Transcript A study of Thin Film Shape Memory Alloys

A study of Thin Film Shape
Memory Alloys
By Rajlakshmi Purkayastha
Metallurgical and Materials Science
IIT Bombay
Shape Memory Alloys
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Shape memory alloys (SMAs) are metals that
"remember" their original shapes.
SMAs are useful for such things as actuators which are
materials that "change shape, stiffness, position, natural
frequency, and other mechanical characteristics in
response to temperature or electromagnetic fields"
The potential uses for SMAs especially as actuators
have broadened the spectrum of many scientific fields.
The study of the history and development of SMAs can
provide an insight into a material involved in cutting-edge
technology.
The diverse applications for these metals have made
them increasingly important and visible to the world.
Shape Memory Effect
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The Shape Memory effect refers to
the fact that after a sample of SMA
has been deformed from its
"original" conformation, it regains
its original geometry by itself during
heating (one-way effect) or, at
higher ambient temperatures,
simply during unloading (pseudoelasticity or superelasticity).
SME is the ability to return to a
previously defined shape through
appropriate thermal procedure after
being severely deformed.
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These extraordinary properties  Those crystal structures
are due to a temperatureare known as martensite
dependent martensitic phase
and austenite.
transformation from a lowsymmetry to a highly symmetric
crystallographic structure.
Thin Films
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Shape memory alloys (SMA’s) possess an array of
desirable properties:
high power to weight (or force to volume) ratio, thus the
ability to recover large transformation stress and strain
upon heating and cooling,
peudoelasticity (or superelasticity),
high damping capacity,
good chemical resistance and biocompatibility, etc.
This attracted much attention to the research of SMAs
as smart (or intelligent) and functional materials.
More recently, thin film SMA has been recognized as a
promising and high performance material in the field of
micro-electro-mechanical system (MEMS) applications,
since it can be patterned with standard lithography
techniques and fabricated in batch process.
Why thin films?
• Thin films have a wide variety of applications. Thus
an analysis of the stress – strain behaviour of
these films becomes very important to analyse,
and predict their beahaviour under different
conditions.
• The main factors that affect the behaviour of
these films are:
1. Temperature
2. Materials Used
3. Stress Applied
Temperature Dependance
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The transformation from Austenite to Martensite and viceversa is temperature Dependant
The temperatures at which the SMA changes its
crystallographic structure are characteristic of the alloy,
and can be tuned by varying the elemental ratios.
Typically, Ms denotes the temperature at which the
structure starts to change from
austenite to martensite upon
cooling. Mf is the temperature
at which the transition is finished.
Accordingly, As and Af are the
temperatures at which the
reverse transformation from
martensite to austenite start
and finish, respectively.
Materials Used
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Each Material is different in terms of:
Transformation Temperature
Youngs Modulus and other Material constants
The most common and important alloy used is
Nitinol – composed of Nickel and Titanium
Other examples include Indium thallium ,
Nickel Aluminium Magnesuim
Stress Applied
Stress can be applied in two Directions – biaxial
stress or in only one direction – uniaxial stress
 Rather than measuring the stress applied, it is
easier to measure the strain caused, and
calculate stress using Hookes law
Stress = [elastic coeff]*strain
 Phase transformations can also be brought
about by the application of stress rather than
change in temperature
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Mechanics of thin films deposited on
 The major three reasons for biaxial
substrate
stresses in thin films are thermal
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strains, growth strains, and epitaxial
strains.
In general thermal strains are the result
of thermal expansion coefficient
mismatch between the substrate and
the thin film equation (1) shows this
effect
(1) ε=−∆α∗∆T
The film is free to move in the z
direction only, as it is constrained by
the substrate in the y and x direction
The transformation and subsequent
stress equations are modelled on this
fact.
Thus a solution to a very real problem,
thermally induced stress, is sought.
The problem
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When austenite transforms
into martensite it transforms
into different variants
The number of variants
depends on the crystal
structure. For example a
tetragonal system has three
variants whilst a monoclinic
(check) has 12 variants.
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variants have different
orientations.
What we have to do is analyse
and detemine the volume
fractions of each of the variants
formed in a phase
transformation , due to applied
stress in a thin film.
Methodology
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The problem is solved using Energy
Minimisation
We know that the film will transform in such a
way so as to minimise the total stress energy.
We therefore calculate the total energy due to
the formation of these variants and then use a
standard optimisation routine with set boundary
conditions to calculate the volume fraction of
each variant.
The details….
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In this problem we are analysing a biaxial
stress applied to a thin film of In-23%Th, i.e. an
Indium-Thallium Shape Memory Alloy
It is a Cubic-Tetragonal transformation.
There are three variants of Martensite formed
Assumptions
We assume that there is no change in
temperature during the transformation
We assume that all of the austenite is
transformed into Martensite i.e. there is no
retained austenite.
All material constants are kept fixed throughout
the transformation.
Basic Equations
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t=
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(I – U1)*v1 + (I – U2)*v2 + (I-U3)*v3
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 strain applied
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t = strain due to phase transformation
Ui = Transformation Matrix in the ‘i’th Direction
I = Identity Matrix
Now, we assume that the body is allowed to
move freely in the ‘z’ direction as it
perpendicular to both of the applied stresses
and is normal to the plane of the substrate.
With this in mind, the applied strain in the zdirection is taken to be 
* , and is calculated
using the stress relationship.
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Therefore ,
* 0 0
app = [ 
0 
0 0
0 0 
0 ]
Thus, = 
app – 
t
E=


where,  = stress, = strain, E = energy

=
*C , where C = stress transformation Coefficient Matrix
Thus, E = 
*C*
Boundary conditions are:
v1 + v2 + v3 = 1
v1,v2,v3 
0
v1,v2,v3 
1
Results
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On running the
optimisation routine
for equal strains
applied in both x and
y direction, we get
the following graph
The graph shows
that the energy
clearly reaches a
minimum at v2 = v3
= 0.5, when v1 = 0.0
• The adjacent graph
shows the variation of
the minimum. The
minimum keeps on
shifting to the right as
the value of v1
increases, but is always
reached at a point
where v2 = v3.
•Also, note that as v1
increases, the total
energy increases,
making the system
more unstable. Thus
the system is most
stable at minimum v1
i.e. v1 = 0.0
...when unequal strains are
applied…
•The variants are formed in
different ratios in order to
accommodate the varying
strain.
•The equations used are
the same, but will change
for energy calculation due
to symmetry
considerations
The two graphs below show the change in the variants
due to the unequal stress. Again, V1 is varied from
0.1 to 1.0
The variation of Energy
The variation of Energy
with V2 shows that all the
with V3 shows that energy
variant forms tends to be
minimisation occurs at
V2
V3 -> 0.
Conclusion
In an ideal situation, the maximum variant
fraction formed depends on the direction in
which the stress is applied.
 When strains are equally applied, then
fractions of V2 and V3 are the same
 When unequal strains are applied, the
maximum variant formed is in the direction
of the larger strain.
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Acknowledgement
I would sincerely like to thank my guide for
this project, Prof Prita Pant, for giving me
this opportunity to work with her. It has
been a very good learning experience,
indeed.
I would also like to mention the IRCC
Summer Fellowship Programme, under
whose aegis this work was carried out
Thank You
Basic Equations…
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t = (I – U1)*v1 + (I – U2)*v2 + (I-U3)*v3

strain applied
t = strain due to phase transformation
Ui = Transformation Matrix in the ‘i’th Direction
I = Identity Matrix
Now, we assume that the body is allowed to
move freely in the ‘z’ direction as it perpendicular
to both of the applied stresses and is normal to
the plane of the substrate.
With this in mind, the applied strain in the zdirection is taken to be * , and is calculated
using the stress relationship.
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Therefore ,app = [ * 0 0
0 0 0
0 0 0 ]
Thus,  = app – t
E =  where,  = stress,  = strain, E = energy
= *C , where C = stress transformation Coefficient
Matrix
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Thus, E = *C*
Boundary conditions are:
v1 + v 2 + v 3 = 1
v1,v2,v3 0
v1,v2,v3 1