giant landslide in Tibet 2000 Shear strength 5.1 General Shear strength 5.1 General

granary foundation failure on clay Shear strength 5.1 General Shear strength 5.1 General

Foundation soil liquefaction caused by earthquake Shear strength 5.1 General

5.1 General Embankment Foundation Retaining wall Key problem: shear strength Shear strength 5.1 General Shear strength

Translation slip Collapse Rotational slip flow slide 5.1 General Shear strength

5.2 Coulomb law τ f = c + σ’ tan φ τ f = shear strength c = cohesion φ = angle of internal friction σ1σ1 σ 1 major principle stress σ3σ3 σ 3 Minor principle stress Confining stress σnσn τfτf  f =c+  tan  粘土 c   ff Shear strength

Consider the following situation: A normal stress is applied vertically and held constant A shear stress is then applied until failure 33 33 11 11    33 11  dl dlcos  dlsin  Shear strength

  O 11 33 1/2(  1 +  3 )  22 A( ,  ) For any given normal stress, there will be one value of shear stress If the normal stress is increased, the shear stress will typically increase in sands and stay the same in clays Shear strength

5.3 Mohr–Coulomb failure criterion 33 11 c   f f 2 f2 f A   cctg  1/2(  1 +  3 ) For sandy soil: c=0 Shear strength

 f f 2 f2 f 33 11 c  A   cctg  1/2(  1 +  3 )  max In the case represented by the figures in this chapter, in which it is assumed that the vertical direction is the direction of the major principal stress, the planes on which the stresses are most critical make an angle π/4 −  / 2 with the vertical direction. Thus it can be expected that sliding failure will occur in planes that are somewhat steeper than 45. If for instance  = 30, which is a normal value for sands, failure will occur by sliding along planes that make an angle of 30 with the vertical direction. Shear strength

Soil Normal stress σ n Shear stress σ 3 (1) Direct shear test 5.4 Shear test Shear strength

 Direct shear test is Quick and Inexpensive  Shortcoming is that it fails the soil on a designated plane which may not be the weakest one Shear strength

Shear stress Shear displacement Peak Strength Residual Strength The discussion thus far have referenced failure of the soil. Failure is indicated by excessive strain with little to no increase (even decrease) in stress. After failure, the soil strength does not go to 0 The soil retains residual strength Shear strength

φ Shear stress normal stress Typical plot for clays - drained condition Overconsolidated OCR >1 normallyconsolidated OCR=1 c Shear strength

(2) Triaxial shear test  3 3  3 3  3 3  3 3  3 3  3 3 △△ △△ The test is designed to as closely as possible mimic actual field or “in situ” conditions of the soil. Shear strength

Triaxial tests are run by: saturating the soil applying the confining stress (called σ3) Then applying the vertical stress (sometimes called the deviator stress) until failure 3 main types of triaxial tests:  Consolidated – Drained  Consolidated – Undrained  Unconsolidated - Undrained Shear strength

The specimen is saturated Confining stress (σ3) is applied This squeezes the sample causing volume decrease Drain lines kept open and must wait for full consolidation (u = 0) to continue with test Once full consolidation is achieved, normal stress applied to failure with drain lines still open Normal stress applied very slowly allowing full drainage and full consolidation of sample during test (u = 0) =’=’  f=ff=f Shear strength

(3) Unconfined compression test ququ ququ The specimen is not placed in the cell Specimen is open to air with a σ 3 of 0 Test is similar to concrete compression test, except with soil (cohesive – why?) Applicable in most practical situations – foundations for example. Drawing Mohrs circle with σ 3 at 0 and the failure (normal) stress σ 3 defining the 2 nd point of the circle – often called q u in this special case c becomes ½ of the failure stress Shear strength

加压 框架 量表 量力环 升降 螺杆 无侧限压缩仪 ququ ququ Shear strength unconfined compression apparatus

Shear strength unconfined compression apparatus Shear strength unconfined compression apparatus

  ququ cucu  u =0 Shear strength unconfined compression apparatus

sensitivity High sensitivity S t >4 Middle sensitivity 2< S t ≤4 Low sensitivity 1<S t ≤2 the effects of disturbance of soil constitutive property on soil strength Shear strength

(4) Vane shear test This test is used for the in-situ determination of the undrained strength of intact, fully saturated clays; the test is not suitable for other types of soil. In particular, this test is very suitable for soft clays, the shear strength of which may be significantly altered by the sampling process and subsequent handling. Generally, this test is only used in clays having undrained strengths less than 100 kN/m2. This test may not give reliable results if the clay contains sand or silt laminations. Shear strength

Example Shear strength

Triaxial tests rarely run The unconfined test is very common In most cases, clays considered φ = 0 and c is used as the strength Sands are considered c = 0 and φ is the strength parameter Direct shear test gives us good enough data for sand / clay mixes (soils with both c and φ) Tables showing N value vs strength very commonly used (page 567 for clays for example). Remark Shear strength