The shear strength of soil can be defined as its maximum resistance to shear stress just before the failure.
Almost all problems related to soil engineering are related
to the shear strength of the soil. It is the primary factor studied to
understand soil-structure interactions and for stability analysis during the
design of foundations, retaining walls, stability of earth slopes, etc.
This article series is exclusively created to explain the basic concept of the shear strength of soils along with their characteristics and theories employed in soil engineering.
The soil shear strength is a measure of soil resistance
against shear stresses. Shear stresses in the soil are caused due to direct
tension or direct compression.
Soil subjected to direct tension results in shear stresses
which would result in tension failure in the form of cracks and fissures. While
the soil under direct compression results in shear stresses that might result
in shear failure.
Hence, direct tension shear stress is not relevant compared
to shear stresses due to direct compression. So we can say, the shear failure
of a soil mass occurs when the shear stresses induced due to the applied
compressive loads exceed the shear strength of the soil.
When we say the shear failure of soil, it implies the failure of the soil due to the relative movements of the particles and not the breaking of the particles.
We now understood that shear strength is a complex
parameter of soil which is in turn a complex function of several contributing
factors, that includes:
These three parameters differ from soil to soil. The shear strength of clayey soil is not similar to that of sandy soil. Granular soil possesses shear strength due to its frictional resistance and particle interlocking. While in the case of cohesive soils, shear strength is due to cohesion between the soil particle.
The behavior of soil under the action of stress is studied
to develop a relation to easily understand the failure mechanism of soil
structure under varied load action. To study this, we consider the soil mass block
B subjected to a stress system.
Note: Stress systems in most studies are considered to be
three-dimensional. As for soil engineering problems, the stress acting in the
third direction is not relevant, the stress system is considered to be
two-dimensional.
Figure-1 |
Consider a block B subjected to a normal force Pn
acting normally on the surface MN and a force Fa acting in a tangential
direction as shown in Figure 2.
The normal force Pn remains constant while the
value of Fa is increasing from a value of zero to a value until it
slides block B.
The action of Fa force results in an equal and
opposite force Fr, which is caused due to the friction between the contact
surface between the block and MN. The force Fr balances the Fa and
keeps the block steady.
The forces Pn and Fr result in a
resultant force R that makes an angle δ with
the normal to the plane MN, as explained in the force diagram (Figure 2 (b)).
The angle δ is
called the angle of obliquity.
Block B starts to slide when Fa increases, and
the value of δ reaches a maximum
value of δm.
In the experiment conducted, block B and surface MN may or
may not be made out of the same material. If they are made of the same
material, then δm is
equal to the angle of friction (φ).
Then tan φ is the angle of
friction.
Then, the applied force Fa is the shearing force
and the force of friction Fr is the shearing resistance. If so,
The maximum shearing resistance that the materials are capable of developing is called shear strength.
The above experiment can be performed for a higher value of
normal force (Pn) which will result in a higher value of Fa.
Hence, from numerous experiments, it is studied that, Fa is directly proportional
to the normal force Pn.
Fa α Pn
Fa = Pn tan φ ( EQ.1)
If the contact area of block B is ‘A’, then,
shear strength s = force/area = Fa/A = [ Pn/A ]
tan φ [EQ.2]
[ from EQ.1]
normal stress σ
= normal force/ area = Pn /A
After substituting in EQ.2
s = σ tan φ [EQ.3]
The fundamental shear strength property of soil was first
recognized by Coulomb and he expressed the shear strength of soil as a linear
function of normal stress:
s= c + σ. tan φ [EQ.4]
The shear strength equation s = σ tan φ, is applied for soil with friction alone i.e.
granular soils. All soils are not purely granular. They will have an additional
strength contributed by the property of cohesion between soil particles.
Hence, the strength of soil is
contributed by two components. One is due to the cohesion between the soil particles (c) and the other one is due to
the friction between them (φ).
In the above equation, S is
the shear strength, c is the
cohesion, φ is the angle of shearing resistance and σ is the
normal stress. The equation expresses the assumption that cohesion c is
independent of the normal pressure σ acting on the plane of failure.
Figure-2: Coulomb's Law |
The c and φ are empirical parameters present in Coulomb’s equation.
These empirical parameters are
dependent on the history of the soil, the initial condition of the soil
(saturated or unsaturated), the permeability characteristics of the soil, and
the drainage conditions allowed in the soil.
For cohesion less soil c=0; then Coulomb’s
equation becomes s = σ tan φ.
The shear strength of the soil is
determined by laboratory methods and field methods. These test methods
determine the shear strength parameters c
and φ of the soils and thereby
calculate the shear strength of the soil at the site.
Read More: Terzaghi's Bearing Capacity Equation | Bearing Capacity of Soil for Foundation
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