Elastomer stretched in tension and elongation

[tone999]

Homework Statement

A strip of elastomer was stretched in tension and elongation and was held constant. After 10 minutes the tensile stress in the specimen dropped by 12%. Assuming that the elastomer behaves in accordance with the maxwell model:

(i) Calculate the relaxation time (to the nearest whole number)
(ii) Show that it takes 22 minutes for the stress to drop to 75% of its final value.

Homework Equations

The Attempt at a Solution

I take it we have to use the maxwell model equation here?

sigma (t) = sigma (0) e ^ -t/tau

 

[Redbelly98]

The Attempt at a Solution

I take it we have to use the maxwell model equation here?

sigma (t) = sigma (0) e ^ -t/tau

I think so. Based on the information in the problem statement, what is the ratio,

sigma(t) / sigma(0),​

when t=10 minutes?

 

[tomcue]

(i) To calculate the relaxation time, we can use the given information that after 10 minutes, the tensile stress dropped by 12%. This means that the final stress is 88% of the initial stress, or 0.88*sigma(0). Plugging this into the maxwell model equation, we get:

0.88*sigma(0) = sigma(0) e ^ -10/tau

Solving for tau, we get tau = 11.7 minutes, which can be rounded to 12 minutes.

(ii) To show that it takes 22 minutes for the stress to drop to 75% of its final value, we can use the same equation and plug in the given values. We know that after 10 minutes, the stress is at 88% of its initial value, so we can use this as our new initial stress, or sigma(0). We also know that we want to find the time it takes for the stress to drop to 75% of this value, or 0.75*0.88*sigma(0). Plugging this into the equation, we get:

0.75*0.88*sigma(0) = 0.88*sigma(0) e ^ -t/12

Solving for t, we get t = 22 minutes, which confirms that it takes 22 minutes for the stress to drop to 75% of its final value. This makes sense, as the relaxation time is the time it takes for the stress to drop to 37% of its initial value, and 75% is almost double that value.