# Mechanism Research of Arch Dam Abutment Forces during Overload

Authors: Yu Xia; Chuangdi Li; Xiaolian Zhao; Zhongqing Zhang
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

This paper presents research on the abutment forces of a double-curvature arch dam during overload based on numerical calculation results obtained through finite element method by Ansys. Results show that, with an increase in elevation, the abutment forces and bending moment of the arch dam increase first and then decrease from the bottom to the top of the dam. Abutment forces and bending moment reach their maximum at the middle or middle-down portion of the dam. The distributions of abutment forces and moment do not change during overload. The magnitude of each arch layer’s forces and moment increases linearly during overload. This result indicates that each arch layer transmits bearing loads to the rocks of the left and right banks steadily. This research explains the operating mechanism of an arch dam under normal and overload conditions. It provides a simple method to calculate the distribution of forcesF
x and F
y and a new method to calculate the overload factor of an arch dam through the estimation of arch layers based on the redistribution characteristic of arch abutment forces.


---

## Body

## 1. Introduction

The development of arch dams has a long history that dates back to 1st century BC [1]. Relative uniformity was achieved in the 20th century after several designs and techniques were developed. The first known arch dam, Glanum Dam, was built by the Romans in France [2]. Arch dam is a type of dam curved in the shape of an arch, with the top of the arch pointing back into the reservoir. Thus, the force of the water against it, known as hydrostatic pressure, presses against the arch. An arch dam is most suitable for narrow gorges with stable rocks. Considering that arch dams are thinner than any other dam type, they require much less construction materials, which make them economical and practical in remote areas. Arch dams are built all over the world because they are safe and involve minimal cost [3]. China has the most number of arch dams [4]. Many numerical methods are currently applied to the analysis of the structure of arch dams. Some examples of these methods are finite element method [5–7], discrete element method [8, 9], block element method [10, 11], discontinuous deformation analysis [12], fast Lagrange analysis of continua [13], and interface element method [14]. Self-adapt element [15–17], meshless [18], and extended finite element methods [19–21] are utilized to simulate the development of cracks in the structure analysis of arch dams. The arch dam is a highly statically indeterminate structure. Through an arching action, arch layers transmit upstream water pressure to bank rocks on two sides. Arch layers play an important role in the safe operation of such dams, which transfers the loads to two banks. It makes large areas of arch dam body under compression through arch layer. This type of structure can make full use of compression strength of concrete. Limit equilibrium method is usually employed in the safety evaluation of arch dams [22, 23]. Little attention is paid to abutment forces in arch dam safety analysis. The mechanical characteristics of a structure can be determined through research on arch abutment forces during overload. It shows that thrust angles at different elevations increase during overload through abutments force analysis [24, 25]. Studying abutment force would thus provide a comprehensive understanding of the overload mechanism of arch dams.


## 2. Concrete Cracking Simulation of Arch Dam during Overload [26]

### 2.1. Concrete Cracking Mode

Before the failure of concrete under tensile condition, a linear relationship exists between stress and strain. The stiffness matrix is(1)
D
c
c
k
=
E
1
+
ν
1
-
2
ν
·
1
-
ν
ν
ν
0
0
0
ν
1
-
ν
ν
0
0
0
ν
ν
1
-
ν
0
0
0
0
0
0
1
-
2
ν
2
0
0
0
0
0
0
1
-
2
ν
2
0
0
0
0
0
0
1
-
2
ν
2
,where E is Young’s modulus for concrete and ν is Poisson’s ratio for concrete.The following conditions apply to cracks in one direction only.If principal stress in one direction is greater than the failure tensile stress, tensile failure occurs. After developing cracks, the concrete becomes an orthotropic material. Given that the stiffness and shear stiffness reduce the normal plane, the stress-strain matrix will change. After the destruction of the concrete, the presence of a crack at an integration point is represented through modification of the stress-strain relations by introducing a plane of weakness in a direction normal to the crack face. A shear transfer coefficientβ
t is introduced to represent a shear strength reduction factor for subsequent loads that induce sliding (shear) across the crack face. The stress-strain relationship is built in the direction of the failure surface and the direction perpendicular to it. When principal stress is greater than the tensile breaking stress in only one direction, the stress and strain of the new matrix are(2)
D
c
c
k
=
E
1
+
ν
·
R
t
1
-
ν
E
0
0
0
0
0
0
1
1
-
ν
ν
1
-
ν
0
0
0
0
ν
1
-
ν
1
1
-
ν
0
0
0
0
0
0
β
t
2
0
0
0
0
0
0
1
2
0
0
0
0
0
0
β
t
2
,where the superscript c
k signifies that the stress-strain relations refer to a coordinate system parallel to principal stress directions with the x
c
k axis perpendicular to the crack face. R
t works with adaptive descent and diminishes to 0.0 as the solution converges.In Figure1, f
t is uniaxial tensile cracking stress and T
c is the multiplier for the amount of tensile stress relaxation.Figure 1
Strength of cracked condition.If the crack closes, then all compressive stresses normal to the crack plane are transmitted across the crack. Only shear transfer coefficientβ
c for a closed crack is introduced. D
c
c
k can then be expressed as(3)
D
c
c
k
=
E
1
+
ν
1
-
2
ν
·
1
-
ν
ν
ν
0
0
0
ν
1
-
ν
ν
0
0
0
ν
ν
1
-
ν
0
0
0
0
0
0
β
c
1
-
2
ν
2
0
0
0
0
0
0
1
-
2
ν
2
0
0
0
0
0
0
β
c
1
-
2
ν
2
.The stress-strain relations for concrete that has cracked in two directions are(4)
D
c
c
k
=
E
R
t
E
0
0
0
0
0
0
R
t
E
0
0
0
0
0
0
1
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
.If both directions reclose(5)
D
c
c
k
=
E
1
+
ν
1
-
2
ν
1
-
ν
ν
ν
0
0
0
ν
1
-
ν
ν
0
0
0
ν
ν
1
-
ν
0
0
0
0
0
0
β
c
1
-
2
ν
2
0
0
0
0
0
0
β
c
1
-
2
ν
2
0
0
0
0
0
0
β
c
1
-
2
ν
2
.The stress-strain relations for concrete that has cracked in all three directions are(6)
D
c
c
k
=
E
R
t
E
0
0
0
0
0
0
R
t
E
0
0
0
0
0
0
R
t
E
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
.If the crack is closed again in three directions, formula (5) will be selected. β
c and β
t values have the following relationship:(7)
1
>
β
c
>
β
t
>
0
.


### 2.2. Distinguishing between Crack Opening and Closing

The open or closed status of integration point cracking is based on strain valueε
c
k
c
k, which is called the crack strain. For the case of a possible crack in the x direction, this strain is evaluated as(8)
ε
c
k
c
k
=
ε
x
c
k
+
ν
1
-
ν
ε
y
c
k
+
ε
z
c
k
if
no
cracking
has
occurred
ε
x
c
k
+
ν
ε
z
c
k
if
y
direction
has
cracked
ε
x
c
k
if
y
and
z
direction
have
cracked
,where ε
x
c
k,  ε
y
c
k,  ε
z
c
k are three normal component strains in crack orientation.Vector{
ε
c
k
} is computed by(9)
ε
c
k
=
T
c
k
ε
′
,where {
ε
′
} is modified total strain (in element coordinates). {
ε
′
} is defined as(10)
ε
′
=
ε
n
-
1
el
+
Δ
ε
n
-
Δ
ε
n
th
-
Δ
ε
n
pl
,where n is the substep number, ε
n
-
1
el is the elastic strain from the previous substep, Δ
ε
n is the total strain increment (based on {
Δ
u
n
}, the displacement increment over the substep), Δ
ε
n
th is the thermal strain increment, and Δ
ε
n
pl is the plastic strain increment.Ifε
c
k
c
k is less than zero, the associated crack is assumed to be closed.Ifε
c
k
c
k is greater than or equal to zero, the associated crack is assumed to be open. When cracking first occurs at an integration point, the crack is assumed to be open for the next iteration.


## 2.1. Concrete Cracking Mode

Before the failure of concrete under tensile condition, a linear relationship exists between stress and strain. The stiffness matrix is(1)
D
c
c
k
=
E
1
+
ν
1
-
2
ν
·
1
-
ν
ν
ν
0
0
0
ν
1
-
ν
ν
0
0
0
ν
ν
1
-
ν
0
0
0
0
0
0
1
-
2
ν
2
0
0
0
0
0
0
1
-
2
ν
2
0
0
0
0
0
0
1
-
2
ν
2
,where E is Young’s modulus for concrete and ν is Poisson’s ratio for concrete.The following conditions apply to cracks in one direction only.If principal stress in one direction is greater than the failure tensile stress, tensile failure occurs. After developing cracks, the concrete becomes an orthotropic material. Given that the stiffness and shear stiffness reduce the normal plane, the stress-strain matrix will change. After the destruction of the concrete, the presence of a crack at an integration point is represented through modification of the stress-strain relations by introducing a plane of weakness in a direction normal to the crack face. A shear transfer coefficientβ
t is introduced to represent a shear strength reduction factor for subsequent loads that induce sliding (shear) across the crack face. The stress-strain relationship is built in the direction of the failure surface and the direction perpendicular to it. When principal stress is greater than the tensile breaking stress in only one direction, the stress and strain of the new matrix are(2)
D
c
c
k
=
E
1
+
ν
·
R
t
1
-
ν
E
0
0
0
0
0
0
1
1
-
ν
ν
1
-
ν
0
0
0
0
ν
1
-
ν
1
1
-
ν
0
0
0
0
0
0
β
t
2
0
0
0
0
0
0
1
2
0
0
0
0
0
0
β
t
2
,where the superscript c
k signifies that the stress-strain relations refer to a coordinate system parallel to principal stress directions with the x
c
k axis perpendicular to the crack face. R
t works with adaptive descent and diminishes to 0.0 as the solution converges.In Figure1, f
t is uniaxial tensile cracking stress and T
c is the multiplier for the amount of tensile stress relaxation.Figure 1
Strength of cracked condition.If the crack closes, then all compressive stresses normal to the crack plane are transmitted across the crack. Only shear transfer coefficientβ
c for a closed crack is introduced. D
c
c
k can then be expressed as(3)
D
c
c
k
=
E
1
+
ν
1
-
2
ν
·
1
-
ν
ν
ν
0
0
0
ν
1
-
ν
ν
0
0
0
ν
ν
1
-
ν
0
0
0
0
0
0
β
c
1
-
2
ν
2
0
0
0
0
0
0
1
-
2
ν
2
0
0
0
0
0
0
β
c
1
-
2
ν
2
.The stress-strain relations for concrete that has cracked in two directions are(4)
D
c
c
k
=
E
R
t
E
0
0
0
0
0
0
R
t
E
0
0
0
0
0
0
1
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
.If both directions reclose(5)
D
c
c
k
=
E
1
+
ν
1
-
2
ν
1
-
ν
ν
ν
0
0
0
ν
1
-
ν
ν
0
0
0
ν
ν
1
-
ν
0
0
0
0
0
0
β
c
1
-
2
ν
2
0
0
0
0
0
0
β
c
1
-
2
ν
2
0
0
0
0
0
0
β
c
1
-
2
ν
2
.The stress-strain relations for concrete that has cracked in all three directions are(6)
D
c
c
k
=
E
R
t
E
0
0
0
0
0
0
R
t
E
0
0
0
0
0
0
R
t
E
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
0
0
0
0
0
0
β
t
2
1
+
ν
.If the crack is closed again in three directions, formula (5) will be selected. β
c and β
t values have the following relationship:(7)
1
>
β
c
>
β
t
>
0
.


## 2.2. Distinguishing between Crack Opening and Closing

The open or closed status of integration point cracking is based on strain valueε
c
k
c
k, which is called the crack strain. For the case of a possible crack in the x direction, this strain is evaluated as(8)
ε
c
k
c
k
=
ε
x
c
k
+
ν
1
-
ν
ε
y
c
k
+
ε
z
c
k
if
no
cracking
has
occurred
ε
x
c
k
+
ν
ε
z
c
k
if
y
direction
has
cracked
ε
x
c
k
if
y
and
z
direction
have
cracked
,where ε
x
c
k,  ε
y
c
k,  ε
z
c
k are three normal component strains in crack orientation.Vector{
ε
c
k
} is computed by(9)
ε
c
k
=
T
c
k
ε
′
,where {
ε
′
} is modified total strain (in element coordinates). {
ε
′
} is defined as(10)
ε
′
=
ε
n
-
1
el
+
Δ
ε
n
-
Δ
ε
n
th
-
Δ
ε
n
pl
,where n is the substep number, ε
n
-
1
el is the elastic strain from the previous substep, Δ
ε
n is the total strain increment (based on {
Δ
u
n
}, the displacement increment over the substep), Δ
ε
n
th is the thermal strain increment, and Δ
ε
n
pl is the plastic strain increment.Ifε
c
k
c
k is less than zero, the associated crack is assumed to be closed.Ifε
c
k
c
k is greater than or equal to zero, the associated crack is assumed to be open. When cracking first occurs at an integration point, the crack is assumed to be open for the next iteration.


## 3. Numerical Model and Parameters

Many constitutive models of concrete exist at present. In 3D finite element numerical calculations, the most preferred model is D-P criterion mainly because model parameters can be easily obtained with this model. D-P criterion has a smooth yield surface and is easy to use. This model can be easily utilized in elastic-plastic analyses for concrete and other materials with similar properties. The Drucker-Prager yield function is(11)
F
=
α
I
1
+
J
2
-
k
=
0
,where α and k are material parameters, I
1 is the first invariant of the stress tensor, and J
2 is the partial stress tensor for the second invariant. They can be calculated as(12)
I
1
=
σ
x
+
σ
y
+
σ
z
,
J
2
=
1
6
σ
x
-
σ
y
2
+
σ
y
-
σ
z
2
+
σ
z
-
σ
x
2
+
τ
x
y
2
+
τ
y
z
2
+
τ
z
x
2
,
α
=
2
sin
⁡
ϕ
3
3
-
sin
⁡
ϕ
,
k
=
6
c
′
cos
⁡
⁡
ϕ
3
3
-
sin
⁡
ϕ
,where σ
x
2, σ
y
2, σ
z
2, τ
x
y
2, τ
y
z
2, and τ
x
z
x
2 are stress components; and ϕ
=
t
g
-
1
f
′.


### 3.1. Finite Element Model

The coordination of the numerical model is shown in Figure2. The x-axis positive direction of the model is along the right banks. The positive direction of y-axis is downstream along the river, and the positive direction of z-axis is along the dam’s increasing height. The range of the numerical model is as follows: 230 m width on the two sides, 270 m depth under the riverbed, 150 m length upstream, and 360 m length downstream. The bottom of the numerical model is subjected to a three-direction constraint, the boundary of the left and right banks is subjected to an x direction constraint, and the boundary of the upstream and downstream is subjected to a y direction constraint. The model has 28,296 elements and 33,830 nodes in total. It will take much more time in numerical calculation with the elastic damage model for computer. So this model is used only in elements of dam body, the elastic-plastic model is used in other parts elements.Figure 2
Finite element model.


### 3.2. Parameters and Loads

The loads of each case include water pressure, temperature load, and dam gravity. The overload process was implemented by gradually adjusting the density of water and keeping the water level and other loads unchanged. Overload factorK is defined as K
=
P
/
P
0, where P is the current water density of the dam and P
0 is the normal water density. In Case 1, the value of water density is 9.8 kg/m3. In Case 2, the magnitude of water density is twice the normal value (19.6 kg/m3) and so on. The relevant parameters are listed in Table 1.Table 1
Material parameters.
Location
Elastic modulus (Gpa)
Bulk density (KN/m3)
Poisson’s ratio
Cohesion (Mpa)
Friction angle (°)
Coefficient of thermal expansion (10−6)
Dam body
18
2350
0.167
2
56
4.87
Abutment
14
2730
0.25
0.9
45
Base
11
2700
0.27
0.7
42
Unconformity
3
2660
0.3
0.35
35


## 3.1. Finite Element Model

The coordination of the numerical model is shown in Figure2. The x-axis positive direction of the model is along the right banks. The positive direction of y-axis is downstream along the river, and the positive direction of z-axis is along the dam’s increasing height. The range of the numerical model is as follows: 230 m width on the two sides, 270 m depth under the riverbed, 150 m length upstream, and 360 m length downstream. The bottom of the numerical model is subjected to a three-direction constraint, the boundary of the left and right banks is subjected to an x direction constraint, and the boundary of the upstream and downstream is subjected to a y direction constraint. The model has 28,296 elements and 33,830 nodes in total. It will take much more time in numerical calculation with the elastic damage model for computer. So this model is used only in elements of dam body, the elastic-plastic model is used in other parts elements.Figure 2
Finite element model.


## 3.2. Parameters and Loads

The loads of each case include water pressure, temperature load, and dam gravity. The overload process was implemented by gradually adjusting the density of water and keeping the water level and other loads unchanged. Overload factorK is defined as K
=
P
/
P
0, where P is the current water density of the dam and P
0 is the normal water density. In Case 1, the value of water density is 9.8 kg/m3. In Case 2, the magnitude of water density is twice the normal value (19.6 kg/m3) and so on. The relevant parameters are listed in Table 1.Table 1
Material parameters.
Location
Elastic modulus (Gpa)
Bulk density (KN/m3)
Poisson’s ratio
Cohesion (Mpa)
Friction angle (°)
Coefficient of thermal expansion (10−6)
Dam body
18
2350
0.167
2
56
4.87
Abutment
14
2730
0.25
0.9
45
Base
11
2700
0.27
0.7
42
Unconformity
3
2660
0.3
0.35
35


## 4. Calculation Results

The abutment forces were obtained through the integration of abutment element nodes. The directions of abutment forces are similar to those in the coordination of the numerical model. The direction of the bending moment complies with the right-hand rule. The following is an analysis of abutmentF
x, F
y, and bending moment. Abutment forces F
x, F
y and bending moment are discussed under normal loads and overloads separately. To express the results clearly, analysis is first performed on the distribution characteristic of abutment forces F
x, F
y and bending moment under normal loads. Then, the abutment forces changing rule during overload are discussed. A simple method is then provided to determine the distribution of abutment forces F
x and F
y. Arch layers 1 to 8 are arch layers from the bottom to the top of the dam. The calculated cases are shown below. The finial failure pattern is shown in Figure 3.Figure 3
Typical damage pattern of dam body (Case6: the black zone is the damage area of dam body.).Case 1.
It is as follows: 1 time water load + gravity + temperature load (Table2).Table 2
Results of Case1.
Location
Left bank
Right bank
F
x
(N)
F
y (N)
Mz (Nm)
F
x (N)
F
y (N)
Mz (NM)
Arch layer 1
64505100
167958000
−7709350000
−67725400
166091000
7369330000
Arch layer 2
141117000
240493000
−18486000000
−149254000
235289000
18784000000
Arch layer 3
175053000
248301000
−25307600000
−182556000
242942000
26715700000
Arch layer 4
181693000
225146000
−28966100000
−183487000
223801000
29565200000
Arch layer 5
165873000
188499000
−29136600000
−164688000
189537000
29105300000
Arch layer 6
130615000
142144000
−25264700000
−129284000
143358000
25159300000
Arch layer 7
81000100
86200100
−17439100000
−79221500
87835000
16947200000
Arch layer 8
20970900
22194700
−5013750000
−20239300
22862400
4735740000Case 2.
It is as follows: 2 times water load + gravity + temperature load (Table3).Table 3
Results of Case2.
Location
Left bank
Right bank
F
x (N)
F
y (N)
Mz (NM)
F
x (N)
F
y (N)
Mz (NM)
Arch layer 1
138468000
347820000
−16028900000
−144378000
344538000
15335700000
Arch layer 2
283616000
482485000
−37084800000
−299808000
472144000
37679300000
Arch layer 3
350106000
496601000
−50615200000
−365112000
485884000
53431400000
Arch layer 4
363385000
450292000
−57932300000
−366974000
447602000
59130300000
Arch layer 5
331747000
376998000
−58273100000
−329376000
379073000
58210600000
Arch layer 6
261230000
284287000
−50529500000
−258569000
286717000
50318600000
Arch layer 7
162000000
172400000
−34878100000
−158443000
175670000
33894400000
Arch layer 8
41941800
44389400
−10027500000
−40478500
45724800
9471470000Case 3.
It is as follows: 3 times water load + gravity + temperature load (Table4).Table 4
Results of Case3.
Location
Left bank
Right bank
F
x (N)
F
y (N)
Mz (NM)
F
x (N)
F
y (N)
Mz (NM)
Arch layer 1
212430000
527681000
−24348500000
−221030000
522985000
23302000000
Arch layer 2
426115000
724477000
−55683500000
−450362000
708998000
56574500000
Arch layer 3
525158000
744902000
−75922900000
−547669000
728827000
80147000000
Arch layer 4
545078000
675438000
−86898400000
−550461000
671402000
88695500000
Arch layer 5
497620000
565496000
−87409700000
−494064000
568610000
87315900000
Arch layer 6
391846000
426431000
−75794200000
−387853000
430075000
75477800000
Arch layer 7
243000000
258600000
−52317200000
−237665000
263505000
50841600000
Arch layer 8
62912700
66584100
−15041200000
−60717800
68587300
14207200000Case 4.
It is as follows: 4 times water load + gravity + temperature load (Table5).Table 5
Results of Case4.
Location
Left bank
Right bank
F
x (N)
F
y (N)
Mz (NM)
F
x (N)
F
y (N)
Mz (NM)
Arch layer 1
286393000
707543000
−32668100000
−297683000
701432000
31268300000
Arch layer 2
568613000
966468000
−74282300000
−600917000
945852000
75469800000
Arch layer 3
700211000
993203000
−101230000000
−730225000
971769000
106863000000
Arch layer 4
726770000
900584000
−115865000000
−733947000
895203000
118261000000
Arch layer 5
663494000
753995000
−116546000000
−658752000
758147000
116421000000
Arch layer 6
522461000
568574000
−101059000000
−517138000
573434000
100637000000
Arch layer 7
324000000
344801000
−69756300000
−316886000
351340000
67788800000
Arch layer 8
83883700
88778800
−20055000000
−80957100
91449700
18942900000Case 5.
It is as follows: 5 times water load + gravity + temperature load (Table6).Table 6
Results of Case5.
Location
Left bank
Right bank
F
x (N)
F
y (N)
Mz (NM)
F
x (N)
F
y (N)
Mz (NM)
Arch layer 1
360356000
887405000
−40987600000
−374335000
879878000
39234600000
Arch layer 2
711112000
1208460000
−92881100000
−751471000
1182710000
94365000000
Arch layer 3
875264000
1241500000
−126538000000
−912781000
1214710000
133578000000
Arch layer 4
908463000
1125730000
−144831000000
−917434000
1119000000
147826000000
Arch layer 5
829367000
942494000
−145683000000
−823440000
947683000
145527000000
Arch layer 6
653076000
710718000
−126324000000
−646422000
716792000
125796000000
Arch layer 7
405000000
431001000
−87195300000
−396108000
439175000
84736000000
Arch layer 8
104855000
110973000
−25068700000
−101196000
114312000
23678700000Case 6.
It is as follows: 6 times water load + gravity + temperature load (Table7).Table 7
Results of Case6.
Location
Left bank
Right bank
F
x (N)
F
y (N)
Mz (NM)
F
x (N)
F
y (N)
Mz (NM)
Arch layer 1
434318000
1067270000
−49307200000
−450988000
1058330000
47200900000
Arch layer 2
853611000
1450450000
−111480000000
−902026000
1419560000
113260000000
Arch layer 3
1050320000
1489800000
−151846000000
−1095340000
1457650000
160294000000
Arch layer 4
1090160000
1350880000
−173797000000
−1100920000
1342800000
177391000000
Arch layer 5
995240000
1130990000
−174819000000
−988128000
1137220000
174632000000
Arch layer 6
783691000
852862000
−151588000000
−775706000
860150000
150956000000
Arch layer 7
486000000
517201000
−104634000000
−475329000
527010000
101683000000
Arch layer 8
125825000
133168000
−30082500000
−121436000
137175000
28414400000


### 4.1. Distribution of Abutment Forces under Normal Loads

The abutment force ofF
x (Figure 4) increases gradually from the bottom to the middle of the dam as the elevation increases. It reaches the maximum at the middle and then decreases from the middle to the top of the arch dam. The maximum of F
x is at approximately 80 m of the height of the dam. The left and right banks F
x are symmetrical because of the symmetry of the dam body shape.Figure 4
Abutment forceF
x in Case 1.With regard to the distribution of forceF
y, force gradually increases as elevation increases; the maximum is reached at the middle-down portion of the arch dam (about 60 m high, Figure 5). The left and right banks F
y are almost similar because of the symmetry of the arch dam. The distribution of F
x and F
y has a similar feature to that from [24, 25].Figure 5
Abutment forceF
y in Case 1.The relation between bending moment and increasing elevation is similar to that inF
x. The only difference is that the maximum bending moment is at approximately 100 m of the dam height (Figure 6). The abutment bending moment of the left and right banks is also symmetrical. The results of abutment forces show that the arch dam transmits loads to rocks on the two sides through an arching action. Most of the loads are transferred to the rocks in the middle and downside of the two banks. The rocks in this location have an important function in the safety of the arch dam. Bending moment is calculated by force multiplied with distance. Compared with distance, the magnitude of force is much larger; thus, the distribution of bending moment mainly reflects the feature of force. Such is the reason why bending moment (Figure 6) appears somewhat similar to abutment force F
x (Figure 4).Figure 6
Abutment bending moment in Case1.


### 4.2. AbutmentF
x and F
y and Bending Moment of Each Arch Layer during Overload

The above calculation and analysis are results of the arch dam being subjected to normal loads. The following presents the results during overload. Analysis of the arch dam during overload would improve the understanding of the overloading mechanism of the arch dam. The feature of each arch layer during overload is shown (Figures7, 8, and 9). The F
x line of each arch layer during overload is straight (Figure 7). F
x is the force that dam abutment applies on the left and right bank rocks. Each arch layer transfers the excess loads to the rocks in the valley. This condition can explain why arch dams usually have a comparatively high tolerance for overloading. The arch plays a key role in overload transfer. In Figures 7, 8, and 9, the x axis is the overload factor. The overload process ensues by gradually adjusting the density of water and keeping the water level and other loads unchanged. The gradient of the lines signifies the transfer coefficient of each arch layer and shows the capacity of an arch layer to transmit overloads to bank rocks. As shown in Figures 7 to 9, the line of each arch layer is straight; thus, the transfer coefficient is a constant. Arch layer 4 has the largest coefficient followed by arch layers 3, 5, 2, 6, 7, 1, and 8. The arch layer of the middle and lower portions of the dam has a large transfer capacity because the middle and lower arch layers have a large central angle and suitable arc length. The rule for F
y force during overload is similar to that for F
x (Figure 8). For F
y, arch layer 3 has the largest coefficient followed by arch layers 2, 4, 5, 1, 6, 7, and 8. For bending moment, the size of the transfer coefficient (from large to small) is 4, 5, 3, 6, 2, 7, 1, and 8.Figure 7
Arch abutmentF
x during overload.Figure 8
Arch abutmentF
y during overload.Figure 9
Arch abutment moment during overload.The arch layers located in top and bottom portions of arch dam usually have a smaller transfer coefficient, and the arc lengths of these arch layers are either too long or too short. The axes of these arch layers are approximately straight lines, which goes against the role that an arch layer is playing. Arch layers located in the middle and lower portions of arch dam have relatively suitable arc length and central angle, which make them have a larger transfer coefficient. The central angle of arch layer is mainly affecting the magnitude ofF
x and F
y. Engineers usually hope that arch dam can transfer loads to two banks as much as possible and a larger transfer coefficient of arch layer is always preferred. Therefore, much attention should be specially paid to the determination of arc length and central angel of each arch layer during its designing process.


### 4.3. Distribution of AbutmentF
x and F
y and Bending Moment in Different Cases

The following presents the results on the distribution of abutment forces in different cases. Abutment forceF
x and bending moment have an obvious symmetry (Figures 10 and 12); the F
y of the left bank is almost similar to that of the right bank. The distribution of F
x, F
y, and bending moment does not change in different cases. The different abutment forces increase linearly under conditions of overload. Once the distribution of abutment forces in the normal case is determined, the distribution in different overload cases can also be determined. The unchanged distribution provides a good suggestion to arch dam designs. Engineers can modify the arch dam shape according to the distribution of abutment forces. The arch should not transmit a large force to weak rocks in a valley. The distribution also helps evaluate the abutment stability of arch dams. Abutment forces differ at different elevations. For an arch layer with larger abutment forces, more abutment forces are required during overload; hence, the rock mass at this location should be stable (Figures 10 and 11).Figure 10
F
x distribution in each case.Figure 11
F
y distribution in each case.Figure 12
Bending moment distribution in each case.


### 4.4. Simple Method to Determine the Distribution of Abutment Forces

Unchanged distribution is helpful to arch dam designers. The use of finite element method in the calculation is inconvenient and complex. For convenience, a simple method is necessary. Vertical force contributes little to the distribution of abutment forces becauseF
x and F
y are horizontal forces. Temperature load is much smaller than water load. Thus, the main factor that affects abutment force distribution is water load. Arch layers differ in terms of the length of the arch, magnitude of water load, height of the arch, and central angle. The first three factors affect the magnitude of force on the upstream face, and the last one affects the value of F
x and F
y. A simple formula to calculate the results is provided below:(13)
F
x
=
p
1
+
p
2
2
l
1
+
l
2
2
h
sin
⁡
α
,
F
y
=
p
1
+
p
2
2
l
1
+
l
2
2
h
cos
⁡
⁡
α
,where p
1 and p
2 are the water pressure of the upstream face, l
1 and l
2 are arch layer length, h is arch layer height, and α is the central angle of an arch layer. All parameters are shown in Figure 13.Figure 13
Arch layer.F
x and F
y distributions were obtained with the simple method (Table 8). The results are shown in Figures 14 and 15. The distribution is similar to that obtained with finite element method.Table 8
Abutment forces by simple method.
Arch layer height (m)
A (m2)
P (m)
Central angle (°)
Results of the simple method
Left bank
Right bank
Left bank
Right bank
F
x (N)
F
y (N)
F
x (N)
F
y (N)
20
2079.656
150
35.3456
34.6478
1768550126
2493602506
1737279556
2515489103
40
3439.683
130
39.9615
41.6339
2814539436
3358818546
2910159962
3276318277
60
4403.907
110
42.6571
43.8137
3216889748
3491351666
3285379568
3426981088
80
5171.987
90
44.4953
44.4745
3197065764
3253891718
3194603029
3256309300
100
5865.691
70
44.9981
45.0003
2845206930
2845395713
2844183315
2846419324
120
6506.773
50
45.0002
44.7193
2254489490
2254473811
2242513203
2266387238
140
7123.782
30
44.9966
43.6255
1480870720
1481046523
1444423162
1516614386
160
7873.98
10
44.867
42.6669
544153765
547120155
522759993.2
567596488.5Figure 14
F
x calculated with the simple method.Figure 15
F
y calculated with the simple method.The above mentioned method is an approximate means to calculateF
x and F
y; the absolute magnitudes of F
x and F
y are different from that obtained from finite element method. The simple method mainly aims to determine the distribution characteristic. To compare the accuracy of results obtained from the simple method and those from finite element method, each result of the two methods is utilized to divide the maximum value obtained through the two methods. The results are shown in Figures 16 and 17. The results of the simple method are consistent with those of finite element method.Figure 16
Comparison ofF
x obtained with the simple method and FEM.Figure 17
Comparison ofF
y obtained with the simple method and FEM.


## 4.1. Distribution of Abutment Forces under Normal Loads

The abutment force ofF
x (Figure 4) increases gradually from the bottom to the middle of the dam as the elevation increases. It reaches the maximum at the middle and then decreases from the middle to the top of the arch dam. The maximum of F
x is at approximately 80 m of the height of the dam. The left and right banks F
x are symmetrical because of the symmetry of the dam body shape.Figure 4
Abutment forceF
x in Case 1.With regard to the distribution of forceF
y, force gradually increases as elevation increases; the maximum is reached at the middle-down portion of the arch dam (about 60 m high, Figure 5). The left and right banks F
y are almost similar because of the symmetry of the arch dam. The distribution of F
x and F
y has a similar feature to that from [24, 25].Figure 5
Abutment forceF
y in Case 1.The relation between bending moment and increasing elevation is similar to that inF
x. The only difference is that the maximum bending moment is at approximately 100 m of the dam height (Figure 6). The abutment bending moment of the left and right banks is also symmetrical. The results of abutment forces show that the arch dam transmits loads to rocks on the two sides through an arching action. Most of the loads are transferred to the rocks in the middle and downside of the two banks. The rocks in this location have an important function in the safety of the arch dam. Bending moment is calculated by force multiplied with distance. Compared with distance, the magnitude of force is much larger; thus, the distribution of bending moment mainly reflects the feature of force. Such is the reason why bending moment (Figure 6) appears somewhat similar to abutment force F
x (Figure 4).Figure 6
Abutment bending moment in Case1.


## 4.2. AbutmentF
x and F
y and Bending Moment of Each Arch Layer during Overload

The above calculation and analysis are results of the arch dam being subjected to normal loads. The following presents the results during overload. Analysis of the arch dam during overload would improve the understanding of the overloading mechanism of the arch dam. The feature of each arch layer during overload is shown (Figures7, 8, and 9). The F
x line of each arch layer during overload is straight (Figure 7). F
x is the force that dam abutment applies on the left and right bank rocks. Each arch layer transfers the excess loads to the rocks in the valley. This condition can explain why arch dams usually have a comparatively high tolerance for overloading. The arch plays a key role in overload transfer. In Figures 7, 8, and 9, the x axis is the overload factor. The overload process ensues by gradually adjusting the density of water and keeping the water level and other loads unchanged. The gradient of the lines signifies the transfer coefficient of each arch layer and shows the capacity of an arch layer to transmit overloads to bank rocks. As shown in Figures 7 to 9, the line of each arch layer is straight; thus, the transfer coefficient is a constant. Arch layer 4 has the largest coefficient followed by arch layers 3, 5, 2, 6, 7, 1, and 8. The arch layer of the middle and lower portions of the dam has a large transfer capacity because the middle and lower arch layers have a large central angle and suitable arc length. The rule for F
y force during overload is similar to that for F
x (Figure 8). For F
y, arch layer 3 has the largest coefficient followed by arch layers 2, 4, 5, 1, 6, 7, and 8. For bending moment, the size of the transfer coefficient (from large to small) is 4, 5, 3, 6, 2, 7, 1, and 8.Figure 7
Arch abutmentF
x during overload.Figure 8
Arch abutmentF
y during overload.Figure 9
Arch abutment moment during overload.The arch layers located in top and bottom portions of arch dam usually have a smaller transfer coefficient, and the arc lengths of these arch layers are either too long or too short. The axes of these arch layers are approximately straight lines, which goes against the role that an arch layer is playing. Arch layers located in the middle and lower portions of arch dam have relatively suitable arc length and central angle, which make them have a larger transfer coefficient. The central angle of arch layer is mainly affecting the magnitude ofF
x and F
y. Engineers usually hope that arch dam can transfer loads to two banks as much as possible and a larger transfer coefficient of arch layer is always preferred. Therefore, much attention should be specially paid to the determination of arc length and central angel of each arch layer during its designing process.


## 4.3. Distribution of AbutmentF
x and F
y and Bending Moment in Different Cases

The following presents the results on the distribution of abutment forces in different cases. Abutment forceF
x and bending moment have an obvious symmetry (Figures 10 and 12); the F
y of the left bank is almost similar to that of the right bank. The distribution of F
x, F
y, and bending moment does not change in different cases. The different abutment forces increase linearly under conditions of overload. Once the distribution of abutment forces in the normal case is determined, the distribution in different overload cases can also be determined. The unchanged distribution provides a good suggestion to arch dam designs. Engineers can modify the arch dam shape according to the distribution of abutment forces. The arch should not transmit a large force to weak rocks in a valley. The distribution also helps evaluate the abutment stability of arch dams. Abutment forces differ at different elevations. For an arch layer with larger abutment forces, more abutment forces are required during overload; hence, the rock mass at this location should be stable (Figures 10 and 11).Figure 10
F
x distribution in each case.Figure 11
F
y distribution in each case.Figure 12
Bending moment distribution in each case.


## 4.4. Simple Method to Determine the Distribution of Abutment Forces

Unchanged distribution is helpful to arch dam designers. The use of finite element method in the calculation is inconvenient and complex. For convenience, a simple method is necessary. Vertical force contributes little to the distribution of abutment forces becauseF
x and F
y are horizontal forces. Temperature load is much smaller than water load. Thus, the main factor that affects abutment force distribution is water load. Arch layers differ in terms of the length of the arch, magnitude of water load, height of the arch, and central angle. The first three factors affect the magnitude of force on the upstream face, and the last one affects the value of F
x and F
y. A simple formula to calculate the results is provided below:(13)
F
x
=
p
1
+
p
2
2
l
1
+
l
2
2
h
sin
⁡
α
,
F
y
=
p
1
+
p
2
2
l
1
+
l
2
2
h
cos
⁡
⁡
α
,where p
1 and p
2 are the water pressure of the upstream face, l
1 and l
2 are arch layer length, h is arch layer height, and α is the central angle of an arch layer. All parameters are shown in Figure 13.Figure 13
Arch layer.F
x and F
y distributions were obtained with the simple method (Table 8). The results are shown in Figures 14 and 15. The distribution is similar to that obtained with finite element method.Table 8
Abutment forces by simple method.
Arch layer height (m)
A (m2)
P (m)
Central angle (°)
Results of the simple method
Left bank
Right bank
Left bank
Right bank
F
x (N)
F
y (N)
F
x (N)
F
y (N)
20
2079.656
150
35.3456
34.6478
1768550126
2493602506
1737279556
2515489103
40
3439.683
130
39.9615
41.6339
2814539436
3358818546
2910159962
3276318277
60
4403.907
110
42.6571
43.8137
3216889748
3491351666
3285379568
3426981088
80
5171.987
90
44.4953
44.4745
3197065764
3253891718
3194603029
3256309300
100
5865.691
70
44.9981
45.0003
2845206930
2845395713
2844183315
2846419324
120
6506.773
50
45.0002
44.7193
2254489490
2254473811
2242513203
2266387238
140
7123.782
30
44.9966
43.6255
1480870720
1481046523
1444423162
1516614386
160
7873.98
10
44.867
42.6669
544153765
547120155
522759993.2
567596488.5Figure 14
F
x calculated with the simple method.Figure 15
F
y calculated with the simple method.The above mentioned method is an approximate means to calculateF
x and F
y; the absolute magnitudes of F
x and F
y are different from that obtained from finite element method. The simple method mainly aims to determine the distribution characteristic. To compare the accuracy of results obtained from the simple method and those from finite element method, each result of the two methods is utilized to divide the maximum value obtained through the two methods. The results are shown in Figures 16 and 17. The results of the simple method are consistent with those of finite element method.Figure 16
Comparison ofF
x obtained with the simple method and FEM.Figure 17
Comparison ofF
y obtained with the simple method and FEM.


## 5. Discussions

The distribution of abutment forces is determined by the special body shape of an arch dam. The deeper the dam is, the greater the pressure is. The distribution of water pressure forms a triangle. The force acting on an upstream layer is equal to the product of water pressure and the acting area. The upstream area of different arch layers differs. Usually, the arch layer area changes from small to large from the bottom to the top of the arch dam. Conversely, water pressure changes from large to small from the bottom to the top. Thus, the product of water pressure and area increases from the bottom to the middle section of the arch dam and decreases from the middle to the top section of the dam.Each arch layer has different left and right central angles. This condition results in the different magnitudes ofF
x and F
y. The linear increase in arch abutment force during overload indicates that the arch transmits excess loads to the rocks of the two banks. To guarantee the safety of an arch dam, the arching action of each arch layer should be ensured and sufficient force storage of rocks should be provided at different elevations. The former depends on the ideal design of arch layers. The latter requires the selection of a proper arch dam site that has a large quantity of strong rock mass on both banks. At the least, arch layers with large arch abutment forces should not be placed in an area where the two banks of a valley cannot offer a sufficiently large quantity of strong rock mass. Thus, distribution can provide valuable information to arch dam design. It also reminds us that the safety evaluation of arch dams may be conducted from two aspects. First, estimate whether the two banks of the valley offer sufficient rock mass in the evaluation of each arch layer according to the distribution of abutment forces because different abutment forces are required at different elevations. Second, ascertain the design of an arch layer to ensure that the arch layer is sufficiently strong because various arch layers bear different load magnitudes. The distribution of abutment forces indicates that arch layers with large abutment forces are usually located in the middle and lower sections of the arch dam. These parts of the dam and rocks bear a large force. Hence, the dam body or the rock mass of the two banks should be sufficiently strong.


## 6. Conclusions

(1) Abutment forceF
x increases from the bottom to the middle section of the arch dam. It reaches the maximum at the middle of the dam and then decreases from the middle to the top of the dam. F
y and bending moment have a similar distribution feature under normal loads. The only difference is that the maximum of F
x, F
y, and bending moment is at different heights of the dam. The F
x and bending moment of the left and right banks exhibit good symmetry. The distribution of F
x, F
y, and bending moment does not change during overload.(2) With regard to the distribution of abutment force along the elevation, attention should be paid to the design process. On the one hand, this condition would ensure that both sides of the bank can provide sufficient force to meet the force requirements at different elevations. Given that each abutment along the elevation requires a different magnitude of force, bank rocks must offer sufficient force. On the other hand, the arch layer should be designed in accordance with the load because arch layers in different elevations bear different magnitudes of load.(3) The left and right bank rocks must have sufficient force storage to ensure the safety of the arch dam during overload. Different arch layers require different magnitudes of rock force. Abutment force increases linearly during overload. The arch layer with the maximum abutment forces will require a large rock force during overload.(4) A simple method to ascertain the distribution characteristic is provided through formula (13). This simple method can only get a precise distribution characteristic of F
x and F
y, and the absolute magnitude of F
x and F
y got by simple method is different from that by finite element method. This simple method can help to choose a suitable dam site. By the distribution of abutment forces got with simple method, engineers will know which layer will bear the largest force. They must choose a better site where the rock is stable enough at the elevation of the layer that has the largest abutment forces. Second, it helps to design a stronger arch dam. The layer that has the largest abutment forces usually bears large water pressure. So, it reminds engineers that dam body at this elevation should be built strong enough.(5) The safe operation of an arch dam relies mainly on the arch. The arch transfers loads to the rocks of the bank. Arch layers play a key role in the safe operation of arch dams. If an arch layer is destroyed under normal loads or overloads, the dam would fail. The overload capacity of an arch dam can be ascertained through the estimation of arch layers. An arch dam’s overload capacity is determined by the weakest arch layer. The overload factor of an arch dam can be simply defined as(14)
K
=
min
⁡
⁡
N
i
F
i
,where N
i is the maximum value of the bearing capacity of all arch layers. F
i is the capacity of an arch layer in normal loads, and i is the number of different arch layers. An arch layer’s overload factor can be evaluated from two aspects. One is the overload capacity of an arch layer, and the other is the force storage of the bank rocks. The overload factor is the minimum of the two aspects mentioned above.



---

*Source: 721602-2015-03-09.xml*