By ASHUTOSH KUMAR

ROLL NO. 15CE64R08GEOTECHNICAL ENGINEERING

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PROBLEM STATEMENTThe failure mode of cemented backfill pillars are represented as a Mohr-Coulomb material. The pillar is three-dimensional in geometry, measuring 75 m in height by 27 min length and 15 m in width. The material properties of the fill are as follows:density (ρ) 2100 kg/m3bulk modulus (K) 110 MPashear modulus (G) 37MPacohesion (c) 0.1 MPafriction angle (φ) 35◦tensile strength ( c/tanφ ) 0.14

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MODELING PROCEDUREThe analysis is made along a two-dimensional longitudinal section through the pillar center. A sliding interface is used along one of the ore body-sand fill contacts to allow downward settling of the sand during collapse.The boundary at the fill-ore contact, which is to be excavated, is given a roller boundary, and the right-hand boundary is fixed in the x- and y-directions; the top is free (Figure 3.1).The modeling sequence is as follows. First, the gravity stresses are allowed to develop in the Sand fill , and forces equilibrate across the interface. This is done elastically so that the fill will not yield. At equilibrium, displacements are reset, cohesion is set to the proper value, and vertical retreat mining is simulated by removing the x-direction fix along the left-face boundary in small increments that simulate the blast height (6 m). Bothx-displacement and x-velocity histories are used to evaluate whether the system is coming to equilibrium at each step. Excavation is continued until active collapse of the pillar occurs.

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FLAC CODE ;Project Record Tree export ;... State: init.sav .... config ;Mining Example Problem ;Stability of cemented sand pillars gri 17,25 Mohr-Coulomb model, 30:1 sand:cement ratio, no tension cutoff m mo ;sand prop s=37e6 bu=110e6 de=2100 coh=1e10 fri=35 tens=1.43e10 ;rock prop s=2.292e10 b=3.056e10 d=2700 fric 35 coh 1e7 i 17 j 1 25 ;pillar 75m high by 15m wide gen 0,0 0,75 17,75 17,0 ;create interface mod null i=16 ini x add -1 i=17

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• ;declare interface int 1 as from 16,26 to 16,1 bs from 17,26 to 17,1 int 1 coh=0 fric=0 kn=1e9 ks=1e9 tbond=0 set grav=9.8 ;boundary conditions fix y j=1 fix x i=1 fix x i=18 his unbal his syy i=8 j=1 ;step to initial gravity equilibrium solve save init.sav ;... State: step1.sav .... ;large strain mode on set large ;turn down cohesion and tension prop coh=0.1e6 tens=0.14e6 ;turn up friction for interface

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• int 1 fric=35 ;fix y at right face boundary fix y i=18 reset displacements ini xdis=0 ydis=0 his xdis i=1 j=3 his xdis i=1 j=5 his xdis i=1 j=7 his xdis i=1 j=9 his xdis i=1 j=11 his xdis i=1 j=13 his xdis i=1 j=15 his xdis i=1 j=17 his xdis i=1 j=19 set sratio 1e-4 ;simulate vertical retreat mining in adjacent slope by ;freeing xfix on node points ; 6 m blast height

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free x i=1 j=4,5solvesave step1.sav;... State: step2.sav ....; 12 m blast heightini xdis=0 ydis=0free x i=1 j=6,7solvesave step2.sav;... State: step2del.sav ....;delete zones that failed in tensionmodel null i 1 j 4 6solvesave step2del.sav;... State: step3.sav ....; 18 m blast heightini xdis=0 ydis=0free x i=1 j=8,9

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model null i 1 j 7solvesave step3.sav;... State: step4.sav ....;... State: step3.sav ....; 18 m blast heightini xdis=0 ydis=0free x i=1 j=8,9model null i 1 j 7solvesave step3.sav;... State: step4.sav ....; 24 m blast height; these final steps will require a greater number of timesteps; to determine potential instability and deformation patternini xdis=0 ydis=0

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free x i=1 j=10,11model null i 1 j 8solvesave step4.sav;... State: step5.sav ....; 30 m blast heightini xdis=0 ydis=0free x i=1 j=12,13step 4000save step5.sav;*** plot commands ****;plot name: Displacement Vectorplot hold bound displacement;plot name: Plasticity Indicatorplot hold bound plasticity;plot name: History of x-displacementplot hold history 7 line

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; 30 m blast height ini xdis=0 ydis=0 free x i=1 j=12,13 step 4000 save step5.sav ;*** plot commands **** ;plot name: Displacement Vector plot hold bound displacement ;plot name: Plasticity Indicator plot hold bound plasticity ;plot name: History of x-displacement plot hold history 7 line

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OUTPUT: Displacement vectors in the model at 6m

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Displacement vectors in the model at 12mHere a small localized displacement is visible.

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Displacement vectors in the model at 18mHere clearly the lines are on On verge of failure.

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Displacement vectors in the model at 24mHere, collapse of the Pillar begins to occur.The failure is shown by The large downward Movement.

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Displacement vectors in the model at 30mHere, the pillar is totally collapses. The zones nearestto the left- face boundary failin tension.

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The deformed boundary and plasticity states at collapse of the pillar

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Result The model comes to equilibrium for the first three blast heights (6 m, 12 m and 18 m).Shown displacement vector at different blast heights from fig 3.2 to fig. 3.7 .Note that after excavation of the 12 m blast height, the zones nearest to the left-face boundary fail in tension. These zones are deleted in order to remove elements that will produce a bad zone geometry as the excavation continues. The model is still in equilibrium after these zones are deleted; first three stages (6 m, 12 m and 18 m blast heights), the model can be brought to equilibrium after each excavation is made by using the SOLVE command.At the fourth blast height (24 m), collapse of the pillar begins to occur. Finally the deformed boundary and plasticity states at collapse of the pillar is shown at output.

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THANK YOU…

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