Soil Strength. Performance Evaluation of Constructed Facilities Fall Prof. Mesut Pervizpour Office: KH #203 Ph: x4046

Transcription

1 ENGR-627 Performance Evaluation of Constructed Facilities, Lecture # 4 Performance Evaluation of Constructed Facilities Fall 2004 Prof. Mesut Pervizpour Office: KH #203 Ph: x Soil Strength 2

2 Soil Strength Shear Strength of Soil (τ): Internal resistance of soil / unit area. MOHR-COULOMB Failure Criteria: Theory of rupture for materials failure under combined σ and τ any stress state that combined effect reaches the failure plane Along the failure plane τ f = f(σ) Failure envelope is a curved line approximated by linear relationship Mohr-Coulomb failure criteria: τ f = c + σ tanφ In terms of effective parameters: τ f = c + σ tanφ Cohesion c τ Mohr s failure envelope φ: internal friction angle Mohr-Coulomb failure criteria σ 3 Soil Strength Inclination of the Plane of Failure Caused by Shear: Failure when shear stress on a plane reaches τ f (line) determine inclination (θ) of failure plane with major & minor principal plane τ h σ 1 A θ D E σ 1 B F C f φ σ 1 > τ f = c + σ tanφ d g c e 2θ b O a σ 1 σ fgh failure plane s = c + σ tanφ ab major principal plane ad failure plane θ to 2θ angled Angle bad = 2θ = 90 + φ θ = 45 + φ/2 σ 1 = tan 2 (45+φ/2) + 2c tan(45+φ/2) Similarly for effective parameters. Shear failure for saturated soils: τ f = c + σ tanφ 4

3 Soil Strength Shear Strength Parameters in Laboratory: Unconfined Compression Test of Saturated Clay: A type of unconsolidated-undrained triaxial test For clayey samples (Cohesive) = 0 (confining pressure) Axial load (σ 1 ) applied to fail the sample (relatively rapid) At failure f = 0 and σ 1f = major principal stress Therefore undrained shear strength is independent of confining pressure τ f = σ 1 / 2 = q u / 2 = C u or S u q u : unconfined compressive strength, c u (S u ): undrained shear strength C u or S u τ σ 1 σ 1 φ = 0 Total stress Mohr s Circle at failure σ 1 = q u σ 5 Soil Strength Direct Shear Test (stress or strain controlled): Specimen is square or circular Box splits horizontally in halves Normal force is applied on top shear box Shear forces is applied to move one half of the box relative to the other (to fail specimen) Stress Controlled: Shear force applied in equal increments until failure Failure plane is predetermined (horizontal) Horizontal deformation & H is measured under each load. Strain Controlled: Constant rate of shear displacement Restraining shear force is measured Volume change ( H) (Advantage: gives ultimate & residual shear strength) Shear Force Loading plate τ Sample Normal force τ Porous Stone Shear Box Shear Stress H Dense sand Loose sand Expansion Peak shear strength τ f Compression Ultimate shear strength τ f Shear Displacement Dense sand Shear Displacement Loose sand 6

4 Soil Strength Direct Shear Test (continued): Repeat Direct Shear under several normal stresses. Plot the normal stress vs. shear stress values. τ f τ f = σ tan φ Dry sand c = 0 for dry sand and σ = σ φ = tan -1 (τ f / σ) φ σ 7 Soil Strength Drained Direct Shear Test on Saturated Sand & Clay: Test conducted on saturated sample at slow rate of loading allowing excess pore water to dissipate. For sand (k is high pwp dissipates quickly) Therefore φ under drained conditions ~ same For clay (k is low under load consolidation takes time, therefore load needs to be applied very slow). τ f c OC clay φ φ τ f = c + σ tan φ NC clay, c=0 τ f = σ tan φ σ General Comments on Direct Shear Test: Failure is not along the weakest plane (forced at horizontal plane) Represents angle of friction between soil and foundation material: τ f = c a + σ tan δ C a : adhesion δ: angle of friction between soil and foundation material 8

5 Soil Strength Triaxial Shear Test: cap Reliable method for determination of shear strength parameters. σ 1 membrane σ 1 Porous stone Axial stress (deviator stress) is applied to cause failure (shear) by vertical loading. Load vs. deformation readings are recorded. Three general types of triaxial test are: 1. Consolidated drained test (CD) 2. Consolidated undrained test (CU) 3. Unconsolidated undrained test (UU) : confining pressure applied all around sample (air/water/glycerine) σ d Porous stone σ 1 = + σ σ d d 9 Soil Strength Triaxial Shear Test: Consolidated-drained test: Specimen is subjected to confining stress all around. As a result the pwp of the sample increases by u c. If the valve is opened at this point the u c will dissipate and sample will consolidate ( V decreases under ) u c B = σ 3 Skempton s pwp parameter (B~1.0 for saturated soils) σ d u d = 0 σ d End of consolidation stage u c = 0. Application of deviator stress ( σ d ): For drained test σ d is increased slowly, while the drainage valve is kept open, & any excess pwp generated by σ d is allowed to dissipate. ( V can be measured by measuring amount outflow-water, since S=100%) CD test excess pwp completely dissipated = 10

6 Soil Strength Triaxial Shear Test: Consolidated-drained test (Continued): At failure (Axial stress) σ 1 = σ 1 = + ( σ d ) f σ 1 major principal stress at failure minor principal stress at failure Conduct other triaxial (CD) tests under different (confining) pressure and obtain the corresponding σ 1 at failure and plot the Mohr s circle for each test. τ σ 1 θ = 45 + φ / 2 B τ f = σ tanφ φ Total and Effective Stress Failure Envelope for OC clays φ 1 σ 1 A c O = 2θ σ 1 = σ 1 2θ σ ( σ d ) f ( σd)f 11 Soil Strength Triaxial Shear Test: Consolidated-undrained test (CU): Consolidation of S=100% sample under (confining stress) & allow u c to dissipate. Drainage valve is closed after complete consolidation (u c = 0) Deviator stress ( σ d ) is applied and increased to failure. u d is developed (due to no drainage). σ d u d 0 σ d End of consolidation stage u c = 0 (and close valves). A σ u = d Skempton s pwp parameter d Loose sand & NC clay Dense sand & OC clay u d increases with strain u d increases with strain up to a certain point and drops & becomes negative (due to dilatation of soil) 12

7 Soil Strength Triaxial Shear Test: Consolidated-undrained test (Continued): Total and Effective principal stresses are not the same. At failure measure ( σ d ) f and ( u d ) f Major principal stress at failure is obtained as: Total: + ( σ d ) f = σ 1 Effective: σ 1 -( u d ) f = σ 1 Minor principal stress at failure is obtained as: Total: Effective: -( u d ) f = τ Note: σ 1 - = σ 1 - ' Mohr s Circle for CU Test: τ f = σ tanφ φ Effective Stress Failure Envelope τ f = σ tanφ cu φ cu Total Stress Failure Envelope A O σ 1 σ 1 B σ ( u d ) f ( u d ) f 13 Soil Strength Triaxial Shear Test: Consolidated-undrained test (Continued): For OC Clay: φ cu τ τ f = σ tanφ cu for OC clays c cu φ 1cu τ f = c cu + σ tanφ 1cu O σ 1 σ 1 A B σ A = A f = ( ud ) f ( σ d ) f for NC clay for OC clay 14

8 Soil Strength Triaxial Shear Test: Unonsolidated-undrained test (UU): Drainage in both stages is not allowed. Therefore application of u c = B And application of σ d u d = Ặ σ d u = u c + u d u = B + Ặ σ d = B + Ặ (σ 1 - ) It can be seen that tests conducted with different results in the same ( σ d ) f, resulting in mohr s circle with same radius. τ Effective φ φ = 0 Failure envelope C u σ 1 σ 1 σ 1 σ σ 1 =[ + ( σ d ) f ] ( u d ) f = σ 1 -( u d ) f = -( u d ) f Example: by u c = = + - u c = ( σ d ) f will be the same. 15 Soil Strength General Comments on Triaxial Tests: Failure plane not predetermined Field strength function of rate of application of load and drainage Granular soil drained shear strength parameters NC Clay Under footing Undrained conditions Excavation in OC Clay Drained case (more critical) Control of stress states are possible in Triaxial test 16

9 Soil Strength Undrained Cohesion of NC and OC Deposits: NC clay undrained shear strength c u or S u increase with effective overburden pressure Skempton (1957) c u / σ = (PI) {PI: in %} Ladd for OC clas (1977) (c u /σ ) OC / (c u /σ ) NC = (OCR) Soil Stresses At A Point Due to Poisson s effect lateral flow (creep) ε x = µ ε z K Ratio of lateral to vertical stress: 0.0 µ 0.5 K = σ h / σ v h σ v = γ t h z x K f Maximum strength failure line σ h K 0 < 1 NC soils K 0 < 1 Slightly OC soils OCR < 3 K 0 > 1 Highly OC soils OCR > 3 18

10 General Comments CD CU Long-term Stability (earth embankments & cut slopes) Soil initially fully consolidated, then rapid loading (slopes in earth dams after rapid drawdown) UU End of construction stability of saturated clays, load rapidly & no drainage (Bearing capacity on soft clays) 19 Slope Stability 20

11 Slope Stability Slope Stability: The engineering assessment Of the stability of natural and man-made Slopes as influenced by natural or induced Changes to their environment. Studied by analytical (closed-form) or numerical (approximate) methods. Both methods are simplification of actual Geological, mechanical and other aspects. The stability of a slope depends on its ability to sustain the effects of load increases or environmental changes. Pre-failure analysis: to assess safety of slope and its intended performance. Post-failure analysis: study of failure and processes causing it. 21 Slope Stability Slope Stability analysis (continued): Determination of shear stress developed on the most likely rupture surface and comparing to shear strength of soil. Likely rupture surface: is the critical surface with minimum factor of safety. Steepened Slope to Wall To increase space 22

12 Slope Stability The effective evaluation of slope stability requires: Site characterization (geological hydrological conditions) Groundwater conditions (pore pressure model) Geotechnical parameters (strength, deformation, drainage) Mechanisms of movement ( kinematics potential failure modes) 23 Landslide Components 24

13 Landslide Components Varnes (1978), Morgenstern (1985) 25 Rotational Slides 26

14 Slope Stability Components of Slopes Facing Crest Toe Slope angle Foundation Reinforcement Reinforced fill Retained Fill Foundation 27 Slope Stability Possible Failure Modes of Slopes Local failure Surficial failure Slope failure Global failure 28

15 Slope Stability Typical Surfical Failure: Shallow failure surface up to 1.2 m (4ft) Failure mechanisms: Poor compaction Low overburden stress Loss of cohesion Saturation Original ground Seepage forces surface Slip Surface Slide Mass 29 Slope Stability Analytical Solutions Limit Equilibrium: Widely applied analytical technique, where force (or moment) equilibrium conditions are determined based on statics. The analyses is based on material strength, rather than stress-strain relationships. A Factor of Safety, is defined as a tool of evaluating the slope stability with limit equilibrium approach. resisting forces shear strength of material FS = = driving forces shear stress required for equilibrium Where FS > 1.0 represents a stable slope and FS < 1.0 stands for failure. Required values: Limit Equilibrium: FS = 1.0 Under Static Loads: FS Under Seismic Loads: FS

16 Slope Stability Limit Equilibrium: Overall measure of the amount by which the strength of the soil would have to fall short of the values described by c and φ in order for the slope to fail. resisting forces shear strength of material FS = = driving forces shear stress required for equilibrium c + σ tanφ τ f FS = = τ τ eq d τ f : Average Shear strength of soil τ d : Shear stress developed on potential surface 31 Slope Stability Limit Equilibrium (continued): Fundamentals of limit equilibrium method (Morgenstern, 1995): Slip mechanism results in slope failure Resisting forces required to equilibriate disturbing mechanisms are found from static solution The shear resistance required for equilibrium is compared with available shear strength in terms of Factor of Safety The mechanism corresponding to the lowest FS is found by iteration 32

17 Slope Stability Stability of Infinite Slopes without Seepage (Surficial slope stability): Soil Shear Strength: τ f = c + σ tanφ Pore water pressure: u = 0 Failing along AB at a depth H Static equilibrium of forces on the block. Assume F on ab and cd are equal. Along line AB: Developed resistance: τ f = c d + σ tanφ d = c d + γ H cos 2 β tanφ d Driving force due to weight: τ d = γ H cosβsinβ Factor of Safety: 2c tanφ FS = + γ H sin 2β tan β For c = 0: tanφ FS = tan β FS = 1 H = H cr A H β β F a b Forces: N a = γ L H cosβ T a = γ L H sinβ σ = γ L H cos β / (L/cosβ) = γ H cos 2 β τ= γ L H sinβ / (L/cosβ) = γ H cosβsinβ N r = γ L H cosβ T r = γ L H sinβ N a T r β L W T a R β N r 33 c d F B Slope Stability Stability of Infinite Slopes with Seepage (Surficial slope stability): Soil Shear Strength: τ f = c + σ tanφ GWT at surface, pore pressure u=γ w h= γ w Hcos 2 β Failing along AB at a depth H Static equilibrium of forces on the block. Assume F on ab and cd are equal. Along line AB: Developed resistance: τ f = c d + σ tanφ d = c d +(σ-u) tanφ d = c d +(γ sat - γ w ) H cos 2 β tanφ d Driving force due to weight: τ d = γ H cosβsinβ Factor of Safety: β FS 2c' γ ' = γ H sat sin 2β + γ sat For c = 0: γ ' FS = γ sat tanφ' tan β FS = 1 H = H cr tanφ' tan β A H β F Forces: N a = γ sat L H cosβ T a = γ sat L H sinβ σ = γ sat L H cos β / (L/cosβ) = γ sat H cos 2 β τ= γ sat L H sinβ / (L/cosβ) = γ sat H cosβsinβ N r = γ sat L H cosβ T r = γ sat L H sinβ a b N a T r β L W T a R β N r h= Hcos 2 β c d F SEEPAGE Equipotential line 34 B

18 Slope Stability Slope Stability with Plane Surface: AC Trial failure place B C H N a W θ T a T r A β θ Factor of Safety: 2csin β + H γ FS = Hγ sin sin( β θ ) ( β θ ) sinθ cosθ tanφ For c = 0: tanφ FS = tan β 35 Slope Stability Modes of Failure of Finite Slopes: Shallow slope failure Slope failure Base failure 36

19 Slope Stability Circular surface Slip circle analysis (φ = 0): Circular slip surfaces are found to be the most critical in slopes with homogeneous soil. There are two analytical, statically determinate, methods used for FS: the circular arc (φ=0) and the friction circle method. M r culr resisting moment FS = = = Circular failure surface in φ=0 soil is defined by its undrained strength, c u. M d Wx driving moment M FS = M r d 2 cur θ = W l W l W 1 W 2 l 2 l 1 37 Slope Stability Circular surface Friction circle (φ, c soil): Trial circle through toe. The friction circle method attempts to satisfy the requirement of complete equilibrium by assuming that the direction of the resultant of the normal and frictional component of strength mobilized along the failure surface corresponds to a line that forms a tangent to the friction circle with radius: Procedure (Abramson et al 1996 more detailed) C parallel to ab R f = R sinφ m P passes through intersection W-C P makes φ m with line through center of friction circle, & tangent to FC U often taken 0 Force polygon determine C Critical circle developed cohesion is maximum For FS = 1, the critical height: C / (γ H cr ) = f(α, β, θ, φ ) = m (stability No.) β P φ m φ > 3 deg critical circles all toe circles 38

20 Slope Stability Method of Slices (limit equilibrium): Soil divided to vertical slices, width of each can vary. The previous methods do not depend on the distribution of the effective normal stresses along the failure surface. The contribution is accounted for by dividing the failing slope mass into smaller slices and treating each individual slice as a unique sliding block. Non-circular: Circular: The discretization of the slip surface to elements results in two force components acting on each: Normal and Shear forces. The other unknown is the location of line of action of the normal force for each element. However the equilibrium conditions: ΣF x =0, ΣF y =0, ΣM=0 No. of unknowns = No. of slices * 3 Therefore assumptions should be made. 39 Slope Stability Circular surface (Bishop method): Soil divided to vertical slices, width of each can vary. Can be applied to layered soil, with different properties. Find minimum FS by several trials. ΣM 0 = 0 FS n ( c' li + Wi cosαi tanφ' ) i= 1 = n ( Wi sinαi ) i= 1 40

21 Slope Stability Search for Minimum Factor of Safety: Minimum FS values for the failure surface for every center is obtained, and recorded by the center of rotation, the contours indicate the location of the center with minimum overall FS. 41 Slope Stability Slope Stability with Seepage (u 0): Obtain the average pwp at the bottom of the slice using the phreatic line. Total pwp for the slice is u n L n Phreatic surface H h z Seepage β FS modified (from Bishop method) for pore pressure: FS n [ c' li + ( Wi ui li ) cosαi tanφ' ] i= 1 = n ( Wi sinαi ) i= 1 42

22 Lateral Earth Pressure 43 Lateral Earth Pressure Lateral Earth Pressure Coefficient: H σ z σ x P=(1/2)K γ H 2 1/3 H K=σ x /σ z σ x = Kσ z = KγH 44

23 Lateral Earth Pressure Lateral Earth Pressure Coefficient at Rest: Relationship between σ z and σ x at a given depth (at rest means no shear). K o : Coefficient of earth pressure at rest, K o = σ x / σ z H Rigid Wall No movement σ z σ x K=σ x /σ z P=(1/2)K γ H 2 1/3 H σ x = Kσ z = KγH For coarse-grained soils: K o = 1 - sinφ (ok for loose sand) For fine grained NC soils: K o = m - sinφ m: 1 for NC cohesionless or cohesive m: 0.95 OCR > 2 Massarch (1979) K o = (PI% / 100) For OC clays: K o = K o (NC) (OCR) (1/2) Or K o = (1 - sinφ ) OCR sinφ 45 Lateral Earth Pressure Coefficient of Active Lateral Earth Pressure: Wall moves away from the soil (pushed out). Movement H σ z σ x K a =σ x /σ z 46

24 Lateral Earth Pressure Wall Movement Required to Reach the Active Condition: Soil Type Dense sand Horizontal movement required to reach the active state H Loose sand H Stiff clay H Soft clay H (From CGS, 1992) 47 Lateral Earth Pressure Coefficient of Passive Lateral Earth Pressure: Wall moves towards the soil (pressed in). Movement H σ z σ x K p =σ x /σ z 48

25 Lateral Earth Pressure Wall Movement Required to Reach the Passive Condition: Soil Type Dense sand Horizontal movement required to reach the passive state H Loose sand H Stiff clay H Soft clay H (From CGS, 1992) 49 Lateral Earth Pressure In Summary: 1. If the wall moves away from the fill (soil) pressure will decrease and reach to active state. (σ h = K a σ v ) 2. If the wall moves towards the fill (soil) pressure will increase and reach to passive case. (σ h = K p σ v ) 3. More deformation is generally required to achieve passive case than the active case. K p K o K a Movement away From backfill Movement towards backfill 50

26 Lateral Earth Pressure Classical Lateral Earth Pressure Theories: Coulomb s Earth Pressure Theory (1776) Rankine s Earth Pressure Theory (1857) 51 Lateral Earth Pressure Rankine s Earth Pressure Theory: Assumptions: The soil is homogeneous and isotropic Frictionless wall Failure surfaces are planar The ground surface is planar The wall is infinitely long (plane strain condition) At the active or passive state (plastic equilibrium, every point in soil about to fail) The resultant on the back of the wall is at angle parallel to ground surface 52

27 Lateral Earth Pressure Rankine s Earth Pressure Theory: Attainment of Rankine s Active State Attainment of Rankine s Passive State 53 Lateral Earth Pressure Rankine s Earth Pressure Theory Force Diagram: C β W β P A θ T N Rankine s Earth Pressure Theory - Force Equilibrium P β N W θ T 54

28 Lateral Earth Pressure Rankine s Theory Critical Angle of Failure Plane: Critical angle of failure plane: The angle (θ) when the thrust (P) reaches the maximum value for the condition or the minimum value for the passive condition At the active state: θ critical = 45 o + φ / 2 At the passive state: θ critical = 45 o - φ / 2 55 Lateral Earth Pressure Rankine s Theory Earth Pressure Distribution (c =0): β H H/3 β P = (1/2) K γ H 2 P H = P cosβ = (1/2) K γ H 2 cosβ β σ = K σ z = K γ H 56

29 Lateral Earth Pressure Rankine s Theory Coefficient of Active Earth Pressure: For β φ : K a cos β = cos β cos β cos φ' 2 2 cos β cos φ' For β = φ : K a = tan 45 φ 2 2 o ' Rankine s Theory Coefficient of Passive Earth Pressure: For β φ : K p cos β + = cos β 2 2 cos β cos φ' 2 2 cos β cos φ' For β = φ : K p 2 o = tan 45 + φ' 2 57