For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to $$ \sigma(t)=\sigma(0) \exp \left(-\frac{t}{\tau}\right) $$ where \(\sigma(t)\) and \(\sigma(0)\) represent the time-dependent and initial (i.e., time = 0 ) stresses, respectively, and \(t\) and \(\tau\) denote elapsed time and the relaxation time, respectively; \(\tau\) is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation \(15.10\) was suddenly pulled in tension to a measured strain of \(0.5 ;\) the stress necessary to maintain this constant strain was measured as a function of time. Determine \(E_{r}(10)\) for this material if the initial stress level was \(3.5 \mathrm{MPa}\) (500 psi), which dropped to \(0.5 \mathrm{MPa}\) (70 psi) after \(30 \mathrm{~s}\).

Step by step solution

01

Find the relaxation time, \(\tau\)

To find \(\tau\), we need to use the provided initial stress level, \(\sigma(0) = 3.5\,\text{MPa}\), and the stress level after 30 seconds, \(\sigma(30) = 0.5\,\text{MPa}\). Plug these values into the stress decay equation and solve for \(\tau\): $$ 0.5\,\text{MPa} = 3.5\,\text{MPa} \exp \left(-\frac{30\,\text{s}}{\tau}\right) $$ Divide both sides by \(3.5\,\text{MPa}\): $$ \frac{1}{7} = \exp \left(-\frac{30\,\text{s}}{\tau}\right) $$ Now, take the natural logarithm of both sides: $$ \ln{\frac{1}{7}} = -\frac{30\,\text{s}}{\tau} $$ Rearrange and solve for \(\tau\): $$ \tau = -\frac{30\,\text{s}}{\ln{\frac{1}{7}}} $$ $$ \tau \approx 8.595\,\text{s} $$

02

Determine the stress at \(t=10\) seconds

Now that we have found the relaxation time, we can calculate the stress at \(t=10\) seconds. Use the stress decay equation with \(\sigma(0) = 3.5\,\text{MPa}\) and \(\tau \approx 8.595\,\text{s}\): $$ \sigma(10) = 3.5\,\text{MPa} \exp \left(-\frac{10\,\text{s}}{8.595\,\text{s}}\right) $$ $$ \sigma(10) \approx 1.287\,\text{MPa} $$

03

Calculate the relaxation modulus, \(E_r(10)\)

The relaxation modulus, \(E_r(10)\), can be determined by dividing the stress at \(t=10\) seconds, \(\sigma(10)\), by the given strain, \(0.5\): $$ E_{r}(10) = \frac{\sigma(10)}{\epsilon} $$ $$ E_{r}(10) = \frac{1.287\,\text{MPa}}{0.5} $$ $$ E_{r}(10) \approx 2.574\,\text{MPa} $$ So the relaxation modulus of the material at \(t=10\) seconds is approximately \(2.574\,\text{MPa}\).

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