《Dredging Engineering-Part 2 soil 疏浚工程 土质篇.docx》由会员分享，可在线阅读，更多相关《Dredging Engineering-Part 2 soil 疏浚工程 土质篇.docx（27页珍藏版）》请在三一办公上搜索。
1、Part 2 Dredging Process. 1Chapter 1 Basic Concepts of Soil Mechanics. 11.1 Soil Classification. 11.2 Soil Properties. 11.2.1 Phase Relationships of Soil.21.2.2 Compaction and/or Consistency. 21.2.3 Permeability. 21.3 Elementary Soil Tests. 71.3.1 Permeability Test. 71.3.2 Mohr Circle. 81.3.3 Case St
2、udy. 101.3.4 Triaxial Shear Test-Mohr Failure Envelop. 121.3.5 Direct Shear Test-Coulombs Law (1776). 131.3.6 Standard Penetration Test-SPT. 161.4 Passive and Active FailureRankines Theory of Lateral Earth Pressure. 171.4.1 Active Stress State. 181.4.2 Passive Stress State. 201.5 Stresses in the Sub
3、soil. 201.5.1 General. 201.5.2 Determination of Vertical Stresses. 221.5.3 Overstressed or Understressed Pore Water in a Clay Layer. 241.6 Consolidation. 25Acknowledgements. 25Part 2 Dredging ProcessChapter 1 Basic Concepts of Soil Mechanics1.1 Soil ClassificationThe object of soil classification is
4、 to divide the soil into groups so that all the soils in a particular group have similar characteristics by which they may be identified. In practice, there are many different classifications are used by different institutions, most of these classification systems are based on the outcome of the par
5、ticle size distribution analyses and the results of the plots on the plasticity chart of Casagrande. Table 1-1 is the internationally most accepted standardization of particle-size-ranges. Table 1-1 Internationally most accepted standardization of particle-size-ranges.fine-grainedcoarse-grainedClayS
6、iltsandgravelstoneColloidsfine26medium620Coarse2060fine60200medium200600coarse6002mmfine26medium620coarse2060cobbles60200boulders200m (micron)mmTable 1-2 shows the British Soil Classification System for Engineering purpose (BS 5930, 1981). The name and descriptive letters used in this system are exp
7、lained in table 1-3.1.2 Soil Properties1.2.1 Phase Relationships of SoilSoil is a multiphase system, containing three distinct phases: solid, liquid (water) and gas (air). In reality the pores in between the solid particles are filled with water and/or gas, figure 1-1(a); however, in order to define
8、 the phase relationships, an element of soil is schematized in figure 1-1(b).1.2.2 Compaction and/or ConsistencyFor natural deposits of soils the compaction of the encountered layer in-situ is an important feature. The compaction of an in-situ sand layer depends on largely on the particle size distr
9、ibution (figure 1-2), the mode of sedimentation and the stress history. For clays the compaction more or less depends on the same features but is somewhat more complicated because for clays the interaction between clay particles depends also on the electrical charges on the surface of the particles.
10、 Therefore it is common practice to distinguish the notation compaction for granular soils, which is expressed in the relative density Dr and the notation consistency for cohesive soils, which is expressed in the unstrained shear strength Cu. See table 1-4.1.2.3 PermeabilityThe permeability of the s
11、oil determines the rate of ingress of water into the soil, either by gravitational flow or by diffusion. Since the individual sand particles are not bonded together by hydrated water and since the particles are relatively large, there is an easy access of water into the pores. Therefore sand has a g
12、ood permeability (permeability varies between k=10-2 mm/s and k=10-4 mm/s). However, clays are almost impermeable (permeability varies between k=10-6 mm/s and k=10-8 mm/s). The two main reasons for this very low permeability are: (1) the particles are very small i.e. of colloidal size; (2) the water
13、 surrounding the clay particles is partly hydrated and/or electrically bonded to the particle and therefore the flow of water entering the pores is very difficult.Permeability is an important soil property that varies considerably for different types of soils, as shown in table 1-5. Relations betwee
14、n soil type and soil parameters are listed in table 1-6.Table 1-2 British Soil Classification System for Engineering Purposes Table 1-3 Names and descriptive symbols for grading and plasticity characteristics, used in the British Soil Classification System for Engineering purposes.Component Main ter
15、mSymbol Qualifying Symbol Coarse Gravel Sand GS Well graded Poorly gradedPoorly graded, uniformPoorly gap gradedWPPuPg Fine Fine soil, finesSilt (M-soil)Clay FMC Low plasticityIntermediate plasticityHigh plasticityVery high plasticityExtremely high plasticityLIHVE Organic Peat Pt Organic O Fig. 1-2
16、Particle size distributionTable 1-4 Compaction of granular soils Consistency of cohesive soilsclassificationDrClassificationCukN/m2Very looseLooseMedium denseDenseVery dense01515-3535-6565-8585-100Very softSoftFirm or mediumStiffVery stiff or hard150Table 1-5Table 1-6 Relation between soil type and
17、soil parameters to NEN 6740.1.3 Elementary Soil Tests1.3.1 Permeability TestAlready in the middle of the 19th century (1850s) Darcy experimentally found his famous law, when he studied the flow properties of water through a sand-filter-bed. For his experiments he used a set-up similar to that shown
18、in Fig 1-3, with which he could vary the length (L) of the sample and the water pressures at the top and at the bottom.Darcy experimentally found that the arte of flow (Q) is proportional to the fall of head:And inversely proportional to the length (L) of the sample: Since Q denotes the total discha
19、rge m3/s passing through the cross-sectional area A m2 of the sample container, it will be clear that the specific rate of the flow or specific discharge (q) has the dimension of a velocity m/s and also represents the velocity v of the free water in the tube where there is no sand. The real (average
20、) velocity with which the water passes the pore channels inside the sand sample is called the seepage velocity (Vs) and can be defined by using the principle of continuity. The discharge of water passing the total area (A) above the sand sample per unit time must also pass the reduced average area (
21、Ap) of the pore channels in the same time: or The constant (k) used in the formula above is the coefficient of permeability which has the dimension of a velocity mm/s and is an important soil property which varies considerably for different types of soils, as shown in Table 1-5. It should be noted t
22、hat k values strictly apply to water ( 200C). All other liquids will have other values for the permeation constant, depending on the viscosity.Fig. 1-3 Determination of the coefficient of permeability 1.3.2 Mohr CircleIn almost all soil mechanics and foundation engineering problems, where concentrat
23、ed loads are applied, the horizontal and vertical stress components will not be principal stresses: i.e. also shear stresses will act on the horizontal and vertical planes. Fig 1-4 Determination of forces acting on an inclined planeAs a matter of fact speaking of the “stress state” in a point of the
24、 subsoil is meaningless, without mentioning the orientation of the plane of interest on which the forces act. Therefore the complete stress-state in a point of the subsoil can only be defined by the stress-tensor.Of course in a given situation the stress-tensor in any point in the subsoil can be def
25、ined in a three-dimensional orthogonal coordinate system. However, for reasons of symmetry, in most cases the resultant of forces acting on a particle of soil lies in a vertical plane. Therefore in most situations it will suffice to consider the stresses in this plane using a two-dimensional orthogo
26、nal coordinate system with a horizontal (x) axis and a vertical (z) axis.If the normal stresses and shear stresses acting on a small element of soil in x- and z-direction are given, then the normal stress and shear stress acting on any other plane with given inclination can be determined (Fig. 1-4):
27、In case the principal planes are orientated parallel to the x- and z-direction then no shear forces act on these principal planes and the normal stresses acting on the principal planes become principal stresses, for example, Therefore in this special case, Since, from a mathematical point of view, t
28、hese two equations represent a circle, Mohr proposed to plot this circle, known as Mohr circle in the - diagram, see Fig. 1-5From this figure it can be observed:1. 1 = major principal stress; 2. (2) 3 = minor principal stress; 3. The center (C) of the ciecle lies on the -axis at a distance 1. OC=p=(
29、 1+ 3)/2 4. The radius of the circle is q=( 1- 3)/2; 5. ( 1- 3) is called the diviator stress or the stress difference; 6. A is called the stress point representing the resulting stress r? acting on the plane with an inclination (between the direction of? 1 and the normal on this plane); 7. This res
30、ulting stress r can be resolved in a normal stress and a shear stress and the angle ? between r and the normal to the plane can be found in the diagram, since 8. tan = / =AB/OB, =AOB; 9. Note, in this special case, the pole (P), also called the origin of planes coincides with the position of 3. The
31、pole P is a point on the Mohr circle, with the following property: a line through the pole P and any point A of the Mohr circle will be parallel to the plane on which the stress pair given by point A acts; 10. The maximum shear stress max =q=( 1- 3)/2 will be obtained in a plane with an orientation
32、=450.Fig. 1-6 The Mohr circle construction1.3.3 Case StudyIf in a point in the subsoil the normal stresses and shear stresses acting on two orthogonal planes with a known orientation, for example relative to the horizontal, are given, Fig. 1-7, then it is possible to determine the magnitude and dire
33、ction of the principal stresses with the aid of the Mohr diagram.This is illustrated in the example below:Fig. 1-7 Mohr circle construction to find the magnitude and direction of 1 and 3Graphical solution:1. Plot the stress point A (25,15) and B(65,15); 2. Draw the AB to find the center C(45,0) of t
34、he circle; 3. Erect of the Mohr circle through Aand B; 4. Read magnitude of 1= 70 kN/m2 and 3= 20 kN/m2; 5. Find the pole P by drawing a line through B parallel to BB (or through A parallel to AA); 6. Find the direction of the principal planes by drawing lines through the pole P1 and P3; 7. The dire
35、ction of the principal stresses acts perpendicular to the principal planes. In this case the direction of 1 is 450 relative to the horizontal. Analytical solution:1)The sum of the normal stresses is constant:(Note this is the coordinate (45,0) of the center C of the circle)2)The radius of the circle
36、 is defined by:3) Combining 1) and 2):4) To find the direction of 1 use the stress pair in which is largest =65kN/m2 and =+15kN/m2: 2=36.8700 and therefore =18.4350.So, relative to the direction of 1 the direction of is (positive) anti-clockwise (see sign. convention figure 1-4).The direction of rel
37、ative to the horizontal is given by, the figure 1-7, tan =2/1=2 =63.4350The direction of relative to 1 was =18.4350The direction of 1 relative to the horizontal is (-)=45.00001.3.4 Triaxial Shear Test-Mohr Failure EnvelopThe most reliable shear test is the triaxial direst stress test (Fig. 1-8). A c
38、ylindrical soil sample with a length of at least twice its diameter is wrapped in a rubber membrane and placed in a triaxial chamber. A specific lateral pressure, called chamber pressure, is applied by means of water within the chamber. The chamber pressure is kept constant during each test. A verti
39、cal load is then applied at the top of the sample and steadily and very slowly increased until the sample fails in shear along a diagonal plane. The Mohr circles of failure stresses for a series of such tests using different values of confining pressure are plotted as shown in Fig. 1-8.Fig. 1-8 Triaxial shear test, Mohr failure envelop (after Sower, Fu Hua Chen)1. The stress is uniformly distributed on t
