SHEAR STRENGTH OF SOIL
One of the most important and the most controversial engineering properties of soil is its shear strength or ability to resist resist sliding along alon g internal interna l surfaces within a mass. The stability stabilit y of a cut, the slope of an a n earth dam, the foundations of structures, the natural slopes of hillsides and other structures built on soil depend upon the shearing resistance offered by the soil along the probable surfaces of slippage. There is hardly a problem in the field of engineering which does not involve the shear properties propertie s of the soil in i n some manner or the other.
SHEAR STRENGTH OF SOIL
SHEAR STRENGTH OF SOIL
BASIC CONCEPT OF SHEARING RESISTANCE AND SHEARING STRENGTH
The basic concept of shearing resistance and shearing strength can be made clear by studying first firs t the basic principles of friction between solid bodies. Consider a prismatic block B resting on a plane pla ne surface MN surface MN as shown in Fig. 8. Block B is subjected to the force P n which acts at right angles to the surface MN surface MN,, and the force F a that acts tangentially to the plane. The normal force Pn remains constant whereas F a gradually increases from zero to a value which will produce p roduce sliding.
SHEAR STRENGTH OF SOIL
If the tangential force F a is relatively small, block B will remain at rest, and the applied horizontal force force will be balanced by an equal and opposite force force Fr on the plane of contact. This resisting resisti ng force is developed as a result of roughness characteristics of the bottom of block B and plane surface surfa ce MN. MN. The angle δ formed by the resultant R of the two forces F r and pn with the normal to the plane MN plane MN is known as the angle of obliquity.
SHEAR STRENGTH OF SOIL
If the applied horizontal force Fa is gradually increased, the resisting force F r will likewise increase, always always being equal in magnitude and opposite in direction to the applied force. Block B will start start sliding along the plane when the force force Fa reaches a value which will will increase the angle of obliquity to a certain maximum maximu m value δm. If block B and plane surface MN surface MN are made of the same material, material , the angle δm. is equal to ϕ which is termed termed the angle of friction, and the value tan ϕ is termed the coefficient of friction. If block Band plane surface MN surface MN are made of dissimilar materials, materials, the angle δ is termed the angle of wall friction.
SHEAR STRENGTH OF SOIL
The applied horizontal force Fa on block B is a shearing force and the developed force is friction or shearing resistance. The maximum shearing sheari ng resistance which the materials are capable of developing is called the shearing strength. If block Band plane surface MN surface MN are made of dissimilar materials, materials, the angle δ is termed the angle of wall friction. The applied horizontal force Fa on block B is a shearing force and the developed force is friction or shearing resistance. The maximum shearing sheari ng resistance which the materials are capable of developing is called the shearing strength.
SHEAR STRENGTH OF SOIL
If another experiment is conducted on the same block with a higher normal load P n the shearing force F a will correspondin correspondingly gly be greater greater.. A series of such experiments would show that the shearing force force F a is proportional to the normal load P n that is Fa =Pn tanϕ If A If A is the overall overall contact area of block B on plane surface MN, the relationship relationship may be written as
Shear strength,s(τ) strength,s(τ) = =
SHEAR STRENGTH OF SOIL
Or s(τ) =σ tanϕ (purely for granular soils) . However However,, soils soil s which are not purely granular granul ar exhibit exhibit an additional strength which is due to the cohesion between the particles It is, therefore, necessary ne cessary to separate separate the shearing strengths of such soils into two components, one due to the cohesion between the the soil particles and the th e other due to the friction between them.
THE COULOMB EQUA EQUATI TION ON
The fundamental shear strength equation proposed by the French engineer Coulomb (1776) is s(τ) = c +σ tan ϕ This equation expresses the assumption that the cohesion c is independent independe nt of the normal pressure a acting on the plane of failure. At zero normal pressure, pressure, the the shear strength of the soil is expressed as s=c According According to above equation, the cohesion of a soil is defined as the shearing strength at zero normal pressure on the plane of rupture.
THE COULOMB EQUA EQUATI TION ON
Normal pressure,
σ
THE COULOMB EQUA EQUATI TION ON
In Coulomb's equation c and ϕ are empirical parameters, the values of which for any soil depend upon several factors; the most important of these thes e are are : 1. The past history of the soil. 2. The initial state of the soil, i.e., whether it is saturated or unsaturated. 3. The permeability permeabil ity characteristics of the soil. 4. The conditions of drainage drainage allowed to take place during the test.
THE COULOMB EQUA EQUATI TION ON
Since c and ϕ in Coulomb's Equation depend upon many factors, c is termed as apparent cohesion and ϕ the angle of shearing resistance. For cohesionless soil soi l c = 0, then Coulomb's equation becomes s(τ) = σ tanϕ The relationship between be tween the various parameters parameters of Coulomb's equation is shown diagrammatically in Fig. 8.1
METHODS OF DETERMINING SHEAR STRENGTH PARAMET ARAMETERS ERS
The shear strength parameters c and ϕ of soils either in the undisturbed undisturbed or remolded remolded states states may be determined by any of the following methods: 1. Laboratory methods (a) Direct or box shear test (b) Triaxial compression test 2. Field method: (a) V (a) Vane ane shear test (b) Penetration test or by any other indirect methods
SHEAR SHE AR TEST APP APPARA ARATUS TUS
1. Shear Box Test Normal load Top platen Load cell to measure Shear Force
Motor drive Soil
Porous plates
Rollers Figure 8(a) Constant rate of strain shear box
Figure 8(b) Strain controlled shear apparatus
Direct Di rect Sh Shear ear Test
The original form of apparatus apparat us for the direct application applic ation of shear force is the shear box. The box shear test, though simple in principle, principl e, has certain shortcomings which will be discussed later on. The apparatus apparat us consists of a square brass box split horizontally horizonta lly at the level of the center of the soil sample, which is held between between metal grilles grilles and porous porous stones as shown in Fig. 8 (a). Vertical ertical load is applied to the sample as shown in the figure fig ure and is held constant during a test. A gradually increasing horizontal load is applied applie d to the lower part of the box until the t he sample fails in shear. shear.
Direct Di rect Sh Shear ear Test
The shear load at failure is divided by the crosscrosssectional area of the sample to give the ultimate shearing strength. The vertical load divided divide d by the area of the sample gives the applied vertical vertica l stress σ. σ. The test may be repeated with a few more samples having the same initial conditions as the first sample. Each sample is tested with a different vertical vertical load
Direct Di rect Sh Shear ear Test
The horizontal load is applied applie d at a constant rate rate of strain. The lower half of the box is mounted on rollers and is pushed forward at a uniform rate by a motorized gearing arrangement. The upper half of the t he box bears against a steel proving proving ring, the th e deformation of which is shown on the dial gauge indicating the shearing force.
Direct Di rect Sh Shear ear Test
To measure the volume change during consolidation and during the shearing process another dial gauge is mounted to show the vertical movement of the top platen. The horizontal displacement displacement of the th e bottom of the box may also be measured measured by another dial gauge which which is is not shown in the figure. Figure 8. (b) (b ) shows a photograph of strain controlled direct shear she ar test apparatus.
Direct Di rect Sh Shear ear Test
Advantages Advantages of Direct Shear Tests 1. The direct shear machine is simple and fast to operate. 2. A thinner soil sample is used in the direct shear test thus facilitating facilitat ing drainage of the pore water quickly from a saturated specimen. 3. Direct shear requirement is much less expensive as compared to triaxial equipment
Triaxial Compression Test Proving ring
Rubber membrane
8 (c) Diagrammatic layout
Triaxial Compression Test
Figure 8(d)Triaxial apparatus
Triaxial Compression Test
A diagrammatic diagrammatic layout layout of a triaxial test apparatus apparatus is is shown in Fig. 8 (c). In the triaxial triaxial compression test, three or more identical identic al samples of soil are subjected subjected to uniformly distributed f luid pressure pressure around the cylindrical surface. The sample is sealed in a watertight watertight rubber membrane.
Triaxial Compression Test
Then axial load l oad is applied to the soil sample until it fails. Although only compressiv compressivee load is applied to to the soil sample, it fails by shear on internal faces. It is possible to determine the shear strength of the soil from the applied loads at failure. Figure 8. (d) gives a photograph photograph of a triaxial triaxial test apparatus.
Comparison and Contr Contrast ast of Direct and Triaxial Shear Tests
Direct shear tests tes ts are generally suitable for cohesionless soils except fine sand and silt sil t whereas the triaxial test is suitable for all types of soils soil s and tests. Undrained and consolidated undrained u ndrained tests test s on clay clay samples can be made with the box-shear apparatus.
Advantages of the triaxial over the Advantages direct shear test
1. The stress distribution distrib ution across the soil sample is more uniform in a triaxial t riaxial test as compared compared to a direct shear test. 2. The measurement of volume changes is more accurate in the triaxial test. tes t. 3. The complete state of stress is known known at all stages during the triaxial test, whereas only the stresses at failure are known in the direct shear test.
Advantages of the triaxial over the Advantages direct shear test
4. In the case of triaxial shear, shear, the sample fails along a plane on which the combination of normal stress and the shear stress gives the maximum angle of obliquity obliquit y of the resultant with the normal, whereas in the case of direct shear, shear, the sample is sheared only on one plane which is the horizontal plane which need not be the plane of actual failure. 5. Pore water pressures can be measured in the case of triaxial shear shear tests whereas it is not possible in direct shear tests. 6. The triaxial machine is more adaptable.
STRESS CONDITION AT A POINT IN A SOIL MASS
Through every point in a stressed body there are three planes at right angles to each other which are are unique as compared compared to all the other planes passing pass ing through the point, because they the y are are subjected only to normal stresses with no accompanying accompanying shearing stresses acting on the planes. These three planes are called principal called principal planes, planes, and the normal stresses stress es acting on these planes are principal are principal stresses. Ordinarily the three principal princi pal stresses stress es at a point differ in magnitude.
STRESS CONDITION AT A POINT IN A SOIL MASS
They may be designated as the major principal stress σ1 the intermediate principal stress σ2 , and the minor principal stress σ3. Principal Principa l stresses at a point in a stressed body are important important because, once on ce they are evaluated, the stresses on any other plane through the point can be determined. Many problems in foundation engineering can be approximated by considering only two-dimensional stress conditions. The influence inf luence of the intermediate principal stress σ2 on failure may be considered considered as not very significant. signif icant.
g . wo- mens ona Demonstration Demonstr ation of the Exist Existence ence of Principal Planes
wo- mens ona emons ra on of the Exist Existence ence of Principal Planes
Consider the body (Fig. 9(a)) is subjected to a system of forces such as F1, F2• F3 . and F4. whose magnitudes magnitude s and lines of action act ion are known. Consider a small prismatic element P. The stresses acting on this element in the directions parallel to the arbitrarily chosen axes x axes x and y are shown shown in Fig. 9(b). Consider a plane AA plane AA through the element, making an angle condition of the α with the x-axis. The equilibrium condition element may be analyzed by considering the stresses acting on the faces of the triangle ECD (shaded) which is shown to an enlarged scale in Fig 9(c). The normal and shearing sheari ng stresses on the faces of the t he triangle are also shown.
wo- mens ona emons ra on of the Exist Existence ence of Principal Planes
The unit stress in compression and in shear on the face ED are designated as σand τ respecti respectively vely.. Expressions for σ and τ may be obtained by applying the principles of statics for theequilibrium condition of the body. The sum of all the forces forces in the x-direction is σxdx tan α+τxy dx+τ dx+τ dx sec αcos α – σ dx sec αsin α=0 (9.1)
wo- mens ona emons ra on of the Exist Existence ence of Principal Planes
The sum of all the th e forces forces in the the y-direction is σ y dx+τ dx+τ xy dx tan α –τ dx sec α sin α – σ dx sec α cos α =0 (9.2) Solving Eqs. Eqs. (9.1) and (9.2) for σ and τ , , we have +
σ=
1
2
+
− 2
cos2α +τxy sin2α
τ =2 − − τxy cos2α
(9.3) (9.3) (9.4) (9.4)
By definition, definition, a principal plane pl ane is one on which which the shearing stress is equal to zero. zero.
wo- mens ona emons ra on of the Exist Existence ence of Principal Planes
Therefore, when τ is made equal to zero in Eq. (9.4), (9.4) , the orientation orientati on of the principal planes is defined defined by by the relationship 2
tan 2α =
−
(9.5) (9.5)
Equation (9.5) (9 .5) indicates that there are two principal planes through the point P in Fig. 9(a) 9( a) and that they are at right angles angl es to each other other.. By differentiating Eq. (9.3) with respect to α and equatin equating g to zero, we have
wo- mens ona emons ra on of the Exist Existence ence of Principal Planes
= -σ σ sin2α+σ sin2α+2τ cos2α=0 y x xy 2
tan 2α =
−
or (9.6) (9.6)
Equation (9.6) indicates the orientation orienta tion of the planes on which the normal normal stresses stresses σ are maximum and minimum. This orientation orienta tion coincides with Eq. (9.5). Therefore, it follows that the principal principa l planes are also planes on which which the normal stresses are maximum and minimum.
