Earth Pressures by the Coulomb Approach

Source Material Provide By : D. A. Cameron Rock & Soils 2009

Dr. I Wayan Sengara

Lateral Earth Pressure

Retaining structures such as retaining walls, basement walls, and bulkheads commonly are used in foundation engineering.

These structures are also subjected to lateral loads, coming from either a soil mass, or from seismic forces operating upon these soil masses.

Lateral Earth Pressure

Some of the examples of these lateral loads covered in this lecture include:

1. Soil stability problems;

2. The temporary of permanent bracing of excavations;

3. The design of retaining walls for highways (abutments) or buildings (walls), and

4. The various types of anchorages that transfer loads from the walls to foundations.

Lateral Earth Pressure

Lateral Earth Pressure

Lateral Earth Pressure

In order to be able to design the structures that can carry these lateral loads, a theory of how these lateral loads behave is reviewed. Specifically, this lecture will present,

1. Define the different types of lateral pressures: at rest, active and passive earth conditions,

2. Distinguish between the theories developed by Charles Coulomb and William Rankine.

Coulomb (1736-1806) Charles-Augustine de

Coulomb (1736-1806) was a military engineer and a famous French physicist that discovered the force between two electrical charges.

Less known was his development of the first thoroughly analytical study of lateral earth pressures which he published in 1776.

That theory remains the standard choice of analysis for lateral forces upon structures in soils.

Rankine (1820-1872)

William J.M. Rankine (1820-1872), the famous Scot engineer and physicist is best known as one of the founders of the science of thermodynamics.

He held the Queen Victoria Chair of civil engineering at the University of Glasgow.

In soil mechanics, he simplified Coulomb’s theory for cases when the surface of the backfill is horizontal, the friction between the wall and the backfill is negligible and the retaining wall is vertical.

COULOMB APPROACH

Upper bound theorem

(Rankine is a lower bound)

Plausible collapse mechanism

assumed

Equilibrium of forces, moments or

energy

Predicted force > collapse load

NOTE: when an upper bound and a

lower bound agree = the true

solution!

COULOMB APPROACH

HANDLES: Irregular backfill surfaces Sloping backs of walls Sloping backfill Surface surcharges Wall friction

Interface friction angle,

Although failure surfaces were

known to be curved, a planar

approximation was adopted

(OK for Active earth pressures)

Influence of Friction

Fa

ilure

line

Pa

Active

Pp

Failu

re lin

e

Passive

Coulomb Wedge Analysis

Case 1: Uniform soil, c = 0,

vertical back of retaining wall

- no wall friction

W

N

T

Pa

a R

FORCES

W = weight of soil wedge

needs support

N = force normal to sliding plane from underlying supporting soil

T = tangential force along sliding plane = Ntan

R = resultant reaction from supporting soil forces, T & N

Pa = maxm reaction from wall required for

equilibrium for “critical” wedge angle, a

VECTOR DIAGRAMSCase 1: no wall friction

W

Pa

R

90 - a +

a -

a

N

R

90Sliding

plane

SOLUTIONBut

a

2

tanθ2

HW

0.322

0.324

0.326

0.328

0.33

0.332

0.334

50 55 60 65 70

trial angle, a (degrees)

2Pa/( H

2)

)tan(θtanθ2HP aa

2

a

)Wtan(θP aa

Note: for this simple

problem, Rankine and

Coulomb expressions

coincide – failure plane is at

(45 + /2) = 60

i.e. where the maximum

value of Pa is obtained

AND Ka = 0.333 = 2Pa/(H2)

for this simple problem

Case 2: Uniform soil, c = 0,

vertical backed retaining wall,

wall friction (tan)

R

W

N

T

a

Pa

VECTOR DIAGRAMS

Case 2: with wall friction

RW

Pa

90 - a +

a -

SOLUTION

(as before)

From the sine rule:

)θsin(90

W

)sin(θ

P

aa

a

))tan(θθsin(90

)sin(θ

2

HP

aa

a2

a

a

2

tanθ2

HW

0.286

0.288

0.29

0.292

0.294

0.296

0.298

50 55 60 65

trial angle, a (degrees)

2Pa/( H

2)

4 flatter failure plane

‘Ka’ 11% less

Influence of = 2/3 = 20- simple case study 2

Consider - - -

natural soil

natural soil

slidi

ng s

urfa

ce?backfill

tension crack

FORCES

natural soil

R

Pa

W

cwh

cl

R

Pa

W

cwh

cl

FORCE VECTOR DIAGRAM

Consider water in crack

natural soil

natural soil

slidi

ng s

urfa

ce?backfill

tension crack

FORCES

natural soil

R

Pa

W

cwh

clU = 0.5wz2

z

Pa

W

cwh

cl

FORCE VECTOR DIAGRAM

U

R

SEEPAGE FORCES

R

Pa

W

cwh

clU

So what’s Coulomb method all about? Must appreciate all the forces acting on

the potentially unstable wedge of backfill soil, including shear against the wall

The forces includes a base reaction, the magnitude of which is unknown

Do know the direction which is useful in finding the active thrust – force vectors

Numerous trial wedges are needed to find the maximum thrust analagous to finding the critical

sliding surface in slope stability

OVERALL SUMMARY

Earth pressures are needed for design

of retaining walls & excavations

3 major states: at rest, Active &

Passive

Earth pressure coefficients are based

on effective stresses

Water pressures are important

Cohesion leads to potential cracked

zone for Active earth pressure state

Coulomb a superior approach as wall

interface forces considered

Masonry Gravity walls

Crib gravitywall(concrete,timber)

Gabiongravitywall

DESIGN OF RETAINING WALLS

Generally rely on mass - needs to be substantial

Concrete Cantilever Walls

Flexural strength BUT also mass of soil above footing, which

moves with the wall

Active thrust

Dead Weights

Design Requirements

a) The wall is structurally sound

flexure, compression & tension

b) The “foundation” can carry the loads

will not fail - “bearing capacity”

c) The wall is safe against overturning

and translation

toppling or sliding

d) Settlement and tilting are minimized

remains “serviceable”

Design Steps

Determine:

1. earth pressures

2. resultant thrust behind wall

3. soil reactions at base of wall (footing)

4. location of resultant soil reaction on

base

take moments of all forces about

toe of wall

Consider the loadssubscript w = “wall”

N

T

“toe”

x

Tw

Pw

H

B

W

xw

y

and the Reactions

EXAMPLE

• Evaluate earth pressures

• Pw and Tw, horizontal and vertical

components of Pa

• Soil reaction: N = (W + Tw)

• Moments of all forces about toe

of wall to find x

(Nx + Pwy) = (TwB + Wxw)

i.e. overturning = restoring moments

“Moment equilibrium”

Design Steps, cont’d.

Checks

1. Base reaction is in MIDDLE third?

If not, then widen footing to avoid “lift

off”

2. Resistance to potential overturning?

1.5moment goverturnin

moment restoring

3. Bearing capacity of soil

foundation?

4. Base sliding resistance?

EXAMPLE cont’d5. 0.67B x 0.33B?

6. Overturning?

7. Bearing capacity = function of

soil shear strength parameters

(later in the course)

8. Base sliding?

1.5yPNx

BTWx

w

ww

NtanBc T where1.5,P

Ta

w

Adhesion & wall friction

Adhesion, ca, between soil foundation and

wall footing

─ natural soil and concrete?

─ sliding resistance of interface for zero

normal force

Wall or interface friction, tan

Typically assume:

─ ca = 0.67c'

─ tan = 0.67tan'

but can vary with roughness of footing

material

Soil base reactionsN Centreline

x

B/2

Average pressure, N/(LB)

+ Pressure due to moment, Ne

m 1LLB

6Ne2

e =(B/2 –x)

Point of lift off at zero contact pressure

6

Be

i.e._when

0B

6e1

B

N

or

0B

6Ne

B

N2

i.e. “the middle third rule”

Design Steps, cont’d.

Finally, check:

9. Global stability

Slope stability

10. Shear & moment capacity of

wall & base

Designers of Retaining Walls:

AS 4678 – 2002

Some Features:

1) 3 levels of classification of structure

2) characteristic strengths factored for

ultimate or serviceability limit state

3) reduction factors depend on how well

controlled the backfill compaction is

4) Load factors 1.25DL,1.25'h, 1.5LL

5) Minm. LL of 5 kPa on backfill surface

6) good information on reinforced soil

systems

SMITH’S EXERCISES

7.2 Gravity retaining wall: Check FoS for sliding and overturning Answers 3.53 and 1.85

2.5 m

1.8 m

4.0 m

1.0 m

c = 0, = 35,

= 18 kN/m3

wall = 23.5 kN/m3

Passive pressure at front of wall?