Homework

A numerical method for solving equilibrium problems of masonry-like solids

Description
A numerical method for solving equilibrium problems of masonry-like solids
Categories
Published
of 19
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
  Meccanica 29: 175-193, 1994. (D 1994 KIuwer Academic Publishers. Printed in the Netherlands. A Numerical Method for Solving Equilibrium Problems of Masonry-like Solids MASSIMILIANO LUCCHESI 1 CRISTINA PADOVANI 2 and ANDREA PAGNI 2 1Universitgt di Chieti, Facolt?t di Arehitettura, Viale Pindaro 42, 65100 Pescara, Italy 2 stituto CNUCE-CNR, Via S. Maria 36, 56100 Pisa, Italy (Received: 25 January 1993; accepted n revised form: 25 May 1993) Abstract. This paper proposes a numerical method for the solution of equilibrium problems of solids which do not support tension. Some boundary-value problems are solved numerically and the solution obtained is compared to the exact one. Sommario. In questo lavoro viene proposto un metodo numerico per la soluzione di problemi di equilibrio di solidi non resistenti a trazione. Vengono successivamente isolti numericamente alcuni problemi di equilibrio e la soluzione ottenuta b confrontata con quella esatta. Key words: Masonry-like materials; finite element method. 1. Introduction The masonry-like materials considered in this paper are characterized by the constitutive hypothesis that the total strain E can be split into the sum of an elastic part E ~ and an inelastic part Ea: E=E~+E e, with E ~ positive semi-definite; and that the stress tensor T is negative semi-definite, depends linearly on E ~ and is orthogonal to E . Since E ~ only depends on the current strain, masonry-like materials are non-linear elastic materials although E is infinitesimal. A solution of the constitutive equation defined in this way exists and is unique [1]; it has been explicitly calculated for isotropic and transversely isotropic materials [2], [3]. The determination of the solution of the equilibrium problem for solids which do not support tension is, in general, very complicated. Some rather restrictive conditions have been found, which guarantee the existence of the solution [4], [5]. Moreover, in general the uniqueness of the stress is guaranteed, but not the uniqueness of the displacement or the strain [6]. The solution can be determined explicitly only in very simple cases [6], [7], [8]; consequently, it is necessary to use numerical techniques in the applications. Numerical techniques based on the finite element method have been proposed in some previous papers: in [9] the equilibrium problem for masonry-like solids was solved by mini- mizing the complementary energy under suitable constraints on the stress. In [10] an iterative procedure was analysed, leading to the progressive reduction of tractions inside the structure. Finally, in [11] the displacement field was determined with the secant matrix method. The numerical method proposed in this paper uses the fact that a masonry-like material is a non-linear elastic material for which, at least in the isotropic case, it is possible to calculate explicitly the derivative of the stress with respect to the total strain. The knowledge of this derivative allows one to calculate the tangent matrix and to determine the displacements by solving a non-linear system, obtained with the discretization into finite elements via the  176 M. Lucchesi et al. Newton-Raphson method [12]. For the sake of simplicity, the study is limited to plane problems. The algorithm, implemented in the finite element code NOSA [13], [14], is used to solve numerically some equilibrium problems; the numerical solution is compared with the exact one. 2. The Constitutive Equation of Masonry-like Materials In this section, after briefly recalling the main properties of the constitutive equation of masonry-like materials, we calculate explicitly the derivative of the stress with respect to the strain. Let us state some notations. Let 12 be a three-dimensional linear space and let Lin be the linear space of all linear applications of ~; into 12, equipped with the linear product A. B = tr(ATB), A, B E Lin, with A T the transpose of A. Let us indicate with Sym, Sym + and Sym- the subsets of Lin constituted by symmetric, symmetric positive semi-definite and symmetric negative semi- definite tensors, respectively. Let us assume that the infinitesimal strain tensor E is the sum of an elastic part E c and of an inelastic part E~: E=E c+E ~, E ~CSym +, (2.1) and that the Cauchy stress tensor T depends linearly and isotropically on E ~, T = 2#E ~ + Atr(E~)I, (2.2) where # and A, the Lam~ moduli of the material, satisfy the inequalities # > 0, 2# + 3A > 0. (2.3) Moreover, let us suppose that T E Sym-, T .E ~ = 0. (2.4) It is known that, given E E Sym and the elastic moduli # and A verifying (2.3), tensors T and E ~ exist and are unique in satisfying (2.1), (2.2) and (2.4). Moreover, T, E and E ~ are coaxial by virtue of (2.1), (2.2) and (2.4), and this property allows one to calculate explicitly the tensors E ~, E and T associated with a given strain tensor E. Let us treat in greater detail the case of plane strain; afterwards, we shall briefly describe the changes needed for plane stress. Let {gl, g2, g3 } be an orthonormal basis of V constituted by eigenvectors of E, such that the eigenvalue e3 = g3 • Eg3 of E is equal to zero; it can be proven that a3 = g3 • E~g3 is equal to zero [2]. Let us put ~ = ),/#1 and indicate with el, e2 and a l, a2 the eigenvalues of E and E ~ respectively; the constitutive relations (2.1), (2.2) and (2.4) become 2 (2(el - al) + c~(el -t- e2 - al - a2)).al = 0 (2(e2 -- a2) + oe(el + e2 -- al -- a2)) • a2 = 0 1 Here we assume that ), _> 0 and, consequently, hat c~ _> 0. 2 From the equalities e3 = a3 = 0, we obtain t3 = g3- Tg3 = a(tl + t2)/2(1 + c~) _< 0.  al >_0, a2_>0, 2(el - al) q- c~(el q- e2 - al - a2) _< 0 2(e2 -- a2) q- o~(el -4- e2 -- al -- a2) < 0. A Numerical Method for Masonry 177 (2.5) Let us indicate with the same symbols E and E ~ the restrictions of E and E = to the two- dimensional linear subspace of P orthogonal to the vector g3. The calculation of al and a2 satisfying (2.5) requires the definition of the following subsets of Sym: S~ = {E E Sym; o~e2 q- (2 + o~)e 1 ~ 0, oeel q- (2 + o0e 2 ~ 0}, S2={EESym; ea_>0, e2>0}, $3 = {lg E Sym; el < 0, c~el + (2+ a)e2 > 0}, where we suppose the eigenvalues el and e2 ordered in such a way that el _< e2. The principal components of E = can be calculated, as is known, from the relations if E E ,51 then al = 0, a2 = 0, ifE E S2then al= el, a2= e2, iflgES3then al =0, a2=e2+ ~el. (2.6) if E E ,51 then tl = #{2el + c~(el + e2)}, t2 = #{2e2 + a(el ÷ e2)}, if E E ,52 then tl =0, t2=0, (2.7) if E E $3 then tl = 90el t2 = 0, where 4#(1 a) 2+a In order to calculate the derivative of T with respect to E, let us begin by observing that from the coaxiality of E and T and the fact that the eigenvalues of T only depend on the eigenvalues of E, it easily follows that the non-linear function 'I?=T(E) is isotropic. By virtue of a well-known representation theorem, two scalar functions/3o and/3t exist of principal invariants of E, II(E) = trig = et + e2, /2(E) = det(E) = ere2, 3 As we did for E and E a, we indicate with T the restriction of the stress tensor to the two-dimensional subspace of V orthogonal to g3. We observe that in S3, the eigenvalues el and e2 are distinct and different from zero, in particular in this region E is invertible. By virtue of (2.1), (2.6) and (2.2), the principal components tl and t2 of stress tensor T 3 are  178 M. Lucchesi et al. such that T =/3oi +/31E. In view of (2.7), we have (2.8) if E E S1 then/30 = AI1, /31 : 2#, if E E $2 then/30 = 0, /J 1 = 0; (2.9) in $3 the pair flo,/~1 is the unique solution of the linear system /30 +/31el = ~el /30 + file2 0, therefore ele2 /30 = ~0 e2-- el if E E ,53 then el (2.10) /31 = --~ e2 - el The eigenvalues el and e 2 are the roots of the characteristic polynomial A2 _/I(E), ~ .~_/2('p, ) : 0; then we can write I1-C-412 /1 + C-412 el = ~ e2= 2 2 It is convenient to express el and e2 in terms of the invariants I1 and I 3 = E • E = 12 - 2/2, instead of the principal invariants I1 and/2. Thus C1 = (2.11) 2 'e2= 2 From these relations we easily obtain the expressions of/3o and/31 as functions of I1 and/3, in the region ,93 /3o = -~ I2 - ~r3 /31 = $ 1 I1_ Zl (2.12) Now we are able, with the help of (2.9) and (2.12), to calculate the derivative DET of T with respect to E. If EE S1 DET = 2#I + AI ® I; (2.13)  A Numerical Method or Masonry 179 if EE $2 DET = @, (2.14) where 1I and @ are respectively the fourth-order identity tensor and the fourth-order null tensor. In the region $3, differentiating (2.8) we have DET = I ® DEflO + E ® DEft1 +/3]]I. (2.15) Taking into account the fact that fl0 and ~1 are functions of invariants I1 and I3 of E and using the well-known expression of their derivatives with respect to E, DEII(E) = I, DE/3(E) = 2E, (2.16) we obtain /1 (3/3 - I2)I - 2/3E Z DE/~O 2 (2/3 - r2)3n ' 131 - [1E DEft1 = -~ (2/3 - 12) 3/2. (2.17) Finally, by virtue of (2.17) and (2.15), we obtain the derivative of T with respect to E in the region 83 ~/1(3/3 - l 2) c21-3 (I @ g + E @ I) DET = 2(2/3 - ~5 I Q I- (2/3 - i2)3/2 -~ (2/3 - 12)3/2 E @ F~ -~- -~ 1 g~ 3 _ f2 We observe that DET is a symmetric fourth-order tensor; this result is in agreement with the fact that the material is hyperelastic, and the potential I (aI 2+213), = o /3 - - r2), EE81, E 82, E 83, calculated with the help of (2.8), (2.9) and (2.12), is a function of class C 2 in the interior of each of the regions 81, $2 and 83. Now we briefly present the calculation of the derivative of T with respect to E for plane stress. Let { gl, g2, g3} be the basis of V with respect to which the eigenvalue t3 =g3 Tg3 of T vanishes; let us again indicate with el, e2, e3 and al, a2, a3 the eigenvalues of E and E ~ respectively. In view of (2.1) and (2.2) we have o~ e3 -- a3 -- 2 +~ (al + a2 - el - e2); (2.19)
Search
Related Search
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks