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by: Mason Hackett


Mason Hackett
GPA 3.91


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This 32 page Class Notes was uploaded by Mason Hackett on Wednesday September 9, 2015. The Class Notes belongs to CEE 521 at University of Washington taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/191982/cee-521-university-of-washington in Civil and Environmental Engineering at University of Washington.

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Date Created: 09/09/15
P w u u m q e 39 Lecture 5 Erosion and Piping wvw a Mn 3 9 ag Hm lllnrllllL mtlx g u x llIII FM gtIV W quot0139 y N gt I ll llll 1P 439 a u c P AWE T2 M 1v Mm no Ems Inf 35 no u u dgtr J Egt u u av Eaxykw t f s rq m u 350 Sank 3 ti F9 5amp3 m zgz Ax M Six Pr g 13 m t rmwxeNx Am mf Ci mixsrf 3 d Fsr wfrw C j 13 ii gwg JIVVSS vsg i v 343 3B N tn wt m g 1333 g3 23 33 333 1m J 2 Kg 0933 x33 33 Aislinglav pd NE 3 mm 13x33 SK d g out Qx vx Q S t u33gt Ax 3 Q t ud 3 xx 3 43 Aug 51 i xxwxxg u BMW h w 4 1 A mo M 3k 95 x h m It u HM viii m WP n 33gt Si 36 Qwamp N W lt gt 3 d5qu imna m m ii an u 0 x m u 3 v w der ETC ki bn Tn lt u S NW SS we ygvv gxk t X Lrex vvm lv e 1 E 3 Kg mumsoxg no u Si oatEVA vn In H Ew vgmbr 32 K ESL qusvn U 3 n Q n x Mr ltSL gyro NQ Fka 3 a YKT 7W Ha mm xu 3 Kayer xmkk he ltSt Qt 9m No rm 3 1 meg S W wgo gn S mg 32 g i g Va Fawn 3 K it 31 Q25 Nth is fex 9 6x sttfq 9 E F 3ch SS 3 5 gt3 1r Tit vAlrrvag 1amp1st n 2w a a u 19 MHPXEltPgt m n rpoxrwu ww h emw DKw m I ii 2w v xv o W 993qu w E mh WNMMQWO w wk HS 5 t ox mi 3 96 up y E 3 I EL 5 t 3 55 A SF 9 g PKNpS Ex e g mec em EN kngx 2 5 43 mvax 3 x5 gwg 8an SF D kmv i nv mw Wmmmmh 38 Ru 9 hm us r k f LS s x lt38 a Fm XE x kg guru NM 5 K a g a eks amp fwfxq V383 KWE F 914 P th3 ngf 1 ltlt 2 s n n w k E 3 u 82 mm rs EX E ES 31 5 c5th nk i Sign EEC 3amp3 ana WEIMA o vd JW my ff do a a Paved AA lam124 fw Atlanta7amp4 MIA070 WV M 3 WM Mn A JigLu flaw 1017 77w M79 MM W 71 W 5 Wave Eul a ajc m W Wft39Im wanmm WAXLAA39 a A LaxM 45014 W 39 54 1247110 EMAwhee m w 7 Wf Maiam M n LA a 0214 I f WW H A 4 014L013 adj M Hm WP 1x1 apazme Table 29 Safe values for the weighted creep ratio Safe Weighted Material creep ratio Very ne sand or silt 85 Fine sand 70 Medium sand 60 Coarse sand 50 Fme gravel 4 0 Coarse gravel 30 Soft 0 medium clay 20 30 Hard clay 18 Hard pan 16 from Das B V Advanced Soil Mechanics McGraw Hill Book Company Inc 1983 Virginia Tech The Charles E Via Jr Department of Civil Engineering CE 5564 seepage and Earth Structures TABLE 1 Examples of the Consequences of Uncontrolled Seepage Category 1 Failures caused by migration of particles to free exits or into coarse openings Category 2 Failures caused by uncontrolled saturation and seepage forces 1 Piping failures of dams levees and reservoirs caused by Lack of lter protection Poor compaction along conduits in foundation trenches etc Gopher holes rotted roots rotted wood etc Filters or drains with pores so large soil can wash through Open seams or joints in rocks in darn foundations or abutments Openwork gravel and other coarse strata in foundations or abutments Cracks in rigid drains reservoir linings dam cores etc caused by earth movements or other causes Miscellaneous mammade or natural imperfections 2 Clogging of coarse drains includ ing French drains as 5 quotP 3 1 Most landslides including those in highway or other cut slopes reservoir slopes etc caused by saturation 2 Deterioration and failure of road beds caused by insuf cient structural drainage 3 Highway and other ll foundation failures caused by trapped ground water 4 Earth embankment and founda tion failures caused by excess pore pressures 5 Retaining wall failures caused by unrelieved hydrostatic pressures 6 Canal linings basement and spillway slabs uplifted by unrelieved pressures 7 Drydock failures caused by unrelieved uplift pressures 8 Dam and slope failures caused by excessive seepage forces or uplift pressures 9 Most liquefaction failures of dams and slopes caused by earthquake shocks From Cedergren HR i967 Seepage Drainage and Flow Nets John Wiley and Sons New York 6 mm Problems Associated with Natural Formations The foundations and abutments of dams usually are stable under the in uence of the natural ground water ow but the impoundment of water in a reser39 voir greatly changes the groundwater regime and may lead to internal erosion and piping Potential seats ofinternal erosion and pipingin natural forma tions include joints in rocks beds of openwork gravel gopher holes and cavities left by rotting roots or other buried organic matter Internal erosion channels can form entirely within soil formations They can also form along con tacts between rigid structural members and adjacent or underlying erodihle materials Unless adequate care is taken a layer ofloose uncompacted material may exist at the Contact Also ifa foundation settles nonuniformly under a rigid structure J01quot example as the result ofvibrations from over ow weirs or con duitsopenings can develop Concentration of low toward loose zones or openings increases the pro bability of piping For these reasons a large propor from Cedergren H EL Seepag e Control in Earth Damsquot Embankment Dam Engineering Ronald C Hirschfeld and Steve J Ponies eds John Wiley 8c Sons Inc 1973 Iquot I y l w In W t PREVENTION OF PIPING FAILURES 39 25 tion of the seepage failures of earth dams originate along outlet conduits or spillway retaining walls or under spillway slabs The danger is greatest in soft erodible sandstones and other soft rocks and in noncohesive soils Some of the piping problems that can occur in natural formations are illustrated by the following examples Example 1 Figure 4a shows a section through a small earth dam in a Western state that failed in 1965 by subsurface erosion Boils and sink holes appeared at the downstream toe intermittently over a period of 5 or 6 years and were lled and covered with a number of truckloads of 3in drain rockquot which the maintenance men believed to be u univer sal material suitable for all drainage problems The downstream toe was under water much of the time hence a substantial loss of foundation soil through the drain rock may have taken place undetected over a number of years Then suddenly one night the dam failedonly a few hours after it had been inspected and reported quotsafe or continued usequot Water surface Homogeneous dam Qcccccccccccc Ccccc IntEl39bedded sand and silt This dam probably failed by the progressive develop ment of underground quotpipesquot or channels which nally reached the reservoir bottom leading to the very rapid failure This failure might have been avoided if a suitable graded filter had been placed over the toublesome areas instead of the coarse rock that was used The use of the 3in rock certainly did little to hold the erodible foundation soil in place and gave those responsible for the maintenance of the darn a false sense of security The piping ratio 05 of lterD35 of soil for the drain rock with respect to the founda tion soil was at least 100 Fig 4b Grading limits of suitable materials for a twodayer graded lter are shown in Fig 4b as dashed lines Example 2 Figure 5 is a cross section through an earth dam in the state of Washington that39 was constructed on a loess line windblown silt founda tion overlying a cemented but very porous congloa merate layer which in turn was underlain by jointed but tight bedrock When the reservoir was rst filled large sink holes 10 to 20 t across formed in 3 5n quotdrain rockquot placed progressive piping from ght to left Clay 6 Suitable coarse filter l T sol l Suitable line filter lllll Des l amino l f oxxm a so i 5 at m 20 o o 0 l s t Y 1 l v 5 339 13 g 4 a 16 30 so 100 200 Sievesize or Fig 4 Small cam that failed by piping through foundation to Cross section simpti ed to Typical grain size curves 5 H SEEPAGE CONTROL lN EARTH DAMS Hard very porous Flow of conglomerate mud Fig 5 Dam that failed by piping of toess through porous rock in foundation 28 Maximum water surface Large sink holes Jointed bedrock the bottom of the reservoir and mud iluwt tl out on the ground several hundred yards ihiwnstrvmn from the darn Continuous open channels in the conglomerate were discovered during the investigation of the fail ure These openings had diameters OH to lin and were considerably larger than the largest particles in the overlying loess When the loess became sutur ated it disintegrated into mud and flowed freely through the openings in the conglomerate Recommendations for Preventing Piping in Natural Formations Field exploration and geologic mapping for earth dam projeCts should be suf ciently thorough to dis close important soil and rock formations that could cause failure by internal piping or heave All new dams and reservoirs should be carefully observed and monitored after being put into service to detect the development of unsafe conditions If seepage quantities increase for no apparent reason or if other unexplained changes in seepage conditions develop adequate protective measures should be taken such as lowering the reservoir or placing weighted lters over areas where seepage discharges Problems Associated with Construction of Embankments and Drains In the preceding paragraphs two examples of dams which experienced severe piping troubles in founda tions are described Foundations must be accepted essentially as provided by nature but dam builders have substantial control over earth embankments drainage zones and other munmudc features Nt vcf theless serious flaws can be built into dams and their drains if the necessary criteria and precautions are not followed Even though a speci cation for an tmliankmcttl zone or a drain is written to satisi Eq l it is possible for piping to occur if the specil cation is not faithfully adhered to during CDnSU Ut tion The following examples emphasize the impo tance of writing adequate speci cations and enforcing them Example I Figure 6a is a cross section through 400fthigh earth dam that was built with grade lter drains As shown in section AA Fig 6b eac drain contained a core of coarse aggregate curve x Fig 62 and outer protective ne lters curve 3 Pi 6c The impervious zone of the dam was construi ted of a pebbly clayey silt curve I which had matrix represented by curve 2 When pebbly rock or gravelly skipgraded soils such as that represente by curve I are to be protected by lters it is impo tant that the grain sizes of the lter material l based on the gradation of the matrix of the soil to protected rather than on the whole gradation It seen that the ne lters in this dam curve 3 provic protection to the soil matrix curve 2 and that tl grading curve of the coarse lter curve 4 satis i Eq 1 with respect to the ne lters As the construction ofthis dam got under way it contractor had di hculties in placing a coarse tilt that everywhere ful lled the speci cations Inspe tors discovered pockets of ne lter with no materi coarser than in slightly outside of speci catic limits being placed directly adjacent to nests 8indiameter boulders curve 5 in the coarse lte The ratio of the D15 size of the nests of boulders the Dag size of the adjacent ne lter was appro imatcly Bin03 in 27 By visual inspection alon as may be seen in Fig 7 it was apparent that tl basic lter criteria were not being ful lled at the locations To solve the problem the unsatisfacto portions of the lters were replaced with materia that satis ed the speci cations Tl Fine filter Coarse filter I W F39me filter b 80 Percent passing Cum 5 l l 6 x quot 4 8 16 30 mo 200 Sieve size 1 Fig 6 Dam in which filter specifications were not me because of size segregation in filter during placement 6 Cross section b SecticniAA c Grain size curves Curve 1 pebbly core material Curve 2 matrix of core Curve 3 fine filter Curve 4 coarse filter Curve 5 nests of boulders in coarse filter quot M l 5112 28 3M I m The severe conditions that developed on this job were caused largely by careless handling and place ment methods that produced segregation in both of the lter zones But segregation in the coarse lter was aggravated by a speci cation that permitted large grain sizes in this zone Severe problems from segregation as in this example usually can he mini mized by 1 restricting the maximum size in the coarse layer of a twoolayer graded lter of this kind to not larger than 1 to 2m diameter 2 requiring lter materials to be handled and placed soon after being thoroughly sprinkled thoroughly dampened aggregates are not prone to segregation and 3 requiring lter materials to he placed with spreader boxes or other equipment that minimizes segrega tion It is important that lter materials be placed in at layers because placement in piles or windrows increases segregation Example 2 Figure 8a is a cross section through a zoned earth dam which was to be constructed with an impervious upstream zone zone I and a much more pervious downstream zone zone 2 The ranges of grain size of the available borrow materials for the two zones are shown in Fig 8b The largest 05 size of the zone 2 material was less than 5 times the 17x3 size of the zone 1 material however if there were Maximum water etevatton SEEPAGE CONTROL lN EARTH DAMS segregated coarse pockets in zone 2 this ratio could be of the order of 40 Recognizing that a wide range of grain sizes in a material such as zone 2 is con ducive to segregation the designer arranged for a test ll Even under the relatively close control that was exercised over the construction of the test ll some pockets of segregated boulders occurred hence the designer decided to modify the design to include a transition between zone 1 and zone 2 This example illustrates the importance of provid ing adequate transitions between ne39grained zones and coarsegrained zones in dams An adequate transition for the dam in Fig 8 was obtained by limiting the maximum size of particles in the rst l2 lit ofzone 2 to 2 in Example 3 Figure 9a is a cross section through an earth dam on a strati ed sand and gravel founda tion The gradation of a local beach sand proposed for the internal drain is given in Fig 96 together with the gradations of the impervious zone In ad dition Fig 9b gives gradations of layers of openwork gravel that were exposed when the cuto 39trench was dug in the foundation It is seen that the beach sand curve 2 contains few particles larger than the No 8 sieve and is a cohesionless material and is there fore very susceptible to piping Further the openings Percent passing g 8 Barn in which fitter criteria were not met because of size segregation in downstream pervious zone a Cross section ill 8 16 30 50 700200 0 Grain size curves 5TI4 mm aa 39 PREVENTlON OF PIPING FAlLURES 29 Maximum mtef elevation Sand and gravel with openwork layers Teamquot Zone impervious 0395quot lt Openwork J39Diii wkgq ii lt39 V39 Sandstone and shale a Percent passing 3quot 12 38 4 8 l 16 30 50 300 200 Sieve size In Fig 9 Earth dam with cutoff trench in foundation containing openwork gravel a Cross section through dam and foundation bl Grain size curves Curve 1 zone 1 Curve 2 beach sand Curve 3 mixed foundation soil Curves 4 openwork layers in foundation in the layers of openwork gravel in the foundation curve 4 are so large that the beach sand could wash freely through Obviously the beach sand would have been a very dangerous material to use under these conditions Because of their susceptibility to piping the do signer should avoid constructing ltirs nl39 mutvriuls that contain only sand sizes Cunsiderubly gruntr security against piping can in ublulnctl liy using lter materials containing reasonable tnrmnlugus ul39 moderate sized particles from l to l in size which will retard piping through cracks or other unantici pazed openings Water surface Presumed mains Example 4 Figure 10 gives a cross section through a levee in Oregon that collapsed as a result of piping of ne sand and silt into openings in drainpipes The drain located about midway between the centerline and the downstream toe was a perforated metal pipe surrounded by wellgmded lter material After a period of unusually high river levels portions of llu river side ol the levee partially collapsed and a consirlvrublc amount of soil discharged from the lmin pipv The river stage dropped before the levee could collapse Completely Excavation into damaged sections showed that the gravel lter had been inadvertently omitted in some areas allowing ne Filter inadvertently omitted Fig 10 Levee that partia y conaoseo due to piping through openings in unprotected drain pipe Ell 30 SEEPAGE CONTROL IN EARTH DAMS sand and silt to be in direct contact with the per forated pipe Let us 39 and that Recommendations for Preventing Piping in Embankments and around Drains i r r are needed in the design and construction of lters and drains in earth dams and levees 1 N S 1 m 57 I Do not allow materials larger than 1 to 2 in in size in the coarser layer of a twolayer drain Do not permit extremely wide ranges of particle sizes in a lter layer Broadly graded mixtures with maximum sizes of several inches or more tend to segregate during placement creating conditions conducive to internal piping Require lter materials to be placed with spreader boxes or other equipment that does not induce segregation Require lter materials to be well saturated at the time of placement and compaction Most dry aggregates tend to segregate badly during handling and placing Do not permit lters to become contaminated with nes that might be dropped from tires of construction equipment washed down slopes by rainstorms or transported by other acciden tal means Whenever possible avoid the use of lter layers that contain only sand sizes since such materials have little resistance to washing through accidental coarse pockets holes in pipes open joints in rock formations openwork gravel seams and other large openings Require careful thorough inspection of the work R m Prep Ornama drags Hi Finite Difference Solu ns of Seepage Problems DEFINITION The nite difference method is a numerical approach to solving partial differen tial equations such as those governing the twovdimensional steady state ow of a uid through a porous medium In the case of con ned problems with simple ge ometry and boundary conditions the nite difference method can easily be imple mented in spreadsheet programs The method can be applied to multiple layers and anisotropic cases FINITE DIFFERENCE SOLUTION OF SEEPAGE PROBLEMS Seepage Theory When waiter ows steadily through a twodimensional porous soil with an aniso tropic permeability kxg ky the distribution of total head 1x y within the satu rated soil obeys the following partial differential equation 32 82h 1639r Era 953 0 1 Equation 1 becomes Laplace s equation in the case of isotropic permeability k Icy 32h 32h 2 3x2 922 O In the case of con ned seepage problems the total head or the uid velocity is prescribed on the boundaries in mathematical terms the boundary conditions are prescribed in either total head or gradients of total head 21 8 Chap 45 Finite Difference Solutions of Seepage Problems Principles of Finite Differences Discretization of function derivatives As shown in Fig 1 a contin uous function fx may be de ned in terms of discrete values f corresponding to values x spaced along the x axis Assuming that the function f is differentiable the function may be expanded by using a Taylor expansion about x if L 12 2 l 3 fxAx fxdxxAx2 dx2xAx 32 dx3xAx 3 Equation 3 may be written for x xi df 1 dzf 1 y g f ELM 4 Fix 2 sz i ix 3 39Ax3 4 9 3421 21 i 3 5 3 dx Aquot2 deiAx 31153 5 The rst order differential may be approximated from discrete values by subtract ing Eq 4 from Eq 5 39 The secondorder derivative may be approximated by adding Eqs 4 and 5 mfi1gtlquot139fi 12fi dx2 A2 7 i Equations 6 and 7 are secondorder approximations of the rst and second order derivatives The errors between the exact and approximate differentials are proportional to Axl When Ax tends toward zero the approximated differential converges quadratically toward its exact values D 39 39 of i J39 p L Equations 6 and 7 also apply to functions of two variables x and y such as the two dimensional dis tribution of total head over a spatial region As shown in Fig 2 the twodimen sional space is discretized with a grid of points the coordinates of which are x X Kid Xi nix xa Figure 1 Discrete representation of a continuous function f Finite Difference Solution of Seepage Problems il Figure 2 Discrete representation a twodimensional region 219 gr 3 3 49 41 l iJ iJ 11j i il x Dig Figure 3 Nodes contributing to Eq 8 denoted by i and j Curved boundaries have to be approximated with straight seg ments in order to be described with points If Ax and Ay are the nodes spacing in the x and y directions respectively the discretized form of Eq 1 at point i j is k k X32 hm1 1 hilj quot 21 3322 hiit 139 hunt 2h 0 3 As shown in Fig 3 only the values of h at the nodes surrounding the node ij contribute to Eq amp When Ax Ay Eq 8 becomes 1 hci 2m a hi1j her Ilu1 liq r 9 where on kkWhen Ax Ay and kt k a 1 Eq 8 becomes hr i hm iii 11 hm inn1 10 Boundary conditions In con ned seepage either the total head or the total ow is precribed on the boundaries For prescribed ow boundaries we con sider only impervious boundaries and exclude prescribed ux boundaries For an impervious boundary the seepage velocity is tangential to the boundary that is ah 20 an n where n is the coordinate normal to the boundary as shown in Fig 4 In the case of a horizontal surface n a y and Eq 11 becomes 8ft 12 a o lt quotHm rst order 39 39 is L 39 39 39 a ctitious node out side the seepage domain see Fig 4 Using Eq 6 at node 139 jgwe obtain 8h z hm hirl 0 13 3 220 Chap 45 Finite Difference Solutions of Seepage Problems Impervious i 1 boundary a J m in 31 Figure 4 Impervious boundary conditions Therefore hid U4 The value of total head at the ctitious node i j 1 is eliminated by combining Eqs 10 and 13 hm i hi1j hsu 2M4 14 In summary for a horizontal impervious boundary it is not necessary to define ctitious nodes however it is necessary to replace Eq 10 by Eq 14 The coef cient 2 in Eq 14 applies to the internal nodes not to the nodes on the boundary Thus Eq 14 may easily be generalized to a vertical boundary Figure 5 gives ad ditional relations for the total head at grid points on inclined boundaries and at various types of corner boundaries In all these cases the sum of the coef cients is equal to 1 Interfaces The partial differential equations Eqs 1 or 2 do not hold on an interface between soils of different permeability because the permeability and the hydraulic gradient are not continuous there in the case of the horizontal in u n11 l I 41 94 m in mu I a hij i hi1 ham 0 MP A 1 1 MM hiLi i quotMJ 139 i by g iJl x x I quot0 ij Rig Cl 14 1 l 5 hr f3 hmi hue 5 hm I l 5 hii 33 hag x 3 any Figure 5 Relations for Gamers and 45 inclined surfaces of imper vious boundaries l mammalian Finite Difference Solution of Seepage Problems 221 terface shown in Fig 6 the discharge velocity can only be de ned on each side of the interface hh1 hrlh v k JLE L and v 2 k2 15 where v and v are the ycomponent of the discharge velocity in medium of r permeability k1 and k2 respectively Due to the conservation of ux of water quott across the interface ire v v Eq 15 becomes ki 52 1 hi kl k2 hiJ l k1 kz hAil 16 In the case of a vertical interface Eq 16 becomes kl k2 E hi m hi 1i m hH lJ 17 It can be shown that the discharge velocity changes direction abruptly at the in terface and that its angle 11 of incidence and angle a of emergence are related i through 39 tan or k1 4 tan a k 18 Seepage flow The total quantity q of ow per unit of time may be cal culated from the discrete values of total head without drawing a ow net q is ob tained for any area A that cuts the ow completely q z vxnx vynydA 19 A where n and ny x and y components of a unit vector normal to surface A and vx and vy z x and y components of seepage velocity Fig 7 If the surface A is vertical then ah qI vdA a kxEdA 20 A A Area m C 3 I I ml 1 O i WA l I 3 w la 2 2 x t dx dy l r I I I I l l l l n d d I M i m Figure 6 lnterface between soils of different Figure 7 Flow lines and calculation of total quantity permeability of seepage from discrete values of total head 222 0 Chap 4395 Finite Difference Solutions of Seepage Problems The area A of Eq 20 may be selected arbitrarily provided that it blocks the ow completely In the case of the vertical section shown in Fig 7 m lt n Eq 20 may be integrated by using a trapezoidal rule n l Ax Ax q 7m Mr 2 m 7 21 j3ml where kX Vu 27quot hidLi iii u 22 Finally the total quantity of seepage is 1 k n q 7f 1i1m quot hi 1m 2 2 hi1i hilj hiLn quot hem 23 jmt Stream function and flow lines 39By de nition the stream function WOW is 39 e 2 i v 8y and v ax 24 The quantity of seepage dq through the small element with sides dx and dy in Fig7 is 41 dq vrdy vydx 3y dy ax dx dw 25 Using Eq 23 the quantity of seepage Aq between two nodes 12139 and i j 1 is Li kx 39 A4 739 de 393 knL hi1j hi1i1 hi lil 26 Lfl W l u Wm The values of v are usually set equal to zero along one of the ow lines on the external boundary After the calculation39of total head 12 the values of WU in the interior are calculated with Eq 26 by moving away from the ow line where him0 The stream function is constant on ow lines To draw a ow net with equipotential and ow lines it is useful to introduce the modi ed stream function wild Viik lhe ow net can be obtainedby superimposing the contour lines of h and wall for identical value of contour interval Solutions of Finite Difference Problems The values of the total head at the grid points may be found by using either a di rect method or an iterative method These methods will be illustrated by consid ering the example in Fig 8 which has no direct relation to a seepage problem 39 ie Laplace equation holds inside the square region 01 by 01 The function hxy is prescribed on the boundary It is equal to zero on the left bottom and Finite Difference Solution of Seepage Problems 223 lt u hxl 4500xlx 9 News hlt1ygto 10 X 14 24 34 44 it Figure 8 Example of boundary value problem right boundaries It is equal to hx 45001 x on the top boundary The problem is to nd the distribution of h inside the square region As shown in Fig 8 the region is coarsely discretized with a grid with 16 nodes lire value of h is known at the 12 nodes at the boundaries There are only four unknown values hm 1123 h3z and hm Direct method There are only two unknowns 11 and hm owing to the symmetry about the line x g which implies that hzz 532 and has has 27 These two unknowns km and has are found by solving the two linear equations I27 2 g 1000 0 hz3 hm and hm g 0 0 1112 Hz 28 The matrix equation corresponding to Eq 28 is 3 1h221000 29 1 3 hz 3 0 Its solutions are hm 375 and h2 3 125 They are found by forming and solving a matrix equation which is a lengthy operation for more complicated grids Relaxation method The relaxation method is one of the solution methods for nite difference equations which is the most suited to spreadsheet calculations In the relaxation method the unknowns are initially assigned an at binary value Then new values are calculated from old ones by iteratively using Eq 28 until their nal values satisfy Eq 28 within a speci ed error tolerance For instance the problem of Fig 8 can be solved by relaxation as shown in Fig9 and 10 Nodes 11 12 are represented by cells A1 A2 in respec tively The speci ed values of h are entered in cells A1 B1 C1 D1 A2 A3 A4 B4 C4 D4 D3 and D2 Equations 27 and 28 are de ned in cells 82 B3 C2 and C3 where the function h is unknown 1 I As shown in Fig 10 the relaxation solution gradually converges toward the exact solution within 100 iterations The iterative calculations are activated by Options Calculation and by clicking on the iteration box The number of itera 224 Chap 45 Finite Difference Solutions of Seepage Problems A l B I C T D 1 o 1000 1000 o 2 0 BlB3A2CZ4 82 O 3 0 82B4A3C34 83 0 4 O 0 0 0 Figure 9 Formulas used for solving Eqs 27 and 28 by relaxation A B C l D l 0 1000 1000 0 2 O 250 250 0 3 0 250 250 0 v T o 0 o 0 l iteration A B C D 1 0 1000 1000 0 T 0 3752 3751 0 3 0 1251 1251 0 4 0 0 O 0 S itemions A B C D I 1 0 1000 1000 0 2 0 375 375 0 3 0 125 125 0 4 0 0 0 0 100 iterations Figure 10 Results of relaxation calculation after 1 5 and 100 iter ations tions and the error tolerance can also be de ned in the Calculation dialog box When the iteration option is not activated the error message Cannot resolve cir cular references should be displayed indicating that the formulas of Fig 9 are referring to each other s values APPLICATION TO SEEPAGE PROBLEMS Figure 11 de nes a seepage problem with a sheetpile wall As shown in Fig 123 only the left half of the problem will be analyzed owing to the symmetry about the sheet pile wall39l he total head is h 6 m on AB Owing to the problem sym metry h z 3 m on CD 111 Fig 123 the equipotential lines AB and CD where the total head is constant are dashed The flow lines AED and BC which are fol lowed by the water are solid Figure 13 shows the spreadsheet representation of the seepage problem of Fig 123 The nite difference nodes are evonly spaced every 2 m in the x and y directions There is a total of 91 nodes 13 and 7 nodes in the x and y directions respectively Figure 14 shows the formulas used in Fig 12 The prescribed total head 11 6 m is copied into cell range MMZ while I 3 m is copied into range M5948 The formulas for vertical impervious boundaries are entered in cell A3 and cop ied into range A4A7 Those for right vertical boundaries are entered in cells M3 and M4 Equation 14 for horizontal impervious boundaries is entered in cell BS and copied to C818 Eq 15 for a comer boundary is entered in cell A8 and Eq 10 is entered in B3 and copied into range B32L7 The iterative calculations are turned on by using Options Calculations The Application to Seepage Probtems E Shubpilc wall 6m 39 t 110m V Damm 225 6m E k Im exvious Figure 11 De nition of seepage problem with a sheetopile wan Equipi ff l f lf l ii m 1 a r B A New line 3m uipotential line 1 4 m Eau39ich cEanB 91 o 39 E Figure 12 Boundary conditions for a origins seepage probfem and b compfementary seepage problem I a l u I E F i l x I J i K i L I M Tote head m Upstream head m S Downstream head m 3 g 600 b 600 600 500 600 600 600 600 50Q 500 d 592 592 591 590 588 585 581 575 568 559 548 536 528 H 585 585 583 580 576 571 563 553 539 521 497 468 44 w 579 578 576 572 567 559 548 533 514 488 452 397 300 w 574 574 571 566 559 549 536 519 495 465 424 370 300 571 570 568 562 555 544 529 510 484 452 410 859 300 570 569 566 561 553 542 27 7 4 B 448 408 356 352 a 50 1 Quantity of ow per unit of rm and unit of aermeabiiiy 32543 Figure 13 Value of total head after 100 iterations results of the calculations after 100 iterations are shown in Fig 13 The error after 100 iterations is less than 0001 m As shown in Figs 13 and 14 the total quantity of seepage q divided by permeabiiity k is calculated in cell E9 by adapting Eq 23 for the horizontal line FG passing at 4 m depth Line FG cuts and blocks he ow completely Excel has several two and threedimensional capabilities to represent the distribution of total head gure 15 shows a twmdimensional contour plat and Fig 16 shows a threedimensional surface plot To get those plats select the range 9323 SF1 zF1 25551 uA2Alt2Bs4 m 39 39 A3A52844 83A455C44 camncsmam Mmmz BS aamswsmsy z 04asce054 A5A72 Bsl4 asmswucew cmamcuosw 39quot F1 D3C4D5E4V4 04050655l4 DEC6D7EGI4 3 it a A6A823987l4 s A788 I I 12 n ABCB23987 I4 BB082C7 l4 p r G8E82 07I4 J 3 Sst J223J4K34 J3l4J5K44 J4IsJ6K54 J516J7K64 J617J6K714 8K8239J74 K xFl K2J3K4L3l4 K3J4K5L44 K4J5K6L5M K5J6K7L64 K6J7K8L7I J8L8239K7I4 L Fv L2K3L4M34 aL5K6L7M64 L8K39IL8M7I4 K8M8239L7M H 1 Guam of ow a unit of me and unit 01 permeablllt a A3A5M3 MS239SUMBSL3 v239SUM 851L5 4 Figure 14 Formulas used in Fig 13 M Fan M2M4239L314 M3M52 L44 J1 J1 vJ J1 Application to Seepage Problems 227 I 300650 I 350400 400 450 E 450500 El 500 550 Cl 550600 Figure 15 Twodimensional contour representation Figure 16 Threevdimensional surface representation of total head for seepage problem of Fig 11 of total head for seepage problem of Fig 11 A22M2 with the mouse and select the appropriate threedimensional chart type by using the Chart Wizard Change the scale of the third axis to select the con tour values InVerse the second axis to display the contour in the right upward di rection Add the contour values by using the Insert Legend option As shown in Fig 17 the equipotential lines which are the lines along which the total head is a constant can be drawn using the threedimensional chart type without the hlh ing option The water pressure 4 is related to the total head 11 through uvwh ygt 39 V 30 where y is the water unit weight and y is the elevation with respect to the datum The distribution of water pressure which corresponds to the total head in Fig 15 is shown in Fig 17 Figure 18 shows the formulas that are used to calculate the water pressure from the total head and the vertical mesh spacing As shown in Fig 17 the water pressure which is hydrostatic away from the pile becomes lower in the vicinity of the pile owing to the water ow 39 as c1505 st quotsome I woe12m Dt251HSOO msoemn I Figure 17 Distribution of water pressure kPa in seepage problem of Fig 11 228 Chap 45 Finite Difference Solutions of Seepage Problems s i F L o l 0 Water pressure kPa Vertical mesh spacin m 2 Ti zeis39A2ROWA1139ROWA51l G10 9s F2ROWFl lROWA1139SGSW 98 62ROW31i ROWASI 139G10 1 2 9e 39 l V In new zm mm quot1 I out a o t eta3w t a ss 39 39 y t w I Maeu w mu wow 29 DOWsAsnrsom 1 4 9 w on w ao e Yn up wuwxu 291 v 1 auemy BquotARampIR quot quot w v II anew n u nu quotv ouch 94 w 1 waiw 839A7ROWA16HOWA11 GSlO 9839F7BOWF16u v ww In new 9o w 1quotumu 17 94 1139G10 98 F3ROWF eve3 0 9839 SA511 SGS IM Figure 18 Formulas used to calculate the water pressure kPa cf Fig 17 from the total head of Fig 134 Figure 19 Flow net for seepage problem of Fig 11 As shown in Fig 19 the ow lines which represent the water trajectory can be obtained b E 26 They can also be obtained by using the same method as for the equxpotential lines but by solving the complementary seepage problem Fig 20 In the complementary seepage problem boundary ow lines become 1 1 39 L J wh 39x 39 39 39 39 become ow lines The total head hcy is also replaced by the modi ed ow function w39wx yk which must also obey the Laplace equation ie Eq 2 Therefore the ow lines of the initial problem are transformed into prescribed ul value lines It is conven ient to set V equal to zero on one of those lines and qk on the other line where q is the total seepage ow calculated using Eq 23 Figure 20 shows the value of w as calculated by the formulas of Fig 21 As shown in Fig 19 the ow lines where p is constant can be plotted by using two dimcnsional contours The ow not is obtained by manually superimposing the twodimensional contours of I and V with identical interval values along the third axis One can verify that the ow lines intersect the equipotential lines at right angles and that these lines form curvilinear squares The ratio between the number Nf of flow channels and the number Nd of equipotential drops should also be equal to 40 M where q is the seepage ow calculated from Eq 23 and Ah is the total head drop The nite difference technique described earlier can be applied to solve many practical seepage problems Some examples of seepage problems are given in the exercises In the case of thin sheet piles as shown in Fig 22 an extra col umn of nodes must he inserted at the location of the sheet pile As shown in Fig 23 this additional column is required to have different total head on the front and back of the sheet pile Beneath the sheet pile the nodes are set to have the same total head As shown in Fig 24 this additional row unfortunately distorts the flow net in the vicinity of the sheet pile This distortion was removed in N N U A 1 B l c l D 1 E I F 1 G l H l x I 1 I K I 1 1 M 31 Ftow tines compiementafz problem Upstream value m 3932543 Downstream vaiue m 0 30 325 317 308 298 286 271 252 quot 229 200 163 117 062 000 A 325 317 309 299 287 273 255 232 204 167 121 065 000 23 325 318 310 301 291 278 262 241 215 181 136 077 000 23 325 319 313 306 297 287 273 256 234 205 166 106 000 31 325 321 317 312 305 298 289 277 261 241 215 183 153 33 325 323 321 818 315 311 306 300 292 282 270 256 248 6 325 325 825 325 325 325 325 325 325 325 325 325 325 Figure 20 Values of 1 after 100 iterations A l u 1 c 1 n 1 x L 1 M AL 1019 A20620239821l4 8200202lt214 czo5202390214 J20L202 K214 K20M202 L214 L19 3L 11319 eaom21mzmczn4 620B210220214 awzowzuazmeeum K201J2HK224L21V4 uL20K21L221M214 usst 2 11319 5quotquot quot M NH quot quotW K21u9mmw 9214 L21K22L23M22l4 L19 330st a822A23824023gt4 quot WM K22J23K24L234 L22K23L24M234 L1 13594519 B23A24825624l4 scza3240250244 023ca40255244 K23J24K25L244 L23K24L25M244 M23M252 L244 25 3311319 Cm MI K24J25K26L254 L24K25L26M25M M24M26239L254 25 11519 nsnms 3311519 H19 H 19 H19 H19 Figure 21 Formulas used in Fig 20 LIMITATIONS REVIEW QUESTIONS Chap 4 5 Finite Difference Solutions of Seepage Problems Figure 22 Seepage problem with a cofferdam and a sheet pile wall Fig 25 after replotting the numerical results of Fig 24 with a more advanced contouring program Figures 26 and 27 show the distributions of water pressure on the bottom surface of the cofferdam and on the front and back of the sheet pile wall The present method is limited to con ned seepage problems for which the boundary conditions have known positions In its present form it does not apply to uncon ned seepage problems such as those in earth dams where the free sur face is unde ned The determination of the position of unknown boundaries with nite difference is possible but requires that additional equations be solved One of the major limitations of the nite difference method is the dif culty encountered in describing curved boundary conditions and complicated layer ge ometries For this reason another numerical technique referred to as the finite ele ment method is often preferred Seepage problems of in nite siZe such as cofferdams on soil strata extending to infinity are also dif cult to analyze by using a grid of nite size In this case the in nite size can be approximated by taking a length equal to three to six times the stratum thickness It is recommended this length be varied in order to assess its effects on the solution of the seepage problem own What is the purpose of the nite difference method How is it applied to con ned seepage problems What is the partial differential equation that controls the distribution of to tal head for anisotropic and isotropic permeability What is the principal numerical technique used to solve the equations of ti nite difference methods Why does the grid spacing control the accuracy of the solution of a seepage problem with nite difference How do you represent curved boundaries in nite difference methods What is the main limitation of nite difference when dealing with seepage problems 8 A L 1 N I o z AA I AL 35651 was use 7 7 39asLsu L1 w A2A423933M aL2K3L4M3M mamaLam oamwz awawanoqm Y3AA32 Z4I4 aAA4ABSM2223213 mAL2AL4239AK3I4 A8A5239B44 ummuuhmm sM3M52 L44 AKNNNSW O It soaN4oaN4 ZBV4ZSAA4M aAA324AA5AB44 AL3AL52 AK4V4 nA4AE2 B 4 L4K5L6M54 nM4M8239L54 N4N62 05M O4N506P4 Z4Y526AA5I4 AA4Z5AA6ABSI4 uALAAL6239AK54 MA6A9239854 L5KEL7M6M emsmnmem unsmnroam n n 7 uAL5AL72 AK64 4A6A102 87M aL6K7LBM74 N7L7M8o74 M7 OsN7OBP74 sZS f7ZBAA74 AA6Z7AA8AB7I4 queummmmm HA5A92aa4 uL7K8L9MeI4 M7L8M9OB4 M3 39quot 39 39 mm xAL7AL92 AKB4 1oA6A1002 894 LBK9L10M94 nM8L9M10094 3M9 a m quot uAL8AL10239AK94 M Lava mmmmz mm me mow1m209y4 YmAA1o239zs4 M5 N 39 Figure 23 Formulas used In soiving the seepage probiem of Fig 22 232 Chap 45 39 Finite Difference Solutions of Seepage Problems Figure 24 Flow net of the seepage emblem in Fig 22 VIIIquot3amp0 O o aaanvo ma t EllIII Figure 25 Flow net of Fig 24 redrawn with a more sophisticated contouring package Distance m Water pressure L39Pa o 5 10 15 o 40 80 120 50 0 45 l e e40 2 g 35 E E 3 a 5 5 a g g u a so 4 3 25 5 20 Figure 26 Distribution of water pressure along the Figure 27 Distribution of water pressure on the horizonta surface of the cofferdam of Fig 22 from and back of the sheet pile of Fig 22 EXERCISES Generalize Eq 9 in the case of normvenly spaced nodes Calculate the hydraulic gradient vector in terms of discrete head valuee End the distribution of total head and aw net for one of the problems a to f in Exercise 2 of Chapter 44 Find the distribution of total head and ow net for one of the problems g to i show beiow Find the distribution of pressures on one of the structures sheetpile wall andor cofferdam de ned in Exercises 3 and 4 Plot the water pressure applied to the structure for one of the probiems 0f Exercisas 3 and 4 a yami 2quot Squot


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