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olga55 [171]
3 years ago
10

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.
Physics
1 answer:
lesya [120]3 years ago
6 0

Answer:

E_r(6)=4.35614\ MPa

Explanation:

\epsilon = Strain = 0.49

\sigma _0 = 3.1 MPa

At t = Time = 32 s \sigma = 0.41 MPa

\tau = Time-independent constant

Stress relation with time

\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)

at t = 32 s

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

The time independent constant is 16.0787 s

E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}

At t = 6

\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}

From the first equation

\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

E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa

E_r(6)=4.35614\ MPa

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