Active earth pressure analysis based on normal stress distribution function along failure surface in soil obeying nonlinear failure criterion

Abstract

In this paper, the normal stress distribution function along the failure surface in soil obeying general nonlinear failure criterion is first derived, and then a method for calculating the static and seismic active earth pressure for soils obeying nonlinear failure criteria is proposed. The method proposed yields not only the active earth pressure and its action point, but also the failure surface and the stress distribution along it. Results of the calculated examples by the proposed method are more reasonable than those reported in the literature. The influence of the parameters of the nonlinear failure criterion on the active earth pressures is also investigated.

Appendix: Derivation of Eq. (11)

Equation (8) can be rearranged into a matrix form:

$$ [K_{p} ]\left\{ {p_{x} } \right\} + 2R[K_{\theta } ]\left\{ {\theta_{x} } \right\} + \gamma \left\{ {K_{\gamma } } \right\} = \left\{ 0 \right\} $$

(39)

where,

$$ \begin{aligned} [K_{p} ] = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right] + \sin \varphi_{\text{t}} \left[ {\begin{array}{*{20}c} {\cos 2\theta } & {\sin 2\theta } \\ {\sin 2\theta } & { - \cos 2\theta } \\ \end{array} } \right] ,\quad \, [K_{\theta } ] = \left[ {\begin{array}{*{20}c} { - \sin 2\theta } & {\cos 2\theta } \\ {\cos 2\theta } & {\sin 2\theta } \\ \end{array} } \right] ,\quad \, \left\{ {K_{\gamma } } \right\} = \left\{ {\begin{array}{*{20}c} {\sin \varepsilon } \\ {\cos \varepsilon } \\ \end{array} } \right\} \hfill \\ \left\{ {p_{x} } \right\} = \left\{ {\begin{array}{*{20}c} {\frac{\partial p}{\partial x}} \\ {\frac{\partial p}{\partial y}} \\ \end{array} } \right\},\quad \, \left\{ {\theta_{x} } \right\} = \left\{ {\begin{array}{*{20}c} {\frac{\partial \theta }{\partial x}} \\ {\frac{\partial \theta }{\partial y}} \\ \end{array} } \right\},\quad \, \left\{ {p_{s} } \right\} = \left\{ {\begin{array}{*{20}c} {\frac{\partial p}{{\partial s_{\alpha } }}} \\ {\frac{\partial p}{{\partial s_{\beta } }}} \\ \end{array} } \right\},\quad \, \left\{ {\theta_{s} } \right\} = \left\{ {\begin{array}{*{20}c} {\frac{\partial \theta }{{\partial s_{\alpha } }}} \\ {\frac{\partial \theta }{{\partial s_{\beta } }}} \\ \end{array} } \right\},\quad \, \left\{ 0 \right\} = \left\{ {\begin{array}{*{20}c} 0 \\ 0 \\ \end{array} } \right\} \hfill \\ \end{aligned} $$

(40)

According to the chain rule, the directional derivatives {p _s } and {θ _s } can be formulated as

$$ \left\{ {p_{s} } \right\} = [J]\left\{ {p_{x} } \right\},\quad \left\{ {\theta_{s} } \right\} = [J]\left\{ {\theta_{x} } \right\}\quad {\text{with}}\quad [J] = \left[ {\begin{array}{*{20}c} {\cos \theta_{\alpha } } & {\sin \theta_{\alpha } } \\ {\cos \theta_{\beta } } & {\sin \theta_{\beta } } \\ \end{array} } \right] $$

(41)

where, θ _α and θ _β are the inclination angles of the α and β line, respectively.Therefore, the derivatives {p _x } and {θ _x } can be expressed as:

$$ \left\{ {p_{x} } \right\} = [J]^{ - 1} \left\{ {p_{s} } \right\} ,\quad \left\{ {\theta_{x} } \right\} = [J]^{ - 1} \left\{ {\theta_{s} } \right\}{\text{ and }}[J]^{ - 1} = \frac{1}{{\sin (\theta_{\beta } - \theta_{\alpha } )}}\left[ {\begin{array}{*{20}c} {\sin \theta_{\beta } } & { - \sin \theta_{\alpha } } \\ { - \cos \theta_{\beta } } & {\cos \theta_{\alpha } } \\ \end{array} } \right] $$

(42)

Substituting Eq. (42) for {p _x } and {θ _x } into Eq. (39), the latter then becomes

$$ [K_{p} ][J]^{ - 1} \left\{ {p_{s} } \right\} + 2R[K_{\theta } ][J]^{ - 1} \left\{ {\theta_{s} } \right\} + \gamma \left\{ {K_{\gamma } } \right\} = \left\{ 0 \right\} $$

(43)

Left multiply Eq. (43) by ([J][K _θ ]⁻¹) = ([J][K _θ ]), the inverse of ([K _θ ][J]⁻¹), then we have

$$ [J][K][J]^{ - 1} \left\{ {p_{s} } \right\} + 2R\left\{ {\theta_{s} } \right\} + \gamma [J][K_{\theta } ]\left\{ {K_{\gamma } } \right\} = \left\{ 0 \right\} $$

(44)

where,

$$ [K] = [K_{\theta } ][K_{p} ] = [K_{\theta } ] + \sin \varphi_{\text{t}} \left[ {\begin{array}{*{20}c} 0 & { - 1} \\ 1 & 0 \\ \end{array} } \right] $$

(45)

In order to decouple the differential equations, we need to make the matrix $[J][K][J]^{ - 1}$ a diagonal one by choosing suitable direction angles.

$$ \begin{aligned} & [J][K][J]^{ - 1} = [J][K_{\theta } ][J]^{ - 1} + \sin \varphi_{\text{t}} [J]\left[ {\begin{array}{*{20}c} 0 & { - 1} \\ 1 & 0 \\ \end{array} } \right][J]^{ - 1} \\ & = \frac{1}{{\sin (\theta_{\beta } - \theta_{\alpha } )}}\left[ {\begin{array}{*{20}c} { - \cos (\theta_{\alpha } + \theta_{\beta } - 2\theta ) + \sin \varphi_{\text{t}} \cos (\theta_{\alpha } - \theta_{\beta } )} & {\cos (2\theta_{\alpha } - 2\theta ) - \sin \varphi_{\text{t}} } \\ { - \cos (2\theta_{\beta } - 2\theta ) + \sin \varphi_{\text{t}} } & {\cos (\theta_{\alpha } + \theta_{\beta } - 2\theta ) - \sin \varphi_{\text{t}} \cos (\theta_{\alpha } - \theta_{\beta } )} \\ \end{array} } \right] \\ \end{aligned} $$

(46)

Let now the non-diagonal elements are zero

$$ \left\{ {\begin{array}{*{20}c} {\cos (2\theta_{\alpha } - 2\theta ) - \sin \varphi_{\text{t}} = 0} \\ { - \cos (2\theta_{\beta } - 2\theta ) + \sin \varphi_{\text{t}} = 0} \\ \end{array} } \right. $$

(47)

which leads to the following two different inclination angles and consequently gives the characteristic line as shown in Fig. 3.

$$ \left\{ {\begin{array}{*{20}c} {\theta_{\alpha } = \theta - \mu } \\ {\theta_{\beta } = \theta + \mu } \\ \end{array} } \right. $$

(48)

Substituting Eq. (48) and Eq. (46) in turn into Eq. (44), we can get the Eq. (11).