§1.Soil Physical Characteristics and Classification

§1.Soil Physical Characteristics and Classification
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 §1.Soil Physical Characteristics and Classification ◆　Soil Formation ◆　 Composition and Structure of Soil ◆　Soil Model Expressions----Phase Relationships 　　　 ◆　Relative density of the granular soils ◆　Consistency of cohesive soils ◆　Soil Compaction ◆　Classification

§1.1 Soil Formation ♦ Residual soil
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 §1.1 Soil Formation To the civil engineer, Soil is any uncemented or weakly accumulation of mineral particles formed by the weathering of rocks, the void space between the particles containing water and /or air. Weak cementation can be due to carbonates or oxides precipitated between the particles or due to organic matter. ♦ Residual soil If the products of weathering remain at their original location they constitute a residual soil. ♦ Transported soil If the products are transported and deposited in a different location they constitute a transported soil, the agents of transportation being gravity, wind, water and glacier.

Physical Process Weathering Chemical Process
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 Temperature Water Ice/glacier Wind Disintegration (freezing/thawing) Physical Process (Single Grain) Weathering Air Water CO2 Liquid (acid, oxygen, etc) Chemical Process (Crystalline particles)

§ 1.2 Composition and Structure of Soil
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 § 1.2 Composition and Structure of Soil 1.2.1 Particle Size Analysis ♦ Particle sizes vary considerably from those measured in microns (clays) to those measured in meters (boulders)---a difference of one million! Most of natural soils are composite soils and the distribution of these sizes gives very useful information about the engineering behavior of the soil. ♦ The distribution is determined by separated the particles using two processes—sieving and sedimentation Sieving ♦ Sieves separate particles in the range between 75mm and 60μm(gravel and sand) and sedimentation separates particles less than 60μm(silt and clay). Particles greater than 75mm are cobble size and not usually include in the tests. They are removed before testing and an estimate made or their proportion.

Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1
1.2.3 Sedimentation Test ♦ This test procedure is based on Stockes’ Law, which governs the velocity at which spherical particles settle in a suspension: the larger the particles the greater is the settling velocity and vice versa. The law does not apply to particles smaller than mm.The size of a particles is given as the diameter of a sphere which would settle at the same velocity as the particle. The soil sample is pretreated with hydrogen peroxide to remove any organic material. The sample is then made up as a suspension in distilled water to which a deflocculating agent has been added to ensure that all particles settle individually. ♦ The suspension is placed in a sedimentation tube. From Stoke’ Law it is possible to calculate the time ,t,for particles of a certain size D(the equivalent settling diameter) ,to settle a specified depth in the suspension. If, after a the calculated time t, a sample of the suspension is drawn off with a pipette at the specified depth below the surface, the sample will contain only particles smaller than the size D at a concentration unchanged from that at the start of

1.2.4 Particle size distribution curves
sedimentation. If pipette samples are taken at the specified depth at times corresponding to other chosen particle sizes the particle size distribution can be determined from the weights of the residues. Stokes’ Law d-颗粒 Gs-土粒比重 G w-温度为T ℃时水的比重 ρ-4时水的密度 1.2.4 Particle size distribution curves ♦ The particle size distribution of a soil is presented as a curve on a semilogarithmic plot, the ordinates being the percentage by weight of particles smaller than the size given by the abscissa. The flatter the distribution curves the larger of the range of particle sizes in the soil; the steeper the curve the smaller the size range. ♦ D10 The size such that 10% of the particles are smaller than that size is denoted by D10 . D10 is defined as effective size. Other sizes such as D30 and D60 (Limited Size)can be defined in a similar way. ♦ The general slope and shape of the distribution curve can be described by means of the coefficient of uniformity (Cu) and the coefficient of curvature (Cc), defined as follow:

Particle size distribution curves
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 Particle size distribution curves

♦ The higher the value of the coefficient of coefficient of uniformity the larger the range of particles sizes in the soil. A well graded soil has a coefficient of curvature between 1 and 3. 1.2.5 Structure of Soil ♦ The nature of each individual particle in a soil is derived from the minerals it contains, its size and its shape. These are affected by the original rocks from which the particle was eroded, the degree of abrasion and comminution during erosion and transportation and decomposition and transportation and disintegration due to chemical and mechanical weathering. The various clay minerals are formed by the stacking of combinations of the basic sheet structures with different forms of bonding between the combined sheets. ♦ Structure of clay minerals ♦ Illite(伊里石) has a basic structure consisting of a sheet of alumina octahedrons between and combined with two sheets of by magnesium and iron and in the tetrahedral sheet there is partial substitution of silicon by aluminum. The combined sheets are linked together by fairly weak bonding due to potassium ions held between them.

♠ Kaolinite（高岭石） consists of a structure based on a single sheet of silica tetrahedrons combined with a single sheet of alumina octahedrons. There is very limited isomorphous substitution. The combined silica-alumina sheets are held together fairly tightly by hydrogen bonding: a kaolinite particle may consist of over one hundred stacks. Montmorillonite（蒙脱石） has the same basic structure as illite. In the octahedral sheets is occupied by water molecules and cations other than potassium. There is a very weak bond between the combined sheets due to these ions. Considerable swelling of montmorillonite can occur due to additional water being adsorbed between the combined sheets.

§1.3 Soil Model Expressions----Phase Relationships
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 §1.3 Soil Model Expressions----Phase Relationships ♦ Soils can be of either two-phase or three-phase composition. In a completely dry soil there are two phases, namely solid soil particles and pore air. A fully saturated soil is also two–phase, being composed of solid particles and pore water. A partially saturated soil is three-phase, being composed of solid soil particles, pore water and pore air .The components of a soil can be represented by a phase diagram as show in Fig.1.1a. (a) Fig.1.1a. (b)

♦ The specific gravity of the solid soil particle (G s) is given by
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 ♦ The specific gravity of the solid soil particle (G s) is given by The bulk density (ρ) of a soil is the ratio of the total mass to the total volume The unit weight(γ) of a soil is the ratio of the total weight(a force) to the total volume, i.e: The water content (w，％), or moisture content (m) is the ration of the mass of water to the mass of solid in the soil

TEST: The water content is determined by weighing a sample of the soil then drying the sample in an oven at a temperature of ℃and reweighing. Drying should continue until the differences between successive weightings at four-hourly intervals are not greater than 0.1% of the original mass of the sample. A drying period of 24h is normally adequate for most soils. The degree of saturation (Sr，％) is the ration of the volume of water to the total volume of void space The degree of saturation can range between the limits of zero for a completely dry soil and 1(or 100%) for a fully saturated soil. The void ratio (e) is the ratio of the volume of voids to the volume of solids

♦ The void ratio and the porosity are inter-related as follows
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 ♦ The porosity (n) in the ratio of the volume of voids to the total volume of the soil ♦ The void ratio and the porosity are inter-related as follows

♦ From the definition of the void ratio, if the volume of solids is 1 unit then the volume of voids is e units. The mass of solids is then Gsρw and, from the definition of water content, the mass of water is w Gsρw. The volume of water is thus w Gs. these volumes and massed are represented in Fig.1.1b ♦ The following relationships can now obtained. ♦ The degree of saturation can be expressed as ♦ In the case of a fully saturated soil, S r=1,hence

♦ The bulk density of a soil can be expressed as:
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 ♦ The bulk density of a soil can be expressed as: (1.11) Or from Equation : (1.12) ♦ For a fully saturated soil (S r=1) (1.13) ♦ For a completely dry soil (S r=0) (1.14) ♦ Equations similar to (1.11) to (1.14) apply in the case of unit weights, for example:

♦ where γw is the unit weight of water. Convenient units are KN/m3, the unit weight of water being 9.8KN/m3. ♦ When a soil in-situ is fully saturated the solid soil particles(volume1 unit weight Gsγw)are subjected to upthrust(γw) . Hence the buoyant unit weight (γ‘)is given by : (1.18) i .e: (1.19)

Example 1 A sample of soil in natural state , given V=210㎝3 ，W=0.0035KN，After oven Drying the soil had a weight of KN. If the particle specific gravity was 2.67, compute itsγ,ω, e, n, Sr,γd ,γsat V’= 210㎝3 =0.21×10-3m3, W=3.5N=0.0035KN Method 1

Method 2

Example 2 For a sample of soil withγs=27kN/m3, Given the volume was 12Cm3，the weight was 0.2N, After oven drying the soil had a weight of 0.165N. Determine the e and Sr in natural state. Method 1

Method 2

Exercise 1 In its natural condition a soil sample has a mass of 2290g and a volume of 1.15×10-3m3.After being completely dried in an oven the mass of the sample is 2035g. The value of Gs for the soil is Determine the bulk density, unit weight, water content, void ratio, porosity, degree of saturation . Bulk density, Unit weight, Water content,

Solution Part(a): the moist unit weight is
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 Example P.5 For a soil in natural state, given e=0.8, and Gs=2.68. (a) Determine the moist unit weight, dry unit weight, and degree of saturation. ( b) If the soil is made completely saturated by adding water,what would its moisture content be at that time? Also find the saturated unit weight. Solution Part(a): the moist unit weight is

1-4. A saturated soil sample weighs 0. 4N and its volume is 21. 5 cm3
1-4. A saturated soil sample weighs 0.4N and its volume is 21.5 cm3. The weight and the volume are 0.33N and 15.7 cm3after being non-completely (partly) dried in an oven for a period. The corresponding degree of saturation is 75%. Determine the water content w, void ration e and dry unit weight γd before drying. 1-4. 某饱和土样重0.4N ，体积为21.5 cm3。放入烘箱内烘一段时间后取出，称得其重量为0.33N ，体积减小至15.7 cm3，饱和度为75% 。试求该土样烘烤前的含水量 w、孔隙比 e 及干容重 γd 。

解：设烘一段时间后，孔隙体积为Vv2，孔隙水所占体积为Vw2，则：
在烘后状态：在烘前状态：联立求解得： Vv2=4.8cm3，Vw2 =3.6cm3

§1.4 Relative density of the granular soils
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 §1.4 Relative density of the granular soils ♦ The relative density is a term generally used to describe the degree of the granular soils. Relative density is defined as Where e max=maximum possible void ratio e min=minimum possible void ratio e =void ratio in natural state of soil

Example Medium dense 1/3< Dr ≤2/3 Dense 2/3< Dr ≤1
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 ♦ According to the equation, the relative density of a soil in it’s densest state (e= e min ) is 100%and the relative density of in it’s densest state (e= e max) is 0. The state of packing can be classified by the Dr : Loose < Dr ≤1/3 Medium dense 1/3< Dr ≤2/3 Dense /3< Dr ≤1 Example A sample of sand, given γ=14.7 kN/m3, w=13%, γdmin=12kN/m3 , γdmax=16.6kN/m3 . Estimate its compaction state.

Solution: So, it is loose.
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 Solution: So, it is loose.

♦ The equation can also be expressed as
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 ♦ The equation can also be expressed as Where the γ dmax ,γdmin ,γd are the maximum, minimum, and natural-state dry unit weights of the soil. ♦ The γdmax ,γdmin ,γd can be acquired by the experiments. Then the relative density can be determined.

§1.5 Consistency of cohesive soils
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 §1.5 Consistency of cohesive soils 1.5.1 Atterberg limits ♦ Atterberg limits are developed for describing the limit consistency of fine-grained soils on the basis of moisture content. These limits are the liquid limit, the plastic limit, and the shrinkage limit. ♦ The liquid limit LL is defined as the moisture content, in percent, at which the soil changes from a liquid state to a plastic state. The moisture contents(in percent)at which the soil changes from a plastic to a semisolid state and from a semisolid state to a solid state are defined, respectively, as the plastic limit Lpand the shrinkage limit Ls . Liquid State Plastic State Semisolid State Solid State Moisture contents Liquid plastic shrinkage limit LL limit Lp limit Ls

TEST—— ♦ The Atterberg limits of cohesive soil depend on several factors, such as several factors, such as the amount and type of clay minerals and type of absorbed cation. ♦ The difference between the liquid limit and the plastic limit of a soil is defined as the plasticity index PI: PI=LL-PL Where LL is the liquid limit and PL is the plastic limit 粘性土的分类《建筑地基基础设计规范》 GB 塑性指数Ip 土的名称 Ip>17 粘土 10<Ip≤17 粉质粘土注:塑性指数由相应于76g圆锥体沉入土样中深度为10mm时测定的液限计算而得。

1.5.2 Liquidity Index ♦ Where w n is the natural moisture content.
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 ♦ The relative consistency of a cohesive soil can be defined by a ratio called the liquidity index LI. It is defined as 1.5.2 Liquidity Index ♦ Where w n is the natural moisture content. If w n =LL, then LI=1. If w n =PL,then LI= 0. thus for a natural deposit which is in a plastic state, the value of LI varies between 1 and 0. A natural soil deposit with w n > LL will have a LI greater than 1.In an undisturbed state, these soils may be stable; however, a sudden shock may transform them into a liquid state. Such soils are called sensitive clays.

粘性土的状态,可按下表分为坚硬、硬塑、可塑、软塑、流塑。粘性土的状态《建筑地基基础设计规范》 GB50007-2002
粘性土的状态《建筑地基基础设计规范》 GB 液性指数IL 状态 IL≤0 坚硬 0.75< IL≤1 软塑 0< IL≤0.25 硬塑 IL>1 流塑 0.25< IL≤0.75 可塑

1.5.3 Activity ♦ Based on laboratory test results for several soils, Skempton made the observation that, for a given soil, the plasticity index is directly proportional to the percent of clay-size fraction (i.e. percent by weight finer than 0.002mm in size). With this observation , Skempton defined a parameter A called activity as Where C is the percent of clay-size fraction, by weight. ♦ Activity has been used as an index property to determine the swelling potential of expansive clays.

§1.6 Soil Compaction ♦ The degree of compaction of a soil is measured in terms of dry density, i.e. the mass of solids only per unit volume of soil. It is apparent that the dry density is given by Where ρis the bulk density of the soil ,w is the water content. ♦ The dry density of a given soil after compaction depends on the water content and the energy supplied by the compaction equipment. ♦ The compaction characteristics of a soil can be assessed by means of standard laboratory tests. The soil is compacted in a cylindrical mould using a standard compactive effort. After compaction, the bulk density and water content of the soil are determined and the dry density calculated. For a given soil the compaction process is repeated at least five times, the water

content of the sample being increased each time. Dry density is plotted against water content and a curve of the form shown in Fig below is obtained: ♦ The curve shows that for a particular method of compaction there is a particular value of water content, known as the optimum water content (w opt) at which a maximum value of dry density is obtained. At low values of water content most soils tend to be stiff and are difficult to compact. As the water content is increased the soil becomes more workable, facilitating compaction and result in high dry densities. At high water contents, however, the dry density decreases with increasing water content, an increasing proportion of the soil volume being occupied by water.

§1.7 Classification 1.7.1 The British Soil Classification System
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 §1.7 Classification 1.7.1 The British Soil Classification System ♦ The British Soil Classification System is shown in detail in Table 1.4. Reference should also be made to the plasticity chart(Fig.1.6).The soil groups in the classification are denoted by group symbols composed of main and qualifying descriptive letters having the meanings given in Table1.5.

1.7.2The Unified Soil Classification System
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 1.7.2The Unified Soil Classification System ♦ In the Unified Soil Classification System, developed in the United States, the group symbols consist of a primary and a secondary descriptive letter. The letters and their meaning are given in Table 1.6.The Unified Soil Classification System, including the laboratory classification criteria, is detailed in Table 1.7 and the associated plasticity chart is shown in Fig1.7.Classification may be based on either laboratory or field test procedures. Soil exhibiting the characteristics of two groups should be given a boundary classification denoted by dual symbol connected by a hyphen.

1.7.3 Chinese classification
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 1.7.3 Chinese classification ◈《建筑地基基础设计规范》 GB ◈《公路土工试验规程》 JTJ

1.7.3 Chinese classification
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 1.7.3 Chinese classification A Gravels ♦ The gravels are those which more than 50% of coarse material is of gravel size(coarser than 2mm). ♦ The classification of the gravels 土的名称颗粒形状颗粒级配漂石圆形及亚圆形为主粒径大于200mm的颗粒超过全重块石棱角形为主％卵石圆形及亚圆形为主粒径大于20mm地颗粒超过全重碎石棱角形为主％圆砾圆形及亚圆形为主粒径大于2mm的颗粒超过全重角砾棱角形为主％

B Sands The classification of the sands
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1 B Sands ♦ The sands are those which more than 50% of coarse material is of sand size(finer than 2mm and coarser than 0.075mm). The classification of the sands 土的名称颗粒级配砾砂粒径大于2 mm的颗粒占全重的25～50％粗砂粒径大于0。5mm的颗粒超过全重50％中砂粒径大于0。25mm的颗粒超过全重50％细砂粒径大于0。075mm的颗粒超过全重85％粉砂粒径大于0。075mm的颗粒超过全重50％

C Clays ♦ The clays are often classified by the the soils’ age of sediment or the plasticity index PI. The classification of the clays according to the age of sediment 土的名称沉积年代工程性质老粘性土第四纪更新世及较高的强度和较低的压缩性其以前沉积的粘性土一般粘性土第四纪全新世（文化期以前）工程性质变化大新近沉积的粘性土文化期以来沉积的粘性土欠固结，强度较低 The classification of the clays according to the PI 土的名称粘质粉土粉质粘土粘土塑性指数 PI≤ ＜PI≤ PI＞17

D Granular soils ♦ The granular soils are those which more than 50% of the material is finer than 0.075mm. The granular soils are classified by the Plasticity chart. The Plasticity chart was advanced by A.Casagrande in 1948: Fig-The line A,B,C divided the chart into six sectors .We can know which class the soil belong to by the value of the PI and LL which can get from the experiment. lineA:PI=0.73(LL-20) lineB:LL=50 lineC:LL=28 SLM：粉质低液限砂土 CL:低液限粘土 CLM:粉质低液限粘土 ML,MI:粉土 CIM:粉质中液限粘 CI:中液限粘土 CH:高液限粘土 CV:很高液限粘土

Stress due to soil weight Contact stress Stress due to loading
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 §2 stress in soil Stress due to soil weight Contact stress Stress due to loading

§2.2 Stress due to Self-weight
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 §2.2 Stress due to Self-weight General situation

Three Cases: -For Many Layers of soil, the vertical stress due to self-weight of soil is given as following.

-With uniform surcharge on infinite land surface
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 -With uniform surcharge on infinite land surface

Effective Vertical Stress due to Self-Weight of Soil
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Effective Vertical Stress due to Self-Weight of Soil - Consider a soil mass having a horizontal surface and with the water table at surface level. The total vertical stress (i.e. the total normal stress on a horizontal plane) at depth z is equal to the weight of all material (solids + water) per unit area above that depth ,i.e. σv = γsat z - The pore water pressure at any depth will be hydrostatic since the void space between the solid particles is continuous, therefore at depth z: u = γw z - Hence the effective vertical stress at depth z will be: σ’v=σv- u =( γsat - γw)z= γ’z where γ’ is the buoyant unit weight of the soil.

-Under water Table General, for sand below water table, the γ’ is used; but for clay below water table, it is difficult to determine which one(γ’ or γsat) is suitable. We often choose the buoyant unit weight when the index of liquid LI>=1; the saturated unit weight when the index of liquid LI<=0. When 0<LI<1, the disadvantageous one is choosen.

Example w=15.6% e=0.57 γs=26.6kN/m3 w=22% wL=32% wp=23% γs=27.3kN/m3
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Example The stratum’s conditions and the related physical characteristics parameters of a foundation are shown in Fig below. Calculate the stress due to self-weight at a,b,c. Draw the stress distribution. w=15.6% e=0.57 γs=26.6kN/m3 w=22% wL=32% wp=23% γs=27.3kN/m3

For silver sand, the buoyant unit weight (γ‘)is used :
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 For silver sand, the buoyant unit weight (γ‘)is used : For saturated clay, So, the clay can be considered the watertight; then the saturated unit weight (γsat)is used :

b σz(upper)=γ’1h1=9.9×2=19.8kPa
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 a σz=0 b σz(upper)=γ’1h1=9.9×2=19.8kPa σz(Down)=γ’1h1+ γw(h1+hw)=9.9×2+10×(2+1.2)=51.8kPa c σz=γ’1h1+ γw(h1+hw)+ γsat2h2 = 9.9×2+10×(2+1.2)+20.8×3=114.2kPa

Exercise 2-2 w=8% e=0.7 γs=26.5kN/m3 e=1.5 γs=27.2kN/m3
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Exercise 2-2 The stratum’s conditions and the related physical characteristics parameters of a foundation are shown in Fig below. Calculate the stress due to self-weight at 10m depth. Draw the stress distribution. Note: For saturated clay, both cases (watertight and non-watertight) need to consider. w=8% e=0.7 γs=26.5kN/m3 e=1.5 γs=27.2kN/m3

§ 2.3 Contact stress 2.3.1 Concept of Contact pressure
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 § 2.3 Contact stress 2.3.1 Concept of Contact pressure ♦ A foundation is the interface between a structural load and the ground. The stress p applied by a structure to a foundation is often assumed to be uniform. The actual pressure then applied by the foundation to the soil is a reaction, called the contact pressure p and its distribution beneath the foundation may be far from uniform. ♦ This distribution depends mainly on: · stiffness of the foundation. i.e. flexible → stiff → rigid. · compressibility or stiffness of the soil. · loading conditions – uniform or point loading. 2.3.2 Contact pressure – uniform loading The effects of the stiffness of the foundation (flexible or rigid) and the compressibility of the soil (clay or sand) are illustrated in Figure below..

Figure The distribution of contact pressure
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Figure The distribution of contact pressure

2.3.3 Stiffness of foundation
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.3.3 Stiffness of foundation A flexible foundation has no resistance to deflection and will deform or bend into a dish-shaped profile when stresses are applied. An earth embankment would comprise a flexible structure and foundation. A stiff foundation provides some resistance to bending and well deform into a flatter dish-shape so that differential settlements are smaller. This forms the basis of design for a raft foundation placed beneath the whole of a structure. A rigid foundation has infinite stiffness and will not deform or bend, so it moves downwards uniformly. This would apply to a thick, relatively small reinforced concrete pad foundation.

2.3.4 Stiffness of soil The stiffness of a clay will be the same under all parts of the foundation so for a flexible foundation a fairly uniform contact pressure distribution is obtained with a dish-shaped (sagging) settlement profile. For a rigid foundation the dish-shaped settlement profile must be flattened out so the contact pressure beneath the centre of the foundation is reduced and beneath the edges of the foundation it is increased. Theoretically, the contact pressure increases to a very high value at the edges although yielding of the soil would occur in practice , leading to some redistribution of stress. The stiffness of a sand increase as the confining pressures around it increase so beneath the centre of the foundation the stiffness will be smaller. A flexible foundation of the sand will, therefore, produce greater strains at the edges than in the centre so the settlement profile will be dish-shaped but upside-down (hogging) with a fairly uniform contact pressure. For a rigid foundation this settlement profile must be flattened out so the contact pressure beneath the centre would be increased and beneath the edges it would be decreased.

2.3.5 Contact pressure-point loading
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.3.5 Contact pressure-point loading pressure distribution similar to the distribution produced by a uniform pressure on a clay is obtained. This suggests that a point load at the centre of a rigid foundation is comparable to a uniform pressure. An analysis for contact pressure beneath a circular raft with a point load W at its centre resting on the surface of an incompressible soil (such as clay) has been provided by Borowicka,1939. This shows that the contact pressure distribution is non-uniform irrespective of the stiffness of the raft or foundation. For a flexible foundation the contact pressure is concentrated beneath the point load which is to be expected and for a stiff foundation it is more uniform. For a rigid foundation the stresses beneath the edges are very considerably increased and a

2.3.6 Stress distribution -Central Loading P=P/F
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.3.6 Stress distribution The stresses that already exist within the ground due to self-weight of the soil are discussed in Chapter 4. When a load or pressure from a foundation or structure is applied at the surface of the soil this pressure is distributes throughout the soil and the original normal stresses and shear stresses ate altered. For most civil engineering applications the changes in vertical stress are required so the methods given below are for increases in vertical stress only. -Central Loading P=P/F

-Eccentric Loading (0<e<=B/6)
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 -Eccentric Loading (0<e<=B/6)

(e>B/6)

pa=p-γh 2.3.7 Additional pressure on the bottom of foundation
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.3.7 Additional pressure on the bottom of foundation pa=p-γh Where pa-additional pressure; p-contact pressure; γ-natural unit weight of soil( bouyant unit weight if below water surface); h- buried depth of the foundation.

Exercise 2-3 A load of 1200kN as well as a moment of 1500 kNm is carried on a rectangular foundation (axb=6x4m) at a shallow depth in a soil mass as shown in Fig below. Determine the maximum & minimum contact stresses on the bottom of foundation. If the stress is tensional, Where does the width of foundation extend assuming the stress at right tip is zero and the position & value of loading, the length of the foundation a do not change? Calculate the contact stress after the extension.

§ 2.4 Stress due to loading 2.4.1 Stresses beneath point load（点荷载）
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 § 2.4 Stress due to loading 2.4.1 Stresses beneath point load（点荷载） Boussinesq published in 1885 a solution for the stresses beneath a point load on the surface of a material which had the following properties: semi-infinite – this means infinite below the surface therefore providing no boundaries of the material apart from the surface homogeneous – the same properties at all locations isotropic –the same properties in all directions elastic –a linear stress-strain relationship.

Linear elastic assumption
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Linear elastic assumption

Properties Substitute Z/R for Cos β,
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Substitute Z/R for Cos β, Properties

Bulbs of pressure(压力泡)
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Bulbs of pressure(压力泡) Isoline(等值线)——Lines or Contours of equal stress increase

2.4.2 Stresses due to line load（线荷载）
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.4.2 Stresses due to line load（线荷载）

-polar coordinates

2.4.3 Stresses due to uniform vertical loading
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.4.3 Stresses due to uniform vertical loading on an infinite strip （条形荷载）

-polar coordinates

-Principal stress

Properties

(1) Isoline(等值线) (2) Direction of principal stress (3) The value of maximum shear stress

→

2.4.4 Stresses due to Linearly increasing vertical
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.4.4 Stresses due to Linearly increasing vertical loading on an infinite Strip

2.4.5 Stresses due to a uniform loaded rectangular area
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2.4.5 Stresses due to a uniform loaded rectangular area -central point method（中点法）

-Corner method（角点法）

Principle of superposition
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Principle of superposition For stresses beneath points other than the corner of the loaded area the principle of superposition should be used, as described in Fig below. →

Stresses beneath flexible area of any shape
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Stresses beneath flexible area of any shape Stresses beneath flexible area of any shape (Figure 5.9) Newmark (1942) devised charts to obtain the vertical stress at any depth, beneath any point (inside or outside) of an irregular shape. Use of the charts is explained in Figure 2.9.

Stresses beneath a flexible rectangle
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Stresses beneath a flexible rectangle ♦ Stresses beneath a flexible rectangle (figure 2.7) The vertical stressσ y at a depth z beneath the corner of a flexible rectangle supporting a uniform pressure q has been determined using: σ v= qI and influence factors I. given by Giroud (1970) are presented in Figure 5.7. they are for an infinite soil thickness. These curves are equivalent to the commonly used charts of Fadum (1948) but are easier to use.

Stresses beneath a rigid rectangle
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Stresses beneath a rigid rectangle

A point load of P is applied at the ground surface.
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Exercise 2-4 A point load of P is applied at the ground surface. Calculate the stress and plot the variation: Vertical stress due to load P along vertical line(r=0) at the different depths Z=1,2,3,4,5m. Vertical stress due to load P from different distances (r=0,1,2,3,4,5m, respectively) along horizontal surface (at the depth Z=2m). Vertical stress due to load P along vertical line(r=1.5m) at the different depths Z=0,0.5,1,2,3,4,5,6m.

Exercise 2-5 A uniform load of p(=20kpa) is applied at the ground surface as shown in Fig below. Calculate the stress due to load p beneath point A at 5m depth.

§3 permeability in soil 土的渗透性
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 §3 permeability in soil 土的渗透性土的渗透性 Permeability 有效应力 Effective stress

3.1 土的渗透性 Permeability 1 达西定律Darcy Law
This states that discharge velocity, v of water is proportional to the hydraulic gradient, i v=q/A= k i

where : K=Darcy coefficient of permeability ,m/s The hydraulic gradient i is the ratio of the head loss h over a distance L The discharge velocity v is defined as the quantity of water , q percolating through a cross-sectional area A in unit time .This is not the sane as the velocity of the water percolating through the soil which is known as the seepage velocity .

Constant head permeameter
Falling head permeameter ←

2 粘性土的渗透性 Permeability of Clay
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 2 粘性土的渗透性 Permeability of Clay

3 渗透性力 Force of permeability

4 流砂与管涌 Running Sand, Heaving or Piping
解水头差：h= =4m 流径长L=2.5+2d 细砂的浮容重: 临界水力梯度考虑安全系数后,实有的水力梯度为:

5 临界水力梯度 Critical hydraulic gradient

Exercise 3-1 For the seepage situations shown in Fig below, the length of the sand sample L=25cm and water head difference h=20cm. Calculate the Force of permeability applied on sand If γs=26.8kN/m3, e=0.72, Determine whether will running sand occur? If the running sand occurs, calculate the necessary water head difference.

3.2 有效应力Effective stress 1.The Principle of Effective Stress
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 3.2 有效应力Effective stress 1.The Principle of Effective Stress ♦ The importance of the forces transmitted through the soil skeleton from particle to particle was recognized in 1923 when Terzaghi presented the principle of effective stress, an intuitive relationship based on experimental data. The principle applies only to fully saturated soils and relates the following three stresses: (1) 总应力 the total normal stress (σ) (2) 孔隙水压力 the pore water pressure (u) (3) 有效应力 the effective normal stress (σ’ )

总应力 the total normal stress (σ) on a plane within the soil mass, being the force per unit area transmitted in a normal direction across the plane, imaging the soil to be a solid (single-phase) material. ←

孔隙水压力 the pore water pressure (u), being the pressure of the water filling the void space between the solid particles. ←

有效应力 the effective normal stress (σ’ ) on the plane, representing the stress transmitted through the soil skeleton only. ←

-Un-Saturated soil The Principle of Effective Stress Bishop,1959
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 ♦ The relationship is : σ = σ’ +u The Principle of Effective Stress -Un-Saturated soil Bishop,1959 For Saturated soil: x=1 For completely dried soil x=0

Skempton A W, 1961 (For Completely Saturated soil) (For Partly Saturated soil)

-静水条件下的有效应力hydrostatic
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 -静水条件下的有效应力hydrostatic Total stress at a-a plane pore water pressure effective normal stress at a-a plane

-自上而下的稳定渗流Seepage to the down
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 -自上而下的稳定渗流Seepage to the down Total stress at a-a plane pore water pressure effective normal stress at a-a plane i=h/h2

-自下而上的稳定渗流Seepage to the upper
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 -自下而上的稳定渗流Seepage to the upper Total stress at a-a plane pore water pressure effective normal stress at a-a plane i=h/h2

Example For sand, e=0.6, Sr=35% (above water table) γs=27kN/m3,
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 2 Example For sand, e=0.6, Sr=35% (above water table) γs=27kN/m3, For clay, γsat=21kN/m3 Calculate the total stress, pore water pressure and the effective stress in 9m depth. Plot the distribution.

§4 Compressibility, consolidation and settlement 土的压缩、固结与地基沉降
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 §4 Compressibility, consolidation and settlement 土的压缩、固结与地基沉降 Compressibility /土的压缩性 Consolidation /土的固结 Settlement /地基沉降我爱你，小姐

Introduction / 概述 Compressibility and consolidation can be distinguished as: ·compressibility –volume changes in a soil when subjected to pressure –giving AMOUNTS of settlement ·consolidation -rate of volume change with time –giving TIME to produce an amount of settlement required These are distinct from: 1. compaction which is the expulsion of air from a soil by applying compaction energy. 2. immediate or undrained settlement which is the resultant deformation of a soil under applied stresses without any volume change taking place

4.1 Compressibility/土的压缩性
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 4.1 Compressibility/土的压缩性 4.1.1 Consolidation Test and compression characteristics/土的固结试验与压缩特性 1 Consolidation test Assumption: Load distribution-uniform Stress distribution(in different height)-the same Lateral deformation-0 The area of the sample section-unchangeable Solid soil-uncompressible oedometer

oedometer

2 Compression curve/压缩曲线
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 2 Compression curve/压缩曲线

3 compression coefficient av/压缩系数

av1-2<0.1Mpa-1, Low compressibility /低压缩性土 0.1≤av1-2<0.5Mpa-1, Middle compressibility/中压缩性土 av1-2 ≥ 0.5Mpa-1, High compressibility /高压缩性土 ←

4 compression index Cc/压缩指数

p>pc, lack consolidation（欠固结） P=pc, normal consolidation （正常固结）
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 Stress History Effect— p>pc, lack consolidation（欠固结） P=pc, normal consolidation （正常固结） P<pc, overconsolidation （超固结）

Determine pc 　　 ←

4.1.2 compressible deformation calculation/单向压缩量的计算 Only compression in vertical，Deformation due to void volume decrease

4.1.3 Effective stress plot The soil is described as normally consolidated when its state exists on the steeper line (1 and 4) and the effective stress can only be increased with subsequent reduction in volume because decreasing the stress takes the soil state away from the normally consolidated line. This line is often referred to as the virgin compression curve or line as any change of effective stress along it will be for the first and only time whereas any number of unload/reload paths could be followed. The soil is described as overconsolidated when it occurs on the flatter portions( 2 and 3) and volume changes can increase or decrease with changes in effective stress.

4.1.4 reloading curves The past geological and stress history has brought the soil to its present condition. If a sample is taken from the ground with no moisture content change and loaded in a laboratory consolidation apparatus a reloading curve can be obtained by plotting the void ratio produced after the soil has consolidated to a new equilibrium for each change in effective stress. The shape of the reloading curve (as well as geological information about the soil at the site) will help to determine whether the soil is normally consolidated or overconsolidated. If the soil is normally consolidated the reloading curve will continue on the virgin compression line from its present condition and will follow the straight line on a log σ＇plot. The reloading curve for an overconsolidated clay will have two portions, one commencing from its present condition and following a flatter path until it reaches the virgin compression line at the preconsolidation pressure, followed by a steeper line corresponding to the virgin compression curve

→ 4.1.5 overconsolidation ratio OCR
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 4.1.5 overconsolidation ratio OCR The overconsolidation ratio (OCR) defined as: For a normally consolidated clay the present effective stress is also the previous maximum so the OCR=1. for a heavily overconsolidated clay the OCR may be 4 or more therefore this type of soil has been subjected to a much greater stress in the past Compared to its present condition. The significance of Pc＇for an overconsolidated clay is that if stresses are kept below this value then settlements can be expected to be small but if the applied stresses due to loading exceed this value then large settlements will occur as consolidation will take place along the virgin compression line. →

4.1.6 Deformation modulus/变形模量 Coefficient of earth pressure at rest/ 静止侧压力系数 Poisson’s ratio/泊松比 μ Modulus of deformation/变形模量E0

Modulus of compression/压缩模量 Es

4.2 Consolidation - Terzaghi theory of one-D consolidation 饱和粘性土太沙基单向固结理论 4.2.1 Introduction/引言 The theory considers the rate at which water is squeezed out of an element of soil and can be used to determine the rates of: volume change of the soil with time settlements at the surface of the soil with time pore pressure dissipation with time t=0, u=p0, σ’=0 t=t1, u+ σ’=p0 t=∞, u=0, σ’=p0 u=f(t,x,y,z)

4.2.2 Assumptions/基本假定 1. Compression and flow are one-dimensional, i.e. they are vertical only. Whereas significant horizontal flow can occur in layered deposits. 2. Darcy’s law is valid at all hydraulic gradients but deviation may occur at low hydraulic gradients. 3. k and m v remain constant. However, they both usually decrease during consolidation. 4. No secondary compression or creep occurs. If this occurs the void ratio-effective stress relation-ship is not solely dependent on the consolidation process. 5. The load is applied instantaneously and over the whole of the soil layer. However, loads are applied over a construction period and usually do not extend over a wide area in relation to the thickness of the consolidating deposit.

4.2.3 Differential equation of consolidation/固结微分方程

4.2.4 Solution of the consolidation equation/固结微分方程求解
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 Solution of the consolidation equation/固结微分方程求解 The basic differential equation of consolidation (equation 3.1) gives the relationship between three values. where: c v is the coefficient of consolidation m v is the coefficient of volume compressibility k is the coefficient of permeability General solution Initial condition and boundary conditions t=0, 0≤Z≤2H, u=σz=p0, 0<t< ∞, Z=0,u=0 Z=2H, u=0 t=∞, 0≤Z≤2H, u=0

Case: triangular distribution
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 Special solution Time factor Case: triangular distribution

4.2.5 Degree of consolidation/固结度
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 4.2.5 Degree of consolidation/固结度 -average degree of consolidation -different cases Rectangular Distribution

triangular distribution

Example 已知:H=10m,大面积荷载p0=0.12Mpa,e0=1.0,av=0.3Mpa-1,K=1.8cm/年求单面及双面排水条件下: (1)加荷一年的沉降量 (2)沉降达14cm所需时间 Solution

an half-closed layer/单面排水 a open layer/双面排水

Tv=0.524, t=4.37(years) for an half-close layer H=10m t=1.09(years) for a open layer H=5m

Type Type Type Type Type 0-2

The Degree of Consolidation Type 0-1 Type Type Type 0-1

4.3 Settlement/地基沉降计算 4.3.1 Classification of foundation settlements/地基沉降分类 Settlements produced by applied stresses originate from: consolidation settlement/主固结沉降, Sc secondary compression次固结沉降, Ss immediate settlement/瞬时沉降, Sd amounts of each must be determined to give the total settlement, S=Sd+Sc+Ss

-Compression layer/压缩层: General,
4.3.2 Principle and procedure of settlement using delamination total Method 分层总和法 -Formula/计算公式: -Compression layer/压缩层: General, There is soft layer below/有软卧层

-Thickness of every layer/分层厚: <0.4B (B- Width of foundation/基础厚度) -Additional contact pressure/附加压力/基地净压力: -Calculating procedure/计算步骤: To delaminate (interface, water table) /分层（土层交界面/地下水位） (hi<0.4B)

Calculate Stress due to self weight/计算地基各层自重应力 Calculate stress due to load/ 计算地基各层附加应力 Determine the compression thickness/ 确定压缩层厚度 Calculate average σ0σz / 计算地基各层自重应力及附加应力的平均值 Calculate e0i,e1i 根据e~p曲线得到各层e0i,e1i Calculate/计算各层 Si Calculate/计算总沉降 S

§4 Compressibility, consolidation and settlement 土的压缩、固结与地基沉降
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 §4 Compressibility, consolidation and settlement 土的压缩、固结与地基沉降 Compressibility /土的压缩性 Consolidation /土的固结 Settlement /地基沉降我爱你，小姐

Introduction / 概述 Compressibility and consolidation can be distinguished as: ·compressibility –volume changes in a soil when subjected to pressure –giving AMOUNTS of settlement ·consolidation -rate of volume change with time –giving TIME to produce an amount of settlement required These are distinct from: 1. compaction which is the expulsion of air from a soil by applying compaction energy. 2. immediate or undrained settlement which is the resultant deformation of a soil under applied stresses without any volume change taking place

4.1 Compressibility/土的压缩性
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 4.1 Compressibility/土的压缩性 4.1.1 Consolidation Test and compression characteristics/土的固结试验与压缩特性 1 Consolidation test Assumption: Load distribution-uniform Stress distribution(in different height)-the same Lateral deformation-0 The area of the sample section-unchangeable Solid soil-uncompressible oedometer

oedometer

2 Compression curve/压缩曲线
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 2 Compression curve/压缩曲线

3 compression coefficient av/压缩系数

av1-2<0.1Mpa-1, Low compressibility /低压缩性土 0.1≤av1-2<0.5Mpa-1, Middle compressibility/中压缩性土 av1-2 ≥ 0.5Mpa-1, High compressibility /高压缩性土 ←

4 compression index Cc/压缩指数

p>pc, lack consolidation（欠固结） P=pc, normal consolidation （正常固结）
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 Stress History Effect— p>pc, lack consolidation（欠固结） P=pc, normal consolidation （正常固结） P<pc, overconsolidation （超固结）

Determine pc 　　 ←

4.1.2 compressible deformation calculation/单向压缩量的计算 Only compression in vertical，Deformation due to void volume decrease

4.1.3 Effective stress plot The soil is described as normally consolidated when its state exists on the steeper line (1 and 4) and the effective stress can only be increased with subsequent reduction in volume because decreasing the stress takes the soil state away from the normally consolidated line. This line is often referred to as the virgin compression curve or line as any change of effective stress along it will be for the first and only time whereas any number of unload/reload paths could be followed. The soil is described as overconsolidated when it occurs on the flatter portions( 2 and 3) and volume changes can increase or decrease with changes in effective stress.

4.1.4 reloading curves The past geological and stress history has brought the soil to its present condition. If a sample is taken from the ground with no moisture content change and loaded in a laboratory consolidation apparatus a reloading curve can be obtained by plotting the void ratio produced after the soil has consolidated to a new equilibrium for each change in effective stress. The shape of the reloading curve (as well as geological information about the soil at the site) will help to determine whether the soil is normally consolidated or overconsolidated. If the soil is normally consolidated the reloading curve will continue on the virgin compression line from its present condition and will follow the straight line on a log σ＇plot. The reloading curve for an overconsolidated clay will have two portions, one commencing from its present condition and following a flatter path until it reaches the virgin compression line at the preconsolidation pressure, followed by a steeper line corresponding to the virgin compression curve

→ 4.1.5 overconsolidation ratio OCR
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 4.1.5 overconsolidation ratio OCR The overconsolidation ratio (OCR) defined as: For a normally consolidated clay the present effective stress is also the previous maximum so the OCR=1. for a heavily overconsolidated clay the OCR may be 4 or more therefore this type of soil has been subjected to a much greater stress in the past Compared to its present condition. The significance of Pc＇for an overconsolidated clay is that if stresses are kept below this value then settlements can be expected to be small but if the applied stresses due to loading exceed this value then large settlements will occur as consolidation will take place along the virgin compression line. →

4.1.6 Deformation modulus/变形模量 Coefficient of earth pressure at rest/ 静止侧压力系数 Poisson’s ratio/泊松比 μ Modulus of deformation/变形模量E0

Modulus of compression/压缩模量 Es

4.2 Consolidation - Terzaghi theory of one-D consolidation 饱和粘性土太沙基单向固结理论 4.2.1 Introduction/引言 The theory considers the rate at which water is squeezed out of an element of soil and can be used to determine the rates of: volume change of the soil with time settlements at the surface of the soil with time pore pressure dissipation with time t=0, u=p0, σ’=0 t=t1, u+ σ’=p0 t=∞, u=0, σ’=p0 u=f(t,x,y,z)

4.2.2 Assumptions/基本假定 1. Compression and flow are one-dimensional, i.e. they are vertical only. Whereas significant horizontal flow can occur in layered deposits. 2. Darcy’s law is valid at all hydraulic gradients but deviation may occur at low hydraulic gradients. 3. k and m v remain constant. However, they both usually decrease during consolidation. 4. No secondary compression or creep occurs. If this occurs the void ratio-effective stress relation-ship is not solely dependent on the consolidation process. 5. The load is applied instantaneously and over the whole of the soil layer. However, loads are applied over a construction period and usually do not extend over a wide area in relation to the thickness of the consolidating deposit.

4.2.3 Differential equation of consolidation/固结微分方程

4.2.4 Solution of the consolidation equation/固结微分方程求解
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 Solution of the consolidation equation/固结微分方程求解 The basic differential equation of consolidation (equation 3.1) gives the relationship between three values. where: c v is the coefficient of consolidation m v is the coefficient of volume compressibility k is the coefficient of permeability General solution Initial condition and boundary conditions t=0, 0≤Z≤2H, u=σz=p0, 0<t< ∞, Z=0,u=0 Z=2H, u=0 t=∞, 0≤Z≤2H, u=0

Case: triangular distribution
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 Special solution Time factor Case: triangular distribution

4.2.5 Degree of consolidation/固结度
Civil Engineering Department of Shanghai University Soil Mechanics Chapter 4 4.2.5 Degree of consolidation/固结度 -average degree of consolidation -different cases Rectangular Distribution

triangular distribution

Example 已知:H=10m,大面积荷载p0=0.12Mpa,e0=1.0,av=0.3Mpa-1,K=1.8cm/年求单面及双面排水条件下: (1)加荷一年的沉降量 (2)沉降达14cm所需时间 Solution

an half-closed layer/单面排水 a open layer/双面排水

Tv=0.524, t=4.37(years) for an half-close layer H=10m t=1.09(years) for a open layer H=5m

Type Type Type Type Type 0-2

The Degree of Consolidation Type 0-1 Type Type Type 0-1

4.3 Settlement/地基沉降计算 4.3.1 Classification of foundation settlements/地基沉降分类 Settlements produced by applied stresses originate from: consolidation settlement/主固结沉降, Sc secondary compression次固结沉降, Ss immediate settlement/瞬时沉降, Sd amounts of each must be determined to give the total settlement, S=Sd+Sc+Ss

-Compression layer/压缩层: General,
4.3.2 Principle and procedure of settlement using delamination total Method 分层总和法 -Formula/计算公式: -Compression layer/压缩层: General, There is soft layer below/有软卧层

-Thickness of every layer/分层厚: <0.4B (B- Width of foundation/基础厚度) -Additional contact pressure/附加压力/基地净压力: -Calculating procedure/计算步骤: To delaminate (interface, water table) /分层（土层交界面/地下水位） (hi<0.4B)

Calculate Stress due to self weight/计算地基各层自重应力 Calculate stress due to load/ 计算地基各层附加应力 Determine the compression thickness/ 确定压缩层厚度 Calculate average σ0σz / 计算地基各层自重应力及附加应力的平均值 Calculate e0i,e1i 根据e~p曲线得到各层e0i,e1i Calculate/计算各层 Si Calculate/计算总沉降 S

《建筑地基基础设计规范》计算地基沉降地基沉降计算的理论公式法

《建筑地基基础设计规范》（GBJ 7-89）所推荐的地基最终沉降量计算方法是另一种形式的分层总和法。它也采用侧限条件的压缩性指标，并运用了平均附加应力系数计算，还规定了地基沉降计算深度的标准以及提出了地基的沉降计算经验系数，使得计算成果接近于实测值。　　1、计算公式　　规范推荐的地基最终沉降量s(mm)的计算公式如下: ψ --沉降计算经验系数，根据地区沉降观测资料及经验确定,一般在之间。 n --地基沉降计算深度范围内所划分的土层数，其分层厚度取法同前面按分层总和法 Po--对应于荷载标准值时的基础底面附加压力(kPa) Esi--基础底面下第i层土的压缩模量， (MPa)； Zi和Zi-1--基础底面至第i层土、第i-1层土底面的距离(m)； ai和 ai-1--基础底面的计算点至第i层土、第i-1层土底面范围内平均附加应力系数　

　　2、计算深度zn的确定　　《建筑地基基础设计规范》用符号表示地基沉降计算深度，并规定应满足下列条件(包括考虑相邻荷载的影响)：　式中Sn为最后一层厚为Dz的土层的压缩量。Dz的取值查表。按上式所确定的沉降计算深度下如有较软土层时，尚应向下继续计算，直至软弱土层中所取规定厚度Δz的计算沉降量满足上式为止。　　当无相邻荷载影响，基础宽度在1～50m范围内时，基础中点的地基沉降计算深度，规范规定，也可按下列简化公式计算

§6 Earth Pressure and Retaining Wall 土压力理论与挡土墙设计
Introduction/引言 Rankine’s earth pressure theory / 朗肯土压力理论 Coulomb’s earth pressure theory / 库仑土压力理论 Discuss / 讨论 Retaining wall design / 挡土墙设计

6.1 Introduction/引言静止土压力 -Coefficient of earth pressure at rest 被动土压力主动土压力

6.2 Rankine Earth Pressure theory /朗肯土压力理论
6.2.1 Limit state/极限状态 -Active limit equilibrium/主动极限平衡 -Passive limit equilibrium/被动极限平衡

6.2.2 Rankine active earth pressure/主动土压力
In Non-Cohesive soils (soil surface horizontal)/地表水平无粘性土

In Cohesive soils (soil surface horizontal)/地表水平粘性土

For sloping soil surface(β) /地表倾斜

6.2.3 Rankine passive earth pressure/被动土压力
In Non-Cohesive soils (soil surface horizontal) /地表水平无粘性土

In Cohesive soils (soil surface horizontal)/ /地表水平粘性土

6.3 Coulomb’s Earth Pressure theory / 库仑土压力理论
6.3.1Basic concept -Assuming -Active earth pressure -Passive earth pressure

Active Case/主动情形 General solution

If i=0,α=0,δ=0, Then

Complex boundary conditions/复杂边界条件

-Coefficient of active earth pressure /主动土压力系数λa

-Distribution of active earth pressure /土压力的分布

-Example A retaining wall of embankment is 6m high. And the relative parameters are c=0,φ=30。,α=14 。 00’,δ=φ/3, as well as shown in Fig. below. Calculate the active earth pressure and draw the distribution.

-Solution ①Assuming sliding surface is BC

So, The Sliding surface agrees with the assumption.
②Verifying sliding surface location So, The Sliding surface agrees with the assumption. ③Calculate the Coefficient of active earth pressure λa ④Calculate the active earth pressure Ea and its distribution

⑤ Draw the distribution of the active earth pressure

Passive Case

6.4 Discuss 6.4.1 Equivalent internal friction angle If i=0,α=0,δ=0,
Then 6.4 Discuss Equivalent internal friction angle

Several-layer soils

6.4.3 wall back in polygonal line

graphical method

6.4.5 Earth pressure of abutments

§7 Stability of slopes Introduction
Stability analysis of granular materials The method of slices Development for the analytical method

7.1 Introduction

7.2 Stability analysis of granular materials

7.3 The method of slices -Concept Fellenius,1927 Bishop -Method

-Critical surface and solution to Kmin
The center of the critical arc φ=0 φ=0　　　　　　　　　　　 φ≠0

Table of the values of angles α，β

The radius of the critical arc
Solution to Kmin

Pressure of Retaining Walls
10th IACMAG 7.4 Development for the analytical methods —— Numerical Analysis to Earth Pressure of Retaining Walls Prof. M Zhang, A A Javadi University of Bradford, UK, Shanghai University

Outline Introduction Formation of earth pressure
10th IACMAG Outline Introduction Formation of earth pressure Principle for calculation based on numerical analysis Active sliding earth pressure Passive earth pressure Example Conclusions

Some drawbacks in traditional theories of earth pressure
To ignore the stress/strain state. Do not calculate the displacement of the wall, though they always claim that it plays a very important role. Based on the limit equilibrium theory, but they give no evidence that limit state takes place in the fracture surface. Pay no attention to the difference of earth pressure calculation between the fill retaining wall and the cutting one. A new method- a combination of numerical analysis with limit equilibrium is proposed.

Formation of earth pressure
To clarify: real cause and actual process of formation of earth pressure Relationship: earth pressure – stress state – construction sequence

Division of earth pressure
Deforming earth pressure ( In some areas, there exists shear failure, but no significant sliding surface occurs.) Sliding/limit earth pressure ( After sliding surface occurs )

Principle for calculation based on numerical analysis
The stress state and the deformation are the most important factors that cause the formation of the earth pressure. Until limit state occurs, the retaining wall has to be subjected to a deforming earth pressure that can be obtained by finite element analysis. As the shear failure makes a fracture surface (or plastic band), the wall will be subjected to a sliding earth pressure. The entire earth pressure activity is usually the mixture of the active sliding one and the passive one.

Fig. 1 Slice method for irregular sliding surface and
10th IACMAG Fig. 1 Slice method for irregular sliding surface and sliding body obtained by the tracking analysis

Fig. 2 Forces acted on slice i and the static equilibrium of slice n
10th IACMAG Fig. 2 Forces acted on slice i and the static equilibrium of slice n

10th IACMAG

10th IACMAG Fig. 3. Static equilibrium of slice n about limit passive earth pressure

10th IACMAG

Table 1. The parameters of materials
10th IACMAG Table 1. The parameters of materials

(a) Displacement vectors
10th IACMAG (a) Displacement vectors

(b) Principal stress vectors
10th IACMAG (b) Principal stress vectors

(c) The distribution maximum principal stress 1.
10th IACMAG (c) The distribution maximum principal stress 1. Fig. 4. Results of a fill retaining wall

10th IACMAG Conclusions The traditional theories of earth pressure for retaining walls, based on the limit equilibrium method, are incorrectly assumed that the earth pressure is independent of the stress/strain state in the soil around the wall. Furthermore, it is not calculated, although the soil deformation is emphasized. The proposed method to calculate the earth pressure, based on numerical analysis, can overcome the weakness of the traditional theories.

Ch 8 Bearing Capacity Part / 地基承载力
8.1 Introduction A foundation is that part of a structure which transmits loads directly to the underlying soil. The ultimate bearing capacity (qu) is defined as the pressure which would cause shear failure of the supporting soil immediately below and adjacent to a foundation. A foundation must satisfy two fundamental requirements: (1) the factor of safety against shear failure of the supporting soil must be adequate, a value between 2 and 3 normally being specified; (2) the settlement of the foundation should be tolerable and, in particular, differential settlement should not cause any unacceptable damage of the structure. The allowable bearing capacity (qa) is defined as the maximum pressure which may be applied to the soil such that the above two requirements are satisfied.

8.2 Types of shear failure and P-S Curve / 地基破坏形式与P-S曲线

8.3 Ultimate Bearing Capacity of Shallow Foundations / 浅基础地基极限承载力

§1.Soil Physical Characteristics and Classification

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§1.Soil Physical Characteristics and Classification

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