Question

The behavior of a polymer is described by a Maxwell model consisting of a spring element of modulus \( 10^{10} \) Pa in series with a dashpot of viscosity \( 10^{12} \) Pa.s. In the solid, 50 s after the sudden application of a fixed strain of 1%, the stress (rounded off to 2 decimal places) will be \(\underline{\hspace{2cm}}\) \( \times 10^7 \) Pa.

Hint:

For Maxwell models, use the formula \( \sigma(t) = \frac{\epsilon}{\eta} \left(1 - e^{-\frac{t}{\tau}}\right) \) to compute the stress at different times.

Answer 1

Correct Answer: 6

For a Maxwell model, the stress \( \sigma(t) \) as a function of time is given by:
\[ \sigma(t) = \frac{\epsilon}{\eta} \left(1 - e^{-\frac{t}{\tau}}\right) \] where:
- \( \epsilon = 1% = 0.01 \),
- \( \eta = 10^{12} \, \text{Pa.s} \),
- \( \tau = \frac{\eta}{G} = \frac{10^{12}}{10^{10}} = 10^2 \, \text{s}. \) Substituting the values at \( t = 50 \, \text{s} \), we get:
\[ \sigma(50) = \frac{0.01}{10^{12}} \left(1 - e^{-\frac{50}{100}}\right) \] Approximating \( e^{-\frac{50}{100}} \approx 0.6065 \), we get:
\[ \sigma(50) \approx 0.01 \times 10^{10} \times (1 - 0.6065) = 0.01 \times 10^{10} \times 0.3935 = 3.935 \times 10^7 \, \text{Pa}. \] Thus, the stress is approximately \( 3.94 \times 10^7 \, \text{Pa} \).