EAT 212 SOIL MECHANICS

Transcription

1 EAT 212 SOIL MECHANICS Chapter 4: SHEAR STRENGTH OF SOIL PREPARED BY SHAMILAH

2 CONTENT Shear failure in soil Drained and Undrained condition Mohr-coulomb failure Shear strength of saturated sands and gravels Shear strength of saturated clays Shear strength evaluation

3 Strength of different materials Steel Concrete Soil Tensile strength Compressive strength Shear strength Complex behavior Presence of pore water 3

4 Shear Failure In Soil The shear strength of a soil can be described as its maximum resistance to shearing stresses. When this strength is exceeded failure occurs along planes in the soil mass called slip surfaces. The shear strength of soil has two components :- a) Frictional strength, ф b) Cohesive, c

5

6 Influencing Factors on Shear Strength Soil Composition - mineralogy, grain size and grain size distribution, shape of particles, pore fluid type and content, ions on grain and in pore fluid. Initial State - State can be describe by terms such as: loose, dense, over-consolidated, normally consolidated, stiff, soft, etc. Structure - Refers to the arrangement of particles within the soil mass; the manner in which the particles are packed or distributed. Features such as layers, voids, pockets, cementation, etc, are part of the structure.

7 shear failure of soil shear failure of soil TYPES OF SOIL FAILURE shear failure of soil Sliding failure of soil Shallow slope stability failure shear failure of soil shear failure of soil

8 Drained condition Occurs when there is no change in pore water pressure due to external loading Pore water pressure can drain out of the soil easily, causing volumetric strains in the soil

9 Undrained condition Occurs when the pore water pressure is unable to drain out of the soil Rate of loading is much quicker than the rate at which the pore water pressure is able to drain out of the soil The tendency of soil to change volume is suppressed Clays have low hydraulic conductivity, hence most often assumed to be under undrained during loading or construction period in short term, and the shear strength must be analyzed accordingly.

10 Stress increment Isotropic confinement stress Excess pore pressure DRAINED UNDRAINED

11 Mohr-coulomb failure criterion f is the maximum shear stress the soil can take without failure, under normal stress of.

12 Mohr-coulomb failure criterion (cont )

13 Peak and Residual-Strength Envelopes for clay

14 Graphical representation of Mohr- Coulomb failure criteria Normal Case

15 Granular (non-cohesive) soil Failure envelopes of cohesion less soil

16 Saturated, plastic clays

17 Shear stress, τ vs. effective stress, σ Nearly linear Slightly non-linear Hence, use idealized linear function a) τ intercept of the line is c b) slope of the line is φ

18 Shear strength of saturated sands & gravel Under typical static loading Little / no excess pore water pressure ( k high) Changes in the state of stress of soil cause the voids to expand / contract, water easily flows in / out Introduction Under dynamic loads (e.g earthquak e) Hence, u = hydrostatic pore water pressure & s analyses may be based on drained condition and effective stress Sand may not drain quickly enough to dissipate all excess pore water pressure Hence, evaluate s under undrained condition

19 Determining c & φ Saturated sands and gravels, c = 0 Determine φ by conducting field / lab test Typical values of φ φ increase as unit weight increase Gravels stronger than sands Well graded soils stronger than poorly graded soils Presence of silt decrease the φ (because particles are smoother, lower coefficient of friction)

20 Value of σ In most design prob, as the state of stress of soil change, the shear and normal stresses also change. Hence σ at beginning of loading and at the end is different Example of situation : External load, P is applied Vertical total stress, σ z increase Pore water pressure, u constant (rapid drainage) Vertical effective stress, σ z increase at same rate as σ z S also increase Shear stress, τ also will induces

21 Shear strength of saturated clays Evaluation of shear strength of clay is more difficult than sand and gravel: Greater volume changes during loading Low k, undrained cond, excess pore water pressure (short term) Excess pore water pressure dissipated, drain condition (long term) Generally weaker than sand & gravel, more often cause problems.

22 When performing shear strength analysis, need to assess drainage condition 3 possibilities: CASE I : Shear strength under drained condition CASE II : Shear strength under undrained condition with positive excess pore water pressures CASE III: Shear strength under undrained condition with negative excess pore water pressures

23 CASE I Easiest to evaluate Two situations Simply evaluate c & φ using test Shear strength can be determined, s = c + σ tan ф Loading very low w.r.t drainage, no excess pore water pressure generated Long time after loading, excess pore water pressure have dissipated

24 +ve because increase in mean normal stress Cannot compute values of σ at failure or s using total stress Mohr coulomb failure s = c T + σ tan ф T CASE II If soil truly saturated and undrained ф T = 0. Shear strength is undrained shear strength, s u s = c T Called as ф = 0 analysis Hence, s u = c T

25 -ve because decrease in mean normal stress CASE III Gradually dissipated but it cause corresponding loss in shear strength also Hence will affect factor of safety, F F = s/τ

26 Shear strength evaluation

27 DIRECT SHEAR TEST Top Platen Normal load Motor Drive Load cell to measure shear force Porous plate Rollers

28 DIRECT SHEAR TEST (cont ) Only slow drained tests are performed in this test. Shearing rate for clays must be chosen to prevent excess pore pressures building up. For sands and gravels tests can be performed quickly. Tests on sands and gravels are usually performed dry. Water does not significantly affect the (drained) strength. If there are no excess pore pressures and as the pore pressure is approximately zero the total and effective stresses will be identical.

29 Advantages of Direct Shear Test Easy and quick test for sands and gravels Large deformations can be achieved by reversing shear direction. This is useful for determining the residual strength of a soil. Large samples may be tested in large shear boxes. Small samples may give misleading results due to imperfections (fractures and fissures) or the lack of them. Samples may be sheared along predetermined planes. This is useful when the shear strengths along fissures or other selected planes are required.

30 Disadvantages of Direct Shear Test Non-uniform deformations and stresses in the specimen. The stress-strain behavior cannot be determined. The estimated stress may not be those acting on the shear plane. There is no means of estimation pore pressures so effective stresses cannot be determined from undrained tests Undrained strengths are unreliable because it is impossible to prevent localized drainage without high shearing rates. In practice shear box tests are used to get quick and crude estimates of failure parameters.

31 Direct shear test Analysis of test results Normal stress Area of Normal force (P) cross section of the sample Shear resistance developed at thesliding surface (S) Shear stress Area of cross section of the sample Note: Cross-sectional area of the sample changes with the horizontal displacement 31

32 Change in height of the sample Compression Expansion Shear stress, Direct shear tests on sands Stress-strain relationship f f Dense sand/ OC clay Loose sand/ NC clay Shear displacement Dense sand/oc Clay Shear displacement Loose sand/nc Clay 32

33 Shear stress, Direct shear tests on sands How to determine strength parameters c and f Normal stress = 3 Normal stress = 2 f3 f2 f1 Normal stress = 1 Shear displacement Shear stress at failure, f f Mohr Coulomb failure envelope Normal stress, 33

34 Direct shear tests on clays In case of clay, horizontal displacement should be applied at a very slow rate to allow dissipation of pore water pressure (therefore, one test would take several days to finish) Failure envelopes for clay from drained direct shear tests Shear stress at failure, f f Normally consolidated clay (c = 0) Normal force, Overconsolidated clay (c 0) 34

35 Example 12.1 Direct shear tests were performed on a dry, sandy soil. The size of the specimen was 50mm x 50mm x 19mm. Test results as follows: Test no. Normal force (N) Normal stress σ = σ (kn/m3) Shear force at failure (N) Shear stress at failure ζ (kn/m3)

36 Exercise A drained shear box test was carried out on a sandy clay and yielded the following results: Area of shear plane = 60 mm x 60mm Determine the apparent cohesion and angle of friction for the soil.

37 Example 12.2 Following are the results of four drained direct shear tests on an overconsolidated clay Diameter of specimen = 50mm Height of specimen = 25mm Test no. Normal force (N) Shear force at failure, S peak (N) Residual Shear force S residual (N) Determine the relationships for peak shear strength (ζ f ) and residual shear strength (ζ r )

38 Triaxial Test Deviator Load Cell water O-ring Seals Confining cylinder Rubber Membrane Porous Filter Disc Cell pressure Pore Pressure and volume change

39 Triaxial Test (cont ) Depending on whether drainage is allowed or not during i) initial isotropic cell pressure application and ii) shearing There are 3 types of triaxial tests i) consolidated drained (CD) test ii) Consolidated undrained (CU) test iii) Unconsolidated Undrained (UU) test

40 Consolidated Drained (CD) Test (slow test) Drainage valves OPEN during consolidation as well as shearing phases. Complete sample drainage is achieved prior to application of the vertical load. The load is applied at such a slow strain rate that particle readjustments in the specimen do not induce any excess pore pressure. Since there is no excess pore pressure total stresses will equal effective stresses. Very slow shearing to avoid build-up of pore pressure Can give the value of c and φ c and φ can be used for analyzing fully drained situations for long term stability of very slow loading)

41 Example 12.3 A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results are as follows : σ 3 = 276 kn/m 2 (Δσ d ) f = 276kN/m 2 Determine a) Angle of friction, φ b) Angle ϴ that the failure plane makes with the major principal plane

42 Exercise A drained triaxial compression test carried out on three specimens of the same soil yielded the following results : Test no Cell pressure (kpa) Deviator stress at failure (kpa) Draw the shear strength envelope and determine the peak strength parameters, c and ф p, assuming that the pore pressure remains constant during the axial loading stage

43 Example 12.4 Refer example 12.3 A) Find the normal stress σ and the shear stress f on the failure plane B) Determine the effective normal stress on the plane of maximum shear stress Solution refer page 453

44 Example 12.5 The equation of the effective stress failure envelope for normally consolidated clayey soil is f = σ tan 30. A drained triaxial test was conducted with the same soil at a chamberconfining pressure of 69 kn/m2. Calculate the deviator stress at failure Solution refer page 454

45 Example 12.6 The results of two drained triaxial test on a saturated clay follow : Specimen I : σ 3 = 70kN/m 2 Specimen II (Δσ d ) f = 130kN/m 2 σ 3 = 160kN/m 2 (Δσ d ) f = 223.5kN/m 3

46 Consolidated Undrained (CU) Test Apply σ 3 and wait until the soil consolidates Drainage valves OPEN during consolidation phase but Closed during the shearing phase. (Drainage and consolidation is allowed to take place during the application of the confining pressure σ 3 ). Loading does not commence until the sample cases to drain (or consolidate). This test can simulates long-term as well as shortterm shear strength for cohesive soils if pore water pressure is measured during the shearing phase. Pore pressure develops during shear to measure σ Also gives c and ɸ

47 Exercise The following results were obtained from undrained tests on specimens of a saturated normally consolidated clay. Determine : Cell pressure (kpa) Ultimate deviator stress (kpa) Ultimate pore pressure (kpa) (a) the effective stress parameters c and ф c (b) The critical state parameter M

48 Exercise The following results were obtained from undrained tests on specimens of an overconsolidated clay : Cell pressure (kpa) Deviator stress at failure (kpa) Deviator pore pressure (kpa) Determine the effective stress parameters c and φ p

49 Unconsolidated Undrained (UU) Test (quick test) Pore pressure develops during shear. Pore pressure is not measured, thus σ is unknown Analyse in terms of σ can gives the values of C u and φ u σ 3 and Δσ are applied fast so the soil does not have time to consolidate The test is performed with the drain valve closed for all phases of the test ( water is not allowed to drain) For this test, ф = ф = 0

50 CD Test CU Test UU Test

51 Advantages of Triaxial Test Easy to control drainage Useful stress-strain data Can consolidate sample hydrostatically Can simulate various loading conditions

52 Disadvantages of Triaxial Test Apparatus more complicated than other types of tests Drained tests on fine grained soils must be sheared very slowly

53 Unconfined compression test Axial compressive load applied to specimen until it fails Cross-sectional area at failure : A f = A 0 / 1-ε f where A f = cross-sectional area of failure A 0 = initial cross-sectional (πd 2 /4) ε f = axial strain at failure Figure : Loading and failure mode in an unconfined compression test.

54 No lateral confinement, so σ 3 = 0 Unconfined compressive strength : q u = P f / A f where P f = normal load at failure A f = cross-sectional area at failure Undrained shear strength : s u = P f / 2A f

55 Vane Shear Test Vane shear test is commonly used to measure the shear strength and sensitivity of clay This test is useful when the soil is soft and its water content is nearer to liquid limit such as clay. Suitable for determination of insitu undrained shear strength of non-fissured fully saturated clay Handle to apply torque Torque and angular displacement scale Laboratory vane test Sample Vane Classification of soil based on sensitivity

56 Advantages of Vane Shear Test Very rapid and inexpensive

57 Disadvantages of Vane Shear Test Not applicable to soils with fissures, silt seams, varves, other defects, or less than 100% saturation Sample disturbance not systematically accounted for

58 Worked Example

59 Worked Example Using data given, compute shear strength, s on horizontal and vertical planes at Point A,B and C

60 Point A vertical Solution vertical effective stress, σ z = H u σ z = [(17.0 x 3) + (17.5 x 1.1)] (9.8 x 1.1) = 59.5 kpa s = c + σ tan ф = 10 + (59.5 tan 28 o ) = 41.6 kpa horizontal horizontal effective stress, σ x = K σ z σ x = (0.54)(59.5) = 32.1 kpa s = c + σ tan ф = 10 + (32.1 tan 28 o ) =27.1 kpa USING SIMILAR STEPS, COMPUTE FOR POINT B AND C!!

61 Solution (cont ) POINT B Vertical : s = 57.2 kpa Horizontal : s = 35.5 kpa POINT C Vertical : s = 68.1 kpa Horizontal : s = 54.4 kpa

62 THANK YOU FOR YOUR ATTENTION