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 timedependent and initial (i.e., time \(=0\) ) stresses, respectively, and \(t\) and \(\tau\) denote elapsed time and the relaxation time; \(\tau\) is a time independent constant characteristic of the material. A specimen of some viscoelastic polymer with the stress relaxation that 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. Deter\(\operatorname{mine} E_{r}(10)\) for this material if the initial stress level was 3.5 MPa \((500\) psi), which dropped to \(0.5 \mathrm{MPa}(70 \mathrm{psi})\) after \(30 \mathrm{s}\).

Step by step solution

01

Determine the relaxation time τ

To determine the relaxation time constant τ, we will use the given values of stress and their corresponding times. We will start with the equation for stress decay as a function of time: $$\sigma(t) = \sigma(0) \exp\left(-\frac{t}{\tau}\right)$$ Now, use the given initial stress at time \(t = 0\), which is \(\sigma(0) = 3.5 \mathrm{MPa}\), and the stress after 30 seconds, which is \(\sigma(30) = 0.5 \mathrm{MPa}\). Plug these values into the equation: $$0.5 = 3.5 \exp\left(-\frac{30}{\tau}\right)$$ We can rearrange it and solve for \(\tau\): $$\frac{0.5}{3.5} = \exp\left(-\frac{30}{\tau}\right)$$ $$\ln\left(\frac{0.5}{3.5}\right) = -\frac{30}{\tau}$$ $$\tau = -\frac{30}{\ln\left(\frac{0.5}{3.5}\right)}$$ Now, calculate the value of \(\tau\): $$\tau \approx 14.582$$ So, the relaxation time constant for this material is approximately 14.582 seconds.

02

Find the relaxed modulus Er(10)

To find the relaxed modulus \(E_{r}(10)\) at a time of 10 seconds, we first need to find the stress at that time using the stress decay equation: $$\sigma(10) = \sigma(0) \exp\left(-\frac{10}{\tau}\right)$$ Plug in the known values \(\sigma(0) = 3.5 \mathrm{MPa}\) and \(\tau \approx 14.582\) seconds: $$\sigma(10) = 3.5 \exp\left(-\frac{10}{14.582}\right)$$ Calculate the value of \(\sigma(10)\): $$\sigma(10) \approx 2.093 \mathrm{MPa}$$ Now we can find the relaxed modulus \(E_{r}(10)\) using the equation: $$E_{r}(10) = \frac{\sigma(10)}{\epsilon}$$ where \(\epsilon\) is the applied strain, which is given as 0.5. Plug in the known values: $$E_{r}(10) = \frac{2.093}{0.5}$$ Calculate the value of \(E_{r}(10)\): $$E_{r}(10) \approx 4.186 \mathrm{MPa}$$ The relaxed modulus of the viscoelastic polymer at a time of 10 seconds is approximately 4.186 MPa.

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