﻿ 斜向荷载作用下岩质地基基础应力分布
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 哈尔滨工程大学学报  2018, Vol. 39 Issue (9): 1526-1531  DOI: 10.11990/jheu.201704101 0

### 引用本文

LI Ke, YIN Ke, YAO Shunyu. Stress distribution of rock foundation under the action of slant load[J]. Journal of Harbin Engineering University, 2018, 39(9), 1526-1531. DOI: 10.11990/jheu.201704101.

### 文章历史

1. 重庆大学 土木工程学院, 重庆 400044;
2. 重庆市高新工程勘察设计院有限公司, 重庆 400047

Stress distribution of rock foundation under the action of slant load
LI Ke1,2, YIN Ke1, YAO Shunyu1
1. School of Civil Engineering, Chongqing University, Chongqing 400044, China;
2. Chongqing Hi-new Geotechnical Investigation and Design Institute Co., Ltd., Chongqing 400047, China
Abstract: The paper discussed the problem that implicit mechanical mechanism exists in the rock-embedded foundation of rock & earth mass support structure containing slant support or slant stress transfer structure. By carrying out rock foundation model test under the condition of acting slant load and utilizing the analysis method of three-dimensional FE model, the stress distribution law and destruction mode of rock foundation under the condition of acting slant load were obtained. By introducing the equivalent isotropic medium theory for solving the internal stress of Goodman layered rock mass, the boundary of the compressive-stress core of foundation was solved; on basis of Southwell basic equation, Hankel integral transformation was adopted to deduce the theoretical algorithm of the main parameter kn and ks in the calculation process of the boundary of the compressive-stress core. By comparing with theoretical calculation and the three-dimensional FE analysis conclusion, the paper verified the theoretical algorithm on the boundary of the compressive-stress core and the parameter kn and ks proposed on basis of the "equivalent isotropic medium" theory, obtained the stress distribution characteristics and the its boundary solution method of rock foundation under the condition of acting slant load, providing a theoretical basis for the subsequent research and design of the internal stress of rock foundation.
Keywords: slant load    rock foundation    stress distribution    compressive-stress core    normal stiffness    shear stiffness

Fox[1]对双层系统接触面受竖向均布荷载下的基底应力进行了试验，得到了不同地基强度比值和不同荷载宽度与埋深比值下的应力比。Whiteman等[2]通过试验得到了半无限体上矩形刚性基础的竖向位移近似解。Muki[3]给出了轴对称圆形弹性地基在半无限空间下的应力分布计算方法。Schiffman等[4]给出了轴对称椭圆形弹性地基在半无限空间下的应力分布。云天铨[5]运用线载荷积分法求解了水平刚性岩基上在受垂直集中力作用下的Fredholm第一类积分方程，并进行了数值验证。文丕华[6]提出的求解方法是在Southwell对称问题基本方程基础上应用Hankle变换对受圆形均布荷载的基底应力求解。在模型试验方面，阴可等[7]对岩石地基上扩展基础的基底反力试验数据进行了分析；艾智勇等[8]对分层地基上矩形刚性基础的基底反力、沉降和倾斜进行了计算；阴可等[9]对岩石地基上扩展基础的受力特性进行了分析。上述方法未对嵌固于岩体中的矩形基础应力分布规律和计算方法进行研究，也未考虑斜向荷载情况。

2 斜向荷载作用下基础模型试验

 Download: 图 4 应变片大值分布区域图 Fig. 4 Large strain gauge value distribution area figures

3 基础三维有限元分析

4 基础压应力核心区的理论计算

 $\frac{\sigma }{E}S + \frac{\sigma }{{{k_n}}} = \frac{\sigma }{{{E_{nt}}}}S$ (1)
 $\frac{1}{E} + \frac{1}{{{k_n}S}} = \frac{1}{{{E_n}}}$ (2)
 $\frac{\tau }{G}S + \frac{\tau }{{{k_s}}} = \frac{\tau }{{{G_{nt}}}}S$ (3)
 $\frac{1}{G} + \frac{1}{{{k_s}S}} = \frac{1}{{{G_{nt}}}}$ (4)

 ${k_n} = \frac{\sigma }{{\Delta v}}$ (5)
 ${k_s} = \frac{\tau }{{\Delta \omega }}$ (6)
 Download: 图 7 试验计算kn数据图 Fig. 7 Test calculation kn used data parameters

 ${\sigma _r} = \frac{{h\left( {X\cos \beta + Yg\sin \beta } \right)}}{{{\rm{ \mathsf{ π} }}r\left( {{{\left( {{{\cos }^2}\beta - g{{\sin }^2}\beta } \right)}^2} + {h^2}{{\sin }^2}\beta {{\cos }^2}\beta } \right)}}$ (7)

 $g = {\left[ {1 + \frac{E}{{\left( {1 - {v^2}} \right){k_n}S}}} \right]^{1/2}}$ (8)
 $h = {\left\{ {\frac{E}{{1 - {\nu ^2}}}\left[ {\frac{{2\left( {1 + \nu } \right)}}{E} + \frac{1}{{{k_s}S}}} \right] + 2\left( {g - \frac{\nu }{{1 - \nu }}} \right)} \right\}^{1/2}}$ (9)

Southwell基本方程：

 $\left\{ \begin{array}{l} {\sigma _r} + {\sigma _\theta } = \frac{1}{r}\frac{{{\partial ^3}\lambda }}{{\partial r\partial {z^2}}}\\ {\sigma _r} - {\sigma _\theta } = r\frac{\partial }{{\partial r}}\frac{1}{{{r^2}}}\left( {\frac{{{\partial ^2}\lambda }}{{\partial {z^2}}} - 2\mu {\vartheta ^2}\lambda } \right)\\ {\sigma _z} = - \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {\frac{{{\partial ^2}\lambda }}{{\partial {z^2}}} - \left( {1 + \mu } \right){\vartheta ^2}\lambda } \right]\\ {\tau _{\gamma z}} = \frac{1}{r}\frac{\partial }{{\partial z}}\left[ {\frac{{{\partial ^2}\lambda }}{{\partial {z^2}}} - \left( {1 + \mu } \right){\vartheta ^2}\lambda } \right] \end{array} \right.$ (10)

 ${\vartheta ^2}\left( {{\vartheta ^2}\lambda } \right) = 0$ (11)
 ${\vartheta ^2} = \frac{{{\partial ^2}}}{{\partial {r^2}}} + \frac{1}{r}\left( {\frac{\partial }{{\partial r}} + \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right)$ (12)

 $\left\{ \begin{array}{l} \phi \left( {r,z} \right) = {\vartheta ^2}\lambda \\ \psi \left( {r,z} \right) = \frac{{{\partial ^2}\lambda }}{{\partial {z^2}}} - \left( {1 + \mu } \right){\vartheta ^2}\lambda \end{array} \right.$ (13)

 ${\vartheta ^2}\phi \left( {r,z} \right) = 0$ (14)
 ${\vartheta ^2}\psi \left( {r,z} \right) = \frac{{{\partial ^2}\phi }}{{\partial {z^2}}}$ (15)

 $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over \phi } = {H_1}\left[ {\phi \left( {r,z} \right);r \to \xi } \right] = \int_0^\infty {\xi \phi \left( {\xi ,z} \right){J_1}\left( {r\xi } \right){\rm{d}}\xi }$ (16)

 $\begin{array}{*{20}{c}} {{\sigma _z} = - \frac{1}{2}\int_0^\infty {\xi {J_0}\left( {\xi r} \right)\left\{ {\exp \left( {\xi z} \right)\left( {2C + \xi zA} \right) + } \right.} }\\ {\left. {\exp \left( { - \xi z} \right)\left[ {2D - \xi zB} \right]} \right\}{\rm{d}}\xi } \end{array}$ (17)
 $\begin{array}{*{20}{c}} {{\tau _z} = \frac{1}{2}\int_0^\infty {\xi {J_1}\left( {\xi r} \right)\left\{ {\exp \left( {\xi z} \right) \cdot \left[ {2C + \left( {1 + \xi z} \right)A} \right] - } \right.} }\\ {\left. {\exp \left( { - \xi z} \right)\left[ {2D + \left( {1 - \xi z} \right)B} \right]} \right\}{\rm{d}}\xi } \end{array}$ (18)
 $\begin{array}{*{20}{c}} {u = \frac{{1 + u}}{{2E}}\int_0^\infty {{J_1}\left( {\xi r} \right)\left\{ {\exp \left( {\xi z} \right)\left[ {2C + \left( {2 - 2u + \xi z} \right)A} \right]} \right. + } }\\ {\left. {\exp \left( { - \xi z} \right)\left[ {2D + \left( {2 - 2u - \xi z} \right)B} \right]} \right\}{\rm{d}}\xi } \end{array}$ (19)
 $\begin{array}{*{20}{c}} {w = - \frac{{1 + u}}{{2E}}\int_0^\infty {{J_0}\left( {\xi r} \right)\left\{ {\exp \left( {\xi z} \right)\left[ {2C - \left( {1 - 2u - \xi z} \right)A} \right]} \right. - } }\\ {\left. {\exp \left( { - \xi z} \right)\left[ {2D - \left( {1 - 2u + \xi z} \right)B} \right]} \right\}{\rm{d}}\xi } \end{array}$ (20)

 ${\sigma _z} = - \int_0^\infty {\xi {J_0}\left( {\xi r} \right)\left[ {2C{\rm{ch}}\left( {\xi z} \right) + \xi zA{\rm{sh}}\left( {\xi z} \right)} \right]{\rm{d}}\xi }$ (21)
 $\begin{array}{l} {\tau _z} = \int_0^\infty {\xi {J_1}\left( {\xi r} \right)\left\{ {2C{\rm{sh}}\left( {\xi z} \right) + A\left[ {{\rm{sh}}\left( {\xi z} \right) + } \right.} \right.} \\ \;\;\;\;\;\;\left. {\left. {\xi z{\rm{ch}}\left( {\xi z} \right)} \right]} \right\}{\rm{d}}\xi \end{array}$ (22)
 $\begin{array}{l} u = \frac{{1 + u}}{{2E}}\int_0^\infty {{J_1}\left( {\xi r} \right)\left\{ {2C{\rm{ch}}\left( {\xi z} \right) + } \right.} \\ \;\;\;\;\;\;\left. {A\left[ {2\left( {1 - \mu } \right){\rm{ch}}\left( {\xi z} \right) + \xi z{\rm{sh}}\left( {\xi z} \right)} \right]} \right\}{\rm{d}}\xi \end{array}$ (23)
 $\begin{array}{l} w = - \frac{{1 + u}}{E}\int_0^\infty {{J_0}\left( {\xi r} \right)\left\{ {2C{\rm{sh}}\left( {\xi z} \right) + } \right.} \\ \;\;\;\;\;\;\left. {A\left[ {\xi z{\rm{ch}}\left( {\xi z} \right) - \left( {1 - 2\mu } \right){\rm{sh}}\left( {\xi z} \right)} \right]} \right\}{\rm{d}}\xi \end{array}$ (24)