Chapter Three Lateral Earth Pressure PDF

2024-10-07 CIVL 4111 Geotechnical Engineering Chapter Three Lateral Earth Pressure By: Hany El Naggar, Ph.D., P.Eng. Professor, Department of Civil and Resource Engineering Dalhousie University...

2024-10-07 CIVL 4111 Geotechnical Engineering Chapter Three Lateral Earth Pressure By: Hany El Naggar, Ph.D., P.Eng. Professor, Department of Civil and Resource Engineering Dalhousie University 1 1 Lateral earth pressure  Lateral earth pressure is the pressure that soil exerts in the horizontal plane. Retaining walls, sheet-pile walls, basement walls, braced excavations and tunnels are common examples of structures which must be designed to withstand the pressure due to the retained soil or fill.  Retaining walls are structures that support backfill and allow for a change of grade. For instance a retaining wall can be used to retain fill along a slope or it can be used to support a cut into a slope as shown the Figure below. Retaining wall to support a fill Retaining wall to support a cut + DQ \(O1 DJJDU3K' 3(QJ  2 1  2024-10-07 Types of Earth-Retaining Structures Earth-retaining structures may be broadly classified as retaining walls and sheetpile walls.  Retaining walls may be further classified as: i. Gravity retaining walls  usually of masonry or mass concrete. - Gravity walls depend on their weight for stability + DQ \(O1 DJJDU3K' 3(QJ  3 ii. Cantilever walls Usually of reinforced concrete. iii. Counterfort walls  R.C. Cantilever walls have a vertical or inclined stem monolithic with a base slab. These are considered suitable up to a height of 7.5 m. A vertical or inclined stem is used in counterfort walls, supported by the base slab as well as by counterforts with which it is monolithic. + DQ \(O1 DJJDU3K' 3(QJ  4 2  2024-10-07  Sheetpile walls may be further classified as cantilever sheetpile walls and anchored sheet pile walls, also called ‘bulkheads’.  The other types of retaining walls, as well as sheet-pile walls, are known as ‘flexible walls’. + DQ \(O1 DJJDU3K' 3(QJ  5  Cantilever sheet pile walls are held in the ground by the passive resistance of the soil both in front of and behind them. Anchored sheet pile wall or bulkhead is fixed at its base as a cantilever wall but supported by tie-rods near the top, sometimes using two rows of ties and properly anchored to a deadman. + DQ \(O1 DJJDU3K' 3(QJ  6 3  2024-10-07 Common uses of retaining walls + DQ \(O1 DJJDU3K' 3(QJ  7 + DQ \(O1 DJJDU3K' 3(QJ  8 4  2024-10-07 General assumptions for estimating the magnitude of earth pressures o Since, length of the wall >> its width, conditions of plane strain are assumed, i.e. strains in the longitudinal direction of the structure are assumed to be zero. o It is assumed that the stress–strain behaviour of the soil can be represented by the rigid–perfectly plastic idealization. In which both yielding and shear failure occur at the same state of stress. o To simplify the analysis, the limit theorems of plasticity can be used to calculate lower and upper bounds to the true collapse load. In certain cases, the theorems produce the same result which would then be the exact value of the collapse load. + DQ \(O1 DJJDU3K' 3(QJ  9 The limit theorems can be summarized as follows Lower bound theorem If maximum state of stress, which at no point exceeds the failure criterion for the soil and is in equilibrium with a system of external loads (which includes the self-weight of the soil), then collapse cannot occur; the external load system thus constitutes a lower bound to the true collapse load. Upper bound theorem If a mechanism of plastic collapse is assumed and the work done by a system of external loads is equal to the dissipation of energy by the internal stresses, then collapse must occur; the external load system thus constitutes an upper bound to the true collapse load. + DQ \(O1 DJJDU3K' 3(QJ  10 5  2024-10-07  In the lower bound approach, the conditions of equilibrium and yield (as per the Mohr–Coulomb failure criterion) are satisfied without consideration of the mode of deformation.  In the upper bound approach, a slip surface is assumed and the work done by the external forces is equated to the loss of energy by the stresses acting along the slip surface, without consideration of equilibrium.  The true collapse load lies somewhere between the lower and upper bounds. + DQ \(O1 DJJDU3K' 3(QJ  11  For the soil cut shown in the Figure below, the soil is at different elevations on either side of the structure, thus some force is required to support the soil. This force may be provided by: o friction at the base (gravity retaining walls) o founding the wall into the ground (sheet retaining walls) o anchors and struts o external loads F a) At rest + DQ \(O1 DJJDU3K' 3(QJ  12 6  2024-10-07  If the force is too small the soil behind the wall will reach a state of failure with the wall moving away from the soil (active failure). If the force is too large the soil will reach another state of failure with the wall moving into the soil (passive failure). F F b) Active Earth Pressure c) Passive Earth Pressure + DQ \(O1 DJJDU3K' 3(QJ  13  There are 3 states of lateral earth pressure Ko = At Rest Ka = Active Earth Pressure (wall moves away from soil) Kp = Passive Earth Pressure (wall moves into soil; passive is more like a resistance) $ FWLYH F 3 DVVLYH + DQ \(O1 DJJDU3K' 3(QJ  14 7  2024-10-07 At Rest Earth Pressure  At rest earth pressure occur when there is no wall rotation such as in a braced wall (basement wall for example) Ko can be calculated as follows: Ko = 1 – sin φ normally consolidated soils Ko = (1 – sin φ) (OCR)1/2 for OC soils  Distribution of lateral earth pressure at rest on a wall (u=0) + DQ \(O1 DJJDU3K' 3(QJ  15  Distribution of lateral earth pressure at rest for partially submerged soil + DQ \(O1 DJJDU3K' 3(QJ  16 8  2024-10-07 Example 1 The 6.50 m –high retaining wall shown in the Figure below is restrained from yielding or moving. Draw the lateral earth pressure distribution. Also, calculate the magnitude and the location of the lateral force Po per unit length of the wall. Assume that the OCR is 2. c'= 0 4.00 φ’= 30° γ= 15.50 kN/m3 GWT c'= 0 2.50 φ’= 30° γ= 19.00 kN/m3 + DQ \(O1 DJJDU3K' 3(QJ  17 Solution: + DQ \(O1 DJJDU3K' 3(QJ  18 9  2024-10-07 Rankine’s theory Rankine's theory, developed in 1857 is a stress field solution that predicts active and passive earth pressure. It assumes that failure will occur when the maximum principal stress at any point reaches a value equal to the failure criterion of the soil, i.e., lower bound solution. Active Earth Pressure  Active earth pressure occurs when the wall tilts away from the soil (a typical free standing retaining wall).  In Rankine’s theory the wall is assumed to be frictionless. The normal stress acting on the wall will therefore be a principal stress. If the wall is vertical and the soil surface horizontal, the vertical and horizontal stresses throughout the retained soil mass will be principal stresses. + DQ \(O1 DJJDU3K' 3(QJ  19  In the figure shown below: v’ = z Initially, there is no lateral movement h’ = Ko v’ = Ko z As the wall moves away from the soil, v’ remains the same; and h’ decreases till failure occurs. + DQ \(O1 DJJDU3K' 3(QJ  20 10  2024-10-07 For cohesionless soils 𝟏 𝐬𝐢𝐧 𝝋 𝝋 𝑲𝐚 𝒕𝒂𝒏𝟐 𝟒𝟓 𝟏 𝐬𝐢𝐧 𝝋 𝟐 2 θ ∴𝑝 𝜎 𝐾 p θ 90-φ v 2 2 )DLOX UH: HGJH 𝒑𝒂 + DQ \(O1 DJJDU3K' 3(QJ  21 For the case of Cohesive soils For the case of drained conditions use effective stress analysis: ∴ 𝒑𝒂 𝝈 𝒗 𝑲𝒂 𝟐𝒄 𝑲𝒂 where, 𝝈′𝒗 is the effective vertical stress  For the case of undrained conditions use total stress analysis with undrained parameters, 𝑐 , 𝜑 0. ∴ 𝒑𝒂 𝝈𝒗 𝟐𝒄𝒖 , and 𝜽 𝟒𝟓∘ where, 𝝈𝒗 is the total vertical stress + DQ \(O1 DJJDU3K' 3(QJ  22 11  2024-10-07 Passive Earth Pressure  The resistance offered by the soil to an object pushing against it is referred to as "passive pressure”.  In the figure shown below: v’ = z Initially, there is no lateral movement. h’ = Ko v’ = Ko z As the wall moves towards the soil, v’ remains the same; and h’ increases till failure occurs. + DQ \(O1 DJJDU3K' 3(QJ  23 For cohesionless soils τ 𝟏 𝐬𝐢𝐧 𝝋 𝑲𝐩 tan 45 B State of passive failure 𝟏 𝐬𝐢𝐧 𝝋 At rest case, before wall moves 𝟏 𝝈 𝝈𝒗 𝟐 𝒉 ∴ 𝒑𝒑 𝝈 𝒗 𝑲𝒑 A ϕ' 90- ϕ' σh σv C= Centre of circle σh' σ' 𝟏 𝝈 𝝈𝒗 𝟐 𝒉 )DLOX UH: HGJH 𝒑𝒑 + DQ \(O1 DJJDU3K' 3(QJ  24 12  2024-10-07 For the case of Cohesive soils For the case of drained conditions use effective stress analysis: τ B ∴ 𝒑𝒑 𝝈 𝒗 𝑲𝒑 𝟐𝒄 𝑲𝒑 ϕ' ϕ' 2ϴ A ϴ where, 𝝈′𝒗 is the effective vertical stress ϕ' c' σ3' C= Centre σ1' σ' ϴ c' cot ϕ' of circle 𝟏 𝝈 𝝈𝟑 𝟐 𝟏  For the case of undrained conditions use total stress analysis with undrained parameters, 𝑐 , 𝜑 0. ∴ 𝒑𝒑 𝝈𝒗 𝟐𝒄𝒖 where, 𝝈𝒗 is the total vertical stress + DQ \(O1 DJJDU3K' 3(QJ  25  Example of earth pressure distribution Pa and Pp are the resultant active and passive thrusts on the wall + DQ \(O1 DJJDU3K' 3(QJ  26 13  2024-10-07 Depth of tension crack  The figure below shows that the active and passive pressures increase linearly with depth, moreover, when c = 0, triangular distributions are obtained in each case. + DQ \(O1 DJJDU3K' 3(QJ  27  When c is greater than zero, the value of pa (lateral earth pressure) is zero at a particular depth zo. This means that in the active case the soil is in a state of tension between the surface and depth zo. 0 𝜎 𝐾 2𝑐 𝐾 at depth = zo  𝜎 𝛾′𝑧 0 𝛾′𝑧 𝐾 2𝑐 𝐾 2𝑐 𝑧 𝛾′ 𝐾  In practice, the active earth pressure above zo in the tension zone cannot be relied upon to act on the wall, since cracks are likely to develop within the tension zone and should be neglected. + DQ \(O1 DJJDU3K' 3(QJ  28 14  2024-10-07 Surface Surcharge  A surcharge load results from forces that are applied along the surface of the backfill behind the wall. These forces apply additional lateral force along the back of the wall.  If a uniformly distributed surcharge pressure of q per unit area acts over the entire surface of the soil mass (very large area), the vertical stress 𝜎 at any depth is increased to 𝜎 𝑞  Resulting in an additional pressure of Kaq in the active case or Kpq in the passive case, both distributions being constant with depth.  If the loaded area adjacent to the wall is not extensive (acting over a very large area), the resulting additional pressure will not be constant with depth. + DQ \(O1 DJJDU3K' 3(QJ  29 Sloping surface  Till now we applied Rankine theory to cases in which the soil surface is horizontal.  For cases in which the soil surface slopes at a constant angle 𝛽 to the horizontal. It is assumed that the active and passive pressures act in a direction parallel to the sloping surface.  For such cases: 𝒄𝒐𝒔 𝜷 𝒄𝒐𝒔𝟐 𝜷 𝒄𝒐𝒔𝟐 𝝋 𝒄𝒐𝒔 𝜷 𝒄𝒐𝒔𝟐 𝜷 𝒄𝒐𝒔𝟐 𝝋 𝑲𝒂𝒊 and, 𝑲𝒑𝒊 𝒄𝒐𝒔 𝜷 𝒄𝒐𝒔𝟐 𝜷 𝒄𝒐𝒔𝟐 𝝋 𝒄𝒐𝒔 𝜷 𝒄𝒐𝒔𝟐 𝜷 𝒄𝒐𝒔𝟐 𝝋 𝟏 𝟏 𝑷𝑨 𝑲 𝜸𝑯𝟐 𝟐 𝒂𝒊 𝑷𝑷 𝑲 𝜸𝑯𝟐 𝟐 𝑷𝒊 30 15  2024-10-07 COULOMB’S THEORY OF EARTH PRESSURE  Coulomb’s theory (1776), which is an upper-bound solution, considers the stability, as a whole, of the wedge of soil between a retaining wall and an assumed failure plane.  The force between the wedge and the wall surface is determined by considering the equilibrium of forces acting on the wedge.  Using Coulomb’s theory, we can account for the effect of the friction between the wall and the soil. The angle of friction between the wall and the soil is called δ.  Main assumptions of this theory are: o The wedge is a free body; o The failure plane is straight line. 31 Active Earth Pressure For cohesionless soils  For the general case of inclined wall and sloping backfill and accounting for the friction between the wall and the adjacent soil, Ka can be calculated considering the equilibrium of the free-body diagram of the failure wedge as shown below: + DQ \(O1 DJJDU3K' 3(QJ  32 16  2024-10-07 We get: 𝑃 𝐾 𝛾𝐻 in the direction shown in the figure above Where, 𝐾 Note: 𝐾 in the direction shown in the figure above  The active thrust acts at H/3 from the base of the wall, inclined at an angle of δ above the normal to the wall. + DQ \(O1 DJJDU3K' 3(QJ  33 Passive Earth Pressure For cohesionless soils  For the general case of inclined wall and sloping backfill and accounting for the friction between the wall and the adjacent soil, Kp can be calculated considering the equilibrium of the free-body diagram of the failure wedge. + DQ \(O1 DJJDU3K' 3(QJ  34 17  2024-10-07 We get: 𝑃 𝐾 𝛾𝐻 in the direction shown in the figure above Where, 𝐾 Note: 𝐾 in the direction shown in the figure above + DQ \(O1 DJJDU3K' 3(QJ  35  Due to the presence of the wall friction, the passive failure plane is not linear, it takes the form of a log-spiral curves. Consequently, the above equation overestimates 𝐾. However, for values of 𝛿 𝜑′, the error is small, and the above cases can be safely used. For 𝛿 𝜑′, it is recommended that passive pressure coefficient, 𝐾 , should be corrected using the correction factors given in the Table 1(after Caquot and Kerisel, 1948). + DQ \(O1 DJJDU3K' 3(QJ  36 18  2024-10-07 Table 1: Correction factors to be applied to Kp to account for log-spiral slip surface (𝛽 0. 𝜹 𝝓′ 𝝓′ =0.7 =0.6 =0.5 =0.4 =0.3 =0.2 =0.1 =0.0 15 0.96 0.93 0.91 0.88 0.85 0.83 0.80 0.78 20 0.94 0.90 0.86 0.82 0.79 0.75 0.72 0.68 25 0.91 0.86 0.81 0.76 0.71 0.67 0.62 0.57 30 0.88 0.81 0.75 0.69 0.63 0.57 0.52 0.47 35 0.84 0.75 0.67 0.60 0.54 0.48 0.42 0.36 40 0.78 0.68 0.59 0.51 0.44 0.38 0.32 0.26 37 Some general and practical comments  Strains required to mobilize full active and passive pressures: Experimental results:  To mobilize full active earth pressure: Low strains are required to achieve full active pressures. Generally, these strains are commonly achieved in practice. 𝚫 𝑯 % = 0.25% for dense sands 𝚫 𝑯 % = 1.00% for loose sands  To mobilize full passive earth pressure: Large strains are required to mobilize passive earth pressures. 𝚫 𝑯 % = 2 - 4 % for dense sands 𝚫 𝑯 % = 10 – 15 % for loose sands + DQ \(O1 DJJDU3K' 3(QJ  38 19  2024-10-07 Because of the large strains required to mobilize passive pressures, it is not common to mobilize the full passive resistance in the soil. Therefore, a factor of safety is normally applied to the passive resistance. + DQ \(O1 DJJDU3K' 3(QJ  39 Example 2 For the figure shown below, plot the pressure diagrams and find the resultant force F and its location. + DQ \(O1 DJJDU3K' 3(QJ  40 20  2024-10-07 Solution: 41 42 21  2024-10-07 Design of Retaining Walls  A retaining wall is a structure designed and constructed to resist the lateral pressure of soil when there is a desired change in ground elevation that exceeds the angle of repose of the soil.  As mentioned before, earth-retaining structures may be broadly classified as retaining walls and sheet pile walls.  The active pressure increases on the retaining wall proportionally from zero at the upper grade level (or the value of q if there is surface surcharge) to a maximum value at the lowest depth of the wall.  The total pressure or thrust may be assumed to be acting through the centroid of the lateral stress distribution pattern. + DQ \(O1 DJJDU3K' 3(QJ  43 Design Approaches + DQ \(O1 DJJDU3K' 3(QJ  44 22  2024-10-07 Table 2: Global factor of safety commonly used for foundation design (CFEM, 2006) FOS Sliding 1.5 Overturning 2.0 Bearing capacity 3.0 7DEOH* HRWHFKQLFDOUHVLVWDQFHIDFWRUVIRU/6'  & )(0 WK(GLWLRQ  Geotechnical Resistance NBCC CHBDC (2019) (2015) Passive resistance ------- 0.5 Sliding (drained) 0.8 0.8 Sliding (drained) 0.6 0.8 Bearing capacity 0.5 0.5 + DQ \(O1 DJJDU3K' 3(QJ  45 Table 4: Load factors for LSD. Load Factor (𝜶) Horizontal active thrust, Dead load 1.25 Live load 1.5 Vertical component of PA 1.0 Hydrostatic forces 1.10 + DQ \(O1 DJJDU3K' 3(QJ  46 23  2024-10-07 For example, for the Working Stress Design  The designing process of a typical retaining wall involves the following:  Establish all design parameters based upon applicable codes (NBCC, 2015; CHBDC, 2019; CFEM, 2006).  Compute all applied loads (active, passive or at rest)  Calculate the vertical (Rv) and horizontal (Rh) components of the resultant force, R, and its location.  Check safety against sliding along the base (FOS > 1.5)  Check safety against overturning about toe (FOS > 2)  Check bearing capacity (FOS > 3) + DQ \(O1 DJJDU3K' 3(QJ  47 Safety check against sliding along the base  For the sliding check ignore the passive resistance xb 𝐹𝑂𝑆 x1 Wb Where, Rv and Rh = the vertical and horizontal W1 PA Xf components of the resultant. Wf W2 ya δ = soil-concrete friction angle  0.5 – 0.7  PP yp Rh x2 Safety check against overturning about toe R Rv Xr 𝑃 𝑦 ∑𝑊𝑥 𝐹𝑂𝑆 𝑃𝑦 Where, Wi = weight of block i, xi = horizontal distance of the centroid of block i from toe + DQ \(O1 DJJDU3K' 3(QJ  48 24  2024-10-07 Safety check of bearing capacity  determine the stress distribution at the base of the wall, and check that the stresses does not exceed the allowable bearing capacity of the soil (we will know how to calculate the bearing capacity next chapter). Where, 𝑹𝑽 𝟔𝒆 𝑷𝒎𝒂𝒙 𝟏 𝑩 𝑩 𝑹𝑽 𝟔𝒆 𝑷𝒎𝒊𝒏 𝑩 𝟏 𝑩 To ensure that the stresses are always compression (𝑷𝒎𝒊𝒏 𝟎), the resultant must act within the middle third of the base (i.e., 𝒆 𝑩⁄ ). 𝟔 + DQ \(O1 DJJDU3K' 3(QJ  49 Table 5: Angles and coefficients of friction and adhesion between dissimilar materials 50 25  2024-10-07 Example 3 Check the stability of the retaining wall shown in the Figure below. Hint: use only 1/2 of the passive resistance + DQ \(O1 DJJDU3K' 3(QJ  51 Solution: 52 26  2024-10-07 53 54 27  2024-10-07 Sheet Pile Walls It is made of a row of interlocking, continuous vertical pile segments driven to form an essentially straight wall as shown in the figure below. A sheet pile wall (cantilever or anchored) sustains a difference in soil surface elevation from one side to the other. The change in soil surface elevations may be produced by excavation, backfilling, or a combination of both. + DQ \(O1 DJJDU3K' 3(QJ  55 Cantilever walls  In the case of a cantilevered wall, sheet piling is driven to a sufficient depth into the ground to become fixed as a vertical cantilever in resisting the lateral active earth pressure. This type of wall is suitable for moderate height up to 3-5 m.  Walls designed as cantilevers usually undergo large lateral deflections and are readily affected by scour and erosion in front of the wall.  Lateral support for a cantilevered wall comes from passive pressure exerted on the embedded portion of the wall. 56 28  2024-10-07 57 58 29  2024-10-07 59  The mode of failure is by rotation about a point O near the lower end of the wall as shown in the Figure below.  At failure, passive resistance acts in front of the wall above O and behind the wall below O, as shown in Figure b below.  The design is generally based on the simplification shown in Figure c, it is being assumed that the passive force below point O is represented by the concentrated force R acting at a point C, slightly below O, at a depth d below the lower elevation.  So, to determine the depth d, equate the moments about point C to zero (apply FOS to the restoring moments). + DQ \(O1 DJJDU3K' 3(QJ  60 30  2024-10-07 Example 4 A cantilever sheet pile wall is to be used to retain the sides of an excavation 2.5 m deep in sand as shown in the figure below. The unit weight of the sand is 18 kN/m3 above the GWT, and 20 kN/m3 below it. Assuming that there is no friction between the wall and the soil determine the embedment depth using a FOS of 2. Solution  For 𝜑 37°  𝐾 0.25 and 𝐾 4 Also, reduce the soil level in front of the wall by 10%  0.25 m, thus the new retained height will be 2.75 m, whereas the depth of GWT below the lower level will be 0.75 m only. In addition, surface surcharge of 10 kPa is assumed to act behind the wall. See the figure in the next slide for details 61 62 31  2024-10-07 63 Anchored sheet pile walls  It is assumed that the depth of embedment is not enough to provide fixity, thus the tip of the wall is reasonably assumed to be free (i.e., SF and BM =0).  The designing process anchored sheet pile walls involves the following:  Establish all design parameters based upon applicable codes (NBCC, 2005; CHBDC, 2000; CFEM, 2006).  Compute the active thrust and the mobilized passive resistance ( ).  Take moments about the anchorage point to determine d.  Take ∑ 𝐹 0 to determine T.  Find the value of the maximum bending moment (at the zero-shear location) + DQ \(O1 DJJDU3K' 3(QJ  64 32  2024-10-07 Design of anchors The rods supporting the SPW are normally anchored to concrete blocks behind the wall. These anchors develop there resistance primarily from the mobilized passive resistance. While the friction along their base is usually ignored.  The anchor must be placed far enough behind the wall to ensure that the active failure surface of the wall does not intersect with the passive failure plane of the anchor, so that full passive resistance can be mobilized. For length of anchor, L, along the wall: The net passive resistance of the anchor ≥ force in the tie rod  Also, assume that the surface 𝟏 𝑲𝒑 𝟏 𝟐 𝑭𝑶𝑺 𝜸𝒅𝟐𝒂 𝑲𝒂 𝒒 𝒅𝒂 𝑲 𝜸𝒅𝟐𝒂 𝟐 𝒂 𝑳 𝑭𝑶𝑺 𝑻 𝑺 surcharge is present only behind the anchor and that there is no surface Where, L = anchor length along the wall, surcharge in front of the anchor. S = spacing between tie rods, T = force in the tie rod. + DQ \(O1 DJJDU3K' 3(QJ 65 Example An anchored sheet pile wall is to be used to retain the sides of an excavation 8 m deep in sand as shown in the figure. The unit weight of the sand is 18 kN/m3 above the GWT, and 20 kN/m3 below it. Assuming that there is no friction between the wall and the soil, determine: a) The embedment depth using a FOS of 2. b) The force in each tie, knowing that the ties are spaced 2.4 m apart. Solution  For 𝜑 37°  𝐾 0.25 and 𝐾 4 + DQ \(O1 DJJDU3K' 3(QJ  66 33  2024-10-07 H D H D  HS 3 D H D  3 D H D  HS H D H S 3 D 3 3 3 D H S H D + DQ \(O1 DJJDU3K' 3(QJ  67 68 34  2024-10-07 69 Braced Excavation  In many construction projects deep excavations must be made before the structure can be built. Temporary earth retaining structures are used to allow the sides of excavation to be cut vertical or near vertical. Sheet pilling or H-piling with timber logging are commonly used.  In this case stability is provided by struts extending across the excavation.  Neither Rankine nor Coulomb can be used in braced excavation because the mode of failure is different. Thus, we need to use empirical design procedures. + DQ \(O1 DJJDU3K' 3(QJ  70 35  2024-10-07 For the case of excavation in sand + DQ \(O1 DJJDU3K' 3(QJ  71 For the case of excavation in clay The lateral earth pressure in the case of clay depends on the stability number, N, which under the undrained condition depends on soil cohesion and is defined as: 𝑁 If N < 4 (i.e., most of the clay is in the If N > 4 (i.e., plastic zone exist near the base elastic equilibrium state) stiff to hard clays of the excavation) 0.2 for short durations; 0.4 for long durations m  1.0 for most clays; 0.4 for very soft, 0.3 is most commonly used. normally consolidated clays (cu < 20 kPa) 72 36  2024-10-07 Example The Figure below shows a bracing system for an open cut in sand. Determine the strut loads. Assume that ɣ=18 kN/m3, φ’= 30˚, and δ = 10˚. + DQ \(O1 DJJDU3K' 3(QJ  73 + DQ \(O1 DJJDU3K' 3(QJ  74 37  2024-10-07 + DQ \(O1 DJJDU3K' 3(QJ  75 38