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 Β© 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 layerβs 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 πΉ
π₯and πΉ
π¦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.
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 [5β7], 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 [15β17], meshless [18], andextended finite element methods [19β21] 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
[π·ππ
π]
=πΈ
(1 + ]) (1 β 2])
β
[[[[[[[[[[[[[
[
(1 β ]) ] ] 0 0 0] (1 β ]) ] 0 0 0] ] (1 β ]) 0 0 0
0 0 0(1 β 2])
20 0
0 0 0 0(1 β 2])
20
0 0 0 0 0(1 β 2])
2
]]]]]]]]]]]]]
]
,
(1)
where πΈ is Youngβs modulus for concrete and ] is Poissonβsratio 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 π½
π‘is
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
[π·ππ
π] =
πΈ
1 + ]
β
[[[[[[[[[[[[[[[[[[
[
π π‘
(1 β ])πΈ
0 0 0 0 0
01
(1 β ])]
(1 β ])0 0 0
0]
(1 β ])1
(1 β ])0 0 0
0 0 0π½π‘
20 0
0 0 0 01
20
0 0 0 0 0π½π‘
2
]]]]]]]]]]]]]]]]]]
]
,
(2)
where the superscript ππ signifies that the stress-strain rela-tions refer to a coordinate system parallel to principal stress
π
ft
Tcft
Rt
6ππ0
11E
πckck
Figure 1: Strength of cracked condition.
directions with the π₯ππaxis perpendicular to the crack face.
π π‘works with adaptive descent and diminishes to 0.0 as the
solution converges.In Figure 1, π
π‘is uniaxial tensile cracking stress and π
πis
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 π½
πfor a closed crack is introduced. [π·ππ
π]
can then be expressed as
[π·ππ
π]
=πΈ
(1 + ]) (1 β 2])
β
[[[[[[[[[[[[[
[
(1 β ]) ] ] 0 0 0] (1 β ]) ] 0 0 0] ] (1 β ]) 0 0 0
0 0 0 π½π
(1 β 2])2
0 0
0 0 0 0(1 β 2])
20
0 0 0 0 0 π½π
(1 β 2])2
]]]]]]]]]]]]]
]
.
(3)
The stress-strain relations for concrete that has cracked in twodirections are
[π·ππ
π] = πΈ
[[[[[[[[[[[[[[[[[[
[
π π‘
πΈ0 0 0 0 0
0π π‘
πΈ0 0 0 0
0 0 1 0 0 0
0 0 0π½π‘
2 (1 + ])0 0
0 0 0 0π½π‘
2 (1 + ])0
0 0 0 0 0π½π‘
2 (1 + ])
]]]]]]]]]]]]]]]]]]
]
.
(4)
If both directions reclose
Mathematical Problems in Engineering 3
[π·ππ
π] =
πΈ
(1 + ]) (1 β 2])
[[[[[[[[[[[[[[
[
(1 β ]) ] ] 0 0 0
] (1 β ]) ] 0 0 0
] ] (1 β ]) 0 0 0
0 0 0 π½π
(1 β 2])2
0 0
0 0 0 0 π½π
(1 β 2])2
0
0 0 0 0 0 π½π
(1 β 2])2
]]]]]]]]]]]]]]
]
. (5)
The stress-strain relations for concrete that has cracked in allthree directions are
[π·ππ
π] = πΈ
[[[[[[[[[[[[[[[[[[[
[
π π‘
πΈ0 0 0 0 0
0π π‘
πΈ0 0 0 0
0 0π π‘
πΈ0 0 0
0 0 0π½π‘
2 (1 + ])0 0
0 0 0 0π½π‘
2 (1 + ])0
0 0 0 0 0π½π‘
2 (1 + ])
]]]]]]]]]]]]]]]]]]]
]
.
(6)
If the crack is closed again in three directions, formula (5) willbe selected. π½
πand π½
π‘values have the following relationship:
1 > π½π> π½π‘> 0. (7)
2.2. Distinguishing between Crack Opening and Closing. Theopen or closed status of integration point cracking is based onstrain value πππ
ππ, which is called the crack strain. For the case
of a possible crack in the π₯ direction, this strain is evaluatedas
πππ
ππ=
{{{{{{{
{{{{{{{
{
πππ
π₯+
]1 β ]
(πππ
π¦+ πππ
π§)
if no cracking has occurred
πππ
π₯+ ]ππππ§
if π¦ direction has cracked
πππ
π₯if π¦ and π§ direction have cracked,
(8)
where ππππ₯, ππππ¦, ππππ§
are three normal component strains incrack orientation.
Vector {πππ} is computed by
{πππ
} = [πππ
] {π
} , (9)
where {π} is modified total strain (in element coordinates).{π
} is defined as
{π
} = {πelπβ1
} + {Ξππ} β {Ξπ
thπ} β {Ξπ
plπ} , (10)
where π is the substep number, {πelπβ1
} is the elastic strainfrom the previous substep, {Ξπ
π} is the total strain increment
(based on {Ξπ’π}, the displacement increment over the sub-
step), {Ξπthπ} is the thermal strain increment, and {Ξπpl
π} is the
plastic strain increment.If πππππis less than zero, the associated crack is assumed to
be closed.If πππππis 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
πΉ = πΌπΌ1+ βπ½2β π = 0, (11)
where πΌ and π are material parameters, πΌ1is the first invariant
of the stress tensor, and π½2is the partial stress tensor for the
second invariant. They can be calculated as
πΌ1= ππ₯+ ππ¦+ ππ§,
π½2=
1
6[(ππ₯β ππ¦)2
+ (ππ¦β ππ§)2
+ (ππ§β ππ₯)2
]
+ π2
π₯π¦+ π2
π¦π§+ π2
π§π₯,
πΌ =2 sinπ
β3 (3 β sinπ),
π =6π cosπ
β3 (3 β sinπ),
(12)
where π2π₯, π2π¦, π2π§, π2π₯π¦, π2π¦π§, and π2
π₯π§π₯are stress components; and
π = π‘πβ1
π.
4 Mathematical Problems in Engineering
Table 1: Material parameters.
Location Elastic modulus(Gpa)Bulk density(KN/m3)
Poissonβsratio
Cohesion(Mpa)
Friction angle(β)
Coefficient of thermalexpansion (10β6)
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 π₯-axis positive direction ofthe model is along the right banks. The positive directionof π¦-axis is downstream along the river, and the positivedirection of π§-axis is along the damβs 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 anπ₯ direction constraint, and the boundary of the upstreamand downstream is subjected to a π¦ 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 πΎ is defined as πΎ = π/π
0,
where π is the current water density of the dam and π0is 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 abutmentπΉπ₯, πΉπ¦, and bending moment. Abutment forces πΉ
π₯, πΉπ¦and
bending moment are discussed under normal loads andoverloads separately. To express the results clearly, analysis isfirst performed on the distribution characteristic of abutmentforces πΉ
π₯, πΉπ¦and 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 πΉ
π₯and πΉ
π¦. 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).
Case 3. It is as follows: 3 times water load + gravity +temperature load (Table 4).
Case 4. It is as follows: 4 times water load + gravity +temperature load (Table 5).
Case 5. It is as follows: 5 times water load + gravity +temperature load (Table 6).
Case 6. It is as follows: 6 times water load + gravity +temperature load (Table 7).
4.1. Distribution of Abutment Forces under Normal Loads.The abutment force of πΉ
π₯(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 πΉ
π₯is at approximately 80m of the height of the
dam. The left and right banks πΉπ₯are symmetrical because of
the symmetry of the dam body shape.With regard to the distribution of forceπΉ
π¦, 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 πΉ
π¦are almost similar
because of the symmetry of the arch dam.The distribution ofπΉπ₯and πΉ
π¦has a similar feature to that from [24, 25].
Mathematical Problems in Engineering 5
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 bankπΉπ₯(N) πΉ
π¦(N) Mz (Nm) πΉ
π₯(N) πΉ
π¦(N) Mz (NM)
Arch layer 1 64505100 167958000 β7709350000 β67725400 166091000 7369330000Arch layer 2 141117000 240493000 β18486000000 β149254000 235289000 18784000000Arch layer 3 175053000 248301000 β25307600000 β182556000 242942000 26715700000Arch layer 4 181693000 225146000 β28966100000 β183487000 223801000 29565200000Arch layer 5 165873000 188499000 β29136600000 β164688000 189537000 29105300000Arch layer 6 130615000 142144000 β25264700000 β129284000 143358000 25159300000Arch layer 7 81000100 86200100 β17439100000 β79221500 87835000 16947200000Arch layer 8 20970900 22194700 β5013750000 β20239300 22862400 4735740000
Table 3: Results of Case 2.
Location Left bank Right bankπΉπ₯(N) πΉ
π¦(N) Mz (NM) πΉ
π₯(N) πΉ
π¦(N) Mz (NM)
Arch layer 1 138468000 347820000 β16028900000 β144378000 344538000 15335700000Arch layer 2 283616000 482485000 β37084800000 β299808000 472144000 37679300000Arch layer 3 350106000 496601000 β50615200000 β365112000 485884000 53431400000Arch layer 4 363385000 450292000 β57932300000 β366974000 447602000 59130300000Arch layer 5 331747000 376998000 β58273100000 β329376000 379073000 58210600000Arch layer 6 261230000 284287000 β50529500000 β258569000 286717000 50318600000Arch layer 7 162000000 172400000 β34878100000 β158443000 175670000 33894400000Arch layer 8 41941800 44389400 β10027500000 β40478500 45724800 9471470000
Table 4: Results of Case 3.
Location Left bank Right bankπΉπ₯(N) πΉ
π¦(N) Mz (NM) πΉ
π₯(N) πΉ
π¦(N) Mz (NM)
Arch layer 1 212430000 527681000 β24348500000 β221030000 522985000 23302000000Arch layer 2 426115000 724477000 β55683500000 β450362000 708998000 56574500000Arch layer 3 525158000 744902000 β75922900000 β547669000 728827000 80147000000Arch layer 4 545078000 675438000 β86898400000 β550461000 671402000 88695500000Arch layer 5 497620000 565496000 β87409700000 β494064000 568610000 87315900000Arch layer 6 391846000 426431000 β75794200000 β387853000 430075000 75477800000Arch layer 7 243000000 258600000 β52317200000 β237665000 263505000 50841600000Arch layer 8 62912700 66584100 β15041200000 β60717800 68587300 14207200000
The relation between bending moment and increasingelevation is similar to that in πΉ
π₯. 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
6 Mathematical Problems in Engineering
Table 5: Results of Case 4.
Location Left bank Right bankπΉπ₯(N) πΉ
π¦(N) Mz (NM) πΉ
π₯(N) πΉ
π¦(N) Mz (NM)
Arch layer 1 286393000 707543000 β32668100000 β297683000 701432000 31268300000Arch layer 2 568613000 966468000 β74282300000 β600917000 945852000 75469800000Arch layer 3 700211000 993203000 β101230000000 β730225000 971769000 106863000000Arch layer 4 726770000 900584000 β115865000000 β733947000 895203000 118261000000Arch layer 5 663494000 753995000 β116546000000 β658752000 758147000 116421000000Arch layer 6 522461000 568574000 β101059000000 β517138000 573434000 100637000000Arch layer 7 324000000 344801000 β69756300000 β316886000 351340000 67788800000Arch layer 8 83883700 88778800 β20055000000 β80957100 91449700 18942900000
Table 6: Results of Case 5.
Location Left bank Right bankπΉπ₯(N) πΉ
π¦(N) Mz (NM) πΉ
π₯(N) πΉ
π¦(N) Mz (NM)
Arch layer 1 360356000 887405000 β40987600000 β374335000 879878000 39234600000Arch layer 2 711112000 1208460000 β92881100000 β751471000 1182710000 94365000000Arch layer 3 875264000 1241500000 β126538000000 β912781000 1214710000 133578000000Arch layer 4 908463000 1125730000 β144831000000 β917434000 1119000000 147826000000Arch layer 5 829367000 942494000 β145683000000 β823440000 947683000 145527000000Arch layer 6 653076000 710718000 β126324000000 β646422000 716792000 125796000000Arch layer 7 405000000 431001000 β87195300000 β396108000 439175000 84736000000Arch layer 8 104855000 110973000 β25068700000 β101196000 114312000 23678700000
Table 7: Results of Case 6.
Location Left bank Right bankπΉπ₯(N) πΉ
π¦(N) Mz (NM) πΉ
π₯(N) πΉ
π¦(N) Mz (NM)
Arch layer 1 434318000 1067270000 β49307200000 β450988000 1058330000 47200900000Arch layer 2 853611000 1450450000 β111480000000 β902026000 1419560000 113260000000Arch layer 3 1050320000 1489800000 β151846000000 β1095340000 1457650000 160294000000Arch layer 4 1090160000 1350880000 β173797000000 β1100920000 1342800000 177391000000Arch layer 5 995240000 1130990000 β174819000000 β988128000 1137220000 174632000000Arch layer 6 783691000 852862000 β151588000000 β775706000 860150000 150956000000Arch layer 7 486000000 517201000 β104634000000 β475329000 527010000 101683000000Arch layer 8 125825000 133168000 β30082500000 β121436000 137175000 28414400000
0 0.5 1 1.5 20
20406080
100120140160180
Dam
hei
ght (
m)
Left bankRight bank
β2 β1.5 β1 β0.5Γ108Abutment force Fx (N)
Figure 4: Abutment force πΉπ₯in Case 1.
0 0.5 1 1.5 2 2.50
20406080
100120140160180
Dam
hei
ght (
m)
Left bankRight bank
Γ108Abutment force Fy (N)
Figure 5: Abutment force πΉπ¦in Case 1.
Mathematical Problems in Engineering 7
0 1 2 30
20406080
100120140160180
Dam
hei
ght
β2β3 β1
Γ1010Abutment moment (NΒ·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 πΉ
π₯(Figure 4).
4.2. Abutment πΉπ₯and πΉ
π¦and 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 πΉ
π₯line of each arch layer
during overload is straight (Figure 7). πΉπ₯is 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 π₯axis 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 πΉ
π¦force during overload is similar
to that for πΉπ₯(Figure 8). For πΉ
π¦, 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
Arch ring 5Arch ring 6Arch ring 7Arch ring 8
Γ108
Abut
men
t for
ceFx
Overload factor, K
Figure 7: Arch abutment πΉπ₯during overload.
0 1 2 3 4 5 6 7 80
2
4
6
8
10
12
14
16
Arch ring 1Arch ring 2Arch ring 3Arch ring 4
Arch ring 5Arch ring 6Arch ring 7Arch ring 8
Γ108
Abut
men
t for
ceFy
Overload factor, K
Figure 8: Arch abutment πΉπ¦during 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 πΉ
π₯and πΉ
π¦.
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 πΉπ₯and πΉπ¦and BendingMoment
in Different Cases. The following presents the results on thedistribution of abutment forces in different cases. Abutmentforce πΉ
π₯and bending moment have an obvious symmetry
(Figures 10 and 12); the πΉπ¦of the left bank is almost similar
8 Mathematical Problems in Engineering
Arch ring 1Arch ring 2Arch ring 3Arch ring 4
Arch ring 5Arch ring 6Arch ring 7Arch ring 8
0 1 2 3 4 5 6 7 80
0.5
1
1.5
2
Abut
men
t mom
ent
Γ1011
Overload factor, K
Figure 9: Arch abutment moment during overload.
β1.5 β1 β0.5 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
Case 4 right bankCase 4 left bankCase 5 right bankCase 5 left bankCase 6 right bankCase 6 left bank
Γ109Abutment force Fx (N)
Figure 10: πΉπ₯distribution in each case.
to that of the right bank. The distribution of πΉπ₯, πΉπ¦, 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).
Case 1 right bankCase 1 left bankCase 2 right bankCase 2 left bankCase 3 right bankCase 3 left bank
Case 4 right bankCase 4 left bankCase 5 right bankCase 5 left bankCase 6 right bankCase 6 left bank
0 0.5 1 1.5 20
20406080
100120140160180
Dam
hei
ght (
m)
Γ109Abutment force Fy (N)
Figure 11: πΉπ¦distribution in each case.
β1β2
Abutment moment (NΒ·m)
Case 1 right bankCase 1 left bankCase 2 right bankCase 2 left bankCase 3 right bankCase 3 left bank
Case 4 right bankCase 4 left bankCase 5 right bankCase 5 left bankCase 6 right bankCase 6 left bank
0 1 2 30
20406080
100120140160180
Dam
hei
ght (
m)
Γ1011
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 πΉ
π₯and πΉ
π¦are
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
Mathematical Problems in Engineering 9
Table 8: Abutment forces by simple method.
Arch layerheight (m) π΄ (m
2) π (m)Central angle (β) Results of the simple method
Left bank Right bank Left bank Right bankπΉπ₯(N) πΉ
π¦(N) πΉ
π₯(N) πΉ
π¦(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 πΉπ₯and πΉ
π¦. A simple formula to calculate the results is
provided below:
πΉπ₯= (
π1+ π2
2)
(π1+ π2)
2β sinπΌ,
πΉπ¦= (
π1+ π2
2)
(π1+ π2)
2β cosπΌ,
(13)
where π1and π
2are the water pressure of the upstream face,
π1and π2are arch layer length, β is arch layer height, and πΌ is
the central angle of an arch layer. All parameters are shownin Figure 13.
πΉπ₯and πΉ
π¦distributions 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 πΉ
π₯and πΉ
π¦; the absolute magnitudes of πΉ
π₯and πΉ
π¦
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
πΌ πΌ
x
y
zh
Figure 13: Arch layer.
β1β2β3Γ109
0 1 2 30
20406080
100120140160180
Dam
hei
ght (
m)
Left bankRight bank
Abutment force Fx (N)
Figure 14: πΉπ₯calculated with the simple method.
Each arch layer has different left and right central angles.This condition results in the different magnitudes of πΉ
π₯
and πΉπ¦. 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
10 Mathematical Problems in Engineering
Γ109
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: πΉπ¦calculated with the simple method.
β1 β0.5 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 πΉπ₯obtained 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 πΉπ¦obtained with the simple method and
FEM.
6. Conclusions
(1) Abutment force πΉπ₯increases 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. πΉ
π¦and bending moment have a similar
distribution feature under normal loads. The only differenceis that the maximum of πΉ
π₯, πΉπ¦, and bending moment is at
different heights of the dam. The πΉπ₯and bending moment
of the left and right banks exhibit good symmetry. Thedistribution of πΉ
π₯, πΉπ¦, 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 πΉ
π₯and
πΉπ¦, and the absolute magnitude of πΉ
π₯and πΉ
π¦got 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
Mathematical Problems in Engineering 11
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 damβs overload capacity is determined by the weakestarch layer. The overload factor of an arch dam can be simplydefined as
πΎ = min{ππ
πΉπ
} , (14)
whereππis the maximum value of the bearing capacity of all
arch layers. πΉπis the capacity of an arch layer in normal loads,
and π is the number of different arch layers. An arch layerβsoverload 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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