Engineering materials

Transcription

1 1 Engineering materials Lecture 2 Imperfections and defects Response of materials to stress

2 2 Crystalline Imperfections (4.4) No crystal is perfect. Imperfections affect mechanical properties, chemical properties and electrical properties. Imperfections can be classified as Zero dimension point deffects. One dimension / line deffects (dislocations). Two dimension deffects (grain boundaries). Three dimension deffects (voids).

3 3 Point Defects Vacancy Vacancy is formed due to a missing atom. Vacancy is formed (one in atoms) during crystallization or mobility of atoms. Energy of formation is 1 ev. Mobility of vacancy results in cluster of vacancies. Also caused due to plastic defor- -mation, rapid cooling or particle bombardment.

4 4 Point Defects - Interstitial Atom in a crystal, sometimes, occupies interstitial site. This does not occur naturally and can be induced by irradiation. This defects caused structural distortion.

5 5 Interstitial Solid Solution Solute atoms fit in between the voids (interstices) of solvent atoms. Solvent atoms in this case should be much larger than solute atoms. Example:- between 912 and C, interstitial solid solution of carbon in γ iron (FCC) is formed. A maximum of 2.8% of carbon can dissolve interstitially in iron. Iron atoms r00.129nm Carbon atoms r=0.075nm

6 6 Substitutional Solid Solution Solute atoms substitute for parent solvent atom in a crystal lattice. The structure remains unchanged. Lattice might get slightly distorted due to change in diameter of the atoms. Solute percentage in solvent can vary from fraction of a percentage to 100% Solvent atoms 4-15 Solute atoms

7 7 Line Defects (Dislocations) Lattice distortions are centered around a line. Formed during Solidification Permanent Deformation Vacancy condensation Different types of line defects are Edge dislocation Screw dislocation Mixed dislocation

8 8 Edge Dislocation Created by insertion of extra half planes of atoms. Positive edge dislocation Negative edge dislocation Burgers vector Burgers vector Shows displacement of atoms (slip). b Figure 4.18

9 9 Screw Dislocation Created due to shear stresses applied to regions of a perfect crystal separated by cutting plane. Distortion of lattice in form of a spiral ramp. Burgers vector is parallel to dislocation line.

10 10 Mixed Dislocation Most crystal have components of both edge and screw dislocation. Dislocation, since have irregular atomic arrangement will appear as dark lines when observed in electron microscope. Dislocation structure of iron deformed 14% at C

11 11 Grain Boundaries Grain boundaries separate grains. Formed due to simultaneously growing crystals meeting each other. Width = 2-5 atomic diameters. Some atoms in grain boundaries have higher energy. Restrict plastic flow and prevent dislocation movement. 3D view of grains Grain Boundaries In 1018 steel

12 12 Planar Defects Grain boundaries, twins, low/high angle boundaries, twists and stacking faults Free surface is also a defect : Bonded to atoms on only one side and hence has higher state of energy Highly reactive Nanomaterials have small clusters of atoms and hence are highly reactive.

13 13 Twin Boundaries Twin: A region in which mirror image pf structure exists across a boundary. Formed during plastic deformation and recrystallization. Strengthens the metal. Twin Plane Twin

14 14 Other Planar Defects Small angle tilt boundary: Array of edge dislocations tilts two regions of a crystal by < 10 0 Stacking faults: Piling up faults during recrystallization due to collapsing. Example: ABCABAACBABC FCC fault Volume defects: Cluster of point defects join to form 3-D void.

15 15 Material Properties Anisotropy ( 非均向性 ): The measured material properties are dependent of the direction and are functions of the symmetry of crystal structure. Isotropy ( 均向性 ): The measured material properties are independent of the direction. Homogeneity ( 均質性 ): The measured material properties are independent of the position. In-homogeneity ( 非均質性 ): The measured material properties are dependent of the position.

16 16 Stress Stress = σ ij lim Δ A 0 ΔF ΔA Normal stress σ 11, σ 22, σ 33 Shear stress σ 12, σ 13, σ 23 σ σ σ σ = σ σ σ ij σ σ σ Symmetric

17 17 Strain Strain: Normal strain ε 11, ε 22, ε 33 ε ij 1 ui = + 2 x j u x i j u 1, u 2, u 3 displacement Shear strain ε 12, ε 13, ε 23 ε ε ε ε = ε ε ε ij ε ε ε Symmetric

18 18 Stress and Strain Metals (materials) undergo deformation under uniaxial tensile force. Elastic deformation: Metal (material) returns to its original dimension after tensile force is removed. Plastic deformation: The metal (material) is deformed to such an extent such that it cannot return to its original dimension 6-10

19 19 Linear elasticity Linearity: linear behavior between input (force) and output (displacement). Elasticity: elastic behavior Consider a body that is loaded by changes in applied forces, displacements or temperatures. If removal of the loading returns the body immediately to its initial state, the material response is called elastic.

20 20 Typical stress-strain curves Ref: Landis, 2001.

21 21 Uniform tension or compression For simple tension, the tensile strain is proportional to the tensile stress σ 11 = σ, σ 22 = σ 33 = σ 12 = σ 13 = σ 23 = 0 σ= E ε E: Young s modulus (elastic modulus)

22 22 Mechanical Properties Modulus of elasticity (E) : Stress and strain are linearly related in elastic region. (Hooks law) E = σ (Stress) ε (Strain) Strain Δσ Δσ E = Δε Higher the bonding strength, higher is the modulus of elasticity. Δε Stress Linear portion of the stress strain curve Examples: Modulus of Elasticity of steel is 207 Gpa. Modulus of elasticity of Aluminum is 76Gpa

23 23 Engineering Stress and Strain Engineering stress σ = (Nominal stress) F (uniaxial tensile force) A 0 (Original cross-sectional area) A 0 Δl Units of Stress are PSI or N/M 2 (Pascals) 1 PSI = 6.89 x 10 3 Pa 0 A 0 Change in length Engineering strain = ε = Original length (Nominal strain) Δ 0 = = 0 0 Units of strain are in/in or m/m.

24 24 Pure shear σ 12 = σ 21 = τ, σ 11 = σ 22 = σ 33 = σ 13 = σ 23 = 0 τ = G γ G: shear modulus Shear angle u x u y γ = + y x Ref: Lai, 1993.

25 25 Shear Stress and Shear Strain Shear stress τ = S (Shear force) A (Area of shear force application) Shear strain γ = Amount of shear displacement Distance h over which shear acts. Elastic Modulus G = τ / γ

26 26 Poisson s Ratio Poisons ratio = ν = ε ( lateral) ε ( longitudinal) = ε y ε z Theoretical value 1 ν w 0 w Usually poisons ratio ranges from 0 to 0.5 Example: Stainless steel 0.28 Copper 0.33

27 27 Hydrostatic pressure σ 11 = σ 22 = σ 33 = p, σ 12 = σ 13 = σ 23 = 0 p Dilatation: Δ= V V V 0 0 p p p ( ε )( ε )( ε ) Δ= Δ= ε11 + ε22 + ε33 = KD Pressure-Dilatation relationship: p Hydrostatic pressure Κ: bulk modulus

28 28 Relations among Material s Properties Among the material parameters: E, G, κ and ν, only two of them are independent. E = 9KG 3K + G K = E 31 2 ( ν ) ν = 3K 2G 23 G E 21 ( K + G) ( + ν ) = -1 < ν 0.5

29 29 Response of Material to Stress For isotropic and linear elastic materials, ε 1 = E σ ν σ + σ ε 1 = E σ ν σ + σ ε 1 = E σ ν σ + σ 1 ε12 = σ12 2G 1 ε13 = σ13 2G 1 ε23 = σ23 2G [ ( )] [ ( )] [ ( )] For example, hydrostatic pressure σ 11 = σ 22 = σ 33 = -p, σ 12 = σ 13 = σ 23 = 0 ε 11 = ε 22 = ε 33 = ε 12 = ε 13 = ε 23 = 0 -(1-2ν)p E Dilatation: Δ = ε11 + ε22 + ε33 3(1 2 ν ) p = E K p = = Δ E 31 2 ( ν )

30 30 Shear distortion Shear deformation only changes shape when deformation is small Infinitesimal shear does not change volume

31 31 Tensile test Strength of materials can be tested by pulling the metal to failure. Load Cell Specimen Extensometer 6-14 Force data is obtained from Load cell Strain data is obtained from Extensometer.

32 32 Engineering Stress and Strain Engineering stress σ = (Nominal stress) F (uniaxial tensile force) A 0 (Original cross-sectional area) A 0 Δl Units of Stress are PSI or N/M 2 (Pascals) 1 PSI = 6.89 x 10 3 Pa 0 A 0 Change in length Engineering strain = ε = Original length (Nominal strain) Δ 0 = = 0 0 Units of strain are in/in or m/m.

33 33 Tension tests Load-displacement Stress-strain

34 34 Tensile Test (Cont) σ UTS Figure 5.21 Commonly used Test specimen Typical Stress-strain curve 6-15

35 35 Yield Strength Yield strength is strength at which metal or alloy show significant amount of plastic deformation. (or 0.2% proof stress) 0.2% offset yield strength is that strength at which 0.2% plastic deformation takes place. Construction line, starting at 0.2% strain and parallel to elastic region is drawn to fiend 0.2% offset yield strength. σ 0.2%

36 36 Ultimate tensile strength 6-19 Ultimate tensile strength (UTS) is the maximum strength reached by the engineering stress strain curve. Necking starts after UTS is reached. More ductile the metal is, more is the necking before failure. S T R E S S M p a Stress increases till failure. Drop in stress strain curve is due to stress calculation based on original area. Al 2024-Tempered Necking Point Al 2024-Annealed Strain

37 37 Percent Elongation Percent elongation is a measure of ductility of a material. It is the elongation of the metal before fracture expressed as percentage of original length. % Elongation = Final length initial Length Initial Length Measured using a caliper fitting the fractured metal together. Example:- Percent elongation of pure aluminum is 35% For 7076-T6 aluminum alloy it is 11% 6-20

38 38 Percent Reduction in Area Percent reduction area is also a measure of ductility. The diameter of fractured end of specimen is measured using caliper. % Reduction Area = Initial area Final area initial area Percent reduction in area in metals decreases in case of presence of porosity Stress-strain curves of different metals

39 39 Stress-strain curve During plastic flow, the total volume of the specimen is conserved: matter is just flowing from place to place. σ UTS E ε f

40 40 Strain Energy The strain energy of a deformed material per unit volume is given by the total area under the stress-strain curve Modulus of Resilience U r = σ y 2 2E U 0 * = ε σ d ε =elastic strain energy + plastic strain energy

41 41 True Stress True Strain True stress and true strain are based upon instantaneous cross-sectional area and length. True Stress = σ t = F A i (instantaneous area) i d li A0 True Strain = ε t = = ln = ln l 0 0 A i True stress is always greater than engineering stress.

42 42 True Stress-True Strain for Plastic Flow During plastic flow, the total volume of the specimen is conserved: matter is just flowing from place to place. Assume volume is constant A o L o =AL Ao A L = = 1+ε L o F F Ao Ao σ t = = = σ = σ(1 + ε) A A A A o L L ε d L A t d ε = t = = = = + L A ε L L ln ln ln(1 ) 0

43 43 Necking in True Stress-True Strain Curve Necking starts at Where the ultimate strength occurs in the engineering stress-strain curve That is at dσ 0 dε = F = F max df =0 F=σ t A df = σ da + Adσ = t t 0 dσ da dl t σ = t A = L = dε t dσ d t t ε = σ t is the condition for necking in true stress-stain curve

44 44 Hardness and Hardness Testing Hardness is a measure of the resistance of a metal to permanent (plastic) deformation. General procedure: Press the indenter that is harder than the metal Into metal surface. Rockwell hardness tester Withdraw the indenter Measure hardness by measuring depth or width of indentation. Figure 6.27

45 45 Hardness Tests Hardness= Force Total surface area of indent

46 46 Hardness True Hardness H = F/A projected = 3 σ y Vicker s Hardness : Hr = F/A inside surface area of the indent

47 47 Plastic Deformation in Single Crystals Plastic deformation of single crystal results in step markings on surface slip bands. Atoms on specific crystallographic planes (slip planes) slip to cause slip bands. Slip bands of zinc Figure 6.28

48 48 Slip Bands and Slip Planes Slip bands in ductile metals are uniform (occurs in many slip planes). Slip occurs in many slip planes within slip bands. Slip planes are about 200A thick and are offset by about 2000A Slip lines Figure 6.30 Slip bands

49 Plastic Deformation by the Slip Mechanism 49

50 50 Slip Mechanism During shear, atoms do not slide over each other simultaneously. The slip occurs due to movement of dislocations. Dislocation density : Sum of length of dislocation line per unit volume Wall of high dislocation density ρ = 1 ds V ρ=10 10 ~10 12 cm/cm 3 for highly deformed crystals Dislocation cell structure in lightly deformed Aluminum

51 51 Slip in Crystals Slip occurs in densely or close packed planes. Lower shear stress is required for slip to occur in densely packed planes. If slip on close packed planes is restricted, then less dense planes become operative. Less energy is required to move atoms along denser planes. Close packed plane 6-32 Non-close-packed plane

52 52 Slip Systems Slip systems are combination of slip planes and slip direction. Each crystal has a number of characteristic slip systems. In FCC crystal, slip takes place in {111} octahedral planes and <110> directions. 4 (111) type planes and 3 [110] type directions. 4 x 3 = 12 slip systems Table 5.3

53 53 Slip Planes and Directions for FCC Crystals

54 54 Slip Systems in BCC Crystal BCC crystals are not close packed. The slip predominantly occurs in {110} planes that has highest atomic density.

55 55 Glide force f Glide force f is the force acting on a dislocation External work = (τ L 1 L 2 )b Internal work = (f L 1 ) L 2 Equivalence between external and internal work: f = τ b Glide force f is defined per unit length of dislocation line

56 56 Critical Resolved Shear Stress Critical resolved shear stress is the stress required to cause slip in pure metal single crystal. Depends upon Crystal Structure Atomic bonding characteristics Temperature Orientation of slip planes relative to shear stress Slip begins when shear stress in slip plane in slip direction reaches critical resolved shear stress. This is equivalent to yield stress. Example :- Zn HCP % pure 0.18MPa Ti HCP 99.99% pure 13.7 MPa Ti HCP 99.9% pure 90.1 MPa

57 57 Schmid s Law 6-37 The relationship between uniaxial stress action on a single cylinder of pure metal single crystal and resulting resolved shear stress produced on a slip system is give by τ r = Shear Force Shear Area Fr Fcos Φ F = = = cos λ cos Φ A A /cosλ A = σ cos λ cos Φ Normal to Slip plane A 1 =Area of Slip plane σ = F A 0 Φ λ τ r = Slip directi F r A 1

58 58 Example Calculate the resolved shear stress on the (111)[011] slip system of an unit cell in a FCC nickel single crystal if a stress of 13.7 MPa is applied in the [001]direction cosλ = (0, 1,1) (0, 0,1) 1 = cosφ = (1,1,1) (0, 0,1) 1 = τ r = 13.7 = 5.6MPa 2 3

59 59 Theoretical Shear Strength (Prediction) d hkl assume 2π x τ = k sin b for small deformation 2π x τ k and b τ = Gγ = x G d hkl 2π k = b τ max = G d hkl Gb 2π dhkl τ G k = max 1 Gb 2π dhkl 1 ~ 10 20

60 60 Theoretical Tensile Strength (Prediction) Interatomic bonding force 2πε σ = k sin λ σ σ=nf=f/r o 2 ε=0 2πε σ = k sin λ ε= r-r o r o ε assume 2πε σ = k sin λ dσ dε ε = 0 E o 2πk = = Eo k = for λ=2 λ π σ max E o σ max 1 1 π E Interatomic force typically is negligible at r=2r 0 ~ 10 20

61 61 Uniaxial Tension Test of Polycrystals F Shear stress τ F Crosssection A θ Fsinθ F τ = = sinθ cosθ = σ sinθ cosθ A/cosθ A Maximum shear stress occurs at θ = 45 o τ = max σ 2 So that yielding occurs at σ y = 2τ y