Mechanical Properties

Transcription

1 Mechanical Properties Elastic deformation Plastic deformation Fracture

2 II. Stable Plastic Deformation S s y For a typical ductile metal: I. Elastic deformation II. Stable plastic deformation III. Unstable deformation IV. Fracture S 0.2% e e p e e Stress-strain relation is non-linear Strain is non-recoverable Relaxation is elastic permanent plastic strain Strain is uniform and stable - work hardening e e p Plasticity initiates at yield strength, s y In ductile material, s y is not obvious s y is usually defined by 0.2% offset strain Yield strength = stress that produces a plastic strain of 0.2%

3 Forms of the Engineering Stress-Strain Curve s u S sy e u e T S I II III e Ductile metal e Yield point s u s u S s e Brittle solid e Elastomer

4 True Stress and Strain True strain is defined from its differential dε = dl L = da A V = constant (plastic deformation): V = A 0 L 0 = AL dv = 0 = AdL + LdA ε = L dl L ε = ln L L = ln A 0 A 0 L 0 ε = ln ΔL +1 ε = ln( 1+ e) L 0 (L/L 0 ) applies when strain is uniform (A 0 /A) can be used for non-uniform strain dl L = da A e = engineering strain e = ΔL L 0 = L L 0 1 True stress σ = P A = P A 0 = s L A 0 A L 0 σ = s(1+ e)

5 Stress-Strain Relations Stress-strain relations σ = s(1+ e) ε = ln(1+ e) s = σ exp( ε) e = exp(ε) 1 stresses equal for small strain S s u sy e u e T Why use engineering curve? Clearly shows tensile strength s u is an important design value No difference in E, s y =σ y I II III e

6 Plastic Deformation: Engineering Significance Design: yield strength Almost all structures operate well below yield Design: ultimate strength Plastic instability limits the capability of ductile materials Manufacture: formability Plasticity is used to form materials into complex shapes Service: failure resistance and failure analysis Plasticity provides margin against fracture Deformation patterns record load history

7 Plastic Deformation Mechanisms Dislocation plasticity (our focus in this course) Dislocation motion causes shear The dominant mechanism of plasticity Diffusion Atoms diffuse to regions of high stress Most important in high temperature creep ( diffusional creep ) Structural phase transformation Structural transformations can cause shape change (martensite) TRIP steel ( transformation-induced plasticity ) Grain boundary sliding Groups of grains tumble by sliding on boundaries Superplasticity : rapid creep to deformations of 1000% or more

8 Example: Indentation of an Al Grain

9 Indentation Deformation Sequence b b = a 2 <110 >

10 The Force on a Dislocation Dislocation bounds an area that has slipped Material is sheared by b when it moves normal to its line The force on the dislocation is F = τb τ is the shear stress in the direction of b Local F is normal to the line, whatever the dislocation shape Assume a set of dislocations, b, move an average <δx> If ρ is the dislocation density (line length/unit volume) The strain is γ = ρb<δx>

11 Plastic Deformation in Shear Dislocations cause shear Shear stress required to drive plasticity Hydrostatic stress does not cause plastic deformation Yield stress in tension (σ y ) is determined by critical shear stress (τ c )

12 The Critical Resolved Shear Stress How does tension (σ) cause shear (τ) Assume tensile stress, σ, along axis Assume glide plane normal at angle, θ Assume Burgers vector at angle, ϕ The resolved shear stress is τ = P b A θ = P cos(ϕ) A cos(θ) = σ cos(θ)cos(ϕ) τ σ /2 Dislocations move when τ = τ c τ c = critical resolved shear stress Yield strength, σ y, is such that τ τ c on the most favorable plane

13 Yield under Tension Yield strength, σ y, is such that τ τ c on the most favorable plane σ y = σ y 2τ c τ c cos(θ)cos(ϕ) Note τ c is a material property, not σ y elongation slip σ σ Tensile deformation by slip The slip planes are angled to the bar Slip causes elongation as shown Many slip planes uniform elongation of the bar

14 Deformation of a Polycrystal P P Plastic deformation of a polycrystal Many grains in all orientations Slip in the grain best aligned Causes Incremental plastic deformation High stress is adjacent grains Deformed regions grow Gradual increase in ε p Eventual large-scale plasticity σ y σ σ y 3τ c Results Gradual yielding Measured by 0.2% offset Yield exceeds minimum } 0.2% ε σ y = k T τ c (k T = Taylor factor ~ 3)

15 Microstructural Control of the Strength Add things that inhibit dislocation glide Microstructural mechanisms: Crystal structure Make lattice resist dislocation motion (Peierls-Nabarro stress) Refine grain size σ y d -1/2 Introduce obstacles into grains Solute atoms Other dislocations (work hardening) Precipitates

16 Inherent Strength: The Peierls-Nabarro Stress Lattice resistance to glide τ p = 2G 1 ν exp 2πd b(1 ν) - d = distance between slip planes Inherently hard materials: - High G - Small ν - Large (d/b) (complex structure) - Diamond - Si - SiO 2

17 Grain Refinement d Grain boundaries resist glide Slip planes not continuous Defect absorption at boundaries Hall-Petch relation Ex: iron σ = σ 0 + Kd -1/2 Control of grain size Recrystallization in metals Recrystallize and quench Fine-grained powder in ceramics

18 Ductility Lost at Nanograin Size Decreasing grain size σ y increases ε u vanishes

19 Obstacle Hardening Random distribution of obstacles: τ c = 2T L s b β 3 / 2 3 / 2 c = Gbβ c n Force on dislocation: f = τb Dislocation bows between obstacles: R = T/τb ~ Gb/2τ Obstacle experiences force: f = 2Tcos(ψ/2) = 2Tβ When β = β c, dislocation cuts through T = line tension n = obstacle density β = cos ψ 2

20 Obstacle Hardening Solute atoms: β c ~ 0.01 n = c (concentration) Random distribution of obstacles: τ c = 2T L s b β 3 / 2 3 / 2 c = Gbβ c n Dislocations: β c ~ 0.1 n = ρ (dislocation density) Precipitates β c ~ 0.7 n = (1/L s ) (obstacle spacing)

21 Solute Hardening compression tension σ y interstitial substitutional Solute have misfit strain Interacts with dislocation field Large solute repels compression Small solute repels tension Strength proportional to misfit Interstitial solutes have greatest effect C, N in iron Strength proportional to c 1/2 when c is small c 2/3 for larger c (solutes overlap) At high T, solutes are mobile Strength lost c

22 Dislocation Hardening compression tension compression tension Dislocations are Crystallographic obstacles Elastic obstacles Forest dislocations harden σ y = αgb ρ α ~ 3β c 3 / 2 ~ 0.3 Work hardening Dislocations multiply during strain dσ dε = αgb dρ 2 ρ dε

23 Work Hardening S Dislocations are Crystallographic obstacles Elastic obstacles Forest dislocations harden e p e e e σ y = αgb ρ α ~ 3β c 3 / 2 ~ Work hardening Dislocations multiply during strain dσ dε = αgb dρ 2 ρ dε

24 Precipitation Hardening Create precipitates τ c = Gbβ c 3 / 2 n β c ~ Age hardening Hardens because β c increases Softens when β c = max (obstacle impenetrable) n decreases under coarsening