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 . 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,

= +

where,

c is the cohesion

is the effective normal stress

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,

0 = +

= 00 +

where,

u is the pore water pressure

0 is the coefficient of at-rest earth pressure

LATERAL EARTH PRESSURE AT REST (No Lateral Movement)

For normally consolidated soil (Jaky, 1944)

0 = 1 For overconsolidated soil (Mayne and Kulhawy, 1982)

0 = (1 )

The total force, 0 , per unit length of the wall can now be obtained from the area of the pressure diagram as,

0 = 1 + 2 = 0 +1

220

The location of the line of action of the resultant force, 0, can be obtained by taking the moment about the bottom of the wall. Thus,

=1

2

+ 23

0

Note: If the surcharge = 0 and the pore water pressure = 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. Rankines

2. Coulombs

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

RANKINES LATERAL EARTH PRESSURE

RANKINES THEORY

z 'v

'h

A

B

Unit weight of soil =

Assumptions:

Vertical frictionless wall

Dry homogeneous soil

Horizontal backfill

' tan ' c' f

RANKINES LATERAL EARTH PRESSURE

ACTIVE EARTH PRESSURE

z

'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 Mohrs

circle in the subsequent

slide.

Unit weight of soil =

' tan ' c' f

'v

RANKINES LATERAL EARTH PRESSURE

o Koo 'a c'

Based on the diagram:

'sin 1

'sin - 1 )

2

' - (45 tan

'

' 2

0

aaK

aa0

2

0

K2c' - K '

)2

' - (45 tan 2c' - )

2

' - (45tan' '

a

' ' tan ' c' f

ACTIVE EARTH PRESSURE

is the Rankine active earth pressure coefficient

RANKINES LATERAL EARTH PRESSURE

ACTIVE EARTH PRESSURE

RANKINES LATERAL EARTH PRESSURE

aa0 K2c' - K'

z

aK2c' -aK2c' -

K' a0

ACTIVE EARTH PRESSURE DISTRIBUTION

a

cK

cz

'2

RANKINES LATERAL EARTH PRESSURE

z 'v

'h

A

B

If the wall is pushed into the

soil mass, the principal stress

h will increase. On the verge of failure the stress

condition on the soil

element can be expressed

by Mohrs circle b.

The lateral earth pressure,

p, which is the major principal stress, is called

Rankines passive earth

pressure.

Unit weight of soil =

' tan ' c' f

PASSIVE EARTH PRESSURE

RANKINES LATERAL EARTH PRESSURE

Shea

r st

ress

Normal stress

' tan ' c' f

C

D

D

O A

'p Koo

b

a

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

is the Rankine passive

earth pressure coefficient

RANKINES LATERAL EARTH PRESSURE

PASSIVE EARTH PRESSURE

RANKINES LATERAL EARTH PRESSURE

For cohesionless soil,

z

K z ppK2c'

ppv K z K p

PASSIVE EARTH PRESSURE DISTRIBUTION

RANKINES LATERAL EARTH PRESSURE

SPECIAL CASES

Submergence:

Inclined backfill:

Inclined but smooth back face of wall:

RANKINES LATERAL EARTH PRESSURE

SPECIAL CASES

Inclined backfill with c- soil:

RANKINES 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

= 15 kN/m3

= 26o

c = 8 kN/m2

q = 10 kN/m2

H = 4 m

Figure 12.22a Frictionless retaining wall AB

RANKINES LATERAL EARTH PRESSURE

SOLUTION

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

Ka =1 sin

1 + sin=1 si n( 26)

1 + si n( 26)= .

a = Kav 2c Ka

v = z

At z = 0 m;

= q = 10 kN/m2

= 0.39(10) 2(8)( 0.39) = . kN/m2

At z = 4 m;

= 10 + 15(4) = 60 kN/m2

= 0.39(60) 2(8)( 0.39) = . kN/m2

RANKINES LATERAL EARTH PRESSURE

SOLUTION

A

B

= 15 kN/m3

= 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)

RANKINES 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; = . m, 4 = . m

Pa =1

217.31 2.96 = . 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 = . m

C

Pa

z=0.99 m

RANKINES LATERAL EARTH PRESSURE

SOLUTION

b) The passive force Pp.

Kp =1 + sin

1 sin=1 + si n( 26)

1 si n( 26)= .

At z = 0 m;

= q = 10 kN/m2

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

At z = 4 m;

= 10 + 15(4) = 60 kN/m2

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

p = Kpv + 2c Kp

v = z

RANKINES LATERAL EARTH PRESSURE

SOLUTION

A

B

= 15 kN/m3

= 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

RANKINES LATERAL EARTH PRESSURE

SOLUTION

(c)

From Figure 12.22c,

Pp = 51.2 4 +1

2153.6 4 = 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 = . m

Pp

z=1.6 m

RANKINES 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.

COULOMBS LATERAL EARTH PRESSURE

ACTIVE EARTH PRESSURE

COULOMBS LATERAL EARTH PRESSURE

PASSIVE EARTH PRESSURE

COULOMBS LATERAL EARTH PRESSURE

where,

is the angle the back face is inclined with the horizontal

is the inclination of backfill with the horizontal

is the wall friction angle

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 /3. 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 Coulombs active earth pressure coefficient.

4.2 Compute the Coulombs active force per unit length of wall.

4.3 Compute the location of the Coulombs 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 /3. The angle of internal friction of soil is 35, the wall friction angle is 20, and soil cohesion is 0. 5.1 Compute the Coulombs passive earth pressure coefficient.

5.2 Compute the Coulombs passive force per unit length of wall

perpendicular to the wall.

5.3 Compute the location of the Coulombs passive force from the bottom

of the wall.