Civil Engineering Department: Foundation Engineering (ECIV 4052)
Part 4: Lateral Earth Pressure and Earth-Retaining
Structures
Chapter 12: Lateral Earth Pressure
Introduction
Vertical or near-vertical slopes of soil are supported by retaining walls,
cantilever sheetpile walls, sheet-pile bulkheads, braced cuts, and other,
similar structures. The proper design of those structures requires an
estimation of lateral earth pressure, which is a function of several factors,
such as:
(a)
(b)
(c)
(d)
the type and amount of wall movement,
the shear strength parameters of the soil,
the unit weight of the soil, and
the drainage conditions in the backfill.
The following Figure shows a retaining wall of height H. For similar types
of backfill:
a. The wall may be restrained from moving (Figure a).
The lateral earth pressure on the wall at any depth is called the atrest earth pressure.
b. The wall may tilt away from the soil that is retained (Figure b).
With sufficient wall tilt, a triangular soil wedge behind the wall will
fail. The lateral pressure for this condition is referred to as active
earth pressure.
c. The wall may be pushed into the soil that is retained (Figure c).
With sufficient wall movement, a soil wedge will fail. The lateral
pressure for this condition is referred to as passive earth pressure.
Engr. Yasser M. Almadhoun
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Lateral Earth Pressure at Rest
Consider a vertical wall of height H, as shown in Figure 12.3, retaining a
soil having a unit weight of g. A uniformly distributed load, q/unit area, is
also applied at the ground surface.
At any depth z below the ground surface, the vertical subsurface stress is:
If the wall is at rest and is not allowed to move at all, either away from the
soil mass or into the soil mass (i.e., there is zero horizontal strain), the
lateral pressure at a depth z is:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
For normally consolidated soil, the relation for Ko (Jaky, 1944) is:
For overconsolidated soil, the at-rest earth pressure coefficient may be
expressed as:
The total force, Po, per unit length of the wall given in Figure 12.3a can
now be obtained from the area of the pressure diagram given in Figure
12.3b and is:
The location of the line of action of the resultant force, Po, can be obtained
by taking the moment about the bottom of the wall. Thus:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
If the water table is located at a depth z, H, the at-rest pressure diagram
shown in Figure 12.3b will have to be somewhat modified, as shown in
Figure 12.4. If the effective unit weight of soil below the water table equals
’ (i.e., sat − w), then:
Hence, the total force per unit length of the wall can be determined from
the area of the pressure diagram. Specifically:
So,
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Active Pressure
Rankine Active Earth Pressure
The Rankine active earth pressure calculations are based on the assumption
that the wall is frictionless. The lateral earth pressure involves walls that
do not yield at all. However, if a wall tends to move away from the soil a
distance Δx, as shown in the following Figure, the soil pressure on the wall
at any depth will decrease. For a wall that is frictionless, the horizontal
stress, σ’h, at depth z will equal Koσ’o (=Koz) when Δx is zero. However,
with Δx > 0, σ’h will be less than Koσ’o.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Rankine active-pressure coefficient
∅′
1 − sin ∅′
𝐾𝑎 = tan (45 − ) =
2
1 + sin ∅′
2
The pressure distribution shows that at z = 0 the active pressure equals
−2𝑐’√𝑘𝑎 , indicating a tensile stress that decreases with depth and becomes
zero at a depth z = zc, or:
The depth zc is usually referred to as the depth of tensile crack, because
the tensile stress in the soil will eventually cause a crack along the soilwall interface. Thus, the total Rankine active force per unit length of the
wall before the tensile crack occurs is:
After the tensile crack appears, the force per unit length on the wall will
be caused only by the pressure distribution between depths z = zc and z =
H, as shown by the hatched area in the previous Figure. This force may be
expressed as:
Or:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
However, it is important to realize that the active earth pressure condition
will be reached only if the wall is allowed to “yield” sufficiently. The
necessary amount of outward displacement of the wall is as given as under:
Soil type
Wall movement for
passive condition, Δx
granular
cohesive
0.001H − 0.004H
.01H − 0.04H
If there exists a surcharge load acting downward on the top surface of the
backfill:
The Rankine active stress at depth z can be calculated as follows:
𝜎ℎ(𝑎𝑐𝑡𝑖𝑣𝑒) = (𝑞 + 𝛾𝐻)𝐾𝑎 − 2𝑐√𝐾𝑎
The Rankine active force per unit length of the wall at depth z can be
calculated as follows:
1
𝑃𝑎 = (𝑞𝐻 + 𝛾𝐻2 ) 𝐾𝑎 − 2𝑐√𝐾𝑎 𝐻
2
Example 12.2
See example 2.2 in textbook, page 602.
Example 12.3
See example 12.3 in textbook, page 604.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
A Generalized Case for Rankine Active Pressure – Granular Backfill
In the previous section, the relationship was developed for Rankine active
pressure for a retaining wall with a vertical back and a horizontal backfill.
That can be extended to general cases of frictionless walls with inclined
backs and inclined backfills.
The previous Figure shows a retaining wall whose back is inclined at an
angle  with the vertical. The granular backfill is inclined at an angle α
with the horizontal. The active force Pa for unit length of the wall then can
be calculated as:
Where Ka can be found from the Table 12.1 or using this equation:
And the horizontal and vertical Rankine active forces (Pa(h) and Pa(v)
respectively) for unit length of the wall is:
𝑃𝑎(ℎ) = 𝑃𝑎 cos(𝛽𝑎 + 𝜃)
𝑃𝑎(𝑣) = 𝑃𝑎 sin(𝛽𝑎 + 𝜃)
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Example 12.4
See example 12.4 in textbook, page 612.
Granular Backfill with Vertical Back Face of Wall
If the backfill of a frictionless retaining wall is a granular soil (c = 0) and
rises at an angle α with respect to the horizontal,
the active earth-pressure coefficient may be expressed in the form (or can
be found from Table 12.3):
At any depth z, the Rankine active pressure may be expressed as:
Also, the total force per unit length of the wall is:
Note that, in this case, the direction of the resultant force Pa is inclined at
an angle α with the horizontal and intersects the wall at a distance H/3
from the base of the wall.
Engr. Yasser M. Almadhoun
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Rankine Active Pressure with Vertical Wall Backface and Inclined c’ϕ’ Soil Backfill
For a frictionless retaining wall with a vertical back face ( = 0) and
inclined backfill of c’− ϕ’ soil (see the following Figure) at an angle α with
the horizontal, the active pressure at any depth z can be given as:
where:
Some values of K’a are given in Table 12.4.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
For a problem of this type, the depth of tensile crack is given as:
Example 12.5
See example 12.5 in textbook, page 613.
Coulomb’s Active Earth Pressure
To apply Coulomb’s active earth pressure theory, let us consider a retaining
wall with its back face inclined at an angle 𝛽 with the horizontal, as shown
in Figure 12.12a. The backfill is a granular soil that slopes at an angle α
with the horizontal.
Also, let 𝛿′ be the angle of friction between the soil and the wall (i.e., the
angle of wall friction).
To find the active force, consider a possible soil failure wedge ABC1. The
forces acting on this wedge (per unit length at right angles to the cross
section shown) are as follows:
1. The weight of the wedge, W.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
2. The resultant, R, of the normal and resisting shear forces along the
surface, BC1. The force R will be inclined at an angle ϕ’ to the
normal drawn to BC1.
3. The active force per unit length of the wall, Pa, which will be
inclined at an angle 𝛿′ to the normal drawn to the back face of the
wall.
The maximum value of Pa thus determined is Coulomb’s active force (see
top part of Figure 12.12a), which may be expressed as:
where,
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
The values of the active earth pressure coefficient, Ka, for a vertical
retaining wall (𝛽 = 90°) with horizontal backfill (𝛼 = 0°) are given in
Table 12.5.
Note that the line of action of the resultant force (Pa) will act at a distance
H/3 above the base of the wall and will be inclined at an angle 𝛿′ to the
normal drawn to the back of the wall.
In the actual design of retaining walls, the value of the wall friction angle
𝛿′ is assumed to be between ϕ’/2 and 2/3ϕ’. The active earth pressure
1
2
2
3
coefficients for various values of ϕ’, α, and 𝛽 with 𝛿 ′ = ∅′ 𝑎𝑛𝑑 ∅′ are
respectively given in Tables 12.6 and 12.7.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
If a uniform surcharge of intensity q is located above the backfill, as shown
in Figure 12.13, the active force, Pa, can be calculated as:
where:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Example 12.6
See example 12.6 in textbook, page 620.
Example 12.7
See example 12.7 in textbook, page 621.
Passive Pressure
Rankine Passive Earth Pressure
The horizontal stress, σ’h, at this point is referred to as the Rankine passive
pressure, or σ’h = σ’p.
For Mohr’s circle, the major principal stress is σ’p, and the minor principal
stress is σ’o:
Rankine active-pressure coefficient
∅′
1 + sin ∅′
𝐾𝑎 = tan (45 + ) =
2
1 − sin ∅′
2
Then, we have:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
The passive pressure diagram for the wall shown in the following Figure.
Note that:
The passive force per unit length of the wall can be determined from the
area of the pressure diagram, or:
The approximate magnitudes of the wall movements, Δx, required to
develop failure under passive conditions are as follows:
If the backfill behind the wall is a granular soil (i.e., c’ = 0), then, the
passive force per unit length of the wall will be:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Example 12.13
See example 12.13 in textbook, page 636.
Rankine Passive Earth Pressure ─ Vertical Backface and Inclined
Backfill
Granular Soil
For a frictionless vertical retaining wall (as the following Figure) with a
granular backfill (c’ = 0), the Rankine passive pressure at any depth can be
determined in a manner similar to that done in the case of active pressure
in a preceeding section.
The pressure is:
and the passive force is:
where:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
As in the case of the active force, the resultant force, Pp, is inclined at an
angle α with the horizontal and intersects the wall at a distance H/3 from
the bottom of the wall. The values of Kp (the passive earth pressure
coefficient) for various values of α and ϕ’ are given in Table 12.9.
C’– ϕ’ Soil
If the backfill of the frictionless vertical retaining wall is a c’– ϕ’ soil, then:
where:
The variation of K’p with ϕ’, α, and c’/z is given in Table 12.10.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Coulomb’s Passive Earth Pressure
To understand the determination of Coulomb’s passive force, Pp, consider
the wall shown in Figure 12.21a.
As in the case of active pressure, Coulomb assumed that the potential
failure surface in soil is a plane. For a trial failure wedge of soil, such as
ABC1, the forces per unit length of the wall acting on the wedge are
1. The weight of the wedge, W,
2. The resultant, R, of the normal and shear forces on the plane BC1,
and
3. The passive force, Pp.
The minimum value of Pp in this diagram is Coulomb’s passive force,
mathematically expressed as:
where,
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
The values of the passive pressure coefficient, Kp, for various values of ϕ’
and 𝛿′ are given in Table 12.11 (𝛽 = 90°, 𝛼 = 0°).
Note that the resultant passive force, Pp, will act at a distance H/3 from the
bottom of the wall and will be inclined at an angle 𝛿′ to the normal drawn
to the back face of the wall.
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Problems
Problem (1)
Consider the retaining wall shown in the Figure below. Calculate the
Coulomb’s active force per unit length of the wall and the location of the
line of action of that resultant at which it acts at the retaining wall.
Solution:
̅
𝒁
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
1
𝑃𝑎 = 𝛾𝑒𝑞 𝐾𝑎 𝐻2
2
sin 𝛽
2𝑞
𝑖𝑓 𝛾𝑒𝑞 = [
] ( ) , 𝑡ℎ𝑒𝑛:
sin(𝛽 + 𝛼) 𝐻
1
sin 𝛽
2𝑞
𝑃𝑎 = (𝛾 + [
] ( )) 𝐾𝑎 𝐻2 … … … 𝑒𝑞𝑛 (1)
2
sin(𝛽 + 𝛼) 𝐻
1
sin 𝛽
𝑃𝑎 = 𝐾𝑎 𝛾𝐻2 + 𝐾𝑎 𝐻𝑞 [
]
2
sin(𝛽 + 𝛼)
… … … 𝑒𝑞𝑛 (2)
𝑃𝑎 = 𝑃𝑎(1) + 𝑃𝑎(2)
… … … 𝑒𝑞𝑛 (3)
Ka can be found using the following equation (equation 12.26 in
textbook):
sin2 (𝛽 + ∅′)
𝐾𝑎 =
2
sin(∅′
𝛿′) sin(∅′
+
− 𝛼)
sin2 𝛽 sin2 (𝛽 − 𝛿′) [1 + √
]
′
′
sin(∅ − 𝛿′) sin(∅ + 𝛼)
𝐾𝑎 =
sin2 (85° + 30°)
sin(30° + 20°) sin(∅′ − 5°)
sin2 85° sin2 (85° − 20°) [1 + √
]
sin(30° − 20°) sin(30° + 5°)
2
𝐾𝑎 = 0.3578
Or even from Table 12.6 in textbook:
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Now, substitute into equation (1):
1
sin 𝛽
2𝑞
𝑃𝑎 = (𝛾 + [
] ( )) 𝐾𝑎 𝐻2 … … … 𝑒𝑞𝑛 (1)
2
sin(𝛽 + 𝛼) 𝐻
1
sin 85°
2×96
𝑃𝑎 = (18 + [
](
)) (0.3578)(6.1)2
2
sin(85° + 5°)
6.1
= 328.5 𝐾𝑁/𝑚′
To determine the location of the line of action of the resultant at which it
acts at the retaining wall:
Recall equations (2) & (3):
1
sin 𝛽
𝑃𝑎 = 𝐾𝑎 𝛾𝐻2 + 𝐾𝑎 𝐻𝑞 [
]
2
sin(𝛽 + 𝛼)
… … … 𝑒𝑞𝑛 (2)
𝑃𝑎 = 𝑃𝑎(1) + 𝑃𝑎(2)
… … … 𝑒𝑞𝑛 (3)
𝐻
𝐻
𝑃𝑎(1) + 𝑃𝑎(2)
3
2
𝐻 1
𝐻
sin 𝛽
𝑧̅𝑃𝑎 = ( 𝐾𝑎 𝛾𝐻2 ) + (𝐾𝑎 𝐻𝑞 [
])
3 2
2
sin(𝛽 + 𝛼)
𝑧̅𝑃𝑎 =
𝑧̅(328.5) =
6.1 1
( ×0.3578×18×6.12 )
3 2
6.1
sin 85°
+
(0.3578×6.1×96× [
])
2
sin(85° + 5°)
𝑧̅ = 2.68 𝑚
Engr. Yasser M. Almadhoun
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
Problem (2)
Consider the retaining wall shown in the Figure below. Calculate the
Rankine passive force per unit length of the wall and the location of the
line of action of that resultant at which it acts on the retaining wall.
Solution:
Rankine passive lateral earth pressure coefficients:
𝐾𝑝(1)
∅1′
30°
= tan (45° + ) = tan2 (45° +
) = 3.0
2
2
𝐾𝑝(2)
∅′2
26°
= tan (45° + ) = tan2 (45° +
) = 2.56
2
2
2
2
Rankine passive lateral earth pressures:
*at h = 0.00 m
σv = 𝛾𝐻 = 0.00 𝐾𝑁/𝑚2
*at h = 2.00 m (just before the GWT):
𝜎𝑣 = 𝛾𝐻 = 2.00×15.72 = 31.44 𝐾𝑁/𝑚2
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Civil Engineering Department: Foundation Engineering (ECIV 4052)
𝜎𝑎 = 𝜎𝑣 ×𝐾𝑝(1) = 31.44×3.00 = 94.32 𝐾𝑁/𝑚2
*at h = 2.00 m (just after the GWT):
𝜎𝑣 = 𝛾𝐻 = 2.00×15.72 = 31.44 𝐾𝑁/𝑚2
𝜎𝑎 = 𝜎𝑣 ×𝐾𝑝(2) + 2.00𝑐√𝐾𝑝(2) = 31.44×2.56 + 2.00(10.00)√2.56
= 112.40 𝐾𝑁/𝑚2
*at h = 3.00 m:
𝜎𝑣 = 𝛾𝐻 = 2.00×15.72 + 1.00×(18.86 − 9.81) = 40.49 𝐾𝑁/𝑚2
𝜎𝑎 = 𝜎𝑣 ×𝐾𝑝(2) + 2.00𝑐√𝐾𝑝(2) = 40.49×2.56 + 2.00(10.00)√2.56
= 135.66 𝐾𝑁/𝑚2
𝑢 = 𝛾𝑤 𝐻 = 9.81×1.00 = 9.81 𝐾𝑁/𝑚2
Rankine passive forces:
1
1
𝑃𝑝(1) = 𝐻𝜎𝑎 = (2.00)(94.32) = 94.32 𝐾𝑁
2
2
𝑃𝑝(2) = 𝐻𝜎𝑎 = (1.00)(112.4) = 112.40 𝐾𝑁
1
1
𝑃𝑝(3) = 𝐻𝜎𝑎 = (1.00)(135.66 − 112.40) = 11.63 𝐾𝑁
2
2
1
1
𝑃𝑝(3) = 𝐻𝜎𝑎 = (1.00)(9.81) = 4.91 𝐾𝑁
2
2
Rankine passive resultant:
𝑃𝑝 = 𝑃𝑝(1) + 𝑃𝑝(2) + 𝑃𝑝(3) + 𝑃𝑝(4)
𝑃𝑝 = 94.32 + 112.40 + 11.63 + 4.91 = 𝑐. 26 𝐾𝑁
Location of the Rankine passive resultant:
𝑧1
𝑧2
𝑧3
𝑧4
𝑧̅×𝑃𝑝 = ×𝑃𝑝(1) + ×𝑃𝑝(2) + ×𝑃𝑝(3) + ×𝑃𝑝(4)
3
2
3
3
2.00
1.00
𝑧̅×(223.26) = (1.00 +
) ×(94.32) + (
) ×(112.40)
3
2
1.00
1.00
+(
) ×(11.63) + (
) ×(4.91)
3
3
𝑧̅ = 0.980531 𝑚
Engr. Yasser M. Almadhoun
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