Question

A stress of 2 MPa is required to stretch a 2 cm strip of excised porcine skin to 5 cm. After 2 hours in the same stretched position, the relaxation time was 1.45 hours. Assume the properties of skin do not vary appreciably during the experiment. What is the new exerted stress

Answer 1

502.2 kPa

Step-by-step explanation:

Stress at any instance, t can be given by

`\sigma (t)=\epsilon_o Ee^(-(t/\tau))`

Strain,
`\epsilon=\frac {\triangle l}{l}=\frac {5-2}{2}=1.5`

`E=\frac {Stress}{Strain}=\frac {2}{1.5}=1.333333\approx 1.33`

Therefore, after 2 hours where t=2 then

`\sigma (2)=1.5* 1.33 Mpa e^(-(2/1.45))\approx 502.2 KPa`