A Lower Bound Estimate of the Bearing Capacity of Foundations on Inherently Anisotropic Sands Implementing the Fabric Tensor

Abstract

An estimate of the bearing capacity of inherently anisotropic sand deposits has been provided by the finite element and linear programming of the lower bound limit analysis. Both strip and circular foundations are considered. The inherent anisotropy can often be reasonably assumed for sands as they are often deposited into layers rendering them a transversely isotropic material. The anisotropy can be described by an internal measure like the fabric tensor. Although determination of the fabric tensor is often a difficult task, for transversely isotropic materials, it was found that no direct measurement on this tensor is required which makes its applications more appealing. In contrast to previous studies on the same topic by assuming a directional dependence of the friction angle, in the present study such an assumption is no longer made and the direct use of the fabric tensor has been made which seems to be logical and more flexible.

Appendix I—Derivation of Equations for Anisotropic Parameters

Parameters $\overline{\eta }$ and $A$ can be found in two standard triaxial tests (${\sigma }_{2}={\sigma }_{3}$) performed at two different orientations, preferably along material axes. According to definitions made by Pietruszczak and Pakdel (2022) for anisotropic materials, the stress state at failure in a transversely isotropic material should satisfy the following equation:

$$\frac{{\sigma }_{1}-{\sigma }_{3}}{{\sigma }_{1}+{\sigma }_{3}}=\overline{\eta }\left[1-A\left(1-3{\zeta }^{2}\right)\right]$$

Let us assume that two tests are conducted along material axes with measured stress state at failure. Therefore:

$$\frac{{\sigma }_{1}-{\sigma }_{3}}{{\sigma }_{1}+{\sigma }_{3}}=\overline{\eta }\left[1-A\left(1-3{\zeta }^{2}\right)\right]={\eta }^{90}=\mathrm{sin}{\phi }^{90} ,\quad {\varvec{n}}={\left[\begin{array}{ccc}1& 0& 0\end{array}\right]}^{T}, {\zeta }^{2}={\sigma }_{1}^{2}/({\sigma }_{1}^{2}+2{\sigma }_{3}^{2})$$

$$\frac{{\sigma }_{1}-{\sigma }_{3}}{{\sigma }_{1}+{\sigma }_{3}}=\overline{\eta }\left[1-A\left(1-3{\zeta }^{2}\right)\right]={\eta }^{00}=\mathrm{sin}{\phi }^{00},\quad {\varvec{n}}={\left[\begin{array}{ccc}0& 0& 1\end{array}\right]}^{T}, {\zeta }^{2}={\sigma }_{3}^{2}/({\sigma }_{1}^{2}+2{\sigma }_{3}^{2})$$

In these two equations, ${\eta }^{90}$ and ${\eta }^{00}$ are shear strength parameters along material axes inclined at $90$ and $0$ degrees with respect to horizon, respectively and stress states are at failure. Therefore, we have a system of two equations into two unknowns, $A$ and $\overline{\eta }$ which after elimination of $\overline{\eta }$ will be reduced to:

$$\frac{1-A\left(1-3{\sigma }_{1}^{2}/({\sigma }_{1}^{2}+2{\sigma }_{3}^{2})\right)}{1-A\left(1-3{\sigma }_{3}^{2}/({\sigma }_{1}^{2}+2{\sigma }_{3}^{2})\right)}=\frac{{\eta }^{90}}{{\eta }^{00}}={\beta }_{\phi }$$

Therefore, by defining $\xi =\left({\sigma }_{1}^{2}-{\sigma }_{3}^{2}\right)/\left({\sigma }_{1}^{2}+2{\sigma }_{3}^{2}\right)$:

$$A=\frac{{\beta }_{\phi }-1}{\xi \left({\beta }_{\phi }+2\right)}, \overline{\eta }=\frac{\mathrm{sin}{\phi }^{90}}{1+2A\xi }=\frac{\mathrm{sin}{\phi }^{00}}{1-A\xi }$$