**5.2. A representative volume unit based mathematical model for investigating the magnetic field dependent sensing capabilities**

#### *5.2.1. Theoretical approach*

From the Dipole Model a Representative Volume Unit (RVU) is derived. A RVU consists of two neighbouring hemispheres and the surrounding polymer matrix, which can be regarded as the minimum volume element in the conventional MRE. Fig. 23 shows the position of RVU in carbonyl iron particle chains and a longitudinal section of the unit is shown in Fig. 24.

**Figure 22.** Resistance versus magnetic field (anisotropic MRE Gr 21.95%)

**Figure 23.** RVU in carbonyl iron chains

**Figure 21.** Resistance versus load (anisotropic MRE Gr 23.81%)

2

2.5

3

3.5

**Resistance(k Ohm)**

4

4.5

2.51 kΩ at 10 N for the sample with graphite weight fraction 23.81%.

**the magnetic field dependent sensing capabilities** 

*5.2.1. Theoretical approach* 

shown in Fig. 24.

43.5 kΩ at a 440 mT magnetic field. This trend is shown in detail in Fig. 22.

As shown in the plots similar trends are observed for the three samples. Specifically, in a fixed magnetic field, when the external load increases from 0 to 10 N, the resistance reduces in all three samples. With small loads, the resistance changes significantly but it decreases slowly when the load is more than 5 N. According to the absolute values, the sample with higher graphite weight fraction shows the higher electrical conductivity and the smaller decline in resistance. For instance, the resistance of the sample with graphite weight fraction 20% drops from 1862 kΩ at 0 N to 942 kΩ at 10 N, whereas a decline from 4.18 kΩ at 0 N to

0 2 4 6 810 **Load(N)**

experimental 13mT experimental 34.1mT experimental 60.5mT experimental 89.1mT experimental 117.8mT experimental 147.3mT experimental 180.7mT

Besides, the resistance of each sample at a fixed external load decreases with increase in magnetic field intensity. Considering the sample with 21.95% graphite weight fraction as an example, at 5 N external force, the resistance 55.4 kΩ without the magnetic field decreases to

**5.2. A representative volume unit based mathematical model for investigating** 

From the Dipole Model a Representative Volume Unit (RVU) is derived. A RVU consists of two neighbouring hemispheres and the surrounding polymer matrix, which can be regarded as the minimum volume element in the conventional MRE. Fig. 23 shows the position of RVU in carbonyl iron particle chains and a longitudinal section of the unit is

**Figure 24.** The longitudinal section of RVU

The RVU is a model for ideal anisotropic MREs which is supposed to have a chain structure. Given the iron particle volume fraction φ in the RVU, the area of cross section can be expressed as

$$S\_r = \frac{V\_r}{h\_r} = \frac{V\_s}{\phi h\_r} = \frac{4\pi r\_p^{\,^3}}{\Im\phi \left(2r\_p + h\right)} = \frac{2\pi r\_p^{\,^2}}{\Im\phi} \tag{1}$$

where φ is the iron particle volume fraction, *rp* is the particle radius, *h* is the particle distance, *hr* is the distance between two half spheres, *Vs* is the particle volume, *Vr* is volume between two particles, and *Sr* is the cross section area.

In the RVU, the conductivity of the iron particles is much higher than that of the polymer, therefore the electric potential drops within the particles can be neglected. Due to system geometry and electrical properties, most of the current flowing through the RVU concentrates on the small area between the two adjacent hemispheres. Specifically, given the intensity of current flowing through the polymer is *E*, the intensity of the local electric field is

$$E\_{loc} = \frac{h\_r E}{h} = \frac{(2r\_p + h)E}{h} = 2r\_p \frac{E}{h} \tag{2}$$

Owing to the magnetic attraction during MRE preparation, the insulating polymer film between two neighbouring iron particles (here described by *h*) is very thin, and it is across this film that the electrical field induced tunnel current can occur. The Fowler-Nordheim equation [19-21] can be used to express the tunnel current. In addition, the iron particles dispersed in the polymer matrix contribute to the conductivity of the polymer and then the total current density is the sum of the tunnel density *jt* and the conduction density of the polymer *jc*

According to the Fowler-Nordheim equation, the relation between the tunnel density *jt* and electric field intensity *E* is

$$j\_t = \alpha E\_{\rm loc}^2 \exp\left(-\frac{\beta}{E\_{\rm loc}}\right) \tag{3}$$

in which *α* and *β* are constants determining the tunnel current. So the total current density *j* is

$$j = j\_t + j\_c = \alpha E\_{\rm loc}^2 \exp\left(-\frac{\beta}{E\_{\rm loc}}\right) + \sigma\_f E\_{\rm loc} \tag{4}$$

in which *σf* is the conductivity of the polymer film.

272 Smart Actuation and Sensing Systems – Recent Advances and Future Challenges

**Figure 24.** The longitudinal section of RVU

Given the iron particle volume fraction

two particles, and *Sr* is the cross section area.

*r*

*S*

expressed as

where φ

The RVU is a model for ideal anisotropic MREs which is supposed to have a chain structure.

( )

is the iron particle volume fraction, *rp* is the particle radius, *h* is the particle distance,

*p p r s*

π

*V V r r*

φ

*hr* is the distance between two half spheres, *Vs* is the particle volume, *Vr* is volume between

In the RVU, the conductivity of the iron particles is much higher than that of the polymer, therefore the electric potential drops within the particles can be neglected. Due to system geometry and electrical properties, most of the current flowing through the RVU concentrates on the small area between the two adjacent hemispheres. Specifically, given the intensity of current flowing through the polymer is *E*, the intensity of the local electric field is

> (2 ) <sup>2</sup> *<sup>p</sup> <sup>r</sup> loc p h E r hE <sup>E</sup> E r hh h*

Owing to the magnetic attraction during MRE preparation, the insulating polymer film between two neighbouring iron particles (here described by *h*) is very thin, and it is across

== = ≈ <sup>+</sup>

3 2 3

3 2 4 2

 π

φ

<sup>+</sup> == ≈ (2)

in the RVU, the area of cross section can be

(1)

φ

*r r p*

φ

*h h r h*

The total current density *j* is for the current flowing through the small area between the tips of two adjacent iron particles, however, the overall density of RVU *jr* should be derived from its entire cross section. So from the total density of RVU *jr* and the electric field intensity *E*, the global conductivity of typical MRE *<sup>r</sup>* σcan be represented as

$$\sigma\_r = \frac{\dot{j}\_r}{E} = \frac{\pi \cdot r\_i^2 \cdot \dot{j}}{S\_r \cdot E} = 3\phi \cdot r\_i^2 \left[ \frac{2\alpha}{h^2} E \exp\left(-\frac{h\beta}{2r\_p E}\right) + \frac{\sigma\_f}{r\_p h} \right] \tag{5}$$

in which *ri* is the radius of the circular section across which the tunnelling current flow, as shown in Fig. 24.

MRE also exhibits piezoresistivity. When a MRE sample is compressed, its conductivity increases. This phenomenon is explained by two factors, one of which is the increments of the conductive area induced by the deformation of MREs and the other is the reduction of the thickness of the polymer membrane between the two adjacent iron particles. Because of the large ratio of *ri/h*, it is difficult to compress the membrane further and thus the increment of the conductive area is the significant reason for the conductivity increasing.

From the Hertz Theory [22-24], when the initial pressure applied on MRE is <sup>0</sup> σ through a constant loading, corresponding to which there is an initial contact area radius *ri0*

$$r\_{i0} = \left[\frac{3\pi\sigma\_0 \left(1 - \nu^2\right)}{2E\_p}\right]^{1/3} \cdot r\_p \tag{6}$$

Where νis the Poisson's ratio, *Ep* is the Young's Modulus.

So the radius *ri* increases along with the increment of pressure

$$r\_i = r\_{i0} + r\_p \left( \left( \sigma\_0 + \sigma \right)^{1/3} - \sigma\_0^{-1/3} \right) \cdot \left( \frac{3\pi \left( 1 - \nu^2 \right)}{2E\_p} \right)^{1/3} \tag{7}$$

The magnetic field also contributes to the electrical resistance of MREs. When the external magnetic field is applied to MREs, the carbonyl iron particles are attracted by the poles of magnetic field, with closer magnetic pole providing more powerful magnetic force than the other pole. So the attraction from the farther magnetic pole can be neglected.

For the two iron particles in each RVU, the magnetic attraction from the pole applied to the farther particle compresses the thin film between the two adjacent iron particles. Similar to the piezoresistivity, the increment of the conductive area is the main cause for the conductivity increasing.

Thus, the radius *ri* can be updated as

$$r\_i = r\_{i0} + r\_p \left( \left( \sigma\_0 + \sigma\_1 + \sigma\_2 \right)^{1/3} - \sigma\_0^{-1/3} \right) \cdot \left( \frac{3\pi \left( 1 - \nu^2 \right)}{2E\_p} \right)^{1/3} \tag{8}$$

in which, 1 σ is the stress due to pressure, 2 σis the stress from the magnetic attraction.

So the dependence of the conductivity of MREs on electric field intensity and the stress due to pressure is

$$\sigma\_m = 3\phi \cdot \left[ \frac{2\alpha}{h^2} \operatorname{Exp} \left( -\frac{h\beta}{2r\_p E} \right) + \frac{\sigma\_f}{r\_p h} \right] \cdot \left[ r\_{i0} + r\_p \left( \left( \sigma\_0 + \sigma\_1 + \sigma\_2 \right)^{1/3} - \sigma\_0^{-1/3} \right) \cdot \left[ \frac{3\pi \left( 1 - \nu^2 \right)}{2E\_p} \right] \right]^2 \tag{9}$$

When the initial condition <sup>0</sup> σ and *ri0* are set, Apart from E andσ , the other parameters in this equation are all constants. So the conductivity of MREs *<sup>m</sup>* σ is dependent on the intensity of the electric field *E* and the stress due to pressureσ.

In response to the effects of graphite and carbonyl iron on the resistance model, two parameters *<sup>g</sup>* λ and *<sup>i</sup>* λ are introduced to show the effects of graphite volume fraction *<sup>g</sup>* φ and carbonyl iron volume fraction *<sup>i</sup>* ϕ to the conductivity of new MREs. Two parameters *<sup>g</sup>* λ and *<sup>i</sup>* λ show the contribution of the carbonyl iron particles and graphite powder to the resistance respectively. The higher are values for *<sup>g</sup>* λ and *<sup>i</sup>* λ , the less resistance the Gr-MREs have.

In the RVU, the iron particle volume fraction is set as φ = 0.4, which means that the volume of two hemispheres is 40% of the whole volume of RVU.

The shape of Gr MRE samples is fixed. The thickness *l* is *l*= 1mm=0.001m, the diameter *D* is *D*=0.021m, so the cross section area is *A*=π· (*D*/2)2= 0.000346 m2.

274 Smart Actuation and Sensing Systems – Recent Advances and Future Challenges

So the radius *ri* increases along with the increment of pressure

*rr r*

is the Poisson's ratio, *Ep* is the Young's Modulus.

00 0

σσ

other pole. So the attraction from the farther magnetic pole can be neglected.

0 012 0

σσσ

2 *ii p*

σ

2 0 012 0 3 1 <sup>2</sup> 3 exp 2 2

*f*

σ

this equation are all constants. So the conductivity of MREs *<sup>m</sup>*

ϕ

*<sup>h</sup> <sup>E</sup> r r*

2 *ii p*

( ) ( ) ( ) 1/3 <sup>2</sup> 1/3 1/3

 σ

The magnetic field also contributes to the electrical resistance of MREs. When the external magnetic field is applied to MREs, the carbonyl iron particles are attracted by the poles of magnetic field, with closer magnetic pole providing more powerful magnetic force than the

For the two iron particles in each RVU, the magnetic attraction from the pole applied to the farther particle compresses the thin film between the two adjacent iron particles. Similar to the piezoresistivity, the increment of the conductive area is the main cause for the

> ( ) ( ) ( ) 1/3 <sup>2</sup> 1/3 1/3

*p p p*

σσσ

 σ

<sup>−</sup> =+ ++ − ⋅

So the dependence of the conductivity of MREs on electric field intensity and the stress due

*h rE rh E*

and *ri0* are set, Apart from E and

In response to the effects of graphite and carbonyl iron on the resistance model, two

show the contribution of the carbonyl iron particles and graphite powder to the

λ

 <sup>−</sup> =⋅ − + ⋅ + ++ − ⋅ 

<sup>−</sup> =+ + − ⋅

3 1

π ν

*p*

(7)

(8)

2

(9)

φ

λ

π ν

, the less resistance the Gr-MREs

= 0.4, which means that the volume

, the other parameters in

is dependent on the

*E*

3 1

π ν

*p*

is the stress from the magnetic attraction.

( ) ( ) ( ) <sup>2</sup>

1/3 1/3

 σ

σ

to the conductivity of new MREs. Two parameters *<sup>g</sup>*

σ.

are introduced to show the effects of graphite volume fraction *<sup>g</sup>*

 and *<sup>i</sup>* λ

φ

σ

*E*

Where ν

conductivity increasing.

in which, 1

to pressure is

σ

parameters *<sup>g</sup>*

and *<sup>i</sup>* λ

have.

σ

 φ

Thus, the radius *ri* can be updated as

α

When the initial condition <sup>0</sup>

λ

 and *<sup>i</sup>* λ

and carbonyl iron volume fraction *<sup>i</sup>*

*rr r*

is the stress due to pressure, 2

*m i p*

 β

σ

resistance respectively. The higher are values for *<sup>g</sup>*

In the RVU, the iron particle volume fraction is set as

of two hemispheres is 40% of the whole volume of RVU.

intensity of the electric field *E* and the stress due to pressure

$$\text{In the tunneling equation} \left[ \frac{2\alpha}{h^2} E \exp\left( -\frac{h\beta}{2r\_pE} \right) + \frac{\sigma\_f}{r\_p h} \right], \text{ a and } \beta \text{ are pre-exponential and } \varepsilon$$

exponential terms of the standard Fowler-Nordheim Tunneling which are both constants. In this case, value *α* is set as 2, value *β* is set as 1. *h* is the height of the RVU as two times of iron particle's radius. So *h*=0.000004 m. The iron particle's radius *rp* is 0.000002 m. *σf* is the conductivity of the polymer film namely silicone rubber. Because of the high resistance of silicone rubber, the value of conductivity of silicone rubber *σf* is set as 1\*10-10. The electric field *E* is from the function file of the multimeter used in the test. The value of *E* is 9V. So the

whole equation can be calculated as <sup>12</sup> 2 2 exp 2.01 \* 10 2 *f p p <sup>h</sup> <sup>E</sup> h rE rh* α β σ − +≈ . At the normal

condition, the MRE Young's modulus Ep is set as 4000 Pa. The Poisson's ratio of particles ν is set as 0.1. Substitute these parameters into Equation (9), the final resistance of Gr-MREs is

$$R\_g = \frac{4.78 \, ^\circ 10^{-13} \times \lambda\_g \lambda\_i}{10^{-8} + 0.000002 \left( \left( 2000 + \sigma\_1 + \sigma\_2 \right)^{1/3} - 2000^{1/3} \right) \times 0.001166} \tag{10}$$

where *σ1* is the compressive stress from the external force and *σ2* is the compressive stress from the external magnetic field [25]. In this study, the weight of plastic plate is 10 g and the weight increment is 110 g. Thus, the force increment is 1 N. The stress, *σ1*, is then calculated by dividing the force with the area A (0.000346 m2), and shown in Table 2.


**Table 2.** Relation of weight, load and compressive stress

The anisotropic MRE with graphite weight fraction 21.95% (contains 10g iron particles, 3g silicone rubber, 3g silicone oil and 4.5g graphite powder) can be used as an example. Table 3 shows the volume fractions of all the ingredients for this sample.


**Table 3.** Volume fractions of all the ingredients (anisotropic MRE Gr 21.95%)

The factors showing the efforts of graphite and iron particles to the resistance in Gr-MREs are *λg* and *λi* respectively and *λg* and *λi* depend on the graphite volume fraction *<sup>g</sup>* φ and iron particles volume fraction *<sup>i</sup>* φ . For each sample there are different values for *λg* and *λi*. Fig. 25 shows the product of *λg* and *λi* versus Gr-MREs (Gr 20%, 21.95% and 23.81%). The data is from the ratio of experimental result and theoretical prediction.

If we set exp( ) *<sup>g</sup> <sup>g</sup>* λ = ⋅ *a* φ and exp( ) *i i* λ = ⋅ *b* φ and substitute the data in Table 4 and Fig. 8 to *λ<sup>g</sup>* and *λi*, the parameters *λg* and *λi* can be obtained and so the parameter *a* and *b* were derived as -65 and 250, respectively. So the relationship of *λg* and *λi* to *<sup>g</sup>* φ and *<sup>i</sup>* φ are 65\* *<sup>g</sup> <sup>g</sup> e* ϕ λ <sup>−</sup> = , 250 *<sup>i</sup> <sup>i</sup> e* ϕ λ=


**Table 4.** Volume fractions of iron and graphite (anisotropic MRE Gr 20%, 21.95% & 23.81%)

After having determined all parameters, Equation 10 is ready to calculate the final resistance *Rg* of the anisotropic MRE with graphite weight fraction 20%.
