Bulk and shear characterization of bituminous mixtures in the linear viscoelastic domain

Abstract

Traffic loads and temperature variations produce three-dimensional stress–strain fields inside road pavements, and therefore the characterization of bituminous mixtures in different deformation modes is important for prediction of the performance of pavement structures. This paper presents a methodology for the bulk and shear characterization of bituminous mixtures in the linear viscoelastic domain, under the hypothesis of material isotropy, by means of uniaxial harmonic tests with the measurement of axial and transverse strains. The theoretical approach was based on the application of the elastic–viscoelastic correspondence principle and was validated by performing tension–compression tests at selected frequencies and temperatures on asphalt concrete specimens characterized by different volumetric properties. The results showed that since uniaxial tests induced both volume and shape variations, the simultaneous measurement of the complex bulk and shear moduli was possible. The validity of the time–temperature superposition principle was also verified for both deformation modes, allowing the construction of master curves for the bulk and shear moduli. The results also showed that the total dissipated energy could be decomposed into its volumetric and deviatoric fractions with excellent accuracy.

Appendix

The amplitude and phase of the bulk and shear strain phasors defined by Eqs. (13a), (13b) can be obtained by simple phasor addition.

Given the two phasors

$$\begin{aligned} A^{*} =& a_{0} \exp(\mathrm{j}\omega+ \theta_{a}) = a_{0}\cos\theta_{a} + \mathrm{j} a_{0}\sin \theta_{a}, \\ B^{*} =& b_{0} \exp(\mathrm{j}\omega+ \theta_{b}) = b_{0}\cos\theta_{b} + \mathrm{j} b_{0}\sin \theta_{b}, \end{aligned}$$

their sum is

$$ C^{*} = A^{*} + B^{*} = c_{0} \exp(\mathrm{j}\omega+ \theta_{c}) = c_{0}\cos\theta _{c} + \mathrm{j} c_{0}\sin\theta_{c}, $$

where

$$\begin{aligned} c_{0}^{2} =& (a_{0}\cos\theta_{a} + b_{0}\cos\theta_{b})^{2} + (a_{0}\sin \theta_{a} + b_{0}\sin\theta_{b})^{2}, \\ \tan\theta_{c} =& \frac{a_{0}\sin\theta_{a} + b_{0}\sin\theta _{b}}{a_{0}\cos\theta_{a} + b_{0}\cos\theta_{b}}. \end{aligned}$$

Hence, for the absolute values of $\varepsilon ^{*}_{p}$ and $\varepsilon ^{*}_{q}$ defined by Eqs. (13a), (13b), we have:

$$\begin{aligned} \varepsilon _{p,0}^{2} =& \bigl[\varepsilon _{1,0}+2 \varepsilon _{2,0}\cos(-\delta _{2})\bigr]^{2} + \bigl[2 \varepsilon _{2,0}\sin(-\delta_{2})\bigr]^{2} \\ =& \varepsilon _{1,0}^2 + 4\varepsilon _{2,0}^2 + 4 \varepsilon _{1,0}\varepsilon _{2,0} \cos\delta_2 \\ =& \varepsilon _{1,0}^2 + 4\varepsilon _{2,0}^2 - 4 \varepsilon _{1,0}\varepsilon _{2,0} \cos\delta_\nu \\ =& \varepsilon _{1,0}^2 \bigl( 1 + 4\nu_{0}^{2} - 4\nu_{0}\cos\delta_\nu \bigr), \\ \varepsilon _{q,0}^{2} =& \biggl[ \frac{2}{3} \varepsilon _{1,0}-\frac {2}{3}\varepsilon _{2,0}\cos(- \delta_{2}) \biggr]^{2} + \biggl[ -\frac {2}{3} \varepsilon _{2,0}\sin(-\delta_{2}) \biggr]^{2} \\ =& \frac{4}{9} \bigl[ \varepsilon _{1,0}^2 + \varepsilon _{2,0}^2 - 2 \varepsilon _{2,0} \varepsilon _{1,0} \cos\delta_2 \bigr] \\ =& \frac{4}{9} \bigl[ \varepsilon _{1,0}^2 + \varepsilon _{2,0}^2 + 2 \varepsilon _{2,0} \varepsilon _{1,0} \cos\delta_\nu \bigr] \\ =& \frac{4}{9} \varepsilon _{1,0}^2 \bigl( 1 + \nu_{0}^2 + 2 \nu _{0} \cos \delta_\nu \bigr), \end{aligned}$$

whereas, for the phases, we have:

$$\begin{aligned} \tan\delta_{\varepsilon p} =& \frac{2\varepsilon _{2,0}\sin(-\delta _{2})}{\varepsilon _{1,0}+2\varepsilon _{2,0}\cos(-\delta_{2})} \\ =& \frac{-2\varepsilon _{2,0}\sin\delta_2}{\varepsilon _{1,0}+2\varepsilon _{2,0}\cos\delta_2} = \frac{2\nu_{0}\sin\delta_\nu}{1-2\nu_{0}\cos\delta_\nu} \\ =& \tan\delta_{\nu}\frac{2\nu_{0}\cos\delta_\nu}{1-2\nu_{0}\cos \delta_\nu}, \\ \tan(-\delta_{\varepsilon q}) =& \frac{-\varepsilon _{2,0}\sin(-\delta _{2})}{\varepsilon _{1,0}-\varepsilon _{2,0}\cos\delta_{2}} \\ =& \frac{\varepsilon _{2,0}\sin\delta_2}{\varepsilon _{1,0}-\varepsilon _{2,0}\cos \delta_2} = \frac{-\varepsilon _{2,0}\sin\delta_\nu}{ \varepsilon _{1,0}+\varepsilon _{2,0}\cos\delta_\nu} = \frac{-\nu_{0}\sin\delta_\nu}{1+\nu_{0}\cos\delta_\nu} \\ =& -\tan\delta_{\nu}\frac{\nu_{0}\cos\delta_\nu}{1+\nu_{0}\cos \delta_\nu}. \end{aligned}$$