Viscoelastic modeling of wood in the process of formation to clarify the hygrothermal recovery behavior of tension wood

Abstract

To explain the hygrothermal recovery (HTR) behavior of tension wood (TW) from the physical and chemical point of view in relation to the time, species and microfibril angle, a theoretical discussion using an analytical one-dimensional viscoelastic modeling was made. The chosen model includes an elastic element, a deformation mechanism and two viscoelastic elements called also as Kelvin–Voigt model. In this analysis, a top-down approach between the model and the experimental data was introduced to find the realistic parameters for the model. It enables us to fit the model to the HTR experimental data for three wood species: konara oak (Quercus serrata Murray), urihada maple (Acer rufinerve Siebold et Zucc.) and keyaki wood (Zelkova serrata Makino). The fitted experimental data show that the two compliances of the two viscoelastic elements are the most important parameters that explain the evolution of TW longitudinal strain during the thermal treatment.

Appendix

First case: $\epsilon = 0$

During the two first steps of the modeling, the stress is null ($\epsilon = 0$). On a step time ${\Delta }t=t_i-t_{i-1}$, we can write that:

$$\begin{aligned} \epsilon = S_0 \sigma + \alpha + \sum _{k} \epsilon ^V_k. \end{aligned}$$

(18)

From this Eq. 18, it is possible to write ${\Delta }\sigma$:

$$\begin{aligned} {\Delta }\epsilon = S_0 {\Delta }\sigma + {\Delta }\alpha + \sum _{k} {\Delta }\epsilon ^V_k. \end{aligned}$$

(19)

In this case, ${\Delta }\epsilon =0$, so we obtain:

$$\begin{aligned} {\Delta }\sigma = S_0^{-1} \left( {\Delta }\alpha +\sum _{k}{\Delta }\epsilon ^V_k \right) \end{aligned}$$

(20)

with ${\Delta }\epsilon ^V_k$ from Eq. 15. So the equation will become:

$$\begin{aligned} \begin{aligned}{\Delta }\sigma &= \left[ S_0 + \sum _k \left[ \left[ \exp \left( - \tau _k^{-1}t_i \right) \left[ \exp \left( - \tau _k^{-1}t_{i-1} \right) \right] ^{-1} - 1 \right] \frac{\tau _k S_k}{{\Delta }t} \right] \right] ^{-1} \\&\quad \times \left[ - {\Delta }\alpha - \sum _k \left[ \exp \left( - \tau _k^{-1}t_i \right) \left[ \exp \left( - \tau _k^{-1}t_{i-1} \right) \right] ^{-1} - 1 \right] \right] \left[ \epsilon ^V_k(t_{i-1}) - S_k \sigma (t_{i-1}) \right] . \end{aligned} \end{aligned}$$

(21)

Using Eq. 20, we can write that $\sigma (t_i)=\sigma (t_{i-1}) + {\Delta }\sigma$.

Second case: $\sigma = 0$

During the fourth step of the modeling ${\Delta }\sigma =0$, Eq. 19 becomes:

$$\begin{aligned} {\Delta }\epsilon = {\Delta }\alpha + \sum _{k} {\Delta }\epsilon ^V_k \end{aligned}$$

(22)

with ${\Delta }\epsilon ^V_k$ from Eq. 15.

Using equation 22, we can write that $\epsilon (t_i)=\epsilon (t_{i-1}) + {\Delta }\epsilon$.