## Thin-Walled Cylindrical Vessel

### Thin-Walled Cylindrical Vessel

The geometry of a vessel has a lot to do with it's compliance.  For the next example, we'll try to figure out the compliance of a thin-walled tube, reminiscent of a straight segment of artery.  We start with a free body diagram as before ( as always ):

Imagine that somewhere along the tube, the ends are actually sealed. Then for an axial force balance:

$$\Large \sigma_z [2 \pi r_0\, h] = \Delta p [\pi r_0^2]$$

$$\Large \sigma_z = \Delta p \frac{r_0}{2 h}$$

In the first of these 2 equations we have a stress, either $$\sigma_z$$ or $$p$$, that acts on (multiplies) a surface area (within the square brackets) to obtain a force ( tension ).  Once again, the tension is of no particular interest, but we could always figure it out if we know the stress.  In the second equation, we have solved for the axial stress in the wall.  It's the same that we obtained for the sphere; the geometry is essentially identical.

For the tangential stress shown as $$\sigma_{\theta}$$ (also called the "hoop stress"), the result is somewhat different:

$$\Large \sigma_{\theta} [2 h l] = \Delta p [2 r_0 l]$$

$$\Large \sigma_{\theta} = \Delta p \frac{r_0}{h}$$

where $$l$$ is the length of the vessel under consideration. The tangential stress is twice the axial stress.  Geometry matters! To figure out the compliance of the vessel due to radial distention only ( thereby stretching the vessel circumferentially ) we have a problem similar to the sphere above.  Here's the strain:

$$\Large \epsilon_{\theta} = \frac{2\pi \Delta r}{2\pi r_0} = \frac{\Delta r}{r_0}$$

Here's the strain substituted in place of the stress:

$$\Large \sigma_{\theta} = E \epsilon_{\theta} =\Delta p \frac{r_0}{h}$$

$$\Large \sigma_{\theta} = E \frac{\Delta r}{r_0} =\Delta p \frac{r_0}{h}$$

Rearranging:

$$\Large \frac{\Delta r}{\Delta p} = \frac{r_0^2}{E h}$$

Here's the volume of the cylinder:

$$\Large V_c = l\pi r^2$$

$$\Large \frac{dV_c}{dr} = 2l\pi r$$

Consequently:

$$\Large \frac{dV}{dp} = \frac{dV_c}{dr} \frac{\Delta r}{\Delta p} = (2l\pi r_0) \frac{r_0^2}{E h}$$

$$\Large \frac{dV}{dp} = 2 \pi l\frac{r_0^3}{E h} = 2 V \frac{r_0}{Eh}$$

$$\Large \frac{dV}{dp}/V = 2 \frac{r_0}{Eh}$$

That's the compliance of the vessel. we call this a "fully constrained" vessel since we haven't allowed it to change dimension in the axial direction at all.  It makes sense for a cylindrical geometry to talk about the compliance per unit of length which is really just the rate of change in cross sectional area with respect to pressure.

$$\Large \frac{dA}{dp} = 2 \pi \frac{r_0^3}{E h} = 2 A \frac{r_0}{Eh}$$

$$\Large \frac{dA}{dp}/A = 2 \frac{r_0}{Eh}$$

### Thin-Walled Cylindrical Vessel 2

We'll embellish the previous results somewhat by considering what might happen if strain occurs in more than one direction.  It turns out that deforming a material in one direction also deforms it in others.  This fact is expressed through another material property called Poisson's ratio:

$$\Large \nu \equiv - \frac{\epsilon_y}{\epsilon_x} = - \frac{\epsilon_z}{\epsilon_x}$$

$$\Large \epsilon_y = -\nu \epsilon_x$$

The first of these 2 equations depicts a definition of Poisson's ratio, $$\nu$$, as a ratio of strains.  If the material is isotropic ( properties are not directionally dependent ) then we just have 1 value for $$\nu$$.  If the material is incompressible ( does not change volume ), as is typically the case for biological tissue, then it will turn out that the value of $$\nu = 0.5$$.  That's the usual situation for biological tissue (it contains a lot of water which is incompressible).  It means that if we stretch a piece of tissue in one direction, it will contract in the other 2 to maintain its volume!

Since strain for a Hookian material can be written as $$\epsilon = \sigma/E$$, a more extensive version of Hooke's law for circumstances where stresses are present in multiple directions appears as follows:

$$\Large \epsilon_x = \frac{1}{E} [\sigma_x - \nu (\sigma_y +\sigma_z)]$$

$$\Large \epsilon_y = \frac{1}{E} [\sigma_y - \nu (\sigma_z +\sigma_x)]$$

$$\Large \epsilon_z = \frac{1}{E} [\sigma_z - \nu (\sigma_x +\sigma_y)]$$

This shows us that the strain ( deformation ) in one direction also depends on stresses applied in other directions.  The above equations can get significantly more complicated when we have an anisotropic material in which case values for both $$\nu$$ and $$E$$ can depend on directional orientation within the material.

Applying these findings to the thin-walled vessel complicates things a little. We still have stresses as before:

$$\Large \sigma_z = \Delta p \frac{r_0}{2 h}$$

$$\Large \sigma_{\theta} = \Delta p \frac{r_0}{h}$$

The deformations however, are due to stresses in more than one direction:

$$\Large \epsilon_{\theta} = \frac{1}{E} [\sigma_{\theta} - \nu \sigma_{z}]$$

$$\Large \epsilon_{z} = \frac{1}{E} [\sigma_{z} - \nu \sigma_{\theta}]$$

The definitions for the  infinitesimal (small) strains:

$$\Large \epsilon_{\theta} = \frac{\Delta r}{r_0}$$

$$\Large \epsilon_{z} = \frac{\Delta l}{l_0}$$

And substituting everything back into the previous equations:

$$\Large \frac{\Delta r}{r_0} = \frac{\Delta p}{hE} [r_0 - \nu \frac{r_0}{2}]=r_0 \frac{\Delta p}{hE} [1 - \frac{\nu}{2}]$$

$$\Large \frac{\Delta l}{l_0} = \frac{\Delta p}{hE} [\frac{r_0}{2} - \nu r_0] =r_0 \frac{\Delta p}{hE} [\frac{1}{2} - \nu]$$

With $$\nu = 0.5$$, we have that the longitudinal strain the equation disappears and:

$$\Large \frac{dr}{dp} = \frac{3r_0^2}{4hE}$$

$$\Large \frac{dV}{dp} = \frac{dV}{dr} \frac{dr}{dp}= \left[2\pi r_0 l\right] \left[\frac{3r_0^2}{4hE}\right] = \frac{3\pi r_0^3 l}{2hE} = \frac{3Vr_0}{2hE}$$

$$\Large \frac{dV}{dp}/V = \frac{3r_0}{2hE}$$

These are the compliance and volume distensibility. Also:

$$\Large \frac{dA}{dp} = \frac{3\pi r_0^3}{2hE} = \frac{3 r_0 A}{2hE}$$

$$\Large \frac{dA}{dp}/A = \frac{3 r_0 }{2hE}$$

These are the compliance per unit length and the area distensibility.