Lateral Earth Pressure 93
Available Factor of safety for gross passive resistance ( Σ M
A
= 0)
FS =
P p
P pm
=
776
( 5165 .
3 13 )
=
776
397
= 1 .
95
Force in each tie rod
2 T = 2 (578.6 – 776) = 395 kN (distance 2 m)
Reinforced soil wall, introduced by Vidal, is an alternative to conventional walls.
This wall consists of a compacted soil mass, reinforcing element, usually in the form of metal strips, geogrids, or geotextiles, and facing unit. It provides a flexible structure that is more tolerable to settlement or movement and it required minimal amount of material and equipment for construction. Thus it is particularly useful for sites with poor ground near the surface or sites with limited access where conventional wall is difficult to construct. Figure 3.16 shows the typical reinforced soil wall.
FACING UNITS
REINFORCING STRIP
H
SELECTED FILL
COMMON
FILL
Figure 3.16 Parts of a reinforced soil wall
The stability analysis of reinforced soil wall is performed in two parts: i.e. the external or overall stability and the internal stability analysis.

94 Introduction to Geotechnical Engineering Part 1
The overall stability analysis is conducted by assuming that the wall behaves as a unity like a gravity wall. The stability is provided by the weight of the reinforced soil part of the wall. The external stability checks include the stability against sliding on the base, overturning, bearing capacity failure and deep seated failure analysis.
The internal stability of reinforced soil wall resulted from frictional forces developed between the soil and the reinforcement. The analysis involves determining tensile and pullout resistance of the reinforcing element and the integrity of the facing element. This analysis require an assumption on the failure plane, distribution of lateral earth pressure, and empirical relationship for soil-skin friction. Depending on the flexibility of the reinforcing element, there are two methods of analyzing the internal stability of the wall. Figure 3.17 shows the methods used for internal stability analysis of the reinforced soil wall.
For wall with flexible reinforcement such as geotextile, the wall is assumed to move around its toe and the soil is in active condition along the height of the wall.
In this case the Coulomb failure wedge is assumed. The internal stability analysis for this type of wall is evaluated using the Tie Back wedge method.
On the other hand, if the reinforcement is relatively stiff, research showed that the failure wedge will form along the locust of maximum tensile force developed in the reinforcing element as shown in Figure 3.17b. In this method, the movement of the wall started at the lower part, thus the soil in the upper part is still in “atrest” condition. Study shows that the active condition is only achieved at depth of
6m below the top of the wall. The method is known as the Coherent Gravity methods.
Tensile force acting on the reinforcement due to transfer of lateral stress from the soil to the reinforcing element is given by:
T = σ v
S
V
S
H
= γ z K S
V
S
H
(3.40) where z is the depth of the reinforcing element, S
V
and S
H
are the vertical and horizontal distance between two reinfocring element, and K is the lateral earth pressure coefficient. For flexible reinforcement K is equal to K reinforcement K varies from K o
at the surface to K a resistance of the reinforcing element is a
, while for rigid
at depth of 6 m. The
T
R
= b t f y
(3.41)

Lateral Earth Pressure 95 where b and t are the width and the thickness of the reinforcement, while f y tensile strength of the material.
is the
H
H
45+ φ ’/ 2
Le (i) ith layer
B
(a) Tie-back wedge method
0.3H
S v
Le (i)
H
2 ith layer
H
2
B
(b) coherent gravity method
Figure 3.17 Assumption made on the failure wedge and the length of anchorage

96 Introduction to Geotechnical Engineering Part 1
The tensile force acting on the element also acts as the pull-out force. The frictional resistance at the interface between the soil and the reinforcement is:
R = 2 b L e
γ z f* (3.42) where L e
is the effective length i.e. the length of the element beyond the failure line (see Figure 3.17), and f* is the friction developed between the soil and the reinforcing element.
The friction ( f* ) depends on the mechanism at the interface between the soil and the reinforcement which is affected also by the overburden pressure on the reinforcing element. For metal strip, there is a dilatancy effect that results in the shear strength developed at the interface. The maximum friction occur near the top of the wall and it decreases linearly as the depth increases. Research showed that the maximum friction for metal strip (reinforced earth wall) is 1.5 at the surface and decrease linearly to tan δ at depth of 6 m. For geogrids, the friction depends on the geogrid’s apperture size and the size of soil particles. For geotextile, the friction occurs only between the soil and the reinforcement, thus f * = tan tan
δ
φ
where
δ ranges between 0.5 – 0.8 φ .
Example Problem 3.14
A 10 m high retaining wall reinforced with galvanized steel strips in a granular backfill is to be constructed. The unit weight of the granular backfill is 16.5 kN/m 3 , while its internal friction angle is 36 o foundation soil are c ’ kPa, and φ ’ = 28
. The shear strength parameters of the o . The width of galvanized strip is 7 cm while its thickness is 5 mm, and the vertical and horizontal center to center distance between the strips ( S
V
= S
H
) are equal to 1 m. The tensile strength of the steel ( f y
) is 250 MPa. (a) Check the external stability of the wall assuming the bearing capacity of the foundation soil is 300 kPa and the base friction angle is
24 o , (b) Evaluate the internal stability of the wall if the lifespan of the structure is
50 years and the corrosion rate is 0.02 mm/year.

Lateral Earth Pressure
H = 10 m
W
97
B = 12 m
Figure P3.14
Solution
External stability analysis (for B = 12 m) assuming active condition prevails along the height of wall (flexibel reinforcement)
K a
= tan 2 ( 45–
φ
) = 0.26
2
P a
= ½ γ
1
K a
H 2 = ½ ×16.5×0.26×(10) 2 = 214.5 kN/m 2
W = γ B H = 16.5×12×10 = 1980 kN/m ’
Check for overturning
M
RO
= ½ W B = ½ × 1980×12 = 11880 kN-m/m ’
M
O
= P a
H /3 = 214.5 × 10/3 = 715 kN-m/m ’
FS overturning
=
M
M
RO
O
=
11880
715
=
16.6 > 3 Æ OK
Check for sliding along the base
Assume the friction angle at the base is 2/3 φ = 24 o
FS sliding
=
W tan δ b
P a
=
1980 tan 24 o
214 .
5
=
4.1 > 3 Æ OK

98 Introduction to Geotechnical Engineering Part 1
Check for bearing capacity e =
B
2
−
M
RO
V
− M
O =
12
2
B e <
6
−
11880 −
1980
715
0.36 < 2 m. Æ OK
=
0.36
Assume ultimate bearing capacity is 300 kPa
P =
V
B
⎜ 1
6
B e
=
1980
12
⎜ 1 ±
6 × 0
12
.
36
= 81.67 (1±0.18)
P max
= 96.37 kPa < BC soil P min
= 67 kPa > 0 Æ OK
Internal stability check
The check should be done for each reinforcement layer. To simplify the calculation we will have to develop a table (Table P 3.14)
Table P3.14
No z
(m)
L e
(m)
T (kN) R (kN)
2bL e
γ z f*
T
R b t f y
FS tensile
FS pullout
1 0.5 0.3975 1.405 9 3.2794
2 1.5 0.3725 1.215 9 9.2194
3
2.5 0.3475 1.025 9 14.3344
4
3.5 0.3225 0.835 9 18.6244
5
4.5 0.2975 0.645 9 22.0894
6 5.5 0.2725 0.455 9.3 24.7294
7 6.5 0.26 0.36 9.9
27.885
8 7.5 0.26 0.36 32.175
9
10
11.1
11.7
36.465
40.755
3.7422 70.0 18.70 1.1411
1.2177
18.711 70.0
4.88 1.3053
26.1954 70.0
3.75 1.4065
33.6798 70.0
3.17 1.5247
42.5363 70.0
2.81 1.7200
53.5135 70.0
2.51 1.9190
65.4885 70.0
2.18 2.0353
78.4615 70.0
1.92 2.1516
92.4323 70.0
1.72 2.2680
Note: For metal strip reinforcement,
K
L e
varies from K o
at the surface to K a
at depth of 6m will follow the coherent gravity methods f* varies from 1.5 at the surface to tan δ at depth of 6m
K a
= K a
= tan 2 ( 45–
φ
) = 0.26 K o
= 1sin 36 = 0.41
2 tan δ = tan 20 = 0.36
Thickness of metal strip after 50 years = 5 – (0.02 x 50) = 4 mm

Lateral Earth Pressure 99
From Table 3.14
The minimum factor of safety against tensile breakage = 1.72 ( at z = 9.5 m)
The minimum factor of safety against pullout = 1.14 (at z = 0.5m)
3.1.
A wall is required to retain a cut on a cohesionless soil which has a unit weight of 18.4 kN/m 3 and effective internal friction angle φ ’ of 25 o . The height of the cut is 3 m. If the wall is very rigid and the soil is in “at rest” condition (a) Draw the diagram of lateral pressure behind the wall (b)
Calculate the magnitude and point of application of the thrust force on a wall
3.2.
An excavation was to be made on a cohesive soil with unit weight kN/m 3 ; c’ = 10 kPa, and ϕ ' = 10 o
γ b
= 19.2
. Calculate the depth of unsupported excavation. Plot the active earth pressure diagram if the excavation was to be made up to 9 m deep, and calculate the resulting force and point of application.
3.3.
A vertical wall retaining a 3 m high dry sand with unit weight 18 kN/m
φ ’ = 36 o
3 , and internal friction angle . What is the thrust force working on the wall if uniform surcharge load of 10 kN/m 2 was placed on the surface of the soil behind the wall. Active condition prevails.
3.4.
A retaining wall as shown in figure Q3.4 retains soil for a depth of 12 m.
The soil mass consists of two layers with soil properties given in the figure.
Water table coincide the interface between layer 1 and 2. (a) determine the magnitude and point of application of the active thrust; (b) determine also the hydrostatic pressure working on the back of the wall.
7.0 m
Figure Q3.4
5.0 m c’ = 0
φ ’
= 28 o
γ b
= 18 kN/m 3
γ c’ = 0
φ ’
= 34 o sat
= 20 kN/m 3

100
5.4m
Introduction to Geotechnical Engineering Part 1
3.5.
A concrete wall is designed to retain a cohesionless soil with the following properties: γ = 20.4 kN/m 3 ; φ ’ = 35 o ; c’ = 0, δ = 20 backfill soil form an angle of 10 o to the horizontal. o . Determine the active thrust force acting on the wall if the height of the wall is 4.6 m and the
3.6.
A vertical wall is retaining a 5 m high soil fill with unit weight of 20 kN/m 3 , φ ’ = 32 o , δ = 20 o . A line load P = 75 kN/m is working on the soil surface parallel to the wall. What is (a) the distance between the line load and the back of the wall so the load will not give additional stress to the wall, (b) the thrust force on the wall if the line load is working 2 m behind the wall.
3.7.
A gravity wall 8 m high is retaining a soil with horizontal surface. The saturated unit weight of the sand is18.5 kN/m 3 and the internal friction angle φ ’ = 30 o . The wall friction angle δ = 20 o . Ground water table is at the surface and there is no drainage to dry the soil quickly. Calculate the total force applied to the back of the wall.
3.8.
For the retaining wall shown in Figure Q3.8, calculate the factor of safety against overturning, sliding, and bearing capacity failure. Use the Rankine equation to compute the earth pressures.
0.3m 18 o
Backfill soil
γ = 18 kN/m 3
φ ’
= 35 o
0.9m
0.6m
Figure Q3.8 0.45m Coefficient of base friction
= 0.58
0.9m 1.65m
3.9.
A retaining wall shown in Figure Q3.10 is to be constructed of concrete with unit weight of 23.5 kN/m properties: γ = 18 kN/m 3 , c’
3 . The backfill soil has the following
= 0, and φ soil and the concrete wall is given by
’
δ
= 30
= 20 o o
. The friction between the
. The coefficient of base friction is 0.55 and the soil’s ultimate bearing capacity is 620 kN/m 2 .

Lateral Earth Pressure 101
0.60 m 1.20 m
15 o
6.0 m
0.60 m
0.50 m 95 o
3 m
Figure Q3.9
3.10.
A sheet-pile wall is used to retain a cut to a depth of 6.5 m. An anchor was placed at depth of 1.65 m below the ground surface (see figure Q3.10). Use the free earth support method to calculate the depth of embedment of the sheet pile for a factor of safety with respect to passive pressure of 2.5.
Estimate the force on the anchor. q = 10 kN/m 2
6.5 m
1.65 m
Loose sand
γ = 16.5 kN/m 3
φ ’
= 23 o d ?
Dense sand
γ = 18.5 kN/m 3
φ ’
= 33 o
Figure Q3.10
3.11
A 6-m high reinforced earth wall is to be constructed using granular soil with unit weight of 17.2 kN/m 3 , and angle of internal friction of 34 o as backfill material. The steel strip of width 75 mm will be used for reinforcement. Following the size of facing unit, the vertical and horizontal spacing of the strips are 0.3 m and 1.0 m respectively. The allowable stress of the steel is 140,000 kN/m 2 . Determine (a) the total length of strip required for a factor of safety against pullout 1.5, and (b) the required thickness of strip.

102 Introduction to Geotechnical Engineering Part 1
3.12 A quay wall as shown in Figure Q3.12 is to be constructed using anchored sheet piling as shown in figure below. The unit weight of the soil is
17kN/m
20kN/m
3
3
above ground water table, and the saturated unit weight is
below. The shear strength parameters are c ’ = 0: and ϕ ’
= 36 o . For a factor of safety of 2.0 with respect to passive resistance, determine the required depth of embedment and the force in each tie if they are spaced at 2 m. Design a continuous anchor to support the ties.
6.40 m
A
1.50 m
T
GWT
2.40 m b d a d
Figure Q 3.12
3.13 A sheet pile wall as shown in Figure Q3.13 is retaining a soil with kN/m 3 , γ sat
= 21 kN/m 3 , c’ = 0, and φ ’ = 33 o
γ = 18
. Using a factor of safety 2.0 for passive resistance, (a) determine the required depth of embedment and the force in the tie rod, and (b) design the size and the length of tie rod if the strength of the steel rod is 140,000 kN/m 2 .
4.00 m
A
1.00 m
T
GWT
3.00 m d
Figure Q 3.13

If a structure is placed on soil surface, then the soil will undergo an elastic and plastic deformation. In engineering practice, the deformation or reduction in the soil volume is seen as settlement or heave depending on weather the load is increased or decreased. There are three types of settlement i.e. the immediate settlement ( S i
), primary consolidation settlement ( S settlement or creep ( S s
). Total settlement is: c
), and secondary compression
S = S i
+ S c
+ S s
(4.1)
In general, immediate settlement is caused by the elastic deformation of the soil mass or by the rearrangement of clay particles when the applied load causes the air to be expelled from the voids. This process occurs immediately following a stress change.
Primary consolidation settlement occurs when the applied stress and subsequently the excess pore pressure has caused the water to dissipate from the voids in saturated soil. The process is time-dependent. Excess pore pressures are set up immediately following a total stress change (undrained) and will gradually transfer the total stresses increase to the effective stresses at a rate proportional to the soil permeability (drained). In sandy, free draining soils, primary consolidation occurs almost instantaneously with relatively small settlements, but in low permeability clays the process can take several years to complete and lead to significant settlements.
Secondary compression, also known as secondary consolidation takes place when the excess pore water pressure is fully dissipated. In this case, the settlement is caused by the deformation of the soil skeleton. This process is also time dependent and the process is slower as compared to the primary consolidation.
103

104 Introduction to Geotechnical Engineering Part 1
The magnitude and rate of settlement is a function of soil type. In dry soil subjected to vibration, the settlement was caused only by rearrangement of soil particles. In granular soil or sand, water dissipates very quickly from void because of its high permeability, therefore only the immediate settlement is considered while the primary consolidation may be insignificant and could be neglected.
Primary consolidation makes an important part in the compressibility of fine grained soil, especially clay. However, the immediate settlement cannot be neglected and should be taken into account in the settlement analysis.
Secondary consolidation may play an important role in the compressibility of organic soil. The proportion of the secondary and the primary consolidation parts should be observed through a laboratory consolidation test.
Generally, immediate settlement can be estimated based on elasticity theory even though the calculation involves some assumptions such as homogeneity and isotropic which are not actually representative of natural soil properties. For an elastic media, and the stress strain relationship is given as:
S i
=
P
E
H
(4.2) where P is the applied pressure per unit area of the soil surface, H is the thickness of the compressible layer, and E is the bulk elastic modulus of the soil.
The calculation of elastic settlement ( S i uniform pressure q o
) of a foundation of width B carrying a
on the surface of a compressible soil layer with Young’s modulus E and Poisson’s ratio ν for the condition shown in Figure 4.1 is based on the formula given by Harr (1966)
S i
= q o
E
B (
1 − ν 2
)
α (4.3)
The formula was developed for a perfectly flexible foundation resting on the surface ( D f
= 0) of a compressible layer of infinite depth ( H = ∞ ). The influence factor α depends on the shape and rigidity of the foundation and the values are given in Table 4.1. If foundation is found on soil layer of definite thickness above a rigid substratum (rock), then Table 4.2 may be used for α parameter. The values

Compressibility & Consolidation 105 of Youngs modulus ( E ) and Poisson’s ratio ( in Table 4.3.
ν ) for various types of soil are given q o D f
Foundation B x L
Soil:
ν = Poisson’s ratio
E = Young’s Modulus
Rigid foundation
Flexible foundation
H
Figure 4.1 Illustration of settlement of footing on elastic medium
Table 4.1 Influence factors for vertical displacement under footing on soil layer of infinite depth
Shape of footing base
α
Center Edge/sides Corner Average
Flexible foundations:
Circle 1.00
Square 1.12
Rectangular L/B = 2
Rectangular L/B = 5
1.53
2.10
1.12
1.68
0.76
1.05
1.30
1.82
Rectangular L/B = 10 2.56 2.10 1.28 2.24
Rigid foundations:
Rectangular L/B = 2
Rectangular L/B = 5
Rectangular L/B = 10
1.12
1.60
2.00
0.79
0.82
0.79
0.82
1.12
1.60
2.00
1.12
1.60
2.00
1.12
1.60
2.00

106 Introduction to Geotechnical Engineering Part 1
Table 4.2 Influence factors for vertical displacement under footing on soil layer of limited depth
Shape of footing base
α
Flexible foundations:
H/B =1 H/B = 2 H/B = 5 H/B = 10 H/B = ∞
Square L/B = 1
Rectangular L/B = 2
Rectangular L/B = 5
Rectangular L/B = 10
0.15
0.12
0.10
0.04
0.29
0.29
0.27
0.26
0.44
0.52
0.55
0.54
0.48
0.64
0.76
0.77
0.56
0.76
1.05
1.28
Rectangular L/B = ∞ 0.04
Rigid foundations:
0.52 0.73 -
Table 4.3 Values of Modulus and Poisson’s ratio
Soil Type Cohesion (kPa) Young’s modulus E (MPa) Poisson’s ratio
ν
Soft clay
Medium to stiff clay
Very Stiff to hard clay
Sandy Clay
Silty Sand
Loose sand
Dense sand
Silt
Example Problem 4.1
< 25
25 – 100
> 100 kPa
2.5 - 15
15 – 50
50 – 100
25 – 250
5 – 20
10 – 25
50 – 81
2 - 20
0.40 – 0.50
0.45 – 0.50
0.45 – 0.50
0.20 – 0.40
A flexible foundation of size 4 × 2 m carrying a uniform load of 100 kPa is lying on a compressible (medium soft) clay layer of infinite thickness. Determine the average immediate settlement under this foundation.
Solution
For a medium soft clay layer: E s
is estimated at 40 MPa, ν = 0.45 (Table 4.3)
For flexible foundation L/B = 2, then α = 1.30
S i
=
B q o
E
(
1 − ν 2
)
α =
2 × 100 × ( 1 − 0 .
45 ) 2
40 , 000
× 1 .
30
= 5 .
2 mm

Compressibility & Consolidation 107
Example Problem 4.2
A foundation of size 4 × 2 m carrying a uniform load of 100 kPa is embedded at a depth of 1 m on a compressible clay layer of 5 m thick. A hard stratum lies below the clay layer. Determine the average immediate settlement under the corner of this foundation if the undrained modulus of the clay is 40 MN/m 2 .
Solution
Foundation is embedded at 1 m, then the thickness of compressible layer below the footing is 4 m. Then:
H
B
= 4/2 = 2,
L
B
= 2 Table 4.2, α = 0.29
Immediate settlement
S i
=
B q o
E
(
1 − ν 2
)
α =
2 × 100 × ( 1 − 0 .
5 2
40,000
) × 0 .
29
= 10 mm
Elastic settlement of footing on saturated clay
The preceding equations can be used for the calculation of elastic settlement of footing on saturated clay layer, by replacing the modulus and Poisson’s ratio by undrained modulus E between 500 – 1500 c u and Poisson’s ratio u where c u may be used for high plasticity clay.
ν = 0.5. The values of E u
frequently lie is the undrained cohesion of soil with lower value
Example Problem 4.3
A raft foundation of size 32 m × 18 m is transmitting a uniform contact pressure of
240 kPa at a depth of 2 m. Determine the immediate settlement that is likely to occur under the center of foundation, assuming it is flexible. E u
= 45 MPa, ν =
0.5, and γ = 20 kN/m 3 .
Solution
For
L
B
= 32/18 = 1.78, then use Table 4.1 for settlement under the center of foundation, interpolate between
L L
B
= 1 and
B
= 2

108 Introduction to Geotechnical Engineering Part 1
L
= 2; α = 1.52
B
L
B
= 1; α = 1.12 Æ α = 1.12 +
( 1 .
78
( 2
−
− 1 )
1 )
(1.52 –1.12) = 1.12 + 0.31 = 1.43
Net contact pressure for raft foundation q o
= 240 – (2 × 20) = 200 kPa
Immediate settlement
S i
=
B q o
E
(
1 − ν 2
)
α =
18 × 200 × ( 1 − 0 .
5 2
45,000
) × 1 .
43
= 88 mm
Elastic settlement of footing based on Schmertmann method
Elastic settlement of granular soils can also be evaluated by use of a semiempirical strain influence factor proposed by Schmertmann and Hartman (1978).
According to this method, the elastic settlement is given by
S i
= C
1
C
2
(q o
− q)
2 B
0
I z
E s
∆ z (4.4) where I z
is the strain influence factor, C
1 foundation embedment, C
2
is a correction factor for the depth of
is a correction factor to account for creep in soil, q o
is the contact pressure at the level of the foundation, and q is the weight of excavated soil.
The values of C
1
and C
2 can be obtained from:
C
1
= 1 − 0 .
5
⎛
⎜⎜ q o q
− q
⎞
⎟⎟ (4.5a)
C
2
= 1 + 0. 2 log (time in years/0. 1) (4.5b)
The variation of the strain influence factor with depth below the foundation is shown in Figure 4.2.
An approximate variation of Young's modulus with depth is required in order to use the equation. This can be estimated based on the type of soil encountered and

Compressibility & Consolidation 109 use Table 4.3 for estimation of Young’s modulus or use the results of in-situ testing such as standard penetration numbers (SPT, N ) and cone penetration resistances (CPT, q c
). For the results of cone penetration test, Schmertman (1978) suggested to use correlation E s
E s
= 2.5 q c
for square or circular foundation, and
= 766 N (kN/m 2 ) was suggested.
E s
= 3.5 q c for strip footing, while for the results of standard penetration test, an empirical
The soil layer can be divided into several layers up to a depth of z = 2 B or 4 B depending on the shape of foundation base, and the immediate settlement of each layer can be estimated. The sum of the settlement of all layers is equal to S i
.
Base of footing
0
I z
B/2
B
Square and circular footing
Strip footing
B = width of footing
2B
4B
I z
Square and circular footing
I
I z z
= 0.1 at z = 0
= 0.5 at z = 0.5B
= 0.0 at z = 2B
Strip footing
(rectangular with L/B >10)
I z
I z
I z
= 0.2 at z = 0
= 0.5 at z = B
= 0.0 at z = 4B
Necessary interpolation can be made for 1<
L
B
< 10
Figure 4.2 Variation of the strain influence factor with depth
Example Problem 4.4
A shallow foundation of size 3 × 3 m in plan is constructed at a depth 1m in a deposit of sandy soil. The foundation is carrying a column load of 1500 kN. Unit

110 Introduction to Geotechnical Engineering Part 1 weight of the soil below foundation is 18 kN/m 3 . Determine the settlement of the foundation after 5 years after construction using Schmertmann (strain influence factor) method if the Young’s modulus of soil can be estimated from the results of
SPT test shown below:
Table P4.4
Depth (m) 1 2 3 4 5 6 7 8
Corrected N 6 9 10 8 12 13 17 23
Solution
Draw the strain influence diagram for foundation of width B = 3m together with the variation of N values with depth
5
N SPT (corrected)
10 15 20
2
3
0
0
1
6
7
4
5
Extend Table 4.4 to calculate the strain influence factor
Depth
(m)
∆ z
(m)
Avg N E s
= 766 N
(kPa)
Avg
I z
0 – 1.5
1.5 – 4.0
4.0 – 6.0
1.5
2.5
2
7.5
10
14
5,745
7,660
10,724
0.3
0.375
0.125
S i
= C
1
C
2
(q o
− q)
2
B
0
I z
Es
∆ z q o
=
Q
B × B
=
1500
3 × 3
= 167 kN/m 2 .
I z
E s
∆ z (m 3 /kN)
0.000082
0.000122
0.000023
Σ = 0.000227

Compressibility & Consolidation 111
q D f
= 18 × 1 = 18 kN/m 3
C
1
= 1 − 0 .
5
⎛
⎝ q o q
− q
⎞
⎟⎟
= 1 − 0 .
5
⎡
⎢
18
167 − 18
⎤
⎥
= 0 .
94
C
2
= 1 + 0.2 log (time in years/0. 1) = 1 + 0.2 log (5/0.1) = 1.34
Hence
S i
= 0.94 ×1.34 ×(167 – 18) × 0.000249 = 0.047 m = 47 mm
When a saturated soil is loaded, then the increase in total stress will be first carried by the water in the soil. The excess pore water pressure will cause the water to dissipate gradually, and the load will be shared by the water and the soil skeleton.
The process will continue until the excess pore water pressure has completely dissipated and all stress is carried by the skeleton. This process is called primary consolidation and will results in the deformation of the soil.
The time required for the water to dissipate from the soil depends on the permeability of the soil itself. In granular soil (sand), the process is very fast and hardly noticeable due to its high permeability. On the other hand, the consolidation process may take years in clay soil.
Consolidation settlement is the vertical displacement of the soil surface corresponding to the volume change at any stage of the consolidation process. In this case, the settlement will result if a structure is built over a layer of saturated clay or if water table is lowered permanently in a stratum overlaying a saturated clay layer. The reverse of settlement, called heaving, will occur if there is a load reduction for example due to excavation.
The amount of settlement and the time needed for the consolidation process can be predicted based on laboratory test i.e. oedometer test which will be explained in this text.
One of the major disadvantages of the conventional oedometer test is the length of time required for completion. A typical test can take up to two weeks to finish.

112 Introduction to Geotechnical Engineering Part 1
Several alternate methods have been proposed to reduce the testing period, among them: the Constant rate of strain (CRS) where the vertical deformation ∆ h is applied at a constant rate, the Constant rate of loading: in which the applied total stress σ is increased at a constant rate and the Consolidation test with back pressure control: in which the back pressure µ
B is initially equal to the pore pressure µ
A in the sample and is steadily reduced to a constant final value. The description of the alternate procedures for the consolidation test can be found in
Head (1986).
The amount of settlement in-situ can be measured by putting settlement plates on the ground or by recording the levels of some reference points. A more precise way to record the progress of consolidation in-situ is by installing piezometers to record the change of pore water pressure with time.
The conventional consolidation or oedometer test (also called 1-dimensional consolidation test) is based on the following assumptions: a) The clay layer is saturated b) The soil is laterally confined c) The flow of water is one-dimensional d) The permeability of the soil is constant e) The consolidation process can be measured by oedometer
Figure 4.3 shows the cross-sectional diagram of the consolidation cell. As shown in the diagram, the soil specimen is in the form of a disc (with a certain diameter and thickness), placed in between two porous stones and held inside a metal ring.
The ring imposes a condition of no lateral strain on the specimen and the porous stones allow a 1-dimensional flow of water from the soil specimen. The specimen is trimmed to fit the consolidation ring and placed on top of the bottom porous stone. The test procedure has been standardized in BS 1377 (Part 5), and ASTM
D 2435-90.
Once set, the specimen will be loaded in a series of pressure; each being doubled the previous value. The initial weight depends on the soil type and the depth where the soil is taken from the field. Each pressure is normally maintained for 24 hours. During this time, deformation of specimen will be observed in specified time (e.g. ¼, ½, 1, 2, 4, 9, 15, 30, 60, 120, 1440 minutes).

Compressibility & Consolidation 113
Applied pressure
∆ h Porous stones ho Soil sample
Consolidation ring
Figure 4.3 Cross-sectional diagram of consolidation cell.
The test results are presented by plotting the readings or vertical strain against time for each load increment (Figure 4.4). This graph is required to observe the time rate of consolidation, and subsequently the time required for a certain degree of consolidation to take place.
0.636
0.635
0.634
0.633
0.632
0.631
0 20 40 60 80 100 120 elapsed time (minutes)
Figure 4.4 Dial reading or sample height vs time curve for a certain loading increment
The second graph is the void ratio ( e ) at the end of each increment period against the corresponding load increment. There are two types of plot: the e – p ’ curve
(Figure 4.5) and the e – log p ’ curve (Figure 4.6). These graphs are needed to obtain the coefficient of volume compressibility ( m v
( C c
) and the compression index
), and to calculate the magnitude of the settlement.

114 Introduction to Geotechnical Engineering Part 1
0.75
0.70
0.65
0.60
0.55
0.50
0.45
0.40
1
0.75
0.70
0.65
0.60
0.55
0.50
0.45
0.40
0 a v
200 400 600 800 1000 1200 1400 1600 1800 pressure (p)
Figure 4.5 e-p’ curve from oedometer test
Cr
Cc
10000 10 100 pressure (p)
1000
Figure 4.6 e-log p’ curve from oedometer test

Compressibility & Consolidation 115
In order to construct the e – p ’ and the e – log p ’ curves, it is necessary to calculate the void ratio ( e ) at the end of each loading increment. This can be done by assuming that the change in void ratio is proportional to the change in volume and thus proportional to the change in the height of specimen (dial readings).
The initial void ratio ( e specimen ( H o thickness of solids ( H s o
) can be calculated based on the initial thickness of the
) where e o
= (H o
-H s
)/H s
. In this case, we need the equivalent
) which can be calculated based on the dry weight ( W the soil specimen, area ( A ), and specific gravity of the solids ( G s
). s
) of
H s
=
A
W s
G s
γ w
(4.6)
Load increment P
1 as much as ∆ H increment P
1 of load increment P causes the change in the thickness of the specimen (measured)
1
. Then the change in void ratio
2
, the void ratio e
2
= e
1
– ∆ e
2
∆ e
1
, the void ratio e
1
will be equal to ( e o
–
=
∆
, etc. e
1
∆ H
1
/H s
. At the end of load
). Subsequently, at the end
a v
, m
From Figure 4.5 ( e – p v
, C c
, C r
’ curve), we can obtain the coefficient of compressibility
( a v
) and the coefficient of volume compressibility (the volume change per unit volume per unit increase in effective stress) defined as m v a v
=
∆
∆ e
σ
= e
1
σ '
1
−
− e
2
σ '
2
(4.7) m v
=
1 a
+ v e o
=
1 +
1 e o
⎛
⎜⎜ e
σ '
1 o
− e
1
− σ '
2
⎞
⎟⎟ (4.8) or in terms of sample height m v
=
1
H o
⎛
⎜⎜
H o
σ '
1
− H
1
− σ '
2
⎞
⎟⎟ (4.9)
As can be seen from the graph (Figure 4.5), the value of m only true for a certain range of stress. v
is not constant and is

116 Introduction to Geotechnical Engineering Part 1
From Figure 4.6 ( e – log p ’ curve), we can obtain the compression index ( C c
) as the slope of the linear portion of the e- log p ’ curve.
C c
=
∆ e
∆ log σ
= e
1 log (
−
σ e
2
′
σ ′
2
1
)
(4.10)
In most instances after a certain loading increment, the test will be continued by unloading (expansion) and reloading (recompression) of the specimen, and this part will also be shown in the graphs (both e – p ’ and e – log p ’ ). The plots show that the soil that has been consolidated will be less compressible then the original condition. The recompression index ( C r
) can be obtained from the slope of recompression part of the graph. The value of recompression index is much less than the value of C c
, usually in the range of (0.05 – 0.1) C c
(Leonards, 1962).
c
The soil that has been loaded and unloaded will be less compressible when it is reloaded again. Thus, it is necessary to estimate the pre consolidation pressure
(the stress carried by soil in the past, σ c
’ ) because consolidation settlement will not usually be great when the applied load remains below the pre consolidation pressure.
Cassagrande (1936) proposed an empirical construction to obtain the preconsolidation pressure by using the e – log p ’ curve obtained from the oedometer test (Figure 4.6). Procedure for obtaining the pre consolidation pressure can refer to Figure 4.7.
Over-consolidation is judged if pre consolidation pressure pressure ( σ o
σ c
’ obtained from e-log p curve based on oedometer test is greater than the currently existing overburden
’ ). In this case, the over-consolidation ratio (OCR) is defined as
OCR =
σ
σ
'
' o c
(4.11)
Among the mechanism causing pre-consolidation are: the changes in total stress due to excavation or removal of structure, the changes in pore water pressure due to the rise of ground water elevation, artesian pressure, pumping or dewatering, and the changes in soil structure due to secondary pressure, environmental changes, climate, creep, etc.

Compressibility & Consolidation 117
0.75
0.70
B
0.65
0.60
0.55
0.50
0.45
A
α
α
0.40
1 10 100
σ ' c
1000 10000 effective consolidation pressure (p')
Figure 4.7 Determination of pre consolidation pressure ( σ ’ c
) from the e-log p’ curve
1. Choose by eye the point of maximum curvature on the consolidation curve
(point A)
2. Draw a horizontal line from point A
3. Draw a line tangent to the curve at point A
4. Bisect the angle made by step 2 and 3
5. Extent the straight line portion of the virgin compression curve up to where it meets the bisector line obtained in step 4. The point of intersection of these two lines is the approximate value of the pre-consolidation pressure
(point B)
Field Compression Index ( C cf
)
Research has shown that a slight disturbance introduced to the soil sample during sampling and preparation of the specimen will influence to the test results. An increase in the degree of soil disturbance will results in a slight decrease in the virgin compression line. Thus the compression index obtained from the results of oedometer test will be slightly lower than the compression index in the field
(Figure 4.8)

118 Introduction to Geotechnical Engineering Part 1
Schmertmann (1953) suggested a procedure to obtain a field virgin compression line based on the e – log p ’ curve (refer to Figure 4.8).
1. Produce two horizontal lines at e o
and 0.42 e o
2. Produce a vertical line at pressure equal to σ o
’ or o
at point 2.
σ c
’ . This line should intersect the horizontal line at e o
at point 1.
3. Plot compression line ( C c
) based on the oedometer test. The line will intersect the horizontal line at 0.42 e
3. Draw a line from point 1 to point 2. The slope of this line is the field compression index ( C cf
).
0.75
0.70
0.65
0.60
0.55
0.50
0.45
0.40
e o
1
C c
C cf
0.35
0.30
σ ' c
2
0.42 e o
0.25
1 10 100 1000 effective consolidation pressure (p')
10000
Figure 4.8 Procedure to determine the field consolidation index C cf
Consolidation settlement is calculated based on the value of either the coefficient of volume compressibility ( m v
) or the compression indices ( C c and C r
). Consider a layer of saturated clay of thickness H as shown in Figure 4.9. Due to construction, the total vertical stress in an elemental layer of thickness d z at depth z is increased by ∆σ ' .

Compressibility & Consolidation 119
Additional stress ( ∆σ ' ) will results in the increase of stress corresponds to σ '
1
– and a decrease in void ratio corresponds to ∆ e = e o
– e
1
.
σ ' o
The change in volume per unit volume of clay soil can be written in terms of void ratio as
∆
V
V o
=
1
∆
+ e e o
= e
1 o
+
− e e o
1 (4.12)
S c z
∆σ ' e o d z
H e
1
∆σ '
σ ' o
σ '
1
S c e o
σ
1
' e
1
σ o
'
⎞
⎟⎟
⎛
⎜⎜
σ
1
'
1 +
− σ e o o
'
Figure 4.9 Consolidation settlement
By knowing the ratio of the change in void ratio to the change in the effective stress in e – p ’ curve (Figure 4.5), then
= ∆ H =
⎛
⎜⎜
−
−
⎞
⎟⎟
H (4.13)
S c
=
S c a v
=
⎝
⎜⎜
⎛
1 +
1 e o m v
⎠
⎟⎟
⎞
∆σ ' H
(
σ
1
' − σ o
'
)
H (4.14)
(4.15) or
S c
=
⎝
⎜⎜
⎛ ∆
1 + e e o
⎠
⎟⎟
⎞
H (4.16)

120 Introduction to Geotechnical Engineering Part 1
By using the e – log p ’ curve (Figure 4.6), the change in void ratio can be written as
∆ e = C c log
σ
1
'
σ o
'
(4.17) and the settlement of a normally consolidated clay due to change of stress ∆σ ’ is given as
S c
= C c 1
H
+ e o log
σ ' o
σ
+
' o
∆σ
(4.18)
Consolidation settlement will not usually be great when the applied load or part of the applied load remains below the pre consolidation pressure. Thus, it is necessary to check if the pre consolidation pressure obtained from laboratory test
( σ ' c
) is greater then the existing overburden pressure ( σ ' o
). If σ ' c
> σ ' o
, then check if the added stress will increase the existing pressure beyond the pre-consolidation pressure.
Figure 4.10 shows the possible loading conditions for an over consolidated soil.
Thus, the following equations will be used for the settlement calculations for condition a and b respectively.
S c
= C r 1
H
+ e o log
σ ' o
σ
+
' o
∆σ '
(4.19)
S c
= C r 1
H
+ e o log
σ
σ '
' c o
+ C c 1
H
+ e o log
σ ' o
σ
+
' c
∆σ
(4.20)
Keep in mind that the e o used in the second part of equation 4.20
is the one corresponds to the pre consolidation pressure ( σ ' c
)

Compressibility & Consolidation e e o e f
σ ' o
σ ' o
+
∆σ
σ ' c
∆σ
∆ e
121
(a)
σ ' o σ ' c
σ ' o
+
∆σ
σ ' e e o ∆ e
1
∆ e
2 e f ∆σ
σ '
(b)
Figure 4.10 Possible loading conditions for an over consolidated soil.
Example Problem 4.5
A sand layer of thickness 10 m overlays a clay layer of 8 m thick as shown in
Figure P4.5. A sand layer exists at the bottom of the clay. Given m v
= 0.83 m 2 /MN,
(a) calculate the consolidation settlement due to the decrease of ground water level from soil surface to a depth of 4 m; (b) What is the settlement of the sand layer if given C c
= 0.50 and e o
= 0.95 (the soil is normally consolidated).

122 Introduction to Geotechnical Engineering Part 1
0
-4
-10 sand
γ sat
= γ b
= 18.5 kN/m 3 clay
-14
γ sat
= 20 kN/m 3
-18
Initial GWT
Final GWT
H = 8m sand
Figure P4.5
Solution a.
Find the increase in effective stress at the middle of the clay layer due to the decrease in ground water table initial condition: σ ’ o
σ ’ o
σ ’ o
= (10×18.5) + (4×20) – (14×9.8)
= 185 + 80 – 137.2
= 127.8 kN/m 2 final condition: σ ’
1
σ ’
σ ’
= (10×18.5) + (4×20) – (10×9.8)
1
1
= 185 + 80 – 98
= 167 kN/m 2
Change in stress ∆σ = σ ’
1
– σ ’ o
= 167 – 127.8 = 39.2 kN/m 2
Thus, S c
= m v
∆σ H
= 0.83 m 2 /MN×0.0392 MN/m 2 ×8 m
= 0.26 m
= 26 cm or 260 mm. b.
Use formula S c
= C c
H
1 + e o log
σ ' o
+
σ ' o
∆σ
S c
= 0.50
1 +
8
0.95
log
167
127.8
=
= 0.24 m = 240 mm.

Compressibility & Consolidation 123
Example Problem 4.6
A clay layer of thickness 3 m is overlain by a sand layer of thickness 6 m. The clay layer overlays another layer of sand. A footing of size 4×4 m was constructed at depth of 1.5 m on the sand layer giving an additional stress of 12 kN/m 2 . Based on laboratory test, the clay layer has an initial void ratio of 1.03, pre consolidation pressure of 140 kPa, compression index C c
of 0.275 and C r
of 0.0575. Calculate the consolidation settlement of the clay layer due to footing load.
0
-2 GWT
4 × 4 m sand γ sat
= γ b
= 18.3 kN/m 3
-6
-9 clay γ sat
= 19 kN/m 3 sand
H = 3m
Figure P4.6
Solution
Find the initial stress at the middle of clay layer σ ' o
σ ’
σ ’
Given σ
Given ∆σ = 12 kN/m
S c
’ c
= 140 kN/m 2
=
=
C r 1 o o
= (2×18.3) + (4 × (18.3–9.8)) + (1.5×9.2)
= 84.36 kN/m 2
H
+
0 .
0575 e o
1
Æ σ ’ log
+
2
3
σ
1 .
03
' o
σ
+
'
σ log o
∆
’ o
σ
< σ ’ c
Æ over-consolidated clay o
+ ∆σ = 84.36 + 12 = 96.36 < σ ’
'
96 .
36
84 .
36 c
S c
= 0.005 m = 5 mm

124 Introduction to Geotechnical Engineering Part 1
Example Problem 4.7
Granular fill is placed on an extended area to consolidate a soft clay layer of 8 m thick. The unit weight of fill ( γ f
) is 20 kN/m 3 and the average thickness of the fill is 3.5 m. Ground water level coincide the ground surface. The saturated unit weight of the clay layer is 18 kN/m 3 . The results of consolidation test is shown in
Table P4.7. (a) Construct the e – log p ’ curve and (b) use the curve to evaluate compression index C pressure ( σ ’ fill. c
, recompression index C r
, and the pre consolidation c
) (c) Calculate the settlement of the clay layer due to the weight of the
Table P4.7
Stress
(kPa)
Void ratio
0 5 10 20 40 80 160 320 640 160 40 5
2.855 2.802 2.793 2.769 2.631 2.301 1.939 1.576 1.314 1.375 1.464 1.589
Solution
(a) see Figure P4.7
(b) Plot the e – log p
C c
C r
=
=
2 .
68 − log
1 .
15 log
(
−
’ curve (Figure P4.7) to get
1 .
53
400
1
40
.
13
)
( )
= 1.15 σ
C
’ c
, C r
, and c
= 38 kPa
σ ’ c
= 0.21 Void ratio corresponds to σ ’ c is 2.68
3
2.8
2.6
2.4
2.2
2
1.8
1.6
1.4
1.2
1
C r
σ ’ c
C c
1000 10 pressure (kPa)
100
Figure P4.7

Compressibility & Consolidation 125
(c ) Find the initial stress at the middle of clay layer
Given
σ
σ ’
’ o
= 4 × (18–9.8) = 32.8 kN/m c
= 38 kN/m 2 σ ’ o
< σ ’ c
2
σ ’ o
Æ the soi is over-consolidated
Increase in stress at depth of 4 m due to the weight of fill
Surcharge load due to fill is 3.5 × 20 = 70 kN/m 2
σ ’ o
+ ∆ σ = 32.8 + 70 = 102.8 kN/m 2 > σ ’ c
Find settlement
S c
= C r 1
H
+ e o log
σ
σ
'
' c o
+ C c 1
H
+ e o log
σ ' o
σ
+
' c
∆σ
S c
= 0 .
21
1 +
8
2 .
855 log
38
32 .
8
+ 1 .
15
1 +
8
2 .
68 log
102 .
8
38
S c
= 0.0375 + 1.41 = 1.45 m
Consolidation is a result of gradual dissipation of excess pore water pressure from a clay layer. This process may take a long time to complete, while the design life of a structure is limited. Thus, it is important to know how fast a structure will settle under an applied load or how much the structure will settle at a certain time after construction.
The calculation of the rate of consolidation is based on Terzaghi 1-D consolidation theory which involves the following assumptions:
1.
Compression and flow of pore water pressure are vertical only, (onedimensional).
2.
The compressible layer is saturated, homogenous and isotropic

126 Introduction to Geotechnical Engineering Part 1
3.
Soil particles and water is incompressible
4.
Darcy’s law is valid at all hydraulic gradient
5.
Strain due to external load is small and within the range of elasticity
6.
The coefficient m v
, k and hence C v
remain constant throughout the consolidation process
7.
There is a unique relationship, independent of time, between void ratio and effective stress
The derivation of the Terzaghi equation in one part considers the volume of water flowing out of a compressible soil element. Since the water is assumed incompressible, by continuity the volume change in the element must be the difference in flow in and out of the element in a differential time ∂ t .
The other part is obtained by relating the volume change or change in void ratio of the soil skeleton to the change in effective stress by means of the coefficient of compressibility a v determined in the oedometer test. From the effective stress principle, we can equate the change in effective stress ( pressure ( ∆ u e
). As before, u is a function of both z and t.
∆σ ’ ) to the change in pore
Putting together both parts of the equation:
C v
∂ 2
∂ z u e
2
=
∂ u
∂ t e
(4.21)
Where u e
is the excess pore water pressure at time t and depth z , and C v the coefficient of rate of consolidation (m 2 /year or m 2 is called
/sec) which contains the material properties that govern the consolidation process.
C v
=
γ k w
1 + a v e o = m v k
γ w
(4.22)
Equation 4.21 is known as Terzaghi one-dimensional consolidation equation. The solution to the equation is made based on the assumption that (1) there exist a twoway flow of water, and (2) Initial pore water pressure is equal to the load increment and (3) the final pore pressure is equal to zero. These conditions can be explained as:

Compressibility & Consolidation
for z = 2 H Æ u = 0
for ∆ u = u i = ∆σ ’ = ( σ
2
’ – σ
1
’ )
for , ∆ u = 0
127 uniform load d h h sand water table d= H/2 z dz a u’ z u o
= ∆σ
∆σ ' z b clay H dx d= H/2 c excess pore pressure distribution d
Figure 4.11 Distribution of excess pore pressure in a clay layer subjected to a uniform increase in loading
The general solution given by Taylor (1948) is in terms of a Fourier series expansion of the form:
µ =
(
σ
2
' − σ
1
'
) n = n
∑
∞
= 0 f
1
(Z)f
2
(T v
) (4.23) where Z and T v
are non-dimensional parameters, for which Z is geometry factor, which is equal to z/H , and T v
is Time factor, in which
T v
=
C
H v d
2 t
= m v k
γ w
H t d
2
(4.24)

128 Introduction to Geotechnical Engineering Part 1
Degree of Consolidation
The consolidation process is related to the change in void ratio. The progress of consolidation at time t and at any depth z can be related to the void ratio at that time ( e ) and the void ratio at the end of consolidation process ( e f
). The degree of consolidation at a specific depth can be expressed in the form of void ratio as
U z
= e o e o
−
− e e f
(4.25) or the change in effective stress and therefore the change in pore water pressure
U z
=
σ '
σ '
1
−
−
σ ' o
σ ' o
=
σ ' − σ ' o
∆σ
= u i u i
− u
= 1 − u u i
(4.26)
From the equation above, we can see that initially U = 0, and become 100% when void ratio is equal to e f
.
U z
= 1 − n
= 0 f
1
(Z) f
2
(T) (4.27)
In most cases, we are not interested in how much a given point in a layer has consolidated. Of more practical interest is the average degree of consolidation of the entire layer. This value, denoted as U , is a measure of how much the entire layer has consolidated. Thus it can be directly related to total settlement of the layer at a given time after loading.
U avg
=
S ct
S f
(4.28)
The relationship between the average degree of consolidation U avg in equation 4.28 and Figure 4.12.
and T v
is given
For U < 60% Æ T = ( π /4) U 2 = ( π /4) (U%/100) 2 (4.28)
For U > 60% Æ T = -0.933 log (1-U) – 0.085 = 1.781 – 0.933 log (100 – U%)

Compressibility & Consolidation 129
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0 10 20 30 40 50 60 70 80 90 100
Percent consolidation (%)
Figure 4.12 Time factor T v for degree of consolidation
Example Problem 4.8
The settlement analysis on Example problem 4.7 (assuming that the clay layer is drained at its top and bottom surfaces) shows that 15 cm of settlement will occur in 1 year. For this case, (a) calculate the coefficient of consolidation of the clay layer, and (b) estimate the time required to reach 50% and 90% consolidation of the layer.
Solution
From problem 4.7, we get Final settlement S cf
= 1.45 m = 145 cm a. For settlement of 15 cm Æ U avg
=
S ct
S f
=
15
145
≈ 0.1 ≈ 10%
Use Figure 4.12 Æ T v
= 0.0077
Drainage at top and bottom Æ H d
= ½ H = 8/2 = 4 m
T v
=
C v
H d
2 t
C v
=
T v
H 2 d t
=
0 .
0077
1
× 4 2
= 0.123 m 2 /year

130 Introduction to Geotechnical Engineering Part 1 c.
For 50% consolidation, T v
= 0.196 t =
T v
H
C v
2 d
=
0 .
196 ×
0 .
123
4 2
= 25 years d.
For 90% consolidation, T v
= 0.848 t =
T v
H
C v
2 d
=
0 .
848
0 .
×
123
4 2
= 110 years
Example Problem 4.9
A 3 m depth of fill (unit weight 20 kN/m 3 ) is to be dumped on the surface over an extensive area covering an 8 m depth of sand deposit overlying a 6 m depth of clay layer (Figure P4.9). Impermeable stratum exists below the clay layer. Ground water table is at 2 m below the ground surface. The unit weight of sand above water table is 17 kN/m 3 and below water table is 19 kN/m 3 . The unit weight of clay is 20 kN/m 3 . The relationship between void ratio and effective stress for the clay can be represented by the equation: e = 0.88–0.32log
coefficient of consolidation C v
is 1.26 m
σ '
, while the
2
100
/year. Calculate (a) the final settlement of the area due to consolidation of the clay, (b) the settlement after a period of 2.5 years, (c) the time to reach 90% consolidation
Solution
Since there is only 1 free draining layer, then
Based on the given equation: e
H d
= H = 6 m.
= 0.88–0.32 log
σ '
100
; we get C c
= 0.32 a.
Settlement of the clay layer, assuming that the soil is normally consolidated
S c
= C c 1
H
+ e o log
σ ' o
σ
+
' o
∆σ '
At the center of clay layer, overburden pressure:
σ ’ o
= (17 × 2) + (19–9.8) × 6 + (20–9.8) × 3 = 119.8 kN/m 2
Void ratio corresponds to the overburden pressure: e o
= 0.88 – 0.32 log (1.198) = 0.88 – 0.025 = 0.855
Stress change due to fill

Compressibility & Consolidation 131
∆σ = γ f
Η f
= 20 × 3 = 60 kN/m 2
S c
= 0 .
32
1 +
6
0 .
855 log
119 .
8 +
119 .
8
60
= 0.182 m = 18.2 cm
Soil profile:
0
-2 q o
GWT sand
γ b
= 17 kN/m 3 ; γ sat
= 19 kN/m 3
-8 clay γ sat
= 20 kN/m 3
-14 impermeable layer
Figure P 4.9
b. Settlement after 2.5 years
T v
=
C v
(H d t
) 2
=
1 .
26
6
×
2
2 .
5
= 0 .
0875
H = 6m for T v
= 0.0875 Æ use equation 4.28 Æ get U = 0.335
Therefore settlement after 2.5 years is U × S c
= 0.335 x 182 = 61 mm. c. Time U = 90% Æ T v
= 0.848 t =
H d
C v
2 T v =
6 2 × 0 .
848
1 .
26
= 24 years
Example Problem 4.10
Refer to Example problem 4.9. What if below the clay layer, there is a sand layer which is permeable?

132 Introduction to Geotechnical Engineering Part 1
Solution
In this case, the final settlement S c
will still be 182 mm
Due to the existence of permeable layer at the bottom of the clay layer, then H d
= H /2 = 3 m the settlement after 2.5 years, t = 2.5
T v
=
C v
(H d t
) 2
=
1 .
26 ×
3 2
2 .
5
= 0 .
35 for T v
= 0.35 Æ use equation 4.28 Æ get U = 0.65
Therefore settlement after 2.5 years is 0.65 × 182 = 118 mm.
Time to reach 90% consolidation
U = 90% Æ T v
= 0.848 t =
H d
C v
2 T v =
3 2 ×
1 .
0 .
26
848
= 6 years
Comment:
The existence of permeable layer below clay layer improves the time for consolidation by 4 times.
Determination of the coefficient of Rate of Consolidation ( C v
)
The value of C v for a particular pressure increment in oedometer test can be determined by “curve fitting” methods. There are two methods commonly used to determine the value of C v
i.e the logarithmic time methods (Cassagrande, 1940), and the square root time methods (Taylor, 1942). These empirical procedures were developed to fit approximately the observed laboratory test data to the
Terzaghi theory of consolidation. The procedures will be outlined in Figure 4.13 and 4.14 respectively.

Compressibility & Consolidation 133
The Cassagrande methods use the plot of dial readings versus the logarithmic of time (log t ). The idea is to find the reading at t
50 or the time for 50% consolidation.
The procedure is as follows (refer to Figure 4.13).
6.0
5.8
5.6
5.4
5.2
6.8
6.6
6.4
6.2
5.0
4.8
4.6
4.4
4.2
4.0
3.8
D
0
A
D
100 t
1 x
B t
2 t
50 x
D
50
C t p
Primary consolidation
1 10 100 time (minutes)
1000 10000
Figure 4.13 Determination of C v
by Cassagrande method
1. Plot a graph relating dial reading (mm) versus log time
2. Produce a straight line for primary consolidation and secondary consolidation part of the graph. The two lines will meet at point C.
3. The ordinate of point C is D
100
= the deformation corresponds to U = 100%
4. Choose t
1
(point A), t reading is equal to x.
2
= 4 t
1
(point B). The difference in the dial
5. An equal distance x set off above point A fixes the point D
0
= the deformation corresponds to U =0%. Notes that Ro is not essentially equal to the initial reading may be due to small compression of air within the sample. between D
100
is called the primary consolidation.
7. A point corresponding to U = 50% can be located midway between D
0
D
100
. The value of T v
corresponds to U = 50% is 0.196.
and

134 Introduction to Geotechnical Engineering Part 1
C v
0 .
196 H d
2
8. Thus
= t
50 where H d
is half the thickness of specimen for a particular pressure increment.
The square root of time methods developed by Taylor is based on the similarity of the shapes of experimental and theoretical curves when plotted versus the square root of t . Refer to Figure 4.14 for procedure to find C v time methods:
using the square root of
6.8
6.6
6.4
6.2
6.0
5.8
5.6
5.4
5.2
5.0
4.8
4.6
4.4
4.2
4.0
0 d
D
0
G
0.15
d
D
90
5 t
90
P Q
10 time (minutes)
15 20
Figure 4.14 Determination of C v
by Taylor method
1. Extent the straight line part of the curve to intersect the ordinate ( t = 0) at point D. The point shows the initial reading ( line with the abscissa is P.
D o
). The intersection of this

Compressibility & Consolidation 135
2. Take point Q such that OQ = 1.15 OP.
3. The intersection of line DQ and the curve is called point G
4. Draw horizontal line from G to the ordinate ( D
90
). The point shows the value of √ t
90
. The value of T v
corresponds to U = 90% is 0.848.
5. Thus C v
=
0 .
848 t
90
H d
2
Determination of the Coefficient of Permeability ( k )
It has been mentioned in the previous part that the rate of consolidation is a function of permeability of the soil. Thus, we can indirectly obtain the coefficient of permeability of the soil based on the results of oedometer test as: k =
C v
1
γ w
+ e o a v =
C v
γ w m v
(4.29)
So far, we have discussed a phenomenon that the compression of soil is only due to the elastic deformation and the dissipation of pore water pressure during primary consolidation. In actual condition, we can see that the compression does not cease when the excess pore water pressure has dissipated to zero but continues at a gradually decreasing rate under constant effective stress. Thus, it is common to differentiate the two processes as primary and secondary compression.
Secondary compression, also referred as creep, is thought to be due to the gradual readjustment of the clay particles into a more stable configuration following the structural disturbance caused by the decrease in void ratio, especially if the clay is laterally confined.
In general, both primary and secondary compressions take place simultaneously.
However, it is assumed that the secondary compression is negligible during primary compression, and is identified after primary consolidation is completed.
The rate of secondary compression in the oedometer test can be defined by the slope ( C
α
) of the final part of the dial reading versus log time curve (Figure 4.15).

136 Introduction to Geotechnical Engineering Part 1
C
α
=
∆ e
∆ log t
=
∆ e log t f t p
(4.30) where ∆ e is the change of void ratio from t
1
to t
2
, in which the completion of primary consolidation, while secondary consolidation settlement is required ( t f t
2 t
1
denotes the time of
is the time for which the
). This estimate is based on assumptions that C
α
is independent of time, thickness of compressible layer or applied pressure. Research showed that the ratio of C
α
/C c varies from 0.025 to 0.1 for normally consolidated soil.
is almost constant and
3.0
2.5
2.0
Primary consolidation
1.5
1.0
0.5
t p
C
α
Secondary consolidation
0.0
1 10 100 1000 10000 100000 time (minutes)
S s
= C
α ε
=
1
C
α
+ e o
Figure 4.15 Void ratio vs time for primary and secondary settlement
The settlement due to secondary consolidation ( S s
) is therefore
H log t t f p
H log t t f p
(4.31) where H is the initial thickness of the layer. The amount of secondary consolidation settlement may be quite significant for highly compressible clay and organic soils.

Compressibility & Consolidation 137
Example Problem 4.11
The results of consolidation test showed that the primary consolidation settlement of 5 m thick clay layer can be completed in 1 year. If the C
αε
of the soil is estimated as 0.005 for the appropriate stress change, estimate the secondary consolidation settlement for the following 20 years.
Solution
Given t p
= 1 year, t
C
αε
= 0.005,
H = 5 m f
= 1 + 20 = 21 year
S s
= C
α ε
H log t t f p
= 0.005 ×5× log
21
1
= 0.033 m = 33 mm
4.1
A 6 m square footing carrying a net pressure of 160 kN/m 2 is located at a depth of 2 m in a deposit of saturated clay 17 m thick. A firm stratum lies immediately below the clay. Calculate immediate settlement below footing if the undrained Young’s modulus E u of the clay is 55 MPa.
4.2
A foundation 4 × 2 m in size carrying a uniform pressure of 150 kN/ m 2 undrained Young’s modulus E u is 40 MN/m 2 . The layer is underlain by a second clay layer 8 m thick for which the undrained Young’s modulus
75 MN/m 2
E
. A firm stratum lies below the second layer of clay. Determine the average immediate settlement below the foundation. u
is placed at a depth of 1 m in a deposit of clay 5 m thick for which the
4.3
A footing 3 m square carries a net foundation pressure of 130 kN/m depth of 1.2 m in deep deposit of sand. The unit weight of sand is 16 kN/m 3
2 at a
. Determine the elastic settlement of footing using Schmertmann’s method one year after the construction if the variation of cone penetration resistance ( q c
) with depth is given as in table Q4.3
Table Q4.3
Depth
(m) q c
(MN/m
2
)
1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4
5.0 6.0
7.0 8.0
3.2 2.1 2.8 2.3 6.1 5.0 6.6 4.5 5.5 10.4 9.9 12.9 14.8

138 Introduction to Geotechnical Engineering Part 1
4.4
A shallow foundation of size 3 × 3 m in plan is placed on a deposit of sandy soil. The actual variation of Young’s modulus with depth determined using
SPT numbers are shown in Table Q4.4 below. Using the strain influence factor method, estimate the elastic settlement of the foundation after 5 years of construction.
Depth (m) 2.0 3.5 4.5 5.5 7.0 8.0 9.0
SPT N value 10 13 12 14 20 22 21
4.5
For the data in the table below, plot the e – log p ’ curve and determine the
C c
, C r
, pre-consolidation pressure
Initial void ratio is 2.60.
σ ’ c
, and the field compression index C cf
.
Effective stress
(kPa)
Void ratio e
1 3 4 10 21 41 82 22 6
2.56 2.52 2.44 2.08 1.75 1.47 1.22 1.27 1.37
4.6
Given a void ratio versus pressure data shown in the table Q4.6. Use the data to plot the e – log p ’ curve to (a) determine compression index ( C c
), (b) evaluate the pre consolidation pressure σ ’ c
, and (c) determine the field compression index C cf
Table Q4.6
Effective stress (kPa)
25 50 100 200 400
Void ratio e
1.115 1.080 1.025 0.965 0.900
4.7 The compression readings in Table Q4.7 were obtained in an oedometer test on a specimen of saturated clay ( G s
= 2.73). The initial thickness of the specimen was 19.0 mm and at the end of the test, the water content was
19.8%

Compressibility & Consolidation 139
Table
Pressure (kN/m
2
) 54 107 214 429 858 1716 3432 0
Dial gage readings
4.747 4.493 4.108 3.449 2.608 1.676 0.737 1.480
(after 24 hr, mm) a.
Plot the e – log p b.
determine m
’ curve and determine the pre-consolidation pressure v
for the stress increments 100 – 200 kN/m 2 and between
1000– 1500 kN/m c.
Calculate C c
2
for the increment of 1000– 1500 kN/m 2
4.8
The data in Table Q 4.8 was obtained from an oedometer test for a clay sample under 100 kN/m 2 pressure. At the end of the test the thickness of the sample was reduced to 17.53 min and the moisture content was 24.7 %.
Assuming the specific gravity of the soil sample equals to 2.70. Plot an appropriate graph and determine the coefficient of consolidation, C v
(in unit of m 2 /year) using Taylor's and Cassagrande’s method.
Table Q 4.8
Time 0.25
∆
H
0.233 0.302 0.390 0.551 0.706 0.970 1.127 1.225 1.290 1.321 1.482
(mm)
4.9
4 m high embankment is placed on the soft clay to pre-consolidate 5m thick saturated soft clay with saturated unit weight of 20 kN/m 3 . The natural moisture content of the soft clay is 37 %, specific gravity is 2.60 and the void ratio after loading is given in Table Q4.9. A layer of dense sand exists below the clay layer.
Effective stress (kPa) 25.5 42 64 98 150
Void ratio e 1.0 0.95 0.9 0.85 0.80
Assume the top width of the embankment is 10 m, the slope is 1 V to 2 H, and the unit weight of the fill material is 19 kN/m 3 , (a) calculate the overburden pressure and additional stress due to embankment loading at the middle of clay layer, (b) use the stresses obtained in question (a) as the maximum and minimum pressure to find the coefficient of volume change,

140 Introduction to Geotechnical Engineering Part 1 m v
. (c ) calculate the consolidation settlement using m consolidation, C v
, for the soft clay is 2.5 m 2 /year. v
and C c
. (d) calculate the time taken to achieve 90 % consolidation if the coefficient of
4.10 A shallow foundation of size 10 x 10m is carrying a uniform pressure of
200kN/m 2 . The foundation was placed at 1 m below the soil surface. The foundation soil is saturated clay with a unit weight of 20.5 kN/m 3 overlain by a loose sand of 5 m thick. The thickness of the clay layer is 25 m and the groundwater table coincide the interface between the sand and the clay layer. Below the clay layer, there is a layer of dense sand. The result of oedometer test on the clay soil showed that the initial void ratio is 1.10, and the compression index C c is 0.4. (a) Use Fadum chart to find the stress below the center of the foundation at the middle of clay layer, (b) Estimate the consolidation settlement of the foundation. (c) Estimate the time needed to reach 50% consolidation if coefficient of consolidation C v
= 2.5m
2 /yr.
4.11
Given a void ratio versus pressure data shown in Table Q4.11. The initial void ratio is 0.725.
Plot the e-log p curve to (a) determine compression index ( C c
) and recompression index ( C r
), (b) evaluate the pre consolidation pressure, ( σ ’ c
) (c) determine the field compression index C cf
.
Table Q4.11
Pressur e
(kPa)
25 50 100 200 400 800 1600 400 100 25
Void
0.708 0.691 0.670 0.632 0.574 0.510 0.445 0.460 0.492 0.530 ratio (e)
4.12
The consolidation test shown in Table Q4.11 is representative of a saturated clay layer 12 m thick exist below a sand deposit (4 m thick) with ground water table is at the surface of clay layer. Unit weight of sand is 17 kN/m 3 , while the saturated unit weight of the clay is 20 kN/m 3 . Based on this data,
(a) compute the overburden pressure at the middle of clay layer and calculate OCR. If there is an additional pressure of 56 kPa at the middle of the clay layer, (b) calculate the consolidation settlement of the soil due to this load. If the clay layer is drained in both directions, and the time rate of consolidation C v
is 8.0 x 10 -8 m 2 /sec, (c) find the average degree or percent consolidation for the clay 5 year after loading.

Compressibility & Consolidation 141
4.13
A foundation slab carrying a uniform loaf of 60 kPa is supported on a bed of compact sand which extends to a depth of 6 m below the base of foundation.
Under the sand there is a stratum of normally consolidated clay of 4.8 m thick (Figure Q4.13a). Ground water table is at 2 m depth. The size of the slab is 100 m x 100 m. The bulk and saturated unit weight of the sand layer are 18.5 and 19.6 kN/m clay is 18.8 kN/m 3
3 respectively, while the average unit weight of the
. Determine (a) the initial overburden pressure and additional stress in the middle of clay layer, compression index C c
Figure Q4.13b; (b) the coefficient of volume compressibility m v range 100 – 160 kN/m 2
from
for pressure
; (c) the ultimate settlement of the clay layer; and (d) the time (in years) for the consolidation to reach half the ultimate settlement. q o
= 60 kPa
Slab, 100 m x 100 m
Dense sand, γ b
= 18.5 kN/m 3
γ sat
= 19.6 kN/m 3 6 m
2 m
Clay, γ = 18.84 kN/m 3 ; C v
= 0.05 mm 2 /s
4.8 m
Figure Q4.13a
0.94
0.92
0.9
0.88
0.86
Figure Q4.13b
0.84
10
Impermeable shale
100 pressure (kPa)
1000

142 Introduction to Geotechnical Engineering Part 1
4.14
A compressible layer 10.0 m thick has an initial void ratio in-situ of 1.026.
Test and computations show that the final void ratio of the clay layer after construction of a structure is 0.978. Determine the estimated primary consolidation settlement of the structure.
4.15
A laboratory consolidation test was performed on a clay specimen, which was drained on both top and bottom. The time for 50% consolidation was
6.2 min, and the specimen thickness at that time was 1.88 cm. Two points on the consolidation line have coordinates ( p
1
, e
1
) and ( p
2
, e
2
) of (50 kN/m 2 ,
1.167) and (100 kN/m 2 , 1.108) respectively. Find the coefficient of permeability of the clay for the given loading range.
4.16
A foundation is to be built on a sand deposit underlain by highly compressible organic clay 5.0 m thick. Primary consolidation of the clay layer is estimated to complete in 10 year. What is the final settlement due to secondary consolidation if the design life of the structure is 50 years. The coefficient of secondary consolidation C
αε
is estimated as 0.001.

Stability analysis of a slope is important in the field of geotechnical engineering.
The stability of a slope should be evaluated as soon as soil movement in slope is identified to cause problems. Slope failure is actually the movement of mass on slope.
Okamura (2001) categorized the mass movement on slope into two types i.e.: the landslide and the slope failure. Slope failure is a mass movement on a slope of not less than 20 o which occur very fast, sometimes without any signs at all. The failure is usually triggered by rainfall and the affected area is small (1-100 hectare). Landslide, on the other hand, involves an extensive area with a slope angle less than 20 o , and the movement is slow and gradual.
Some factors have been identified to cause slope failures i.e.: slope inclination, additional load or fill height, excessive pore water pressure, and loss of shear strength due to weathering, liquefaction, and water (infiltration and seepage).
There are two types of slope in general: natural and man-made slope. Natural slopes were formed by a series of long-term and short term natural processes such as the formation of hills and valleys. Man-made slopes are of two types: excavated or cut slope and embankment or fill slope. Analysis of related slopes must be carried out when dealing with the constructing for example railways, highways, canals and excavations. This is also true for designing embankments, earth dams, dumping or waste fill.
The principal of slope stability analysis is very simple: that sliding will occur if the shear stress developed in the soil exceeds the corresponding shear resistance of the soil. The not so simple part is to find location where the failure started. Sliding may occur along any of a number of possible surfaces. Furthermore the shear strength of the soil generally varies with location and throughout time. It also
143

144 Introduction to Geotechnical Engineering Part 1 varies with soil moisture and some other factors. It is therefore common in practice to impose a factor of safety in slope stability analysis. Factor of safety 1.25 to 1.4 is considered appropriate for natural slope and routine cut and fills, but for the design of dam, higher factor of safety is required.
There are three types of slope failure based on the shape of failure surface i.e: wedge failure, circular and non-circular failures, and translational failures. Wedge failure usually occurs on a weak plane or weak joint due to external forces. This type of failure should be evaluated using 3-dimensional analysis. Circular slips are associated with homogeneous soil conditions while non-circular slips are associated with non-homogeneous conditions. Translational and compound slip failures occur where the form of failure is influenced by the presence of weak layer. In this case, the failure surface tends to be plane and roughly parallel to the slope surface
The first thing to do in the slope stability analysis is to find the most dangerous or the most critical surface. After defining the shape and location of the failure surface, the limit equilibrium methods can be applied for slope stability analysis by imposing several assumptions i.e: (1) the failure occurs in two-dimension, (2) rigid block movement taking place on the failure surface itself, and (3) uniform shear stress is mobilized over the whole length of the failure surface.
In general, there are two methods of limit equilibrium analysis i.e.: Linear methods which is relatively simple, and the Non-linear methods or Method of Slices.
Linear methods can be used for stability analysis of infinite slopes, slope in dry cohesionless soil ( c = 0), slope in cohesive soil where undrained condition prevail
( φ u
= 0), and wedge failure. Method of slices is necessary for irregular slope geometry, non-uniform soil condition, and slope analyses involving the consideration of seepage in soil.
In limit equilibrium method, the shear strength at the time of failure τ f is compared to the stress mobilized at that plane τ m
. Then the factor of safety is
f
FS =
m
≥ 1 (5.1)

Slope stability 145
Theoretically if the factor of safety is equal to 1, then the slope is in critical condition. At the time of failure, the shear strength of the soil is fully mobilized along the failure plane. The shear strength is represented by the Mohr-Coulomb criteria where for total stress analysis,
τ f
= c u
(5.2a) and for effective stress analysis
τ f
= c ’ + σ ’ tan φ ’ (5.2b)
In relation to the shear strength parameters to be used in the slope stability analysis, it is important to know the time when the most critical state occurs. If the most critical condition is predicted at the end of its construction period, i.e. short term, then the total stress concept is more appropriate. On the other hand if the critical condition is predicted in a long term situation, then the effective stress concept is to be applied. This depends on the type of slopes in question and on the type of soil.
For example, the most critical condition for excavated slope is long term, while the most critical condition for embankment is short term. For slope made of low permeability soil such as clay, the undrained assumption may be more appropriate since the dissipation of pore water pressure will take a long time. For highly permeable soil such as sand where drainage takes place almost immediately, we use effective stress parameter c ’ , and φ ’ .
In relation to the type of failure surface, the subsequent parts will deal with slope stability analysis for infinite slope which is related to the translational type of failure, the analysis of finite slope with planar surface of failure, and circular failure for undrained conditions. Evaluation of slope stability for a more complex situation should be analyzed using methods of slices.
Infinite Slope Analysis
The condition for infinite slope failure can be seen in Figure 5.1. In this case the factor of safety is assumed to be constant along the failure plane where the shear strength is

146 Introduction to Geotechnical Engineering Part 1
1
τ f
= c ’ + ( σ – µ ) tan φ ’ (5.3)
GWT
T m z cos 2 β
β
W
Flow net z mz z
µ
T or
N or
Figure 5.1 Translational failure on infinite slope
The expressions for σ and µ based on Figure 5.1 are
σ = W cos β
σ = {(1-m) γ b
µ = m z γ
+ m w cos 2
γ
β sat
} z cos 2 β (5.4)
(5.5)
The shear stress causing slope failure is
τ m
= W sin β
τ m
= {(1-m) γ b
+ m γ sat
} z sin
Substitute the above expressions to get factor of safety
β cos β
FS = c ' + ( σ −
τ m
µ ) tan φ '
(5.6)
FS =
γ sat c z cos 2 β sin β
+
γ
γ sat
' tan tan
φ '
β
(5.7)

Slope stability 147
For effective stress analysis where c ’ = 0,
FS =
γ '
γ sat tan tan
φ
β
'
And for the case where water table is far below the failure plane ( m = 0)
FS = tan tan
φ
β
'
(5.8)
(5.9)
Note that when c ’ = 0, then factor of safety is independent of the height of the slope. The slope will be stable as long as slope angle β is less than the internal friction angle φ ’ .
If both cohesion and angle of internal friction angle is not zero, then the critical condition ( FS = 1) will be achieved when: z = z cr
=
γ ' cos 2 c
β tan φ
(5.10)
For a total stress analysis, the shear strength parameters c u zero value of m
Example Problem 5.1
and φ u
are used with a
A long natural slope in a fissured over consolidated clay is inclined at 12 o to the horizontal. The water table is roughly at the surface of the slope. A slip has developed on a plane parallel to the surface at depth of 5 m. The saturated unit weight of the clay is 20 kN/m and φ ’
= 26
Solution: o
3 and the peak strength parameters are
. Determine the factor of safety along the slip plane.
c
’
= 10 kN/m 2
For β = 12 o , z = 5 m, m = 1 (water table is at the surface)
γ sat
= 20 kN/m 3 , c
’
= 10 kN/m 2 φ ’
= 26 o

148 Introduction to Geotechnical Engineering Part 1
σ = { ( 1 – m ) γ + m γ sat
} z cos 2 β
= m γ sat
z cos 2 β
= 1×20×5×cos 2 12 = 95.5 kN/m 2
τ m
= {(1 – m) γ + m γ sat
} z sin β cos β
= m γ sat
z sin β cos β
= 1×20×5×sin12×cos 12 = 20.3 kN/m 2
µ = m z γ w
cos 2 β
= 1×5×9.8×cos 2 12 = 46.8 kN/m 2
τ f
= c ’ + ( σ – µ ) tan φ ’
= 10 + (95.5 – 46.8) tan 26 o
= 10 + 48.7×tan 26 = 33.8 kN/m 2
FS =
τ
τ f =
33 .
8
= 1 .
66 m
20 .
3
Finite Slope with Linear Failure plane
For an assumption of linear failure plane, the soil mass is divided into the upper soil mass that will slide on the surface of the lower mass. The plane of weakness is assumed to pass through the toe of the fill as shown in Figure 5.2. In this figure, line AC is the trial failure plane.
The weight of soil (ABC) is:
W =
1
2
γ L h where γ = the unit weight of the soil mass,
H
L = sin θ
(5.11)
(5.12a)
H h = L sin ( β – θ ) = sin
sin ( β – θ ) (5.12b)

Slope stability 149 h
C
H
B
N =W cos θ
W sin θ
R s
L
A
β
θ
Figure 5.2 Slope with planar failure plane
Substituting equations 5.12 a and b into equation 5.11 gives:
W =
1
2
γ LH sin ( β sin
− θ
β
)
(5.13)
The force that will cause the failure is
T = W sin θ and the resistance to sliding is given by
R s
= c d
L + W cos θ tan φ d
(5.14)
(5.15) where c d
is the developed cohesion ( c d
= c
FS
C coefficient of friction (tan φ d
= tan
FS
φ
φ
), and tan φ d
is the developed
). The factor of safety will be
FS =
W
R s sin θ
= c L + W
W cos θ sin θ tan φ
(5.16)

150 Introduction to Geotechnical Engineering Part 1
Critical condition prevails when T is equal to R s equation 5.16, we get
.
By substituting FS equal to 1 in c d
=
1
2
γ H
⎡
⎢ sin
(
β − θ
) sin β sin ( θ cos φ d
− φ d
) ⎤
⎥ (5.17)
The critical angle for θ can be determined by equating the first derivative of c d
in equation 5.17 with respect to θ to zero, and solving for θ . The result is
θ =
β + φ d
2
(5.18)
Substituting θ into equation 5.17 yields: c d
=
γ H
4 ⎣
⎢
⎡ 1 − cos sin β
(
β − φ d
) cos φ d
⎦
⎥
⎤
(5.19)
Solving for H and replacing c d by c and tan ϕ d
by tan ϕ , gives
H cr
=
4 c
γ
⎡
⎢
1 sin
−
β cos cos
β
φ
(
− φ
)
⎤
⎥ (5.20) where H is the safe depth of cut and β is the angle of cut surface from horizontal.
Same principal valid for condition where a slope consists of two layers where the upper layer is assumed to slide along the interface between the two layers (Figure
5.3). h
H
D
θ
B
N= W cos θ
R s
W
T = W sin θ
L
β
A
Figure 5.3 Slope with linear failure surface
C

Slope stability 151
Example Problem 5.2
A slope as in Figure P5.2 is cut through two soil strata. The height of the slope is
5 m, and the height of the upper stratum is 3 m. The lower stratum is highly permeable cohesive soil. The angle of the slope is assumed failure plane
’
θ is 30 o
= 7 kPa, and φ ’ = 25
β = 45 o , and the angle of the the two soil is c kN/m 3
. The shear strength parameters at the interface of o . The unit weight of the upper layer is 16.5
. Determine the factor of safety against sliding.
B
C
Solution
5 m
3 m
A 45
ο
L =
D sin θ
=
3 sin 30
= 6 m h =
D sin β
sin ( β – θ ) = 1.1 m
30
ο
Soil 1
Soil 2
W =
1
2
γ L h = ½ × 16.5× 6 × 1.1 = 54.4 kN/m ’
Driving force T = W sin θ = 54.4 sin θ = 27.2
kN/m'
Resisting Force R s
= cL + W cos θ tan φ = 7 × 6 + 54.45×cos θ ×tan φ = 64 kN/m ’
=
64
27 .
2
= 2.35
Example Problem 5.3
A vertical cut is to be made through a soil mass. The soil has the following properties: γ = 16.5 kN/m 3 , c ’ = 25 kPa, and φ ' = 21 o . Evaluate the safe depth of the cut using a factor of safety = 2.

152 Introduction to Geotechnical Engineering Part 1
Solution
β = 90 o
For FS = 2, c =
25
2
= 12.5 kPa; φ = tan -1 ( tan 21
) = 11 o
2
H cr
=
4 c
γ ⎣
⎢
⎡
1 sin
−
β cos
( cos
β −
φ
φ
)
⎦
⎥
⎤
=
4 × 12 .
5
16 .
5 ⎣
⎢
⎡ sin
1 −
90 cos
(
× cos 11
90 − 11
)
⎦
⎥
⎤
= 3.6 m
Example Problem 5.4
A 2 m vertical cut is required for construction of a trench. The unit weight of the soil is 19.0 kN/m 3 , while its strength parameters are c ’ = 20.2 kN/m
Calculate the factor of safety of the cut with respect to both c ’ and
2 and φ ' = 28
φ ' parameters. o .
Solution:
Use equation 5.20, H cr
=
4 c
γ ⎣
⎢
⎡ sin
1 −
β cos cos φ
(
β − φ
)
⎦
⎥
⎤
Try FS
φ
= 1.0 Æ tan φ d
= tan φ
1
For vertical wall, β = 90
= tan 2 8
= 0.532
1
Substituting φ by φ d
in equation 5.20, get c d
=
γ H
4 ⎣
⎢
⎡ 1 − cos sin β
(
β cos
− φ
φ d d
)
⎦
⎥
⎤
=
19 × 2
4 ⎣
⎢
⎡ 1 − cos sin 90
(
90 cos
− 28
28
)
⎦
⎥
⎤
=
5.70 kN/m 2 c
FS c
= c d
=
20 .
2
5 .
70
= 3 .
54
Try FS
φ
= 2.0 Æ tan φ d
= tan φ
2
= tan 2 8
2
= 0.2659 Æ φ d
= 14.89
o
Substituting φ by φ d
in equation 5.20, get

Slope stability 153 c d
=
γ H
4 ⎣
⎢
⎡ 1 − cos sin β
(
β cos
− φ
φ d d
)
⎦
⎥
⎤
=
19 x 2
4
⎡
⎢
1 − cos sin 90
(
90 − 14 .
89 cos 14 .
89
) ⎤
⎥
=
7.30 kN/m 2 c
FS c
= c d
=
20 .
2
7 .
30
= 2 .
77
Try FS
φ
= 3.0 Æ tan φ d
= tan φ
3
= tan 2 8
3
= 0.1772 Æ φ d
= 10.05
o
Substituting φ by φ d
in equation 5.20, get c d
=
γ H
4 ⎣
⎢
⎡ 1 − cos sin β
(
β cos
− φ
φ d d
)
⎦
⎥
⎤
=
19 x 2
4
⎡
⎢
1 − cos sin 90
90 − 10 .
05
cos 10 .
05
⎤
⎥
=
7.96 kN/m 2 c
FS c
= c d
=
20 .
2
7 .
96
= 2 .
54
Plot FS
φ
vs FS c
in a graph to get the actual factor of safety = 2.60
FS c
5.0
4.0
3.0
2.0
1.0
0.0
0.0
1.0
2.0
FS
φ
3.0
4.0
5.0

154 Introduction to Geotechnical Engineering Part 1
Slope in dry cohesionless soil
When the slope angle β of a pile of dry sand exceeds the internal friction angle φ , the slope tends to fail by sliding in a downhill direction parallel to the slope. Thus the greatest angle for a free standing dry cohesionless soil is at the angle of internal friction, while the angle of which the dry sand slope fail is called the angle of repose. The factor of safety for slopes in dry cohesionless soil is therefore:
FS = tan tan
φ
β
(5.16)
Finite Slope with Circular Failure plane
In homogeneous soils relatively unaffected by faults or bedding, deep seated shear failure surfaces tend to form in a circular, rotational manner. Figure 5.4 shows the various types of circular failure planes. Toe and slope circle develop in slope made up of cohesionless soil and it is generally shallower that that of cohesive soil. On the other hand, deep seated failure may occur in cohesive soil. This failure plane sometimes reached interface between the soil and the stronger layer underneath the soil. The slope with larger inclination will have a shallow failure plane as compared to those with small slope angle.
The shape and location of the failure plane should be determined in advance before a slope stability analysis can be made. For circular failure plane, we can find the most critical failure plane by considering a series of slip circles of different radii and centre of rotation, and find the minimum factor of safety. This is a tedious task if we do not have any guidance to start with. Some charts and tables have been prepared to be used for starting point. Figure 5.5 and 5.6 show two methods used to get the first estimate of the most critical failure plane for slope in homogeneous soil.

Slope stability
O
155
O a. Toe circle
O b. Slope circle c. Base circle
Figure 5.4 Types of circular failure surfaces

156 Introduction to Geotechnical Engineering Part 1
β
11.32
18.43
26.57
33.79
45
60
α
25
25
25
26
28
29
δ
35
35
35
35
37
40
θ
α
β
δ
Figure 5.5 First approximation of the circular failure plane
5
4
3
Y c
X c
β
2
1
Y c
/H
X c
/H
0
0 10 20 30 40 slope angle ( β )
50 60
Figure 5.6 Chart for finding the location of center of the failure circle
H

Slope stability 157
Slope in Homogeneous Cohesive soil (
= 0 analysis)
Consider a slope in saturated cohesive soil with a potential circular failure surface centered at O with radius r , and arc length L a
as shown in Figure 5.7. The moment caused by the movement of the soil mass above the assumed failure plane is W d where W is the weight of the soil mass and d is the distance between the center of gravity of the soil mass and point O.
θ
R
B
R d
W
A
L a
θ
R y c a. No tension crack
B
R d
W
P w z c
Hydrostatic pressure in tension crack
L a
τ b. with tension crack c u
Figure 5.7 Circular failure planes
The shear strength mobilized along the failure surface to resist the moment is: m
=
τ f
FS
=
FS
(5.21) which yield in a force resistance of

158 Introduction to Geotechnical Engineering Part 1
R s
= τ m
L a
= c u
FS
L
Equating moments about point O (see figure 5.7) a
W d = c u
FS
L a
R
Therefore
FS = c u
W
L a d
R
(5.22)
(5.23)
(5.24)
An appearance of tension cracks at the head of a slide suggests strongly that instability is imminent. Tension cracks are sometimes considered in slope stability calculations, and sometimes they are considered to be full of water. If this is the case, then hydrostatic forces will develop. If the tension cracks do exist, then L a equation 5.24 is shortened and hydrostatic force will act normal to the crack if the crack is filled with water. In this case, the L a will reduce to L a
’ , and the moment causing failure will increase by the moment due to pore water pressure P w
Thus, equation 5.24 becomes
in
y c
.
FS =
W c u d
L a
'
+ P w
R y c
(5.25)
Example Problem 5.5
An excavation was made to depth of 6 m forming an angle 35 cohesive with c u
= 25 kN/m 2 and φ o . The soil is
= 0. Unit weight of the soil is 20 kN/m 3 . (a) use Figure 5.5 to get the first approximation of failure plane for the slope; (b) find the factor of safety of the slope if the weight of sliding mass is 583 kN and the center of gravity of the sliding mass is at a horizontal distance 3.92 m from point
O. The radius of the circle is 10 m .
Solution:
From figure 5.5, with β = 35 o
α = 26 o , δ = 35 o , then θ = 180 – α – β – δ = 180 – 26 – 35 – 35 = 84 o
Thus FS = c u
W
L a d
R
=
25 ×
{
2 × π × 10 2 × 84 /360
}
583 × 3.92
= 1.57

Slope stability 159
Example Problem 5.6
The slope of an embankment is 1 vertical : 2 horizontal and the vertical height is
10m (Figure P5.6). The soil has an undrained cohesion of 35 kPa and a unit weight of 18 kN/m 3 . A failure plane is assumed with a radius of 22.83 m. The weight of the moving mass is 3150 kN. A tension crack exists on the surface of the slope and it is filled with water. Determine the factor of safety against shear failure along the trial arc shown in the figure.
O
83.1
o y c
= 12.2 m d = 7.3 m
P w z w
=3.3m
H = 10 m
W
Figure P5.6
Solution
Slope 1V: 2 H Æ β = tan -1 (½) = 26.5
o
Soil is in undrained condition c u
= 35 kPa
Use formula FS = c u
L a
R
W d
But tensile crack exist, then
FS =
W c u d +
L a
'
P w
R y c
Here P w
= ½ γ w
z w
2
L a
=
θ
360
2 π R
= 54.35 kPa
=
83 .
360
1
× 2 π × 22 .
83 = 33.11

160 Introduction to Geotechnical Engineering Part 1
35
FS =
3150 ×
× 33 .
7 .
3 5
11
+
× 22 .
83
4 .
35 × 12 .
2
=
26444
22932
=
1.15
There are many charts available for evaluating the factor of safety of a slope, each based on various assumptions and specific use. The most popular one is the
Taylor’s stability number which will be discussed in the following. Other charts include the Janbu stability charts for slope in cohesive soils, Morgenstein graphs for rapid drawdown in dam, the Bischop and Morgenstain charts for effective stress analysis, and many others.
Taylor’s chart (Figure 5.8) can be used to evaluate the stability of slope in a homogeneous cohesive soil with a hard stratum at depth of n d
H from the surface of the slope. The factor of safety can be calculated as:
FS =
N s c
(5.26)
γ H
Where N s
is the Taylor’s stability number obtained from the chart.
Example Problem 5.7
A cutting in a cohesive soil forms a slope angle of 35 undrained cohesion c u
is 40 kN/m 2 o and a height of 8 m. The
, and unit weight γ = 18 kN/m 3 . Using Taylor stability charts determine the factor of safety against shear failure for (a) a hard stratum exists well below the slope, and (b) a hard stratum exists at 4 m below the toe
Solution a.
c u
= 40 kN/m 2 ,
γ = 18 kN/m
φ
3 , n d from chart N s
= 5.52 u
= 0 is infinity
β = 35 o
FS =
N
γ s
H
c x 40
18 x 8
= 1.53 b.
If hard stratum exist at 4 m below the toe, then n d
H = 8 + 4 = 12

Slope stability 161
For n d
= n d
H
H
=
12
8
= 1.5 Æ N s
= 6.0
FS =
N
γ s
H
c
=
6.0
× 40
18 × 8
= 1.65
Thus, the presence of hard layer forced a formation of smaller circle, which results in higher factor of safety. n d
H H β
Figure 5.8 Taylors stability number

162 Introduction to Geotechnical Engineering Part 1
Example Problem 5.8
Determine the factor of safety in terms of effective strength for the slope with
40
10 o o and height 10 m, if the properties of the soil are as follows: c
’
, γ = 17 kN/m 3
= 20 kN/m 2 , φ
β =
’ =
Solution
From Taylor’s chart: for φ =10 o , β = 40 o , N s
= 10.7
FS =
N
γ s
H
c
=
10.7
x 20
17 x 10
= 1.26
Example Problem 5.9
Determine the safety factor for a slope with a height of 5.5 m, and slope angle 40
The unit weight of the soil forming the slope is 16.5 kN/m 3 , while the shear strength parameter of the soil is c ’ = 20 kPa and φ ’ = 10 o o .
Solution
For β = 40 o , φ ’ = 10
FS c
= o , H = 5.5 m, and
N
γ s
H
=
10.7
x 20
16.5
x 5.5
γ =16.5 kN/m
= 2.36
3
Use FS
φ
= 1 Æ N s
from chart = 10.7, then using c ’ = 20 kPa
c
Use FS
φ
= 1.8 Æ φ d
= tan − 1 tan 10
1.8
= 5.6
N s
from chart Æ for for
φ
φ d
= 5 ; N s
= 8 d
= 10 ; N s
= 10.7
Interpolate for φ d
= 5.6 ; N s
= 8.324 then using c ’ = 20 kPa

Slope stability 163
FS c
=
N s
c
γ H
=
8.324
x 20
16.5
x 5.5
= 1.83
Then FS
φ
Example problem 5.10
= FS c
= 1.8
Determine the safety factor for a submerged fill with a height of 30 m and slope angle 30 o . The saturated unit weight of the soil is 19 kN/m strength parameters of the soil are: c ’ = 35 kPa and φ ’ = 5 o .
3 , while the shear
Solution:
Use FS
φ
= 1
From Taylor’s chart for φ =5 o , β = 30 o Æ N s
= 9.25
FS c
=
N s
γ '
c
H
=
9.25
x 35
( 19 9.8) x 30
= 1.17
Most of the times, the slope stability cannot be analyzed using the simple methods presented earlier especially for slope with both cohesion and internal friction angle
φ > 0. This analysis is generally done by using slices methods. In this method, the potential failure surface is assumed to be circular or non-circular.
Figure 5.9 shows a failure plane forming a circular arc with center O and radius R.
The soil mass (ABCD) above the failure surface (AC) is divided by vertical planes into a series of slices of width b. The base of each slice is assumed to be a straight line. For any slice, the inclination of the base to the horizontal line is α i
and the height (measured at the centerline) is h i
.

164 Introduction to Geotechnical Engineering Part 1
O
D C
A
B
1
2
3
4
5
6
7
8
9
E i-1
X i-1
α b
W i forces acting on a slice
X i
E i
T i
N i
-U
Figure 5.9 Slices methods applied to a slope
As shown in the Figure, the forces acting on slice i are:
1.
The total weight of the slice, W = γ bh i
2.
The total normal force on the base: the effective normal force N the boundary water force U = µ l i center of the base and l is the length of the base
’ = σ ’ l i
and
. where µ is the pore water pressure at the
3.
The shear force on the base, T = τ m l i
4.
The total normal forces on the sides, E i
and E i-1
5.
The shear forces on the sides, X i and X i-1
6.
Any external forces working on the slope must be included in the analysis.
The factor of safety in all slices is assumed to be the same, implying mutual support between the slices. In this case the forces acting between slices are denoted by E and X .
We can see that there are many unknown forces involved in the equation of equilibrium for a slice. However if we assume that there is no pore water pressure and the inter-slice forces E and X is equal to zero, then the factor of safety can be

Slope stability 165 evaluated based only on the weight of the slice which can be resolved into two components T and N .
In circular failure plane, the overall factor of safety can be evaluated by examining the sliding and resistance moment about point O. As usual, the factor of safety is defined as the ratio of the available shear strength ( τ f
) to the shear stress acting on the plane ( τ m
)
FS =
f m
(5.27)
The component T which is parallel to the base of the slice tends to cause sliding.
Taking moment about O, the sum of the moments of the shear forces T on the failure arc AC must be equal to the moment due to the weight of the soil mass
ABCD.
TR =
W R sin α (5.28)
Resistance to this sliding force is provided by the cohesion and internal friction of the soil. The cohesion force is equal to the product of the cohesion of the soil and the length of the slices curved base ( L a
). The friction force is the normal components of the weight ( N ) multiplied by the friction coefficient which is equal to tan φ . For analysis in terms of effective stress the general factor of safety is given as:
FS =
∑ c' i l i
∑
+
W
∑ sin
N'
α tan φ '
(5.29a) or if c constant, then FS = c' L a
+
∑
∑
W
N' sin tan φ '
α
(5.29b)
Example Problem 5.11
Calculate factor of safety for a slope with the given failure plane in term of effective strength concept. The slope angle is 45 strength parameter of soil in layer A are c kN/m 3 , while for layer B, c ’ = 0 kN/m 2 , φ ’
’ = 17.1 kN/m
= 34 o o , height is 15.24 m. The shear
2 , φ ’ = 18
, and γ = 19.2 kN/m 3 . o , γ = 17.6

166 Introduction to Geotechnical Engineering Part 1
H = 15.24 m
Figure P5.11
O
4
3
2
1
Soil B
Soil A
Solution :
The factor of safety ( FS ) can be calculated using Equation 5.29:
FS =
∑ c' i l i
∑
+
W
∑
N' sin α
For case of no ground water table; N ’ = W cos α tan φ '
The soil mass in divided into 4 slices as shown in Figure P5.11. The weight of each slice is computed based on the shape/geometry.
W = crossectional area x unit weight = b h γ b h
Length of circular arc on soil A is 9 m (based on scaled graph).
Each component of the formula for calculating FS for each slice is tabulated
(Table P5.11)

Slope stability 167
Table P5.11
Slice
No
Soil at the base
W
(kN)
φ ’ c ’
(kPa) l i
(m)
W cos
α tan
φ
W sin
α
1 A 297 18 17.1 9 45.6 264.2
17.1
FS =
17 .
1 × 9 +
1340
2470
= 1.96
Fellenius (Swedish) Method
Fellenius assumed that the resultant of the inter-slice forces is zero, but the porewater pressure exists at the base of the slice, then
N ’ = W cos α – µ l (5.30)
Where µ is the pore-water pressure and l is the length of the base of a slice. Hence, substituting equation 3.30 into equation 3.29, the factor of safety in terms of effective stress is given by:
FS =
∑ [ c' l +
(
W cos
∑
W
α sin
−
α
µ l
) tan φ '
]
(5.31)
The components W cos α and W sin α can be determined graphically while angle α can be calculated or measured. For analysis in terms of total stress parameter or φ u
= 0, then
FS = c u
∑
W
L a sin α
(5.32)
Example Problem 5.12
Determine the Factor of Safety of the slope with an assumed failure plane shown in Figure P5.12 in terms of effective stress using Felenius slices methods. The

168 Introduction to Geotechnical Engineering Part 1 slope is 1V:1½ H. The unit weight of the soil is 20 kN/m 3 parameters are figure. c ’ = 10 kN/m 2 and φ ’ = 29
and the shear strength o . Ground water level is shown in the
Solution:
The factor of safety ( FS ) can be calculated using Equation 5.31:
FS =
∑ [ c' l +
(
W
∑ cos
W
α sin
−
α
µ l
) tan φ '
]
Radius of the circular failure plane is 9.5 m, then arc length
L a
= (86.5/360) × 2 × π ×9.50 = 14.35 m
The soil mass is divided into 8 slices of 1.5 m wide. The weight of each slice is given by W i
= γ b h i
The pore water pressure at the center of the base of each slice is
µ = γ w
h w where h w
is the vertical distance of the center point below the water table.
O
86.5
o
R = 9.50 m
3.15 m
6.00 m
2.5 m
3
4
6
5
1.5 m
7 8
WT
1
2
Figure P5.12

Slope stability b
Table P 5.12
Slice W cos
α h w
α h
W sin
α µ l
µ
l
No (kN/m’) (kN/m’) (kN/m
2
) (m) (kN/m)
1 22.5 -4.5 5.9 1.55 9.1
2 54.0 -3.0 11.8 1.50 17.7
3 81.0 12.0 16.2 1.55 25.1
4 97.5 30.0 18.1 1.60 29.0
5 103.5 52.5 17.1 1.70 29.1
6 93.0 70.5 11.3 1.95 22.0
7 57.0 67.5 0 2.35 0
8 16.5 28.5 0 2.15 0
169
FS =
( 10 × 14 .
35 ) + ( 525
254
− 132 ) × 0 .
554
= 1.42
Bischop (Routine) Method
Bischop assumed that the resultant of vertical inter-slice forces ( X i
–X to zero, but the sum of the horizontal inter-slice force ( E i
–E is not equal to the soil resistance at the base of the slices, but i-1
) is equal i-1
) is not zero. Then T
T =
1
FS
( c' l + N'
Resolving forces in the vertical direction, then: tan φ '
)
(5.33)

170 Introduction to Geotechnical Engineering Part 1
W = N' cos α + µ l cos α + c' l
FS sin α +
N'
FS
N' =
⎜
⎝
⎛
W − c' l
FS cos α sin
+
α − tan φ '
µ l cos α
⎞
⎠ sin α
FS tan φ ' sin α (5.34)
(5.35)
Substituting the length of the base of the slice ( l ) by b sec α , and after some rearrangement, we obtain:
FS =
∑
W
1 sin α
∑
⎢
⎢
⎣
⎡
⎢
{ c' b +
(
W − µb
) tan φ '
}
1 + sec tan α
α tan
FS
φ '
⎥
⎥
⎦
⎤
⎥
(5.36)
µ
µ by defining r u
=
γ h
= W then: b
FS =
∑
W
1 sin α
∑
⎢
⎢
⎣
⎡
⎢
{ c' b + W
(
1 − r u
) tan φ '
}
1 + sec α tan α tan φ '
FS
⎥
⎥
⎦
⎤
⎥
(5.37) and by defining m
α
= cos α 1 tan α tan φ
FS
FS =
∑
W
1 sin α
∑
⎡
⎢
{ c' b + W
(
1 − r u
) tan φ '
} 1 m
α
⎤
⎥ (5.38)
Since FS appears in both sides of the equation, we have to use trial and error to find the factor of safety. Janbu, Bjerrum and Kjaernsli (1956) suggested the use of a chart to help the trial and error process. However, currently with the help of spreadsheet program, equation 5.37 can be solved easily.

Slope stability 171
Example Problem 5.13
Determine the factor of safety in terms of effective stress of the slope shown in
Figure P5.13 with respect to the trial circle shown. The shear strength parameters of the upper layer are kN/m 3 c
1
’
= 25 kN/m
. For the lower layer c that the pore pressure ratio r u
2
’
= 7 kN/m
= 0.3.
2 and
2
φ
;
1
’
φ
= 12
2
’ o while the unit weight
= 25 o ; γ
2
= 19.5 kN/m
γ
1
= 18
3 . Assume
7m
O
R = 13.9 m
23.7
o
4 m 79.9
o
B x 10
9
5 m 8
6 h
1
7
2m
4
5
Soil 1
Soil 2
3 m h
2
3
A
2
1
Figure P5.13
Solution
The slip mass is divided into a convenient number of slices of width b . The average height of each slice is measured off a scale drawing of the cross section in two components h
1
and h
2
. The weight of each slice is therefore: angle
W = ( γ
1
h
1
+
α is evaluated for each slice by using α
γ
2
h
2
)b
= arc sin x
R

172 Introduction to Geotechnical Engineering Part 1
Use formula 5.37 and tabulate the calculation in spreadsheet (Table P5.13)
FS =
∑
W
1 sin α
∑
⎢
⎢
⎣
⎡
⎢
{ c' b + W
(
1 − r u
) tan φ '
}
1 + sec tan α
α tan
FS
φ '
⎥
⎥
⎦
⎤
⎥
Table P5.13
Slices
Measured Measured Measured Measured h
1 h
2
W
x b
α
W sin
α
1 0
2 0
3 0
4 0.55
5 2
6 3.3
7 4.7
8 5
9 5
0.6 5.67
2.1 4
3.9 2
5.0 0
4.8 2
4.4 4
3.5 6
2.5 8 2
0.7 10 2
2
2
2
2
2
2
2
23.4 24.1 - 9.552
81.9 16.7 -23.585
152.1 8.3 -21.901
214.8 0 0
259.2 8.3 37.322
290.4
16.7 83.829
305.7
25.6 132.052
277.5 35.2 159.827
207.3 46.0 149.244
10 2.9 0 11.8 2.4 125.28 58.2
613.465
Table P5.13 (continued)
Slices
Given Given c
’ φ c
’ b tan
φ ' sec
α
1+ (tan α tan ϕ ’)/FS
W (1-r u
) tan
φ '
14 7.638
2 7 25
3 7 25
4 7 25
5 7 25
14
14
14
14
0.466
0.466
0.466
0.466
0.945 26.733
0.962 49.648
1 70.114
0.962
84.607
6 7 25
7 7 25
8 7 25
9 7 25
14
14
14
14
0.466
0.466
0.466
0.466
0.945 94.791
0.950 99.785
0.981 90.580
1.057 67.666
18.640
Table P5.13 was obtained from calculation for FS 1.333 after some trial and error, and the final FS obtained was 1.333.
Example Problem 5.14
Determine the long term stability of the cutting slope given in Figure P5.14 for the given failure surface. The unit weight of the soil above water table is 19.5 kN/m 3

Slope stability 173 while the unit weight below water table is 20 kN/m 3 parameters are c
’
= 18 kN/m 2 and φ ’
= 23 o .
. The shear strength
O
79 o
R = 29.50 m
16.25 m
9.5 m
1
WT
12.0 m
6
5
1:2.5
4
3
2
Figure P5.14
Solution
The slip mass is divided into 6 slices of width b. The average height of each slice is measured off a scale drawing of the cross section. The weight of each slice is and the pore water pressures are:
W = γ h b and µ = γ w
h w angle α is evaluated for each slice by using α = arc sin x
R
Use formula 5.37 and tabulate the calculation in spreadsheet to get the factor of safety by iteration process. In this case the FS at the right hand side of Equation
5.37 were tried for several values until FS at the left hand side is equal to the trial
FS value. Table P5.14 represent the tabulation of Factor of safety for assumed.
FS =
∑
W
1 sin α
∑
⎢
⎢
⎣
⎡
⎢
{ c' b + W
(
1 − r u
) tan φ '
}
1 + sec α tan α tan φ '
FS
⎥
⎥
⎦
⎤
⎥

174 Introduction to Geotechnical Engineering Part 1
Table 5.14
Slices c
’ b
+(W-
µ
)tan
φ
’
α sin
α
W
µ
W sin
α
(6)
0 51.3
37 56.3
62 50.4
66 44.4
49 38.0
27.8
190.5 267.7 sec
α
1+(tan
α tan
φ
’)/FS
(7)
(6)/(7)
1.099 56.7
1.000 56.3
0.962 48.4
0.971 43.1
1.010 38.3
1.087 30.2
272.8
Table P5.14 was obtained from calculation for FS = 1.4 after some trial and error,
272 .
8 and the final FS obtained was
190 .
5
=1.432 (close enough)
Other Methods of Slices
Due to repetitive nature of the calculations and the need to select the most critical failure surface, the method of slices is particularly suitable for solution by computer. In this case, more complex geometry and soil strata can be introduced.
Many computer programs have been developed based on limit equilibrium methods and available for use. However, it is quite easy to develop your own program for some simple problems.
Besides the Fellenius and Bishop’s method described above, there are many other methods of slices have been developed by other researchers. These methods differ in term of the assumptions made on inter-slices forces and the shape of the failure surface. Other methods of slices are introduced by Janbu (Janbu simplified and
Janbu rigorous methods), Lowe and Karafiath, Spencer, and Morgenstein and
Price. Table 5.1 lists the methods of slope stability analysis frequently used in practice, their assumption of slip surface and the corresponding equilibrium analysis. Discussion on these methods can be found elsewhere.

Slope stability 175
Table 5.1. Slices methods of analysis frequently used in practice.
Method
Ordinary method of slices (Fellenius,
1927)
Bishops Modified
(Bishop, 1955)
Force equilibrium
Does not satisfy horizontal or vertical forces equilibrium
Satisfy vertical force but not horizontal force equilibrium
Moment
Shape of slip surface equilibrium
Yes. Circular circular may have numerical problems.
Janbu’s simplified method
(Janbu, 1956) frequent numerical problems than other methods
Morgenstern and
Price (Morgenstern and Price, 1965)
Spencer’s Method
(Spencer, 1967)
Yes. Permits side forces to be varied
Yes. Side forces are assumed to be parallel
As mentioned in the beginning of this chapter, the stability of a slope should be evaluated as soon as some type of movement such as development of surface crack is identified on the slope. If the analysis shows that the slope is indeed in critical stage, then some effort should be done to stabilize the slope.
There are many methods available for slope stabilization, most falls into the category of prevention or initial treatment of failed slopes. The methods can be grouped into (1) reducing the load, (2) increasing the shear resistance, and (3) drainage system. Alternative methods include the so called bioengineering and soil cover system.
Reducing the load causing soil movement include re-grading or changing the geometry of the slope includes: reducing the height of the slope, compacting slopes to a flatter angle, creating stepping slope, and putting an additional load
(berm) at the toe. This method has an immediate effect on improving the stability of the slope, but unlikely to become effective with time. This method is popular for initial treatment for slope failure
Traditional methods of controlling slope failures have relied on structural practices like rip-rap, retaining wall, buttress wall, and sheet pile. A more recent

176 Introduction to Geotechnical Engineering Part 1 development is the use of gabions, anchor, soil nailing, use of geotextiles and geogrids, wire meshes, and shotcrete.
The method can be grouped into passive and active scheme. Passive schemes such as the retaining wall and piling system only take effect on further movement of the slope, which of course may not be desirable as slope stabilization.
Furthermore, these methods incurred a high initial cost and are not environmental friendly. Active scheme, for example soil nailing and anchor adds the strength directly into the soil through friction at the interface. Adjustment of building position, if possible, will also have the benefit to the stability of slope.
Surface drainage is very common as slope stabilization method because it can be used to regulate the amount of water that can infiltrate into the soil. Application of drainage grids just below the surface may have immediate effect in high permeability soils but may have to take more time to function in fine grained soils.
The drainage is best used as a short term stabilizing methods, due to the fact that , in the long term, the drains need much maintenance and repair which is often difficult to perform and expensive. Other types of drainage include the deep drain, relieve well or trench at the toe, cutting trench, and vertical drains are used to reduce pore water pressure in soil.
Other approaches are the so called the bioengineering methods and the soil cover system. In bioengineering method, vegetation is used to help stabilize the slope indirectly through the root system and its effect on the soil moisture regime. By covering over a slope with grass or any other cover system, we immediately reduce the amount of water which can infiltrate it. This method is often used in conjunction with more effective and long-term methods. The method is actually inexpensive and simple but may take some time to gain the stabilizing function.
Selection of the stabilization methods will very much depends the slope condition, the type of soil that make up the slope, the urgency, and the cost. Generally in looking at failing slopes, we need to choose a primary stabilization method i.e. one which will immediately take effect in stopping the slide. In deciding the method to use, the consideration usually we given to option to re-grade the slope, then installing surface and blanket drain, and last consideration would be the application of reinforcement. If time is not an utmost important factor, the vegetation may worth serious consideration.

Slope stability 177
5.1
An infinite slope is inclined at 20 o . A plane of failure has developed at depth of 8 m from the surface. The unit weight of the soil is 18.6 kN/m 3
While its shear strength parameters are given as c ’ = 18 kN/m 2 and φ ’ = 25 o
.
.
Calculate factor of safety (a) if there is no water table, (b) if water table exists at the surface, given c ’ = 0; γ sat
= 20 kN/m 2
5.2 A slope has an identified failure plane at an angle
β is 50
.
θ of 30 o . The slope angle o and the height of the slope is 3 m. Calculate the safety factor if the shear strength parameter of the soil are unit weight of the soil 16 kN/m 3 c ’ = 28.8 kN/m 2 and φ ’ = 10 o . The
5.3 A canal is planned on a soil with shear strength parameters c and φ ’ = 28 o that FS = 3 with respect to both c ’ and φ ’ .
’ = 20.2 kN/m 2
. If the slope of the cut is 50 o , what is the height of the cut so
5.4
The most critical slip surface is the one for which the calculated factor of safety has the lowest value. The minimum factor of safety is clearly the criterion required for design. The accuracy of values of factor of safety depends on a process of trial and error, using a reasonable number of trial circles and a thoughtful search pattern. The slope shown in Figure Q5.4 is in homogeneous undrained condition with cohesion c unit weight 18.5 kN/m 3 u
= 40 kN/m 2 , and bulk
. (a) Using the charts provided in this chapter, determine the first trial coordinate for the slope, (b) Estimate of the factor of safety of the slope.
H = 5 m
Slope
1V: 2H
Figure Q5.4
5.5
A cut is to be made in saturated clay to a depth of 9 m. The unit weight of the soil is 19 kN/m 3 . Shear strength parameters of the soil are c u
= 30 kN/m 2 and φ u
= 0. A hard stratum exists at depth of 11 m below the original soil surface (see Figure Q5.5). (a) By using Taylor’s chart, determine the

178 Introduction to Geotechnical Engineering Part 1 critical angle β for the cut slope, (b) Determine the slope angle for the factor of safety 1.2.
Figure Q5.5
9 m
11 m
β
12 m c u
= 30 kPa
5.6 A cutting in a saturated clay is inclined at a slope of 1 vertical :1½ horizontal and has a vertical height of 10 m (Figure Q5.6). The bulk unit weight of the soil is 18.5 kN/m 3 , and its undrained cohesion is 40 kPa.
Determine the factor of safety against immediate shear failure along slip circle shown in the figure (a) ignoring the tension crack, (b) assuming the tension crack empty of water, (c) assuming that the tension crack filled with water.
5.0 m
θ
O
6.7 m R y c
= 9.58 m d
2.7m
P w z o
=4.32 m
10.0 m
W
A
Figure Q5.6

Slope stability 179
5.7
The stability of a slope shown in Figure Q5.7 is to be analyzed by the method of slices for the given failure surface. The length of the failure surface is 11 m. The internal friction angle of the soil is 10 o , and the cohesion is 20 kPa. Calculate the factor of safety of the slope along the given failure surface. The data for each slice are tabulated as in Table Q5.7
O
Figure Q5.7
Table Q5.7
1
2
3
4
5
6
α
7
8
Trial failure surface
Slice
Number
Weight
(kN/m)
1 2 3 4 5 6 7 8
Base angle (
α
)
5.8
It is said that the Fellenius method gives a conservative factor of safety (i.e. smaller than) in comparison to results of stability analysis by Bischop
Modified Method. State the reasons for the above statement with respect to assumptions made in the formulation of the Fellenius and Modified
Bischop’s methods.
5.9 A stability analysis was made for a 2 m high embankment built on soft foundation soil (Figure Q5.9) using Fellenius method of slices. The inclination of the slope is 1 vertical : 1 ½ horizontal. The laboratory tests show that the soil parameters for embankment soil are: c
35 o and and γ
γ b
= 20 kN/m 3 sat
= 14 kN/m
, while for the foundation soil: c u
’ = 0 kN/m
= 5 kN/m 2 ; φ
2 , u
φ ’ =
= 0 o existing foundation soil. Calculate the factor of safety of the slope for the
3 . The groundwater level is at the ground surface of given failure surface.

180
O
Introduction to Geotechnical Engineering Part 1
7.0 m
6.5 m
1
2 3
4
5
6
1V: 1.5
7
8
2.0 m
2.5 m
Figure Q5.9
5.10
A cutting of depth 15 m is to be formed in a fine soil with a slope 1 vertical:
2.5 horizontal. The soil has a unit weight of 20 kN/m 3 and undrained shear strength parameter c u kN/m 2 and φ ’ = 28 o
= 90 kPa or drained shear strength parameters c ’ = 10
. The average pore pressure ratio r u is 0.40. Determine the factors of safety: (a) immediate after excavation, and (b) long time after excavation
5.11
A test trench with vertical sides failed when excavated to a depth of 6 m and the tension crack is seen to be 2 m deep. Laboratory tests on the clay gave constant situation. c u
= 40 kPa, and unit weight γ = 18 kN/m 3 . Comment on this
5.12
A wide cutting 8 m deep with side slopes of 30 o inclination to the horizontal is made rapidly through a horizontal deposit of soft saturated uniform clay.
The clay, which overlies rock, has a unit weight of 18 kN/m 3 and thickness of 12 m. On completion of the excavation, the cutting slope failed by landslipping. Using Taylor’s chart, estimate the average undrained shear strength of the clay. What is the depth of excavation for a factor of safety of
2.0

Slope stability 181
5.13
As part of a new motorway, a permanent cutting 15.0 m deep is to be made with sides sloping at 1 V to 3 H in a stiff clay. The bulk unit weigh of the clay is 19.0 kN/m kN/m 2 and φ ’
3 , and the peak shear strength parameters are c
= 25 o . The average pore pressure coefficient
Calculate the factor of safety of the slope in the long term. r u
’ = 11
= 0.3.
5.14
A stability analysis was made for slope of dumping of mine waste material using simplified Bischop method of slices. The height of the slope is 30 m and the average surface slope is 36 waste material has a unit weight of 17.6 kN/m parameters c ’ = 0, and φ ’ = 32 o o . Laboratory test results shows that the
3 and shear strength
. The simplified cross-sesction and failure plane are shown in Figure Q5.14. (a) Redraw Figure Q5.14 using an appropriate scale, (b) Calculate the factor of safety of the slope (c) What is the factor of safety if the average surface slope is reduced to 30 o .
1 2
3
4
5
10
9
7
Mine waste material
6
8
Original ground
Trial failure circle
Figure Q5.14
5.15
(a) State the slope stabilization methods that you are aware of. (b) List the factors to consider for the selection of slope stabilization method for a certain situation, (c) If you find that the slope in Question 5.14 is unsafe, then you need to plan a slope stabilization. What is the most suitable slope stabilization method for this situation?

As described in Chapter 1, soil is defined as a collection of solid particles surrounded by void which is filled with water and air. The individual solid particle and water is incompressible, while air is highly compressible. Soil volume would change due to rearrangement of solid particles into a new position. Thus, the deformation of soil depends on the structural arrangement of the solid particles.
In dry or partly saturated soil, deformation takes place because air is compressed from the voids. The rearrangement of soil particles due to compression of air can happen by itself or through compaction process. When all the voids are filled with water, the deformation will occur only if water can dissipate from the voids due to applied load. This process of dissipation of excess pore water due to applied load is called consolidation.
There are two types of stress in soil. The stress due to soil’s own weight, normally called the in-situ stress or overburden pressure, and the change of stress due to applied load. This chapter will discuss the basic theory and methods to calculate both stresses.
In-situ stress in saturated soil presents in two situations i.e.: the total normal stress and the effective stress. The normal stress is the total stress feels by the soil element as if it is a solid material. Effective stress is the stress transmitted through inter-granular particle contacts within the soil (Figure 2.1). The concept of effective stress is important because shear stress can be resisted only by soil particles by means of forces developed at the inter-particle contacts. If the soil is fully saturated, then the water filing the voids can also withstand the increase in pressure.
27

28 Introduction to Geotechnical Engineering Part 1
A
Figure 2.1 Representation of stresses in soil
If a force P is applied on a soil element with a cross-sectional area A , it will be resisted in part by water and the inter-granular forces. The inter-granular force may be split into two components: normal ( N normal stress is then:
’ ) and tangential (T). The effective
σ ' =
A
N '
(2.1) and the total stress is
P
A
=
∑
=
A
If water pressure is working on all area A, then the stress is
P =
N ' + µ A or
A
N '
P
+ µ
Thus
(2.2)
(2.3)
σ = σ ’ + µ (2.4)
Total Stress = Effective Stress + Pore water pressure

Stresses in Soil 29
Application of the concept of effective stress is very wide including the evaluation of shear strength characteristics of soil, evaluation of consolidation problems, and the evaluation of quick and uplift condition. A general application of effective stress is the determination of in-situ stress in soil.
As described above, stress in saturated soil is the combination of effective stress and pore water pressure. Thus the stress in a soil element at any depth below ground surface due to its own weight is given as:
σ o
= σ ’ o
+ µ (2.5)
Figure 2.2 illustrates the distribution of pore-water pressure, effective stress, and total stress in a soil mass having a horizontal surface with water level above the soil surface.
H
1 z
X H Y
µ = (H
1
+H) γ w
; σ o
’ = H γ ’ ; σ o
= H
1
γ w
+ H γ sat
Figure 2.2 Illustration of in-situ stresses in soil
The effective stress along
XY is as follows:
Pore-water pressure:
Effective stress:
µ = (H
1
+ z) γ w
σ ’ o
= z ( γ sat
– γ w
) = z
Total stress: σ o
= H
1
γ w
+ z γ sat
γ ’
The above condition occurs if the soil is in a fully saturated condition. If the soil is not saturated, then the pore-pressure is the combination of pore-water and pore-air pressures.

30 Introduction to Geotechnical Engineering Part 1
Example Problem 2.1
A soil profile consists of a surface layer of loose sand 3.5 m thick overlying a layer of stiff clay as shown in Figure P2.1. Unit weight of sand is 16.5 kN/m saturated unit weight of stiff clay is 18.5 kN/m 3
3 while the
. Draw the distribution of total, effective and pore water pressure in the soil and calculate the stresses at depth of
– 3.5 and – 5.5 m from surface. Take γ w
= 9.8 kN/m 3 .
Solution:
Total pwp Effective
Loose sand
γ b
= 16.5 kN/m 3
-3.5m
GWT
Stiff clay
γ sat
= 18.5 kN/m 3
-5.5m
Figure P2.1
At depth of – 3.5 m (GWT)
Total stress:
Pore-water pressure:
Effective stress:
σ o
= 16.5 × 3.5 = 57.75 kN/m 3
µ = 0
σ ’ o
= 57.75 – 0 = 57.75 kN/m 3
At depth of –5.5 m
Total stress:
Pore-water pressure:
Effective stress:
σ o
= (16.5 × 3.5) + (18.5 × 2) = 94.75 kN/m 3
µ = 2 × 9.8 = 19.6 kN/m 3
σ ’ o
= 94.75 – 19.6 = 75.15 kN/m 3

Stresses in Soil 31
Example Problem 2.2
Calculate the effective stress at 5 m below a river bed which consist of sand with saturated unit weight of 20 kN/m 3 . The depth of the water in the river was 2 m.
Take γ w
= 9.8 kN/m 3 .
Solution:
Figure P2.2
At point A
-2 m
5m
River
A
Sand
γ sat
= 20 kN/m 3
Total stress:
Pore water pressure
Effective stress
σ o
= 2 × γ w
+ 5 x γ sat
= 119.6 kN/m 3
µ = γ w
(2 + 5) = 68.6 kN/m 3
σ ’ o
= σ o
– µ
= 119.6 – 68.6
= 51 kN/m 3
The effective stress at point A is 51kN/m 3
Besides the stress due to its own weight, a soil element will carry an additional stress when it is loaded. In saturated soil, this additional stress is initially carried by water and as the water is forced to dissipate from the voids, the stress will be gradually transferred to soil skeleton or solid particles. This load will eventually induce a change in effective stress in soil.
If the applied load covers large areas such as soil fill on an extended construction site or a shopping mall, then the additional stress at any depth would be equal to the applied load itself. However, if the load is transferred by structure through a

32 Introduction to Geotechnical Engineering Part 1 footing of limited size, the applied stress would dissipate rapidly with depth and distance. Figure 2.3 shows the distribution of externally applied load in soil.
P
1
P
2
Q d
1
∆σ d
2
Figure 2.3 Stress distribution below a foundation
∆σ
Simple Method
Many methods have been used to calculate stress in a soil element at depth due to applied load. One of the simplest methods is the so called 2:1 distribution of load.
The load distribution under rectangular or strip foundation is shown in Figure 2.4.
P
σ z
B
B + z
1
2 z
B
B + z
L
L + z z
Figure 2.4 2.:1 distribution of load under strip and rectangular foundation

Stresses in Soil 33
Based on this distribution, the stress at depth z under a rectangular footing with dimensions B × L is
∆ σ z
=
(
B + z
Q
)(
L + z
) (2.6) and for continuous footing of width B
∆ σ z
=
(
B
Q
+ z
)
1
(2.7)
Other methods are generally based on elasticity assumptions i.e. assuming linear relationship between stress and strain.
Example Problem 2.3
For a rectangular footing 3 × 4 m in size carrying a column load 1500 KN, determine the vertical stress at depth of 2 m using a 2:1 load distribution.
Solution
For Q = 1500 kN, B = 3, L = 4, and z = 2 m
σ z
=
(
B + z
Q
)(
L + z
)
=
(3 +
1500
2) × (4 + 2)
= 50 kN/m 2
Note that the stress obtained using the 2:1 distribution is the average stress
Boussinesq & Westergaard Elastic Solutions
Point Load
In 1912 Westergaard developed a theory of stress distribution in soil by assuming that the soil consists of thin layers. For stress at depth z and distance r from a point load on soil surface (Figure 2.5), Westergaard gives a formulation:

34 Introduction to Geotechnical Engineering Part 1
σ z
=
2 π z 2
⎧
⎪⎩
(
Q
2 −
(
(
(
1
2
−
−
2 ν
2 ν
)
)
1
2 ν
−
)
2 ν
+
)
( )
2
⎫
⎪⎭
3
2
(2.8) where ν is Poisson’s ratio while other parameters are defined in Figure 2.5. If
Poisson’s ratio is equal to 0, or the soil is confined in lateral direction, then the above equation can be written as:
Q/ π
σ z
=
(2.9) z 2
⎡
⎢
⎛ +
⎝
1 2 ( r z
) 2
⎤
⎥
3
2
Boussinesq (1885) theory is based on assumption that the soil is homogenous, isotropic, linearly elastic semi-infinite medium for a point load act perpendicular to soil surface. Boussinesq formula for stress at a point under concentrated load is:
σ z
= z 2
⎝
1
3
+
Q r z
2 π
2
5 / 2
(2.10) with all parameters defined in Figure 2.5
Q z
∆σ z
∆σ x r
∆σ
θ
Figure 2.5 Stress distribution in soil due to point load.

Stresses in Soil 35
Simplified solution for Westergaard and Boussinesq equation is given in the form of chart by Taylor (Figure 2.6). Using this Figure, the stress at depth z and horizontal distance r is
∆ σ z
= z
Q
2
N (2.11)
0.5
0.4
0.3
0.2
0.1
0
0.0
0.5
1.0
1.5
2.0
Boussinesq
Westergaard
2.5
3.0
Figure 2.6 Stress distribution under a point load
The figure shows that Westergaard solution gives lower stress as compared to
Boussinesq solution. Other comparisons of stress in soil calculated using
Boussinesq and Westergaard equations were made by Duncan and Buchignani (see
Holtz and Kovacs, 1981) for continuous footing as well as rectangular footing also shows similar trends. Based on these studies, we can conclude that Boussinesq formula gives a more conservative value, and therefore more widely used.
Example Problem 2.4
Calculate the stress in a soil element at depth of 4 m due to a column load of 300 kN at point A and B by using (a) Boussinesq formula, (b) Figure 2.6 to find the stress according to Boussinesq and Westergaard formulas. Point A is directly below the load ( r = 0) and point B is at a distance of 1.5 m from point A ( Figure
P2.4).

36 Introduction to Geotechnical Engineering Part 1
Solution a.
Using Boussinesq formula
∆σ z
=
3 Q
2 π z 2
⎝
⎜ 1 ( r z
) 2
⎠
⎟
⎞ 5 / 2
At point A; r z
= 0
∆σ
A
=
2 × π ×
3 × 300
( )
2 × ( 1 + 0 2 )
5
2
= 8 .
95
kPa
Q = 300 kN
A
1.5 m
B
4 m
At point B; r z
= 1.5/4 = 0.375
∆σ
B
=
2 × π
3 × 300
×
( )
2 × ( 1 + 0 .
375 2 )
5
2
= 6 .
45 kPa
Figure P2.4 b.
Using Figure 2.6
Q
∆ σ z
=
2
N z
For r
= 0 z
Boussinesq N
B
= 0.48 ∆ σ
A
∆ σ
A
= 300 × 0.48/4 2
= 300 × 0.32/4 2
= 9 kPa
= 6 kPa
For r
= 0.375 z
Boussinesq N
B
= 0.31 ∆ σ
B
∆ σ
B
= 300 × 0.31/4 2
= 300 × 0.19/4 2
= 5.81 kPa
= 3.56 kPa
Line load
Stresses due to line load can be found by integrating Boussinesq equation (2.9) for point load. The stress under a line load is given by

Stresses in Soil 37
∆σ z
=
π
2 Q
(z 2 z 3
+ r 2 ) 4
(2.12) where Q is the line load in kN/m ’
Example Problem 2.5
A long concrete wall fence induced a line load of Q = 18 kN/m’ on a soil (Figure
P2.5). Calculate the stress induced at distance 1 m from the fence at a depth of 2m below the base.
Solution
Use equation 2.10
Q = 18 kN/m’ ∆σ z
=
π z
Q z r
3
(
2
2 + 2
4
)
=
2 × 18 × (2) 3
π (2 2 + 1 2 ) 4
Figure P2.5 r
A z
= 0.1467 kN/m 2
Load distributed on circular area
The integration of Boussinesq equation for stress distribution under circular footing can be calculated as:
∆σ z
= ∆ q o
I z
(2.13) where I z
is influence factor taken from a chart in Figure 2.7 which is called the bulb method.

38 Introduction to Geotechnical Engineering Part 1 a z r
Figure 2.7 Influence factor for stress distribution under circular load
Example Problem 2.6
A circular footing with diameter 3m is carrying a column load of 300 kN. Use figure 2.7 to find the stress in soil due to the uniformly distributed load at dept of 4 m below the foundation base for point A below the center, and point B below the

Stresses in Soil 39 sides of the foundation. Compare your results with Example Problem 2.4 in which the load is assumed as a point load.
Solution
Contact pressure ∆ q = column contact load area
=
300
π
4
3 2
= 42.44
kPa
Radius of the footing = 1.5 m
For point A, below the center of the footing r = 0 z = 4 m; a = 1.5 m z a
=
4
1 .
5
= 2.67, r a
= 0, then I z
= 0.19
∆ σ
A
= ∆ q I z
= 42.44 × 0.19 = 8.06 kPa
(assuming point load and using Boussinesq equation, ∆ σ
A
= 9 kPa)
For point B, below the side of the footing
I z
= 0.14
∆ σ
B
= ∆ q I z
= 42.44 0.14 = 5.94 kPa
(assuming point load and using Boussinesq equation, ∆ σ
B
= 5.81 kPa)
The stress obtained by assuming the uniform load is more distributed than that obtained by using Boussinesq formula assuming that the load is applied at a point.
Example Problem 2.7
An oil tank is carrying a uniform load of 117 kPa. Determine the vertical stresses at depth 2 m below (a) the center (b) the sides of the tank (c) point C at distance 4 m from the center. The diameter of the tank is 3.9 m.

40 Introduction to Geotechnical Engineering Part 1
Solution
Radius of the tank a =
3 .
9
2
= 1.95 a.
Below the center of the tank,
∆ σ z
= ∆ q I z
= z a
= 117 × 0.63 = 74 kPa
2
1 .
95
= 1.02, r a
= 0, then I z
= 0.63 b. Below the sides of the tank,
∆ σ z
= ∆ q I z
= z a
= 117 × 0.33 = 39 kPa
1 .
2
95
= 1.02, r a
= 1.0, then I z
= 0.33 c. Below point C, z
=
2
= 1.02, r a 1 .
95 a
∆ σ z
= ∆ q I z
= 117 × 0.045 = 5.2 kPa
Uniform load on rectangular footing
= 4/0.95= 2.04, then I z
= 0.045
Similarly, the stress distribution due to load applied on rectangular footing can be found using Equation
∆σ z
= ∆ q o
I z
(2.14) for which ∆σ z and ∆ q o
as defined previously and developed by Fadum (1948).
I z is obtained from Figure 2.8
Example Problem 2.8
For a rectangular foundation of size 3 × 4 m carrying a uniform load of 200 kPa, determine the vertical stress at depth of 2 m below (a) the edge of the foundation, and (b) center of foundation using Figure 2.8.
C
3 m
Figure P 2.8
E
4 m

Stresses in Soil z
σ z mz q nz
41
Figure 2.8 Influence factor for stress distribution under rectangular footing

42 Introduction to Geotechnical Engineering Part 1
Solution
For point E: B = 2 m ; m =
B
= 2/2 = 1 z
L
L = 3 m ; n = = 3/2 = 1.5 z
From figure 2.8, I z
= 0.195
∆σ
E
= 2 ∆ q I z
= 2 × 200 × 0.195 = 78 kPa
For point C: B = 2 m ; m =
B z
= 2/2 = 1
L = 1.5 m ; n =
L
= 1.5/2 = 0.75 z
From figure 2.8, I z
= 0.165
∆σ c
= 4 ∆ q I z
= 4 × 200 × 0.165 = 132 kPa
Example Problem 2.9
A plan of a building shows areas ABCD and CEFG carrying uniform pressure of
140 kN/m 2 and 70 kN/m 2 respectively (Figure P2.9). Determine the vertical stress
10 m below point X and E
Solution
Arrange the points in question so that they are at the corner of each rectangular with a certain load (see Figure P2.9)
For Point X
4 x Area XKAM with
1 x Area XFEI with
∆ q = + 140 kPa
∆ q = + 70 kPa
1 x Area XGCI with ∆ q = –70 kPa
For Point E
1x Area EBAN with ∆ q = + 140 kPa
1 x Area ECDN with ∆ q = –140 kPa
1 x Area ECGF with ∆ q = + 70 kPa

Stresses in Soil 43
60
B
M
A
140 kPa
X
K
40
C
G
D
70 kPa
E
F
N
30
20
Prepare a Table for each point to get the influence factor I and the vertical stress
Point X
Area B(m) L n = L/z
∆ q
(kN/m
2
)
I z
XKAM
XFEI
XGCJ
10
30
10
30
30
30
1.0
3.0
1.0
3.0
3.0
3.0
+140
+70
-70
0.203
0.244
0.203
∆σ
X
= 4 × 140 × 0.203 + 70 × [0.244 –0.203] = 111.61 kN/m 2
Point E
Area B(m) L (m) m =B/z n = L/z
∆ q
(kN/m
2
)
I z
EBAN
ECDN
40
20
60
60
4.0
20
6.0
6.0
ECGF 20 30 20 3.0
∆σ
E
= 140 [0.248 – 0.240] + 70 × [0.240] = 17.92 kN/m 2
+140
-140
+70
0.248
0.240
0.240
Example Problem 2.10
A strip footing of width 2.5m induced a 100kN/m ’ column load on a soil.
Calculate the maximum vertical stress at 2, 4, 6, 8 and 10 m below the base of the footing using Figure 2.8.

44 Introduction to Geotechnical Engineering Part 1
Solution
Contact pressure at the base of footing is
For strip footing, we use n = ∞ ; and m =
∆ q
B z
=
=
100
2 .
5
2 .
5 z
= 40 kN/m 2
Maximum vertical pressure occurs along the centerline of the footing, thus
∆σ z
= 2 ∆ q o
I z
(kN/m 2 ) z (m) n I z
∆σ z
= 2 ∆ q o
I z
(kN/m
2
)
18
13
9.76
7.76
6.40 10 0.250 0.08
Embankment Loading
The influence chart to be used for calculating the stress distribution under trapezoidal loading representing a long embankment was developed by Osterberg
(Figure 2.9). The size of the embankment is represented only by the half of the top and bottom width of the embankment ( b and a + b respectively). The chart can only be used to calculate the stress at the centerline of the embankment and since the chart only represents half of the embankment, the stress is:
∆σ z
= 2 ∆ q I z
(2.15) where ∆ q is the load due to the weight of embankment. Assuming that the embankment is very large, then ∆ q = h γ f
, with h is the height of the embankment and γ f
is the unit weight of the fill.

Stresses in Soil 45 a b q o
σ z
= 2I
σ z z
Figure 2.9 Influence factor for stress distribution under embankment loading
Example Problem 2.11
An embankment (Figure P2.11) is to be constructed on foundation soil. Calculate the additional stress 10 m below the center of the embankment using Osterberg chart if the height of the embankment is 3.5m, and the unit weight of the compacted fill is 20kN/m 3 .

46 Introduction to Geotechnical Engineering Part 1 a = 5 m b = 7 m
h = 3.5 m
Figure P 2.11
Solution
For the embankment ∆ q = h γ f
= 20 ×3.5 = 70 kN/m 2
At z = 10 m, a/z = 0.5 b/z = 0.7 From Osterberg chart, we get I z
= 0.40
The stress below the center of the embankment is
∆σ z
= 2 ∆ q I z
= 2 × 0.40
× 70 = 56 kN/m 2
Load on Irregular Shaped Area
The stress distribution under irregular shaped footing can also be found using equation 2.11. The chart for influence factor was developed by Newmark (1942).
In this case, the influence factor I z
is the product of influence value ( I ) given in the graph and the number of block covered by the scaled diagram of the foundation base (N).
∆σ z
= ∆ q I z
= ∆ q I N (2.16)
Figure 2.10 shows Newmark chart for an influence factor of 0.001. The scale QQ is the depth of the point in question. The foundation shape can be drawn in a transparent paper. The point in question is placed over the center of the chart
(point O). Then count the number of box covered by your foundation base ( N ).

Stresses in Soil 47
Q Influence value 0.001 Q
Scale
Figure 2.10 Newmark influence chart for stress on irregular shaped foundation
Example Problem 2.12
A rectangular foundation 6 × 3 m carries a uniform pressure of 250 kN/m 2 is placed near the surface of a soil mass. Determine the vertical stress at a depth of
3 m (a) below the corner of the foundation (point C), (b) below one edge of the

48 Introduction to Geotechnical Engineering Part 1 foundation (point E), (c) below the center of foundation (point O), and (d) below a point on the center line 1.5 m outside the long edge of the foundation (point D).
Use Newmark chart.
6 m
C
E 3 m
O
1.5 m
D
Figure P 2.12
Solution
Do the following steps:
Scale the line QQ on Figure 2.10 to be equal to 3 m (depth in question)
Draw the base of the foundation according to this scale
Put the points in questions on the center of Newmark chart
Count the number of boxes covered by the graph (foundation base) a.
For point C N = 196 b.
For point E N = 268 c.
For point O N = 472 d.
For point D N = 74
∆σ
∆σ z
= 250 × 0.001 × 196 = 49 kPa z
= 250 × 0.001 × 268 = 67 kPa
∆σ
∆σ z
= 250 × 0.001 × 472 = 118 kPa z
= 250 × 0.001 × 74 = 18.5 kPa
Example Problem 2.13
An L shaped foundation shown in Figure P 2.13 is carrying a uniform load of 70 kPa. Determine the vertical stress 15 m below point A by using Newmark chart.

Stresses in Soil 49
10 m
10 m
Figure P 2.13
Solution
A
30 m
15 m
Do the following steps:
Draw the base of the foundation by scaling z = 15 m = QQ
Put point A on top of point O in Newmark chart
The number of boxes covered by the foundation base on Newmark chart is
N = 458
∆σ z
= 70 × 0.001
× 458 = 32 kPa
2.1
A layer of saturated clay 4 m thick is overlain by sand 5 m thick. Water table being 3 m below ground surface. The saturated unit weight of the sand is 19 kN/m 3 and of the clay is 20 kN/m 3 , the unit weight of sand above ground water table is 17 kN/m 3 . Plot the total and effective stress against depths.
2.2
A 5m depth of sand overlies 6 m layer of clay, water table being at the surface. The permeability of clay is very low. The saturated unit weight of sand is 19 kN/m 3 and that of clay is 20 kN/m with unit weight of 20 kN/m 3
3 . A 4m depth of fill material
is placed on the surface over an extensive area. Determine the effective stress at the center of clay layer. (a)

50 Introduction to Geotechnical Engineering Part 1 immediately after the fill has been placed, (b) many years after the fill is placed
2.3
A soil profile consists of 3 m depth of gravel with saturated unit weight of
22 kN/m 3 lies above a 10 m depth of clay with unit weight of 20 kN/m 3 .
Initially the water table exists at the soil surface. Gravel layer has undergone a drying process due to application of a surcharge load of 30 kPa. Draw the effective stress diagram for (a) initial condition, and (b) for final condition if the dry unit weight of the gravel is 17 kN/m 3 .
2.4
A layer of sand with bulk unit weight of 16 kN/m 3 , saturated unit weight of
19 kN/m 3 and thickness of 9 m overlays a clay layer with low permeability.
The unit weight of the clay is 20 kN/m 3 . Ground water level is initially at
6m below the ground surface. In all of a sudden, the groundwater table raised to a depth of 3 m below ground surface, and become constant at this position. Determine the effective stress at depth of 8 and 12 m below the ground surface (a) initially, (b) right after the raise of groundwater table, and (c ) years after the raise of ground water table.
2.5
A concentrated load of 1000 kN is applied to the ground surface. What is the vertical stress increment due to the load at a point 5 m below the ground surface at a horizontal distance of 3 m from the line of the concentrated load.
2.6
Three point loads, 5000 kN, 7500 kN, and 6000 kN act in a line 5 m apart near the surface of a soil mass. Calculate the vertical stress at a depth 3 m vertically below the center (7500 kN) load.
2.7
A line load of 150 kN/m ’ acts 2 m behind the back surface of an earth retaining structure of 4 m high. Calculate the vertical stress increment at the base of the wall due to the line load.
2.8
A circular area carrying a uniformly distributed load of 100 kPa is applied to the ground surface. The radius of the circular area is 3 m. Calculate the vertical stress increment due to this load at a point 6 m below (a) the center of the circular area, (b) the edge of the circular area, and (c) at a horizontal distance of 5 m from the center of the circular area.
2.9
A 2 × 2 m footing is carrying a column load of 1500 kN. Calculate the vertical stress at depth of 5 m below the center of the foundation base assuming (a) the load is distributed uniformly over the footing base (use
Figure 2.8), (b) the load as a point load. Compare the answer.

Stresses in Soil 51
2.10
A shallow foundation 25 × 18 m carries a uniform load of 175 kPa.
Determine the vertical stress increment at a point 12 m below the mid-point of one of the longer sides using Newmark chart.
2.11
A 5 × 10 m area is loaded by a uniform load of 100kPa. (a) Calculate the stress at depth 5 m below point A shown in Figure Q2.11, (b) Calculate stress at point A if the left part of the foundation carry an additional load
100 kPa. Use Fadum method (Figure 2.8) and Newmark method (Figure
2.10). Compare the results.
A
5 m
5 m
Figure Q2.11
5 m 10 m
2.12
A wall footing of 1 m wide applies a contact pressure of 295 kN/m length of wall to a soil with unit weight of 18 kN/m 3 . (a) Determine the maximum vertical stress increment due to the wall footing at depth of 3 m from the foundation base, (b) calculate the total vertical stress at that point.
2.13
Calculate the vertical stress below the centerline of the embankment shown in Figure Q2.13 at depth of 3 and 6 m below the foundation base. The unit weight of the fill material is 20 kN/m 3 , and the height of the fill is 3 m.
⎣
3m 2:1
3m
Figure Q2.13
3 m
2.14
A foundation of size 6 × 10 m carrying a uniformly distributed load 600 kPa is shown in Figure Q2.14. (a) Determine the stress at depth of 5 m below points A, B, C, and D.

52 Introduction to Geotechnical Engineering Part 1
5m 5m
3 m
3 m
3 m
D
Figure Q2.14
2.15
A long strip footing of 3.5 m wide is to transmit a contact pressure of 225 kPa. The footing rests on a 5 m deep sand layer underlain by clay layer of 4 m thick. Determine the maximum embedment for the footing so that the increase in vertical stress at the center of the clay layer does not exceed 75 kPa. Assume the groundwater table is well below the footing and the unit weight of sand is 19 kN/m 3 .

Analysis and determination of lateral or horizontal pressure in soil are necessary for the design of walls and other earth retaining structures such as diaphragm wall, abutment etc. The lateral earth pressure at a point in soil is the portion of the vertical pressure at that point, and the ratio of the horizontal effective stress to the vertical effective stress at any point is defined as the coefficient of lateral earth pressure K.
The value of the coefficient of lateral earth pressure and the distribution of lateral earth pressure on the wall depends upon several factors such as: the physical characteristics of soil, time dependent shear strength characteristics of the soil, soil-structure interaction, and deformation characteristics of the material. It is also influenced by the type and the geometry of the structure which govern the mechanism of wall movement. Furthermore, any external loads applied on the wall will also influence the lateral earth pressure on the wall.
In relation to the soil-structure interaction and deformation characteristics, the lateral earth pressure is categorized into three conditions as shown in Figure 3.1.
σ h
= K o
σ v
σ v z
σ h
= K a
σ v
σ h
= K p
σ v
(a)”at-rest” (b) active
Figure 3.1 Three conditions of lateral earth pressure
53
(c) passive

54 Introduction to Geotechnical Engineering Part 1
Earth pressure “at rest” refers to condition in which the soil is prevented from lateral movement by the surrounding soil or by an unyielding wall. In this case, the coefficient of lateral earth pressure “at rest” is identified as K o
.
In actual condition where a soil element exists behind a retaining wall, it will reach either an active or passive conditions due to wall movements. An active condition occurs when the wall moves away from soil resulting in the lowering of the earth surface. If the wall movement is enough to develop shear failure of the soil, then a soil wedge will be formed and move forward and downward. The earth pressure developed at this state of failure is known as the active earth pressure σ a
= σ v
’ K a
.
On the other hand, if wall moves towards the soil, then the lateral earth pressure will increase until it overcomes the shear strength of the soil. A soil wedge will be formed and moves backward into the soil. The earth pressure developed at this state of failure is known as the passive earth pressure σ p
= σ v
’ K p
.
Table 3.1 shows the wall movement required to reach the active and passive condition
Table 3.1 Wall movement required to reach active and passive conditions
σ ’ h
Passive
“At-rest”
Active
Soil Type
Dense Cohesionless
Loose Cohesionless
Stiff Cohesive
Soft Cohesive
∆ La
H
∆ Lp
H
Wall movement
Active condition
∆ La
H
0.001
0.004
0.010
0.020
Passive condition
∆ Lp
H
0.020
0.060
0.020
0.040

Lateral Earth Pressure 55
To analyze the lateral earth pressure at rest, we need to consider an element of soil at depth z as shown in Figure 3.2a. The soil element is confined by the surrounding soil, so it cannot deform laterally due to vertical effective pressure
σ v
’. In this case, a lateral pressure was developed and it is defined as
σ h
’ = σ v
’ K o
(3.1) or
K o
=
h
'
v
'
Based on experiments, the typical value of K o is about 0.40 – 0.50 for sand, 0.35 –
0.70 for normally consolidated clay, and 0.50 – 3.00 for over-consolidated soil.
General formula for coefficient of lateral earth pressure “at rest” for sand is given by Jaky (1944):
K o
= 1 – sin φ ’ (3.2) where ϕ ’ is the effective angle of internal friction of the soil. Another formula is proposed by Brooker and Ireland (1965) for normally consolidated clay:
K o
= 0.95 – sin φ ’ (3.3)
Empirical correlation between the coefficient of lateral earth pressure K o plasticity of normally consolidated clay is proposed by Alpan (1967):
and the
K o
= 0. 19 + 0.233 log (PI) (3.4) where PI is the plasticity index of the soil. Other empirical correlations were proposed by Mayne and Kulhawy (1982):
K o
= 0.4 + 0.007 (PI) for 0 < PI < 40 (3.5a)
K o
= 0.64 + 0.001 (PI) for 40 < PI < 80 (3.5b)
Distribution of lateral earth pressure at the wall is similar to the distribution of vertical pressure with depth, so the total horizontal force applied at the wall shown in Figure 3.2a is:
P o
= ½ K o
γ b
H 2 (3.6)

56 Introduction to Geotechnical Engineering Part 1
If ground water table exists in the soil behind the wall as shown in Figure 3.2b, then the distribution of lateral earth pressure will be affected by the pore water pressure. We knew that water pressure is equal in all direction, thus the horizontal pressure due to water is equal to the vertical pressure.
H
H
H w
K o
γ b
H w
P o
= ½ K o
γ b
H 2
H/3
K o
γ b
H
(a) No ground water table
σ v
σ h
K o
γ b
H w z
GWT
K o
[ γ b
H w
+ γ ’(H-H w
)]
γ w
(H-H w
)
K o
[ γ b
H w
+ γ ’(H-H w
) ]+ γ w
(H-H w
)
(b) Ground water table exists at depth H w
Figure 3.2 Distribution of lateral earth pressure on wall
The vertical and horizontal pressures above water table at depth z are:
σ v
’ = γ b
z
σ h
’ = K o
σ v
’ = K o
γ b
z
The pore-water pressure developed at depth z below water table (refer to Figure
3.2b) is

Lateral Earth Pressure 57
µ = γ w
(z – H w
) (3.7)
The total vertical pressures at depth z below water table is then
σ v
= γ b
H w
+ γ sat
(z-H w
) and the effective vertical pressure at that depth is
σ v
’ = γ b
H w
+ γ sat
(z – H w
) – γ w
(z – H w
) or σ v
’ = γ b
H w
+ γ ’ (z – H w
) (3.8)
The effective lateral or horizontal pressure at depth z below water table is then equal to:
σ h
’ = K o
σ v
’ = K o
( γ b
H w
+ γ ’ (z – H w
)) (3.9) and the total horizontal stress is
σ h
= K o
σ v
= K o
( γ b
H w
+ γ ’ (z – H w
)) + γ w
(z – H w
) (3.10)
The stress distribution is shown in Figure 3.2b. The total force applied to the wall taking H
2
= H-H w
, then
P o
= ½ K o
γ b
H w
2 + K o
γ b
H w
H
2
+ ½ K o
γ ’ H
2
2
+ ½ γ w
H
2
2 (3.11)
Example Problem 3.1
A retaining structure is supporting a 6 m high excavation as shown in Figure P3.1.
The wall is very rigid so that the soil behind the wall is in “at rest” condition. The properties of the soil are shown in the figure. (a) Sketch lateral pressure diagram on the Figure and (b) determine the lateral earth pressure “at rest” P o
Determine the hydrostatic force P w surface.
and (c)
. Ground water table is at 3.5 m below ground
Solution
Use Jaky’s formula to determine K o

58 Introduction to Geotechnical Engineering Part 1
For both layers Æ K o
= 1 – sin φ ’ = 1 – sin 25 o = 0.577 a. Sketch of lateral earth pressure diagram is given in the Figure P3.1
3.5 m
Fill sand,
φ ’ = 25 o c’ = 0
γ b
= 18 kN/m
3
18x3.5x 0.577 = 36.35
2.5 m
Silt,
φ ’ = 25 o c’ = 0
γ sat
= 20 kN/m
3
36.35 + (20-9.8) 2.5 x 0.577 = 51.06
9.8 x 2.5 = 24.5
Figure P3.1 b.
Lateral earth pressure (see diagram)
P o
P o
= ½ × 36.35 × 3.5 + 36.35 × 2.5 + ½ × 14.71 × 2.5
= 63.61 + 90.88 + 18.39
P o
= 172.88 kN c.
Hydrostatic pressure (see diagram)
P
P w w
= ½ × 24.5 × 2.5
= 30.63 kN
T otal horizontal pressure on wall = P o
+ P w
= 172.88 + 30.63 = 203.51 kN
Horizontal Backfill
Rankine’s theory (1857) considers stress condition in a homogeneous isotropic soil with horizontal surface bounded by a smooth wall extending to a semi-infinite depth (Figure 3.3). The theory also assumed that the critical surface is plane (2-D case) forming a certain angle with horizontal.
Vertical and horizontal stresses at any point in the soil mass are the major and minor principal stresses σ
1
= σ z
and σ
3
= σ x
.
If no movement is allowed, then σ x

Lateral Earth Pressure 59
= K o
σ v
. This is shown as circle a in Figure 3.4. If the wall moves sufficiently away from the soil, plastic equilibrium of the soil is reached. At this state, all points in soil has reached the state of failure or an active condition, and σ x decreases to a minimum value ( σ
3.4. a
). This condition is shown in as circle b in Figure
Active case Passive case
σ z
σ x z
Figure 3.3 Stress condition at a soil element behind retaining wall
A
τ c
O
Failure envelope
D
σ a
’
σ o
’ b a
C active condition
σ v
’
“at rest” condition
Failure envelope
σ
φ
P a movement
Shear failure plane
45 + φ /2
Figure 3.4 Rankine’s active earth pressure

60 Introduction to Geotechnical Engineering Part 1
From figure 3.4, sin φ =
CD
AC
=
CD
AO + OC where CD = ½ ( σ v
– σ a
) ; AO = c cot φ ; and OC = ½ ( σ v
+ σ a sin φ = c cot
1
2
φ
( σ
+ v
1
2
−
(
σ
σ a v
)
− σ a
)
) , then
Thus
σ a
= σ v
'
⎡
⎢ (
(
1
1
−
+ sin sin
φ
φ )
)
⎤
⎥
− 2 c
⎡
⎢ (
1
( cos φ )
+ sin φ )
⎤
⎥
From geometry
1 − sin φ
1 + sin φ
= tan 2 ( 45 −
φ
2
) cos φ
1 + sin φ
= tan ( 45 −
φ
2
)
(3.12)
Then
σ ’ a
= σ ’ v tan 2 ( 45–
φ
2
) – 2 c tan ( 45–
φ
2
) (3.13)
For cohesionless soil (c = 0), the equation becomes
σ ’ a
= σ ’ v tan 2 ( 45–
φ
) (3.14)
2
Thus the Rankine’s coefficient of active lateral earth pressure
K a
=
σ
σ a
′ v
′
= tan 2 ( 45 −
φ
2
) (3.15) and the failure plane is assumed to form an angle ( 45+
φ
2
) with horizontal

Lateral Earth Pressure 61
Passive condition is described in Figure 3.5. Initial condition when the soil is in
“at rest” state, is shown as circle a . If the wall moves into the soil mass, the minor principal stress σ x
increases until the soil reaches failure (represented by circle b ).
Failure envelope
φ
Passive condition
σ p
’ σ c σ o
’ a
σ v
’
“at rest” condition
Failure envelope b movement
Pp
Shear failure plane
45 - φ /2
Figure 3.5 Rankine’s passive earth pressure
Using similar steps for active earth pressure, we get
σ ’ p
= σ ’ v
tan 2 ( 45+
φ
2
) + 2 c tan ( 45+
φ
2
) (3.16)
For cohesionless soil (c = 0),

62 Introduction to Geotechnical Engineering Part 1
σ ’ p
= σ ’ v
tan 2 ( 45+
φ
) (3.17)
2
The failure plane for passive condition is assumed to form an angle ( 45–
φ
2
)with horizontal, and the Rankine’s passive lateral earth pressure coefficient is
K p
=
σ
σ ′ v
′ p = tan 2 ( 45 +
φ
2
) (3.18)
Sloping Backfill
Rankine also developed the equation for coefficient of lateral earth pressure for sloping backfill. The Mohr circle for stress condition in soil behind a wall with sloping backfill is shown in Figure 3.6. In this case the derivation of the formula is made for cohesionless soil.
In figure 3.6 the plane is sloping at an angle of β , so that
σ
1
= σ ’ v
= OA
σ
3
= σ ’ a
= OB
Thus when c = 0 K a
=
σ ' a
σ ' v
=
OB
OA
=
OD −
OD +
AD
AD
OD = OC cos β
AD = OC 2 sin 2 φ − OC 2 sin 2 β = OC sin 2 φ − sin 2 β
K a
=
OC cos β
OC cos β
− OC
+ OC sin 2 φ − sin 2 β sin 2 φ − sin 2 β
(3.19)
From trigonometry sin 2 φ − sin 2 β = cos 2 β − cos 2 φ

Lateral Earth Pressure
Then
63
K a
= cos β − cos β + cos 2 β − cos 2 φ cos 2 β − cos 2 φ
(3.20)
Similarly for K p
, we get
K p
= cos β + cos β − cos 2 β − cos 2 φ cos 2 β − cos 2 φ
(3.21)
β
β
P
P
P av
= P a
sin ah
= P a
cos
β
β
τ
β
B’
F
Failure envelope
φ
A
D
C
O σ
G
Figure 3.6 Wall with sloping backfill

64 Introduction to Geotechnical Engineering Part 1
Thus, the active and passive pressure acting parallel to the slope are:
σ ' a
= K a
γ z cos β
σ ' p
= K p
γ z cos β
(3.22)
(3.23)
And the total active thrust and passive resistance on a vertical wall surface of height H are:
P a
= ½ K a
γ H 2 cos β
P p
= ½ K p
γ H 2 cos β
(3.24)
(3.25)
Example Problem 3.2
A retaining structure is supporting a 5.3 m high excavation. This wall has moved sufficiently to develop active condition. The properties of the soil behind retaining wall are c ’ = 0 ; φ ’ = 30 o ; γ b
= 20.4 kN/m 3 , γ sat
= 22.0 kN/m 3 . Compute the lateral pressure distribution acting on this wall if ground water table exist at 2.5 m below ground surface. Assume that the back of the wall is smooth so that the wall friction angle δ = 0.
Solution
Use Rankine’s theory K a
= tan 2 (45 – 30/2) = 0.333
2.5 m Sand ,
φ ’ = 30 o c
γ
γ
’ = 0 b
= 20.4 kN/m 3 sar
= 22 kN/m 3
20.4x2.5x 0.333 = 17
2.8 m
17 + (22-9.8) 2.8 x 0.333 = 28.38
9.8 x 2.8 = 27.45
Figure P3.2

Lateral Earth Pressure 65 a.
Lateral earth pressure (see diagram)
P a
P a
P a
= ½ × 17 × 2.5 + 17 × 2.8 + ½ ×11.38×2.8
= 21.25 + 47.6 + 15.93
= 84.78 kN b.
Hydrostatic pressure (see diagram)
P
P w w
= ½ × 27.45 × 2.8
= 38.43 kN
Total horizontal pressure on wall = P a
Example Problem 3.3
+ P w
= 84.78 + 38.43 = 123.2 kN
A 6 m tall cantilever wall retain a soil with the following properties: c
30 o ; γ = 19.2 kN/m 3
’ = 0; φ ’ =
. The ground surface behind the wall is inclined at a slope of
3H : 1V. Assume that the wall moves sufficiently to develop active condition.
Determine the normal and shear forces acting on the back of this wall using
Rankine theory
Solution
3H : 1V slope Æ β = tan − 1 ( 1 / 3 ) = 18 o
H
1
P a v
3
P a
P a H
K a
= cos β − cos β + cos 2 cos 2
β − cos 2 φ
β − cos 2 φ
= 0.415
P a
=½ γ H 2 K a cos β
P a
= ½ (19.2) (6) 2 (0.415) cos 18
P a H
P a V
= P a
cos 18 o = 129.7 kN/m
= P a
sin 18 o = 42.1 kN/m o = 136 kN/m
Figure P3.3

66 Introduction to Geotechnical Engineering Part 1
Cohesive Soils
The application of Rankine’s theory outlined above is for cohesionless soil. The generalized lateral earth pressure distribution for cohesive soil based on Rankine’s theory is illustrated in Figure 3.7a and b. For active condition, the presence of cohesion reduces the active earth pressure and it is negative for the upper part of the wall, the depth at which the active earth pressure becomes zero is: z o
=
γ
2 c
K a
(3.22)
The active and passive forces are given by:
P a
=
1
2
H σ ' v
K a
− H 2
( c K a
)
=
1
2
γ H
P p
=
1
2
H σ ' v
K p
+ H
(
2 c K p
)
=
1
2
γ
2 K a
H 2
− H
(
2 c K a
)
K a
+ H
(3.23)
(
2 c K a
)
(3.24)
Z o
–
=
P a
σ ' v
K a
2c V K a a) Active earth pressure
σ ' v
K a
–
2c V K a
+ =
P p
σ ' v
K p
2c V K p
σ ' v
K p
+ 2c V K p b) Passive earth pressure
Figure 3.7 Active and passive earth pressure for cohesive soil

Lateral Earth Pressure 67
Example Problem 3.4
A retaining structure is supporting a 6 m high excavation as shown in Figure P3.4.
The wall moves sufficiently that active condition prevail behind the wall. Fill of dry sand of 1 m height was placed in front of the wall in order to reduce the movement. The dry unit weight of the sand fill was 18.5 kN/m 3 . (a) determine the active thrust force P a
working on the wall, and (b) determine the passive force induced by the sand fill in front of the wall, (c) if the passive thrust were 100% mobilized, is the 1 m sand fill sufficient to retain the movement of the wall (use horizontal force equilibrium).
Solution
Rankine’s K a
= tan 2 ( 45–
φ
2
K p
=
1
K a
= 2.46
) = 0.406
5 m
γ b
= 20 kN/m c’ = 10 kPa
3
φ ’
= 25 o
P a2
P a1
1 m
Fill sand,
φ ’ = 25 o
γ b
= 18.5 kN/m 3 P p
σ ’ p
= 18.5 ×1 ×2.46 = 45.51
-2 ×10 × √ 0.406 =-12.74
σ ’ a
=20 ×6 ×0.406=48.72
Figure P3.4 a.
Active force (see diagram)
P a
= ½ × 48.72 × 6 – 12.74 × 6 = 146.16 – 76.44 = 69.72kN b.
Passive force (see diagram)
P a
P p
= ½ × 45.51 × 1 = 27.76 kN
> P p
Æ The sand fill is not sufficient to resist wall movement

68 Introduction to Geotechnical Engineering Part 1
Example Problem 3.5
Determine the height of unsupported excavation in clay soil with c = 40 kN/m 2 and
γ = 16.3 kN/m 3 .
Solution
For φ = 0 ; K a
σ h
’
=
= 1 (Rankine)
σ v
’
K a
– 2 c K a
= γ z K a
− 2 c K a
Critical condition when σ h
’
= 0
γ z .
K a
= 2 c K a z =
γ
2 c
K a
=
2 × 40
16.3
1
= 4.90m
The maximum height for unsupported excavation is 4.90 m
COULOMB’S THEORY
One of the important assumptions made in Rankines’s theory is that the wall is smooth so that there is no friction between the soil and the wall. Coulomb’s theory developed in 1776 takes into account the friction. It also assumes that the failure occurs in the form of wedge passing through the toe of the wall. Resultant of the active earth pressure acts at the center of gravity of the wedge at an angle of wall friction ( δ ) with horizontal. The basics of this theory apply for cohesionless soil.
Figure 3.8 shows a wall retaining cohesionless soil, where soil surface forms an angle β with horizontal. Line AB is the predicted failure plane in which the following forces are acting: (a) weight of failed soil mass, (b) resultant normal and shear forces F that form an angle φ with the normal line to the failure plane, and
(c) the active earth pressure form an angle wall.
δ with the normal line to the back of the

Lateral Earth Pressure 69
C
A β
α − β
H W D
W
δ
P a
φ
F
θ
B
α
Figure 3.8 Coulomb’s active earth pressure: Failure wedge and force polygon
C
P a
F
A
β
α − β
H P p
δ
F
W D
W
φ
F
Pp
θ
B
α
Figure 3.9 Coulomb’s passive earth pressure: Failure wedge and force polygon

70 Introduction to Geotechnical Engineering Part 1
From figure 3.8, we can see that sin ( 180 − ( θ
W
− δ ) −
(
α − φ )
)
=
P a sin( α − φ )
P a
= sin
W sin
( 180 − ( θ
( α
− δ )
− φ
−
)
(
α − φ )
) the weight of failed soil mass is
W =
γ H 2
2 sin sin 2
(
θ
θ
+ α sin
) (
(
α −
θ
β
)
+ β
)
Thus,
P a
=
γ H
2 sin 2 θ
2 sin sin
(
α
(
θ
−
−
β
α
)
) ( sin
[
θ
180
+
−
β
(
θ
) (
− δ
α −
) (
φ
α
)
− φ
) ] (3.25)
We can see that the only variable in Equation 3.25 is obtained when is found when α
P a
reaches maximum. In this case,
= 45+
φ
2
Substituting α = 45+
P a where
= ½ γ H 2
φ
2
∂ P a
∂ α
α . Critical value for into Equation 3.25, we get:
K a
α is
should be equal to 0. This
K a
= sin 2 θ sin
( θ − δ ) sin
⎢
⎣
⎡
1 +
2
( θ + sin sin
φ
(
( φ
θ
)
+
−
δ
δ
) (
) (
φ
θ
−
+
β
β )
) ⎤
⎥
⎦
(3.26)
Similar derivation can be made for coefficient of passive earth pressure based on
Figure 3.9 yielding in:
K p
= sin 2 θ sin
(
θ + δ
) sin
⎢
⎣
⎡
1 +
2
( θ − sin
φ
(
)
+ δ ) ( φ + β ) sin
(
φ
θ + δ sin
) (
θ + β
)
⎤
⎥
⎦
(3.27)
Note that for β = 0, θ = 90, and δ = 0, Equation 3.26 and 3.27 become:

Lateral Earth Pressure 71
K a
=
1
1
−
+ sin sin
φ
φ
= tan 2 ( 45 −
φ
2
)
K p
= cos
1 +
φ sin φ
= tan ( 45 −
φ
2
) which are equal to the Rankine’s active and passive earth pressure coefficients.
Example Problem 3.6
A retaining wall as shown in Figure P3.6 is designed to retain a 9 m excavation
Calculate the active thrust on the wall and its point of application using Coulomb’s theory if the wall friction angle is 25
= 30 o , c = 0, and unit weight γ o . The properties of the soil behind wall are:
= 17.6 kN/m 3 .
φ
δ
P a
9 m
Figure P3.6
Solution
For φ = 30 o , δ = 25 o , β = 0 o , and θ = 90 o
K a
= sin 2 90 sin
(
90 − 25
) sin
⎢
⎣
⎡
1
2
+
(
90 + sin sin
30
)
(
(
30
90
+
−
25
25
) (
) (
30
90
−
+
0
0
)
) ⎤
⎥
⎦
= 0.296
P a
= ½ γ H 2 K a
= ½ × 17.6 × (9) 2 × 0.296 = 211 kN/m ’

72 Introduction to Geotechnical Engineering Part 1
LATERAL PRESSURE EXERTED BY APPLIED LOAD
There are various types of load that may be working directly or indirectly either on the soil surface behind the retaining wall or on the wall itself. These loads can be idealized in four types of load: uniform load, strip load parallel to the wall, line load parallel to wall, and point load. The additional stress acting on a wall due to these loads can be evaluated using the formulas derived from elastic theory.
It should be noted that these load have influence on wall only if it acts above the failure wedge. Thus, we should define the failure wedge before we consider the effect of the applied load to the structure.
Uniform Surcharge Load
Additional pressure exerted against a retaining wall due to uniform surcharge load q o
applied on the soil surface (Figure 3.10a) can be computed as
Strip load parallel to wall
∆σ x
= q o
H K a
(3.28)
Lateral stress due to strip load parallel to the wall (Figure 3.10b) can be calculated using Scott’s formula:
∆ σ x
= q
π
(
α − sin α cos
(
α + 2 β
) )
(3.29a) with α and β as given in the figure. Equation 3.29a can be used if the wall is flexible. If the wall is rigid, then
∆
x
=
2
q (
− sin
cos
(
+ 2
) )
(3.29b)
Line load Parallel to Wall
Pressure distribution behind retaining wall due to a line load Q (kN/m) at a distance x = mH from the wall is shown in Figure 3.10c. The lateral earth pressure can be calculated as:

Lateral Earth Pressure 73
∆ σ x
=
π
2 Q
H
⎡
⎢
⎣
( m 2 m 2 n 2
+ n 2
2
)
⎤
⎥
⎥ for flexible wall ( 3.30a)
∆ σ x
=
π
4 Q
H
⎡
⎢
⎣ m 2 n 2
( m 2 + n 2
2
)
⎥
⎤
⎥
⎦ for rigid wall (3.30b) where y = nH is the depth of the point from the soil surface.
The total thrust force on the structure is given by:
P x
=
1
∫
0
σ x h dn =
2 Q
π ⎣
⎢
⎡
( m 2
1
+ 1
)
⎤
⎦
⎥ (3.31) q o x = mH
Q z=nH
H ∆σ x
= q o
H K a
H/2
H ∆σ x a) Uniform surcharge load q c) Line load x = mH
Q z=nH
z
∆σ z
X
α β
∆σ x b. Strip load
H
∆σ x d. Point load
Figure 3.10 Lateral stresses due to applied load

74 Introduction to Geotechnical Engineering Part 1
Point Load
Lateral stress acting on wall due to point load Q on the soil surface as shown in
Figure 3.10d. In this case, soil is assumed as elastic material with Poisson’s ratio
ν = 0.5.
∆ σ x
=
Q
2 x
⎛
⎜⎜
(x 2
3 x 2
+ z 2 z
) 5 / 2
⎞
⎟⎟
(3.32)
By substituting x = mH , dan z = nH in equation 3.30, then
∆ σ x
=
Q
2 xH 2
2 m n
( m 2 + n 2
5
)
/ 2
⎟
(3.33)
Example Problem 3.7
A sheet pile wall is needed for normalization of a river bank. The depth of soil to be retained is 3.5 m and sheet piles of 5.5m length will be used. The soil profile consists of a surface layer of loose sand 3.5 m thick overlying a layer of stiff clay.
Unit weight of sand is 16.5 kN/m 3
φ ’ = 23 o . The saturated unit weight of stiff clay is 18.5 kN/m strength parameters are c ’
, and the soil strength parameters are
= 10 kN/m 2 and φ ’ = 33 o
3 c ’ = 0 and
while the shear
. A uniform load q = 10 kN/m 2 is estimated to work on the surface of the soil as the wall is constructed. Draw the active pressures on the diaphragm wall.
Solution
The sheet pile wall has a smooth surface, thus the friction between the wall and the soil can be neglected and Rankine’s theory can be used to calculate the coefficient of lateral earth pressure. Assume w
= 9.8 kN/m 3 ayer 1 loose sand,
Layer 2 c ’ = 0, stiff clay c ’
φ ’ = 23 o
= 10 kPa,
γ
Æ
φ ’
K a1
= 33
= tan o Æ
2
K
( 45– a2
φ
2
= tan
) = 0.438
2 ( 45–
φ
2
)= 0.295
At z = 0 m σ a
= q o
K a1
= 10 × 0.438 = 4.38 kN/m 2
At z = 3.5 m σ a
= q o
K a1
+ γ
1
×3.5 × K a1
= 4.38 + 25.3 = 29.68 kN/m 2 (above)

Lateral Earth Pressure
σ a
= q o
K a2
+ γ
1
×3.5× K a2
- 2 c √ K a2
= 2.95 + 17.04 – 10.86 = 9.13 kN/m 2 (below)
At z = 5.5 m σ a
= q o
K a2
+ γ
1
×3.5
× K a2
+ γ '
2
×2 × K a2
- 2 c √ K
= 2.95 + 17.04 + 5.13 – 10.86 = 14.26 kN/m 2 a2
The pressure distribution diagram is shown in Figure P3.7. q o
= 10 kN/m 2
4.38
75
3.5 m
Loose sand, c’ = 0
φ ’ = 23 o
γ = 16.5 kN/m 3
GWT 9.13
29.68
2.0 m
Stiff clay, c’ = 10 kPa,
γ
φ ' = 33 sat
= 18.5 kN/m 3 o
14.26
Figure P3.7
Earth retaining walls are commonly used to support soils and structures to maintain a difference in elevation of the ground surface through excavation or by earth work above existing ground level. There is a large variety of retaining walls to choose from and a number of factors govern the selection.
Retaining walls are normally grouped into gravity and cantilever walls, and sheet pile. Other types of retaining structures are anchored wall, counter-fort wall, soil nailing, and reinforced soil wall.
Gravity walls are the most common when construction is taking place above ground level. Stability is provided through their weight alone or by a combination of the weight of the structure and the soil resting on the heel of the wall. The

76 Introduction to Geotechnical Engineering Part 1 choice of material for gravity walls are concrete, masonry or selected stone.
Gabions have been widely applied to river and tidal works.
Sheet pile walls rely on the resistance of the soil in front of the wall. The stability can be enhanced through the use of external support systems such as anchored tie rods, props, and bored piles.
Reinforced soil walls rely on the weight of reinforced soil and the anchorage provided by the reinforcement for their stability. Reinforced soil structures tend to be flexible and have a large tolerance for movement; therefore they are ideal for sites with poor ground near the surface. Reinforcing elements ranges from metallic strips to geotextiles. Construction times are generally shorter and material requirements are minimal when compared to conventional structures. This type of wall is suitable for site with limited access, but less likely to be selected when strips interfere with boundaries and obstructions.
The first things to consider when selecting the types of retaining wall include the proposed height, topography, the soil and groundwater conditions and the nature of project. Other factors are related to the construction including the availability of materials (e.g. backfill) and special equipment, construction space available, design life and maintenance requirements, environmental factors, appearance, and cost implications. The final decision will not normally be influenced by a single factor but be based more on a number of the factors given above.
Once the type of retaining wall is selected, the following step is to estimate the magnitude, the distribution, and the location of active earth pressure acting on the wall. The dimensions of the wall are determined based on the numerical calculations necessary to ensure both external and internal stability of the wall.
Conventional retaining walls are generally free standing in which the wall stand on or a little below the soil surface. The stability of these walls depends primarily on the weight of the wall material and any soil resting above its heel. Gravity and semi-gravity walls, cantilever walls and counter fort walls are among the form of free standing wall (Figure 3.12). Gravity and semi gravity wall is only suitable for wall less than 3m high, while cantilever or counter fort wall can stand an excavation or fill up to 8 m high.

Lateral Earth Pressure 77
Active earth pressure is used to design the free-standing wall. The next step is to assume the size of the wall. Normally the required wall height is known, other geometry should be assumed. As a preliminary estimate, the base of the wall B is assumed to be 2/3 of the height of the wall.
Stability analyses against overturning, sliding horizontally, and bearing capacity of foundation soil are performed for this geometry. If any of these conditions is not safe, then the assumed wall size must be modified.
Toe Heel
(a) gravity wall (b) cantilever type wall
Figure 3.12 Types of free standing retaining wall
To check the stability of the wall against overturning, we have to define point O at the toe of the wall. Then we need to calculate the overturning moment ( M
O resisting moment ( M
Ro
) and
) at point O. The factor of safety against overturning is
FS =
M
RO ≥ 2.5 (3.33)
M
O
Sliding on the base might occur due to the sum of horizontal forces working at the back of the wall. The resistance is provided by the weight of the wall and any soil resting on its heel and the base friction. The factor of safety against sliding on the base is:
FS =
∑
V
∑ tan
H
δ
≥ 2.0 (3.34) where δ is the angle of friction between the foundation base and the soil. The resulting pressure ( P ) below the wall base is calculated by taking into account the

78 Introduction to Geotechnical Engineering Part 1 moment working on the center of gravity of the wall base (defined as point C). It is calculated as
P =
V
B
±
M
C
I y
(3.35) where M
C
= V e in which e is the load eccentricity, I is the moment of inertia of the foundation base, and y is maximum possible eccentricity =
B
. In the case of
2 two-dimensional analysis, the foundation of the wall is a strip footing of width B and length 1 . Taking moment inertia of a rectangular equation 3.35 can be written as:
I =
1
12 l B 3 and y =
B
2
,
P =
V
B ⎝
⎜
6
B e
⎠
⎟
⎞
(3.36)
Maximum pressure ( P max
) should be less than the allowable bearing capacity of the soil, while minimum pressure ( P min that soil cannot take tension.
) should be greater than 0 because we know
Gravity or cantilever walls are not designed to withstand large additional pressure due to water existing on the back of the wall. Thus, it is important to make sure that water is prevented from accumulating in the fill soil behind the wall. One method of preventing water from accumulating behind a wall is to provide some types of drainage on the surface. It is also desirable to use a highly pervious soil such as sand, gravel, or crushed stone as backfill material so that any water that infiltrate into the soil will be drained quickly. The other method is to drain the backfill soil by installing weep holes through the wall or perforated drain pipe placed longitudinally along the back of the wall.
Example Problem 3.8
A gravity wall is designed to retain a cut along a highway project as shown in
Figure P3.8. The wall is made of unreinforced concrete with backfill soil is free draining medium sand with
20 o , and unit weight 19.2 kN/m 3 bearing capacity of 550 kN/m 2
φ ’ = 34
γ c
= 22 kN/m 3 . The
δ =
. The wall is resting on stiff clay deposit with
. Evaluate the stability of the wall against overturning, sliding and bearing capacity failure. o , wall friction angle
Solution

Lateral Earth Pressure 79
The back of the wall is not smooth and
In this case θ = 90 o , β = 0.
K a
= sin 2 90 sin
(
90 − 20
) sin 2
⎢
⎣
⎡
1 +
δ = 20 o , then use Coulomb method
(
90 + sin sin
34
(
(
34
90
)
+
−
20
20
) (
) (
34
90
−
+
0
0
)
) ⎤
⎥
⎦
= 0.3767
To check the stability of the wall against overturning, sliding on the base, and bearing capacity failure, we can prepare Table P 3.8 by referring to Figure P3.8.
1 m
1 m
W
1
W
3
W
2
P av
P ah
6 m
Free draining medium sand
φ
δ
γ
’= 34
= 20 o o b
= 19.2kN/m
3
P ah
2 m
Figure P3.8
3 m
Stiff clay
No Force
Moment Arms
(about O)
M
O
4/3
2.5
3 2×1×22 = 44
4 ½ ×19.2× (6) 2 ×0.3767×cos20 o = 122.30
1
2
5 ½ × 19.2 × 6 2 × 0.3767 × sin20 o = 44.5
Stability against overturning
FS =
M
M
RO o
=
147 + 330 + 44
244 .
6
+ 133 .
5
Stability against sliding on the base
3
= 2 .
68 > 2 .
50
+147
+330
+44
-244.60
+133.50

80 Introduction to Geotechnical Engineering Part 1
FS =
V tan
H
Bearing capacity failure e =
M
V
O −
B
2
δ
=
=
( 110 + 132 + 44 +
122 .
3
44 .
5 ) tan 20 o
= 0 .
98 < 2
147
(
+ 330
110 +
+ 44
132 +
+ 133 .
5
44 +
−
44 .
244
5 ).
.
6
−
3
2
= 1 .
24 − 1 .
50 = 0.26 m
B /6 = 3/6 = 1.5 m
Check P max
and P min
Æ e < B /6
P =
V
B ⎝
⎜
6
B e
⎠
⎟
⎞
=
330
3
.
5 ⎛
⎝
1 ±
6 × 0
3
.
26
= 110 .
17 ± 57 .
29
P max
= 167.5< Q all
and P min
= 53.1 > 0
The wall is safe against overturning and bearing capacity failure but unsafe against sliding. We may need to provide heel to increase the passive resistance against sliding.
Example Problem 3.9
A gravity wall as shown in Figure P3.9 is retaining a backfill with unit weight of
18 kN/m 3 and shear strength parameters c ’ = 0, and the soil and wall is represented by an angle δ = 25 o
φ ’ = 38 o . The friction between
. Evaluate the stability of the wall against overturning, sliding and bearing capacity failure if the unit weight of wall material is 23.5 kN/m 3

Lateral Earth Pressure 81
0.50 m 0.70 m
20 o
2
Pa
6.0 m
0.50 m 35 o
0.75 m
4
3
O
2.75 m
1
100 o
Figure P3.9
Solution
For φ ’ = 38 o , δ = 25 o , β = 20 o and θ = 100 o
K a
= sin 2 100 sin
(
100 − 25 sin
)
2
⎢
⎣
⎡
1 +
(
100 + sin sin
38
)
(
38
(
100
+
−
25
25
) (
) (
38
100
−
+
0
)
0
)
⎤
⎥
⎦
= 0.39
P a
= ½ × 0.39 × 18 × (6) 2 = 126 kN/m acts at 1 /
3
height and forming an angle of 25 + (100– 90) = 35
Calculation of force and pressure on wall o to horizontal.
No Force
W
1
½ × 1.05 × 6 × 23.5 = 74
Moment Arms
(about O)
2.05
M omen
+151.7
1.35
+133.25 W
2
0.7 × 6 × 23.5 = 98.7
W
3
½ × 0.5 × 5.25 × 23.5 = 30.8
W
4
1 × 0.75 × 23.5 = 17.6
P h
126 cos 35 o = 103.2
P v
126 sin 35 o = 72.3
Σ V = 293.4 ; Σ H = 103.2
0.833
0.50
+25.66
+8.81
2 -206.40
2.40
+173.52
Σ M
O
= 206.40 ; Σ M
RO
= 484.13

82 Introduction to Geotechnical Engineering Part 1
Stability against overturning
FS =
M
M
Ro o
=
484
206
Stability against sliding on the base
.
13
.
4
= 2 .
34
FS =
V tan
H
Bearing capacity failure e =
M
V
−
B
2
δ
=
=
293 .
4
484 .
tan
103 .
2
13 −
293
B/6 = 2.75/6 = 0.458 m
25 o
206
.
4
=
.
4
1 .
33
−
2
< 2.0
.
75
2
=
P =
V
B ⎝
⎜
6 e
B ⎠
⎟
⎞
=
293.4
2.75
⎝
⎜
⎛
1 ±
6.
× 0.43
2.75
⎠
⎟
⎞
0 .
43 m
Æ e < B/6
=
106.7 (1 ± 0.06)
P max
= 113.1 kN/m 2 < σ s
P min
= 100.3 kN/m 2 > 0
Example Problem 3.10
A cantilever type wall as shown in Figure P3.10 is required to retain an excavation. The shear strength parameters for the soil are kN/m 3 the base of wall is 30 o c ’ = 0, φ ’ = 40 o , γ = 17
. Water table is located far below the base of the wall. The friction angle at
. Evaluate the stability of the wall against overturning, sliding and bearing capacity failure. (Take the unit weight of concrete γ c
= 23.5 kN/m 3 and use Rankine theory to calculate lateral stress)

Lateral Earth Pressure 83 q = 40 kN/m 2
W
4
30 cm
W
3
540 cm (1)
(2)
W
1
W
2
40 cm
O
300 cm
175 cm
Figure P3.10
Solution:
For φ ' = 40 o ; δ = 0 Æ K a
= tan 2 (45
–
20
) = 0.22
2
Calculation of force and pressure on wall is tabulated in table P3.10
Table P 3.10
No
W
1
W
2
W
3
W
4
P
1
P
2
Force
5 × 0.3 × 23.5 = 35.3
Arms (about O) Moment
1.10
+38.83
3 × 0.4 × 23.5 = 28.2 1.50
+42.30
5 × 1.75 ×17 = 148.8 2.125
+316.20
1.75 × 40 = 70.0
0.22 × 40 × 5.40 = 47.5
½ × 0.22 × 17 ×5.40
2 = 54.6
2.125
2.70
1.80
+148.75
-128.20
-98.30
Stability against overturning
FS =
MRo
Mo
=
38 .
83 + 42 .
30 +
128 .
20
316 .
20
+ 98 .
30 .
+ 148 .
75
= 2 .
40

84 Introduction to Geotechnical Engineering Part 1
Stability against sliding on the base
FS =
V tan
H
Bearing capacity failure e =
M
V
−
B
2
=
38 .
83 + 42 .
30
δ
+
=
( 35 .
3 +
( 35 .
3
316 .
2 +
28 .
2
+
148 .
75
+
28 .
2 +
148 .
8
148 .
8
47 .
5
−
+
+
+
54 .
6
128 .
20
70 )
70 )
− tan 30 o
98 .
30
−
3
2
=
=
1 .
60
1 .
13 − 1 .
50 = 0 .
37 m
B /6 = 3/6 = 0.5 m Æ e < B/6
P =
V
B
1
6 e
B
=
282.3
3 ⎝
1
P max
= 163.7 kN/m 2 < σ s
6.0.37
3
=
94.1 (1 ± 0.74)
P min
= 24.5 kN/m 2 > 0
Cantilever sheet pile wall is recommended for walls of moderate height of up to 5 m for flexible steel sheet piles, or 12 m for stiffer reinforced concrete sections.
However, deflections should always be checked especially at the head of the wall.
The wall acts as a wide cantilever beam above the dredge line. The aim of the stability analysis of cantilever sheet pile wall is to find the depth of embedment of the sheet pile. The basic principles for estimating net lateral earth pressure distribution on a cantilever sheet pile wall are explained in Figure 3.13.
The mode of failure by the rotation is assumed as rotation about a point O near the lower end of the wall as shown in Figure 3.13. Active earth pressure acts behind the wall and consequently passive resistance acts in front of the wall above point O and behind the wall below point O. The idealized pressure distribution is shown in
Figure 3.13b. Factor of safety is applied to the consideration of passive pressure of the wall and to the depth of embedment.

Lateral Earth Pressure 85 h active active
O d passive passive
R
(a) (b) (c)
Figure 3.13 Cantilever sheet pile wall
In the design of sheet-pile, it is common practice to assume that water exists at the same level in front and behind the wall, thus hydrostatic pressures at any depth from both sides of the wall cancels each other. In this case, we consider only the lateral soil pressure. If there is a difference in the water level, then there are some methods to consider the presence of water pressure on the wall (see Craig, 2004).
Example Problem 3.11
A cantilever sheet pile wall is required to support 5 m depth of soil with strength parameters c ’ = 0, φ ’ = 35 o , and γ = 20.8 kN/m 3 . Determine the length of pile required for a safety factor of 1.5 applied to passive pressure. Use horizontal equilibrium only.
Solution
The problem is shown in Figure P3.11

86 Introduction to Geotechnical Engineering Part 1
H = 5 m
P a
P p
Figure P 3.11
Assume that the back of sheet pile is smooth, then use Rankine’s formula to calculate K a
and K p
K a
= tan 2
K p
= tan 2
( 45–
φ
( 45+
2
φ
2
) = 0.271
)= 3.690
Active force: P a
= ½ γ ( H + d ) 2 K a
= ½ × 20.8
×( H + d ) 2 ×0.271 =
P a
= 2.818 (5 + d ) 2
Passive force P p
= ½ γ z 2 K a
= ½ × 20.8
×z 2 ×3.69 =
P p
= 38.38 z 2
Horizontal equilibrium is reached when P a
= P p
Apply factor of safety 1.5 with respect to passive resistance
P a
= 1.5
P p
2.818 (5 + d ) 2 = 1.5 ×38.38 d 2
Get d = 3.85 m.
Take the length of embedment = 4 m

Lateral Earth Pressure 87
Example Problem 3.12
The sides of an excavation 3.00 m in sand are to be supported by a cantilever sheet pile wall. The water table is 1.5 m below the bottom of the excavation. The sand has a bulk unit weight of 17 kN/m 3 , and saturated unit weight of 20 kN/m effective friction angle of the sand is 36
3 . The o . Determine the depth of penetration of the piling below excavation to give a factor of safety of 2.0 with respect to passive resistance.
Solution:
For φ ’= 36 o K a
= tan 2 ( 45–
φ
) = 0.26, and K p
= tan 2 ( 45+
2
φ
) = 3.85; with FS = 2, K p
= 1.925
2
γ ’ = 20 – 9.8 = 10.2 kN/m 3
The pressure distribution is shown in Figure P3.11. Hydrostatic pressure on the two sides of wall balances.
Consider moments about X (per m), assuming d > 0.
3 m
1.5 m (1)
(4)
(6)
(5) d
(2) (3)
R
Figure P3.11
X
To simplify, develop a table for calculation of depth of embedment based on the free-body of the pressure diagram shown in Figure 3.11:

88 Introduction to Geotechnical Engineering Part 1
No Forces Arms (m) Moment at X (kN m)
1. ½ × 0.26 × 17 × 4.5
2 = 44.8
2 0.26 × 17 × 4.5 × d = 19.9 d
4 - ½× ½×3.85 ×17 × 1.5
2 = -36.8
1.5+d d/2 d+0.5
67.2 + 44.8d
9.95 d 2
3 ½ × 0.26 × 10.2× d 2 = 1.33 d 2 d 3
-36.8 d – 18.4
5 - ½ × 3.85× 17 × 1.5 × d = -49.1 d
6 -½ × ½ ×3.85 ×10.2 × d × d = -9.82 d 2 d/3 -3.27
Taking moment at X equal to 0, we get the equation
-2.83 - 14.6 d 2 + 8 d + 48.8 = 0
By trial and error, we get d = 1.79 m and applying FS = 1.2, depth of penetration = 1.2 x (1.79+1.50) = 3.95m
Substituting d = 1.79 m to the horizontal force equilibrium
44.8 + 19.9
d + 1.33 d 2 – 36.8 – 49.1 d – 9.82 d 2 = R
Get force resultant R = 71.5 kN
Over additional depth of 20% ( d = 3.95 – 1.5 = 2.45 m)
P p
– P a
= ( K p
γ 4.5) + ( K a
γ 1.5) + ( K p
-K a
) γ ’ (2.45)
= (3.85 × 17 × 4.5) + (0.26 × 17 ×1.5) + (3.85–0.26) 10.2 × 2.45
= 365.5 kN (> R )
When the height of soil to be retained is large or when the calculated depth of embedment for cantilever sheet pile wall is too large as compared to its height, then it is time to consider a tie back to reduce working moment on the wall (Figure
3.14). Tie rods are normally resisted by an anchorage system such as beams, plates or concrete blocks at some distance behind the retaining wall and beyond the failure wedge (Figure 3.15). Wall of this type are used extensively in waterfront construction and in support for deep excavation.
Instead of taking moment at point O below the dredge line, the analysis of anchored sheet pile wall is performed assuming that the wall rotates at the point of

Lateral Earth Pressure 89 application of the anchorage (point A in Figure 3.14). The force per unit length resisted by the tie rod ( T ) is the difference in the horizontal forces applied to the wall.
T T
A h
Bending moment
Active Active
Passive Passive
R
Figure 3.14 Anchored sheet pile wall (free earth support methods)
Tie rod
Interlocking sheet-pile
Anchorage
Figure 3.15 Sheet pile wall with Tie Rod
If T is the tie rod force per unit length of wall, s is the spacing, and FS is the factor of safety, then based on the active and passive pressure applied to the anchorage, an equation for tie rod force is:
T =
γ
2 d a
2
FS s l (
K p
− K a
)
(3.37) and the required depth of the anchorage as d a
2 =
γ L
2 FS
(
K p
− s T
K a
) (3.38)

90 Introduction to Geotechnical Engineering Part 1 if s = l , then d a
=
γ
2
(
K
FS p
−
T
K a
) (3.39)
Example Problem 3.12
A sheet-pile wall is used to retain a cut to a depth of 6 m. An anchor was placed at depth of 1.5 m below the ground surface (see figure P3.12). Use the free earth support method to calculate (a) the depth of embedment of the sheet pile for a factor of safety with respect to passive pressure of 2, (b) the force on the anchor if the distance between two anchors is 1 m.
Solution
Since the wall is smooth, then we use Rankine’s formula for calculation of earth pressure coefficient
For
For
φ
K p
φ
= 30
= 35 o K a o K a
K p
= tan 2
= tan
= tan 2
= tan
2
2
( 45–
( 45+
φ
2
φ
( 45+
( 45–
φ
2
2
φ
2
) = 0.3333
) =3
) = 0.271
) =3.69
6 m
1.5 m d ?
Figure P 3.12
(5)
Loose sand
γ b
= 17 kN/m 3
φ ’
= 30 o
Dense sand
γ b
= 19 kN/m 3
φ ’
= 35 o q = 10 kN/m 2
Tie rod
(1)
(3)
(2)
(4)

Lateral Earth Pressure 91
Produce table to calculate depth of embedment d
Force per m (kN) Arms (m) Moment (A)
(1) 10×6×0.33 = 19.8
(2) ½ × 17 × 6 2 ×0.33 = 101
(3) (10 +17× 6) d ×0.271 = 30.35 d
(4) ½×19 × d 2 × 0.271 = 2.574 d 2
(5) –(½ × ½ × 19 × d 2 ×3.69) = – 17.524 d 2
1.5
2.5
+29.7
+252.5
4.5 + d/2 +136.575 d + 15.175
4.5 + 2d/3 +11.583 d 2 + 1.716 d 3
4.5 + 2d/3 – 78.86 d 2 – 11.68 d 3
0 0
From the table, taking Σ M
A
= 0,
–9.96 – 67.28 d 2 +136.575 d +297.38 = 0
By trial and error, get d =3 m Æ take d = 3 m.
Subsituting d into the force equilibrium
∑ Force = 0
19.8 + 101 + 30.35
d + 2.574 d 2 – 17.524 d 2 – T = 0
T = 77.3 kN/m ’
The force in each tie = 1 m × 77.3 = 77.3 kN
For a continuous anchor and factor of safety = 1.5, the required depth of d a
=
γ
2
(
K
FS p
−
T
K a
) =
17
2 × 1 .
5 × 77 .
3
( 3 .
69 − 0 .
271 )
= 2 m
The anchor is centered 1.5 m below the surface
Then width of the anchorage b = 2 × (2– 1.5) = 1 m
Example Problem 3.13
The soil in both sides of the anchored sheet pile wall detailed in Figure P3.13 has a bulk unit weight of 18 kN/m 3 and saturated unit weight of 21 kN/m 3 strength parameters of the soil are c ’ = 0, and φ ’ = 36
. The shear o . (a) Determine the factor of safety of the wall with respect to gross passive resistance and (b) calculate the force in each tie rod.

92 Introduction to Geotechnical Engineering Part 1
6.00 m
A
4.50 m
10 kN/m 2
1.50 m
Tie rod, 2 m spacing
GWT
(2)
(1)
(3)
6.00 m
(5)
(4)
Figure P3.13
Solution
Use Rankine’s formula for calculation of earth pressure coefficient
For φ ’ = 36 o K
K
γ a p
=
=
’ = γ tan tan 2
2 ( 45–
( 45+
φ
φ
2
2
)= 0.26
) =3.85 sat
= 21– 9.8 = 11.2 kN/m 3
Pressure due to water can be ignored because water table at both sides of the wall is at the same level.
Consider moment about anchor point A and tabulate the calculation
Force per m (kN)
(1) 10 x 0.26 x 15 = 39
(2) ½ x 18 x 6 2 x 0.26 = 84.24
(3) 18 x 6 x 10.5 x 0.26 = 294.84
(4) ½ x 11.2 x 10.5
2 x 0.26 = 160.52
(5) -½ x 11.2 x 6 2
Tie -T
x 3.85 = -776
Σ F = 578.6-776
Arms (m)
6.0
2.5
9.75
11.5
13
0
Moment (A)
+234
+210.6
+2874.69
+1846
-10090
0
Σ M = 5165.3-10090