Find Radial & Circumferential Strain in Incompressible Vessel

[chusifer]

here's the question...the inner radius of a blood vessel with circular cylindrical cross section is distended during pressure elevation from radius a to a+∆a. assume the wall of the vessel is incompressible and the length of the vessel is constant. find the radial and circumferential infinitesimal strain, [tex]\epsilon_i_j[/tex] throughout the wall.

ok...so the circumferential strain is defined as [tex]\epsilon_i_j = \lambda_\theta - 1[/tex]. and the radial strain is [tex]\epsilon_r = \lambda_r - 1[/tex]. and the stress ratio, lambda is computed by the final length/initial length right? so is the radial strain just a+∆a/a - 1? and there's no value for the z axis since the vessel is imcompressible...so how would i get the stress ratio for the circumferential strain? many thanks...

 

[anathema]

Hi there,

Thank you for your question. In order to calculate the radial and circumferential infinitesimal strain throughout the wall of the blood vessel, we need to first define the parameters involved. In this case, we have the initial radius of the vessel, a, and the change in radius, ∆a, due to pressure elevation. We also have the length of the vessel, which is constant.

To calculate the radial infinitesimal strain, we use the formula \epsilon_r = \frac{\Delta a}{a}, where a is the initial radius and ∆a is the change in radius. This is because the strain is defined as the change in length divided by the initial length. Therefore, the radial infinitesimal strain throughout the wall would be \epsilon_r = \frac{a+\Delta a}{a} - 1.

To calculate the circumferential infinitesimal strain, we use the formula \epsilon_\theta = \frac{\Delta l}{l}, where l is the initial length of the vessel and ∆l is the change in length due to pressure elevation. However, in this case, the length of the vessel is constant, so we do not have a value for ∆l. Instead, we can use a different formula for the circumferential strain, which is \epsilon_\theta = \lambda_\theta - 1, where \lambda_\theta is the stress ratio in the circumferential direction. The stress ratio is calculated as the final length divided by the initial length, so in this case, it would be \lambda_\theta = \frac{a+\Delta a}{a}. Therefore, the circumferential infinitesimal strain throughout the wall would be \epsilon_\theta = \frac{a+\Delta a}{a} - 1.

I hope this helps to clarify the calculations for the radial and circumferential infinitesimal strain in this scenario. Please let me know if you have any further questions. Thank you.

 

[wais]

To find the radial and circumferential strain in an incompressible vessel, we can use the equations \epsilon_i_j = \lambda_\theta - 1 and \epsilon_r = \lambda_r - 1, where \lambda_\theta and \lambda_r are the stress ratios in the circumferential and radial directions, respectively.

Since the vessel is assumed to be incompressible, the length of the vessel remains constant. Therefore, the stress ratio in the z-direction is equal to 1, and we can focus on the stress ratios in the circumferential and radial directions.

To find the stress ratio in the radial direction, we can use the equation \lambda_r = \frac{a+\Delta a}{a}, where a is the initial radius and a+\Delta a is the final radius. This results in a radial strain of \epsilon_r = \frac{a+\Delta a}{a} - 1.

To find the stress ratio in the circumferential direction, we can use the equation \lambda_\theta = \frac{2\pi(a+\Delta a)}{2\pi a}, where 2\pi a is the initial circumference and 2\pi(a+\Delta a) is the final circumference. This results in a circumferential strain of \epsilon_\theta = \frac{2\pi(a+\Delta a)}{2\pi a} - 1.

Overall, the infinitesimal strain throughout the wall would be given by \epsilon_i_j = \begin{bmatrix} \epsilon_r & 0 & 0 \\ 0 & \epsilon_\theta & 0 \\ 0 & 0 & 0 \end{bmatrix}.

I hope this helps clarify the calculation process. Please let me know if you have any further questions.