CE466.3 Modeling of Earth Structures. Contents

Transcription

1 1 CE466.3 Modeling of Earth Structures Earth Retaining Structures Part One Contents Introduction Definition of key terms Lateral earth pressure Basic concepts Coulomb s earth pressure theory

2 2 Introduction Earth retaining structures are ubiquitous in a manmade environment. These structures have the distinction of being the first to be analyzed using the concepts of mechanics. In this lecture and the next two lectures, we will analyze the stability of a few typical earth retaining structures. Only the ultimate limit state will be considered in such a stability analysis. The learning objectives of these lectures are: Understand and determine the lateral earth pressures Understand the forces that make an earth retaining structure unstable Acquire the ability to analyze the stability of simple earth retaining structures It will help you a great deal if you have a firm grasp of the following concepts: Static equilibrium Effective stresses and seepage Mohr s circle Shear strength of soils Planar flow of water through soils

3 3 Definition of key terms Active earth pressure coefficient (K a ): It is the ratio of lateral and vertical principal effective stresses when an earth retaining structure moves away (by a small amount) from the retained soil. Passive earth pressure coefficient (K p ): It is the ratio of lateral and vertical principal effective stresses when an earth retaining structure is forced against a soil mass. Gravity retaining wall: It is a massive concrete or stone masonry wall that relies on its mass to resist the lateral forces from the retained soil mass. Flexible retaining wall: It is a long, slender wall that relies on passive resistance and anchors or props for its stability. Mechanically stabilized earth: It is a gravity type retaining wall in which the soil is reinforced by thin reinforcing elements (steel, fabric, fibers, etc.).

4 4 Lateral earth pressure Basic concepts We are going to consider the lateral earth pressures on a vertical wall that retains a soil mass on one side. We will develop a basic understanding of lateral earth pressures using a generic φ and make the following assumptions: The retaining wall is vertical. The interface between the wall and the soil is frictionless. The soil surface is horizontal and no shear stresses act on horizontal and vertical boundaries. 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.

5 5 Lateral earth pressure (continued ) Consider the wall shown in the figure on the top right. If no movement of the wall takes place, the vertical and horizontal effective stresses at rest on elements A and B are: σ z = σ = γ 1 z τ σ = σ = K σ = K γ z x where K 0 is the lateral earth pressure at rest. Mohr s circle for the at-rest stress state is shown in the figure on the bottom right. σ [Please label the figure during the lecture.]

6 6 Lateral earth pressure (continued ) Let us now consider a rotation about the bottom of the wall that is sufficient to produce slip planes in the soil mass behind and in front of the wall as shown in the figure below.

7 7 Lateral earth pressure (continued ) Much larger rotation is required to produce slip planes in front of the wall as compared to the back of the wall as shown in the figure below. The soil mass at the back of the wall is assisting in producing the failure while the soil mass in the front of the wall is resisting failure.

8 8 τ Lateral earth pressure (continued ) Let us examine the changes in effective stresses that act on elements A and B. σ [Please label the figure during the lecture.] The vertical effective stress does not change for both the elements but the lateral effective stress acting on element Areducesand lateral effective stress acting on element B increases. We can now plot two additional Mohr s circles: one to represent the stress state of element A and the other to represent the stress state of element B.

9 9 Lateral earth pressure (continued ) [Please fill in the equations during the lecture.] For element B, the lateral effective stress must be greater than the vertical effective stress to reach the failure state. The ratio of lateral principal effective stress to vertical principal effective stress for element A is given by: Similarly, the ratio of lateral principal effective stress to vertical principal effective stress for element B is given by: where K p is called the passive lateral earth pressure coefficient. Therefore, where K a is called the active lateral earth pressure coefficient.

10 10 Rankine stress states [Please fill in the equations during the lecture.] The stress states of soil elements A and B are called the Rankine active state and Rankine passive state (named after Scottish engineer Rankine who proposed this theory in 1857). Each of these Rankine states is associated with a family of slip planes shown in the figure on page 6. For the Rankine active state, the slip planes are oriented at For the Rankine passive state, the slip planes are oriented at with respect to the horizontal plane. The lateral earth pressure for the Rankine active state is and for the Rankine passive state is with respect to the horizontal plane.

11 11 Variation of lateral earth pressure with depth Equations for earth pressure for Rankine active and passive earth pressure varies linearly with depth as shown in the figure below.

12 12 Lateral stresses due to groundwater [Please fill in the equations during the lecture.] The lateral earth force is the area of the lateral stress diagram, which, for the Rankine active state, is and, for the Rankine passive state, is DO NOT multiply the pore water pressure with a coefficient of earth pressure. If the groundwater table is located at a height h w from the bottom of the wall, the hydrostatic pore water pressure is If groundwater is present, you need to add the pore water pressure to the lateral earth pressure. and, the hydrostatic force is

13 13 Lateral stresses due to surcharge [Please fill in the equations during the lecture.] Surcharge loading also imposes lateral earth pressures on retaining walls. A uniform surface stress, q s, will transmit a uniform active lateral earth pressure of The active and passive lateral earth pressures due to the soil, groundwater and the uniform surface stresses are then and a uniform passive earth pressure of and provided the entire surface stress has become effective, i.e. it is carried only by the soil skeleton and not by the pore water. where u is the hydrostatic pressure and the subscripts a and p denote the active and passive states, respectively.

14 14 Lateral earth pressure Essential points The lateral earth pressures on retaining walls are related directly to the vertical effective stress through the active (K a ), at-rest (K 0 ) and passive (K p ) earth pressure coefficients. (K a ) < (K 0 ) < (K p ). 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. The lateral earth pressure coefficients derived thus far are valid only for a smooth, rigid vertical wall supporting a homogenous soil mass with a horizontal surface. The lateral earth pressure coefficients must be applied only to the effective stresses.

15 15 Coulomb 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 θ to the horizontal. The critical slip plane is the one which gives the maximum lateral pressure on the wall. Let us consider a vertical, frictionless wall of height H, supporting a soil mass with a horizontal surface as shown in the figure on the right. We are going to assume a dry, homogenous soil mass with an angle of internal friction φ. Since the soil is dry, γ = γ.

16 16 Coulomb s earth pressure theory (continued ) [Please fill in the equations during the lecture.] As we did for a slope stability analysis, let us consider a freebody diagram for the sliding soil mass as shown in the figure on the right and solve for the lateral force acting on the wall as follows: The weight of the sliding soil mass is Solving for P a, we get At limit equilibrium,

17 17 Coulomb s earth pressure theory (continued ) [Please fill in the equations during the lecture.] To get the maximum thrust, we differentiate the equation for P a with respect to θ: which leads to Substituting this value of θ into the equation for P a, we obtain This is the same result obtained earlier from considering Mohr s circle. A limit equilibrium method always gives a solution that is usually greater than the true solution. This solution is called an upper bound solution. On the other hand, a solution from the consideration of static equilibrium and stresses that do not exceed the failure stresses, is usually smaller than the true solution. Such a solution is called a lower bound solution. Rankine stress state solution is one such lower bound solution.

18 18 Coulomb s earth pressure theory (continued ) If the upper bound solution and the lower bound solution are in agreement, we have a true solution. This is certainly the case with the estimation of lateral earth pressures. The solution obtained using Coulomb s theory (an upper bound solution) is the same as that obtained from Rankine s theory (a lower bound solution). Therefore, for a smooth, rigid vertical wall, the true values of active and passive earth pressure can be obtained using However, in practice, the retaining walls are neither smooth nor fully rigid. Therefore, the above expressions need to be modified to take into account the friction between the soil and the retaining wall.

19 1 CE466.3 Modeling of Earth Structures Earth Retaining Structures Part Two Contents Lateral earth pressure coefficients for rough walls Lateral earth pressures for a total stress analysis Types of retaining walls and their modes of failure Stability of rigid retaining walls

20 2 Lateral earth pressure for rough walls Poncelet (1840) used Coulomb s limit equilibrium approach to obtain the active and passive earth pressure coefficients for cases where Wall friction (δ) is present Wall face inclined at an angle η to the vertical, and The backfill is sloping at an angle β to the horizontal Referring to the figure on the top right, Poncelet (1840) obtained the following expressions for active earth pressure coefficient: K ac = cos 2 η cos ( η + δ ) cos ( φ η) sin( φ + δ ) sin( φ β ) cos( η + δ ) cos( η β ) 2

21 3 Lateral earth pressure for rough walls (continued ) Similarly, Poncelet (1840) obtained the following expression for the passive earth pressure coefficient: K pc = cos 2 η cos ( η δ ) cos 1 ( φ + η) sin( φ + δ ) sin( φ + β ) cos( η δ ) cos( η β ) In the above two expressions, the subscript C denotes Coulomb. Note that K pc 1/K ac. The inclination of the slip plane to the horizontal is θ 1 sinφ cosδ tan = ± tanφ cosφ sin ( φ + δ ) where the positive sign refers to the active state (θ a ) and the negative sign refers to the passive state (θ p ). 2 2

22 4 Lateral earth pressure for rough walls (continued ) Wall friction causes the slip planes in both the active and the passive states to be curved. The curvature of the slip surface for the active state is small in comparison with that for the passive state. The implication of the curved slip surface is that the values of K ac and K pc obtained from Poncelet s equations are not accurate. For the active case, the error is negligibly small. However, the value of the passive earth pressure coefficient is overestimated. For the passive state, the error is small only if δ < φ /3. In practice, however, δ is generally greater than φ /3. Several investigators have tried to find the values of K a and K p using nonplanar slip surfaces. Caquot and Kerisel (1948) used a logarithmic spiral. Packshaw (1969) used a circular arc. The values of K a and K p proposed by Caquot and Kerisel (1948) are generally used in practice. Factors that can be applied to K pc to correct it for a logarithmic spiral slip surface are given in Table 1 on the next page.

23 5 Correction factors for nonplanar slip surface [after Budhu, 2000]

24 6 Rough walls Sign convention for δ The sign convention for the inclination of δ are shown in the figure on the right. Appropriate sign must be used when determining the values of K ac and K pc. The direction of the frictional force on the wall depends on whether the soil moves upwards or downwards relative to the wall. In general, the active soil wedge moves downwards (δ +ve) and the passive soil wedge moves upwards (δ ve) relative to the wall as shown in the figure on the left.

25 7 Rough walls Essential points The presence of wall friction causes the slip planes to curve which leads to an overestimation of passive earth pressure using Coulomb s analysis. The equations for active and passive earth pressure coefficient for rough walls proposed by Poncelet (1840) can be used provided a correction factor given in Table 1 is applied to the equation for passive earth pressure coefficient. No correction is necessary for the active earth pressure coefficient obtained from Poncelet s equation. The resultant active and passive forces are inclined at an angle δ with respect to the normal to the wall face. Appropriate sign of δ must be considered when using Poncelet s equations for the estimation of earth pressure coefficient. For rough walls, K pc 1/K ac.

26 8 Lateral earth pressures for a total stress analysis Figure on the right shows a smooth, vertical, rigid wall supporting a homogenous soil mass under undrained conditions. Using limit equilibrium, we will once again assume that for the active state, a slip plane is formed at an angle θ to the horizontal. The forces acting on this soil wedge are also shown in the figure on the right. Using statics, we obtain the sum of forces along the slip plane as: [Please fill in the equations during the lecture.] T is given by W is given by

27 9 Total stress analysis (continued ) [Please fill in the equations during the lecture.] Substituting the expressions for T and W in the equation for equilibrium along the slip plane, we get By substituting θ = 45 into the expression for P a, we get the maximum active lateral force as: To find the maximum value of P a, we differentiate the above equation with respect to θ and equate the derivative to zero: which gives If we assume a uniform distribution of stresses on the slip plane, then the active lateral stress is Let us examine the above equation. In case of an unsupported excavation

28 10 Total stress analysis (continued ) As we already know, z cr is the maximum depth of a tension crack and that if the tension crack is filled with water, z cr increases to For an unsupported excavation, the active lateral force is also zero. Therefore, If the excavation is filled with water, then We have two possible unsupported depths as given by the equations for z cr and H cr. The correct solution lies somewhere in between these two critical depths. In design practice, a value of and solving for H 0, we obtain: is used for unsupported excavation in fine-grained soils.

29 11 Total stress analysis (continued ) [Please fill in the equations during the lecture.] The passive lateral earth force for a total stress analysis can be written as and the passive lateral pressure is given by Please keep in mind that we are unable to use the coefficients of active and passive earth pressures in case of a total stress analysis. The reason for this is that a total stress analysis is formulated in terms of total stresses whereas the earth pressure coefficients are formulated in terms of effective stresses. We can certainly use these coefficients if we know the variation of pore water pressure with depth. Then, we will be able to estimate the variation of effective stresses with depth. However, estimation of variation of pore water pressure is not straightforward as the groundwater condition need not be hydrostatic.

30 12 Types of retaining walls There are two general classes of retaining walls rigid and flexible. Rigid retaining walls consist of concrete walls that rely on gravity for their stability. Four different rigid retaining walls are shown in the figure on the left.

31 13 Flexible retaining walls Flexible retaining walls consist of slender members of either steel or concrete or wood and rely on passive soil resistance, props or anchors for stability. Three different types of flexible retaining walls are shown in the figure below.

32 14 Failure modes for rigid retaining walls

33 15 Failure modes for flexible retaining walls

34 16 Rigid retaining walls Translational stability [Please fill in the equations during the lecture.] A rigid retaining wall must have adequate resistance against translation, i.e. the sliding resistance of the base of the wall must be greater than the resultant lateral force pushing against the wall. The factor of safety against translation is defined as where T is the sliding resistance at the base and P ax is the lateral force pushing against the wall. For an ESA, T can be expressed in terms of R z the resultant vertical force acting on the base of the wall and φ b the interfacial frictional angle between the base of the wall and the soil: If the base of the wall rests on a fine-grained soil, TSA is applicable. For a TSA, T can be expressed in terms of s w the adhesion between the base of the wall and the soil and B the horizontal width of the base of the wall:

35 17 Translational stability (continued ) [Please fill in the equations during the lecture.] Typical sets of forces acting on gravity and cantilever rigid retaining walls are shown in the figure below. Using statics, we obtain the factor of safety against translation for an ESA as:

36 18 Translational stability (continued ) [Please fill in the equations during the lecture.] In the above equation, θ b is the inclination of the base of the wall with the horizontal. θ b is positive if the base of the wall dips towards the backfill. If θ b is equal to zero, For a TSA, the factor of safety is: If θ b is equal to zero, The embedment of rigid retaining walls is generally small and therefore the passive lateral force is not taken into account. If the base resistance is found to be inadequate, the width B of the wall can be increased. For cantilever walls, a shear key can be constructed to provide additional base resistance against translational sliding.

37 19 Rigid retaining walls Rotational stability A rigid retaining wall must have adequate resistance against rotation (usually about the toe). The rotation of the wall about its toe is satisfied if the resultant vertical force lies within the middle third of the base. Referring to the figure on the right and taking moments about the toe, the resultant vertical force at the base is located at [Please fill in the equations during the lecture.] The wall is safe against rotation if In the above equation,

38 20 Rigid retaining walls Other failure modes [Please fill in the equations during the lecture.] A rigid retaining wall must have a sufficient margin of safety against soil bearing capacity failure. The maximum pressure imposed on the soil at the base of the wall must not exceed the allowable soil bearing capacity; that is where σ max is the maximum vertical stress imposed by the wall and q a is the allowable soil bearing capacity. A rigid retaining wall must not fail by deep-seated failure whereby the sliding mass includes both the wall and the soil. The stability against a deep-seated failure can be checked by using one of the many slope stability methods discussed in the previous lectures. Seepage forces, if left unchecked, can alter the stability of a rigid retaining wall significantly. The hydraulic gradient should generally be less than the critical hydraulic gradient:

39 1 CE466.3 Modeling of Earth Structures Earth Retaining Structures Part Three Contents Mechanically stabilized earth (MSE) walls Basic concept Components and construction of MSE walls Stability of MSE walls

40 2 Mechanically stabilized earth (MSE) The concept of mechanically stabilized earth (MSE) was accidentally discovered by Henri Vidal in 1967 while playing with his son on a beach. He discovered that he could build sand hills at an angle greater than the angle of internal friction by reinforcing the sand using strips of palm leaves. He called the composite material Terre Armeé or Reinforced Earth. The horizontal reinforcement mobilizes friction at its interfaces with the sand and therefore, imposes lateral restraint on the sand mass. This effect is shown using Mohr s circle of stress in the above figure. For the MSE, the Mohr s circle is smaller and is further away from the failure line.

41 3 An MSE Wall

42 4 Components of a MSE wall Front Panel Backfill Strip Reinforcement

43 5 Construction stages of a MSE wall Installing the front facing panels Attaching strip reinforcement to the facing panel Compacting the backfill

44 6 Examples of MSE walls - Canada Toronto Vancouver

45 7 Stability of a MSE wall A MSE wall has to satisfy two stability criteria: Internal stability External or overall stability The external stability of a MSE wall is determined by using the same procedure as used for a gravity retaining wall with a vertical face. The internal stability depends on The tensile strength of the reinforcing material, and The slip at the soil-reinforcement interface. Tensile failure of the reinforcement at any level leads to progressive collapse of the wall. Slip at the soil-reinforcement interface leads to redistribution of stresses and progressive deformation of the wall. Two methods of analysis are used to determine the internal stability of a MSE wall. One method is analogous to treating the MSE wall as an anchored flexible retaining wall and is generally used for reinforcing material with high extensibility such as geotextiles and geogrids. The other method is the coherent gravity method and is used for reinforcement of low extensibility such as metal strips.

46 8 Stability of a geotexile reinforced MSE wall For such a wall, Rankine active earth pressure theory is used with the active slip plane inclined at to the horizontal as shown in the figure on the right. The frictional resistance develops over an effective length L e, outside the active failure zone. At a depth z, the frictional resistance developed on both surfaces of the reinforcing material is where w is the width of the reinforcing material, σ z is the vertical effective stress, q s is the surcharge and φ i is the angle of friction at the soil reinforcement interface.

47 9 Geotextile reinforced MSE wall (continued ) Consider a layer of reinforcement at a depth z. The tensile force in the reinforcement is given by where S z and S y are the spacing in Z and Y directions, respectively, and T is the tensile force per unit length of the wall. For geotextiles and geogrids, one would normally consider one unit length and one unit width, so S y = 1 and w = 1. By setting T = P r, we can find the effective length of the reinforcement required for limit equilibrium (FS = 1). To find design effective length, a factor of safety (FS) t is applied on the tensile force T. Solving for L e, we get (FS) t usually ranges from 1.3 to 1.5. The total length of reinforcement is

48 10 Geotextile reinforced MSE wall (continued ) Because L R is zero at the base of the wall, the calculated length of the reinforcement at the base is often the shortest. This calculated length, while adequate for internal stability, is often inadequate for translation or bearing capacity (external stability). The adequacy of the calculated length for translation or bearing capacity can be determined as follows: 1. Calculate the maximum lateral active force, P ax : where K ac is the active lateral earth pressure coefficient using Poncelet s method with wall friction. If you neglect the interface friction, you may use K ar the Rankine active earth pressure coefficient. 2. Calculate length of the reinforcement required for translation. For short term loading in clays, the base resistance is where L b is the length of the reinforcement at the base and s w is the adhesion stress.

49 11 Geotextile reinforced MSE wall (continued ) 3. For external stability against translation where (FS) T is a factor of safety against translation with a usual range of 1.5 to 3. Therefore, the required length at the base against translation for short-term loading in clays is then 4. For long-term loading, where W i is the weight of the soil layer i, n is the number of layers and φ b is the effective interfacial friction angle between the reinforcement and the soil at the base. 5. Assuming a uniform soil weight throughout the height of the wall, and the reinforcement length at the base required to prevent translation under long-term loading is

50 12 Geotextile reinforced MSE wall (continued ) The procedure for analysis of a geotextile reinforced MSE wall is as follows: 1. Calculate the allowable tensile strength per unit width of the reinforcement: where T A is the allowable tensile strength, T U is the ultimate tensile strength, FS is a factor of safety and the subscripts have the following meaning: ID Installation damage CR Creep CD Chemical degradation BD Biological degradation 2. Calculate the vertical spacing at different wall heights using where (FS) sp is another factor of safety (between 1.3 and 1.5). It is customary to check the minimum vertical spacing (at the base of the wall) and then check the vertical spacing required at one-third and two-third wall heights.

51 13 Geotextile reinforced MSE wall (continued ) 3. Check the adequacy of the length of the reinforcement from translation stability pointof-view. 4. Determine the total length of reinforcement at different levels: where 5. Check for external stability against bearing capacity failure and overturning. Let s solve an example to understand the analysis procedure better. Design a 4 m MSE wall using a geotextile as the reinforcement. The backfill would be a compacted, coarse-grained soil with φ cs = 30 and γ sat = 18 kn/m 3. The surcharge is 15 kpa. The geotextile has ultimate tensile strength of 58.5 kn/m and the soil-geotextile interface friction angle is 20. The native soil is clay with parameters γ sat = 18.5 kn/m 3, φ cs = 28, φ b = 2φ cs /3 and s u = 60 kpa. The example will be solved during the lecture. Please use the space provided to note down the solution.

52 14 Geotextile reinforced MSE wall An example Solution: