Foundation Analysis LATERAL EARTH PRESSURE

INTRODUCTION

Vertical or near-vertical slopes of soil are supported by

retaining walls, cantilever sheet-pile walls, sheet-pile

bulkheads, braced cuts, and other similar structures. The

proper design of these 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.

INTRODUCTION

Lateral earth pressure is a function of wall movement (or

relative lateral movement in the backfill soil).

LATERAL EARTH PRESSURE AT REST

Consider a vertical wall of height H, as shown in Figure 7.3,

retaining a soil having a unit weight of γ. A uniformly

distributed load, q/unit area, is also applied at the ground

surface.

(No Lateral Movement)

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)

The shear strength of the soil is,

𝜏 = 𝑐′ + 𝜎′ 𝑡𝑎𝑛∅′

where,

c’ is the cohesion

σ’ is the effective normal stress

ϕ’ is the effective angle of friction

At any depth z below the ground surface, the vertical

subsurface stress and lateral earth pressure are expressed as,

𝜎0′ = 𝑞 + 𝛾𝑧

𝜎ℎ = 𝐾0𝜎0′ + 𝑢

where,

u is the pore water pressure

𝐾0 is the coefficient of at-rest earth pressure

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)

For normally consolidated soil (Jaky, 1944)

𝐾0 = 1 − 𝑠𝑖𝑛∅′ For overconsolidated soil (Mayne and Kulhawy, 1982)

𝐾0 = (1 − 𝑠𝑖𝑛∅′)𝑂𝐶𝑅𝑠𝑖𝑛∅′ The total force, 𝑃0 , per unit length of the wall can now be obtained from the area of the pressure diagram as,

𝑃0 = 𝑃1 + 𝑃2 = 𝑞𝐾0𝐻 +1

2𝛾𝐻2𝐾0

The location of the line of action of the resultant force, 𝑃0, can be obtained by taking the moment about the bottom of the wall. Thus,

𝑧 =𝑃1

𝐻2

+ 𝑃2𝐻3

𝑃0

Note: If the surcharge 𝑞 = 0 and the pore water pressure 𝑢 = 0, the pressure diagram will be a triangle.

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)

If the water table is located at a depth z < H, the at-rest

pressure diagram will have to be somewhat modified.

PROBLEM SET 10

1. For the retaining wall shown in the figure below, determine

the lateral earth force at rest per unit length of the wall. Also determine the location of the resultant force. Assume OCR = 1.

LATERAL EARTH PRESSURE

ACTIVE AND PASSIVE

Based on assumptions of the intervening forces and the failure

mode, different theories have been developed.

These theories differ only in terms of the coefficient of lateral

earth pressure but operate with similar stress/pressure

equations.

The three widely-accepted theories are the following: 1. Rankine’s

2. Coulomb’s

3. Log-spiral

LATERAL EARTH PRESSURE

ACTIVE AND PASSIVE

Below are the comparison of the three theories and their

applicability.

Method Failure

Mode Wall Friction

Active Case Passive Case

based on experimentation and actual

failure observations

Rankine planar no wall friction poor estimate poor estimate

Coulomb planar considered good estimate (less) poor

estimate

Log-spiral curved considered better

estimate

better

estimate

RANKINE’S LATERAL EARTH PRESSURE

RANKINE’S THEORY

z σ'v

σ'h

A

B

Unit weight of soil = γ

Assumptions:

Vertical frictionless wall

Dry homogeneous soil

Horizontal backfill

' tan ' c' f

RANKINE’S LATERAL EARTH PRESSURE

ACTIVE EARTH PRESSURE

z

σ'h

A

B

If wall AB is allowed to move

away from the soil mass

gradually, horizontal stress

will decrease.

Plastic equilibrium in soil

refers to the condition

where every point in a soil

mass is on the verge of

failure.

This is represented by Mohr’s

circle in the subsequent

slide.

Unit weight of soil = γ

' tan ' c' f

σ'v

RANKINE’S LATERAL EARTH PRESSURE

σ‘o Koσ’o σ'a

c'

Based on the diagram:

'sin 1

'sin - 1 )

2

' - (45 tan

'

' 2

0

a

aK

aa0

2

0

K2c' - K '

)2

' - (45 tan 2c' - )

2

' - (45tan' '

a

ø' ' tan ' c' f

ACTIVE EARTH PRESSURE

𝐾𝑎 is the Rankine active

earth pressure coefficient

RANKINE’S LATERAL EARTH PRESSURE

ACTIVE EARTH PRESSURE

RANKINE’S LATERAL EARTH PRESSURE

aa0 K2c' - K'

z

aK2c' -aK2c' -

K' a0

ACTIVE EARTH PRESSURE DISTRIBUTION

a

cK

cz

'2

RANKINE’S LATERAL EARTH PRESSURE

z σ'v

σ'h

A

B

If the wall is pushed into the

soil mass, the principal stress

σ’h will increase. On the

verge of failure the stress

condition on the soil

element can be expressed

by Mohr’s circle b.

The lateral earth pressure,

σ’p, which is the major

principal stress, is called

Rankine’s passive earth

pressure.

Unit weight of soil = γ

' tan ' c' f

PASSIVE EARTH PRESSURE

RANKINE’S LATERAL EARTH PRESSURE

Shea

r st

ress

Normal stress

' tan ' c' f

C

D

D’

O A

σ'p Koσ’o

b

a

σ‘o ' c'

pp0

2

0

K2c' K'

)2

' (45 tan 2c' )

2

' (45 tan' '

p

'sin 1

'sin 1 )

2

' (45 tan

'

'2

0

p

pK

PASSIVE EARTH PRESSURE

𝐾𝑝 is the Rankine passive

earth pressure coefficient

RANKINE’S LATERAL EARTH PRESSURE

PASSIVE EARTH PRESSURE

RANKINE’S LATERAL EARTH PRESSURE

For cohesionless soil,

z

K z ppK2c'

ppv K z K p

PASSIVE EARTH PRESSURE DISTRIBUTION

RANKINE’S LATERAL EARTH PRESSURE

SPECIAL CASES

Submergence:

Inclined backfill:

Inclined but smooth back face of wall:

RANKINE’S LATERAL EARTH PRESSURE

SPECIAL CASES

Inclined backfill with c’-ϕ’ soil:

RANKINE’S LATERAL EARTH PRESSURE

ILLUSTRATIVE PROBLEM

A frictionless retaining wall is shown in Figure 12.22a. Determine:

a) The active force Pa after the tensile crack occurs.

b) The passive force Pp.

z

A

B

γ = 15 kN/m3

ø’ = 26o

c’ = 8 kN/m2

q = 10 kN/m2

H = 4 m

Figure 12.22a – Frictionless retaining wall AB

RANKINE’S LATERAL EARTH PRESSURE

SOLUTION

a) The active force Pa after the tensile crack occurs.

Ka =1 − sinØ′

1 + sinØ′=1 − si n( 26)

1 + si n( 26)= 𝟎. 𝟑𝟗

σa′ = Kaσv′ − 2c′ Ka

σv′ = γ′z

At z = 0 m;

𝜎𝑣′= q = 10 kN/m2

𝜎𝑎′= 0.39(10) – 2(8)( 0.39) = −𝟔. 𝟎𝟗 kN/m2

At z = 4 m;

𝜎𝑣′= 10 + 15(4) = 60 kN/m2

𝜎𝑎′= 0.39(60) – 2(8)( 0.39) = 𝟏𝟕. 𝟑𝟏 kN/m2

RANKINE’S LATERAL EARTH PRESSURE

SOLUTION

A

B

γ = 15 kN/m3

ø’ = 26o

c’ = 8 kN/m2

q = 10 kN/m2

H = 4 m

Figure 12.22 – (a) Frictionless retaining wall AB, and (b) active pressure distribution diagram

-6.09 kN/m2

17.32 kN/m2

y = 1.04 m

4 – y = 2.96 m

(a) (b)

RANKINE’S LATERAL EARTH PRESSURE

SOLUTION

-6.09 kN/m2

17.32 kN/m2

y = 1.04 m

4 – y = 2.96 m

(b)

From Figure 12.22b,

6.09

y=17.31

4 − y; 𝑦 = 𝟏. 𝟎𝟒 m, 4 − 𝑦 = 𝟐. 𝟗𝟔 m

Pa =1

217.31 2.96 = 𝟐𝟓. 𝟔𝟐 kN/m

The active force Pa after the tensile crack occurs is equal to the area of the active pressure distribution diagram below point C, or

and the location is located at

z =1

32.96 = 𝟎. 𝟗𝟗 m

C

Pa

z=0.99 m

RANKINE’S LATERAL EARTH PRESSURE

SOLUTION

b) The passive force Pp.

Kp =1 + sinØ′

1 − sinØ′=1 + si n( 26)

1 − si n( 26)= 𝟐. 𝟓𝟔

At z = 0 m;

𝜎𝑣′= q = 10 kN/m2

𝜎𝑝′= 2.56(10) + 2(8)( 2.56) = 𝟓𝟏. 𝟐 kN/m2

At z = 4 m;

𝜎𝑣 ’ = 10 + 15(4) = 60 kN/m2

𝜎𝑝’ = 2.56(60) + 2(8)( 2.56) = 𝟐𝟎𝟒. 𝟖 kN/m2

σp′ = Kpσv′ + 2c′ Kp

σv′ = γ′z

RANKINE’S LATERAL EARTH PRESSURE

SOLUTION

A

B

γ = 15 kN/m3

ø’ = 26o

c’ = 8 kN/m2

q = 10 kN/m2

H = 4 m

Figure 12.22 – (a) Frictionless retaining wall AB, and (c) passive pressure distribution diagram

(a) (c)

51.2 kN/m2

51.2 kN/m2 153.6 kN/m2

RANKINE’S LATERAL EARTH PRESSURE

SOLUTION

(c)

From Figure 12.22c,

Pp = 51.2 4 +1

2153.6 4 = 𝟓𝟏𝟐 kN/m

The passive force Pp is equal to the area of the passive pressure distribution, or

and the location is calculated by taking summation moment at the base, or

512 z = 51.2 44

2+

1

2153.6 (4)

1

3(4)

51.2 kN/m2

51.2 kN/m2 153.6 kN/m2 z = 𝟏. 𝟔 m

Pp

z=1.6 m

RANKINE’S LATERAL EARTH PRESSURE

FINAL ANSWERS

a) The active force Pa after the tensile crack

occurs has a magnitude of 25.62 kN per unit

length of the frictionless retaining wall and is

acting at 0.99 meters above the base.

b) The passive force Pp has a magnitude of 512 kN

per unit length of the frictionless retaining wall

and is acting at 1.6 meters above the base.

PROBLEM SET 10

2. Assume that the retaining wall shown in the figure below

can yield sufficiently to develop an active state. Determine

the Rankine active force per unit length of the wall and the

location of the resultant line of action.

PROBLEM SET 10

3. Assume that the retaining wall shown in the figure below

can yield sufficiently to develop passive state. Determine the

Rankine passive force per unit length of the wall.

COULOMB’S LATERAL EARTH PRESSURE

ACTIVE EARTH PRESSURE

COULOMB’S LATERAL EARTH PRESSURE

PASSIVE EARTH PRESSURE

COULOMB’S LATERAL EARTH PRESSURE

where,

β is the angle the back face is inclined with the horizontal

α is the inclination of backfill with the horizontal

δ is the wall friction angle

ϕ is the angle of internal friction

PROBLEM SET 10

4. A retaining wall shown below has a height of 4.5 m. The unit

weight of soil is 16.5 𝑘𝑁/𝑚3. The angle of internal friction of soil

is 36°, the wall friction angle is 24°, and soil cohesion is 0. The

wall is supporting a horizontal backfill. 4.1 Compute the Coulomb’s active earth pressure coefficient.

4.2 Compute the Coulomb’s active force per unit length of wall.

4.3 Compute the location of the Coulomb’s active force from the bottom of

the wall.

PROBLEM SET 10

5. A vertical retaining wall has a height of 4 m and is

supporting a horizontal backfill. The unit weight of soil is 16.5 𝑘𝑁/𝑚3. The angle of internal friction of soil is 35°, the wall

friction angle is 20°, and soil cohesion is 0. 5.1 Compute the Coulomb’s passive earth pressure coefficient.

5.2 Compute the Coulomb’s passive force per unit length of wall

perpendicular to the wall.

5.3 Compute the location of the Coulomb’s passive force from the bottom

of the wall.