Find Radial & Circumferential Strain in Incompressible Vessel

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To find the radial and circumferential strain in an incompressible blood vessel, the relevant equations are \epsilon_i_j = \lambda_\theta - 1 for circumferential strain and \epsilon_r = \lambda_r - 1 for radial strain. The radial strain is calculated as \epsilon_r = (a + ∆a)/a - 1, where 'a' is the initial radius. For circumferential strain, the stress ratio is determined using \lambda_\theta = (2π(a + ∆a))/(2πa), leading to \epsilon_\theta = (2π(a + ∆a))/(2πa) - 1. The overall infinitesimal strain tensor throughout the vessel wall is represented as \epsilon_i_j = [\epsilon_r, 0, 0; 0, \epsilon_\theta, 0; 0, 0, 0].
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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, \epsilon_i_j throughout the wall.

ok...so the circumferential strain is defined as \epsilon_i_j = \lambda_\theta - 1. and the radial strain is \epsilon_r = \lambda_r - 1. 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...
 
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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.
 
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