Masonry Constructions: Mechanical Models and Numerical by Massimiliano Lucchesi, Cristina Padovani, Giuseppe

By Massimiliano Lucchesi, Cristina Padovani, Giuseppe Pasquinelli, Nicola Zani

This monograph to begin with offers an in depth description of the constitutive equation of masonry-like fabrics, in actual fact starting up its most crucial positive factors. It then is going directly to supply a numerical approach to unravel the equilibrium challenge of masonry solids.

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1. 120) where ω t = σ t /μ and ω c = σ c /μ. 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 . 110), we obtain the principal components of Ef , Ec and T. 137) C if E ∈ C9 ∪ I9,10 then ef1 = 0, ef2 = 0, α 1+α t ω c − 2+3α ω, ef3 = e3 − 2+3α 2+α α ω c + 2(2+3α) ωt , ec1 = e1 + 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 ef3 = e3 + 2(1+α) e2 − 4(1+α) ω t − 4(1+α) ωc , 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) t3 = σ t .

8. 32) DE ψ(E) = T(E). Proof. 32). 34) for each E1 , E2 ∈ Sym. 13, this allows concluding that function ψ is convex. 6. 1 Isotropic Materials This subsection deals with materials for which the elasticity tensor C is isotropic (cf. 85)). 2) can be calculated explicitly. We recall that stress function T is said to be isotropic if, for every E ∈ Sym and Q ∈ Orth, it holds that T(QEQT ) = QT(E)QT . 36) holds, where coefficients βi are functions of the principal invariants of E [52]. 9. 2) is isotropic, then stress function T is isotropic.

36) holds, where coefficients βi are functions of the principal invariants of E [52]. 9. 2) is isotropic, then stress function T is isotropic. Proof. 38) for each T∗∗ ∈ Sym− . 39) (T(QEQT ) − T∗ )·(QEQT − C−1 [T(QEQT )]) ≥ 0 for each T∗ ∈ Sym− . 35) follows. 2) is isotropic, if its elasticity tensor C is isotropic. 86), the Lam´e moduli. 40) hold and we have C−1 [A] = λ 1 A− (trA)I. 10. If C is isotropic, then E, T, Ee and Ef are coaxial. Proof. 5. 41) it follows that C−1 is isotropic as well. 19, for A ∈ Sym, AC−1 [A] = C−1 [A]A.

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