Answer to Question #225569 in Mechanical Engineering for john luvis

Question

An artery with dimensions OD = 20 mm and ID = 17 mm is subjected to pressures ranging from 80 mmHg to 130 mmHg.

(a) Determine the diameter of the artery of the systolic (highest) and diastolic (lowest) points. The material in the artery follows the relationship σ = kε 2 , with k = 25 MPa.

(b) Plot the pressure–radius curve due to the internal pressure from the stress–strain response of the material.

Answer 1

Part a

Even though the radial strain and the circumferential strain physically represent two different things, they are numerically equal.

First, we need to calculate the hoop stress and then use the material relation, we can calculate the hoop strain, and from the Houston, we can calculate the changing radius.

$Hoop \space stress = \frac{Pd}{2t}\\ \sigma_d= \frac{10665*17}{2*1.5}=60439.5 Pa\\ Hoop \space stress \space at \space systolic , \sigma_s= \frac{17331.9*17}{2*1.5}=98241.1 Pa\\ Stress \space strain \space relation, \sigma = 25*10^6 \varepsilon^2\\ then \space \varepsilon_d; 60349.5= 25*10^6* \varepsilon_d^2\\ \varepsilon_d = \sqrt{2.41398*10^{-3}}=0.049132\\ \varepsilon_d= \frac{Change \space in \space diameter}{Diameter}\\ 0.049132= \frac{Change \space in \space diameter}{17}\\ Change \space in \space diameter = 0.835 \\ Final \space diameter = 17+0.835=17.835 mm\\$

Similarly for systolic

$\varepsilon_s; 98241.1= 25*10^6 *\varepsilon_s^2\\ \varepsilon_s= \sqrt{39296*10^{-3}}\\ \varepsilon_s = 0.06268 = \frac{Diameter \space change }{Original \space diameter}\\ Diameter \space change = 17*0.06268=1.06577 mm\\ Final \space diameter = 18.6577 mm$

Part b

The relation between strain and pressure

$\varepsilon= \sqrt{\frac{Pd}{2t}* \frac{1}{25*10^6}}\\ Radius = (1+ \varepsilon)R\\ Final \space Radius = (1+\sqrt{\frac{P*17}{2*1.5}* \frac{1}{25*10^6}})*8.5\\$