Chapter 8. Deformation and Strengthening Mechanisms

Transcription

1 Chapter 8 Deformation and Strengthening Mechanisms

2 Chapter 8 Deformation

3 Deformation and Strengthening 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?

4 Dislocations and Plastic Deformation In crystalline solids, the onset of plastic deformation is caused by the movement of dislocations.

5 Dislocations & Materials Classes Metals: Disl. motion easier -non-directional bonding -close-packed directions for slip. electron cloud ion cores Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding Ionic Ceramics (NaCl): Motion hard -need to avoid ++ and -- neighbors

6 Dislocation Motion_1 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!

7 Dislocation Motion_2

8 Dislocation Motion_3 Dislocation moves along slip plane in slip direction perpendicular to edge dislocation line Slip direction same direction as Burgers vector Edge dislocation Screw dislocation

9 Stress Concentration at Dislocation_1

10 Stress Concentration at Dislocations_2 Dislocation_2

11 Deformation Mechanisms Plastic flow of crystalline materials usually occurs by slip along certain crystal planes and in certain crystal directions. A slip plane and direction is called a slip system. Rule for slip 1. Slip always occurs in closed packed direction - in direction of shortest lattice translation vector. 2. Slip usually occurs on the smoothest and most widely spaced (or densely packed) plane, but this rule is not always obeyed.

12 Slip System

13 Slip Planes and Directions b(fcc) = a/2<110> b(bcc) = a/2<111> b(hcp) = a/2<1120>

14 Dislocations & Crystal Structure Structure: close-packed planes & directions are preferred. close-packed directions close-packed plane (bottom) close-packed plane (top) Comparison among crystal structures: FCC: many close-packed planes/directions; HCP: only one plane, 3 directions; BCC: none view onto two close-packed planes. Results of tensile testing. tensile direction Mg (HCP) Al (FCC)

15 Dislocation Density_1 ρ line length per unit Volume (cm/cm 3 ) L

16 Dislocation Density_2 Typical Dislocation Densities heavily cold worked metals /cm 2 deformed metals /cm 2 n n : # of dislocations L annealed metals /cm 2 Semiconductors /cm 2 whiskers 0 L

17 Plastic Flow and Dislocation Motion v : average velocity of dislocation. shear strain increment during the time increment Average shear strain rate is, but n/(lh) is the dislocations density ρ m : mobile dislocation density Taylor - Orowan Relation for electrical conductivity

18 Slip in Single Crystals Dislocations can move by a shear stress. In σ - ε curve, we learned strength of the material characterized by tensile yield strength Shear or tensile? Compression 1) Dislocations move in response to shear. 2) Applied tensile stress in one direction acts like a shear stress on any plane angled to that direction.

19 Schmid Law s s s s Single crystal yields when F: Applied load

20 ex Ex: : Deformation of Single Crystal_1 =60 =35 a) Will the single crystal yield? b) If not, what stress is needed? crss = 3000 psi cos cos 6500 psi = 6500 psi (6500 psi) (cos35 )(cos60 ) (6500 psi) (0.41) 2662 psi crss 3000 psi So the applied stress of 6500 psi will not cause the crystal to yield.

21 ex Ex: : Deformation of Single Crystal_2 What stress is necessary (i.e., what is the yield stress, s y )? crss 3000 psi y cos cos y (0.41) y crss coscos 3000 psi psi So for deformation to occur the applied stress must be greater than or equal to the yield stress y 7325 psi

22 Critical Resolved Shear Stress Condition for dislocation motion: Crystal orientation can make it easy or hard to move disl. R cos cos σ σ typically 10-4 G to 10-2 G σ R = 0 l =90 R = s/2 l =45 f =45 R = 0 f =90 R maximum at = = 45º

23 Single Crystal Slip

24 Disl. Motion in Polycrystals 300 m

25 Twin Deformation

26 Four Strengthening Mechanisms Grain Size Hardening Solid Solution Hardening Precipitation Hardening Strain Hardening

27 Strengthening Strategy 1 : Grain Size Grain boundaries are barriers to slip. Barrier "strength" increases with misorientation. Smaller grain size: more barriers to slip. slip plane grain A grain B grain boundary Hall-Petch Equation: yield o k y d 1/2

28 Grain Size Hardening 70wt%Cu-30wt%Zn brass alloy yield o k y d 1/2 Data: yield(mpa) grain size, d (mm) x ky [grain size (mm)] mm

29 Grain Size Hardening (Kinetics) more specifically σ σ K y 0 = σ : y 0 +K y average d -0.5 σ : complex parameter y Hall - Petch relation of a single grain

30 Strengthening Strategy 2 : 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 shear at A and B that opposes disl motion to the right. Impurity generates local shear at C and D that opposes disl motion to the right.

31 Strengthening by Alloying_1 small impurities tend to concentrate at dislocations reduce mobility of dislocation increase strength

32 Strengthening by Alloying_2 large impurities concentrate at dislocations on low density side

33 Ex ex : Solid Solution Strengthening in Copper Tensile strength & yield strength increase w/wt% Ni. Tensile strength (MPa) Yield strength (MPa) wt. %Ni, (Concentration C) Adapted from Fig (a) and (b), Callister 6e. wt. %Ni, (Concentration C) Empirical relation: y ~C 1/2 Alloying increases σ y and TS.

34 Strengthening Strategy 3 : Precipitation Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Side View precipitate Large shear stress needed to move dislocation toward precipitate and shear it. Top View Unslipped part of slip plane S Slipped part of slip plane Dislocation advances but precipitates act as pinning sites with spacing S. Result: y~ 1 S

35 Application : Precipitation Strengthening Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates formed by alloying. 1.5mm

36 Strengthening Strategy 4 : Cold Work (%CW) Room temperature deformation. Common forming operations change the cross sectional area: -Forging force -Rolling Ao die blank -Drawing Ao die die force Ad Ad tensile force Adapted from Fig. 11.7, Callister 6e. Ao force Ao -Extrusion container ram billet container roll roll Ad die holder extrusion Ad die %CW A o A d A o x100

37 Strain Hardening (Work Hardening) Increase the density of dislocation increase yield strength W.H

38 Dislocations during Cold Work Ti alloy after cold working: Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. 0.9 m

39 Anisotropy in Deformation 1. Cylinder of Tantalum machined from a rolled plate: 2. Fire cylinder at a target. 3. Deformed cylinder side view rolling direction The noncircular end view shows: anisotropic deformation of rolled material. end view plate thickness direction

40 Anisotropy in σ yield Can be induced by rolling a polycrystalline metal -before rolling -after rolling 235 mm -isotropic since grains are approx. spherical & randomly oriented. rolling direction -anisotropic since rolling affects grain orientation and shape.

41 Result of Cold Work_1 Dislocation density (ρ d ) goes up: Carefully prepared sample: ρ d ~ 10 3 mm/mm 3 Heavily deformed sample: ρ d ~ mm/mm 3 Ways of measuring dislocation density: 40mm Volume, V length, l 1 Area, A dislocation pit length, l 2 length, l 3 d l 1 l 2 l 3 V OR d N A N dislocation pits (revealed by etching)

42 Result of Cold Work_2 Dislocation density Carefully grown single crystal 10 3 mm -2 Deforming sample increases density mm -2 Heat treatment reduces density mm -2 Yield stress increases as ρ d increases: σ y1 σ y0 large hardening small hardening

43 Impact of Cold Work As cold work is increased Yield strength (σ y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases.

44 Dislocation Dislocation Trapping Dislocation generate stress. This traps other dislocations. Red dislocation generates shear at pts A and B that opposes motion of green disl. from left to right. A B

45 Impact of Cold Work Yield strength (σ y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. Stress % cold work Strain

46 Cold Work Analysis What is the tensile strength & ductility after cold working? 2 ro -rd CW = r % 2 o 2 x100 = 35.6% Do=15.2mm Copper Cold work -----> Dd=12.2mm yield strength (MPa) 700 tensile strength (MPa) ductility (%EL) MPa Cu % Cold Work y=300mpa MPa Cu % Cold Work TS=340MPa % 0 0 Cu % Cold Work %EL=7%

47 Concepts for Microstructural Control Soft while fabrication, Strong under service! casting forging annealing - heating to a suitable T followed by cooling at a moderate rate in order to reduce hardness, improve machinability cold working hot working : materials worked above recrystallization T recrystallization : 0.3T m for pure metal 0.5T m for alloys ( ~ up to 0.7 T m )

48 Effect of Heating After after %CW 1 hour treatment at T anneal... decreases TS and increases %EL. Effects of cold work are reversed! tensile strength (MPa) annealing temperature (ºC) tensile strength ductility ductility (%EL) 3 annealing stages to discuss...

49 The Relation between Mechanical Properties and Microstructural Control cold worked and recovered strain E by dislocation VS surface E

50 Recovery Annihilation reduces dislocation density. Scenario 1 extra half-plane of atoms atoms diffuse to regions of tension extra half-plane of atoms Disl. annhilate and form a perfect atomic plane. Scenario 2 annihilation 3. Climbed disl. can now move on new slip plane 2. grey atoms leave by vacancy diffusion allowing disl. to climb 1. dislocation blocked; can t move to the right 4. opposite dislocations meet and annihilate Obstacle dislocation t R

51 σ-ε Behavior vs Temperature Results for polycrystalline iron: Stress (MPa) C -100C 25C Strain s y and TS decrease with increasing test temperature. %EL increases with increasing test temperature. Why? Vacancies help dislocations past obstacles. 2. vacancies replace atoms on the disl. half plane disl. glides past obstacle 1. disl. trapped by obstacle obstacle

52 Recrystalization Temperature, T R T R = recrystallization temperature point of highest rate of property change or temperature at which the crystallization reaches completion in 1hr T m => T R T m (K) Due to diffusion annealing time T R = f(t) shorter annealing time => higher T R Higher %CW => lower T R strain hardening Pure metals => lower T R due to easier movement of grain boundary

53 Recrystalization New crystals are formed that: --have a small disl. density --are small --consume cold-worked crystals. 0.6 mm 0.6 mm 33% cold worked brass New crystals nucleate after 3 sec. at 580C.

54 Further Recrystalization All cold-worked crystals are consumed. 0.6 mm 0.6 mm After 4 seconds After 8 seconds

55 Grain Growth_1 T R = recrystallization temperature T R º

56 Grain Growth_2 At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced. 0.6 mm 0.6 mm After 8 s, 580C Empirical Relation: exponent typ. ~ 2 grain diam. at time t. After 15 min, 580C d n d o n Kt coefficient dependent on material and T. elapsed time

57 Coldwork Calculations A cylindrical rod of brass originally 0.40 in (10.2 mm) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 55,000 psi (380 MPa) and a ductility of at least 15 %EL are desired. Further more, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.

58 Coldwork Calculations Solution If we directly draw to the final diameter what happens? 8% 43 x x x 100 % D D x A A A A A CW o f o f o f o D o = 0.40 in Brass Cold Work D f = 0.30 in Coldwork Calculations Solution

59 Coldwork Calculation Solution : Cont._ For %CW = 43.8% y = 420 MPa TS = 540 MPa > 380 MPa %EL = 6 < 15

60 Coldwork Calculation Solution : Cont._ For TS > 380 Mpa For %EL < 15 > 12 %CW < 27 %CW our working range is limited to %CW = 12-27

61 Coldwork Calculation Solution : Recyrstalization Cold draw-anneal-cold draw again For objective we need a cold work of %CW We ll use %CW = 20 Diameter after first cold draw (before 2nd cold draw)? must be calculated as follows: 2 2 D 2 D 2 % CW % CW f 1 f x D D02 D D f f 2 D Intermediate diameter = % CW 100 D f 1 D % CW D in

62 Coldwork Calculations Solution Summary: 1.Cold work D 01 = 0.40 in D f1 = in %CW x Anneal above D 02 = D f1 3.Cold work D 02 = in D f2 =0.30 in % CW 2 = x 100 = 20 Therefore, meets all requirements 2 σ y =340 MPa TS=400 MPa %EL=24

63 Summary Dislocations are observed primarily in metals and alloys. Here, 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.