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Braced cut in deep excavation

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  • 1. Under the guidance ofProf. Kalyan Kumar ChattopadhyaySubmitted By:Yogesh Kr PandeyExam roll: 110904034
  • 2. IntroductionAn excavation supported by suitable bracing system are calledbraced cut. These excavation support systems are used to,• Minimize the excavation area,• Keep the sides of deep excavations stable, and• Ensure that movements of soil will not cause damage toneighboring structures or to utilities in the surroundingground.
  • 3. The design of braced cuts involves two distinct butinterrelated features, namely Stability of excavation, ground movement, control ofwater into the excavation, effect of adjoiningstructures and so on. Design of structural elements i.e sheet pile, struts oranchors and so forth.
  • 4. Type I use of soldier beams•Soldier beam is driven into the ground before excavation and is avertical steel or timber beam.•Laggings, which are horizontal timber planks, are placed betweensoldier beams as the excavation proceeds.•When the excavation reaches the desired depth, wales and struts(horizontal steel beams) are installed. The struts are horizontalcompression members.
  • 5. Type II: Use of Sheet Piles•Interlocking sheet piles are driven in to the soil before excavation.•Wales and struts are inserted immediately after excavation reachesthe appropriate depth.
  • 6. DIFFERENT TYPES OFSHEETING AND BRACINGSYSTEMS
  • 7.  Vertical Timber Sheeting: Vertical timber sheeting consisting of planks about8 to 10 cm thick are driven around the boundary of the proposed excavation tosome depth below the base of the excavation. The soil between the sheeting isthen excavated. The sheeting is held in place by a system of wales and struts.The wales are horizontal beams running parallel to the excavation wall. Thewales are supported by horizontal struts which extend from side to side of theexcavation. However, if the excavations are relatively wide, it becomeseconomical to support the wales by inclined struts, known as rakers. Forinclined struts to be successful, it is essential that the soil at the base of theexcavation be strong enough to provide adequate reaction. If the soil can betemporarily support itself an excavation of limited depth without an externalsupport, the timber sheeting can be installed in the open or in a partiallycompleted excavation. Vertical timber sheeting is economical up to a depth of4 to 6 m.
  • 8. VERTICAL TIMBER SHEETING
  • 9. Steel Sheet Pile: In this method, the steel sheet piles are drivenalong the sides of the proposed excavation. As the soil is excavatedfrom the enclosure, wales and struts are placed. The wales aremade of steel. The struts may be of steel or wood. As theexcavation progresses, another set of wales and struts is inserted.The process is continued till the excavation is complete. It isrecommended that the sheet piles should be driven several metersbelow the bottom of excavation to prevent local heaves. If the widthof a deep excavation is large, inclined bracing may be used.Steel sheet pile
  • 10. Soldier Beams: Soldier beams are H-piles which are driven at aspacing of 1.5 to 2.5 m around the boundary of the proposed excavation. Asthe excavation proceeds, horizontal timber planks called laggings are placedbetween the soldier beams. When the excavation advances to a suitabledepth, wales and struts are inserted. The lagging is properly wedgedbetween the pile flanges or behind the back flange.Soldier Beam
  • 11. Tie Backs: In this method, no bracing in the form of struts orinclined rakers is provided. Therefore, there is no hindrance to theconstruction activity to be carried out inside the excavated area. Thetie back is a rod or a cable connected to the sheeting or lagging onone side and anchored into soil (or rock) outside the excavation area.Inclined holes are drilled into the soil (or rock), and the hole isconcreted. An enlargement or a bell is usually formed at the end ofthe hole. Each tie back is generally prestressed the depth ofexcavation is increased further to cope with the increased tension.
  • 12. Use of Slurry Trenches: An alternative to use of sheeting andbracing system, which is being increasingly used these days, is theconstruction of slurry trenches around the area to be excavated and iskept filled with heavy, viscous slurry of a bentonite clay-water mixture.The slurry stabilizes the walls of the trench, and thus the excavation canbe done without sheeting and bracing. Concrete is then placed througha tremie. Concrete displaces the slurry. Reinforcement can also beplaced before concreting, if required. Generally, the exterior walls of theexcavation are constructed in a slurry trench.Slurry Trench
  • 13. Lateral Earth Pressure Distribution
  • 14. Lateral earth pressure is the pressure that soil exerts against astructure in a sideways, mainly horizontal direction. Since most opencuts are excavated in stages within the boundaries of sheet pilewalls or walls consisting of soldier piles and laggings and sincestruts are inserted progressively as the excavation proceeds, thewalls are likely to deform (as shown in figure below). Little inwardmovement can occur at the top of the cut after the first strut isinsertedTypical pattern of deformation of vertical wall (Braced cuts)
  • 15. EARTH PRESSURE DISTRIBUTION• In Sand• In Clay
  • 16. In SandFollowing figures shows various recommendations for earthpressure distribution behind sheeting This pressure, pa may beexpressed as0.8γHKaTerzaghi and Peck’s Peck’s earthpressure distribution for loosesand
  • 17. 0.8γHKaTerzaghi and Peck’s earth pressuredistribution for dense sand0.8γHKaTschebotarioff’s Peck’s earth pressuredistributionHH
  • 18. 0.65γHKaPeck, Hansen and Thornburn’sPeck’s earth pressure distributionfor moist and dry sandsHWhere,, γ= unit weightH= height of the cutKa= Rankine’s active pressure coefficient.
  • 19. Cuts in ClayThe given figures represent the different earth pressure distributionrecommendations for clay. In clay braced cuts becomes unstabledue to bottom heave .To ensure the stability of braced systemγH/cb must be kept less than 6, where γH/cb is the undrained shearstrength of soil below base or excavation level.For plastic clay by Peck
  • 20. HNeutral earth pressure ratiomethod by TschebotarioffH Peck, Hanson and Thornburn’sdiagram when γH/c ≤ 4
  • 21. HPeck, Hanson and Thornburn’sdiagram when γH/c > 4As the most probable value of any individual strut load is about25 percent lower than the maximum as obtained fromPeck, Hanson and Thornburn’s earth pressure distributiontheories, so among all the given earth pressure distributionprofiles, Peck, Hanson and Thornburn’s earth pressuredistribution theories are most widely and popularly used
  • 22. Pressure envelope for cuts in layered soilSometimes, layers of both sand and clay are encountered when a bracedcut is being constructed. In this case, Peck (1943) proposed that anequivalent value of cohesion should be determined according to theformula,cav = [γsKsHs2tanФs’ + (H-HS)n’qu]The average unit weight of the layers may be expressed as,γa H =[ γs Hs + ( H - HS ) γc ]
  • 23. Where,H = total height of cutγs = unit weight of sandKs = lateral earth pressure coefficient for sand layer ( =1 )Hs = height of sand layerФs’= effective angle of friction of sandqu= unconfined compression strength of clayn’= a coefficient of progressive failure (ranging from 0.5 to 1.0average value 0.75)γc = saturated unit weight of clay layer•Once the average values of cohesion and unit weight aredetermined, the pressure envelopes in clay can be used todesign the cuts
  • 24. Similarly, when several clay layers are encountered in the cut, theaverage undrained cohesion becomesCav = (c1H1 + c2H2 + ...... + cnHn)The average unit weight is now,γa =[ γ1 H1 + γ 2H2 + ... + γn Hn ]Where,c1,c2,...., cn = undrained cohesion in layer 1,2,...,nH1 , H2 ,..., Hn = thickness of layers 1, 2, ... , n
  • 25. Limitations of the Pressure EnvelopeWhen using the pressure envelope just described, following points are tobe noted: The pressure envelopes are sometimes referred to a apparent pressureenvelope. However, the actual pressure distribution is a function of theconstruction sequence and relative flexibility of wall. They apply to excavation having depths greater than about 6m They are based on the assumption that water table is below the bottomof the cut. Sand is assumed to be drained with zero pore water pressure. Clay is assumed to be undrained and pore water pressure is notconsidered
  • 26. LOADS ON BRACESTributary Area MethodEquivalent Beam Method
  • 27. Tributary Area MethodThe load on a strut is equal to the load resultingfrom pressure distribution over the tributary area over that strut.For e.g Strut load PB in the fig. is the load on the tributary area 1-2-3-4.
  • 28. Equivalent Beam Method:In this method entire depth is split into segments ofsimply supported beams and reactions can then be determined bystandard process.
  • 29. STABILITY OF BRACED CUTS
  • 30. Heaving in Clay SoilThe danger of heaving is greater if the bottom of the cut issoft clay. Even in a soft clay bottom, two types of failure arepossible. They areCase 1: When the clay below the cut is homogeneous atleast up to a depth equal 0.7 B where B is the width of thecut.Case 2: When a hard stratum is met within a depth equal to0.7 B.
  • 31. Case:1 The anchorage load block ofsoil a b c d in Fig. (a) of width(assumed) at the level of thebottom of the cut per unitlength may be expressed as The vertical pressure q perunit length of ahorizontal, ba, is
  • 32. The bearing capacity qu per unit area at level ab isqu = Ncc = 5.7cWhere, Nc =5.7The factor of safety against heaving isBecause of the geometrical condition, it has been found that the widthcannot exceed 0.7 B . Substituting this value for,For excavations of limited length L, the factor of safety can be modified to
  • 33. Where, B’=T (thickness of clay below the base of excavation) or B/ (whicheveris smaller)In 2000, Chang suggested few revisions, with hismodification, equation takes the formB’=T if T B/B’ = B/ if T> B/B’’= B’
  • 34. Case: 2 Replacing 0.75B by D in Eq, thefactor of safety is represented by For a cut in soft clay with aconstant value of cu below thebottom of the cut, D in Eq.becomes large, and Fsapproaches the value
  • 35. Ns is termed as Stability Number. The stability number is auseful indicator of potential soil movements. The soilmovement is smaller for smaller values of Ns.
  • 36. Heaving in Cohesionless SoilA bottom failure in cohesionless soils may occur because of a piping, orquick, condition if the hydraulic gradient h/L is too large. A flow netanalysis may be used to estimate when a quick condition may occur.Possible remedies are to drive the piling deeper to increase the length ofthe flow path L of Fig or to reduce the hydraulic head h by less pumpingfrom inside the cell. In a few cases it may be possible to use a surchargeinside the cell.Fig. (a) condition for piping or quick, conditions ;(b) conditions for blow in
  • 37. In Fig. 16(b ) the bottom of the excavation may blow in if thepressure head hw indicated by the piezometer is too great, as follows(SF = 1.0):γwhw= γshsThis equation is slightly conservative, since the shear, or walladhesion, on the walls of the cofferdam is neglected. On the otherhand, if there are soil defects in the impervious layer, the blow-inmay be local; therefore, in the absence of better data, the equalityas given should be used. The safety factor is defined as
  • 38. BJERRUM AND EIDE (1956) METHOD OF ANALYSISThis method of analysis is applicable in the caseswhere the braced cuts are rectangular, square orcircular in plan or the depth of excavation exceeds thewidth of the cut.In this analysis the braced cut is visualizedas a deep footing whose depth and horizontaldimensions are identical to those at the bottom of thebraced cut. The theory of Skempton for computing Nc(bearing capacity factor) for different shapes of footingis made use of.
  • 39. Figure gives values of Nc as a function of H/B forlong, circular or square footings. For rectangularfootings, the value of Nc may be computed by theexpression:
  • 40. NC (rect) = (0.84 + 0.16B/L) NC (sq)Where,L = length of excavation.B = width of excavationThe factor of safety for bottom heave may be expressed asWhere,q, is the uniform surcharge load.
  • 41. Design of Various Components of BracingStruts: The strut is a compression member whose load-carryingcapacity depends upon slenderness ratio, l/r. The effective length ‘l’ ofthe member can be reduced by providing vertical and horizontalsupports at intermediate points. The load carried by a strut can bedetermined from the pressure envelope. The struts should have aminimum vertical spacing of about 2.5 m. In the case of braced cuts inclayey soils, the depth of the first strut below the ground surfaceshould be less than the depth of tensile crack (Zc), which is equal to ,Zc= (2c/ γ)While calculating the load carried by various struts, it is generally assumedthat the sheet piles (or soldier beams) are hinged at all the strut levels expectfor the top and bottom struts.
  • 42. Determination of strut load ;(a) section and plan of cut;(b) Method for determining strut loads
  • 43. Steps Draw the pressure envelope of the braced cut and also show theproposed strut level. Strut levels are marked A, B, C and D. Determine the reactions for two simple cantilever beams (top andbottom) and all the simple beams between. These reactions areA, B1,B2,C1,C2 and D. The strut loads may be calculated asPA= (A) (s)PB= (B1+B2) (s)Pc= (C1+C2) (s)PD= (D) (s)Where PA,PB,PC,PD are the loads to be taken by individual struts atlevels A,B,C and D respectively Knowing the strut loads at each level and intermediate bracingcondition allows selections for the proper selection from the steelconstruction manual.
  • 44. WALESThey are considered as horizontal beams pinned at strut levels. Themaximum bending moment will depend upon the span “s” and loads on thestruts. As the strut loads are different at various levels, maximum bendingmoment would also be different.At level A, Mmax : (A)(s2)/8At level B, Mmax : (B1 +B2)(s2)/8At level C, Mmax : (C1 + C2)(s2)/8At level D, Mmax : (D)(s2)/8once the maximum bending moment has been computed, the section modulusis computed as,S= (Mmax)/( σall)σall = Allowable bearing stress.
  • 45. SHEET PILESSheet piles act as vertical plates supported at strut levels.The maximum bending moments in various sections such asA, B, C, D is determined. Once the maximum bending momentshave been computed, the section modulus of the sheet pile can becomputed and the section chosen.
  • 46. Design of Braced Sheeting in Cuts
  • 47. EARTH PRESSURE DISTRIBUTION BEHIND SHEETINGFOR DRY OR MOIST SAND FOR SOFT TO MEDIUM CLAY FOR STIFF CLAY
  • 48. Design of Braced Sheeting in CutsSheet piles are used to retain the sides of thecuts in sands and clays. The sheet piles are keptin position by wales and struts. The first bracelocation should not exceed the depth of thepotential tension cracks.245tan20chSince the formation of cracks will increase thelateral pressure against the sheeting and if thecracks are filled with water, the pressure will beincreased even more. The sheeting of a cut isflexible and is restrained against deflection at thefirst series of struts. The deflection, therefore, islikely to be as shown in Fig. (a). The pressuredistribution on sheet pile walls to retain sandy soiland clay soil are shown in Figs. (b) and (c)respectively.abc
  • 49. Design1. The sheet pile is considered as continuous beamsupported on wales either cantilevered attop, fixed, partially fixed, hinged, orcantilevered at the bottom depending upon theamount of penetration below the excavationline.2. Bending moment and shearing force diagramare then obtained using moment distributionmethod.3. Section of the sheet pile is then designed in theconventional way for the maximum bendingmoment.A fast way of designing sheeting is to assumeconditions as shown in Fig. (d). The top istreated as a cantilever beam including the firsttwo struts. The remaining spans between strutsare considered as simple beams with a hinge orcantilever at the bottom.Struts are designed as columns subjected to anaxial force. The wales as continuous membersor simply supported members pinned at thed
  • 50. EXAMPLE
  • 51. #1 A long trench is excavated in medium dense sand for the foundation of amulti-storey building. The sides of the trench are supported with sheet pilewalls fixed in place by struts and wales as shown in figure below. The soilproperties are:γ = 18.5KN/m3; c=0 ; Ф = 38oDetermine:(a) The pressure distribution on the walls with respect to depth.(b) Strut loads. The struts are placed horizontally at distances L =4 m centre to centre.(c) The maximum bending moment for determining the pile wallsection.(d) The maximum bending moments for determining the section ofthe wales.
  • 52. (a) For a braced cut in sand use the apparent pressure envelope givenin Fig. 20.28 b. The equation for pa ispa = 0.65 γ H KA = 0.65 x 18.5 x 8 tan2 (45 - 38/2) = 23 kN/m2b) Strut loadsThe reactions at the ends of struts A, B and C are represented by RA, RBand Rc respectivelyFor reaction RA , take moments about BRA x3 = 4x23x4/2or RA = 184/3 = 61.33 kNRB1 = 23 x 4 - 61.33 = 30.67 kNDue to the symmetry of the load distribution,RB1 = RB2 = 30.67 kN, and RA = Rc = 61.33 kN.Now the strut loads are (for L = 4 m)Strut A, PA = 61.33 x 4 = 245 kNStrut B, PB = (RB1 + RB2) x 4 = 61.34 x 4 = 245 kNStrut C, Pc = 245 kN
  • 53. (c)Moment of the pile wall sectionTo determine moments at different points it is necessary to draw a diagramshowing the shear force distribution.Consider sections DB1 and B2E of the wall in Fig. (b). The distribution ofthe shear forces are shown in Fig. (c) along with the points of zero shear.The moments at different points may be determined as followsMA = 0.5 x 1 x 23 = 11.5 kN- mMc = 0.5x 1 x 23 = 1 1.5 kN- mMm = 0.5 x 1.33 x 30.67 = 20.4 kN- mMn =0.5 x 1.33 x 30.67 = 20.4 kN- mThe maximum moment Mmax = 20.4kN-m. A suitable section of sheet pilecan be determined as per standard practice.(d) Maximum moment for walesThe bending moment equation for wales isMmax = (RL2)/8Where R = maximum strut load = 245 kNL = spacing of struts = 4 mMmax = (245 x 42)/8 = 490 kN-mA suitable section for the wales can be determined as per standardpractice.
  • 54. #2 The cross section of a long braced cut is shown in Figurea. Draw the earth-pressure envelope.b. Determine the strut loads at levels A, B, and C.c. Determine the section modulus of the sheet pile section required.d. Determine a design section modulus for the wales at level B.e. Calculate the factor of safety against heave (L= 20m, T= 1.5m and q= 0)(Note: The struts are placed at 3 m, centre to centre, in the plan.) Use σall = 170 x103 kN/m2γ = 18 kN/m2; c= 35 kN/m2 and H= 7m
  • 55. (a) Given: γ = 18 kN/m2; c= 35 kN/m2 and H= 7m. So ,Thus , the pressure envelope will be like the one as shown in previous figureand the maximum pressure intensity Pa,Pa = 0.3 γ H = (0.3) (18) (7) = 37.8 kN/m2(b)To calculate the strut load, examine fig. b, taking moments abountB1, we have MB1 = 0,RA x 2.5 – 0.5 x 37.8 x 1.75 x (1.75 + 1.75/3 ) – 1.75 x 37.8 x (1.75/2) = 0RA = 54.02 kN/mAlso vertical forces = 0 . Thus,0.5 x 1.75x 37.8 + 37.8 x 1.75 = RA + RB1Therefore, RB1 = 45.2 kN/mDue to symmetry,RB2 = 45.2kN/mRC = 54.02 kN/m
  • 56. Hence the horizontal strut loads ,Pa = RA x Horizontal spacing(s) = 54.02 x 3 = 162.06 kNPb = (RB1+ RB2 ) x s = (45.02 +45.02) x 3 = 271.2 kNPc =RC x s = 54.02 x 3 = 162.06kN(c) Location of the point of maximum moment, i.e. shear force is zero,37.8(x) = 45.2Therefore, x= 1.196m from BMoment at A= 0.5x1x (37.8/1.75 x1) x 0.33 = 3.6 kN-m/m of wallMoment at E (point of maximum moment)= 45.2x 1.196 – 37.8 x 1.196x 1.196/2 = 27.03kN-m/m of wall.Therefore section modulus of sheet pileS= (Mmax / σall) = (27.03)/(170x 103)= 15.9 x 10-5m3/m of wall
  • 57. (d) the reaction at level B has been calculated in part b. Hence,Mmax =(RB1 + RB2)S2 / 8 = (45.2 + 45.2)32 / 8 = 101.7 kN-mAnd section modulus s = = (Mmax / σall) = (101.7)/(170x 103)= 0.598x 10-3 m3(e)Factor of safety against heave is given by the equation,Where,L=20m; c=35kN/m2 ;Nc = 5.7 ; γ = 18 kN/m2 ; H= 7m ; B=3m ;q=0B/ = 3/ = 2.12mB’=T = 1.5mB’’= = 1.5x = 2.12m
  • 58. Braced cut showing arrangement of sheet piles, wales and struts
  • 59. The wall is supported by “rakers,” or inclined struts. Thebottom ends of the rakers are raced against the centralpart of the building foundation slab.

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