Question

For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to where σ(t) and σ(0) represent the time-dependent and initial (i.e., time = 0) stresses, respectively, and t and τ denote elapsed time and relaxation time, respectively; τ is a time-dependent constant characteristic of the material. A specimen of some viscoelastic polymer, the stress relaxation of which obeys this equation, was suddenly pulled in tension to a measured strain of 0.69; the stress necessary to maintain this constant strain was measured as function of time. Determine Er(10) for this material (in MPa) if the initial stre

Answer 1

Answer:

Hello your question lacks some vital parts here is the complete question

Determine Er(10) for this material (in Mpa) If the initial stress level was 2.76 Mpa (400psi) which dropped to 1.72 Mpa(250psi) after 60s

Answer: Er(10) for the material  = 2.55 / 0.69 = 3.695 ≈ 3.70 Mpa

Explanation:

The equation of stress decay for viscoelastic polymers

$\alpha (t) = \alpha (0)exp(-\frac{t}{T} )$    equation 1

$\alpha (t) = time - dependent stress \\ \alpha (0) = initial time stresses$            

t = elapsed time = 60s

T = relaxation time

$\alpha (t) = 1.72 MPa \\ \alpha (0) = 2.76MPa$

input the given values into the equation

1.72 = 2.76 exp ( - $\frac{60}{T})$

exp ( - 60/T ) = 1.72/2.76  therefore T = 126.87S

Considering 10s for t  

$\alpha (10) = (2.76) exp(- \frac{10}{126.87})$

taking Ln of both sides of the equation

Ln $\alpha (10) = ln(2.76) - \frac{10}{126.87}$

therefore $\alpha (10) = 2.55 Mpa$ = 370 psi

To determine the Er(10) for the material in (Mpa) we apply the relaxation modulus equation

Er(t) = $\frac{\alpha(t) }{Eo}$

where Eo = strain level = 0.69

Er(t) = E(10)

$\alpha (t)$ = 2.55 Mpa

input this variables into equation

Er(10) for the material  = 2.55 / 0.69 = 3.695 ≈ 3.70 Mpa