A wheat dough specimen is modeled as a linear Maxwell viscoelastic element consisting of a Hookean spring with modulus $E = 25.0\text{ kPa}$ in series with a Newtonian dashpot with viscosity $\eta = 300.0\text{ kPa}\cdot\text{s}$. If an instantaneous constant strain $\varepsilon_0$ is applied at $t = 0$, the percentage of initial relaxation stress remaining after $t = 18.0\text{ seconds}$ is ________ % (round off to two decimal places).
Answer Key:
22.31
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Maxwell Constitutive Equation**
For a Maxwell model under constant strain ($\dot{\varepsilon} = 0$ for $t > 0$):
$\frac{\dot{\sigma}}{E} + \frac{\sigma}{\eta} = 0 \implies \frac{d\sigma}{\sigma} = -\frac{E}{\eta} dt$
Integrating from $t = 0$ (where $\sigma = \sigma_0 = E \varepsilon_0$) to $t$:
$\sigma(t) = \sigma_0 \exp\left(-\frac{t}{\tau_r}\right)$
where the relaxation time is:
$\tau_r = \frac{\eta}{E} = \frac{300.0\text{ kPa}\cdot\text{s}}{25.0\text{ kPa}} = 12.0\text{ s}$
At $t = 18.0\text{ s}$:
$\frac{\sigma(18)}{\sigma_0} = \exp\left(-\frac{18.0}{12.0}\right) = \exp(-1.5) \approx 0.22313$
Percentage of initial stress remaining:
$0.22313 \times 100\% = 22.31\%$
**Method 2: Dimensionless Relaxation Ratio**
$\text{Stress Ratio} = \exp(-t/\tau_r) = e^{-18/12} = e^{-1.5} = 0.22313 \implies 22.31\%$