research article mechanism research of arch dam ...arch dams. e arch dam is a highly statically...

Research ArticleMechanism Research of Arch Dam AbutmentForces during Overload

Yu Xia,1 Chuangdi Li,1 Xiaolian Zhao,2 and Zhongqing Zhang3

1College of Civil and Architecture Engineering, Guangxi University of Science and Technology, Liuzhou 545006, China2College of Materials Science and Engineering, Guangxi University, Nanning 530004, China3College of Civil and Architecture Engineering, Guangxi University, Nanning 530004, China

Correspondence should be addressed to Yu Xia; [email protected]

Received 5 September 2014; Revised 8 February 2015; Accepted 18 February 2015

Academic Editor: Paolo Lonetti

Copyright © 2015 Yu Xia et al. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper presents research on the abutment forces of a double-curvature arch damduring overload based onnumerical calculationresults obtained through finite element method by Ansys. Results show that, with an increase in elevation, the abutment forces andbending moment of the arch dam increase first and then decrease from the bottom to the top of the dam. Abutment forces andbendingmoment reach theirmaximum at themiddle ormiddle-down portion of the dam.The distributions of abutment forces andmoment 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 researchexplains the operating mechanism of an arch dam under normal and overload conditions. It provides a simple method to calculatethe distribution of forces 𝐹

𝑥and 𝐹

𝑦and 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.

1. Introduction

The development of arch dams has a long history that datesback to 1st century BC [1]. Relative uniformity was achievedin the 20th century after several designs and techniques weredeveloped. The first known arch dam, Glanum Dam, wasbuilt by the Romans in France [2]. Arch dam is a type ofdam curved in the shape of an arch, with the top of the archpointing back into the reservoir. Thus, the force of the wateragainst it, known as hydrostatic pressure, presses against thearch. An arch dam is most suitable for narrow gorges withstable rocks. Considering that arch dams are thinner thanany other dam type, they require much less constructionmaterials, which make them economical and practical inremote areas. Arch dams are built all over the world becausethey are safe and involve minimal cost [3]. China has themost number of arch dams [4]. Many numerical methodsare currently applied to the analysis of the structure of archdams. Some examples of these methods are finite elementmethod [5–7], discrete element method [8, 9], block elementmethod [10, 11], discontinuous deformation analysis [12], fast

Lagrange analysis of continua [13], and interface elementmethod [14]. Self-adapt element [15–17], meshless [18], andextended finite element methods [19–21] are utilized tosimulate the development of cracks in the structure analysis ofarch dams. The arch dam is a highly statically indeterminatestructure. Through an arching action, arch layers transmitupstream water pressure to bank rocks on two sides. Archlayers play an important role in the safe operation of suchdams, which transfers the loads to two banks. It makes largeareas of arch dam body under compression through archlayer.This type of structure can make full use of compressionstrength of concrete. Limit equilibrium method is usuallyemployed in the safety evaluation of arch dams [22, 23].Little attention is paid to abutment forces in arch dam safetyanalysis. The mechanical characteristics of a structure can bedetermined through research on arch abutment forces duringoverload. It shows that thrust angles at different elevationsincrease during overload through abutments force analysis[24, 25]. Studying abutment force would thus provide acomprehensive understanding of the overload mechanism ofarch dams.

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015, Article ID 721602, 12 pageshttp://dx.doi.org/10.1155/2015/721602

2 Mathematical Problems in Engineering

2. Concrete Cracking Simulation of Arch Damduring Overload [26]

2.1. Concrete Cracking Mode. Before the failure of concreteunder tensile condition, a linear relationship exists betweenstress and strain. The stiffness matrix is

[𝐷𝑐𝑘

𝑐]

=𝐸

(1 + ]) (1 − 2])

⋅

[[[[[[[[[[[[[

[

(1 − ]) ] ] 0 0 0] (1 − ]) ] 0 0 0] ] (1 − ]) 0 0 0

0 0 0(1 − 2])

20 0

0 0 0 0(1 − 2])

20

0 0 0 0 0(1 − 2])

2

]]]]]]]]]]]]]

]

,

(1)

where 𝐸 is Young’s modulus for concrete and ] is Poisson’sratio for concrete.

The following conditions apply to cracks in one directiononly.

If principal stress in one direction is greater than thefailure tensile stress, tensile failure occurs. After developingcracks, the concrete becomes an orthotropic material. Giventhat the stiffness and shear stiffness reduce the normal plane,the stress-strain matrix will change. After the destructionof the concrete, the presence of a crack at an integrationpoint is represented throughmodification of the stress-strainrelations by introducing a plane of weakness in a directionnormal to the crack face. A shear transfer coefficient 𝛽

𝑡is

introduced to represent a shear strength reduction factor forsubsequent loads that induce sliding (shear) across the crackface. The stress-strain relationship is built in the directionof the failure surface and the direction perpendicular to it.When principal stress is greater than the tensile breakingstress in only one direction, the stress and strain of the newmatrix are

[𝐷𝑐𝑘

𝑐] =

𝐸

1 + ]

⋅

[[[[[[[[[[[[[[[[[[

[

𝑅𝑡

(1 − ])𝐸

0 0 0 0 0

01

(1 − ])]

(1 − ])0 0 0

0]

(1 − ])1

(1 − ])0 0 0

0 0 0𝛽𝑡

20 0

0 0 0 01

20

0 0 0 0 0𝛽𝑡

2

]]]]]]]]]]]]]]]]]]

]

,

(2)

where the superscript 𝑐𝑘 signifies that the stress-strain rela-tions refer to a coordinate system parallel to principal stress

𝜎

ft

Tcft

Rt

6𝜀𝜀0

11E

𝜀ckck

Figure 1: Strength of cracked condition.

directions with the 𝑥𝑐𝑘axis perpendicular to the crack face.

𝑅𝑡works with adaptive descent and diminishes to 0.0 as the

solution converges.In Figure 1, 𝑓

𝑡is uniaxial tensile cracking stress and 𝑇

𝑐is

the multiplier for the amount of tensile stress relaxation.If the crack closes, then all compressive stresses normal to

the crack plane are transmitted across the crack. Only sheartransfer coefficient 𝛽

𝑐for a closed crack is introduced. [𝐷𝑐𝑘

𝑐]

can then be expressed as

[𝐷𝑐𝑘

𝑐]

=𝐸

(1 + ]) (1 − 2])

⋅

[[[[[[[[[[[[[

[

(1 − ]) ] ] 0 0 0] (1 − ]) ] 0 0 0] ] (1 − ]) 0 0 0

0 0 0 𝛽𝑐

(1 − 2])2

0 0

0 0 0 0(1 − 2])

20

0 0 0 0 0 𝛽𝑐

(1 − 2])2

]]]]]]]]]]]]]

]

.

(3)

The stress-strain relations for concrete that has cracked in twodirections are

[𝐷𝑐𝑘

𝑐] = 𝐸

[[[[[[[[[[[[[[[[[[

[

𝑅𝑡

𝐸0 0 0 0 0

0𝑅𝑡

𝐸0 0 0 0

0 0 1 0 0 0

0 0 0𝛽𝑡

2 (1 + ])0 0

0 0 0 0𝛽𝑡

2 (1 + ])0

0 0 0 0 0𝛽𝑡

2 (1 + ])

]]]]]]]]]]]]]]]]]]

]

.

(4)

If both directions reclose

Mathematical Problems in Engineering 3

[𝐷𝑐𝑘

𝑐] =

𝐸

(1 + ]) (1 − 2])

[[[[[[[[[[[[[[

[

(1 − ]) ] ] 0 0 0

] (1 − ]) ] 0 0 0

] ] (1 − ]) 0 0 0

0 0 0 𝛽𝑐

(1 − 2])2

0 0

0 0 0 0 𝛽𝑐

(1 − 2])2

0

0 0 0 0 0 𝛽𝑐

(1 − 2])2

]]]]]]]]]]]]]]

]

. (5)

The stress-strain relations for concrete that has cracked in allthree directions are

[𝐷𝑐𝑘

𝑐] = 𝐸

[[[[[[[[[[[[[[[[[[[

[

𝑅𝑡

𝐸0 0 0 0 0

0𝑅𝑡

𝐸0 0 0 0

0 0𝑅𝑡

𝐸0 0 0

0 0 0𝛽𝑡

2 (1 + ])0 0

0 0 0 0𝛽𝑡

2 (1 + ])0

0 0 0 0 0𝛽𝑡

2 (1 + ])

]]]]]]]]]]]]]]]]]]]

]

.

(6)

If the crack is closed again in three directions, formula (5) willbe selected. 𝛽

𝑐and 𝛽

𝑡values have the following relationship:

1 > 𝛽𝑐> 𝛽𝑡> 0. (7)

2.2. Distinguishing between Crack Opening and Closing. Theopen or closed status of integration point cracking is based onstrain value 𝜀𝑐𝑘

𝑐𝑘, which is called the crack strain. For the case

of a possible crack in the 𝑥 direction, this strain is evaluatedas

𝜀𝑐𝑘

𝑐𝑘=

{{{{{{{

{{{{{{{

{

𝜀𝑐𝑘

𝑥+

]1 − ]

(𝜀𝑐𝑘

𝑦+ 𝜀𝑐𝑘

𝑧)

if no cracking has occurred

𝜀𝑐𝑘

𝑥+ ]𝜀𝑐𝑘𝑧

if 𝑦 direction has cracked

𝜀𝑐𝑘

𝑥if 𝑦 and 𝑧 direction have cracked,

(8)

where 𝜀𝑐𝑘𝑥, 𝜀𝑐𝑘𝑦, 𝜀𝑐𝑘𝑧

are three normal component strains incrack orientation.

Vector {𝜀𝑐𝑘} is computed by

{𝜀𝑐𝑘

} = [𝑇𝑐𝑘

] {𝜀

} , (9)

where {𝜀} is modified total strain (in element coordinates).{𝜀

} is defined as

{𝜀

} = {𝜀el𝑛−1

} + {Δ𝜀𝑛} − {Δ𝜀

th𝑛} − {Δ𝜀

pl𝑛} , (10)

where 𝑛 is the substep number, {𝜀el𝑛−1

} is the elastic strainfrom the previous substep, {Δ𝜀

𝑛} is the total strain increment

(based on {Δ𝑢𝑛}, the displacement increment over the sub-

step), {Δ𝜀th𝑛} is the thermal strain increment, and {Δ𝜀pl

𝑛} is the

plastic strain increment.If 𝜀𝑐𝑘𝑐𝑘is less than zero, the associated crack is assumed to

be closed.If 𝜀𝑐𝑘𝑐𝑘is greater than or equal to zero, the associated crack

is assumed to be open. When cracking first occurs at anintegration point, the crack is assumed to be open for the nextiteration.

3. Numerical Model and Parameters

Many constitutive models of concrete exist at present. In3D finite element numerical calculations, the most preferredmodel is D-P criterion mainly because model parameterscan be easily obtained with this model. D-P criterion hasa smooth yield surface and is easy to use. This model canbe easily utilized in elastic-plastic analyses for concrete andother materials with similar properties. The Drucker-Prageryield function is

𝐹 = 𝛼𝐼1+ √𝐽2− 𝑘 = 0, (11)

where 𝛼 and 𝑘 are material parameters, 𝐼1is the first invariant

of the stress tensor, and 𝐽2is the partial stress tensor for the

second invariant. They can be calculated as

𝐼1= 𝜎𝑥+ 𝜎𝑦+ 𝜎𝑧,

𝐽2=

1

6[(𝜎𝑥− 𝜎𝑦)2

+ (𝜎𝑦− 𝜎𝑧)2

+ (𝜎𝑧− 𝜎𝑥)2

]

+ 𝜏2

𝑥𝑦+ 𝜏2

𝑦𝑧+ 𝜏2

𝑧𝑥,

𝛼 =2 sin𝜙

√3 (3 − sin𝜙),

𝑘 =6𝑐 cos𝜙

√3 (3 − sin𝜙),

(12)

where 𝜎2𝑥, 𝜎2𝑦, 𝜎2𝑧, 𝜏2𝑥𝑦, 𝜏2𝑦𝑧, and 𝜏2

𝑥𝑧𝑥are stress components; and

𝜙 = 𝑡𝑔−1

𝑓.


Table 1: Material parameters.

Location Elastic modulus(Gpa)Bulk density(KN/m3)

Poisson’sratio

Cohesion(Mpa)

Friction angle(∘)

Coefficient of thermalexpansion (10−6)

Dam body 18 2350 0.167 2 56 4.87Abutment 14 2730 0.25 0.9 45Base 11 2700 0.27 0.7 42Unconformity 3 2660 0.3 0.35 35

3.1. Finite Element Model. The coordination of the numericalmodel is shown in Figure 2. The 𝑥-axis positive direction ofthe model is along the right banks. The positive directionof 𝑦-axis is downstream along the river, and the positivedirection of 𝑧-axis is along the dam’s increasing height. Therange of the numerical model is as follows: 230m width onthe two sides, 270m depth under the riverbed, 150m lengthupstream, and 360m length downstream. The bottom of thenumerical model is subjected to a three-direction constraint,the boundary of the left and right banks is subjected to an𝑥 direction constraint, and the boundary of the upstreamand downstream is subjected to a 𝑦 direction constraint. Themodel has 28,296 elements and 33,830 nodes in total. It willtakemuchmore time in numerical calculationwith the elasticdamage model for computer. So this model is used only inelements of dam body, the elastic-plastic model is used inother parts elements.

3.2. Parameters and Loads. The loads of each case includewater pressure, temperature load, and dam gravity. Theoverload process was implemented by gradually adjustingthe density of water and keeping the water level and otherloads unchanged. Overload factor 𝐾 is defined as 𝐾 = 𝑃/𝑃

0,

where 𝑃 is the current water density of the dam and 𝑃0is the

normal water density. In Case 1, the value of water densityis 9.8 kg/m3. In Case 2, the magnitude of water density istwice the normal value (19.6 kg/m3) and so on. The relevantparameters are listed in Table 1.

4. Calculation Results

Theabutment forces were obtained through the integration ofabutment element nodes. The directions of abutment forcesare similar to those in the coordination of the numericalmodel. The direction of the bending moment complies withthe right-hand rule. The following is an analysis of abutment𝐹𝑥, 𝐹𝑦, and bending moment. Abutment forces 𝐹

𝑥, 𝐹𝑦and

bending moment are discussed under normal loads andoverloads separately. To express the results clearly, analysis isfirst performed on the distribution characteristic of abutmentforces 𝐹

𝑥, 𝐹𝑦and bending moment under normal loads.

Then, the abutment forces changing rule during overload arediscussed. A simple method is then provided to determinethe distribution of abutment forces 𝐹

𝑥and 𝐹

𝑦. Arch layers 1

to 8 are arch layers from the bottom to the top of the dam.Thecalculated cases are shown below. The finial failure pattern isshown in Figure 3.

x

yz

Figure 2: Finite element model.

Case 1. It is as follows: 1 time water load + gravity +temperature load (Table 2).

Case 2. It is as follows: 2 times water load + gravity +temperature load (Table 3).





4.1. Distribution of Abutment Forces under Normal Loads.The abutment force of 𝐹

𝑥(Figure 4) increases gradually

from the bottom to the middle of the dam as the elevationincreases. It reaches the maximum at the middle and thendecreases from the middle to the top of the arch dam. Themaximum of 𝐹

𝑥is at approximately 80m of the height of the

dam. The left and right banks 𝐹𝑥are symmetrical because of

the symmetry of the dam body shape.With regard to the distribution of force𝐹

𝑦, force gradually

increases as elevation increases; the maximum is reachedat the middle-down portion of the arch dam (about 60mhigh, Figure 5).The left and right banks 𝐹

𝑦are almost similar

because of the symmetry of the arch dam.The distribution of𝐹𝑥and 𝐹

𝑦has a similar feature to that from [24, 25].


Upstream Downstream

Figure 3: Typical damage pattern of dam body (Case 6: the black zone is the damage area of dam body.).

Table 2: Results of Case 1.

Location Left bank Right bank𝐹𝑥(N) 𝐹

𝑦(N) Mz (Nm) 𝐹

𝑥(N) 𝐹

𝑦(N) Mz (NM)

Arch layer 1 64505100 167958000 −7709350000 −67725400 166091000 7369330000Arch layer 2 141117000 240493000 −18486000000 −149254000 235289000 18784000000Arch layer 3 175053000 248301000 −25307600000 −182556000 242942000 26715700000Arch layer 4 181693000 225146000 −28966100000 −183487000 223801000 29565200000Arch layer 5 165873000 188499000 −29136600000 −164688000 189537000 29105300000Arch layer 6 130615000 142144000 −25264700000 −129284000 143358000 25159300000Arch layer 7 81000100 86200100 −17439100000 −79221500 87835000 16947200000Arch layer 8 20970900 22194700 −5013750000 −20239300 22862400 4735740000



𝑦(N) Mz (NM) 𝐹

𝑥(N) 𝐹

𝑦(N) Mz (NM)





𝑥(N) 𝐹

𝑦(N) Mz (NM)


The relation between bending moment and increasingelevation is similar to that in 𝐹

𝑥. The only difference is that

the maximum bending moment is at approximately 100m ofthe dam height (Figure 6). The abutment bending momentof the left and right banks is also symmetrical. The results of

abutment forces show that the arch dam transmits loads torocks on the two sides through an arching action. Most of theloads are transferred to the rocks in themiddle and downsideof the two banks.The rocks in this location have an importantfunction in the safety of the arch dam. Bending moment is





𝑥(N) 𝐹

𝑦(N) Mz (NM)





𝑥(N) 𝐹

𝑦(N) Mz (NM)





𝑥(N) 𝐹

𝑦(N) Mz (NM)


0 0.5 1 1.5 20

20406080

100120140160180

Dam

hei

ght (

m)

Left bankRight bank

−2 −1.5 −1 −0.5×108Abutment force Fx (N)

Figure 4: Abutment force 𝐹𝑥in Case 1.

0 0.5 1 1.5 2 2.50

20406080

100120140160180

Dam

hei

ght (

m)

Left bankRight bank

×108Abutment force Fy (N)

Figure 5: Abutment force 𝐹𝑦in Case 1.


0 1 2 30

20406080

100120140160180

Dam

hei

ght

−2−3 −1

×1010Abutment moment (N·m)

Left bankRight bank

Figure 6: Abutment bending moment in Case 1.

calculated by force multiplied with distance. Compared withdistance, the magnitude of force is much larger; thus, thedistribution of bending moment mainly reflects the featureof force. Such is the reason why bending moment (Figure 6)appears somewhat similar to abutment force 𝐹

𝑥(Figure 4).

4.2. Abutment 𝐹𝑥and 𝐹

𝑦and Bending Moment of Each Arch

Layer during Overload. The above calculation and analysisare results of the arch dam being subjected to normalloads. The following presents the results during overload.Analysis of the arch dam during overload would improvethe understanding of the overloading mechanism of thearch dam. The feature of each arch layer during overload isshown (Figures 7, 8, and 9). The 𝐹

𝑥line of each arch layer

during overload is straight (Figure 7). 𝐹𝑥is the force that

dam abutment applies on the left and right bank rocks. Eacharch layer transfers the excess loads to the rocks in the valley.This condition can explain why arch dams usually have acomparatively high tolerance for overloading. The arch playsa key role in overload transfer. In Figures 7, 8, and 9, the 𝑥axis is the overload factor. The overload process ensues bygradually adjusting the density of water and keeping thewaterlevel and other loads unchanged. The gradient of the linessignifies the transfer coefficient of each arch layer and showsthe capacity of an arch layer to transmit overloads to bankrocks. As shown in Figures 7 to 9, the line of each arch layer isstraight; thus, the transfer coefficient is a constant. Arch layer4 has the largest coefficient followed by arch layers 3, 5, 2, 6,7, 1, and 8. The arch layer of the middle and lower portionsof the dam has a large transfer capacity because the middleand lower arch layers have a large central angle and suitablearc length. The rule for 𝐹

𝑦force during overload is similar

to that for 𝐹𝑥(Figure 8). For 𝐹

𝑦, arch layer 3 has the largest

coefficient followed by arch layers 2, 4, 5, 1, 6, 7, and 8. Forbending moment, the size of the transfer coefficient (fromlarge to small) is 4, 5, 3, 6, 2, 7, 1, and 8.

The arch layers located in top and bottomportions of archdam usually have a smaller transfer coefficient, and the arclengths of these arch layers are either too long or too short.The axes of these arch layers are approximately straight lines,

0 1 2 3 4 5 6 7 80

2

4

6

8

10

12

Arch ring 1Arch ring 2Arch ring 3Arch ring 4


×108

Abut

men

t for

ceFx

Overload factor, K

Figure 7: Arch abutment 𝐹𝑥during overload.

0 1 2 3 4 5 6 7 80

2

4

6

8

10

12

14

16



×108

Abut

men

t for

ceFy

Overload factor, K

Figure 8: Arch abutment 𝐹𝑦during overload.

which goes against the role that an arch layer is playing. Archlayers located in the middle and lower portions of arch damhave relatively suitable arc length and central angle, whichmake themhave a larger transfer coefficient.The central angleof arch layer is mainly affecting the magnitude of 𝐹

𝑥and 𝐹

𝑦.

Engineers usually hope that arch dam can transfer loads totwo banks asmuch as possible and a larger transfer coefficientof arch layer is always preferred. Therefore, much attentionshould be specially paid to the determination of arc lengthand central angel of each arch layer during its designingprocess.

4.3. Distribution of Abutment 𝐹𝑥and 𝐹𝑦and BendingMoment

in Different Cases. The following presents the results on thedistribution of abutment forces in different cases. Abutmentforce 𝐹

𝑥and bending moment have an obvious symmetry

(Figures 10 and 12); the 𝐹𝑦of the left bank is almost similar




0 1 2 3 4 5 6 7 80

0.5

1

1.5

2

Abut

men

t mom

ent

×1011

Overload factor, K

Figure 9: Arch abutment moment during overload.

−1.5 −1 −0.5 0 0.5 1 1.5 20

20406080

100120140160180

Dam

hei

ght (

m)

Case 1 right bankCase 1 left bankCase 2 right bankCase 2 left bankCase 3 right bankCase 3 left bank


×109Abutment force Fx (N)

Figure 10: 𝐹𝑥distribution in each case.

to that of the right bank. The distribution of 𝐹𝑥, 𝐹𝑦, and

bending moment does not change in different cases. Thedifferent abutment forces increase linearly under conditionsof overload. Once the distribution of abutment forces inthe normal case is determined, the distribution in differentoverload cases can also be determined. The unchangeddistribution provides a good suggestion to arch dam designs.Engineers can modify the arch dam shape according to thedistribution of abutment forces.The arch should not transmita large force to weak rocks in a valley. The distribution alsohelps evaluate the abutment stability of arch dams. Abutmentforces differ at different elevations. For an arch layer withlarger abutment forces, more abutment forces are requiredduring overload; hence, the rock mass at this location shouldbe stable (Figures 10 and 11).



0 0.5 1 1.5 20

20406080

100120140160180

Dam

hei

ght (

m)

×109Abutment force Fy (N)

Figure 11: 𝐹𝑦distribution in each case.

−1−2

Abutment moment (N·m)



0 1 2 30

20406080

100120140160180

Dam

hei

ght (

m)

×1011

Figure 12: Bending moment distribution in each case.

4.4. Simple Method to Determine the Distribution of Abut-ment Forces. Unchanged distribution is helpful to arch damdesigners.The use of finite elementmethod in the calculationis inconvenient and complex. For convenience, a simplemethod is necessary. Vertical force contributes little to thedistribution of abutment forces because 𝐹

𝑥and 𝐹

𝑦are

horizontal forces. Temperature load is much smaller thanwater load. Thus, the main factor that affects abutment forcedistribution is water load. Arch layers differ in terms of thelength of the arch,magnitude ofwater load, height of the arch,and central angle. The first three factors affect the magnitudeof force on the upstream face, and the last one affects the value


Table 8: Abutment forces by simple method.

Arch layerheight (m) 𝐴 (m

2) 𝑃 (m)Central angle (∘) Results of the simple method

Left bank Right bank Left bank Right bank𝐹𝑥(N) 𝐹

𝑦(N) 𝐹

𝑥(N) 𝐹

𝑦(N)

20 2079.656 150 35.3456 34.6478 1768550126 2493602506 1737279556 251548910340 3439.683 130 39.9615 41.6339 2814539436 3358818546 2910159962 327631827760 4403.907 110 42.6571 43.8137 3216889748 3491351666 3285379568 342698108880 5171.987 90 44.4953 44.4745 3197065764 3253891718 3194603029 3256309300100 5865.691 70 44.9981 45.0003 2845206930 2845395713 2844183315 2846419324120 6506.773 50 45.0002 44.7193 2254489490 2254473811 2242513203 2266387238140 7123.782 30 44.9966 43.6255 1480870720 1481046523 1444423162 1516614386160 7873.98 10 44.867 42.6669 544153765 547120155 522759993.2 567596488.5

of 𝐹𝑥and 𝐹

𝑦. A simple formula to calculate the results is

provided below:

𝐹𝑥= (

𝑝1+ 𝑝2

2)

(𝑙1+ 𝑙2)

2ℎ sin𝛼,

𝐹𝑦= (

𝑝1+ 𝑝2

2)

(𝑙1+ 𝑙2)

2ℎ cos𝛼,

(13)

where 𝑝1and 𝑝

2are the water pressure of the upstream face,

𝑙1and 𝑙2are arch layer length, ℎ is arch layer height, and 𝛼 is

the central angle of an arch layer. All parameters are shownin Figure 13.

𝐹𝑥and 𝐹

𝑦distributions were obtained with the simple

method (Table 8). The results are shown in Figures 14 and 15.The distribution is similar to that obtainedwith finite elementmethod.

The above mentioned method is an approximate meansto calculate 𝐹

𝑥and 𝐹

𝑦; the absolute magnitudes of 𝐹

𝑥and 𝐹

𝑦

are different from that obtained from finite element method.The simplemethodmainly aims to determine the distributioncharacteristic. To compare the accuracy of results obtainedfrom the simple method and those from finite elementmethod, each result of the two methods is utilized to dividethe maximum value obtained through the two methods. Theresults are shown in Figures 16 and 17. The results of thesimple method are consistent with those of finite elementmethod.

5. Discussions

The distribution of abutment forces is determined by thespecial body shape of an arch dam. The deeper the dam is,the greater the pressure is. The distribution of water pressureforms a triangle. The force acting on an upstream layer isequal to the product of water pressure and the acting area.The upstream area of different arch layers differs. Usually, thearch layer area changes from small to large from the bottom tothe top of the arch dam. Conversely, water pressure changesfrom large to small from the bottom to the top. Thus, theproduct of water pressure and area increases from the bottomto the middle section of the arch dam and decreases from themiddle to the top section of the dam.

l1

l2

p2

p1

𝛼 𝛼

x

y

zh

Figure 13: Arch layer.

−1−2−3×109

0 1 2 30

20406080

100120140160180

Dam

hei

ght (

m)

Left bankRight bank

Abutment force Fx (N)

Figure 14: 𝐹𝑥calculated with the simple method.

Each arch layer has different left and right central angles.This condition results in the different magnitudes of 𝐹

𝑥

and 𝐹𝑦. The linear increase in arch abutment force during

overload indicates that the arch transmits excess loads to therocks of the two banks. To guarantee the safety of an archdam, the arching action of each arch layer should be ensuredand sufficient force storage of rocks should be provided atdifferent elevations. The former depends on the ideal designof arch layers. The latter requires the selection of a proper


×109

Left bankRight bank

0 0.5 1 1.5 2 2.5 3 3.50

20406080

100120140160180

Dam

hei

ght (

m)

Abutment force Fy (N)

Figure 15: 𝐹𝑦calculated with the simple method.

−1 −0.5 0 0.5 1 1.50

20406080

100120140160180

Dam

hei

ght (

m)

Left bank by simple methodRight bank by simple method

Left bank by FEMRight bank by FEM

Fx/max(Fx)

Figure 16: Comparison of 𝐹𝑥obtained with the simple method and

FEM.

arch dam site that has a large quantity of strong rock masson both banks. At the least, arch layers with large archabutment forces should not be placed in an area where thetwo banks of a valley cannot offer a sufficiently large quantityof strong rock mass. Thus, distribution can provide valuableinformation to arch dam design. It also reminds us that thesafety evaluation of arch dams may be conducted from twoaspects. First, estimate whether the two banks of the valleyoffer sufficient rock mass in the evaluation of each arch layeraccording to the distribution of abutment forces becausedifferent abutment forces are required at different elevations.Second, ascertain the design of an arch layer to ensure thatthe arch layer is sufficiently strong because various arch layersbear different load magnitudes.The distribution of abutmentforces indicates that arch layers with large abutment forcesare usually located in the middle and lower sections of thearch dam. These parts of the dam and rocks bear a largeforce. Hence, the dam body or the rockmass of the two banksshould be sufficiently strong.

0 0.2 0.4 0.6 0.8 10

20406080

100120140160180

Dam

hei

ght (

m)

Right bank by simple methodRight bank by FEM

Fy/max(Fy)

Figure 17: Comparison of 𝐹𝑦obtained with the simple method and

FEM.

6. Conclusions

(1) Abutment force 𝐹𝑥increases from the bottom to the

middle section of the arch dam. It reaches the maximum atthe middle of the dam and then decreases from the middle tothe top of the dam. 𝐹

𝑦and bending moment have a similar

distribution feature under normal loads. The only differenceis that the maximum of 𝐹

𝑥, 𝐹𝑦, and bending moment is at

different heights of the dam. The 𝐹𝑥and bending moment

of the left and right banks exhibit good symmetry. Thedistribution of 𝐹

𝑥, 𝐹𝑦, 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 designprocess. On the one hand, this condition would ensure thatboth sides of the bank can provide sufficient force to meetthe force requirements at different elevations. Given that eachabutment along the elevation requires a different magnitudeof force, bank rocks must offer sufficient force. On the otherhand, the arch layer should be designed in accordance withthe load because arch layers in different elevations beardifferent magnitudes of load.

(3)The left and right bank rocksmust have sufficient forcestorage to ensure the safety of the arch dam during overload.Different arch layers require different magnitudes of rockforce. Abutment force increases linearly during overload.Thearch layer with the maximum abutment forces will require alarge rock force during overload.

(4) A simple method to ascertain the distribution charac-teristic is provided through formula (13).This simple methodcan only get a precise distribution characteristic of 𝐹

𝑥and

𝐹𝑦, and the absolute magnitude of 𝐹

𝑥and 𝐹

𝑦got by simple

method is different from that by finite element method. Thissimple method can help to choose a suitable dam site. Bythe 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 stableenough at the elevation of the layer that has the largestabutment forces. Second, it helps to design a stronger archdam. The layer that has the largest abutment forces usually


bears large water pressure. So, it reminds engineers that dambody at this elevation should be built strong enough.

(5)The safe operation of an arch dam relies mainly on thearch. The arch transfers loads to the rocks of the bank. Archlayers play a key role in the safe operation of arch dams. Ifan arch layer is destroyed under normal loads or overloads,the dam would fail. The overload capacity of an arch damcan be ascertained through the estimation of arch layers. Anarch dam’s overload capacity is determined by the weakestarch layer. The overload factor of an arch dam can be simplydefined as

𝐾 = min{𝑁𝑖

𝐹𝑖

} , (14)

where𝑁𝑖is the maximum value of the bearing capacity of all

arch layers. 𝐹𝑖is the capacity of an arch layer in normal loads,

and 𝑖 is the number of different arch layers. An arch layer’soverload factor can be evaluated from two aspects. One is theoverload capacity of an arch layer, and the other is the forcestorage of the bank rocks.Theoverload factor is theminimumof the two aspects mentioned above.

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper.

Acknowledgments

This study is supported by the National Natural ScienceFoundation of China (51468005, 51469005), Guangxi Nat-ural Science Foundation in China (2013GXNSFBA019247),Science Plan Projects of Guangxi Education Departmentin China (201203YB130), and the Doctor Foundation ofGuangxi University of Science and Technology in China(11Z02).

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