THEORETICAL BACKGROUND OF MASONRY MODEL

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In the reference of figure 8, the significance of nonlinearity is more convincing if we consider the resultant base shear results. In the figure 8, the horizontal axis is showing the pier positions and the vertical axis is representing the resultant shear of each pier divided by the total shear force. In the left pier, the base shear result of nonlinear masonry model is almost half of that of linear masonry model. Contrarily, in the right pier, the resultant base shear of nonlinear is almost twice that of linear masonry model. Also, overall shear force distribution is quite different. The linear masonry model shows symmetric force distribution about the mid pier. In the nonlinear masonry model, however, the right pier has the largest shear force results. From this consideration, it should be noted that the shear forces after crack are shared not by elastic stiffness but by the strength capacity as suggested by Magenes(2006). Nonlinear vs. Linear Analysis of Masonry [ Both cases have the same homogenization ] V/Vtotal [%] 55 50 45 40 35 30 25 20 15 10 5 0 Left Pier Mid Pier Right Pier Nonlinear Linear Fig.8 Base shear force distribution THEORETICAL BACKGROUND OF MASONRY MODEL page 10 / 20
APPENDIX I ORTHOTROPIC PROPERTIES OF MASONRY BASED ON STRAIN ENERGY RULE Orthotropic material properties of masonry can be derived employing a strain energy concept and the details are given in the following. It is noted that homogenization is performed between brick and bed joint first. Similar details can also be obtained when brick and head joint are homogenized first. Referring to Fig. 1, volume fraction of brick and bed joint can be described as h tbj µ b = ; µ bj = h+ t h+ t bj bj A.1 where subscript b and bj represent the brick and bed joint, respectively. If the brick and bed joint are homogenized in the beginning, the following stress/strain components in the sense of volume averaging can be established; { xx, yy, zz , xy, yz, zx} T { xx, yy, zz, xy, yz, zx} σ = σ σ σ τ τ τ ε = ε ε ε γ γ γ T A.2 where, σ ε 2 1 = ∑∫ σ dV A.3 xx xxi i V Vi i= 1 2 1 = ∑∫ ε dV A.4 xx xxi i V Vi i= 1 and i=1 for brick, i=2 for bed joint. For each strain component, THEORETICAL BACKGROUND OF MASONRY MODEL page 11 / 20
Page 1 and 2: THEORETICAL BACKGROUND OF MASONRY M
Page 3 and 4: ε = ⎡ ⎣ C⎤ ⎦ σ (2) where,
Page 5 and 6: 3. Criteria for Failure for Constit
Page 7 and 8: easons. Fig.4 Example of in-plane d
Page 9: Fig.6 Cracked and deformed shape at
Page 13 and 14: ε ε ε γ γ γ xxi xx yyi yy yyi
Page 15 and 16: where, µν b b χb = 1 − ν 2 2
Page 17 and 18: eq ( + ) µ eq ν yx ν yzνzx λeq
Page 19 and 20: S S S hj xy hj zx hj,21 2 1 ν ⎜

THEORETICAL BACKGROUND OF MASONRY MODEL

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