## Thick-Walled Spherical Vessel

The video suggests something about the way deformation occurs in a thick-walled spherical vessel.  The video starts with the vessel fully distended and lines indicating the radial coordinate are equally spaced.  As the vessel deforms (contracts), the internal dimension decreases much more dramatically than the outer dimension; i.e. strain is more pronounced at the inner surface.  The video was made intentionally with a very thick wall and large deformations so that the difference between the inner and outer radii would be immediately apparent to the naked eye.  You should be able to see the pronounced gradation of deformation as a function of the radial coordinate.  Both radial and azimuthal ("fractional shortening") deformation are more pronounced at the inner wall and we will see that stresses ( forces ) in the wall also follow this distribution ( in a Hookean material anyway ).   The video shown is from a deforming prolate ellipsoidal structure (not a sphere);  it demonstrates the principals described qualitatively but is not quantitatively representative of a spherical vessel.

This page includes the determination of stresses and strains resulting from pressurization of a thick-walled spherical vessel whose wall is comprised of a Hookean material, one that has a linear stress-strain relationship.  What does this have to do with a heart?  Well, more than the "Law of LaPlace" so often cited in cardiology texts.    As a matter of fact, this Law that's attributed to LaPlace was actually developed by Thomas Young ca 1804 (apparently).

We're talking about a hollow sphere whose wall is made up of a uniform material that deforms linearly when a stress is applied (Hooke's law, Hookean material). The vessel is deformed by a change in internal pressure ( $$p_i$$), outer (exterior) pressure ( $$p_o$$), or both.  When this occurs, each material element in the wall can be displaced radially; only.  How much displacement occurs may depend on where the material element originated, it's original radial coordinate.

The figure below may give you an idea of what we're up to.  Suppose the left side of the figure represents a (very) thick-walled sphere, sliced in half so we can watch the wall deform.  This particular sphere has a wall thickness that's twice its internal radius.  Material layers of the wall are also seen (concentric circles), and these have been placed equidistant from each other; it's $$\Delta r$$ that's constant from one layer to the next in the left hand figure.  On the right side, the sphere has undergone a (realistic) deformation; this would occur by increasing the internal pressure, decreasing the external pressure, or some combination.

This would be a very large deformation in the engineering science world; the radius has increased by a factor of 2 and you probably wouldn't want this happening to a bridge or an airplane.  It's the sort of thing that occurs all the time with the heart however.  NOTICE that all the concentric layers of material are thinner (radial deformation or radial strain), but that this thinning isn't uniform.  The layers closer to the chamber (endocardial?) obviously have been compressed more than those near the external layer (epicardial?).  Also note that internal circumference has changed dramatically (circumferential strain) in comparison with the external.  So there's direct visual evidence that the deformation is graded; both the radial ("compression") and circumferential ("elongation") deformations are clearly more pronounced near the endocardium.

So why does this occur?  And exactly how does it occur?  I'm going to introduce some math that will explain exactly how these figures were drawn.  The first thing to understand is that the material making up the sphere  is incompressible (cardiac tissue also  to a very good approximation).   This was employed mathematically to draw the figure by specifying that each and every layer of the sphere retains its original volume subsequent to the deformation. Suppose that the inner radius of the sphere on the left side is $$r_{i,1}$$ and that outer radius is $$r_{o,1}$$ using $$i$$ and $$o$$ to stand for "inner" and "outer"; subscript $$1$$ refers to the initial conformation of the sphere.  Then the volume of material in the wall, $$V_w$$ is equal to the difference in volume between the inner and outer spheres:

$$\large V_w = \frac{4}{3} \pi (r^3_{o,1} - r^3_{i,1})$$

With an incompressible material, the volume of wall material is the same after the sphere is deformed:

$$\large V_w = \frac{4}{3} \pi (r^3_{o,1} - r^3_{i,1}) = \frac{4}{3} \pi (r^3_{o,2} - r^3_{i,2})$$

where subscript $$2$$ refers to the sphere in the deformed state.  Now the wall volume must remain constant regardless of the initial choice of radial coordinate.  If the initial "outer" coordinate is arbitrarily called $$r_1$$, then we have a simple prescription for determining what the radial coordinate will be in terms of the deformation of the inner wall.   The wall volume equation is:

$$\large V_w(r_1) = \frac{4}{3} \pi (r^3_{1} - r^3_{i,1}) = \frac{4}{3} \pi (r^3_{2} - r^3_{i,2})$$

This equation is readily solved for the radial coordinate in the deformed state, $$r_2$$, given the initial coordinate, $$r_1$$ and a description of the deformation at the inner wall ( $$r_{i,1}$$ and $$r_{i,2}$$ ).

$$\Large r_2 = \sqrt[3]{r^3_{1} - r^3_{i,1} + r^3_{i,2}}$$

The figure above is nothing but a graphic representation of this formula with the prescribed deformation noted previously.  The volume of each concentric sphere remains the same after the deformation as does the volume of the wall in total.

This concept means that each layer of the sphere deforms differently.  Suppose we have a contracting sphere for a left ventricle.  For a given fractional shortening (as usually defined), we will see a different fractional shortening for each of the layers (radial coordinate), assuming we could track them (speckles?).  The figure below shows fractional shortening of several spheres, each with FS at the "endocardium" arbitrarily set to 0.35, but showing FS as it depends on initial radial coordinate.  The spheres differ in their relative wall thickness ( RWT ) in "diastole" with RWT  defined as the diastolic wall thickness divided by internal radius.   The plot shows how the FS varies with relative radial coordinate that ranges from 0 at the "endocardium" (inner wall) to 1.0 at the "epicardium" (outer wall).

As you can see, the fractional shortening varies greatly, at wall locations other than the inner wall, and depends on the RWT.  For a very thick walled heart, we don't expect the epicardial surface to be moving much, even if the endocardium exhibits a normal or even enhanced FS.  Because of this phenomenon, some have suggested determining FS at the midwall as a more just (accurate? meaningful? predictive? Jazzy?) measure of LV deformation.   My feeling is simply that we need to understand the phenomenon and hopefully find better ways to evaluate heart function.  By the way: If you measure the motion of the midwall (defined as the point midway between epicardium and epicardium), have you actually tracked the motion of a piece of tissue?  (No.  The tissue that's midway in diastole is not midway in systole.)

If you can't get your head wrapped around "fractional shortening" determined at various values of the radial coordinate, this is actually the same thing as circumferential strain (except that with the sphere getting smaller, we would say that the strain is negative).  For simple elongations and contractions, strain ($$\large \epsilon$$ ) is defined as:

$$\Large \epsilon = \frac{l_f - l_0}{l_0} = \frac{\Delta l}{l_0}$$

where $$\large l_f$$ is the final length and $$\large l_0$$ is the initial length.  For our sphere the initial length, the circumference ( $$\large c_1$$  ), depends on initial radial coordinate ( $$\large r_1$$ ).

$$\Large l_0 = c_1 = 2\pi r_1$$

After the deformation, there's a new radius that we've already figured out in terms of the initial and deformed inner radius, $$\large r_{i,1}$$, and $$\large r_{i,2}$$ respectively:

$$\Large l_f = c_2 = 2 \pi r_2 = 2 \pi \sqrt[3]{r^3_{1} - r^3_{i,1} + r^3_{i,2}}$$

A mathematical expression for the circumferential strain follows readily:

$$\Large \epsilon = \frac{c_2 - c_1}{c_1} = \frac{ \sqrt[3]{r_1^3 + r_{i,2}^3 - r_{i,1}^3}}{r_1} - 1$$

Although the sphere is an obvious oversimplification of a cardiac chamber, we'll always see this kind of geometric dependence of the strain.  The complexity of the heart will result in deviations from an oversimplified formula, yet the basic dependencies will persist.  You will see greater strains near the endocardium than the epicardium and this is not a biological property of the heart but a physical one of incompressible materials in general.  The thicker the heart wall, the greater the differences will be between endo- and epicardium.  For folks doing strain imaging, I'm guessing this is all common knowledge.

Wall Stress in the Sphere

The foregoing shows how the thick-walled sphere deforms, but doesn't tell us directly about the stresses (forces) in the wall.  For that we need to know a relationship between force and deformation, what's called a constitutive law (or relationship).  We're going to assume a linear relationship between stress and strain that's prescribed by Hooke's law.

$$\Large \epsilon = \frac{\sigma}{E}$$

where $$\epsilon$$ is strain, $$\sigma$$ is stress, and $$E$$ is the proportionality constant having physical units of stress/strain (Young's modulus of elasticity).  Heart muscle doesn't behave like this, but we'll gain some insights by solving a simpler problem.   Actually the problem isn't that simple!  The solution is detailed at the bottom of the page but takes a bit of work and some calculus.  In this section we're just going to use the result of that work.  Here are equations for the stresses, displacements, and strains in terms of the internal ($$\large p_i$$ ) and external pressures  ( $$\large p_o$$ ); radii are as previously defined.

$$\Large \sigma_{rr} =\frac{ p_o r_o^3}{r^3} \frac{r^3-r_i^3}{r_i^3-r_o^3} + \frac{ p_i r_i^3}{r^3} \frac{r_o^3-r^3}{r_i^3-r_o^3}$$ Radial stress

$$\Large \sigma_{tt} =\frac{ p_o r_o^3}{2r^3} \frac{2r^3+r_i^3}{r_i^3-r_o^3} - \frac{ p_i r_i^3}{2r^3} \frac{2r^3+r_o^3}{r_i^3-r_o^3}$$ Circumferential stress

$$\Large u_r = \frac{3 ( p_i - p_o) r_i^3 r_o^3}{4 E r^2 (r_o^3-r_i^3)}$$  Radial displacement as a function of initial radius, $$r$$

$$\Large u_r(r_i) = \frac{3 (p_i-p_o) r_i r_o^3}{4 E (r_o^3 - r_i^3)}$$ Radial displacement of the inner surface

$$\Large \epsilon_{rr} = u_r'(r) = -\frac{3}{2E} \frac{p_i-p_o}{r^3} \frac{r_i^3 r_o^3}{r_o^3-r_i^3}$$ Radial strain

$$\Large \epsilon_{tt} = \frac{u(r)}{r} = \frac{3}{4E} \frac{p_i - p_o}{r^3} \frac{r_i^3 r_o^3}{r_o^3 - r_i^3}$$ Circumferential strain

So that was the "answer" (in case that wasn't obvious).  The answer isn't a number or a set of numbers, but a set of functions, relationships that tell us how stresses, deformations, and displacements are related to the internal and external applied pressures for the sphere as defined.  Unlike the thin-walled sphere, wall stress with a thick wall depends on (initial) radial coordinate, $$r$$.  We have a circumferential stress, $$\large \sigma_{tt}$$, that stretches the wall and increases the radius of the sphere if the internal pressure is greater than external.  We also have a radial stress, $$\large \sigma_{rr}$$, that ends up compressing the wall in the radial direction ( if $$\large p_i > p_o$$).

For the thin-walled sphere, we had just one value for circumferential stress (it was actually an average).  For the thick-walled case, we get a function that depends on initial radius.  Since we saw at the top of the page that the inner wall undergoes greater deformation than the outer, it won't be too surprising that the stress is greater there (both circumferential and radial).

The following figure suggests the circumferential stress distribution in a (very) thick-walled spherical vessel with a wall-thickness to radius ratio of 2.0.  The stress is concentrated near the inner ("endocardial") wall.  Note that although the thickness of the vessel may be exaggerated for cardiology discussions, the stress distribution is not!  The graphic is derived from the above formula and faithfully represents how the stress is distributed (for the oversimplified model).  It's the wall thickness that causes this rather drastic variation with radial coordinate. (not the magnitude of the distending pressure).

Just to make sure you don't take a bad hop on me, be sure you understand that the Law of LaPlace is not going to tell you the stress at the various locations in the wall. The LaPlace relation suggests that the stress would be greater at a greater radius, but that doesn't apply to the thick-walled sphere.

The figure above depicts the distribution of relative circumferential stress (actual / mean) for a thick-walled sphere as a function of relative radial coordinate; 0 is the inner surface and 1 is the outer.  The stress distribution is plotted for three different values of the relative wall thickness ( wall thickness divided by the inner radius ). The wall material is assumed to be isotropic and homogeneous.  While a thicker wall results in lower mean stress, it also results in stress concentration at the inner wall.  The Laplace relationship predicts larger circumferential stress as radius increases, but the law of Laplace has nothing to do with this situation.  DOWNLOAD an Excel spreadsheet if you want to play with the numbers yourself.

Compliance of the Thick Walled Sphere

## It turns out that we can compute the pressure-volume relationship and the compliance of the structure also.  The Internal radius after the structure has deformed Is equal to the initial radius plus the displacement radius.  The volume as a function of distending pressure:

$$\Large V(p) = \frac{4}{3} \pi \left[ r_{i} + u_r(r_i)\right]^3 = \frac{4}{3} \pi \left[ r_{i} +\frac{3 (p_i-p_o) r_i r_o^3}{4 E (r_o^3 - r_i^3)} \right]^3$$

The compliance is the rate of change of volume with respect to pressure, $$\large \frac{dV}{dp}$$:

$$\Large \frac{dV}{dp} = C = \frac{4}{3} \pi r_i^3 \frac{9 r_o^3}{4 E (r_o^3 - r_i^3)} \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2 = \frac{9}{4} V_0 \frac{1}{E} \frac{r_o^3}{r_o^3 - r_i^3} \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2$$

$$\large V_0$$ has been substituted for the initial (unstressed) volume of the sphere.  The compliance is not a constant, as we had for the thin-walled sphere with small deformations but depends on $$\large \Delta p$$ the distending pressure ( $$p_i - p_o$$ ) that appears in the square brackets.  Look closely and you'll see that the quantity in square brackets is nondimensional – it's a pure number that modifies the result depending on how distended the sphere is; the compliance increases as the sphere is distended.  The $$\large \frac{9}{4}$$ at the beginning is also a pure number that's due to the shape of the vessel (a sphere).  $$\large \frac{1}{E}$$ is due only to the wall material. We also see a fraction, $$\large \frac{r_o^3}{r_o^3 - r_i^3}$$, that takes the place of $$\large \frac{r_0}{h}$$ of the thin-walled sphere.  We will always see these 4 components in describing the compliance of a vessel: 1) something about the shape, 2) something about the size, 3) something about the mechanical properties of the material and, 4) something about how much material there is surrounding the vessel.  Of particular interest to us as cardiologists is the fact that the compliance increases with the size of the structure, in this case literally $$\large V_0$$.  The heart and blood vessels increase their compliance by enlarging (dilatative remodeling, whether they want to or not).  As was noted for the thin-walled structure, engineers often normalize the compliance for the size of the structure, the volume distensibility.   For the thick-walled sphere this is obviously:

$$\Large \frac{dV}{dp}/V_0 = \frac{9}{4} \frac{1}{E} \frac{r_o^3}{r_o^3 - r_i^3} \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2$$

As to why the vessel changes its compliance with distention, this is due to changing geometry (at least).  When the sphere distends it is of course larger, and the wall also becomes thinner.  So changing volume alters at least 2 of the essential ingredients of the compliance formula. For the actual heart chambers, we'll also have changes in properties of the material as it changes length (it's nonlinear), and likely changes in chamber shape as well.

## The Prolate Ellipsoid

Obviously the LV is much more complicated than a Hookean sphere in the way it deforms.  Yet the example on this page affords insight for more complicated situations.  We are always going to have greater strains at the inner wall of a convex pressure vessel.

The video above suggests the deformation of a ellipsoid of revolution with parameters selected to be similar to the left ventricle.  The virtual structure is shown with its volume changing sinusoidally and has no relation to the time course of volume change of the actual LV.  It's important for you to understand however that the video was made by prescribing the deformation.  There was no solution of Navier-Cauchy equations ( no equation of motion or compatibility equations were solved ); I can't tell you the wall stresses, the shape of the structure as it contracts/distends, or how large the vessel will be for a particular distending pressure.  That turns out to be intractable for a paper and pencil type of solution, even for this relatively simple problem. What the video does depict is the behavior of incompressible tissue for a more complicated geometry.  Like the simpler sphere, the material grid superimposed on the structure depicts a result where each material element maintains a constant volume during the deformation.  This is quite a bit more complicated than the sphere which is essentially 1 dimensional in terms of the solution difficulty.  The elipsoid is still only 2-dimensional (3D structure but with an axis of symmetry ) but the curvature of the structure is different depending on the coordinate (e.g. radial versus azimuthal) and location. The video changes orientation halfway through so that you can observe the appearance in a "short axis" view as well (at the level of the equator), similar to the video at the top of the webpage.  The structure is the result of employing an prolate spheroidal coordinate system which seems well suited to a simplistic LV geometry.

## Deformation of a Thick-Walled Sphere

Engineers will recognize this as a standard approach to a continuum mechanics problem.  (I'm not so good at solid mechanics so I compiled this from several sources; hope it's correct.)

The spherical coordinate system (image from Wikipedia)

The Navier-Cauchy equations in spherical coordinates (The equation of motion or force balance equation):

$$\Large \frac{\partial \sigma_{rr}}{\partial r} + \frac{1}{r} \frac{\partial \sigma_{r \theta} }{\partial \theta} + \frac{1}{r\sin \theta } \frac{\partial \sigma_{r \phi} }{\partial \phi} + \frac{1}{r} \large \left[ 2 \sigma_{rr} - \sigma_{\theta \theta} - \sigma_{\phi \phi} + \sigma_{r \theta} \cot \theta \right] + F_r = \Large \rho \frac{\partial^2 u_r }{\partial t^2}$$

$$\Large \frac{\partial \sigma_{r \theta} }{\partial r} + \frac{1}{r} \frac{\partial \sigma_{\theta \theta} }{\partial \theta} + \frac{1}{r \sin \theta} \frac{\partial \sigma_{\theta \phi}}{\partial \phi} + \frac{1}{r} \large \left[ (\sigma_{\theta \theta} - \sigma_{\phi \phi}) \cot \theta + 3 \sigma_{r \theta} \right] + F_{\theta} = \Large \rho \frac{\partial^2 u_{\theta} }{\partial t^2}$$

$$\Large \frac{\partial \sigma_{r \phi} }{\partial r} + \frac{1}{r} \frac{\partial \sigma_{\theta \phi} }{\partial \theta} + \frac{1}{r \sin \theta} \frac{\partial \sigma_{\phi \phi} }{\partial \phi} + \frac{1}{r} \large \left[ 2 \sigma_{\theta \phi} \cot \theta + 3 \sigma_{r \phi} \right] + F_{\phi} = \Large \rho \frac{\partial^2 u_{\phi} }{\partial t^2}$$

Infinitesmal strain tensor in spherical coordinates.

$$\Large \epsilon_{rr} = \frac{\partial u_r }{\partial r}$$

$$\Large \epsilon_{\theta \theta} = \frac{1}{r} \left[ \frac{\partial u_{\theta} }{\partial \theta} + u_{r} \right]$$

$$\Large \epsilon_{\phi \phi} = \frac{1}{r \sin \theta} \left[ \frac{\partial u_{\phi} }{\partial \phi} + \large u_{r} \sin \theta + u_{\theta} \cos \theta \right]$$

$$\Large \epsilon_{r \theta} \frac{1}{2} \left[ \frac{1}{r} \frac{\partial u_{r} }{\partial \theta} +\frac{\partial u_{\theta} }{\partial r } - \frac{u_{\theta}}{r} \right]$$

$$\Large \epsilon_{\theta \phi} = \frac{1}{2 r} \left[ \frac{1}{\sin \theta} \frac{\partial u_{\theta} }{\partial \phi} + \left( \frac{\partial u_{\phi} }{\partial \theta} - \large u_{\phi} \cot \theta \right) \right]$$

$$\Large \epsilon_{r \phi} = \frac{1}{2} \left[ \frac{1}{r \sin \theta} \frac{\partial u_{r} }{\partial \phi } + \frac{\partial u_{\phi}}{\partial r } - \frac{u_{\phi}}{r} \right]$$

NOTE: The solution is being applied to a large (finite) strain situation.  My understanding is that this is OK since there are no rotations or shears. If you know differently, please fill me in (This email address is being protected from spambots. You need JavaScript enabled to view it.)

## Solution

$$\Large u_r(r)$$ is the radial displacement field, the difference between the final location and the original one of each particle (element).   As suggested by the symbol, this can depend on initial radial position, but doesn't changes with other aspects of starting position in the wall.

In terms of the displacement field, the strains that occur in the wall appear as:

$$\Large \epsilon_{rr} = \frac{dr [1+u_r'(r)] - dr}{dr} = u_r'(r)$$

$$\Large \epsilon_{\theta \theta} = \frac{[r+u_r(r)] d\theta - r d\theta}{r d\theta} = \frac{u_r(r)}{r}$$

$$\Large \epsilon_{\phi \phi} = \frac{[r+u_r(r)] \cos(\theta) d\phi - r \cos (\theta) d\phi}{r \cos (\theta) d\phi} = \frac{u_r(r)}{r}$$

So even though there is only radial displacement, strains can occur in multiple directions.  $$\large \epsilon_{rr}$$ has to do with thickening or thinning of the wall.  $$\large \epsilon_{\theta \theta}$$ and $$\large \epsilon_{\phi \phi}$$ have to do with stretching the wall i.e. to accommodate a larger volume.

Because of the symmetry of the sphere, the circumferential stresses and strains can be reduced from the $$\large \theta$$ and $$\large \phi$$ angular coordinates by defining a tangential direction:

$$\Large \epsilon_{\theta \theta } = \epsilon_{\phi \phi} = \epsilon_{tt}$$  Tangent, i.e. circumferential strain

$$\Large \sigma_{\theta \theta} = \sigma_{\phi \phi} = \sigma_{tt}$$ Tangent, i.e. circumferential stress

It's assumed that no shear stresses occur.

$$\Large \sigma_{r \theta} = \sigma_{r \phi} = \sigma_{\theta \phi} = 0$$

$$\Large \frac{\partial \sigma_{rr}}{\partial \theta} = \frac{\partial \sigma_{\theta \theta}}{\partial \theta} = \frac{\partial \sigma_{\phi \phi}}{\partial \theta} = \frac{\partial \sigma_{rr}}{\partial \phi} = \frac{\partial \sigma_{\theta \theta}}{\partial \phi} = \frac{\partial \sigma_{\phi \phi}}{\partial \phi} = 0$$

With the problem simplifications noted, the equation of motion reduces to:

$$\Large 2 \sigma_{rr} + r \frac{\partial \sigma_{rr}}{\partial r} - \sigma_{\theta \theta} - \sigma_{\phi \phi} = 2 \sigma_{rr} + r \frac{\partial \sigma_{rr}}{\partial r} - 2 \sigma_{tt} = 0$$

Hooke's Law

$$\Large \epsilon_{rr} = \frac{1}{E} \left[ \sigma_{rr} - 2 \nu \sigma_{tt} \right]$$

$$\Large \epsilon_{tt} = \frac{1}{E} \left[\sigma_{tt} - \nu(\sigma_{rr} + \sigma_{tt}) \right]$$

Then the following

$$\Large u_r'(r) = \frac{1}{E} (\sigma_{rr} - 2 \nu \sigma_{tt})$$ Hooke's law with strain substitution.

$$\Large \frac{u_r(r)}{r} = \frac{1}{E} \left[ \sigma_{tt} - \nu ( \sigma_{rr} + \sigma_{tt}) \right]$$ Hooke's law with strain substitution

$$\Large 2 \sigma_{rr}(r) + r \sigma_{rr}'(r) - 2 \sigma_{tt} = 0$$ Equation of motion

can be reduced to a single ordinary differential equation:

$$\Large 2 u'(r) - 2 \frac{u(r)}{r} + r u''(r) = 0$$

The radial subscript on $$u$$ has been dropped since there is only radial displacement.  The solution is:

$$\Large u(r) = A r + B \frac{1}{r^2}$$  The solution

$$\Large \sigma_{tt} = E \frac{A}{1-2\nu} + E \frac{B}{1+\nu} \frac{1}{r^3}$$

$$\Large \sigma_{rr} = E \frac{A}{1-2\nu} - 2E \frac{B}{1+\nu} \frac{1}{r^3}$$

With $$\large r_i$$, the inner radius of the sphere, $$\large r_o$$ the outer radius, $$\large p_i$$ the internal pressure, $$\large p_o$$ the external pressure:

$$\Large \sigma_{rr}(r_i) = -p_i$$

$$\Large \sigma_{rr}(r_o) = -p_o$$

$$\Large A = \frac{1 - 2\nu}{E} \frac{p_i r_i^3 - p_o r_o^3}{r_o^3-r_i^3}$$

$$\Large B = \frac{1+\nu}{E}\frac{(p_i-p_o) r_i^3 r_o^3}{r_o^3-r_i^3}$$

( with $$\large \nu = 0.5$$, i.e. incompressible)

$$\Large \sigma_{rr} =\frac{ p_o r_o^3}{r^3} \frac{r^3-r_i^3}{r_i^3-r_o^3} + \frac{ p_i r_i^3}{r^3} \frac{r_o^3-r^3}{r_i^3-r_o^3}$$

$$\Large \sigma_{tt} =\frac{ p_o r_o^3}{2r^3} \frac{2r^3+r_i^3}{r_i^3-r_o^3} - \frac{ p_i r_i^3}{2r^3} \frac{2r^3+r_o^3}{r_i^3-r_o^3}$$

$$\Large u_r = \frac{3 ( p_i - p_o) r_i^3 r_o^3}{4 E r^2 (r_o^3-r_i^3)}$$  Radial displacement as a function of initial radius, $$r$$

$$\Large u_r(r_i) = \frac{3 (p_i-p_o) r_i r_o^3}{4 E (r_o^3 - r_i^3)}$$ Radial displacement of the inner surface

$$\Large \epsilon_{rr} = u_r'(r) = -\frac{3}{2E} \frac{p_i-p_o}{r^3} \frac{r_i^3 r_o^3}{r_o^3-r_i^3}$$

$$\Large \epsilon_{tt} = \frac{u(r)}{r} = \frac{3}{4E} \frac{p_i - p_o}{r^3} \frac{r_i^3 r_o^3}{r_o^3 - r_i^3}$$

Substituting $$\large \Delta p = p_i-p_o$$

$$\Large V = \frac{4}{3} \pi \left[ r_i + u_r(r_i) \right]^3 = \frac{4}{3} \pi r_i^3 \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^3$$  This is the "Pressure-Volume Relationship" or PVR.  However we're used to seeing this as an expression of pressure in terms of volume, so the formula needs to be "inverted":

$$\Large \Delta p = \frac{2}{3} \sqrt[3]{\frac{6}{\pi}} E \left[ V^{1/3} - V_0^{1/3} \right] \frac{r_o^3 - r_i^3}{r_i r_o^3}$$

where $$\large V_0 = \frac{4}{3} \pi r_i^3$$

$$\Large \frac{dV}{dp} = 4 \pi r_i^3 \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2 \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} = \frac{3 \pi r_i^3 r_o^3}{ E (r_o^3 - r_i^3)} \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2$$   This is the Compliance of the structure - nonlinear in this case even though the material is linear.  I prefer to write this in such a way that the 4 intrinsic contributors to compliance are evident:

$$\Large \frac{dV}{dp} = C = \frac{4}{3} \pi r_i^3 \frac{9 r_o^3}{4 E (r_o^3 - r_i^3)} \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2 = \left\{ \frac{9}{4} \right\} \left\{ V_0 \right\} \left\{ \frac{r_o^3}{r_o^3-r_i^3} \right\} \left\{ \frac{1}{E} \right\} \left[1 + \Delta p \frac{3 r_o^3}{4 E (r_o^3 - r_i^3)} \right]^2$$

In the latter we have 4 quantities in curly brackets.  A pure number ($$9/4$$) has only to do with the shape of the vessel (a sphere); $$\large V_0$$ is the intrinsic size of the vessel;  $$\large 1/E$$ has to do with the material properties (only); $$\large \frac{r_o^3}{r_o^3 - r_i^3}$$ is a fraction (pure number) that has to do with the (relative) amount of material incorporated into the vessel.  The square bracketed term is actually a pure number that varies with the distending pressure, $$\large \Delta p$$.  It describes the nonlinear aspect of how the compliance varies with distending pressure.