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
[tex]\alpha (t) = \alpha (0)exp(-\frac{t}{T} )[/tex] equation 1
[tex]\alpha (t) = time - dependent stress \\ \alpha (0) = initial time stresses\\[/tex]
t = elapsed time = 60s
T = relaxation time
[tex]\alpha (t) = 1.72 MPa \\ \alpha (0) = 2.76MPa[/tex]
input the given values into the equation
1.72 = 2.76 exp ( - [tex]\frac{60}{T})[/tex]
exp ( - 60/T ) = 1.72/2.76 therefore T = 126.87S
Considering 10s for t
[tex]\alpha (10) = (2.76) exp(- \frac{10}{126.87})[/tex]
taking Ln of both sides of the equation
Ln [tex]\alpha (10) = ln(2.76) - \frac{10}{126.87}[/tex]
therefore [tex]\alpha (10) = 2.55 Mpa[/tex] = 370 psi
To determine the Er(10) for the material in (Mpa) we apply the relaxation modulus equation
Er(t) = [tex]\frac{\alpha(t) }{Eo}[/tex]
where Eo = strain level = 0.69
Er(t) = E(10)
[tex]\alpha (t)[/tex] = 2.55 Mpa
input this variables into equation
Er(10) for the material = 2.55 / 0.69 = 3.695 ≈ 3.70 Mpa
