# Application of Base Force Element Method to Mesomechanics Analysis for Concrete

Authors: Yijiang Peng; Yao Wang; Qing Guo; Junhua Ni
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

The base force element method (BFEM) on damage mechanics is used to analyze the compressive strength, the size effects of compressive strength, and fracture process of concrete at mesolevel. The concrete is taken as three-phase composites consisting of coarse aggregate, hardened cement mortar, and interfacial transition zone (ITZ) on mesolevel. The random aggregate model is used to simulate the mesostructure of concrete. The mechanical properties and fracture process of concrete under uniaxial compression loading are simulated using the BFEM on damage mechanics. The simulation results agree with the test results. This analysis method is the new way for investigating fracture mechanism and numerical simulation of mechanical properties for concrete.


---

## Body

## 1. Introduction

Concrete is considered as heterogeneous composites whose mechanical performance is much related with the microstructure of material. The composite behavior of concrete is exceedingly complex and up to now many details such as strain softening, microcrack propagation, failure mechanisms, and size effects are still far from being fully understood. Since it is difficult to look inside concrete to observe the actual crack propagation or to experimentally determine the microscopic stress field, it has become obvious that further progress based exclusively on experimental studies will be limited [1].In order to overcome this defect, the concept of numerical concrete was presented by Wittmann et al. [2] based on micromechanics. Subsequently, some scholars did some creative works in this field and made a number of models. Among them, the two important models are the lattice model and the random aggregate model. For example, Schlangen et al. [3, 4] applied the lattice model to simulate the failure mechanism of concrete. Liu and Wang [5] adopted the random aggregate model to simulate cracking process of concrete using FEM. Peng et al. [6] adopted the random aggregate model to simulate the mechanical properties of rolled compacted concrete on mesolevel using FEM. Du et al. [7–11] researched a mesoelement equivalent method for the simulation of macromechanical properties of concrete and a mesoscale analysis method for the simulation of nonlinear damage and failure behavior of reinforced concrete members.The finite element method (FEM) is one of the most important numerical methods developed from 1950s, and it has been the most popular and widely used numerical analysis tool for problems in engineering and science. Over the past 50 years, numerous efforts techniques have been proposed for developing finite element models [12–15] and some other improvement and alternative methods have been proposed and developed, such as boundary element methods [16, 17] and meshless methods [18, 19]. In recent years, a new type of finite element method, the base force element method (BFEM), has been developed by Peng et al. [20–25] based on the concept of the base forces by Gao [26]. Further, the base force element method (BFEM) on potential energy principle was used to analyze recycled aggregate concrete on mesolevel [27].In this paper, the base force element method (BFEM) on damage mechanics is used to analyze the size effect on compressive strength for concrete at mesolevel.


## 2. Basic Formula of Base Force Theory

### 2.1. Base Forces

Consider a two-dimensional material domain.x
α
(
α
=
1,2
) denote the Lagrangian coordinates. Let Q denote the position vector of a point after deformation. The tangent basis vectors can be written in the form
(1)
Q
1
=
∂
Q
∂
x
1
,
Q
2
=
∂
Q
∂
x
2
.The position vector of a pointQ can be written as
(2)
Q
=
x
1
Q
1
+
x
2
Q
2
.In order to describe the stress state at a pointQ, a parallelogram with the edges d
x
1
Q
1
,
d
x
2
Q
2 is shown in Figure 1. Let d
T
α denote the force acting on the α edge. We calculate the limit
(3)
t
α
=
lim
⁡
d
x
α
→
0
d
T
α
d
x
α
+
1
,
where we promise 3
=
1 for indexes. Quantities t
α
(
α
=
1,2
) are called the base forces at point Q in the two-dimensional coordinate system x
α.Figure 1
The base forces.In order to further explain the meaning oft
α, let us compare t
α with the stress vector σ
α which represent the forces per unit area in the deformed body. That is,
(4)
t
α
=
A
Q
σ
α
,
where A
Q is called base area A
Q
=
|
Q
1
×
Q
2
|.The base forcest
α can also be understood as stress flux.According to the definitions of various stress tensors, the relation between the base forces and various stress tensors can be given. For example, the Cauchy stress is(5)
σ
=
1
A
Q
t
α
⊗
Q
α
,
where ⊗ is the dyadic symbol and the summation rule is implied.When the body force per unit mass is zero, the equilibrium equation can be written as(6)
∂
t
α
∂
x
α
=
0
.


### 2.2. Conjugate Variable of Base Forces

LetP denote the position vector of a point before deformation; the displacement u of a point is
(7)
u
=
Q
-
P
.The gradient of displacementu
α can be written as
(8)
u
α
=
∂
u
∂
x
α
=
Q
α
-
P
α
.Further, the elastic law can be given as follows:(9)
t
α
=
ρ
A
Q
∂
W
∂
u
α
,
in which W is the strain energy density and ρ is the mass density after deformation.Equation (9) expresses the t
α by strain energy directly. Thus, u
α is just the conjugate variable of t
α. It can be seen that the mechanics problem can be completely established by means of t
α and u
α.For the small deformation case, the Green strainε can be written as
(10)
ε
=
1
2
(
u
α
⊗
P
α
+
P
α
⊗
u
α
)
,
where P
α is the conjugate of P
α.


## 2.1. Base Forces

Consider a two-dimensional material domain.x
α
(
α
=
1,2
) denote the Lagrangian coordinates. Let Q denote the position vector of a point after deformation. The tangent basis vectors can be written in the form
(1)
Q
1
=
∂
Q
∂
x
1
,
Q
2
=
∂
Q
∂
x
2
.The position vector of a pointQ can be written as
(2)
Q
=
x
1
Q
1
+
x
2
Q
2
.In order to describe the stress state at a pointQ, a parallelogram with the edges d
x
1
Q
1
,
d
x
2
Q
2 is shown in Figure 1. Let d
T
α denote the force acting on the α edge. We calculate the limit
(3)
t
α
=
lim
⁡
d
x
α
→
0
d
T
α
d
x
α
+
1
,
where we promise 3
=
1 for indexes. Quantities t
α
(
α
=
1,2
) are called the base forces at point Q in the two-dimensional coordinate system x
α.Figure 1
The base forces.In order to further explain the meaning oft
α, let us compare t
α with the stress vector σ
α which represent the forces per unit area in the deformed body. That is,
(4)
t
α
=
A
Q
σ
α
,
where A
Q is called base area A
Q
=
|
Q
1
×
Q
2
|.The base forcest
α can also be understood as stress flux.According to the definitions of various stress tensors, the relation between the base forces and various stress tensors can be given. For example, the Cauchy stress is(5)
σ
=
1
A
Q
t
α
⊗
Q
α
,
where ⊗ is the dyadic symbol and the summation rule is implied.When the body force per unit mass is zero, the equilibrium equation can be written as(6)
∂
t
α
∂
x
α
=
0
.


## 2.2. Conjugate Variable of Base Forces

LetP denote the position vector of a point before deformation; the displacement u of a point is
(7)
u
=
Q
-
P
.The gradient of displacementu
α can be written as
(8)
u
α
=
∂
u
∂
x
α
=
Q
α
-
P
α
.Further, the elastic law can be given as follows:(9)
t
α
=
ρ
A
Q
∂
W
∂
u
α
,
in which W is the strain energy density and ρ is the mass density after deformation.Equation (9) expresses the t
α by strain energy directly. Thus, u
α is just the conjugate variable of t
α. It can be seen that the mechanics problem can be completely established by means of t
α and u
α.For the small deformation case, the Green strainε can be written as
(10)
ε
=
1
2
(
u
α
⊗
P
α
+
P
α
⊗
u
α
)
,
where P
α is the conjugate of P
α.


## 3. Model of the BFEM

### 3.1. Stiffness Matrix

Consider a triangular element as shown in Figure2; the stiffness matrix of a base force element can be obtained [27] as
(11)
K
I
J
=
E
2
A
(
1
+
v
)
[
2
v
1
-
2
v
m
I
⊗
m
J
+
m
I
J
U
+
m
J
⊗
m
I
]
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
(
I
=
1,2
,
3
;
J
=
1,2
,
3
)
,
in which E is Young’s modulus, v is Poisson’s ratio, A is the area of an element, U is the unit tensor U
=
P
α
⊗
P
α
=
P
α
⊗
P
α, m
I
J
=
m
I
·
m
J, and m
I can be calculated from
(12)
m
I
=
m
α
I
P
α
=
1
2
(
L
I
J
n
I
J
+
L
I
K
n
I
K
)
,
where L
I
J and L
I
K are the lengths of edges I
J and I
K of an element and n
I
J and n
I
K denote the external normals of edges I
J and I
K, respectively.Figure 2
A triangular element.Further, the stiffness matrix of a base force element can be written as(13)
K
I
J
=
E
2
A
(
1
+
v
)
[
2
v
1
-
2
v
m
α
I
P
α
⊗
m
β
J
P
β
+
(
m
α
I
P
α
·
m
β
J
P
β
)
U
+
m
β
J
P
β
⊗
m
α
I
P
α
2
v
1
-
2
v
]
,
where the summation rule is implied.For the Cartesian coordinate system, we introduce the unit basic vectorse
α
(
α
=
1,2
) and have
(14)
P
α
=
e
α
,
P
β
=
e
β
,
U
=
e
γ
⊗
e
γ
or
(15)
P
1
=
e
x
,
P
2
=
e
y
;
P
1
=
e
x
,
P
2
=
e
y
,
U
=
e
x
⊗
e
x
+
e
y
⊗
e
y
.Substitute (15) into (13) and consider e
x
·
e
x
=
1, e
x
·
e
y
=
0; we can obtain
(16)
K
I
J
=
E
2
A
(
1
+
v
)
[
2
v
1
-
2
v
(
m
x
I
m
x
J
e
x
⊗
e
x
+
m
x
I
m
y
J
e
x
⊗
e
y
=
E
2
A
(
1
+
v
)
nn
+
m
y
I
m
x
J
e
y
⊗
e
x
+
m
y
I
m
y
J
e
y
⊗
e
y
)
=
E
2
A
(
1
+
v
)
+
(
m
x
I
m
x
J
+
m
y
I
m
y
J
)
(
e
x
⊗
e
x
+
e
y
⊗
e
y
)
=
E
2
A
(
1
+
v
)
+
(
m
x
I
m
x
J
e
x
⊗
e
x
+
m
x
I
m
y
J
e
y
⊗
e
x
=
E
2
A
(
1
+
v
)
+
m
y
I
m
x
J
e
x
⊗
e
y
+
m
y
I
m
y
J
e
y
⊗
e
y
)
2
v
1
-
2
v
]
or
(17)
[
K
I
J
]
e
=
E
2
A
(
1
+
v
)
×
[
2
-
2
v
1
-
2
v
m
x
I
m
x
J
+
m
y
I
m
y
J
2
v
1
-
2
v
m
x
I
m
y
J
+
m
y
I
m
x
J
2
v
1
-
2
v
m
y
I
m
x
J
+
m
x
I
m
y
J
2
-
2
v
1
-
2
v
m
y
I
m
y
J
+
m
x
I
m
x
J
]
.For the plane stress problem, it is necessary to replaceE
/
(
1
-
v
2
) by E and v
/
(
1
-
v
) by v in (16) and (17).Further, the stiffness matrix of a base force element for the plane stress problem can be written as(18)
[
K
I
J
]
e
=
E
A
(
1
-
v
2
)
×
[
m
x
I
m
x
J
+
1
-
v
2
m
y
I
m
y
J
v
m
x
I
m
y
J
+
1
-
v
2
m
y
I
m
x
J
v
m
y
I
m
x
J
+
1
-
v
2
m
x
I
m
y
J
m
y
I
m
y
J
+
1
-
v
2
m
x
I
m
x
J
]
,
in which m
I and m
J can be calculated from
(19)
m
I
=
m
i
I
e
i
=
1
2
(
L
I
J
n
I
J
+
L
L
I
n
L
I
)
=
1
2
(
L
I
J
n
i
I
J
e
i
+
L
L
I
n
i
L
I
e
i
)
=
1
2
(
L
I
J
n
i
I
J
+
L
L
I
n
i
L
I
)
e
i
,
m
J
=
m
i
J
e
i
=
1
2
(
L
J
K
n
J
K
+
L
I
J
n
I
J
)
=
1
2
(
L
J
K
n
i
J
K
e
i
+
L
I
J
n
i
I
J
e
i
)
=
1
2
(
L
J
K
n
i
J
K
+
L
I
J
n
i
I
J
)
e
i
or
(20)
{
m
x
I
m
y
I
}
=
1
2
(
L
I
J
{
n
x
I
J
n
y
I
J
}
+
L
L
I
{
n
x
L
I
n
y
L
I
}
)
=
1
2
(
L
I
J
{
y
J
-
y
I
L
I
J
-
x
J
-
x
I
L
I
J
}
+
L
L
I
{
y
I
-
y
L
L
L
I
-
x
I
-
x
L
L
L
I
}
)
=
1
2
{
y
J
-
y
L
x
L
-
x
J
}
,
{
m
x
J
m
y
J
}
=
1
2
(
L
J
K
{
n
x
J
K
n
y
J
K
}
+
L
I
J
{
n
x
I
J
n
y
I
J
}
)
=
1
2
(
L
J
K
{
y
K
-
y
J
L
J
K
-
x
K
-
x
J
L
J
K
}
+
L
I
J
{
y
J
-
y
I
L
I
J
-
x
J
-
x
I
L
I
J
}
)
=
1
2
{
y
K
-
y
I
x
I
-
x
K
}
.


### 3.2. Strain Tensor of an Element

When the element is small enough, the real strainε can be replaced by the average strain ε
-. We can obtain the average strain ε
- in the element as
(21)
ε
-
=
1
A
∫
A
ε
d
A
,
in which A is the area of an element.Substituting (10) into (21), we can easily derive the following equation:
(22)
ε
-
=
1
2
A
(
u
I
⊗
m
I
+
m
I
⊗
u
I
)
,
where I
=
(
1,2
,
3
) is the local code number for the nodes of a triangular element and the summation rule is implied.Substituteu
I
=
u
I
i
e
i and m
I
=
m
j
I
e
j into (22); we can obtain the explicit expression of the tensor formula as
(23)
ε
-
=
1
2
A
∑
I
=
1
3
[
(
2
u
I
x
m
x
I
)
e
x
⊗
e
x
+
(
u
I
x
m
y
I
+
u
I
y
m
x
I
)
e
x
⊗
e
y
nn
+
(
u
I
x
m
y
I
+
u
I
y
m
x
I
)
e
y
⊗
e
x
+
(
2
u
I
y
m
y
I
)
e
y
⊗
e
y
]
or
(24)
ε
-
x
=
1
A
∑
I
=
1
3
(
u
I
x
m
x
I
)
,
ε
-
y
=
1
A
∑
I
=
1
3
(
u
I
y
m
y
I
)
,
γ
-
x
y
=
1
A
∑
I
=
1
3
(
u
I
x
m
y
I
+
u
I
y
m
x
I
)
.


### 3.3. Stress Tensor of an Element

When the element is small enough, the real stressσ can be replaced by the average stress σ
-. According to the generalized Hooke law, the stress component expressions of an element can be obtained for the plane stress problem as follows:
(25)
σ
x
=
E
1
-
v
2
(
ε
x
+
v
ε
y
)
,
σ
y
=
E
1
-
v
2
(
ε
y
+
v
ε
x
)
,
τ
x
y
=
E
2
(
1
+
v
)
γ
x
y
.For the plane strain problem, it is necessary to replaceE by E
/
(
1
-
v
2
) and v by v
/
(
1
-
v
) in (25).


## 3.1. Stiffness Matrix

Consider a triangular element as shown in Figure2; the stiffness matrix of a base force element can be obtained [27] as
(11)
K
I
J
=
E
2
A
(
1
+
v
)
[
2
v
1
-
2
v
m
I
⊗
m
J
+
m
I
J
U
+
m
J
⊗
m
I
]
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
(
I
=
1,2
,
3
;
J
=
1,2
,
3
)
,
in which E is Young’s modulus, v is Poisson’s ratio, A is the area of an element, U is the unit tensor U
=
P
α
⊗
P
α
=
P
α
⊗
P
α, m
I
J
=
m
I
·
m
J, and m
I can be calculated from
(12)
m
I
=
m
α
I
P
α
=
1
2
(
L
I
J
n
I
J
+
L
I
K
n
I
K
)
,
where L
I
J and L
I
K are the lengths of edges I
J and I
K of an element and n
I
J and n
I
K denote the external normals of edges I
J and I
K, respectively.Figure 2
A triangular element.Further, the stiffness matrix of a base force element can be written as(13)
K
I
J
=
E
2
A
(
1
+
v
)
[
2
v
1
-
2
v
m
α
I
P
α
⊗
m
β
J
P
β
+
(
m
α
I
P
α
·
m
β
J
P
β
)
U
+
m
β
J
P
β
⊗
m
α
I
P
α
2
v
1
-
2
v
]
,
where the summation rule is implied.For the Cartesian coordinate system, we introduce the unit basic vectorse
α
(
α
=
1,2
) and have
(14)
P
α
=
e
α
,
P
β
=
e
β
,
U
=
e
γ
⊗
e
γ
or
(15)
P
1
=
e
x
,
P
2
=
e
y
;
P
1
=
e
x
,
P
2
=
e
y
,
U
=
e
x
⊗
e
x
+
e
y
⊗
e
y
.Substitute (15) into (13) and consider e
x
·
e
x
=
1, e
x
·
e
y
=
0; we can obtain
(16)
K
I
J
=
E
2
A
(
1
+
v
)
[
2
v
1
-
2
v
(
m
x
I
m
x
J
e
x
⊗
e
x
+
m
x
I
m
y
J
e
x
⊗
e
y
=
E
2
A
(
1
+
v
)
nn
+
m
y
I
m
x
J
e
y
⊗
e
x
+
m
y
I
m
y
J
e
y
⊗
e
y
)
=
E
2
A
(
1
+
v
)
+
(
m
x
I
m
x
J
+
m
y
I
m
y
J
)
(
e
x
⊗
e
x
+
e
y
⊗
e
y
)
=
E
2
A
(
1
+
v
)
+
(
m
x
I
m
x
J
e
x
⊗
e
x
+
m
x
I
m
y
J
e
y
⊗
e
x
=
E
2
A
(
1
+
v
)
+
m
y
I
m
x
J
e
x
⊗
e
y
+
m
y
I
m
y
J
e
y
⊗
e
y
)
2
v
1
-
2
v
]
or
(17)
[
K
I
J
]
e
=
E
2
A
(
1
+
v
)
×
[
2
-
2
v
1
-
2
v
m
x
I
m
x
J
+
m
y
I
m
y
J
2
v
1
-
2
v
m
x
I
m
y
J
+
m
y
I
m
x
J
2
v
1
-
2
v
m
y
I
m
x
J
+
m
x
I
m
y
J
2
-
2
v
1
-
2
v
m
y
I
m
y
J
+
m
x
I
m
x
J
]
.For the plane stress problem, it is necessary to replaceE
/
(
1
-
v
2
) by E and v
/
(
1
-
v
) by v in (16) and (17).Further, the stiffness matrix of a base force element for the plane stress problem can be written as(18)
[
K
I
J
]
e
=
E
A
(
1
-
v
2
)
×
[
m
x
I
m
x
J
+
1
-
v
2
m
y
I
m
y
J
v
m
x
I
m
y
J
+
1
-
v
2
m
y
I
m
x
J
v
m
y
I
m
x
J
+
1
-
v
2
m
x
I
m
y
J
m
y
I
m
y
J
+
1
-
v
2
m
x
I
m
x
J
]
,
in which m
I and m
J can be calculated from
(19)
m
I
=
m
i
I
e
i
=
1
2
(
L
I
J
n
I
J
+
L
L
I
n
L
I
)
=
1
2
(
L
I
J
n
i
I
J
e
i
+
L
L
I
n
i
L
I
e
i
)
=
1
2
(
L
I
J
n
i
I
J
+
L
L
I
n
i
L
I
)
e
i
,
m
J
=
m
i
J
e
i
=
1
2
(
L
J
K
n
J
K
+
L
I
J
n
I
J
)
=
1
2
(
L
J
K
n
i
J
K
e
i
+
L
I
J
n
i
I
J
e
i
)
=
1
2
(
L
J
K
n
i
J
K
+
L
I
J
n
i
I
J
)
e
i
or
(20)
{
m
x
I
m
y
I
}
=
1
2
(
L
I
J
{
n
x
I
J
n
y
I
J
}
+
L
L
I
{
n
x
L
I
n
y
L
I
}
)
=
1
2
(
L
I
J
{
y
J
-
y
I
L
I
J
-
x
J
-
x
I
L
I
J
}
+
L
L
I
{
y
I
-
y
L
L
L
I
-
x
I
-
x
L
L
L
I
}
)
=
1
2
{
y
J
-
y
L
x
L
-
x
J
}
,
{
m
x
J
m
y
J
}
=
1
2
(
L
J
K
{
n
x
J
K
n
y
J
K
}
+
L
I
J
{
n
x
I
J
n
y
I
J
}
)
=
1
2
(
L
J
K
{
y
K
-
y
J
L
J
K
-
x
K
-
x
J
L
J
K
}
+
L
I
J
{
y
J
-
y
I
L
I
J
-
x
J
-
x
I
L
I
J
}
)
=
1
2
{
y
K
-
y
I
x
I
-
x
K
}
.


## 3.2. Strain Tensor of an Element

When the element is small enough, the real strainε can be replaced by the average strain ε
-. We can obtain the average strain ε
- in the element as
(21)
ε
-
=
1
A
∫
A
ε
d
A
,
in which A is the area of an element.Substituting (10) into (21), we can easily derive the following equation:
(22)
ε
-
=
1
2
A
(
u
I
⊗
m
I
+
m
I
⊗
u
I
)
,
where I
=
(
1,2
,
3
) is the local code number for the nodes of a triangular element and the summation rule is implied.Substituteu
I
=
u
I
i
e
i and m
I
=
m
j
I
e
j into (22); we can obtain the explicit expression of the tensor formula as
(23)
ε
-
=
1
2
A
∑
I
=
1
3
[
(
2
u
I
x
m
x
I
)
e
x
⊗
e
x
+
(
u
I
x
m
y
I
+
u
I
y
m
x
I
)
e
x
⊗
e
y
nn
+
(
u
I
x
m
y
I
+
u
I
y
m
x
I
)
e
y
⊗
e
x
+
(
2
u
I
y
m
y
I
)
e
y
⊗
e
y
]
or
(24)
ε
-
x
=
1
A
∑
I
=
1
3
(
u
I
x
m
x
I
)
,
ε
-
y
=
1
A
∑
I
=
1
3
(
u
I
y
m
y
I
)
,
γ
-
x
y
=
1
A
∑
I
=
1
3
(
u
I
x
m
y
I
+
u
I
y
m
x
I
)
.


## 3.3. Stress Tensor of an Element

When the element is small enough, the real stressσ can be replaced by the average stress σ
-. According to the generalized Hooke law, the stress component expressions of an element can be obtained for the plane stress problem as follows:
(25)
σ
x
=
E
1
-
v
2
(
ε
x
+
v
ε
y
)
,
σ
y
=
E
1
-
v
2
(
ε
y
+
v
ε
x
)
,
τ
x
y
=
E
2
(
1
+
v
)
γ
x
y
.For the plane strain problem, it is necessary to replaceE by E
/
(
1
-
v
2
) and v by v
/
(
1
-
v
) in (25).


## 4. Random Aggregate Model

Based on the Fuller grading curve, Walraven and Reinhardt [28] put the three-dimensional grading curve into the probability of any point which located in the sectional plane of specimens, and its expression is as follows:
(26)
P
c
(
D
<
D
0
)
=
P
k
(
1.065
(
D
0
D
max
⁡
)
1
/
2
-
0.053
(
D
0
D
max
⁡
)
4
-
0.012
(
D
0
D
max
⁡
)
6
-
0.0045
(
D
0
D
max
⁡
)
8
-
0.0025
(
D
0
D
max
⁡
)
10
)
,
where P
k is the volume percentage of aggregate volume among the specimens, in general P
k
=
0.75, D
0 is the diameter of sieve pore, and D
max
⁡ is the maximum aggregate size.According to (26), the numbers of coarse aggregate particles with various sizes can be obtained. By Monte Carlo method, random to create the centroid coordinates of all kinds of coarse aggregate particles, namely, to generate random aggregate model.According to the projection method, we dissect the specimens of concrete with different phases of materials. Then, the phase of coarse aggregate, the phase of hardened cement, and the phase of interfacial transition zone (ITZ) can be judged by a computer code as in Figure3.Figure 3
Attribute recognition figure.


## 5. Damage Model of Materials

Components of concrete such as coarse aggregate, cement mortar, and interfacial transition zone (ITZ) are basically quasibrittle material, whose failure patterns are mainly brittle failure.In this paper, according to the characteristics of concrete on mesostructure, the damage degradation of concrete is described by the bilinear damage model, and the failure principal is the criterion of maximum tensile stress. Damage constitutive model is defined asE
~
=
E
0
(
1
-
D
)
(
0
≤
D
≤
1
), shown in Figure 4, where E
0 is the initial Young modulus; the damage factor D can be expressed as follows:
(27)
D
=
{
0
ε
<
ε
0
1
-
η
-
λ
η
-
1
ε
0
ε
+
1
-
λ
η
-
1
ε
0
<
ε
≤
ε
r
1
-
λ
ε
0
ε
ε
r
<
ε
≤
ε
u
1
ε
>
ε
u
,
where f
t is the tensile strength of material, the residual tensile strength is defined as f
t
r
=
λ
f
t, the residual strength coefficient λ ranges from 0 to 1, the residual strain is ε
r
=
η
ε
0, ε
0 is the peak strain, η is the residual strain coefficient, the ultimate strain is defined as ε
u
=
ξ
ε
0, where ξ is ultimate strain coefficient, and ε is principal tensile strain of element.Figure 4
Bilinear damage model.


## 6. Numerical Example

According to the test results obtained from the experiment, material parameters of concrete are selected, which is completely coherent with model material used in the experiment. Material parameters of numerical simulation are shown in Table1. The concrete specimens were loaded by displacement steps.Table 1
Material parameters of uniaxial compressive tests.
Materials
Elastic modulus/GPa
Poisson’s ratio
Tensile strength/MPa
λ
η
ξ
Coarse aggregate
70
0.16
10
0.1
5
10
Cement mortar
30
0.22
4
0.1
4
10
Interfacial transition zone
15
0.2
2.5
0.1
3
10


### 6.1. Compressive Strength of Concrete Specimen

For the size150
mm
×
150
mm
×
150
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The random aggregate model is generated as in Figure 5.Figure 5
Random aggregate model with150
mm
×
150
mm
×
150
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure6. The compressive strengths of the three specimens were 24.28 MPa, 24.40 MPa, and 24.41 MPa. The uniaxial compressive strength average of the specimen group is 24.36 MPa. The result BFEM on mesodamage analysis for concrete is consistent with the test results [29].Figure 6
Stress-strain curve of specimen with150
mm
×
150
mm
×
150
mm.


### 6.2. Size Effects of Compressive Strength

In order to study the size effects of compressive strength for concrete specimen, we used three different sizes of specimens in the numerical simulations, which are100
mm
×
100
mm
×
100
mm, 150
mm
×
150
mm
×
150
mm, and 300
mm
×
300
mm
×
300
mm.


#### 6.2.1. Compressive Strength of 100 mm × 100 mm × 100 mm Specimens

For the size100
mm
×
100
mm
×
100
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The random aggregate model is as in Figure 7.Figure 7
Random aggregate model with100
mm
×
100
mm
×
100
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure8. The compressive strengths of the four specimens were 25.85 MPa, 25.33 MPa, and 25.67 MPa. The uniaxial compressive strength average of the specimen group is 25.62 MPa.Figure 8
Stress-strain curve of concrete with100
mm
×
100
mm
×
100
mm.


#### 6.2.2. Compressive Strength of 300 mm × 300 mm × 300 mm Specimens

For the size300
mm
×
300
mm
×
300
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The generated random aggregate model is as in Figure 9.Figure 9
Random aggregate model with300
mm
×
300
mm
×
300
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure10. The compressive strengths of the four specimens were 22.59 MPa, 22.72 MPa, and 22.70 MPa. The uniaxial compressive strength average of the specimen group is 22.67 MPa.Figure 10
Stress-strain curve of specimen with300
mm
×
300
mm
×
300
mm.The results of150
mm
×
150
mm
×
150
mm specimens are shown in Figure 6. The size effects of mechanical properties of concrete under uniaxial compression loading are shown in Table 2.Table 2
Different sizes of concrete compressive strength under uniaxial compression loading.
Different sizes of concrete
Compressive strength/MPa
100 mm × 100 mm × 100 mm
25.62
150 mm × 150 mm × 150 mm
24.36
300 mm × 300 mm × 300 mm
22.67


### 6.3. Fracture Process of Concrete

In the calculation, we use the incremental method and the displacement loading steps. According to the first strength theory, the element causes damage when the first principal stress exceeds the allowable value of the element. All the damaged elements were represented by the black color.The propagation process of cracks of the specimen of concrete with100
mm
×
100
mm
×
100
mm by uniaxial compression is shown in Figure 11.Propagation process of cracks of the specimen with100
mm
×
100
mm
×
100
mm.
(a)
Specimen 1 with100
mm
×
100
mm
×
100
mm
(b)
Specimen 2 with100
mm
×
100
mm
×
100
mm
(c)
Specimen 3 with100
mm
×
100
mm
×
100
mmThe propagation process of cracks of the specimen with150
mm
×
150
mm
×
150
mm by uniaxial compression is shown in Figure 12.Propagation process of cracks of the specimen with150
mm
×
150
mm
×
150
mm.
(a)
Specimen 1 with150
mm
×
150
mm
×
150
mm
(b)
Specimen 2 with150
mm
×
150
mm
×
150
mm
(c)
Specimen 3 with150
mm
×
150
mm
×
150
mmThe propagation process of cracks of the specimen with300
mm
×
300
mm
×
300
mm by uniaxial compression is shown in Figure 13.Propagation process of cracks of the RAC specimen with300
mm
×
300
mm
×
300
mm.
(a)
Specimen 1 with300
mm
×
300
mm
×
300
mm
(b)
Specimen 2 with300
mm
×
300
mm
×
300
mm
(c)
Specimen 3 with300
mm
×
300
mm
×
300
mm


## 6.1. Compressive Strength of Concrete Specimen

For the size150
mm
×
150
mm
×
150
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The random aggregate model is generated as in Figure 5.Figure 5
Random aggregate model with150
mm
×
150
mm
×
150
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure6. The compressive strengths of the three specimens were 24.28 MPa, 24.40 MPa, and 24.41 MPa. The uniaxial compressive strength average of the specimen group is 24.36 MPa. The result BFEM on mesodamage analysis for concrete is consistent with the test results [29].Figure 6
Stress-strain curve of specimen with150
mm
×
150
mm
×
150
mm.


## 6.2. Size Effects of Compressive Strength

In order to study the size effects of compressive strength for concrete specimen, we used three different sizes of specimens in the numerical simulations, which are100
mm
×
100
mm
×
100
mm, 150
mm
×
150
mm
×
150
mm, and 300
mm
×
300
mm
×
300
mm.


### 6.2.1. Compressive Strength of 100 mm × 100 mm × 100 mm Specimens

For the size100
mm
×
100
mm
×
100
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The random aggregate model is as in Figure 7.Figure 7
Random aggregate model with100
mm
×
100
mm
×
100
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure8. The compressive strengths of the four specimens were 25.85 MPa, 25.33 MPa, and 25.67 MPa. The uniaxial compressive strength average of the specimen group is 25.62 MPa.Figure 8
Stress-strain curve of concrete with100
mm
×
100
mm
×
100
mm.


### 6.2.2. Compressive Strength of 300 mm × 300 mm × 300 mm Specimens

For the size300
mm
×
300
mm
×
300
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The generated random aggregate model is as in Figure 9.Figure 9
Random aggregate model with300
mm
×
300
mm
×
300
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure10. The compressive strengths of the four specimens were 22.59 MPa, 22.72 MPa, and 22.70 MPa. The uniaxial compressive strength average of the specimen group is 22.67 MPa.Figure 10
Stress-strain curve of specimen with300
mm
×
300
mm
×
300
mm.The results of150
mm
×
150
mm
×
150
mm specimens are shown in Figure 6. The size effects of mechanical properties of concrete under uniaxial compression loading are shown in Table 2.Table 2
Different sizes of concrete compressive strength under uniaxial compression loading.
Different sizes of concrete
Compressive strength/MPa
100 mm × 100 mm × 100 mm
25.62
150 mm × 150 mm × 150 mm
24.36
300 mm × 300 mm × 300 mm
22.67


## 6.2.1. Compressive Strength of 100 mm × 100 mm × 100 mm Specimens

For the size100
mm
×
100
mm
×
100
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The random aggregate model is as in Figure 7.Figure 7
Random aggregate model with100
mm
×
100
mm
×
100
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure8. The compressive strengths of the four specimens were 25.85 MPa, 25.33 MPa, and 25.67 MPa. The uniaxial compressive strength average of the specimen group is 25.62 MPa.Figure 8
Stress-strain curve of concrete with100
mm
×
100
mm
×
100
mm.


## 6.2.2. Compressive Strength of 300 mm × 300 mm × 300 mm Specimens

For the size300
mm
×
300
mm
×
300
mm of compression specimen, the numbers of coarse aggregate particles can be obtained according to (26). The generated random aggregate model is as in Figure 9.Figure 9
Random aggregate model with300
mm
×
300
mm
×
300
mm.The uniaxial compressive stress-strain curve of concrete is got as shown in Figure10. The compressive strengths of the four specimens were 22.59 MPa, 22.72 MPa, and 22.70 MPa. The uniaxial compressive strength average of the specimen group is 22.67 MPa.Figure 10
Stress-strain curve of specimen with300
mm
×
300
mm
×
300
mm.The results of150
mm
×
150
mm
×
150
mm specimens are shown in Figure 6. The size effects of mechanical properties of concrete under uniaxial compression loading are shown in Table 2.Table 2
Different sizes of concrete compressive strength under uniaxial compression loading.
Different sizes of concrete
Compressive strength/MPa
100 mm × 100 mm × 100 mm
25.62
150 mm × 150 mm × 150 mm
24.36
300 mm × 300 mm × 300 mm
22.67


## 6.3. Fracture Process of Concrete

In the calculation, we use the incremental method and the displacement loading steps. According to the first strength theory, the element causes damage when the first principal stress exceeds the allowable value of the element. All the damaged elements were represented by the black color.The propagation process of cracks of the specimen of concrete with100
mm
×
100
mm
×
100
mm by uniaxial compression is shown in Figure 11.Propagation process of cracks of the specimen with100
mm
×
100
mm
×
100
mm.
(a)
Specimen 1 with100
mm
×
100
mm
×
100
mm
(b)
Specimen 2 with100
mm
×
100
mm
×
100
mm
(c)
Specimen 3 with100
mm
×
100
mm
×
100
mmThe propagation process of cracks of the specimen with150
mm
×
150
mm
×
150
mm by uniaxial compression is shown in Figure 12.Propagation process of cracks of the specimen with150
mm
×
150
mm
×
150
mm.
(a)
Specimen 1 with150
mm
×
150
mm
×
150
mm
(b)
Specimen 2 with150
mm
×
150
mm
×
150
mm
(c)
Specimen 3 with150
mm
×
150
mm
×
150
mmThe propagation process of cracks of the specimen with300
mm
×
300
mm
×
300
mm by uniaxial compression is shown in Figure 13.Propagation process of cracks of the RAC specimen with300
mm
×
300
mm
×
300
mm.
(a)
Specimen 1 with300
mm
×
300
mm
×
300
mm
(b)
Specimen 2 with300
mm
×
300
mm
×
300
mm
(c)
Specimen 3 with300
mm
×
300
mm
×
300
mm


## 7. Conclusions

In this paper, the base force element method (BFEM) on damage mechanics is used to analyze the size effect on compressive strength for concrete at mesolevel. The characteristics of the BFEM compared with the traditional FEM are as follows. The expression of the stiffness matrix is a precise expression, and it is not necessary to introduce Gauss’ integral for calculating the stiffness coefficient at a point. This expression of stiffness matrix can be used for calculating the stiffness of various elements with a unified method. This expression of stiffness matrix can be used in any coordinate system. The method of constructing the stiffness matrix does not regulate the introduction of interpolation. The model of the base force element method was used to analyze the damage problem for concrete and was used to analyze the relationships of mesostructure and macroscopic mechanical performance of concrete in this paper. The concrete is taken as three-phase composites consisting of coarse aggregate, hardened cement mortar, and interfacial transition zone (ITZ) on mesolevel. The random aggregate model is used for the numerical simulation of uniaxial compressive performance of concrete. The results by the BFEM show that the uniaxial compressive strengths of specimens are approximately coincident with the experiment results and the size effect of specimens is in agreement with the common rule. The following conclusions can be drawn.(1)
It shows that the BFEM with the mesodamage model is feasible and effective to study failure process and mechanical parameter of concrete.
(2)
When specimens are under the same conditions, the compressive strength of concrete decreases with increasing specimen size.
(3)
Macrofailure of concrete apparently lags behind the growth of microdamage; that is, macrofailure can be considered as the accumulation of microdamage.
(4)
The interface is the weakest part of concrete, and the selection of mechanical parameters of interface plays an important role on the numerical results of uniaxial compression for specimens. More future studies should be done on researching the mechanical property of the interface.



---

*Source: 528590-2014-08-04.xml*