LATERAL EARTH PRESSURE

TOPIC 2:

BY: ENGR AZHANI ZUKRI

BAA3513 SEMI2013/14

Learning Outcomes

Introduction to retaining walls and

earth pressures

Lateral earth pressure – Theory of

elasticity

Rankine’s theory of earth pressure

(horizontal and sloping backfill)

Coulomb’s theory of earth pressure

Braced cut BAA3513 SEMI2013/14

Introduction

Lateral earth pressure estimation

are crucial to structures such as

retaining walls, sheet piles walls,

buried pipes, basement walls,

braced excavation, cofferdams

and others.

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Retaining walls

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What are Factors that

should be considered when

selecting the type of

retaining wall?

RETAINING WALL

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Type of retaining wall…

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Earth pressures

There are 3 condition that need to be consider

i) At rest pressure

ii) Active pressure

iii) Passive pressure

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Three condition of lateral earth pressure

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At rest pressure

Refer to condition in which soil is prevented from lateral movement by the surrounding soil or by unyielding wall.

Happen if the wall AB is static (not move either to left or right)

The soil mass in a state of static equilibrium

Coefficient of lateral earth pressure at rest is identified as Ko.

Coefficient of earth pressure at rest, Ko = 1 – sin

Lateral soil pressure, o = KoH

Resultant force per unit length of wall,

Po = ½ KoH2

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BAA 3513 GEOTECHNICAL ENGINEERING

Coefficient of lateral earth pressure at rest

The ratio of the horizontal to the vertical stress at any point is defined as the

coefficient K0.

Jaky, 1944

Sherif, Fang &

Sherif, 1984

Mayne &

Kulhawy, 1982

For a dense sand, compacted and back fill :

Recommended by Mayne & Kulhawy (1982) because of over-consolidated:

For fine grained, normally consolidated soil , suggested by Massrsch (1979),

For over-consolidated clay:

Type of soil Ko

Loose sand 0.45 – 0.6

Dense sand 0.3 – 0.5

NC clay 0.5 – 0.7

OC clay 1.0 – 4.0

Compacted clay 0.7 – 2.0

Range of values for Ko

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Typical value for K0

Total Force/unit length of wall =Po

If z H₁ ….

z > H₁ …… ,,,,,, = sat - w

So the total LEP:

The force/unit length;

At rest Lateral earth

pressures

Let’s try examples given…

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A retaining structure is supporting a 6m high

excavation as shown in figure 1. The wall is very

rigid so that the soil behind the wall is in “at rest”

condition.

A) Sketch lateral earth pressure diagram

B) Determine the lateral earth pressure at rest Po

C) Determine the hydrostatic force Pw

Let’s try examples given… BAA3513 SEMI2013/14

DAA3513 GEOTECHNICAL ENGINEERING

Figure 1 (example)

Fill sand

c’ = 0

d = 18 kN/m3

c’ = 0

sat = 20 kN/m3

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Active lateral earth pressure

Happen if the wall moves away from the soil

The soil tend to experienced shear failure which resulted a sliding soil wedge that move forward and downward

The earth pressure exerted on the wall at this state of failure is known as active earth pressure, Pa

There are several method that can be use to calculate the active / passive lateral earth pressure

i) Rankine’s approach

ii) Coulomb’s approach

iii) Culmann’s method

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Passive pressure

Happen if the wall moves toward the soil

The soil tend to experienced shear failure

which resulted a sliding soil wedge that

move backward and upward

The earth pressure exerted on the wall at

this state of failure is known as passive

earth pressure, Pp

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coefficient of passive earth pressure, Kp

Pp = ½ KpH2

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Rankine active pressure

The effective pressure on the vertical plane is rankine’s active earth pressure,,, if we derive it in term of , z, c, from above figure, we found: For cohesionless soil: The ratio of is called coeff. of rankine’s active earth pressure = Ka

Rankine passive pressure

Generalized case for Rankine active/passive pressure

The inclined back fill (granular soil)

Generalized case for Rankine active/passive pressure

The inclined back fill (granular soil)

Generalized case for Rankine active pressure

The pressure work in direction :

So, active force (Pa),

The location & direction of resultant Pa is shown

For wall vertical backface = 0, as shown

Generalized case for Rankine passive pressure

For wall vertical backface = 0, as shown

The inclination of as shown on

So, the Passive Force:

The location &direction of Pa along with the failure wedge is shown in Fig.13.11b,

for wall with vertical backface = 0,

For THE CASE of = = , Kp(R) shown

Lateral Rankine passive pressure distribution

Backfill – cohesionless

soil with hor. Surface

ACTIVE CASE (Fig. 13.13a),

PASSIVE CASE (Fig. 13.13b)

Backfill–Partially Submerged

Soil Supporting a Surcharge,

ACTIVE CASE

Backfill–Partially Submerged

Soil Supporting a Surcharge,

PASSIVE CASE (Fig. 13.15),

Backfill – cohesive soil with hor. Backfill ACTIVE CASE (Fig. 13.16), For =0 condition

Backfill – cohesive soil with hor. Backfill ACTIVE CASE (Fig. 13.16), (cont) For common practical, we have take the tensile crack into account, so For =0 condition,

Backfill – cohesive soil with hor. Backfill PASSIVE CASE (Fig. 13.17), For =0 condition, Kp = 1,

For the retaining wall shown in Fig. 13.19a, determine the force /unit length of the wall for rankine’s active state, and find the location of the resultant.

For c = 0,,,

So for layer 1 and 2,

At z = 0, ₀ = 0

z = 3 (bottom of upper layer), ₀ = 3*16 = 48 kN/m²

= 1/3*48 = 16 kN/m²

At z = 3 ( in the lower layer) ₀ = 3*16 = 48 kN/m² and

At z = 6, ₀ = 3*16 +3(18-9.81) = 72.57 kN/m² and

The variation of a with z ; LEP due to pore water:

at, z = 0, u = 0

z = 3, u = 0

z = 6, u = 3*9.81 = 29.43 kN/m² , see Fig 13.19c

Pa = ½*3*16 + 3*13 + ½ *3*36.1

= 24 + 39 + 54.15

= 117.15 kN/m²,

Take M bottom of the wall,

Pa z = Pa1*z1 +Pa2*Z2 + Pa3*Z3

z = (Pa1*z1 +Pa2*Z2 + Pa3*Z3)/ Pa

= 24*(3+3/3) + 39*(3/2) + 54.15*(3/3)}/117.15

= 1.78 m

Rankine’s Pressure for c and soil –inclined back fill

Mazindrani & Ganjali (1977):

Active pressure =

a = zKa(R) = zKa(R) Cos

Ka(R) = Rankine’s Active E P Coefficient

zKa(R) = KaR) / Cos

Passive pressure =

p = zKp(R) = zKp(R) Cos

Kp(R) = Rankine’s Passtive E P Coefficient

Kp(R) = Kp(R) / Cos

Ka(R), Kp(R) =iation

See Tables 13.4 & 13.5 for variation Ka(R), Kp(R)

Rankine’s theory

Assume no frictions between wall and soil

Coefficient of active earth pressure, Ka

245tan2 o

coscoscos

coscoscoscos

Reminder..some

book use

instead of

*** Normally (i) and (ii)

use for horizontal

backfill while (iii) use

for sloping backfill ii)

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Lateral soil pressure,

a = KaH

Pa = ½ KaH2

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William J.M. Rankine

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Work example

Calculate the resultant active thrust on

a vertical smooth retaining wall of

height 5.4m. The water table is well

below the base of the wall.

Soil properties :

= 30, c = 0, = 20kN/m3

Let’s try this example…

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Solution….

Calculate Ka using appropriate equation.

Calculate the lateral soil pressure, a = KaH.

Find the resultant active thrust, Pa = ½ KaH2

Determine the point of action at a height of

H/3 above the base .

245tan2 o

BAA3513 SEMI2013/14

Work example

Sebuah tembok penahan tegak setinggi 4m

menyokong tanah tak jeleket. Cerun

permukaan kambus balik ialah 12. Tentukan

magnitud tekanan aktif semeter larian ke

atas tembok dengan menggunakan teori

Rankine.

Diberi : = 25, = 17.8kN/m3

Let’s try this too…

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Solution….

Calculate Ka

Calculate Pa = ½ KaH2

coscoscos

coscoscoscos

BAA3513 SEMI2013/14

Coefficient of passive earth pressure, Kp

*** Normally (i) and (ii)

use for horizontal

backfill while (iii) use

for sloping backfill ii)

245tan2 o

coscoscos

coscoscoscos

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Lateral soil pressure, p = KpH

Pp = ½ KpH2

Redo the previous work example and let’s

try this example No 2 and 3…

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For multi layer

For cohesive soil

Lateral soil pressure, may increase (for passive) or decrease (for

active). The value may be determine using the following equation, z :

paKc /2

If ground water table encounter in soil

Lateral soil pressure, will increase. The value may be determine

using the following equation :

If extra surcharge imposed on top of the soil

Lateral soil pressure, will increase and its

depend to the surcharge value.

Also applicable if imposed by UDL and point

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Let’s try more examples…

Don’t worry…

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COULOMB'S EARTH PRESSURE THEORY FOR SAND

FOR ACTIVE STATE

Coulomb made the following assumptions in the development of his theory: 1. The soil is isotropic and homogeneous 2. The rupture surface is a plane surface 3. The failure wedge is a rigid body 4. The pressure surface is a plane surface 5. There is wall friction on the pressure surface 6. Failure is two-dimensional and 7. The soil is cohesionless

FOR ACTIVE STATE

Consider the figure 1. AB is the pressure face 2. The backfill surface BE is a plane inclined at an angle with the horizontal 3. is the angle made by the pressure face AB with the horizontal 4. H is the height of the wall 5. AC is the assumed rupture plane surface, and 6. is the angle made by the surface AC with the horizontal

FOR ACTIVE STATE

Area of wedge ABC = A = 1/2 AC x BD where BD is drawn perpendicular to AC. From the law of sines, we have

FOR ACTIVE STATE

The various forces that are acting on the wedge are shown. As the pressure face AB moves away from the backfill, there will be sliding of the soil mass along the wall from B towards A. The direction of the shear stress is in the direction from A towards B. If Pn is the total normal reaction of the soil pressure acting on face AB, the resultant of Pn and the shearing stress is the active pressure Pa making an angle with the normal. Since the shearing stress acts upwards, the resulting Pa dips below the normal. The angle for this condition is considered positive.

FOR ACTIVE STATE

The weight of the wedge ABC is W. If Wn is the normal component of the weight of wedge W on plane AC, the resultant of the normal Wn and the shearing stress is the reaction R. This makes an angle with the normal since the rupture takes place within the soil itself. Statical equilibrium requires that the three forces Pa, W, and R meet at a point. Thus, a polygon of forces can be established

FOR ACTIVE STATE

From the polygon,

FOR ACTIVE STATE

the only variable is and all the other terms for a given case are constants. Substituting for W, we have The maximum value of Pa may be written as

FOR ACTIVE STATE

COULOMB'S EARTH PRESSURE THEORY

FOR ACTIVE STATE

where KA is the active earth pressure coefficient. The total normal component Pn of the earth pressure on the back of the wall

FOR ACTIVE STATE

FOR PASSIVE STATE

and thus

FOR PASSIVE STATE

Also from Rankine’s passive pressure coefficient

Coulomb’s theory

Assume that failure occurs in the form of a wedge

Assume frictions occurs between wall and soil

Apply for cohesionless soil

coefficient of active earth pressure, Ka

Pa = ½ KaH2

BAA3513 SEMI2013/14

FOR ACTIVE STATE

Coulomb made the following assumptions in the development of his theory: 1. The soil is isotropic and homogeneous 2. The rupture surface is a plane surface 3. The failure wedge is a rigid body 4. The pressure surface is a plane surface 5. There is wall friction on the pressure surface 6. Failure is two-dimensional and 7. The soil is cohesionless

FOR ACTIVE STATE

Consider the figure 1. AB is the pressure face 2. The backfill surface BE is a plane inclined at an angle with the horizontal 3. is the angle made by the pressure face AB with the horizontal 4. H is the height of the wall 5. AC is the assumed rupture plane surface, and 6. is the angle made by the surface AC with the horizontal

FOR ACTIVE STATE

Area of wedge ABC = A = 1/2 AC x BD where BD is drawn perpendicular to AC. From the law of sines, we have

FOR ACTIVE STATE

The various forces that are acting on the wedge are shown. As the pressure face AB moves away from the backfill, there will be sliding of the soil mass along the wall from B towards A. The direction of the shear stress is in the direction from A towards B. If Pn is the total normal reaction of the soil pressure acting on face AB, the resultant of Pn and the shearing stress is the active pressure Pa making an angle with the normal. Since the shearing stress acts upwards, the resulting Pa dips below the normal. The angle for this condition is considered positive.

FOR ACTIVE STATE

The weight of the wedge ABC is W. If Wn is the normal component of the weight of wedge W on plane AC, the resultant of the normal Wn and the shearing stress is the reaction R. This makes an angle with the normal since the rupture takes place within the soil itself. Statical equilibrium requires that the three forces Pa, W, and R meet at a point. Thus, a polygon of forces can be established

FOR ACTIVE STATE

From the polygon,

FOR ACTIVE STATE

the only variable is and all the other terms for a given case are constants. Substituting for W, we have The maximum value of Pa may be written as

FOR ACTIVE STATE

COULOMB'S EARTH PRESSURE THEORY

FOR ACTIVE STATE

where KA is the active earth pressure coefficient. The total normal component Pn of the earth pressure on the back of the wall

FOR ACTIVE STATE

FOR PASSIVE STATE

and thus

FOR PASSIVE STATE

Also from Rankine’s passive pressure coefficient

Work example

Sebuah tembok penahan tegak setinggi 4m

menyokong tanah tak jelekit. Cerun

permukaan kambus balik ialah, = 12.

Tentukan magnitud tekanan aktif dan pasif

semeter larian (kN/m) ke atas tembok

dengan menggunakan teori Coulomb.

Diberi : = 25, = 17.8kN/m3, = 15

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Solution

Calculate Ka

Calculate Pa = ½ KaH2

(ans : 62.562kN/m)

Redo the solution steps for the passive pressure

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CULLMAN LATERAL EARTH PRESSURE

Culmann's (1875) method is the same as the trial wedge method. In Culmann's method, the force polygons are constructed directly on the -line AE taking AE as the load line. The procedure is as follows: AB is the retaining wall drawn to a suitable scale. The various steps in the construction of the pressure locus are: 1. Draw -line AE at an angle to the horizontal. 2. Lay off on AE distances, AV, A1, A2, A3, etc. to a suitable scale to represent the weights of wedges ABV, A51, AS2, AS3, etc. respectively.

3. Draw lines parallel to AD from points V, 1, 2, 3 to intersect assumed rupture lines AV, Al, A2, A3 at points V", I',2', 3', etc. respectively. 4. Join points V, 1', 2' 3' etc. by a smooth curve which is the pressure locus. 5. Select point C‘ on the pressure locus such that the tangent to the curve at this point is parallel to the -line AE. 6. Draw C‘ parallel to the pressure line AD. The magnitude of C‘ in its natural units gives the active pressure Pa. 7. Join AC" and produce to meet the surface of the backfill at C. AC is the rupture line. For the plane backfill surface, the point of application of Pa is at a height of H/3 from the base of the wall.

Braced cut

Sometimes earth cuts are retained by braced

sheeting rather that the rigid walls which

considered previously

Sheeting were commonly made of wood or

Sheeting are normally driven vertically

Usage : to retain earth temporarily during a

construction project BAA3513 SEMI2013/14

Strut : a horizontal brace providing lateral support to resist earth pressure behind the sheeting

Wale : a continuous horizontal (longitudinal) member extending along a sheeting’s face to provide intermediate sheeting support between strut

Braced cut cannot ordinarily be analyzed by ordinary theories (refer text book for further explanation)

However, the theory will not be discussed further

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DON’T FORGET TO DO YOUR EXERCISES !!!!

THE END

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