Mechanism Research of Arch Dam Abutment Forces during Overload

This paper presents research on the abutment forces of a double-curvature arch damduring overload based onnumerical 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 bendingmoment reach theirmaximum at themiddle ormiddle-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 forces F 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.


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][6][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][16][17], meshless [18], and extended finite element methods [19][20][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.

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 where  is Young's modulus for concrete and ] 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 coefficient   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 where the superscript  signifies that the stress-strain relations refer to a coordinate system parallel to principal stress directions with the   axis perpendicular to the crack face.  works with adaptive descent and diminishes to 0.0 as the solution converges.
In Figure 1,   is uniaxial tensile cracking stress and   is 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 shear transfer coefficient   for a closed crack is introduced.[    ] can then be expressed as The stress-strain relations for concrete that has cracked in two directions are If both directions reclose The stress-strain relations for concrete that has cracked in all three directions are If the crack is closed again in three directions, formula (5) will be selected.  and   values have the following relationship: 1 >   >   > 0. (7)

Distinguishing between Crack Opening and Closing.
The open or closed status of integration point cracking is based on strain value    , which is called the crack strain.For the case of a possible crack in the  direction, this strain is evaluated as where    ,    ,    are three normal component strains in crack orientation.
Vector {  } is computed by where {  } is modified total strain (in element coordinates).{  } is defined as where  is the substep number, { el −1 } is the elastic strain from the previous substep, {Δ  } is the total strain increment (based on {Δ  }, the displacement increment over the substep), {Δ th  } is the thermal strain increment, and {Δ pl  } is the plastic strain increment.
If    is less than zero, the associated crack is assumed to be closed.
If    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.

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 where  and  are material parameters,  1 is the first invariant of the stress tensor, and  2 is the partial stress tensor for the second invariant.They can be calculated as where   2. The -axis positive direction of the model is along the right banks.The positive direction of -axis is downstream along the river, and the positive direction of -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  direction constraint, and the boundary of the upstream and downstream is subjected to a  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.

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 factor  is defined as  = / 0 , where  is the current water density of the dam and  0 is the normal water density.In Case 1, the value of water density is 9.8 kg/m 3 .In Case 2, the magnitude of water density is twice the normal value (19.6 kg/m 3 ) and so on.The relevant parameters are listed in Table 1.

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 abutment   ,   , and bending moment.Abutment forces   ,   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   ,   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   and   .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.  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).

Distribution of Abutment Forces under Normal Loads.
The abutment force of   (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   is at approximately 80 m of the height of the dam.The left and right banks   are symmetrical because of the symmetry of the dam body shape.
With regard to the distribution of force   , 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   are almost similar because of the symmetry of the arch dam.The distribution of   and   has a similar feature to that from [24,25].The relation between bending moment and increasing elevation is similar to that in   .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   (Figure 4).

Abutment 𝐹 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 (Figures 7, 8, and 9).The   line of each arch layer during overload is straight (Figure 7).  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  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   force during overload is similar to that for   (Figure 8).For   , 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.
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 of   and   .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.to that of the right bank.The distribution of   ,   , 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).

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 because   and   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 where  1 and  2 are the water pressure of the upstream face,  1 and  2 are arch layer length, ℎ is arch layer height, and  is the central angle of an arch layer.All parameters are shown in Figure 13.  and   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.
The above mentioned method is an approximate means to calculate   and   ; the absolute magnitudes of   and   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.

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.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.

Conclusions
(1) Abutment force   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.  and bending moment have a similar distribution feature under normal loads.The only difference is that the maximum of   ,   , and bending moment is at different heights of the dam.The   and bending moment of the left and right banks exhibit good symmetry.The distribution of   ,   , 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   and   , and the absolute magnitude of   and   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 where   is the maximum value of the bearing capacity of all arch layers.  is the capacity of an arch layer in normal loads, and  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.

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

Figure 11 :
Figure 11:   distribution in each case.

Figure 12 :
Figure 12: Bending moment distribution in each case.

Figure 16 :
Figure 16: Comparison of   obtained with the simple method and FEM.

Figure 17 :
Figure 17: Comparison of   obtained with the simple method and FEM.

Table 2 :
Results of Case 1.

Table 3 :
Results of Case 2.

Table 4 :
Results of Case 3.

Table 5 :
Results of Case 4.

Table 6 :
Results of Case 5.

Table 7 :
Results of Case 6.

Table 8 :
Abutment forces by simple method.