Chapter 12: Lateral Earth Pressure - site.iugaza.edu.pssite.iugaza.edu.ps/ymadhoun/files/2016/09/Chapter-12.pdf · Civil Engineering Department: Foundation Engineering (ECIV 4052)

Civil Engineering Department: Foundation Engineering (ECIV 4052)

Engr. Yasser M. Almadhoun Page 1

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) the type and amount of wall movement,

(b) the shear strength parameters of the soil,

(c) the unit weight of the soil, and

(d) 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 at-

rest 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.



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:



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:



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,



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 (=Koz) when Δx is zero. However,

with Δx > 0, σ’h will be less than Koσ’o.



Rankine active-pressure coefficient

𝐾𝑎 = tan2 (45 −∅′

2) =

1 − sin ∅′

1 + sin ∅′

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 soil-

wall 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:



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 0.001H − 0.004H

cohesive .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:

𝑃𝑎 = (𝑞𝐻 +1

2𝛾𝐻2) 𝐾𝑎 − 2𝑐√𝐾𝑎𝐻

Example 12.2

See example 2.2 in textbook, page 602.

Example 12.3




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)



Example 12.4


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.



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.



For a problem of this type, the depth of tensile crack is given as:

Example 12.5


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.



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,



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

coefficients for various values of ϕ’, α, and 𝛽 with 𝛿′ =1

2∅′𝑎𝑛𝑑

2

3∅′ are

respectively given in Tables 12.6 and 12.7.



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:



Example 12.6


Example 12.7


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

𝐾𝑎 = tan2 (45 +∅′

2) =

1 + sin ∅′

1 − sin ∅′

Then, we have:



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:



Example 12.13


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:



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.



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,



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.



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:

�̅�



𝑃𝑎 =1

2𝛾𝑒𝑞𝐾𝑎𝐻2

𝑖𝑓 𝛾𝑒𝑞 = [sin 𝛽

sin(𝛽 + 𝛼)] (

2𝑞

𝐻) , 𝑡ℎ𝑒𝑛:

𝑃𝑎 =1

2(𝛾 + [

sin 𝛽

sin(𝛽 + 𝛼)] (

2𝑞

𝐻)) 𝐾𝑎𝐻2 … … … 𝑒𝑞𝑛 (1)

𝑃𝑎 =1

2𝐾𝑎𝛾𝐻2 + 𝐾𝑎𝐻𝑞 [

sin 𝛽

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):

𝐾𝑎 =sin2(𝛽 + ∅′)

sin2 𝛽 sin2(𝛽 − 𝛿′) [1 + √sin(∅′ + 𝛿′) sin(∅′ − 𝛼)sin(∅′ − 𝛿′) sin(∅′ + 𝛼)

]

2

𝐾𝑎 =sin2(85° + 30°)

sin2 85° sin2(85° − 20°) [1 + √sin(30° + 20°) sin(∅′ − 5°)

sin(30° − 20°) sin(30° + 5°)]

2

𝐾𝑎 = 0.3578

Or even from Table 12.6 in textbook:



Now, substitute into equation (1):

𝑃𝑎 =1

2(𝛾 + [

sin 𝛽

sin(𝛽 + 𝛼)] (

2𝑞

𝐻)) 𝐾𝑎𝐻2 … … … 𝑒𝑞𝑛 (1)

𝑃𝑎 =1

2(18 + [

sin 85°

sin(85° + 5°)] (

2×96

6.1)) (0.3578)(6.1)2

= 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):

𝑃𝑎 =1

2𝐾𝑎𝛾𝐻2 + 𝐾𝑎𝐻𝑞 [

sin 𝛽

sin(𝛽 + 𝛼)] … … … 𝑒𝑞𝑛 (2)

𝑃𝑎 = 𝑃𝑎(1) + 𝑃𝑎(2) … … … 𝑒𝑞𝑛 (3)

𝑧̅𝑃𝑎 =𝐻

3𝑃𝑎(1) +

𝐻

2𝑃𝑎(2)

𝑧̅𝑃𝑎 =𝐻

3(

1

2𝐾𝑎𝛾𝐻2) +

𝐻

2(𝐾𝑎𝐻𝑞 [

sin 𝛽

sin(𝛽 + 𝛼)])

𝑧̅(328.5) =6.1

3(

1

2×0.3578×18×6.12)

+6.1

2(0.3578×6.1×96× [

sin 85°

sin(85° + 5°)])

𝑧̅ = 2.68 𝑚



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) = tan2(45° +∅1

′

2) = tan2(45° +

30°

2) = 3.0

𝐾𝑝(2) = tan2(45° +∅2

′

2) = tan2(45° +

26°

2) = 2.56

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



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 = 𝛾𝑤𝐻 = 9.81×1.00 = 9.81 𝐾𝑁/𝑚2

Rankine passive forces:

𝑃𝑝(1) =1

2𝐻𝜎𝑎 =

1

2(2.00)(94.32) = 94.32 𝐾𝑁

𝑃𝑝(2) = 𝐻𝜎𝑎 = (1.00)(112.4) = 112.40 𝐾𝑁

𝑃𝑝(3) =1

2𝐻𝜎𝑎 =

1

2(1.00)(135.66 − 112.40) = 11.63 𝐾𝑁

𝑃𝑝(3) =1

2𝐻𝜎𝑎 =

1

2(1.00)(9.81) = 4.91 𝐾𝑁

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:

𝑧̅×𝑃𝑝 =𝑧1

3×𝑃𝑝(1) +

𝑧2

2×𝑃𝑝(2) +

𝑧3

3×𝑃𝑝(3) +

𝑧4

3×𝑃𝑝(4)

𝑧̅×(223.26) = (1.00 +2.00

3) ×(94.32) + (

1.00

2) ×(112.40)

+ (1.00

3) ×(11.63) + (

1.00

3) ×(4.91)

𝑧̅ = 0.980531 𝑚

Documents

Chapter 12: Lateral Earth Pressure - site.iugaza.edu.pssite.iugaza.edu.ps/ymadhoun/files/2016/09/Chapter-12.pdf · Civil Engineering Department: Foundation Engineering (ECIV 4052)