02 mecc a

191 views
121 views

Published on

Published in: Technology, Business
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
191
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
5
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

02 mecc a

  1. 1. Materiali per l’IngegneriaM.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. 2. Materiali per l’IngegneriaM.Ferraris2σ (STRESS) and ε (STRAIN) : definitionσ (STRESS) = F (force, Newton, N) / A (surface, m2)σ = F/A (N/m2= Pa) (N/mm2= MPa)DEFORMATION or ε (STRAIN)(adimensional)ε = (l – l0)/l0 = ∆l/l0sample length = linitial sample length = l0tensile stress, compressive stress, bending stress, shear stress
  3. 3. Materiali per l’IngegneriaM.Ferraris3Tensile and compressionstress/strainA0FFll0 l0FFlA0Tension CompressionStress, σ = F/A0Strain, ε = (l – l0)/l0= ∆l/l0
  4. 4. Materiali per l’IngegneriaM.Ferraris4ShearA0FFθ (Shear stress), τ = F/A0(Shear strain), γ = tan θ
  5. 5. Materiali per l’IngegneriaM.Ferraris5TorsionTTφTorsion τ = T/A0Shear strain, γ = tan φAo
  6. 6. Materiali per l’IngegneriaM.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. 7. Materiali per l’IngegneriaM.Ferraris7For most of materials, low stresses correspond toproportionally low deformations (directproportionality)σ= E ε (Hooke law, E=Young modulus orelastic modulus)σ=F/Aε=∆l/loF/A=E ∆l/lo
  8. 8. Materiali per l’IngegneriaM.Ferraris8Tensile test curve on a metalTensile strengthElastic fieldPlastic fieldσ =Eε
  9. 9. Materiali per l’IngegneriaM.Ferraris9Elastic behaviour• Small stresses corresponds to smalldeformation having a direct proportionalitybetween them.• Small deformations are immediately recoveredwhen stress is removed• E = σ / ε = [N/mm2] = [MPa]Young modulus or Elastic Modulus E(Hooke law)
  10. 10. Materiali per l’IngegneriaM.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. 11. Materiali per l’IngegneriaM.Ferraris11CallisterHow atomicbonds areresponsible ofmaterialspropertiesCondon-Morse curves
  12. 12. Materiali per l’IngegneriaM.Ferraris12Elastic modulus and bond strengthE is directlyproportional to(dF/dr)roStrong bonds:ceramicsWeak bonds: polymers
  13. 13. Materiali per l’IngegneriaM.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. 14. Materiali per l’IngegneriaM.Ferraris14g Only the elastic field
  15. 15. Materiali per l’IngegneriaM.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. 16. Materiali per l’IngegneriaM.Ferraris16How to measure the elasticmodulussteelAlstressstrain
  17. 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. 18. Materiali per l’IngegneriaM.Ferraris18How to measure the elasticmodulus– Drawbacks of tensile tests:Sample preparationDestructive testUnsuitable for brittle materials• Measure of the elastic modulus by Tensile,compression or bending tests• Non destructive tests
  19. 19. Materiali per l’IngegneriaM.Ferraris19Elastic modulus by sonic test• E= 0.9465 * (Γν2L4ρ/t2)∀Γ = costant depending on material shape∀ν = sonic wave frequency;• L= sample length;∀ρ= material density;• t= sample thickness.•
  20. 20. 20Elastic modulus vs TWsteelAlElastic modulus, GPa
  21. 21. Materiali per l’IngegneriaM.Ferraris21Elastic modulus of a steel vs T
  22. 22. Materiali per l’IngegneriaM.Ferraris22∆lz/2l0x∆lx/2l0zσσzyzxεεεευ−=−=Es. 6.2 pag.124Poisson coefficient (ν)Metals: about 0.33Ceramics: 0.17 - 0.27Polymers: 0.33 - 0.5
  23. 23. Materiali per l’IngegneriaM.Ferraris23Non linear elastic behaviour• Some polymers, cast iron, concrete,….• E calculated locallyStrainStressσ1σ2ε1 ε2Secant modulus at σ1 = σ1/ε1Tangent modulus at σ2 = dσ/dε
  24. 24. Materiali per l’IngegneriaM.Ferraris24summaryLinear elasticBehaviour(most materials)Non linear elasticbehaviour(some polymers,cast iron,concrete,…)Anelastic behaviourTime is required torecover deformationafter stress removal(some polymers)
  25. 25. Materiali per l’IngegneriaM.Ferraris25Plastic deformation and chemical bond• When applied stresses increases and correspondingdeformations are higher than about 0.002, the Hookelaw doesn’t 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. 26. Materiali per l’IngegneriaM.Ferraris26σ/ε for metalsfractureTensilestrengthUpperlowerσystrainstressstrictionElastic field Plastic fieldStrain at fractureMetals follow oneof the two curveswww.matcoinc.com/images/sem1a.jpg
  27. 27. Materiali per l’IngegneriaM.Ferraris27
  28. 28. Materiali per l’IngegneriaM.Ferraris28Yield strength at 0.2 %strain (0.002)
  29. 29. Materiali per l’IngegneriaM.Ferraris29Yield strength for differentmaterialsσ 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. 30. Materiali per l’IngegneriaM.Ferraris30Dislocationsexplain the 3orders ofmagnitutebetweentheorethicalandexperimentalyield strengthYield strength and dislocations
  31. 31. Materiali per l’IngegneriaM.Ferraris31Withoutdislocations, theyield strengthwould be aboutE/8 (GPa)
  32. 32. Materiali per l’IngegneriaM.Ferraris32
  33. 33. Materiali per l’IngegneriaM.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. 34. Materiali per l’IngegneriaM.Ferraris34Es: Cu, Al, Ag, AuEs: Fe, W, CrEs: Mg, Ti, ZnHexagonal compactCCB
  35. 35. Materiali per l’IngegneriaM.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 <110>
  36. 36. Materiali per l’IngegneriaM.Ferraris36• Plastic deformation of grains in a stainless steel after laminationfrom CallisterAfter deformation
  37. 37. Materiali per l’IngegneriaM.Ferraris37Stress/strain curve for AlTensile strength (TS)
  38. 38. Materiali per l’IngegneriaM.Ferraris38If we stop loading here a residual plastic deformation (ε res)is still present
  39. 39. Materiali per l’IngegneriaM.Ferraris39Plastic(residual,permanent)deformation
  40. 40. Materiali per l’IngegneriaM.Ferraris40
  41. 41. Materiali per l’IngegneriaM.Ferraris41Ductile fractureStriction Formation of cavitiesCavities coalescence’Propagationhttp://web.umr.edu/~be120/lessons/intro/tension/testing_st/fracture.gifCoarse surface due toplastic deformation
  42. 42. Materiali per l’IngegneriaM.Ferraris42Tensile strengthbehaviour fordifferentmaterials
  43. 43. Materiali per l’IngegneriaM.Ferraris43Stress/strain curves for brittlematerialsceramics, glasses;brittle alloys, brittlepolymers, brittlecompositesNo (or limited) plasticdeformation
  44. 44. Materiali per l’IngegneriaM.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. 45. Materiali per l’IngegneriaM.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. 46. Materiali per l’IngegneriaM.Ferraris46Tensile strength (σ) and % εdeformation at fracture• Material σ (MPa) ε (%)• 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. 47. Materiali per l’IngegneriaM.Ferraris47How to modify plasticproperties of materials• Make dislocation motionmore difficult = increasematerials mechanicalstrength• Cold working• Grain size• precipitates• second phases• Monocrystals• Solid solutionsgrain boundarygrain boundaryσy=σ0+kyd−12Hall-Petch lawd = grain diameter (ave)σy = yield strengthMaterials related costantGrain size modification by thermal treatment
  48. 48. Materiali per l’IngegneriaM.Ferraris48Solid solutionsDislocations motion slowedTensile stressesCompression stresses• Metallic solidsolutions(alloys) havesuperiormechanicalstrength thanpure metals
  49. 49. Materiali per l’IngegneriaM.Ferraris49Role of alloying elements onmechanical properties
  50. 50. Materiali per l’IngegneriaM.Ferraris50
  51. 51. Materiali per l’IngegneriaM.Ferraris51
  52. 52. Materiali per l’IngegneriaM.Ferraris52COLD WORKING OF STEELS•REVERSIBLE PROCESS:•COLDWORKING/ANNEALING

×