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`Earth Retaining Structures· Earth retaining structures are common in a manmade environment. · These structures were amongst the first to be analyzed using the concepts of mechanics. · One of the most popular earth retaining structure is a retaining wall. · In this lecture, we'll learn the theory of earth pressure that is used to analyze the stability of retaining walls. · It will help you a great deal if you revised the following important concepts:­ Static equilibrium, Effective Stress, Mohr's Stress Circle and Shear Strength of Soils2Lateral Earth Pressure ­ Theories of Rankine and CoulombLecture No. 15 November 12, 2002Definition of Key Terms· Active earth pressure coefficient (Ka): It is the ratio of horizontal and vertical principal effective stresses when a retaining wall moves away (by a small amount) from the retained soil. · Passive earth pressure coefficient (Kp): It is the ratio of horizontal and vertical principal effective stresses when a retaining wall is forced against a soil mass. · Coefficient of earth pressure at rest (K0): It is the ratio of horizontal and vertical principal effective stresses when the retaining wall does not move at all, i.e. it is &quot;at rest&quot;.3Lateral Earth Pressure ­ Basic Concepts· We will consider the lateral pressure on a vertical wall that retains soil on one side. · First, we will consider a drained case, i.e. the shear strength of the soil is governed by its angle of friction . · In addition, we will make the following assumptions:­ The interface between the wall and the soil is frictionless. ­ The soil surface is horizontal and there are no shear stresses on horizontal and vertical planes, i.e. the horizontal and vertical stresses are principal stresses. ­ The wall is rigid and extends to an infinite depth in a dry, homogenous, isotropic soil mass. ­ The soil is loose and initially in an at-rest state.4Basic Concepts (Continued..)· Consider the wall shown in the figure on the top right. If no movement of the wall takes place, the soil is at rest and the vertical and horizontal effective stresses acting on elements A and B are:Active Failure· If the wall is moved away from element A and towards element B, the effective horizontal stress in element A will reduce but the effective vertical stress will remain constant. · Therefore, the Mohr's circle will expand as shown in the figure at the bottom right. · When the Mohr's circle touches the failure surface, element A will undergo active failure.5   =  1 = z z  x =   = K 0 1 = K 0 z 3K0 ­ Coefficient of Earth Pressure at rest · Mohr's circle for the at-rest stress state is shown in the figure on the bottom right.'x'z''z 'xf'6· Since the soil is not at failure, the Mohr's circle stays within the failure surface boundaries.Active Earth Pressure Coefficient (Ka)· Considering triangle OFC in the figure on the right:sin = (  -   ) (  +  xf ) zf xf zf· Rearranging the above equation, we can obtain an expression for the active earth pressure coefficient (Ka) as:O( zf -  xf ) 2F C'zf'xf' 1 - sin   xf =  1 + sin  = K a     zf( zf +  xf ) 27·Since the Mohr's circle at Passive Failure  wall is moved towards B, its effective horizontal stress will 'xf 'x increase but 'zf the effective ' vertical stress will remain constant. ·Hence, the Mohr's circle will first contract and then expand as shown in the figure. 8Passive FailurePassive Earth Pressure Coefficient (Kp)· Considering triangle OFC in the figure on the right:Slip Planes for Active and Passive FailureFElement AElement Bp = 45 - 2sin = ( xf -   ) (  +  xf ) zf zf· The above equation can be rearranged O to obtain an expression for the passive earth pressure coefficient   xf (Kp) as: zf[w.r.t. horizontal]( xf -  zf ) 2'xC(90 + ) (90 ­ )'xf'zf'O Pole for active failurep'zf a'Pole for passive failure( zf +  xf ) 2 1 + sin  1 =  1 - sin  = Kp = K    a9a = 45 + [w.r.t. horizontal] 210Slip Planes (Continued..)Required Horizontal Movement· Much larger rotation is required to produce slip planes for passive failure as compared to the active failure as shown in · The soil mass at the back of the wall is the figure assisting in failure while the soil mass on the left. in the front of the wall is resisting 12 failure.· There are always two sets of slip planes ­ one for positive shear stress and the other for negative shear stress.11Variation of Lateral Earth Pressure with DepthLateral Stresses due to Groundwater· If groundwater is present, the pore water pressure must be added to the lateral z earth pressure. zw Pw · DO NOT multiply the pore water pressure with a zw/3 coefficient of earth pressure. wzw · If the groundwater table is located at a height zw from · The lateral force P w the bottom of the wall, the due to groundwater hydrostatic pore water is given by: pressure is given by:· Since the vertical effective stress varies linearly with depth, the active, the lateral earth pressure also vary linearly with depth. · Only in case of a surcharge (figure(d)), the lateral earth pressure is constant with depth. · A vertical wall retaining saturated soil on one side must resist both lateral earth pressure and hydrostatic pressure. 13u = w (z - z w )Pw = 1 wz w 2214Lateral Stresses due to Surcharge· Surface stresses, such as a qs uniform surcharge qs as shown in the figure on the right, also impose lateral Pq z earth pressures on retaining walls. z/2 · Depending on whether the retained soil is under active, at-rest or passive condition, Kqs a uniform surcharge qs will · The lateral force Pq impose a uniform lateral due to the surcharge earth pressure of Kqs where is given by: K is the earth pressure coefficient. Pq = K  qs  z15Lateral Earth Pressure ­ An Example· Determine the active lateral qs = 50 kPa earth pressures on the frictionless wall shown in the figure. The retained soil sat = 20 kN/m3 carries a uniform surcharge of ' = 30° 5m 50 kPa at the surface. Calculate the resultant lateral force per meter length on the wall and its location from the base of the wall. The groundwater within the retained soil is at hydrostatic condition. [This example will be solved during the class.]16Lateral Earth Pressure ­ Key Points· The lateral earth pressures on retaining walls are related directly to the vertical effective stress through the active (Ka) , at-rest (K0) and passive (Kp) earth pressure coefficients. · Ka &lt; K0 &lt; Kp and Ka = 1/Kp. · Only a small movement of the wall away from the soil is required to mobilize full active earth pressure in the soil mass. · Substantial movement of the wall towards the soil is required to mobilize full passive earth pressure in the soil mass. · Rankine's earth pressure coefficients are only applicable to a frictionless, vertical, rigid wall retaining a homogenous soil with a horizontal ground surface.17Coulomb's Earth Pressure Theory· Coulomb (1776) proposed that a condition of limit equilibrium exists in the soil mass retained behind a vertical wall and that the retained soil mass will slip along a plane inclined at an angle  to the horizontal. · The critical slip plane is the one which gives the maximum lateral pressure on the wall. · Limit equilibrium describes the state of a soil mass that is on the verge of failure, i.e. the applied stresses are equal to the available strength along the slip plane.18Coulomb's Theory (Continued..) · Let us consider a vertical, frictionless wall of height H, supporting a dry, homogenous soil mass with an angle of internal friction , as shown in the figure on the top right. · The forces acting on the soil wedge above the slip plane are shown in the figure on the bottom right. · For a dry soil, ' = .19Coulomb's Theory (Continued..) · Consider the force equilibrium of the soil wedge in the x- and zdirections:F Fx= Pa + Tcos - Nsin = 0 = W - Tsin - Ncos = 0 Pa = 1 H2cot  tan( -  ) 0 2zwhere T = N.tan and W = ½H02cot N.tan cot ½· Solving for Pa, we get:· Differentiating the above expression w.r.t.  and equating the derivative to zero, we can obtain critical value of  that gives maximum Pa: = cr = 45o +  2   Pa = 1 H2 tan2  45o -  = 1 K a H2 0 0 2 2 2 20Coulomb's Theory (Continued..) · The expression for lateral force due to active earth pressure is the same as that obtained earlier using stress equilibrium considerations (Mohr's circle). · The solution obtained using a limit equilibrium analysis always results in a failure load that is greater than the true failure load. This solution is called an upper bound or an unsafe solution. · The main reason for this is that the soil will always be able to choose a failure mechanism that is more efficient than the assumed failure mechanism (shape and location of slip plane). · The accuracy of a limit equilibrium analysis depends on how realistic the chosen mechanism is.21Coulomb's Theory (Continued..) · On the other hand, an analysis from the consideration of static equilibrium of stresses usually results in a failure load smaller than the true failure load. Such a solution is called a lower bound or safe solution. · For the fortuitous case when both these analyses give the same solution, we have a true solution. · In general, Coulomb's earth pressure theory gives an upper bound estimate and Rankine's theory gives a lower bound estimate of lateral earth pressure. · For a vertical, frictionless, rigid wall retaining a horizontal homogenous soil mass, both these theories give the same solution.22Rough Wall and Sloping Backfill· Poncelet (1840) used Coulomb's limit equilibrium approach to obtain the active and passive earth pressure coefficients for cases where­ wall friction  is present ­ wall face inclined at an angle  to the vertical, and ­ the backfill is sloping at an angle  to the horizontalRough Wall and Sloping Backfill (Continued..) · Following expressions for active and passive earth pressure coefficients were obtained by Poncelet:K aC = cos2 ( - )  sin( +  )sin( -  )  cos2cos( +  )1 +  cos( +  )cos( -  )   cos2 ( + )  sin( +  )sin( +  )  cos cos( -  )1 +  cos( -  )cos( -  )  2 2 2KpC =· Following Coulomb's approach, Poncelet also used a linear slip plane inclined at an angle  with respect to the horizontal.23· Unlike the Rankine earth pressure coefficients, KaC is not equal to 1/KpC.24Rough Wall and Sloping Backfill (Continued..) · The critical inclination of the slip plane w.r.t. the horizontal is given by: 1  = tan   cos -1Error due to Curvature of Slip Plane· The curvature of the slip surface for the active state is small in comparison with that for the passive state as shown  in the figure. · For the active condition, the error is negligibly small. · However, the value of the passive earth pressure coefficient is overestimated.  · For the passive condition, the error is small only if &lt;/3. In practice, however,  is generally greater than /3. sin  cos ± tan   sin( +  ) Pa· In the above expression, positive sign refers to the active condition (a) and the negative sign refers to the passive condition (p). · Please note that the presence of wall friction results in a curved slip plane for both the active and the passive condition and therefore, Poncelet's coefficients are not 100% accurate.Active ConditionPpPassive Condition2625`

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