A Micro-Mechanical Model For Homogenisation of Masonry: International Journal of Solids and Structures June 2002
A Micro-Mechanical Model For Homogenisation of Masonry: International Journal of Solids and Structures June 2002
A Micro-Mechanical Model For Homogenisation of Masonry: International Journal of Solids and Structures June 2002
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of Masonry
A. ZUCCHINI 1
ENEA, INN.FIS.MACO, C.R.E. “E.Clementel”, v.Don Fiammell,2, I – 40129 Bologna, Italy.
P.B. LOURENÇO *
University of Minho, Department of Civil Engineering, Azurém, P – 4800-058 Guimarães, Portugal
1
E-mail: zucchini@indy.bologna.enea.it,Ph: +39 0516098256, Fax: +39 0516098062
*
E-mail: pbl@eng.uminho.pt, Ph: + 351 253 510200, Fax: + 351 253 510217
ABSTRACT
Masonry is a composite material made of units (brick, blocks, etc.) and mortar. For
periodic arrangements of the units, the homogenisation techniques represent a powerful tool
for structural analysis. The main problem pending is the errors introduced in the
homogenisation process when large difference in stiffness are expected for the two
components. This issue is obvious in the case of non-linear analysis, where the tangent
stiffness of one component or the tangent stiffness of the two components tends to zero with
The paper itself does not concentrate on the issue of non-linear homogenisation. But as
the accuracy of the model is assessed for an increasing ratio between the stiffness of the two
components, the benefits of adopting the proposed method for non-linear analysis are
demonstrated. Therefore, the proposed model represents a major step in the application of
The micro-mechanical model presented has been derived from the actual deformations
of the basic cell and includes additional internal deformation modes, with regard to the
standard two-step homogenisation procedure. These mechanisms, which result from the
staggered alignment of the units in the composite, are of capital importance for the global
response. For the proposed model, it is shown that, up to a stiffness ratio of one thousand, the
maximum error in the calculation of the homogenised Young’s moduli is lower than five
percent. It is also shown that the anisotropic failure surface obtained from the homogenised
KEYWORDS
1
1. INTRODUCTION
experiments and the actual geometry of both components, viz. units (e.g.
the behaviour of masonry structures, see e.g. Lofti and Shing (1994), and
Lourenço and Rots (1997). Nevertheless, the representation of each unit and
terms of macro or average stresses and strains so that the material can be
that can be used confidently in the analyses. It is stressed that the results are
limited to the conditions under which the data are obtained. New materials
describing the behaviour of the composite from the geometry and behaviour
of the representative volume element (or basic cell, see Fig. 1), grants us a
predictive capability.
in terms of averaged stresses and strains from the geometry and constitutive
and efficient solution, Besdo (1985) and Mühlhaus (1993), possesses some
and true discontinuum behaviour. The step towards the practical application
basic cell, i.e. to carry out a single step homogenisation, with adequate
interface has not yet been accounted for by researchers. The complexity of
the masonry basic cell implies a numerical solution of the problem, which
has been obtained using the finite element method. The theory was thus
not, actually, to carry out analysis at the structural level. In fact, the rigorous
the complex masonry basic cell implies solving the problem for all possible
3
macroscopic loading histories, since the superposition principle does not
structural level, Pande et al. (1989), Maier et al. (1991) and Pietruszczak and
head (or vertical) and bed (or horizontal) joints being introduced
stressed that the unit-mortar interface has not been accounted for by the
cited researchers.
account for the regular offset of vertical mortar joints belonging to two
consecutive layered unit courses. Moreover, the final result depends on the
1
“Engineering” is used here not in the sense that it is empiric or practical but in the sense
4
Nevertheless, this simplified homogenisation approach has been used by
several authors and performs very satisfactorily in the case of linear elastic
analysis, Anthoine (1995) and Lourenço (1997). For the case of non-linear
analysis, where the ratio between the stiffness of unit and mortar becomes
errors and should not be used. Lourenço (1997) has shown that large errors
can occur in the standard two step homogenisation technique if there are
(1995,1997) has shown that the standard two step homogenisation technique
does not take into account the arrangement of the units in the sense that
different bond patterns (running bond and stack bond for example) may lead
2
“Stretcher bond” represents the typical arrangement of masonry units in a wall, with an offset half unit
for the vertical mortar joints belonging to two consecutive masonry courses. The “stacked bond”
arrangement, in which the vertical joints run continously through all the courses, is not allowed for
and more important for increasing unit/mortar stiffness ratios. At this stage,
der Pluijm (1999) for the analysis of masonry subjected to flexural bending
and by Lopez et al. (1999) for the non-linear analysis of masonry walls
6
distribution of deformations over units and mortar, compared with the
(internal) stresses of units and mortar deviate from the average (external)
occur on the boundaries of the basic cell, detailed finite element calculations
have been carried out for different loading conditions. For a clear discussion
system was defined, where the x-axis is the parallel to the bed joints, the y-
axis is parallel to the head joints and the z-axis is normal to the masonry
plane, see Fig. 2. This figure also shows the components considered in the
present paper. The cross joint is defined as the mortar piece of the bed joint
The unit dimensions are 210 × 100 × 52 mm3 and the mortar thickness is 10
mm. The assumption that the units are stiffer than the joints is usually made
better understand the deformational behaviour of the mortar, the units are
considered infinitely stiff (for this purpose, the adopted ratio between unit
basic cell under compression along the axes x, y and z, and under shear in
the planes xy, xz and yz. Loading is applied with adequate tying of the nodes
7
in the boundaries, making use of the symmetry and antisymmetry conditions
appropriate to each load case. Therefore, the resulting loading might not be
Fig. 4a demonstrates that, for compression along the x-axis, the unit
and the bed joint are mostly subjected to normal stresses, the bed joint is
strongly distorted in shear and the cross joint is subjected to a mixed shear /
normal stress action. While the cross joint effect can be neglected if the
cross joint is small compared to the basic cell, the shear of the bed joint
Fig. 4b demonstrates that, for compression along the y-axis, the unit
and the bed joint are mostly subjected to normal stresses, and the head and
cross joints are subjected to a mixed shear / normal stress action. These
model, have small influence on the overall behaviour of the basic cell and
which neglects such effects, leads to almost exact results (errors smaller
leads to almost exact results (errors smaller than 0.2% for ratios unit /
mostly subjected to shear stresses, the bed joint is strongly distorted in the
normal direction (tension) and the cross joint is subjected to a mixed shear /
will feature normal compression in the bed joint. While the cross joint effect
can be neglected if the cross joint is small compared to the basic cell, the
model.
The cell components are mostly subjected to shear stresses, with unit and
head joint deformed in the horizontal plane, while the bed joint is distorted
also in the vertical plane. Therefore the shear stress σ yz cannot be neglected
in a micro-mechanical model.
Fig.4f. All cell components are mainly distorted by shear in the vertical
overall effects.
3.1 General
Lourenço (1997) has shown that large errors can occur in the standard
9
this paper overcomes this limitation by a more detailed simulation of the
neglects some deformation mechanisms of the bed joint, that become more
and more important for increasing unit/mortar stiffness ratios, such as:
• vertical normal stress in the bed joint, when the basic cell is loaded with
in-plane shear;
• in-plane shear of the bed joint, when the basic cell is loaded with an
• out-of-plane shear σ yz of the bed joint, when the basic cell is loaded
head joints.
the elastic response of the basic cell to a generic load can be determined by
studying six basic loading conditions: three cases of normal stress and three
cases of simple shear. In the present formulation, for each loading case and
each basic cell component, suitably chosen components of the stress and
strain tensors are assumed to be of relevance for the stress-strain state of the
basic cell, while all the others are neglected, see Fig. 6 and Fig. 8 for
10
examples. Equilibrium is, of course, ensured for all loading cases. The
order" effects. The unknown internal stresses and strains can be found from
with a few ingenious assumptions on the cross joint behaviour and on the
kinematics of the basic cell deformation, see Fig. 5 for the adopted
are then easily derived from the internal stresses and strains, by forcing the
same, meaning that both systems must contain the same strain energy.
orthotropic material can be derived from the elastic strains of the basic cell
given axis (x, y or z). All other stresses vanish on the boundary. Fig. 9 shows
this case all shear stresses and strains inside the basic cell are neglected,
except the in-plane shear stress and strain (σxy and εxy) in the bed joint and in
the unit. Non-zero stresses and strains are assumed to be constant in each
basic cell component, except the normal stress σxx in the unit, which must
be a linear function of x to account for the effect of the shear σxy in the bed
11
With these hypotheses, the following relations hold for the stresses at
l −t
(1) Interface brick − head joint σ xx2 = σ xxb − σ 1xy
2h
( 2) Interface brick − bed joints σ byy = σ 1yy
l −t
(3) Right boundary hσ xx2 + 2tσ xx
3
+ h(σ xxb + σ 1xy ) = 2(h + t)σ xx
0
2h
( 4) Upper boundary lσ byy + tσ 2yy = (l + t)σ 0yy
(5) Front boundary 2 thσ zz2 + 2(l − t)tσ 1zz + 2lhσ zzb + 4t 2 σ zz3 = [ 2th + 2(l + t)t + 2lh]σ zz0
where l is half of the unit length, h is half of the unit height and t is half of
the bed joint width. Unit, bed joint, head joint and cross joint variables are
3, respectively. σ xxb and ε xxb are the mean value of the normal stress
σ xx and normal strain ε xx in the unit. σ xx0 , σ yy0 , σ zz0 are the uniform
respectively in the x-, y- and z-direction. The equilibrium of the unit (Fig. 7)
yields:
where we assume that the shear acts only on the bed-unit interface (l-t).
12
If σ bxx is linear in x, its mean value in the mid-unit (equal to the mean value
σ bxx1 + σ bxx2
(11) σ xx
b
=
2
l−t
σ xx
b1
= σ xx
b
− σ1xy
(12) 2h
l −t
σ bxx2 = σ xx
b
+ σ1xy
2h
which have been used in Eqs. (1,3). The couple required for the momentum
equilibrium of one fourth of the unit in the basic cell (Fig. 6) derives from
the neighbouring cell along y-axis. The symmetric unit quarter of the cell
above (Fig. 7) reacts at the centre line of the unit with a couple due to a self-
In Eqs.(1-9) the unknown stresses and strains in the cross joint can be
E2 2 E3 1 E1 1
(13) ε 3yy = ε yy σ zz3 = σ zz ε xx
3
= ε xx
E3 E1 E3
series with the bed joint in the x-direction, connected in series with the head
joint in the y-direction and connected in parallel with the bed joint in the z-
It can be noted that the stress-strain state in the cross joint does not play a
13
major role in the problem, because of its usually small volume ratio, so finer
unknowns related to the cross joint. Further coupling with the nine elastic
stress-strain relations in the unit, head joint and bed joint, namely,
ε xxk =
1 k
Ek
[ ]
σ xx − ν k (σ yyk + σ zzk )
(15) ε yyk =
1 k
Ek
[
σ yy − ν k (σ xxk + σ zzk ) , ] k = b,1,2,
ε zzk =
1 k
Ek
[ ]
σ zz − ν k (σ xxk + σ yyk )
yields a linear system of 18 equations. The unknowns are the six normal
stresses and strains of the three components (unit, head joint and bed joint)
and the shear stress and shear strain in the bed joint, amounting to a total of
20 unknowns.
The equations can be derived introducing the shear deformation of the bed
joint: the elastic mismatch between the normal x strains in the unit and in
the head joint is responsible for shear in the bed joint because of the
in Fig. 9 (where only the horizontal displacements have been magnified for
l−t
to ε xx = ε xx + σ xy
b2 b 1
, but usually the second term in the right-hand side can be
2hE b
neglected.
14
1 ∆x 2 − ∆xb ε xx2 × t − ε xx
b2
× t ε xx2 − ε xxb
(16) ε 1xy = × = ≅
2 2t 4t 4
This relation holds in the hypothesis that the bed joint does not slip on the
where i represents the three orthogonal directions associated with the axis x,
y or z. The shear stress in the unit can be found by means of the internal
equilibrium equation :
∂σ xxb ∂σ xy ∂σ xzb
b
(19) + + = 0,
∂x ∂y ∂z
which leads to :
y
(20) σ xyb = σ 1xy 1 −
h
4
It is noted that an explicit symbolic solution does exist and has been obtained.
Nevertheless, the complexity of the solution precludes its use for practical purposes. The
system of twenty equations can be easily reduced to a system of nine equations, which
15
The homogenised Young's moduli and Poisson's coefficients of the
σ ii0 ε
(21) Ei = , ν ij = jj , i, j = x, y, z
ε ii ε ii i
where
l − t + 2tE1 / E3
ε xx = ε 1xx
l +t
ε zz = ε zzb
and the subscript i in the Young’s modulus E and the Poisson’s ratio
calculation ( ) i indicates that the values are calculated for uniaxial loading
basic cell with simple in-plane shear by means of suitable load and
displacement fields. All external loads are zero on the basic cell boundary,
except a uniform shear stress σ xy0 applied on the upper and lower face, and
the equilibrium reactions σ xy on the left and right face. In this case the
model neglects all stresses (and corresponding strains), except the in-plane
shear in each basic cell component and the normal vertical component
σ 1yy in the bed joint. Non-zero stress and strain components are assumed to
16
must be a linear function of x to account for the effect of the normal stress
σ 1yy in the bed joint. The deformation of the basic cell is approximated as
shown in Fig. 10, with the bed joint in traction. Note that in the
neighbouring basic cells (along x-axis) the bed joint is in compression, due
l 1
(23) Interface brick − head joints σ xy2 = σ xyb + σ yy
2h
Interface brick − bed joints σ xyb = σ 1xy
The normal strain ε1yy can be derived from the geometric considerations in
Fig. 10, where all the geometric quantities can be defined. Neglecting
t
∆y + ∆y
2t ′ − 2t l ∆y ∆y
(24) ε 1yy ≅ ≅ , ε xy2 − ε xyb = +
2t 2t 2t 2l
which lead to :
and, introducing the linear elastic relation between stress and strain, finally:
17
yield the shear stresses in the basic cell components:
lE1 + 4hGb
σ xy2 =σ xy
0
= kσ xy
0
l Gb
2
lE1 + 4hGb + E1 − 1
l + t G2
t+l t 2
(28) σ 1xy =σ xy
b
=σ xy
0
− σ xy
l l
2h
σ 1yy = (σ xy2 − σ xyb )
l
The shear strains of the basic cell components and of the homogenised
material, according to the deformation shown in Fig. 10, are related by the
strain-displacement relations :
1 ∆x 1 ∆y 1 ∆x 2 ∆y
ε 1xy ≅ − , ε xy2 ≅ +
2 2t l 2 h t
(29) ∆x 1
+ ∆x 2
1 ∆x 2 ∆y
≅
2
ε xyb ≅ − , ε xy
2 h l (
2 h+t )
which lead to
1 b h t2
ε xy = (lε xy + tε xy )
2
+ tε xy + (ε xy − ε xy )
1 2 b
h + t l + t
(30)
l +t
The shear strains ε xyi in the above equation can be calculated from the shear
σ xy l (t + l )(t + h )
0
G xy = =
tl (t + h ) (t + l − kt )(lh − t 2 ) t (t + l )(t + l − kt )
( 31)
2ε xy
k + +
G2 Gb G1
18
To calculate the homogenised shear modulus Gxz, simple out-of-plane
shear conditions in the xz-plane are imposed to the basic cell. Right and left
faces are loaded with a uniform shear σ 0xz , while all other boundary stresses
are zero, except the equilibrium reactions σ xz on front and rear face. Only
out-of-plane shear stresses σ xz in each basic cell component and σ1yz in the
bed joint (and corresponding strains) are taken into account in the model,
while all others are neglected. Non-zero stress and strain components are
in Fig. 11, where one side has been fixed for the purpose of graphical
clarity. The shear strain ε1yz , with geometric considerations, can be found to
be:
t
∆z 2 − ∆zb
1 ∆z l
(32) ε 1yz = ≅
2 2t 4t
(l − t ) 1
Interface brick − head joint σ xz2 = σ xzb − σ yz
2h
(l − t ) 1
(33) Right boundary h σ xzb + σ yz + 2t σ xz3 + hσ xz2 = 2(t + h)σ xz0
2h
Interface cross − bed joints σ xz = σ 1xz
3
19
and of the kinematic relations
1 ∆z1 1 ∆z 2 1 ∆z b
( 35) ε 1xz ≅ , ε xz2 ≅ , ε xzb ≅
2 l 2 t 2 l
Eqs.(32,33) yield :
t+h
ε 1xz = ε xzb =σ xz0
2(tG1 + hGb )
4hGb + (l − t )G1
(36) ε xz2 = ε xzb
4hG2 + (l − t )G1
1 2
ε 1yz = (ε xz − ε xzb )
2
(37) G xz =
σ xz0 σ0
= xz 2
t +1
=
(t + l )(tG1 + hGb )
(t + h ) t 4hGb + (l − t )G1 + l
2ε xz 2 tε xz + lε xzb
4hG2 + (l − t )G1
(in the plane yz) by means of appropriate boundary conditions. The external
load is a uniform shear stress σ 0yz applied on upper and lower face of the
basic cell, while equilibrium reactions σ yz act on front and rear face, where
the boundary condition uy=0 is imposed. Only the shear stresses σ yz (and
corresponding strains) are taken into account in the model. It can be argued,
from the deformation shown in Fig. 12 (where one side has been fixed for
yields:
Gb b Gb 0 l +t
(40) ε 1yz = ε yz = σ yz
G1 G1 2(lGb + tG2 )
tε 1yz + hε yzb
(41) ε yz =
t+h
4. ELASTIC RESULTS
The model described in the previous section has been applied to a real
masonry basic cell and compared with the results of an accurate finite
In fact, the analytical model needs material data for the components and this
type of data, at least for the mortar, always result from debatable
composite level (the curing conditions of mortar inside the composite are
specimens have been cured outside the composite). In the finite element
21
analysis and the analytical model, the properties of the components can be
The same elastic properties have been adopted for the bed joint, head
stiffness ratios between mortar and unit are considered. This allows to
assess the performance of the model for inelastic behaviour. In fact, non-
differences between the components. In the limit, the ratio between the
components has no stiffness left. The unit dimensions are 210 × 100 × 52
mm3 and the mortar thickness is 10 mm. The material properties of the unit
are kept constant, whereas the properties of the mortar are varied. For the
unit, the Young's modulus Eb is 20 GPa and the Poisson's ratio νb is 0.15.
problem. Note that the ratio Eb / Em tends to infinity when softening of the
22
The elastic properties of the homogenised material, calculated by means
6%. The thinner curves in Fig. 14 ("simplified model") give the results of a
simplified model (Ex only), derived from the model presented in the paper,
Sec.3.1, have not been taken into account. The simplified model therefore
neglects the main effects due to the misalignment of the units in the
masonry wall and coincides with the full model when the units are aligned
in the wall. The simplified model is, therefore, closer to the standard two-
the elastic Young’s modulus along the x direction. Directions y and z are not
shown in the picture for the sake of clarity of the picture. Less pronounced
the x direction and the running bond reduces largely the influence of the
For large ratios Eb / Em the simplified model predicts value of Ex, vxz and
Gxz much smaller than the actual values obtained by FE analysis. The large
and increasing errors of the simplified model on these variables (up to 50%)
mechanisms of the bed joint contribute significantly to the overall basic cell
23
behaviour. In the proposed micro-mechanical model the in-plane shear
resistance of the bed joint ( σ 1xy ) is responsible for the increased stiffness in
the x-direction (up to 50%), which could not be accounted for only with the
normal stresses in the unit and in the mortar. This increase of the stiffness
stress in the bed joint ( σ 1yy ) contributes to the in-plane shear stiffness, while
the out-of-plane shear ( σ 1yz ) can double (for very large ratios Eb / Em) the
shear resistance of the basic cell to a shear load σ xz0 calculated with the
simplified model.
local crushing are the object of a long-going debate among researchers, see
van Mier (1998) for a discussion. For masonry under uniaxial compression,
and the unit is in a mixed uniaxial compression - biaxial tension, see e.g.
by the tensile failure of the unit, induced by the expansion effect of mortar,
24
(voids, inclusions, etc.) is also a key issue. A discussion on these aspects is
out of the scope of the present paper and will not be carried out.
The sole objective of this section is to demonstrate that the shape of the
is not the issue here. Currently, a research project being carried out at
the homogenised material properties of the basic cell, but also stresses and
mortar and unit, the stress distribution for an arbitrary loading case can be
presented in Chapter 3.
Then, the failure load for the homogenised cell results from reaching the
failure criteria of any of the two components. For the purpose of this
section, the simplest failure criteria can be considered for the unit and
mortar. Assuming that both materials are isotropic, the Rankine yield
surface has been assumed to describe the tensile behaviour, while the classic
von Mises criteria has been adopted to describe the compression behaviour,
Rankine: σ kp = σ tk
25
(43)
von Mises stress, and σ tk , σ ck are the tensile and compressive strengths of
dimensional stress states, which is not the case here. On the contrary, it can
compression regime for plane stress problems, as adopted here. It has been
Fig. 16 shows the resulting failure surfaces in the plane stress space
coincide with the material axes, i.e. only in-plane normal stresses σ 1 , σ 2
and no shearing are applied on the cell faces. The material and geometric
parameters for unit and mortar, which are defined in the picture, aim at
model, the principal directions in the bed joint do not coincide with the
material axes even in the absence of shear loading, due to presence of shear
in the bed joint. The intersection of all failure surfaces (the thicker line in
Fig. 16a which is reproduced in Fig. 16b) is the failure surface of the
can be noted that the plot of the yield stress in the unit of Fig. 16a is not a
perfect ellipse (check top and bottom parts): actually it is the intersection
vary linearly with x in the unit. For a given stress path, the failure loads and
material properties and above all on the relative material strengths of the
different cell components. Note that the direction of the maximum principal
stress in each component does not correspond always to the same material
direction, but does change with the load ratio σ 1 / σ 2 . Additionally, the
compression) is the cause of a tensile stress state of the unit in the direction
z.
According to the proposed model, Fig. 16b shows that, for the selected
failure of the cell for decreasing σ 1 / σ 2 ratios: tensile failure in the bed joint
(for very high ratios), compressive failure in the head joint and compressive
failure in the bed joint. Again, it is believed that these conclusions are
27
debatable and more research is needed on the issue of compressive failure of
masonry.
given in Fig. 17. The agreement in the actual values is misleading as the
agreement is found in the shape of the yield surface, indicating that the
It is stressed that the present work is, at this stage, mostly fundamental
analysts, but are not yet fully developed. The aim of this section is only to
adequate experimental values to assess the model, and the actual simplicity
of the model, the analytical results presented seem of value to the authors.
Finally, it must be stressed that failure by tension of the head joints will
not imply necessarily the failure of the composite system in the macroscale,
as adopted in this paper. For the simplified approach used here, this seems
the most reasonable assumption (i.e. if the weakest link fails, the system
28
fails). The issue of actual non-linear behaviour of the components with
of failure is a tricky issue for a composite material such masonry. The well-
tension, in the case of compression parallel to the bed joints, see Dhanasekar
et al. (1985).
6. CONCLUSIONS
joints being introduced successively, in which very large errors occur for
large differences between the unit and mortar stiffness, Lourenço et al.
(1998).
Finally, it is shown that the anisotropic failure surface obtained from the
unit and mortar, seems to, qualitatively, reproduce well the experimental
29
assessment of the model cannot be addressed at this stage, due to the
Acknowledgements
and MURST.
(FCT).
7. REFERENCES
Netherlands.
30
Bati, S.B., Ranocchiai, G., Rovero, L., 1999. A micromechanical
32, 22-30.
619.
Dhanasekar, M., Page, A.W. and Kleeman, P.W., 1985. The failure of
brick masonry under biaxial stresses, Proc. Intsn. Civ. Engrs., Part 2, 79,
295-313.
Kingdom.
Lopez, J., Oller, S., Oñate, E., Lubliner, J., 1999. A homogeneous
Materials 1, 273-294.
123(7), 660-668.
31
Lourenço, P.B., Rots, J.G., Blaauwendraad, J., 1998. Continuum
Maier, G., Papa, E., Nappi, A., 1991. On damage and failure of unit
Mühlhaus, H.-B., 1993. Continuum models for layered soil and blocky
26-31.
van der Pluijm, R., 1999. Out of plane bending of masonry: Behaviour
Netherlands.
32
Urbanski, A., Szarlinski, J., Kordecki, Z., 1995. Finite element
33
Fig. 1 – Basic cell for masonry and objective of homogenisation
34
y
z x
Unit
Bed joint
Head joint
Cross joint
y
z x Basic cell (R.V.E.)
Fig. 2 – Definition of (a) masonry axes and (b) masonry components considered in the analysis: unit, head joint, bed joint and cross joint
35
y
x
z
Fig. 3 – Finite element mesh for the basic cell adopted in the analyses
36
y
z x
(a) X direction
37
y
x
z
(b) Y Direction
38
y
x
z
(c) Z Direction
39
y
x
z
(d) XY Direction
40
y
x
(e) XZ Direction
41
y
x
(f) YZ Direction
Fig. 4 – Deformed configuration resulting from the finite element analysis on the basic cell: (a) compression x, (b) compression y, (c)
compression z, (d) shear xy, (e) shear xz and (f) shear yz.
42
t l
Head Joint
Unit h
Cross Joint
y 2 b
3 1
Bed Joint 2t
x 1 3
z
b 2
Cross Joint
Unit h
Head Joint
l t
Fig. 5 – Adopted geometry symbols.
43
Cross joint (3)
(a) (b)
σ 1xy σ xyb
σ xx3 σ xx3 σ 1xx σ 1xx σ xx3
σ xyb σ 1xy
(c)
Fig. 6 – Normal stress loading parallel to the x axis: (a) equivalent homogenised cell; (b) assumed deformation behaviour; (c) assumed involved stress components
44
σ
σ bxx2
σ xx
b
σ bxx1 x
l/2 l 2l
σ xyb σ xyb
σ xxb1 σ xxb1
σ xyb σ xyb
σ byy
σ xxb1 σ xxb 2
σ xyb
Fig. 7 – Normal stress loading parallel to the x axis: unit equilibrium (couple moment equal to self-equilibrating vertical stress distribution)
45
σ yy0
Cross joint (3)
Unit (b)
Head joint (2)
(a) (b)
σ yy2 σ byy
σ yy2 σ byy
σ byy σ yy2
σ byy σ yy2
(c)
Fig. 8 – Normal stress loading parallel to the y axis: (a) equivalent homogenised cell; (b) assumed deformation behaviour; (c) assumed involved stress components
46
∆x2
2t
∆xb
x
t
Fig. 9 – Model assumptions for compression along the x axis
47
h
∆y
α ∆y
y t ∆x1 ∆x2
h
l t
48
t l
α ∆zb
∆z ∆z2
49
h
2t
h y
z
∆zb ∆z1 ∆zb
50
Titolo:
/disk6/zucchini/OFFICE/PLOT/EmortE1.eps
Autore:
xmgr
Anteprima:
L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
Commento:
L'immagine EPS potrà essere stampata con una stampante
PostScript e non con
altri tipi di stampante.
(a)
51
Titolo:
/disk6/zucchini/OFFICE/PLOT/EmortNU1.eps
Autore:
xmgr
Anteprima:
L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
Commento:
L'immagine EPS potrà essere stampata con una stampante
PostScript e non con
altri tipi di stampante.
(b)
52
Titolo:
/disk6/zucchini/OFFICE/PLOT/EmortG1.eps
Autore:
xmgr
Anteprima:
L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
Commento:
L'immagine EPS potrà essere stampata con una stampante
PostScript e non con
altri tipi di stampante.
(c)
Fig. 13 – Comparison between the micro-mechanical model and FEA results for different stiffness ratios: (a) Young’s moduli, (b) Poisson’s ratio
(a)
54
Titolo:
/disk6/zucchini/OFFICE/PLOT/errNU.eps
Autore:
xmgr
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L'immagine EPS non è stata salvata
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PostScript e non con
altri tipi di stampante.
(b)
55
Titolo:
/disk6/zucchini/OFFICE/PLOT/errG.eps
Autore:
xmgr
Anteprima:
L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
Commento:
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PostScript e non con
altri tipi di stampante.
(c)
Fig. 14 – Comparison between the micro-mechanical model and FEA results for different stiffness ratios: (a) Young’s moduli, (b) Poisson’s ratio
56
σ2
σt
−σc
σt σ1
−σc
Fig. 15 – Composite von Mises-Rankine failure criteria in the principal stress space.
57
Titolo:
/usr/people/zucchini/OFFICE/MASONRY/fig12a.eps
Autore:
xmgr
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L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
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PostScript e non con
altri tipi di stampante.
(a)
58
Titolo:
/usr/people/zucchini/OFFICE/MASONRY/fig12b.eps
Autore:
xmgr
Anteprima:
L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
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L'immagine EPS potrà essere stampata con una stampante
PostScript e non con
altri tipi di stampante.
(b)
Fig. 16 – Calculated micro-mechanical failure criterion for masonry under biaxial in-plane loading (principal axes coincident with material axes):
(a) complete failure modes of the unit and mortar and (b) composite masonry failure.
59
Titolo:
/usr/people/zucchini/OFFICE/MASONRY/fig13.eps
Autore:
xmgr
Anteprima:
L'immagine EPS non è stata salvata
con l'anteprima inclusa in essa.
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Fig. 17 – Comparison between micro-mechanical failure criterion and experimental results of Page (1981,1983).
60