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- 1. MT 610Advanced Physical Metallurgy Session : Phase Transformations in Solids IV Materials Technology School of Energy and Materials
- 2. Contents Diffusional transformations Long-range diffusion Short-range diffusion Diffusionless transformations Martensitictransformation Geometric observation Mechanism 2
- 3. Shear transformation Exp. Martensite can be generated by shear on γ Both shears are possible and identical to Bain distortion if disregarded the rigid body rotation. 3
- 4. Shear transformation Shear of cooperative movements of atoms can be in different planes rather than (111)γ plane, depending on alloy composition and transformation temp. Shear does not have to act along the same direction on every parallel atomic plane. 4
- 5. Shear transformation Greninger and Troiano (1949) found that Observed shear plane in Fe-22% Ni-0.8% C was not the {111}γ plane and the shear angle was 10.45°, not 19.5° as predicted by shear mechanism. Theysuggested that another shear had to be added in order to complete the mechanism. 5
- 6. Double shear transformation The first shear isa macroscopic shear that contributes the shape change and change in crystal structure. The second shear is a microscopic shear. Invariant plane Bain distortion has no invariant plane Lattice-invariant shear with Bain distortion 6
- 7. Invariant plane During the martensitic transformation The interface should be an invariant plane Undistorted and unrotated plane Any deformation on the invariant plane will be termed an invariant plane strain. 7
- 8. From the Bain distortion α lattice with bcc can be generated from an fcc γ lattice by Compression about 20% along one principle axis and a simultaneous uniform expansion about 12% along the other two axes perpendicular to the first principle axis 8
- 9. Bain distortion of a sphere Due to the Bain distortion A unit sphere of the parent crystal transforms into an oblate spheroid of the product crystal Contraction about 20% along the one principle axis Expansion about 12% along the other two axes perpendicular to the first principle axis 9
- 10. Bain distortion of a sphere Initial sphere equation of the parent crystal x12 + x2 + x3 = 1 2 2 12% expansion 20% contraction Ellipsoid equation of the transformed crystal (x ) 2 1 + (x ) 2 2 + (x ) 2 3 =1 ( 1.12 ) ( 1.12 ) ( 0.80 ) 2 2 2 10
- 11. Bain distortion of a sphere Due to the lattice deformation x12 + x2 + x3 = 1 2 2 Vectors OA’ and OB’ represent the final position of vectors Vectors OA and OB represent the initial position of the same vectors unchanged in ( x1 ) + ( x2 ) + ( x3 ) = 1 2 2 2 ( 1.12 ) ( 1.12 ) ( 11 ) 2 2 2 magnitude 0.80
- 12. Bain distortion of a sphere Vectors unchanged in magnitude during the lattice deformation Corresponding to the cones AOB and COD and the cones A’OB’ and C’OD’ These vectors are termed unextended lines. A homogeneous strain would result in an undistorted plane of contact between the initial sphere of austenite and the ellipsoid of martensite. 12
- 13. Bain distortion of a sphere Allother vectors not involved in the cones A’OB’ and C’OD’ would be changed in magnitude. Bain distortion would result in no undistorted plane. Hence, there is no invariant plane. Very difficult to obtain a coherent planar interface between the parent and the product crystals only by the Bain distortion. 13
- 14. Bain distortion of a sphere Therefore,Bain distortion has no invariant plane. 14
- 15. Lattice-invariant shear Lattice-invariant shear must be of such magnitude so as to produce an undistorted plane when combined with the Bain distortion. Consider slip or twinning Must not make any change in crystal structure. 15
- 16. Lattice-invariant shear Graphical analysis of a simple shear of slip or twinning of a unit sphere Shear on an equatorial plane K1 as the shear plane d as the shear direction α as shear angle Slip 16
- 17. Lattice-invariant shear As a result of shear on K1 Any vector in the plane AK B is 2 transformed into a vector in the plane AK’2B, which is unchanged with length although rotated relatively to its original position. The plane AK B is the initial Slip 2 position of a plane AK’2B, which remains undistorted as a result of the shear. 17
- 18. Lattice-invariant shear As a result ofshear on K1 The relative positions of the planes AK2B and AK’2B depend on the amount of shear involved. The shear plane itself remains undistorted after shear. Slip Vectors that remain invariant in length (unextended lines) to this shear operation are define as potential habit planes. 18
- 19. Lattice-invariant shear As a result ofshear on K1 The relative positions of the planes AK2B and AK’2B depend on the amount of shear involved. The shear plane itself remains undistorted after shear. Slip Vectors that remain invariant in length (unextended lines) to this shear operation are define as potential habit planes. 19
- 20. Lattice-invariant shear When initial sphere → ellipsoid by lattice deformation using Bain distortion is distorted by simple shear into another ellipsoid + and the lattice is left invariant, The simple shear is termed a lattice-invariant shear. shear 20
- 21. Stereographic projection 21
- 22. Stereographic representationof the Bain distortion Any vector lying on the initial cone AOB with a semiapex of φ moves radially onto the final cone A’OB’ with a semiapex of φ’. Vectors in the cones of unextended lines do not change their length, but only the angle ∆φ. 22
- 23. Stereographic representationof the lattice-invariant shear An unextended line C moves to the final position along the circumference of the great circle defined by d* (dash line). 23
- 24. Stereographic representationof the lattice-invariant shear Vectors in K’2 plane do not change their length due to shear, and the line OC’ in the plane represents the final position of an unextended line. Line OC in K2 plane represents the initial position of OC’. 24
- 25. Requirement for habit plane Both Bain distortion and lattice invariant shear provide an undistorted plane for the habit plane. Additional requirement is that the habit plane be unrotated. A rigid body rotation must be able to return the undistorted plane to its original position before transformation. 25
- 26. 3 important components Bain distortion Lattice invariant shear Rigid body rotation 26
- 27. Bain distortion with slip #1 Vectors b and c are defined by the intersections of the initial Bain cone with K1 plane 1.Apply a complementary shear Vectors b and c become b’ and c’ and still lie in the K1 plane and remain unchanged in both direction and magnitude. They are invariant lines. 27
- 28. Bain distortion with slip #1 Vectors b and c are defined by the intersections of the initial Bain cone with K1 plane 2.Apply a Bain distortion Vectors b’ and c’ become b’’ and c’’ lie on the initial and final Bain cones, respectively, without changing their magnitude. 28
- 29. Bain distortion with slip #1 Complementary shear b and c to b’ and c’ Bain distortion b’ and c’ to b’’ and c’’ Angle btw b and c ≠ angle btw b” and c” Appropriate rotation cannot be applied to return b” and c” to initial positions of b and c. Plane defined by b and c cannot be an invariant plane. 29
- 30. Bain distortion with slip #2 To obtainan invariant plane, must have other extended lines Ifassumed to know the shear angle α, vectors a and d obtained from the intersections of the K2 plane change to a’ and d’ along the great circles. Bain distortion, vectors a’ and d’ become a” and d”, respectively 30
- 31. Bain distortion with slip #2 Through the transformation of the complementary shear and the Bain distortion Sequences of a→a’→a” and sequences d→d’→d” reveal no change in length However, angle btw a & d ≠ angle btw a” & d” Plane defined by a and d cannot be an invariant plane. 31
- 32. Complete transformationprocess Possible invariant planes will depend on the choice of combination of b or c with a or d such as Vectors a and b Vectors a and c Vectors b and d Vectors c and d 32
- 33. Complete transformation process If theinvariant plane is the plane defined by vectors a & c Angle btw a & c = angle btw a’’ & c’’ Let the axis required for rotation be at point u Determine the amount of rotation stereographically by intersection of a great circle bisecting a-a” with another great circle bisecting c-c” 33
- 34. Complete transformationprocess Once a” and c” coincide simultaneously with a and c, respectively Angle btw a & c = angle btw a’’ & c’’ Therefore, orientation relationship btw γ plane (defined by the vectors a and c) and α’ plane (defined by the vectors a” and c”) can be determined for a specific variant of the Bain distortion (B), lattice invariant shear (P), and rotation operation (R). T = BPR 34
- 35. Complete transformationprocess T = BPR Bain distortion (B) Lattice invariant shear (P) Rotation operation (R) 35
- 36. Bain distortion with twinning Twinned martensite can take place by having alternate regions in the parent phase undergo the lattice deformation along different contraction axes, which are initially at right angles to each other. In the first region, contraction occurs along the x3 [ 001] f axis. In the adjacent region, contraction direction can be either x1 [100] f or x2 [ 010] f axis. Two rigid body rotations are also involved in the twinning analysis. 36
- 37. Nucleation and growth It only takes about 10-5 to 10-7 seconds for a plate of martensite to grow to its full size. The nucleation during the martensitic transformation is extremely difficult to study experimentally. Average number of martensite is as large as 104 nuclei/mm3 Number of martensite nuclei can be increased by increasing ∆T prior to Ms. It is too small in term of number of nucleation sites for homogeneous nucleation. 37
- 38. Nucleation and growth Less likely to occur by homogeneous nucleation process, but heterogeneous. Surfaces and grain boundaries are not significantly contributing to nucleation. Most likely types of defect that could produce the observed density of martensite nuclei are dislocations (> 105 dislocation/mm2). C. Zener (1948): movement of partial dislocations during twinning could generate a thin bcc region of lattice from an fcc region. 38
- 39. Nucleation and growth Dissociation of a dislocation into 2 partials is favorable → lower strain energy. r r r To generate b1 = b2 + b3 bcc structure, a a a [ 110] = [ 211] + 121 the requirements are that all 2 6 6 green atoms move (shear) a forward by 12 [ 211] and an additional dilatation to correct lattice spacings. 39
- 40. Nucleation and growth Growth of lath martensite with dimension a > b >> c growing on a {111}γ planes Thickening mechanism would involve the nucleation and glide of transformation dislocations moving on discrete ledges behind the growing front. Due to large misfit between bct and fcc lattice, dislocations could be self-nucleated at the lath interface as the lath moves forward. 40
- 41. Nucleation and growth In medium and high carbon steels, Morphology of martensite turns to change from a lath to a plate-like shape. As carbon concentration decreases, Decrease lath structure Decrease martensitic temperature Increase twinning Increase retained austenite Depending on compositions, the habit plane changes from {111}γ → {225}γ → {259}γ 41
- 42. Effect of pressure to martensite As pressure increases In Fe unary system, the equilibrium temperature decreases In Fe-C binary system, the phase region around γ phase shifts to the left and downward. Similar to adding austenite stabilizer 42
- 43. Effect of alloying element tomartensite Each alloying element will effect the martensitic transformation differently. If initially Hγ = Hα When adding C The ē of C will decrease Hα and cause α to be less stable. ∆H = Hγ – Hα > 0, stabilize the γ When adding X Increase Hα and ∆H < 0, stabilize the α 43
- 44. Effect of external stress tomartensite As martensite prefers to nucleate and grow along the dislocation Expected that an externally applied shear stress will assist and accelerate the generation of dislocations and hence the growth of martensite. An external shear stress can aid martensite nucleation if the external elastic strain components play as a part of the Bain strain. This can also help by raising the M s temperature. 44
- 45. Effect of external stress tomartensite Once the plastic deformation occurs There is an upper limit value of M that the s stress can be applied. The limit temp. of M is called M (highest s d temperature that stress helps to form martensite) Too much plastic deformation will suppress the transformation. 45
- 46. Effect of external stress tomartensite If a tensile stress is applied M temperature can be suppressed to lower s temperature Transformation may be reversed from α’ → γ Presence of large magnetic field may favor the formation of the ferromagnetic phase and therefore raise Ms temp. 46
- 47. Effect of external stress tomartensite Plastic deformation of γ before transformation will assist on increasing number of nucleation sites. Once the transformation occurs Result in very fine plate size of martensite (Called the ausforming process) Combined effect of very fine martensite plates, 1 2 solution hardening of carbon, and 3dislocation hardening Very high strength ausformed steel 47
- 48. Shape-memory alloys (SMA) Unique property of some alloys After being deformed at one temperature, they recover the original undeformed shape when heated to a higher temperature. 48
- 49. Shape-memory alloys (SMA) Unique property of some alloys After being deformed at one temperature, they recover the original undeformed shape when heated to a higher temperature. Fundamental to the shape-memory effect (SME) is the occurrence of a martensitic phase transformation and its subsequent reversal. Alloys: Ni-Ti (called NiTiNOL), Ni-Al, Fe-Pt, Cu-Al-Ni, Cu-Au-Zn, Cu-Zn-(Al,Ga,Sn,Si), Ni-Mn-Ga 49
- 50. SMA Common characteristics Atomicordering transformation from ordered parent phase to ordered martensite phase Thermoelastic martensitic transformation that is crystallographic reversible Martensite phase that forms in a self- accommodating manner (slip or twinning) 50
- 51. SMA Typical plot of property changes versus temp. A hysteresis is usually on the order of 20°C 51
- 52. One-way SMA Sample is cooled from above Af to below Mf → martensite forms Sample has no shape change Sample is deformed below Mf Sample remains deformed until heated. Begin shape recovery at A and complete at A s f No shape change when cooled below Mf Deforming the 52 martensite again will reactivate SME
- 53. Two-way SMA Sample is cooled from above Af to below Mf → martensite forms Sample has no shape change Sample is deformed below Mf Sample remains deformed until heated. Begin shape recovery at A and complete at A s f Returnsto the deformed shape when cooled below Mf 53

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