Homework Problems and Solutions for Chapter 1...Homework Problems and Solutions for Chapter 1 For each of the following materials, provide a simple description of their actuation and

Supplementary information for Fundamentals of Smart Materials © The Royal Society of Chemistry 2020

Homework Problems and Solutions for Chapter 1 For each of the following materials, provide a simple description of their actuation and sensing

characteristics. Are they just actuators, sensors or both?

Homework Problem 1.1: Piezoelectric materials (piezoelectric) Solution to Homework 1.1: Piezoelectric materials are both actuators, energy-harvesters and sensors. The piezoelectric effect describes a reversible electrodynamic relationship in some solid

crystalline structures with embedded dipoles. These crystalline solids possess microscopic

regions containing dipole charges such that under applied mechanical stress they change their

internal arrangements of embedded electrical dipoles generating a voltage across the material

boundaries. Conversely, an applied voltage to the solid crystalline changes the orientation of the

embedded internal dipole charges and generates a deformation or strain in the solid.

Homework Problem 1.2: Piezoresistive materials (piezoresistive) Solution to Homework 1.2: With piezoresistivity as a property of certain materials such as semiconductors for which the materials electrical resistance changes purely due to mechanical

pressure, force, acceleration, strain, and stress. It is the physical property of certain materials,

which has been widely used to convert a mechanical signal into an electrical signal in smart

sensors, accelerometers, tactile sensors, strain gauges, and flow meters and similar devices and

microdevices. The piezoresistive effect is present in semiconductors such as germanium,

amorphous silicon, polycrystalline silicon, silicon carbide, among others.

Homework Problem 1.3: Electrostrictive materials (ER)

Solution to Homework 1.3: Electrostriction is the nonlinear electromechanical coupling in all electrical-nonconductors (dielectric materials). Under the application of an electric field, these

materials show deformation, strain, and stress. All electrostrictive materials exhibit second-order

nonlinear coupling between the elastic strains or stresses and the dielectric terms such as the

strain tensor. For a single uniaxial strain (deformation), the induced strain (deformation) is

directly proportional to the square of the applied electric field (voltage).


Homework Problem 1.4: Fibrous polyacrylonitrile (PAN) artificial muscles Solution to Homework 1.4: Fibrous contractile ionic PAN materials are then introduced as mimicking mammalian muscles. PAN fibers in an active form (PAN or PAN gel modified by

annealing/cross-linking and partial hydrolysis) elongate and contract when immersed in pH

solutions (caustic and acidic solutions, respectively). Activated PAN fibers can also contract and

expand in polyelectrolyte when electrically and ionically activated with cations and anions,

respectively. The change in length for these pH-activated fibers is typically greater than 100%.

However, more than 900% contraction/expansion of PAN nanofibers (less than 1 micron in

diameter) have been observed in our laboratories. PAN muscles present great potential as

artificial muscles for linear actuation. PAN fibers can convert chemical energy directly into

mechanical motion.

Homework Problem 1.5: Giant magnetostrictive materials (MS) Solution to Homework 1.5: Magnetostrictive materials are then introduced in which deformation is observed in ferromagnetic materials when subjected to a magnetic field. The

effect was first identified in 1842 by James Joule when observing a sample of nickel (the Joule

Effect). At a fundamental level, the changes in dimensions result from the interactive coupling

between an applied magnetic field and the magnetization and magnetic moments of the

material’s magnetic dipoles, the material initially under some stress. Reviews of giant

magnetoresistive (GMR) materials are then presented.

Homework Problem 1.6: Giant magneto resistive materials (GMR) Solution to Homework 1.6: Magnetoresistance is defined as the property of a material to change its electrical conductivity or its inverse electrical resistance when an external magnetic

field is applied to it. In 1851, William Thomson (Lord Kelvin), discovered that when pieces of

iron or nickel are placed within an external magnetic field the electrical resistance increases

when the current is in the same direction as the magnetic force, which is aligned with the

magnetic N–S vector and decreases when the current is perpendicular to the direction of the

magnetic force. Lord Kelvin was unable to reduce the electrical resistance of any metal by more

than about 5%.


Homework Problem 1.7: Magnetic gels (MG)

Solution to Homework 1.7: Brief reviews of magnetic gels (ferrogels) are presented by Zrinyi and co-workers. A prelude to the development of ferrogels was a classic paper by Rosenzweig in

1985 on ferrohydrodynamics. A colloidal ferrofluid, or a magnetic fluid, is a colloidal dispersion

of monodomain magnetic particles. Typically, monodomain magnetic particles have typical sizes

of about 10–15 nm, and they are superparamagnetic, in which magnetization can randomly flip

direction under the influence of temperature.

Homework Problem 1.8: Electrorheological fluids (ERFs) Solution to Homework 1.8: Reviews of electrorheological fluids (ERFs) are then given. ERFs belong to a class of smart materials capable of changing from a liquid phase to a much more

viscous liquid and then to an almost solid phase in the presence of a dynamic electric field. They

are essentially colloidal suspensions of highly polarizable particles in a nonpolarizable solvent.

The solid phase of an ERF typically has mechanical properties similar to a solid such as a gel and

can undergo a phase change from liquid to a thick liquid like honey and then solid or in reverse

from a solid transform to a thick liquid and then a thin liquid in a matter of a few milliseconds.

The effect is called the “Winslow effect” after its discoverer Willis M. Winslow, who obtained a

US patent on the effect in 1947 and published an article on it in 1949. The effect is better

described as electric field dependent shear yield stress.

Homework Problem 1.9: Magnetorheological fluids (MRFs)

Solution to Homework 1.9: Magnetorheological fluids (MRFs) are then introduced. MRFs are suspensions of micron-sized magnetic particles such as carbonyl iron powder in a host liquid,

which is usually a type of oil with some additives to minimize particle sedimentation and particle

wear and tear. When the MRF suspension is placed in a magnetic field, the suspended colloidal

particles reconfigure to form chains in the direction of the magnetic flux and make the solution

more solid-like than liquid.

Homework Problem 1.10: Dielectric elastomers (DEs)

Solution to Homework 1.10: In DEs,rubbery elastomers like a silicone rubber sheet are sandwiched between two compliant electrodes, then any applied electric field induces


electrostatic forces (attraction) between the electrodes. Thus, the rubber sheet in between them

can be compressed by the electrostatic forces, which then cause the rubbery sheet to expand

sideways due to the Poisson’s ratio effect. Thus actuation results. In 1880, Röntgen demonstrated

this actuation by using two glasses as dielectrics, and once the opposing surfaces of these glasses

were charged, small thickness changes were observed. Later, electrostatically-induced pressures

acting to compress dielectrics became known as the “Maxwell stress.” It was, however, Perline,

Kornbluh, and Joseph in 1998 who introduced dielectric elastomer technology with compliant

electrodes.

Homework Problem 1.11: Shape memory alloys (SMAs) Solution to Homework 1.11: SMAs are capable of solid-phase transformation from a body-centered tetragonal form called thermoelastic martensite to a face-centered cubic superelastic

called austenite. These materials are named shape memory materials (SMMs) and their thermal

versions called SMAs. These martensitic crystalline structures are capable of returning to their

original shape in the austenite phase, after a large plastic deformation in the martensitic phase

and return to their original shape when heated towards austenitic transformation. These novel

effects are called thermal shape memory and superelasticity (elastic shape memory),

respectively.

Homework Problem 1.12: Magnetic shape memory alloys (MSMAs) Solution to Homework 1.12: Magnetic shape memory alloys (MSMAs), often also referred to as ferromagnetic shape memory alloys (FSMAs), have emerged as an interesting extension of the

class of shape memory materials (SMMs). FSMAs combine the attributes and properties of

ferromagnetism with a reversible martensitic crystalline solid phase transformation. MSM

phenomena were originally suggested by Ullakko, O’Handley, and Kantner and were

demonstrated for a Ni–Mn–Ga alloy in as early as 1996.

Homework Problem 1.13: Shape memory polymers (SMPs)

Solution to Homework 1.13: Shape memory polymers belong to the family of shape–memory materials (SMMs), which can be deformed into a predetermined shape under some imposed

specific fields such as temperature, electric or magnetic, as well as strain and stress. These


shapes can be relaxed back to their original field-free shapes under thermal, electrical, magnetic,

strain, stress, temperature, laser, or environmental stimuli. These transformations are essentially

due to the elastic energy stored in SMMs during the initial deformation. As a member of SMMs,

SMPs are stimuli-sensitive polymers.

Homework Problem 1.14: Smart materials for controlled drug release Solution to Homework 1.14: Smart materials for controlled drug release allow fine-tuning of drug bioavailability, by regulating the amount and the rate at which the drug reaches the

bloodstream, which is critical for the success of the therapy. Some drugs pose important

problems in terms of efficacy and safety (e.g., antitumor drugs, antimicrobials) and suffer

instability problems in the biological environment (e.g., gene materials), and thus the therapeutic

performance of these drugs is improved when they are selectively directed (targeted) from the

bloodstream to the site of action (tissues, cells or cellular structures). Both macro-dosage forms

and nano-delivery systems may notably benefit from stimuli-responsive materials. Drug release

triggering where, when, and how requires detailed knowledge of the changes that the illness

causes in physiological parameters. These changes can be characterized in terms of biomarkers

(e.g., glucose, specific enzymes, or quorum sensing signals in the case of infection) and

physicochemical parameters (pH, ions, temperature, glutathione) that may be exploited as

internal stimuli.

Homework Problem 1.15: Mechanochromic and metamaterials (MC, MM) Solution to Homework 1.15: Mechanochromic materials change their optical properties and, in particular, photoluminescence characteristics if subjected to mechanical loading. Metamaterials

define materials that are not ordinarily produced in nature. Smaller units rather than the

properties of the host material play a fundamental role in material behavior.

Metamaterials are nanocomposite materials made of a periodically repeated micro or nano units

of metals, alloys, and plastics that exhibit properties different from the natural properties of the

participating materials.


Homework Problem 1.16: Ionic polymer–metal nano composites (IPMCs) Solution to Homework 1.16: Ionic polymeric networks contain conjugated ions that can be redistributed by an applied electric field and consequently act as distributed nano actuators,

nanosensors, and energy harvesters. Gel-based and chitosan-based conductor composites have

also been considered as electrically active composite smart materials similar to IPMCs. The basic

mechanism is redistribution of ions within the material due to imposition of a voltage, causing

the material to deform accordingly.

Homework Problem 1.17: Smart ionic liquids (ILs) Solution to Homework 1.17: Ionic solids such as sodium chloride (table salt) have been known for centuries. The very first examples are the ground-breaking endeavours of Sir Humphry Davy

in the synthesis of alkali metals by electrolysis. However, this needs a high temperature as ionic

bonds are strong. Electrolysis of sodium chloride should be conducted at a temperature higher

than 801 °C. Since high temperatures are not technologically favorable, the melting point of such

ionic solids is reduced by weakening the ionic bonds in eutectic mixtures. One of the very first

examples is the pioneering work of Charles Martin Hall in the synthesis of aluminum, which is

still the dominant approach for the exploitation of metallic aluminum. The high melting point of

these ionic liquids is due to the close arrangement of highly charged ions within the lattice.

Homework Problem 1.18: Conductive polymers (CPs) Solution to Homework 1.18: Conductive polymers have the ability to conduct electrical charges, in addition to being flexible, optically active and not difficult to synthesize, and present

a tremendous opportunity for industrial and medical applications. Pioneering work on conductive

polymers reported the observation that the conductivity of polyacetylene increases by millions of

times when it was oxidized by “doping” with iodine vapor. Conductive polymers can conduct

electrical charge because within their molecular network charges can jump between the

molecular chains of the polymer. Conductive polymer molecular structures possess both single

and double chemical bonds, which enhance charge transfer.


Homework Problem 1.19: Liquid crystals and liquid crystal elastomers (LCEs) Solution to Homework 1.19: LCEs can be used as robotic actuators through inducing a nematic–isotropic phase transition in them via a temperature increase, which causes them to

shrink. LCEs have been made electroactive by creating a composite material that consists of

nematic LCEs and a conductive phase such as graphite or conducting polymers that are

distributed within their network structure. The actuation mechanism of these materials involves

phase transition between a nematic (cholesteric, smectic) and isotropic phases over less than a

second. The reverse process is slower, taking about 10 sec, and requires cooling the LCE back to

its initial temperature as the LCE expands back to its original size.

Homework Problem 1.20: Chemomechanical polymers (CMPs) Solution to Homework 1.20: Chemoresponsive gels can be used for many applications, and are increasingly developed also given their possible biocompatibility. Such smart materials can,

depending on suitable chemical components, bind or release for example drugs, pollutants,

catalysts upon interaction with external effectors, and swell or shrink under the influence of

different pH, various chemical compounds, temperature, or light. Most hydrogels are

amorphous, some are semicrystalline mixtures of amorphous and crystalline phases, or are

crystalline. Hydrogels have a water content typically between 80 and 99%, which can change

according to external stimuli; this is the basis of many applications. Natural sources of hydrogels

are for example agarose, chitosan, methylcellulose or hyaluronic acid, but most smart hydrogels

are based on synthetic polymers or rely on chemical modification of natural systems. Synthetic

polymers for gels are usually obtained by copolymerization or cross-linking free-radical

polymerizations, reacting hydrophilic monomers with multifunctional cross-linkers.

Homework Problem 1.21: Nanogels (NG) Solution to Homework 1.21: Smart nanogels are one of the most important innovations that have emerged in the field of nanomedicine and biomedical applications. As recent advances in

the applications of biomaterials, nanogels have emerged as novel candidates for drug delivery,

biosensing, imaging, tissue engineering, and the targeted delivery of bioactives. The present

chapter gives a basic understanding of the hydrogels and introduces the nanoparticle form of

hydrogels known as “nanogels.” Nanogels have the synergistic properties of interpenetrating


networks as well as nanoscale properties such as a small size and high surface-to-volume ratio.

These hybrid materials show high drug loading, are capable of crossing strong barriers, as well

being highly biocompatible. In brief, this chapter describes the basic synthetic methodology and

characterization techniques of nanogels. It also discusses the natural and synthetic polymers

deployed for the synthesis of nanogels.

Homework Problem 1.22: Self-healing materials (SHMs) Solution to Homework 1.22: The self-healing characteristics of these materials, and in particular biomaterials, and the concepts of the self-healing processes in nature and biology are

already well known by scientific communities. One can start by describing their impact and

occurrence in nature, in plants, in animals and human beings. These understandings of self-

healing processes in biology and nature are particularly more advanced in terms of dermatology

and skin repair by scar tissues. The advantage of self-healing materials is that they can treat

materials degradation by initiating a repair mechanism that responds to the incurred damage or

degradation. One good example is self-healing bacterial concrete.

Homework Problem 1.23: Janus particles as smart materials (JPs)

Solution to Homework 1.23: In ancient Roman times, Janus was the god who had two faces (beginnings and endings). In modern science, we have adopted the term to describe particles with

two distinct and usually contrasting sides. These particles have the resemblance of the Taijitu

symbol in ancient Asian philosophy, where Yin and Yang (dark and bright) were used to

describing seemingly opposite forces. It is believed that these two basic elements give rise to

complicated change and transition in the whole world. In the same sense, Janus particles are

defined by their duality, which can take on a variety of forms and create a wide range of new

materials with the simple Janus motif. The possibilities for properties that can be assigned to

each half of the Janus particles are vast (for example, hydrophobicity and charge), and are

limited only by the fabrication capabilities of their creators.


Homework Problems and Solutions for Chapter 2

Homework Problem 2.1: What is a piezo modulus?

Solution to Homework Problem 2.1: The piezo modulus is the ratio of induced electric charge to mechanical stress or of achievable mechanical stress to the applied electric field (T = constant).

For piezo actuators, the piezo modulus is also referred to as deformation coefficient dij.

Homework Problem 2.2: Design a piezo stack of your choice composed of n actuators. Present a solution to its performance in terms of voltage applied and displacement achieved.

Solution to Homework Problem 2.2: The active part of the positioning element consists of a stack of ceramic disks separated by thin metallic electrodes. The maximum operating voltage is

proportional to the thickness of the disks. PI stack actuators are manufactured with layers from

0.02–1 mm in thickness. Piezoceramics are generally brittle and cannot withstand large tensile

forces. Thus, preloaded stacks are favorable and are industrially available. Piezo stack models can

be used for static and dynamic operation. Displacement of a Piezo stack actuator can be estimated

using the following equation:

L d33 × n × U (2.1)

where d33 is the strain coefficient (field and deflection in polarization direction) [m V−1], n is the

number of ceramic layers and U is the operating voltage [V]. Figure 2.2 shows a typical piezo

stack arrangement.


Figure 2.2 Typical piezo stack arrangement.

Homework Problem 2.3: Assuming the host boring bar shown in Figure 2.3 has mode shapes

r(z) (which include the cutout), derive an expression for the modal control forces including the active

and passive components of the actuator. Use the actuator force model:

Fa(t) = Ka(p d33V(t) − L(t)), (2.2)

in which Ka is the actuator stiffness, p is the number of wafers in the stack, d33 is the

piezoelectric constant, V(t) is the input voltage, and L(t) is the axial deformation of the actuator

resulting from bar bending deformations of the beam.

Figure 2.3 Conceptual model of a boring bar with a structurally integrated PZT stack

actuator located at the bar root.


Solution to Homework Problem 2.3: Ignoring shear and rotary inertia effects for simplicity, the dynamic displacement of a simple bar in bending is given by:

1( , ) ( ) ( )r r

ru z t z q t

in which ( )r z are the mass normalized mode shapes and ( )rq t are the modal displacements for a cantilevered bar. The mode shapes reflect only the dynamics of the structure including the material cutouts intended to accommodate actuators. The modal control forces are the projection of the external forces onto the modal sub-space given by the inner product:

0( ) ( ) ( , )

L

r rq t z f z t dz

Where f(z, t) is the external force per unit length. Actuator deformations are approximated geometrically by

1 2( ) [ ( , ) ( , )]y yL t b z t z t in which the beam slope ( , )y z t is obtained from the first spatial derivative of the

displacement given in the above equation. This leads to the alternate expression for the actuator deformations.

' '1 2

1( ) [ ( ( ) ( )) ( )]S S S

sL t b z z q t

The effect of the actuator force can be approximated as moments concentrated at the actuator–bar interfaces. Accordingly, the moment per unit length is given by:

1 2( , ) ( )( ( ) ( ))am z t bF t z z z z

' '33 1 2 1 21

( , ) [ ( ) ( ( ) ( )) ( ) ( ( ) ( ))a S S SS

m z t bK pd V t b z z q t z z z z


' ' ' '1 2 33 1 21

( ) [ ( ) ( )][ ( ) ( ( ) ( )) ( )r a r r S S SS

Q t bK z z pd V t b z z q t

which includes both the active and passive effects of the actuator. The control input V(t) is governed by an appropriate feedback control algorithm. Note that the degree of coupling of the control to a particular mode depends on the first spatial derivative of the mode shape at the interface locations. Therefore, the modal control forces can be maximized by placing the actuators in locations of peak modal strain energy.

Homework Problem 2.4: Define the permittivity, ε

Solution to Homework Problem 2.4: The permittivity, ε, or the relative dielectric coefficient DC is the ratio of the absolute permittivity of the ceramic material to the absolute permittivity in

a vacuum (ε0 = 8.85 × 10−12 F m−1), where the absolute permittivity is a measure of the polarizability in the electrical field. The dependency of the dielectric coefficient from the

orientation of the electric field and the dielectric displacement is symbolized by the

corresponding indices. Examples are:

ε33T DC value in the polarization direction when an electric field is applied in the direction of polarization (direction 3) at a constant mechanical stress (T = 0: "free" permittivity).

ε11S Electrical field and dielectric displacement in the direction of axis 1, at constant deformation (S=0: "clamped" permittivity). The following Figure 2.4 depicts the typical variations in the

piezoelectric coefficients according to the electric field.


Figure 2.4 Typical variations in piezoelectric coefficients according to the electric field.


Homework Problems and Solutions for Chapter 3 Homework Problem 3.1: A piezoresistor is embedded on the top surface of a silicon cantilever near the cantilevered base. The cantilever points in the direction. The piezoresistor is a p-

type doped with resistivity of 7.8 cm. Find the longitudinal gauge factor of the material.

Solution to Homework Problem 3.1: The longitudinal piezoresistive coefficient is given by:

(11 + 12 + 44)/2 = (1/2) × (6.6 − 1.1 + 138.1) × 10−11 = 71.8 × 10−11 Pa−1

Note that Young’s modulus of a single p-type silicon crystal is 168 GPa in the orientation.

Thus, the effective gauge factor GF = 71.8 × 10−11 (1/Pa) × 168 × 109 (Pa) = 120.6

Homework Problem 3.2: A piezoresistor is embedded on the top surface of a silicon cantilever near the cantilevered base. The cantilever points in the direction. The piezoresistor is a n-

type doped with resistivity of 7.8 cm. Find the longitudinal gauge factor of the material.

Solution to Homework Problem 3.2: The longitudinal piezoresistive coefficient is given by:

(11 + 12 + 44)/2 = (1/2) × (−103.2 + 53.4 − 13.6) × 10−11 = 31.7 × 10−11 Pa−1

Note that Young’s modulus of a single p-type silicon crystal is 168 GPa in the orientation.

Thus, the effective gauge factor GF = 31.7 × 10−11 (1/Pa) × 168 × 109 (Pa) = 53.26


Homework Problems and Solutions for Chapter 4 Homework Problem 4.1: An electrostrictive actuator utilizing a 2 mm thick layer of PVDF can be used for applications requiring very short strokes. What is the strain? Solution to Homework Problem 4.1: The relevant strain equation is 𝑠 𝜀 𝜀𝐸 /𝑌, where ε0 is the permittivity of free space, ε is the relative permittivity of PVDF, E is the applied electric field

strength, and Y is the modulus of elasticity. PVDF can exhibit a fairly wide range of relative

permittivity, reaching 12 in some cases. In this case a permittivity of 8 will be used. In a field of

100 MV m−1, a strain of 272 microstrain is achieved, resulting in a stroke length of 0.5 μm.

Homework Problem 4.2: Compare the constitutive equations for electrostrictive materials with those of piezoelectric materials.

Solution to Homework Problem 4.2 Both electrostrictives and piezoelectrics convert electrical energy into mechanical energy by mechanically deforming under an applied field, or voltage.

Their response, however, differs both in magnitude and direction. Piezoelectric materials have a

linear response to an applied voltage while electrostrictive materials have a displacement

proportional to the square of the voltage. This is shown by the constitutive Equations 4.1 and 4.2

for the piezoelectrics and Equations 4.3 and 4.4 for the electrostrictive materials;

Electrostriction is generally defined as a quadratic coupling between strain (Sij) and polarization

(Pm):

Em =mn Pn + 2Qklmn Tkl Pn (4.1) Sij=sPijkl Tkl +dkij Ek + Qijmn Pm Pn (4.2) Where sPijkl is the elastic compliance, Qijkl is the polarization-related electrostriction

coefficient, mn is the inverse of the linear dielectric permittivity, Tkl is the stress and Em is the

electric field. A linear relationship is assumed between the polarization and the electric field. The

strain Sij and the electric flux density Di are expressed as independent variables of the electric

field intensity Ek, El, and stress Tkl by the constitutive relations according to the equations

reported in ref. 8 and 9, where is the elastic compliance under constant electric field, Mijkl is

the electric field-related electrostriction coefficient, and is the linear dielectric permittivity.

S = SE × T + d × E (4.3)


D = T × E + T × E (4.4) Sij = sEijkl Tkl +dkij Ek + mklij Ek El (4.5) Dm = dmkl Tkl +mn En + 2mmnij EnTij (4.6) Em = mn Pn + 2Qklmn Tkl Pn (4.7) Sij = sPijkl Tkl +dkij Ek + Qijmn Pm Pn (4.8)

S = SE × T + m × E2 (4.9) D = T × E + m × T × E (4.10) where S is the strain tensor, SE is the compliance tensor, T is the strain tensor, d the piezoelectric

coefficients, E the electric field, D is the electric displacement, T the dielectric permittivity

tensor at constant stress, and m the electrostrictive coefficients. Equation 4.3 shows that the

strain of an electrostrictive material in the same direction as the electric field is positive,

regardless of the polarity of the field, and will not contract from its original size. On the other

hand, for a piezoelectric material, a negative voltage, or a voltage in the opposite direction of the

poles will induce a contraction. This is shown graphically in Figure 4.1.



Homework Problem 5.1:

Propose a design for a resilient PAN actuator.

Solution to Homework Problem 5.1:

The configuration below (Figure 5.1) shows a fabricated PAN actuator system with some resilient

body, where the dimensions are provided. It can cause the PAN fiber to contract within the flexible

membrane (rubber boots). Once the polarity is changed then the PAN fiber tends to expand and

the compressed flexible membrane will help it expand in a resilient manner. The fabrication of this

unit was completed and is now ready for performance testing. The test results will be reported in

the near future.

Figure 5.1 A resilient PAN actuator system.

First we measured the spring constant of rubber boots by applying predetermined loads. The

measurement gave a spring constant of k = 0.01 kg mm−1. Inside rubber boots the following

components are positioned as can be seen; the PAN muscle bundle, electrodes, and a solution.

Applying electrical currents through the electrodes can carry out the system operation. The inner

electrode (a circular shape) surrounds the PAN muscle bundle and the other is attached to the boot

wall. The clearance between the boot wall and the inner electrode is approximately 15 mm.


Homework Problem 5.2:

Describe the pH-activated contraction and expansion mechanisms of PAN

Solution to Homework Problem 5.2: These are shown in Figure 5.2 below, where elongation

and contraction of a modified PAN fiber in (A) basic and (B) acidic solutions are shown. The advantages of PAN fibers among other electrolyte gels are that PAN fiber gels have good

mechanical properties, which compare to those of biological muscles. The large volume change

of PAN fiber gels also allows the reduction in the size of the gel, which is an important factor in

determining response time. This procedure results in the need to study PAN fibers and make

promising materials for high-performance artificial muscles. PAN fibers can convert chemical

energy directly into mechanical motion. Figure 5.2 depicts a possible explanation for the

contraction and expansion of modified PAN fibers. Based on ion diffusion theory, the response

time of swelling is proportional to the square of the gel fiber diameter. The surface/volume ratio

also affects the response time. Note that such pH-induced contraction–expansion of modified

PAN fibers can also be induced electrically in a chemical cell by electrolysis and production of

H+ and OH− ions.

(A) (B)

Figure 5.2 Elongation and contraction of a modified PAN fiber in (A) basic and (B) acidic solution. Reproduced from Ref. 16 with permission from the Royal Society of Chemistry.


Figure 5.3 also depicts the neutral, pH contracted and pH expanded fiber bundles of PAN (active

PAN muscles):

This alternative setup involves securing the activated muscles to a structure and allowing the

muscles to hang. The solutions are then poured or squirted onto the bundles. This set-up allows

applications to be witnessed, such as the lifting of an object (Figure 5.3). A large dish should be

placed under the structure to catch excess solution.

Figure 5.3 Lifting application of PAN muscles in an alternative experimental set-up. Homework Problem 5.3: Describe the differences between the stress–strain curves of contracted and expanded PAN

fibrous muscles.

Solution to Homework Problem 5.3: By annealing raw PAN fibers (Orlon) at temperatures of

220–240 °C, pyridine rings are formed by cross-linking. The cross-linked PAN fibers can then

be made active by boiling the fibers in 1N NaOH or LiOH solutions (saponification of annealed

PAN fibers.

Annealing temperature and time will determine the degree of cross-linking. Typical Stress–

strain curves for active PAN (swollen and contracted) fiber bundles is also depicted in Figure

5.4, showing the difference in constitutive equations for PAN contraction and expansion.


Figure 5.4 Normal stress–strain relationship for the contracted and expanded state of PAN fibrous muscles.

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Stre

ss (M

Pa)

Strain

Stress vs. strain curve: PAN muscle (50 strands)

Contracted PAN Muscle

Elongated PAN Muscle

Contracted muscle: length under 0 load = 6.9 cm; maximum load applied = 650 gm

(did not break); max length = 24.2 cm. Elongated muscle: length under 0 load = 15.5

cm; maximum load applied = 200 gm (broke under load); max length = 20.3 cm.

Estimated cross-sectional area of one strand = 1.57 × 10−7 m2; for 50 strands = 7.85 ×

10−6 m2.



Homework Problem 6.1: Describe the Matteucci effect.

Solution to Homework Problem 6.1: The Matteucci effect manifests itself as helical anisotropy

of the magnetic susceptibility when a magnetostrictive material is subjected to a twisting torque

or torsional stress. On the other hand, the Wiedemann effect is torsional deformation of the

magnetostrictive material under an applied helical magnetic field.

The design and applications of magnetostrictive materials leads to a number of effects emanating

from the Joule effect such as the Villari effect, the Wiedemann effect and the Matteucci effect. The Joule effect at a fundamental level is the change in dimensions of a ferromagnetic material

due to the interactive coupling between an applied magnetic field and the magnetization and

magnetic moments of the internal domains or magnetic dipoles. The reverse of the Joule effect is

called the Villari effect. Here, when the magnetostrictive material is subjected to mechanical

stress, its magnetic susceptibility changes due to loading.

There is another effect called the ΔE-effect which changes the Young’s Modulus of elasticity E of the GMM as a result of an imposed magnetic field. The inverse Wiedemann effect is called the Matteucci effect. Alternating current fed to a coil creates a longitudinal magnetic field in a

sample, and this, in turn, creates a magnetic flux density in the sample. An additional

magnetostrictive effect is the Barret effect. This effect indicates that the volume of GMMs can

change if placed in a magnetic field. There is also the inverse Barret effect known as the

Nagaoka–Honda effect. In this effect, if the GMM is placed in a hydrostatic pressure

environment which may cause its volume to decrease, then the magnetic state of the GMM

changes accordingly. The most widely employed magnetostrictive effects are the Joule effect and

its inverse, the Villari effect.

Homework Problem 6.2 Describe the Wiedemann effect.

Solution to Homework Problem 6.2: The Wiedemann effect is torsional deformation of the magnetostrictive material under an applied helical magnetic field. The inverse Wiedemann effect

is called the Matteucci effect, which has been previously defined.

Homework Problem 6.3: Describe the ΔE-effect.


Solution to Homework Problem 6.3: In the ΔE-effect, changes in the Young’s modulus of elasticity E of the GMM as a result of an applied magnetic field is observed.

Homework Problem 6.4: Describe the Barret effect.

Solution to Homework Problem 6.4: An additional magnetostrictive effect is the Barret effect.

In this effect, the volume of GMMs can change if placed in a magnetic field.

Homework Problem 6.5: Describe the Nagaoka–Honda effect.

Solution to Homework Problem 6.5: There is also the inverse Barret effect known as the

Nagaoka–Honda effect. In this effect, if the GMM is placed in a hydrostatic pressure

environment which may cause its volume to decrease, then the magnetic state of the GMM

changes accordingly.


Homework Problems and Solutions for Chapter 7 Homework Problem 7.1: Describe the concepts of spin-up and spin-down electrons Solution to Homework Problem 7.1: Electrons possess intrinsic magnetic fields due to their spin. In the absence of magnetization of a ferromagnetic material, half of the conduction

electrons possess a positive spin (spin-up) along the chosen axis, and half will have a negative

spin (spin-down).

If a ferromagnetic material is magnetized along the positive spin direction, a spin-up electron

will travel through the material easier than a spin-down electron.

Homework Problem 7.2: How can traveling electrons in ferromagnetic materials experience less or more collisions with other electrons and atoms?

Solution to Homework Problem 7.2: Note that generally, the resistance of a conductor depends on the magnetization which is controlled by the applied external magnetic field. Note that

electrical resistance of conductors is due to collisions between conduction electrons with other

charge carriers, electrons, ions or atoms. If a perfect metallic crystalline structure is at a

temperature of absolute zero, it will allow an electron to move through it and experience no

collisions so that the crystal will have zero resistance. Due to imperfections in the crystalline

structure, at temperatures above the absolute zero the atoms vibrate out of their lattice locations

and these imperfections and vibrations cause collisions, which increases the resistance of the

metallic crystal. An applied external magnetic field can also increase the resistance of the

material since the magnetic force on the moving charges will tend to increase the number of

collisions between charges. This dependence of resistance on the magnetic field is called the

magnetoresistivity effect. Note that changes in resistance are directly proportional and related to the strength of the imposed external magnetic field.

The magnetic force developed disrupts the normal flow of electrons, causing more collisions

with atoms and other electrons, thus increasing the resistance of the wire. Based on quantum

mechanical principles, if conduction electrons moving through a magnetoresistive medium have

inherent spins in the same direction as the imposed magnetic field, they experience fewer


collisions with other entities (electrons, ions, atoms, etc.) than electrons with inherent spins in the

opposite direction of the imposed magnetic field.

Homework Problem 7.3: How are the electrons scattered in GMR materials?

Solution to Homework Problem 7.3: Note that each layer acts like a spin valve in the sense that its magnetization direction determines whether it allows spin-up or spin-down electrons to pass

through or get scattered back and not go through the layers. In the case of parallel-aligned

magnetic layers, the resistivity for one spin channel is low, leading to low total resistance.

Homework Problem 7.4: Present a simple model for GMR effects.

Solution to Homework Problem 7.4: The GMR effect occurs in granular mixtures or powders of ferromagnetic and non-magnetic metals. Here, the spin-dependent scattering of electrons

occurs at the surface and in the bulk of the grains. The granular mixture of ferromagnetic and

non-ferromagnetic metals form powder-like ferromagnetic nanoclusters of about 10 nm in size

within particles of about 10 nm in diameter embedded in a non-magnetic metallic powder,

forming a superlattice.



Homework Problem 8.1: Are shape changes in magnetic gels reversible?

Solution to Homework problem 8.1: Yes, they are. Magnetic gels or ferrogels belong to the general family of magnetostrictive materials, producing strain when exposed to a magnetic field.

One may also embed magnetic coils within these materials to be able to also electrically control

the deformation of ferrogels. The reversible swelling and deswelling of polymeric magnetic gels

are possible. Such swelling and deswelling can also be induced chemically (pH Muscles) or

electrically (EAP). Note that magnetic gels are considered a member of the smart materials

family and in ways are similar to soft silicon rubber magnetic composites used in our daily life as

soft magnetic stickers. However, magnetic gels are softer and more stretchable and

maneuverable in a magnetic field and can sustain soft actuation at the micro and nano levels.

Ferrogels are chemically cross-linked polymer networks swollen by a colloidal ferrofluid.

Ferrofluids are also called magnetic fluids and are colloidal dispersions of monodomain

magnetic particles. Typically single domain colloidal magnetic particles have typical dimensions

of about 10–15 nm. Furthermore, they are superparamagnetic, in which magnetization can

randomly flip direction under the influence of the temperature field.

Homework Problem 8.2: If the engineering stress for Neo–Hookean magnetic gels is given by: 2

11 1 1( )eng

n G

Show that true stress is given by: 2 111 1 1( )G

Solution to Homework Problem 8.2: Note from equation 8.15 that

2 2 111 1 1 1 1 1 1

1

( ) ( )U G G


Note thatis the true stress or axial force applied to the ferrogel rod divided by the deformed

cross-section of the rod, which in this case is related to the initial undeformed cross section by

the product of and which is equal to 𝜆 𝜆 𝜆 . Thus, the uniaxial engineering stress 1

11 1 11eng

n in the ferrogels can be expressed as:

2 zzn G Here, the nominal stress n is defined as the ratio of the equilibrium elastic force to the

undeformed cross-sectional area of the sample, and z is the stretch in the z-direction.

Homework Problem 8.3: Describe the configuration of uniaxial magnetic gels, which will lead to a positive definite stretch.

Solution to Homework problem 8.3: Consider a vertically suspended cylindrical magnetic gel intended to pull a weight upward. Initially, due to the weight of the strand, the gel is preloaded

with its own weight Mgg, where Mg is the mass of the magnetic gel strand.

Thus, the gel is under a passive stretch, m, due to its weight. The nominal stress due to this

passive stretch can be written as:

0

g

nM g

a

where Mg is the mass of the magnetic gel, g denotes the gravitational acceleration and a0 denotes

the initial undeformed cross-section of the magnetic gel strand. Let us rewrite this equation in the

form of a cubic equation to solve for M :

0123 MM

0

gn M g

G a G

Here 0a denotes the undeformed cross-sectional area of the gel at rest. The above equations show us how the passive stretch M depends on the applied mechanical stress and the shear modulus G. We can show that the passive stretch 1M . Note that the solution to equation 8.21 is:


3 2 1 1M M

Thus, in the presence of a load or stress, the magnetic gel tends to elongate, i.e., ( 1M ). Homework Problem 8.4: Show by a Taylor series expansion of the Langevin distribution function 1cothL given in equation 8.24 that it is approximately L() /3. Solution to Homework problem 8.4: Under an applied magnetic field, the magnetic gel strand

will develop a new total stretch HM , as well as magnetization, M. The magnetization, M, of a

magnetic gel is defined by the Langevin function based on the assumption that the magnetization

of individual colloidal particles in the magnetic gel is approximately equal to the saturation

magnetization of the pure ferromagnetic material. Thus, the magnetization, M, of magnetic gels,

in an applied magnetic field can be expressed as:

1thcoMLMM smsm

where m stands for the volume fraction of the magnetic particles in the whole gel, and of the

Langevin function L() is defined as:

0

B

mHk T

In this equation m stands for the magnetic moment of embedded particles, Bk is the Boltzmann

constant and T stands for the ferrogel temperature. It establishes that the magnetization of a

ferrogel is proportional to the volume fraction of embedded magnetic particles and their

saturation magnetization. Assuming a linear relationship between the magnetization, m, and

magnetic field strength, H, and homogeneous deformation for the ferrogel sample, one obtains a

linear relationship between the magnetization, M, and magnetic field strength, H, such that this

latter assumption is valid at small magnetic field intensities. In this case, a Taylor series

expansion of the Langevin distribution function in Eq. 8.24 yields: L () /3.


Homework Problem 8.5: Find the relationship between the magnetization M and magnetic field strength H.

Solution to Homework problem 8.5: The above conclusion of L () /3 leads to a linear

relationship between the magnetization M and field intensity H or M = H, where denotes the initial susceptibility, which is defined as:

3m S BmMk T

Homework Problem 8.6: Show that for a magnetic gel fiber with a total stretch of M,H

M H M H h m M HH H, , ,3 2 2 2 1 0

where hH and mH are the magnetic field strengths at the bottom and the top of a rod-like ferrogel

fiber, respectively.

Solution to Homework problem 8.6: Note that for ferrogels M m k TS B/ 3 = 0.338. The magnetic

force can be determined using Equation 8.1. If we assume the total stress to be mechanical stress

plus magnetic stress, then one can derive the following equations:

M H M H h m M HH H, , ,3 2 2 2 1 0 where hH and mH are the magnetic field strengths at the bottom and the top of a rod-like ferrogel

fiber, respectively (see Figure 8.4).


S N

z = z -h

z = z

mfM

Hm

H h

Ferrogel

z = 0

m

Force measuring unit

Figure 8.4 Schematic diagram of the experimental set-up to study the magnetoelastic properties of the cylindrical form of ferrogels. Reproduced from Ref. 19 with permission from the Royal

Society of Chemistry.

The parameter can be considered as the stimulation coefficient defined as:

02G

This equation establishes that elongation occurs, i.e., , 1M H M , if we suspend a ferrogel fiber in a nonhomogeneous magnetic field in such a way that h mH H . On the other hand, when the field is turned off, the stress is lifted. In the opposite case, when Hh < Hm work is released when the magnetic field is applied, i.e., M > 1, but M H,


Homework Problems and Solutions for Chapter 9 Homework Problem 9.1: Design an ERF cylindrical brake system.

Solution to Homework Problem 9.1: In order to design an ERF brake system note that an ERF requires a channel width of 1 mm or so and a voltage of 1 kV. Let us assume the radius of the

housing is about 100 mm. The width of the plate and housing section will be set as 2 and 4mm,

respectively. As discussed before for the ERF clutch, one of the common applications of ERFs is

in smart ERF brakes, as shown below:

(a) (b)

Possible design of an (a) ERF brake and (b) a model of a simple ERF brake system.

In automatic transmissions, the torque converter takes the place of the brake found on

standard shift vehicles. If we use an ERF fluid instead of oil in the torque converter, by applying

an electric field, we can improve the Torque converter performance. For an electrorheological

brake to be used in a passenger car, component dimensions similar to existing clutch plates

should be chosen. This brake system will use six plate sections in the assembly.

Homework Problem 9.2: List possible carrier fluids for ERFs.

Solution to Homework Problem 9.2: An ERF is typically a suspension consisting of electrically

polarizable nano or micro non-conducting particles of a few microns (~4 m) in diameter well

dispersed in a non-conducting (insulating), low-viscosity medium like a colloidal solution. The

particles are randomly distributed initially before an electric field is applied. A simple ERF can

be made by mixing cornflour in a light vegetable oil or silicone oil/cross-linked polyurethane


particles in silicone oil. The particles are 4 microns in diameter and contain dissolved metal ions

for fast polarization and the electrorheological effect. ERFs have a typical response time of a few

milliseconds with reversible liquid–solid transformation under a dynamic electric field. After an

ERF is subjected to an imposed electric field, the suspending particles will be polarized. ERFs

may be in particulate form or colloidal form depending on the size of particles suspended in the

dielectric fluid environment. A liquid that consists of a suspension of very fine semi-conducting

particles in an electrically-insulating fluid (usually silicone oil) is an ERF suspension. When an

electric current is put through this liquid, its viscosity changes from the consistency of a liquid to

that of a gel in milliseconds and reverts when the current is removed.

Homework Problem 9.3: List possible particulate materials to be suspended in ERF carriers.

Solution to Homework Problem 9.3: Similar to in the previous problem, possible particulate

materials to be suspended in ERF carriers are electrically polarizable nano or micro non-conducting

particles of a few microns (~4 m) in diameter well dispersed in a non-conducting medium like a

colloidal solution. The particles are 4 microns in diameter and contain dissolved particles for fast

polarization and the electrorheological effect. ERFs may be in particulate form or colloidal form

depending on the size of particles suspended in the dielectric fluid environment. A liquid that

consists of a suspension of very fine semi-conducting particles in an electrically-insulating fluid

(usually silicone oil) is an ERF suspension. When an electric current is put through this liquid, its

viscosity changes from the consistency of a liquid to that of a gel in milliseconds and reverts

when the current is removed.


Homework Problems and Solutions for Chapter 10 Homework Problem 10.1: The shear stress curve of a magnetorheological (MR) fluid and its Bingham plastic model with respect to applied magnetic fields are presented in Figure 10.4.

Using the rheological properties of the MR fluid given in this problem, please answer the

following questions.

Figure 10.4 Shear stress curve of an MR fluid and its Bingham plastic model with respect to applied magnetic field.

0 2000 4000 6000 8000 100000

10

20

30

40

500 kA/m50 kA/m100 kA/m150 kA/m200 kA/m250 kA/m

Bingham plastic model

𝜏 0.15 Pa s 𝛾 at 0 kA m 𝜏 0.13 Pa s 𝛾 2183 Pa at 50 kA m 𝜏 0.12 Pa s 𝛾 7604 Pa at 100 kA m 𝜏 0.12 Pa s 𝛾 15 776 Pa at 150 kA m 𝜏 0.13 Pa s 𝛾 26 478 Pa at 200 kA m 𝜏 0.13 Pa s 𝛾 39 566 Pa at 250 kA m


(a) The yield stress of the MR fluid can be a function of magnetic field, such as 𝜏 𝛼𝐻 . Identify the values of 𝛼 and 𝛽 to be fitted with the measured yield stress by a curve-fitting method. (b) Plot the measured yield stress data and the identified yield stress curve as a function of applied

magnetic field.

(c) Create a log-log plot for the non-dimensional Bingham number versus the shear rate. Use the

identified yield stress values at the magnetic field inputs of 0, 50, 150, and 250 kA m-1.

(d) Create a log–log plot for the apparent viscosities of the MR fluid at the magnetic field inputs

of 0, 50, 150, and 250 kA m-1.

Solution to Homework Problem 10.1

(a) 𝛼=1.91 and 𝛽 1.8 (b)

(c) The non-dimensional Bingham number is given in eqn (10. 11):

𝐵𝑖 𝜏𝜂𝛾

Then, 𝐵𝑖 . .. .

0 50 100 150 200 250Magnetic field input (kA/m)

0

5

10

15

20

25

30

35

40

Yiel

d st

ress

(kPa

)

measuredcurve-fitting ( y=1.91 H

1.8)


(d) The apparent viscosity is given in eqn (10.2)

𝜂 𝜏𝛾 𝜂

Then,

𝜂 0.15 𝑃𝑎 𝑠 at H 0 kA/m

𝜂 0.13 2,183 𝛾 Pa s at H 50 kA m 1

𝜂 0.12 15,776 𝛾 Pa s at H 150 kA m 1

𝜂 0.13 39,566 𝛾 Pa s at H 250 kA m 1

10-2 100 102 104

Shear rate (s-1)

10-2

100

102

104

106

108

Bing

ham

num

ber (

-)

H=0 kA/mH=50 kA/mH=150 kA/mH=250 kA/m

10-2 100 102 104

Shear rate (s-1)

10-2

100

102

104

106

108

Appa

rent

vis

cosi

ty (P

a s)

H=0 kA/mH=50 kA/mH=150 kA/mH=250 kA/m


Homework Problem 10.2: The mudline, 𝑧 of an MR fluid contained in a stationary tube was measured at 2 hour intervals and results were listed in Table 10.1. Using the sedimentation data

given in this problem, please answer the following questions.

Table 10.1: Measured mudline data

Time (h) Mudline (mm) Time (h) Mudline (mm) Time (h) Mudline (mm)

0 200.0 14 187.41 28 186.05

2 198.0 16 187.21 30 185.85

4 196.0 18 187.02 32 185.66

6 194.0 20 186.82 34 185.37

8 192.0 22 186.63 36 185.26

10 190.0 24 186.44 38 185.17

12 188.0 26 186.24 40 185.08

Note: The origin of the mudline, 𝑧 was the bottom of the tube. (a) Plot the measured mudline versus time.

(b) The measured mudline can be approximated by using two piecewise functions (the first

segment at 0 ≤ t ≤12 h is a linear function and the second segment at 12 ≤ t ≤ 40 h is a second-

degree polynomial function). Please find two piecewise functions.

(c) Obtain the mudline velocity, 𝑧 using the identified two piecewise functions. (d) Using the mudline velocity obtained from (c), predict the time when the sedimentation process

has completed.


Solution to Homework Problem 10.2

(a)

(b) Piecewise function #1

𝑧 200 1 𝑡 mm at 0 𝑡 12 h Piecewise function #2

𝑧 189.5913 0.1622 𝑡 0.0012 𝑡 mm at 12 𝑡 40 h (c) Piecewise linear function #1

𝑧 1 mm h at 0 𝑡 12 h Piecewise linear function #2

𝑧 0.1622 0.0024 𝑡 mm h at 12 𝑡 40 h (d) If the mudline velocity of the piecewise linear function #2 becomes zero, then the sedimentation process

will be complete. Then,

𝑧 0.1622 0.0024 𝑡 0 𝑡 67.5833 h

0 10 20 30 40Time (hour)

180

185

190

195

200

Mud

line,

zM

(mm

)

measured datapiecewise function #1piecewise function #2



Homework Problem 11.1: Derive equation 11.13 from the generalized constitutive equations for rubber elasticity.

Solution to Homework problem 11.1: Taking the assumption of dielectric elastomers as an incompressible, isotropic, and homogeneous hyper-elastic Neo-Hookean model, the true stress

formula can be presented as:

ij i ijj

W p

So that

11 11

W p

, 22 22

W p

, 33 33

W p

Note that W is a function of I1 and I2 and thus by chain rule of differentiation, one can find the

following expressions for stresses in DEAs. Note that in this case due to incompressibility,

1 2 3 1 , and thus, for a uniform plate of DEA, ( 1/2) ( 1/2)2 1 3 and thus:

2 2 2 2 2 21 1 2 3 2 1 2 3( 3) ( 3)W C C

or simply: 2 1 21 3 3 2 3 3( 2 3) ( 2 3)W C C

Thus, it can be deduced from equations (10–12) that:

33 22 3 23 2

W W

, 22 11 2 12 1

W W

(11.13)

Since in this case, the expression for the engineering Maxwell stress is derived

such that:

2 1 233 3 2 1 3 3 2 3 3

3 2

2 ( ) 2 ( )W W C C

For Mooney–Rivlin-type DEAs, note thatis the true stress or axial force applied to the DEA

divided by the deformed cross section of the plate, which in this case is related to the initial

unreformed cross section by the product of and which due to incompressibility (Equation


5), is equal to 1 11 2 3 . Thus, the engineering stress 1 133 3 33 33

engn

in the dielectric elastomer actuator in the z-direction can be expressed as:

1 21 22( )n C C (11.15) where the nominal stress n is defined as the ratio of the equilibrium elastic force to the undeformed cross-sectional area of the sample and is the stretch in the z-direction.

Homework Problem 11.2:Design a helical dielectric elastomer actuator Solution to Homework Problem 11.2: You may consider a geometry of an actuator, of thickness = 0.8 mm, width = 8 mm, length = 100 mm, with a modulus of = 0.01 GPa (softer then rubber).

Assuming the material is isochoric and acts in a Neo–Hookean manner, this will accurately model

the actuator for the lower end of the stress. This actuator will compress, as seen in Figure 11.4.

Figure 11.4 Helical dielectric elastomer actuator, (a) undeformed and (b) deformed.

Note that:

E = (V/l) and the Maxwell stress where is the Maxwell pressure, is the

vacuum permittivity, r is the dielectric constant of the polymer, is the voltage, and is the

modulus of elasticity. A Neo–Hookean model can be determined in the following way:

where is the stretch or L/L0 where L and L0 are the current and initial thicknesses of the DEA that determine the change in length from 0 to 4000 V.


Homework Problems and Solutions for Chapter 12 Homework Problem 12.1: Describe the design of parallel spring-loaded fibrous SMA actuators, as depicted in Figure 12.7.

Solution to Homework Problem 12.1: There are many applications in connection with shape–memory materials. These include couplings, seals, electrical connectors, virtual two-way

actuators, non-biased safety devices, thermal interrupter, eyeglass frames, cellular phone

antennas, and home appliances. There are also many scientific experiments that utilize shape–

memory materials as experimental means to perform other experiments.

Adaptive structures, structural damping, high force devices, jet engines and other aeronautical

applications including active backend rotors of helicopters. Refer to work by Shahinpoor for

fibrous parallel spring-loaded shape–memory alloy (SMA) robotic linear actuators (Figure 12.7),

and to work by Shahinpoor and Martinez for the applications of SMAs as thaw sensors. The design

is composed of bundles of SMA wires that stretch under the expansive load of a helical spring

encompassing wires which are all encased inside a foldable outer wall or cylindrical mantle.

Figure 12.7 Fibrous parallel spring-loaded SMA robotic linear actuators with a resilient

helical spring.


Homework Problem 12.2: Describe how SMAs can be used as thaw sensors. Solution to Homework Problem 12.2: In order to create a thaw sensor (refrozen food sensor) one can use one-way SMAs. Here, recall that once the SMA is activated by raising its

temperature above the thaw temperature, also above the solid phase transformation temperature,

it will make a solid phase transformation to the austenite phase and can turn a dial indicator to

red, indicating a thaw event. If the food is refrozen by lowering the temperature, the SMA does

not automatically revert back to Martensite and thus the thaw sensor still indicates red, meaning

that the food was thawed and refrozen.

Homework Problem 12.3: What are the differences between the Tanaka and Brinson models of SMAs?

Solution to Homework Problem 12.3: The Tanaka formulation maintains that the thermomechanical behavior of the SMAs can be fully expressed by three major state variables:

strain , temperature T, and martensite fraction . The constitutive equation in this model relating

the state variable stress ( ), strain ( ), and temperature ( ) in terms of martensitic volume

fraction ( ) is:

))(()())(( 0000 TTE (12.1)

where ),,,( 0000 T represent the initial state or original condition of the material. In this equation

is the module of elasticity and assumed to be a linear function of the martensite volume fraction:

)()( AMA EEEE (12.2)

and is called phase transformation coefficient and is given by equation 12.3:

)()( EL (12.3)

where is the maximum recoverable strain or residual strain. The kinetics equations describing

the martensite volume fraction as an exponential function of stress and temperature are:

))(exp(1 MsMMA bTMa for fMT and )()( fMsM MTCMTC (12.4)

Based on experimental data, the two material constants CA and CM, which are called stress-

influence coefficients, show the influence of stress on the transition transformations. Note that:

T

E

L


))(exp( AsAAM bTAa for sAT and )()( sAfA ATCATC (12.5)

where Aa , Ma , Ab , and Mb are materials constants in terms of transition temperatures As, Af , Ms,

and Mf. Figure 12.4 depicts the two materials constants CA and CM in relation to the slopes of

martensitic and austenitic transformations and stress temperature variations proposed in Tanaka’s

model.

Figure 12.4 Critical stress–temperature profiles used in the Tanaka model.

Compared to the Brinson Model, since the shape–memory effect (SME) at a lower temperature is caused by the conversion between stress-induced martensite and temperature-induced martensite, these models cannot be implemented due to the detwinning of martensite that is responsible for the SME. This problem was solved by Brinson and his model. In this model, the martensite volume fraction ( ) is separated into stress-induced ( s ) and temperature-induced martensite fractions (

T ):

S T (12.8)

The first form of the constitutive equation in this model is:

0 0 0 0 0 0 0 0( ) ( ) ( ) ( ) ( ) ( ) ( )s s s s T T T TE E T T (12.9)

but it was shown by Brinson and Huang [11] that the constitutive Equation 12.9 could be reduced

to the simplified form of:

)())(( 00 TTE sL (12.10)

where L is the residual strain when = 0.

Homework Problem 12.4: Describe Brinson’s constitutive equations for SMAs.

Solution to Homework Problem 12.4: Note that in the Brinson model, the shape–memory effect


(SME) at a lower temperature is caused by the conversion between the stress-induced martensite

and temperature-induced martensite, these models cannot be implemented due to the detwinning

of martensite that is responsible for the SME. This problem was solved by Brinson and his model.

In this model, the martensite volume fraction ( ) is separated into stress-induced ( s ) and

temperature-induced martensite fraction ( T ):

S T (12.8)

The first form of the constitutive equation in this model is:

0 0 0 0 0 0 0 0( ) ( ) ( ) ( ) ( ) ( ) ( )s s s s T T T TE E T T (12.9)

but it was shown by Brinson and Huang [11] that the constitutive Equation 12.9 could be reduced

to the simplified form of:

)())(( 00 TTE sL (12.10)

where L is the residual strain when = 0.

The variation in the critical stresses with temperature for transformation consistent with the

separation of into two components is shown schematically in Figure 12.5. The evolution

equations for calculation of the martensite fractions as a function of temperature and stress can

now be represented according to Figure 12.5 as follows:

Conversion to detwinned martensite: For sMT & )()( sM

crfsM

crs MTCMTC

)(1

,2

1)]([cos

21

00

00

00ss

s

TTT

ssM

crfcr

fcrs

ss MTC

(12.11)


Figure 12.5 Critical stress–temperature profiles used in the Brinson model.

For 𝑇 𝑀 𝑎𝑛𝑑 𝜎 𝜎 𝜎

21

)(cos2

1 00 scrfcr

fcrs

ss

,

Tss

s

TTT

)(1 00

00 (12.12)

where, if sf MTM and 0TT then ̀1)(cos21 0

fMT

T MTa

otherwise 0 T

Conversion to Austenite: For sAT and )()( sAfA ATCATC

)(),(,1cos2 00

000

0

00

0

TTT

sss

AsA C

ATa (12.13)

This completes our Brinson formulation of constitutive equations for SMAs. Homework Problem 12.5: Design a large motion SMA actuator.

Solution to Homework Problem 12.5: In order to design a large motion SMA actuator one needs to accumulate small deformation of multiple connected wires to generate large motions for SMAs.

A possible design is depicted below. As one can see in part a, the SMA wires are rolled around

pulleys in such a way that when the temperature is raised to the austenite transformation the wires

contract and move the central actuator by a large amount. If the end point is also resiliently

stretched, once the temperature decreases to the martensite phase the spring will pull the wires

back to their original length, as shown in part b of the figure.

𝜎 𝜎

Detwinned Martensite

Pure Austenite

Stress

Mf As Af

CM

CA

Twinned Martensite

Temperature Ms


Linear Actuator: Pulleys and wire of the actuator:

Part (a) of the large motion SMA actuator, unextended

Spring Extended:

Part (b) of the large motion SMA actuator, extended.


Homework Problem 12.6: Medical Applications

Solution to Homework Problem 12.6: Some Medical Applications of SMAs are:

Dental devices

Orthodontic wires, dental implants, endodontic files

Orthopedic devices

Bone anchors, reamers, staples, spinal correction devices

Surgical devices

Laparoscopic surgical instruments, organ bag springs, needles and trocars, malleable surgical

instruments

Cardiovascular devices

Stents, guidewires, vena cava filter, atherectomy devices, catheters, arterial septal defect

devices, clips

Gastroenterology–urology devices

Stents, catheter re-enforcement, endoscopic instruments, snares and baskets

Radiology devices – lesion localizer

Oncology devices – localized therapy


Homework Problems and Solutions for Chapter 13 Homework Problem 13.1: Present a comparison between magnetic shape–memory alloys, thermal shape–memory alloys, magnetostrictive materials, and piezoelectric materials

Solution to Homework Problem, 13.1: Magnetic shape–memory alloys (MSMAs) or materials (MSMMs), are often also referred to as ferromagnetic shape memory alloys (FSMAs). FSMAs

combine the attributes and properties of ferromagnetism with reversible martensitic–austenitic

crystalline solid phase transformations involving twining similar to thermal shape–memory

alloys like SMAs. The magnetic field-induced reorientation of the twin microstructures of a

magnetic shape–memory alloy are similar to those of SMAs. It was reported that the magnetic

SME possesses a much more rapid response in actuation compared with rather a slow response

of thermal SMAs. Here, the magnetic field controls the reorientation of the twin variants in

MSMAs analogously as the twin variants are controlled by the stress in classical thermal SMAs.

Magnetostrictive materials rely on magnetic dipole reorientation due to changing the imposed

magnetic field and thus are mechanically different from MSMAs. Piezoelectric materials also

respond to changing electric dipole orientation due to an imposed dynamic electric field.

Homework Problem 13.2: Describe the properties of MSM linear actuators using Ni–Mn–Ga actuating elements.

Solution to Homework Problem 13.2: The MSM effect has demonstrated that certain SMMs that are also ferromagnetic can show very large dimensional changes up to 10% under the

application of an external magnetic field. These strains occur within the low-temperature

(martensitic) phase. MSMAs are a family of materials in the larger family of Heusler materials.

These are ordered intermetallic materials with the generic formula X2YZ with X and Y being 3D

elements and Z a group IIIA–VA element. FSM materials combine the properties of

ferromagnetism with those of a reversible martensitic transformation. These materials, such as

the MSMA Ni2MnGa with a cubic Heusler structure in the high-temperature austenitic phase is

capable of undergoing a cubic-to-tetragonal martensitic transformation to exhibit MSM effects.

Note that similar to SMAs with twinning and detwinning, when they are subjected to one


externally applied magnetic field, the twins will have a favorable orientation and grow at the

expense of the other twins. This observation causes a shape change in the MSMAs. Compared to

SMAs such as Nitinol, in which temperature is the main driving force in addition to stress and

strain, FSMAs transform back to their original configuration when exposed to a specific magnetic

field for the intended shape memory. MSM actuators generally have a very fast response time of

less than 0.2 ms and can operate at high frequencies and large mechanical stroke. Hysteresis

issues in connection with MSM actuation are also of interest in discussing MSM actuation.

Hysteresis in MSM actuators occurs between the strain and the actuator input current as well as

between the strain and stress. It is clear that hysteresis generates losses in the MSM material and

makes the control of some positioning system applications fairly complicated. On the other hand,

due to its energy dissipation characteristics, hysteresis increases the vibrational damping

capability of the MSM materials. It has been shown that the strain of the MSM actuators depends

on the load it has to work against.


Homework Problems and Solutions for Chapter 14 Homework Problem 14.1: Discuss the types of modeling of SMPs.

Solution to Homework Problem 14.1: Three types of modeling exist for SMPs:

1. Modeling based on storage deformation

2. Modeling based on phase transition

3. Modeling based on viscoelasticity

At a given temperature T, the total Helmholtz energy of the SMP is Ht, which is a sum of the

Helmholtz energy of the rubbery phase Hr plus the Helmholtz energy of the glassy phase Hg such

that

Ht = fr(T) Hr + fg(T) Hg (14.8)

where fr (T)= fa(T) and fg (T) = ff (T) are volume fractions of the rubbery (active phase) and glassy

phase (frozen phase), respectively, such that fr(T) + fg(T) = 1. Furthermore, simple constitutive

equations in terms of the total stress, st, and total strain, e, of the SMP can be postulated to be

given in terms of stress in the rubbery or active phase sr = sa, stress in the glassy or frozen phase

sg = sf such that:

st = fr(T) sr + fg(T) sg = fa(T)sa + ff (T)sf = (1- ff (T)) sa + ff (T) sf (14.9)

Homework Problem 14.2: How is the total strain et related to the strains of the rubbery phase ef and the strain of the glassy phase eg?

Solution to Homework Problem 14.2: Note that in terms of the total strain and corresponding strain of the rubbery phase, ea, and the glassy phase, ef:

et = fr(T) er + fg(T) eg = fa(T)ea + ff (T)ef = (1− ff (T)) ea + ff (T) ef (14.10)

Note from the above equations that the volume fraction of the frozen or glassy phase is given by

ff (T)=1 −{1/ [1+Cf (Th − T)n]. (14.11)

where, Th is a reference temperature, and Cf and n are fitting parameters determined by

appropriate experiments.


Homework Problems and Solutions for Chapter 15 Homework Problem 15.1: Propose three polymers that could be useful to prepare controlled drug release systems responsive to:

a) an increase in glutathione concentration

b) matrix metalloproteases

c) the presence of a certain antigen

Solution to Homework Problem 15.1: a) A polymer with SS links between monomers or that has been cross-linked with reagents containing S–S bonds, such as N,N′-bis(acryloyl)cystamine.

Glutathione may induce the transformation of disulfide bonds into thiol groups.

b) A polymer grafted to peptides that can be degraded by matrix metalloproteases, such as the

(Gly–Pro–Val–Gly–Leu–Ile–Gly–Lys) octapeptide.

c) A polymer grafted to an adenosine triphosphate (ATP) aptamer that interacts with other

polymer chains that have a derivative of the ATP molecule (e.g., adenosine-50-carboxylic acid,

denoted as adenosine COOH). These two polymers form a strong network able to encapsulate

drugs. By a competitive displacement reaction, the interaction becomes weaker in the presence of physiological ATP molecules, and the cargo can be released.

Homework Problem 15.2: Design an imprinted network loaded with an antimicrobial agent that

exhibits self-regulated release because of pH-responsiveness. Solution to Homework Problem 15.2: A hydrogel can be designed to collapse at pH 7.4 but swollen at pH 4.0 using monomers bearing amino groups such as N-(3-aminopropyl)acrylamide,

and a cross-linker such as N,N′-methylenebis(acrylamide). N-(3-aminopropyl) acrylamide can

also interact with antimicrobial agents that have carboxylic acid groups, such as amoxicillin, and

therefore it can work as a functional monomer to interact with template amoxicillin molecules

during hydrogel synthesis. A common functional monomer:template mol ratio to obtain

imprinted networks is 4:1. After hydrogel synthesis, the hydrogels are expected to release minor

amounts of amoxicillin at pH 7.4 (i.e., in the absence of infection because the network shrinks),

but the release can be triggered at acidic pH because of the protonation of the amino groups of

the N-(3-aminopropyl) acrylamide moieties. This causes the swelling of the network and the

weakening of the interactions with amoxicillin molecules.


Homework Problem 15.3: Propose a controlled drug release system that combines photothermal therapy and photo-triggered drug release.

Solution to Homework Problem 15.3: PEGylated polyamidoamine (PAMAM) dendrimers have been developed to integrate gold nanoparticles for photothermal therapy, with high

payloads of chemotherapy agents in a hydrophobic inner space. Also, the decoration of

implantable medical devices with gold nanoparticles offer promise for localized treatments. Gold nanoparticles (AuNPs) can be used for photothermal therapy because they absorb light and release

energy, causing local heating. AuNPs can be decorated with drugs that are bound to the nanoparticles

using spacers with photo-cleavable bonds. Irradiation of these drug-decorated AuNP systems with light of

adequate wavelength may cause both an increase in temperature and the release of the drug from the

surface of the AuNPs.

Homework Problem 15.4: Sketch the drug release rate versus the application/cessation of a stimulus for (a) an activation-modulated system, and (b) a feedback regulated system.

Solution to Homework Problem 15.4: In case (a), the system does not release the drug until the stimulus appears. Once the release is triggered it cannot be stopped. In case (b), the release can

be switched on and off when the stimulus appears/disappears.

Homework Problem 15.5: Sketch the swelling behavior of (a) a pH-responsive network made of polymers with acidic groups with pKa values equal to 5.5, and (b) a temperature-responsive

network made of polymers showing a lower critical solution temperature (LCST) of 34 ºC.

Please indicate for each case, the interval of pH or temperature where the release of a drug

encapsulated in the network would take place.


Solution to Homework Problem 15.5:


Homework Problems and Solutions for Chapter 16 Homework Problem 16.1: What are the basic characteristics of mechanochromic materials? Solution to Homework Problem 16.1: Mechanochromic materials change their optical properties, and in particular, photoluminescence characteristics, if subjected to mechanical

loading or interactions with their environment. Chemical and physical molecular changes across

various length scales and the rearrangement of molecular chemical bonds to modifications in

molecular arrangements in the nanometer regime generally trigger mechanochromic

characteristics.

Homework Problem 16.2: What are the basic microstructural arrangements in mechanochromic materials?

Solution to Homework Problem 16.2: These basic microstructural arrangements originate in the rearrangement of chemical and molecular bonds to changes in molecular domains by

approximately hundreds of nanometers. Such mechanochromic effects and characteristics will

further improve the understanding of strain/stress distributions in solids.

Homework Problem 16.3: Where do the basic microstructural arrangements originate from in terms of chemical and physical molecular changes across various length scales, such as the

rearrangement of molecular chemical bonds to modifications in molecular arran

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Homework Problems and Solutions for Chapter 1...Homework Problems and Solutions for Chapter 1 For each of the following materials, provide a simple description of their actuation and