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where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed time and the relaxation time; τ is a time-independent constant characteristic of the material. A specimen of some viscoelastic polymer with the stress relaxation that obeys equation above was suddenly pulled in a tension to a measured strain of 0.49; the stress necessary to maintain this constant strain was measured as a function of time. Determine Er(6) for this material if the initial stress level was 3.1 MPa (440 psi), which dropped to 0.41 MPa (59 psi) after 32 s.

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Answer:

[tex]E_r(6)=4.35614\ MPa[/tex]

Explanation:

[tex]\epsilon[/tex] = Strain = 0.49

[tex]\sigma _0[/tex] = 3.1 MPa

At t = Time = 32 s [tex]\sigma[/tex] = 0.41 MPa

[tex]\tau[/tex] = Time-independent constant

Stress relation with time

[tex]\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)[/tex]

at t = 32 s

[tex]0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s[/tex]

The time independent constant is 16.0787 s

[tex]E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}[/tex]

At t = 6

[tex]\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}[/tex]

From the first equation

[tex]\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451[/tex]

[tex]E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa[/tex]

[tex]E_r(6)=4.35614\ MPa[/tex]

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