Question

Stress relaxation: A stress of 1.5 MPa was required to stretch a 3-cm long aorta strip to 3.4 cm. After 2 hours in the same stretched position, the strip exerted a stress of 1.2 MPa. Assume the mechanical property of the aorta did not vary appreciably during the experiment.What is the relaxation time, assuming a simple exponential decay model?

a) 1 hour : 30 minutes
b) 2 hours : 15 minutes
c) 3 hours : 45 minutes
d) 4 hours : 30 minutes

Answer 1

The relaxation time can be calculated using the formula Δσ/σ₀ = e⁽⁻ᵗ/τ⁾. Substituting the given values, we find that the relaxation time is approximately 1 hour : 30 minutes.

The answer is option ⇒A

Step-by-step explanation:

To calculate the relaxation time, we can use the formula:

Δσ/σ₀= e⁽⁻ᵗ/τ⁾

Where:

• Δσ is the decrease in stress
• σ₀ is the initial stress
• t is the time
• τ is the relaxation time

Given that the initial stress is 1.5 MPa and decreases to 1.2 MPa after 2 hours, we can rearrange the formula to solve for τ:

τ = -t / ln(Δσ / σ₀)

Substituting the values into the formula:

τ = -2 / ln(1.2 / 1.5)

Calculating this, we find that τ is approximately 1 hour : 30 minutes.

Therefore, the correct answer is option a) 1 hour : 30 minutes.