4. Two Stable phases at different temperature
Fig 1. Different phases of an SMA
5. SMA’s Phase Transition
Fig 2. Martensite Fraction v.s. Temperature
Ms : Austensite -> Martensite Start Temperature
Mf : Austensite -> Martensite Finish Temperature
As : Martensite -> Austensite Start Temperature
Af : Martensite -> Austensite Finish Temperature
A
A
M
M
Hysteresis size =
½ (As – Af + Ms - Mf)
6. How SMA works ? One path-loading
Fig 3. Shape Memory Effect of an SMA.
M
D-M
A
7. Example about # of Variants of Martensite [KB03]
Fig 4. Example of many “cubic-tetragonal” martensite variants.
8. How SMA works ? One path-loading
M
D-M
A
T-M
Fig 5. Fig 6. Loading path.
9. Austenite directly to detwinned martensite
Fig 7. Temperature-induced phase transformation with applied load.
D-M
A
10. Austenite directly to detwinned martensite
M
D-M
A
Fig 8. Fig 9. Thermomechanical
loading
12. Summary: Shape memory alloy (SMA) phases and crystal structures
Fig 12. How SMA works.
13. ① Maximum recoverable strain
② Thermal/Stress Hysteresis size
③ Shift of transition temperatures
④ Other fatigue and plasticity problems and other factors, e.g. expenses…
What SMA’s pratical properties we care about ?
Fig 13. SMA hysteresis & shift temp.
14. SMA
facing
challenges!
• High
expenses;
• Fa5gue
Problem;
• Large
temperature/stress
hysteresis
• Narrow
temperature
range
of
opera5on
• Reliability
15. • Since
the
crystal
laCce
of
the
martensi5c
phase
has
lower
symmetry
than
that
of
the
parent
austeni5c
phase,
several
variants
of
martensite
can
be
formed
from
the
same
parent
phase
crystal.
• Parent
and
product
phases
coexist
during
the
phase
transforma5on,
since
it
is
a
first
order
transi5on,
and
as
a
result
there
exists
an
invariant
plane
(relates
to
middle
eigenvalue
is
1),
which
separates
the
parent
and
product
phases.
Summary: Shape memory alloy (SMA) phases and crystal structures
17. QUj -Ui = a⊗n
• Nature Materials, (April 2006;
Vol 5; Page 286-290)
• Combinatorial search of
thermoelastic shape-memory
alloys with extremely small
hysteresis width
• Ni-Ti-Cu & Ni-Ti-Pb
New findings: extremely small hysteresis width when λ2 è 1
Fig 14.
18. QUj -Ui = a⊗n
• Adv. Funct. Mater. (2010), 20,
1917–1923
• Identification of Quaternary
Shape Memory Alloys with
Near-Zero Thermal Hysteresis
and Unprecedented Functional
Stability
New findings: extremely small hysteresis width when λ2 è 1
Fig 15.
19. Conditions of compatibility for twinned martensite
Definition. (Compatibility condition) Two positive-definite symmetric
stretch tensors Ui and Uj are compatible if:
, where Q is a rotation, n is the normal direction of interface, a and Q
are to be decided.
Result 1 [KB Result 5.1]
Given F and G as positive definite tensors, rotation Q, vector a ≠ 0, |n|=1,
s.t.
iff: (1) C = G-TFTFG-1≠Identity
(2) eigenvalues of C satisfy: λ1 ≤
λ2 =1 ≤
λ3
And there are exactly two solutions given as follow: (k=±1, ρ is chosen to let |n|=1)
∃
QF -G = a⊗n
QUj -Ui = a⊗n
a= ρ
λ3
1− λ1( )
λ3
− λ1
ˆe1
+k
λ1
λ3
−1( )
λ3
− λ1
ˆe3
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
; n=
λ3
− λ1
ρ λ3
− λ1
− 1− λ1
GT
ˆe1
+k λ3
−1GT
ˆe3( )
20. Conditions of compatibility for twinned martensite
Definition. (Compatibility condition) Two positive-definite symmetric
stretch tensors Ui and Uj are compatible if:
, where Q is a rotation, n is the normal direction of interface, a and Q
are to be decided.
Result 2 (Mallard’s Law)[KB Result 5.2]
Given F and G as positive definite tensors, (i) F=Q’FR for some rotation Q’ and some
180° rotation R with axis ê; (ii)FTF≠GTG, then one rotation Q, vector a ≠ 0, |n|=1,
s.t.
And there are exactly two solutions given as follow: (ρ is chosen to let |n|=1)
QF -G = a⊗n
QUj -Ui = a⊗n
(Type Ι) a= 2
G−T
ˆe
|G−T
ˆe|
− Gˆe
⎛
⎝
⎜
⎞
⎠
⎟ ; n= ˆe
∃
(Type ΙΙ) a= ρGˆe; n=
2
ρ
ˆe−
GT
Gˆe
|Gˆe|2
⎛
⎝
⎜
⎞
⎠
⎟
Need
to
sa5sfy
some
condi5ons;
Usually
there
are
TWO
solu5ons
for
each
pair
of
{F,G}
;
22. Austenite-Martensite Interface
QUi -I = a⊗n
QUj -Ui = a⊗n;
Q'(λQUj +(1-λ)Ui ) =I+b⊗m
(★)
(★★)
R'(Ui +λa⊗n) =I+b⊗m
Need to check middle eigenvalue of is 1.
Which is equivalent to check:
Order of g(λ) ≤ 6, actually it’s at most quadratic in λ
and it’s symmetric with 1/2. so it has form:
And g(λ) has a root in (0,1) ç g(0)g(1/2) ≤
0. and use this get one condition;
Another condition is that from 1 is the middle eigenvalue ç (λ1-1)(1-λ3) ≥
0
Gλ = (Ui +λa⊗n)(Ui +λn⊗ a)
g λ( )= det Ui
+ λn⊗a( ) Ui
+ λa⊗n( )−I⎡
⎣
⎤
⎦ = 0
g λ( )= β λ −1/ 2( )
2
+ η
23. Austenite-Martensite Interface
Result 3 [KB Result 7.1]
Given Ui and vector a, n that satisfy the twinning equation (★), we can obtain a
solution to the aust.-martensite interface equation (★★), using following procedure:
(Step 1) Calculate:
The austenite-martensite interface eq has a solution iff: δ ≤ -2 and η ≥ 0;
(Step 2) Calculate λ (VOlUME fraction for martensites)
(Step 3) Calculate C and find C’s three eigenvalues and corresponding eigenvectors.
And ρ is chosen to make |m|=1 and k = ±1.
b = ρ
λ3
(1− λ1
)
λ3
− λ1
ˆe1
+k
λ1
(λ3
−1)
λ3
− λ1
ˆe3
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
m=
λ3
− λ1
ρ λ3
− λ1
− 1− λ1
ˆe1
+k λ3
−1ˆe3( )
Need
to
sa5sfy
some
condi5ons;
Usually
there
are
Four
solu5ons
for
each
pair
of
{Ui,
Uj}
;
QUi -I = a⊗n
QUj -Ui = a⊗n;
Q'(λQUj +(1-λ)Ui ) =I+b⊗m
(★)
(★★)
δ = a⋅Ui
Ui
2
−I( )
−1
n; η= tr Ui
2
( )−det Ui
2
( )− 2 +
| a|2
2δ
λ* =
1
2
1− 1+
2
δ
⎛
⎝
⎜
⎞
⎠
⎟ λ = λ* or (1-λ*)
C = Ui
+ λn⊗a( ) Ui
+ λa⊗n( )
24. Austenite-Martensite Interface
QUi -I = a⊗n
QUj -Ui = a⊗n;
Q'(λQUj +(1-λ)Ui ) =I+b⊗m
(★)
(★★)
R'(Ui +λa⊗n) =I+b⊗m
What if
Order of g(λ) < 2, β=0; g(λ) has a root in (0,1),
Now, λ is free only if belongs to (0,1).
Another condition is that from 1 is the middle eigenvalue ç (λ1-1)(1-λ3) ≥
0
g λ( )= det Ui
+ λn⊗a( ) Ui
+ λa⊗n( )−I⎡
⎣
⎤
⎦ = β λ −1/ 2( )
2
+ η
g λ( )= η= constant ≡ 0
25. Cofactor conditions
• Under certain denegeracy conditions on the input data U, a, n, there can be
additional solutions of (★★), and these conditions called cofactor conditions:
• Simplified equivalent form: (Study of the cofactor condition. JMPS 2566-2587(2013))
QUi -I = a⊗n
QUj -Ui = a⊗n;
Q'(λQUj +(1-λ)Ui ) =I+b⊗m
(★)
(★★)
λ2
=1
a⋅Ucof U2
−I( )n= 0
trU2
−detU2
−
| a|2
|n|2
4
− 2 ≥ 0
λ2
=1
XI
:=|U-1
ˆe|=1 for Type I twin
XII
:=|Uˆe|=1 for Type II twin
-‐1/2
β
ß
27. Conditions to minimize hysteresis
• Conditions:
• Geometrical explanations of these conditions:
1) det U = 1 means no volume change
2) middle eigenvalue is 1 means there is an invariant plane btw Aust. and Mart.
3) cofactor conditions imply infinite # of compatible interfaces btw Aust. and Mart.
Objective in this group meeting talk:
--- Minimization of hysteresis of transformation
det U( )=1
λ2
=1
a⋅Ucof U2
−I( )n= 0
trU2
−detU2
−
| a|2
|n|2
4
− 2 ≥ 0
det U( )=1
λ2
=1
XI
:=|U-1
ˆe|=1 for Type I twin
XII
:=|Uˆe|=1 for Type II twin
or
28. A simple transition layer
C−I = f ⊗m
Cv = Av
Cw = Bw
We can check there is solution for C:
C = I+ f ⊗m; f =b+
ε
α
λ 1-λ( )a
Using linear elasticity theory, we can see
the C region’s energy:
Area of C region:
Energy:
εα
2m⋅n⊥
E = Area
µ
2
1
2
CA−1
− I( )+ CA−1
− I( )
T
( )
2
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝⎜
⎞
⎠⎟
=
εαw
2m⋅n⊥
µλ2
4
a⋅cλ( )2
+ | a |2
| cλ |2
( )⎛
⎝⎜
⎞
⎠⎟
where cλ = A−T
n +
ε
α
1− λ( )A−T
m
min
α
E ⇒ εwhµλ2
1− λ( )ξ
Fig 17.
29. A simple transition layer
Where ξ is geometric factor related with
m, n, A, a;
And it’s can be changed largely as for
various twin systems for Ti50Ni50-xPdx,
x~11:
From 2000 ~ 160000
E =
2κwhl
ε
+ εwhµλ2
1− λ( )ξ
+ϕ A,θ( )whl +ϕ I,θ( )wh L − l( )
min
α
E ⇒ εwhµλ2
1− λ( )ξ
Introduce facial energy per unit area κ:
min
ε
E = 2whλ 2κµl 1− λ( )ξ
+ whl ϕ A,θ( )−ϕ I,θ( )( )+ const
max
ε, l
E =
2λ2
whκµ 1− λ( )ξ
ϕ I,θ( )−ϕ A,θ( )( )
with lc =
2λ2
κµ 1− λ( )ξ
ϕ A,θ( )−ϕ I,θ( )( )
Fig 17.
30. A simple transition layer
ϕ A,θ( )−ϕ I,θ( )= L
θc −θ
θc
L = θc
∂ϕ I θc( ),θc( )
∂θ
−
∂ϕ A θc( ),θc( )
∂θ
⎛
⎝
⎜
⎞
⎠
⎟
Do Tayor expansion for φ near θc:
Let’s identify hysteresis size
H = 2 θc −θ( )
=
2λθc
L
2κµ 1− λ( )ξ
lc
min
ε
E = 2whλ 2κµl 1− λ( )ξ
+ whl ϕ A,θ( )−ϕ I,θ( )( )+ const
min
ε, l
E =
2λ2
whκµ 1− λ( )ξ
ϕ I,θ( )−ϕ A,θ( )( )
with lc =
2λ2
κµ 1− λ( )ξ
ϕ A,θ( )−ϕ I,θ( )( )
Fig 17.
31. General Case
H = 2 θc −θ( )
=
2λθc
L
2κµ 1− λ( )ξ
lc
Some Gamma-Convergence Problem
Fig 18.
33. • Nature, (Oct 3, 2013; Vol 502; Page 85-88)
• Enhanced reversibility and unusual microstructure of a
phase-transforming material
• Zn45AuxCu(55-x) (20 ≤
x
≤30) (Cofactor conditions satisfied)
Theory driven to find –or- create new materials
36. Why Riverine microstructure is possible?
a. Planar phase boundary (transition layer);
b. Planar phase boundary without Trans-L;
c. A triple junction formed by Aust. & type I
Mart. twin pair;
d. (c)‘s 2D projection;
e. A quad junction formed by four variants;
f. (e)’s 2D projection;
g. Curved phase boundary and riverine
microstructure.
Fig 21.
38. References
1. [KB] Bhattacharya K. Microstructure of martensite: why it forms and how it gives rise
to the shape-memory effect[M]. Oxford University Press, 2003.
2. Song Y, Chen X, Dabade V, et al. Enhanced reversibility and unusual microstructure of
a phase-transforming material[J]. Nature, 2013, 502(7469): 85-88.
3. Chen X, Srivastava V, Dabade V, et al. Study of the cofactor conditions: Conditions of
supercompatibility between phases[J]. Journal of the Mechanics and Physics of
Solids, 2013, 61(12): 2566-2587.
4. Zhang Z, James R D, Müller S. Energy barriers and hysteresis in martensitic phase
transformations[J]. Acta Materialia, 2009, 57(15): 4332-4352.
5. James R D, Zhang Z. A way to search for multiferroic materials with “unlikely”
combinations of physical properties[M]//Magnetism and structure in functional
materials. Springer Berlin Heidelberg, 2005: 159-175.
6. Cui J, Chu Y S, Famodu O O, et al. Combinatorial search of thermoelastic shape-
memory alloys with extremely small hysteresis width[J]. Nature materials, 2006, 5(4):
286-290.
7. Zarnetta R, Takahashi R, Young M L, et al. Identification of Quaternary Shape Memory
Alloys with Near‐Zero Thermal Hysteresis and Unprecedented Functional Stability[J].
Advanced Functional Materials, 2010, 20(12): 1917-1923.
Thanks Gal for help me understand one Shu’s paper!