土中应力（本书重点内容之一,理论讲解并进行课堂课后练习）.doc

蚁袈膀蒄螃螁肆蒃蒃羆羂肀薅蝿袈聿蚇羄膇肈莇螇肃膇葿羃罿膆薂螆袅膅螄薈芃膅蒃袄腿膄薆蚇肅膃蚈袂羁膂莈蚅袇芁蒀袀膆芀薂蚃肂艿蚅衿羈艿蒄蚂羄芈薇羇袀芇虿螀腿芆荿羅肅芅蒁螈羀莄薃羄袆莃蚆螆膅莃莅蕿膁莂薇螅肇莁蚀蚈羃莀荿袃衿荿蒂蚆膈莈薄袁肄蒇蚆蚄羀蒇莆袀袆蒆蒈蚂芄蒅蚁袈膀蒄螃螁肆蒃蒃羆羂肀薅蝿袈聿蚇羄膇肈莇螇肃膇葿羃罿膆薂螆袅膅螄薈芃膅蒃袄腿膄薆蚇肅膃蚈袂羁膂莈蚅袇芁蒀袀膆芀薂蚃肂艿蚅衿羈艿蒄蚂羄芈薇羇袀芇虿螀腿芆荿羅肅芅蒁螈羀莄薃羄袆莃蚆螆膅莃莅蕿膁莂薇螅肇莁蚀蚈羃莀荿袃衿荿蒂蚆膈莈薄袁肄蒇蚆蚄羀蒇莆袀袆蒆蒈蚂芄蒅蚁袈膀蒄螃螁肆蒃蒃羆羂肀薅蝿袈聿蚇羄膇肈莇螇肃膇葿羃罿膆薂螆袅膅螄薈芃膅蒃袄腿膄薆蚇肅膃蚈袂羁膂莈蚅袇芁蒀袀膆芀薂蚃肂艿蚅衿羈艿蒄蚂羄芈薇羇袀芇虿螀腿芆荿羅肅芅蒁螈羀莄薃羄袆莃蚆螆膅莃莅蕿膁莂薇螅肇莁蚀蚈羃莀荿袃衿荿蒂蚆膈莈薄袁肄蒇蚆蚄羀蒇莆袀袆蒆蒈蚂芄蒅蚁袈膀蒄螃螁肆蒃蒃羆羂肀薅蝿袈聿蚇羄膇肈莇螇肃膇葿羃罿膆薂螆袅膅螄薈芃 第3章 土中应力（本书重点内容之一，理论讲解并进行课堂课后练习）3.1 概述土体在自身重力、建筑物荷载、交通荷载或其他因素（如地下水渗流、地震等）的作用下，均可产生土中应力。土中应力将引起土体或地基的变形，使土工建筑物（如路堤、土坝等）或建筑物（如房屋、桥梁、涵洞等）发生沉降、倾斜以及水平位移。土体或地基的变形过大时，影响建筑物等的正常使用。而应力过大时，又会导致土体的强度破坏，使土工建筑物发生土坡失稳或地基建筑物地基的承载力不足而发生失稳。因此在研究土的变形，强度及稳定时，都必须掌握土中的应力状态。土应力的计算和分布规律是土力学的基本内容之一。Stresses in the ground will be induced by self-weight，load from structures and so on。If the shear stresses induced in the ground are within the allowable range of the soil strength， the soil is stable。Conversely if the shear stresses induced in a certain localized area in the ground are in excess of the allowable range of the soil strength，failure of the soil will take place there，and this may result in the slippage of the global ground，leading to an overturning of the structure。On other hand， if the deformation of the ground exceeds allowable values，the structure may also be damaged and lose its service function although the soil has not failed yet。Therefore，in order to ensure the safety and the normal service function of the structure， the stress distribution patterns and the deformations induced in the ground under various loading conditions must be studied。基本内容：掌握土中三种应力（自重应力、基底压力以及各种荷载条件下的土中附加应力）计算方法。学习基本要求：1. 掌握土中自重应力计算； 2. 掌握基底压力和基底附加压力分布与计算； 3. 掌握圆形面积均布荷载、矩形面积均布荷载、 矩形 面积三角形分布荷载以及条形载等条件下的土中竖向附加应力计算方法； 4. 了解地基中其他应力分量的计算公式。 应力方向说明 在用摩尔圆进行分析时，法向应力仍以压为正，剪 应力方向的符号规定则与材料力学 相反。材料力学中规定剪应力以顺 时针方向为正，土力学中则规定剪 应力以逆时针方向为正。3.2 土中自重应力计算3.2.1 均质土自重应力计算在深度z处平面上，土体因自身重力产生的竖向应力cz（称竖向自重应力）等于单位面积上土柱体的重力W，如下图所示。在深度z处土的自重应力为：Consider an element of soil at a given depth z meters below the ground surface，as shown in the below figure，the vertical stress induced by self-weight is， 式中： g 为土的重度，KN/m3 ； A 为土柱体的截面积，m2。K0：静止土压力系数，：泊松比In equation，g is the bulk unit weight ，A is the cross section area of the soil column。K0：coefficient of lateral pressure， ：Poissons ratio 从上式可知，自重应力随深度z线性增加，呈三角形分布图形。From the above formula，we can see that the vertical stress induced by self-weight increases linearly with the depth。3.2.2 成层土自重应力计算（calculation the vertical stresses of layered soils）地基土通常为成层土。当地基为成层土体时，设各土层的厚度为hi，重度为gi，则在深度z处土的自重应力计算公式为:In general，the ground consists of multi layer。Consider the thickness and the bulk unit weight of each layer are hi and gi，respectively，the vertical effective overburden pressure at the bottom level of the nth soil layer is given by 3.2.3 地下水位升降时的土中自重应力当计算地下水位以下土的自重应力时，应根据土的性质确定是否需要考虑水的浮力作用。通常认为水下的砂性土是应该考虑浮力作用的。粘性土则视其物理状态而定，一般认为，若水下的粘性土其液性指数IL 1，则土处于流动状态，土颗粒之间存在着大量自由水，可认为土体受到水浮力作用；若IL0，则土处于固体状态，土中自由水受到土颗粒间结合水膜的阻碍不能传递静水压力，故认为土体不受水的浮力作用；若0IL1，土处于塑性状态，土颗粒是否受到水的浮力作用就较难肯定，在工程实践中一般均按土体受到水浮力作用来考虑。若地下水位以下的土受到水的浮力作用，则水下部分土的重度按有效重度g计算，其计算方法同成层土体情况。1122z0原地下水位变动后地下水位1122z0原地下水位变动后地下水位If the calculation point is below the ground water table，the buoyant unit weight should be used for the calculation of the soil column，because of the buoyant effect of the water on the soil。For the clays，when IL 1，the soil is in liquid state，the buoyancy should be considered，when IL0，the soil is in solid state，it is not necessary to consider the buoyancy。If 0IL1 ，it is very difficult to decide the buoyant effect，in most engineering practices，the buoyancy is considered。 If there is buoyant effect，the buoyant bulk unit weight should be used for the soils below the groundwater level。 当地基中存在隔水层时，隔水层面以下土的自重应考虑其上的静水压力作用。If there is a staunch stratum， the hydraulic pressure due to the upside water should be considered for the soils under the staunch stratum。 i 第i层土的天然重度，对地下水位以下的土取有效重度，取i ；hw 地下水到隔水层的距离(m)。i：natural bulk unit weight for i layer，the effective bulk unit weight is used for soils under groundwater level；hw3.3基底压力(Contact pressure between the foundation and the ground)3.3.1 基本概念(basic concept) 建筑物荷载通过基础传递给地基的压力称基底压力（地基反力）。也就是作用于基础底面土层单位面积的压力，单位为kPa。 Contact pressure is the intensity of loading transmitted from the underside of a foundation to the ground soil. 基底压力分布及其影响因素(Influence factors on the magnitude and the distribution of contact pressure)相对刚度、地基土的性质、基础大小、形状和埋深、作用在基础上的荷载大小、分布和性质等The magnitude and the distribution pattern of the contact pressure depend on many factors such as the magnitude and the distribution of the structure load applied, the rigidity, size, shape and embedment depth of foundation and the soil properties.当荷载较小时，基底压力分布形状如图a，接近于弹性理论解；荷载增大后，基底压力呈马鞍形(图b)；荷载再增大时，边缘塑性破坏区逐渐扩大，所增加的荷载必须靠基底中部力的增大来平衡，基底压力图形可变为抛物线型(图d)以至倒钟形分布(图c)。刚性基础放在砂土地基表面时，由于砂颗粒之间无粘结力，其基底压力分布更易发展成图d所示的抛物线形；而在粘性土地基表面上的刚性基础，其基底压力分布易成图b所示的马鞍形。When the structure load applied is very small, the distribution of contact pressure distribute as shown in figure (a). The distribution of contact pressure will develop from (a) to (b), (c), (d) when the structure load increases. For a low rigidity or for a flexible foundation, the magnitude and the distribution patter of the contact pressure is the same as those of the load applied on the foundation. This is because the foundation is compatible to the deformation of the ground soil.For a rigid foundation that cannot be compatible to the ground deformation due to significant difference in rigidity, the distribution of the contact pressure varies with magnitude of applied load, the embedment depth of foundation, the properties of the ground soil, etc. When the applied load is relatively small, higher contact pressure will be imposed on the edge of the foundation and lower contact pressure will be imposed on the center of the foundation as shown in figure (a). When the load is increased gradually to failure load, the distribution curve of the contact pressure develops from (a) to (b), (c), (d), finally becomes higher on the center and lower on the edge of the foundation as shown in figure (d). 根据弹性理论中圣维南原理，在总荷载保持定值的前提下，地表下一定深度处，基底压力分布对土中应力分布的影响并不显著，而只决定于荷载合力的大小和作用点位置。因此，除了在基础设计中，对于面积较大的片筏基础、箱形基础等需要考虑基底压力的分布形状的影响外，对于具有一定刚度以及尺寸较小的柱下单独基础和墙下条形基础等，其基底压力可近似地按直线分布的图形计算，即可以采用材料力学计算方法进行简化计算。Empirically, when the width of the rigidity foundation is not two wide and the applied load is relatively small, the contact pressure distribution follows approximately a linear distribution assumption. The error induced between the assumption and the reality would be acceptable, according to St. Venants principle. Introduced below is the simplified calculation method normally used in the engineering practice for computing the contact pressure based on the linear distribution assumption.3.3.2 基底压力的简化计算基底压力分布是很复杂的，一般并非线形分布。当基础有一定刚度且基底尺寸较小时，工程上常将基底压力假定为线形分布，应用材料力学理论进行简化计算。Distribution of the contact pressure is very complicated, in general, it isnt linearly distributed. When the size and the rigidity is relatively is small, it is usually assumed as linearly distributed in engineering practice, and calculate the contact pressure by the theory of material mechanics. 中心荷载下的基底压力(Contact pressure due to vertical centric load) A vertical centric load is applied on the rectangular foundation. According to the linear distribution assumption, the value of the contact pressure is 式中 p 作用任基础上的竖向力设计值(kN)； G 基础自重设计值及其上回填土重标准值的总重(kN)； G=gGAd ,gG 其中为基础及回填土之平均重度，一般取20kN/m3， 但在地下水位以下部分应扣去浮力，即取10kN/m3；d 基础埋深，必须从设计地 面或室内外平均设计地面算起(m)，A 基底面积(m 2)，对矩形基础Alb，l和b分别为其的长 和宽 。对于荷载沿长度方向均匀分布的条形基础，取单位长度进行基底平均压力设计值p(kPa)计算，A改为b(m)，而F及G则为基础截面内的相应值(kN/m)。 Where lowercase p represents the contact pressure (kPa); capital P represents the vertical toad on the underside of the foundation: and A represents the area of the foundation; G : total load of self-weight and the refill soil; gG: average bulk unit weight of the foundation and the refill soil, it is assumed as 20 KN/m3 usually, but the parts under water level should be considered as 10 KN/m3偏心荷载下的基底压力计算(Contact pressure due to vertical eccentric load) When a eccentric load is applied to a rectangular foundation, the contact pressure at any arbitrary point can be calculated using the formula of eccentric compression in mechanics of materials, as given byWhere p represents the contact pressure at an arbitrary point: represents the moment of the eccentric load about X-X axis (ey is the offsetting of the eccentric load line to the X-X axis); represents the moment of the eccentric load about Y-Y axis (ex is the offsetting of the eccentric load line to the Y-Y axis); Ix=lb3/12 represents the moment of inertia of the area of the underside of the foundation about X-X axis; Iy=l3b/12 represents the moment of inertia of the area of the underside of the foundation about Y-Y axis; If the load is applied on the major axis, e.g., the maximum and the minimum contact pressure on both sides of the underside of the rectangular foundation under vertical eccentric load are given as exeyXYlb 式中 : pmax ， pmin 分别为基础底面边缘的最大、最小压力设计值(kPa)(maximum and minimum contact pressure)； R =F+G 竖向荷载的合力，(kN) (total vertical load)；M 作用于基础底面形心的力矩设 计值(kN .m)(moment about X or Y axis), W 基础底面的抵抗矩(moment of resistance)(m3) 当el/6时，基底压力分布图呈梯形图(b)( el/6, distribution curve of contact pressure is trapezoidal； 当e=l/6时，则呈三角形图(c)( e=l/6, distribution curve of contact pressure is triangular)； 当el/6时，按计算结果，距偏心荷载较远的基底边缘反力为负值，即p min0 图(c)，根据偏心荷载与基底反力平衡，并通过3角形中心，( el/6, p min0)在双向偏心情况下(for two side eccentric load)， 3.3.3 基底附加压力（additional contact pressure）一般情况下，建筑物建造前天然土层在自重作用下的变形早已结束。因此，只有基底附加压力才能引起地基的附加应力和变形。 For a natural soil layer, before building, in general, the displacement due to self-weight has been finished. Only the contact pressure increase of underside may cause the stress increase and the displacement of ground. 基底附加压力是基础底面处地基土在初始应力基础上增加的压力。该处的初始应力为基础底面处土的自重应力cd，现有压力为基底压力p，所以基底附加压力p0等于基底压力p与自重应力cd的差，即： The contact pressure increase is the 3.3.4 桥台前后填土引起的基底附加应力(self-study) 本节作为学生自学内容，要求理解由于桥头路基填土等引起的附加竖向应力机制，并掌握附加竖向应力的计算方法。3.4 地基附加应力（本章难点，课堂讲解理论公式推导，并要求课堂自行推导，课后作业加深理解）土中的附加应力是由建筑物荷载所引起的应力增量，（即土在初始应力基础上增加的应力）。假设地基土是均匀、连续、各向同性的半无限空间线形弹性体，一般采用将基底附加压力当作作用在弹性半无限体表面上的局部荷载，用弹性理论求解的方法计算 Soil stress increase is the stress increment caused by structure load. Assuming the ground soil as a continuous homogeneous isotropic and semi-infinite (half space) elastic material, the stress increase can then be calculated busing the basic formulae of the elasticity theory. 竖向集中力下的地基附加应力 在均匀的、各向同性的半无限弹性体表面作用一竖向集中力P时，半无限体内任意点M的应力可由布西奈斯克解计算，如下图所示。工程中常用的竖向正应力sz及地表上距集中力为r处的竖向位移w（沉降）可表示成如下形式： when a vertical concentrated load P is applied on the surface of an elastic half-space as shown in the figure. At an arbitrary point M in the interior of the elastic body can be solved using elasticity theory, as given byK土的竖向附加应力系数，是r/z的函数。上式可写为 对实际工程中普遍存在的分布荷载作用时的土中应力计算，如下方法处理：当基础底面的形状或基底下的荷载分布不规则时，可以把分布荷载分割为许多集中力，然后用布西奈斯克公式和叠加原理计算土中应力。当基础底面的形状及分布荷载都是有规律时，则可以通过积分求解得相应的土中应力。In engineering practice, in many cases, the load is distributed. When the distribution is not regular, the load can be divided in to some concentrated load, then using the boussinqs equation to solve the stress increase in the soil ground. If the distribution is regular, the stress increase can be solved be integrating the distributed load.3.4.2 矩形荷载和圆形荷载下的地基附加应力均布矩形荷载在地基表面作用一分布于矩形面积(lb)上的均布荷载p，计算矩形面积中点下深度z处M点的竖向应力z值，可从下式解得When a vertical uniform load p is applied to the underside of a rectangular foundation, the vertical stress increase under the corners of the foundation can be calculated by integrating the basic equation with respect to the whole rectangular area, as shown in the above figure. If the vertical uniform load intensity on the foundation underside is p, the acting force dp on the infinitesimal area dxdy is pdxdy and it can be taken granted as a concentrated point load. There, the vertical stress increase at depth z under the foundation corner induced by the vertical uniform load, is given by Where 矩形面积均布荷载作用下土中任意点下的附加应力计算可利用下式和叠加原理求解，此法称为角点法。 For a point inside or outside the range of the foundation underside, the vertical stress increase can be calculated by using the above equation and the principle of superposition. L=6m4.5m1.5m1.5m2mEOADC1 求解点在荷载面边缘2 求解点在荷载面内3 求解点在荷载面边缘外4 求解点在荷载面角点外侧 Example The area of a rectangular foundation is 2mx6m, as shown in the figure. A vertical uniform load of 300 kPa is applied on the foundation underside. Try to find the vertical stress increase at a depth z of 2m under points A,E,C,D,E and O.SolutionStep 1 in order to obtain the vertical stress increase under point A, the foundation underside is divided into two rectangular of equal area of 2mx3m,such that point A is under the corners of two rectangular of 3m in length L1, and 2m in width B. with L1/B=3/2=1.5 and z/b=2/2=1, and from table we get . Therefore, the vertical stress increase under point A is Step 2 in order to obtain the vertical stress increase under point E, the foundation underside is divided into two rectangular of equal area of 1mx6m,such that point E is under the corners of two rectangular of 6m in length L, and 1m in width B. with L1/B=6/1=6 and z/B=2/1=2, and from table we get . Therefore, the vertical stress increase under point A is Step 3 point c is right under the corner of a rectangular area of 6m in length Land 2m in width B. with L1/B=6/2=3 and z/B=2/2=1, and from table we get . Therefore, the vertical stress increase under point A is Step 4 in order to obtain the vertical stress increase under point D, the foundation underside is divided into four rectangular by drawing two sublines pass through point D and parallel to the longer and shorter sides of the foundation respectively. The stress increase coefficients at point D for each rectangular can be obtained from table as listed in the following table.IdentityLBzL/BZ/B11.50.52.03.04.00.0621.51.52.01.01.330.13834.51.52.03.01.330.17744.50.52.09.04.00.076Step 5 similarly, in order to obtain the vertical stress increase under point O, the foundation underside is divided into four rectangular of equal area by drawing two sublines pass through point O and parallel to the longer and shorter sides of the foundation, respectively. With L1/B=3/1=1 and z/B=2/1=2, and from table we get . Therefore, the vertical stress increase under point A is 。当地基表面作用矩形面积(lb)三角形分布荷载时，为计算荷载为零的角点下的竖向应力值sz1，可将坐标原点取在荷载为零的角点上，相应的竖向应力值sz可用下式计算：When triangularly distributed load is applied to a rectangular foundation underside, at an arbitrary depth z under the corner where the load intensity is zero can be calculated by the following equation。 （1）荷载强度为零的角点下（at the corner where load intensity is zero）公式内 （2）荷载最大角点下 （at the corner where load intensity is maximum） m=l/b， n=z/b， b is the length of one side of the underside in the loading variation direction and l is the length of another side。公式内所含可由p80，表3-8查得（can be obtained from table 3-8）。均布的圆形荷载 （a vertical uniform load is applied to the underside of a circular foundation）为了计算圆形面积上作用均布荷载p0时土中任一点M(r,z)的竖向正应力，可采用原点设在圆心O的极坐标(如3-22图)，由公式在圆面积范围内积分求得： m=r0/z公式内所含可由p81，表3-9查得。3.4.3 线荷载和条形荷载下地基附加应力（stress increase in plane problem） 线荷载（line load） dyxdxP0R1MxyroxR2R1yM（x，0，z）z （a）线荷载 （b）条形荷载下条形分布荷载下土中应力状计算属于平面应变问题，对路堤、堤坝以及长宽比lb10的条形基础均可视作平面应变问题进行处理。（When lb10， it can be considered as a plane problem）条形荷载下铅直应力（vertical stress increase under a strip foundation underside due to vertical uniform load）Example A retaining wall of 6m wide is embedded into the ground at a depth of 1.5m. A vertical line load p=2400kN/m is acting on a point at a distance of 3.2m to the wall front toe A. The wall back is subjected to a horizontal thrust H=400kN/m, whose application point is 2.4m to the wall back toe, as shown in the figure. The unit weight of the ground soil is . Try to find the vertical stress increase at a point 7.2m below the center of the foundation. (with no consideration of the vertical stress increase due to the fill at the wall back)SolutionStep 1 Find the offsetting e: assuming the horizontal distance between the application point of the resultant force and the wall front toe A is x and equating moment induced by the resultant force with respect to A to that induced by each force component yields,2400x=2400x3.2-400x2.4then x=3.2-(400x2.4)/2400=2.8Therefore, the offsetting of the resultant force isE=B/2-x=0.2m Step 2 Find the contact pressure: the vertical pressure on the foun