## Thick-Walled Cylindrical Vessel

The video suggests something about the way deformation occurs in a thick-walled tubular 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;  it demonstrates the principals described qualitatively but is not quantitatively representative of a cylindrical vessel.

For the thick-walled cylindrical vessel,  look at the before ( left ) and after ( right ) figures of the end-on vessel below.  To motivate the next discussion, how do you think I determined how to draw this schematic?

The thing that's the same between the left and right figures is the amount of of tissue, i.e. the cross-sectional area of the vascular wall itself (in gray). Assuming the tissue is incompressible ( a good assumption ) this is what would happen to a thick-walled vessel if it were distended from the configuration on the left to the one on the right (I've assumed that the length of the vessel didn't change.)  The outer diameter changes a little; the inner diameter changes a lot.  As a matter of fact, this is a rather large deformation with the inner diameter changing by a factor of 2; the outer diameter changes much less. Why does this occur? Suppose the inner radius of the figure on the left is $$r_{1,a}$$ and the outer radius is $$r_{1,b}$$; The vessel has a length $$l_1$$.  Then the volume of material in the wall is simply determined for the radial geometry.

$$\Large V = \pi l_1[r_{1,b}^2 - r_{1,a}^2]$$

We can be more general than that. Whatever radial coordinate you choose, $$r_1$$, the volume of tissue from the inner wall out to that coordinate is given by the same formula:

$$\Large V_1 = \pi l_1[r_{1}^2 - r_{1,a}^2]$$

Likewise the volume of tissue in the deformed state on the right, with inner radius $$r_{2,a}$$ and length $$l_2$$, is given by a similar formula:

$$\Large V_2 = \pi l_2[r_{2}^2 - r_{2,a}^2]$$

If the volume of each/every element of tissue remains the same, then the radial coordinate in the deformed state ($$2$$) is determined from the original state ($$1$$).

$$\Large \pi l_1[r_{1}^2 - r_{1,a}^2] = \pi l_2[r_{2}^2 - r_{2,a}^2]$$

$$\Large r_2^2 = \frac{l_1}{l_2} [r_1^2-r_{1,a}^2]+r_{2,a}^2$$

$$\Large r_2 = \sqrt{\frac{l_1}{l_2} [r_1^2-r_{1,a}^2]+r_{2,a}^2}$$

Observe the above figure closely.  In the original undeformed state (left), the grid shows an even (linear) progression of the radial grid lines.  In the deformed state (right), the radial grid lines are unevenly spaced. There is obviously greater circumferential deformation at the inner wall; radial deformation is greater there also.

### Boundary Conditions

In this article we are going to derive the stresses and strains on a thick-walled cylindrical tube using the principles of Newton's laws as they are applied in solid mechanics.  It's assumed that you haven't necessarily been exposed to this type of endeavor.  Unfortunately this is the type of math problem that is typically done in a solid mechanics course after you've had some background and experience.   Don't fret if this seems like a difficult problem -- we're just trying to get some insights into how materials behave ( and consequently why this isn't a very easy endeavor applied to something as complicated as a heart ).    Before proceeding allow me to confess that I'm simply paraphrasing the development given in a document at an Iowa State University website (author not stated).  This will give any readers a chance to try to struggle through the logic of solving a solid mechanics problem using the pencil and paper route.

When solving a mathematical problem of this complexity "boundary conditions" describe the circumstances of the problem.  We solve mathematical problems involving partial differential equations inside a region of interest called the "domain".  To do that, we have to describe exactly ( mathematically ) what is happening at all the boundaries of the domain.  For our problem we have a cylindrical tube of inner radius $$r_a$$, outer radius $$r_b$$, and length $$l$$.  In all that follows, $$\sigma$$ is the symbol used to represent mechanical stress ( force per unit area ).  $$\sigma_{xy}$$ should be read as "the stress on the $$x$$ face in the $$y$$ direction".  Only our problem is best suited to a cylindrical coordinate system where the directions are radial ($$r$$), azimuthal ($$\theta$$), and longitudinal ($$z$$).  While the radial and azimuthal gridlines are always perpendicular to each other, the directions are not constant.  This is an example of a curvilinear coordinate system.

The initial part of the problem includes a specification of the pressures at the inner ($$p_a$$) and outer walls ($$p_b$$) of the tube.  These constitute compressive stresses in the radial direction and so appear as negative quantities in the problem specification as shown.

$$\Large \sigma_{rr}(r,\theta,z)|_{r=r_a} = -p_a$$

$$\Large \sigma_{rr}(r,\theta,z)|_{r=r_b} = -p_b$$

The notation $$|r=r_a$$ is used to indicate where on the surface of the domain the pressure specification is being made. Writing $$\sigma_{rr}(r,\theta,z)$$ with the independent variables in parentheses indicates explicitly that the stress might vary in all directions; we will simplify the specification shortly.

$$\Large \sigma_{r\theta}(r,\theta,z)|_{r=r_a} = \sigma_{rz}(r,\theta,z)|_{r=r_a} = 0$$

$$\Large \sigma_{r\theta}(r,\theta,z)|_{r=r_b} = \sigma_{rz}(r,\theta,z)|_{r=r_b} = 0$$

$$\Large \sigma_{zz}(r,\theta,z)|_{z=0} = \sigma_{zr}(r,\theta,z)|_{z=0} = \sigma_{z\theta}(r,\theta,z)|_{z=0} = 0$$

$$\Large \sigma_{zz}(r,\theta,z)|_{z=l} = \sigma_{zr}(r,\theta,z)|_{z=l} = \sigma_{z\theta}(r,\theta,z)|_{z=l} = 0$$

The previous 4 lines state explicitly that shear stresses at all the boundary surfaces are being set to 0.  Shear stresses are indicated whenever the 2 subscripts of the stress are not the same, $$\sigma_{rz}$$ for example (" stress on the $$r$$ surface in the $$z$$ direction "). Shear stresses are generated in a blood vessel at the luminal surface when fluid ( blood ) flows through the vessel.  They also occur when the blood vessel moves but is restrained by the surrounding connective tissue.  So this might not be a realistic specification for some purposes; the purpose  of a model is always a key issue since the solution will not exhibit any behavior not included in the model.  There is an indication in the above that the stresses at the ends of the vessel are also being set to 0 at both ends of the vessel ($$\sigma_{zz} = 0$$).  The simplified boundary condition statements are shown below; the situation has been reduced to one where only radial and azimuthal stresses are present and these depend only on the radial coordinate $$r$$.

$$\Large \sigma_{rr} = \sigma_{rr}(r)$$

$$\Large \sigma_{\theta \theta} = \sigma_{\theta \theta} (r)$$

$$\Large \sigma_{r \theta} = \sigma_{rz} = \sigma_{z \theta} = \sigma_{zz} = 0$$

$$\Large \sigma_{rr}(r)|_{r=r_a} = -p_a$$

$$\Large \sigma_{rr}(r)|_{r=r_b} = -p_b$$

### Equations of Motion

The next thing we do is figure out the equations that describe the motion of the material everywhere inside the domain.  The starting place for such a derivation is a free body diagram (below) of a chunk of material. In this case, it consists of an elemental piece from the polar coordinate system. The equations are derived from Newton's laws of motion and are a direct mathematical statement that $$\textbf{F}=m\textbf{a}$$ everywhere inside the domain.  A scrupulous summation of the stresses acting on the surfaces of an elemental volume in a polar coordinate system yields the following:

$$\Large \frac{\partial \sigma_{rr}}{\partial r } + \frac{\sigma_{rr}-\sigma_{\theta \theta}}{r } + \frac{1}{r} \frac{\partial \sigma_{r\theta}}{\partial \theta } + \frac{\partial \sigma_{rz}}{\partial z } + F_r = \rho \frac{\partial^2 u_r }{\partial t^2}$$

$$\Large \frac{\partial \sigma_{r\theta}}{\partial r } + \frac{2 \sigma_{r\theta}}{r} + \frac{1}{r}\frac{\partial \sigma_{\theta \theta}}{\partial \theta } + \frac{\partial \sigma_{\theta z}}{\partial z } + F_{\theta}= \rho \frac{\partial^2 u_{\theta} }{\partial t^2}$$

$$\Large \frac{\partial \sigma_{zr}}{\partial r } + \frac{\sigma_{zr}}{r} + \frac{1}{r} \frac{\partial \sigma_{z \theta}}{\partial \theta } + \frac{\partial \sigma_{z z}}{\partial z } + F_z = \rho \frac{\partial^2 u_z }{\partial t^2}$$

These are the Navier-Cauchy equations (in polar coordinates).   Clinicians I've encountered will typically throw up their hands on sight of this jumble, but there are engineers all over the world that use these equations on a daily basis. (Furthermore you are looking at a work of beauty and simplicity; you should be able to appreciate the sound of a Stradivarius violin even if you can't play one.)  There are journals, societies, and conferences dedicated to the application and solution of these.  The derivation of the N-C equations is no simple matter, but there are many engineering texts that stick it in as a matter of course and engineering students are sometimes asked to derive them as part of a homework set ( at least they were when I was in school ).  What we have here are 3 equations (1 for each of the $$r$$, $$\theta$$, and $$z$$ directions) stating that force equals mass multiplied by acceleration.  The left-hand side contains terms having to do with the stresses ($$\sigma$$) acting on the surface of the elemental volume, and also a body force, $$\textbf{F}$$, which has been subscripted to indicate components in each of the 3 directions.  Body forces are ones that seem to reach out to affect material elements from beyond the domain; the most obvious one from everyday experience is gravity, but an accelerating frame of reference is another example where an apparent body force may be present.  The right-hand side of the equation is here written in terms of the displacement ( distance from original position ), $$u_r$$, $$u_r$$, and $$u_{\theta}$$.  Shown is the second derivative with respect to time of each displacement multiplied by the density ($$\rho$$) of the element.  Hence the right-hand side is the mass ( per unit of volume ) multiplied by acceleration.

The next group of equations define the engineering strains ($$\epsilon$$) in terms of the displacements:  $$u_r$$, $$u_{\theta}$$ and $$u_z$$.  Displacement means that an element in the original material has moved to a new location; it's displaced in the $$r$$, $$\theta$$, and $$z$$ directions by the amount indicated by the displacement vector, $$\mathbold{u}$$.

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

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

$$\Large \epsilon_{zz} = \frac{\partial u_{z} }{\partial z}$$

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

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

$$\Large \epsilon_{zr} = \frac{1}{2} \left[ \frac{\partial u_{r} }{\partial z} + \frac{\partial u_{z} }{\partial r} \right]$$

Note that these equations include no assumptions whatsoever as to the type of material under study.  This is just a rigorous statement of Newton's law for a continuous material ( hence the restrictions are that the continuum hypothesis is applicable and that the chunk of material is moving nowhere near the speed of light relative to us ).

$$\LARGE \frac{\partial \sigma_{rr}}{\partial r } + \frac{\sigma_{rr}-\sigma_{\theta \theta}}{r } = 0$$

$$\Large \epsilon_{rr} = \frac{1}{E}(\sigma_{rr}-\nu \sigma_{\theta \theta})$$

$$\Large \epsilon_{\theta \theta} = \frac{1}{E}(\sigma_{\theta \theta}-\nu \sigma_{rr})$$

$$\Large \epsilon_{zz} = \frac{-\nu}{E}(\sigma_{rr}+ \sigma_{\theta \theta})$$

$$\Large \epsilon_{r\theta} = \frac{\sigma_{r\theta}}{G} = 0$$

$$\Large \epsilon_{z\theta} = \frac{\sigma_{z\theta}}{G} = 0$$

$$\Large \epsilon_{rz} = \frac{\sigma_{rz}}{G} = 0$$

$$\Large \sigma_{rr} = \frac{E}{1-\nu^2} (\epsilon_{rr}+\nu\epsilon_{\theta \theta})$$

$$\Large \sigma_{\theta \theta} = \frac{E}{1-\nu^2} (\epsilon_{\theta \theta}+\nu\epsilon_{rr})$$

$$\Large \epsilon_{zz} = \frac{-\nu}{E} (\sigma_{rr}+\sigma_{\theta \theta})$$

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

$$\Large \epsilon_{\theta \theta} = \frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta} + \frac{u_r}{r}$$

$$\Large \epsilon_{zz} = \frac{\partial u_z}{\partial z}$$

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

$$\Large \epsilon_{rz} = \frac{1}{2} \left[\frac{\partial u_z}{\partial r} + \frac{\partial u_r}{\partial z} \right]$$

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

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

$$\Large \epsilon_{\theta \theta} = \frac{u_r}{r}$$

$$\Large \epsilon_{zz} = \frac{\partial u_z}{\partial z}$$

$$\Large \epsilon_{r\theta} = \epsilon_{\theta z} = \epsilon_{rz} = 0$$

$$\Large \sigma_{rr} = \frac{E}{1-\nu^2} \left[ \frac{d u_r}{dr} +\nu\frac{u_r}{r}\right]$$

$$\Large \sigma_{\theta \theta} = \frac{E}{1-\nu^2} \left[ \frac{u_r}{r} +\nu\frac{du_r}{dr}\right]$$

$$\Large \epsilon_{zz} = \frac{du_z}{dz} = \frac{-\nu}{E}(\sigma_{rr}+\sigma_{\theta \theta})$$

$$\Large \frac{d^2 u_r}{dr^2} + \frac{1}{r} \frac{du_r}{dr} - \frac{u_r}{r^2} = 0$$

$$\Large \frac{d}{dr} \left[ \frac{1}{r} \frac{d}{dr} (r u_r)\right] =0$$

$$\Large \frac{d}{dr} (r u_r) = C'_1 r$$

$$\Large u_r = \frac{C'_1}{2} r + \frac{C_2}{r} = C_1 r + \frac{C_2}{r}$$

$$\Large \sigma_{rr} = \frac{E}{1-\nu} C_1 - \frac{E}{1+\nu} \frac{C_2}{r^2}$$

$$\Large \sigma_{\theta \theta} = \frac{E}{1-\nu} C_1 + \frac{E}{1+\nu} \frac{C_2}{r^2}$$

$$\Large \epsilon_{zz} = \frac{-2\nu}{1-\nu} C_1$$

$$\Large A \equiv \frac{E}{1-\nu} C_1$$

$$\Large B \equiv \frac{E}{1+\nu} C_2$$

$$\Large \sigma_{rr} = A - \frac{B}{r^2}$$

$$\Large \sigma_{\theta \theta} = A + \frac{B}{r^2}$$

$$\Large \epsilon_{zz} = \frac{du_z}{dz} = \frac{-2\nu A}{E}$$

$$\Large A - \frac{B}{r_a^2} = -p_a$$

$$\Large A - \frac{B}{r_b^2} = -p_b$$

$$\Large A = \frac{r_a^2 p_a-r_b^2 p_b}{r_b^2-r_a^2}$$

$$\Large B = \frac{r_a^2 r_b^2(p_a-p_b)}{r_b^2-r_a^2}$$

$$\Large \sigma_{rr} = \frac{p_a r_a^2}{r_b^2-r_a^2} \left[ 1 - \frac{r_b^2}{r^2} \right] - \frac{p_b r_b^2}{r_b^2-r_a^2} \left[ 1 - \frac{r_a^2}{r^2} \right]$$

$$\Large \sigma_{\theta \theta} = \frac{p_a r_a^2}{r_b^2-r_a^2} \left[ 1 + \frac{r_b^2}{r^2} \right] - \frac{p_b r_b^2}{r_b^2-r_a^2} \left[ 1 + \frac{r_a^2}{r^2} \right]$$

$$\Large \epsilon_{zz} = \frac{du_z}{dz} = \frac{-2\nu}{E} \left[ \frac{r_a^2 p_a - r_b^2 p_b}{r_b^2-r_a^2} \right] \rightarrow u_z = \frac{-2\nu}{E} \left[ \frac{r_a^2 p_a - r_b^2 p_b}{r_b^2-r_a^2} \right] z + C$$

$$\Large u_r = \frac{1-\nu}{E} \left[ \frac{r_a^2 p_a - r_b^2 p_b}{r_b^2-r_a^2} \right] r+ \frac{1+\nu}{E} \left[\frac{r_b^2 r_a^2 (p_a-p_b)}{r_b^2-r_a^2} \right] \frac{1}{r}$$