Masonry Constructions: Mechanical Models and Numerical Applications (Lecture Notes in Applied and Computational Mechanics, 39) 9783540791102, 9783540791119, 3540791108

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Lecture Notes in Applied and Computational Mechanics Volume 39 Series Editors Prof. Dr.-Ing. Friedrich Pfeiffer Prof. Dr.-Ing. Peter Wriggers

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Masonry Constructions: Mechanical Models and Numerical Applications

Massimiliano Lucchesi Cristina Padovani Giuseppe Pasquinelli Nicola Zani •

•

With 78 Figures

•

Massimiliano Lucchesi Dipartimento di Costruzioni Università di Firenze Piazza Brunelleschi 6 50121 Firenze Italy [email protected]

Cristina Padovani Istituto di Scienza e Tecnologie dell’Informazione “A. Faedo” Consiglio Nazionale delle Ricerche Via G. Moruzzi 1 56124 Pisa Italy [email protected]

Giuseppe Pasquinelli Istituto di Scienza e Tecnologie dell’Informazione “A. Faedo” Consiglio Nazionale delle Ricerche Via G. Moruzzi 1 56124 Pisa Italy [email protected]

Nicola Zani Dipartimento di Costruzioni Università di Firenze Piazza Brunelleschi 6 50121 Firenze Italy [email protected]

ISBN

978-3-540-79110-2

ISSN

1613-7736

e-ISBN

978-3-540-79111-9

Library of Congress Control Number: 2007941933 © 2008 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other ways, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: WMX Design GmbH, Heidelberg Printed on acid-free paper 9 8 7 8 6 5 4 3 2 1 0 springer.com

To Anna, Tommaso, Lisa, Enrico and Orlando

Preface

Many historically and artistically important masonry buildings of the world’s architectural heritage are in dire need of maintenance and restoration. In order to optimize such operations in terms of cost-effectiveness, architectural impact and static effectiveness, accurate models of the structural behavior of masonry constructions are invaluable. The ultimate aim of such modeling is to obtain important information, such as the stress field, and to estimate the extent of cracking and its evolution when the structure is subjected to variations in both boundary and loading conditions. Although masonry has been used in building for centuries, it is only recently that constitutive models and calculation techniques have been available that enable realistic description of the static behavior of structures made of this heterogeneous material whose response to tension is fundamentally different from that to compression. Important insights on the mechanical behavior of masonry arches and vaults come from as far back as Leonardo [10], Hooke [58], Poleni [92] and many other authors (see [47], [9] and [10] for detailed references). Castigliano, in his famous paper on the Mosca bridge [23], and Signorini, in his studies on masonry beams [97], [98], showed both the possibility and necessity of taking into account the weak tensile strength of masonry material. Subsequently, the success attained in applying linear-elastic calculations to iron and reinforced concrete structures elicited so much enthusiasm that the models sometimes came to be employed, in a rather arbitrary fashion, to the study of masonry constructions as well. This not only hampered research efforts to further the work begun in 19th century by Castigliano on developing proper calculation techniques for masonry structures, but also contributed to breeding scepticism regarding the very possibility of modeling the static behavior of masonry buildings. For many years the fundamental studies on materials not withstanding tension remained without practical application, until the sixties, when they were resumed in order to determine the collapse load of relatively simple masonry structures [61], [54].

VIII

Preface

More recently, the availability of ever more powerful computers and sophisticated numerical techniques for solving nonlinear problems has reopened the debate over the possibility and advisability of modeling restoration work on historical building with the aim of evaluating its effects on the structures’ statics before its implementation. Many authors have proposed determining the collapse load for masonry structures through rigid blocks models with different kind of interfaces [62], [64], and have applied these models to the study of vaults as well [86]. Elasticplastic constitutive equations are widely adopted to model both the block units and mortar, as well as taking into account the anisotropy of masonry (see e.g. [65] and [66]). A comparison among elastic-plastic constitutive models adopting different yield criteria (Galileo-Rankine, Drucker-Prager, etc.) can be found in [45], and an application to the study of vaults is presented in [101]. Moreover, many proposed models use homogenization techniques to take the masonry’s texture into account, even if their application is generally limited to the study of panels [81], [82], [104]. A constitutive equation widely adopted to model the behavior of masonry materials views them as nonlinear elastic materials with zero tensile strength and infinite compressive strength [41], [42] and [38]. Many authors have contributed to a better understanding of this constitutive equation and to detailing its capabilities [97], [98], [54], [56], [90], [28], [37], [93], [94] and [95]. This equation, which is known as the masonry-like or no-tension model, can, at least in certain aspects, realistically describe the mechanical behavior of masonry [12], [13], [76]. Despite the relative simplicity of the constitutive equation of masonry-like materials, explicit solution of equilibrium problems of any practical interest is nonetheless very difficult [7], [8], [80]. Therefore, in order to study real problems, it is necessary to resort to numerical methods. To this end, suitable numerical techniques have been developed [68], [1], [96]. The constitutive model has been generalized in order to take into account both masonry’s bounded compressive strength and its weak tensile strength, and both the model and the associated numerical method studied have then been applied to the analysis of some buildings of historical and architectural interest [77], [75], [67], [78], [13], [12], [14], [74]. In this book we shall firstly provide a detailed description of the constitutive equation of the masonry-like materials, clearly setting out its most important features. In order to make the approach easier to follow, we shall first present the case of infinite compressive strength and zero tensile strength, and only afterwards address the modifications necessary to take into account the material’s bounded compressive strength and weak tensile strength. Lastly, the approach to masonry-like materials under non-isothermal conditions is dealt with. The main properties of the constitutive equations presented here are proved for the general anisotropic case, though the stress tensor as function of the strain tensor has been calculated for the isotropic case only. Since the numerical techniques for determining an approximated solution to the

Preface

IX

equilibrium problem make use of the tangent stiffness matrix, the derivative of the stress with respect to strain is calculated explicitly. These issues are taken up in Chapter 2 using some basic concepts of tensor algebra and analysis, which are summarized in Chapter 1. This chapter is devoted to some basic results on the gradient and divergence of vector and tensor fields, as well as some less standard questions such as differentiation of eigenvalues and eigenvectors of a symmetric tensor with respect to the tensor itself. Chapter 3 describes the equilibrium problem of solids made of a masonrylike material and presents a proof of the uniqueness of the solution in terms of stress. This result has been obtained in [4] and [46] under much more general hypotheses than those considered here. In fact, we preferred to formulate the results in such a way that their understanding did not call particular skills on functional analysis. Moreover, different variational formulations of the equilibrium problem are dealt with, and the equivalence between some of them proved. Chapter 4 briefly describes the numerical method used to solve the equilibrium problem and implemented in COMES-NOSA, a finite element code developed at the Istituto di Scienza e Tecnologie dell’Informazione ”A. Faedo” of the Italian National Research Council (CNR). Chapter 5 is devoted to masonry arches and vaults. Although the limit analysis of masonry structures is not dealt with in this book, some basic results are recalled, as they are used later to interpret the results of some numerical analyses. The maximum modulus eccentricity surface (m.m.e.s) is then defined for masonry vaults; in our opinion, this can play a role analogous to that of line of thrust for arches and thus allows concise, effective rendering of the results of the finite element analyses, as well an evaluation of the safety of vaults. In Chapter 6 we first present some simple equilibrium problems whose explicit solutions are known and compare the explicit solutions with the numerical ones obtained via the COMES-NOSA code. We then consider some different masonry structures. Although explicit solutions to the equilibrium problem are not available for such cases, we explicitly determine the value of the collapse load. Then, for the sake of comparison, the collapse load is calculated via numerical analysis by progressively increasing the load until convergence is not longer attainable. Finally, Chapter 7 presents some applications. The first three cases deal with some important monuments with the aim of modeling eventual strengthening operations and assessing their effects on their static behavior. The fourth example, of a completely different nature, is a study of a ladle employed in the iron and steel industry. The analysis aims to determine the behavior of the refractory material lining the metallic vessel, which must hold molten steel. Four appendices supplement the book. Some results concerning the constitutive equation of masonry-like materials in the two-dimensional case are listed in Appendix A and a brief description of the numerical method for the solution of the equilibrium problem in the non-isothermal case is given

X

Preface

in Appendix B. Appendix C contains some information about the finite element code COMES-NOSA and lastly, Appendix D, written by Stefano Secchi, presents a graphical tool which can be used to process both input and output data for COMES-NOSA. This work has been developed within the framework of Lagrange Laboratory, a European research group comprising CNRS, CNR, the Universities of ´ Rome ”Tor Vergata”, Calabria, Cassino, Pavia and Salerno, Ecole Polytechnique, University of Montpellier II, ENPC, LCPC and ENPTE. Research on masonry-like materials and implementation in the COMESNOSA code of the subroutines necessary to the analysis of masonry constructions have been made possible through the funding of the CNR (progetto finalizzato Beni Culturali, progetto COMES - Network for Computational Solid Mechanics) and of the Italian Ministry of the University and Research (Fondo Speciale per la Ricerca di Interesse Strategico ”Diagnostica e salvaguardia di manufatti architettonici con particolare riferimento agli effetti derivanti da eventi sismici e altre calamit`a naturali”). These financial supports are gratefully acknowledged. The COMES-NOSA code and graphical tool described in Appendix D may be freely downloaded, together with the associated user manuals, from the Website http://www.isti.cnr.it/comesnosa. We wish to thank Silvia Degl’Innocenti and Andrea Pagni for their support in creating the Website and compiling the software documentation and Anthony Cafazzo for the care he took to improve the quality of the book’s English.

Firenze and Pisa February, 2008

Massimiliano Lucchesi Cristina Padovani Giuseppe Pasquinelli Nicola Zani

Contents

1

Elements of Tensor Algebra and Analysis . . . . . . . . . . . . . . . . . . 1 1.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Finite-Dimensional Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Linear Complementarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4 Vectors and Second-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 Gradient and Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.6 Higher-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.7 Derivatives of Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . 15

2

The Constitutive Equations of Masonry-Like Materials . . . . 2.1 Masonry-Like Materials with Zero Tensile Strength and Infinite Compressive Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Isotropic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 The Two-Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Masonry-Like Materials with Small Tensile Strength and Bounded Compressive Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Masonry-Like Materials Under Non-Isothermal Conditions . . . . 2.4 The Derivative of the Stress Function . . . . . . . . . . . . . . . . . . . . . .

19 19 25 30 34 42 45

3

Equilibrium of Masonry Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.1 The Equilibrium Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.2 Variational Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4

The Numerical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Algorithm for Solution of the Equilibrium Problem . . . . . . . . . . 4.2 Fracture Strain Tensor and Cracked Regions . . . . . . . . . . . . . . . . 4.3 The Finite Element Code COMES-NOSA . . . . . . . . . . . . . . . . . .

59 59 63 66

5

Masonry Arches, Vaults and Domes . . . . . . . . . . . . . . . . . . . . . . . 5.1 Shells and Masonry Vaults . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Maximum Modulus Eccentricity Surface . . . . . . . . . . . . . . . . 5.3 The Limit Analysis of Masonry Arches and Vaults . . . . . . . . . . .

67 67 68 71

XII

Contents

6

Comparison Between Explicit and Numerical Solutions . . . . 6.1 The Circular Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Spherical Container . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Trapezoidal Panel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Mosca Bridge in Turin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The Circular Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 The Circular Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 The Spherical Dome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 76 79 82 89 92 95 98

7

Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 7.1 The Medici Arsenal in Pisa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 7.2 The Church of San Pietro in Vinculis in Pisa . . . . . . . . . . . . . . . . 110 7.3 The Dome of the Church of S. Maria Maddalena in Morano Calabro . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 7.4 The Ladle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 7.4.1 Analysis at the End of Stage CT2 . . . . . . . . . . . . . . . . . . . 126 7.4.2 Analysis at the End of Stage CT4 . . . . . . . . . . . . . . . . . . . 127

A

The Constitutive Equation of Masonry-Like Materials: the Two-Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 A.1 Masonry-Like Materials with Small Tensile Strength and Bounded Compressive Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 A.1.1 Plane Strain State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 A.1.2 Plane Stress State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 A.2 Masonry-Like Materials under Non-Isothermal Conditions . . . . 136 A.3 The Derivative of the Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

B

Algorithm for the Solution of the Equilibrium Problem: Non-Isothermal Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

C

Flow-Chart and Element Library of COMES-NOSA . . . . . . . 145

D

The GiD2Nosa Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

1 Elements of Tensor Algebra and Analysis

The reader is assumed to have some familiarity with the elements of linear algebra. Here we recall some results of tensor algebra and tensor analysis that will be used later on. For further details see e.g. [51], [52], [19] and [53].

1.1 Notations Italic boldface minuscules a, b, u, v,...: vectors and vector fields; x, y, z, ...: points of space. Italic boldface majuscules A, B,...: (second-order) tensors and tensor fields. Symbol Name A, C B b β C E E Ee Ef Ec E I I Lin Orth R s S

Fourth-order tensors Body Body force Thermal expansion Elasticity tensor Young’s modulus Infinitesimal strain tensor Elastic part of the strain Fracture strain Crushing strain Three-dimensional Euclidean space Second-order identity tensor Fourth-order identity tensor Space of second-order tensors Set of orthogonal tensors Set of real numbers surface force n−dimensional real vector space

2

1 Elements of Tensor Algebra and Analysis

Skw Sym Sym+ Sym− T  T  BC T  NI T u V λ μ σt σc ψ ϑ ϑ0 ⊗ ()· ()T ∇ div tr

Space of skew-symmetric tensors Space of symmetric tensors Set of positive semidefinite symmetric tensors Set of negative semidefinite symmetric tensors Cauchy stress tensor Stress function for masonry-like materials Stress function for masonry-like materials with bounded compressive strength Stress function for masonry-like materials under non-isothermal conditions Displacement vector Three-dimensional real vector space Lam´e modulus Lam´e modulus Tensile strength Compressive strength Strain energy density Absolute temperature Reference temperature Tensor product Time derivative Transpose Gradient Divergence Trace

1.2 Finite-Dimensional Vector Spaces Let S be a real vector space; the elements of S are called vectors. An inner product (or scalar product) on S is a function ( , ) defined on S × S with values in R such that 1. (a, b) = (b, a) for each a, b ∈ S (symmetry), 2. (α1 a1 + α2 a2 , b) = α1 (a1 , b) + α2 (a2 , b) for each a1 , a2 , b ∈ S, and α1 , α2 ∈ R (bilinearity), 3. (a, a) ≥ 0 for each a ∈ S and (a, a) = 0 if and only if a = 0 (positiveness).  For a ∈ S, a = (a, a) is the norm or length of a. Moreover, for a, b ∈ S, the Schwarz inequality |(a, b)| ≤ a b (1.1) and the parallelogram law

1.2 Finite-Dimensional Vector Spaces 2

2

2

a + b + a − b = 2 a + 2 b

2

3

(1.2)

both hold. Vectors a, b ∈ S are orthogonal if (a, b) = 0. Vectors u1 , ..., uk are orthonormal if  1, i = j, (1.3) (ui , uj ) = δij = 0, i = j. Orthonormal vectors are linearly independent, that is α1 u1 + ... + αk uk = 0 implies

α1 = ... = αk = 0.

(1.4)

Throughout the chapter we consider vector spaces S with an inner product and finite dimension n, and indicate by {e1 , ..., en } an orthonormal basis of S, i.e., a set of orthonormal vectors generating S. In other words, vectors e1 , ..., en satisfy (1.3), and moreover, for each u ∈ S, there exist unique n real numbers β1 , ..., βn such that u=

n 

βi ei .

(1.5)

i=1

A vector space S with inner product is a metric space with the distance d(u, v) = ||u − v||, for every u, v ∈ S.

(1.6)

A sequence of vectors {vk }k∈N is a Cauchy sequence if vm − vp  → 0, as m, p → ∞. S is a complete metric space if for every Cauchy sequence {vk }k∈N in S, there is a (unique) vector v ∈ S such that {vk }k∈N converges to v as k → ∞, that is, vk − v → 0, as k → ∞. As is well known, a real finitedimensional vector space with inner product is complete. Let A be a subset of S. A is closed if v ∈ A whenever {vk }k∈N converges to v, with {vk }k∈N ⊂ A; A is open if its complement S \ A is closed, and A is convex if u, v ∈ S implies that tu + (1 − t)v ∈ A, for every t ∈ [0, 1]. A is a cone with vertex u if {u + t(u − u) | t ≥ 0} ⊂ A, whenever u ∈ A. If A is a subset of S and u0 ∈ S, the distance of u0 from A is the number dist(u0 , A) = inf ||u0 − v||. v∈A

(1.7)

Theorem 1.1. (Minimum Norm Theorem) Let S be a finite-dimensional real vector space equipped with the inner product ( , ) and K ⊂ S a non-empty closed convex set. For each f ∈ S there exists a unique u ∈ K satisfying the following equivalent conditions (i) (1.8) f − u = min f − v = dist(f , K). v∈K

(ii) (f − u, v − u) ≤ 0

for each v ∈ K.

(1.9)

4

1 Elements of Tensor Algebra and Analysis

The variational inequality (1.9) is the Euler’s inequality for the minimum problem (1.8) [100]. We put u = PK f and we call u the projection of f onto the closed convex set K.

Fig. 1.1. Convex Set

Proof. First of all, we shall prove that there exists u ∈ K satisfying (1.8), then we shall prove the equivalence of (1.8) and (1.9) and lastly the uniqueness of u ∈ K that satisfies (1.9). If f ∈ K, then u = f ; if f ∈ / K, we put d = dist(f , K). Then, there exists a sequence {uk }k∈N ⊂ K such that dk = uk − f  → d, as k → ∞. {uk }k∈N is a Cauchy sequence; in fact, by applying the parallelogram law (1.2) with a = f − um , b = f − up , we arrive at ||um − up ||2 ≤ 2d2m + 2d2p − ||2f − um − up ||2 .

(1.10)

Taking the convexity of K into account, we have ||2f − um − up ||2 = 4||f −

um + up 2 || ≥ 4d2 , 2

(1.11)

and then, from (1.10) we obtain ||um − up ||2 ≤ 2d2m + 2d2p − 4d2 ,

(1.12)

||um − up || → 0,

(1.13)

from which we get as m, p → ∞.

Thus, um → u ∈ K as m → ∞ and d = u − f  . Now, we have to prove the equivalence of (1.8) and (1.9). Let u ∈ K satisfy (1.8), for each w ∈ K we have v = (1 − t)u + tw ∈ K

for t ∈ [0, 1]

(1.14)

and then, f − u ≤ f − (1 − t)u − tw = f − u − t(w − u) .

(1.15)

Consequently, for t ∈ (0, 1], 2

2

f − u ≤ f − u − 2t(f − u, w − u) + t2 w − u

2

(1.16)

1.2 Finite-Dimensional Vector Spaces

5

2

and 2(f − u, w − u) ≤ t w − u , which leads to (1.9) as t → 0. Vice versa, let u ∈ K satisfy (1.9); we have 2

u − f  − v − f  2

= 2(f − u, v − u) − u − v ≤ 0

2

for all v ∈ K,

(1.17)

from which (1.8) follows. In order to prove the uniqueness of u, let u1 ,u2 ∈ K satisfy (1.9). We have (f − u1 , v − u1 ) ≤ 0

for each v ∈ K,

(1.18)

(f − u2 , v − u2 ) ≤ 0

for each v ∈ K.

(1.19)

Putting v = u2 in (1.18) and v = u1 in (1.19), by addition we get ||u2 − u1 ||2 ≤ 0,

(1.20)

and then u1 = u2 .  Let us now introduce the concepts of continuity and differentiability and set forth some basic results. Let X , X1 , X2 and Y denote four finitedimensional normed vector spaces, and D and G open subsets of X and X1 , respectively. Let g : X → Y be a function between X and Y, g is linear if g(αu + βv) = αg(u) + βg(v), for each u, v ∈ X and for each α, β real numbers. Let f : D → Y be a function between D and Y. f is continuous at v ∈ D if {f (vk )}k∈N converges to f (v) whenever {vk }k∈N converges to v, with {vk }k∈N ⊂ D, as k → ∞. Function f is continuous in D if is continuous at every v ∈ D. Proposition 1.2. Function f is continuous in D if and only if for each closed (open) subset C of Y the inverse image of C f −1 (C) = {v ∈ X | f (v) ∈ C}

(1.21)

is closed (open) in D [40]. Let g be a function defined in a neighborhood of zero in X and have values in Y. We say that g(h) approaches zero faster than h, and write g(h) = o(h) as h → 0, if ||g(h)|| lim = 01 . (1.22) h=0, h→0 ||h|| Function f is said to be (Fr´echet) differentiable at v ∈ D if there exists a linear transformation Df (v) : X → Y such that 1

In other words, for each k > 0 there exists k > 0 such that ||g(h)|| < k||h|| whenever ||h|| < k .

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1 Elements of Tensor Algebra and Analysis

f (v + h) = f (v) + Df (v)[h] + o(h)

as h → 0.

(1.23)

We call Df (v) the derivative of f at v. If Df (v) exists, it is unique; in fact for each h, we have 1 d Df (v)[h] = lim (f (v + h) − f (v)) = f (v + h)|=0 . →o d

(1.24)

f is said to be differentiable in D if it is differentiable at each v ∈ D. Function f is of class C 1 , or smooth in D, if f is differentiable and Df is continuous in D; f is of class C 2 if f is of class C 1 and Df is of class C 1 . Proposition 1.3. Let f : D → Y be differentiable at v ∈ D; then f is continuous at v [40]. Let us consider the general bilinear product operation π : X1 × X2 → Y which assigns the product π(f0 , g0 ) ∈ Y to each f0 ∈ X1 and g0 ∈ X2 . Within this framework, the product p = π(f , g) of two functions f : D → X1 and g : D → X2 is the function p : D → Y defined by p(v) = π(f (v), g(v)),

for all v ∈ D.

(1.25)

For the proofs of the next two propositions, refer to [52]. Proposition 1.4. (product rule) Let f and g be differentiable at v ∈ D; then their product p = π(f , g) is differentiable at v and Dp(v)[h] = π(Df (v)[h], g(v)) + π(f (v), Dg(v)[h])

(1.26)

for all h ∈ X . Let f : D → X1 and g : G → Y, with the range of f contained in G. Proposition 1.5. (chain rule) If f is differentiable at v ∈ D and g is differentiable at y = f (v), then the composition c = g ◦ f is differentiable at v and Dc(v)[h] = Dg(f (v))[Df (v)[h]] (1.27) for all h ∈ X .

1.3 Linear Complementarity Now, let us we briefly recall the linear complementarity problem, which will be used in the next section. Let Rn be the n-dimensional Euclidean space whose elements are the n-tuples x = (x1 , ..., xn ) of real numbers, equipped with the inner product n  xi yi . (1.28) x·y = i=1

1.4 Vectors and Second-Order Tensors

7

Given x ∈ Rn , the inequality x ≥ 0 means that xi ≥ 0 for i = 1, ..., n. Rn×n denotes the space of all n × n real matrices A = [Aij ], i, j = 1, ..., n. Matrix A is positive definite if x · Ax > 0 for every x ∈ Rn , x = 0. Given q ∈ Rn and a matrix M ∈ Rn×n , the linear complementarity problem is to find z, w ∈ Rn such that ⎧ ⎨ w ≥ 0, z ≥ 0, w = q + M z, (1.29) ⎩ z · w = 0. For the proof of the next proposition, the reader is referred to [30]. Proposition 1.6. If matrix M is positive definite, then problem (1.29) has a unique solution for all q ∈ Rn .

1.4 Vectors and Second-Order Tensors Let E be a three-dimensional Euclidean space and V a three-dimensional real vector space associated with E and with inner product u · v. Let {e1 , e2 , e3 } be an orthonormal basis of V, so that the components of a vector u ∈ V are ui = u · ei , i = 1, 2, 3. Moreover, we have

3 3

  u·v = ui vi and u =  u2i . (1.30) i=1

i=1

Let Lin = {A : V → V | A(αu + βv) = αAu + βAv, for each α, β ∈ R, u, v ∈ V}

(1.31)

be the vector space of all linear transformations of V, which will be called (second-order) tensors. For u, v ∈ V the tensor product u ⊗ v is the element of Lin defined by the relation, (u ⊗ v)w = (w · v)u for each w ∈ V. For A ∈ Lin we have A=

3 

Aij ei ⊗ ej ,

(1.32)

i,j=1

with Aij = ei · Aej the components of A. The trace tr is the linear functional in Lin such that tr(u ⊗ v) = u · v, 3 3 Aij ei ⊗ ej , trA = Aii . The for every u, v ∈ V. Then, for each A = i,j=1

i=1

transpose AT of A is the unique tensor such that AT u · v = u · Av for all

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1 Elements of Tensor Algebra and Analysis

u, v ∈ V, and A ∈ Lin is symmetric if AT = A. Lin has the natural inner product (1.33) A · B = tr(ABT ), A, B ∈ Lin,

so that A = tr(AAT ). If A, B ∈ Lin, then the product AB ∈ Lin is defined by (AB)u = A(Bu) 3 for all u ∈ V, (AB)ij = Aik Bkj . The determinant of a tensor A, denoted k=1

by det A, is the determinant of the 3 × 3 real matrix of the components Aij of A . A is invertible if Au = 0 only for u = 0, and then A−1 , the inverse of A, is the unique tensor for which A−1 A = AA−1 = I, where I ∈ Lin is the identity tensor. It is an easy matter to prove that A is invertible if and only if det A = 0. Because the following relations  0, j = k, (1.34) (ei ⊗ ej )(ek ⊗ el ) = ei ⊗ el , j = k,  (ei ⊗ ej )·(ek ⊗ el ) = (ei · ek )(ej · el ) =

0, i = k, j = l, 1, i = k, j = l,

(1.35)

hold, in view of (1.32) and (1.35), {ei ⊗ ej }i,j=1,2,3 is an orthonormal basis of Lin. Moreover, for A ∈ Lin and u, v ∈ V, we have A(u ⊗ v) = (Au) ⊗ v,

(u ⊗ v)A = u ⊗ (AT v).

(1.36)

Now, Orth = {Q ∈ Lin | QQT = QT Q = I}

(1.37)

denotes the subset of Lin of all orthogonal tensors, Sym = {A ∈ Lin | A = AT }

(1.38)

the subspace of Lin of all symmetric tensors, and Skw = {W ∈ Lin | W = −WT }

(1.39)

the subspace of Lin of all skew-symmetric tensors. Each A ∈ Lin is uniquely decomposed into the sum of (A + AT )/2 ∈ Sym and (A − AT )/2 ∈ Skw. Therefore, since Sym∩Skw = {0}, Lin is the direct sum of Sym and Skw. A real number a is an eigenvalue of A if there exists u0 ∈ V, u0 = 0, such that Au0 = au0 , and each vector u = 0, such that Au = au, is an eigenvector of A corresponding to a. Moreover, for a an eigenvalue of A, M(a) = {u ∈ V | Au = au} is a subspace of V whose dimension is equal to the multiplicity of a. Lastly, a is an eigenvalue of A if and only if A − aI is not invertible, and therefore a is a root of the characteristic polynomial of A,

1.4 Vectors and Second-Order Tensors

p(a) = det(A − aI) = −a3 + I1 (A)a2 − I2 (A)a + I3 (A),

9

(1.40)

where I1 (A) = trA, I2 (A) =

(1.41)

1 [(trA)2 − tr(A2 )], 2

(1.42)

and I3 (A) = det A,

(1.43)

are the principal invariants of A. The eigenvalues of a tensor are also called principal components (of the tensor); correspondingly, the direction of an eigenvector is called the principal direction. Theorem 1.7. (Spectral theorem) Let A be a symmetric tensor. Then, there exist an orthonormal basis of V constituted by eigenvectors g1 , g2 , g3 of A and three eigenvalues a1 , a2 , a3 of A, Agi = ai gi , such that A=

3 

i = 1, 2, 3,

ai gi ⊗ gi .

(1.44)

(1.45)

i=1

Relation (1.45) is the spectral representation of A. In particular, if a1 = a2 , a2 = a3 , then A = a1 g1 ⊗ g1 + a2 (I − g1 ⊗ g1 ), and if a1 = a2 = a3 = a, then A = aI (for the proof see [53]). The set σ(A) of the ordered eigenvalues of A ∈ Sym (a1 ≤ a2 ≤ a3 ) is the spectrum of A. It should be noted that in view of this ordering, a1 , a2 and a3 can be univocally expressed as functions of the principal invariants (1.41), (1.42) and (1.43) (see [59] for their explicit expressions). For C = {x ∈ R3 | x1 ≤ x2 ≤ x3 }, it can be proved that the function σ : Sym → C, A → σ(A), which associates the corresponding spectrum to each tensor A, is continuous [99]. Tensors A and B ∈ Sym are said to be coaxial if they have the same eigenvectors and they are said to commute if AB = BA. Proposition 1.8. Tensors A, B ∈ Sym commute if and only if they are coaxial. Proof. Under the assumption that A and B are coaxial, let g1 , g2 , g3 be a common basis of eigenvectors A=

3  i=1

ai gi ⊗ gi ,

B=

3 

bi gi ⊗ gi .

(1.46)

i=1

Thus, in light of (1.34) we have AB = BA. Vice versa, if we suppose that AB = BA, we can distinguish the following cases.

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1 Elements of Tensor Algebra and Analysis

(i) If A = aI, coaxiality is evident. (ii) If A has three distinct eigenvalues a1 , a2 , a3 , let us consider its spectral representation 3  A= ai gi ⊗ gi , (1.47) i=1

and put B=

3 

Bij gi ⊗ gj .

(1.48)

ai Bij (gi ⊗ gj − gj ⊗ gi );

(1.49)

i,j=1

Then, we have 0 = AB − BA =

3  i,j=1 i=j

and since the skew-symmetric tensors gi ⊗gj −gj ⊗gi are linearly independent, from (1.49) we obtain (1.50) (a1 − a2 )B12 = 0, (a1 − a3 )B13 = 0,

(1.51)

(a2 − a3 )B23 = 0,

(1.52)

whence it follows that B12 = B13 = B23 = 0. Thus, g1 , g2 , g3 are also eigenvectors of B. (iii) Instead, if A has two distinct eigenvalues a1 = a2 = a3 , A = a1 g1 ⊗ g1 + a2 (I − g1 ⊗ g1 ),

B=

3 

Bij gi ⊗ gj ,

(1.53)

i,j=1

with g2 and g3 orthogonal to the subspace generated by g1 . By once again considering the condition 0 = AB − BA, we see that B12 = B13 = 0, and hence Bg1 = b1 g1 . Denoting by f2 and f3 the other two eigenvectors of B such that {g1 , f2 , f3 } is an orthonormal basis of V, we conclude that {g1 , f2 , f3 } is a basis of eigenvectors for both B and A.  We denote by Sym+ = {A ∈ Sym | v · Av ≥ 0 for each v ∈ V}

(1.54)

the subset of Sym made up of all positive semidefinite symmetric tensors and by Sym− = {A ∈ Sym | v · Av ≤ 0 for each v ∈ V} (1.55) the subset of Sym of negative semidefinite symmetric tensors. A ∈ Sym+ (Sym− ) if and only if its eigenvalues are non-negative (non-positive).

1.4 Vectors and Second-Order Tensors

11

Note that sets Sym+ and Sym− are closed convex cones of Sym. In fact, if A ∈ Sym+ (Sym− ), then tA ∈ Sym+ (Sym− ) for each t ≥ 0 and if A1 , A2 ∈ Sym+ (Sym− ), then tA1 + (1 − t)A2 ∈ Sym+ (Sym− ) for each t ∈ [0, 1]. Moreover, Sym+ (Sym− ) is closed because it is the inverse image of the closed set {x ∈ R3 | 0 ≤ x1 ≤ x2 ≤ x3 } ({x ∈ R3 | x1 ≤ x2 ≤ x3 ≤ 0}) by means of the continuous function associating the corresponding spectrum to each tensor. In the following proposition we collect some simple results. Proposition 1.9. 1. Let A ∈ Sym− (Sym+ ), if there exists u ∈ V such that u · Au = 0, then Au = 0. 2. Let A, B ∈ Sym. Then A · B ≥ 0 for each B ∈ Sym+ (Sym− ) implies that A ∈ Sym+ (Sym− ). 3. Let A ∈ Sym+ . If B ∈ Sym+ (Sym− ), then A · B ≥ 0 (≤ 0). 4. Let A ∈ Sym+ , B ∈ Sym+ (Sym− ). If A · B = 0, then AB = BA = 0. The following proposition is known as square-root theorem, its proof can be found in [52]. Proposition 1.10. For each C ∈ Sym+ , with det C = 0, there exists a unique invertible tensor U ∈ Sym+ , such that U2 = C.

(1.56)

Given A ∈ Sym, B ∈ Sym+ with det B = 0, we say that a real number a is a generalized eigenvalue of A if there exists u ∈ V, u = 0, such that Au = aBu;

(1.57)

u is a generalized eigenvector and (1.57) is called generalized eigenvalue problem [5]. Vectors u1 , u2 , u3 are B-orthonormal if ui · Buj = δij ,

(1.58)

with δij defined in (1.3). Theorem 1.11. If A ∈ Sym, B ∈ Sym+ with det B = 0, then the generalized eigenvalues a1 , a2 , a3 of generalized eigenvalue problem (1.57) are real (numbers), and there exist B-orthonormal generalized eigenvectors g1 , g2 , g3 which constitute a basis of V. Proof. In view of proposition 1.10, B = U2 , with U ∈ Sym+ and det U = 0, whence the generalized eigenvalue problem (1.57) becomes  = av, Av

(1.59)

 = U−1 AU−1 and v = Uu. From theorem 1.7 it follows that tenwhere A  has real eigenvalues a1 , a2 , a3 and that there exists an orthonormal sor A  Thus, a1 , a2 , a3 and basis of V constituted by eigenvectors v1 , v2 , v3 of A.

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1 Elements of Tensor Algebra and Analysis

u1 = U−1 v1 , u2 = U−1 v2 , u3 = U−1 v3 are respectively the eigenvalues and the corresponding eigenvectors of (1.57). Relations δij = vi · vj = Uui · Uuj = ui · U2 uj = ui · Buj ,

(1.60)

allow us to conclude that u1 , u2 , u3 are B-orthonormal.  For each vector u, u = 0 the ratio q(u) =

u · Au , u · Bu

(1.61)

is the Rayleigh quotient corresponding to the generalized eigenvalue problem (1.57) [5]. Proposition 1.12. Let a1 ≤ a2 ≤ a3

(1.62)

be the ordered generalized eigenvalues of problem (1.57). The Rayleigh quotient q defined in (1.61) satisfies the inequalities a1 ≤ q(u) ≤ a3 ,

for every u ∈ V, u = 0.

(1.63)

Proof. Let u1 , u2 , u3 be the B-orthonormal eigenvectors constituting a basis of V; for each u we have u = u1 u1 + u2 u2 + u3 u3 , and q(u) =

(1.64)

u21 a1 + u22 a2 + u23 a3 . u21 + u22 + u23

(1.65)

By taking (1.62) into account, (1.63) follows directly from (1.65).  A function f defined on Sym with values in R is convex if f ((1 − t)E1 + tE2 ) ≤ (1 − t)f (E1 ) + tf (E2 ),

(1.66)

for each t ∈ [0, 1], E1 , E2 ∈ Sym. In view of (1.23), f is differentiable in E1 ∈ Sym if f (E1 ) = f (E2 )+Df (E2 )·(E1 −E2 )+o(E1 −E2 ),

as E1 −E2 → 0. (1.67)

Let us now add the following proposition ([24]). Proposition 1.13. Let f :Sym→ R be differentiable in Sym; f is convex if and only if f (E1 ) ≥ f (E2 ) + Df (E2 )·(E1 − E2 ), (1.68) for all E1 , E2 ∈ Sym.

1.5 Gradient and Divergence

13

A function A :Sym→Sym is monotone if (A(E1 ) − A(E2 ))·(E1 − E2 ) ≥ 0,

for each E1 , E2 ∈ Sym.

(1.69)

The following proposition links convex functions and monotone functions. Proposition 1.14. Let f :Sym→ R be differentiable and convex in Sym. Then Df is monotone, (Df (E1 ) − Df (E2 ))·(E1 − E2 ) ≥ 0,

for each E1 , E2 ∈ Sym.

(1.70)

Proof. From proposition 1.13, it follows that inequalities (1.68) and f (E2 ) ≥ f (E1 ) + Df (E1 )·(E2 − E1 )

(1.71)

hold, and their sum yields (1.70). 

1.5 Gradient and Divergence Let D be an open set in the Euclidean space E. A function on D is called scalar, vector, or tensor field when its values are respectively scalars, vectors, or tensors. If v is a vector field on D of class C 1 , then Dv(x) is a linear transformation from V into V and hence a second order tensor. We use the notation ∇v(x) for Dv(x) and write ∇v(x)h = Dv(x)[h],

h ∈ V.

(1.72)

The tensor ∇v(x) is the gradient of v at x. Given a vector field v on D of class C 1 , the scalar field div v = tr∇v,

(1.73)

is called the divergence of v. For S, a tensor field of class C 1 , divS is the unique vector field such that (divS) · a = div(ST a)

(1.74)

for every vector a. Proposition 1.15. Let v and S be fields of class C 1 . Then [52] div(ST v) = S · ∇v + v · divS.

(1.75)

R is said to be a regular region if R is a connected closed set of the Euclidean space E whose boundary ∂R is piece-wise C 1 [52], [51], and we denote by n the outward unit normal field on ∂R.

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1 Elements of Tensor Algebra and Analysis

Proposition 1.16. (Divergence theorem). Let R be a bounded regular region, and let v : R → V and S : R →Lin be fields of class C 1 . Then [52]   divv dv = v · n da, (1.76) R

∂R



 divS dv =

R

Sn da.

(1.77)

∂R

Proposition 1.17. Let R be a bounded regular region and let v : R → V and S : R →Lin be fields of class C 1 . Then    u · divS dv + ∇u · S dv = Sn · u da. (1.78) R

R

∂R

Proof. The result follows from the divergence theorem and (1.75). 

1.6 Higher-Order Tensors Throughout this book a third-order tensor F is a linear transformation from Lin into V. In particular, given u, v, w ∈ V, u ⊗ v ⊗ w denotes the third-order tensor defined by u ⊗ v ⊗ w[H] = (v ⊗ w · H)u,

H ∈ Lin.

(1.79)

A fourth-order tensor A is instead a linear transformation from Lin into itself. We denote by Lin the space of all fourth-order tensors and I as the unit tensor defined by I[H] = H for every H ∈ Lin. The tensor product A ⊗ B of two (second-order) tensors A and B is the fourth-order tensor defined by A ⊗ B[H] = (B · H)A,

H ∈ Lin.

(1.80)

The components of A with respect to the orthonormal basis {ei ⊗ ej ⊗ ek ⊗ el }i,j,k,l=1,2,3 of Lin are Aijkl = (ei ⊗ ej )·A[ek ⊗ el ]. The major transposition of a fourth-order tensor A is the fourth-order tensor AT satisfying AT [H]·K = A[K]·H, for each H, K ∈ Lin. A is said to have a major symmetry if AT = A. In indices this means that Aijkl = Aklij . The symmetry in the first pair of indices (Aijkl = Ajikl ) means that A takes symmetric values, A[H]T = A[H],

H ∈ Lin,

(1.81)

while the symmetry in the second pair of indices (Aijkl = Aijlk ) means that A vanishes on Skw, A[W] = 0, W ∈ Skw, (1.82) or equivalently,

1.7 Derivatives of Eigenvalues and Eigenvectors

A[HT ] = A[H],

H ∈ Lin.

15

(1.83)

Relations (1.81) and (1.83) express the so-called minor symmetries of the fourth-order tensor A. Let G be a subset of all orthogonal tensors Q (see (1.37)). The fourth-order tensor A :Sym→Sym is invariant under G if A[QEQT ] = QA[E]QT , for each E ∈ Sym, Q ∈ G;

(1.84)

A is said to be isotropic if it is invariant under Orth, A[QEQT ] = QA[E]QT , for each E ∈ Sym, Q ∈ Orth.

(1.85)

The next proposition provides a representation formula for the isotropic fourth-order tensors [52]. Proposition 1.18. The fourth-order tensor A :Sym→Sym is isotropic if and only if there exist two real numbers μ and λ such that A[E] = 2μE + λ (trE)I, for every E ∈ Sym.

(1.86)

Corollary 1.19. If A is isotropic, then EA[E] = A[E]E,

for each E ∈ Sym.

(1.87)

Moreover, given E, F ∈Sym, FA[E] = A[E]F

is and only if

FE = EF.

(1.88)

1.7 Derivatives of Eigenvalues and Eigenvectors The derivatives of the eigenvalues and eigenvectors of a symmetric tensor play a determining role in formulating the numerical method presented in Chapter 4. The main results are summed up by proposition 1.20. Let Sym* stand for the subset of Sym of all symmetric tensors having distinct eigenvalues. Given A ∈ Sym*, let a1 < a2 < a3 be its eigenvalues, and g1 , g2 , g3 a triad of corresponding normalized eigenvectors. For the sake of convenience, we put G11 = g1 ⊗ g1 , G22 = g2 ⊗ g2 , G33 = g3 ⊗ g3 , 1 1 G12 = √ (g1 ⊗ g2 + g2 ⊗ g1 ), G13 = √ (g1 ⊗ g3 + g3 ⊗ g1 ), 2 2 1 G23 = √ (g2 ⊗ g3 + g3 ⊗ g2 ). 2

(1.89) (1.90) (1.91)

For i = 1, 2, 3, function ai : Sym*→ R, which associates to each A the eigenvalue ai (A), and Gii : Sym*→Sym, given in (1.89), are uniquely determined

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1 Elements of Tensor Algebra and Analysis

by virtue of the fact that the eigenvalues of A are distinct. Moreover, tensors Gij (i = j) are determined, except for their sign, and Gij ⊗ Gij are therefore uniquely determined. We denote by DA ai and DA Gii the derivatives of ai and Gii with respect to A. Proposition 1.20. Under the foregoing hypotheses, the following relations hold DA a1 = G11 , (1.92) DA a2 = G22 ,

(1.93)

DA a3 = G33 ,

(1.94)

1 1 G12 ⊗ G12 + G13 ⊗ G13 , a1 − a2 a1 − a3 1 1 = G12 ⊗ G12 + G23 ⊗ G23 , a2 − a1 a2 − a3 1 1 = G13 ⊗ G13 + G23 ⊗ G23 . a3 − a1 a3 − a2

DA G11 =

(1.95)

DA G22

(1.96)

DA G33

(1.97)

Proof. Here we limit ourselves to proving equations (1.92) and (1.95), as the other relations follow through analogous proofs. For A ∈ Sym* and H ∈ Sym fixed, let us consider ∈ R; let a1 ( ) and g1 ( ) be the smallest eigenvalue and the corresponding eigenvector of A + H, such that g1 ( ) · g1 > 0, (A + H)g1 ( ) = a1 ( )g1 ( )

(1.98)

Within and error of order o( ), we can put ·

·

a1 ( ) = a1 + a1 (0) and g1 ( ) = g1 + g1 (0),

(1.99)

where a1 = a1 (0), g1 = g1 (0) and the superimposed dot denotes differentiation with respect to . By substituting (1.99) in (1.98) we obtain ·

·

·

Ag1 (0) + Hg1 = a1 (0)g1 + a1 g1 (0).

(1.100)

·

Since g1 ( ) · g1 ( ) = 1, then g1 (0) · g1 = 0; thus, if we multiply (1.100) by g1 , we get · a1 (0) = g1 · Hg1 = g1 ⊗ g1 ·H. (1.101) Because of (1.24), for every H ∈ Sym we can write ·

a1 (0) =

d a1 (A + H)|=0 = DA a1 (A)·H, d

(1.102)

and then, by virtue of (1.101), we obtain (1.92). In order to calculate the derivative of G1 , we must calculate the derivative of g1 . To this end, by substituting (1.101) into (1.100), we obtain

1.7 Derivatives of Eigenvalues and Eigenvectors ·

·

Ag1 (0) + Hg1 = (g1 ⊗ g1 ·H)g1 + a1 g1 (0).

17

(1.103)

·

Since g1 and g1 (0) are orthogonal, we can write ·

g1 (0) = χg2 + ξg3 ,

(1.104)

where χ and ξ are real numbers which depend on A. By substituting (1.104) into (1.103), the relation χ(a2 − a1 )g2 + ξ(a3 − a1 )g3 = (g1 ⊗ g1 ·H)g1 − Hg1

(1.105)

follows. Multiplying (1.105) by g2 and g3 , we obtain respectively χ=

1 g1 ⊗ g2 ·H, a1 − a2

(1.106)

ξ=

1 g1 ⊗ g3 ·H. a1 − a3

(1.107)

Thus, from (1.104), (1.106) and (1.107), by virtue of the symmetry of H, and recalling (1.24), we have ·

g1 (0) =

d g1 (A + H)|=0 = DA g1 (A)[H] d

=

1 (g2 ⊗ g1 ⊗ g2 + g2 ⊗ g2 ⊗ g1 )[H] 2(a1 − a2 )

+

1 (g3 ⊗ g1 ⊗ g3 + g3 ⊗ g3 ⊗ g1 )[H]. 2(a1 − a3 )

(1.108)

In view of (1.79), the desired result follows from the product rule DA G11 [H] = DA g1 [H] ⊗ g1 + g1 ⊗ DA g1 [H]. 

(1.109)

2 The Constitutive Equations of Masonry-Like Materials

The purpose of this section is to characterize the mechanical behavior of so-called masonry-like or no-tension materials. In particular, we deal with a class of nonlinear elastic materials that are incapable of withstanding tensile stresses and behave like a linear elastic material when subjected to compressive stresses. To this end, we begin in section 2.1 by assuming that the stress must be negative semidefinite and that the strain is the sum of two parts: the former depends linearly on the stress, the latter is orthogonal to the stress and positive semidefinite. The material is moreover allowed to extend freely in directions of zero stress. Later on this constitutive equation will be generalized in order to take the material’s bounded compressive strength into account.

2.1 Masonry-Like Materials with Zero Tensile Strength and Infinite Compressive Strength We shall denote by E = (∇u + ∇uT )/2 ∈ Sym the infinitesimal strain tensor, with u the displacement field, and T ∈ Sym the Cauchy stress tensor. Definition 2.1. A masonry-like material is an elastic material whose stress  : Sym→Sym, function T  T = T(E),

E ∈ Sym

satisfies the following properties ⎧ T ∈ Sym− , ⎪ ⎪ ⎪ ⎪ ⎨ E = Ee + Ef , Ef ∈ Sym+ , ⎪ ⎪ T = C[Ee ], ⎪ ⎪ ⎩ f E ·T = 0.

(2.1)

(2.2)

20

2 The Constitutive Equations of Masonry-Like Materials

Ee and Ef are respectively the elastic and fracture1 parts of the strain. As in the linear theory of elasticity, we call C the elasticity tensor, which is assumed to have both major and minor symmetry (cf. (1.81), (1.83)), and to be positive definite, in the sense that A · C[A] > 0 for each A ∈ Sym, A = 0.

(2.3)

With the aim of proving the existence and uniqueness of the solution to the constitutive equation (2.2), it is convenient to characterize it further. For every E ∈ Sym, let us consider the problem of finding T ∈ Sym− such that the variational inequality (T − T∗ )·(E − C−1 [T]) ≥ 0 for each T∗ ∈ Sym−

(2.4)

holds. Proposition 2.2. For every E ∈ Sym, the triplet (T, Ee , Ef ) of elements of Sym is a solution to the constitutive equation (2.2) if and only if (T, Ee , Ef ) satisfies (2.2)2 , (2.2)4 , and T belongs to Sym− and is a solution to the variational inequality (2.4). Proof. For Ef = E − C−1 [T], let us prove that the inequality (T − T∗ )·Ef ≥ 0 for each T∗ ∈ Sym−

(2.5)

is equivalent to the conditions Ef ∈ Sym+ ,

Ef ·T = 0.

(2.6)

If conditions (2.6) hold, then, in view of proposition 1.9(3), (T − T∗ )·Ef = −T∗ ·Ef ≥ 0.

(2.7)

Vice versa, if (2.5) is satisfied, for T∗ = 0, we have T · Ef ≥ 0. Moreover, for T∗ = 2T we have T · Ef ≤ 0, and (2.6)2 is proved. Thus, (2.5) reduces to the inequality (2.8) T∗ ·Ef ≤ 0 for each T∗ ∈ Sym− , from which, with the help of proposition 1.9(2), (2.6)1 follows.  Proposition 2.3. For every E ∈ Sym, there exits a unique T ∈ Sym− satisfying the variational inequality (2.4). Proof. To prove the proposition, first of all let us show that (2.4) characterizes T as the C−1 -orthogonal projection of C[E] onto the closed convex set Sym− . To this end, consider the inner product on Sym defined by 1

In [68] and [70] Ef was called inelastic part of the strain tensor E and was denoted by Ea .

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

(A, B)C−1 = C−1 [A]·B.

21

(2.9)

In virtue of the minimum norm theorem 1.1, given C[E] ∈ Sym, there is a unique T ∈ Sym− such that (T − T∗ , C[E] − T)C−1 ≥ 0 for each T∗ ∈ Sym− .

(2.10)

From the definition of the inner product (A, B)C−1 , we obtain (T−T∗ , C[E]− T)C−1 = (T − T∗ )·(E − C−1 [T]), from which the thesis follows directly.  From the two foregoing propositions, it follows that the solution to constitutive equation (2.2) both exists and is unique. Proposition 2.4. For each E ∈ Sym, there exists a unique triplet (T, Ee , Ef ) of tensors of Sym that satisfies (2.2). The constitutive equation (2.2) and, equivalently (2.4), are the formalization of a natural and intuitive characterization of masonry-like materials. Tensor C[E], which for a linear elastic material coincides with the stress tensor associated to strain E, is not generally negative semidefinite. Consequently, it is not an admissible stress for a masonry-like material. We therefore consider the ”closest” negative semidefinite tensor to C[E], in a suitable norm, and assume it to be the stress associated to E. More precisely, from propositions 2.3 and 2.2 and the minimum norm theorem 1.1, it follows that, given  E ∈ Sym, the stress T = T(E) associated to E is the tensor of Sym− having the minimum distance from C[E], with respect to the norm induced by the inner product ( , )C−1 defined in (2.9). That is to say, T is the projection PSym− (C[E]) of C[E] onto Sym− with respect to the inner product ( , )C−1 . This characterizes T as the unique minimum point of the functional J(T∗ ) = C−1 [C[E]−T∗ ]·(C[E]−T∗ ) defined on Sym− , whose Euler’s inequality is right (2.4). It should be noted that, in view of inequality (2.5), fracture strain Ef is an element of the normal cone to Sym− at T, N (T) = {A ∈ Sym | (T − T∗ )·A ≥ 0 for each T∗ ∈ Sym− }.

(2.11)

Since a masonry-like material is characterized by the unilateral constraint on the stress which forces the stress T to be negative semidefinite, the fracture strain can be interpreted as the reactive part of the strain, satisfying the normality assumption (2.5), as prescribed by the theory of constrained materials [38]. In particular, when T belongs to the interior of Sym− , then the normal cone reduces to the null element, N (T) = {0}, and accordingly we have Ef = 0.

22

2 The Constitutive Equations of Masonry-Like Materials

Fig. 2.1. The normal cone to Sym− at T.

Proposition 2.5. Tensors T and Ef satisfying (2.2) are coaxial. Proof. Since T ∈ Sym− , Ef ∈ Sym+ and Ef ·T = 0, the result follows from propositions 1.8 and 1.9.  In particular, in view of proposition 1.9, we have Ef T = TEf = 0.

(2.12)

In addition, note that, since T ∈ Sym− , if Tn · n = 0, then in view of proposition 1.9, Tn = 0, that is, n is an eigenvector of T corresponding to the null eigenvalue. The following proposition sums up some properties of the nonlinear stress  defined by the constitutive equation (2.2), or equivalently, by the function T variational inequality (2.4) [38].  exhibits the following properties: Proposition 2.6. Stress function T  is not injective and therefore not invertible; indeed, (i) T E ∈ Sym+ Moreover

C[E] ∈ Sym−

 if and only if T(E) = 0. if and only if

 T(E) = C[E].

(2.13) (2.14)

 is positively homogeneous of degree one, (ii) T   T(βE) = β T(E),

for each β ≥ 0, E ∈ Sym.

(2.15)

 is monotone, (iii) T  2 ))·(E1 − E2 ) ≥ κ ||T(E  1 ) − T(E  2 )||2 ,  1 ) − T(E (T(E for each E1 , E2 ∈ Sym, with

(2.16)

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

κ=

inf

A∈ Sym, ||A||=1

A·C−1 [A],

23

(2.17)

 is Lipschitz continuous, (iv) T  1 ) − T(E  2 )|| ≤ κ−1 ||E1 − E2 ||, ||T(E

(2.18)

for each E1 , E2 ∈ Sym. Proof. (i): (2.13) and (2.14) are immediate consequences of (2.2). (ii): From the variational inequality (2.4), by taking into account that Sym− is a cone, and putting T∗∗ = βT∗ , it follows that   ≥ 0 for each T∗∗ ∈ Sym− . (β T(E) − T∗∗ )·(βE − C−1 [β T(E)])

(2.19)

 inequality On the other hand, from the definition of T,   ≥ 0 for each T∗∗ ∈ Sym− (T(βE) − T∗ )·(βE − C−1 [T(βE)])

(2.20)

follows. By comparing (2.19) and (2.20), and bearing in mind the uniqueness of the solution to (2.4), we get (2.15).  1 ), T2 = T(E  2 ). From (iii): In order to prove (2.16), let us put T1 = T(E (2.4) we obtain the following inequalities (T1 − T2 )·(E1 − C−1 [T1 ]) ≥ 0

(2.21)

(T2 − T1 )·(E2 − C−1 [T2 ]) ≥ 0

(2.22)

(T1 − T2 )·(E1 − E2 ) ≥ (T1 − T2 )·C−1 [T1 − T2 ].

(2.23)

which summed, yield

Then, from (2.23) and (2.17), we get (2.16). (iv): From the property (2.16), by using the Schwarz inequality (1.1), we obtain  2 )||2 ≤ (T(E  1 ) − T(E  2 ))·(E1 − E2 )  1 ) − T(E κ ||T(E  2 )|| ||E1 − E2 ||,  1 ) − T(E ≤ ||T(E

(2.24)

whence, we directly arrive at (2.18).  Now, we intend to verify that the material defined by the constitutive equation (2.2) is hyperelastic, that is, there exists a differentiable function ψ, called strain energy density, defined on Sym with values in R such that its  derivative with respect to E coincides with T(E). In fact, we shall prove that the function 1 ψ(E) = T(E) · E, for each E ∈ Sym, (2.25) 2 is a strain energy density. Let us start by proving the following preliminary result [38].

24

2 The Constitutive Equations of Masonry-Like Materials

Proposition 2.7. Let ψ be the function defined in (2.25). The following inequalities hold for any E1 , E2 ∈ Sym 1   2 )||2 ≤ ψ(E1 ) − ψ(E2 ) − T(E  2 ) · (E1 − E2 ) κ||T(E1 ) − T(E 2 ≤

1 −1 κ ||E1 − E2 ||2 , 2

(2.26)

where κ is given in (2.17).  1 ), T2 = T(E  2 ), from (2.25) the following identities Proof. For T1 = T(E hold ψ(E1 ) − ψ(E2 ) = T2 ·(E1 − E2 ) − T2 ·E1 1 1 + T1 ·E1 + T2 ·E2 = T2 ·(E1 − E2 ) − T2 ·(C−1 [T1 ] + Ef1 ) 2 2 1 1 + T1 ·C−1 [T1 ] + T2 ·C−1 [T2 ] 2 2 1 = T2 ·(E1 − E2 ) + (T1 − T2 )·C−1 [T1 − T2 ] − T2 ·Ef1 . 2

(2.27)

Since T2 ·Ef1 ≤ 0, then ψ(E1 ) − ψ(E2 ) − T2 ·(E1 − E2 ) ≥

1 (T1 − T2 )·C−1 [T1 − T2 ]. 2

(2.28)

The former inequality in (2.26) comes from the definition of κ given in (2.17). In order to prove the latter inequality, we only have to interchange E1 and E2 in the former inequality, rewrite it in the form ψ(E1 ) − ψ(E2 ) − T2 ·(E1 − E2 ) 1 ≤ (T1 − T2 )·(E1 − E2 ) − κ||T1 − T2 ||2 2 and use the inequality (T1 − T2 )·(E1 − E2 ) ≤

(2.29)

1 κ||T1 − T2 ||2 2

1 + κ−1 ||E1 − E2 ||2 . (2.30) 2 The algebraic relation (2.30) is proved by the following chain of inequalities 1 1 κ||T1 − T2 ||2 + κ−1 ||E1 − E2 ||2 2 2 =

1 −1 2 κ (κ||T1 − T2 || − ||E1 − E2 ||) + ||T1 − T2 || ||E1 − E2 || 2 ≥ ||T1 − T2 || ||E1 − E2 || ≥ (T1 − T2 )·(E1 − E2 ).  (2.31)

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

25

We are now in a position to prove the desired result. Proposition 2.8. Function ψ defined in (2.25) is continuously differentiable, convex and  DE ψ(E) = T(E). (2.32) Proof. From (2.26) with E1 = E and E2 = E + H, we deduce that  ψ(E + H) − ψ(E) − T(E) · H = o(H),

as H → 0,

(2.33)

thus proving (2.32). From the first inequality in (2.26), by taking (2.32) into account, we obtain ψ(E1 ) − ψ(E2 ) − DE ψ(E2 ) · (E1 − E2 ) ≥ 0,

(2.34)

for each E1 , E2 ∈ Sym. By virtue of proposition 1.13, this allows concluding that function ψ is convex.   is monotone, We point out that from proposition 1.14, it follows that T as already proved in proposition 2.6. 2.1.1 Isotropic Materials This subsection deals with materials for which the elasticity tensor C is isotropic (cf. (1.85)). As we shall see, for this class of materials the solution to the constitutive equation (2.2) can be calculated explicitly.  is said to be isotropic if, for every E ∈ We recall that stress function T Sym and Q ∈ Orth, it holds that T T   ) = QT(E)Q . T(QEQ

(2.35)

 is isotropic then the following representation Moreover, if T  T = T(E) = β0 I + β1 E + β2 E2 ,

(2.36)

holds, where coefficients βi are functions of the principal invariants of E [52]. Proposition 2.9. If the elasticity tensor C in (2.2) is isotropic, then stress  is isotropic. function T Proof. Taking into account that Sym− is invariant under Orth, i.e., if A ∈ Sym− , then QAQT ∈ Sym− , for each Q ∈ Orth, and that the trace is isotropic, i.e., tr(QAQT ) = trA,

for each Q ∈ Orth, A ∈ Sym,

from (2.4), in view of the isotropy of C, we obtain

(2.37)

26

2 The Constitutive Equations of Masonry-Like Materials T T   (QT(E)Q − T∗∗ )·(QEQT − C−1 [QT(E)Q ]) ≥ 0

(2.38)

T  ) is defined by the infor each T∗∗ ∈ Sym− . On the other hand T(QEQ equality T T   ) − T∗ )·(QEQT − C−1 [T(QEQ )]) ≥ 0 (2.39) (T(QEQ

for each T∗ ∈ Sym− . From the uniqueness of the solution to (2.4), by comparing (2.38) and (2.39), (2.35) follows.  We say that a masonry-like material with constitutive equation (2.2) is isotropic, if its elasticity tensor C is isotropic. In this case, as for linear elastic materials, we call the two real numbers λ and μ defined in (1.86), the Lam´e moduli. In view of condition (2.3), the inequalities μ > 0,

2μ + 3λ > 0

(2.40)

hold and we have C−1 [A] =

λ 1 A− (trA)I. 2μ 2μ(2μ + 3λ)

(2.41)

Proposition 2.10. If C is isotropic, then E, T, Ee and Ef are coaxial. Proof. T and Ef are coaxial in view of proposition 2.5. Since C is isotropic, from (2.41) it follows that C−1 is isotropic as well. Moreover, in view of the corollary 1.19, for A ∈ Sym, AC−1 [A] = C−1 [A]A. From both the isotropy of C−1 and the coaxiality of T and Ef , it follows that ET = Ef T + Ee T = TEf + C−1 [T]T = TEf + TC−1 [T] = TE;

(2.42)

EEf = Ee Ef + (Ef )2 = C−1 [T]Ef + (Ef )2 = Ef C−1 [T] + (Ef )2 = Ef E.

(2.43)

Analogously, it is a simple matter to prove that Ee E = EEe ,

Ee Ef = Ef Ee

(2.44)

and then, the thesis is a consequence of proposition 1.8.  Because C is isotropic, in view of the preceding proposition, system (2.2) can be rewritten as a linear complementarity problem, thereby allowing its solution to be calculated explicitly. Let {q1 , q2 , q3 } be orthonormal eigenvectors of E, which, in view of proposition 2.10, are eigenvectors of T and Ef as well. Moreover, let (e1 , e2 , e3 ), (ef1 , ef2 , ef3 ), and (t1 , t2 , t3 ) be the eigenvalues of E, Ef and T, respectively. Then, by virtue of (1.86), system (2.2) is equivalent to

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

⎧ ⎪ t1 = μ[2(e1 − ef1 ) + α(e1 + e2 + e3 − ef1 − ef2 − ef3 )] ⎪ ⎪ ⎪ ⎪ t2 = μ[2(e2 − ef2 ) + α(e1 + e2 + e3 − ef1 − ef2 − ef3 )] ⎪ ⎪ ⎨ t3 = μ[2(e3 − ef3 ) + α(e1 + e2 + e3 − ef1 − ef2 − ef3 )] ⎪ t1 ≤ 0, t2 ≤ 0, t3 ≤ 0 ⎪ ⎪ ⎪ ⎪ ef1 ≥ 0, ef2 ≥ 0, ef3 ≥ 0 ⎪ ⎪ ⎩ f t1 e1 = t2 ef2 = t3 ef3 = 0

27

(2.45)

with α = λ/μ (in the following we assume λ ≥ 0, so that we have α ≥ 0). Putting e = (e1 , e2 , e3 ), ef = (ef1 , ef2 , ef3 ) and t = (t1 , t2 , t3 ), (2.45) is a linear complementarity problem of the type (1.29), where M = D, with

q = −De,

w = −t,

z = ef ,

(2.46)

⎡

⎤ 2+α α α ⎦. 2+α α D = μ⎣α α α 2+α

(2.47)

Since by virtue of (2.40), D is positive definite, proposition 1.6 guarantees that given e, there exist unique ef and t satisfying (2.45). This result provides an alternative proof of the existence and uniqueness of the solution to the constitutive equation (2.2), for isotropic C. Now let us proceed to explicitly calculating the eigenvalues (ef1 , ef2 , ef3 ), and (t1 , t2 , t3 ) as functions of (e1 , e2 , e3 ). To this aim, we suppose that the eigenvalues of E are ordered in such a way that e1 ≤ e2 ≤ e3 .

(2.48)

In view of (2.48), the solution of (2.45) takes one of the following forms t < 0,

ef = 0,

(2.49)

t = 0,

ef > 0,

(2.50)

t1 < 0, t2 = t3 = 0, t1 < 0, t2 < 0, t3 = 0,

ef1 = 0, ef2 > 0, ef3 > 0, ef1

=

ef2

= 0, ,

ef3

> 0,

(2.51) (2.52)

depending on the values of e1 , e2 and e3 . In conformity with (2.49)-(2.52), it is natural to define the following subsets of Sym, R1 = {E ∈ Sym | 2e3 + α trE < 0},

(2.53)

R2 = {E ∈ Sym | e1 > 0},

(2.54)

R3 = {E ∈ Sym | e1 < 0, αe1 + 2(1 + α)e2 > 0},

(2.55)

R4 = {E ∈ Sym | αe1 + 2(1 + α)e2 < 0, 2e3 + α trE > 0},

(2.56)

28

2 The Constitutive Equations of Masonry-Like Materials

where trE = e1 + e2 + e3 . Moreover, we define the following interfaces between regions R1 , R4 ; R2 , R3 and R3 , R4 , respectively, as R I1,4 = {E ∈ Sym | g14 (E) = 2e3 + α trE = 0},

(2.57)

R = {E ∈ Sym | g23 (E) = e1 = 0}, I2,3

(2.58)

R I3,4 = {E ∈ Sym | g34 (E) = αe1 + 2(1 + α)e2 = 0}.

(2.59)

Regions Ri characterize the different types of behavior exhibited by the material. In R1 the material is compressed in all directions and behaves like a linear elastic material. On the contrary, in R2 it is subjected to a positive semidefinite strain, and the stress is zero. Regions R3 and R4 present mixed behavior; in particular, they respectively contain two directions and one direction along which the stress is zero and the material can fracture orthogonally to these directions. For later use, we observe that from (2.55) and (2.56), it clearly follows e1 = e2 and e2 = e3 , respectively in R3 and R4 . By solving system (2.45) we obtain the eigenvalues of Ef and T, R then ef1 = 0, if E ∈ R1 ∪ I1,4 ef2 = 0, ef3 = 0, t1 = μ[(2 + α)e1 + α(e2 + e3 )], t2 = μ[(2 + α)e2 + α(e1 + e3 )], t3 = μ[(2 + α)e3 + α(e1 + e2 )];

(2.60)

R if E ∈ R2 ∪ I2,3 then ef1 = e1 , ef2 = e2 , ef3 = e3 , t1 = 0, t2 = 0, t3 = 0;

(2.61)

R if E ∈ R3 ∪ I3,4 then ef1 = 0, α ef2 = e2 + 2(1+α) e1 , f α e3 = e3 + 2(1+α) e1 , t1 = E e1 , t2 = 0, t3 = 0;

(2.62)

if E ∈ R4 then ef1 = 0, ef2 = 0, α (e1 + e2 ), ef3 = e3 + 2+α 2μ t1 = 2+α [2(1 + α)e1 + αe2 ], 2μ [αe1 + 2(1 + α)e2 ], t2 = 2+α t3 = 0, where E = μ(2μ + 3λ)/(μ + λ) is Young’s modulus.

(2.63)

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

Therefore, given a symmetric tensor E =

3

29

ei qi ⊗ qi and having deter-

i=1

mined the region to which E belongs, the solution to the constitutive equation (2.2) is given by 3  Ef = efi qi ⊗ qi , (2.64) i=1

 T = T(E) =

3 

t i qi ⊗ q i ,

(2.65)

i=1

where efi , ti are the functions of (e1 , e2 , e3 ) given in (2.60)-(2.63). It should be noted that in view of (1.36), relation (2.65) is consistent with the isotropy  expressed by (2.35). of T Proposition 2.11. The coefficients β0 , β1 and β2 in the representation formula (2.36) are given by R then β0 = λ trE, if E ∈ R1 ∪ I1,4 β1 = 2μ, β2 = 0,

(2.66)

R if E ∈ R2 ∪ I2,3 then β0 = 0, β1 = 0, β2 = 0,

(2.67)

R if E ∈ R3 ∪ I3,4 then β0 = E

β1 = β2 = if E ∈ R4 then β0 = β1 = β2 =

e1 e2 e3 (e3 −e1 )(e2 −e1 ) , 1 (e2 +e3 ) −E (e3e−e , 1 )(e2 −e1 ) e1 E (e3 −e1 )(e2 −e1 ) ,

2μ −(2+3α)e1 e2 e3 +αe3 [e1 (e3 −e1 )+e2 (e3 −e2 )] , 2+α (e3 −e2 )(e3 −e1 ) 2 2 2 2 2μ α(e1 +e2 +e3 )+2e3 +(2+3α)e1 e2 , 2+α (e3 −e2 )(e3 −e1 ) 2μ 2e3 +α(e1 +e2 +e3 ) − 2+α (e3 −e2 )(e3 −e1 ) .

(2.68)

(2.69)

Proof. By substituting (2.65) into (2.36), we obtain the linear system ⎧ ⎨ β0 + e1 β1 + e21 β2 = t1 , β0 + e2 β1 + e22 β2 = t2 , (2.70) ⎩ β0 + e3 β1 + e23 β2 = t3 . This can be solved in the different regions Ri by replacing t1 , t2 , t3 with their values given in (2.60)-(2.63) and explicitly calculating the coefficients (2.66)(2.69).  Note that, in view of (2.48), e1 , e2 , e3 and then β0 , β1 , β2 can be expressed as functions of the principal invariants of E.

30

2 The Constitutive Equations of Masonry-Like Materials

Explicit knowledge of the solution to the constitutive equation allows us to calculate the strain energy density defined in (2.25). In particular, it holds that R then if E ∈ R1 ∪ I1,4 ψ(E) = μ||E||2 +

λ (trE)2 , 2

(2.71)

R if E ∈ R2 ∪ I2,3 then

ψ(E) = 0, if E ∈ R3 ∪

R I3,4

(2.72)

then 1 2 Ee , 2 1

(2.73)

2μ [(1 + α)(e21 + e22 ) + αe1 e2 ]. 2+α

(2.74)

ψ(E) = if E ∈ R4 then ψ(E) =

2.1.2 The Two-Dimensional Case Firstly, let us consider the case in which the infinitesimal strain tensor E has a null eigenvalue. We assume that q1 , q2 , q3 is an orthonormal basis of V constituted by eigenvectors of E with Eq3 = 0. Proposition 2.12. Let C be isotropic; for E = the solution to system (2.2), with Ef =

3 i=1

3

ei qi ⊗qi , let (T, Ee , Ef ) be

i=1

efi qi ⊗ qi . Then, e3 = q3 · Eq3 = 0

implies ef3 = q3 · Ef q3 = 0. Proof. The ab absurdo proof of the proposition is as follows. Let us assume that ef3 > 0. From the orthogonality of Ef and T, we obtain the condition t3 = 0 which, together with (2.45)3 yields ef3 =

α (e1 + e2 − ef1 − ef2 ). 2+α

(2.75)

If α = 0, we immediately obtain ef3 = 0. On the other hand, if α > 0, then the quantity ef3 in (2.75) is positive, if and only if e1 + e2 − ef1 − ef2 > 0.

(2.76)

Now, by substituting (2.75) into the expressions of t1 and t2 in (2.45)1 and (2.45)2 , respectively, and summing, we obtain 0 ≥ t1 + t2 = 2μ

2 + 3α (e1 + e2 − ef1 − ef2 ) 2+α

which is incompatible with (2.76). 

(2.77)

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

31

In particular, under the hypotheses of the preceding proposition, we have t3 = α(t1 + t2 )/(2(1 + α)). Let us indicate with the same symbols E, Ef and T, the restrictions of E, f E and T to the two-dimensional subspace of V orthogonal to q3 . Calculation of ef1 , ef2 , t1 and t2 satisfying (2.45) requires definition of the following subsets of Sym, shown in Figure 2.2, S1 = {E ∈ Sym | αe1 + (2 + α)e2 < 0},

(2.78)

S2 = {E ∈ Sym | e1 > 0},

(2.79)

S3 = {E ∈ Sym | e1 < 0, αe1 + (2 + α)e2 > 0},

(2.80)

with e1 ≤ e2 . Moreover, we define the following interfaces between regions S1 , S3 and S2 , S3 , respectively as S I1,3 = {E ∈ Sym | αe1 + (2 + α)e2 = 0},

(2.81)

S = {E ∈ Sym | e1 = 0}, I2,3

(2.82)

Note that in S3 , e1 and e2 are distinct and different from zero. Under these circumstances, from system (2.45) we deduce S if E ∈ S1 ∪ I1,3 then ef1 = 0, ef2 = 0, t1 = μ[(2 + α)e1 + αe2 ], t2 = μ[(2 + α)e2 + αe1 ];

(2.83)

Fig. 2.2. Subdivision of the half-plane e1 ≤ e2 into the regions S1 , S2 and S3

32

2 The Constitutive Equations of Masonry-Like Materials S if E ∈ S2 ∪ I2,3 then ef1 = e1 , ef2 = e2 , t1 = 0, t2 = 0;

(2.84)

if E ∈ S3 then ef1 = 0, α e1 , ef2 = e2 + 2+α 1+α t1 = 4μ 2+α e1 , t2 = 0.

(2.85)

 is an isotropic function of E, we have Since T  T = T(E) = β0 I + β1 E;

(2.86)

and β0 and β1 can be obtained from (2.70), with β2 = 0, S then β0 = λ(e1 + e2 ), if E ∈ S1 ∪ I1,3 β1 = 2μ,

(2.87)

S if E ∈ S2 ∪ I2,3 then β0 = 0, β1 = 0,

(2.88)

e 1 e2 if E ∈ S3 then β0 = 4μ 1+α 2+α e2 −e1 , e1 β1 = −4μ 1+α 2+α e2 −e1 .

(2.89)

Lastly, from (2.83)-(2.85) we deduce the following expressions for the strain energy density. S then If E ∈ S1 ∪ I1,3 ψ(E) = μ||E||2 +

λ (trE)2 , 2

(2.90)

S then if E ∈ S2 ∪ I2,3

ψ(E) = 0,

(2.91)

if E ∈ S3 then

1+α 2 e . (2.92) 2+α 1 Now, let us consider the case in which T has a null eigenvalue. For q1 , q2 , q3 , orthonormal eigenvectors of T, with Tq3 = 0, from (2.45)3 we get α (ef + ef2 − e1 − e2 ). e3 − ef3 = (2.93) 2+α 1 ψ(E) = 2μ

Moreover, as ef3 (2.45)6 , it is assumed to be equal to zero. Putting T1 = {E ∈ Sym | αe1 + 2(1 + α)e2 < 0},

(2.94)

T2 = {E ∈ Sym | e1 > 0},

(2.95)

2.1 Masonry-Like Materials with Zero Tensile & Infinite Compressive Strength

we have

33

T3 = {E ∈ Sym | e1 < 0, αe1 + 2(1 + α)e2 > 0},

(2.96)

T I1,3 = {E ∈ Sym | αe1 + 2(1 + α)e2 = 0},

(2.97)

T = {E ∈ Sym | e1 = 0}, I2,3

(2.98)

T if E ∈ T1 ∪ I1,3 then ef1 = 0, ef2 = 0, 2μ [2(1 + α)e1 + αe2 ], t1 = 2+α 2μ t2 = 2+α [2(1 + α)e2 + αe1 ];

(2.99)

T if E ∈ T2 ∪ I2,3 then ef1 = e1 , ef2 = e2 , t1 = 0, t2 = 0;

if E ∈ T3 then ef1 = 0, ef2 = e2 + t1 = Ee1 , t2 = 0.

α 2(1+α) e1 ,

(2.100)

(2.101)

In this case the coefficients of the representation (2.86) are 2λ T if E ∈ T1 ∪ I1,3 then β0 = 2μ+λ (e1 + e2 ), β1 = 2μ,

(2.102)

T if E ∈ T2 ∪ I2,3 then β0 = 0, β1 = 0,

(2.103)

e2 if E ∈ T3 then β0 = E ee21−e , 1 1 β1 = −E e2e−e , 1

(2.104)

and the strain energy density has the following expressions, T then if E ∈ T1 ∪ I1,3 ψ(E) = μ||E||2 +

λ (trE)2 , 2+α

(2.105)

T if E ∈ T2 ∪ I2,3 then

ψ(E) = 0, if E ∈ T3 then ψ(E) =

1 2 Ee . 2 1

(2.106) (2.107)

34

2 The Constitutive Equations of Masonry-Like Materials

2.2 Masonry-Like Materials with Small Tensile Strength and Bounded Compressive Strength In many applications masonry materials can be considered infinitely resistant to compression because the stresses in the structure are definitely less than the maximum compressive strength [56]. However, in some cases it is important to have a constitutive model that takes into account the limited compressive strength of the material, since re-distribution of the stresses in the structure, which occurs when the compressive strength is reached in large parts of it, may be significant and may sometimes considerably modify its behavior. The reasons for introducing low tensile strength stem from different considerations. Indeed, it is well known that, although some masonry components may have non-negligible tensile strength, the frequent presence of fractures, which often occur during a building’s construction, suggest viewing masonry materials as not withstanding tension at all. On the other hand, in some circumstances, providing for low tensile strength can facilitate the numerical analysis without significantly modifying the results. In this section the constitutive equation presented in the preceding section is generalized by allowing the material to possess some, albeit low, tensile strength and setting a limit to its compressive strength. We shall limit ourselves to the particular case in which the dependence of the stress on the elastic strain is isotropic. Let C be an isotropic tensor, C[A] = 2μA+λtr(A)I, A ∈Sym, where λ and μ satisfy the inequalities (2.40). Definition 2.13. Let σ t and σ c be two positive material constants representing the maximum resistance to tension and compression, respectively. A masonry-like material with bounded compressive strength (and bounded tensile  BC : Sym→Sym, strength) is an elastic material whose stress function T  BC (E), T=T

E ∈ Sym

satisfies the following conditions ⎧ ⎪ T − σ t I ∈ Sym− , ⎪ ⎪ ⎪ ⎪ ⎪ T + σ c I ∈ Sym+ , ⎪ ⎪ ⎪ ⎪ ⎪ E = Ee + Ef + Ec , ⎪ ⎪ ⎪ ⎨ T = C[Ee ], ⎪ Ef ∈ Sym+ , ⎪ ⎪ ⎪ ⎪ ⎪ Ec ∈ Sym− , ⎪ ⎪ ⎪ ⎪ ⎪ Ef ·Ec = 0, ⎪ ⎪ ⎪ ⎩ (T − σ t I)·Ef = (T + σ c I)·Ec = 0.

(2.108)

(2.109)

2.2 Masonry-Like Materials with Small Tensile & Bounded Compressive Strength

Once again we shall call C the elasticity tensor and λ and μ the Lam´e moduli. Ee , Ef and Ec are respectively the elastic, fracture and crushing parts of the strain. When the crushing stress σ c goes to infinity, and tensile stress σ t goes to zero, relations (2.109) reduce to the constitutive equation of the masonry-like material described in Section 2.1. A material whose behavior is described by (2.109) will be also called a BCS material, for the sake of brevity. Proposition 2.14. Tensors E, Ef , Ec , Ee , T, T − σ t I and T + σ c I are coaxial. Proof. Because C is isotropic, the proof is analogous to that of proposition 2.10.  Due to the preceding proposition, the constitutive equation (2.109) can be rewritten with respect to the basis {q1 , q2 , q3 } of the eigenvectors of E. Let (e1 , e2 , e3 ), (ef1 , ef2 , ef3 ), (ec1 , ec2 , ec3 ) and (t1 , t2 , t3 ) be the eigenvalues of E, Ef , Ec and T, respectively. It is an easy matter to show that the constitutive equation (2.109) is equivalent to the system ⎧ ⎪ t = μ{2(e1 − ef1 − ec1 ) + α[tr(E) − tr(Ef ) − tr(Ec )]} ⎪ ⎪ 1 ⎪ ⎪ ⎪ t2 = μ{2(e2 − ef2 − ec2 ) + α[tr(E) − tr(Ef ) − tr(Ec )]} ⎪ ⎪ ⎪ ⎪ ⎪ t3 = μ{2(e3 − ef3 − ec3 ) + α[tr(E) − tr(Ef ) − tr(Ec )]} ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (t1 − σ t )ef1 = 0 ⎪ ⎪ ⎪ ⎪ ⎪ (t2 − σ t )ef2 = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (t3 − σ t )ef3 = 0 ⎪ ⎪ ⎨ (t1 + σ c )ec1 = 0 c c ⎪ ⎪ ⎪ (t2 + σ )e2 = 0 ⎪ ⎪ ⎪ ⎪ (t3 + σ c )ec3 = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ef1 ≥ 0, ef2 ≥ 0, ef3 ≥ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ec1 ≤ 0, ec2 ≤ 0, ec3 ≤ 0 ⎪ ⎪ ⎪ ⎪ ⎪ t1 − σ t ≤ 0, t2 − σ t ≤ 0, t3 − σ t ≤ 0 ⎪ ⎪ ⎪ ⎪ t + σ c ≥ 0, t + σ c ≥ 0, t + σ c ≥ 0 ⎪ ⎪ 1 2 3 ⎪ ⎪ f ⎩ f c f c c e1 e1 = e2 e2 = e3 e3 = 0

(2.110)

where tr(E) = e1 + e2 + e3 , tr(Ef ) = ef1 + ef2 + ef3 , tr(Ec ) = ec1 + ec2 + ec3 and α = λ/μ. Given the elastic moduli λ and μ and the material constants σ t and σ c , the principal components (ef1 , ef2 , ef3 ), (ec1 , ec2 , ec3 ) and (t1 , t2 , t3 ) satisfying (2.110) can be calculated as functions of the eigenvalues e1 , e2 , e3 of E, which are

35

36

2 The Constitutive Equations of Masonry-Like Materials

presumed to be ordered so that e1 ≤ e2 ≤ e3 . The solution to (2.110) can be calculated explicitly by using a procedure similar to that used in subsection 2.1.1. To this end, we define the following subsets of Sym, C1 = {E ∈ Sym | 2e3 + α tr(E) − ω t < 0,

2e1 + α tr(E) + ω c > 0}, (2.111)

C2 = {E ∈ Sym | 2e1 + α tr(E) + ω c < 0, 2αe2 +4(1+α)e3 −αω c −(2+α)ω t < 0,

2(1+α)e2 +αe3 +ω c > 0}, (2.112)

C3 = {E ∈ Sym | 2(1 + α)e2 + αe3 + ω c < 0, (2 + 3α)e3 − αω c − (1 + α)ω t < 0,

(2 + 3α)e3 + ω c > 0},

C4 = {E ∈ Sym | (2 + 3α)e3 + ω c < 0},

(2.113) (2.114)

C5 = {E ∈ Sym | 2e3 + α tr(E) − ω t > 0, 2(1+α)e2 +αe1 −ω t < 0,

4(1+α)e1 +2αe2 +αω t +(2+α)ω c > 0}, (2.115)

C6 = {E ∈ Sym | 2(1 + α)e2 + αe1 − ω t > 0, (2 + 3α)e1 − ω t < 0,

(2 + 3α)e1 + αω t + (1 + α)ω c > 0},

C7 = {E ∈ Sym | (2 + 3α)e1 − ω t > 0},

(2.116) (2.117)

C8 = {E ∈ Sym | 2(2 + 3α)e2 − αω c − (2 + α)ω t > 0, (2 + 3α)e1 + αω t + (1 + α)ω c < 0},

(2.118)

C9 = {E ∈ Sym | 2(2 + 3α)e2 + αω t + (2 + α)ω c < 0, (2 + 3α)e3 − αω c − (1 + α)ω t > 0},

(2.119)

C10 = {E ∈ Sym | 4(1 + α)e1 + 2αe2 + αω t + (2 + α)ω c < 0, 4(1 + α)e3 + 2αe2 − αω c − (2 + α)ω t > 0, 2(2 + 3α)e2 − αω c − (2 + α)ω t < 0, 2(2 + 3α)e2 + αω t + (2 + α)ω c > 0},

(2.120)

where ω t = σ t /μ and ω c = σ c /μ. In addition, just as for the case of masonrylike materials with zero tensile strength and infinite compressive strength, we define the following interfaces

2.2 Masonry-Like Materials with Small Tensile & Bounded Compressive Strength C I1,2 = {E ∈ Sym | 2e1 + α tr(E) + ω c = 0},

(2.121)

C = {E ∈ Sym | 2e3 + α tr(E) − ω t = 0}, I1,5

(2.122)

C = {E ∈ Sym | 2(1 + α)e2 + αe3 + ω c = 0}, I2,3

(2.123)

C = {E ∈ Sym | (2 + 3α)e3 + ω c = 0}, I3,4

(2.124)

C = {E ∈ Sym | αe1 + 2(1 + α)e2 − ω t = 0}, I5,6

(2.125)

C = {E ∈ Sym | (2 + 3α)e1 − ω t = 0}, I6,7

(2.126)

C = {E ∈ Sym | (2 + 3α)e1 + αω t + (1 + α)ω c = 0}, I6,8

(2.127)

C = {E ∈ Sym | 2(2 + 3α)e2 − αω c − (2 + α)ω t = 0}, I8,10

(2.128)

C = {E ∈ Sym | 2(2 + 3α)e2 + αω t + (2 + α)ω c = 0}, I9,10

(2.129)

It is easy to prove that in regions C2 , C6 and C8 we have e1 < e2 ≤ e3 and that in C3 , C5 and C9 , e1 ≤ e2 < e3 . Lastly, the eigenvalues of E are distinct in C10 . By solving system (2.110), we obtain the principal components of Ef , Ec and T. C C If E ∈ C1 ∪ I1,2 ∪ I1,3 then ef1 = 0, ef2 = 0, ef3 = 0, ec1 = 0, ec2 = 0, ec3 = 0, t1 = μ[(2 + α)e1 + α(e2 + e3 )], t2 = μ[(2 + α)e2 + α(e1 + e3 )], t3 = μ[(2 + α)e3 + α(e1 + e2 )];

(2.130)

C then ef1 = 0, if E ∈ C2 ∪ I2,3 ef2 = 0, ef3 = 0, α 1 ec1 = e1 + 2+α (e2 + e3 ) + 2+α ωc , c e2 = 0, ec3 = 0, t1 = −σ c , α [2(e2 + e3 ) − ω c ]}, t2 = μ{2e2 + 2+α α t3 = μ{2e3 + 2+α [2(e2 + e3 ) − ω c ]};

(2.131)

37

38

2 The Constitutive Equations of Masonry-Like Materials C if E ∈ C3 ∪ I3,4 then ef1 = 0, ef2 = 0, ef3 = 0, α 1 ec1 = e1 + 2(1+α) e3 + 2(1+α) ωc , α 1 c e2 = e2 + 2(1+α) e3 + 2(1+α) ω c , ec3 = 0, t1 = −σ c , t2 = −σ c , μ [(2 + 3α)e3 − αω c ]; t3 = 1+α

if E ∈ C4 then ef1 = 0, ef2 = 0, ef3 = 0, 1 ec1 = e1 + 2+3α ωc , 1 ωc , ec2 = e2 + 2+3α 1 c e3 = e3 + 2+3α ω c , t1 = −σ c , t2 = −σ c , t3 = −σ c ;

(2.133)

C then ef1 = 0, if E ∈ C5 ∪ I5,6 ef2 = 0, α 1 (e1 + e2 ) − 2+α ωt , ef3 = e3 + 2+α c e1 = 0, ec2 = 0, ec3 = 0, μ [4(1 + α)e1 + 2αe2 + αω t ], t1 = 2+α μ t2 = 2+α [4(1 + α)e2 + 2αe1 + αω t ], t3 = σ t ; C C ∪ I6,8 then ef1 = 0, if E ∈ C6 ∪ I6,7 ef2 = e2 +

α 2(1+α) e1 α 2(1+α) e1

−

(2.132)

(2.134)

ωt 2(1+α) , ωt 2(1+α) ,

− ef3 = e3 + c e1 = 0, ec2 = 0, ec3 = 0, μ [(2 + 3α)e1 + αω t ], t1 = 1+α t t2 = σ , t3 = σ t ;

(2.135)

2.2 Masonry-Like Materials with Small Tensile & Bounded Compressive Strength

if E ∈ C7 then ef1 = e1 − ef2 = e2 − ef3 = e3 − ec1 = 0, ec2 = 0, ec3 = 0, t1 = σ t , t2 = σ t , t3 = σ t ;

1 t 2+3α ω , 1 t 2+3α ω , 1 t 2+3α ω ,

(2.136)

C then ef1 = 0, if E ∈ C8 ∪ I8,10

ef2 = e2 − ef3 ec1 ec2 ec3

= e3 −

= e1 + = 0, = 0, t1 = −σ c , t2 = σ t , t3 = σ t ;

αω c 2(2+3α) αω c 2(2+3α) (1+α)ω c 2+3α

− − +

2+α t 2(2+3α) ω , 2+α t 2(2+3α) ω , α t 2+3α ω ,

(2.137)

C then ef1 = 0, if E ∈ C9 ∪ I9,10

ef2 = 0, α 1+α t ef3 = e3 − 2+3α ω c − 2+3α ω, 2+α α c c ωt , e1 = e1 + 2(2+3α) ω + 2(2+3α) 2+α α c c e2 = e2 + 2(2+3α) ω + 2(2+3α) ω t , ec3 = 0, t1 = −σ c , t2 = −σ c , t3 = σ t ;

if E ∈ C10 then ef1 = 0, ef2 = 0, α 2+α α e2 − 4(1+α) ω t − 4(1+α) ωc , ef3 = e3 + 2(1+α) α 2+α α c c e1 = e1 + 2(1+α) e2 + 4(1+α) ω + 4(1+α) ω t , ec2 = 0, ec3 = 0, t1 = −σ c , μ [2(2 + 3α)e2 + α(ω t − ω c )], t2 = 2(1+α) t t3 = σ . We are now in a position to prove the following theorem.

(2.138)

(2.139)

39

40

2 The Constitutive Equations of Masonry-Like Materials

Theorem 2.15. Given the Lam´e moduli of the material and the constants σ t and σ c , for each E ∈ Sym the constitutive equation (2.109) has a unique solution (T, Ee , Ef , Ec ). Proof. As far as the existence of the solution is concerned, formulae (2.130)(2.139) provide an explicit solution as E varies in Sym. In order to prove its uniqueness, let (T1 , Ee1 , Ef1 , Ec1 ) and (T2 , Ee2 , Ef2 , Ec2 ) be two different solutions. For the elastic parts, we have Ee1 = E − Ef1 − Ec1 and Ee2 = E − Ef2 − Ec2 and thus (2.140) C−1 [T1 − T2 ] = Ee1 − Ee2 = Ec2 − Ec1 + Ef2 − Ef1 . By virtue of (2.109)1 , (2.109)2 and (2.109)8 , the positive definiteness of C and proposition 1.9, we can write 0 ≤ (T1 − T2 )·C−1 [T1 − T2 ] = (T1 − T2 )·(Ec2 − Ec1 ) + (T1 − T2 )·(Ef2 − Ef1 ) = (T1 − σ t I)·Ef2 + (T2 − σ t I)·Ef1 +(T1 + σ c I)·Ec2 + (T2 + σ c I)·Ec1 ≤ 0. Ef2

(2.141)

Ef1

Consequently, T1 = T2 , Ee1 = Ee2 and +Ec2 = +Ec1 . On the other hand, by using (2.109)7 and proposition 1.9, we obtain 0 ≤ (Ec2 − Ec1 )·(Ef1 − Ef2 ) = Ec2 ·Ef1 + Ec1 ·Ef2 ≤ 0, and finally, Ec2 − Ec1 = Ef1 − Ef2 = 0. 

Therefore, given a symmetric tensor E =

3

ei qi ⊗ qi and having deter-

i=1

mined the region to which E belongs, the solution to the constitutive equation (2.109) is 3   BC (E) = T=T t i qi ⊗ q i , (2.142) i=1

Ef =

3  i=1

efi qi ⊗ qi ,

Ec =

3 

eci qi ⊗ qi ,

(2.143)

i=1

where efi , eci , ti are functions of ei , given in (2.130)-(2.139). In particular,  if σ t = 0 and σ c = ∞.  BC = T T As in the case of a masonry-like material with zero tensile strength and infinite compressive strength, the solution to the constitutive equation (2.109) can be characterized as the solution of a suitable variational inequality. To this end, put Sym(σ Sym(σ

t

t

,σ c )

,σ c )

= {A ∈ Sym | A − σ t I ∈ Sym− , A + σ c I ∈ Sym+ };

is a closed, convex subset of Sym, which is not a cone.

(2.144)

2.2 Masonry-Like Materials with Small Tensile & Bounded Compressive Strength

For E ∈ Sym, let us consider the problem of finding T ∈ Sym(σ that the variational inequality (T − T∗ )·(E − C−1 [T]) ≥ 0

for each T∗ ∈ Sym(σ

t

t

,σ c )

,σ c )

such

(2.145)

holds. Proposition 2.16. For every E ∈ Sym, there exists a unique T ∈ Sym(σ which satisfies the variational inequality (2.145).

t

,σ c )

Proof. Inequality (2.145) characterizes T as the C−1 -orthogonal projection of t c C[E] onto the closed convex set Sym(σ ,σ ) , whose existence and uniqueness is guaranteed by the minimum norm theorem 1.1.  Proposition 2.17. For E ∈ Sym, (T, Ee , Ef , Ec ) is the solution to the cont c stitutive equation (2.109) if and only if T ∈ Sym(σ ,σ ) and satisfies (2.145). Proof. Let (T, Ee , Ef , Ec ) be the solution to (2.109); conditions (2.109)1 and t c t c (2.109)2 imply that T ∈ Sym(σ ,σ ) . For T∗ ∈ Sym(σ ,σ ) , from (2.109)8 , (2.109)4 and proposition 1.9, it follows that (T − T∗ )·Ef = [(T − σ t I) − (T∗ − σ t I)]·Ef = −(T∗ − σ t I)·Ef ≥ 0.

(2.146)

Analogously, from (2.109)8 , (2.109)3 and proposition 1.9, we have (T − T∗ )·Ec = [(T + σ c I) − (T∗ + σ c I)]·Ec = −(T∗ + σ c I)·Ec ≥ 0.

(2.147)

Thus, by adding (2.146) and (2.147), bearing in mind that E + E = E − C−1 [T], (2.145) follows. Conversely, from propositions 2.15 and 2.16, it follows that if T ∈ t c Sym(σ ,σ ) is the solution to (2.145), then T satisfies (2.109).  f

c

For a BCS material with constitutive equation (2.108), we define the strain energy density as 1 C[Ee ] · Ee + T · (Ef + Ec ), for each E ∈ Sym. 2  BC is monotone, Proposition 2.18. Stress function T ψBC (E) =

 BC (E1 ) − T  BC (E2 ))·(E1 − E2 ) ≥ κ||T  BC (E1 ) − T  BC (E2 )||2 , (T

(2.148)

(2.149)

for each E1 , E2 ∈ Sym, and Lipschitz continuous,  BC (E1 ) − T  BC (E2 )|| ≤ κ−1 ||E1 − E2 ||, ||T

(2.150)

for each E1 , E2 ∈Sym. Moreover, function ψBC in (2.148) is continuously differentiable, convex and  BC (E). DE ψBC (E) = T

(2.151)

41

42

2 The Constitutive Equations of Masonry-Like Materials

Proof. The proof of analogous results for is the solution of the

(2.149), (2.150) and (2.151) is similar to the proof of  in Section 2.1 and is based on the fact that T  BC (E) T variational inequality (2.145). 

The two-dimensional case will be dealt with in the Appendix.

2.3 Masonry-Like Materials Under Non-Isothermal Conditions There are many problems in which the presence of thermal dilatation must be taken into account. Consider, for example, the influence of thermal variations on the stress field in masonry bridges [49], or the thermo-mechanical behavior of the refractory materials used in the iron and steel industry [60], and, lastly, geological problems connected with the presence of volcanic calderas, such as that at Pozzuoli [29]. In the last two cases, the thermal variation during the thermo-mechanical process under examination can be so high that the dependence of the material constants on temperature cannot be ignored. In this section we recall the constitutive equation for isotropic masonry-like materials with zero tensile strength and infinite compressive strength under non-isothermal conditions developed in [71]. Let ϑ ∈ [ϑ1 , ϑ2 ], with ϑ1 > 0, be the absolute temperature and ϑ0 ∈ [ϑ1 , ϑ2 ] the reference temperature. In view of the target applications, no limitations are placed on the range of temperature variation. We assume that the thermal dilatation due to the temperature variation ϑ−ϑ0 is the spherical tensor β(ϑ)I, where β(ϑ) is a material function of the temperature called thermal expansion, with β(ϑ0 ) = 0. Moreover, we assume that there exist δ ∈ [0, 1) such that ||E − β(ϑ)I|| ≤ δ,

for each ϑ ∈ [ϑ1 , ϑ2 ].

(2.152)

Condition (2.152) is equivalent to requiring that the norm of the deviatoric part of E and the scalar function β(ϑ) − tr(E)/3 be O(δ)2 . Let E(ϑ) and ν(ϑ) be temperature-dependent functions such that E(ϑ) > 0,

0 ≤ ν(ϑ) < 1/2,

and let us set γ(ϑ) =

2

for each ϑ ∈ [ϑ1 , ϑ2 ],

ν(ϑ) . 1 − 2ν(ϑ)

(2.153) (2.154)

Given a mapping B from a neighbourhood of 0 in R into a vector space with norm || ||, we write B(δ) = O(δ) if there exist k > 0 and k > 0 such that ||B(δ)|| < kδ whenever |δ| < k .

2.3 Masonry-Like Materials Under Non-Isothermal Conditions

43

By generalizing the constitutive equation described in section 2.1, we present a nonlinear elastic constitutive equation which associate a negative semidefinite stress T to each strain E − β(ϑ)I. Let C(ϑ) be the positive definite fourth-order tensor C(ϑ) =

E(ϑ) (I + γ(ϑ)I ⊗ I). 1 + ν(ϑ)

(2.155)

Definition 2.19. A masonry-like material under non-isothermal conditions  N I : Sym×[ϑ1 , ϑ2 ] →Sym, is an elastic material whose stress function T  N I (E, ϑ), T=T satisfies the system

(E, ϑ) ∈ Sym × [ϑ1 , ϑ2 ]

⎧ − ⎪ ⎪ T ∈ Sym , e ⎪ ⎪ ⎨ E − β(ϑ)I = E + Ef , Ef ·T = 0, ⎪ ⎪ T = C(ϑ)[Ee ], ⎪ ⎪ ⎩ f E ∈ Sym+ .

(2.156)

(2.157)

As in the isothermal case, T is the projection of C(ϑ)[E − β(ϑ)I] onto Sym− with respect to the inner product (A, B) = C(ϑ)−1 [A]·B in Sym, and Ef = E − β(ϑ)I − C(ϑ)−1 [T] belongs to the normal cone N (T) to Sym− at T defined in (2.11). Once again denoting {q1 , q2 , q3 } as an orthonormal basis constituted by eigenvectors of E, and with (e1 , e2 , e3 ) the eigenvalues of E, we can show that for each (E, ϑ) ∈ Sym ×[ϑ1 , ϑ2 ], the unique solution (T, Ef ) to (2.157) is given by T = C(ϑ)[E − β(ϑ)I], (2.158) Ef = 0,

(2.159)

T = 0,

(2.160)

E = E − β(ϑ)I,

(2.161)

T = E(ϑ)(e1 − β(ϑ))q1 ⊗ q1 ,

(2.162)

W for (E, ϑ) ∈ W1 ∪ I1,4 ; f

W for (E, ϑ) ∈ W2 ∪ I2,3 ;

Ef = [e2 − β(ϑ) + ν(ϑ)(e1 − β(ϑ))]q2 ⊗ q2 +[e3 − β(ϑ) + ν(ϑ)(e1 − β(ϑ))]q3 ⊗ q3 , for (E, ϑ) ∈ W3 ∪ T=

(2.163)

W I3,4 ;

E(ϑ) {[e1 − β(ϑ) + ν(ϑ)(e2 − β(ϑ))]q1 ⊗ q1 1 − ν 2 (ϑ) +[e2 − β(ϑ) + ν(ϑ)(e1 − β(ϑ))]q2 ⊗ q2 },

(2.164)

44

2 The Constitutive Equations of Masonry-Like Materials

Ef =

1 [e3 − β(ϑ) 1 − ν(ϑ)

+ν(ϑ)(e1 + e2 − e3 − β(ϑ))]q3 ⊗ q3 ,

(2.165)

for (E, ϑ) ∈ W4 ; where W1 = {(E, ϑ) | e3 − β(ϑ) + γ(ϑ)(tr(E) − 3β(ϑ)) < 0}, W2 = {(E, ϑ) | e1 − β(ϑ) > 0},

(2.166) (2.167)

W3 = {(E, ϑ) | e1 − β(ϑ) < 0, ν(ϑ)(e1 − β(ϑ)) + e2 − β(ϑ) > 0},

(2.168)

W4 = {(E, ϑ) | ν(ϑ)(e1 − β(ϑ)) + e2 − β(ϑ) < 0, e3 − β(ϑ) + γ(ϑ)(tr(E) − 3β(ϑ)) > 0}, W = {(E, ϑ) | e3 − β(ϑ) + γ(ϑ)(tr(E) − 3β(ϑ)) = 0}, I1,4 W I2,3

= {(E, ϑ) | e1 − β(ϑ) = 0},

W I3,4 = {(E, ϑ) | ν(ϑ)(e1 − β(ϑ)) + e2 − β(ϑ) = 0}.

(2.169) (2.170) (2.171) (2.172)

It is an easy matter to verify that in the absence of temperature variations, the material characterized by the constitutive equation (2.157) conforms to the isothermal masonry-like material dealt with in section 2.1. For a masonry-like material under non-isothermal conditions we define the strain energy density 1 TN I (E, ϑ) · (E − β(ϑ)I). 2 is continuously differentiable, convex and ψN I (E, ϑ) =

Function ψN I

 N I (E, ϑ). DE ψN I (E, ϑ) = T

(2.173)

(2.174)

The reader is referred to [71] for a more detailed description of the behavior of masonry-like materials under non-isothermal conditions. Starting from the explicit expression of the stress as function of strain and temperature, the paper deduces the free energy, internal energy, entropy and enthalpy, and defines the specific heat at constant strain. By assuming the classical Fourier hypothesis for heat flux, the material presented therein is characterized completely by five functions of the temperature: Young’s modulus, Poisson’s ratio, thermal expansion, conductivity and specific heat. Indeed, when these material functions are known, the thermodynamic potentials (and consequently the thermo-mechanical behavior) of the material can be determined. Just as in the linear case, the basic equations of the thermoelastic theory for no-tension materials are: the strain-displacement relation, the equilibrium equation, the constitutive equations for stress and heat flux, and the equilibrium energy equation. The system obtained is coupled because the temperature coefficient and the coefficient of the derivative of temperature with respect to time in

2.4 The Derivative of the Stress Function

45

the energy equation depend on strain and strain rate. However, without any hypothesis on the range of variation of temperature, if we assume small values for the displacement gradient, the thermal expansion and its derivative with respect to temperature, strain rate and temperature rate are small, then the thermoelastic equilibrium equations are uncoupled and can be integrated separately. It is worth noting that if the further temperature condition ϑ = ϑ0 + O(δ)

(2.175)

holds, with ϑ0 the reference temperature, we then have β(ϑ) = β(ϑ0 ) + β  (ϑ0 )(ϑ − ϑ0 ) + o(δ)

(2.176)

and thus, by taking into account that β(ϑ0 ) = 0, we obtain that in a neighborhood of ϑ = ϑ0 β(ϑ) = β  (ϑ0 )(ϑ − ϑ0 ), (2.177) within an error of order o(δ). The quantity α = β  (ϑ0 )

(2.178)

is the linear coefficient of thermal expansion. In this case, functions E and ν must be presumed to be temperature-independent and coincident with their value at ϑ0 [22].

2.4 The Derivative of the Stress Function  T  BC and T  N I with In this section we calculate the derivative of functions T, respect to E, under the assumption that C is isotropic. These derivatives will be used in the next chapter, dealing with numerical solution of the equilibrium problem via the finite elements method. In particular, knowing the explicit  DE T  BC and DE T  BN I allows us to calculate for each expressions of DE T, strain tensor E the tangent stiffness matrix used in the numerical procedure described in chapter 4.  given in (2.65) is differentiable in each Proposition 2.20. Stress function T region Ri , i = 1, ..., 4. Proof. With the help of the results given in subsection 1.7 we are in a position  in the four regions Ri . Let e1 ≤ e2 ≤ e3 and q1 , q2 , q3 to calculate DE T be the eigenvalues and the eigenvectors of E, respectively. We consider the orthonormal basis of Sym,

46

2 The Constitutive Equations of Masonry-Like Materials

O11 = q1 ⊗ q1 , O22 = q2 ⊗ q2 , O33 = q3 ⊗ q3 , 1 O12 = √ (q1 ⊗ q2 + q2 ⊗ q1 ), 2 1 O13 = √ (q1 ⊗ q3 + q3 ⊗ q1 ), 2 1 O23 = √ (q2 ⊗ q3 + q3 ⊗ q2 ), 2

(2.179) (2.180) (2.181) (2.182)

and (2.65) then becomes  T = T(E) =

3 

ti Oii .

(2.183)

i=1

 for E beFrom (2.60) and (2.61), it follows that the expression of DE T  when E longing to R1 and R2 can be calculated easily. Calculation of DE T belongs to the two other regions is slightly more complex and requires differentiating expression (2.183) with respect to E by using proposition 1.20. As  when E ∈ R3 , where e1 < e2 ≤ e3 . a single example, we shall calculate DE T Let us begin by supposing that e1 < e2 < e3 ; from equations (2.183), (2.62) and (1.92)-(1.97), using the relation = DE T

3 

(DE ti Oii + ti DE Oii ),

(2.184)

i=1

we obtain

 = E(O11 ⊗ O11 − DE T

e1 O12 ⊗ O12 e2 − e1

e1 O13 ⊗ O13 ), e3 − e1 which is well-defined also when e2 = e3 .  −

E ∈ R3 ,

(2.185)

 in the four regions Ri , Let us then summarize the expressions of DE T  = T1 = 2μI + λI ⊗ I, DE T  = T2 = O, DE T

E ∈ R1 ,

E ∈ R2 ,  = T3 = E(O11 ⊗ O11 − e1 O12 ⊗ O12 DE T e2 − e1 e1 − O13 ⊗ O13 ), E ∈ R3 , e3 − e1

(2.186) (2.187)

(2.188)

 = T4 = 2μO12 ⊗ O12 − 2μ 2(1 + α)e1 + αe2 O13 ⊗ O13 DE T 2+α e3 − e1

2.4 The Derivative of the Stress Function

−

47

2μ 2(1 + α)e2 + αe1 O23 ⊗ O23 2+α e3 − e2

μ(2 + 3α) (O11 + O22 ) ⊗ (O11 + O22 ) 2+α +μ(O11 − O22 ) ⊗ (O11 − O22 ), E ∈ R4 , +

(2.189)

where I and O are the fourth-order identity tensor and the fourth-order null tensor, respectively. It is worthwhile noting that the expressions given in  in the three regions (2.187)-(2.189) are the spectral representations of DE T R2 , R2 and R4 . Moreover, it can be easily verified that the eigenvalues of  are non-negative, a result which conforms with convexity of the strain DE T energy density defined in (2.25), already proved in Section 2.1. R R R , I2,3 and I3,4 respectively defined in (2.57), (2.58) On the interfaces I1,4  and (2.59), tensor DE T(E) does not exist and, in view of the Lipschitz con it is possible to define the subgradient of T  constituted by the tinuity of T, convex combination S(E) of tensors Ti [33], [26] S(E) = {ξT3 + (1 − ξ)T4 | ξ ∈ [0, 1]},

R E ∈ I3,4 ,

(2.190)

S(E) = {ξT2 + (1 − ξ)T3 | ξ ∈ [0, 1]},

R E ∈ I2,3 ,

(2.191)

S(E) = {ξT1 + (1 − ξ)T4 | ξ ∈ [0, 1]},

R I14 .

(2.192)

E∈

 is differentiable, DE T  has Proposition 2.21. For each E ∈ Sym in which T the following properties   = T(E), (2.193) DE T(E)[E] e   ] = T(E), DE T(E)[E

  DE T(βE) = DE T(E),

f  DE T(E)[E ] = 0,

(2.194)

for each β ≥ 0,

(2.195)

T   . = DE T(E) DE T(E)

(2.196)

Proof. From  + H) = T(E)   T(E + DE T(E)[H] + o(H)

as H → 0,

(2.197)

putting H = hE, h ≥ 0, and taking (2.15) into account, we get    (1 + h)T(E) = T(E) + hDE T(E)[E] + o(h)

as h → 0,

(2.198)

whence (2.193) follows directly. The proof of (2.195) is similar and is also based on (2.15). Relations (2.194) and the major symmetry in (2.196) follow directly from the (2.186)-(2.189).  R   , of DE T(E) across the interfaces I3,4 In addition, the jump [DE T(E)] R and I1,4 satisfies the conditions given in [33], which express the absence of tangential discontinuity of the derivative of the stress with respect to the strain, as stated in the following proposition. R I2,3

48

2 The Constitutive Equations of Masonry-Like Materials

 Proposition 2.22. Tensor DE T(E) satisfies the jump conditions  =− [DE T(E)]

μ ∇g34 (E) ⊗ ∇g34 (E), (1 + α)(2 + α)

 [DE T(E)] = E ∇g23 (E) ⊗ ∇g23 (E), μ  ∇g14 (E) ⊗ ∇g14 (E), [DE T(E)] = 2+α

R E ∈ I3,4 ,

R E ∈ I2,3 , R E ∈ I1,4

(2.199) (2.200) (2.201)

where g34 , g23 and g14 are given in (2.59), (2.58) and (2.57), respectively. In particular, it holds that  [DE T(E)][E] = 0, (2.202) R R R or E ∈ I2,3 , or E ∈ I1,4 . for E ∈ I3,4

Proof. The proof is based on simple calculations involving relations (2.186)(2.189) and the derivative with respect to E of functions g34 , g23 and g14 .   BC given in (2.142) is differentiable in each Proposition 2.23. Function T  N I given in (2.156) is differentiable in each region Ci , i = 1, ..., 10. Function T region Wi , i = 1, ..., 4.  N I can be calculated by  BC and T Proof. The derivatives of stress functions T following a procedure analogous to that sketched out for proposition 2.20.  In the case of material having constitutive equation (2.109), the expressions  BC in the ten regions Ci , are of DE T  BC = 2μI + λI ⊗ I, DE T  BC = DE T +

E ∈ C1 ,

(2.203)

2μ ω c + 2(1 + α)e2 + αe3 O12 ⊗ O12 2+α e2 − e1

2μ ω c + 2(1 + α)e3 + αe2 O13 ⊗ O13 + 2μO23 ⊗ O23 2+α e3 − e1 +

μ(2 + 3α) (O22 + O33 ) ⊗ (O22 + O33 ) 2+α

+μ(O22 − O33 ) ⊗ (O22 − O33 ),  BC = EO33 ⊗ O33 + DE T +

E ∈ C2 ,

(2.204)

μ ω c + (2 + 3α)e3 O13 ⊗ O13 1+α e3 − e1

μ ω c + (2 + 3α)e3 O23 ⊗ O23 , 1+α e3 − e2  BC = O, DE T

E ∈ C4 ,

E ∈ C3 ,

(2.205) (2.206)

2.4 The Derivative of the Stress Function

 BC = 2μO12 ⊗ O12 + DE T +

2μ ω t − 2(1 + α)e1 − αe2 O13 ⊗ O13 2+α e3 − e1

2μ ω t − 2(1 + α)e2 − αe1 O23 ⊗ O23 2+α e3 − e2

+

μ(2 + 3α) (O11 + O22 ) ⊗ (O11 + O22 ) 2+α

+μ(O11 − O22 ) ⊗ (O11 − O22 ),  BC = EO11 ⊗ O11 + DE T +

49

E ∈ C5 ,

μ ω − (2 + 3α)e1 O12 ⊗ O12 1+α e2 − e1

μ ω t − (2 + 3α)e1 O13 ⊗ O13 , 1+α e3 − e1  BC = O, DE T

t

(2.207)

t

c

t

E ∈ C6 ,

E ∈ C7 ,

(2.208) (2.209)

c

 BC = σ + σ O12 ⊗ O12 + σ + σ O13 ⊗ O13 , DE T e2 − e1 e3 − e1

E ∈ C8 ,

(2.210)

t c t c  BC = σ + σ O13 ⊗ O13 + σ + σ O23 ⊗ O23 , DE T e3 − e1 e3 − e2

E ∈ C9 ,

(2.211)

 BC = EO22 ⊗ O22 + DE T

αω t + (2 + α)ω c + 2(2 + 3α)e2 μ O12 ⊗ O12 2(1 + α) e2 − e1 +

+

σt + σc O13 ⊗ O13 e3 − e1

αω c + (2 + α)ω t − 2(2 + 3α)e2 μ O23 ⊗ O23 , 2(1 + α) e3 − e2

E ∈ C10 .

(2.212)

 N I (E, ϑ) with respect to E, we have As for the derivative of T  NI = DE T

ν(ϑ) E(ϑ) {I + I ⊗ I}, 1 + ν(ϑ) 1 − 2ν(ϑ)  N I = O, DE T

(E, ϑ) ∈ W1 ,

(E, ϑ) ∈ W2 ,

(2.213) (2.214)

 N I = E(ϑ)(O11 ⊗ O11 − e1 − β(ϑ) O12 ⊗ O12 DE T e2 − e1 −

e1 − β(ϑ) O13 ⊗ O13 ), e3 − e1

 NI = DE T +

(E, ϑ) ∈ W3 ,

E(ϑ) ((O11 − O22 ) ⊗ (O11 − O22 ) 2(1 + ν(ϑ))

1 + ν(ϑ) (O11 + O22 ) ⊗ (O11 + O22 ) + 2O12 ⊗ O12 1 − ν(ϑ)

(2.215)

50

2 The Constitutive Equations of Masonry-Like Materials

−2 −2

e1 + ν(ϑ)e2 − β(ϑ)(1 + ν(ϑ)) O13 ⊗ O13 (1 − ν(ϑ))(e3 − e1 )

e2 + ν(ϑ)e1 − β(ϑ)(1 + ν(ϑ)) O23 ⊗ O23 ), (1 − ν(ϑ))(e3 − e2 )

(E, ϑ) ∈ W4 .

(2.216)

 N I for the plane cases are summarized  T  BC and T The derivatives of T, in the Appendix.

3 Equilibrium of Masonry Bodies

3.1 The Equilibrium Problem The equilibrium problem for masonry-like solids (with infinite compressive strength and zero tensile strength) was studied in the 1980s, though the existence of a solution was proved solely for a restricted class of loads [4], [46]. On the other hand, the uniqueness of the solution is guaranteed only in terms of stress, in the sense that different displacement and strain fields may correspond to the same stress field. In this chapter we shall consider the equilibrium problem of a continuous body made of a material with constitutive equation (2.1), and shall prove some basic results that also hold for the case of materials with constitutive equation (2.142) and (2.156). For the sake of simplicity, the hypotheses on the smoothness of the solution considered here are much less general than those in [4] and [46]. The body is identified with its reference configuration B ⊂ E. B is a bounded regular region having boundary ∂B, with outward unit normal n (cf. section 1.5) Let ∂Bu and ∂Bs be two disjoint subsets of the boundary ∂B of B, such that their union covers ∂B. We assume given body forces b on B, surface forces  s on ∂Bs and surface  on ∂Bu and we call the triplet ( u,  s, b) the load. We moreover displacement u assume that the fourth-order tensor C in (2.2), which may depend on the  is continuous on place, is of class C 1 on B, and that b is continuous on B, u s is piece-wise smooth on ∂Bs [51]. ∂Bu , and  A vector field u : B → V is an admissible displacement field provided u is of class C 2 . We denote by Au the set of all admissible displacements. A symmetric tensor field T : B →Sym− is an admissible stress field if T is of class C 1 . The associated surface force field is the vector field s defined by s = Tn.

(3.1)

We denote by AT the set of all admissible stress fields. Given the load ( u,  s, b), the equilibrium problem for a body made of a material with constitutive equation (2.1) is to find a triplet (u, E, T), with

52

3 Equilibrium of Masonry Bodies

u ∈ Au and T ∈ AT , that satisfies the strain-displacement relation E= the constitutive equation

1 (∇u + ∇uT ), 2

(3.2)

 T = T(E),

(3.3)

 defined in (2.1), the equilibrium equation with T divT + b = 0 on B,

(3.4)

 on ∂Bu , u=u

(3.5)

Tn =  s on ∂Bs .

(3.6)

the displacement condition

and the traction condition

Such a triplet, if it exists, is a solution to the equilibrium problem. We say that the load ( u,  s, b) is admissible if the corresponding equilibrium problem has at least one solution. Proposition 3.1. If ( u,  s, b) is an admissible load, and (u1 , E1 , T1 ) and (u2 , E2 , T2 ) are two solutions to the corresponding equilibrium problem, then T1 (x) = T2 (x) for each x ∈ B. Proof. The triplet (u, E, T), with u = u1 − u2 , E = E1 − E2 , T = T1 − T2 satisfies equation (3.2). It furthermore satisfies (3.4), (3.5) and (3.6) with  = 0 and  b = 0, u s = 0. Thus, in agreement with the hypothesis of smoothness of the solutions, simple application of relation (1.78) proves that  T·E dV = 0. (3.7) B

 1 (x)),  is monotone, since T1 (x) = T(E Bearing in mind that stress function T  2 (x)), from (3.7) we obtain T2 (x) = T(E  ||T||2 dV = 0, (3.8) B

whence, by the continuity of T, we get T = 0 and then T1 = T2 .  From the preceding proposition, by taking into account that the elastic strain is Ee = C−1 [T], we arrive at Ee1 = Ee2 . In general, both the displacement and inelastic strain fields are not unique, as on the contrary happens for linear

3.2 Variational Principles

53

elastic materials. Indeed, for these latter materials the function linking the strain tensor and the Cauchy stress tensor is strongly monotone, that is (C[E1 ] − C[E2 ])·(E1 − E2 ) ≥ ξ||E1 − E2 ||2 , for each E1 , E2 ∈ Sym, with ξ =

inf

(3.9)

A·C[A] (compare (3.9) and

A∈ Sym, ||A||=1

(2.16)). Given the load ( u,  s, b), the stress field T is statically admissible if T is admissible and satisfies both the equilibrium equation (3.4) and the traction condition (3.6). Load ( u,  s, b) is compatible, if there exists at least one statically admissible stress field T. Clearly, the compatibility of the load is a necessary condition for its admissibility. The necessary conditions for a load to be compatible are presented in [38]. It is interesting to highlight that by virtue of property (2.15) of stress  if (u, E, T) is a solution to the equilibrium problem (3.2)-(3.6) function T,  = 0, then (βu, βE, βT) is a solution to (3.2)-(3.6), where in (3.4) and with u (3.6) loads b and  s have been respectively replaced by βb and β s, with β ≥ 0.  BC has no such property. It is a simple matter to verify that T Let us now consider no-tension solids in the presence of thermal variations. The equations governing their thermo-mechanical equilibrium have been set out in [71]. These equations are the strain-displacement relation, the equilibrium equation, the constitutive equations for stress and heat flux, and the equilibrium energy equation. The system obtained in [71] is coupled because the temperature coefficient and the coefficient of the derivative of temperature with respect to time in the energy equation depend on strain and strain rate. ·

·

Here we assume that E, β(ϑ), β  (ϑ), E and ϑ are O(δ), just as E − β(ϑ)I in (2.152). Thus, the thermoelastic equilibrium equations are uncoupled and can be integrated separately. In other words, the temperature field is obtained by integrating the equation of heat conduction and used as a thermal load in the subsequent resolution of the equilibrium problem. In this circumstance, by proceeding as in proposition 3.1, it is possible to prove that the solution is unique in terms of stress [71].

3.2 Variational Principles Besides the approach described in the preceding section, we also consider an alternative formulation of the equilibrium problem, called the weak formulation. In addition, we establish the two minimum principles of the statics of bodies made of masonry-like materials: the principle of minimum potential energy and the principle of minimum complementary energy. These generalize the classical principles of linear elastostatics [51]. In this section we investigate the relations among minimum problems and the two formulations of the equilibrium problem.

54

3 Equilibrium of Masonry Bodies

Hereinafter, the problem described in section 3.1 will be referred as the strong formulation of the equilibrium problem and its solution as the strong solution. Let E and T be continuous symmetric tensor fields on B, with T having negative semidefinite values. We define the strain energy by  ψ(E) dv (3.10) U1 (E) = B

where ψ is the strain-energy density defined in (2.25), and the stress energy by  1 U2 (T) = T · C−1 [T] dv. (3.11) 2 B  For T = T(E), we have U1 (E) = U2 (T);

(3.12)

in fact, from (2.25) and (2.2) we get ψ(E) =

1 1 1 1 E · T = (Ee + Ef ) · T = Ee · T = T · C−1 [T]. 2 2 2 2

(3.13)

By a kinematically admissible state we mean a triplet (u, E, T), with u an admissible displacement field and T an admissible stress field, that satisfies the strain-displacement relation (3.2), the constitutive equation (2.1) and the displacement boundary condition (3.5). We denote AK as the set of all kinematically admissible states and Φ1 as the functional defined on AK ,    Φ1 (u, E, T) = U1 (E) − b · u dv − s · u da, (3.14) B

∂Bs

with (u, E, T) ∈ AK . Φ1 is called the potential energy. Proposition 3.2. (Principle of minimum potential energy) If (u1 , E1 , T1 ) is a solution to the strong equilibrium problem, then (u1 , E1 , T1 ) is a minimum point for Φ1 , (3.15) Φ1 (u1 , E1 , T1 ) ≤ Φ1 (u2 , E2 , T2 ) for every (u2 , E2 , T2 ) ∈ AK , and equality holds only if T1 = T2 . Proof. From the equality E2 · T2 − E1 · T1 = Ee2 · T2 − Ee1 · T1 and the symmetries of C, we obtain  1 U1 (E2 ) − U1 (E1 ) = (Ee − Ee1 ) · C[Ee2 − Ee1 ] dv+ 2 B 2

(3.16)

3.2 Variational Principles

55

 1 2



B

T1 · (Ee2 − Ee1 ) dv =

(Ee2 − Ee1 ) · C[Ee2 − Ee1 ] dv+   T1 · (E2 − E1 ) dv − T1 · Ef2 dv, B

B

(3.17)

B

where the last step follows from the relations   T1 · (E2 − E1 ) dv = T1 · (Ee2 + Ef2 − Ee1 − Ef1 ) dv = B

B

 B

T1 ·

(Ee2

−

Ee1 )

 dv + B

T1 · Ef2 dv.

The identity (1.75) with the divergence theorem (1.76) yields  1 U1 (E2 ) − U1 (E1 ) = (Ee − Ee1 ) · C[Ee2 − Ee1 ] dv+ 2 B 2  − T1 · Ef2 dv+ B    divT1 · (u2 − u1 ) dv. s · (u2 − u1 ) da −

(3.18)

(3.19)

B

∂Bs

From (3.14), in view of (3.19) and the fact that T1 satisfies the equilibrium equation (3.4), we get

1 2

 B

Φ1 (u2 , E2 , T2 ) − Φ1 (u1 , E1 , T1 ) =  (Ee2 − Ee1 ) · C[Ee2 − Ee1 ] dv − T1 · Ef2 dv ≥ 0,

(3.20)

B

by virtue of the positive definiteness of C, the negative semidefiniteness of T1 and the positive semidefiniteness of Ef2 . Thus, (3.15) is proved. In addition, in view of (3.20), Φ1 (u2 , E2 , T2 ) = Φ1 (u1 , E1 , T1 ), if and only if Ee2 = Ee1 , that is, if and only if T2 = T1 .  Let SV and SW be the sets of all vector fields u of class C 1 on B such that,  on ∂Bu . SV and SW are known as the space of respectively, u = 0 and u = u variations and the set of weakly admissible displacements, respectively.  and  Given b, u s as above, the weak equilibrium problem for a body made of a material with constitutive equation (2.1) is to find a triplet (u, E, T) where u ∈ SW , E satisfies (3.2) and T defined by (3.3) is such that     b · u∗ dv = T · E∗ dv (3.21) s · u∗ da + B

∂Bs ∗

∗

1 ∗ 2 (∇u

B

∗T

for each u ∈ SV and E = + ∇u solution to the equilibrium problem.

). (u, E, T) is said to be a weak

56

3 Equilibrium of Masonry Bodies

Proposition 3.3. Let (u, E, T) be a minimum point for the functional Φ1 defined in (3.14). Then, (u, E, T) is a weak solution to the equilibrium problem. Proof. We aim to show that (u, E, T) satisfies (3.21) for each u∗ ∈ SV . For   = T(E + E∗ ) u∗ ∈ SV and ∈ R, we put E∗ = (∇u∗ + ∇u∗T )/2 and T( )   (in particular T(0) = T(E) = T), and consider the kinematically admissible  Thus, we have state (u + u∗ , E + E∗ , T( )).  0 ≤ Φ1 (u + u∗ , E + E∗ , T( )) − Φ1 (u, E, T) =    1 ∗ ∗   s · (u + u∗ ) da− (E + E ) · T( ) dv − b · (u + u ) dv − 2 B B ∂Bs    1  E · T dv + b · u dv + s · u da = 2 B B ∂Bs  1 ∗  (E∗ · T + E · DE T(E)[E ]) dv− { 2 B    s · u∗ da} + o( ) = b · u∗ dv −  { B

B

∗

E · T dv −

∂Bs

 B

∗



b · u dv −

 s · u∗ da} + o( ),

(3.22)

∂Bs

 is differentiable almost as → 0. In (3.22) we have used the fact that T  in everywhere in Sym (cf. proposition 2.20), the major symmetry of DE T(E) (2.196) and relation (2.193). Hence, (3.21) follows from the arbitrariness of .  Now, we are in a position to prove that every weak solution to the equilibrium problem is a strong solution, provided that it is sufficiently smooth. To this end, let us consider triplets (u, E, T), where u is a weakly admissible displacement, E satisfies (3.2) and T is given by (3.3). Proposition 3.4. Let (u, E, T) be a triplet that satisfies (3.21); if u is of class C 2 on B, then (u, E, T) is a strong solution to equilibrium problem (3.2)-(3.6). Proof. Since T satisfies the constitutive equation (2.1), then T is symmetric. In view of (1.75) we have T · ∇u∗ = div(TT u∗ ) − u∗ · divT; from (3.21), by applying the divergence theorem (1.76), we get     s · u∗ da + b · u∗ dv = T · ∇u∗ dv = ∂Bs

B

B

3.2 Variational Principles

 B

div(Tu∗ ) dv −



∗

 

Tn · u da −

B

∂B



B

Tn · u∗ da −

u∗ · divT dv = u∗ · divT dv =



∂Bs

57

B

u∗ · divT dv,

(3.23)

where the last step follows from the fact that u∗ belongs to SW . Lastly, (3.23) gives   ( s − Tn) · u∗ da +

∂Bs

B

(b + divT) · u∗ dv = 0

(3.24)

and from the arbitrariness of u∗ we obtain (3.4) and (3.6).  Let AS be the set of all statically admissible stress fields, and Φ2 the functional defined on AS   da. Φ2 (T) = U2 (T) − s·u (3.25) ∂Bu

Φ2 is called the complementary energy. Proposition 3.5. (Principle of minimum complementary energy) Let T1 be a stress field corresponding to a solution of the strong problem. Then Φ2 (T1 ) ≤ Φ2 (T2 ),

(3.26)

for every T2 ∈ AS , and equality holds only if T1 = T2 . Proof. The proof is analogous to that of proposition 3.2. In fact, it is an easy matter to prove that  1 Φ2 (T2 ) − Φ2 (T1 ) = (T2 − T1 ) · C−1 [T2 − T1 ]dv− 2 B  T2 · Ef1 dv ≥ 0, (3.27) B

−1

because C is positive definite, T2 is negative semidefinite and Ef1 is positive semidefinite. Lastly, Φ2 (T2 ) = Φ2 (T1 ), if and only if T2 = T1 . 

4 The Numerical Method

In chapter 3 we proved that, given the load ( u,  s, b), the set of all weak solutions to the equilibrium problem is exactly the set of all minimizers of the potential energy (propositions 3.2, 3.3 and 3.4). We moreover showed that each strong solution is a minimizer for the complementary energy (proposition 3.5). From such results, it follows that a numerical solution to the equilibrium problem of masonry-like solids can be calculated via different methods. The most common approach is based on a displacement formulation of the finite element method applied to the weak formulation (3.21) of the equilibrium problem [1], [68], [70], [76], [96]. In [32] the equilibrium problem is instead solved by minimizing the complementary energy functional on a suitable set of stress fields. Another possible approach is to minimize the potential energy on the set of all kinematically admissible states [2]. It is assumed that the reader is familiar with the finite element method, otherwise, detailed descriptions can be found in [25], [18], [6], [20], [21], [84], [85] and [57].

4.1 Algorithm for Solution of the Equilibrium Problem In order to solve the equilibrium problem for masonry-like materials by using the finite element method, we are often obliged, for numerical reasons, to assign the load incrementally. To this end, although the material being considered is hyperelastic, we must assign a loading process and consider the associated incremental equilibrium problem. We then intend to prove that the numerical solution obtained by using an incremental procedure is independent of the particular loading process chosen. Instead, it depends solely on the final assigned load, provided that the loading process considered is admissible in the sense specified as follows. A loading process (τ ), with τ ∈ [0, τ ], is a function triplet (τ ) = ( u(., τ ),  s(., τ ), b(., τ )),

0 ≤ τ ≤ τ,

(4.1)

60

4 The Numerical Method

 and  with b, u s defined respectively on B × [0, τ ], ∂Bu × [0, τ ] and ∂Bs × [0, τ ], piece-wise differentiable with respect to τ , and such that (0) = (0, 0, 0). We say that a loading process  is admissible if (i) for every τ , (τ ) is an admissible load, (ii) at least one of the curves τ → (u(., τ ), E(., τ ), T(., τ )) of solutions to the equilibrium problem (3.2)-(3.6), corresponding to (τ ) is regular, i.e., piece-wise differentiable with respect to τ . Let  be an admissible loading process, and (u(., τ ), E(., τ ), T(., τ )) a corresponding regular curve of solutions. By differentiating (3.2)-(3.6) with respect to τ , we obtain, for τ ∈ [0, τ ], the following incremental equilibrium problem ·

E=

1 · ·T (∇u + ∇u ), 2

·

·

 T = DE T(E(τ ))[E], ·

·

divT + b = 0 on B, ·

·

 on ∂Bu , u=u ·

·

(4.2) (4.3) (4.4) (4.5)

s on ∂Bs , Tn = 

(4.6)

u(x, 0) = 0, E(x, 0) = 0, T(x, 0) = 0 on B,

(4.7)

and where the dot · denotes the derivative with respect to τ . Note that (4.3) is  well-defined, since in light of proposition 2.20, E → T(E) is differentiable almost everywhere in Sym. Thus, (u(., τ ), E(., τ ), T(., τ )) is a solution to the incremental equilibrium problem (4.2)-(4.6). Vice versa, if a regular curve τ → (u(., τ ), E(., τ ), T(., τ )) solves (4.2)-(4.6), then for each 0 ≤ τ ≤ τ , it is a solution to the equilibrium problem corresponding to ( u(., τ ),  s(., τ ), b(., τ )), as we deduce by integrating (4.2)-(4.6) on [0, τ ] and taking into account (4.7). Hence it follows that (i) The solution to the incremental problem, if it exists, is unique in terms of stress; (ii) If 1 and 2 are two admissible processes on [0, τ ], such that 1 (τ ) = 2 (τ ), and (u1 (., τ ), E1 (., τ ), T1 (., τ )) and (u2 (., τ ), E2 (., τ ), T2 (., τ )) are two solutions to the incremental problem respectively corresponding to 1 and 2 , then (4.8) T1 (x, τ ) = T2 (x, τ ), for each x ∈ B. This last result guarantees that the solution to the incremental problem does not depend on the loading process, at least with regard to the stress. In dealing with the equilibrium problem in presence of thermal variations, the loading process includes the temperature as well. Just as in the isothermal

4.1 Algorithm for Solution of the Equilibrium Problem

61

case, we can prove that the stress field obtained by solving the incremental equilibrium problem is independent of the loading process. Now we assume that body B is subjected to the loading process (τ ) in (4.1). Our aim is to numerically determine a triplet (u(., τ ), E(., τ ), T(., τ )) that satisfies the equilibrium problem (3.2), (3.4), (2.1), (3.5) and (3.6) for each τ ∈ [0, τ ]. The non-isothermal case is dealt with in the Appendix. Let w be a displacement field such that w = 0 on ∂Bu . From (3.4) and (3.6) it follows that at every τ , the following equilibrium equation must be satisfied     s · w dA + b · w dV, T · ∇w dV = (4.9) B

B

∂Bs

 with T = T(E) (cf. (3.21)). By differentiating (4.9) with respect to τ, we obtain the incremental equilibrium equation  ·  ·   s · w dA DE T(E)[ E] · ∇w dV = B



∂Bs ·

b · w dV,

+

(4.10)

B

where the dot denotes the derivative with respect to τ. As usual, the body B is discretized into p finite elements B e , for a total number of q nodes [57]. The coordinates and displacements of a point belonging to an element are expressed in terms of coordinates and displacements of the nodes of the element, via suitable shape functions. Thus, taking into account the arbitrariness of w, the finite element method allows us to transform the equilibrium equation (4.9) into the nonlinear equilibrium system [K{u}]{u} = {f }

(4.11)

and the incremental equilibrium equation (4.10) into the nonlinear evolution system ·

·

[KT ]{u} = {f },

(4.12)

·

where {u} and {u} are the arrays of the nodal displacements and velocities, ·

{f } and {f } are the vectors of nodal forces and their time derivatives, [K] is the stiffness matrix and [KT ] the tangent stiffness matrix. The 3q × 3q matrix [KT ] stems from the assembly of the p elemental matrices [KTe ] obtained from the relation  · ·e e e  E] · ∇w dV (4.13) {c } · [KT ]{u } = DE T(E)[ Be ·e

·

where {ce } and {u } are the vectors of the nodal values of fields w and u, ·

corresponding to the finite element e. Vector {f } in (4.12) is the assembly of ·e

the p elemental vectors {f } calculated from the relation

62

4 The Numerical Method ·e



·



 s · w dA +

{ce } · {f } = ∂Be ∩∂B

s

·

b · w dV.

(4.14)

Be

We assume that equation (4.9) holds in correspondence of τ , and that the body is therefore in equilibrium. We then assign a load increment {Δf } defined by the relations  ( s(τ + Δτ ) −  s(τ )) · w dA {ce } · {Δf e } =  +

∂Be ∩∂Bs

(b(τ + Δτ ) − b(τ )) · w dV.

(4.15)

Be

After assembly, we then solve the nonlinear system [KT ]{Δu} = {Δf }

(4.16)

by following the Newton-Raphson iterative procedure described below.  necessary for calculation of The matrix of the components of tensor DE T, the tangent stiffness matrix, is explicitly calculated in Section 2.4 for threedimensional and in Appendix for the two-dimensional case. In order to briefly recall the iterative procedure aimed at determining the displacement increment for any load increment, let us consider the following quantities related to the i − th iteration of the j − th load increment, vector of nodal displacements, {u(i,j) }  D({u(i,j) }) matrix of the components of DE T, (i,j) })] tangent stiffness matrix, [KT ({u nodal equivalent of the assigned incremental loads if i = 0, {f (i,j) } nodal equivalent of the residual loads if i ≥ 1, (i,j) eG vector of the engineering components of the total strain, (i,j) aG vector of the engineering components of the fracture strain, (i,j) tG vector of the engineering components of the stress, where the subscript G indicates the Gauss integration point at which these quantities are calculated. In the first iteration of the first load increment, {u(0,1) } is nil, and D({u(0,1) }) coincides with the matrix of the material’s elastic moduli. Let us suppose that during the j − th load increment, we have calculated the displacement {u(i,j) }, the tangent stiffness matrix [KT ({u(i,j) })] and the nodal equivalent loads {f (i,j) } corresponding to the i − th iteration; we must therefore solve the linear system [KT ({u(i,j) })]{Δu(i,j) } = {f (i,j) }

(4.17)

in order to determine the displacement {u(i+1,j) } = {u(i,j) } + {Δu(i,j) } relative to the (i + 1) − th iteration.

4.2 Fracture Strain Tensor and Cracked Regions

63

Then, for every Gauss point of every element we calculate the total strain associated with the displacement {u(i+1,j) }, and its eigenvalues which (i+1,j) (i+1,j) and the stress tG , as are needed to calculate the fracture strain aG per formulae (2.60)-(2.63). We moreover calculate matrix D({u(i+1,j) }) which may be used as needed in the next iteration or in the next load increment. In order to avoid the presence of null pivot during solution of the linear system (4.17), at the Gauss points at which all stress components are zero, it is convenient to calculate  = O (see equations D({u(i+1,j) }), rather than by starting with relation DE T (2.186)-(2.189)), by using the elastic moduli matrix multiplied by a sufficiently small quantity . Using this numerical expedient makes solving the system easier, and consequently allows calculating a solution to the equilibrium problem. A slightly different technique has been proposed in [1]. Lastly, we calculated the vector of the residual loads given by

(i+1,j) eG

{f (i+1,j) } =

j 

{f (0,k) } − [K{u(i+1,j) }]{u(i+1,j) }

(4.18)

k=1

and perform the convergence check ||{f (i+1,j) }|| ≤ ξc , j ||{f (0,k) }||

(4.19)

k=1

where ξc is a suitable parameter assigned by the user, and

j

{f (0,k) } is the

k=1

nodal equivalent of the overall loads, including the reaction forces. If convergence is reached, we proceed with the next load increment; otherwise, we repeat all operations beginning with solution of system (4.17). The convergence check guarantees that equilibrium is reached. Therefore, the stress field calculated is admissible (it satisfies the constitutive equation) and, within error ξc , equilibrated with the assigned loads.

4.2 Fracture Strain Tensor and Cracked Regions Tensor Ef is called fracture strain because, if it is non-null in any region of a masonry-like solids, then we can expect fractures to be present in that region. In the three-dimensional case, determining the cracks can be complex and their rendering require sophisticated graphic tools. Nevertheless, knowing the principal components t1 ≤ t2 ≤ t3 of the stress, and ef1 ≤ ef2 ≤ ef3 of the fracture strain in each point of the structure can provide information on any eventual crack surfaces, bearing in mind that in a masonry-like material the principal directions of T and Ef coincide. At every point P of the structure,

64

4 The Numerical Method

the following four situations can occur. If the three principal stresses are negative, and the principal fracture strains equal to zero (t1 ≤ t2 ≤ t3 < 0, ef1 = ef2 = ef3 = 0), the masonry behaves as a linear elastic material (the total strain at P belongs to R1 in (2.53)). If the principal stresses are zero and the fracture strains non-zero (t1 = t2 = t3 = 0, 0 < ef1 ≤ ef2 ≤ ef3 ), the material is completely cracked (the total strain at P belongs to R2 in (2.54)). If the principal stress t3 along the principal direction q3 is zero, and ef3 is different from zero (t1 ≤ t2 < t3 = 0, ef1 = ef2 = 0 < ef3 ), then the plane orthogonal to q3 is a fracture plane (Figure 4.1). Such a situation occurs when the total strain at P belongs to R4 , as defined in (2.56).

Fig. 4.1. Fracture plane orthogonal to the direction q3 corresponding to the principal stress t3 = 0.

Lastly, if both principal stresses t2 and t3 are zero (t1 < t2 = t3 = 0, ef1 = 0 < ef2 ≤ ef3 ), the material exhibits two fracture planes, one orthogonal to q2 , the other orthogonal to q3 , both shown in 4.2. This case takes place when the total strain at P belongs to R3 in (2.55). In the plane cases it is possible to obtain information about the direction of the factures from an analysis of the components of Ef . Let us assume that in a fixed point of the solid, one (and only one) eigenvalue of T is zero, or, equivalently, that E belongs to the region S3 defined in (2.80) (or T3 in (2.96)). In this case, Ef has one eigenvalue equal to zero and one that is positive, and the characteristic direction of the latter coincides with the characteristic direction corresponding to the null eigenvalue of T. Therefore, fractures can open in this direction. f f f , E22 and E12 , be the components of the fracture strain Ef with Let E11 respect to a fixed orthonormal basis {e1 , e2 }. The eigenvalues ef1 and ef2 of Ef are

4.2 Fracture Strain Tensor and Cracked Regions

65

Fig. 4.2. planes orthogonal to the directions q2 and q3 corresponding to the principal stresses t2 = 0 and t3 = 0, respectively.

f Fig. 4.3. Behavior of the cracks as the sign of E12 varies.

ef1 ef2

=

=

f f + E22 − E11

f f + E22 + E11



f f 2 f 2 − E22 ) + 4(E12 ) (E11

2 f f 2 f 2 (E11 − E22 ) + 4(E12 )

2

,

(4.20)

.

(4.21)

Let us assume now that 0 = ef1 < ef2 . Since we may limit ourselves to the f = 0, the components q1 , q2 of the eigenvector corresponding to ef2 case of E12 satisfy the condition Ef q2 = 22 q . (4.22) f 1 E12

66

4 The Numerical Method

f f Because E22 > 0, we can conclude that as the sign of E12 varies, the fractures exhibit the behavior shown in Figure 4.3. By applying this criterion for each point of the structure under examination, it is possible to draw the corresponding fracture curve.

4.3 The Finite Element Code COMES-NOSA The finite element code NOSA (Nonlinear Structural Analysis) has been developed at the Mechanics of Materials and Structures Laboratory of ISTI-CNR with the aim of testing new material constitutive models. It has moreover been applied to checking the algorithms used for integrating the equations of motion, as well as to other numerical techniques for solving structural engineering problems. The development of the code began in 1980 and has continued over the ensuing years along the research lines of the Laboratory. In 1996, the Laboratory became a member of the Multi-centre Network for Computational Solid Mechanics (COMES). Since then, a large part of the laboratory’s development activities have been conducted within this framework, giving rise to the COMES-NOSA code [34]. The code includes in its material library all the constitutive models described in chapter 2, as well as the numerical techniques described in section 4.1 [68], [70], [76]. It allows for studying the statics of masonry solids and modeling restoration and reinforcement operations on constructions of particular architectural interest [89], [79]. Moreover, the code has been extended to perform nonlinear heat conduction analysis on solids, even in non-stationary cases, with boundary conditions concerning temperature and thermal fluxes. Today, the code enables thermomechanical analysis of no-tension solids whose mechanical characteristics depend on temperature, in the presence of thermal loads [71], [88]. The code has been used to study the San Nicol`o Cathedral in Noto [77], the scenic arch of the Goldoni’s theatre in Livorno [75], [67], the Medici Arsenal in Pisa [78], the Buti bell tower [13], the Pisa church of San Pietro in Vinculis [12], [11], and the church of Santa Maria Maddalena in Morano Calabro [14] and [74]. These studies proved that although the masonry-like model does not, at least in its original formulation, take into account some of the material’s actual features, such as anisotropy, it does enable conducting realistic analysis of the complex, large-size structures frequently encountered in real applications. The flow-chart and element library of COMES-NOSA are presented in the Appendix C.

5 Masonry Arches, Vaults and Domes

The numerical method proposed in this chapter allows for analyzing any kind of vault (masonry spire, fan vault, incomplete domes, etc.), subjected to any type of static load, problems which can be difficult to solve using classical methods [55]. In addition, it is possible to determine the collapse multiplier for different types of loads.

5.1 Shells and Masonry Vaults Consider the vault element with thickness h shown in Figure 5.1; let η1 and η2 be an orthogonal coordinate system, not necessarily the principal one, defined on the mean surface S, with ζ ∈ [−h/2, h/2] as the coordinate in the normal direction n. We denote by g1 and g2 the unit tangent vectors to η1 and η2 axis, respectively, and assume that g1 and g2 are linearly independent. We assume that for each p ∈ S and ζ ∈ [−h/2, h/2], the stress tensor T satisfies the condition T(p, ζ)n(p) = 0. (5.1) As far as the constitutive equation is concerned, we assume that the tensile strength is null and the compressive strength is infinite. Thus, in view of the assumption (5.1), we adopt the relations (2.99)-(2.101) obtained in subsection 2.1.2. The equilibrium problem of masonry vaults and domes is solved by using quadrilateral, eight-node shell elements based on the Love-Kirchhoff hypothesis [83], [50]. As described in section 4.2, knowing fracture strain Ef allows one to garner information on the crack distribution. To this end, the vault can be considered to be made up of the layers Sζ = {p | p = p + ζn, p ∈ S, n = n(p)},

ζ ∈ [−h/2, h/2].

(5.2)

and the criterion of section 4.2 can be applied for each point of the surface Sζ , making it possible to draw the corresponding fracture curves.

68

5 Masonry Arches, Vaults and Domes

Fig. 5.1. Vault element.

5.2 The Maximum Modulus Eccentricity Surface For each point p = p + ζn, with p belonging to the mean surface S, we denote by g(γ) = (cos γ, sin γ) the unit tangent vector to S at p, where γ ∈ [−π/2, π/2) is the angle formed by g(γ) with the direction of η1 . We set σ(p, ζ, γ) = g(γ) · T(p, ζ)g(γ);

(5.3)

thus, the relations h/2 N (p, γ) =

h/2 σ(p, ζ, γ) dζ,

M (p, γ) =

−h/2

σ(p, ζ, γ)ζ dζ,

(5.4)

−h/2

respectively define the normal force and the bending moment per unit length corresponding to g(γ). Because T(p, ζ) is negative semidefinite, N (p, γ) is non-positive for every γ. Moreover, we have    h/2      |M (p, γ)| =  σ(p, ζ, γ)ζ dζ    −h/2  h/2 ≤

|σ(p, ζ, γ)ζ| dζ ≤ −h/2

h |N (p, γ)|, 2

(5.5)

where the next to last step is justified by the fact that σ(p, ζ, γ) ≤ 0 for each γ and ζ. We define tensors M(p) and N(p) as follows  h/2  h/2 M(p) = T(p, ζ)ζ dζ, N(p) = T(p, ζ) dζ, (5.6) −h/2

−h/2

5.2 The Maximum Modulus Eccentricity Surface

69

and call their components internal forces. Since M (p, γ) = g(γ) · M(p)g(γ),

(5.7)

N (p, γ) = g(γ) · N(p)g(γ),

(5.8)

it is a simple matter to show that inequality (5.5) is equivalent to the conditions g(γ) · (M(p) −

h h N(p))g(γ) ≥ 0, g(γ) · (M(p) + N(p))g(γ) ≤ 0, (5.9) 2 2

for each γ ∈ [−π/2, π/2), which in turn, express the positive and negative semidefiniteness of M(p) − hN(p)/2 and M(p) + hN(p)/2. Now let us consider a subset S of S where M and N are continuous and N = 0. For each γ ∈ [−π/2, π/2), let us define the corresponding eccentricity e(p, γ) =

M (p, γ) . N (p, γ)

(5.10)

Proposition 5.1. Let p ∈ S . If there exists γ1 such that N (p, γ1 ) = 0, then e(p, γ) is a constant function of γ. Proof. If N (p, γ1 ) = 0, then, from the negative semidefiniteness of N(p), the relation (5.11) N(p)g(γ1 ) = 0 follows. Moreover, using the preceding relations in (5.9), we arrive at M(p)g(γ1 ) = 0.

(5.12)

Thus, a vector g(γ2 ) orthogonal to g(γ1 ) is a common eigenvector of M(p) and N(p). We now denote by m and n < 0 the corresponding eigenvalues. By setting g(γ) = β1 g(γ1 ) + β2 g(γ2 ), with β12 + β22 = 1, we obtain e(p, γ) =

m (β1 g(γ1 ) + β2 g(γ2 )) · M(p)(β2 g(γ2 )) = , (β1 g(γ1 ) + β2 g(γ2 )) · N(p)(β2 g(γ2 )) n

(5.13)

and thus e(p, γ) does not depend on γ.  In view of proposition 5.1 function e(p, γ) can be extended by continuity to γ = γ1 , putting e(p, γ1 ) = m/n. Therefore, for each p ∈ S , e(p, γ) depends continuously on γ. For each point p of S , let γ0 ∈ [−π/2, π/2) be the value of γ (not necessarily unique) for which the function |e(p, γ)| = |M (p, γ)/N (p, γ)| reaches its maximum value. The quantity e(p) =

M (p, γ0 ) N (p, γ0 )

(5.14)

70

5 Masonry Arches, Vaults and Domes

is the maximum modulus eccentricity at point p. The surface M = {p | p = p + e(p)n, p ∈ S , n = n(p)}

(5.15)

obtained by translating points p belonging to S by the value e(p) along the unit normal vector n to S is the maximum modulus eccentricity surface (m.m.e.s.). The only points p where M is not defined are those for which two values γ0 and γ1 exist that maximize the function |e(p, γ)| with e(p, γ0 ) = −e(p, γ1 ). Proposition 5.2. Let p ∈ S . The maximum modulus eccentricity defined in (5.14) has the expression ⎧ ⎨ ω1 if det N(p) = 0 and |ω1 | ≥ |ω2 |, e(p) = ω2 if det N(p) = 0 and |ω1 | ≤ |ω2 |, (5.16) ⎩m if det N(p) = 0, n where tr(N−1 M)(det N) −



(tr(N−1 M))2 (det N)2 − 4(det N)(det M) , 2 det N (5.17)  tr(N−1 M)(det N) + (tr(N−1 M))2 (det N)2 − 4(det N)(det M) ω2 = . 2 det N (5.18) ω1 =

Proof. With the aim of determining the maximum and the minimum of function e(p, γ), as γ varies in [−π/2, π/2), we can limit ourselves to the case in which det N(p) = 0. In fact, as already pointed out, if det N(p) = 0, then e(p, γ) is the constant function m/n. Due to (5.10), (5.7) and (5.8), the quantity e(p, γ) is the Rayleigh quotient (cf. (1.61)) corresponding to the generalized eigenvalue problem M(p)f = ωN(p)f ,

(5.19)

whose eigenvalues are ω1 and ω2 given in (5.17) and (5.18) with eigenvectors f1 and f2 . In view of proposition 1.12, e(p, γ) belongs to the interval [ω1 , ω2 ], and we can therefore conclude that the minimum and the maximum of e(p, γ) are ω1 and ω2 , and (5.16) is thus proved.  In the particular case in which the vault’s geometry and loads possess axial symmetry, then the eccentricity can attain its maximum modulus only in the direction of parallel or meridians [73]. Note that, in view of equality (5.5), the m.m.e.s. corresponding to a negative semidefinite stress field is entirely contained within the vault. The m.m.e.s. allows concise, effective rendering of the results of finite element analyses, as well as an evaluation the safety of vaults, as we shall see in the next section.

5.3 The Limit Analysis of Masonry Arches and Vaults

71

Let us consider the barrel vault of Figure 5.2, subjected to boundary conditions and load distribution which are independent of axial coordinate z. We choose g1 and g2 as the unit vectors in the circumferential and longitudinal direction, respectively, and assume that the normal force N22 (p) = g2 ·N(p)g2 is zero. In view of proposition 5.1, e(p, γ) = M11 (p)/N11 (p), for each γ, where M11 (p) = g1 · M(p)g1 , N11 (p) = g1 · N(p)g1 . In particular, an arch can be viewed as the portion of a barrel vault between two planes with equations z = z1 and z = z2 . The intersection of the maximum modulus eccentricity surface corresponding to the vault with a vertical plane is known as the arch’s line of thrust [56].

Fig. 5.2. Barrel vault.

5.3 The Limit Analysis of Masonry Arches and Vaults The methods of the limit analysis, usually employed for structures made of ideally plastic materials, have been extended to the study of masonry arches in [61], [54], [56], [62], [17], [35], [36], [27] and [28]. Such extension is based on the assumption that masonry has zero tensile strength and infinite compressive strength and, moreover, that sliding failure can not occur. From these hypotheses it follows that if a thrust line can be found, which is in equilibrium with external loading, and which lies everywhere within the masonry of the arch ring, then the arch is safe [56]. Subsequently, [39] showed that the classical static and kinematic theorems of limit analysis, proved for ideally plastic materials [43], hold for a larger class of materials which also includes masonry-like ones. Characterization of the collapse and proof of the limit analysis theorems performed in [39] are based on two hypotheses peculiar to both no-tension and elastic plastic materials. In fact, both these materials types are characterized by the existence of a convex set of admissible stresses and an

72

5 Masonry Arches, Vaults and Domes

orthogonality property linking admissible stresses and inelastic strains (i.e., fracture strain for no-tension materials and plastic strain for elastic plastic materials). In [72] a method is proposed for determining the collapse load in the case of circular arches subjected to their own weight and a vertical point load applied at a point of the extrados. The maximum modulus eccentricity surface, introduced in section 5.2 plays for masonry vaults a role analogous to the line of thrust for arches: if it is possible to find a m.m.e.s. equilibrated with the load and entirely contained within the thickness of the vault, then the vault is safe. By following Heyman’s suggestion for arches [56], it is possible to define a ”geometrical” factor of safety for vaults as well. Specifically, for each p ∈ S, we put  ω2 − ω1 , if det N(p) = 0, h(p) = (5.20) 2|m/n|, if det N(p) = 0, 

and determine

h(p) da,

v=

(5.21)

S

which is the volume of the smallest vault able to contain the m.m.e.s. Then we determine the volume V of the vault and set ϕv = Alternatively, we can put

v V −v =1− . V V

(5.22)

2 | e(p)| , h(p) p∈S

(5.23)

e = sup

where h(p) is the thickness of the vault, and consider ϕs = 1 − e.

(5.24)

Both ϕv and ϕs can be used to assess the safety of vaults. However, it should be noted that coefficient ϕs may turn out to be very conservative. Indeed, in light of the definition in (5.23), the presence of only one point at which the m.m.e.s. comes near the boundary of the vault will yield a ϕs approaching zero. In order to obtain an estimate of a vault’s safety, it can be important to know the value of the collapse load and the corresponding mechanism for some types of loads. Explicit calculation of the collapse load and mechanisms is possible only in a few, particularly simple, structures, such as arches [72], toroidal tunnels [73], circular plates and spherical domes [74]. In more complex cases, an incremental finite element analysis is required. The value of the collapse load is then obtained by progressively increasing the load assigned to the structure until it is no longer possible to determine an admissible equilibrated solution.

5.3 The Limit Analysis of Masonry Arches and Vaults

73

Knowing of the m.m.e.s. corresponding to the last equilibrated load may be very useful in determining the collapse mechanism [39]. In fact, if the m.m.e.s. is tangent to the extrados or intrados along a path, such a path can be considered the site of cylindrical hinges. The corresponding rotational axis coincides with the direction orthogonal to the direction for which the absolute value of the eccentricity is maximum. Therefore, under the hypothesis that the masonry has infinite compressive strength and sliding failure cannot occur, as the load increases, vault collapse occurs when the m.m.e.s. is tangent to the intrados or extrados along paths, in such a way as to determine a hinge distribution sufficient to render the vault a kinematically undetermined structure. Finally, it should be pointed out that in the model presented herein, collapse of a masonry vault is due to the formation of hinges on its extrados and intrados. However, other collapse mechanisms can be considered, for example by removing the ”non-sliding” assumption [62] or assuming a bounded compressive strength for the masonry [63].

6 Comparison Between Explicit and Numerical Solutions

This chapter is aimed at assessing the effectiveness of proposed numerical method. To this end, comparisons between explicit and numerical solutions are presented for some equilibrium problems. Some simple solutions to the equilibrium problem of masonry-like solids have been collected, and the exact solutions then compared with the corresponding numerical results obtained via the finite element code COMES-NOSA. Firstly, we analyze a circular ring and a spherical container, both subjected to uniform radial pressures pe and pi acting, respectively, on the outer and inner boundary, and explicitly calculate the solution (u, E, T) of the equilibrium problem. Next, we determine the explicit solution to the equilibrium problem for trapezoidal panels made of a no-tension material; the panels are clamped at their bases and subjected to normal and tangential loads distributed on their tops, under the hypothesis of plane stress and the absence of body forces. In all these cases the solution is unique also in terms of displacement and strain. In the following, ν and E are respectively Poisson’s ratio and Young’s modulus of the material. For all the numerical calculations we put ν = 0.1 and E = 5000 MPa. We assume that σ t = 0, 0 < σ c < +∞ for the circular ring, and σ c = ∞ for the other cases. Moreover, we suppose that pe and pi satisfy the compatibility conditions pe ≤ σ c and pi ≤ σ c . A stress field T that is in equilibrium with the assigned loads and satisfies the constitutive conditions T − σ t I ∈Sym− and T + σ c I ∈Sym+ (cf. (2.109)1 ) and (2.109)2 ) is said to be statically admissible (cf. definition of statically admissible stress field in section 3.2). The last three examples of this section deal with the limit analysis of simple masonry structures. They are subjected to a load c(λ) = c0 + λc1 , where c0 and c1 are respectively the permanent and variable part of the load, and λ ≥ 0 is the load multiplier. In order to determine the collapse multiplier λc for the three examples considered, we proceed as follows. For λ > 0 we determine internal forces fields that are admissible (i.e., satisfying condition (5.5)) and equilibrated

76

6 Comparison Between Explicit and Numerical Solutions

with the load c(λ), and then find the multiplier λ = λc such that there is a collapse mechanism corresponding to the internal forces equilibrated with c(λ). These are called collapse internal forces [72], [73], [74]. Once the collapse load has been explicitly determined, for the sake of comparison, we conduct a finite element analysis by increasing the variable load until a value λs of λ beyond which it is impossible to obtain a numerical solution to the equilibrium problem. The following examples show that in order to obtain a good approximation of the collapse load, it is necessary to use sufficiently fine meshes as well as small load increments.

6.1 The Circular Ring The circular ring shown in Figure 6.1 has inner radius a, outer radius b and is subjected to a state of plane strain as a consequence of the action of two uniform radial pressures pe and pi acting, respectively, on the outer and inner boundary [8].

Fig. 6.1. The circular ring.

Due to the symmetry of both geometry and loads, the shear components of the stress are zero, and in view of proposition 2.14, the principal components of the strain, fracture strain and crushing strain tensors are the radial and circumferential ones. We refer to [8] for more details on the procedure adopted to calculate the explicit solution to the equilibrium problem. Here we keep pi fixed and summarize the different solutions as pe varies. In particular, the following situations can occur.

6.1 The Circular Ring

77

1. Pressure pe satisfies the inequality pe


(6.15)

For the numerical analysis, the following values of the constants have been chosen a = 1 m, b = 1.5 m, pi = 0.1 MPa, pe = 0.23 MPa, σ c = 0.5 MPa, In this case the ratio (σ c − pe )/(σ c − pi ) = 0.675 lies within the interval (a/b, (a2 + b2 )/(2b2 )) = (0.667, 0.722) and the transition radius is approximately rc = 1.28 m. By reasons of symmetry, only a quarter of the circular ring has been studied, and this was discretized into 1000 plane strain fournode elements. Figure 6.2, 6.3 and 6.4 show the behavior of the radial stress, circumferential stress and circumferential crushing strain. The black line represents the exact solution, while the red one the numerical results. The circular ring is then subjected to a loading process, with pi = 0.1 MPa and pe increasing from pe0 = 0.0667 MPa to pef = 0.2333 MPa. Figure 6.5

6.2 The Spherical Container

79

shows the behavior of the radius r∗ separating the region where the inelastic strain Ef +Ec is non-zero from the region in which Ef +Ec = 0. In accordance with equations (6.3) and (6.11), the expression of r∗ is

r

∗



pe pi







pe = 1.5 1.5 − pi r



∗

 pe p2e ≤ 0.722, (6.16) 2.25 2 − 1 , for 0.0667 ≤ pi pi



 pe pe ≤ 2.111, = 1, for 0.722 ≤ pi pi     pe pe ∗ r = 0.375[1.5 5 − pi pi



pe − 2.25 5 − pi

2 − 16], for 2.111 ≤

pe ≤ 2.333. pi

(6.17)

(6.18)

Fig. 6.2. Radial stress σr (MPa) vs. r (m).

6.2 The Spherical Container Let us now consider a spherical container made of a masonry-like material; it has inner radius a and outer radius b, and is subjected to two uniform radial pressures pe and pi acting on the external and internal boundary, respectively [76]. Let {O, r, θ, ϕ} be a spherical reference system, with origin O coinciding with the centre of the container. As in the preceding example, due to the symmetry in both geometry and loads, the only displacement component different from zero is the radial one, and no shear component is allowed for stress and strain. In [8] it has been shown that if

80

6 Comparison Between Explicit and Numerical Solutions

Fig. 6.3. Circumferential stress σθ (MPa) vs. r (m).

Fig. 6.4. Circumferential crushing strain cθ vs. r (m).

pe 2a3 + b3 ≥ , pi 3b3

(6.19)

the solution coincides with that of a linear elastic material, and the stress has principal components σr (r) =

a3 b3 (pe − pi ) 1 pi a3 − pe b3 + , b3 − a3 r3 b3 − a3

σθ (r) = σϕ (r) = −

a3 b3 (pe − pi ) 1 pi a3 − pe b3 + . 3 3 3 2(b − a ) r b3 − a3

If the ratio pe /pi satisfies the inequalities

(6.20) (6.21)

6.2 The Spherical Container

81

Fig. 6.5. Transition radius r∗ (m) vs. pe /pi .

a2 pe 2a3 + b3 ≤ ≤ , 2 b pi 3b3 

we have σr (r) =

2

− ar2 pi , 2 − a3 pi

 σθ (r) = σϕ (r) =

2 rr03

0, a2 3 pi



+

1 r02

r0 r3

−

(6.22)

 a ≤ r ≤ r0 , , r0 ≤ r ≤ b;

1 r02

 a ≤ r ≤ r0 , , r0 ≤ r ≤ b,

(6.23)

(6.24)

where r0 is the sole real root of the cubic equation, 2a2 pi r3 − 3b3 pe r2 + a2 b3 pi = 0

(6.25)

that belongs to [a, b] and separates the region of non-zero circumferential fracture strain from the region where it is zero. As pe /pi varies from (2a3 + b3 )/(3b3 ) to a2 /b2 , radius r0 correspondingly varies from a to b. The radial displacement and the circumferential fracture strain are ⎧ ! ⎨ pi a2 1 − (1 − ν) 1 , a ≤ r ≤ r0 , E r r0 ! (6.26) u(r) = 3 2 ⎩ pi a2 (2ν − 1)r + (1 + ν) r02 , r0 ≤ r ≤ b, 3E r r 0



and εfθ (r)

= εfϕ (r) =

1−ν a2 E r pi

0,



1 r

−

1 r0



, a ≤ r ≤ r0 , r0 ≤ r ≤ b.

(6.27)

For values of pe /pi less than a2 /b2 , there are no statically admissible stress fields.

82

6 Comparison Between Explicit and Numerical Solutions

The container was discretized with 1000 axisymmetric eight-node elements, using the following values for the material properties and pressures: a = 1 m, b = 1.5 m, pi = 0.1 MPa, pe = 0.046 MPa With this choice of parameters, in view of (6.25), we get r0 = 1.24 m. Figures 6.6, 6.7, 6.8 and 6.9 show the radial stress, the circumferential stress, the circumferential fracture strain and the radial displacement, respectively; the black line represents the exact solution, the red one the numerical solution.

Fig. 6.6. Radial stress σr (MPa) vs. r (m).

6.3 The Trapezoidal Panel We propose to determine the explicit solution to the equilibrium problem for panels made of a no-tension material. A trapezoidal panel with minor base 2b, major base 2B and height h is clamped at its base and subjected to normal and tangential loads distributed on its top, under the hypothesis of plane strain and in absence of body force (Figure 6.10). The method used to determine the solution is described in [80]; herein we limit ourselves to a concise presentation of the results. Firstly, in the spirit of the semi-inverse method, we assume that one of the principal stresses is null, whereas the other is strictly negative, so that we can write for the stress tensor   2   κ κ σx τxy , (6.28) = σy T= τxy σy κ 1

6.3 The Trapezoidal Panel

83

Fig. 6.7. Circumferential stress σθ (MPa) vs. r (m).

Fig. 6.8. Circumferential fracture strain εfθ vs. r (m).

with σy < 0, and κ the cotangent of the angle between the active isostatic line (the one corresponding to the non-null principal stress) and the x axis. From (6.28) we deduce the system of equilibrium equations  ∂κ κ ∂κ ∂x + ∂y = 0, (6.29) ∂σy ∂σy κ ∂x + σy ∂κ ∂x + ∂y = 0, which formally coincide with the one-dimensional Euler’s equations for the flow of an isentropic gas when the pressure gradient vanishes [31]. (6.29)1 coincides with the inviscid Burger’s equation [105], which has already been derived in [41], and (6.29)2 is a linear equation. With p and q the horizontal and vertical loads, respectively, and f = q/p, the solution to system (6.29) is implicitly defined by the relations κ(x, y) = f (x − κy),

σy (x, y) = −

p(x − κy) , 1 + yf  (x − κy)

(6.30)

84

6 Comparison Between Explicit and Numerical Solutions

Fig. 6.9. Radial displacement u (m) vs. r (m).

Fig. 6.10. The trapezoidal panel.

where f  denotes the derivative of f [105]. Once the stress is determined, the elastic part of the strain is obtained from  2  κ − ν κ(1 + ν) σy e , (6.31) E = E κ(1 + ν) 1 − νκ2 where E and ν are respectively Young’s modulus and Poisson’s ratio of the material. For the fracture strain, from (6.28) and (2.12), we deduce

6.3 The Trapezoidal Panel

 Ef = a

1 −κ −κ κ2

85

 ,

(6.32)

where, in the light of (2.2)3 , a must be a scalar non-negative field, a(x, y) ≥ 0 for each (x, y) ∈ Ω.

(6.33)

By requiring that the total strain E = Ee + Ef satisfy the compatibility condition, after some manipulations, we obtain that a must satisfy the parabolic partial differential equation ∂a ∂(aκ) 1 ∂(aκ2 ) + 2 = (σy (1 + κ2 )), +2 2 ∂x ∂x∂y ∂y E

(6.34)

where  is the Laplace operator. It should be noted that, for no-tension materials, stress compatibility does not mean that the trace of the stress must be harmonic, as in the linear elastic case (in the absence of body forces), but calls for the existence of a non-negative function a which satisfies (6.34). In [80], an explicit representation formula is derived for the solution to (6.34), from which a can be fully determined with the help of the displacement boundary conditions. If the determined scalar field a is non-negative everywhere, we can try to complete the solution to then equilibrium problem by integrating the equation (∇u + ∇uT ) = 2E, in order to determine a displacement field u, which satisfy the boundary conditions. Otherwise, we must conclude that, for the given loads, there is no solution of the form (6.28). Referring to Figure 6.10, for n ≥ 0 an integer, P > 0 and t cn = −t

dx , (1 + x2 )n

(6.35)

with t = tan φ, let us choose p=

P y 2n−1 P xy 2(n−1) , q = . cn (x2 + y 2 )2 cn (x2 + y 2 )2

(6.36)

It is an easy matter to verify that κ(x, y) =

x , y

σy = −

P y 2n−1 cn (x2 + y 2 )2

(6.37)

are solutions to the equilibrium system (6.29). Moreover, in view of (6.28), the other components of the stress tensor are σx = −

P x2 y 2n−3 , cn (x2 + y 2 )2

τxy = −

P xy 2(n−1) . cn (x2 + y 2 )2

(6.38)

86

6 Comparison Between Explicit and Numerical Solutions

It can be seen that, for every n, on each horizontal plane y = y0 of the panel, we have ty0 ty0 σy (x, y)dx = −P, τxy (x, y)dx = 0, (6.39) −ty0

−ty0

as is easily deduced from (6.37)2 , (6.38)2 and (6.36). Now we propose to determine the strain and the displacement fields satisfying the boundary conditions at the base of the panel. To this aim, we observe that the determined stress field, when expressed in polar coordinates with the origin in O, has components σr (r, θ) = −

P sin2n−3 (θ) , cn (t)r

σθ = τrθ = 0.

(6.40)

Let us put r sin θ ; h with a procedure similar to that used in [80], Sect. 5, we obtain s=

a(r, θ) =

P sin2n−3 θ { − 2[2(n2 − 3n + 2) cos2 θ + 2 − n] ln s+ Ercn (t)

[4s(n − 1)2 − 4n2 + 8n − 3] cos2 θ + 2(n − 1)(1 − s)}, ur (r, θ) = − uθ (r, θ) =

(6.41)

P sin2n−3 θ ln s , Ecn (t)

P sin2(n−2) θ cos θ [(3 − 2n) ln s + 2(n − 1)(1 − s)], Ecn (t)

(6.42) (6.43) (6.44)

which solve our equilibrium problem if and only if the following conditions a(r, θ) ≥ 0, for

π π − φ ≤ θ ≤ + φ and y0 ≤ r sin θ ≤ h 2 2

(6.45)

are satisfied. For n = 0, inequality (6.45) can be proved as in the example shown in [80]. For n = 1 (Figure 6.11) in view of (6.36), (6.40), (6.42), (6.43) and (6.44), we have P , (6.46) σr (r, θ) = − 2r arctan t sin θ P (−2 ln s + cos2 θ) a(r, θ) = , (6.47) 2Er arctan t sin θ P ln s ur (r, θ) = − , (6.48) 2E arctan t sin θ P cos θ ln s uθ (r, θ) = . (6.49) 2E arctan t sin2 θ

6.3 The Trapezoidal Panel

87

Fig. 6.11. Behaviour of loads p and q for n = 1.

Due to inequalities

y0 ≤ s ≤ 1, (6.50) h a satisfies (6.45), and then relations (6.46)-(6.49) define a solution for the equilibrium problem for every value of φ. For n = 2 (Figure 6.12) from (6.36) and (6.40), we obtain σr (r, θ) = −

P sin θ , r(sin φ cos φ + φ)

(6.51)

Fig. 6.12. Behaviour of loads p and q for n = 1.

which is the well-known stress field proposed by Michell for a symmetric wedge with amplitude 2φ, subjected to a concentrated force at its apex in the direction parallel to the symmetry axis [102].

88

6 Comparison Between Explicit and Numerical Solutions

Moreover, from (6.42)-(6.44) we deduce a(r, θ) =

P sin θ[(4s − 3) cos2 θ + 2(1 − s)] , Er(sin φ cos φ + φ)

ur (r, θ) = − uθ (r, θ) =

P sin θ ln s , E(sin φ cos φ + φ)

P cos θ[ − ln s + 2(s − 1)] . E(sin φ cos φ + φ)

(6.52) (6.53) (6.54)

Now we propose to verify that relations (6.51)-(6.54) are a solution to the equilibrium problem, i.e., that a satisfies (6.45), but only for √ 3  ). (6.55) φ ≤ φ = arccos ( 3 To this end, we observe that a(

h P sin2 θ cos2 θ , θ) = sin θ Eh(sin φ cos φ + φ)

(6.56)

and we then have a(h/ sin θ, θ) ≥ 0 for each θ. Thus, a sufficient condition for the validity of (6.45) is that P sin θ(1 − 3 sin2 θ) ∂a = ≤ 0, ∂r Er2 (sin φ cos φ + φ) hold everywhere, which means √ √ 3 3 ) ≤ θ ≤ π − arcsin ( ). arcsin ( 3 3

(6.57)

(6.58)

On the other hand, as the denominator √of a is always positive, the numerator takes negative values when θ < arcsin( 3/3) and r goes to 0. Therefore, (6.58) is also a necessary condition for the validity of (6.45). Finally, we observe that relation (6.58) is verified everywhere if and only if φ ≤ φ holds. For n > 2, there is no value of t for which function a defined in (6.42), verifies (6.45). In fact, we have   2P (n − 1)(h − r) π r a(r, ) = (n − 2) ln + (6.59) 2 Ercn (t) h h and then, because r ≤ h, inequality a(r, π/2) < 0 holds for some value of r. So, (6.40), (6.42), (6.43) and (6.44) define a solution of the equilibrium problem√for n = 0 and n = 1 for each value of t, and for n = 2 only when 0 ≤ t ≤ 2 holds. Finally, we observe that of all the stress fields defined by (6.40), the only one characterized by a harmonic trace of stress is that proposed by Michell. However, it should be noted that in order to satisfy the boundary condition

6.4 The Mosca Bridge in Turin

89

of this equilibrium problem, also the stress field proposed by Michell requires the presence of a non-zero fracture strain (cf. (6.50)). One half of the panel was discretized with 3200 four-nodes plane strain elements, using the following set of parameter values: b = 0.5 m, B = 1.5 m, h = 1. m, p = 10 MPa, q = 10 MPa, E = 5 · 107 MPa. Figures 6.13, 6.14 and 6.15 show the vertical displacement v, the stress component σxx and the component fxx of the fracture strain vs. y for x = −b. Figures 6.16, 6.17 and 6.18 show the horizontal displacement u, the vertical displacement v and the component fxx of the fracture strain vs. x for y = −h/2. The black lines represent the exact solution, the red ones the numerical solution.

6.4 The Mosca Bridge in Turin The Mosca bridge over the Doria Riparia river in Turin was constructed in 1827. It consists of 93 voussoirs made of Malanaggio granite, has a span of 45 m, an intrados rise of 5.5 m and a thickness varying from 2 m at the

Fig. 6.13. Vertical displacement v (m) vs. y (m).

Fig. 6.14. Stress component σxx (MPa) vs. y (m).

90

6 Comparison Between Explicit and Numerical Solutions

Fig. 6.15. Fracture strain component fxx vs. y (m).

Fig. 6.16. Horizontal displacement u (m) vs. x (m).

Fig. 6.17. Vertical displacement v (m)vs. x (m).

6.4 The Mosca Bridge in Turin

91

Fig. 6.18. Fracture strain component fxx vs. x (m).

springing to 1.5 m at the crown (Figure 6.19). Mortar has been interposed between the granite blocks only in the first 11 joints at the springings and the 22 joints nearest the crown. The bridge has been studied by Castigliano [23] with the goal of verifying the advantageous effects of mortar joints on the behavior of the line of thrust. In fact, Castigliano proved that when the mortar is taken into account, the line of thrust is contained entirely within the middle-third, while considering the arch ring as a monolithic body leads to a value of the eccentricity e at the springing of 0.556 m, as opposed to a middle-six value of 0.33 m, corresponding to an opening of 0.682 m at the extrados. In order to obtain this last result, Castigliano considered a linear elastic material and used an iterative procedure aimed at progressively eliminating tensile stresses. Our goal is to determine the line of thrust by using the finite element code COMES-NOSA under the assumptions of infinite compressive strength and zero tensile strength. The arch ring is considered to be a monolith and discretized into 800 four-node elements, arranged in eight longitudinal lines, each consisting of 100 elements. The amount of loads is the same as that considered by Castigliano in his study, but rather than concentrating the total load in 12 points, we have assigned the weight of the arch as a body force and distributed the rest of the load (the permanent load and the overload) along the extrados. Figure 6.20 shows the behavior of the line of thrust; it is contained entirely within the middle-third, except in the region delimited by the springing and the normal section at 2.87 m from the springing, that is, approximately the region of the first six voussoirs. In particular, the eccentricity value at the springing is 0.616 m, with a corresponding opening of 0.848 m at the extrados. The horizontal component of thrust at the abutments is 3.963 · 106 N/m; this is 17% higher than Castigliano’s findings (3.2728 · 106 N/m). The greatest compression stress occurs at the intrados in the springing and holds about 7.6 MPa. The region characterized by the openings is illustrated in Figure 6.21, where the isostatic lines are also drawn.

92

6 Comparison Between Explicit and Numerical Solutions

6.5 The Circular Arch In the polar reference system {O, r, θ}, consider the circular arch A with radius R and thickness h. The arch is clamped at the springings and subjected to its own weight c0 and a vertical point load λc1 applied at the keystone. Our aim is to determine the value of the collapse multiplier λc (see also [72]). The weight c0 = |c0 | of the arch, with density δ = c0 /πR, is considered to be uniformly distributed along the center line. Let N , T and M be the normal force, the shear force and the bending moment, respectively. By reason of symmetry, we can limit ourselves to consider the half-arch 0 ≤ θ ≤ π/2. The equilibrium equations for the circular arch are [103] N (θ) = −δR sin θ − T (θ) =

1 d2 M , R dθ2

1 dM , R dθ

d3 M dM = −2δR2 cos θ, + dθ3 dθ whose general integrals are

Fig. 6.19. Mosca’s bridge.

Fig. 6.20. The line of thrust for Mosca’ bridge.

(6.60) (6.61) (6.62)

6.5 The Circular Arch

93

Fig. 6.21. The isostatic lines near the springing.

  K1 N (θ) = −δRθ cos θ + δR + sin θ + R   K1 T (θ) = −δRθ sin θ + δR + cos θ − R

K2 cos θ, R

(6.63)

K2 sin θ, R

(6.64)

M (θ) = δR2 θ cos θ + K1 sin θ + K2 cos θ + K3 .

(6.65)

Since force λc1 must be equilibrated by the shear force, from (6.64) we get the expression for K2 , 1 K2 = − δπR2 (1 + p), (6.66) 2 where p = λ/c0 is the ratio between the value of the point load and the weight of the arch. As we are interested in determining collapse internal forces, we assume that two hinges form at the extrados of the arch, one at the crown and the other at the springing, e(0) =

h M (0) =− , N (0) 2

π M (π/2) h e( ) = =− . 2 N (π/2) 2

(6.67)

From (6.63) and (6.65), by taking into account conditions (6.67), we obtain " # 1 2 2t K1 = − δR π(1 + p) + , (6.68) 2 1+t and K3 =

1 δπR2 (1 + p)(1 + t), 2

(6.69)

with t = h/2R. From (6.63), (6.64) and (6.65), bearing (6.68), (6.66) and (6.69) in mind, we arrive at a determination of the internal forces

94

6 Comparison Between Explicit and Numerical Solutions

N (θ) =

2t δR {[2 − π(1 + p) − ] sin θ 2 1+t +[2θ − π(1 + p)] cos θ},

(6.70)

δR {[π(1 + p) − 2θ] sin θ 2 2t ] cos θ}, +[2 − π(1 + p) − 1+t δR2 2t M (θ) = {−[π(1 + p) + ] sin θ 2 1+t +[2θ − π(1 + p)] cos θ + π(1 + p)(1 + t)}. T (θ) =

(6.71)

(6.72)

Therefore, the eccentricity, which depends on t, θ and p, is e(t, θ, p) = R

−π(1 + p)(sin θ + cos θ − 1 − t) −

2t 1+t

sin θ + 2θ cos θ

−π(1 + p)(sin θ + cos θ) + 2(sin θ + θ cos θ)

.

(6.73)

Setting t0  0.0537 and t∞  0.1716, it is an easy matter to show that the following three situations can occur. (i) Circular arches with thickness h ≤ 2Rt0 are unable to sustain any positive load; indeed, 2Rt0 is the minimum thickness necessary in order for the arch to be in equilibrium with its own weight alone. (ii) For t0 < t < t∞ , there is a unique pc (t), such that a unique θ(t) ∈ (0, π/2) exists, for which e(t, θ(t), pc (t)) = h/2; and λc = pc (t)c0 is the collapse multiplier. (iii) No positive load λc1 is able to cause collapse of an arch with thickness h ≥ 2Rt∞ . Figure 6.22 shows the behavior, as parameter t varies, of both the ratio pc (black line) between the collapse multiplier and the weight c0 of the arch, and the ratio sc (red line) between the horizontal thrust at the springing at the instant of collapse and c0 .

Fig. 6.22. Ratios pc (black line) and sc (red line) vs. t.

One half of the arch with R = 2 m, h = 0.4 m and specific weight 20000 N/m3 was discretized with 512 beam elements (see Appendix C). The load

6.6 The Circular Plate

95

multiplier λ is increased incrementally up to the value λs = 8250 N, beyond which it is no longer possible to get convergence. The collapse multiplier is λc = 8319 N, as obtained from Figure 6.22. Figure 6.23 shows the line of thrust derived from the stress field calculated numerically at the instant of collapse.

Fig. 6.23. The line of thrust at the collapse instant.

6.6 The Circular Plate In the cylindrical reference system {O, r, θ, z}, consider the circular plate P with radius R and thickness h, constrained along the lateral surface {r = R} in such a way that rotation and vertical displacements are prevented (Figure 6.24). The plate is subjected to a horizontal pressure c0 , uniformly distributed on the lateral surface, and a vertical pressure λc1 , uniformly distributed on the circle of the extrados 0 ≤ r ≤ b, with 0 < b ≤ R. For the sake of simplicity, we have ignored weight, but the corresponding generalization presents no difficulties. Once again, our aim is to determine the collapse multiplier λc . To this end, for λ > 0 we determine a statically admissible internal forces field equilibrated with the load c0 + λc1 . Let Q, Nr , Nθ , Mr and Mθ be respectively the shear, radial and circumferential normal forces, and the radial and circumferential bending moment per unit length. They must satisfy the equilibrium equations [103] dQ 1 + Q = λc1 , dr r

(6.74)

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6 Comparison Between Explicit and Numerical Solutions

Fig. 6.24. Circular plate.

1 1 + Nr − Nθ = 0, r r 1 1 Mr − Mθ + Q = 0, r r  1, 0 ≤ r ≤ b , c1 (r) = 0, b < r ≤ R

dNr dr dMr + dr where

(6.75) (6.76)

(6.77)

and the boundary conditions Q(R) =

λb2 , 2R

Nr (R) = −c0 h,

(6.78)

with c0 = |c0 |. Since the internal forces to be determined must be collapse internal forces, by setting er = Mr /Nr , eθ = Mθ /Nθ , we require that they satisfy the further boundary conditions h h (6.79) er (0) = − , er (R) = , 2 2 and that the equality h (6.80) eθ (r) = − 2 hold for 0 ≤ r ≤ R. Lastly, in order for internal forces to be statically admissible, we require that the inequalities Nr (r) ≤ 0,

Nθ (r) ≤ 0,

−

h h ≤ er (r) ≤ 2 2

(6.81)

be satisfied for 0 ≤ r ≤ R. From (6.74), (6.77) and (6.78)1 we can immediately deduce the shear,  λr 2 2, 0 ≤ r ≤ b, Q(r) = λb (6.82) 2r , b < r ≤ R.

6.6 The Circular Plate

97

Equations (6.75), (6.76) and (6.80) are insufficient to determine the remaining four internal forces. A further condition is needed and can be obtained by simply assuming that the circumferential normal force is constant, for 0 ≤ r ≤ R,

Nθ (r) = κ

(6.83)

with κ ≤ 0. From (6.80), we then deduce h Mθ (r) = − κ 2

for 0 ≤ r ≤ R.

(6.84)

By integrating (6.75) with the help of (6.78)2 , we obtain Nr (r) = −

R (c0 h + κ) + κ r

(6.85)

and from the admissibility conditions (6.81)1 and (6.81)2 , we obtain −c0 h ≤ κ ≤ 0. Lastly, by integrating (6.76) with boundary condition (6.79)1 , we arrive at the expression of Mr which depends on λ. The boundary condition (6.79)2 is satisfied for λ = λc with λc =

6c0 Rh2 . b2 (3R − 2b)

(6.86)

This, in turn, allows us to determine the final expression for Mr , Mr (r) = −

1 [hR(c0 h + κ) − hκr 2r

2c0 h2 R r3 ], 2 b (3R − 2b)

0 < r ≤ b,

1 [hR(c0 h + κ) − hκr 2r 2c0 h2 b2 R − (3r − 2b)], b ≤ r ≤ R. (3R − 2b)3

(6.87)

Mr (r) =

(6.88)

It is a simple matter to verify that the admissibility condition (6.81)3 is satisfied. In view of (6.79) and (6.80), the maximum modulus eccentricity surface corresponding to the determined internal forces is a plane region coinciding with the extrados for 0 ≤ r < R and undefined for r = R. In fact, for 0 ≤ r ≤ R the eccentricity eθ is −h/2, and the absolute value of the radial eccentricity er is less than h/2. For r = R, we have er = h/2. The value of λc given by (6.86) is the collapse multiplier. In fact, the displacement field suggested by the shape of the m.m.e.s and shown in Figure 6.25 is a collapse mechanism. Each section θ = constant rotates around the absolute center of rotation P of the extrados. This rotation is possible because, in view of (6.79)2 , the sliding clamp shown in Figure 6.24 has released,

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6 Comparison Between Explicit and Numerical Solutions

Fig. 6.25. The plate collapse mechanism.

as depicted in Figure 6.25a. From (6.80), it follows that each radius of the extrados is the axis of a cylindrical hinge. These hinges allow the surface of the extrados to form the lateral surface of a cone (see Figure 6.25b). One quarter of the plate, with R = 0.5 m and h = 0.01 m has been discretized with the shell elements described in chapter 5. After applying the horizontal pressure c0 = 5.0 · 104 Pa, we progressively increased the vertical pressure λ, which is uniformly distributed throughout the extrados (b = R), by assigning successive load increments up to the value λs , beyond which it is impossible to obtain a numerical solution to the plate equilibrium problem. By using the stress field corresponding to the last load increment, the maximum modulus eccentricity surface has been determined (in Figure 6.26, the thickness of the plate has been magnified ten times in order to better illustrate the surface). For the numerical example presented here, relation (6.86) with b = R yields λc = 6c0 h2 /R2 = 120 Pa. We have verified that using a less refined mesh yields a higher value of λs ; in particular, with 900 and 4000 elements, we get λs = 125.8 Pa and λs = 120 Pa, respectively.

6.7 The Spherical Dome In the spherical reference system {O, r, θ, ϕ}, consider the spherical dome D, with mean radius R and thickness h. The dome is clamped at the springings and subjected to its own weight c0 and a point load λc1 applied at the keystone. Our aim is once again to determine the value of the collapse multiplier λc .

6.7 The Spherical Dome

99

Fig. 6.26. The m.m.e.s. at the instant of collapse.

We begin by writing the equilibrium equations for the dome under the assumption that the circumferential normal force and bending moment vanish. Denoting q as the weight of the vault per unit area, N and M as the meridional normal force and bending moment per unit length, respectively, and Q as the shear per unit length, we can write [103] dN (ϕ) − Q(ϕ) + Rq sin2 ϕ = 0, dϕ

(6.89)

dQ(ϕ) + N (ϕ) + Rq sin ϕ cos ϕ = 0, dϕ

(6.90)

dM (ϕ) − RQ(ϕ) = 0, dϕ

(6.91)

where N (ϕ) = N (ϕ) sin ϕ, Q(ϕ) = Q(ϕ) sin ϕ, M (ϕ) = M (ϕ) sin ϕ. Moreover, by imposing vertical force equilibrium in the spherical bowl with amplitude ϕ, we obtain (6.92) 2πRN (ϕ) + 2πRQ(ϕ) cos ϕ + T (ϕ) = 0, where T (ϕ) = 2πR2 q(1 − cos ϕ) + λ is the total load acting on the bowl. By taking Q(ϕ) from (6.92) and substituting into (6.89), we obtain dN (ϕ) Rq(p − cos3 ϕ) + N (ϕ) tan ϕ + = 0, dϕ cos ϕ

(6.93)

where we have put p=1+

λ , 2πR2 q

with p ≥ 1.

(6.94)

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6 Comparison Between Explicit and Numerical Solutions

The solution to (6.93) in the interval [0, π/2] is N (ϕ) = N (0) cos ϕ + Rq sin ϕ(cos ϕ − p). On the other hand, by imposing horizontal force equilibrium on a slice of amplitude Δθ, we obtain RΔθ[N (0) + Q(π/2)] = 0, from which we deduce π (6.95) N (ϕ) = −Q( ) cos ϕ + Rq sin ϕ(cos ϕ − p). 2 From the above relation, with the help of (6.92), we can easily derive the expression for shear, Q(ϕ) = −Q(π/2) − Rq cos ϕ(p − cos ϕ). As Q(ϕ) is now known, we can integrate equation (6.91), thereby obtaining π π M (ϕ) = {M ( ) − RQ( ) cos ϕ 2 2 2ϕ − π + sin 2ϕ + p(1 − sin ϕ)]}. (6.96) 4 Since we are interested in determining collapse internal forces as in example 6.5, we assume that two hinges form at the extrados, one at the crown, the other at the springing, +qR2 [

e(0) =

M (0) h =− , 2 N (0)

M ( π2 ) π h e( ) = =− . 2 2 N ( π2 )

(6.97)

Using these relations and setting t = h/(2R), we obtain π π ], Q( ) = Rq[p − 2 4(1 + t)

π M ( ) = R2 qpt, 2

(6.98)

which, in turn, enable us to arrive at the expressions for the normal force and bending moment. In fact, from (6.96), (6.97) and (6.98), we obtain N (ϕ) = −Rq{[p −

π ] cos ϕ + sin ϕ(p − cos ϕ)}, 4(1 + t)

M (ϕ) = R2 q{ − [p −

(6.99)

2ϕ − π + sin 2ϕ π ] cos ϕ + 4(1 + t) 4

+p(1 + t − sin ϕ)}.

(6.100)

Therefore, the eccentricity, which depends on ratio t, angle ϕ and load parameter p, is e(t, ϕ, p) = −R

p(1 + t − sin ϕ) − [p − [p −

π 4(1+t) ] cos ϕ

π 4(1+t) ] cos ϕ

+

2ϕ−π+sin 2ϕ 4

+ sin ϕ(p − cos ϕ)

.

(6.101)

Since the circumferential normal force is zero, the eccentricity is independent of the direction in the tangent plane. Thus, the maximum modulus eccentricity surface is the surface whose distance from the mean surface of the vault, measured along the radial direction, is e(t, ϕ, p). Let t0 be the minimum value of t necessary for the vault to be in equilibrium with its own weight. Simple

6.7 The Spherical Dome

101

calculation reveal that for t ≥ t0 , ∂e/∂p > 0 for each ϕ ∈ [0, π/2) and p ≥ 1. Therefore, there is a unique pc such that a unique ϕ ∈ (0, π/2) exists, for which the maximum modulus eccentricity surface meets the vault intrados (see Figure 6.27), i.e., e(t, ϕ, pc ) = h/2. To verify that λc = 2πR2 q(pc − 1) is the collapse multiplier (compare (6.94)), it is sufficient to determine a corresponding mechanism. Figure 6.27 suggests considering the vault to be a kinematically undetermined structure made up of slices of infinitesimal amplitude, each in turn made up of two bodies {0 ≤ ϕ ≤ ϕ} and {ϕ ≤ ϕ ≤ π/2} linked by the hinge at point C. The absolute centers of rotation of the first and second body are the points A and B, respectively. h . Figure 6.28 shows the behavior of the ratio λc /c0 as function of t = 2R In particular, t0  0.019 and, as t approaches the value of about 0.17, the collapse load grows infinitely. One quarter-vault with R = 1 m, h = 0.16 m and specific weight 20000 N/m3 , has been discretized with 3200 shell elements. The load multiplier λ is increased incrementally up to the value λs = 6200 N, beyond which it is no longer possible to obtain convergence. For the case at hand, the collapse multiplier is λc = 6230 N, as obtained from (6.94) and (6.101). Figure 6.29 shows a comparison of the eccentricity derived from the stress field calculated numerically at the instant of collapse (dotted line) with the eccentricity obtained from relation (6.101) (continuous line).

Fig. 6.27. M.m.e.s. corresponding to the collapse multiplier and associated collapse mechanism.

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6 Comparison Between Explicit and Numerical Solutions

Fig. 6.28. λc /c0 as function of t.

Fig. 6.29. Eccentricity e as function of z = R cos ϕ at collapse.

7 Applications

The examples reported in the following, some of which were commissioned to the Mechanics of Materials and Structures Laboratory by public agencies, serve to highlight the potentials of the numerical method described in the previous sections in real applications. The results demonstrate that numerical analysis make a significant contribution to advancing the knowledge of the mechanical behavior of age-old masonry structures. In addition, the method provides for modeling several possible strengthening strategies and determining the response of the structure to each. Consequently, it can guide the choice of the most suitable strengthening operations for a specific structure. The analyses have been conducted by using the finite element code COMES NOSA, adopting the value ξc = 0.1% for the convergence criterion (4.19). The results are described briefly, with an emphasis on the more significant aspects. For greater details, the reader is referred to the bibliography. A great many results of the numerical analyses have been visualized by using the graphic code MSC Mentat .

7.1 The Medici Arsenal in Pisa At the request of the University of Pisa, in 1997 a study was conducted of the Medici arsenal. The aim of the study was to model some restorative operations to improve the structure’s seismic resistance, evaluate their effectiveness and aid in making the best choices in order to guarantee the safety and functionality of this historic building. The arsenal dates back to the 13th century, thought was completely modified during the 16th century under the de Medici government. The building, which was arranged in eight aisles (Figure 7.1), had a simple enough structure: pillared arches, a saddle roof with wooden counterlath set on purlins and rafters and roof covering in keeping with traditional Tuscan architecture. The arches were devoid of chains, so as not to obstruct construction of the galleys above; the thrust of the transverse arches was supported by the aisle

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7 Applications

abutting on the church of San Vito on one side and a small additional aisle set scarp-buttress fashion, on the other. The inner arcades were completely free, just as the overlooking arcades were also open to the river.

Fig. 7.1. View of the Medici arsenal overlooking the Arno.

After the building complex had outlived its usefulness as an armory, it suffered many losses and underwent many modifications. In the late 18th century, external buffering of the arches was performed, seven bays adjacent to San Vito were demolished, and the roofs of the last three aisles to the west raised to bring them to the same height as the others. Still more serious damage was inflicted during the last war, with loss of the western aisle, including the section flanking it.

Fig. 7.2. Series of arches making up the current structure of the arsenal.

7.1 The Medici Arsenal in Pisa

105

The structure is made up of a set of pillars surmounted by full ogival arches (Figure 7.2). The series of parallel arches on the building’s main fa¸cade (W-E direction) have an average height of about 8.7 m at the keystone, while those running parallel to the aisles’ axes are about 4.7 m in height and are for the most part closed in by masonry walls from the third aisle onward. The one exception to this is along the direction of the “roadway” that crosses the entire arsenal from west to east. In this direction, the series of arches in the first four aisles (beginning from the west) form an angle of 15◦ with those in the N-S direction (Figure 7.3). The arches of the fifth and the sixth aisles, on the other hand, are orthogonal to each other; this is owing to the fact that each series of arches arranged in the W-E is rotated counterclockwise in correspondence to the fifth pillar. Currently the structure is devoid of any sort of buttress able to sustain the thrust of the arches. On the building’s western side, an attempt was made to compensate, at least partially, for this situation by chaining the arches of the last aisle. Recently, restoration work has been initiated by the Region of Tuscany’s Office for the Arts and Environment and, as part of the re-roofing operations, the filling material has been substituted with reinforced concrete beams above each series of arches along both the W-E and N-S directions. In this study the effects of seismic actions directed parallel to the building’s main fa¸cade have been determined (earlier studies had, in fact, revealed that there are no static problems with regard to actions parallel to the aisles’ axes.) The part of structure analyzed is represented by three series of arches parallel to the building’s main fa¸cade, indicated in Figure 7.3 by the rows of pillars 1a-7a and 1c-7c. This section has been discretized with 2500 shell elements (see chapter 5 and Appendix C), and the masonry modeled via constitutive equation (2.2). The mechanical properties of masonry, its Young’s modulus Em = 45000 daN/cm2 , Poisson’s ratio νm = 0.12 and ultimate compressive stress σmr = 100 daN/cm2 , have been deduced through laboratory tests performed on sample materials at the Constructions Department of the University of Florence. The reinforced concrete beams, as well as the steel beams making up the retaining structures and chains have been assumed to be linear elastic. As for the load conditions, the structure is subjected to its own weight, the weight of the roofing, and a horizontal load parallel to the building’s main fa¸cade, which is meant to simulate the action of an earthquake. The structure has been studied under three different conditions: (a) in the presence of retaining structures applied to the eastern and western sides of the building, (b) in the presence of retaining structures as in point (a), together with chains fitted in the W-E direction and connected to the retaining structures and pillars, (c) in the structure’s actual state, neglecting the existing chains on the western aisle.

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7 Applications

Fig. 7.3. The two rows of pillars 1a-7a and 1c-7c delimiting the portion of the building analysed with the NOSA code.

In all cases considered, the bases of the pillars are clamped. In case (b) the chains, set at a height of 4.16 m from ground level, have an extensional stiffness of 1.89 107 daN and are connected to both the retaining structures as well as the individual pillars, to which they are attached by steel plates. The structure’s own weight is applied as a vertical body force calculated on the basis of each material’s corresponding specific weight (1800 daN/m3 for the masonry, 2500 daN/m3 for the reinforced concrete and 7800 daN/m3 for the steel). The overall weight of the roofing is assumed to be 180 daN/m2 . The horizontal load, which simulates the action of an earthquake, is assigned as a body force and, given the structure’s modest dimensions, is assumed constant throughout its height. In each case, the loads have been assigned incrementally. The vertical loads are applied at the first increment, while the horizontal loads are assigned over subsequent increments by pro-

7.1 The Medici Arsenal in Pisa

107

gressively increasing the ratio C between horizontal and vertical loads up to 0.28, with a step of 0.035 for each increment. For each case and every load increment we have determined: 1. the reacting section at the base of each pillar, the maximum compressive stress attained and the mean shear stress; 2. the internal forces in the reinforced concrete beams set in the W-E direction. This in order to determine the stresses in the chains and concrete. Moreover, for case (a) we have studied the behavior of the arches in the W-E direction, as well as that of the masonry walls, the reinforced concrete beams in the N-S direction and the retaining structures. We then determine the excess probabilities of some quantities deemed particularly important for assessing the building’s safety. From the stresses obtained for each pillar via the finite element analysis we have calculated the normal force N , the bending moment M1 acting on the mean plane of the pillar, the bending moment M2 perpendicular to M1 and the shear force T1 parallel to the mean plane. The maximum compressive stress σM at the base of the considered pillars and the mean shear stress τM on the reacting section have been calculated, via an analysis of the combined compressive and bending stress. In case (a), the greatest values of σM are found at the base of pillars 5 and 6. This is owing to the fact that the stress center at the base of these pillars is also eccentric in the direction orthogonal to the mean plane because of the 15◦ rotation of the last two arches with respect to the preceding ones. The highest shear stress values are found at the base of pillar 5. Regarding the reinforced concrete beams, no significant differences were found between the results from beams belonging to different rows of arches. Analogously to case (a), in case (b) (structure fitted with both retaining structures and chains) the highest values of σM are found at the base of pillars 5 and 6. Lastly, in case (c) (structure in its actual state, though without the chains on the western aisle or retaining structures), the horizontal load has been applied only for C = 0.21. The absence of the retaining structures also allows analyzing the data relative to pillars 1 and 7. The highest values of the maximum compressive stress σM occur at the base of pillars 5, 6 and 7; at the base of pillar 7, this reaches a value of 93 daN/cm2 at C = 0.21. Figure 7.4 presents a comparison of the values of σM obtained in cases (a) (continuous line), (b) (dotted line) and (c) (dashed line). The data is relative to pillar 5 in cases (a) and (b), and pillar 7 in the case of (c). The data relating to the seismic risk in Pisa have been drawn from the catalogue compiled by the National Group for Earthquake Defense of the CNR and developed within the framework of the collaboration agreement between the Research Institute on Earthquake Hazard of the CNR and the Tuscan regional Department of the Environment [91]. The graph in Figure 7.5 allows determining the probability that an earthquake with ground acceleration equal to or greater than γg occur in the in-

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7 Applications

Fig. 7.4. Comparison of the σM pattern as a function of C, in cases (a) (continuous line), (b) (dotted line) and (c) (dashed line).

tervals of time ΔT = 50 years (continuous line), 100 years (dotted line) and 500 years (dashed line) for each value of γ (ratio between ground acceleration and gravity, g = 10 m/s2 ).

Fig. 7.5. Probability that an earthquake occur with ground acceleration equal to or exceeding g, for T = 50 years (continuous line), 100 years (dotted curve) and 500 years (dashed curve). Ordinates are in logarithmic scale.

Given the modest height of the building, it appears reasonable to assume that the horizontal acceleration at each elevation does not exceed twice that at the ground. We may therefore set C = 2γ. For each value C0 of C and

7.1 The Medici Arsenal in Pisa

109

for each time interval ΔT , let us indicate by PΔT (C ≥ C0 ) the corresponding probability of exceeding such value. In other words, PΔT (C ≥ C0 ) is the probability that in a period of ΔT years, the ratio between the horizontal and vertical loads acting on the building is equal to or greater than C0 . Knowing the value of PΔT (C ≥ C0 ) with varying C0 , from the results of the numerical analyses, it is easy to deduce the excess probabilities of σM and τM for an interval of ΔT years. Figure 7.6 plots the curve of P100 (σM ≥ σM 0 ) (the excess probability of σM 0 (daN/cm2 ) for a 100-year period) with varying σM 0 , for cases (a) (continuous line), (b) (dotted line) and (c) (dashed line).

Fig. 7.6. Comparison of the behavior of P100 (σM ≥ σM 0 ) as a function of σM 0 in cases (a) (continuous line), (b) (dotted line) and (c) (dashed line). The ordinates are in logarithmic scale.

The analyses conducted underscore the need to carry out consolidation and seismic adaptations on the building. Indeed, in its current state, there is a 3.3% probability that within 100 years an earthquake will occur whose intensity is enough to determine a compressive stress level twice that of the ultimate stress at the base of some of its pillars. Moreover, by simulating two different types of reinforcement measures we have been able to evaluate the improvement that each would bring about in the building’s response to seismic loads. The results highlight the need, at the very least, to fit some sort of retaining structures on both the western and eastern sides of the building, bearing in mind that the structure is currently lacking any buttress whatever able to sustain the thrust of the arches. Such consolidation work could reduce the maximum stress at the base of pillar 5 to 53 daN/cm2 for C = 0.21 and thereby reduce the probability of exceeding half the ultimate stress within 100 years to 8 · 10−4 .

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7 Applications

A further significant finding is that, although chaining the arches can provide significant improvements in terms of increased stability, the effect, as can be concluded from Figures 7.5 and 7.6, is definitely inferior to that obtained by retaining structures alone.

7.2 The Church of San Pietro in Vinculis in Pisa The centuries-old Pisan church of St. Pietro in Vinculis, more commonly referred as ”San Pierino”, was built by the Agostinian order between 1072 and 1118 on the site of a pre-existing church. The structure is typically Romanesque: the inside is divided into three apsidal aisles by semicircular arches. The fa¸cade Figure 7.7) bears three portals and mullioned windows. The mosaic floor overlies a large crypt divided into three aisles with cross vaults and Roman capitals - perhaps the remains of an age-old arcade for merchants, later transformed into a Christian temple, as testified to by fascinating traces of earlier decorations and ancient tombs. This section is devoted to a study conducted on the static behavior of the right longitudinal wall of the church: it is 28 m long, 12 m high and 60 cm thick (Figure 7.8) and made up of eight circular arches of unequal radii, supported by nine columns of varying heights. Five small arched windows are also present in the wall. The structure has been discretized with 7074 plane-stress eightnode elements (Figure 7.9). As for the boundary conditions, the bases of the columns are clamped. The presence of both the front and back walls, which counter the bulging of the arches has been accounted for: their thickness has been assumed to be equal to 60 cm. The longitudinal wall has been studied under the action of both its own weight (specific weight γ = 20000 N/m3 ) and the weight of the roof (a uniformly distributed load of 16973 Pa). Out-of-plane effects, in particular the thrust of the vaults of the right side aisle, have been neglected. As for the mechanical properties, we have taken E = 3.2 · 103 MPa and ν = 0.1. A glance at the deformed structure (Figure 7.10) reveals that the columns and arches located at the ends of the wall are highly deformed, a finding in keeping with the presence of cracks in these regions. Such numerical results are in good agreement with the actual lowering of the keystone (Figure 7.11). The values of the stress σxx (see the reference system in Figure 7.10) are very low throughout (< 0.3 MPa). The values of the stress σyy are low (< 0.3 MPa) in the region of the wall above the arches, while values of about 1.5 MPa are reached in the columns, though these are still lower than the maximum admissible compressive strength. The only stress peaks, about 3.0 MPa (close to the maximum compressive strength), are located in two small regions: the external springing of the arch 8-9 above column 9, and the external sides of the bases of pillars 1 and 9. In agreement with these results, the wall presents no crushing fractures. The shear stress is less than 0.5 MPa and reaches its

7.2 The Church of San Pietro in Vinculis in Pisa

Fig. 7.7. S. Pietro in Vinculis church.

Fig. 7.8. Plan of S. Pietro in Vinculis.

111

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7 Applications

maximum at the springing of arch 1-2, above column 2 as well as at the springing of arch 8-9, above column 9. We also analyzed the consequences of any eventual settling of the pillars, which cannot be excluded considering that pillars rest on a system of vaults constituting the roof of the crypt below. In particular, a settling of 0.01 m assigned to pillars 1 and 3 brings about the fracture strain shown in Figure 7.13 and can explain the presence of the cracking over pillar 2, as shown in Figure 7.12. As expected, the maximum compressive strength under such conditions is attained in correspondence to pillar 2, at whose base it reaches a value of 6.8 MPa.

Fig. 7.9. The wall, finite element mesh.

Fig. 7.10. Deformed configuration of the wall.

7.3 The Dome of the Church of S. Maria Maddalena in Morano Calabro

113

Fig. 7.11. Keystone of arch 8-9.

The analysis has allowed determining the current stress and strain states of the structure in question, as well as assessing its safety. The numerical results are in good agreement with the actual distribution of cracking along the wall and can contribute to a better understanding of the causes of crack formation.

7.3 The Dome of the Church of S. Maria Maddalena in Morano Calabro The available historical information on the church and the operations to which it has been subjected over the centuries is quite scarce. It was probably built during the Middle Ages over an earlier chapel situated outside the town and then enlarged and restored during the 16th and 18th centuries. As for the structural layout, the church is a load-bearing masonry building in a Latin cross plan, with a central nave and two side aisles. The church is about 50 m long, 24 m wide and 33 m high (Figure 7.14). The presbytery zone is delimited by an octagonal cross-section drum supporting a dome and a lantern. The drum is about 11 m in diameter, 6.10 m in height with constant thickness of 110 cm. The 8 m-high dome is equal in diameter to the drum, and its thickness varies from 110 cm at the springing to 70 cm at its top. It is crowned by the 3.4 m-high lantern, 2 m in diameter and about 40 cm thick. A survey of the cracking and its distribution in both the drum and dome was carried out in October 2004. As for the dome, the crack distribution

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7 Applications

Fig. 7.12. Fracture above pillar 2.

is known only for the intrados, as the extrados is completely covered with majolica tiles (Figure 7.15). The dome presents widespread cracking, as well as evident signs of water seepage (Figure 7.16). The intrados bears two sets of fractures, different in both dimensions and behavior. The first set, which contains the larger cracks, involves the ribs connecting the webs of the dome. These cracks extend from the top of the drum up to well beyond the mid-height of the dome (Figures 7.17 and 7.18). The width of the cracks is maximum in correspondence to the dome base and decreases towards its top, where the cracks close and do not affect the last part of the ribs. The second set of cracks is horizontal and located in the webs, in the upper half of the dome. They extend from one rib to the next, inscribing a closed circular curve which involves nearly the entire dome. The width of these openings is much smaller than the first, vertical set of cracks. The drum, instead, is in relatively good condition; the survey revealed the presence of a few cracks with moderate extension. In particular, the cracks are located at the top of the drum and are continuations of those on the dome. They extend onto the drum for a short stretch and stop above the windows. The crack survey of the structures’s intrados is shown in Figures 7.17 and 7.18, where different colors have been used to represent fractures with

7.3 The Dome of the Church of S. Maria Maddalena in Morano Calabro

115

Fig. 7.13. Behaviour of the fracture strain εfxx .

different width: red, blue and green represent cracks of major, medium and minor width, respectively.

Fig. 7.14. Longitudinal section of the church by the nave.

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7 Applications

Fig. 7.15. View of the dome and the drum.

Fig. 7.16. View of the intrados of the dome and the drum.

The structure was discretized with 5496 shell elements. At first, the base of the drum was considered to be clamped, and the effects of the other parts of the church thereby neglected. Further analyses were then carried out with the aim of assessing the effects of vertical settlings of the drum.

7.3 The Dome of the Church of S. Maria Maddalena in Morano Calabro

117

Fig. 7.17. Cracks distribution on the dome intrados.

Fig. 7.18. Sections A-A and B-B of the dome, cracks distribution on the extrados.

118

7 Applications

Since no experimental data are available for the masonry material constituting the structure, we assumed γ = 20000 N/m3 , for the specific weight; E = 3.0 · 109 Pa, for Young’s modulus and ν = 0.1 for Poisson’s ratio. In the following, we describe the results of the analysis of the structure subjected to its own weight. Compressive stresses are less than 13· 105 Pa, a value considered acceptable. Figure 7.19 illustrates the maximum modulus eccentricity surface (see section 5.2) for a web, together with the extrados and the intrados (in blue). Regarding the intrados, results are in agreement with the crack distribution shown in Figures 7.17 and 7.18. Moreover, the factors defined in (5.22) and (5.24) turned out to be ϕv = 0.71 and ϕs = 0.17. With the aim of determining the direction of any eventual cracking, the components of the fracture strain in correspondence to both the intrados and f (p, ζ) = g1 · the extrados were analyzed. For g1 and g2 as in section 5.1, let E11 f f f E (p, ζ)g1 and E22 (p, ζ) = g2 · E (p, ζ)g2 be the components of the fracture f (p, ζ) = strain respectively along the parallels and the meridians, and E12 f f f f is g1 · E (p, ζ)g2 the shear component. The behavior of E11 , E22 and E12 f shown in Figures 7.20, 7.21 and 7.22 for the dome. As for E11 , the highest values are reached in the ribs starting at the top of the drum and continuing f f and E12 are equal to zero, which up to a certain height. In these regions E22 leads to the conclusion that there are fractures along the meridians, starting at the base of the dome and concentrated in the ribs. On the lower part, they continue downward, past the dome onto the upper portion of the drum. f component reaches its maximum value in the upper half of the The E22 dome, at about 6.90 m from the springing, in the centre of the webs, where f f and E12 are negligible. Thus, we can deduce that the fractures in these E11 f , regions have the same directions as parallels. Moreover, near the ribs, E11 f f and E12 are non-zero and the parallels are no longer principal directions E22 for the fracture strain. By applying the criterion described in subsection 4.2 (Figure 4.3), we deduce that here the fractures join the ribs, arranged so as to form an approximately upside down ”U” shape (Figure 7.17). As for the extrados, the numerical results suggest the presence of fracturing along the ribs, higher up than that on the intrados. Over the base of the dome, at about 2 m from the springing, there is another series of fractures similar to those on the intrados: they follow the parallel in the central part of each web and join the ribs, with a similar upside down ”U” trend. The agreement between the crack distribution actually observed in the intrados and the numerical results obtained represents substantial grounds for considering the results for the dome extrados both realistic and reliable, as well. Further numerical analyses were conducted with the aim of assessing the behavior of the structure in the presence of vertical settlings of the base of the drum. These investigations may help to explain the reasons for the asymmetries of the crack distribution shown in Figures 7.17 and 7.18. In order to model sinking of the foundations, three adjacent sides of the base of the

7.4 The Ladle

119

drum are subjected to a vertical piecewise linear displacement. In particular, we impose a constant displacement of 1 cm to the central side, and a linear displacement varying from 1 to 0 to the other two sides. Displacements of the same sort are assigned to the other three opposite sides, with a maximum displacement of 0.5 cm. In this case, compressive stresses are less than 18· 105 Pa. As for the geometrical factors of safety, the worst conditions occur in webs 3 and 7, where ϕv = 0.46 and ϕs = 0. Figures 7.23, 7.24 and 7.25 show f f f , E22 and E12 for the dome intrados. It should be noted the components E11 that the assigned vertical displacement does not lead to the formation of any new fractures, apart from those corresponding to the weight, and these latter re-arrange in a asymmetric way, in agreement with Figures 7.17 and 7.18.

Fig. 7.19. Maximum modulus eccentricity surface in a web.

7.4 The Ladle Thermo-mechanical analyses of structures such as converters, ladles, torpedo ladles, etc. employed in the iron and steel industry are usually performed by modeling the refractory as a linear elastic or elastic-plastic material [3], [15], [16], [44] and [48]. In this section, we describe an example to highlight the good results that can be obtained by using the constitutive equation of masonry-like materials under non-isothermal conditions in the presence of thermal expansion, as presented in section 2.3. In particular, we consider a ladle (Figures

120

7 Applications

f Fig. 7.20. Component E11 of the fracture strain in the intrados of the dome subjected to its own weight.

f Fig. 7.21. Component E22 of the fracture strain in the intrados of the dome subjected to its own weight.

7.4 The Ladle

121

f Fig. 7.22. Component E12 of the fracture strain in the intrados of the dome subjected to its own weight.

f Fig. 7.23. Component E11 of the fracture strain in the intrados of the dome subjected to its own weight and a vertical displacement of the drum.

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7 Applications

f Fig. 7.24. Component E22 of the fracture strain in the intrados of the dome subjected to its own weight and a vertical displacement of the drum.

f Fig. 7.25. Component E12 of the fracture strain in the intrados of the dome subjected to its own weight and a vertical displacement of the drum.

7.4 The Ladle

123

7.26 and 7.27) subjected to its own weight and a high thermal gradient. Figure 7.27 shows the presence of three stiffening rings, two upper (one inner and the other outer) and one lower, immediately below the suspension plate. One quarter of the structure, composed of the metal vessel, protective lining and wear refractory (Figure 7.28), has been discretized with 9272 eight-node, three-dimensional finite elements. We first conducted a thermal analysis to determine the temperature fields in the structure subjected to a standard thermal cycle. Then we considered two temperature distributions. In the first case, the interior of the ladle reaches the temperature of molten steel, and the thermal gradient across the thickness is very high. In the second case, the difference in temperature between the interior and the exterior is smaller: the interior has cooled and the exterior warmed up with respect to the preceding case. In both cases, we conducted a static analysis to calculate displacement, strain and stress fields in the structure and determine whether any cracks are present in the refractory lining. A detailed description of the finite element analysis is given in [69].

Fig. 7.26. Finite element discretisation of the ladle (inner view) with the materials constituting the ladle.

We modelled a thermal cycle made up the following stages:

124

7 Applications

Fig. 7.27. Finite element discretisation of the ladle (outer view).

Fig. 7.28. Metal vessel, protective lining and wear refractory in the ladle.

7.4 The Ladle

125

CT1. During the first six hours, preheating is performed, and the temperature of the inner surface increases linearly over to time, from room temperature up to 1000o C. CT2. Filling of the ladle with molten material is modeled by a temperature increase of the inner surface of 1650o C in 10 minutes. CT3. Subsequently, the temperature of the inner surface is kept constant for 3.5 hours. CT4. Over the next two hours the ladle is assumed to be empty, and its temperature is left to vary by assuming a suitable cooling flux. Figures 7.29 and 7.30 show the temperature distributions reached at the end of stages CT2 and CT4, which have been used in the subsequent mechanical analyses.

Fig. 7.29. Temperature distribution in the ladle at the end of stage CT2.

Figures 7.29 and 7.30 suggest some conclusions. The bottom of the metal vessel does not undergo any significant heating. After the filling stage, the external surface temperature is between 100 and 200o C, and there is a total thermal jump of about 1500o C between the ladles’s interior and exterior. Moreover, 50% of such thermal range is concentrated in a very small thickness of the refractory lining. At the end of the cycle, the temperature of the inner surface falls and exhibits a value of about 800 − 900o C almost everywhere. The external vessel, on the other hand, continues to heat up to 300 − 400o C

126

7 Applications

Fig. 7.30. Temperature distribution in the ladle at the end of stage CT4.

in the upper part, and the thermal jump across the thickness of the ladle is reduced to below 500o C. The static analyses of the ladle were conducted by preventing any axial displacements in the central region of the suspension plate and taking the geometrical symmetries into account. For the refractory we assumed the constitutive equation (2.156), without attempting to model the mortar joints between bricks. Young’s moduli and the coefficients of thermal expansion have been experimentally determined and depend on temperature, while Poisson’s ratio is instead taken as constant. The structure is subjected to its own weight and to the actions of thermal dilatation. 7.4.1 Analysis at the End of Stage CT2 At the end of stage CT2, the maximum value of the displacements is about 1.2 cm, concentrated at the top of the ladle, near the plane part above the suspension region. From the analysis of the eigenvalues 0 ≤ ef1 ≤ ef2 ≤ ef3 of the fracture strain in the inner and outer surface of the refractory lining, we deduce the presence of cracks in the upper part of the ladle, more or less concentric to the internal reinforcement ring. Other cracks are located immediately below the inner surface of the wear refractory, in the region of

7.4 The Ladle

127

maximum thermal gradient; the fracture surfaces are parallel to the surface itself (Figure 7.32) The outer surface of the protective lining against the metallic vessel exhibits two series of crack surfaces (Figures 7.31 and 7.33): the former along the meridians, and the latter along the orthogonal direction. Still in the outer surface of the protective lining, there are some completely cracked regions immediately above and below the suspension plate and in the surface against the upper reinforcement ring. Here the eigenvalues of the fracture strain are positive (Figures 7.31 and 7.33), and the principal stresses nil. From the analysis of the stress components, we deduce that the inner wall of the ladle is subjected to compressive stresses, which reach the maximum of about 35 MPa in the region ref7 of Figure 7.26

Fig. 7.31. Principal fracture strain ef2 in the outer surface of the refractory lining at the end of stage CT2.

7.4.2 Analysis at the End of Stage CT4 As for the displacement field in the structure at the end of stage CT4, the value of the maximum displacement does not substantially change (about 1.4 cm) with respect to stage CT2. Instead, displacements are no longer concentrated

128

7 Applications

Fig. 7.32. Principal fracture strain ef3 in the inner surface of the refractory lining at the end of stage CT2.

at the top of the ladle, and the structure is deformed in a more uniform fashion. By examining the principal fracture strains and the corresponding principal directions, we deduce the presence of a crack distribution qualitatively similar to that found at the end of stage CT2, with the difference that the values of the eigenvalues of the fracture strain are higher and distributed more widely. This means that, with respect to the preceding case, even though fractures follow the same directions, they tend to open and spread throughout the lining. The only exception is represented by the disappearance of the cracked region immediately below the inner wall of the region ref9 of Figure 7.26, due to the strong reduction of the thermal gradient in that region. It is noteworthy that the stresses are generally lower than in stage CT2, especially in those regions characterized by high thermal gradients in stage CT2.

7.4 The Ladle

129

Fig. 7.33. Principal fracture strain ef3 in the outer surface of the refractory lining at the end of stage CT2.

A The Constitutive Equation of Masonry-Like Materials: the Two-Dimensional Case

The states of plane stress and plane strain for masonry-like materials are dealt with in subsection 2.1.2; here we list the expressions of the solutions to the twodimensional version of the constitutive equations of masonry-like materials with bounded compressive strength and masonry-like materials under nonisothermal conditions as well.

A.1 Masonry-Like Materials with Small Tensile Strength and Bounded Compressive Strength We analyze the plane strain state and the plane stress state separately. A.1.1 Plane Strain State First, let us consider the case in which E has a null eigenvalue and assume, as done in subsection 2.1.2, that q1 , q2 , q3 is an orthonormal basis of V constituted by eigenvectors of E with Eq3 = 0 . A result analogous to that proven in proposition 2.12 holds, if e3 = q3 · Eq3 = 0, then ef3 = ec3 = 0 and t3 =

α (t1 + t2 ). 2(1 + α)

(A.1)

Let us designate E, Ef , Ec and T as the restrictions of E, Ef , Ec and T to the two-dimensional subspace of V orthogonal to q3 . Calculation of ef1 , ef2 , ec1 , ec2 , t1 , and t2 which satisfy system (2.110) requires definition of the following sets Fi , i = 1, ..., 6 shown in Figure A.1, F1 = {E ∈ Sym | αe1 + (2 + α)e2 − ω t < 0, (2 + α)e1 + αe2 + ω c > 0},

(A.2)

132

A The Constitutive Equation of Masonry-Like Materials

F2 = {E ∈ Sym | e1 >

ωt }, 2(1 + α)

(A.3)

F3 = {E ∈ Sym | αe1 + (2 + α)e2 − ω t > 0, e1
− }, 2(1 + α) 4(1 + α)

(A.4)

F4 = {E ∈ Sym | (2 + α)e1 + αe2 + ω c < 0, e2 > −

ωc (2 + α)ω t + αω c , e2 < }, 2(1 + α) 4(1 + α)

F5 = {E ∈ Sym | e2 < − F6 = {E ∈ Sym | e1 < − e2 >

ωc }, 2(1 + α)

(A.5) (A.6)

αω t + (2 + α)ω c , 4(1 + α)

(2 + α)ω t + αω c }, 4(1 + α)

(A.7)

and the following interfaces, F I1,3 = {E ∈ Sym | αe1 + (2 + α)e2 − ω t = 0}, F I2,3 = {E ∈ Sym | e1 =

ωt }, 2(1 + α)

F I1,4 = {E ∈ Sym | (2 + α)e1 + αe2 + ω c = 0}, F I3,6 = {E ∈ Sym | e1 = −

(A.9) (A.10)

αω t + (2 + α)ω c }, 4(1 + α)

(A.11)

ωc }, 2(1 + α)

(A.12)

F I4,5 = {E ∈ Sym | e2 = − F I4,6 = {E ∈ Sym | e2 =

(A.8)

(2 + α)ω t + αω c }. 4(1 + α)

(A.13)

We still suppose that the eigenvalues e1 and e2 are ordered in such a way that e1 ≤ e2 , and observe that in F3 , F4 , and F6 , e1 and e2 are distinct. The principal components of Ef , Ec and T are F if E ∈ F1 ∪ I1,3 then ef1 = 0, ef2 = 0, ec1 = 0, ec2 = 0, t1 = μ[(2 + α)e1 + αe2 ], t2 = μ[(2 + α)e2 + αe1 ],

(A.14)

A.1 Masonry-Like Materials with Small Tensile & Bounded Compressive Strength

Fig. A.1. Subdivision of the half-plane e1 ≤ e2 into the regions Fi , i = 1, ..., 6.. F if E ∈ F2 ∪ I2,3 then ef1 = e1 − ef2 = e2 − ec1 = 0, ec2 = 0, t1 = σ t , t2 = σ t ,

1 t 2(1+α) ω , 1 t 2(1+α) ω ,

F then ef1 = 0, if E ∈ F3 ∪ I3,6 α 1 ef2 = e2 + 2+α e1 − 2+α ωt , c e1 = 0, ec2 = 0, α t t1 = 4μ 1+α 2+α e1 + 2+α ω , t t2 = σ ,

(A.15)

(A.16)

133

134

A The Constitutive Equation of Masonry-Like Materials F if E ∈ F4 ∪ I1,4 then ef1 = 0, ef2 = 0, α 1 ec1 = e1 + 2+α e2 + 2+α ωc , c e2 = 0, t1 = −σ c , α c t2 = 4μ 1+α 2+α e2 − 2+α ω ,

(A.17)

F then ef1 = 0, if E ∈ F5 ∪ I4,5 ef2 = 0, 1 ec1 = e1 + 2(1+α) ωc , 1 ec2 = e2 + 2(1+α) ωc , c t1 = −σ , t2 = −σ c ,

(A.18)

F if E ∈ F6 ∪ I4,6 then ef1 = 0, 2+α α ω t − 4(1+α) ωc , ef2 = e2 − 4(1+α) α 2+α ω t + 4(1+α) ωc , ec1 = e1 + 4(1+α) c e2 = 0, t1 = −σ c , t2 = σ t .

(A.19)

From (A.1), in view of the non-negativeness of α, it follows that the eigenvalue t3 of T satisfies the inequalities −σ c ≤ t3 ≤ σ t as well. A.1.2 Plane Stress State Now let us assume t3 = q3 · Eq3 = 0. Then ef3 can be set equal to zero and ec3 , by virtue of the positiveness of σ c , must be equal to zero, so that we have e3 =

α (ef + ef2 + ec1 + ec2 − e1 − e2 ). 2+α 1

(A.20)

In order to calculate ef1 , ef2 , ec1 , ec2 , t1 , and t2 which satisfy system (2.110) we define the following sets G1 = {E ∈ Sym | 2αe1 + 4(1 + α)e2 − (2 + α)ω t < 0, 4(1 + α)e1 + 2αe2 + (2 + α)ω c > 0},

(A.21)

2+α ω t }, 2(2 + 3α)

(A.22)

G2 = {E ∈ Sym | e1 >

G3 = {E ∈ Sym | 2αe1 + 4(1 + α)e2 − (2 + α)ω t > 0, e1
− ωt − ω }, 2(2 + 3α) 2(2 + 3α) 2 + 3α

G4 = {E ∈ Sym | 4(1 + α)e1 + 2αe2 + (2 + α)ω c < 0,

(A.23)

A.1 Masonry-Like Materials with Small Tensile & Bounded Compressive Strength

e2 > −

2+α 1+α t α ω c , e2 < ω + ω c }, 2(2 + 3α) 2 + 3α 2(2 + 3α) G5 = {E ∈ Sym | e2 < −

G6 = {E ∈ Sym | e2 > e1 < −

2+α ω c }, 2(2 + 3α)

(A.24) (A.25)

1+α t α ω + ωc 2 + 3α 2(2 + 3α)

α 1+α c ωt − ω }, 2(2 + 3α) 2 + 3α

(A.26)

and the following interfaces, G I1,3 = {E ∈ Sym | 2αe1 + 4(1 + α)e2 − (2 + α)ω t = 0}, G I2,3 = {E ∈ Sym | e1 =

2+α ω t }, 2(2 + 3α)

G I1,4 = {E ∈ Sym | 4(1 + α)e1 + 2αe2 + (2 + α)ω c = 0}, G = {E ∈ Sym | e1 = − I3,6

α 1+α c ωt − ω }, 2(2 + 3α) 2 + 3α

G I4,5 = {E ∈ Sym | e2 = − G I4,6 = {E ∈ Sym | e2 =

2+α ω c }, 2(2 + 3α)

1+α t α ω + ω c }. 2 + 3α 2(2 + 3α)

(A.27) (A.28) (A.29) (A.30) (A.31) (A.32)

We observe that in G3 , G4 , and G6 the eigenvalues e1 and e2 are distinct. The principal components of Ef , Ec and T are G if E ∈ G1 ∪ I1,3 then ef1 = 0, ef2 = 0, ec1 = 0, ec2 = 0, 2μ [2(1 + α)e1 + αe2 ], t1 = 2+α 2μ t2 = 2+α [2(1 + α)e2 + αe1 ], G then ef1 = e1 − if E ∈ G2 ∪ I2,3 ef2 = e2 − ec1 = 0, ec2 = 0, t1 = σ t , t2 = σ t ,

(A.33)

2+α t 2(2+3α) ω , 2+α t 2(2+3α) ω ,

(A.34)

135

136

A The Constitutive Equation of Masonry-Like Materials G if E ∈ G3 ∪ I3,6 then ef1 = 0, α 2+α ef2 = e2 + 2(1+α) e1 − 4(1+α) ωt , c e1 = 0, ec2 = 0, α t t1 = μ 2+3α 1+α e1 + 2(1+α) ω , t t2 = σ ,

(A.35)

G then ef1 = 0, if E ∈ G4 ∪ I1,4 ef2 = 0, α α e2 + 2(1+α) ωc , ec1 = e1 + 2(1+α) c e2 = 0, t1 = −σ c , α c t2 = μ 2+3α 1+α e2 − 2(1+α) ω ,

(A.36)

F then ef1 = 0, if E ∈ G5 ∪ I4,5 ef2 = 0, 2+α ωc , ec1 = e1 + 2(2+3α) 2+α c e2 = e2 + 2(2+3α) ω c , t1 = −σ c , t2 = −σ c , F then ef1 = 0, if E ∈ G6 ∪ I4,6 1+α t α ef2 = e2 − 2+3α ω − 2(2+3α) ωc , α 1+α c c t e1 = e1 + 2(2+3α) ω + 2+3α ω , ec2 = 0, t1 = −σ c , t2 = σ t .

(A.37)

(A.38)

A.2 Masonry-Like Materials under Non-Isothermal Conditions An analysis of the plane stress state is provided here. Let us assume that t3 = q3 · Tq3 = 0, from (2.157)4 we obtain e3 − β(ϑ) − ef3 =

ν(ϑ) (ef + ef2 − e1 − e2 + 2β(ϑ)), 1 − ν(ϑ) 1

(A.39)

moreover, since by virtue of (2.157)3 , ef3 is arbitrary, it can be assumed to be equal to 0. In order to calculate the eigenvalues ef1 , ef2 , t1 and t2 , we define the following subsets of Sym×[ϑ1 , ϑ2 ], where e1 ≤ e2 , H1 = {(E, ϑ) | γ(ϑ)(e1 − β(ϑ)) + (1 + 2γ(ϑ))(e2 − β(ϑ)) < 0}, H2 = {(E, ϑ) | e1 − β(ϑ) > 0},

(A.40) (A.41)

A.3 The Derivative of the Stress

137

H3 = {(E, ϑ) | e1 − β(ϑ) < 0, γ(ϑ)(e1 − β(ϑ)) + (1 + 2γ(ϑ))(e2 − β(ϑ)) > 0}, H I1,3

= {(E, ϑ) | γ(ϑ)(e1 − β(ϑ)) + (1 + 2γ(ϑ))(e2 − β(ϑ)) = 0}, H I2,3 = {(E, ϑ) | e1 − β(ϑ) = 0}.

(A.42) (A.43) (A.44)

Then, T and Ef have the following expressions H , then If (E, ϑ) ∈ H1 ∪ I1,3 T=

ν(ϑ) E(ϑ) {E − β(ϑ)I + tr(E − β(ϑ)I)I}, 1 + ν(ϑ) 1 − ν(ϑ)

(A.45)

Ef = 0;

(A.46)

T = 0,

(A.47)

E = E − β(ϑ)I;

(A.48)

T = E(ϑ)(e1 − β(ϑ))q1 ⊗ q1 ,

(A.49)

E = [ν(ϑ)e1 + e2 − (1 + ν(ϑ))β(ϑ)]q2 ⊗ q2 .

(A.50)

if (E, ϑ) ∈ H2 ∪

H I2,3 , f

if (E, ϑ) ∈ H3 , f

The solution to the constitutive equation (2.157) for the plane strain state can be obtained in a similar way and is here omitted.

A.3 The Derivative of the Stress We list the expressions of the derivative of the stress for states of plane strain and plane stress. Let e1 < e2 be the eigenvalues of E and q1 and q2 the corresponding eigenvectors, putting O11 = q1 ⊗ q1 ,

(A.51)

O22 = q2 ⊗ q2 ,

(A.52)

1 O12 = √ (q1 ⊗ q2 + q2 ⊗ q1 ), 2

(A.53)

we have DE e1 = O11 , DE O11

1 = O12 ⊗ O12 , e1 − e2

DE e2 = O22 ,

DE O22

1 = O12 ⊗ O12 . e2 − e1

(A.54) (A.55)

 in the three regions For a state of plane strain the derivatives of function T S1 − S3 are  = 2μI + λI ⊗ I, E ∈ S1 , (A.56) DE T

138

A The Constitutive Equation of Masonry-Like Materials

 = O, DE T

E ∈ S2 ,

(A.57)

 = 4μ(1 + α) O11 ⊗ O11 DE T 2+α 4μ(1 + α) e1 − O12 ⊗ O12 , E ∈ S3 , 2 + α e2 − e1

(A.58)

 in the three regions T1 − T3 are and the derivatives of T  = 2μI + 2μα I ⊗ I, DE T 2+α  = O, DE T

E ∈ T1 ,

E ∈ T2 ,

(A.59) (A.60)

 = EO11 ⊗ O11 DE T −

μ(2 + 3α) e1 O12 ⊗ O12 , (1 + α) e2 − e1

E ∈ T3 .

(A.61)

 BC in the six regions F1 − F6 are Moreover, the derivatives of T  BC = 2μI + λI ⊗ I, DE T  BC = O, DE T

E ∈ F1 ,

(A.62)

E ∈ F2 ,

(A.63)

 BC = 4μ(1 + α) O11 ⊗ O11 DE T 2+α +

2μ ω t − 2(1 + α)e1 O12 ⊗ O12 , 2+α e2 − e1

E ∈ F3 ,

4μ(1 + α) O22 ⊗ O22 2+α 2μ ω c + 2(1 + α)e2 + O12 ⊗ O12 , E ∈ F4 , 2+α e2 − e1

(A.64)

 BC = DE T

 BC = O, DE T t

E ∈ F5 ,

c

σ +σ O12 ⊗ O12 , e2 − e1 For a state of plane stress we have  BC = DE T

 BC = 2μI + DE T

2μα I ⊗ I, 2+α

 BC = O, DE T

(A.65) (A.66)

E ∈ F6 .

(A.67)

E ∈ G1 ,

(A.68)

E ∈ G2 ,

(A.69)

 BC = EO1 ⊗ O1 DE T +

(2 + α)ω t − 2(2 + 3α)e1 μ O12 ⊗ O12 , 2(1 + α) e2 − e1

E ∈ G3 ,

(A.70)

A.3 The Derivative of the Stress

139

 BC = EO22 ⊗ O22 DE T c

+

(2 + α)ω + 2(2 + 3α)e2 μ O12 ⊗ O12 , 2(1 + α) e2 − e1  BC = O, DE T  BC = DE T

E ∈ G4 ,

E ∈ G5 ,

σt + σc O12 ⊗ O12 , e2 − e1

(A.71) (A.72)

E ∈ G6 .

(A.73)

 N I in the three regions H1 − H3 are Finally, the derivatives of T  NI = DE T

ν(ϑ) E(ϑ) (I + I ⊗ I), 1 + ν(ϑ) 1 − ν(ϑ)  N I = O, DE T

(E, ϑ) ∈ H1 ,

(E, ϑ) ∈ H2 ,

 N I = E(ϑ)(O11 ⊗ O11 − e1 − β(ϑ) O12 ⊗ O12 ), DE T e2 − e1

(A.74) (A.75)

(E, ϑ) ∈ H3 . (A.76)

B Algorithm for the Solution of the Equilibrium Problem: Non-Isothermal Case

We assume that body B is subjected to the loading process (τ ) = ( u(., τ ),  s(., τ ), b(., τ )),

τ ∈ [0, τ ],

(B.1)

our aim is to numerically determine a triplet (u(τ ), E(τ ), T(τ )) which satisfies the equilibrium problem (3.2), (3.4), (2.156), (3.5) and (3.6) for each τ ∈ [0, τ ]. Let w be a vector field such that w = 0 on ∂Bu . From (3.4) and (3.6) it follows that at every τ , the following equilibrium equation must be satisfied     T · ∇w dV = (B.2) s · w dA + b · w dV. B

B

∂Bs

Since T depends nonlinearly on E, we must also consider the following incremental equilibrium equation  ·  ·   s · w dA DE TN I (E, ϑ)[E] · ∇w dV = B

∂Bf



·



b · w dV −

+ B

·

 N I (E, ϑ)ϑ · ∇w dV, Dϑ T

(B.3)

B

where the dot denotes the derivative with respect to τ. As usual the body B is discretized into p finite elements B e for a total number q of nodes. Coordinates and displacements of a point belonging to an element are expressed in terms of coordinates and displacements of the nodes of the element, via suitable shape functions. Thus, the finite element method allows us to transform the incremental equation (B.3) into the nonlinear evolution system ·

·

[KT ]{u} = {f }, ·

(B.4) ·

where {u} is the vector of the nodal velocities, {f } is the vector of the time derivative of nodal forces and [KT ] is the tangent stiffness matrix. The 3q ×

142

B Algorithm for the Solution of the Equilibrium Problem

3q matrix [KT ] comes from the assembly of the p elemental matrices [KTe ] obtained from the relation  · ·e  N I (E, ϑ)[E] · ∇w dV {ce } · [KTe ]{u } = DE T (B.5) Be ·e

·

with {ce } and {u } the vectors of the nodal values of fields w and u, corre·

sponding to the finite element e. Vector {f } in (B.4) is the assembly of the p ·e

elemental vectors {f } calculated from the relation  · ·e  s · w dA {ce } · {f } = ∂Bse ∩∂Bs





·

b · w dV −

+ Be

·

 N I (E, ϑ)ϑ · ∇w dV. Dϑ T

(B.6)

Be

We assume that equation (B.2) holds in correspondence of τ , and that the body is therefore in equilibrium; subsequently, we assign a load increment {Δf } defined by means of the relations  e e ( s(τ + Δτ ) −  s(τ )) · w dA {c } · {Δf } = ∂Bse ∩∂Bs

 (b(τ + Δτ ) − b(τ )) · w dV

+  −

Be

 N I (E, ϑ2 ) − T  N I (E, ϑ1 )) · ∇w dV, (T

(B.7)

Be

where ϑ2 = ϑ(τ + Δτ ), ϑ1 = ϑ(τ + Δτ ). It is an easy matter to verify that the following equality   N I (E, ϑ2 ) − T  N I (E, ϑ1 )) · ∇w dV − (T Be

 {C(ϑ2 )[ΔEf ] + 3χ(ϑ2 )ΔβI

= Be

−ΔC[E − Ef1 ] + 3Δχβ(ϑ1 )I} · ∇w dV

(B.8)

holds, where we have C(ϑ) =

ν(ϑ) E(ϑ) {I + I ⊗ I}, 1 + ν(ϑ) 1 − 2ν(ϑ)

(B.9)

B Algorithm for the Solution of the Equilibrium Problem

ΔC = C(ϑ2 ) − C(ϑ1 ),

(B.10)

Δβ = β(ϑ2 ) − β(ϑ1 ),

(B.11)

Δχ = χ(ϑ2 ) − χ(ϑ1 ),

(B.12)

3χ(ϑ) =

E(ϑ) , 1 − 2ν(ϑ)

ΔEf = Ef2 − Ef1 , Ef2

143

Ef1

(B.13) (B.14)

and and are the solutions to the constitutive equation (2.157) corresponding to the points (E, ϑ2 ) and (E, ϑ1 ), respectively. We then solve the linear system (B.15) [KT ({u})]{Δu} = {Δf } and follow the iterative procedure described in section 4.1.

C Flow-Chart and Element Library of COMES-NOSA

In the following the flow chart (Figure C.1) and the element library (Table C.2) of COMES-NOSA are shown. For further details the reader is referred to [34]. COMES-NOSA enables determination of the stress field and the presence of any cracking in the masonry structure under examination, as well as the modeling of any potential strengthening and restoration work, such as, for example, the fitting of rods and reinforcement rings. In order to model the interaction between the masonry and the reinforcement applied, the generalized displacement components (degrees of freedom) of the nodes facing each other at the interface must be properly linked. Such links are relations between different degrees of freedom, which can be considered a special case of constraints and are managed in the solver phase [89]. These techniques allow linking elements of the same, as well as of different types, for example three-dimensional and shell elements, and shell and beam elements.

146

C Flow-Chart and Element Library of COMES-NOSA

Fig. C.1. Flow-chart for COMES-NOSA Structural type

Element Interpolating number functions

3D element

1

Quadratic

Plane stress

2

Quadratic

Plane strain

3

Quadratic

Axisymmetric 4 element

Quadratic

Thin shell element

Linear for 8-node displacements, isoparametric quadratic for element rotations

5

Remarks 20-node isoparametric element 8-node isoparametric element 8-node isoparametric element 8-node isoparametric element

C Flow-Chart and Element Library of COMES-NOSA

Plane strain

6

Linear

4-node isoparametric element

Structural type

Element Interpolating Remarks number functions 4-node Axisymmetric isoparametric 7 Linear element element 8-node isoparametric 3D element 8 Linear element 2-node Straight beam isoparametric 9 Linear element element 4-node Thick shell isoparametric 10 Linear element element 8-node Plane heat isoparametric 11 Quadratic transfer element element 4-node Plane heat isoparametric 12 Linear transfer element element 8-node Axisymmetric heat isoparametric 13 Quadratic transfer element element 4-node Axisymmetric heat isoparametric 14 Linear transfer element element 4-node 3D heat isoparametric 15 Linear transfer element element 20 nodes 3D heat isoparametric 16 Quadratic transfer element element 4-node Heat transfer isoparametric 17 Linear shell element element

TABLE C.2. Element library of the COMES-NOSA code.

147

D The GiD2Nosa Interface

The GiD2Nosa interface1 is a system with a set of add-on applications for the COMES-NOSA software. This interface is plugged into the GiD software (http://gid.cimne.upc.es/), developed by CIMNE (International Center for Numerical Methods in Engineering). It provides a set of windows-interfaced tools for managing numerical model in both the pre- and post- processing stages. The geometrical modeling features of the GiD program enable building or importing the geometry of the problem, which is made up of points, lines, surfaces and solids, and automatically generating the corresponding finite element mesh. Several custom dialog boxes allow inserting information on the problem parameters and attributes of the finite elements, loads, materials, boundary conditions, etc., as well as the general settings of the problem and the control parameters for the solver. The result is a text input data file that includes all the necessary information for analysis via the COMES-NOSA solver. Although the resulting data file is complete, fully compatible with the solver input data structure, and ready for processing, some data can obviously be modified manually in conformity with the advanced functionalities of the COMESNOSA program. Before the solution stage, two pre-processor applications are first executed. The first is a reordering algorithm, which optimizes the numbering of the nodes and elements in order to minimize the bandwidth and the wave front of the sparse tangent stiffness matrix. The second procedure reduces the size of the input data file by grouping equal load and restraint conditions. When the model is solved, before preceding to the post-processing stage, another series of applications restores the original numbering of nodes and elements and calculates the stress and strain results with respect to the global reference system.

1

the author of both the appendix and GiD2Nosa interface is Stefano Secchi, Istituto di Ingegneria Biomedica, ISIB-CNR.

150

D The GiD2Nosa Interface

Definition of the input data file and subsequent execution of the applications and solver are always managed automatically through the “calculate” command of the GiD interface. In this way, the user can switch between the pre- and post-processor modes, as well as run the problem solution, all through the GiD program. The post-processing module, which completes the GiD2Nosa interface, utilizes the GiD rendering tools to display the results of the finite elements analysis in the most appropriate way. Thus, it is possible to view stress, strain, displacements and reactions as contours, iso-surfaces, vectors, graphs, and so on. It is moreover possible to select various graphic representations of the results through different rendering modalities, on both the initial and deformed configurations.

Fig. D.1. GiD interface: geometry creation.

D The GiD2Nosa Interface

151

Fig. D.2. GiD interface: automatic mesh generation.

Fig. D.3. Example of dialog box for inputting the problem’s mechanical properties.

152

D The GiD2Nosa Interface

Fig. D.4. Example of dialog box for inputting the problem’s boundary conditions.

Fig. D.5. ”calculate” command, which starts the solving process.

D The GiD2Nosa Interface

153

Fig. D.6. Switching between pre- and post-processor.

Fig. D.7. Example of post-processing, contour bands of vertical displacements on deformed configuration.

154

D The GiD2Nosa Interface

Fig. D.8. Example of post-processing,stress component with iso-line contour.

Fig. D.9. Example of post-processing, reaction vectors.

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Index

Absolute temperature, 42 Admissible displacement field, 51, 54 internal forces field, 75 load, 52, 60 loading process, 60 stress field, 51, 53, 54 Bending moment, 68, 95, 99, 100, 107 Bilinearity of an inner product, 2 Cauchy sequence, 3, 4 Chain rule, 6 Characteristic polynomial, 8 Church of San Pietro in Vinculis, 110 of Santa Maria Maddalena, 113 Circular plate, 72, 95 Circular ring, 75, 76, 78 Closed set, 3–5, 11, 13, 20, 40, 41 Collapse, 73, 101 internal forces, 76, 96, 100 load, 72, 101 mechanism, 72, 73, 76, 97 multiplier, 67, 75, 92, 95, 97, 98, 101 Compatible load, 53 Complementary energy, 57, 59 principle of minimum, 53, 57 Complete metric space, 3 Cone, 3, 11, 23 Constitutive equation for BCS materials, 35, 40, 41, 48, 51 for masonry-like materials, 20–23, 25–27, 29, 30, 51, 52, 54–56

for masonry-like materials under non-isothermal conditions, 42, 44, 51 Continuous function, 5, 6, 9, 11, 51, 54 Convergence check, 63 Convex, 25 function, 12, 13, 41, 44 set, 3, 4, 11, 20, 40, 41 Crushing strain, 35, 76–78 Derivative, 6, 16, 23, 45, 53, 60, 61, 84, 141 of eigenvalues, 15 of eigenvectors, 15 of stress function, 46, 47, 49, 137–139 Determinant, 8 Differentiable function, 5, 6, 12, 13, 23, 25, 41, 44, 45, 47, 56, 60 Direct sum, 8 Displacement condition, 52, 54, 85 Distance, 3 Divergence, 13 theorem, 14, 55, 56 Eccentricity, 69, 70, 73, 91, 97, 100, 101 Eigenvalue, 8–11, 15, 16, 22, 26–28, 30, 32, 35, 37, 43, 45, 47, 63, 64, 69, 70, 132, 134–137 Eigenvector, 8–11, 15, 16, 22, 26, 30, 32, 35, 43, 45, 65, 69, 70, 137 Elastic strain, 20, 35, 40, 52, 84 Elasticity tensor, 20, 25, 26, 35 Equilibrium equation, 52, 53, 55, 61, 63, 83, 141

162

Index

for circular plates, 95 for spherical domes, 99 incremental, 61, 141 problem, 51–53, 59–61, 75, 76, 82, 85, 89 incremental, 60 strong formulation, 54, 56 weak formulation, 53, 55, 59 Euclidean space, 6, 7, 13 Euler equation, 83 inequality, 4, 21 Finite element, 61, 141, 142 code, 66, 75, 91 method, 45, 59, 61, 141 Fourth-order tensor, 14, 15, 51 identity, 14, 47 invariant, 15 null, 47 Fracture strain, 20, 21, 35, 62–64, 76–78, 81, 82, 84, 89, 112, 118 Geometrical factor of safety, 72, 119 Gradient, 13 Inner product, 2, 3, 6–8, 20, 21, 43 Internal forces, 69, 75, 76, 95, 96, 107 Inverse image of a set by a function, 5, 11 of a tensor, 8 Invertible function, 22 tensor, 8 Isostatic line, 83, 91 Isotropic fourth-order tensor, 15, 25–27, 30, 45 material, 26, 42 scalar function, 25 tensor function, 25, 32 Jump of the derivative of stress function, 47 Kinematically admissible state, 54, 56, 59 Ladle, 119 Lam´e moduli, 26, 35, 40 Laplace operator, 85

Length of a vector, 2 Line of thrust, 91 Linear complementarity problem, 6, 7, 26, 27 Linear function, 5 Load, 51 Loading process, 59–61, 78, 141 Maximum modulus eccentricity, 70 surface, 68, 70 for a web, 118 for circular plate, 97, 98 for spherical vault, 100 Medici arsenal, 103 Metric space, 3 complete, 3 Michell problem, 88 Minimum norm theorem, 3, 21, 41 Monotone function, 13, 22, 25, 41, 52 Mosca bridge, 89 Newton-Raphson method, 62 Norm of a vector, 2 Normal cone, 21, 43 Normal force, 68, 95, 97, 99, 100, 107 Open set, 3, 5, 13 Orthonormal basis of a real vector space, 3 of a three-dimensional real vector space, 7, 9, 11, 30 of the space of fourth-order tensors, 14 of the space of second-oder tensors, 8 Outward normal field, 13 Parallelogram law, 2, 4 Plane strain, 30, 76, 82, 131, 137 Plane stress, 32, 131, 134, 136–138 Poisson’s ratio, 75, 84, 105, 118 Positive definite fourth-order tensor, 20, 43, 57 matrix, 7, 27 Positively homogeneous function, 22 Positiveness of an inner product, 2 Potential energy, 54, 59 principle of minimum, 53, 54 Principal invariants, 9, 25, 29 Product

Index of functions, 6 of tensors, 8 rule, 6, 17 Projection, 4, 20, 21, 41, 43 r plate, 95 Rayleigh quotient, 12, 70 Reference temperature, 42, 45 Regular curve of solutions, 60 Regular region, 13, 14, 51 Schwarz inequality, 2, 23 Smooth function, 6 Solution to the constitutive equation of masonry-like materials, 20 to the equilibrium problem, 52 strong, 54, 56, 59 weak, 55, 56, 59 Space of variations, 55 Spectral representation, 9, 10, 47 Spectral theorem, 9 Spectrum, 9, 11 Spherical container, 75, 79 Spherical dome, 72, 98 Statically admissible internal forces field, 95, 96 stress field, 53, 57, 75 Stiffness matrix, 61 Strain energy, 54 energy density, 23, 30, 32, 33, 41, 47, 54 tensor, 19–21, 30, 53, 76, 85 Strain-displacement relation, 52–54 Stress energy, 54 function, 19, 22, 25, 34, 41, 43–45, 48, 52, 53 tensor, 19, 21, 47, 53, 77, 78, 82, 85 Strongly monotone function, 53 Subgradient of stress function, 47 Surface displacement, 51 Surface force, 51 Surface force field, 51 Symmetry, 70, 76, 78, 79, 87 of a fourth-order tensor major, 14, 20, 47, 56

163

minor, 15, 20 of a inner product, 2 of a second-order tensor, 17 Tangent stiffness matrix, 61, 62, 141 Tensor, 7, 9, 11, 13, 14 coaxial, 9, 22, 26, 35 components of, 7, 64, 69, 76, 79, 85, 86, 118, 119 identity, 8 negative semidefinite, 10, 19, 21, 43, 57, 68 orthogonal, 8, 15 positive semidefinite, 10, 19, 28, 57 principal components of, 35, 37, 63, 76–78, 80, 132, 135 product of tensors, 14 of vectors, 7 skew-symmetric, 8, 10 spherical, 42 symmetric, 8, 9, 15, 29, 40, 54, 56 Thermal expansion, 42 linear coefficient of, 45 Third-order tensor, 14 Trace, 7, 25, 85, 88 Traction condition, 52, 53 Transpose, 7 Trapezoidal panel, 75, 82 Variational inequality, 4 Vector field, 13, 51, 55, 141 Vector space, 2, 3, 5, 7 of second-order tensors, 7 Vectors, 2, 13 components, 7 linear independent, 3 of nodal forces, 61 of nodal velocities, 141 of residual loads, 63 orthogonal, 3 orthonormal, 3 sequence of, 3 Weakly admissible displacement, 55 Young’s modulus, 28, 75, 84, 105, 118