**2. Anatomy: Spine analysed as an intrinsically optimal structure**

## **2.1 Spinal vertebral body (an intrinsically efficient load-bearer)**

The spinal vertebral body (VB) geometry resembles a hyperboloid (HP) shell (fig 2) which is loaded by compressive and torsional loadings, portrayed in fig 1 as resolved into component forces along its generators.

Fig. 1. Vertebral body, a hyperboloid (HP) shell formed of 2 sets of generators [3].

### **Stress analysis of the VB under Axial Compression [3]:**

2 Biomedical Science, Engineering and Technology




Finally, we have also shown the application of this concept and methodology to hospital management. There is a considerable (and hitherto under developed) scope for application of Industrial Engineering discipline for effective hospital administration, in the form of how to determine and allocate hospital budget to optimise the functional performances of all the hospital departments. This leads us to what can be termed as the *Hospital Management* 

Herein, we have shown how to formulate a performance index (PFI) for ICU. This index divided by the Resource index gives us the cost-effectiveness index (CEI). The Management strategy is to maintain certain acceptable values of both PFI and CEI for all hospital departments, by judicious allocation of staff to the departments. This enables the determination of the Optimal Resource index (RSI) and hospital budget (HOB) to maintain a balance between PFI and CEI for all the hospital departments. This can constitute the basis

The spinal vertebral body (VB) geometry resembles a hyperboloid (HP) shell (fig 2) which is loaded by compressive and torsional loadings, portrayed in fig 1 as resolved into

**2. Anatomy: Spine analysed as an intrinsically optimal structure** 

Fig. 1. Vertebral body, a hyperboloid (HP) shell formed of 2 sets of generators [3].

**2.1 Spinal vertebral body (an intrinsically efficient load-bearer)** 


elastance (*Eao*, to characterize the LV systemic load).

patients;

*System*.

becoming diabetic;

of Hospital Management.

component forces along its generators.

We now analyse for stresses in the HP shell (generators) due to a vertical compressive force *P*, as shown in figures 3 and 4.Assume that there are two sets of '*n'* number of straight bars placed at equal spacing of (*2πa/n*) measured at the waist circle, to constitute the HP surface, as shown in figure 3 (right). Due to the axi-symmetric nature of the vertical load, no shear stresses are incurred in the shell, i.e. σφθ= *0,* as in figure 3 (left)*.* We then delineate a segment of the HP shell, and consider its force equilibrium (as illustrated in figure 4), to obtain the expressions for stresses *N*φ *and N*θas depicted in figure 4.

$$r\_o = r\_2 \text{ (sum} \text{\textdegree)}$$

$$r\_1 = -\left(\frac{b^2}{a^4}\right)r\_2^3$$

$$\frac{\left(\frac{\nu^2}{a^2} + \nu^2\right)}{a^2} - \frac{z^2}{b^2} = 1$$

$$At \ x = 0, \quad \frac{\left(\nu^2\right)}{a^2} - \frac{z^2}{b^2} = 1$$

$$z = \pm \left(\frac{b}{a}\right) \nu$$

Fig. 2. Geometry of a Hyperboloid (HP) shell**.** In the figure z = b, and y = a. We define tan β = a/b [3].

Fig. 3. Stress Analysis for Vertical Loading: Stresses at the waist section of the VB HP Shell: (a) stress components (b) equivalent straight bars aligned with the generators) placed at equal spacing to take up the stresses. In fig 3 (left) due to axi-symmetric vertical load, no shear stresses are incurred in the shell, i.e. σφθ= *0***.** In fig 3 (b), there are 2 sets of '*n'* number of straight bars placed at equal spacing of (*2πa/n*) measured at the waist circle, to constitute the generators of the HP surface [3].

### **Equilibrium of Forces on a Shell Segment under Vertical load P:**

Fig. 4. Equilibrium of Forces on a Shell Segment: Analysis for stresses *N*φ *and N*θ due to the vertical force *P* [3,2].

Then based on the analysis in Fig 5, we obtain the expression for the equivalent resultant compressive forces *C* in the fibre-generators of the VB HP shell. Thus it is seen that the total axial loading is transmitted into the HP-shell's straight generators as compressive forces. It is to be noted that the value of *C* is independent of dimensions R and a.

Fig. 3. Stress Analysis for Vertical Loading: Stresses at the waist section of the VB HP Shell: (a) stress components (b) equivalent straight bars aligned with the generators) placed at equal spacing to take up the stresses. In fig 3 (left) due to axi-symmetric vertical load, no

of straight bars placed at equal spacing of (*2πa/n*) measured at the waist circle, to constitute

 *0***.** In fig 3 (b), there are 2 sets of '*n'* number

φ *and N*θ

due to the

σφθ=

**Equilibrium of Forces on a Shell Segment under Vertical load P:** 

Fig. 4. Equilibrium of Forces on a Shell Segment: Analysis for stresses *N*

is to be noted that the value of *C* is independent of dimensions R and a.

Then based on the analysis in Fig 5, we obtain the expression for the equivalent resultant compressive forces *C* in the fibre-generators of the VB HP shell. Thus it is seen that the total axial loading is transmitted into the HP-shell's straight generators as compressive forces. It

shear stresses are incurred in the shell, i.e.

*r*

β

*p*

the generators of the HP surface [3].

At the waist (ro= a),

, <sup>2</sup> *<sup>P</sup> <sup>N</sup>*

π

φ

Hence,

θ

*a*

Now since, 1 2

2

⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

*aPP <sup>N</sup>*

vertical force *P* [3,2].

<sup>=</sup> compressive

*N N*

 θ

tan

 π

− =

*r r* φ

2 2

2 2

*b a a*

π

Fig. 5. Equivalent compressive force C in the generators (corresponding to the stressresultants acting on the shell element) equilibrating the applied axial loading [3,2].

It can be noted that the value of C is independent of dimensions R and a.

### **Stress analysis under Torsional loading [3]:**

Next, we analyse the compressive and tensile forces in the HP shell generators when the VB is subjected to pure torsion (*T*).In this case (referring to **fig. 6)**, the normal stress resultants are zero, and we and only have the shear stress-resultant , as given by

$$N\_{\phi} = N\_{\theta} = 0 \qquad \text{and} \quad N\_{\phi\theta} = \mathbf{r}\mathbf{t}$$

The equilibrium of a segment of the shell at a horizontal section at the waist circle (depicted in **figure** 6 **a**) gives the shear stress-resultant as follows:

$$[(\pi \cdot t)(2\pi a)]a = T \text{ , i.e., } N\_{\phi\theta} = \frac{T}{2\pi a^2}$$

τ*)* 

**Stress Analysis for Torsional Loading M** 

Fig. 6. Stress analysis of the vertebral body under torsional loading [3].

Now, we consider an element at the waist circle as shown in **figure** 7. The equivalent compressive force (*FcT*) and tensile force (*FtT*), in the directions aligned to their respective set of shell generators, are given by

$$\begin{aligned} \mathbf{F}\_{cT}^2 = \mathbf{F}\_{IT}^2 = \left(\mathbf{N}\_{\phi\theta} \frac{\pi a}{n}\right)^2 + \left(\mathbf{N}\_{\phi\theta} \frac{\pi b}{n}\right)^2 \\\\ \text{or,} \\\\ \mathbf{F}\_{cT} \end{aligned} \qquad \left| \mathbf{F}\_{cT} \right| = \left| \mathbf{F}\_{IT} \right| = \frac{T}{2na\sin\beta}$$

wherein FcT andFtT are depicted as c and T respectively in figure 7.

Fig. 7. Analysis of equilibrium of a shell element comprising of two intersecting generators: Expressions for tensile forces T and compressive forces C in the generators, indicate that torsion loading is also transmitted as axial compressive and tensile forces through the generators of the VB, which makes it a naturally optimum (high-strength and light-weight) structure [3].

For equilibrium, ( ) 2 () π τ*a ta M* ⋅ =

> <sup>2</sup> 2 *M a t*

<sup>2</sup> 2 *<sup>M</sup> <sup>N</sup>*

π=

*a*

*Stresses in the HP shell* 

*due to torsion M acting on the* 

2 2 2 1/2 2 ( cot ) 2

θ

*na <sup>M</sup> TorC*

θ

θ

2 sin cos 2

*na M ec na*

=

=

<sup>+</sup> <sup>=</sup>

*Ma a*

σφθ*=* τ*)* 

*element (*

*VB* 

2 2

φθ

β

 π

*n n*

⎛ ⎞⎛ ⎞

*na*

σφ*=*σθ*=0 and* 

π=

Now, we consider an element at the waist circle as shown in **figure** 7. The equivalent compressive force (*FcT*) and tensile force (*FtT*), in the directions aligned to their respective set

*a b FF N N*

π

== + ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= =

φθ

*<sup>T</sup> F F*

Fig. 7. Analysis of equilibrium of a shell element comprising of two intersecting generators: Expressions for tensile forces T and compressive forces C in the generators, indicate that torsion loading is also transmitted as axial compressive and tensile forces through the generators of the VB, which makes it a naturally optimum (high-strength and light-weight)

τ

φθ

Fig. 6. Stress analysis of the vertebral body under torsional loading [3].

2 2 *cT tT*

wherein FcT andFtT are depicted as c and T respectively in figure 7.

or, 2 sin *cT tT*

**Stress Analysis for Torsional Loading M** 

*Equilibrium of a shell segment under torsion (M) and shear* 

of shell generators, are given by

*) (or shear stress-*

*stresses (*

*resultant N*

structure [3].

τ

φθ*)*  Thus, a torsional loading on the VB HP shell is taken up by one set of generators being in compression and the other set of generators being in tension.

Equilibrium of a Shell Element Comprising of Two Intersecting Generators: The equivalent compressive forces (*C*) and tensile forces (*T*); in the shell generators (required to equilibrate the applied load), are given by:

$$|T| = |C| = \sqrt{\left(\tau \cdot t\left(\frac{\pi a}{n}\right)\right)^2 + \left(\tau \cdot t\left(\frac{\pi b}{n}\right)\right)^2} = \tau \cdot t\left(\frac{\pi}{n}\right)\sqrt{a^2 + b^2} = \frac{M(a^2 + b^2)^{\frac{1}{2}}}{2na^2}$$

### **2.2 Spinal disc optimal design (to bear loading with minimal deformation and maximal flexibility)**

Fig 8 illustrates how the mechanism of how the spinal disc bears compression without bulging. It is seen that the nucleus pulposus plays the key role in this mechanism, as will be explained further in this section. Hence the absence of it in a denucleated disc causes the disc to bulge.

Fig. 8. Mechanism of how the spinal disc bears compression without bulging.

Fig 9 displays the geometry and the deformation variables of the spinal disc. We now present the elasticity analysis of the disc to first obtain the radial, circumferential and axial stresses in terms of the disc deformations and annulus modulus *E*[4].

We next carry out the stress analysis of the disc under vertical loading *P* (fig 10), to obtain the expressions (equations 10) for the stresses in the disc annulus in terms of the load *P* , pressure p in the nucleus-pulposus, and the disc dimensions (a and b) [4].

We then derive the expressions for the disc axial and radial deformations *<sup>u</sup>* δ and δ*<sup>h</sup>*in terms of nucleus pulposus pressure *p*, the annulus modulus *E* and the disc dimensions, as given by equations (17) and (18). Now the annulus modulus *E* is a function of the stress in the annulus (*k* being the constitutive proportionality constant), and hence of the pressure *p* and the disc dimensions, as shown by equation (21). As a result of this relationship, we finally show that the disc deformations are only functions of *k* and the disc dimensions. This implies that irrespective of the increase in the value of load *P*, the disc deformations remain constant, and only depend on the constitutive property parameter *k*. This is the novelty of the intrinsic design of the spinal disc!

Fig. 9. Geometry and Deformation Variables of the Spinal Disc [4].

### **ANALYSIS**

The equilibrium equation is:

$$
\sigma\_r - \sigma\_\theta + r \frac{d\theta\_r}{dr} = 0 \tag{1}
$$

Strain-displacement relations: *<sup>r</sup>*

$$
\varepsilon\_r = \frac{du}{dr}, \varepsilon\_\theta = \frac{u}{r} \,\,\,\,\,\tag{2}
$$

Constitutive relations:

$$
\sigma\_r = E \in \underline{\mathbf{E}} = E \frac{d\underline{u}}{dr}, \; \sigma\_\theta = E \in \underline{\mathbf{e}}\_\theta = E \frac{\underline{u}}{r} \tag{3}
$$

Substituting eqn (3) into eqn (1), we have:

$$\frac{d^2u}{dr^2} + \frac{1}{r}\frac{du}{dr} - \frac{u}{r^2} = 0 \; \; \; \; u = Ar + \frac{B}{r} \tag{4}$$

Because of the incompressible nucleus pulposus fluid inside:

$$
\pi a^2 h = \pi (a + \mathcal{S}\_u)^2 (h - \mathcal{S}\_h) \tag{5}
$$

the (radial and axial) deformations *u* and *h* are related as:

$$
\delta\_{\rm li} = \frac{\pi a^2 \delta\_{\rm li}}{2\pi ab \hbar} = \left(\frac{a}{2\hbar}\right) \delta\_{\rm li} \tag{6}
$$

Now, we designate:

$$\left.u\right|\_{r=a} = \left.\mathcal{S}\_u, \sigma\_r\right|\_{r=b} = 0\tag{7}$$

Substituting eqn (7) into eqns (3 & 4), we get

$$\begin{aligned} Aa + \frac{B}{a} &= \delta\_u \\ E\left[A - \frac{B}{b^2}\right] &= 0 \end{aligned} \tag{8}$$

 So that

8 Biomedical Science, Engineering and Technology

by equations (17) and (18). Now the annulus modulus *E* is a function of the stress in the annulus (*k* being the constitutive proportionality constant), and hence of the pressure *p* and the disc dimensions, as shown by equation (21). As a result of this relationship, we finally show that the disc deformations are only functions of *k* and the disc dimensions. This implies that irrespective of the increase in the value of load *P*, the disc deformations remain constant, and only depend on the constitutive property parameter *k*. This is the novelty of

> 0 *<sup>r</sup> <sup>r</sup> d r dr*

θ

*r* ∈ =θ

−+ = (1)

*r*

*r*

 θ

, (2)

= ∈= (3)

= + (4)

⎝ ⎠ (6)

θ

*du dr* ∈ = , *<sup>u</sup>*

= ∈= , *<sup>u</sup> E E*

σθ

+ −= , *<sup>B</sup> u Ar*

(5)

*ah a h* =+ −

2 2 ( )( ) *u h*

2 2 *h u h a a ah h*

⎛ ⎞ = = ⎜ ⎟

 δ

 δ

2

π

π δ

 πδ

δ

π

σ σ

*r r*

σ

2

Because of the incompressible nucleus pulposus fluid inside:

the (radial and axial) deformations *u* and *h* are related as:

*du E E dr*

2 2 <sup>1</sup> <sup>0</sup> *d u du u dr r r dr*

the intrinsic design of the spinal disc!

**ANALYSIS** 

The equilibrium equation is:

Constitutive relations:

Strain-displacement relations: *<sup>r</sup>*

Substituting eqn (3) into eqn (1), we have:

Fig. 9. Geometry and Deformation Variables of the Spinal Disc [4].

$$A = \frac{a\mathcal{S}\_u}{a^2 + b^2}, \quad B = \frac{a\mathcal{S}\_u b^2}{a^2 + b^2} \text{ in the 'u' function} \tag{9}$$

Then, substituting A and B into eqns (4 & 3), we obtain:

$$\mathbf{-} \quad \text{for the radial stress,} \qquad \sigma\_I = \frac{\mathrm{Ea}\left(r^2 - b^2\right)\delta u}{(a^2 + b^2)r^2} = \frac{a^2\left(r^2 - b^2\right)}{2h(a^2 + b^2)r^2} \mathrm{E}\delta h \tag{10-a}$$

$$\mathbf{-} \quad \text{for the circumferential stress,} \quad \sigma \theta = \frac{\mathrm{E}a\left(r^2 + b^2\right)\delta\_{\mathrm{H}}}{(a^2 + b^2)r} = \frac{a^2\left(r^2 + b^2\right)}{2\mathrm{H}(a^2 + b^2)r} \mathrm{E}\delta \mathbf{b} \tag{10-b}$$


Once σ*z* is evaluated, δ*h* will become known (from eqn. 10-c) and subsequently σ*r*,σ*<sup>θ</sup>* (from eqns 10-a & 10-b) and *δu*(from eqn 10-c).

### **Stress Analysis for Vertical Loading [4]:**

For a vertically applied force *P*,

$$P = \left(\pi a^2\right)\sigma\_f + \pi (b^2 - a^2)\sigma\_z \tag{11}$$

where *f* is the hydrostatic pressure in the NP fluid, and *z* the axial stresses in the annulus. Because the disc height (*h*) is small, σ*<sup>f</sup>*≈ *constant, and hence* 

$$\left. \sigma\_f = -\sigma\_r \right|\_{r=a} = p \tag{12}$$

Then, based on eqns (12, 10-a & 10-c), we obtain

$$\sigma\_f = -\frac{\left(a^2 - b^2\right)}{2h\left(a^2 + b^2\right)} E \delta\_h = \frac{\left(b^2 - a^2\right)}{2h\left(a^2 + b^2\right)} E\left(\frac{\delta h}{h}\right), \text{ or } \sigma\_f = \frac{\left(b^2 - a^2\right)}{2\left(a^2 + b^2\right)} \sigma\_z = p \tag{13}$$

**Normal stresses** σ**f &** σ**z equilibrating the applied force P** 

Fig. 10. Induced Stresses in the disc annulus and Pressure *p* in the nucleus-pulposus in response to the load *P* [4].

Substituting

$$\sigma \sigma \prime = \frac{\left(b^2 - a^2\right)}{2\left(a^2 + b^2\right)}, \quad \sigma\_z = p \prime \sigma \sigma = E \frac{\delta h}{h} = 2E \frac{\delta u}{a}$$

$$\text{Into}\tag{1}$$

$$P = (\pi a^2)\sigma\gamma + \pi(b^2 - a^2)\sigma\gamma$$

We get: 
$$\delta\_h = \frac{h\sigma\_z}{E} = \frac{h}{E} \left. \frac{P}{\pi(b^2 - a^2)} \right[ \frac{2(a^2 + b^2)}{3a^2 + 2b^2}] \tag{14}$$

$$\mathcal{S}\_{\text{ul}} = \frac{a}{2h} \mathcal{S}\_{\text{h}} = \frac{a}{2E} \quad \frac{P}{\pi (b^2 - a^2)} \left[ \frac{2(a^2 + b^2)}{3a^2 + 2b^2} \right] \tag{15}$$

$$\text{Nucleus-pulposus fluid pressure}, \qquad p = \frac{P}{\pi (3a^2 + 2b^2)}\tag{16}$$

$$\sigma\_z = \frac{2p(a^2 + b^2)}{(b^2 - a^2)} = \frac{2P(a^2 + b^2)}{\pi(b^2 - a^2)(3a^2 + 2b^2)}\tag{16-a}$$

$$\mathcal{S}\_{\text{lt}} = \frac{h\sigma\_z}{E} = \frac{h}{E} \left. \frac{P}{\pi(b^2 - a^2)} \right| \left. \frac{2(a^2 + b^2)}{3a^2 + 2b^2} \right|$$

### **Stress in the Annulus**

$$\text{From,}\\
\qquad \qquad \sigma \theta = \frac{\text{E}a\left(r^2 + b^2\right)\delta\_{\text{H}}}{(a^2 + b^2)r} = \frac{a^2\left(r^2 + b^2\right)}{2\text{h}(a^2 + b^2)r} \text{E}\delta n$$

$$\text{We get : } \begin{aligned} a^2 \left( r^2 + b^2 \right) \mathbf{P} &= \frac{pa^2 \left( r^2 + b^2 \right)}{r^2 \left( b^2 - a^2 \right) \left( 3a^2 + 2b^2 \right)}, \end{aligned} \tag{17}$$

$$\sigma\_{\theta} \left( \mathbf{r} = \mathbf{a} \right) = \frac{p \left( a^2 + b^2 \right)}{\left( b^2 - a^2 \right)}, \text{ in the annulus} \tag{18}$$

wherein, ( ) 2 2 , 3 2 *P p* π *a b* = + pressure in nucleus-pulposus fluid

### **The disc deformations have been obtained as:**

$$\text{Axial deformation}, \qquad \delta\_{\hbar} = \frac{2ph(a^2 + b^2)}{E\_c(b^2 - a^2)} \tag{19}$$

$$\text{Radial deformation}\tag{20}$$

$$\delta\_u = \frac{pa(a^2 + b^2)}{E\_c(b^2 - a^2)}\tag{20}$$

wherein Ec = E - Eo = kσ.

$$\text{Let } E\_c = k \sigma\_\theta (r = a) = k p \frac{\left(a^2 + b^2\right)}{\left(b^2 - a^2\right)} \tag{21}$$

Now, as the magnitude of the load *P* increases, the pressure *p* in nucleus-pulposus fluid also increases. Then, as p increases, so does the modulus Ec according to eqn (21)

$$\dots \text{ : } \frac{p}{E\_c} = \frac{\left(b^2 - a^2\right)}{k\left(a^2 + b^2\right)} = \text{constant} \tag{23}$$

$$\delta \delta\_h = \frac{2ph(a^2 + b^2)}{E\_c(b^2 - a^2)} = \frac{2h}{k} \text{, a constant, and } \begin{aligned} \delta\_\mu = \frac{pa(a^2 + b^2)}{E\_c(b^2 - a^2)} = \frac{2a}{k} \text{, a constant.} \end{aligned}$$

This means that, irrespective of increase in the value of load P, the disc deformations *<sup>u</sup>* δ and *<sup>h</sup>* δ remain constant, and only depend on the constitutive property parameter k. This is the novelty of the intrinsic design of the spinal disc!

10 Biomedical Science, Engineering and Technology

**Normal stresses** σ**f &** σ**z equilibrating the applied force P** 

*<sup>z</sup>* <sup>=</sup> *<sup>p</sup>* , 2 *h u <sup>z</sup> E E*

*b*

*-*σ*z*

*σf*

 σ

2( )

2 2

2( )

22 2 2

− + ⎢⎣ ⎥⎦

22 2 2

− + ⎢⎣ ⎥⎦

( )3 2

σ= =

*f z*

δ

*h a*

 δ *h*

(14)

(15)

(16-a)

(16)

Fig. 10. Induced Stresses in the disc annulus and Pressure *p* in the nucleus-pulposus in

 σ

πσ

π

2 2 ( )3 2 *u h a a P ab h E ba a b*

π

 π

*h h P ab E E ba a b*

<sup>⎡</sup> <sup>+</sup> <sup>⎤</sup> = = <sup>⎢</sup> <sup>⎥</sup>

<sup>⎡</sup> <sup>+</sup> <sup>⎤</sup> = = <sup>⎢</sup> <sup>⎥</sup>

*P*

2 2 2 2 22 22 2 2 2( ) 2( ) ( ) ( )(3 2 ) *<sup>z</sup> pa b Pa b ba ba a b*

π

<sup>+</sup> <sup>+</sup> = = − −+

( ) ( )

−

*b a a b*

2 *f*

We get: 2 2

*<sup>z</sup> <sup>h</sup>*

σ

> δ

> > *p* π*a b* <sup>=</sup> <sup>+</sup>

=

Into 2 22 *P a ba* = +− () ( )

δ

δ

Nucleus-pulposus fluid pressure, 2 2 (3 2 )

σ

σ

2 2 2 2 ,

+

response to the load *P* [4].

Substituting

### **3. Physiology: Mechanism of left ventricle twisting and pressure increase during isovolumic contraction (due to the contraction of the myocardial fibres)**

**Introduction and objective**: The left ventricular (LV) myocardial wall is made up of helically oriented fibers. As the bioelectrical wave propagates along these fibers, it causes concomitant contraction wave propagation. Our LV cylindrical model is illustrated in figure 11. The contraction of the helical oriented myocardial fibers causes active twisting and compression of the LV (as illustrated in fig 11), thereby compressing the blood fluid contained in it. Then due to the very high bulk modulus of blood, this fluid compression results in substantial pressure increase in the LV cavity.

Herein we simulate this phenomenon of LV isovolumic contraction, which causes the intra-LV pressure to rise so rapidly during 0.02-0.06 seconds of isovolumic contraction. Our objective is to determine how the pressure generated during isovolumic contraction, due to by active torsion (with LV twisting) and compression (with LV shortening) caused by the contractile stress in the helically wound myocardial fibers [5].

Fig. 11. Top: Fiber orientation and twisting model of the left ventricle (LV). Bottom: The fluid-filled LV cylindrical shell model: (i) geometry (ii) material property, and (iii) equivalent compression Δ*F* and Δ*T* associated with its internal stress state due to internal pressure rise within it.

**Concept**: In order to simulate the left ventricle deformation during isovolumic contraction, we have modeled it as a pressurized fluid-filled thick-walled cylindrical shell supported by the aorta along its upper edge. The LV cylindrical model consists of an incompressible hyperelastic material with an exponential form of the strain energy function ψ.

The contraction of the myocardial fibers causes active twisting and compression of the left ventricle, thereby compressing the blood fluid contained in it. Then, due to the very high bulk modulus of blood, this compression results in pressure increase in the ventricular cavity. Hence, we simulate this phenomenon by applying and determining equivalent active torque and compression to the LV-cylindrical model incrementally (as ΔF and ΔT), so as to raise the LV pressure by the monitored amounts.

## **Modeling approach:**

12 Biomedical Science, Engineering and Technology

Herein we simulate this phenomenon of LV isovolumic contraction, which causes the intra-LV pressure to rise so rapidly during 0.02-0.06 seconds of isovolumic contraction. Our objective is to determine how the pressure generated during isovolumic contraction, due to by active torsion (with LV twisting) and compression (with LV shortening) caused by the

Fig. 11. Top: Fiber orientation and twisting model of the left ventricle (LV). Bottom: The fluid-filled LV cylindrical shell model: (i) geometry (ii) material property, and (iii) equivalent compression Δ*F* and Δ*T* associated with its internal stress state due to internal

**Concept**: In order to simulate the left ventricle deformation during isovolumic contraction, we have modeled it as a pressurized fluid-filled thick-walled cylindrical shell supported by the aorta along its upper edge. The LV cylindrical model consists of an incompressible

hyperelastic material with an exponential form of the strain energy function ψ.

pressure rise within it.

**3. Physiology: Mechanism of left ventricle twisting and pressure increase during isovolumic contraction (due to the contraction of the myocardial fibres) Introduction and objective**: The left ventricular (LV) myocardial wall is made up of helically oriented fibers. As the bioelectrical wave propagates along these fibers, it causes concomitant contraction wave propagation. Our LV cylindrical model is illustrated in figure 11. The contraction of the helical oriented myocardial fibers causes active twisting and compression of the LV (as illustrated in fig 11), thereby compressing the blood fluid contained in it. Then due to the very high bulk modulus of blood, this fluid compression

results in substantial pressure increase in the LV cavity.

contractile stress in the helically wound myocardial fibers [5].

We monitor LV pressure (*p*), LV volume (V), myocardial volume (*VM*), and wall thickness (h) at time intervals during isovolumic contraction. From the monitored *V, VM,* and *h*, we determine the LV model radius *R* and length *L* and the wall thickness *h.* We also monitor LV twist angle ϕ.

We then invoke blood compressibility to determine *∆V* at subsequent instants , as *∆V = (∆p/K) V,* where *∆p* is the monitored incremental pressure and *K* is the bulk modulus of blood. From the volume strain *∆V/V*, we then determine the model length and radius strains ∆I/L and ∆r/R, and hence the LV dimensions with respect to LV dimensions at the start of isovolumic contraction. This enables us to determine the stretches (strains) (*λz, λr, λθ* and *γ),* and thereafter the Lagrange strain tensor components (*Err, Eθθ, Ezz, Eθz*) in terms of these stretches and the hydrostatic pressure.

Then, we express the LV wall stresses in terms of the strain energy density function *Ψ,* of the Lagrange strain tensor components in cylindrical coordinates and the material constitutive parameters *(bi)*, in which (i) the stretches (*λz, λr, λθ* and *γ)* have been calculated and are known*,* (ii) the hydrostatic pressure and the constitutive parameters *bi* (*i* = 1*,* 2*, . . . ,* 9) are the unknowns.

So now, we substitute the stress expressions *σrr* and *σθθ* into the boundary conditions equations (of equilibrium between the internal pressure and the wall stresses *σrr* and *σθθ*, and between the internal pressure and wall stress *σzz)*, and determine the best values of the constitutive parameters (*bi*) and the hydrostatic pressure to satisfy these equations.

We then go back, and determine the stress expressions. We utilize the stress expressions for *σzz* and *σθz*, to determine the generated values of torsion (Δ*T*) and axial compression (Δ*F*),due to the contraction of the helically wound myocardial fibres.

Finally, we determine the principal stresses and principal angle along the radial coordinate of the LV wall thickness, from which we can interpret the fibre orientations, which can be related to the LV contractility index.

This procedure is carried out at three instants of time from the start of isovolumic contraction, and at 0.02 s, 0.04 s, 0.06 s into the isovolumic contraction phase. Hence, from the monitored LV *∆p* and computed *∆V* at these three instants (with respect to the pressure and volume at t = 0 at the start of isovolumic contraction phase, we determine (i) the time variation of the internally generated torque and axial compression during the isovolumic contraction phase (fig 12), as well as (ii) the time variations of the principal (tensile) stress and the principal angle (taken to be equivalent to the fiber angle) during the isovolumic contraction phase (fig 13).

### **Model Kinematics:**

We model the LV as an incompressible thick-walled cylindrical shell subject to active torsion torque and compression as illustrated in fig 11. The upper end of the LV model is constrained in the long-axial direction to represent the suspension of the left ventricle by the aorta at the base. Now, considering the LV at end-diastole to be in the unloaded reference configuration, the cylindrical model in its undeformed state is represented geometrically in terms of cylindrical coordinates (*R*, Θ, *Z*) by

$$R\_{\mathbf{i}} \le R \le R\_{o'} \quad 0 \le \Theta \le 2\pi \quad 0 \le Z \le L \tag{1}$$

where *Ri*, *Ro* and *L* denote the inner and outer radii, and the length of the undeformed cylinder, respectively.

In terms of cylindrical polar coordinates (*r, θ, z*), the geometry of the deformed LV configuration (with respect to its undeformed state at the previous instant) is given by:

$$r\_l \le r \le r\_{o'} \quad 0 \le \theta \le 2\pi, \quad 0 \le \mathbf{z} \le l \tag{2}$$

where *ri*, *ro* and *l* denote the inner and outer radii, and the length of the deformed cylinder, respectively.

We further consider the incompressible LV model in its reference state to be subjected to twisting, radial and axial deformations in the radial and long-axis directions during isovolumic contraction, such that (also based on incompressibility criterion), the deformations of incompressible LV cylindrical shell can be expressed as

$$r = \sqrt{\frac{R^2 - R\_i^2}{\lambda\_z} + r\_i^2}, \ \theta = \Theta + Z\frac{\phi}{L}, \ z = \lambda\_z Z \tag{3}$$

where *λz* is the constant axial stretch, *ri* is the inner radius in the deformed configuration and φ is the measured angle of twist at the apex of the LV (relative to the base). It can be seen that the twist angle (*θ*) and the axial deformation (z) are zero at the upper end of the LV.

### **Model Dimensions:**

At any instant (t), the geometrical parameters (or dimensions) of the LV cylindrical model (instantaneous radius R and length L, as defined in fig 11) can be determined in terms of the monitored LV volume (V), myocardial volume (*VM*) and wall thickness (*h*), as follows:

$$R\_i = \frac{2Vh \;/\; V\_M + \sqrt{\left(2Vh \;/\; V\_M\right)^2 + 4Vh^2 \;/\; V\_M}}{2} \tag{4-a}$$

$$\mathbf{L} = \mathbf{V} \;/\, \pi \mathbf{R}\_i^{\;2} \tag{4-b}$$

$$\text{Then}\\
\text{r} \tag{4.6} \\
\text{r} \tag{4.7} \\
\text{s} \tag{4-\text{c}} \\
\text{r} \tag{4-\text{c}}$$

These equations will be employed to determine the LV dimensions at the start of isovolumic contraction phase (t = 0). The determination of the dimensions of the deformed LV (due to contraction of the myocardial fibers) at the subsequent instants of the isovolumic contraction phase is indicated in the next subsection. We also utilize the information on the LV twist angle (φ ) during the isovolumic phase, from MRI myocardial tagging. From this information, we can determine the stretches (*λz, λr, λθ* and *γ).*

### **Theoretical Analysis:**

14 Biomedical Science, Engineering and Technology

constrained in the long-axial direction to represent the suspension of the left ventricle by the aorta at the base. Now, considering the LV at end-diastole to be in the unloaded reference configuration, the cylindrical model in its undeformed state is represented geometrically in

*R RR i o* ≤ ≤ ≤Θ≤ ≤ ≤ , 0 2 , 0 Z

where *Ri*, *Ro* and *L* denote the inner and outer radii, and the length of the undeformed

In terms of cylindrical polar coordinates (*r, θ, z*), the geometry of the deformed LV configuration (with respect to its undeformed state at the previous instant) is given by:

> , 0 2 , 0 z *i o r rr* ≤≤ ≤ ≤ ≤≤ θ π

where *ri*, *ro* and *l* denote the inner and outer radii, and the length of the deformed cylinder,

We further consider the incompressible LV model in its reference state to be subjected to twisting, radial and axial deformations in the radial and long-axis directions during isovolumic contraction, such that (also based on incompressibility criterion), the

<sup>2</sup> , , *<sup>i</sup>*

*R R r r Zz Z*

θ

where *λz* is the constant axial stretch, *ri* is the inner radius in the deformed configuration and

At any instant (t), the geometrical parameters (or dimensions) of the LV cylindrical model (instantaneous radius R and length L, as defined in fig 11) can be determined in terms of the monitored LV volume (V), myocardial volume (*VM*) and wall thickness (*h*), as follows:

> ( )<sup>2</sup> <sup>2</sup> 2/ 2/ 4 / 2 *M M M*

> > <sup>2</sup> *LV R* = /π

Then, . *R Rh o i* = + (4-c) These equations will be employed to determine the LV dimensions at the start of isovolumic contraction phase (t = 0). The determination of the dimensions of the deformed LV (due to contraction of the myocardial fibers) at the subsequent instants of the isovolumic contraction phase is indicated in the next subsection. We also utilize the information on the LV twist

) during the isovolumic phase, from MRI myocardial tagging. From this

*Vh V Vh V Vh V*

that the twist angle (*θ*) and the axial deformation (z) are zero at the upper end of the LV.

is the measured angle of twist at the apex of the LV (relative to the base). It can be seen

*i z*

*L* φ

 λ

<sup>−</sup> = + =Θ+ = (3)

+ + <sup>=</sup> (4-a)

*<sup>i</sup>* (4-b)

deformations of incompressible LV cylindrical shell can be expressed as

2 2

*z*

λ

*i*

information, we can determine the stretches (*λz, λr, λθ* and *γ).*

*R*

π

*L* (1)

*l* (2)

terms of cylindrical coordinates (*R*, Θ, *Z*) by

cylinder, respectively.

respectively.

φ

angle (

φ

**Model Dimensions:** 

The strain energy density function suitable for the myocardium material, is given by:

$$\varphi = \mathcal{C}(\exp(\mathcal{Q}) - 1) / \mathcal{Z} \tag{5}$$

wherein Q is a quadratic function of the 3 principal strain-components (to describe 3-d transverse isotropy) in the cylindrical coordinate system, given by:

$$\begin{aligned} Q &= b\_1 \mathbf{E}\_{\theta\theta}^2 + b\_2 \mathbf{E}\_{ZZ}^2 + b\_3 \mathbf{E}\_{RR}^2 + 2b\_4 \mathbf{E}\_{\theta\theta} \mathbf{E}\_{ZZ} + 2b\_5 \mathbf{E}\_{RR} \mathbf{E}\_{ZZ} \\ &+ 2b\_6 \mathbf{E}\_{\theta\theta} \mathbf{E}\_{RR} + 2b\_7 \mathbf{E}\_{\theta Z}^2 + 2b\_8 \mathbf{E}\_{RZ}^2 + 2b\_9 \mathbf{E}\_{\theta R}^2 \end{aligned} \tag{6}$$

wherein bi are non-dimensional material parameters, and Eij are the components of the modified Green-Lagrange strain tensor in cylindrical coordinates (R, Θ, Z). In order to reduce the mathematical complexity of the problem, we assume negligible shear during isovolumic contraction. Thus *ERZ* and *E*θ*<sup>R</sup>* in equation (6) and their corresponding stress components (σ*RZ* and σθ*<sup>R</sup>*) are neglected.

The stress equilibrium equation (in the cylindrical coordinate system) is given by the following equation:

$$\frac{\mathrm{d}\sigma\_{rr}}{\mathrm{d}r} + \frac{\left(\sigma\_{r} - \sigma\_{\theta\theta}\right)}{r} = 0\tag{7}$$

The boundary conditions on the outer and inner surfaces of the LV cylindrical model are given by

$$
\sigma\_{rr}\left(r = r\_o\right) = 0 \quad \lor \quad \sigma\_{rr}\left(r = r\_i\right) = -p \tag{8}
$$

where *p* is the LV pressure acting on the inner surface of the LV model; we employ incremental pressure *∆p* with respect to the LV pressure at t = 0, the start of isovolumic contraction.

By integrating eq (7), the Cauchy radial stress σ*rr* is given by:

$$\sigma\_{rr}\left(\xi\right) = \int\_{\xi}^{r\_o} \left(\sigma\_{rr} - \sigma\_{\theta\theta}\right) \frac{\mathrm{d}r}{r}, \qquad r\_i \le \xi \le r\_o \tag{9}$$

There from, the boundary condition eq (8) of the internal pressure p = -σrr (r=ri) is obtained (by substituting equation 9 into the boundary condition 8), in the form:

$$p\_i = -\int\_{r\_i}^{r\_o} (\sigma\_{rr} - \sigma\_{\theta\theta}) \frac{\mathbf{d}r}{r},\tag{10}$$

Since the valves are closed during isovolumic contraction, we impose another set of boundary conditions (at both the top and bottom of the internal LV surface), giving:

$$
\sigma\_{zz}\pi\left(r\_o^2 - r\_i^2\right) = p\left(\pi r\_i^2\right) \tag{11}
$$

where σ*ZZ* denotes the axial component of the Cauchy stresses. In the analysis, we will employ ∆p with respect to the pressure at t = 0, at the start of isovolumic contraction.

The blood in the left cavity is assumed compressible, and the change in cavity volume (*∆V*) due to the monitored incremental pressure (*∆p*), is given by

$$
\Delta V = \Delta p \, / \,\text{K}; \qquad \text{K} = 2.0 \times 10^9 \, pa \tag{12}
$$

where *K* is the bulk modulus of blood.

### **Analysis and computational procedure:**

The following analysis is carried out at the three time instants *t* (or *j*) = 0.02 secs, 0.04 secs and 0.06 secs (from the start *t* = 0 of the isovolumic contraction phase, from the monitored *Δp* and computed *ΔV* (eq 12) at the three time instants with respect to *p* and *V* at t = 0 ( the start of isovolumic contraction phase).


$$
\Delta l\_{\bar{j}} = \left(1 - \sqrt[3]{1 - \Delta p/K}\right) L\_r \quad \Delta r\_{\bar{i}\bar{j}} = \left(1 - \sqrt[3]{1 - \Delta p/K}\right) R \tag{13}
$$

From equation (13), the incremental quantities Δ*lj* and Δ*rij* can be calculated, and hence:

$$l\_j = L - \Delta l\_j,\text{ and }\ r\_{i\bar{j}} = R - \Delta r\_{i\bar{j}} \tag{14}$$

where *lj* and *rij* are the deformed model length and radius at time t (or *j*) So the wall-thickness *h* can be obtained from:

$$\hbar\_j = \sqrt{\frac{V\_M \left/ \left. l\_j + \pi r\_{ij} \right|^2}{\pi}}} - r\_{ij} \tag{15}$$

Let Δφ denote the relative angle of twist measured at the apex, at each of the 3 stages of isovolumic contraction phase, obtained by magnetic resonance imaging (MRI).

3. Next we determine the stretches in the 3 directions (due to deformed dimensions *l* and *r* with respect to the undeformed dimensions *L* and *R*) as follows

$$\mathcal{A}\_{\mathbb{Z}}\left(r\right) = \frac{1}{L}, \quad \mathcal{A}\_{r}\left(r\right) = \frac{\partial r}{\partial R} = \frac{R}{r\mathcal{A}\_{\mathbb{Z}}}, \quad \mathcal{A}\_{\theta}\left(r\right) = \frac{r\partial\theta}{R\partial\Theta} = \frac{r}{R} \tag{16}$$

We define the twist stretch due to torsion, as

$$\gamma(r) = \frac{r\partial\theta}{\partial z} = \frac{r\phi}{l} \tag{17}$$

wherein φis zero at the top surface of the LV held by the aorta

4. Next we express the components of the Lagrange Green's strain tensor in terms of the stretches and deformations obtained from equations (8-10), as:

$$\mathcal{E}\_{\boldsymbol{m}} = \frac{1}{2} \left( \boldsymbol{\lambda}\_{\boldsymbol{r}}^{2} \cdot \mathbf{1} \right), \; \mathcal{E}\_{\boldsymbol{\theta}\boldsymbol{\theta}} = \frac{1}{2} \left( \boldsymbol{\lambda}\_{\boldsymbol{\theta}}^{2} \cdot \mathbf{1} \right), \; \mathcal{E}\_{\boldsymbol{\varpi}} = \frac{1}{2} \left( \boldsymbol{\lambda}\_{\boldsymbol{z}}^{2} \left( \mathbf{1} + \boldsymbol{\gamma}^{2} \right) \cdot \mathbf{1} \right), \; \mathcal{E}\_{\boldsymbol{\theta}\boldsymbol{z}} = \frac{\mathcal{M}\_{\boldsymbol{z}} \mathcal{X}\_{\boldsymbol{\theta}}}{2} \tag{18}$$

5. Then, by using the strain energy function (eqs 5 and 6), we obtain the expressions of Cauchy stress tensor in terms of the parameters *bi*:

$$
\sigma\_{\theta\theta} = \lambda\_{\theta}^{2} \frac{\partial \Psi}{\partial \mathbf{E}\_{\theta\theta}} + 2\mathcal{M}\_{z}\lambda\_{\theta} \frac{\partial \Psi}{\partial \mathbf{E}\_{\theta z}} + \lambda\_{z}^{2}\lambda\_{z}^{2} \frac{\partial \Psi}{\partial \mathbf{E}\_{zz}} - \overline{p} \qquad \sigma\_{m} = \lambda\_{r}^{2} \frac{\partial \Psi}{\partial \mathbf{E}\_{m}} - \overline{p}
$$

$$
\sigma\_{zz} = \lambda\_{z}^{2} \frac{\partial \Psi}{\partial \mathbf{E}\_{zz}} - \overline{p} \qquad \sigma\_{\theta z} = \lambda\_{z}\lambda\_{\theta} \frac{\partial \Psi}{\partial \mathbf{E}\_{\theta z}} + \mathcal{M}\_{z}^{2} \frac{\partial \Psi}{\partial \mathbf{E}\_{zz}} \tag{19}
$$

wherein ψ is given by equations (5 & 6), and the to-be-determined unknown parameters are the hydrostatic pressure and the constitutive parameters *bi* (in equation 6). *p*


$$
\sigma\_{1,2} = \frac{\sigma\_{zz} + \sigma\_{\theta\theta}}{2} \pm \sqrt{\left(\frac{\sigma\_{zz} - \sigma\_{\theta\theta}}{2}\right)^2 + \sigma\_{\theta z}^2} \tag{20}
$$

$$\tan 2\phi = \frac{2\sigma\_{\theta z}}{\sigma\_{\theta \theta} - \sigma\_{zz}} \tag{21}$$

9. Then, we compute the equivalent active axial force ΔF and the torsional couple ΔT as follows:

$$
\Delta F = 2\pi \int\_{r\_i}^{r\_o} \sigma\_{zz} r \mathrm{d}r, \quad M\_T = 2\pi \int\_{r\_i}^{r\_o} \sigma\_{\theta z} r^2 \mathrm{d}r,\tag{22}
$$

### **Results and comments:**

16 Biomedical Science, Engineering and Technology

The blood in the left cavity is assumed compressible, and the change in cavity volume (*∆V*)

The following analysis is carried out at the three time instants *t* (or *j*) = 0.02 secs, 0.04 secs and 0.06 secs (from the start *t* = 0 of the isovolumic contraction phase, from the monitored *Δp* and computed *ΔV* (eq 12) at the three time instants with respect to *p* and *V* at t = 0 ( the

1. We obtain the increments in LV pressure (*Δp*) from the monitored pressure data, during isovolumic contraction. By taking *K=Δp/ΔV*, we compute the corresponding changes in

2. Left ventricular deformation: For this change *ΔV* in the LV volume, by assuming the ratios of cylindrical-model length and radius strains *Δl/L* and *Δr/R* to be equal during

( ) 11 , <sup>3</sup> *<sup>j</sup>* Δ = − −Δ *l p K L* ( ) 1 13 *ij* Δ = − −Δ *r p K R* (13)

Δ*lj* and Δ

– , and *<sup>j</sup> j ij ij lLl rR r* = Δ = −Δ (14)

*M j ij* / *j ij Vl r h r*

π

3. Next we determine the stretches in the 3 directions (due to deformed dimensions *l* and *r*

( ) *r r*

4. Next we express the components of the Lagrange Green's strain tensor in terms of the

π

denote the relative angle of twist measured at the apex, at each of the 3 stages of

*z*

*z l* θ

*L Rr R R*

*rR r r*

θ

 λ λ

> φ

2

<sup>9</sup> Δ =Δ *V p K K pa* / ; 2.0 10 = × (12)

*rij* can be calculated, and hence:

<sup>+</sup> = − (15)

θ

<sup>∂</sup> = = ∂ (17)

∂ ∂ = == = = ∂ ∂<sup>Θ</sup> (16)

due to the monitored incremental pressure (*∆p*), is given by

isovolumic contraction, we obtain their expressions as:

where *lj* and *rij* are the deformed model length and radius at time t (or *j*)

isovolumic contraction phase, obtained by magnetic resonance imaging (MRI).

( ) ( ) ( ) <sup>1</sup> , , *z r*

*r*

γ

is zero at the top surface of the LV held by the aorta

stretches and deformations obtained from equations (8-10), as:

*rr r*

with respect to the undeformed dimensions *L* and *R*) as follows

λλ

where *K* is the bulk modulus of blood. **Analysis and computational procedure:** 

start of isovolumic contraction phase).

From equation (13), the incremental quantities

So the wall-thickness *h* can be obtained from:

We define the twist stretch due to torsion, as

LV volume *ΔV*.

Let Δφ

wherein

φ


We hence demonstrate that the big increment of internal pressure in LV cavity is caused by the compression of the blood (due to its high bulk modulus) by the internally generated torque and axial force during the isovolumic phase, due to the contraction of the myocardial fibers.

3. The radial variations of the principal stress and principal angle are found to be almost uniform across the wall thickness [5]. The time variation of the principal stress and angle during isovolumic phase are shown in **Fig. 13**. At the end of the isovolumic phase, the magnitude of the tensile principal stress is around 1.75x105pa, which is in good agreement with the isometric tension value of 1.40x105pa achieved under maximal activation.

Fig. 12. Variations of axial force and torque as function of time during isovolumic phase [5].

The notable result from **Fig. 13** is that both the principle stresses and their orientation angle keep changing during the isovolumic phase. It is seen from Figure 13, that the myocardial fiber was oriented 380 at the start and became 330 at the end of the isovolumic phase, as the monitored internal pressure increased during isovolumic phase (due to active torsion and compression, corresponding to active contraction of the helically woven fibers from 380 to 330) from 24 to 44 to 62 mmHg during the isovolumic contraction phase. Observing the values of the principle stresses, it is seen that the LV is acted upon by internally generated (i) active torsion (T) causing its twisting, and (ii) compression (F) to cause compression of the blood and develop the LV pressure rise.

Fig. 13.Variation of principal stresses as functions of time, during isovolumic phase [5].

**In conclusion**: we have indirectly shown that the contraction of the myocardial fibers (i) develops active stresses in the LV wall (and corresponding principal stresses), and (ii) causes LV twist and shortening, resulting in the development of substantial pressure increase during isovolumic contraction.

The tensile principal stress corresponds to the active contractile stress generated in the myocardial fibers, while the angle of the tensile principal stress corresponds to the fiber helical angle, which is in agreement with the experiment data on the fiber angle.

The computational results are shown in Tables 1 to 5, for a sample subject.


Table 1. Pressure-volume and model parameters for a sample subject with *VM*=185ml.

axial force

Fig. 12. Variations of axial force and torque as function of time during isovolumic phase [5]. The notable result from **Fig. 13** is that both the principle stresses and their orientation angle keep changing during the isovolumic phase. It is seen from Figure 13, that the myocardial fiber was oriented 380 at the start and became 330 at the end of the isovolumic phase, as the monitored internal pressure increased during isovolumic phase (due to active torsion and compression, corresponding to active contraction of the helically woven fibers from 380 to 330) from 24 to 44 to 62 mmHg during the isovolumic contraction phase. Observing the values of the principle stresses, it is seen that the LV is acted upon by internally generated (i) active torsion (T) causing its twisting, and (ii) compression (F) to cause compression of the

0.2 0.4 0.6 0.8 1 1.2 1.4

T=ΣΔTi (Nm)

<sup>0</sup> 0.01 0.02 0.03 0.04 0.05 0.06 <sup>0</sup>

Torque

Time (second)

<sup>0</sup> 0.01 0.02 0.03 0.04 0.05 0.06 <sup>33</sup>

Time (second)

 Fig. 13.Variation of principal stresses as functions of time, during isovolumic phase [5].

33.5 34 34.5 35 35.5 36 36.5 37 37.5 38

principle angle (°

)

**In conclusion**: we have indirectly shown that the contraction of the myocardial fibers (i) develops active stresses in the LV wall (and corresponding principal stresses), and (ii) causes LV twist and shortening, resulting in the development of substantial pressure

The tensile principal stress corresponds to the active contractile stress generated in the myocardial fibers, while the angle of the tensile principal stress corresponds to the fiber

helical angle, which is in agreement with the experiment data on the fiber angle. The computational results are shown in Tables 1 to 5, for a sample subject.

blood and develop the LV pressure rise.

principal stress (tension) principal stress (compression)

<sup>0</sup> 0.01 0.02 0.03 0.04 0.05 0.06 -120

Time (second)





Stress (Pa)

0

0.5

<sup>1</sup> x 105



F=ΣΔFi (N)



0

increase during isovolumic contraction.

<sup>0</sup> 0.01 0.02 0.03 0.04 0.05 0.06 -2

Time (second)


Table 2. The parameters of the strain energy function for the sample case shown in Table 1.


Table 3. The stretches calculated from equations (16 & 17) for the sample case shown in Table 1.


Table 4. Radial stresses distributions along LV wall from endocardium to epicardium.


Table 5. Results for the sample subject at different instants

### **4. Clinical evaluation of Physiological systems in terms of Non-dimensional physiological Indices**

Non-dimensional numbers (made up of several phenomenon related terms) are employed to characterise – disturbance phenomena. For example, in a cardiovascular fluid-flow regime, the Reynold'snumber

$$\mathcal{N}\_{\rm re} = \rho \text{VD} \mid \mu \text{ } \tag{1}$$

*(V :* flow velocity, *D:* diameter, μ: fluid viscosity, ρ: fluid density)

parameter b1 b2 b3 b4 b5 b6 b7 b8 b9 value 5946.2278 15690.58158 422.514993 16157.10454 16360.53744 33299.28998 680.7385218 0 0 Table 2. The parameters of the strain energy function for the sample case shown in Table 1.

t λ<sup>z</sup> λ<sup>r</sup> λθ ϒ

0.02 0.9999999958 1.000000008E+00 9.99999996E-01 1.28626948E-01 0.04 0.9999999925 1.000000015E+00 9.99999993E-01 2.56482520E-01 0.06 0.9999999895 1.000000021E+00 9.99999990E-01 3.85688000E-01 Table 3. The stretches calculated from equations (16 & 17) for the sample case shown in

> Endocardium-------------------------------------------------------------------------------------------------------Epicardium 0 1.99E-02 1.02E-01 2.37E-01 4.08E-01 0.591715 0.762765 0.89833 0.98014 1

> > Axis stress (Pa)

Shear stress (Pa)

Principal stresstension (Pa)

+04 3.29E+04 28221.68 -38395.38 37.82

+04 8.21E+04 63836.28 -105547.96 36.18

+05 1.22E+05 81673.18 -180897.35 33.02

(1)

Principal stresscompression (Pa)

Principal direction (°)

0.02 -24.0809 -23.4615 -17.2801 -12.9839 -8.73378 -4.9888 -2.11726 -0.41173 -0.2144 0 0.04 -44.2917 -43.1341 -31.6543 -23.7482 -15.9694 -9.13029 -3.88098 -0.75569 -0.3542 0 0.06 -6.21E+01 -6.05E+01 -5.43E+01 -4.48E+01 -3.39E+01 -2.31E+01 -1.34E+01 -5.75E+00 -1.13E+00 0

Table 4. Radial stresses distributions along LV wall from endocardium to epicardium.

Torque (Nm)

0 0.00 0.00 0.00 0.00 0.00E+00 0.00E+00 0.00 0.00

**4. Clinical evaluation of Physiological systems in terms of Non-dimensional** 

Non-dimensional numbers (made up of several phenomenon related terms) are employed to characterise – disturbance phenomena. For example, in a cardiovascular fluid-flow regime,

> *N VD re* = ρ /μ

(second) Results

Radial stress (Pa)

0.02 3358.33 -2844.53 -13.51 0.38 -1.35E

0.04 5318.61 -5956.38 -46.06 0.89 -4.70E

0.06 4481.83 -8530.94 -102.38 1.30 -1.04E

Table 5. Results for the sample subject at different instants

*(V :* flow velocity, *D:* diameter, μ: fluid viscosity, ρ: fluid density)

Axial force (N)

Circumferential stress (Pa)

**physiological Indices** 

the Reynold'snumber

0

Table 1.

t (second)

t

characterizes turbulent flow, which can occur in the ascending aorta when the aortic valve is stenotic (giving rise to murmurs) and accentuated in the case of anaemia (decreased blood viscosity).

Integration of a number of isolated but related events into one non-dimensional physiological index (NDPI) can help to characterise an abnormal state of a particular physiological system [1].

For utilization of an NDPI diagnostically, we evaluate it in a large patient population and develop its distribution into normal and dysfunctional ranges, as illustrated in figure 14. Then, upon evaluation of the NDPI of a particular subject, we can see if that value falls in the normal or dysfunctional range, and accordingly make the medical diagnosis.

Fig. 14. Illustration of the Distribution of an NDI, showing its normal and dysfunctional ranges which can be employed for diagnostic purpose.
