Stability of dam abutment including seismic loading.pdf

Stability of dam abutment including seismic loadingMorteza Sohrabi Gilani Civil Engineering Department Sharif University of Technology Azadi Ave.- Tehran- Iran Rupert Feldbacher Institute of Hydraulic Engineering and Water Resources Management Graz University of Technology Stremayrgasse 10/II 8010 – Graz Gerald Zenz Institute of Hydraulic Engineering and Water Resources Management Graz University of Technology Stremayrgasse 10/II 8010 – Graz [email protected] [email protected] [email protected] 1. Introduction One of the most important aspects in the stability analysis of arch dams, which has been encountered for many years, is the stability of the abutment. This study is aimed to evaluate within the Tenth Benchmark Workshop on Numerical Analysis of Dams-Theme C, the abutment stability of Luzzone arch dam under static and seismic loadings. At first the three dimensional model of the dam has been transferred for being applicable in the finite element program of Abaqus 6.7. With the FEM the interface forces between concrete dam and wedge are calculated for the required loading cases. The stability analysis of the given wedge is evaluated by Londe method. 2. System Assumption 2.1. Luzzone dam The Luzzone dam is a double curved concrete arch dam which was initially built in the sixties. The dam was heightened within the ninetieths. The total height of the dam is 225 m. Figure 1 shows the Luzzone dam. Fig. 1: Luzzone dam 1 In other words a massless foundation is considered.2. only the stiffness of foundation is considered and density of it is taken as zero. Material properties The material property of the mass concrete dam and foundation are defined as: Concrete of dam: Density ( ρ ) = 2400 kg/m3 `` Poisson ratio (ν ) = 0. The wedge position is shown in figure 2.6 and β =0.167 Modulus of elasticity (E) = 27 GPa Rayleigh damping coefficients: α =0.6 and β =0. To verify about this situation a stability assessment is necessary. 2 . The volume of the wedge has been estimated as 1.2 Modulus of elasticity (E) = 25 GPa Rayleigh damping coefficients: α =0. Wedge definition For the benchmark in the left bank of the dam are two geological joints.2: Finite element model of dam and foundation and Geometry of the wedge 2. With these joints a wedge is defined and has a potential to slide under arch dam and uplift loading.001 Water: Density ( ρ ) = 1000 kg/m3 Fig 3: Material properties of dam and foundation It should be mentioned that in calculating the interface forces between dam and wedge.001 Foundation rock: Density ( ρ ) = 2600 kg/m3 Poisson ratio (ν ) = 0.3.2.92 × 106 m 3 . Fig. three stochastically independent acceleration time histories are used according to the data provided by the formulator. 3. Under the reservoir full condition the hydrostatic pressure is applied to the dam’s upstream surface according to the programs loading definition. and all planes are under full uplift pressure. The added mass per unit area of the upstream wall is given in approximate form by the expression 7 ρ w hw (hw − y ) .direction) = 0.16 g Vertically upwards (Y.1063 g Cross valley direction (Z. The dam-foundation interface is modeled as a joint with a high friction coefficient to reduce relative displacement between dam and foundation to a minimum. In the self-weight condition the dam is considered monolithic and isotropic material behavior is used. only the stiffness of foundation is considered and density of it is taken as zero. According to westergaard. 3 . Calculation Procedure Figure 2 presents the finite element model of the dam. In other words a massless foundation is considered. Loading The static and seismic load cases are considered to calculate the dam-foundation interface forces.4.2.16 g In wedge stability assessment uplift pressure at the wedge interfaces is considered very conservative. 1610 m 1510 m 1385 m Fig 4: cross section of the dam For seismic analysis. For the seismic analysis direct time history approach is used and hydrodynamic pressure is computed by Westergaard’s added mass method. This model is created within Abaqus 6. the hydrodynamic pressures that the water exerts on the dam during an earthquake are the same as if a certain body of water moves back and forth with the dam whiles the remainder of the reservoir is left inactive. These accelerations are scaled according to the peak ground accelerations of these components are: Downstream-upstream (X. where ρ w is the density 8 of water.direction) = 0. It should be mentioned that in calculating the interface forces between dam and wedge.direction) = 0.7 and linear elements (C3D8) are used to define dam and foundation body. The resultant forces transmitted between dam and wedge are computed as the sum of pressure and shear stresses at the wedge dam interface. Tensile forces. eight cases are possible. For this purpose the equilibrium equation is solved with these three existing forces. This force is time dependent and its magnitude and direction will change by time. N3 and S23 again. Figure 5-a. Seismic forces due to applied earthquake: the three components of seismic forces are considered as the max. The reaction of planes: due to applied forces. The applied forces can be categorized as: - - - Weight of the wedge (Ww) The thrust force which is the resultant force at the dam wedge interface. U2 and U3 which are applied to the planes 1. N2 and N3). For this purpose Londe method for stability of rock slopes is used and some simplifying assumptions are made. - Case 2: The reaction force of plane 1 is tensile. Stability of wedge The next step for the analysis is to evaluate is to evaluate the wedge stability. Plane2 and Plane3 in the figure 2).these forces do not change during investigation. three reaction forces will develop on the planes (N1. may and maz which m is the mass of the wedge and ax. Table 1 shows all possibilities. For this wedge geometry and applied forces. The wedge is considered as a rigid body and the geometry of the wedge would not change during application of the forces throughout the investigation. Cohesion and tensile strength are neglected in the contact planes and therefore. In this case to check the movement along the intersection of plane 2 and 3 the force in this direction is calculated. The volume of the wedge is limited by intersections of three planes (Planes Plane1. N2<0 and N3<0). As mentioned before these forces can only be compressive. It is supposed that the moments of the forces have negligible influences and can be ignored. This assumption is conservative as the natural surfaces are generally irregular [2]. which mean that the plane is open. Fy and Fz. - Case 1: All plane reaction forces are compressive: all planes are in contact and the wedge is perfectly stable. it is supposed that the friction between surfaces is the only parameter that can resist sliding. The forces due to uplift pore pressure: U1. N1<0 and N3<0). N2. but the other two reaction forces are compressive (N1>0. Figure 5-b. Then the stability factor can be calculated accordingly: SF = S 23 N 2 tan ϕ 2 + N 3 tan ϕ 3 If the safety factor is less than one the rupture will occur and wedge will move along intersection line of planes 2 and 3. 4 . due to equilibrium condition and calculated plane reaction forces. For wedge stability evaluation at first the three plane reaction forces are to be calculated by solving static equilibrium equations in three direction x.4. This case is similar to case 2. y and z. - Case 3: Sliding along the intersection of plane 1 and 3 (N2>0. In other words plane 1 is open but planes 2 and 3 are in contact yet. 2 and 3 respectively. are not acceptable and will lead to a different sliding mode respectively exclude sliding in the decoupling plane due to tensile forces. So this force can be defined by Fx. ay are az are acceleration time histories which were defined before. The wedge stability safety factor of the dam during the earthquake is plotted in the figure 6. This case is similar to case 5. - Case 5: Sliding in plane 3 (N1>0. In this case the only plane which remains in contact is plane 3. Fig. N2>0 and N3<0. N3>0 and N1<0). The Safety factor reads accordingly: SF = Shear force on plane 3 ( S 3 ) N 3 tan ϕ 3 - Case 6: Sliding in plane 2 (N1>0. 5-a: Normal contact forces Fig.- Case 4: Sliding along the intersection of plane 1 and 3 (N3>0. This case is similar to case 2. The concept of Newmark’s method is used to calculate displacement of the wedge. N1<0 and N2<0). N3>0 and N2<0). the safety factor for a short period of time is less than 1. N2>0 and N3<0). In this case all planes are open and the wedge is obviously freely moving. As shown. which means that the wedge would move during this time period. 5-b: Contact forces for case 2(sliding along the intersection of planes 2 and 3) 5 . This case is similar to case 5. The normal and shear forces of this plane are calculated again by solving the equilibrium equation and ignoring the plane 1 and 2. - Case 8: N1>0. - Case 7: Sliding in plane 1 (N2>0. 3 1. 3 Table 1: all possible movement cases of the wedge 6 .2 2. 3 8 in space Diagram 1. 3 2 intersection of Planes 2. 2. 3 2 4 intersection of Planes 1. 3 7 in plane 1 1 2.3 2. 3 3 5 in plane 3 3 1. 3 1 3 intersection of Planes 1.Case Nature of sliding Contact faces Open Faces 1 No sliding 1. 2 6 in plane 2 2 1. 2. this assumption used normally.0 1. 6. 6: Safety factor of the wedge stability during seismic loading 5.5 1. With the help of FEM the wedge is suggested to be analyzed as deforming body and with this the stability of the abutment. water loading and uplift. Figure 7 shows the calculated displacement for the wedge in the x and y directions. The integration should continue till the velocity in the considered direction vanishes. (For one plane sliding the resultant acceleration is decomposed F − ( N 3 tan ϕ ) y due to the applied forces.0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 Time(s) Fig.5 2. The uplift pressure at the wedge interface is very conservative. Displacement of the wedge To calculate the displacement of the wedge in the first step acceleration of the wedge is calculated in the x. was out of scope of this benchmark.5 0. The calculated displacement is reported in the table 3. as it would never be the case that the entire planes are under full uplift. but for m m two planes sliding the direction is the intersection of corresponding planes which is constant) Displacement of the wedge is the double integration of this computed acceleration. Driving Force − Stabilizing Force and its direction y and z direction.0 SF 2. for example in case 2. a x = Fx − ( N 3 tan ϕ ) x and a y = y . However. The magnitude of the acceleration is mass of the wedge is being defined due to the movement case. Conclusion Under the assumption of a rigid body wedge the analysis is carried out for dead weight.0 0.5 3.Wedge Stability Safety Factor 3. No variation of the earthquake acceleration along the valley is assumed. 7 . 8 0. Russell Michael Gunn.02 13.02 Vy (m/s) 0.7-EF Documentation”. No.04 0.02 1. “Abaqus version 6.06 0 13. Londe.08 time (s) time (s) 12.98 ax (m/s2) 0 time (s) Fig 7: acceleration. “Computational aspects of analysis and design of dams”. published in Geotechnique.98 Vx (m/s) 13.8 0. p. vol.2 -1. N. “effect of earthquakes on dams and embankments”. p.10 13.00 Uy (mm) 2. 8 . 109-129.6 -2 -2 -2.5 time (s) 13.08 12. References 1. Prof. 8. velocity and displacement of the wedge in x and y direction 7.2 1.00 0 0 12.04 13.00 13.06 13.02 13.01 0. Newmark.98 ay (m/s2) 13. 6. 2. 3. M.06 13.98 13.06 13.03 0.06 13. 1965.10 Uy 2. 4.01 13. 1973.4 time (s) Vy 0.01 13. 15. “Analysis of the stability of rock slopes”.8 -1. P.04 13.5 13.6 -1.4 12.02 -0.ax ay 1.03 0. 93-124.00 0 12.5 13.4 -1.4 -2.04 0 13.98 13.8 -0. “Milestones in Engineering”.04 13.06 1 0. 2.98 -0.01 Ux 13.02 0.00 -0.08 -0. Dr. published in “Quarterly Journal of Engineering Geology and Hydrogeology”.10 time (s) Vx Ux (mm) 13.02 0.5 2 2 1. Acknowledgement The support of this work by the research project “Design of Hydraulic Structures” by Pöyry Energy Ltd is gratefully acknowledged. issue 1.2 0.4 0. vol.5 13.00 12. “Tenth benchmark workshop on numerical analysis of dams”.04 0.5 13.02 13.2 -0.04 13.04 1 13.4 13. 00 0.12 .65 23968.1 127.55 ‐3000.64 34312.82 65764.72 6910.52 ‐4858.32 ‐58096.00 0.17 ‐20452.1 57. Yes Yes Yes Yes Yes Yes Yes Loss of Contact Plane Jh J1 J2 No No No No No .85 ‐18618.3 59.6 22.55 44304.12 ‐3086. .5 59.00 0.15 ‐2663. ‐68269.6 59.00 0.42 .00 0. 31.24 ‐8792.19 22488.54 48174.29 75230. 89. No No No No No No No No No No No No No No No No . Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes No No No No No . 12.40 ‐2000. .9 131. .88 .94 ‐58094.85 1056.5 60.07 13.14 ‐10685.95 ‐14115.47 16190.06 ‐58095.90 ‐20813.00 0.86 ‐1366.5 .25 ‐20777.58 ‐17461.7 132.91 23308.85 ‐18620. .80 ‐65403.26 ‐18315.00 Contact Plane Forces J1(MN) J2(MN) 0.42 17493.94 ‐66254.73 12824.44 30347.5 132.94 ‐20152.08 ‐43341.44 ‐10874.95 78216.80 11481.96 31.55 ‐35570.33 .06 17697.24 ‐23248. ‐5814.89 48162.00 0. .6 Beta 31.30 ‐26704.86 ‐26572.21 . 37. .97 12.75 78793.45 16940.25 ‐20776.92 ‐19023.6955.21 Yes No No No No . .58 15936.51 ‐26774. No No No No No No No 9 Results Case Fsw Fsw+hyd 0 0.7 120.26 ‐8511.40 ‐67487. ‐27019.3 Resultant Forces (Dam thrust) FR(MN) 12715.46 24691.24 ‐10681.09 13.10 ‐67722. 0.77 ‐18622.95 66318.50 Alpha 50.30 ‐47437.5 9. 66832.5 60. ‐4175.61 30101.10 78488. 54.90 ‐55629. driving and stabilizing forces .55 585.06 22397.5 60.38 30168.4 132.30 22496. Contact plane forces.26 ‐4980.99 13 13.0 131.00 0.47 .9 .41 0. .00 0.21 52025.65 ‐20776.50 ‐66957.5 71.00 0.02 13.24 13320. .85 ‐16396.18 48157.6 43.6 59. Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes .98 ‐29517.6 60.00 0.90 ‐38231.51 30340.08 13.06 ‐17602.32 ‐24014.41 24906.11 22496.78 . .37 24744.20 22497.00 0.06 ‐20487.46 26162.44 ‐3150.84 ‐20361.6 59. .27 ‐27772.00 .5 59.03 2694.24 30003.72 ‐29190.00 0.5 98.00 0.46 66332.29 ‐29114.71 ‐47437.51 ‐17512.27 ‐74245.6 59.00 0.20 ‐67104.68 ‐20249.93 66117.39 15275.73 ‐6682.03 13.00 0.55 ‐18618.82 55032.1 . .00 0.06 13. . .32 ‐29051.78 ‐25308.00 0.94 . .34 77977.01 0.30 ‐61647.90 12414.5 111.14 25137.46 .53 .1 134.88 ‐10684. 79124.85 ‐26490.00 0.4 133.69 ‐21197.92 . .01 ‐8045.17 ‐60393.56 17440.3 125.00 0.41 ‐2358.38 78084.6 42.48 ‐21815. 14332.50 ‐47441. .63 ‐7630.00 0.01 13.83 66572.11 3574.9 131. 30442.00 ‐29382.88 3592.71 58362.04 13. .69 77773. ‐74669.35 ‐74540.00 ‐1965.30 ‐47439.14 ‐14152. ‐29496.4 132.95 .41 ‐74063. .5 59.4 132.75 ‐73769.5 134.00 0. .70 ‐20171.70 ‐29353.12 ‐72166.00 0.85 ‐2477. 9418.78 54399.3 44.09 55032. .69 77603.34 S(MN) 26928.29 ‐726. 0.00 0.97 2485.64 ‐20627.3 .54 ‐26892.97 31.27 Stability and Driving Force D (MN) ‐7576.00 0.61 11360.5 58.00 0. 7158. 59.81 ‐26667.00 0.87 ‐20070.87 55038.00 0.89 ‐29002.21 30046.61 ‐58094. .8 59.5 132.91 ‐13678.05 13.52 17219.00 0. 0.00 ‐67984.7 50.40 ‐67281.7 55.00 0.02 .38 ‐52829.10 ‐25861.30 .85 65628.10 Resultant Forces (Dam thrust) Fx(MN) Fy(MN) Fz(MN) 4182.84 ‐26909.3 29. .42 19660.96 Jh(MN) ‐38457.9 131.25 19206.2 128.53 ‐74448.4 132.96 36991. .7 130.00 0.6 59.33 11145.95 1736.99 32 Table 2: Dam thrust force.66 .6 59.04 ‐15917.23 9787.74 62793. ‐20469.98 12.18 .56 ‐24983.38 65927.1 16.00 0.00 .82 55035.01 1018.00 0.23 4067.1 13.27 18587. .22 ‐10742. .80 ‐67836.07 ‐73656.38 43664.51 ‐69672.00 ‐26419. .00 ‐10705.62 ‐14013.60 ‐47234. 132.94 31.73 ‐73904. .05 ‐4860.3 60.11 13. ‐20777.00 0.58 ‐20777. No No No No No No No No Yes Yes Yes Yes .70 .97 20668.43 48159.31 63909.71 70835. .4 60.09 ‐29279.11 30248.95 31.51 3707.28 . .98 31. 23 2.32 2. 0 0 28. . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 .06 0.72 1. 2.41 25. . 0 0 0 0 0 0 0 0 0 0 0 0 .29 0.44 31.05 34.23 2.91 2. 2.01 2.04 13. 0 0 0 0 0 0 0 beta 0 0 0 0 0 .01 2.01 0 0 0 0 0 .43 1.41 25. 12.19 2.01 0.52 1.49 29. .96 0.19 2. .01 0.55 0. .33 0.10 0.Results Case Fsw Fsw+hyd 0 0.03 13.81 1. 0 0 0.96 .96 0.72 30.08 13.55 2.01 2.65 1. .23 2. 2.59 0.16 1. .65 1.41 25.96 0.99 32 Ux(mm) 0 0 0 0 0 .01 2. 0 0 0.18 33.23 2.01 2.01 2.82 26.86 1.12 27. . .19 .28 1.23 2.23 2.05 2.36 0. . .41 25.97 12.96 0. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 .02 0.41 25.1 13.04 1.01 2.98 31.07 13.96 0.01 .41 25.96 0.70 0. 0 0 0.01 13.54 2.23 2.96 0.17 0.96 0. 0.81 0. .14 2.19 2.43 1.54 1.95 31.12 0. . 25.80 1.41 25.14 .98 12.42 1.85 33.72 0.23 2.65 1.09 13.23 alpha 0 0 0 0 0 . 31. . .01 0.23 .19 2.94 31.19 2. .53 .05 13.03 1.96 Wedge displacements Uy(mm) Uz(mm) UR 0 0 0 0 0 .99 13 13.96 0.14 2.06 13.97 31.02 .96 31.14 2.02 13.96 0.03 0.96 0.41 Table 3: displacement of the wedge in the x and y direction 10 Factor of safety 3.93 2.41 25.46 34.87 2.65 0.20 2.01 2.11 13.41 .17 2.96 0.33 25.12 .41 25.96 0.16 2. 1.23 2.01 2.