I've been trying to work this out, but I haven't gotten it yet. The problem:
Consider two polymer strands A and B, each held fixed at the top at a distance of 50 cm above the floor. Strand A is 35 cm long, with a diameter of 0.5 mm, and behaves like an ideal Maxwell viscoelastic solid. Strand B is 25 cm long, with a diameter of 0.25 cm, and behaves like an ideal Voight viscoelastic solid. A stress of 10 kPa is applied to each strand at t = 0. The figure below illustrates the arrangement before the stress is applied.Consider two polymer strands A and B, each held fixed at the top at a distance of 50 cm above the floor. Strand A is 35 cm long, with a diameter of 0.5 mm, and behaves like an ideal Maxwell viscoelastic solid. Strand B is 25 cm long, with a diameter of 0.25 cm, and behaves like an ideal Voight viscoelastic solid. A stress of 10 kPa is applied to each strand at t = 0. The figure below illustrates the arrangement before the stress is applied.
a) Explain the principle differences between Maxwell and Voight models for viscoelasticity.
b) What is the length of each strand immediately after applying the stress?
c) Which strand hits the floor first, and after how much elapsed time?
d) How much residual stress remains in that strand when the other strand finally hits the floor?
Some potentially useful equations:
ε(t) = σ/E*(1+(t/τ)) --> maxwell equation, ε = strain, σ = stress, t = time, τ = relaxation time, E = Young's modulus
ε(t) = σ/E*(1- exp(t/τ)) --> voigt equation
σ = E*ε
σ = η* derivation of ε, where η = dynamic viscosity
τ = η*E
σ = E/3([l(f)/l(o)]-[1/[(l(f))/l(o))^2]) where lo = initial length, l(o) = final length.
Part a is easy to answer. For part b, I'm thinking that t = 0 for voigt, and no stretching occurs. For the maxwell, I figured out that σ = F/A (force over area) --> so A = πr^2, and it will deform. I'm completely stumped on c, which is needed for d.
How am I supposed to solve without Young's modulus or the relaxation time? I was thinking of using the f* formula, but I'm not sure that's right.