Classification of soil
The soil classification according to the size of particles.
The soil which diameter of soil less than 0.067 mm called cohesive soil and consists of clay and silt.
The soil diameter of particles more than 0.067 called cohesion less soil and consists of sand, gravel and stone
Soil separate – size limit as:
Stone > 60 mm
|Gravel||Coarse gravel||20 – 60 mm|
|Medium gravel||6 – 20 mm|
|Fine gravel||2 – 6 mm|
|Sand||Coarse sand||0.6 – 2 mm|
|Medium sand||0.2 – 0. 6 mm|
|Fine sand||0.06 – 0.2 mm|
|Silt||Coarse silt||0.02 – 0. 06 mm|
|Medium silt||0.006 – 0.02 mm|
|Fine silt||0.002 – 0.006 mm|
Clay < 0.002
Sketch the classification of soil according to the size of Diameter shown in fig. (1)
The Soil Consists of two kinds of soil
Inorganic Soil which consists of cohesive soil (clay - silt ) and cohesive less soil (sand - gravel - stone)
Properties of Soil:
Between the soil particles there are the internal forces that the soil can be offer to resist failure and sliding a long any plane inside it. One of there is cohesive strength (C) and the other is angle of internal friction (φ).
The cohesive strength appear clearly in cohesive soil, more than the cohesion less soil is called by this forces ..
Cohesive soil à called C – soil and
Cohesion less soil à called φ – soil
But the soil in general called C – φ soil.
Determination of soil properties (shear strength):
The shear strength parameters of soil can be determined in the laboratory primarily by three types of tests. Direct shear test, triaxial test and unconfined compression test.
1. Direct shear test:
This is the oldest and simplest form of shear test. The test equipment consist of a metal shear box in which the sample is placed as shown in fig. (2).
The box is split horizontally into two halves. Normal force on the sample is applied from the top of the shear box by dead weight. Shear force is applied to the side of the top half of the box to cause failure in the soil sample. For the given test. The normal stress can be calculated as:-
|σ = normal stress =||=|
and shear stress can be calculated as :
|τ = shear stress =||=||
The test repeated more than 3 time with different value for force P (normal force). Followed that the difference value for force T (shear force).
Put the results in table and sketch as shown in fig. (3).
Fig (3) – A : The results in table shear stress
The shear strength value can be determined as shown, where …
φ = Angle of internal friction.
C = Cohesive stress or adhesion stress
The equation for the average line obtained from experimental results called coulomb law.
S = C + σ tan φ
S : Shear strength
C : Adhesion stress
φ : Friction angle
σ : Normal stress
1. Triaxial Compression Test:
· Triaxial compression test is one of the most common method for determination the shear strength parameters or C and φ for soil.
· The sample dimensions are 1.5 in and 3 in diameter and length, As shown in fig (4) the sample is encased by a thin rubber membrane and placed inside plastic cylindrical chamber that is usually filled with water which is under pressure, the sample is effected with axial load which caused axial stress. The axial stress increment until the sample fails, and the axial deformation is measured by a dial gauge Δ σ, as shown in fig (5), the soil sample is subjected to an all around confining pressure σ3.
σ3 = Pore water pressure on confining stress
σ1 = Total axial stress at failure
σ3 = σ3 + Cσ1
Fig (5) : Stress Application
σ3 : Pore water pressure on confining stress
σ1 : Total axial stress at failure
σ1 = σ3 + Δ σ
in triaxial test σ1 is the major principle stress and σ3 is the minor stress several test on similar samples can be conducted by varying the confining pressure, with the major and minor principle stress at failure for each envelop can be obtained the following relation show fig. (6) and fig. (7)
Fig. (6) : Mohr's Circle
Fig (7) : The table of the results
As shown in fig (6) the plan of failure inclination Θ with the major principle plane.
|Θ = 45 +|
φ : An angle of internal friction
And the shear strength equation can be written as
S = C + σ tan φ
The triaxial compression test was carried and the results were as follow. Determine the shear strength parameters of soil.
Cell Pressure σ3 KN/m2
Deviator Stress at failure Δ σ
|280||205||130||Δ σ KN/m2|
|580||405||230||σ1 = σ3 + Δ σ|
Fig (8) : From the curve we can measure C and Q
3 . Unconfined Compression Test:
This special type of test used for clay sample as shown in fig (9) , where φ = 0 in that test the confining pressure σ3 is zero, axial load is rapidly applied to cause failure, at failure the minor principal stress σ3 = 0 and the major principal stress is σ1, the relation between stresses, shear and normal as shown in fig. (10). So unconfined cohesive strength is (Cu).
Fig. (9) : Unconfined Compression Test
σ1 = minor stress called unconfined stress qu
Fig. (10) : stresses relations for unconfined compression test.