# 2D Nonlinear Finite Element Analysis of Masonry Bedjoint

2.2 Strain-displacement relations

Strains can be expressed as functions of displacement components at a point in

the three Cartesian coordinate directions. If u,v and w (all functions of location of

point, represented by its x, y, z coordinates) represent the displacement components

along x, y and z directions, then

x

u

x

_{�}

�

�

�

_{y}

v

y

_{�}

�

�

�

_{z}

w

z

_{�}

�

�

�

(2.01)

x

v

y

u

xy

_{�}

�

�

�

�

�

�

_{y}

w

z

v

yz

_{�}

�

�

�

�

�

�

_{z}

u

x

w

zx

_{�}

�

�

�

�

�

�

2.3 Stress-strain relations

Stress-strain relations or the constitutive equations (according to the

generalised Hooke law

_{�}

� E

_{� }) for linearly elastic and isotropic material can be

expressed as

;

)

(

E

z

y

x

x

��

��

�

�

^{�}

�

�

_{G}

xy

xy

�

� �

;

)

(

E

z

y

x

y

��

�

��

�

^{�}

�

�

�

_{G}

yz

yz

�

� �

_{(2.02)}

;

)

(

E

z

y

x

z

�

��

��

�

^{�}

�

�

�

_{G}

zx

zx

�

� �

where

_{)}

1

(

2 �

�

�

^{E}

_{G }is the shear modulus.

These equations can also be written in terms of stresses as function of strains.

� �

� �

_{;}

)

2

1

)(

1

(

1

�

�

��

��

�

�

�

_{�}

�

�

�

�

�

^{z}

y

x

x

E

)

1

(

2 �

�

�

_{�}

�

^{xy}

xy

E

� �

� �

_{;}

)

2

1

)(

1

(

1

�

�

��

�

�

��

�

_{�}

�

�

�

�

�

^{z}

y

y

y

E

)

1

(

2 �

�

�

_{�}

�

^{yz}

yz

E

_{(2.03)}

� �

� �

_{;}

)

2

1

)(

1

(

1

�

�

�

�

��

��

�

_{�}

�

�

�

�

�

^{z}

y

x

z

E

)

1

(

2 �

�

�

_{�}

�

^{zx}

zx

E

These equations are expressed in the form as

� �

_{� �}� �

�

� D

�

^{(2.04)}

2.4 Equilibrium equations

Stress at a point in a component is described by the stress tensor – one normal

stress component and two shear components on each of six sides of a cube around

that point. To reach an equilibrium of a cube, next eighteen stress components must

satisfy the following equilibrium conditions. Forces acting on a cube are F

_{x}, F

_{y }and F

_{z}

and they act along axes X,Y and Z.

;

x

xz

xy

x

_{F}

z

y

x

^{�}

�

�

�

�

�

�

�

� �

�

�

yx

xy

^{�}

� �

;

y

yz

y

yx

_{F}

z

y

x

^{�}

�

�

�

�

�

�

�

� �

�

�

zy

yz

^{�}

� �

^{(2.05)}

;

z

z

zy

zx

_{F}

z

y

x

^{�}

�

�

�

�

�

�

�

�

_{�}

�

�

xz

zx

^{�}

� �

DAMAGE MODELS 23

3 Damage Models

3.1 Isotropic damage model

An isotropic damage model expects that the stiffness degradation is isotropic.

It means that modulus of elasticity corresponding to different directions decreases

proportionally and independently of the loading direction. The damaged stiffness

tensor is expressed as where is a scalar damage variable and is

D � (1 ��)D

e

the elastic stiffness tensor. The damage evolution law is assumed in an explicit form,

connecting the damage variable to the largest previously reached equivalent strain

level.

The equivalent strain is a scalar measure related from the strain tensor. The

shape of the elastic domain in the strain space has got an influence to the choice of

the specific expression for the equivalent strain. There are many supported

equivalent strain definitions, for example: Mazars definition based on norm of

positive part of strain, Rankine criterion of maximum principal stress, energy norm

scaled by Young’s modulus, Modifies Mises definition. All these definitions are

based on the three-dimensional description of strain (and stress). If they are used in a

reduced problem, the strain components from the underlying assumptions are used

in the estimation of equivalent strain.

Attention should be paid to proper regularization, since the growth of damage

usually leads to softening and may cause localization of the dissipative process. The

damage law should be modified according to the element size, in the spirit of the

crack-band approach, if the model is kept local.

�

� D

e

3.2 Anisotropic damage model

The concept of an isotropic damage is appropriate for materials weakened by

gaps and empty spaces, but if the physical source of damage is the initiation and

propagation of micro-cracks, isotropic stiffness degradation can be considered only

as a first imprecise approximation. More sophisticated damage models take into

account the highly oriented nature or cracking, which is obtained in the anisotropic

character of the damaged stiffness or compliance matrices.

There are many anisotropic damage formulations in the literature. In our case

we use a model based on the principle of energy equivalence and on the construction

of the inverse integrity tensor by integration of a scalar over all three-dimensional

directions. This model uses concepts from microplane theory and it is called the

microplane-based damage model – MDM.

Microplane damage model is very advanced technique to capture the failure.

The damage evolution is calculated in each integration point (node) in 21 directions

(on 21 micro planes having a different orientation). Our model is theoretically

dealing only with the tensile stresses. The compressive strength turns out to depend

on the Poisson ratio and its value is too low compared to the tensile strength. The

DAMAGE MODELS 24

MDM is a very advantageous to capture moment of a failure and is a perfect method

to find a peak damage. More information about this technique are explained in [9] or

in M. Jirásek: Comments on microplane theory, Mechanics of Quasi-Brittle Materials

and Structures, Paris, 1999.

DAMAGE LOCALIZATION 25

4 Damage localization

Many materials exhibit softening when subjected to excessive tensile stress. It

means, a stress resistance ability is decreasing with growing deformation. The

softening problem is caused by a progress of material defects and decreasing ability

to carry stresses through the effective area. A problem in a brittle materials is, that

the crack development is too fast and it causes immediate model failure. Any stress

transfer in material practically disappears. Instead of softening process we call it

brittle failure.

Fig. 4.1: Local stress-strain diagram with linear softening

For illustration, we will use a simple material model (Fig. 4.1) for uniaxial

tension. We assume that the material is linearly elastic only until reaching a

deformation _{�}

0

. After that, according to the diagram, a linear softening is initiated

and the stress transfer totally disappears after reaching the deformation . The

maximum stress _{f}

t

, which is also the strength of material, expressed as

where E is the modulus of elasticity.

�

f

f_{t }� E�

0

,

4.1 Numerical solution of tensile test with softening

material

A numerical solution of one-dimensional task by finite element method splits

a whole bar (Fig. 4.2) into N parts – elements. A number of connections are ,

called nodes. Another two nodes are on the ends of the bar. As we know, the method

is based on approximation of a fields with displacement functions, which are

continuous on the whole bar and linear on each element. Every function is specified

by a node displacement values. An element deformation is calculated as a difference

of a node displacement divided by bar length. A deformation of a whole bar is

approximated from steps by constant function. The stresses are obtained from

deformations. An equivalent condition needs to be satisfied – a stresses in every

element are equal.

N �1

1

DAMAGE LOCALIZATION 26

Fig. 4.2: Tensioned bar

As we mentioned before, the material is assumed to be linearly elastic until

reaches a peak of the stress-strain diagram. After that, a behaviour solution is not

clear. In every element a two situations can be distinguished – elastic unloading or

softening. There are together reasonable solutions of unloading or softening a

whole bar and combination in every element. We are looking for a failure situation

and because of this requirement we can eliminate an elastic unloading of the whole

bar.

A process of loading is divided on steps, where every step presents loading

by force or displacement. Force control is not appropriate for our model, it would be

impossible to find out a moment of failure. So, we will use displacement control and

imagine moving with a right side of the bar. In case, that a material characteristic of

elements differs (for example strength of element) and number of steps is adequate,

only one element starts to soften – it will be the weakest element with smallest

strength. In the next step, the element is more softening, the stress in it is decreasing.

This leads to the stress decreasing in other elements, which are still in elastic state

and their deformation is also being reduced. The non-elastic deformation in one

element is localized. It is common in practise to weaken one element on purpose to

help localization. Nevertheless, usually, in a numerical solution, localization is

reached spontaneously. It is caused by a rounding mistakes, which leading to small

deviations from idealised assumptions.

2

N

4.2 Damage work

We will show how much energy (work) is needed to be expended for

complete material damage of unitary volume. During an uniaxial tension and

monotone loading a relation between a stress and deformation can be written

�� s(

�) � ^{�}1 � g^{�}�

�_{�}_{E}_{� }_{4.1}

where equation 4.1 is divided from deformation law (dependence of damage parameter on

deformation → ω=g(ε) ) and Hook’s law → σ=Eε. The variable _{g }is a function taken from

a deformation law. The area ^{g}

t

under the local stress-strain diagram is calculated as

g

t

�

�

�

0

s

�_{�}

�

d�� E

�

�

0

^{� }� g^{�}�

�_{�}_{�}

d�

1 _{4.2}

For example, in a model with linear softening (Fig. 4.1), the area under a local stressstrain

diagram is calculated as

t

f

t

t

^{E}

f

g �

�

�

_{0}

2

1

2

1

_{�}

�

^{4.3}

For a model with exponential softening (Fig. 4.3) an area

_{t}

_{g }can be obtained as

�

�

�

�

�

�

_{�}

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

_{�}

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

_{�}

�

�

�

�

�

�

�

f

f

f

f

f

t

E

E

E

E

E

d

E

d

E

g

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

0

0

0

2

0

0

0

0

2

0

0

0

2

1

2

1

exp

2

1

exp

0

0

0

0

4.4

Fig. 4.3: Local stress-strain diagram

In general, an area under the stress-strain diagram represents a dissipated

energy - the energy spent on an object damage.

4.3 Fracture energy

For explanation, we can imagine a bar tensile experiment with a constant

cross-section. A material has got defined his special characteristics and behaves

linearly elastic until reaching a force-displacement diagram peak. A softening is

concentrated in a damage localized zone and we can denote it as

_{p}

^{L }. A total energy

dissipated during the experiment

_{f}

^{W }is related to the area of the cross-section. Then,

the ratio

^{A}

^{W }

_{f }

^{/ }is dependent only on a material characteristic and symbolizes energy

needed to break the bar. The fracture energy in brittle materials is not dissipated only

in one crack, but is dissipated in many micro-cracks localized in a process zone.

DAMAGE LOCALIZATION 28

If _{G}

f

is the fracture energy then it holds that

G

f

W

f

A

� ^{4.5}

The advantage is, that it is easier to measure the fracture energy then the width of a

localization zone .

L_{s}

L

p

The numerical testing also localizes damage into a certain zone, but a width

symbolizes the width of a finite element and that is why a true width zone

differs from the width L_{s}

. The dissipated volume is

stress-strain diagram for uniaxial tension is

calculations is

W

f

� AG

f

, then

s f

g

AL

g

t

AL_{s}

L

and the area under the

, then a total dissipation obtained with

, which needs to be equal with the true dissipated energy

p

L g

s

f

� G

f

4.6

The energy on the right side

the left side ^{g}

f

G

f

is a material characteristic – constant. The energy on

is related with a selected material model and _{L}_{s}

is the width of a

simulated localization zone. If the condition 4.6 is not valid, the area under a

simulated force-displacement diagram differs with a true area. In case of making the

finite element mesh smoother, the width is changing but not the area .

L

s

g

f

PART II:

CALCULATIONS

CALCULATION WITH CRUSHED BRICK PARTICLES 30

5 Calculation with Crushed Brick Particles

Geometry and loading of the tested specimen is described in on Fig. 6.1. The

geometry represents a unit cell of masonry wall and consists of a bedjoint connecting

two halves of bricks. There were tested two load-cases: pure shear and shear with

precompression _{P }equal to _{0.5 }_{MPa }.

Fig. 6.1: Geometry and loading of the tested specimen

Code

Joint

thickness

Crushed brick

particles

Quality of

crushed

bricks

[mm] coarse/fine/none high/low

Loading

pure shear/

shear + compression

T20_P_S 20 - - pure shear

T20_P_SC 20 - - shear + compression

T30_P_S 30 - - pure shear

T30_P_SC 30 - - shear + compression

T40_P_S 40 - - pure shear

T40_P_SC 40 - - shear + compression

T40_C_HQ_S 40 coarse high pure shear

T40_C_HQ_SC 40 coarse high shear + compression

T40_C_LQ_S 40 coarse low pure shear

T40_C_LQ_SC 40 coarse low shear + compression

T40_F_HQ_S 40 fine high pure shear

T40_F_HQ_SC 40 fine high shear + compression

T40_F_LQ_S 40 fine low pure shear

T40_F_LQ_SC 40 fine low shear + compression

Tab. 6.1: Codes of the samples