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- 1. Under the guidance of Prof. Kalyan Kumar Chattopadhyay Submitted By: Yogesh Kr Pandey Exam roll: 110904034
- 2. Introduction An excavation supported by suitable bracing system are called braced 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 to neighboring structures or to utilities in the surrounding ground.
- 3. The design of braced cuts involves two distinct but interrelated features, namely Stability of excavation, ground movement, control of water into the excavation, effect of adjoining structures and so on. Design of structural elements i.e sheet pile, struts or anchors and so forth.
- 5. Type I use of soldier beams •Soldier beam is driven into the ground before excavation and is a vertical steel or timber beam. •Laggings, which are horizontal timber planks, are placed between soldier beams as the excavation proceeds. •When the excavation reaches the desired depth, wales and struts (horizontal steel beams) are installed. The struts are horizontal compression members.
- 6. 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 reaches the appropriate depth.
- 7. DIFFERENT TYPES OF SHEETING AND BRACING SYSTEMS
- 8. Vertical Timber Sheeting: Vertical timber sheeting consisting of planks about 8 to 10 cm thick are driven around the boundary of the proposed excavation to some depth below the base of the excavation. The soil between the sheeting is then 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. The wales are supported by horizontal struts which extend from side to side of the excavation. However, if the excavations are relatively wide, it becomes economical to support the wales by inclined struts, known as rakers. For inclined struts to be successful, it is essential that the soil at the base of the excavation be strong enough to provide adequate reaction. If the soil can be temporarily support itself an excavation of limited depth without an external support, the timber sheeting can be installed in the open or in a partially completed excavation. Vertical timber sheeting is economical up to a depth of 4 to 6 m.
- 10. Steel Sheet Pile: In this method, the steel sheet piles are driven along the sides of the proposed excavation. As the soil is excavated from the enclosure, wales and struts are placed. The wales are made of steel. The struts may be of steel or wood. As the excavation progresses, another set of wales and struts is inserted. The process is continued till the excavation is complete. It is recommended that the sheet piles should be driven several meters below the bottom of excavation to prevent local heaves. If the width of a deep excavation is large, inclined bracing may be used. Steel sheet pile
- 11. Soldier Beams: Soldier beams are H-piles which are driven at a spacing of 1.5 to 2.5 m around the boundary of the proposed excavation. As the excavation proceeds, horizontal timber planks called laggings are placed between the soldier beams. When the excavation advances to a suitable depth, wales and struts are inserted. The lagging is properly wedged between the pile flanges or behind the back flange. Soldier Beam
- 12. Tie Backs: In this method, no bracing in the form of struts or inclined rakers is provided. Therefore, there is no hindrance to the construction activity to be carried out inside the excavated area. The tie back is a rod or a cable connected to the sheeting or lagging on one side and anchored into soil (or rock) outside the excavation area. Inclined holes are drilled into the soil (or rock), and the hole is concreted. An enlargement or a bell is usually formed at the end of the hole. Each tie back is generally prestressed the depth of excavation is increased further to cope with the increased tension.
- 13. Use of Slurry Trenches: An alternative to use of sheeting and bracing system, which is being increasingly used these days, is the construction of slurry trenches around the area to be excavated and is kept filled with heavy, viscous slurry of a bentonite clay-water mixture. The slurry stabilizes the walls of the trench, and thus the excavation can be done without sheeting and bracing. Concrete is then placed through a tremie. Concrete displaces the slurry. Reinforcement can also be placed before concreting, if required. Generally, the exterior walls of the excavation are constructed in a slurry trench. Slurry Trench
- 14. Lateral Earth Pressure Distribution
- 15. Lateral earth pressure is the pressure that soil exerts against a structure in a sideways, mainly horizontal direction. Since most open cuts are excavated in stages within the boundaries of sheet pile walls or walls consisting of soldier piles and laggings and since struts are inserted progressively as the excavation proceeds, the walls are likely to deform (as shown in figure below). Little inward movement can occur at the top of the cut after the first strut is inserted Typical pattern of deformation of vertical wall (Braced cuts)
- 16. EARTH PRESSURE DISTRIBUTION • In Sand • In Clay
- 17. In Sand Following figures shows various recommendations for earth pressure distribution behind sheeting This pressure, pa may be expressed as 0.8γHKa Terzaghi and Peck’s Peck’s earth pressure distribution for loose sand
- 18. 0.8γHKa Terzaghi and Peck’s earth pressure distribution for dense sand 0.8γHKa Tschebotarioff’s Peck’s earth pressure distribution H H
- 19. 0.65γHKa Peck, Hansen and Thornburn’s Peck’s earth pressure distribution for moist and dry sands H Where,, γ= unit weight H= height of the cut Ka= Rankine’s active pressure coefficient.
- 20. Cuts in Clay The given figures represent the different earth pressure distribution recommendations for clay. In clay braced cuts becomes unstable due to bottom heave .To ensure the stability of braced system γH/cb must be kept less than 6, where γH/cb is the undrained shear strength of soil below base or excavation level. For plastic clay by Peck
- 21. H Neutral earth pressure ratio method by Tschebotarioff H Peck, Hanson and Thornburn’s diagram when γH/c ≤ 4
- 22. H Peck, Hanson and Thornburn’s diagram when γH/c > 4 As the most probable value of any individual strut load is about 25 percent lower than the maximum as obtained from Peck, Hanson and Thornburn’s earth pressure distribution theories, so among all the given earth pressure distribution profiles, Peck, Hanson and Thornburn’s earth pressure distribution theories are most widely and popularly used
- 23. Pressure envelope for cuts in layered soil Sometimes, layers of both sand and clay are encountered when a braced cut is being constructed. In this case, Peck (1943) proposed that an equivalent value of cohesion should be determined according to the formula, cav = [γsKsHs 2tanФs’ + (H-HS)n’qu] The average unit weight of the layers may be expressed as, γa H =[ γs Hs + ( H - HS ) γc ]
- 24. Where, H = total height of cut γs = unit weight of sand Ks = lateral earth pressure coefficient for sand layer ( =1 ) Hs = height of sand layer Фs’= effective angle of friction of sand qu= unconfined compression strength of clay n’= a coefficient of progressive failure (ranging from 0.5 to 1.0 average value 0.75) γc = saturated unit weight of clay layer •Once the average values of cohesion and unit weight are determined, the pressure envelopes in clay can be used to design the cuts
- 25. Similarly, when several clay layers are encountered in the cut, the average undrained cohesion becomes Cav = (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,...,n H1 , H2 ,..., Hn = thickness of layers 1, 2, ... , n
- 26. Limitations of the Pressure Envelope When using the pressure envelope just described, following points are to be noted: The pressure envelopes are sometimes referred to a apparent pressure envelope. However, the actual pressure distribution is a function of the construction 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 bottom of the cut. Sand is assumed to be drained with zero pore water pressure. Clay is assumed to be undrained and pore water pressure is not considered
- 27. LOADS ON BRACES Tributary Area Method Equivalent Beam Method
- 28. Tributary Area Method The load on a strut is equal to the load resulting from 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.
- 29. Equivalent Beam Method: In this method entire depth is split into segments of simply supported beams and reactions can then be determined by standard process.
- 30. STABILITY OF BRACED CUTS
- 31. Heaving in Clay Soil The danger of heaving is greater if the bottom of the cut is soft clay. Even in a soft clay bottom, two types of failure are possible. They are Case 1: When the clay below the cut is homogeneous at least up to a depth equal 0.7 B where B is the width of the cut. Case 2: When a hard stratum is met within a depth equal to 0.7 B.
- 32. Case:1 The anchorage load block of soil a b c d in Fig. (a) of width (assumed) at the level of the bottom of the cut per unit length may be expressed as The vertical pressure q per unit length of a horizontal, ba, is
- 33. The bearing capacity qu per unit area at level ab is qu = Ncc = 5.7c Where, Nc =5.7 The factor of safety against heaving is Because of the geometrical condition, it has been found that the width cannot exceed 0.7 B . Substituting this value for , For excavations of limited length L, the factor of safety can be modified to
- 34. Where, B’=T (thickness of clay below the base of excavation) or B/ (whichever is smaller) In 2000, Chang suggested few revisions, with his modification, equation takes the form B’=T if T B/ B’ = B/ if T> B/ B’’= B’
- 35. Case: 2 Replacing 0.75B by D in Eq, the factor of safety is represented by For a cut in soft clay with a constant value of cu below the bottom of the cut, D in Eq. becomes large, and Fs approaches the value
- 36. Ns is termed as Stability Number. The stability number is a useful indicator of potential soil movements. The soil movement is smaller for smaller values of Ns.
- 37. Heaving in Cohesionless Soil A bottom failure in cohesionless soils may occur because of a piping, or quick, condition if the hydraulic gradient h/L is too large. A flow net analysis may be used to estimate when a quick condition may occur. Possible remedies are to drive the piling deeper to increase the length of the flow path L of Fig or to reduce the hydraulic head h by less pumping from inside the cell. In a few cases it may be possible to use a surcharge inside the cell. Fig. (a) condition for piping or quick, conditions ;(b) conditions for blow in
- 38. In Fig. 16(b ) the bottom of the excavation may blow in if the pressure head hw indicated by the piezometer is too great, as follows (SF = 1.0): γwhw= γshs This equation is slightly conservative, since the shear, or wall adhesion, on the walls of the cofferdam is neglected. On the other hand, if there are soil defects in the impervious layer, the blow-in may be local; therefore, in the absence of better data, the equality as given should be used. The safety factor is defined as
- 39. BJERRUM AND EIDE (1956) METHOD OF ANALYSIS This method of analysis is applicable in the cases where the braced cuts are rectangular, square or circular in plan or the depth of excavation exceeds the width of the cut. In this analysis the braced cut is visualized as a deep footing whose depth and horizontal dimensions are identical to those at the bottom of the braced cut. The theory of Skempton for computing Nc (bearing capacity factor) for different shapes of footing is made use of.
- 40. Figure gives values of Nc as a function of H/B for long, circular or square footings. For rectangular footings, the value of Nc may be computed by the expression:
- 41. NC (rect) = (0.84 + 0.16B/L) NC (sq) Where, L = length of excavation. B = width of excavation The factor of safety for bottom heave may be expressed as Where, q, is the uniform surcharge load.
- 42. Design of Various Components of Bracing Struts: The strut is a compression member whose load-carrying capacity depends upon slenderness ratio, l/r. The effective length ‘l’ of the member can be reduced by providing vertical and horizontal supports at intermediate points. The load carried by a strut can be determined from the pressure envelope. The struts should have a minimum vertical spacing of about 2.5 m. In the case of braced cuts in clayey soils, the depth of the first strut below the ground surface should 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 assumed that the sheet piles (or soldier beams) are hinged at all the strut levels expect for the top and bottom struts.
- 43. Determination of strut load ;(a) section and plan of cut; (b) Method for determining strut loads
- 44. Steps Draw the pressure envelope of the braced cut and also show the proposed strut level. Strut levels are marked A, B, C and D. Determine the reactions for two simple cantilever beams (top and bottom) and all the simple beams between. These reactions are A, B1,B2,C1,C2 and D. The strut loads may be calculated as PA= (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 at levels A,B,C and D respectively Knowing the strut loads at each level and intermediate bracing condition allows selections for the proper selection from the steel construction manual.
- 45. WALES They are considered as horizontal beams pinned at strut levels. The maximum bending moment will depend upon the span “s” and loads on the struts. As the strut loads are different at various levels, maximum bending moment would also be different. At level A, Mmax : (A)(s2)/8 At level B, Mmax : (B1 +B2)(s2)/8 At level C, Mmax : (C1 + C2)(s2)/8 At level D, Mmax : (D)(s2)/8 once the maximum bending moment has been computed, the section modulus is computed as, S= (Mmax)/( σall) σall = Allowable bearing stress.
- 46. SHEET PILES Sheet piles act as vertical plates supported at strut levels. The maximum bending moments in various sections such as A, B, C, D is determined. Once the maximum bending moments have been computed, the section modulus of the sheet pile can be computed and the section chosen.
- 47. Design of Braced Sheeting in Cuts
- 48. EARTH PRESSURE DISTRIBUTION BEHIND SHEETING FOR DRY OR MOIST SAND FOR SOFT TO MEDIUM CLAY FOR STIFF CLAY
- 49. Design of Braced Sheeting in Cuts Sheet piles are used to retain the sides of the cuts in sands and clays. The sheet piles are kept in position by wales and struts. The first brace location should not exceed the depth of the potential tension cracks. 2 45tan 2 0 c h Since the formation of cracks will increase the lateral pressure against the sheeting and if the cracks are filled with water, the pressure will be increased even more. The sheeting of a cut is flexible and is restrained against deflection at the first series of struts. The deflection, therefore, is likely to be as shown in Fig. (a). The pressure distribution on sheet pile walls to retain sandy soil and clay soil are shown in Figs. (b) and (c) respectively. a b c
- 50. Design 1. The sheet pile is considered as continuous beam supported on wales either cantilevered at top, fixed, partially fixed, hinged, or cantilevered at the bottom depending upon the amount of penetration below the excavation line. 2. Bending moment and shearing force diagram are then obtained using moment distribution method. 3. Section of the sheet pile is then designed in the conventional way for the maximum bending moment. A fast way of designing sheeting is to assume conditions as shown in Fig. (d). The top is treated as a cantilever beam including the first two struts. The remaining spans between struts are considered as simple beams with a hinge or cantilever at the bottom. Struts are designed as columns subjected to an axial force. The wales as continuous members or simply supported members pinned at the d
- 51. EXAMPLE
- 52. #1 A long trench is excavated in medium dense sand for the foundation of a multi-storey building. The sides of the trench are supported with sheet pile walls fixed in place by struts and wales as shown in figure below. The soil properties are: γ = 18.5KN/m3; c=0 ; Ф = 38o Determine: (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 wall section. (d) The maximum bending moments for determining the section of the wales.
- 54. (a) For a braced cut in sand use the apparent pressure envelope given in Fig. 20.28 b. The equation for pa is pa = 0.65 γ H KA = 0.65 x 18.5 x 8 tan2 (45 - 38/2) = 23 kN/m2 b) Strut loads The reactions at the ends of struts A, B and C are represented by RA, RB and Rc respectively For reaction RA , take moments about B RA x3 = 4x23x4/2 or RA = 184/3 = 61.33 kN RB1 = 23 x 4 - 61.33 = 30.67 kN Due 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 kN Strut B, PB = (RB1 + RB2) x 4 = 61.34 x 4 = 245 kN Strut C, Pc = 245 kN
- 55. (c)Moment of the pile wall section To determine moments at different points it is necessary to draw a diagram showing the shear force distribution. Consider sections DB1 and B2E of the wall in Fig. (b). The distribution of the shear forces are shown in Fig. (c) along with the points of zero shear. The moments at different points may be determined as follows MA = 0.5 x 1 x 23 = 11.5 kN- m Mc = 0.5x 1 x 23 = 1 1.5 kN- m Mm = 0.5 x 1.33 x 30.67 = 20.4 kN- m Mn =0.5 x 1.33 x 30.67 = 20.4 kN- m The maximum moment Mmax = 20.4kN-m. A suitable section of sheet pile can be determined as per standard practice. (d) Maximum moment for wales The bending moment equation for wales is Mmax = (RL2)/8 Where R = maximum strut load = 245 kN L = spacing of struts = 4 m Mmax = (245 x 42)/8 = 490 kN-m A suitable section for the wales can be determined as per standard practice.
- 56. #2 The cross section of a long braced cut is shown in Figure a. 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 x 103 kN/m2 γ = 18 kN/m2; c= 35 kN/m2 and H= 7m
- 57. (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 figure and 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 abount B1, 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) = 0 RA = 54.02 kN/m Also vertical forces = 0 . Thus, 0.5 x 1.75x 37.8 + 37.8 x 1.75 = RA + RB1 Therefore, RB1 = 45.2 kN/m Due to symmetry, RB2 = 45.2kN/m RC = 54.02 kN/m
- 58. Hence the horizontal strut loads , Pa = RA x Horizontal spacing(s) = 54.02 x 3 = 162.06 kN Pb = (RB1+ RB2 ) x s = (45.02 +45.02) x 3 = 271.2 kN Pc =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) = 45.2 Therefore, x= 1.196m from B Moment at A= 0.5x1x (37.8/1.75 x1) x 0.33 = 3.6 kN-m/m of wall Moment 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 pile S= (Mmax / σall) = (27.03)/(170x 103)= 15.9 x 10-5m3/m of wall
- 59. (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-m And 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=0 B/ = 3/ = 2.12m B’=T = 1.5m B’’= = 1.5x = 2.12m
- 60. Braced cut showing arrangement of sheet piles, wales and struts
- 61. The wall is supported by “rakers,” or inclined struts. The bottom ends of the rakers are raced against the central part of the building foundation slab.