Modeling of the simple shear deformation of sand: effects of principal stress rotation

Abstract

The paper presents a simple constitutive model for the behavior of sands during monotonic simple shear loading. The model is developed specifically to account for the effects of principal stress rotation on the simple shear response of sands. The main feature of the model is the incorporation of two important effects of principal stress on stress–strain response: anisotropy and non-coaxiality. In particular, an anisotropic failure criterion, cross-anisotropic elasticity, and a plastic flow rule and a stress–dilatancy relationship that incorporate the effects of non-coaxiality are adopted in the model. Simulations of published experimental results from direct simple shear and hollow cylindrical torsional simple shear tests on sands show the satisfactory performance of the model. It is envisioned that the model can be valuable in modeling in situ simple shear response of sands and in interpreting simple shear test results.

Appendix: Non-coaxiality angle due to principal stress rotation

The angle of non-coaxiality angle Δ required to determine the non-coaxiality parameter c in the stress–dilatancy relationships given in Eqs. 9 and 10 is evaluated using the flow rule developed by [10] for plastic flow in the (σ _y  − σ _x )/2 versus σ _xy stress rotation plane. This flow rule is shown in Fig. 6. In this figure, the strain increment components (dε _y  − dε _x ) and 2dε _xy have been superimposed on the stress plane. Point A is the current stress point. Neglecting the effect of anisotropy, the failure surface is circular and centered at the origin of the (σ _y  − σ _x )/2 versus σ _xy axes. The distances $\overline{\text{OA}}$ and $\overline{\text{OB}}$ are equal to the current shear stress t and the radius of the circular failure surface, respectively. These distances can be related to the current mean stress s, the mobilized friction angle ϕ and the peak friction angle ϕ _p as follows:

$$ \overline{\text{OA}} = s\,{ \sin }\,\phi ,\quad \overline{\text{OB}} = s\,{ \sin }\,\phi_{\text{p}} $$

(32)

Fig. 6

Non-coaxial flow rule in the (σ _y  − σ _x )/2 versus σ _xy stress rotation plane [10]

On the (σ _y  − σ _x )/2 versus σ _xy stress plane, a stress vector makes an angle equal to 2α from the (σ _y  − σ _x )/2 axis (Eq. 7). Similarly, a stress increment vector makes an angle equal to 2ξ from the (σ _y  − σ _x )/2 axis (Eq. 19). On the (dε _y  − dε _x ) versus 2dε _xy strain increment plane, a strain increment vector has a length equal to the plastic shear strain increment dγ and makes an angle equal to 2β from the (dε _y  − dε _x ) axis. This plastic strain increment direction is evaluated as the normal to the failure surface at the conjugate point A which is the intersection of the failure surface and the stress increment vector extended from the current stress point. This flow rule is based on the experimental observations that plastic flow on the (σ _y  − σ _x )/2 versus σ _xy stress plane is dependent on the stress increment direction [10]. This flow rule should be contrasted with conventional plasticity formulations where the plastic strain increment direction is evaluated at the current stress point independent of the stress increment direction. Details including the experimental verification of the flow rule are given in [10].

From triangle OAB in Fig. 6, the following angles can be obtained:

$$ \angle {\text{OAB}} = \pi - (2\xi - 2\alpha ) $$

(33)

$$ \angle {\text{BOA}} = 2\Updelta $$

(34)

$$ 2\Updelta = \pi - \angle {\text{OAB}} - \angle {\text{ABO}} $$

(35)

Using the law of sines:

$$ {\frac{\text{OA}}{{{ \sin }\,\angle {\text{ABO}}}}} = {\frac{\text{OB}}{{{ \sin }\,\angle {\text{OAB}}}}} $$

(36)

$$ \angle{\text{ABO}} = \sin^{ - 1} \left( {{\frac{{\text{OA}}}{{\text{OB}}}}\sin \angle{\text{OAB}}} \right) $$

(37)

Substituting Eqs. 32 and 33 in Eq. 37:

$$ \angle {\text{ABO}} = { \sin }^{ - 1} \left( {{\frac{{{ \sin }\,\phi }}{{{ \sin }\,\phi_{\text{p}} }}}{ \sin }(\pi + 2\alpha - 2\xi )} \right) $$

(38)

Substituting Eqs. 33 and 38 in Eq. 34 gives:

$$ 2\Updelta = 2\alpha - 2\xi - { \sin }^{ - 1} \left( {{\frac{{{ \sin }\,\phi }}{{{ \sin }\,\phi_{\text{p}} }}}{ \sin }\left( {\pi + 2\alpha - 2\xi } \right)} \right) $$

(39)

Simplifying the above equation results in the expression for Δ given in Eq. 18.