288 14 Further Topics in Hyperplasticity there are a number of other possible forms for compressible large strain elastic- ity. On differentiation, this gives (with no summation over i) ooo lo g ii ii J G J K VO UO U U O (14.47) or 2 log i i G KJ J J O V (14.48) For uniaxial tension 1 O O 23 0V V and 23 O O, 2 2 0lo g lo g J GKJGKJ O O (14.49) so that lo g KJ G J O , and it follows that 22 1 log 1 KJ G G JJ J §· OO V ¨¸ ¨¸ O ©¹ (14.50) which is very similar to the incompressible result. Note that J cannot be ex- pressed analytically in terms of O because of the transcendental nature of lo g KJ G J O . For small strain, it can be shown that the above equations reduce to 1123123 2 2 3 G KV HHH HHH (14.51) which is, as expected, the usual small strain expression in terms of principal strains. 14.6 Fibre-reinforced Material Consider an elastic material as defined in Section 2.1. The free energy is 32 62 ii jj ij ij fK G cc HH HH (14.52) Now consider that embedded within this material is a (dilute) proportion of unidirectional fibres, with a small concentration c (volume of fibres divided by total volume). The fibres have a Young’s modulus f E , and they are aligned in the direction with unit vector i n . We assume that the fibres are bonded to the parent (matrix) material and therefore undergo the same strain in the direction i n . The elongational strain in 14.6 Fibre-reinforced Material 289 this direction is ij i j nnH , so it is straightforward to see that the free energy of the composite material is f 32 622 ii jj ij ij ij i j kl k l cE f KG nnnn cc HH HH HH (14.53) and the stress is f 2 ij kk ij ij kl k l i j KGcEnnnn c V HG H H (14.54) For a given strain, the material simply exhibits a stress that is augmented by a term that depends on the fibres. Now consider the possibility that the matrix material exhibits von Mises type plasticity, as described (in terms of the Gibbs free energy) in Section 5.2.2. The Helmholtz free energy becomes f 32 622 ij ij ij ij ii jj ij i j kl k l cE fK G nnnn cc HD HD HH HH (14.55) which we augment by the constraint 0 kk D so that there is no plastic volumet- ric strain. The dissipation is 2 ij ij dk DD (14.56) Now we obtain f 2 ij kk ij ij ij kl k l i j KG cEnnnn c V HG HD H (14.57) 2 ij ij ij ij G c F HD /G (14.58) 2S ij ij ij kF D (14.59) Equating ij F and ij F immediately gives 0/ , and eliminating the general- ised stresses, we obtain an expression for the yield surface in strain space: 22 420 ij ij ij ij Gk cc HD HD (14.60) In principle we can solve (14.57) for the strains in terms of the stresses and in- ternal variables, and substitute this result in (14.60) to obtain the yield surface in terms of the stress and internal variables. In practice, this is most straightfor- wardly carried out by numerically inverting the expression: f 1 22 3 ij ij ij kl ik jl ij kl i j k l kl GK G cEnnnn §· §· V D GG G G GG H ¨¸ ¨¸ ©¹ ©¹ (14.61) The presence of the internal variables in the expression for the yield surface in terms of stresses means that the composite material exhibits strain hardening. 290 14 Further Topics in Hyperplasticity 14.7 Analysis of Axial and Lateral Pile Capacity The final example, drawn from geotechnical engineering, demonstrates how the continuous hyperplastic approach can be used to analyse simple systems involv- ing soil-structure interaction. Unlike some previous applications, in this case we can give a physical meaning to the internal coordinate K. We consider a pile analysed by the “Winkler” method in which the reaction from the soil is deter- mined by pointwise load-displacement relationships along the pile, but interac- tions among the relationships are not taken into account. The method is other- wise known as the use of “t-z” curves for vertical loading and “p-y” curves for lateral loading. 14.7.1 Rigid Pile under Vertical Loading Consider a rigid pile of length L and radius R in clay with shear modulus G and undrained shear strength c, under purely vertical loading. The resistance to movement is provided by shaft resistance on the side of pile and end bearing at the tip, see Figure 14.8. At each point along the pile, we assume a simple elastic- plastic relationship (the “t-z” relationship) between the vertical displacement w and the shear stress W (these variables are usually called z and t, respectively, in the piling literature). We specify the response at any point along the pile by the following functions, in which it is convenient to work from the Helmholtz free energy: 2 2 G fw R D ] (14.62) D dac (14.63) where ] and a are dimensionless factors (see Fleming et al., 1985) and D is the plastic vertical displacement of the pile shaft. The standard approach gives f G w wR w W D w] (14.64) f G w R w F D W wD ] (14.65) S d ac w F D wD (14.66) The yield function for the shear stress on the side of the pile can be expressed as 0yac F . 14.7 Analysis of Axial and Lateral Pile Capacity 291 The end-bearing stress V on the pile is also specified by an elastic-plastic rela- tionship: 2 2 1 L G fw R D SQ (14.67) cL dNc D (14.68) where the stiffness and end-bearing factors follow the approach taken by Flem- ing et al., (1985). We distinguish L D , the plastic displacement of the pile base, from D (the equivalent for the pile shaft) because we have in effect two parallel plasticity mechanisms in which the onset of plastic strains occurs at a much smaller displacement on the shaft than at the base. The analogy is shown in Figure 14.9. Figure 14.8. Tractions on a pile 292 14 Further Topics in Hyperplasticity Figure 14.9. Conceptual models of shaft and base of pile 14.7 Analysis of Axial and Lateral Pile Capacity 293 It follows that 4 1 L f G w wR w V D wSQ (14.69) 4 1 LL L f G w R w F D V wD S Q (14.70) S Lc L d Nc L w F D wD (14.71) and the yield function at the pile tip can be derived as 0 Lc yNc F . The entire pile response can now be obtained by integrating the energy and dissipation terms over the shaft of the pile and adding the end-bearing term summed over the end area. For convenience, we express the distance down the pile in terms of a dimensionless coordinate zLK , where z is the distance below ground level and L is the length of the pile. To emphasise that it is now a function of K, we rewrite D as ˆ D. Again, we must distinguish here between ˆ 1D , the plastic displacement at the tip relevant to skin friction, and L D , the plastic displacement at the tip of the pile relevant to end bearing. The free en- ergy and dissipation functions are 1 2 2 0 1 2 0 2 ˆ 2 21 2 ˆ 1 L L GGR fwRLdw R GR fd w DSKD ]Q K D Q ³ ³ (14.72) 1 2 0 1 2 0 ˆ 2 ˆ cL cL dac RLd RNc dd R N c DS KS D KS D ³ ³ (14.73) Application of the standard approach now results in the following expression for the vertical load V: 1 0 4 ˆ 2 1 L f GGR VwRLdw wR w DS K D w] Q ³ (14.74) ˆ ˆ ˆ 2 ˆ f G wRL R w F D S wD ] (14.75) ˆ ˆˆ S2 ˆ d ac R L w F D S wD (14.76) 294 14 Further Topics in Hyperplasticity 4 1 LL L f GR w w F D wD Q (14.77) 2 S LcL L d RNc w F S D wD (14.78) It is straightforward to show that these equations correctly define the me- chanical behaviour of the pile, in that the total load is the integral of the shear stress over the surface of the pile, plus the end bearing; each of these terms is determined by a simple elastic-plastic relationship. Note that no additional diffi- culty is caused if the strength c is a function of depth. 14.7.2 Flexible Pile under Vertical Loading The only modification to the energy expressions necessary to accommodate a pile that is flexible rather than rigid is that there is now an additional free en- ergy term due to elastic compression of the pile: 2 1 2 00 ˆˆ 22 L p EA w EA w fdzd zL §· ww §· K ¨¸ ¨¸ wwK ©¹ ©¹ ³³ (14.79) where we have written the vertical displacement w as ˆˆ ww K to emphasise that it is now a function of the distance down the pile. The total free energy therefore becomes 2 1 2 2 0 ˆ 2 ˆ ˆˆ 21 221 L GEAwGR fwRL dw RL §· §· w ¨¸ DS K D ¨¸ ¨¸ ]wKQ ©¹ ©¹ ³ (14.80) and the dissipation expression is unchanged. We consider the possibility of the following externally applied loading on the pile: (a) a vertical load V at the top (as before) and (b) a distributed load ˆ v per unit length along the pile. Later, we shall assume that the latter is zero, but it is convenient to introduce it at this stage. The work done by these external forces is the Frechet differential of the free energy with respect to the displacement ˆ w. (Compare it with the usual equation fV w wH for a continuum, so that Wf VH w wH H ): 1 ˆ 0 ˆˆˆ ˆ 0, w WVw vwLd f w K ³ (14.81) 14.7 Analysis of Axial and Lateral Pile Capacity 295 Carrying out the Frechet derivative, we obtain 11 00 ˆˆ ˆ ˆˆˆ ˆˆ 02 2 ˆˆ 11 1 L GEAww Vw vwLd w w R L d RL GR ww §· §· §· ww K DS K ¨¸ ¨¸ ¨¸ ¨¸ ¨¸ ]wKwK ©¹ ©¹ ©¹ D Q ³³ (14.82) Applying integration by parts to the second term in the integral on the right- hand side, this becomes: 11 2 2 00 1 0 ˆ ˆ ˆˆˆ ˆˆ ˆ 02 ˆ 2 ˆˆˆ 11 1 L GEAw Vw vwLd w w R L w d RL EA w GR www L §· §· w ¨¸ K DS K ¨¸ ¨¸ ¨¸ ] wK ©¹ ©¹ ªº §· w D «» ¨¸ wK Q ©¹ ¬¼ ³³ (14.83) Noting though that ˆ w is arbitrary, we can simply compare coefficients in terms in ˆ w . For 01K (note the strict inequalities), we compare the terms within the integrals to obtain 2 2 ˆ ˆ ˆˆ 02 GEAw vwRL RL §· w D S ¨¸ ¨¸ ] wK ©¹ (14.84) Since the (upward) shear stress on the pile is ˆ ˆ G w R W D ] and the com- pressive force at any point in the pile is ˆ EA w P L §· w ¨¸ wK ©¹ , this simply represents the differential equation for the vertical equilibrium of the pile 1 2 P R L w S W wK . At the top of the pile, we compare terms in ˆ 0w and obtain 0 ˆ EA w V L K §· w ¨¸ wK ©¹ (14.85) which is simply the relationship between the applied vertical load and the com- pressive strain at the top of the pile. Similarly, at the base, we compare terms in ˆ 1w to obtain 1 ˆ 2 ˆ 01 1 L EA w GR w L K §· w D ¨¸ wK Q ©¹ (14.86) which is the relationship between the end-bearing force and the compressive strain at the pile toe. 296 14 Further Topics in Hyperplasticity Finally, if we express f and d in ways similar to those used for the rigid pile, 1 2 0 2 ˆ ˆ 1 1 L GR ffd w K D Q ³ and 1 2 0 ˆ cL ddd RNc KS D ³ , we can obtain the generalised forces: ˆ ˆ ˆ ˆ 2 ˆ f G wRL R w F D S wD ] (14.87) ˆ ˆˆ S2 ˆ d ac R L w F D S wD (14.88) 4 ˆ 1 1 LL L f GR w w F D wD Q (14.89) 2 ˆ S LcL L d RNc w F S D wD (14.90) Combining the above equations yields the following results. For 0 1 K , 2 2 ˆ ˆˆ ˆ 2S2 EA d w G wRLacRL LR d DS DS ] K (14.91) which is the differential equation for the compression of the pile in terms of the mobilised shear stress on the pile shaft (which is in turn given by an elastic- plastic expression). At 1 K , 2 1 ˆ 4 ˆ 1S 1 LcL EA dw GR wRNc Ld K §· D SD ¨¸ KQ ©¹ (14.92) which expresses the elastic-plastic response for the end bearing at the pile tip and its relationship to the strain in the base of the pile. The above equations could have been readily (and more easily) obtained by conventional methods of analysis of the mechanics of the pile, but our purpose here is to demonstrate the consistency and generality of the hyperplasticity ap- proach. The important principle illustrated in this example is that we were able to define the correct mechanical behaviour from a free energy functional that contained differential as well as algebraic terms. Note that the above formalism for describing the pile problem is only one of a number of possible approaches. In particular, we chose to treat ˆ w as if it were an external kinematic variable. An alternative would be to treat ˆ w as an internal variable for 0 1 Kd and just the displacement ˆ 0 w as an external variable. In this latter case, the orthogonality condition results in a vertical equilibrium equation at any point down the pile, and this provides an example which brings physical meaning to the interpretation of orthogonality as an “internal equilib- rium” condition. 14.7 Analysis of Axial and Lateral Pile Capacity 297 14.7.3 Rigid Pile under Lateral Loading Now consider the rigid pile under lateral loading. We shall consider the resis- tance to movement provided solely by shaft resistance ( i. e. we ignore any lateral shear resistance on the tip). At each point along the pile, we assume a simple elastic-plastic relationship (the “ p-y” relationship) between the lateral displace- ment ˆ uuz u L T KT and the normal stress V (called y and p, respectively, in piling literature). We specify the following functions: 2 ˆ ˆ 4 k fu R D (14.93) ˆ h dNc D (14.94) where k is the “modulus of subgrade reaction” and h N is a lateral bearing capacity factor. The internal variable ˆ D now represents the lateral plastic dis- placement of the pile. The standard approach gives ˆ ˆ ˆ 2 f k u uR w V D w (14.95) ˆ ˆ 2 f k u R w F D V wD (14.96) ˆ S h d Nc w F D wD (14.97) The yield function for the stress on the side of the pile can be expressed as 0 h yNc F . Now the entire pile response can be obtained by integrating the energy and dissipation terms over the length of the pile: 11 22 000 ˆ ˆˆ 2 42 L kkL fuLRd uLdfd R KTDK KTDK K ³³³ (14.98) 11 00 ˆ ˆ 2 h dRLNcddd DK K ³³ (14.99) Application of the standard method results in 1 0 ˆ f HkLuLd u w KTDK w ³ (14.100) 1 2 0 ˆ f M kL u L d w KKTDK wT ³ (14.101) [...]... 2 J2 1 aij a ji 2 1 aii a jj 6 1 2 I2 I1 3 1 2 tr a 3 tr a 2 (B.9) 2 2 2 a1 a2 a3 a1 a2 a2 a3 a3 a1 a1 a2 1 aij a jk aki 3 2 a2 a3 1 aij a jk aki 3 1 1 tr a 3 tr a 2 tr a 3 3 1 2a1 a2 a3 2a2 a3 27 1 3 3 3 2 a1 a2 a3 27 2 3 a1 a2 a3 2 aij a ji akk 2 3 tr a 9 2 a2 a3 a1 a3 a1 I3 2 2 aii a jj akk 9 1 I 2 I1 3 2 3 I1 27 a1 2a3 a1 a2 2 a3 a1 a2 1 2a1 a 2a3 (B.10) ... orthogonal if its inverse is equal to its transpose aij a jk ik The tensor has principal values a1 , a2 , a3 , which are the eigenvalues of the matrix, and are the solutions of the cubic equation, a3 a2 I1 aI 2 I3 (B.1) 0 where I1 , I 2 , I3 are the invariants of the tensor, which are I1 I2 aii tr a (B.2) a1 a2 a3 1 1 tr a 2 aij a ji aii a jj 2 2 a1 a2 a2 a3 a3 a1 tr a 2 (B.3) 312 Appendix B Tensors 1 2aij a. .. aii a1 a2 a3 2 2 2 a1 a2 a3 aij a ji 3 3 3 a1 a2 a3 aij a jk aki (B.5) I1 2I 2 2 I1 (B.6) 3 3I 3 3I 2 I1 I1 (B.7) The deviator of a tensor is defined as follows: aij so that I1 tr a 1 akk 3 aij (B.8) ij 0 The second and third invariants of the deviator are also often required and may be written in a variety of forms: I2 1 2 1 3 1 6 I3 J3 1 aij a ji 2 J2 1 aij a ji 2 1 aii a jj 6 1 2 I2 I1 3 1 2 tr a. .. simply a clarification of the terminology used here A variable is a symbol used to represent an unspecified value of a set The value of the variable is one particular member of the set and the range is the set itself For example, the variable x might represent a real number In this case, the range is the set of real numbers from to , and in a particular instance, the value of x may be the number 3.81 A. .. jk aki 3aij a ji akk aii a jj akk 6 1 3 2tr a 3 3tr a 2 tr a tr a det a 6 I3 (B.4) a1 a 2a3 Note that some authors define I 2 with the opposite sign, but we prefer the notation used here; otherwise, J 2 (see below), which plays a major role in the analysis of shear behaviour, is always negative The traces of the powers may alternatively be chosen as defining the three invariants, tr a tr a 2 tr a 3 aii... functionals describing the system behaviour are rather more complex than we have encountered hitherto In particular, they include functions of differentials of continuous kinematic variables The application of hyperplasticity principles to modelling foundation behaviour is finding a number of applications Einav (2005) presents an analysis of the behaviour of piles which follows a rather similar approach... concept of the differential of a function can be extended to that of a functional by using either the Gateaux or Frechet differential, and these are developed as follows where each of the terms of the form 308 Appendix A Functions, Functionals and their Derivatives In the classical calculus of variations, the variation of a functional f u is defined in terms of a variation u of its argument function: f... can derive the entire constitutive response without recourse to any additional ad hoc assumptions This gives the structure of our theories an appealing simplicity, and also opens the door for development of theorems about the behaviour of such materials The approach also allows materials to be simply classified and put into hierarchies We make no extravagant claims of generality for our approach, and... we have attempted to present a unified approach to the constitutive modelling of dissipative materials, based on thermodynamic principles Although we have concentrated on rate-independent “plastic” materials, we have also considered a wider class of problems In the approach we term hyperplasticity, ” once two scalar functions are known (essentially defining the stored energy and the rate of dissipation),... over a3 3 3 a repeated index; thus, for instance, akk a1 1 a2 2 a3 3 and aij a jk aij a jk j 1 Alternatively, we may write aij a jk a 2 The unit tensor, or Kronecker is defined by ij 1, i j ; ij 0, i j A tensor is symmetric if it is equal to its transpose aij a ji and is antisymmetric, or skew-symmetric, if aij a ji The 1 inverse of a tensor or matrix is defined by aij a jk ik , and a tensor or matrix . one of a number of possible approaches. In particular, we chose to treat ˆ w as if it were an external kinematic variable. An alternative would be to treat ˆ w as an internal variable. 14.7 Analysis of Axial and Lateral Pile Capacity 299 Applying integration by parts (a) to the last term in the integral on the left- hand side of the equation and (b) twice to the last term. theory (and more par- ticularly hyperplasticity) is naturally expressed within the language of convex analysis. This approach has already been adopted by some authors (notably Han and Reddy,