62 Groundwater flow

The hydrostatic distribution of pore pressures is valid when the groundwater is at rest. When the groundwater is flowing through the soil the pressure distribution will not be hydrostatic, because then the equations of equilibrium (6.1) are no longer complete. The flow of groundwater through the pore space is accompanied by a friction force between the flowing fluid and the soil skeleton, and this must be taken into account. This friction force (per unit volume) is denoted by f. Then the...

61 Hydrostatics

As already mentioned in earlier chapters, the stress distribution in groundwater at rest follows the rules of hydrostatics. More precise it can be stated that in the absence of flow the stresses in the fluid in a porous medium must satisfy the equations of equilibrium in the form Here it has been assumed that the z-axis is pointing in vertically upward. The quantity yw is the volumetric weight of the water, for which Yw 1 kN m3. It has further been assumed that there are no shear stresses in...

52 The general procedure

It has been indicated in the examples given above how the total stresses, the effective stresses and the pore pressures can be determined on a horizontal plane in a soil consisting of practically horizontal layers. In most cases the best general procedure is that first the total stresses are determined, from the vertical equilibrium of a column of soil. The total stress then is determined by the total weight of the column (particles and water), plus an eventual surcharge caused by a structure....

51 Vertical stresses

In many places on earth the soil consists of practically horizontal layers. If such a soil does not carry a local surface load, and if the groundwater is at rest, the vertical stresses can be determined directly from a consideration of vertical equilibrium. The procedure is illustrated in this chapter. A simple case is a homogeneous layer, completely saturated with water, see Figure 5.1. The pressure in the water is determined by the location of the phreatic surface. This is defined as the...

43 Effective stress

On an element of soil normal stresses as well as shear stresses may act. The simplest case, however, is the case of an isotropic normal stress, see Figure 4.4. It is assumed that the magnitude of this stress, acting in all directions, is a. In the interior of the soil, for instance at a cross section in the center, this stress is transmitted by a pore pressure p in the water, and by stresses in the particles. The stresses in the particles are generated partly by the concentrated forces acting...

42 Pore pressures

Soil is a porous material, consisting of particles that together constitute the grain skeleton. In the pores of the grain skeleton a fluid may be present usually water. The pore structure of all normal soils is such that the pores are mutually connected. The water fills a space of very complex form, but it constitutes a single continuous body. In this water body a pressure may be transmitted, and the water may also flow through the pores. The pressure in the pore water is denoted as the pore...

41 Stresses

As in other materials, stresses may act in soils as a result of an external load and the volumetric weight of the material itself. Soils, however, have a number of properties that distinguish it from other materials. Firstly, a special property is that soils can only transfer compressive normal stresses, and no tensile stresses. Secondly, shear stresses can only be transmitted if they are relatively small, compared to the normal stresses. Furthermore it is characteristic of soils that part of...

Cell Test

An early version of the triaxial test was developed around 1938 by Keverling Buisman, see Figure 23.1. Actually, the triaxial test can be considered to be an improved version of this Dutch cell test. The apparatus consists of a container with a cylindrical glass wall (the cell), with a fixed rubber membrane, in which the sample can be installed. The water pressure in the cell can be controlled. In contrast with the triaxial apparatus, in which the cell pressure also acts on the top of the...

Problems

24.1 On a number of identical soil samples CU-triaxial tests are being performed. The cell pressure is applied, then consolidation is allowed to reduce the pore water pressures to zero, and in the second stage the sample is very quickly brought to failure, undrained. The pore pressures are measured. The results are given in the table (all stresses in kPa). Determine the values of the cohesion c and the friction angle 0. 24.2 What can you say about the coefficients A and B in this case 24.3...

162 Solution

The problem defined by the equations (16.1)-(16.4) can be solved, for instance, by separation of variables, or, even better, by the Laplace transform method. This last method will be used here, without giving the details. The Laplace transform p of the pressure p is defined as The basic principle of the Laplace transform method is that the differential equation (16.1) is multiplied by exp(-st)dt, and then integrated from t 0 to t to. This gives, using partial integration and the initial...

102 Flow under a structure

As an example the flow under a structure will be considered, see Figure10.2. In this case a sluice has been constructed into the soil. It is assumed that the water level on the left side of the sluice is a distance H higher than the water on the right side. At a certain depth the permeable soil rests on an impermeable layer. To restrict the flow under the sluice a sheet pile wall has been installed on the upstream side of the sluice bottom. The flow net for a case like this can be determined...

Llllillillillliililillillillll

I axx 2c, azz 0, axz 0. (41.3) This stress state satisfies the equilibrium conditions and the boundary conditions on the upper surface (zero shear stress and zero normal stress), and it does not violate the yield condition in any point. Actually, in every point of this zone the yield condition is just reached. On a plane inclined at an angle of 45 the stresses Figure 41.2 Prandtl's schematization. are, see also Figure 41.3, aee c, and aer -c. The sign of these stresses can best be verified by...

Po 4 j [

This is the analytical solution of the problem, see Figure 16.2. At a first glance the solution (16.11) may not seem to give much insight, but after some closer inspection many properties of the solution can be obtained from it. It is for instance easy to see that for z h the pressure p 0, which shows that the solution satisfies the boundary condition (16.4). The cosine of each term of the series (16.11) is zero if z h, because cos(n 2) 0, cos(3n 2) 0, cos(5n 2) 0, etc. It can also be verified...

173 Numerical versus analytical solution

As may be evident from this chapter and the previous one, the numerical solution method is simpler than the analytical solution, and perhaps much easier to use. It may be added that the numerical solution method can easier be generalized than the analytical method. It is, for instance, rather simple to develop a numerical solution for the consolidation of a layered soil, with different values for the permeability and the compressibility in the various layers. The analytical solution for such a...

DP

The material property cv and the size h have now been eliminated from the mathematical problem, and the only numerical values in the problem are the numbers 0 and 1. Both Z and P are of the order of magnitude of 1. This will then probably also hold for T, and it can be expected that the process will be finished when T 1. This means that it can be stated that the process will be governed by the factor T cvt h2, as was indeed found in the analytical solution in the previous chapter. The...

A1 Transformation formulas

Suppose that the state of stress in a certain point is described by the stresses Txx, Txy, Tyx and Tyy, see Figure A.1. Following the usual sign convention of applied mechanics a component of stress is considered positive when the force component on a plane whose outward normal vector is directed in a positive coordinate direction, acts in positive direction as well, or when the force component acts in negative direction on a plane whose outward normal vector is directed in negative coordinate...

172 Numerical stability

In Program 17.1 the value of the factor a in the algorithm (17.5) is being assumed to be 0.25. Using this value the program calculates a suggestion for the time step At, and then the user of the program may enter a value for the time step. The user may follow that suggestion, but this is not absolutely necessary, of course. The suggestion is being given because the process is numerically unstable if the value of a is too large. This can easily be verified by running the program and then...

H a ojfc

The integral of this quantity can not be zero, as equation (C.42) states. This means that the assumption a > 1 must be false. Therefore a < 1, and this is just what had to be proved. The theorem means that a statically admissible stress field that does not violate the yield criterion, constitutes a lower bound for the failure load. The real failure load is always larger than the load for that equilibrium system. The load is on the safe side.

Shear Test

The notion that failure of a soil occurs by sliding along a plane on which the shear stress reaches a certain maximum value has lead to the development of shear tests. In such tests a sample is loaded such that it is expected that one part of the sample slides over another part, along a given sliding plane. It is often assumed that the sliding plane is fixed and given by the geometry of the equipment used, but it will appear that the deformation mode may be more complicated.

E1 v

When expressed into the constants K and G this can also be written as zz (K + 4G) zz. (14.15) The elastic coefficient for one dimensional confined compression appears is sometimes denoted as D, the constrained modulus, D K + 4 G --- 3K (-). (14.16) When v 0 it follows that D E if v > 0 D > E. In the extreme case that v 1 the value of D to. Such a material is indeed incompressible. Similar to the considerations in the previous chapter on tangent moduli the logarithmic relationship (14.4) may...

Ij

Is perpendicular to the yield surface, in the point a j. o22 As an example consider a yield surface in the form of an ellipse, see Figure C.1, with tangent plane (in this two-dimensional case this is a - oh)+4o212(O22 - o212) 0, (C.18) Figure C.2 Examples of tangents. in which the superscript 1 indicates that the point is located on the yield surface. In the rightmost point of the yield surface o 2a and o 0. Equation (C.18) then defines the tangent as o21 2a. In the topmost point of the yield...

93 Floatation of a pipe

The second example is concerned with a pipeline in the bottom of the sea (or a circular tunnel under a river), see Figure 9.3. The pipeline is supposed to consist of steel, with a concrete lining, having a diameter 2R and a total weight (above water) G, in kN m. This weight consists of the weight of the steel and the concrete lining, per unit length of the pipe. For the risk of floatation the most dangerous situation will be when the pipe is empty. For the analysis of the stability of the...

Floatation

In the previous chapter it has been seen that under certain conditions the effective stresses in the soil may be reduced to zero, so that the soil looses its coherence, and a structure may fail. Even a small additional load, if it has to be supported by shear stresses, can lead to a calamity. Many examples of failures of this type can be given the bursting of the bottom of excavation pits, and the floatation of basements, tunnels and pipelines. The floatation of structures is discussed in this...

21 Grain size

Soils are usually classified into various types. In many cases these various types also have different mechanical properties. A simple subdivision of soils is on the basis of the grain size of the particles that constitute the soil. Coarse granular material is often denoted as gravel and finer material as sand. In order to have a uniformly applicable terminology it has been agreed internationally to consider particles larger than 2 mm, but smaller than 63 mm as gravel. Larger particles are...

381 Computer program

100 CLS PRINTSheet pile wall in layered soil NN 1000 120 DIM D(20),Z(20),CA(20),CP(20) 130 DIM 140 DIM M(NN),Q(NN),F(NN),P(NN),U(NN) 150 INPUT Depth of anchor (m) DA 160 INPUT Number of layers N 170 Z(0) 0 GW 10 F0R I 1 TO N CLS PRINT Layer I PRINT 180 INPUT Thickness (m) D(I) 190 INPUT Cohesion (kN m2) CC(I) 200 INPUT Active stress coefficient CA(I) 210 INPUT Passive stress coefficient CP(I) 220 INPUT Dry weight left (kN m3) GDL(I) 230 INPUT Saturated weight left (kN m3) ________ GNL(I) 240...

341 Active earth pressure

For the active case (a retreating wall) the procedure is illustrated in Figure 34.1. It is assumed that in case of a displacement of the wall towards the left, a triangular wedge of soil will slide down, along a straight slip plane. The angle of the slope with the vertical direction is denoted by 0. It is also assumed that at the moment of sliding, the weight of the soil wedge is just in equilibrium with the forces on the slip surface and the forces on the wall. For reasons of simplicity it is...

210 Print U 2sqrtpiend

Program 16.2 Degree of consolidation. calculated by the program 16.2. The program also gives an approximate value (U'), see the next section. Figure 16.3 Degree of consolidation. Theoretically speaking the consolidation process takes infinitely long to be completed. For engineering practice, however, it is sufficient if the first (and largest) term in (16.19), the infinite series, is about 0.01. Then 99 of the final deformation has been reached. It can be seen that this is the case if cvt h2...

81 Flow in a vertical plane

Suppose that the flow is restricted to a vertical plane, with a cartesian coordinate system of axes x and z. The z-axis is supposed to be in upward vertical direction, or, in other words, gravity is supposed to act in negative z-direction. The two relevant components of Darcy's law Conservation of mass now requires that no water can be lost or gained from a small element, having dimensions dx and dz in the x, z-plane, see Figure 8.1. In the x-direction water flows through a vertical area of...

C6 Druckers postulate

It has been found, by comparing theoretical results with experimental data, that for metals very good agreement is obtained if the plastic potential g is identified with the yield function f. This is often called Drucker's postulate, It has been attempted to find a theoretical derivation of this property, for instance on the basis of some thermodynamical principle. It has been found later, however, that there is no physical necessity for the validity of Drucker's postulate, other than that it...

461 Circular slip surface

Most methods assume that the soil fails along a circular slip surface, see Figure 46.1. The soil above the slip surface is subdivided into a number of slices, bounded by vertical interfaces. At the slip surface the shear stress is t, which is assumed to be a factor F smaller than the maximum possible shear stress, i.e. The factor F is assumed to be the same for all slices, an assumption that is common to all methods. The equilibrium equation to be used in conjunction with a circular slip...

Aj4 dV f tcuc dS FcUc dVC37

Now assume that for a load t atc and F aF a statically admissible stress field aj has been found, and that all these stresses are inside the yield criterion. Then this load is smaller than the failure load, i.e. The proof (ad absurdum) of this theorem can be given as follows. Let it be assumed that the theorem is false, i.e. assume that a > 1. From the virtual work theorem it follows that From (C.37) and (C.39) it follows that Using Drucker's postulate, which has been assumed to be valid, the...

Sheet Pile Wall In Layered Soil

For a sheet pile wall in a layered soil, the method of analysis is the same as for a wall in homogeneous soil, as considered in the previous chapter. The main difference is that the computation of the horizontal stresses against the wall is . more complicated. The computation can best be performed using a computer program. . In this chapter a simple program is presented, using Blum's method. The complications are that the weight and the properties of the various layers may . .is gii .is gii .3...

C1 Yield surface

The simplest description of plastic deformations is by considering a perfectly plastic material. This is a material that exhibits plastic deformations if (and only if) the stresses satisfy the yield condition. For a perfectly plastic material this yield condition is a function of the stresses only (and not of the deformations, or of the time). This yield condition is written in the form Plastic deformations can occur only if f (aj) 0. Stress states for which Oj) > 0 are impossible, and if f...

Soil Mechanics

Arnold Verruijt Delft University of Technology, 2001 This is the screen version of the book SOIL MECHANICS, used at the Delft University of Technology. It can be read using the Adobe Acrobat Reader. Bookmarks are included to search for a chapter. The book is also available in Dutch, in the file GrondMechBoek.pdf. Exercises and a summary of the material, including graphical illustrations, are contained in the file SOLMEX.ZIP. All software can be downloaded from the website http geo.verruijt.net...

453 Flow parallel to the slope

An interesting problem is the stability of an embankment or dam in which groundwater flows parallel to the slope, in downward direction, see Figure 45.3. This may occur in a dike that is just not high enough to retain the water in a river, so that water flows over the slope. This water penetrates into the dike material, and after some time a flow of groundwater parallel to the slope may be created, as shown in Figure 45.4. If the flow is uniform the pressure distribution must be linear in and...

272 Elasticity

For the analysis of stresses and strains in a homogenous, isotropic linear elastic material various methods have been developed. The general theory can be found in many textbooks on the theory of elasticity. Here, only the basic equations will be given, without giving the details of the derivations. Some of these details, and some derivations of solutions are given in Appendix B. In this chapter, and in the next chapters, the analysis always concerns the calculation of stresses and deformations...

220 Next Kend

Program 16.1 Analytical solution for one dimensional consolidation. This means that the total deformation of the sample is f h f h Ah e dz mvhq + mv pdz. (16.14) The first term on the right hand side is the final deformation, which will be reached when all pore pressures have been reduced to zero. That value will be denoted by AhTO, Immediately after the application of the load the pore pressure p p0, see eq. (16.2). The deformation then is, with (16.14), Aho mv h - -. (16.16) If the water is...

471 Cone Penetration Test

A simple, but very effective method of soil investigation consists of pushing a steel rod into the soil, and then measuring the force during the penetration, as a function of depth. This force consists of the reaction of the soil at the point (the cone resistance), and the friction along the circumference of the rods. The method was developed in the 1930's in the Netherlands. It was mainly intended as an exploration tool, to give an indication of the soil structure, and as a modelling tool for...

474 Soil sampling

For many engineering projects it is very useful to take a sample of the soil, and to investigate its properties in the laboratory. The investigation may be a visual inspection (which indicates the type of materials sand, clay or peat), a chemical analysis, or a mechanical test, such as a compression test or a triaxial test. A simple method to take a sample is to drive a tube into the ground, and then recovering the tube with the soil in it. The tube may be about 1 meter long, see Figure 47.5,...

362 Pore pressures

In the previous sections the soil was assumed to be dry, for simplicity. In general the soil may consist of soil and water, however, and the excavation may even contain free water. Thus the general problem of a sheet pile wall should take into account the presence of groundwater in the soil. Because the failure of soils, as described by the Mohr-Coulomb criterium, for instance, refers to effective stresses, the relations formulated above for the earth pressure coefficients Ka and Kp, should be...

Ni 0 N2 0 N3 2kw N4 6kw

Figure 49.3 Statically indeterminate pile foundation. Figure 49.3 Statically indeterminate pile foundation. The forces in the piles have been considered positive for tension. The equations of equilibrium of the foundation plate are that the sum of the horizontal forces should be -200 kN, the sum of the vertical forces should be -2000 kN, and the sum of the moments with respect to the origin should be -2000 kNm. These equations can be written as Substitution of (49.3) into (49.4) yields the...

Secular Effect

As mentioned in the previous chapter, in a one dimensional compression test on clay, under a constant load, the deformation usually appears to continue practically forever, even if the pore pressures have long been reduced to zero, see Figure 18.1. Similar types of behavior are found in other materials, such as plastics, concrete, etcetera. The phenomenon is usually denoted as creep. For many materials this behavior can be modelled reasonably well by the theories of visco-elasticity or...

242Types of triaxial tests

The procedure in the tests described above, with the results given in Table 24.1, is that in the first stage of the tests, the application of the cell pressure, the soil is free to consolidate, and sufficient time is taken to allow for complete consolidation, i.e. the excess pore pressures are reduced to zero. In the second stage of the test, however, no consolidation is allowed, by closing the tap to the drainage reservoir. Such a test is denoted as a Consolidated Undrained test, or a CU-test....

352 Example

As an example the case of a wall at an inclination of 80 is considered. The slope of the soil is 10 , see Figure 35.2. The soil is sand, with 0 30 , and the friction angle between the wall and the soil is J 20 . The problem is to determine the horizontal component of the force against the wall, in the case of active earth pressure. In this case Table 35.2 gives K 0.438, so that the force on the wall is Q 0.219 Yh2. Its horizontal component is, with (35.2), Qh 0.190Yh2. In the case of passive...

441 Lower bound

In this case it is essential that the weight of the material is taken into account. The equations of equilibrium now are A simple equilibrium system is shown in Figure 44.2, consisting of three zones. On the interfaces between the zones the normal stresses parallel to these interfaces may be discontinuous, without disturbing equilibrium, see Chapter 40. The boundary conditions for the stresses are that the normal stresses and the shear stresses are zero, all along the upper surface. These...

82 Upward flow

A very simple special case of groundwater flow occurs when the water flows in vertical direction only. The solution for this case is h iz, where i is a constant, a measure for the intensity of the flow. Actually i, that is dh dz, is called the gradient. In this case qx 0 and qz -ki. The equation of continuity (8.2) is now indeed satisfied. If the specific discharge is now denoted as q0, the gradient appears to be i q0 k, and h q0z k. Because in general h z + p Yw it now follows that the...

Stress Strain Relations

As stated in previous chapters, the deformations of soils are determined by the effective stresse, which are a measure for the contact forces transmitted between the particles. The soil deformations are a consequence of the local displacements at the level of individual particles. In this chapter some of the main aspects of these deformations will be discussed, and this will lead to qualitative properties of the relations between stress and strain. In later chapters these relations will be...

22 Grain size diagram

The size of the particles in a certain soil can be represented graphically in a grain size diagram, see Figure 2.1. Such a diagram indicates the percentage of the particles smaller than a certain diameter, mea-100 i . i . i . i . . . . . . . i . . . i i . i . sured as a percentage of the weight. A steep slope of the curve in the diagram indicates a uniform soil, a shallow slope of the diagram indicates that the soil contains particles of strongly different grain sizes. For rather coarse...

391 Basic theorems of plasticity theory

In considerations of limit analysis not all the details of the constitutive relations are taken into account, but one aspect is given priority, namely the failure criterion of the material. For soils this may be the Mohr-Coulomb criterion, described by a cohesion c and a friction angle 0. Also, not all the conditions of equilibrium and compatibility equations are taken into account, but only a subset of these equations. The purpose of limit analysis is not to determine the complete field of...

Triaxial Test

The failure of a soil sample under shear could perhaps best be investigated in a laboratory test in which the sample is subjected to pure distorsion, at constant volume. mains constant during the test, The volume could be kept constant by taking care that the isotropic stress a0 1 (< ri + < r2 + a3) re-or, better still, by using a test setup in which the volume change can be measured and controlled very accurately, so that the volume change can be zero. In principle such a test is possible,...

222 Simple shear test

Apart from the difficulty that the state of stress is not completely given in a shear test, the direct shear test suffers from the disadvantage that the deformation is strongly inhomogeneous, because the deformations are concentrated in a zone in the center of the shear box. An improved shear box has been developed by Roscoe in Cambridge (England), in which the deformation is practically homogeneous. The apparatus has been constructed with rotating side walls, so that a uniform shear...

De Josselin de Jong see Figure 222

It seems very likely that in a shear test the horizontal normal stress jxx is smaller than the vertical normal stress jzz. If the sand has been poured into the shear box, and the vertical load has been applied by gradually increasing the load, it seems likely that the horizontal - - < xx stress is smaller than the vertical stress. In an elastic material, for instance, the ratio of horizontal to vertical stress would be jxx jzz v (1 v), where v is Pois-son's ratio, which must be smaller than...

121 Compression and distorsion

In the contact point of two particles a normal force and a shear force can be transmitted, see Figure 12.1. The normal force can only be a compressive force. Tension can not be transmitted, unless the soil particles are glued together. Such soils do exist (e.g. calcareous soils near the coast of Brazil or Australia), but they are not considered here. The magnitude of the shear force that can be transmitted depends upon the magnitude of the normal force. It can be expected that if the ratio of...

164 Approximation for small values of time

If the time parameter cvt h2 is very small, many terms are needed in the analytical solutions to obtain accurate results. That may not be a great disadvantage if the computations are performed by a computer program, but it does not give much insight into the solution. A more convenient approximation can be obtained using a theorem from Laplace transform theory saying that an approximation for small values of t can be obtained by assuming the value of s in the transformed solution as very large....

Flow Towards Wells

For the theoretical analysis of groundwater flow several computational methods are available, analytical or numerical. Studying groundwater flow is of great importance for soil mechanics problems, because the influence of the groundwater on the behavior of a soil structure is very large. Many dramatic accidents have been caused by higher pore water pres. Q0 sures than expected. For this reason the study of groundwater flow requires special attention, much more than given in the few chapters of...

431 Bearing capacity of strip foundation

An important problem of foundation engineering is the computation of the maximum load the bearing capacity of a strip foundation, i.e. a very long foundation, of constant width, at a certain depth below the soil surface. The influence of the depth of the foundation is accounted for by considering a surcharge at the foundation level, to the left and the right of the applied load. For the simplest case, of a strip of infinite length, on weightless soil, the first computations were made by...

13 Why Soil Mechanics

Soil mechanics has become a distinct and separate branch of engineering mechanics because soils have a number of special properties, which distinguish the material from other materials. Its development has also been stimulated, of course, by the wide range of applications of soil engineering in civil engineering, as all structures require a sound foundation and should transfer its loads to the soil. The most important special properties of soils will be described briefly in this chapter. In...

141 Confined compression test

In the confined compression test, or oedometer test a cylindrical soil sample is enclosed in a very stiff steel ring, and loaded through a porous plate at the top, see Figure 14.1. The equipment is usually placed in a somewhat larger container, filled with water. Pore water may be drained from the sample through porous stones at the bottom and the top of the sample. The load is usually applied by a dead weight pressing on the top of the sample. This load can be increased in steps, by adding...

202 Mohrs circle

From the theory of stresses see Appendix A it is known that the stresses acting in a certain point on different planes can be related by analytical formulas, based upon the equilibrium equations. In these formulas the basic variable is the angle of rotation of the plane with respect to the principal directions. These principal directions are the directions in which the shear stress is zero, and in which the normal stresses are maximal or minimal. It is assumed here that the maximum principal...

Vertical Slope In Cohesive Material

A well known and important problem of soil mechanics is the case of a vertical cutoff in a purely cohesive material 0 0 , as occurs when making a vertical excavation, or a vertical slope, see Figure 44.1. The problem to be considered in this chapter is the determination of a lower bound or an upper bound for the maximum possible height hc of the slope, for a material having a constant cohesive strength c, and a constant volumetric weight 7.

122 Unloading and reloading

Because the deformations of soils are mostly due to changes in the particle assembly, by sliding and rolling of particles, it can be expected that after unloading a soil will not return to its original state. Sliding of particles with respect to each other is an irreversible process, in which mechanical energy is dissipated, into heat. It is to be expected that after a full cycle of loading and unloading of a soil a permanent deformation is observed. Tests indeed confirm this. When reloading a...

Boussinesq

In 1885 the French scientist Boussinesq obtained a solution for the stresses and strains in a homogeneous isotropic linear elastic half space, loaded by a vertical point force on the surface, see Figure 28.1. A derivation of this solution is given in Appendix B, see also any textbook on the theory of elasticity for instance S.P. Timoshenko, Theory P of Elasticity, paragraph 123 . The stresses are found to be Figure 28.1 Point load on half space. In these equations r is the cylindrical...

35 Water content

The water content is another useful parameter, especially for clays. It has been used in the previous chapter. By definition the water content w is the ratio of the weight or mass of the water and the solids, It may be noted that this is not a new independent parameter, because For a completely saturated soil S 1 and assuming that pp pw 2.65, it follows that void ratio e is about 2.65 times the water content. A normal value for the porosity is n 0.40. Assuming that pk 2650 kg m3 it then follows...

334 Neutral earth pressure

It has been found that the possible states of stress in a soil may vary between fairly wide limits, especially if the friction angle is large. For a normal sand, with 30 , the smallest value of the horizontal stress is 3 of the vertical stress which usually is known from the surcharge and the weight of the overlying soil , and the largest Hi value is 3 times the vertical stress. In case of a rigid retaining wall, the lateral stress against the wall is unknown, at least from a strictly...

Flamant

In 1892 Flamant obtained the solution for a vertical line load on a homogeneous isotropic linear elastic half space, see Figure 30.1. This is the two dimensional equivalent of Boussinesq's basic problem. It can be considered as the superposition of an infinite number of point loads, uniformly distributed along the y-axis. A derivation is given in Appendix B. . . . . . . . .v. . . . . . . In this case the stresses in the x, z-plane are In these equations r x2 z2. The quantity F has the dimension...

252 Undrained shear strength

For the comparison of drained and undrained calculations, and for the actual calculation in an undrained analysis, it is often necessary to determine the undrained shear strength su of a soil, from the basic shear strength parameters c and 0. This can be done by noting that in a saturated soil there can be practically no volume change in undrained conditions, so that the isotropic effective stress remains constant. Thus the average effective stress remains constant, and this means that the...

432 Inclination factors

Loading by a vertical force and a horizontal load, see Figure 43.2, the bearing capacity is considerably reduced. This can be understood by noting that sliding would occur if the horizontal force approaches the maximum possible shear force on the foundation surface, The formulas should be such that for this limiting value of the shear stress t with respect to the constant value of the vertical stress p the bearing capacity reduces to zero. For cases in which the shear force is smaller than its...

A3 Mohrs circle

The formulas derived above can be represented in a simple graphical form, using Mohr's circle. For this purpose it is most convenient to use the transformation formulas in the form A.4 , but expressed into the principal stresses. The orientation of the x-axis with respect to the direction of the major principal stress is denoted by y, see Figure A.4. The directions of the major and the minor principal stresses are indicated by 1 and 2. The transformation formulas for the transition from the...

44 Archimedes and Terzaghi

The concept of effective stress is so important for soil mechanics that it deserves careful consideration. It may be illuminating, for instance, to note that the concept of effective stress is in agreement with the principle of Archimedes for the upward force on a submerged body. Consider a volume of soil of magnitude V, having a porosity n. The total weight of the particles in that volume is 1 n YPV, in which yp is the volumetric weight of the particle material, which is about 26.5 kN m3....

12 History

Soil mechanics has been developed in the beginning of the 20th century. The need for the analysis of the behavior of soils arose in many countries, often as a result of spectacular accidents, such as landslides and failures of foundations. In the Netherlands the slide of a railway embankment near Weesp, in 1918 see Figure 1.1 gave rise to the first systematic investigation in the field of soil mechanics, by a special commission set up by the government. Many of the basic principles of soil...