Foundation Engineering

Transcription

1 Foundation Engineering Draft Version April 01 S. van Baars

2 PREFACE This book is just a help for the students fort the old course Grundbau. Luxembourg, April 01 Stefan Van Baars

3 CONTENT I SHALLOW FOUNDATIONS Elastic stresses and deformations... 6 Boussinesq Flamant Deformation of layered soil Strip footing Lower bound Upper bound... 6 Prandtl Brinch Hansen Bearing capacity of strip foundation Inclination factors Shape factors Brinch Hansen and codes CPT and undrained shear strength II PILE FOUNDATIONS Cone Penetration Test (CPT) Compression piles Pile types Bearing capacity Tip resistance using Brinch Hansen (Prandtl or Meyerhof) Tip resistance using Koppejan Shaft resistance Tension piles Differences between tension and compression piles Reduction of the cone value q c due to excavation Reduction of cone resistance due to tensile pile force Clump criterion Edge Piles

4 4

5 I Shallow foundations 5

6 1 Elastic stresses and deformations Very important in the calculation of the shallow foundations are the stresses and deformations. As long as we assume that we have non failing soils, behaving rather linea elastic, we can make simple design calculations by hand. Because of equilibrium of moments the stress tensor must be symmetric, xy yz zx yx,, zy. xz (1.1) The equations of equilibrium constitute a set of six equations involving nine stress components. In itself this can never be sufficient for a mathematical solution. The deformations must also be considered before a solution can be contemplated. For a linear elastic material the relation between stresses and strains is given by Hooke s law, 1 xx [ xx ( yy )], E 1 yy [ yy ( xx )], E 1 [ ( xx yy )], E 1 xy xy, E 1 yz yz, E 1 zx zx. E (1.) (1.3) where E is the modulus of elasticity (Young s modulus), and is Poisson s ratio. The equations (1.) and (1.3) add six equations to the system, at the same time introducing six additional variables. The six strains can be related to the three components of the displacement vector, xx yy ux, x u y, y uz, z xy yz zx u u x y ( ), y x u y uz ( ), z y uz ux ( ). x z (1.4) (1.5) 6

7 These are the compatibility equations. In total there are now just as many equations as there are variables, so that the system may be solvable, at least if there are a sufficient number of boundary conditions. The minus sign in the equations has been introduced in geotechnical engineering because a volume decrease leads to a positive strain and a volume increase to a negative strain. For a number of problems solutions of the system of equations can be found in the literature on the theory of elasticity. In soil mechanics the solutions for a half space or a half plane, with a horizontal upper surface, are of special interest. In the next chapters some of the most important solutions for soil mechanics are further discussed. For the solved elastic problems the strains and effective stresses can be calculated for drained or consolidated loads, with the help of two stiffness parameters ( E, or K, G). If these parameters are replaced by two undrained parameters ( Eu, u) the strains and total stresses can be calculated for undrained problems (short loads on clay or peat). When calculating these two parameters, one should realize that an undrained load, because of the incompressibility of water, causes no volume change, so u 0.5. (1.6) The undrained compression modulus becomes infinite: K u Eu. 3(1 ) u (1.7) Besides this the water has no shear stresses, both in the drained as in the undrained state. From this it follows that G u E Eu G. (1 ) (1 ) From which can be derived that Eu (1 u ) E. (1 ) u (1.8) (1.9) Replacing the drained stiffness parameters ( E, ) from the previous chapters by the undrained stiffness parameters ( Eu, u) and calculating these values in the way demonstrated above, the solutions can be used not only for the drained, but also for the undrained case. 7

8 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 -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 of Elasticity, paragraph 13). The stresses are found to be 3Pz R 3, 5 (.1) rr P 3r z 1 5 R R( R z) [ (1 ) ], P 1 R z ( ), R R z R 3P rz R (.) (.3) rz. 5 (.4) In these equations r is the cylindrical coordinate, r x y, (.5) and R is the spherical coordinate, R x y z. (.6) The solution for the displacements is u r Figure -1. Point load on half space. P(1 ) r z z [ 3 (1 )(1 )], (.7) ER R R u 0, (.8) u z P(1 ) z [(1 ) )]. (.9) ER R 8

9 The vertical displacement of the surface is particularly interesting. This is P(1 ) z 0 : uz. (.10) Er For r 0 this tends to infinity. At the point of application of the point load the displacement is infinitely large. This singular behaviour is a consequence of the singularity in the surface load, as in the origin the stress is infinitely large. That the displacement in that point is also infinitely large may not be so surprising. Another interesting quantity is the distribution of the stresses as a function of depth, just below the point load, i.e. for r 0. This is found to be r r Figure -. Vertical normal stress. 3P z 0 :, (.11) P 4 z 0 : rr (1 ). (.1) These stresses decrease with depth, of course. In engineering practice, it is sometimes assumed, as a first approximation, that at a certain depth the stresses are spread over an area that can be found by drawing a line from the load under an angle of about 45º. That would mean that the vertical normal stress at a depth z would be P z, homogeneously over a circle of radius z. That appears to be incorrect (the error is 50 % if r 0 ), but the trend is correct, as the stresses indeed decrease with 1 z. In Figure - the distribution of the vertical normal stress is represented as a function of the cylindrical coordinate r, for two values of the depth z. 9

10 The assumption of linear elastic material behaviour means that the entire problem is linear, as the equations of equilibrium and compatibility are also linear. This implies that the principle of superposition of solutions can be applied. Boussinesq s solution can be used as the starting point of more general types of loading, such as a system of point loads, or a uniform load over a certain given area. As an example consider the case of a uniform load of magnitude p over a circular area, of radius a. The solution for this case can be found by integration over a circular area (S.P. Timoshenko, Theory of Elasticity, paragraph 14), see Figure -3. The stresses along the axis r 0, i.e. just below the load, are found to be r z b 3 0 : p(1 ), 3 (.13) 3 z 1 z r 0 : rr p[(1 ) (1 )], 3 (.14) b b in which b z a. The displacement of the origin is r z u Figure -3. Uniform load over circular area. pa E 0, 0 : z (1 ). (.15) This solution will be used as the basis of a more general case in the next chapter. Another important problem, which was already solved by Boussinesq (see also Timoshenko) is the problem of a half space loaded by a vertical force on a rigid plate. The force is represented by P a p, see Figure -4. The distribution of the normal stresses below the plate is found to be 1 p z 0, 0 r a:. 1 r a (.16) 10

11 This stress distribution is shown in Figure -4. At the edge of the plate the stresses are infinitely large, as a consequence of the constant displacement of the rigid plate. In reality the material near the edge of the plate will probably deform plastically. It can be expected, however, that the real distribution of the stresses below the plate will be of the form shown in the figure, with the largest stresses near the edge. The centre of the plate will subside without much load. The displacement of the plate is z r a u Figure -4. Rigid plate on half space. pa E 0, 0 : z (1 ). (.17) When this is compared with the displacement below a uniform load, see (.15), it appears that the displacement of the rigid plate is somewhat smaller, as could be expected. 11

12 3 Flamant In 189 Flamant obtained the solution for a vertical line load on a homogeneous isotropic linear-elastic half space, see Figure 3-1. 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. In this case the stresses in the x,z-plane, resulting from the line load, are 3 F z F 3 4 cos, r r (3.1) F x z F 4 sin cos, xx r r (3.) F xz F 4 sin cos. Figure 3-1. Flamant s Problem. xz r r (3.3) In these equations r x z. The quantity F has the dimension of a force per unit length, so that Fr has the dimension of a stress. Expressions for the displacements are also known, but these contain singular terms, with a factor ln r. This factor is infinitely large in the origin and at infinity. Therefore these Figure 3-. Strip load. expressions are not so useful. On the basis of Flamant s solution several other solutions may be obtained using the principle of superposition. An example is the case of a uniform load of magnitude p on a strip of width a, see Figure 3-. In this case the stresses are p [( 1 ) sin 1 cos 1 sin cos ], (3.4) 1

13 p xx [( 1 ) sin 1 cos 1 sin cos ], (3.5) p xz [cos cos 1]. (3.6) In the centre of the plane, for x 0, 1. Then the stresses are p x 0 : [ 1 sin 1 cos 1], (3.7) Figure 3-3. Stresses for x=0. p x 0 : xx [ 1 sin 1 cos 1], (3.8) x 0 : xz 0. (3.9) That the shear stress xz 0 for x 0 is a consequence of the symmetry of this case. The stresses xx and are shown in Figure 3-3, as functions of the depth z. Both stresses tend towards zero for z, of course, but the horizontal normal stress appears to tend towards zero much faster than the vertical normal stress. It also appears that at the surface the horizontal stress is equal to the vertical stress. At the surface this vertical stress is equal to the load p, of course, because that is a boundary condition of the problem. Actually, in every point of the surface below the strip load the normal stresses are xx p. 13

14 It may be interesting to further explore the result that the shear stress xz 0 along the axis of symmetry x 0 in the case of a strip load, see Figure 3-. It can be expected that this symmetry also holds for the horizontal displacement, so that ux 0 along the axis x 0. This means that this solution can also be used as the solution of the problem that is obtained by considering the right half of the strip problem only, see Figure 3-4. In this problem the quarter plane x 0, x 0 is supposed to be loaded by a strip load of width a on the surface z 0, and the boundary conditions on the boundary x 0 are that the displacement u 0 and the shear stress xz 0, representing a perfectly smooth and rigid vertical wall. The wall is supposed to extend to an infinite depth, which is impractical. For a smooth rigid wall of finite depth the solution may be considered as a first approximation. The formulas (3.7) and (3.8) can also be written as p a az x 0 : [arctan( ) ], z a z p a az x 0 : xx [arctan( ) ]. z a z Integration of the horizontal stress height h, Figure 3-4. Strip load next to a smooth rigid wall. x (3.10) (3.11) xx from z 0 to z h gives the total force on a wall of a Q pharctan( ). h (3.1) For a very deep wall ( h a) this becomes, because then arctan( a h) a h, h : Q pa pa. (3.13) The quantity pa is the total vertical load F (per unit length perpendicular to the plane of the drawing). It appears that the horizontal reaction in an elastic material is 0.637F. For a very shallow wall ( h a) the total lateral force will be, because then arctan( ah), h 0 : Q ph. (3.14) 14

15 This is in agreement with the observation made earlier that the value of the horizontal stress at the surface, just below the load, is xx p. For a very short wall the horizontal force will be that horizontal stress, multiplied by the length of the wall. Another interesting application of Flamant s solution is shown in Figure 3-5. In the case of a surface load by two parallel line loads it can be expected that at the axis of symmetry ( x 0 ) the horizontal displacement and the shear stress will be zero, because of symmetry. This means that the solution of that problem can also be used as the solution of the problem of a line load at a certain distance from a smooth rigid wall, because the boundary conditions along the wall are that 0 x u and 0 for x 0. By the symmetry of the problem shown zx in the left half of Figure 3-5, these conditions are satisfied by the solution of that problem. The horizontal stresses against the wall are given by equation (3.) for Flamant s basic problem, multiplied by, because there are two line loads and each gives the same stress. In this formula the value of x should be taken as x a, where a is the distance of the force to the wall. The horizontal stress against the wall in this case is xx 4 Fa z ( a z ). (3.15) The distribution of the horizontal stresses against the wall are also shown in Figure 3-5. The maximum value occurs for z 0.577a, and that maximum stress is F xx max a The total force on a wall of depth h can be found again by integration of the horizontal stress over the depth of the wall. This gives F Q 1 a h Figure 3-5. Line load next to a smooth rigid wall.. (3.16) If a 0 this is Q 0.637F. If a increases the value of Q will gradually become smaller. A force at a larger distance from the wall will give smaller stresses against the wall. It should be kept in mind that only the extra forces and stresses resulting from the loads are mentioned. The weight of the soil itself also causes stresses, both in vertical and horizontal direction. 15

16 4 Deformation of layered soil An important problem of soil mechanics practice is the prediction of the settlements of a structure built on the soil. For a homogeneous isotropic linear elastic material the deformations could be calculated using the theory of elasticity. That is a completely consistent theory, leading to expressions for the stresses and the displacements. However, solutions are available only for a half space and a half plane, not for a layered material (at least not in closed form). Moreover, soils exhibit non-linear properties (such as a stiffness increasing with the actual stress), often have anisotropic properties, and in many cases the soil consists of layers of different properties. For such materials the description of the material properties is already a complex problem, let alone the analysis of stresses and deformations. Figure 4-1. Load on layered soil. For these reasons an approximation is often used, based on a semi-elastic analysis. In this approximation it is first postulated that the vertical stresses in the soil, whatever it true properties are, can be approximated by the stresses that can be calculated from linear elastic theory. On the basis of these stresses the deformations are then determined, using the best available description of the relation between stress and strain, which may be non-linear. If the soil is layered, the deformations of each layer are calculated using its own properties, and then the surface displacements are determined by a summation of the deformations of all layers. In this way the different properties of the layers can be taken into account, including a possible increase of the stiffness with depth. The procedure is not completely consistent, because in a soil consisting of layers of different stiffness, the stress distribution will not be the same as in a homogeneous linear elastic material. A partial justification may be that the stresses following from an elastic computation at least satisfy the equilibrium conditions. Also it has been found, by comparison of solutions for layered materials with the solution for a homogeneous material, that the distribution of the vertical stress is not very sensitive to the material properties, provided that the differences in material properties are not very large, i.e. excepting extreme cases such as a very stiff layer on a very soft subsoil. Example The computation method can best be illustrated by considering an example. This example concerns a circular fluid reservoir, having a diameter of 0 meter, which is being constructed on a foundation plate on a layer of fairly soft soil, of 0 meter thickness, see Figure

17 Figure 4-. Reservoir on soft soil. Below the soft soil the soil is a hard layer of sand or rock. The compressibility of the soft soil is about C The pressure of the foundation plate on the soil is 0 kpa, and the additional load by the fluid in the reservoir is 100 kpa. The problem is to determine the settlement caused by the load, in the centre of the reservoir. depth weight found. 1 log h 1 (m) (kpa) (kpa) (kpa) (kpa) kpa (m) load Table 4-1: Computation of settlement. The example has been elaborated in Table 4-1. The soil has been subdivided into 10 layers, of meter thickness each. The first column of the table gives the average depth of each layer. The second column gives the effective stress due to the weight of the soil, assuming that the effective unit weight of the soil is 10 kn/m 3, so that for each meter depth the stress increases by 10 kpa. The third column gives the additional stress due to the weight of the foundation plate. These stresses have been calculated using the formula for the stresses below a uniform circular load, equation (.13), z b 3 p(1 ), 3 (4.1) in which b z a, and a is the radius of the circular area. The fourth column is the sum of the second and third columns. This is considered as the initial stress, before the application of the load, but after the construction of the foundation plate. The fifth column gives the stresses caused by the load, the weight of the fluid in the reservoir. These stresses have also been calculated by the formula (4.1). The sixth column is the sum of the fourth and the fifth column. These are the final effective stresses. The seventh and eight columns contain the 17

18 actual computation of the deformations of each layer, using Terzaghi s logarithmic formula, and the value C. By adding the deformations of the layers the total settlement of the reservoir is obtained, which is found to be 0.10 m. That is not very small, and it might mean that the construction of the reservoir on such a soft soil is not feasible. The procedure described above can easily be extended. It is, for instance, simple to account for different properties in each layer, by using a variable compressibility. The method is also not restricted to circular loads. The method can easily be combined with Newmark s method to calculate the stresses below a load of arbitrary magnitude on an area of arbitrary shape. It is also possible to incorporate creep by adding the formula of Koppejan in the calculation. The method can also be elaborated with little difficulty to a computer program. Such a program may use a numerical form of Newmark s method to determine the stresses, and then calculate the settlements of the loaded area by the method illustrated above. The formula to compute the deformation of each layer may be Terzaghi s formula, but it may also include a time dependent term, to account for creep and consolidation. 18

19 5 Strip footing One of the simplest problems for which lower limits and upper limits can be determined is the case of an infinitely long strip load on a layer of homogeneous cohesive material ( 0 ), see Figure 5-1. Figure 5-1. Strip footing. The weight of the material will be disregarded, at least in this chapter. This means that it is assumed that 0. The problem is a first schematisation of the shallow foundation of a structure, using a long strip foundation, made of concrete, for instance. It will first be attempted to obtain a lower bound for the failure load, using an equilibrium system. Such a system should consist of a field of stresses that satisfies the conditions of equilibrium in all points of the field, that agrees with the given stress distribution on the soil surface, and that does not violate the yield condition in any point. 5.1 Lower bound An elementary solution of the conditions of equilibrium in a certain region is that the stresses in that region are constant, because then all conditions are indeed satisfied. In a two-dimensional field these equilibrium conditions are, in the absence of gravity, x xx x xz z zx z xz. zx 0, 0, (5.1) (5.) (5.3) The main difficulty is to satisfy the boundary condition, because the normal stress is discontinuous along the surface, see Figure 5-1. This difficulty can be surmounted by noting that in a statically admissible field of stresses (an equilibrium system), not all stresses need be continuous. Formally this can be recognized by inspection of the equations of equilibrium, eqs. (5.1) (5.3). All partial derivatives in these equations must exist, which means that the stresses must at least be continuous in the directions in which they have to be differentiated. It follows that the shear stress xz must be continuous in both directions, that the normal stress xx must be continuous in x-direction, and the normal stress must be continuous in z-direction. However, two of the partial derivatives, xx z and x, do not appear in the equations of equilibrium, and therefore no conditions have to be imposed on the continuity of 19

20 these two normal stresses in these directions. This means that xx may be discontinuous in z- direction, and that may be discontinuous in x-direction. Such a discontinuity is shown, for the vertical direction, in Figure 5-. Figure 5-. Stress discontinuity. This figure shows a small element, with all the stresses acting upon its boundaries. The normal stress xx must be continuous in x-direction, because of equilibrium, as can most easily be seen by letting the width of the element approach zero. Then the continuity of the stress can be seen as a consequence of Newton s principle that the reaction must be equal xx to the action. The normal stress, however, may jump across the vertical line, without disturbing equilibrium. In Figure 5- the stress is discontinuous in x-direction. The partial derivative x is infinitely large at the location of the vertical axis, but the element, and all of its parts, are perfectly well in equilibrium. This property of equilibrium systems has been applied by Drucker, one of the originators of the theory of plasticity, to construct equilibrium fields for practical problems. In this method the field is subdivided into regions of simple form, in each of which the stress is constant, so that the equations of equilibrium are automatically satisfied. The various subregions then are connected by requiring that all the stresses transferred on the boundary surfaces are continuous, allowing the normal stresses in the direction of these boundaries to be discontinuous. Figure 5-3. Drucker s equilibrium system. An example is shown in Figure 5-3, for the case of a strip footing. In a vertical strip below the load the stresses are supposed to be c, 4c, and 0. In the two regions xx xz 0

21 to the left and right of this strip the stresses are c, 0, and 0. On the two vertical discontinuity lines only the vertical normal stress is discontinuous. The other stresses are continuous, as required by equilibrium. This field of stresses satisfies all the conditions of equilibrium, and satisfies the boundary conditions on the upper surface. The shear stress zx 0, and the normal stress 0 if x a, and p if x a, where a is the width of the loaded strip. The stress distribution should also satisfy the condition that the yield condition is never violated. This can be checked most conveniently by considering the Mohr circles for this case, as shown in Figure 5-4. xx xz Figure 5-4. Circles of Morh belonging to Drucker s equilibrium system. In order that all circles remain within the yield envelope the value of the load p should be such that p 4c. The stress distribution satisfies all the conditions for a statically admissible stress field, and it can be concluded that p 4c is a lower bound for the failure load. If the true failure load is denoted by p c, it now has been shown that pc 4. c (5.4) It is possible that by considering more than two discontinuity lines slightly higher lower bounds can be found. This will not be investigated here, however. Figure 5-5. Elastic solution. Another method to obtain a statically admissible stress field is to use an elastic solution, when available. Such a solution satisfies the equilibrium equations and the boundary conditions. It also satisfies Hooke s law and the compatibility equations (which is not required for a statically admissible stress field, but not forbidden either). If the stress field is such that the maximum shear stress is not larger than the strength c, a lower bound of the 1

22 failure load has been obtained. For the case of a strip load, see Figure 5-5, the elastic solution has been given in Chapter 3. It can be shown that the maximum shear stress is p sin( 1 ). (5.5) This equation can be derived from the formulas (3.4) (3.6) by noting that xx ( ) xz. (5.6) The maximum value of sin( 1 ) is 1, so that the maximum elastic shear stress is p. If this is taken equal to c, the load is p c. For this value of the load the elastic solution is a statically admissible stress field, and the corresponding load is a lower bound for the failure load, i.e. pc 3.14 c. (5.7) Unfortunately, this is a lower value than the value found before (4c), so that this elastic lower bound does not contribute to a better approximation of the failure load. 5. Upper bound An upper bound for the failure load can be obtained by considering the mechanism shown in Figure 5-6. Figure 5-6. Mechanism 1. This mechanism consists of a displacement field in which half a circle, of radius a, rotates over a small angle, without internal deformations. This half circle slides along the remaining part of the body. The displacement field is compatible, and satisfies the boundary conditions on the displacements (that is very simple: these are only present on the circular slip surface). The load corresponding to this deformation can be determined by examining the moment equilibrium. If the circle rotates, a shear stress occurs at the periphery. If the shear stress is assumed to be maximal, so c, the moment with respect to the axis of rotation of the internal friction stresses at the periphery of the circle equals ca, because the length of the circular arc is a. The eccentricity of the external load is 1 a, so the exerted moment becomes 1 pa. Equating these two moments gives p c.

23 This is an upper bound for the failure load p c, pc 6.8 c. (5.8) A somewhat lower upper bound can be found by choosing the centre of the circle somewhat higher, see Figure 5-7. Figure 5-7. Mechanism. If the angle at the top is, it follows and because cr pa, 1 a Rsin, in which R is the radius of the circle and a the width of the load, 4c p. sin 1 For the previous upper bound is recovered. The smallest value is obtained for , or The centre of the circle then is located at a height 0.49a. The corresponding value of p is 5.5c. This is an upper bound, hence pc 5.5 c. (5.9) It can be concluded at this stage that it has been shown that 4c pc 5.5 c. (5.10) In the next chapter the failure load will be approximated even closer. 3

24 6 Prandtl In 190 Prandtl succeeded in finding a solution for the problem of a strip load on a half plane that is both statically admissible and kinematically admissible. This solution must therefore give the true failure load. The lower bound part of Prandtl s solution, with an equilibrium system of stresses, will be presented in this chapter. The proof that this solution is also kinematically admissible, which is much more difficult, will be omitted here. A complete proof can be found in textbooks on the theory of plasticity. The material is considered to be weightless ( 0 ), and frictionless ( 0 ), so that its only relevant property is the cohesive strength c. That is a great restriction, but it will be relaxed in later chapters. The basic principle of Prandtl s solution of the problem of the determination of the failure load of a half plane carrying a strip load on its surface is a subdivision of the region below the load into three zones, see Figure 6-1, two triangles and a wedge. Figure 6-1. Prandtl s schematisation. In each of these three zones the stress state is assumed to be critical. In that case it follows that the true failure load in the case of a strip load is pc ( ) c 5.14 c. (6.1) 4

25 7 Brinch Hansen Although the computation of limit states for materials with internal friction does not lead to results that are certainly on the safe side or on the unsafe side of the failure load (as they are for cohesive materials), these computations can still be very valuable, because it can be expected that the results will at least give an approximation of the failure load. Engineering calculations are always based upon a number of assumptions and approximations, and engineers have to accept that their results at best are an approximation of reality. The circumstance that it cannot be stated with certainty that a given load is above or below the failure load does not make the results useless. As long as the computations are based upon reasonable assumptions regarding the stresses and deformations, and if they incorporate certain experiences from the field or the laboratory, the results may be considered as giving a useful approximation of the real world. In this process there is also room for engineering judgement, which is a combination of intuition, experience and common sense. Finally, it may be mentioned that the applicability of a certain methodology may be enhanced by its possible agreement with known results for special cases. Figure 7-1. Failure planes in sand. In this chapter the case of a strip footing on cohesive material, considered in chapters 5 and 6, is extended to a general type of shallow foundation, on a soil characterized by its cohesion c, friction angle and volumetric weight. The soil is assumed to be completely homogeneous. Although the formulas were originally intended to be applied to foundation strips of buildings, at a shallow depth below the soil surface, they are also applied to large caisson foundations used in offshore engineering for the foundation of huge oil production platforms. 7.1 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 Prandtl, see Figure 7-, on the basis of the assumption that in a certain region at the soil surface the stresses satisfy the equilibrium conditions and the Mohr-Coulomb failure criterion. In this entire region the soil then is on the verge of yielding. 5

26 Figure 7-. Strip foundation. This analysis is a direct generalization of the problem considered in Chapter 6 of a strip load on the surface of a cohesive material. The foundation pressure is denoted by p. The surcharge q, next to the foundation, is supposed to be given. It can be used to represent the effect of the depth of the foundation (d) below the soil surface. In that case q d, where is the unit weight of the soil. N c N q N N c Table 7-1: Bearing capacity coefficients. N q N Prandtl s solution, which will not be derived in detail here, again uses a subdivision of the soil into three zones, see Figure 7-. In zone I the horizontal stress is supposed to be larger than the vertical stress, which is equal to the surcharge q. This horizontal stress is then the passive lateral stress corresponding to the vertical stress q. In zone III the vertical normal stress is supposed to be the largest stress, and its value is equal to the unknown load p. The 6

27 transition is formed by the wedge shaped zone II (Prandtl s wedge), which is bounded below by a logarithmic spiral. The results of the analysis can be written as p cn, c qnq (7.1) where the coefficients N c en N q are dimensionless constants, for which Prandtl obtained the following expressions, N q 1 sin exp( tan ), 1 sin (7.) Nc ( Nq 1) cot. (7.3) In Table 7-1 the values of N c and N q are given, as a function of the friction angle. In the limiting case 0 the value of Nc, as found in Chapter 6. If c 0 and 0 the bearing capacity must be equal to the surcharge, i.e. p q. Even a layer of mud can support a certain load, provided that it is the same all over its surface. This is expressed by the value N 1 for 0. q Prandtl s formula (7.1) has been extended by Keverling Buisman, Caquot, Terzaghi and Brinch Hansen with various terms, including one for the unit weight of the soil. The complete formula is written in the form p cn qn BN 1 c q, (7.4) where B is the total width of the loaded strip, and is the volumetric weight of the soil. That all effects may be superimposed, as has been assumed in eq. (7.4), has been confirmed by various investigations, but has never been proved rigorously. For the coefficient N various suggestions have been made, on the basis of theoretical analysis or experimental evidence, for instance N ( N q 1) tan. (7.5) There appears to be general agreement on the character of this expression, but various researchers have proposed different values for the constant factor. Brinch Hansen used a factor 3 rather than a factor, probably to avoid an overestimation, and therefore including some safety. In modern engineering it is considered that safety factors should be kept apart from the theoretical formulas, so that it was agreed that the best value of the multiplication factor is. A safety factor must be taken into account explicitly, in the design stage, by reducing the soil strength, or as a load factor. Later the formula (7.4) has been further extended with various correction coefficients, in order to take into account the shape of the loaded area, the inclination of the load, a possible inclined soil surface, and a possible inclined loading area. Most of these effects were assembled into a single formula by Brinch Hansen, p i s cn i s qn i s BN 1 c c c q q q. (7.6) In this equation the coefficients i c and i q are correction factors for a possible inclination of the load (inclination factors), and s c and s q are correction factors for the shape of the loaded area (shape factors). Some other factors may be used (for a sloping soil surface, or a sloping foundation foot), but these are not considered here. 7

28 7. Inclination factors In case of an inclined load, i.e. loading by a vertical force and a horizontal load, see Figure 7-3, 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, t c p tan. (7.7) 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. Figure 7-3. Inclined load. For cases in which the shear force is smaller than its maximum possible value, the correction factors for the inclination of the load are usually expressed as i c t 1, c ptan (7.8) i q i c, (7.9) i i 3 c. (7.10) There is no general agreement on the precise value of these reduction factors, on an international level and even on a national level. Various researchers prefer slightly different values, and even the national standards may (e.g. NEN 6744) may give different values. The formulas given above at least are in agreement with certain special cases. The coefficients approach 0 if the shear stress approaches the maximum value max c n tan. The other extreme case is when the load is vertical ( t 0). Then all factors reduce to 1, as required. 7.3 Shape factors If the shape of the foundation area is not an infinitely long strip, but a rectangular area, of width B and length L (where it is assumed, for definiteness, that the width is the shortest dimension, i,e, L B), the usual correction factors are of the form 8 s s c q B 1 0., (7.11) L B 1 sin, (7.1) L B s (7.13) L

29 There is no international agreement on the precise values of these correction factors either. Some consultants prefer to take sq 1, for all values of, and some use coefficients with slightly different values for the factors 0. and 0.3. It may be noted that for BL 0, the formulas all give a factor 1, in agreement with the basic results for an infinite strip. It should also be remembered that BL 1, by definition. Figure 7-4. Rectangular area. Some justification for values of the coefficients s c and s q larger than 1 is that when loading a rectangular plate, some of the soil surrounding the plate will also deform with the plate, so that a larger area contributes to the bearing capacity of the plate. The coefficient s is smaller than 1 because tests indicate that for a rectangular plate in sand a sliding surface may occur that is less deep than the sliding surface for a long strip. In case of an eccentric resultant force of the load, the width B and the length L may be reduced such that the resulting force does apply in the centre of the reduced area, see Figure 7-5. Part of the foundation plate then does not contribute to the bearing capacity, at least for this loading case. It may, of course, give a contribution to the bearing capacity of other loading cases. As mentioned before, there is no general agreement about the values of many of the correction factors, because the results obtained by researchers in different countries, from theoretical or experimental studies, appear to give different results. Great care is needed when using data from literature. When a certain value has been obtained by one single researcher, and deviates from the results of many others, that value may well be in error. It is also very inconvenient that there is sometimes no agreement about the basic formula (7.4). In some older publications the factor 1 is omitted. Then the values of N are (approximately) half as large, so that the final result is the same, but it may give rise to some confusion when using a formula from one publication, and taking the coefficients from another publication. In this book Terzaghi s original formula has been used, as is common practice internationally. Figure 7-5. Eccentric load. The formulas presented in this chapter have originally been derived for foundations on land, with relatively modest dimensions, say a few square meters. The third term in Brinch Hansen s formula (7.6) then is rather small, because of the small value of B, and it is often omitted. In offshore engineering the development of gravity foundations has meant that production platforms may be founded on huge concrete caissons that are placed on the sea 9

30 bottom, in deep water. The surface area may be up to 80 m 80 m. For the design of such structures the bearing capacity of the foundation is of great importance, and then the third term in the formulas (7.4) or (7.6), which describes the influence of the unit weight of the soil (i.e. the gravity term), is the most important term of all, giving the major contribution to the bearing capacity, especially in sand. This is why considerable attention has been paid to a more accurate definition of this term. It must be emphasized that all the considerations of this chapter are restricted to dry soils, in which there is no difference between effective stresses and total stresses. For saturated soils the formulas should be expressed in terms of effective stresses. Usually this can be accomplished simply by replacing the volumetric weight by the effective volumetric weight s w. That is a simple, but very fundamental adjustment. 7.4 Brinch Hansen and codes The code TGB 1990 (NEN 6744) gives the Brinch Hansen method for determining the maximum bearing force F of a foundation. This method is based on Prandtl s theoretical slip surfaces (see Figure 7-6). One distinguishes between drained and undrained situations. Undrained situations occur in cohesive impermeable soils, in which pore water pressures increase directly after the load is applied. In undrained soil an undrained shear strength c u is used instead of c' in calculations. One also uses = 0. Long-term loads on clay and peat are, of course, calculated as on drained soils! Figure 7-6. Prandtl s slip surfaces. p q H F Figure 7-7. Brinch Hansen s slip surfaces. Brinch Hansen extended Prandtl s formulas to include reduction factors for the influence of a possible shear force H and the relation between the foundation s width B and length L. The following discusses a method which is somewhat simplified, relative to the TGB

31 The maximum bearing force can be approximated by: whereby: F p A max ' max (7.14) p' max c' Ncscic q' Nqsqiq 0,5 B N s i (7.15) In which the factors for the bearing force are: tan Nc Nq 1 cot N 1 sin q 1 sin e N ( N 1) tan In which the factors for the shape of the foundation ( B L ) are: B B sc 1 0. sq 1 sin s B L L L In which the factors for the horizontal load ( B L ) are: For drained soil: For H parallel to L and L /B : in 1 H 1 N 1 F Accot q q ic iq i q For H parallel to B : in 3 3 q q c q 1 H i i i 1 H Nq 1 F Accot F Accot For undrained soil: i c H Af undr for the rest, see above. q Only the part of the foundation slab which has effective stresses underneath is included in the effective width B. The factors for the bearing force are also given in the figure below. Figure 7-8. Bearing force factors as functions of the angle of internal friction. 31

32 where: p' max = maximal average effective stress on the effective foundation area [kpa] A = the effective foundation area [m ] c' = (weighted) cohesion (calculation value) [kpa] q' = effective stress at the depth of but next to foundation surface (calculation value) in = v;z;o f;g. d i. car u (calculation value) i1 [kpa] u = the water pressure [kn/m ] n = the number of horizontal soil layers between the construction depth and the soil cover level; ' = (weighted) effective volumetric weight of the soil below the construction depth (calculation value) [kn/m 3 ] d i = the thickness of layer i [m] car = the characteristic volumetric weight of the soil [kn/m 3 ] for which: * for a soil layer above groundwater level: car = rep ; * for a soil layer under groundwater level: car = sat;rep ; rep = the representative value of the volumetric weight with a natural moisture content [kn/m 3 ] sat;rep = the saturated volumetric weight [kn/m 3 ] f;g = the load factor for a favourable load; c = factor for the influence of cohesion; q = factor for the influence of the soil cover; = factor for the influence of the effective volumetric weight of the soil; ' = (weighted) effective angle of friction (calculation value) [] H = the shear force, i.e.: component of load in the plane of the foundation surface (calculation value) [kn] F = component of the bearing force perpendicular to the foundation surface (calculation value) [kn] f undr = the calculation value of the undrained shear strength = c u [kpa] L = the length of the effective foundation area, for circular slabs: L = B [m] B = the width of the effective foundation area, for circular slabs: L = B [m] For an eccentrically loaded foundation B is approximated by B-e See the following figure: true schematised 3

33 Note: The factors for the bearing force are too conservative. This is because it is assumed that the factors do not influence each other, which is incorrect. The factors for the shape of the foundation and particularly for the horizontal load are not substantiated scientifically but they are based on empirical relations, experiments, calculations, etc. For a horizontal load H one not only has to apply a reduction in the calculation of the vertical bearing force F, but one also has to check if sliding can occur (e.g. using Coulomb). The reduction of the vertical bearing force F as a result of the horizontal load H is considerable. Assuming H/F = 0.30 results in i q = 0.50!! 7.5 CPT and undrained shear strength In the Netherlands the cone penetration test is mainly used as a model test for pile foundations. In the Western parts of the Netherlands the soil usually consists of 10 0 meters of very soft soil layers (clay and peat), on a rather stiff sand layer. This soil structure is very well suited for a pile foundation, of wooden or concrete piles of about 0 cm 40 cm diameter, reaching just into the sand. The weight of the soft soil acts as a surcharge on the sand, which has a considerable cone resistance. The allowable stress on the sand depends upon its friction angle, its cohesion c (usually very small, or zero), and the surcharge q. The dimensions of the foundation pile have very little influence, because this parameter appears only in the third term of Brinch Hansen s formula, which is a small term if the width is less than, say, 1 meter. This means that the maximum pressure for a large pile and the thin pile of a cone penetrometer will be practically the same, so that the allowable pressure on a pile can be determined by simply measuring the cone resistance. This will be elaborated in Chapter 8. The cone penetration test can also be used to determine physical parameters of the soil, especially the shear strength. It can be postulated, for instance, that in clays the cone resistance will be determined mainly by the undrained shear strength of the soil (s u ). In agreement with the analysis of Brinch Hansen the relation will be of the form q c v N, csu (7.16) where v is the local vertical stress caused by the surcharge, and N c is a dimensionless factor. For a circular cone in a cohesive material a cone factor N c of the order of magnitude is usually assumed, on the basis of plasticity calculations for the insertion of a cone into a cohesive material of infinite extent. By measuring the cone resistance q c the undrained shear strength s u can be determined. The results are not very accurate, because of theoretical shortcomings and practical difficulties, but the measurement has the great advantage of being done in situ, on the least disturbed soil. The alternative would be taking a sample, bringing it to a laboratory, and then doing a laboratory test. This process includes many possible sources of disturbance, which can be avoided by doing a test in situ. 33

34 34

35 II Pile foundations 35

36 8 Cone Penetration Test (CPT) The cone penetration test (CPT) is a testing method used to determine the geotechnical engineering properties of soils and the soil stratigraphy. It was initially developed in the 1950s at the Dutch Laboratory for Soil Mechanics in Delft (now Deltares) to investigate soft soils. Based on this history it has also been called the "Dutch cone test". Today, the CPT is one of the most used and accepted in soil methods for soil investigation worldwide. The test is used to determine: soil types, (stratigraphic) soil layering, soil stiffnesses and strengths (empirical) bearing capacities of pile foundations permeability and consolidation parameters, groundwater pressure environmental data (pollution) The test method consists of pushing an instrumented cone, with the tip facing down, into the ground at a controlled rate (between cm/s). The resolution of the CPT in delineating stratigraphic layers is related to the size of the cone tip, with typical cone tips having a crosssectional area of either 10 or 15 cm², corresponding to diameters of 3.6 and 4.4 cm. At least two things are constantly measured: 1. the tip resistance /cone resistance /cone pressure q c [MPa], and. the shaft or friction sleeve resistance s [MPa] Also other items are measured, for example the water pressure, the inclination of the conus, or chemical products like hydrocarbons or aromatics. One parameter is calculated from the first two, which is the friction ratio R f : R f s 100% (8.1) q c Figure 8-1. Left:CPT wagen. Right: CPT cone. 36

37 The friction number is quite helpful in determining the soil type, see Table 8-1, but a checking with a few additional borings remains very important. Soil Type Friction R f cone pressure q c Sand, coarse Sand, medium Sand, fine Sand, silty Sand, clayey Sandy clay or loam Silt Clay, silty Clay Clay, organic (peaty) Peat 0,4% 0,6% 0,8% 1,1% 1,4% 1,8%,%,5% 3,3% 5,0% 8,0% 5-30 MPa 5-10 MPa 0,5- MPa 0-1 MPa Table 8-1: Friction number and cone resistance for different soil types. For example in the sounding (CPT) of Rotterdam in Figure 8-3, the soil types and layering are very clear. There are very soft layers until m: a clear clay layer until -5,5 m; peat until -11,5 m; clay until -16,0 m; clayey peat until -17,5 and first then soft sand. Figure 8- shows the soil layering of the Hubertus Dune in Den Hague, the Netherlands. This is used to plan a bore tunnel (see black lines under dune). Figure 8-. Soil layering of the Hubertus Dune in Den Hague. 37

38 38 Figure 8-3. CPT of Rotterdam, with clearly a layering of clay, peat, clay, mixed clay and peat and finally deep sand.

39 9 Compression piles In many river delta areas, for example the Western part of the Netherlands, the soil consists of layers of soft soil (clay and peat), of 10 0 meter thickness, on a stiff sand layer, of pleistocene origin. The bearing capacity of this sand layer is derived for a large part from its deep location, with the soft layers acting as a surcharge. The properties of the sand itself, a relatively high density, and a high friction angle, also help to give this sand layer a good bearing capacity, of course. The system of soft soils and a deeper stiff sand layer is very suitable for a pile foundation. In this chapter a number of important soil mechanics aspects of such pile foundations is briefly discussed. 9.1 Pile types Piles can be made from wood, steel, (partially) reinforced concrete or prestressed concrete. The installation can be done by driving, vibrating, boring or pressing, but not all installation techniques can be applied to the different pile types. The following distinction can be made between the piles, but it is a strong simplification: Wooden piles (driven or pressed) Steel piles (driven, vibrated or bored) Precast concrete piles (driven) In situ concrete, piles (driven with steel pipe) In situ concrete, piles (bored with steel pipe support) In situ concrete, piles (bored with slurry support) Wooden piles Wooden piles exist already a long time, probably more than thousand years. In the 15 th century piles were used a lot in Amsterdam and also in Venice there is similar old tradition of using wooden piles. The palace on the dam in Amsterdam was built in 1665 on wooden piles. In the Netherlands there are still 5 million wooden piles in use. Figure 9-1. Left: Wooden pile driving in Amsterdam in the 15 th century. Right: Palace on the dam. Wooden piles are cheap, light, easy to transport, easy to cut off, but have two disadvantages: they have a limited bearing capacity and they will rot above the groundwater table and need 39

40 therefore a concrete extension or something similar and should never face a sinking groundwater (not even a leak draining sewage system nearby). Figure 9-. Wooden piles with concrete extension against rotting. Steel piles Steel piles can be found in all sorts: H-profiles, tubes or other profiles. They exist in all lengths, width and wall thicknesses. If there is no closed foot of a tube, and the soil conditions are correct, the piles can be vibrated into the ground, otherwise they will be driven. Tubes can be filled afterwards with concrete to make them stronger and to option tip resistance. The advantage is: very strong and perfect for high horizontal and vertical loads and heavy ground for pile driving. The disadvantages are: expensive and rusting (corrosion). Therefore those piles are used when other are not possible anymore. Figure 9-3. Steel piles. Precast concrete piles Precast concrete piles are mostly used. They can be bought in many sizes up to 0,5 m wide and 40 m long. They are mostly driven into the ground, even for rather stiff sands (cone resistance up to q c = 65 MPa). The advantages are that they are relatively cheap, strong and available in all sizes. The disadvantages are little, only for very high load steel has to used and in case of problems with driving vibrations, in-situ concrete piles can be used. 40

41 Figure 9-4. Precast concrete piles. In situ concrete driven piles One can drive a steel tube with a steel plate at the tip, into the ground, fill it with concrete, lift the tube and lose the plate. In this way a concrete in-situ pile is made. The advantage is that the pile is still driven, so the soil is compacted and there is a higher bearing capacity than bored piles. Also no transport of long piles into urban areas are needed, Disadvantages are: the pile is of a poorer quality than precast piles and the driving still causes vibrations. Figure 9-5. Driven in-situ concrete piles. 41

42 Figure 9-6. Making of driven in-situ concrete piles. In situ concrete bored piles in steel casing Another way of in-situ pile making is boring. A steel casing supports the hole. The advantages are: no vibrations and also possible in complex grounds with stones are hard soils. The disadvantage is the lower bearing capacity due to the looser ground by boring. Figure 9-7. Making of bored in-situ concrete piles. Concrete, in situ concrete, piles, with slurry support There are also bore techniques without a steel casing supporting the ground. Mostly a bentonite slurry is used to create support pressure. Figure 9-8. Bore for making bored in-situ concrete piles. 4

43 Other piles There are many other pile types, for example the Franki pile, a type of bored pile, but with a in the steel tube driven and wider pile tip. Figure 9-9. Franki pile. Or the fundex pile, a drilled pile, which can be installed in small and low areas, perfect for improving a foundation afterwards, but not as good as a driven pile and above all very expensive because it is complex and one loses an expensive steel drill as pile tip. Figure Fundex pile. 43

44 9. Bearing capacity A pile may also have a bearing capacity due to friction along the length of the pile. This is very important for piles in sand layers. In applications in very soft soil (clay layers), the contribution of friction is generally very unreliable, because the soil may be subject to settlements, whereas the pile may be rigid, if it has been installed into a deep sand layer. It may even happen that the subsiding soil exerts a downward friction force on the pile, negative skin friction, which reduces the effective bearing capacity of the pile. Friction is of course very important for tension piles, for which it is the only contributing mechanism. The maximum value of the skin friction can be determined very well using a friction cone, which is a penetration test in which the sleeve friction is also measured. The local values are often very small, however, so that the measured data are not very accurate. For sandy soils the friction therefore is often correlated to the cone resistance. The ultimate bearing capacity of a compression pile is determined by both the tip bearing capacity and the shaft bearing capacity: whereby: Fr ;max Fr ;max: tip Fr ;max: schaft (9.1) F A p (9.) r;max: tip tip r;max; tip L F O p (9.3) r;max: shaft p; avg o r ;max; shaft in which: F r;max = maximum bearing force F r;max;tip = maximum tip resistance force F r;max;shaft = maximum shaft friction force A point = surface area of the tip of the pile p r;max;tip = maximum tip resistance according to the sounding ( = q c ) O p;avg = circumference of the pile shaft L = length of the pile p r;max;shaft = maximum pile shaft friction according to the sounding ( = sqc) Notes In case prefab piles with enlarged feet are used, L may not be larger than the height of the enlarged foot H (see Figure 9-15). A reduced value of maximum tip resistance of 15 MPa is valid (see Figure 9-16). One must not forget to reduce the net bearing capacity of the pile by the value of negative shaft friction. For s see Table 9- in Chapter 9.5 In areas with soft soils, mostly methods based on the Cone Penetretation Tests are used (for example method Koppejan). Another way is to calculate the bearing capacity of the pile with Brinch-Hansen, this is done when there are no Cone Penetretation Tests to use for the previous method. 44

45 9.3 Tip resistance using Brinch Hansen (Prandtl or Meyerhof) For the determination of the bearing capacity of a foundation pile it is possible to use a theoretical analysis, on the basis of Brinch Hansen s general bearing capacity formula. In this analysis the basic parameters are the shear strength of the sand layer (characterized by its cohesion c and its friction angle ), and the weight of the soft layers, which can be taken into account as a surcharge q. The maximum tip resistance can be determined analogously to the bearing capacity of a shallow foundation, which is based on a slide plain according to Prandtl. This entails simply using the -dimensional solution for 3-dimensional collapse (conservative) and simply disregarding the shear strength, not the weight, of the soil above the foundation plane (conservative). The following example shows that the term 0.5 e;d ef B N s is negligible relative to v;z;o;d Ns q q. Therefore the tip resistance can be calculated most simply according to: p N s r;max;tip v;z;o;d q q Figure Slide plains under a pile according to Prandtl. Example Given is a pile that is driven 11 metres into the ground according to Figure The pile is in approximately 1 metre of sand; above the sand layer is 10 metres of clay. Groundwater level is 1 metre below ground level. The following data is given: Sand: = 40 c = 0 w,d = 0 kn/m 3 Clay: w,d = 15 kn/m 3 Pile dimensions: 0.4 x 0.4 m According to Prandtl, because the cohesion is zero the maximum pressure below the tip of the pile will be: pr;max;tip v;z;o;d Nqsqiq 0.5 e;dbef N s i According to Table 7-1 in Chapter 7: N q = 64 and N = 106 The pile is only subjected to axial loads, so: i q = i =1 The shape factors can be derived from (B ef = L ef ): s q = 1+sin 40 = 1.64 s = = 0.7 The initial effective ground pressure is: v;z;o;d kn/m The maximum acceptable tip stress is: p So: r;max;tip kn/m 0.5 e;dbef Ns Ns << v;z;o;d q q 45

46 9.4 Tip resistance using Koppejan In engineering practice a simpler, more practical and more reliable method has been developed, on the basis of a cone penetration test, considering this as a model test. It would be even better to perform a pile loading test on the pile, in which the pile is loaded, for instance by concrete blocks on a steel frame, with a test load approaching its maximum bearing capacity. This is very expensive, however, and the CPT is usually considered reliable enough. Figure 9-1. CPT and pile. In a homogeneous soil it can be assumed that under static conditions the failure load of a long pile is independent, or practically independent of the diameter of the pile. This means that the cone resistance measured in a CPT can be considered to be equal to the bearing capacity of the pile tip (or pile base or pile foot). A possible theoretical foundation behind this statement is that the failure is produced by shear deformations in a zone around the pile, the dimensions of which are determined by the only dimension in the problem, the diameter of the pile. If the pile diameter is taken twice as large, the dimensions of the failure zone around the pile will also be twice as large. The total force (stress times area) then is four times as large, see Figure 9-1. This is also in agreement with the theory behind Brinch Hansen s formula, provided that the third term (representing the weight of the soil below the foundation level, and the width of the foundation) is small. This will be the case if the pile diameter is small compared to its length. In reality the soil around the pile tip usually is not perfectly homogeneous. Very often the soil consists of layers having different properties. For this case practical design formulas have been developed, which take into account the different cone resistance below and above the level of the pile tip. Moreover, in these design formulas the possibility that the failure mode will prefer the weakest soil can be accounted for. In engineering practice the Koppejan formula is often used. In this formula the resistance of the pile is assumed to consist of three contributions, p p p p 1 1 [ ( 1 ) 3]. (9.4) In this equation p 1 is the smallest value of the cone resistance q c below the pile tip, up to a depth of 4d, where d is the diameter of the pile, p is the average cone resistance in that zone, and p 3 is a representative low value of the cone resistance above the pile tip, in a zone up to 8d above the pile tip. In this way a representative average value of the cone resistance around the pile tip is obtained, in which engineering judgement is combined with experience. 46