For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to 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 7.40, was suddenly pulled in a tension to a measured strain of 0.5; the stress necessary to maintain this constant strain was measured as a function of time. Determine Er(8) for this material if the initial stress level was 2.62 MPa (380 psi), which dropped to 1.73 MPa (251 psi) after 64 s.

of the material is 4.98 MPa

Explanation:

Consider the equation of stress decay for viscoelastic polymers

where; σ(t): 1.73 MPa , σ(0): 2.62 MPa , t: 64 s

σ(t) = σ(0)exp(-t/r)

1.73 = 2.62 exp(-64/r)

exp(-64/r) = 1.73 ÷ 2.62

exp(-64/r) = 0.6603

-64 / r = Ln 0.6603

-64 / r = -0.4151

-64 / -0.4151 = r

r = 154.18 s

Consider the equation of stress decay for viscoelastic polymers

σ(t) = σ(0)exp(-t/r)

substitute 8 s for t

σ(8) = (2.62)exp(-8/154.18)

Ln σ(8) = Ln(2.62) - 8/154.18

Ln σ(8) = 0.9632 - 0.051887

Ln σ(8) = 0.911313

σ(8) = 2.49 MPa

Consider the equation relaxation modulus

σ(t) / ε₀

Here, ε₀ is the strain level, which is maintained constant for σ(t) and ε₀ is the strain.

Substitute 0.5 for ε₀

= 2.49 MPa / 0.5

        = 4.98 MPa

Therefore, of the material is 4.98 MPa

answer:

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