lateral earth pres

8/10/2019 Lateral Earth Pres

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- 1 -L iterature Review

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CHAPTER- 2

Lateral Earth Pressures1. Introduction

Lateral earth pressure is the pressure that is exerted by a soil mass on a retaining

structure or a wall. The problem associated with lateral earth pressures and retaining wall

stability is one of the most common in the civil engineering field, and a segment of soil

mechanics that has been receiving widespread attention from engineers for a long time.

Design of structures such as retaining walls, sheet piling, and braced sheeting of pits and

trenches, bulkheads, bridge abutments, basement or pit walls etc., requires a quantitative

estimate of lateral earth pressure. In this chapter we will learn how to determine the

magnitude and direction of lateral earth pressures for which a structure must be designed. In

last part of the chapter the intro to design of a gravity retaining wall will be taken up. By

definition a retaining wall is a structure used for maintaining ground surface at different

elevations on its either side. The soil supported by the retaining wall is called backfill.

In this chapter we will be concentrating on lateral earth pressures experienced by a retraining

wall. To understand the concept behind these earth pressures lets examine the sequence of

retaining wall construction for a road construction project in a hilly area. The sequence is

shown below:

Ideally a vertical cut should be provided to economize on the cost of excavation but soils are

seldom stable at vertical slope and tend to slough down to attain equilibrium. Generally they tend to

stabilize at a gentle slope. The stable slope angle depends on the soil properties and the height of

slope. A gentle cut may not be feasible due to high cost of excavation or due to existence of some

structures which may be pulled down. Under such circumstances soil is cut at a steep slope and is

supported by constructing a retaining wall. A reasonable slope is cut during dry part of the year (step2)

and retaining wall is constructed. The space behind the wall is later filled with soil and proper drainage

is provided to protect it from storm water erosion. The retaining wall supports the backfill and its

thickness at every horizon must be adequate to resist lateral earth pressure so that the wall neither tilts

excessively nor is displaced horizontally away from the soil mass it is supporting. The approach to

design of these walls is as under:-

a. Determine different forces which contribute to wall stabilityb. Determine forces which are trying to destabilize the wall


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c. Compare these forces to ensure that the structure is safe.

Two basic types of soil pressures are evaluated in this chapter, active and passive. If the

soil mass pushes against a retaining wall such as to push it away, the soil becomes the

actuating element and the pressure resulting thereby is known as active pressure. On the

other hand, if the wall pushes against the soil, the resulting pressure is known as passive

pressure. In this case the actuating element is the retaining wall itself.

Analysis of a wide range of earth-retaining structures is primarily based on either ofthe two theories formulated by such persons as Coulomb in 1776 and Rankine in 1857.Furthermore, although some research data and experience indicate that assumptions relatedto pressure distributions on retaining walls, or on the failure surface of backfills, are not quitethose depicted by these early investigators, substantial evidence exists that the analysis anddesign efforts based on their theories give acceptable results for most cases of cohesion-lesstype backfills. The results are significantly less dependable for the more cohesive soils.2. Common Uses of Retaining Walls


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3. Factors Affecting Lateral Earth Pressure

The main factors that affect lateral earth pressure include wall movement, wall friction,

water in the backfill, compaction of the backfill, slope of backfill, seasonal variation and the

nature of the backfill, the interaction between the soil and the retaining structure at the

interface and the imposed loading (e.g., height of backfill, surcharge loads). These are

discussed briefly in the following sections.

a. Wall Moment.Under service conditions a rigid retaining wall possibly can rotate

about its base, its top, its middle or another point: it can translate or there can be

a combination of these modes. It is also possible that wall may not yield at all.

Terzaghi (1934) is believed to be the first to conclude from his large scale earth

pressure tests that the lateral earth pressure distribution behind retaining wall is

associated, with the type and the magnitude of wall movement. Earth pressure

distribution behind a wall rotating about its base is considered as linear (Terzaghi

1934).

b. Wall Friction

A relative movement between a retaining wall and its back fill develops shear

forces between the face of the wall and the backfill. These are termed wall

friction (Lambe and Whitman 1969). The wall friction depends on roughness of

the wall, amount and direction of the wall movements, soil properties of the

backfill and inclination of ground surface behind the wall. The wall friction is also

influenced by the settlement of the structure and the backfill soil (Grandil 1987).

The maximum value of wall friction may not occur simultaneously with the

maximum shearing resistance and the value of wall friction may vary across the

wall (Bowles 1982). Wall friction angle () between 1/3 to 2/3 is a good

approximation for computing lateral earth pressure using Coulombs theory. The

wall friction affects the magnitude and the direction of earth pressure.

c. Water in the Backfill

The presence of water in the wall backfill without compensating water in front

substantially increases total pressure because the coefficient K for water is unity.

The distribution of pressure on a retaining wall due to water is hydrostatic and the

resultant water force acts at one third the water height from wall base. The

effective earth force acts above one third of wall height due to the presence of

water in the backfill. The combined force due to soil and water acts below the

one-third of wall height.

d. Seasonal Variation

Seasonal Variations affect the magnitude and the distribution of earth pressure

mainly because of changes in temperature, freezing of backfill and heavy rainfall.

Field test results show about 30% variation of earth pressure due to seasonal

variations (Terzaghi and peck 1967). There is variation in earth pressure in

summer and winter (Duncan et al, 1990).The warm front face of a wall in the

summer expands relative to the back face and causes the wall to deflect back

against the fill. The result is an increase in earth pressure. In the winter the


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pressure is reversed, the earth pressure also increases in winter due to

expansion of water in the backfill voids due to freezing.

e. Nature of Backfill Soil

The engineering properties of the backfill soil such as(c, , )cohesion, internal

friction angle and unit weight have substantial effect on the magnitude of lateral

earth pressure on retaining structures. Higher the cohesion and internal

friction angle, the lower will be the pressure. On the other hand, higher the

Unit weights, higher will the pressure on the wall.

f. Slope of Backfill

The backfill slope changes the direction of lateral earth pressure. Thus, the

magnitude of horizontal earth force acting on the wall changes with the increase

or decrease in slope.

4. Lateral Earth Pressures at Rest, Active and Passive States

Let there be a mass of soil of density at

depth z, vertical and horizontal stresses

can be given as, v= *z ,

h= k* *z = k* v ,where k= h/ v

k = Coefficient of earth pressure.

Statically value of k is indeterminate. The notation of kFor certain special cases are given below:-

ko= Coefficient of earth pressure at rest which corresponds to zero strain.

ka= Coefficient of active earth pressure which corresponds to shear

failure by expansion of soil.

Kp= Coefficient of passive earth pressure which corresponds to shear

failure by compression of soil.

Figure 13.2 above shows some of the forces acting on a typical gravity retaining wall.

Frictional forces that may be developed on the front and back faces of the retaining wall are

not shown. The lateral force induced by the backfill pushes against the wall with a resultant

pressure Pa. In turn, the retaining wall resists the lateral force of the backfill, thereby

preventing its movement. In this case, it is apparent that the soil becomes the actuating force.

The thrust Pa is the resultant of the active pressure, or simply the active thrust. The

resistance to the active thrust is provided by the frictional force at the bottom of the

wall and by the soil in front of the wall . For the sake of illustration, assume that the wall was

pushed to the left by the active thrust Pa.In this case the soil in front of the wall at its toe

provides a passive resistance to movement. This resistance is known as the passive earth

pressure, with the resultant of this pressure denoted by Pp. The magnitude of the lateral

force varies considerably as the wall undergoes lateral movement resulting in either tilting or

lateral translation, or both. Terzaghi focused on this phenomenon during his classic

experiments in 1929 -1934.

The concept of active and passive earth pressure can be best illustrated by reviewing the

experiments conducted at MIT. In these experiments a glass box was used. A retaining wall

was placed as shown below and steel bars, 6 inch long and inch diameter, were placed on

1

3 244

1z


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either side of the wall up to its top.

Since the top level of bars was the same initially the wall stood vertically i.e., every thing was in

equilibrium, forces from left side of wall were balancing forces from right hand side. Later

some bars from the left side were removed. Since there was no force to balance out forces

from the right hand side, the wall tilted slightly to the left resulting in a decrease in pressure on

the wall. As more rods were removed from the left the pressure on the wall reduced further

and it swayed away from the rods on the right side. Finally as the depth of rod removal on left

side reached a certain magnitude the bars on the right moved one past the other and a zone of

shear failure was formed. A clear cut surface of shear failure could be seen. The wall

experienced the minimum horizontal pressure at this instance. As more rods were removed,

there was no further reduction in pressure on the wall. A similar phenomenon occurs in soils.

Letsconsider soil strata under geostatic state of stress. Initially on any element the horizontal

stresses on left and right side of the element are the same to keep it in equilibrium. Letspass

an imaginary retaining wall of zero volume through this soil. Now we start removing soil from

left of this imaginary wall just like the example above. As above the pressure on the wall will

decrease as the wall tilts to the left due to removal of support. As more soil is removed the tilt

increases and finally the soil behind the wall fails in shear and a zone of shear failure is formed

right behind the wall. There is no further reduction in the lateral pressure experienced by the

wall. This minimum lateral earth pressure experienced by the wall as it moves away from the

soil mass it is supporting is called ACTIVE earth pressure. By definition active earth

pressure is the minimum lateral earth pressure experienced by the wall as

it moves away from the soil mass to which it is supporting.


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Figure 12.2 above depicts the relationship betweenthe earth pressure and the walI movement.

Porepresents the magnitude of pressure when no movement of the retaining wall takes

place; it is commonly referred to as earth pressure at rest.As the wall moves toward the

backfill, the pressure increases, reaching a maximum value of Ppat point C. On the other

hand, if the wall moves away from the backfill, the force on the wall decreases, reaching a

minimum value of Paat point B.

Figures 12.3a and 12.3b show the active case, where the wall moves away from the backfill,

together with the corresponding force polygon. Similarly, the case of passive resistance,

together with its force polygon, is shown in Figs. 12.3c and 12.3d.


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5. Active and Passive Earth Pressures for (c- ) Soils with Horizontal Backfill

a. Active Earth Pressure(c- , soils). Letsconsider a case for wall with horizontal

backfill of c- , soilsas above. Here we assume that backfill is pushing the wall

away from fill. So as we start excavation, the wall yields and lateral stresses 3

will keep decreasing (shown in dotted). Finally the lateral pressure reduce to a

level that the Mohrs circle touches strength envelope. This is limiting conditionsat which pressure can not reduce any further or its Mohrs circle cross strength

envelope, which is impossible. The stresses so induced are plotted on Mohrs

Circle and followed the well known Mohrs/Coulumbs shear strength equation

(=C+ ntan )


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From small triangle, A = C/tan , B = (1+A)/2 and R = (1- A)/2 ------------ (i)

But from fig R = (A+B) sin --------------------------------------------------------------(ii)

Comparing eqn (i) & (ii) we get (1- A)/2 = (A+B) sin --------------------------(iii)By putting values of A & B in eqn (iii) we get,

(1- A)/2 = ((C/tan + (1+A)/2)* sin(1- A)/2 = C/tan *sin+1sin/2+ Asin/2(1- A) = 2C cos + 1sin+ AsinA(1+ sin) = 1(1- sin) - 2C cos A= 1(1- sin)/ (1+ sin) - 2C cos /(1+ sin)PA = 1*ka- 2C kawhere, 1= *z and ka= (1- sin)/ (1+ sin)or Ka= tan

2(45-/2)b. Passive Earth Pressure(c- , soils). Letsconsider a case for wall with horizontal

backfill of c- , soils. Here we assume that backfill is being pushed by the walltowards fill. Therefore lateral earth pressure 3 will keep increasing due topushing of backfill by wall with passive pressure p and thus failure occur due tocompression of soil. The stresses so induced are plotted on Mohrs Circle andfollowed the well known Mohrs/Coulumbs shear strength equation

( =C+ ntan )

From small triangle, A = C/tan , B = (1+p)/2 and R = (p- 1)/2 ------------ (i)But from fig R = (A+B) sin --------------------------------------------------------------(ii)Comparing eqn (i) & (ii) we get (p- 1)/2 = (A+B) sin --------------------------(iii)By putting values of A & B in eqn (iii) we get,(p- 1)/2 = ((C/tan + (1+p)/2)* sin(p- 1)/2 = C/tan * sin+ 1sin/2+ psin/2(p- 1)= 2C cos + 1sin+ psinp(1- sin) = 1(1+ sin) + 2C cos p= 1(1+ sin)/ (1- sin) +2C cos /(1- sin)Pp = 1*kp+ 2C kpwhere, 1= *z and kp= (1+ sin)/ (1- sin)or Kp= tan

2(45+/2)

Alternately

CA

(1+

A)/2

R=(1- A)/2

=c+ ntan

v

h

h

BA

n1

CA

(1+p)/2C/tan

R=(p- 1)/2

=c+ ntan

3 p

BA


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Lets starts with simple case of horizontal ground level and zero wall friction.

Consider an element located at depth Z initially as wall moves awaydecreases and finally the soil fails. The state of stress for soil mass as it moves away from thebackfill from static equilibrium to plastic equilibrium is shown below:

For the circle touching the strength envelope, h is minimum and is equal to Pa.

From ABC

Here

Therefore

1, (Active earth pressure)

& -------2, (Passive earth pressure)

Equation no one is called Bells Equation for active earth pressure for

cohesive soils.

From equation- 1 at ground surface z = 0

at the base of wall

Stress diagram is shown below:

vh K 0 h

2

2

31

31

CotC

Sin

hav P 31 &

22

avav PPCotCSin

Sin

CosC

Sin

SinP

CosCSinSinP

PSinPSinCosC

PSinPSinSin

CosSinC

a

va

avav

avav

1

2

1

1

211

2

2

aava KCKP 2

pvpp KCKP 2

a

aaa

KC

KCZKP

2

02

aaa KCHKP 2

Zv

h = K v (C - )

A C

O

B

C

= K0h v

2

av P

= Pa = Ka h v

v

C/tan

aKC2A

2Z0 Z0

+

H

60.6kn/m

-


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At certain depth zo, Pa= 0

As indicated in the figure up to a depth the soil is under tension. Since soil

cannot sustain tension a tension crack is generally formed at top to a depth of . If the

crack is filled with water then the pressure of this water on wall must be given due

consideration.

From fig it is clear that up to a depth of 2Z0the net pressure on the wall is zero i.e. a cohesive

soil can stand without any lateral support up to a depth,

Therefore, the critical Hcan unsupported vertical cut in clay is given as: 2Z0= Hc

For computation of lateral stress in case of cohesive soils, theve pressure are neglected &

whole +ve pressure below depth z is taken into consideration.

6. Unsupported cuts in (c -) soil

Unsupported excavations would theoretically be possible in (c-) soils if the lateral

pressure (3for the active case) would not exceed the strength of the soil. The general

expression forthe horizontal stress for the active is given as:

(d)

At ground surface, h = 0. Thus,

3= -2cKa (Tension)

This implies the formation of a crack as depicted in Fig. 12.15a given below. The

corresponding pressure distribution based on Eq. (d) is shown in Fig. 12.15 b.

The theoretical depth of the crack htcan be determined by recognizing that, at the

bottom of the crack, 3 = 0 or Pa=0. Thus, from Eq. (d),

0 = htKa-2cka or ht = 2c /ka (12-17)

The theoretical maximum depth Hc of unsupported excavation may be calculated as the

02 aa KCZK

a

a

a

KCZ

K

KCZ

2

02

0

0

aK

CZ

20

aK

C

2

aK

CZ

42 0

a

cK

CH

4

aaa KCHKP 2


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point where the tension forces equal the cohesive strength. Hence, from

Figure 12.15 b, Hc= 2 ht.

Though the above equation provides a theoretical depth to which an excavation may be

made without lateral support, it should be used cautiously. Surface moisture that may enter the

crack may induce hydrostatic stresses or may decrease the shear strength of the soil. Hence,

the unsupported excavation to such depths should be for short duration at best. Even then,judgment reflecting on the potential consequences from unsupported excavation is indeed

warranted.

In general it is advisable to minimize the use of cohesive backfill when ever possible.

With changes in moisture content the pressure induced by highly cohesive soils may change

significantly. If the clay dries out and shrinks then the pressure on the wall will reduce. On the

other hand a dry soil may take up moisture and swell thus exerting tremendous pressure on

the wall.

a. At Rest Earth Pressure. Let us consider the mass of soil shown in Figure 2.1. The

mass of soil bounded by a frictionless wall AB that extends to infinite depth. An element

located at a depth z will be subjected to a vertical pressure vand horizontal pressure h. For

the case considered here, vand hare effective and total pressures and there are no shear

stresses on the vertical and horizontal planes.

If the wallAB is static-that is, if it does not move either to the right to or the left of its initialposition-the soil mass will be in a state of elastic equilibrium. The ratio of the horizontal stressto the vertical stress is the coefficient of earth pressure at rest, Ko, or Ko= h/v-------------(2.1)Since v= *z, so, h= Ko(*z) -----------------------------------------(2.2)

For granular soils, the coefficient of earth pressure at rest can be represented by the

empirical relation (Jaky, 1944), Ko= 1- Sin ------ (2.3)Brooker and Ireland (1965) recommended the following equation for Ko in normallyconsolidated clays, Ko= 0.95- Sin -------------- (2.4)

Also(1) For -soils

(a) For horizontal backfill, ko= (1- sin )(b) For sloping backfill, ko= (1- sin )(1+ sin)=ko(1+ sin)

(2) For c-soils (Brooker & Ireland)(a) For 0


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b. Earth Pressure at Rest for Partially Submerged Soil

Figure 2.3a shows a wall of height H. The ground water table is located at a depth H1below the ground surface. For z H1 the lateral earth pressure at rest can be given ash= Ko (*z). The variation of h with depth is shown by the triangle ACE in Figure-2.3 a.However, for z H1 (Below GWT)

The pressure on the wall is found from the effective stress and pore water pressurecomponents in the following manner:

Effective vertical pressure = /v= H1+ '(z - H1) ------------------- (2.9)

where ' = sat- w= the effective unit weight of soilSo, the effective lateral pressure at rest is/h= Ko

/h= Ko {(H1+ '(z - H1)} -------------------------------------(2.10)

The variation of /hwith depth is shown by CEGB in Figure 2.3 a.Again, the lateral pressure due to pore water is, u = w(z-H1)--- (2.11)

The variation of u with depth is shown in fig 2.3 bHence, the total lateral earth pressure at any depth z H1is equal toh=

/h+u = Ko {(H1+ '(z - H1)}+ w(z - H1) -----------------------(2.12)

The force per unit width of the wall can be found from the sum of areas of the pressure

diagrams in Figures 9.3a and b and is equal to

Po= {(1/2*KoH21)+( KoH1H2)+ ( Ko

/ + w)H22} ----------------(2.13)

(Area ACE) (Area CEFB) (Area EFG&IJK)c. General Comments on Earth Pressure Coefficient at Rest

While designing a wall that may be subjected to lateral earth pressure at rest, care must

be taken in evaluating the value of Ko. Sherif, Fang, Sherif (1984), based on their laboratory

tests, have shown that Jaky's eqn for Ko eqn (2.3), gives good results when the backfill is

H2

KoH1

(a)

GWT

GF

E

B

A

z

Unit weight of soil =

H

Fig-2.3

C

Ko(H1+ /H2)

KoH1

H1

Saturated Unit

weight of soil = sat

KJ

I

wH2

(b)

Ko(H1+/H2)+wH2

H2

H1

=

60.


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loose. However, for a dense sand backfill, eqn. (2.3) grossly underestimates lateral earth

pressure at rest. This is due to the process of compacting backfill. For that reason, they have

recommended the following design relationship.

Ko= (1-sin )+( d/ d(min)-1)*5.5 ------------------------------------(2.14)

where d= actual compacted dry unit weight of the sand behind the wall

d(min)= dry unit weight of the sand in the loosest stateExample:

Data: As shown on fig

Req: Calculate fol:

ko, ka, kp

PO,Pa, Pp

location of pressures.

Solution:

a. Values of k

ko= 1- sin = 1- sin 35o= 0.43

ka= tan2(45-/2) = tan2(45-35o/2) = 0.27

kp= tan2(45+/2)= tan2(45+35o/2)= 3.69

b. Po= * Ko H2= *0.43*125*30

2= 24187 lb/ft

2

Loc of Po= H/3=30/3= 10 ft from base of wall.

c. Pa= * Ka H2= *0.27*125*30

2= 15187 lb/ft

2

Loc of Pa= H/3=30/3= 10 ft from base of wall.

d. Pp= * Kp (H1)2= *3.69*120*10

2= 22140 lb/ft

2

Loc of Pp= H1/3=10/3= 3.33 ft from base of wall.

Example:

Data: As shown on fig

Req: Calculate active force per unit of wall

for Rankins state. Also find location of resultant.

Solution

1. Calculation of Ka

a. For top layer Ka1

Ka1= 1-sin30o/1+sin30o= 0.33

b. For Bottom layer Ka2

Ka2= 1-sin35o/1+sin35

o= 0.27

H

PO

KoH

H

Pa

KaH

Pp

H1

KpH1

= 120 pcf= 35o

H1=10 ft

= 125 pcf= 35o

H = 30 ft

3m

3m= 16kn/m= 30oc=0

= 18kn/m= 35oc=0

3


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2. Calculation of Pa

a. For top layer Pa1

at z=0, v=0 but at z=3m, v1= 16*3=48kn/m2

So a1=Ka1*v1= 16*3* 0.33= 16 kn/m2

and Pa1= *3*16= 24 kn/m2

b. For Bottom layer Pa2

at z=3, v=3*16 =48kn/m2 (just below top layer)

and a2=Ka2* = 0.27*48= 13 kn/m2

at z=6m, /v2= v1+ /v2= 48+3*(18-9.81)=72.6kn/m

2

/a2= Ka2* /v2= 0.27*72.6= 19.7 kn/m

2

c. Pressure due to water-uAt z=0, u=0

At z=3m, u=0At z=6m, u= 3*9.81=29.43kn/m2

d. Total active pressure,

Pa=(*16*3)+13*3+ 19.7-13(3*1/2)+*29.43*3

= 24+39+10+44.14 =117.15 kn/m2

e. Location of pressure By taking moment at the base of wall.

117.15*z= 24(3+3/3)+39*3/2+10*3/3+44.15*3/3=208.15

z= 208.15/117.15= 1.78 m

Example:

An unsupported cut as shown in fig with the given

soil properties, find

a. Stress at top and bottom of cutb. Max depth of potential tension crackc. Max un-sported excavation.

Solution As we know that major & minor principal stresses are decided on the basis of their

magnitudes. Here, a is greater in magnitude, so considering it as major principal stress,hence, stresses would be calculated basing ona.

a. Stresses at Top and BottomStresses at point A (Top)- As h = 0, so a= h Ka- 2ckaa= -2cka= (2*25) tan (45-5

0) = -41.95 kn/m

2

Stresses at point B (Bottom)- As h = 4.2 m,

a= h Ka- 2cka= 18.2*4.2 tan2(45-50)2*25 tan (45-50) = 11.87 kn/m2

b. Max depth of Potential Tension Crack

ht= 2c /ka = (2*25) /18.2tan2(45-50)= 3.27 mc. Max Un sported ExcavationAs we know that Hc= 2 ht,

so Hc= 2 *3.27 = 6.54 m7. Rankines Theory

Coulombs and Rankine are the best-known theories which are used to compute the

magnitude and direction of forces which act on the retaining wall due to the backfill behind it.

These theories are frequently referred to as the classical earth pressure theories. The theory

proposed by Rankine in 1857 is based on the following assumptions:

Frictional forces between backfill and retaining wall are zero. ( the implication of this

assumption is that shear stresses on the vertical and horizontal planes are zero i.e.,

these planes are principal planes)

The back of the wall is vertical,

The surface of the backfill is horizontal.

19.7kn/m2

3

13kn/m

3

16kn/m

29.43kn/m

3

A

B

= 18.2

kn/m3= 10o

H = 4.2 m


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The backfill is a homogeneous, isotropic, dry and cohesion-less material.

The failure surface is plane.

The wall yields about the base and satisfies conditions of plastic equilibrium.

The basic concept behind Rankines theory can be depicted by Mohrs circle. Consider the

element shown in Fig.12.3, subjected to the geostatic stresses shown. The value for 1/could

be approximated as the product of the average unit weight times depth, namely, 1/= h. If

the wall were to move to the left, thereby creating a case of active stress, the value for 1/

would become the major principal stress and at failure the minor principal stress h/ is equal

to the active earth pressure. The corresponding Mohr circle for this case is depicted by circle-I

in Fig. 5. On the other hand, if the wall were to push against the backfill i.e., move towards the

backfill, a case of passive pressure would be developed. The vertical stress would then

become the minor principal stress

3/ and the lateral stress would thus become the major

principal stress which at failure is the passive earth pressure. The Mohr circle for this condition

is depicted by circle- 2 in Fig. 5.

The pressures on retaining wall vary linearly with depth, as indicated by Fig. 12.7.

The corresponding resultant pressures, active and passive, can be calculated for a unit length

of retaining wall as

Pa = 0.5 H ka= 0.5 * H2tan

2(45 - /2)

Pp = 0.5 H kp= 0.5 * H2tan 2(45 + /2)

The corresponding slip planes for the active and passive cases are shown in Figure

12.5c and 12.5d. The or ientat ion of these planes can be best being und erstood u sing

the conc ept of po le. For active case point 3represents pole. A line drawn from this point

to the point representing state of stress on failure plane yields the orientation of failure planes

for the active state. It is to be pointed out that once the lateral stresses in this case drop to the

active value, shear failure along an infinite number of planes behind the wall is imminent and at

the instant the soil is said to be in a state of limit equilibrium. Since the state of stress in backfill

is known for either at rest condition or the active condition, retaining walls are designed for the

limit equilibrium conditions and such design is called limit design.

8. Effects of Surcharge loads

Concentrated or uniformly distributed loads, commonly referred to as surcharge loads, acting

on the surface of the backfill, will:

o Increase the lateral pressure against the, retaining wall.

o Move the point of application of the resultant pressure upward.

a. Effect of UDL Surcharge

Figure 12.17 shows uniformly distributed surcharge q (kN/m2) acting on the surface of the


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backfill. The vertical stress 1on an element within the soil backfill at a depth h is equal to

h + q. Correspondingly, the lateral stress 3for a (c-)soil, is

99.. RRaannkkiinnAAnnaallyyssiissBBaacckkffiillllwwiitthhUUnniiffoorrmmSSuurrcchhaarrggee

If the backfill is horizontal & carries a uniform surcharge q-kn/m2, the vertical stress at

any depth will increase by q. The increase in lateral pressure for active case therefore will be

kaq as shown below:

Example

Find the passive resistance (Pp) of given frictionless

Retaining wall as shown in fig. Also calc the location

of passive resultant.Solution

a. Calculation of K

Kp= 1+sin250/1-sin250= 2.45

2. Calculation of Stresses

a. Passive Resistance/Stresses

at top of layer, z=0, v=q= 12 kn/m2(surcharge)

p= 2.45*12+2*102.45= 60.6 kn/m2

at bottom of layer, z=5m, v=q+h = 12+15*5 =87kn/m2

p= 2.45*87+2*102.45= 244.35 kn/m2

b. Passive Resistance

Pp= 60.6*5+244.3*5*1/2= 303+610.9 = 914 kn/m

c. Location of Resultant. Taking moment of pressure diagram about wall base.

914*z = 303*5/2 + 611*5/3= 757.5+1018.3 = 1775.83 = z=1775.83/914 =1.94m

b. Effect of Point load Surcharge (by Boussinesqs equation)

To determine lateral pressure on wall due to point load

aaa KCHKP 2

z=nH

x=mH

Q

H

q kN/m2

aa KCHK 2

244.3kn/m2

60.6kn/m2

5 m

= 15 kn/m

= 25oc=10 kn/m2

q=12kn/m2

914kn/m


17/36


17

Q placed at surface of the backfill as shown in fig.

If the load Q is placed on the plane of section as shown;

We can substitute y=0 and =0.5 in vertical stress distr

equation developed by Boussinesq. This is modified as

x =Q/2(3x2*z/R

5) -------------(A)

where R= (x2+z2)1/2

Substituting x=mH & z=nH in eqn (A), we get

x =3Q/2H2{m

2n/(m

2+n

2)5/2

}-------------- (B)

The horizontal stresses expressed by eqn (B), does not include restraining effect of wall.

Therefore the eqn has been modified as per real time situation as under:-

x =1.77Q/H2{m

2n

2/(m

2+n

2)3} for m> 0.4-------------- (C)

x =0.28Q/H2{n

2/(0.16+n

2)3} for m 0.4-------------- (D)

For concentrated surcharge loads Q such as may be induced by a continuous footing,

railroad tracks, and the like, running parallel to the wall, it is possible, although rather

laborious, to estimate the increased stresses on the wall based on Boussinesqs equation,

consistent with the theory of elasticity for a semi-infinite homogeneous soil mass. However,

graphical methods given below are more expedient for this purpose.

Experimental data indicate that Boussinesqs formula for lateral stress gives acceptable

results where the wall movement is compatible to soil deformations within the backfill. On the

other hand, if the retaining wall is totally rigid such that the soil deformation is greatly restricted

by the rigid boundary, the horizontal stress approaches a value twice that given by

Boussinesqs equation.This effect becomes less noticeable as the distance of Q relative to

the wall increases.

c. Effect of Point load Surcharge (by graphical method)

The line of action of the active thrust induced by the concentrated surcharge loads Qis

commonly based on empirical procedures. Such a procedure gives acceptable results for

cohesion less backfill; it may be illustrated with the aid of Fig 12.19. When the concentrated

surcharge load is located to the left of point C, the active thrust Pa may be determined by

drawing lines EDand FDparallel to lines AGand AC,respectively. The point of application

of Pais one-third the distance FEfrom point F, as indicated in Fig. 12.19a. But when load Q

is located to the right of the failure plane, as indicated in Fig. 12.19b, line EDis parallel to line

AC. The point of application of Pais one-third the distance EAfrom point F, as shown in

Fig. 12.19b. We noted that the line of action of the resultant thrust moves up the wall as the

load Qapproaches the wall. Furthermore, the lateral thrust as well as the overturning effect

decreases as the load Qmoves away from the wall.


18/36


18

ExampleFind the lateral force Pa due to backfilland

point load 1.5m away from back face of wall as shown in fig.

Solution (for point load)

Here x=1.5m, H= 5m, z= 0-5m (for n= 0.2,0.4,0.6,0.8,1.0)

Divide H in five equal parts for calc pressure per unit of wall.

m= x/H = 1.5/5=0.3 i.e less than 0.4

Therefore we will use

x =0.28Q/H2{n

2/(0.16+n

2)3} for m 0.4

x =0.28*150/52{0.2

2/(0.16+0.2

2)

3} = 1.68*5= 8.4kn (for n= 0.2)

x =0.28*150/52{0.4

2/(0.16+0.4

2)

3} = 1.68*4.88= 8.2kn (for n= 0.4)

x =0.28*150/52{0.6

2/(0.16+0.6

2)

3} = 1.68*2.56= 4.3kn (for n= 0.6)

x =0.28*150/52{0.8

2/(0.16+0.8

2)

3} = 1.68*1.25= 2.1kn (for n= 0.8)

x =0.28*150/52{1

2/(0.16+1

2)3} = 1.68*0.41= 0.69kn (for n= 1.0)

Now lateral force P due to point load will be

P= (0+8.4)*1+ (8.4+8.2)*1+ (8.2+4.3)*1+ (4.3+2.1)*1 + (2.1+0.69)*1= 23.35kn/m of wall ht2 2 2 2 2

For geostatic stresses ka= tan2(45-/2)= 0.626

At top lateral stresses, h= 2*12.50.626=2*12.5*0.7= 19.75kn/m2

At bottom of wall, h= 18*5*0.626-2*12.5*0.79= 36.6 kn/m2

Pressure on wall= 19.75*5+16.85*52/2= 98.75+210.6= 309.35kn/m of wall ht

For resultant Pressure location taking moments at base of wall

Z= (98.75*5/2)+(210.6*1.67)/309.6=246.87+351.7=598.57/309.6=1.93m from base.

d. Effect of line load Surcharge (by Boussinesqs equation)

To determine lateral pressure on wall due to line load

q placed at surface of the backfill as shown in fig.

If the load q is placed on the plane of section as shown;

Eqn (D) is modified as

x =4q/H {m2n/(m

2+n

2)

2} for m> 0.4-------------- (E)

x =0.203q/H {n/(0.16+n2)2} for m 0.4-------------- (F)

ExampleA line load of 50kn/m is located at a distance of 3m from back

face of retaining wall 4.5m high. Determine lateral force P on wall due to line load.

Solution

Here x=3m, H= 6m, z= 0-6m (for n= 0.25,0.5,0.75,1.0)

Divide H in four equal parts of 1.5 m for calc pressure on wall.

m= x/H = 3/6=0.5 i.e more than 0.4

Therefore we will use

x =4q/ H {m2n/(m

2+n

2)2} for m> 0.4

x =4*50/*6 {0.520.25/(0.52+0.252)2} = 10.61*0.64= 6.8kn (for n= 0.25)

x =4*50/*6 {0.520.5/(0.5

2+0.5

2)2} = 10.61*0.5= 5.3kn (for n= 0.5)

x =4*50/*6 {0.520.75/(0.5

2+0.75

2)2} = 10.61*0.28= 3.0kn (for n= 0.75)

z=nH

x=3m

q= 50kn/m

H=6m

z=nH

x=mH

Q=150Kn

H

=18kn/mc=12.5kn/m2

=150

z=nH

x=mH

q= kn/m

H


19/36


19

x =4*50/*6 {0.521.0/(0.5

2+1

2)2} = 10.61*0.16= 1.7kn (for n= 1.0)

Now lateral force P due to point load will be

P= (0+6.8)*1.5+ (6.8+5.3)*1.5+ (5.3+3)*1.5+ (3+1.7)*1.5 = 24kn/m of wall width

2 2 2 2

e. Effect of strip load Surcharge (by Boussinesqs equation)

To determine lateral pressure on wall due to strip load

q/area placed at surface of the backfill as shown in fig.

If the load q is placed at distance m1 from wall back of

of height H with width as m2.The horizontal stress

at depth z on wall due to strip loading is given as

x =q/H(-sin*cos 2), angles & as defined on fig-------------- (G)

for actual soil behavior conditions eqn (G) can be modified as

x =2q/H*{(-sin*cos 2)}, -------------- (H)

The force P per unit width of wall can be obtained by integ eqn (H)For z & H this is given as

P= q/90{H(21)}, where is in deg and given as

1 = tan-1

(m1/H) & 2 = tan-1

((m1+m2)/H)

ExampleA strip loading as shown in fig; m1= 10ft, m2= 5ft

Calculate xat 2/interval of H.

Also determine lateral force P on wall due to strip load.

Solution

Calculation of xby using eqn (H)

x =2q/H*{(-sin*cos 2)}

Calculation of x

z m1 m2 = tan-

(m1+ m2/2)/z = {tan-

(m1+ m2)/z}-{ tan-

(m1)/z} x=2q/z*{(-sin*cos 2)}2 10 5 80.91

3.71

5657lb/ft

4 10 5 72.25

6.9

5225lb/ft

6 10 5 64.36

9.2

4629 lb/ft

8 10 5 57.38

10.60 4000 lb/ft

10 10 5 51.34

11.31

3402lb/ft12 10 5 46.17

11.5

2877 lb/ft

Calculation of PUsing eqn given above as, P= q/90(21) and1 = tan

-1(m1/H) & 2 = tan

-1(m1+m2/H)

1 = tan-1

(m1/H)= tan-1

(10/12)= 39.80

2 = tan-1

((m1+m2)/H))= tan-1

((10+5)/12)= 51.340

Now, P= q/90{H(21)}= 1500/90{12(51.34-39.8)}= 2308 lb/ft of wall width

10. Cohesion less Backfill and Inclined Surface

Let us now consider cohesion less mass with a sloping surface behind a smooth vertical

retaining wall. Assume this condition to be depicted by Fig. 12.8a. The lateral stresses acting

on the vertical faces of the element (i.e., the faces parallel to the wall) are parallel to the

inclined surface. Thus, any such planes experience not only normal but also shear stresses.

These planes, therefore, are no longer principal planes as was the case for horizontal ground

surface.

For the following analysis Rankine made an additional assumption i.e., the vertical stress

acting on any plane and the lateral stress acting on the element are conjugate stresses .

This assumption implies that the lateral earth pressure is parallel to the surface of the backfill

slope. The corresponding resultant pressure on the wall could be determined with the aid of

q= 1500psf

H = 12ft

x

m2m1

q/area

HP


20/36


20

Mohrs circle. Figure l2.8c symbolizes an active state of stress. The magnitude of the vertical

stress is depicted by the distance OC; the lateral stress, acting parallel to the sloped surface, is

represented by the distance OA. Hence, from Fig. 12.8c we have:


21/36


21


22/36


22

Equation 12-9 is Rankines expression for the active lateral pressure at depth H.

Equation 12-10 is Rankines expression for the passive case.

For a given sloped surface and uniform soil properties Kabecomes a constant. Thus,

the intensity of load, or stress, varies linearly with depth. Hence, as before, the total resultant

active force may be given by Eq. 12-9. Again, we noted that the direction of the resultant is

parallel to the sloped surface. For the case of level surface, Eqs. 12-9 and 12-10 reduce to

Eqs. 12-5 and 12-6, respectively.

11. Rankine State for Sloping Backfill (In short)

The coefficient of active earth pressure will be:

Ka= Cos{(Cos- Cos2- Cos2)/ (Cos+ Cos2- Cos2)}

Kp= Cos{(Cos+ Cos2- Cos2)/ (Cos- Cos2- Cos2)}

H


23/36


23

12. Rankin Analysis, R/W with submerged Backfill

At A

At B

13. Active Earth Pressure Partially submerged Backfill

0v

Hsatv

HHKK

HH

wsata

van

wsatv

.baseatHHK

HHKP

wa

wsataHa

A

B

C

Dry

Sat

H1

H2

A

B

H


24/36


24

Assuming that '' is same for dry & saturated granular fill:

Lets assume that Moist > Saturated

Therefore Ka sat> Kamoist.

The shape of pressure diagram for such a case will be

14. Active Earth Pressure against Inclined Back Retaining Wall

Consider a retaining wall with inclined back as above. Rankines analysis can be extended to

solve such problem by computing active earth pressure against a vertical plane C D. Let F 1

be the force due to active earth pressure. The weight & its location of wedge BCD, W can be

determined.

The resultant of F1&W gives the required force R. Similarly if the backfill is inclined

then the active force against plane 'CD' is lot and let it be F1. The resultant of F1& weight W is

R which is required force againstthe wall as shown below.

F1 = Ka H12

15. Coulombs Analysis of lateral Earth Pressures (inclined back & inclined fill)

.

''

''

0''

221

21

21

221

21

1

1

baseatHHKHKP

HKHK

HH

HHH

HHCAt

HKP

HBAt

AAt

wadaha

adah

dv

wsatdv

satdv

daa

dv

v

awamaH

amah

m

wsatmv

satmv

maa

maamv

av

PHHKHKHKHK

HH

HHH

HHCAt

HKPerfaceAt

HKPHBAt

PAAt

22212

2212

21

221

21

12

111

''

int

00

A

B

C

Ka1 mH1 H1

Ka2 mH1

H2

F1

CBA

D

H

W H/3

F1

H/3

R

CBA

D

H


25/36


25

In 1776 Coulomb introduced an expression for determining the active thrust on retaining walls.

Coulomb knew about effect of both cohesion and friction on shear strength of soil. He

observed a number of retaining wall failures and he concluded that when ever a retaining wall

fails a certain mass of soil breaks off in the form a wedge and exerts pressure on wall thus

concluding that the retaining wall shall be designed for the MAXIMUM pressure exerted by this

mass of soil. Although he observed that the failure surface is not plane but for simplicity ofanalysis he assumed that the failure surface is plane and the soil is cohesion less. He did this

in order to simplify somewhat the mathematically complex problem introduced when cohesion

and non-planer sliding surfaces are considered. He did, however, account for the effects of

frictional interaction between the soil backfill and the face of the retaining wall. Due to relative

motion between soil and the wall shear forces develop at interface. In active zone the soil

tends to move down along wall face as the wall moves away from the backfill. This downward

motion imparts a downward drag force on the wall. The magnitude of this shear force depends

on the friction angle between the wall & the soil wand is called angle of wall friction.

Typically for concrete. w = 2/3

In case of passive earth pressure the soil moves up as it is pushed by the wall & in the process

it exerts an upward shear force on the wall. The direction of this frictional force depends on the

direction of motion of soil mass and shall be evaluated carefully.

16. Coulombs Assumptions

a. Backfill is dry, cohesion less isotropic, homogeneous and elastically un-

deformable but breakable.

b. The slip surface is plane which passes through the heel of the wall.

c. The sliding wedge itself acts as a rigid body and the value of earth pressure is

obtained by considering the limiting equilibrium of sliding wedge as a whole.

d. The position and direction of resultant earth pressure are known. The resultant

force acts on the back of the wall at 1/3 rdthe height of the wall form the base and

is inclined at (angle of wall friction) to the normal to the back)

e. Back of the wall is rough and relative movement of the wall and soil takes place

which develops frictional forces that influence the direction of resultant pressure.

According to Coulombs theory, the thrust is induced by the sliding wedge, as shown in Fig.

12.l0a. For this reason, it is sometimes referred to as the sliding wedge analysis. The

corresponding force polygon is shown in Fig. 12.l0b. The development of Coulombs equation

follows from this basic relationship. From Fig. 12.l0b, using the Law of Sines, we have


26/36


26

Explanation of Force Polygon Angles

W = Wt of wedge

F = Resultant of Normal and Shear forces on the plane.

Pa= Active Pressure.

H = Ht of wall.

= Inclination of back face of wall

= Angle of force Pa with normal, friction angle between soil and wall.

= Internal friction angle, or angle of friction with normal.

= Angle of failure plan with horizontal.

i = Slope of backfill with horizontal


27/36


27


28/36


28


29/36


29

Example: A retaining wall as shown in Fig, below with soil properties. Find Paand Pp.

Note: Solve examples 9.1-9.8 Braja M. Das (Principles of Geotechnical Engineering)

17. Coloumbs Solution for R/W with Inclined Back and Inclined Back fill

Consider a RW as shown below the geometry of problem has changed but will we have

two unknown R& Fa, and one known quantity W the weight of wedge as shown.

The vector diagram is shown above.

By applying sine lawto force diagram.

Fa

CBA

D

H

i

W

w

Normal to

inclined surface

R

- i

-

- w

180-( - ) ( - w)

W

R

Fa


30/36


30

Also from sine law:

Putting values of eqn- B in eqn- A, we get

To determine maximum value of Fawe differentiate above equation with respect to and

equate it to zero. The resulting equation for maximum Fais,

18. Culmanns Method

The following graphical procedure was devised by Karl Culmann in (1875). It is used to

determine the magnitude and the location of the resultant earth pressures, both active and

passive, on retaining walls. This method is applicable with acceptable accuracy to cases where

the backfill surface is level or sloped, regular or irregular and where the backfill material is

uniform or stratified. Also, it considers such variables as wall friction, cohesion less soils, and

with some procedural modifications, cohesive soils and surcharge loads, both concentrated

and uniformly distributed. It however, requires that the angle of internal friction of the soil be a

constant for the total backfill. The procedure presented here is limited to cohesion-less soils.

)(180

180

180

&

12

1

180

180

iSinSin

HAD

SinABAD

AB

ADSinAlso

Sin

HAB

AB

HSin

BCADW

ABCFrom

ASin

SinWF

F

Sin

W

Sin

w

a

a

w

)(

)(

)(

2

1

)()(

)(

)(

)(

)()(

2

2 BiSinSin

iSinSinHW

iiiSin

iSin

Sin

HBC

iSin

iSinABBC

iSin

BC

iSin

AB

w

aSin

Sin

iSinSin

iSinSinHF

180)(

)(

2

12

2

2

12

22

)(2

1

iSinSin

iSinSinSinSin

SinHF

w

ww

Maxa


31/36


31

Reference is made to Fig. 12.20 in describing the procedure for determining the active

pressure for a case of cohesion less soil by Cullmans method:-

a. Select a convenient scale to show a representative configuration of retaining wall

and backfill. This should include height and slope of the retaining wall, surface

configuration of the backfill, location and magnitude of concentrated (line)

surcharge loads, uniformly distributed surcharge, and so on.

b. From pointA draw lineAC, which makes an angle of with the horizontal.

c. Draw lineAD at an angle of from lineAC. Figure 12.20 shows the angle to

be the angle between the vertical and the resultant active pressure.

d. Draw raysAB1, AB2, AB3, and so on, that is, assumed failure surfaces.

e. Determine the weight of each wedge, accounting for variations, if the backfill is a

layered system, for variable moisture content, and so on.

f. Select a convenient scale and plot these weights along lineAC. For the distance

fromA to W1 along lineAC equals W1 similarly, from theW1 to W2 along lineAC

equals W2, and so on.

g. From each of the points located on lineAC, draw lines parallel to line AD, to

intersect the corresponding assumed failure surfaces; that is, from W1 will

intersect lineAB1, then from W2 will intersect line AB2and so on.

h. Connect these points of intersection with a smooth line, Culmanns curve.

i. Parallel to lineAC; draw a tangent to Culmanns curve. In Fig. 12.20 point Erepresents a tangent point. More than one tangent is possible if Culmann line isirregular.

j. From the point of tangency, draw line EF parallel to lineAD. The magnitude of

EF, based on the selected scale, represents the active pressure Pa. Severaltangents to the curve are possible; the largest of such values becomes the value

of Pa. The failure surface passes through E andA, as Fig. 12.20.

Surcharge loads and their respective effects on the location of the resultant are accounted

for as described in the previous section. Examples 12.5 and 12.6 further enhance this

explanation.


32/36


32

Figure 12.21 illustrates the procedure for determining the passive resistancevia culmanns

method.

The approach is similar to that for the active pressure, with some notable differences:

(1) LineAC makesan angle of ' ' degrees below rather than above the horizontal;

(2) The reference line makes an angle of with line AC, with measured magnitude as

indicated in Fig. 12.21.

For the assumed sliding wedges, the weights W1, W2, and so on, are plotted along line

AC. From these points, lines are drawn parallel to the reference line to intersect the

corresponding rays, as shown in Fig. 12.21. The Culmann line represents a smooth curve

connecting such points of intersection. A tangent to the Culmann curve parallel to line AC is

drawn, with the resultant earth pressure being the scaled value of line FE, as shown in fig.

Example: A retaining wall with backfill as shown in fig-12-22, with the given data as:-

Ht of wall= 7 m, =30o, line load = 100 kn/m, =18.2 kn/m3, =900

Find the active thrust by Culmans method.

For convenient calculation of wedge weight; the bases for all the wedges are taken same.

Hence, the weight of all the wedges equals 127.4 kN, as shown in Fig. 12.23. The

corresponding points alongAC are shown in Fig. 12.22 for an arbitrary scale of 1 cm = 100 kN.


33/36


33

From these points lines are drawn parallel to lineAD so as to intersect raysAB1, AB2, AB3, and

so on. Note that a similar line is drawn for the line load by connecting these points of

intersection with a smooth curve (Culman curve) and drawing a tangent to this curve parallel to

lineAC, we obtain the Pa, which is equal to the corresponding scaled value FE. The scale for

FE = 1.85 cm for 185 kN. The point where 'Pa' acts is determined as described in the

preceding section and as shown in Fig. 12.24. Line EG is parallel to lineAC, and line GE isparallel to the failure plane. Patherefore acts at one-third distance FE from point E, a total of

4.25 m above pointA.

19. Design of Gravity Retaining Walls

Retaining walls are structures used to provide stability for earth or other material whereconditions disallow the mass to assume its natural slope, and are commonly used to hold back

or support soil banks, coal or ore piles and water.

Retaining walls are classified, based on the method of achieving stability, into six

principal types.


34/36


34

a. Gravity Retaining Walls. The gravity wall depends upon its weight, as the name

implies, for stability. The tensile stresses are avoided by proper proportioning of

the wall.

b. Cantilever Walls. The cantilever wall is a reinforced-concrete wall that utilizescantilever action to retain the mass behind the wall from assuming a natural slope.

Stability of this wall is partially achieved from the weight of soil on the portion of the

base slab.

c. Counter fort Retaining Walls. A counter fort retaining wall is similar to a

cantilever retaining wall, except that it is used where the cantilever is long or for

very high pressure behind the wall and has counter forts, which tie the wall and

breast together, built at intervals along the wall to reduce the bending moments

and shears. As indicated in Fig, the counter fort is behind the wall and subjectedto tensile forces.

d. Buttressed Wall. A buttressed retaining wall is in compression instead of tension

as shown in figure.

e. Crib Walls. These are built-up members of pieces of pre-cast concrete, metal or

timber and are supported by anchor pieces embedded in the soil for stability.

f. Bridge Abutments. These are retaining with wing wall extensions to retain the

approach fill and provide protection against erosion.

Keys

A roach Slabs


35/36


35

20. Gravity Retaining Walls

a. Common Proportions

Retaining wall design proceeds with the selection of tentative dimension, which

are then analyzed for stability and structural requirements and are revised as

required. Since this is trial process, several solutions to the problem may be

obtained all of which are satisfactory. A computer solution greatly simplifies the

work in retaining wall design and provides the only practical means to optimize

the design.

Gravity-wall dimensions may be taken as shown in figure below. Gravity walls,

generally, are trapezoidal-shaped but also may be built with broken backs. The

base and other dimensions should be such that the resultant falls within the

middle one-third of the base. The top width of the stem should be on the order of

0.30 m. If the heel projection is only 100 to 150 mm, the Coulmb equation may

be used for evaluating the lateral earth pressure, with the surface of sliding taken

along the back face of the wall. The Rankine solution may also be used on or

along the back face of the wall. The Rankine solution may also be used on a

section taken through the heel. Because of the massive proportions and resulting

low concrete stresses, low-strength concrete can generally be used for the wall

construction.

A critical section for analysis of tensile flexure stress will occur through the junction ofthe toe portion at the front face of the wall.

b. Stability of Wall. Gravity retaining walls must provide adequate stability againstsliding. The soil in front of the wall provides a passive-earth pressure resistanceas the wall tends to slide into it. If the soil is excavated or eroded after the wall isbuilt, the passive-pressure component is not available and sliding instability mayoccur. If there is certainty of no loss of toe soil, the designer may use the passivepressure in this zone as part of the sliding resistance. Additional sliding stabilitymay be derived from the use of a key beneath the base.

c. Design of Gravity Retaining Wall. A gravity retaining wall is a structure whichresists lateral earth pressure by its weight.

A retaining wall is considered safe, if it is safe against:

(1) Overturning (2) Sliding (3) Bearing failure

The proportioning should be such that no tension should develop anywhere in wall.

0.5 to 0.7 H

H

Minimum

batter

1:48

D to D

D to D

H/B to H/b

0.30 m to H/12

Slope change to

reduce concrete

(b)(a)

Approach Fill

PaPv

a

d

w


36/36

36

Consider a retaining wall as shown in the figure. Let R be the reaction from base actingat distance x from toe of the wall O. The total active force can be split into its horizontal andvertical components Ph and Pv respectively. The passive earth pressure Pp is ignored since it issupporting and less. The wall will be safe against overturning if the resultant passes thoughmiddle third of the base. Taking moment of all forces about O

Rv * x = W*a + Pv*d- Ph*b

x = (W*a + Pv * d - Ph * b) / Rv

Where Rv = W + Pv

The eccentricity of R from the centre of base is given as:

e = B/2 - xWe must ensure that e < B/6 so that there is no overturning.

a. FOS against overturning.

b. FOS against sliding, where = friction angle between wall

base and soil.

c. FOS against bearing capacity

The maximum stress should not exceed bearing capacity of soil.

5.1

Re

bP

dPW

gMomentOverturnin

entsistingMomFOS

h

va

5.1tan

h

v

P

RFOS

B

e

B

Rf

B

e

B

Rf

v

v

61

61

min

max

lateral earth pres

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