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The Boltzmann superposition principle states that the strain response of a viscoelastic material is a superposition of the responses to all previous stress histories. Assuming no stress before time $t=0$, the constitutive equation can be written as:

$\begin{equation} \gamma(t)=\int_0^t J(t-t^\prime)\dot{\sigma}(t^\prime)\mathrm{d}t^\prime \end{equation}$

Applying the Laplace transform to this equation yields:

$\begin{align} \hat{\gamma}(s)&=\hat{J}(s)\hat{\dot{\sigma}}(s)\\ &=\hat{J}(s)\left(s\hat{\sigma}(s)-\sigma(0^{-})\right) \\ &=\hat{J}(s)\left(s\hat{G}(s)\hat{\dot{\gamma}}(s)-\sigma(0^{-})\right)\\ &=\hat{J}(s)\left(s\hat{G}(s)(s\hat{\gamma}(s)-\gamma(0^-))-\sigma(0^{-})\right)\\ &=s^2\hat{J}(s)\hat{\gamma}(s)\hat{G}(s)  \end{align}$

Here, $\hat{f}(s):=\mathcal{L}\lbrace f(t)\rbrace(s)$ denotes the Laplace transform of $f(t)$. Since the stress starts at time $t=0$, both $\gamma(0^-)$ and $\sigma(0^-)$ are zero. The second to third step is derived from the convolution relation $\sigma(t)=\int_0^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$. Cancelling out $\hat{\gamma}(s)$ and rearranging the last equation gives:

$\begin{equation} \frac{1}{s^2}=\hat{J}(s)\hat{G}(s) \end{equation}$

Taking the inverse Laplace transform of this equation leads to the convolution relation:

$\begin{equation} t=\int_0^{t}J(t-t^\prime)G(t^\prime)\mathrm{d}t^\prime \end{equation}$

Substituting $J(t)=J_e + t/\eta$ into the convolution relation and taking the limit as $t\to\infty$ gives:

$\begin{equation} \begin{aligned} \color{blue}{t} &=\int_0^{t}\left(J_e+\frac{t-t^\prime}{\eta}\right)G(t^\prime)\mathrm{d}t^\prime\\ &= J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime + {\color{blue}{\frac{t}{\eta}\int_0^\infty G(t^\prime)\mathrm{d}t^\prime}} – \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime \end{aligned} \end{equation}$

Since $\eta:=\int_0^\infty G(t^\prime)\mathrm{d}t^\prime$, the blue terms cancel out, resulting in:

\begin{equation} J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime = J_e\eta= \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime = \frac{1}{\eta} \int_0^\infty G(t^\prime)t^\prime\mathrm{d}t^\prime \end{equation}

Therefore, we obtain:

$J_e = \frac{1}{\eta^2}\int_0^\infty tG(t)\mathrm{d}t$

$Q.E.D.$