lIngegneriaM.Ferraris30Dislocationsexplain the 3orders ofmagnitutebetweentheorethicalandexperimentalyield strengthYield strength and dislocations 31. Materiali per lIngegneriaM.Ferraris31Withoutdislocations, theyield strengthwould be aboutE/8 (GPa) 32. Materiali per lIngegneriaM.Ferraris32 33. Materiali per lIngegneriaM.Ferraris33Sliding planeShear stressDislocations provide an effective way for bonds to break and re-form in another place, when load is applied.High density sliding planes favoured by dislocation motion 34. Materiali per lIngegneriaM.Ferraris34Es: Cu, Al, Ag, AuEs: Fe, W, CrEs: Mg, Ti, ZnHexagonal compactCCB 35. Materiali per lIngegneriaM.Ferraris35 CCB e CFC have at least 12 sliding planes HCP has a maximum of 6 CFC metals (Cu, Al, Ni, Ag, Au) and CCB (Fe)are ductile, with large plastic deformation HCP metals (Ti, Zn) are more brittleplane (111) for CFC,with three slidingdirections 36. Materiali per lIngegneriaM.Ferraris36 Plastic deformation of grains in a stainless steel after laminationfrom CallisterAfter deformation 37. Materiali per lIngegneriaM.Ferraris37Stress/strain curve for AlTensile strength (TS) 38. Materiali per lIngegneriaM.Ferraris38If we stop loading here a residual plastic deformation ( res)is still present 39. Materiali per lIngegneriaM.Ferraris39Plastic(residual,permanent)deformation 40. Materiali per lIngegneriaM.Ferraris40 41. Materiali per lIngegneriaM.Ferraris41Ductile fractureStriction Formation of cavitiesCavities coalescencePropagationhttp://web.umr.edu/~be120/lessons/intro/tension/testing_st/fracture.gifCoarse surface due toplastic deformation 42. Materiali per lIngegneriaM.Ferraris42Tensile strengthbehaviour fordifferentmaterials 43. Materiali per lIngegneriaM.Ferraris43Stress/strain curves for brittlematerialsceramics, glasses;brittle alloys, brittlepolymers, brittlecompositesNo (or limited) plasticdeformation 44. Materiali per lIngegneriaM.Ferraris44Brittle fracture Without plastic deformation Flat fracture surfaces Along crystalline planes for mono-crystalline materials(cleavage) Intergranular fracture for poly-crystalline materialshttp://www.jwave.vt.edu/crcd/farkas/lectures/Fract1/fig3.gif 45. Materiali per lIngegneriaM.Ferraris45Ductile and brittle fracturesurfaces Ductile fracture: plastic deformation at the crackedge, slow crack propagation Brittle fracture: no plastic deformation at the crackedge, quick crack propagation (catastrophic failure) 46. Materiali per lIngegneriaM.Ferraris46Tensile strength () and % deformation at fracture Material (MPa) (%) steel 400-2500 40-12 Al alloys 300-770 8-16 Al2O3 sint. 200-340 / SiO2Glass 110 / ZrO2 sint. 80 / Polyesthers 40-80 1-2 Nylon 70-160 50-200 Epoxy 30-120 3-6 47. Materiali per lIngegneriaM.Ferraris47How to modify plasticproperties of materials Make dislocation motionmore difficult = increasematerials mechanicalstrength Cold working Grain size precipitates second phases Monocrystals Solid solutionsgrain boundarygrain boundaryy=0+kyd12Hall-Petch lawd = grain diameter (ave)y = yield strengthMaterials related costantGrain size modification by thermal treatment 48. Materiali per lIngegneriaM.Ferraris48Solid solutionsDislocations motion slowedTensile stressesCompression stresses Metallic solidsolutions(alloys) havesuperiormechanicalstrength thanpure metals 49. Materiali per lIngegneriaM.Ferraris49Role of alloying elements onmechanical properties 50. Materiali per lIngegneriaM.Ferraris50 51. Materiali per lIngegneriaM.Ferraris51 52. Materiali per lIngegneriaM.Ferraris52COLD WORKING OF STEELSREVERSIBLE PROCESS:COLDWORKING/ANNEALING