The document discusses the shear strength of soil, detailing its importance in construction, failure modes, and methods of investigation such as unconfined compression and direct shear tests. It outlines the factors influencing shear strength, including cohesion and the angle of internal friction, along with various testing procedures. Examples provide calculations for determining soil cohesion and internal friction based on experimental data.
Shear Strength dueto Sliding Friction
Normal Stress vs. Shear Stress
Normal Stress (s) Shear Stress (f)
10 kPa (=kN/m2
) 6 kPa
20 kPa 12 kPa
30 kPa 18 kPa
9.
Shear Strength dueto Sliding Friction
Normal Stress (s)
ShearStress(s)
Angle of internal friction (f)
φσ
φσ
σ
φ
tan
tan
tan
+=
=
=
cs
s
s
10.
Coulomb Equation
:tan
:
:
:c
:
φ
φ
σ
s
Effective intergranularnormal pressure
(perpendicular to the shear plane)
Shear strength
Cohesion
(Strength gained from the ionic bound between grain particles)
Angle of internal friction
(Strength gained from internal friction resistance)
Coefficient of friction
φσ tan+= cs
11.
Shear strength ofsoil
• Soil cannot resist tension
• Soil can resist compression.
• For excessive compression failure occurs in the form
of shearing along the internal surface within the soil
• Structural Strength of soil = f ( soil’s shear strength)
• Shear Strength
– Soil’s ability to resist sliding
– Important for
• foundation design
• Lateral earth pressure calculations
• Slope stability analysis
12.
Methods of investigatingshear strength
• Unconfined compression test (for cohesive soil)
• Direct shear test
• Triaxial compression test
• Vane test (for soft clay)
• Standard penetration test (for cohesionless soil)
• Penetrometer test
Example
• A clayeysoil subjected to an unconfined compression
test fails at a pressure of 2540 lb/ft2
(i.e., qu = 2540
lb/ft2
). What is cohesion of this clayey soil?
15.
Direct Shear Test
NormalLoad
Sharing Force
Sharing Force
Normal Load
Sharing Force
Sharing Force
Soil Specimen
16.
Direct Shear Test
SoilSpecimen
Normal Load
Sharing Force
Normal Stress
ShearStress
c
Normal Stress = Normal load / the
specimen’s cross-sectional area
Shear stress = Shearing Force / the
specimen’s cross- sectional area
The graph can be used to determine the
given soil’s shear strength for any load
Problem: Shear failure is forced to occur
along a predetermined plane, which is
not necessarily the weakest plane of the
soil specimen tested.
Test with different Normal Load
φ
φ: angle of internal friction
c: cohesion
17.
Example
• A seriesof direct shear tests was performed on a soil
sample. Each test was carried out until the soil
specimen experienced shear failure. The test data
are listed next. What is soil’s cohesion and angle of
internal friction?
Specimen Number Normal Stress (lb/ft2
) Shearing Stress (lb/ft2
)
1 604 1522
2 926 1605
3 1248 1720
Triaxial Compression Test
s3
s3
p∆
Applya vertical load
and increase until the
specimen fails
Use different lateral
pressure, conduct the
same experiment
s3
s3
p∆
22.
Triaxial Compression Test
s3
p∆
p∆s3 : Minor principal
stress
Dp : Deviator stress at
failure axial = (Load at
failure / cross-sectional
area)
s3
s3
s3
23.
Triaxial Compression Test
s3
p∆+=31 σσ
s3 : Minor principal
stress
Dp : Deviator stress at
failure axial = (Load at
failure / cross-sectional
area)
s1 = s3 + Dp; major
principal stress
How to obtain cohesion
and angle of internal
friction?
s3
p∆+= 31 σσ
Example
• Gien
– Triaxialcompression tests on three specimens of a soil sample
were performed. Each test was carried out until the specimen
experienced shear failure. The test data are tabulated as follows:
• Required
– The soil’s cohesion and angle of internal friction
Specimen
Number
Minor Principal Stress
(kips/ft2
)
Deviator Stress at Failure
(kips/ft2
)
1 1.44 5.76
2 2.88 6.85
3 4.32 7.50