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A known standard Kelvin tissue material has relaxation response f(t) = 2(1 – 0.5e-t/3). Find its dynamic modulus |G| under sinusoidal input when w = 2 rad/s

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Answer:

Step-by-step explanation:

The dynamic modulus of the Kelvin material is defined as the ratio of the complex stress response to the complex strain response under sinusoidal loading.

The complex stress response (σ*) and complex strain response (ε*) can be obtained by applying a sinusoidal input of angular frequency ω = 2 rad/s to the relaxation function:

σ*(jω) = G(jω)ε*(jω)

where G(jω) is the complex modulus of the material, and j = √(-1) is the imaginary unit.

The relaxation response of the Kelvin material is given by:

f(t) = 2(1 – 0.5e-t/3)

Taking the Laplace transform of f(t), we get:

F(s) = 2(1/(s + 1/3) - 0.5/(s + 1/3))

Simplifying this expression, we get:

F(s) = 2/(s + 1/3) - 1/(s + 1/3)

= 2/(s + 1/3) - 2/(2s + 2/3)

This can be rewritten as:

F(s) = 2/(s + 1/3) - 2/(s + 1/3 + j√(8)/3) + 2/(s + 1/3 - j√(8)/3)

This is the Laplace transform of a sum of three exponential functions. Each term corresponds to a relaxation mode with relaxation time τ = 3, and the amplitudes of the modes are given by 2, -2/(1 + j√(8)), and -2/(1 - j√(8)).

The complex modulus of the Kelvin material can be obtained by taking the ratio of the stress response to the strain response:

G(jω) = σ*(jω)/ε*(jω)

Substituting ω = 2 rad/s, we get:

G(j2) = σ*(j2)/ε*(j2)

Using the complex stress and strain responses obtained from the Laplace transform of the relaxation function, we get:

G(j2) = [2/(j2 + 1/3) - 2/(j2 + 1/3 + j√(8)/3) + 2/(j2 + 1/3 - j√(8)/3)] / (j2)

Evaluating this expression, we get:

G(j2) = -1.82 - j0.63 MPa

Therefore, the dynamic modulus of the Kelvin material under a sinusoidal input of angular frequency 2 rad/s is |G| = sqrt((-1.82)^2 + (-0.63)^2) = 1.91 MPa.

User Leandro Mora
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