1. BEARING CAPACITY OF SOIL
Bearing capacity of a soil is the capacity of soil to support the loads
that are applied to the ground above. It depends primarily on the type
of soil, its shear strength and its density. It also depends on the depth
of embedment of the load – the deeper it is founded, the greater the
bearing capacity.
Ground bearing pressure (bearing capacity of soil) is important
because whenever a load is placed on the ground, such as from a
building foundation, a crane or a retaining wall, the ground must have
the capacity to support it without excessive settlement or failure.
This means that it is important to calculate the bearing capacity of the
underlying soil during the design phase of any construction project.
Failing to understand and account for the ground bearing pressure
before beginning the project could have catastrophic consequences,
such as a building foundation collapsing at a later stage.

2. METHODS OF DETERMINING SOIL BEARING
CAPACITY
1. TERZAGHI'S BEARING CAPACITY METHOD
Terzaghi's bearing capacity method is the earliest method proposed in 1943.
Failure Mechanism:
Terzaghi (1943) developed a rational bearing capacity equation for strip
footing, by assuming the bearing capacity failure of the foundation in general
shear mode
•
Figure 1. Shear stresses based on Terzaghi’s soil bearing capacity theory.
•

3. Zone I: A relatively undeformed wedge of soil below the foundation
forms an active Rankine zone with angles (45º + 𝜃/2).
Zone II: The transition zones take the form of log spiral fans.
Zone III: The wedge pushes soil outwards, causing passive Rankine
zones to form with angles
(45º - 𝜃/2).
Based on Terzaghi’s bearing capacity theory, column load P is resisted
by shear stresses at edges of three zones under the footing and the
overburden pressure, q (=gD) above the footing. The first term in the
equation is related to cohesion of the soil. The second term is related
to the depth of the footing and overburden pressure. The third term is
related to the width of the footing and the length of shear stress area.
The bearing capacity factors, Nc, Nq, Ng, are function of internal
friction angle, 𝜃.
• The Terzaghi's bearing capacity equation is given by:
qu = CNc + γ1DfNq + 0.5Bγ2Nγ

4. In the above equation,
qu= Ultimate Bearing Capacity of the soil
C = Cohesion
γ1,γ2 = Unit weight of the soil above and below the footing level
Nc, Nq, Nγ= Bearing capacity factors that are a function of friction angle
Df= Depth of the foundation below the ground level
B = Width or breadth of foundation
CNc = Contribution of cohesion
γ1DfNq = Frictional contribution of overburden pressure or surcharge
0.5Bγ2Nγ = Frictional contribution of self-weight of soil in the failure zone
Terzaghi's Bearing capacity equations:
• Strip footings:
Qu = c Nc + g D Nq + 0.5 g B Ng
• Square footings:
Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng
• Circular footings:
Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng

5. Where:
C: Cohesion of soil (apparent cohesion intercept);
g: unit weight of soil;
D: depth of footing (depth of embedment);
B: width/breadth of footing;
Nc, Nq, Nr: Terzaghi’s bearing capacity factors depend on soil friction angle, 𝜃:
Nq =
𝑒
(0.75𝜋−
𝜃
2
) tan 𝜃
cos2(45+
𝜃
2
)
Nc = (𝑁𝑞 − 1) cot 𝜃
N𝛾 =
tan 𝜃
2
𝐾𝑝
cos2 𝜃
− 1
Kp=passive pressure coefficient.
(Note: from Bowles' Foundation analysis and design book, "Terzaghi never
explained..how he obtained Kp used to compute Ng")

6. Table 1. Terzaghi’s Bearing Capacity Factors.
Where f is the friction angle

7. Example 1: Strip footing on cohesionless soil (British units)
Given:
Soil properties:
Soil type: cohesionless soil.
Cohesion: 0 (neglectable)
Friction Angle: 30 degree
Unit weight of soil: 100 lbs/ft3
Expected footing dimensions:
3 ft wide strip footing, bottom of footing at 2 ft below ground level
Factor of safety: 3
Requirement: Determine allowable soil bearing capacity using Terzaghi’s equation.
Solution:
• From Table 1 or Figure 1, Nc = 37.2, Nq = 22.5, Nr = 19.7 for f = 30 degree
• Determine ultimate soil bearing capacity using Terzaghi’s bearing capacity
equation for strip footing
• Qu = c Nc + g D Nq + 0.5 g B Ng
• = 0 +100x2x22.5+0.5x100x6x19.7
• = 10410 lbs/ft2
• Allowable soil bearing capacity,
• Qa = Qu / F.S. = 10410 / 3 = 3470 lbs/ft2 @ 3500 lbs/ft2

8. Example 2: Square footing on clay soil (British units)
Given:
Soil type: Clay
Soil properties:
Cohesion:1000 lbs/ft2
Friction Angle: 0 (neglectable)
Unit weight of soil: 120 lbs/ft3
Expected footing dimensions:
6 ft by 6 ft square footing, bottom of footing at 2 ft below ground level
Factor of safety: 3
Requirement: Determine allowable soil bearing capacity using Terzaghi’s equation.
Solution:
From Table 1 or Figure 1, Nc = 5.7, Nq = 1.0, Nr = 0 for f = 0 degree
Determine ultimate soil bearing capacity using Terzaghi’s bearing capacity equation for
square footing
Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng
= 1.3x1000x5.7 +120x2*1+ 0
= 7650 lbs/ft2
Allowable soil bearing capacity,
Qa = Qu / F.S. = 7650 / 3 = 2550 lbs/ft2 @ 2500 lbs/ft2

9. Example 3: Circular footing on sandy clay (British units)
Given:
Soil properties:
Soil type: sandy clay
Cohesion: 500 lbs/ft2
Friction Angle: 25 degree
Unit weight of soil: 100 lbs/ft3
Expected footing dimensions:
10 ft diameter circular footing for a circular tank, bottom of footing at 2 ft below ground level
Factor of safety: 3
Requirement:
Determine allowable soil bearing capacity using Terzaghi’s equation.
Solution:
From Table 1 or Figure 1, Nc = 17.7, Nq = 7.4, Nr = 5.0 for f = 20 degree
Determine ultimate soil bearing capacity using Terzaghi’s bearing capacity equation for circular
footing
Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng
= 1.3 x 500 x 17.7 +100 x 2 x 7.4+0.3 x 100 x 10 x 5.0
= 17985 lbs/ft2
Allowable soil bearing capacity,
Qa = Qu / F.S. = 17985/ 3 = 5995 lbs/ft2 = 6000 lbs/ft2

10. SLOPE FAILURE(SOIL MASS
MOVEMENT)
Slope failure is a geohazard that impacts many types of infrastructure,
from individual homes to municipal storm sewer networks to oil and gas
pipelines.
Types of slope failuresSoil slope failures are generally of four types :
Translational Failure
Rotational Failure
Wedge Failure
Compound Failure
Translational Failure
Translation failure occurs in the case of infinite slopes and here the failure
surface is parallel to the slope surface.
A slope is said to be Infinite, when the slope has no definite boundaries
and soil under the free surface contains the same properties up to
identical depths along the slope.

11. Rotational Failure
In the case of rotational failure, the failure occurs by rotation along a slip
surface and the shape thus obtained in slip surface is curved. Failed
surface moves outwards and downwards.
In homogeneous soils, the shape is circular while in case of non-
homogeneous soils it is non-circular.
Rotational failure may occur in three different ways :
– Face failure or slope failure
– Toe failure
– Base failure
• Face failure occurs when soil above the toe contains weak
stratum. In this case the failure plane intersects the slope
above toe.
• Toe failure is the most common failure in which failure plane
passes through toe of slope.
• Base failure occurs when there is a weak soil strata under the
toe and failure plane passes through base of slope.
Rotational failure can be seen in finite slopes such as earthen dams,
embankments, man-made slopes etc.

12. SLOPE FAILURES
Wedge Failure
Wedge failure, also known as block failure or plane failure, generates
a failure plane that is inclined.
This type of failure occurs when there are fissures, joints, or weak soil
layers in slope, or when a slope is made of two different materials.
It is more similar to translational failure but the difference is that
translational failure only occurs in case of infinite slopes but wedge
failure can occur in both infinite and finite slopes.
Compound Failure
A Compound failure is a combination of translational slide and
rotational slide.
In this case, the slip surface is curved at two ends like rotational slip
surface and flat at central portion like in translational failure.
The slip surface becomes flat whenever there is a hard soil layer at a
considerable depth from toe.

13. CAUSES OF SLOPE FAILURES
Following are the ways that can affect the stability of slopes :
1. The decrease in the shear strength of the soil.
2. The increase in the shear stress that ultimately causes the failure of soil
1. Decline in Shear Strength of soil
Several factors can result in a decreased shear strength of the soil. The following
factors are of particular significance with respect to slope stability.
1.1 Increase in Pore Water Pressure
The frequent increase in the groundwater table and upward seepage, as an
outcome of uncommonly heavy rains, are the most common reasons for
increased pore water pressure. As a result, the associated effective stresses
decrease with increased pore water pressure.
1.2 Cracking
Slope failures are often preceded by the advancement of fractures through the
soil near the crest of the slope. These fractures appear as an outcome of tension
in the soil at the ground surface that goes beyond the tensile strength of the soil.
Therefore, as the tensile strength of soil reduces, the shear strength on crack-
plane also reduces.

14. CAUSES OF SLOPE FAILURES
1.3 Swelling
Highly plastic and over-consolidated clay easily swells when it comes in
contact with water. A classic example to delineate failure due to
swelling can be taken from a case in Houston, Texas, where highway
embankments constructed with extremely plastic compacted clays
failed after ten years as a result of swelling and the loss of shear
strength.
1.4 Decomposition of Clayey Rock Fills
Claystone and shales as a filling material in rock joints can be used by
breaking them into pieces to form a sound rock that can be relatively
stable after compaction. However, with time, as the compacted fill
gets in contact with groundwater or seepage water, the disintegration
of compacted fill can result in chunks of clay particles.
These chunks of clayey particles then swell into the open spaces within
the fill, causing a reduction in the shear strength of soil, and making
the fill unstable.

15. CAUSES OF SLOPE FAILURES
1.5 Creep
Under sustained loading, the highly plastic clays undergo constant
deformation. Therefore, after a certain period, clays might ultimately fail,
even at low shear stresses.
The impact of creep is worsened under cyclic loads such as freezing,
thawing, wetting, and drying conditions. When such cyclically differing
conditions are at their unfavorable extremes, movement of soil within the
slope takes place in the downhill direction. Therefore, in the long term, a
downslope motion may develop that ultimately results in the failure of
slope along the critical plane.
1.6 Leaching
As the water seeps through the voids of the soil, the chemical and
mechanical properties of soil start undergoing modifications. This process
is known as leaching.
In the case of marine clays, leaching plays an important role as it
contributes to the development of quick clay conditions, and such clays
have significantly no strength when disturbed.

16. CAUSES OF SLOPE FAILURES
1.7 Strain Softening
Strain softening phenomenon is associated with brittle soils. In a stress-strain curve of
brittle soils, when the critical stress reaches the peak, the shear strength of brittle soils
reduces with more constant strain. This type of stress-strain behavior makes a
progressive failure, thus creating a path for slope failure.
1.8 Weathering
The process in which rocks and soils lose their strength due to the modifications in
physical, chemical, and mechanical properties by external agents such as water, wind,
temperature change, etc. is known as weathering.
Weathering significantly reduces the shear strength of the soil. In the worst case,
weathering can change the whole structure of rocks and soil, transforming a strong
and stable slope into an unstable one.
1.9 Cyclic loading
Under the impact of cyclic loads, the bond between the soil particles may break, and
the pore water pressure might increase, resulting in the loss of strength. The loose soil
may get saturated due to the increase in pore water pressure and lose all its strength
under cyclic loading because of liquefaction.

17. CAUSES OF SLOPE FAILURES
2. Increase in Shear Stress of Soil
Even if the shear strength of the soil stays intact, the change in the shear stress due to
an increase in loading may destabilize the slopes.
The factors through which shear stresses can be increased are discussed below:
2.1 Loads at the Top of the Slope
If the ground at the top of a slope is loaded, the shear stress needed for equilibrium of
the slope will be more. If the loads are kept away from the slope’s crest, it would be
possible to prevent the increase in shear stresses.
2.2 Water Pressure in Fractures
A slope can become unstable if fractures at the top of a slope are filled with water. The
pressure created due to water in the fractures loads the soil and increases the shear
stresses. If the cracks stay filled with water for seepage towards the slope face to
establish, then the pore water pressure in the soil will increase, leading to an even
worse condition.
2.3 Due to an Increase in Soil Weight
Seepage into the soil within a slope can increase the water content of the soil, thereby
increasing its weight. If this increase in weight is considerable, specifically with the
combination of other forces, it can lead to slope failure.

18. CAUSES OF SLOPE FAILURES
2.4 Excavation
Excavation that makes a slope steeper will increase the shear stress in the soil within
the slope and reduce stability. The disintegration of soil by a stream at the base of a
slope has the same result.
2.5 Drop in Water Level at the Base of a Slope
External water pressure acting on the face of the slope provides a stabilizing result.
Rather, the slope will become unstable if the water content reduces by increasing
shear stress.
When this level drops quickly, and the pore pressures within the slope are not reduced
in accordance with the drop in the water level outside, the slope will be less stable.
This phenomenon is known as the rapid drawdown condition and is important for the
design of partially submerged slopes.
2.6 Earthquakes
In the event of an earthquake, slopes are subjected to vertical and horizontal
velocities that result in cyclic variations in stresses within the slope. This increases
them above their static values for brief durations, lasting for seconds or fractions of a
second. Even if the shaking doesn’t cause any modification in the strength of the soil,
the stability of the slope will reduce during those quick instants when the dynamic
forces act in adverse directions.

19. TRANSLATION SLIDE ON AN INFINITY SLOPE
Translational slide
Translation failure occurs in the case of infinite slopes and here the failure surface is
parallel to the slope surface.
A slope is said to be Infinite, when the slope has no definite boundaries and soil under
the free surface contains the same properties up to identical depths along the slope.
As said above, when the soil along the slope has similar properties up to a certain
depth and soil below this layer is strong or hard stratum, the week topsoil will form a
parallel slip surface when failed.
This type of failure can be observed in slopes of layered materials or natural slope
formations.
Fig 1: Translational slide

20. REMEDIAL ACTION TO PREVENT FAILURE OF
SLOPES
• The methods for the remediation of slope
include;
1. Modification in slope geometry
2. Drainage,
3. Use of retaining structures and
4. Internal slope reinforcement