Lecture 4 Dislocations & strengthening mechanisms

Transcription

1 Lecture 4 Dislocations & strengthening mechanisms ISSUES TO ADDRESS... Why are dislocations observed primarily in metals and alloys? How are strength and dislocation motion related? How do we increase strength? How can heating change strength and other properties? Edge dislocation and screw dislocation The atom positions around an edge dislocation A screw dislocation within a crystal Plictic deformation corresponds to the motion of large numbers of dislocations 1

2 Dislocations & Materials Classes Metals: Dislocation motion easier. - non-directional bonding - close-packed directions for slip. Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding electron cloud ion cores Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors Dislocation Motion Dislocations & plastic deformation Cubic & hexagonal metals - plastic deformation by plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations). If dislocations don't move, deformation doesn't occur! Adapted from Fig. 7.1, Callister 7e. 2

3 Dislocation Motion Dislocation moves along slip plane in slip direction perpendicular to dislocation line Slip direction same direction as Burgers vector Edge dislocation Screw dislocation Adapted from Fig. 7.2, Callister 7e. Slip system _ Slip plane plane allowing easiest slippage wide interplanar spacings highest planar densities Slip direction direction of movement highest linear densities A {111} <110>Slip system within an FCC unit cell Slip occurs on {111} planes (close-packed) in <110> directions (close-packed). Three <110> slip directions as shown in (b), within the (111) plane. => total of 12 slip systems in FCC Slip system in BCC unit cell: {110} <111> Adapted from Fig. 7.6, Callister 7e. 3

4 Slip system Crystals slip due to a resolved shear stress, t R. Applied tensile stress: s = F/A A slip direction F F Resolved shear stress: t R =Fs/As slip plane normal, n s t R slip direction F S t R A S F s = Fcosl As = A/cosf t R = scosfcosl t R (max) = s(cosfcosl) max Critical resolved shear stress t CRSS : The minimium shear stress required to initiate slip. The applied stress required to initial yielding s y : s y = t CRSS / (cosfcosl) max 4

5 Single Crystal Slip Adapted from Fig. 7.9, Callister 7e. Slip in a zinc single crystal. Adapted from Fig. 7.8, Callister 7e. Slip Motion in Polycrystals s Stronger - grain boundaries pin deformations Slip planes & directions (l, f) change from one crystal to another. t R will vary from one crystal to another. The crystal with the largest t R yields first. Other (less favorably oriented) crystals yield later. 300 mm Slip lines on the surface of a polycrystalline specimen of copper that was polished and subsequently deformed. Adapted from Fig. 7.10, Callister 7e. 5

6 Mechanisms of strengthening in metals Strain hardening (work hardening) Grain boundary strengthening Solid-solution strengthening Precipitation strengthening Strain hardening (work hardening) Strain hardening Within the region between Y.S (stress at 2) and T.S (stress at 3) The region after T.S Neck down of a tensile test specimen within the gage lengh Dislocations in plastically deformed stainless steel necking 6

7 Strain hardening (work hardening) Room temperature deformation (cold work). Common forming operations change the cross sectional area: -Forging Ao die blank force Ad Adapted from Fig. 11.8, Callister 7e. -Rolling Ao roll roll Ad -Drawing Ao die die force Ad tensile force -Extrusion Ao container force ram billet container die holder extrusion die Ad Percent Cold Work: % CW = Ao - A A o d x 100 Dislocations During Cold Work Ti alloy after cold working: Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. 0.9 mm Adapted from Fig. 4.6, Callister 7e. (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.) 7

8 As cold work is increased: Yield strength (sy) increases Tensile strength (TS) increases Ductility (%EL or %AR) decreases Exercise: Compute the tensile strength and ductility (%EL) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm. Cold Work Analysis example problem 7.2 p194 What is the tensile strength & ductility after cold working? p 2 2 r - p % = o r CW d p 2 ro yield strength (MPa) 700 x 100 = 35.6 tensile strength (MPa) 800 Do =15.2mm Copper Cold Work ductility (%EL) 60 Dd =12.2mm MPa 300 Cu % Cold Work s y = 300MPa MPa Cu % Cold Work TS = 340MPa % 0 0 Cu % Cold Work %EL = 7% Adapted from Fig. 7.19, Callister 7e. 8

9 Grain boundary hardening Grain boundaries are barriers to slip. Barrier "strength" increases with Increasing angle of misorientation. Smaller grain size: more barriers to slip. Hall-Petch Equation: Grain boundary hardening Hall Petch (H-P) relationship s y = s + kd s 0 and s y : yield strength k: the Hall Petch slope d: grain size k: constant In analogy, hardness (Hv) can be related to the grain size by H v = H + k d 0 H 1-2 H 0 and k H are constants. 9

10 Homework 1. The following yield strengths were obtained in ferritic steel as a function of grain size. Estimate the two constants in the Petch equation for this material and predict the expected yield strength of the steel in which the grain size is reduced to i micron. grain size (micron) Yield strength (MPa) Solid-Solution strengthening Solid Solutions Impurity atoms distort the lattice & generate stress. Stress can produce a barrier to dislocation motion. Smaller substitutional impurity Larger substitutional impurity A C B D Impurity generates local stress at A and B that opposes dislocation motion to the right. Impurity generates local stress at C and D that opposes dislocation motion to the right. 10

11 Stress Concentration at Dislocations Adapted from Fig. 7.4, Callister 7e. Strengthening by Alloying small impurities tend to concentrate at dislocations reduce mobility of dislocation \ increase strength Tensile lattice strain imposed on host atoms by a small substitutional impurity atom Possible locations of small impurity atoms relative to an edge dislocation such that there is partial cancellation of impuritydislocation lattice strains large impurities concentrate at dislocations on low density side Adapted from Fig. 7.17, Callister 7e. 11

12 Solid Solution Strengthening in Copper Tensile strength & yield strength for copper-nickel alloys increase with wt% Ni. Tensile strength (MPa) wt.% Ni, (Concentration C) Yield strength (MPa) wt.%ni, (Concentration C) Adapted from Fig (a) and (b), Callister 7e. Empirical relation: s ~ C Alloying increases s y and TS. y 1/ 2 Summary Dislocations are observed primarily in metals and alloys. Strength is increased by making dislocation motion difficult. Particular ways to increase strength are to: --decrease grain size --solid solution strengthening --precipitate strengthening --cold work Heating (annealing) can reduce dislocation density and increase grain size. This decreases the strength. 12

13 Lecture 5 Creep p 238 Tensile test at elevated temperatures under a constant load the strain will increase with time Creep curve Creep is defined as time dependent plastic deformation at constant stress and temperature. Creep deformation becomes important only for temperatures greater than about 0.4Tm (Tm = absolite melting temperature) The slope of this curve is the creep rate Questions and problems 8.26 p249 Give the approximate temp. At which creep deformation becomes an important consideration for each of the following metals: Sn, Mo, Fe, gold, Zn and Cr. 13

14 The primary stage: a decreasing strain rate dislocation climb The secondary stage: straight line constant-strain rate Final stage: Strain rate increase due to necking or internal cracking Creep Occurs at elevated temperature, T > 0.4 T m tertiary primary secondary elastic Adapted from Figs. 8.29, Callister 7e. Homework 8.26 p249 14

15 Influence of stress s and temperature T on creep behavior with either increasing stress or temperature: 1. The instantaneous strain at the time of stress application increases; 2. The steady-state creep rate is increased; 3. The rupture lifetime is diminished Diffusion mechanisms in creep Activition energy Q Rate = C exp (-Q/RT) R: gas constant Diagram showing stressdirected flow of vacancies (solid lines) from tensile to compressive grain boundaries and corresponding reverse flow of atoms or ions (dashed lines 15

16 Dislocation climb At elevated temperature Thermally activated atom mobility Creep rate of the secondary stage the steady-state e& = A creep rate: s D de dt n - d p - exp( Q / RT ) 0 s: applied stress d: grain size D 0 : diffusion coefficient Q: activation energy for diffusion T: absolute temperature Herring Nabarro creep, n=1, p=2, (D 0 and Q refer to bulk diffusion) Coble creep (GB diffusion), n=1, p=3, (D 0 and Q refer to GB diffusion) Dislocation climb n=4-7, p=0, (D 0 and Q refer to bulk diffusion) Grain boundary sliding 16

17 Arrhenius plot (ln e& e& = A s ) versus (1/T): n - d p - D exp( Q / RT ) 0 When stress s is a constant: e& = k exp (-Q/RT) ln e& = K + (-Q/R) (1/T) The slop of the curve gives the activition energy for the creep mechanism Creep mechanism the stress exponent n e& = A (log e& s n - d p - D ) versus log s exp( Q / RT ) 0 When temperature is a constant, (log e& e& = ks n )= K + n log s The slop of the curve gives the (n ) value for the creep mechanism 17

18 Creep deformation map Pure silver, Grain size 32 mm, Elastic boundaries determined at a strain rate of 10-8 /sec s: stress, m: shear modulus T: temperature (K), T m melting temperature (K) 18

19 Turbine blade Subjected centrifugal forces Creep map for MAR-M200 nickel base alloy (100mm) The creep rate below sec -1 is usually negligible. 19

20 Larson-Miller plot Give a plot in the form of stress - life time Applied stress versus time to failure at a given temperature Applied stress versus time to a certain strain at a given T Log s - T (C + log t) C is a constant, for most materials C = 20 T temperature (K) t: life time for steady state, t = e / e& e.g. t 0.01 means the time to get a strain e = 0.01 t 0.01 = 0.01 / e& Creep Failure p242 design example 8.2 Failure: along grain boundaries. g.b. cavities Time to rupture, t r T( 20 +logtr ) = L applied stress Estimate rupture time S-590 Iron, T = 800 C, s = 20 ksi data for S-590 Iron L( K-log hr) Stress, ksi T( 20 + logtr ) = L Adapted from Fig. 8.32, Callister 7e. (Fig is from F.R. Larson and J. Miller, Trans. ASME, 74, 765 (1952).) 24x10 3 K-log hr temperature function of applied stress time to failure (rupture) 1073K Ans: t r = 233 hr 20

21 Sample problems (the life time is t 0.01, i.e. the time to a strain of 10-2 ) Larson prediction, problem 4 21

22 homework A common creep requirement is a 1000h creep life to 2% strain at a (shear) stress of 100 MPa (1MPa = 10 6 N/m 2 ) (1) Calculate the creep rate ( e& ) for this creep requirement (2) Referring to the creep maps below (Ni-base superalloy), at what temperatures in the two materials is this requirement met? (3) What is the dominant creep mechanism in the two cases? Grain size d = 100 mm d = 1cm 22