Chapter 5.Earth Dam

8/10/2019 Chapter 5.Earth Dam.

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CHAPTER 5

EARTH DAMS

5.1 INTRODUCTION

Earth dams for the storage of water for irrigation have been built since the earliest

times. These dams were however, limited in height but not necessarily in extent. Earth dams

are now being built to unprecedented heights. Sites which have hitherto been considered unfit

for the construction of darns are now being exploited. Development of soil mechanics, study of

behavior of earth dams, and the development of better construction techniques have all been

helpful in creating confidence to build higher dams with improved designs and more ingenious

details. The result is that the highest dam in the world today is an earth dam. The highest

earth/rockfill dams in the world are Roguni U.S.S.R ( 335 m ) Nurek, U.S.S.R. ( 300 m );

(Fig. 5.1) Mica, Tehri India ( 260 m ) Canada ( 244 m ) and Oroville, U.S.A. ( 235 m ).

Fig. 5.1 Nurek dam U.S.S.R.

In spite of these developments it is difficult to establish mathematical solutions to the

problems of design, and many of its components are still guided by experience or judgment.

For a realistic design of an earth dam it. is necessary that the foundation conditions and

materials of construction are thoroughly investigated. It is also necessary that controlled

methods of construction are used to achieve necessary degree of compaction at predetermined

moisture.

The discussion in the text is limited to design procedure for earth dams which are rolled

fill type of construction. This type of construction is now being used almost entirely for the

construction of earth dams to the exclusion of hydraulic and semi-hydraulic fills. In this type,

the major portion of the embankment is constructed in successive mechanically compacted

layers of 150 mm to 220 mm thickness.


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5.2 Foundation for Earth Dams

The essential requirements of a foundation for an. earth dam are (i) that it provides

stable support for the embankment under all conditions of saturation and loading, and (ii) thatit provides sufficient resistance to seepage to prevent piping and excessive loss of water.

In general foundations may be grouped into three main classes according to their

predominant characteristics.

1. Rock foundations,

2. Pervious foundations, and

3. Impervious foundations.

Rock Foundations

These foundations, including shale generally do not present any problem of bearingstrength. The principal considerations are erosive leakage, excessive loss of water through

joints, fissures, crevices, permeable strata and along fault planes, etc. Grouting is usually done

to treat this type of foundation.

Shale may however present foundation problems specially if they contain joints, faults,

seams filled with soft material and weak layers. Such defects and excess pore water pressure

may control the overall strength of foundation.

FOUNDATION FOR EARTH DAMS

Pervious Foundations

Often the foundations for earth dams consist of recent alluvial deposits composed of

relatively pervious sand and gravels overlying impervious geological formations like rock or

clay. There are two basic problems with which these types of foundations are associated viz. (i)

excessive amount of under seepage, and (ii) piping and boils caused by forces exerted due to

seepage. The treatment which may be provided to control these problems is governed by the

thickness of pervious strata viz, whether the pervious foundation extend to a moderate depth or

to an infinite depth.Loose fine sand or coarse silt deposits in a foundation present one of the most difficult

problems. The difficulty arises not only due to low strength or high compressibility of the loose

sand, but also through a phenomenon known as liquefaction. A certain fine uniform sand in a

loose condition when subjected to sudden applications of shock (as in earth quake) loses all its

shear strength and behaves as though it were a heavy viscous fluid. This phenomenon is

exhibited by uniform sands which are very fine and consists of rounded grains and their

relative density is less than 50%.

Fortunately no failures of this type have occurred. At Obra dam in U.P., the foundations

were thick loose sands and its susceptibility to liquefaction was carefully evaluated by fieldand laboratory investigations of the foundation material and by actual blasting tests. It was

concluded that the foundations were not likely to liquefy.


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Impervious Foundations

Foundations of silt and clay extending to large depths are sufficiently impermeable to

preclude the necessity of providing treatment for under seepage and piping. The main problem

with these, foundations may be excessive pore water pressure and significant deformations.Where the embankments are constructed on foundations consisting of brittle, highly

plastic or over-consolidated clays, serious investigations are required as their presence may

cause excessive deformations. The embankment design in such cases would be controlled by

likely strains in the foundations. If there is silt and clay to large depths, then there is not

much necessity of providing treatment for under seepage and piping. The main problem with

these foundations is of stability for which generally the slopes of the embankments are made

flatteror berms on either side are provided.

If the structure crosses swampy or similar area where the foundation material will be of

plastic nature, the matter would require serious investigations as plastic clays are verydeficient in shear strength.

An approximate method to determine the safety of foundation material against the shear

stress is as follows :

(i) Determine the total horizontal shear under the slope ofdam by the formula,

S =

2

45tan2

hh 122

2

2

1 (5.1)

where, h1 = vertical distance from top of dam down to therigid boundarysuch as rock, the strength of which is great as compared to the

overlying material.

h2 = vertical distance from base of dam to the rigidboundary.

= effective weight per cubic metre of material.1 = equivalent angle of internal friction given by,

tan 1 =1

1

h

tanhC

+ (5.2)

(ii) Calculate the average unit shear by the formula

Sa = S/b

where Sa = average horizontal foundation shear per sq. m.

b = horizontal distance along base from top shoulder

of slope to the toe of dam.

Fig. 5.2 Location of maximum Shear-Definition sketch


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130(iii) The maximum unit shear may be obtained by multiplying the average

unit stress by 1.4 i.e. Smax=1.4 Sa. The location of the point of maximum shear may be taken

0.4 b from upper shoulder of the slope as shown in fig. 5.2.

5.3 Causes of Failures of Earth Dams

Like most other damages to engineering structures, earth dam failures are caused by

improper design frequently based on insufficient investigations, and lack of care in

construction and maintenance.

Failures of earth darns may be grouped into the following basic causes

(a) Hydraulic failures

(b) Seepage failures and

(c) Structural failures

Hydraulic Failures

They account for about one third of the failure of dams and are produced by surface

erosion of the dam by water. They include wash-outs from overtopping (Fig. l-5-3a), wave

erosion of upstream face, scour from the discharge of the spillway etc. and erosion from

rainfall.

Seepage FailuresSeepage of water through the foundation or embankment has been responsible for more

than one third of earth dam failures. Seepage is inevitable in all earth dams and ordinarily it

does no harm. Uncontrolled seepage; may however, cause erosion within the embankment or

in the foundation which may lead to piping (Fig. 5.3 b). Piping is the progressive erosion

which develops through under the dam. It begins at a point of concentrated seepage where the

gradients are sufficiently high to produce erosive velocities. If forces resisting erosion i.e.

cohesion, inter-locking effect, weight of soil particles, action of downstream filter etc. are less

than those which tend to cause, the soil particles are washed away causing piping failure.

Seepage failures are generally caused by (a) pervious foundations, (b) leakage throughembankments, (c) conduit leakage and (d) sloughing.

Pervious Foundations

Presence of strata and lenses of sand or gravel of high permeability or cavities and

fissures in the foundation may permit concentrated flow of water from the reservoir causing

piping. Presence of buried channels under the seat of dam have also been responsible for this

type of failure.


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131

Leakage Through Embankrnents

The following are the common causes of embankment leaks which lead to piping:

(i) Poor construction control which includes insufficient compaction adjacent

to outlet conduits and poor bond between embankment and the foundationor between the successive layers of the embankment.

(ii) Cracking in the embankment or in the conduits caused by foundation

settlement (Fig. 5.3 c),

(iii) Animal burrows

(iv) Shrinkage and dry cracks (Fig. 5.3 d)

(v) Presence of roots, pockets of gravel or boulders in the embankment.

Condui t Leakage

Conduits through the dam have been responsible for nearly one third of the seepagefailure and more than one eighth of all failures. Failures are of two types (i) contact seepage

along the outside of the conduit which develops into piping and (ii) seepage through leaks in

the conduit which may also develop into piping.

Contact seepage along the conduit wall is caused either by a zone of poorly compacted

soil or small gap between the conduit and remainder of the embankment. Seepage through

poorly compacted zones soon develops into piping.

Conduit cracking is caused by differential settlement or by overloading from

embankment.

Sloughing

Failure due to sloughing takes place where downstream portion of the dam becomes

saturated either due to choking of filter toe drain, or due to the presence of highly pervious

layer in the body of the dam. The process begins when a small amount of material at the

downstream toe is eroded and produces a small slide. It leaves a relatively steep face which

becomes saturated by seepage from the reservoir and slumps again, forming a higher and more

unstable face. This process is continued until the remaining portion of the dam is too thin to

withstand the water pressure and complete failure occurs.

Structural Failur es

Structural failures of the embankment or its foundation account for about one fifth of

the total number of failures. Structural failures may result in slides in foundation or

embankment due to various causes as explained below.

Foundation Failur es(Fig. 5.3 e). Faults and seams of weathered rocks, shale, soft clay strata

are responsible for the foundation failure in which the top of the embankment cracks and

subsides and the lower slope moves outward and large mud waves are formed beyond the toe.

Another form of foundation failure occurs because of excessive pore water pressure inconfined seams of silt or sand. Pore water pressure in the confined cohesionless seams, artesian

pressure in the abutments or consolidation of clays interbedded with the sands or silt, reduces


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132the strength of the soil to the extent that it may not be able to resist the shear stresses

induced by the embankment. The movement develops very rapidly without warning. Excess

settlement of foundation may also cause cracking of the embankment (Fig.5.3 c).

Fig. 5.3 Types of failuresEarth dams

Slides in Embankment (Fig 5.3 f and g) An embankment is subjected to shear stresses

imposed by pool fluctuations, seepage or earthquake forces. Embankment slides may occur

when the slopes are too steep for the shear strength of the embankment material to resist the

stresses imposed.Usually the movement develops slowly and is preceded by cracks on the top or the

slope near the top. Failure of this type are usually due to faulty design and construction. In high

dams slope failure may occur during dissipation of pore pressure just after construction. The

upstream slope failure may occur due to sudden drawdown as shown in fig. 5.3 f. The

downstream slope is critical during steady seepage condition.

5.4 Design Criteria of Earth Dams

Based on the experience of failures, the following main design criteria may be laid

down for the safety of an earth dam.1. To prevent hydraulic failures the dam must be so designed that erosion of the

embankment is prevented.

This implies that the following conditions are satisfied.

(a) Spillway capacity is sufficient to pass the peak flow.

(b) Overtopping by wave action at maximum water level is prevented.

(c) The original height of structure is sufficient to maintain the minimum safe

freeboard after settlement has occurred.

(d) Erosion of the embankment due to wave action and surface run-off does not

occur.(e) The crest should be wide enough to withstand wave action and earthquake

shock.


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2. To prevent the seepage failures, the flow of water through the body of the dam

and its foundation must not be sufficiently large in quantity to defeat the purpose of the

structure nor at a pressure sufficiently high to cause piping.

This implies that

(a) Quantity of seepage water through the dam section and foundation should be limited.

(b)

The seepage line should be well within the downstream face of the dam to

prevent sloughing.

(c)

Seepage water through the dam or foundation should not remove any

particle or in other words cause piping. The driving force depends upon the

pressure gradient while the resisting force depends upon the strengthcharacteristics of the boundary material.

(d) There should not be any leakage of water from the upstream to downstream

face. Such leakage may occur through conduits, at joints between earth and

concrete sections or through holes made by aquatic animals.

3. To prevent structural failures, the embankment and its foundation must be stable under all

conditions.

This implies that(i) The upstream and downstream slopes of the embankment should be stable under all

loading conditions to which they may be subjected including earthquake.

(ii) The foundation shear stresses should be within the permissible limits of shear

strength of the material.

5.5 Prevention of Erosion-Embankment Details

Spillway Capacity

It must be calculated and fixed by relevant hydrological studies and flood routing such thatsufficient freeboard is available between the maximum flood level and top of the dam.

Freeboard for wave action

The required allowance for waves is based on the effect of wind of maximum velocity

blowing down the reservoir and setting up a wave splash on the dam face. Various empirical

formulae depending on wind velocity and reservoir fetch have been suggested for computing

wave heights. The MolitorStevenson formulas are normally used which are

hw = 0.032 4271.0763.0 FFV +

where F 32 km. (5.5)


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134where hw = wave height in meters measured between trough and crest.

V = wind velocity in km per hour

F = fetch in kilometers.

On a sloping surface the wave rides along the slope upto a vertical height of 1.5 times the waveheight above the reservoir level hence 1.5 hw is provided as freeboard.

According to U.S.B.R. criteria distinction is to be made between the normal and minimum

freeboards. Normal freeboard is defined as the difference in elevation between the crest of the

dam and normal reservoir water surface. According to the fetch of reservoir the freeboard may

be provided as given in Table 5.1.

Table 5.1

Recommended Values of Freeboards

Fetch in km Normal freeboard Minimum freeboard

in meter in meter

Less than 1.5 1.25 1.00

1.5 1.50 1.25

4.0 1.80 1.50

8.0 2.50 1.80

15.0 3.00 2.20

It is also recommended that freeboard shown in, Table 5.1 be increased by 50 percent

if a smooth pavement is provided as protection on the upstream slope.

Settlement allowance

Settlement of an embankment will be caused by consolidation in the foundation and in

the fill. A settlement allowance of 2% is considered adequate and is generally provided.

However, in case of dams of more than 30m height, an extra 1% allowance is provided to

account forthe settlement due to earthquake.

Upstream slope protection

Surface protection of upstream slope is meant to prevent the destructive wave action.

Usual type of surface protection for the upstream slope is stone rip-rap either dry dumped or

hand placed. When a thin layer is used, hand placed rip-rap may be more economical than

dumped rip-rap.

There are several empirical methods to find out the thickness of rip-rap. These methods

take into account the wave height, embankment slope, weight of average size rock and its

specific gravity.


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135

Hand placed rip-rap

The hand placed rip-rap for upstream protection is very common in this country. The

size of stones used for hand placed rip-rap may be determined with the following formula,

dm= 2.23 C. hw)2(

1.

2

++

sss

w

w

(5.6)

where dm = diameter of stone brought to form a ball, in metre, in the zone of maximum

blow of the wave.

w = unit weight of water in t/m3 = unit weight of stones in t/m3s = slope of embankment.

hw = height of wave in metre.

C = factor depending on the type of protection

For hand placed rip-rap C = 0.54

For rock fill or dumped rip-rap C = 0.80

Average size of stone required dav= dm/O.85 (average shape)

The average weight of the stone can be found out by the formula

Wav=

.)d(6

3

av (5.7)

F il ter below ri p-rap

A layer of filter material consisting of gravel Or crushed rock is always required under

rip-rap to prevent waves from eroding the underlying embankment material. No definite rule

can be given for the minimum necessary thickness of the filter layer. Most filters are

constructed with thickness ranging from 20 cm to 75 cm.

The following factors govern the thickness of the filter layer

(i) Wave action

The less the wave action, the less the need for a thick filter under the rip-rap.

(ii) Gradation of ri p-rap

If the rip-rap is well graded with plenty of quarry fines to fill the larger voids there is less

stress on the filter.

(iii) Plasticity and gradation of embankment materials

If the embankment material is well graded granular soil with a tough clay hinder it needs

less protection against erosion than if it is fine silty sand.


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136

(iv) The cost of fi lter

If the material for the filter is obtained without washing or screening as a pit run natural

gravel and quarry wash, and is not expensive as compared to the cost of average embankment

material, there is no reason not to use maximum filter thickness.The following limits are recommended to satisfy stability criteria for graded uniform filters:

(1)materialbaseofD

filterofD

15

15 = 5 to 40, provided that the filter does not contain more than 5

percent of material finer than 0.075 mm ( I.S. sieve no.-minus

75 micron ).

(2)baseofD

filterofD

85

15 = 5 or less

(3)drainpipeofopeningMax.

filterofD85 = 2 or more

(4) The grain size curve of filter should be roughly parallel to that of the base material.

D15 or D85 is the size at which 15 percent and 85 percent (respectively) of the total soil

particles are smaller. The percentage is by weight as determined by mechanical analysis.

If more than one filter layer is required; the same criterion is followed; the finer filter is

considered as the base material for selection of the coarser filter.

Horizontal filter layers can safely be made thinner than steeply inclined or vertical filters.

Minimum thickness of each horizontal layer is 15 cms for sand and 30 cms for gravel. If the

filter contains excessive fines or coarse material such thatbaseofD

filterofD

15

15 > 4 but < 5 or

baseofD

filterofD

15

15 > 5 but < 6, the thickness of filter layer may be increased by 50 percent.

Design Cr iteria The following criteria of upstream protection is being adopted by

Central Design Directorate of Irrigation Department, Uttar Pradesh.

(i) Upto 5 m height of bound No pitching(ii) Between 5 m and 10 m 0.25 m stone pitching over 0.15 m graded shingle or

spalls.

(iii) Above l0m and upto 15m 0.30 m stone pitching over 0.15 m graded shingle or

spalls.

(iv) Above 15 m and upto 25 m 0.5 m stone pitching over 0.25 m graded shingle or spalls.

(v) Above 25 m and upto 50m 0.5 m stone pitching over 0.3 m grade shingle or spalls.

(vi) Above 50 m and upto 75 0.75 m stone pitching over 0.5 m graded shingle or spalls.

(vii) Above 75 m 1.0 m stone pitching over 0.75 m graded shingle.The entire upstream face should be pitched right upto the top of bund.


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Downstream Slope Protection

The problem of erosion of downstream slopes due to surface runoff may be effectively

controlled by turfing. In area too deficient in rainfall during parts of the year to maintain a

proper cover, berms and other erosion control be applied.

Crest Width

The crest or top width of an earth dam should not be less than 4 m for maintenance

purposes. However, the width depends on several considerations such as (i) nature of emban-

kment material and minimum allowable percolation distance through the embankment at

normal reservoir water level (ii) height and importance of structure (iii) required width to

provide embankment mass for resistance to earthquake shock and (iv) roadway requirements.

Common practice is fairly well represented by the formula,

Bt= 5/3. H

5.6 Seepage Through Dams

Phreatic or Seepage line

The two dimensional flow of fluid through porous soil can be expressed by Laplaces

equation

02

2

2

2

=

+

Graphically, the equation can be represented by two sets of curves that intersect at rightangles. The combined representation of two sets of lines is called a flow net (Fig. 5.4). With

the help of a flow net, the seepage problems can be analyzed at any point within the section of

the embankment.

The seepage or phreatic line may be defined as the line within a dam section below

which there are positive hydrostatic pressures in the dam. On the line itself, the hydrostatic

pressure is zero. Above the line, there will be a zone of capillary situation. The phreatic line

represents the top flow line or the boundary condition for drawing the flow net.

Fig. 5.4 Flow net through earth dam

The location of the phreatic line is necessary in order to draw accurately the flow net. It

is also useful in analyzing stability of the dam. It may be noted that the location of the seepage

line is dependent only on the cross section of the dam. Its position is not influenced by the

permeability of thematerial composing the dam so long as thematerial is homogeneous.


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138According to A. Cassagrande the phreatic line for the homogeneous fill section is a

basic parabola except at the ingress and egress points. The presence of a pervious stratum

below the dam does not influence the position of the phreatic line. For graphical construction

of the phreatic line the following procedure may be adopted.

Stepwise procedure for locating phreatic line

1. With hori zontal drainage fi lter (F ig.5.5)

(i)

The horizontal distance between upstream toe A and the point B where water

surface meets the upstream face is calculated or measured (say L). The point B0is then located

on the water surface at a distance 0.3 L from B.

(ii) The basic parabola has to pass through B0and have its focus at F which is the starting

point of the horizontal

Fig. 5.5 Phreatic line with horizontal filter

drainage. With these points known the basic parabola may be constructed graphically.

(iii) With centre B0 and radius B0F, draw an arc to meet the water line at C. Draw the

vertical line CDwhich is the directrix. Let FD, the focal distance = yo. Bisect the distance FD

to get the point E, the vertex of the parabola. Draw FG parallel to CD and equal to yo.

Knowing Bo,Gand Ethe basic parabola can be drawn.

The focal distance y0 can also be determined on the consideration that if (x y) is one point

on the parabola,

=+ 22 yx x+yo (5.9)

Since the point B0 of coordinates d, h lies on the equation.

yo= dhd22 + (5.10)

(iv) The ingress portion of phreatic line is joined to the base parabola from point B,

keeping the starting end normal to the upstream face.

2. With incl ined discharge face (F ig. 5.6)

For embankments with no drainage measures the base parabola cuts the discharge face at

point G0 at a distance (a + a) along the discharge face from point F, and extends beyond the

limits of the embankments. The actual seepage line meets the discharge face (at point G) at adistance a below the point G0. The value of a can be worked outfrom the curve (5.7) after A.


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139

Cassagrande giving the value of

+

a

a

afor different angles () of discharging face. The

value of (a+ a) can either be measured directly on the face when the parabola has been drawnor its value determined from the equation,

a + a =cos-1

yo (5.11)

3. With rock toe (F ig. 5.8)

The basic parabola may be drawn in a similar way taking F

Fig. 5.6 Phreatic line with inclined discharge face

as focus. As already shown this parabola it self is the seepage line for a horizontal filter. (For a

horizontal filter = 180).For a rock toe, an appropriate value of , measured clockwise from the horizontal base

should be taken and the value of

+

a

a

aread from the curve given in fig. 5.7. The parabola

is corrected at the egress point.

Fig. 5.7 Fig. 5.8

For values of between 30to 180, the distance a measured along the slope from the

toe may be determined as explained above. However, if it is less than 30 , the distance a orthe point of emergence of the phreatic line at the downstream slope may be determined with

the help of Schaffernaks equation.


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140

22

2

sincoscos

hdda = (5.12)

or 22222 cothdhda += (5.13)

where d and h are the coordinates of the initial point B, as explained in fig. 5.5

Quantity of Seepage

Consider earth embankment of homogeneous material given in fig. 5.4.

Flow net through the dam section has been drawn by trial and error method. If h is the total

hydraulic head and Ndis the number of potential drops (9 in fig. 5.4) the potential drop h =

dN

hConsider a field of length l, the field being an approximate square its width is also equal

to l, the hydraulic gradient across the fieldl

h.

The discharge through the field is given by

dN

hKhKl

l

hKq ==

= ..

If Nf is the total number of flow channels ( Nf = 3 in fig. 5.4 ) the seepage per unit width

of embankment is

d

f

N

NKhq= (5.14)

The discharge through a homogeneous embankment with horizontal filter may also be

calculated with the help of equation 5.9.

The equation of base parabola under steady seepage condition may be written as

y =2

00..2 yyx +

For a unit length, y represents the area of flow.

[ ] 20002/12

00 2/2

..

yxyyyxydx

d

dx

dy

ydxdyKq

+=+=

=

Since discharge passing through any vertical plane is the same, at x = 0 we have 1=dx

dyand

y = y0.

q = K y0 (5.15)The equation although is applicable to embankments with horizontal filters but hold

good for determining approximate discharge in all other cases.


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141

Example 5.1

For the homogeneous dam section shown in fig. 5.9, draw the pheratic line if a

horizontal filter of length 10 m is provided. Also determined discharge per m length of dam if

K 10

-5

m / sec.Solution

The angle of discharge face = tan-1/1.5= 33.69'

From Fig. 5.7, =+

a

a

a0.35

Fig. 5.9 Design example 15.2 Dam section

h = 22, L = 22 x 2.5 = 55, 0.3 L = 16.5m

d = l6.5 + (2.5 + 1.5) 3 + 6 + 1.5 x 22 -10

= 57.5 m.

Equation 5.10

dhdy += 220

06.45.57225.5722 =+=

Equation of the parabola is

06.422 +=+ xyx

with the help of this equation, the pheratic line can be drawn.

For example at x = 10, y = 9.88

at x = 20, y = 13.37

Discharge per metre length q = K yo

q = 4.06 x10-5m2/ sec.


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142

Flow net in Anisotropic Soil

Natural deposits made by flowing water and rolled earth dam sections are horizontally

stratified. Such stratifications display much higher horizontal permeability (Kh) than vertical

permeability (Kv).When the horizontal and vertical permeability differ in a dam section, the flow net may be

drawn in the usual manner in a transformed section. To make a transformed section the

horizontal dimensions are multiplied byh

v

K

K while vertical dimensions remain unchanged.

After drawing the flow net in the transformed section, the flow net is restored back in the

original section. The procedure is illustrated in .Fig. 5.10 in case when the horizontal

permeability is 4 times the vertical permeability. It may be noted that while the fields are

approximate squares in the transformed section, they get distorted when restored back in the

original section. The higher the ratio between the horizontal and vertical permeability the more

would be the distortion in the flow net. The flow lines tend to come near the downstream face

of the dam, and for a certain ratio the flow lines may even intersect the section making it

unsafe. The discharges, seepage force, pore water pressure can be determined with help of the

flow net in the original section.

Fig 5.10 Transformation of section having different permeability in horizontal

and vertical directions.

Thequantity of seepage in an anisotropic section can also be determined by using an

effective permeability coefficient (K')in equation 5.15, K' is given by

K' = hvKK (5.16)

5.7 Control of Seepage Through Foundations

Different methods to control seepage of water through foundations are described

below. The suitability of the method of treatment depends primarily on the nature of

foundation.

1. Grouting and grout curtain2. Cut off trenches.

3. Partial cut-off


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1434. Sheet piling cutoff

5. Cement bound curtain cutoff

6. Cast in situ concrete diaphragm

7. Upstream blanket8. Pressure relief wells.

Grouting and grout curtain

Certain materials when infected as grout in the foundation strata acts as a binder and

fills the voids, thus reducing the permeability and increasing its stability. Cement, asphalt, clay

and various chemicals are used as grouting materials.

The choice of grouting material, pattern, depth and sequence of grouting are related to

the foundation conditions, type and height of the dam and objective desired. Cement grouting

is extensively used in rock foundations. In pervious foundations the choice of a suitable groutdepends primarily upon the grain size of the material and its permeability. Table 5.2. indicates

approximate range of grain sizes that can normally be grouted by different types of grout

materials and mixtures.

Table 5.2

Diameter of the material

Type of grout that can be grouted

mm

Cement 0.5 to 1.4

Clay-cement bentomite 0.3 to 0.5

Clay-chemical, bentomite-chemical 0.2 to 0.4

Chemical 0.1 to 0.2

Blanket grouting is done to a depth of 5 to 10 m through holes spaced 3 to 5 m to prevent

piping. Curtain grouting is done to much greater depth to reduce seepage through foundation.

The number of lines and spacing of holes depends upon the nature of foundation and width of

grout curtain which is usually 1/3 to 1/5 of the water head. A single line of grout holes hasbeen adopted in a number of dams. It is however desirable to have two lines of holes.

The pervious zones should be grouted first with coarse grouts with holes spaced apart

followed by finer grouts in further stages with holes spaced closely.

Cut off trenches (Fig. 5.11 a)

The cut-off trenches with side sloping or vertical are excavated below the dams and

filled with well compacted impervious material. These trenches are located well upstream of

the centre line of the dam but within a point where the cover of the impervious embankment

above the trenches is sufficient to resist the percolation at least to the extent offered by thetrench itself. The centre line of the trench is kept parallel to the centre line of the dam. The

trench should be provided up to bed rock or other impervious strata. The bottom width of the


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144trench is governed by the space required for treatment of foundation and type of equipment

used for rolling. Usually minimum 5 m width is adopted. Other method of cut-off may be

economical as compared to deep cutoff trenches. The maximum depth of trench is governed by

economical considerations. For moderate pervious foundations positive cut-off up to hardstratum is provided. The cut-off may be (i) sheet pile (ii) cement bound curtain, (iii) concrete

diaphragm.

Partial cut-off trench (Fig. 5.11 c)

Apartial cut off trench is effective in stratified foundation by intersecting more impervious

layer in the foundation and by increasing the vertical path of seepage. In a uniform pervious

strata the role of partial cut off in reducing percolation is very limited. A cut off going to 80%

of the total depth of pervious

strata reduces the seepage discharge by only 50%. Thus with a partial cut off the reliance is

primarily on the length of the seepage path. Therefore for treatment of deep pervious

foundation where it is not economically possible to provide a positive cut-off, partial cut-off

along with upstream blanket is provided to reduce the discharge and seepage pressures.


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145

Sheet piling cut-off ( Fig. 5.11 b )

Steel sheet piling cut off can be used in silty, sandy and fine

Fig 5.11 Typical foundation treatment and trenches details

gravel foundations. If the foundation strata contains boulders the sheet piles will not easily

penetrate. It is very difficult to make sheet pile cut off water tight. There are always chances of

leakage through the joints and at contact with bed rock.

These disadvantages and limitations of sheet pile cut off have been overcome by use of

cellular sheet pile wall back filled with concrete. Another method of sealing the interlocks is to

drill arrow of boles and fill them with bentonite before driving the sheet pipe line across the

holes.

Cement bound curtain cut off

This type of cut off is used in pervious foundations which do not contain large cobbles

and boulders. The grout is pumped through hollow rotating drill rod which is fixed with a

mixing head on the end. The grout is forced down with vanes provided for in the mixing head

which results in the formation of cylindrical elements of cement impregnated sand and gravel.

A continuous curtain is formed by successive over lapping of such cylinders.


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146

Cast in-situ concrete diaphragm

In this process a trench 5 to 10 m long and up to 1.2 m wide is made by a percussion

tool of the required diameter connected to a rigid tube through which drilling mud is circulated

from the bottom to the top. The mud retains the walls of the trench by hydrostatic pressure as.well as by forming a cake of bentonite along the side of the trench. The excavating tool and

the pipes travel horizontally to and fro over the length of the panels. The vertical percussion

motion of the tool along with the horizontal movement of the assembly excavates a trench

equal in length to the horizontal movement and the excavated material is sucked out through

the rigid tube along with the drilling mud. This drilling mud is then led into large tanks where

the excavated material settles down. The mud is replanished with bentonite and is returned to

the trench. After excavating the trench to the full depth, it is backfilled by tremic concrete.

Alternate panels are first completed, and the intervening panels are subsequently excavated and

backfilled with concrete.This process is suitable for making of cut-offs in sand material and it is for the first

time that this process has been adopted in Obra dam in India (Fig. 5.12)

Fig. 5.12 Cross section of Obra Dam U.P.

Fig. 5.13 Upstream blanket

Upstream Blanket

(i) Advantages

An impervious clay blanket placed upstream of a dam and connected to the impervious

section is a convenient way of effecting moderate reduction in the amount of seepage (Fig.

5.13). The quantity of seepage is some what less than inversely proportional to the total length

of impervious material. The effectiveness of a blanket depends upon the proportionate increase

in number of equipotential drop that results from its addition to the dam. A blanket is

advantageous only when an appreciable length of the path of flow can be affected by a blanket

of reasonable length.


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147

(i i) Length of blanket

Usually the material of the blanket is so tight in relation to pervious stratum that it is not

necessary to consider flow through the blanket in determining the blanket length. The length of

the blanket is given by (Fig. 5.13).

pq

bqpdhKL

.... = (5.17)

in which L = length of upstream blanket in metres.

K = mean horizontal permeability coefficient of the pervious stratum.

h = gross head in metres on impervious upstream blanket.

d = depth of pervious stratum in metres

p = percentage (stated as a decimal) of flow under the dam without a blanket,

to which it is desired to reduce the seepage by means of the blanket. For

example, if the seepage is required to be reduced to 25% of its originalvalue, then p = 0.25.

b = length of impervious portion of base of dam in metres.

q = flow under dam, without a blanket, per metre of dam K. (h/b). d.

(iii) Thickness of blanket

The thickness of the blanket is a function of the relative permeability of the blanket,

permeability of the foundation material and its depth. Theoretically it should vary in thickness

from zero at its upper edge to the maximum where it joins the impervious section of the dam.

The thickness of blanket t at distance h from upstream toe of blanket may bedetermined from the equation,

t= ( K2/K1 ) x b x ( B/d ) (5.18)

where

t = thickness of blanket at point under consideration,

b = distance from point under consideration to upstream toe of blanket,

K1 = average permeability of stratum,

K2 = permeability of blanket,

B = length of blanket from upstream toe to impervious section

d = depth of pervious stratum in meter.For normal condition the thickness of upstream blanket is kept between 1.5 to 3.0 m

and length about 10 times the head of ponded water. In case of fine sand or silty foundations,

the length of blanket is kept 15 times the head.

Pressure Relief Wells

The primary purpose of relief wells is to reduce artesian pressures which otherwise

would cause formation of sand boils and piping. Relief wells also intercept and provide

controlled outlets for seepage which otherwise would emerge uncontrolled downstream of thedam. Theoretically, piping occurs when the uplift pressure at a point at some level in the

foundation near downstream toe reaches, is greater than the combined weight of soil and water


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148above it. If the thickness of impervious layer is equal to the reservoir head, h 1, the uplift

pressure beneath the layer cannot exceed the weight of layer, because the saturated weight of

soils equals approximately twice of water. Thus, if the thickness of the top impervious

structure is equal to the reservoir head, there is no danger against piping. In this situation nofurther treatment of foundation is required. If the thickness of the top impervious stratum is

less than reservoir head, h, but is too thick for treatment by drainage trenches or if the pervious

foundation is stratified, pressure relief wells are required.

Relief wells should be designed to penetrate into the principal pervious strata to obtain

efficient pressure relief, especially where the foundation is stratified. Whereas full penetration

is desirable in case of shallow depth of pervious aquifers (6 m to 9 meter thickness), at least

0.50 percent penetration should be ensured into thicker aquifers. Generally depths of wells

equal to the height of the dam is satisfactory.

Wells should be spaced sufficiently close together (generally 15 metre apart) tointercept seepage and reduce uplift pressures between wells. The wells must offer little

resistance to infiltrator of seepage and discharge there of. The wells must be so designed that

they do not become ineffective due to clogging or corrosion.

Where no control measures are present, relief wells should be so designed that pressure

gradients between wells or downstream from the wells do not exceed 0.5 to 0.6. Where down-

stream berms are provided pressure gradient between wells should not exceed 0.6 to 0.7.

The slots in the well screen must have adequate area and at the same time be of such

size as to prevent movement of filter through the screen after development of the well and shall

satisfy the following criteria.

slotwidth

filterD85 1.2

orerHolediamet

filterMinD85 1.0

The gradation of the filter must also comply with the following criteria

sandDMin

filterDMax

85

85

)(

)(5.0

andsandDMax

filterDMin

15

15

)(

)(4.0

The well screen consists of G.I. pipe 10 cm to 15 cm diameter slotted with 4.75 mm to

6 mm by 50 mm size opening covering about 10 percent of the circumferential area of the pipe.

Vertical slotting is preferable to horizontal slotting. The pipe should be coated with anti-

corrosive paint or regularized after slotting. A type design of pressure relief well is illustrated

in fig. 5.14.


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149

Fig. 5.14 Type design of relief well

In fig. 5.11 appropriate treatments for the following foundation conditions areillustrated.

1. Pervious foundation for shallow depth.

2. pervious foundation extending to a moderate depth.

3. Pervious foundation extending to great depth.

4. Thin impervious layer underlain by pervious foundation.

5. Impervious layer of thickness greater than 3 m and less than the height of the dam.

5.8 Drainage In Earth Dams

Drainage in earth dams is primarily provided to bring the phreatic line well within thedownstream face. A proper drainage system also helps in avoiding heaving and piping by

arresting the soil particles, which may otherwise move by seepage discharge. The drainage

system also reduces the pore water pressure in the downstream portion of the dam and thus the

stability of the downstream face is increased.

Design of drainage arrangement is governed mainly by the height of the dam,

availability of pervious material, and the permeability of the foundation. Different types of

arrangements are shown in fig. 5.15 and short description of each follows.


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150

Horizontal drainage blanket

The horizontal blanket is provided over that portion of the foundation downstream from

the impervious zone of the dam where high upward seepage forces exist. The blanket must be

pervious to drain off effectively and its design should fulfill the usual filter criteria, in order toprevent the movement of particles of the foundation or embankment by seepage discharge.

They may be provided on homogeneous pervious foundations overlain by thin impervious

layers, in order to stabilize the foundation and relieve pressures that may break through the

impervious layer. The requirement of the length of blanket may be determined theoretically by

means of flow net, provided the horizontal and vertical permeability of the foundation are

known. Generally a length three times the height of dam is sufficient. The horizontal blanket is

generally combined with a rock toe (Fig. 5.15 a).

Chimney DrainsThe main disadvantage of the horizontal drainage blankets is to make the earth dam

embankment stratified and consequently more pervious in the horizontal direction. The

inclined or vertical chimney drains are thus provided in many homogeneous dams to intercept

seepage water before it reaches the downstream slope. A number of dams with chimney

arrangements have recently been constructed in this country.( Fig. 5.15)

Fig.5.15 Types of drainage arrangements

Rock Toe

The rock toe consists of sizes usually varying from 15 cm to 20 cm ( Fig. 5.16 ).

Transition filter is required to be provided between the homogeneous fill and the rock toe. The

filter usually comprises


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15130 cm thick layer of fine sand (15% size = 0.09 mm)

45 cm thick layer of coarse sand (15% size 0.81 mm)

60 cm thick layer of gravel (15% size = 7.3 mm).

Fig. 5.16 Rock toe and filter

The capacity of the filter should as a rule be kept twice the discharge calculated by the

Darcys formula.

The minimum required height of rock toe is governed by two factors

(a) The minimum allowable cover on the phreatic line.

(b) Downstream water level.

A minimum cover of 1 m between the seepage line and downstream slope is consideredadequate. The height of rock toe is generally kept between 1/3 to 1/4 the height of the dam.

The top of rock toe must be sufficiently higher than the tail water level to prevent any

wave splashes on the downstream face. Sometimes it may not be economical to raise the level

of rock toe for this purpose. In such cases stone pitching extending to a minimum vertical

height of 1.5 m above the downstream-H.F.L. is provided in continuation of the rock toe.

Generally the discharge face (inward face) of rock toe is kept at slope of 1:1 and the

outer face is a continuation of the downstream slope. In case of high dams, considerable

economy can be affected by providing a berm 3 m to 6 m wide at top of the rock toe and

reducing its outer slope to 1.5:1 to 2:1.

Toe Drains and Drainage Trenches

The purpose of these drains is to collect the seepage from the horizontal blanket which

discharges into the spillway stilling basin or into the river channel below the dam. Toe drains

are also used on impervious foundations to ensure that any seepage that may come through the

foundation or the embankment is collected and the ground water is kept below the surface to

avoid the creation of boggy areas below the dam. The toe drains ( Fig. 5.17 ) may be of

vitrified clay or perforated asphalt dipped corrugated metal pipe. Drainage trenches can be

used to control seepage where the top stratum is thin and pervious foundation is shallow so thatthe trench can be built to penetrate the aquifer substantially. Where the pervious foundation is

deep, a drainage trench of any reasonable depth would attract only a small portion of under


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152seepage, its effect would be local and detrimental because the under seepage would bypass

the trench. A typical design of drainage trench is shown in fig. 5.18

Fig 5.17 Typical toe drain

Fig. 5.18 Details of drainage trench

The filter comprising the drainage layers should be designed in accordance with the

filter criteria.

5.9 Stability of Slopes

Methods of Investigating Stability of Slopes

Every soil mass which has slope at its end is subjected to shear stresses on internal

surfaces in the soil mass, near the slope. This is due to the force of gravity which tries to pull

down the portions of the soil mass, adjoining the slope. If, however, the shearing resistance of

the soil is greater than the shearing stress induced along the most severely stressed or criticalinternal surface, the slope will remain stable; and if on the other hand the shearing resistance of

the soil, at any time after the construction of the slope becomes less than the induced shearing

stress, the portion of soil mass between the slope and the critical internal surface will slide

down along this surface, until, the new slope formed by the sliding mass makes the shearing

stress less than the shearing strength of the soil. (The stability of slopes of earth-structures thus

depends on the shear resistance or strength of the soil.)

The well known method of investigating stability of slope is Swedish method, devised

by Swedish Engineers in 1922 which is simple, and therefore, is more commonly used.

In this method, the curved slip surface is taken to be an arc of circle with a certaincentre. There will be a number of such likely slip circles with their respective centres. It is

necessary to pick up the most dangerous of critical slip circle i.e. that circle along which the


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153soil has the least shear resistance. The centre of this circle is located by trial and error.

Below are shown methods of locating the critical slip circle in the case of various types of

soils, assuming that the soil slope is of homogeneous structure, i.e., it consists of one type of

soilonly.

(i) Cohesive soil

Let PHV ( Fig.5.19 ) be any one slip circle with center O1, x is the distance of the

centroid G ofarea DVHP from the centre O1. The position of G can be obtained in the same

manner as the centroid of an irregular plane area.

Actuating moment which may

= W.x kg metres

cause a slip along surface PHV

Fig. 5.19 Slip circle method

where, W = weight of soil mass of area DVHP and length one metre.

For cohesive soil = 0 and, its shearing resistance depends on cohesion only and is thesame along the entire surface VHP.

The maximum resisting moment mobilized by the soil to prevent the likely Occurrence

of a slip.= cohesive shear resistance developed along VHP

= (L x 1) x c xR kg metres

= cLR kg metres

where L = length of arc VHP in meters

c = unit cohesion in kg/m2

R = radius of likely slip circle in metres

But, L = R where is the angle of arc VHP, in radiansMaximum resisting moment = c R R kg metre

= c R2kg metre


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154Factor of safety against slipping = Maximum resisting moment

along surface VHP actuating moment

W

cR

.

2=

Various other slip circles like the above said one, e.g. QH1V, SH2V, TH3V, etc. are

drawn to scale and the factor of safety is found, as shown above, for each such slip circle. The

slip circle giving the minimum factor of safety is the critical slip circle, because, the failure, if

itoccurs, will occur along this, slip circle. This critical slip circle is located as shown below.

The values of the factor of safety worked out for various slip circles say four, are

plotted as shown in fig. 5.19, and a curve is drawn passing through the tops of the ordinates

representing the four values of the factors of safety corresponding to four slip circles. The

lowest point on this curve is noted and a vertical line is drawn through it, meeting the top of

soil mass at a point say Z; V and Z then mark the two points through which the requiredcritical slip circle will pass.

(ii) c- Soil

The shear strength of such a soil due to cohesion and internal friction varies from point

to point along the slip circle. The equation for shear strength of this soil is, S = c + pntan ,kg/m2 where, c is constant along the slip circle but pnwhich is the pressure normal to the slip

circle due to weight of soil varies from point to point along the slip circle. The method of

locating the critical slip circle is as follows:

The whole soil mass adjoining the slope, is divided into vertical strips or slices asshown in fig. 5.20 (i).

The fifth strip from the left in the figure it taken for the study and is shown on a bigger

scale in fig. Fig. 5.20 (ii). Taking one running metre of the strip, its equilibrium is practically

due to two forces only viz; its weight W5 and its shearing strength S5 along the curved surface

of length l5 inat the bottom of the strip. The weight W5 is equivalent to two forces,

Fig. 5.20 Swedish method for c - soil


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155namely, Wn5which is normal to the curved surface and, the tangential force Wt5 which is

tangential to the curved surface as shown in fig. 5.20. We will, therefore have the equation,

S5 = ( c. l5+ Wn5 tan )kg

where, c.l5= cohesive strength per metre length of the strip and Wn5 tan = frictional strengthper metre length of the strip.The moment of this shear strength is

= R. S5, kg metres

= R ( c.15 + Wn5 tan )kg metresand the moment of shear strength

along the entire slip surface VHP = R ( c.15 + Wn5 tan )= R ( ct5 + W n5 tan )= R (c.L.+ tan Wn5) kg.m

where, L = length of the whole slip surface VHP.The actuating moment along slip surface at bottom of the strip R.. Wt5kg metres

The actuating moment alongthe entire slip surface PHV = R Wt5 kg metres

The factor of safety, against shear = resisting moment

failure, along slip surface PHV actuating moment

or, in general, the factor of safety =t

n

W

WLc

+ tan.

The weight of each strip one metre long, is proportional to its area and this latter can be

found approximately by the trapezoidal formula, or, accurately by a planimeter. Wn and Wtof

each strip can be found graphically by drawing the triangle of forces for each strip as shown in

fig. 5.20 (i).

The factor of safety is found for each of other slip circles and the critical circle is

located as usual. It should be carefully noted, that, the tangential components of the weights of

the first few strips near the toe of the slope, will resist the slipping tendency and these

components must therefore, be taken with their proper sign.

Location of cri tical sli p sur face

The location of critical slip surface by trial and error entails much loss of time. To save

time in locating the centre of critical slip circle for homogeneous sections, Fellenius

construction, for locating the line on which the centre of the critical circle is most likely to lie,

is generally adopted. Various circles with center on this line are tried until the one with

minimum factor of safety is found.

According to the Fellenius construction, the position of the line on which the centre of the

critical circle lies depends only


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156

Fig . 5.21 Fellenius method of locating line of center for critical slip circle

on the height and slopes of the embankment. Referring to fig. 5.21, the values of , and aretaken from Table 5.3 for the embankment slope. From point A and B these angles are drawn to

meet at point 01. Point 02 is then located at horizontal distance of 4.5 H, starting directly below

point A, and at a depth equal to H, the height of the embankment. The line joining 01 and 02

points is the line on which the centre of the critical circle lies. The maximum depth to which

the rupture can occur is limited by the presence of hard stratum underneath.

Table 5.3

Recommended values of and - Fellenius construction

Slope

hor: vertical

1:1 27.5 372:1. 25 32

3:1 25 35

4:1 25 365:1 25 37

Pore Water Pressure

When moist soil mass is loaded without permitting air/or water to escape, part of the load

causes the soil grains to deform elastically or to undergo non-elastic rearrangement but without

significant change in their solid volume. This part of the load is carried on the soil skeleton as

effective stress. The remaining portion of the load is carried by stress in the air and water

contained in the voids and is known as pore-water pressure.

The pore-water pressures below the phreatic line reduces the shearing strength of the soil mass

in accordance with Coulombs equation.

S = c + ( pn u ) tan


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157The practice of accounting for pore pressures in drawdown and steady-seepage

conditions is different from that of end of construction conditions.

Pore Pressures in drawdown and steady conditions:Pore pressures on any slice of a failure are taken to act normally on the arc surface and are

equivalent to the weight of water transmitted by that slice. Consider a slice ( Fig. 5.22 ) below

phreatic line and above dead storage level or tail water level. If h is theaverage height and d its

width, force due to pore pressure.

u = h.w x arc length

= h x w x d sec .The resisting forcedue to pore pressure will, be reduced to u tan .

= ( h d.w. sec ) tan = (h x arc length) x w tan

Fig. 5.22 Pore water pressure

If are length for a small slice be taken approximately equal to the width of the slice, the

contribution of pore pressure becomes (h.b.w.) tan With the above approximation the effect of pore pressures will amount in reducing the

saturated weight W of the soil by the weight of water of the same volume. This would mean

that submerged weight may he taken for saturated weight of soil while calculating the resistingcomponent. Hence a_steady seepage an drawdown conditions, weigh of soil below phreatic

line and above dead storage level is taken as submerged for calculating resisting forces and

saturated for actuating forces.

Pore Pressure just after construction

In this condition the pore pressures are maximum and hence their correct evaluation

from proper consolidation test is advisable. It has been found that the pore pressure generally

varies from 1.1 H to 1.25 H. In the absence of a reliable test it is reasonable to adopt a value of

pore pressures equal to 1.25 H.The effect of pore pressures estimated on above principle is calculated separately as u

tan u being 1.25 H sec , and is deducted from the value of pn tan .


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158

Critical cases for analysis

Reservoir drawdown The upstream slope of the dam is subjected to most adverse

condition during sudden drawdown. When sudden drawdown takes place, the water pressureacting on the upstream slope is removed above the drawdown level while the saturation line

remains higher. Since the drainage is not as rapid as the drawdown, this means that the

resisting forces are reduced in comparison to actuating forces. To take account of this fact in

stability computations, the resisting forces are calculated for submerged weight of the material

below water surface, and the actuating forces are calculated for the saturated weight of the

material below water surface. All materials below drawdown level are submerged, and

therefore, resisting and actuating force below the drawdown level are calculated on the basis of

the submerged weight of the materials.

It has been observed for homogeneous section that drawdown to an intermediate pool levelusually presents a more critical case than complete draw down. In all upstream slope stability

analysis, therefore, such a condition should be tested. For each slip circle, the intermediate

surface of the pool is assumed to intersect the embankment slope directly below the centre of

the circle. This represents the worst condition for assumed failure arc.

Steady Seepage Condition

The effect of seepage through the embankment is to reduce embankment stability by

increasing the actuating forces and decreasing the resisting forces. For the analysis of this

condition, reservoir isassumed to be at thenormal storage level in which case the phreatic lineis assu1riecl to have fully developed. The upstream slope will be subjected to water pressure

from the reservoir while the rest of darn will be subjected to pore pressure from the established

flow net. This condition is critical for the downstream slope only.

End of Construction Condition

Stability at the end of construction is most critical for homogeneous embankments

constructed of plastic materials. Immediately on completion of embankment there would be

construction pore pressure due to consolidation of fill under the embankment load. There

would be no water loads.

Earthquake

There is little experience on the performance of earth dams shaken by earthquakes in

seismically active regions. Practice to account the horizontal acceleration caused by earthquake

is, however, a common practice in stability computations for darns lying in seismic active

regions.

It is recommended that applicable values of seismic coefficient may be taken into

account in stability computation for steady seepage conditions, while half of its values be

adopted in case of sudden drawdown and end of construction conditions.


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159

Factor of Safety

Factor of safety in the present context is known as the ratio of forces resisting

movement to these tending to produce movement.

The minimum required factor of safety in earth dams never exceeds 1.5, which mayseem to be quite a low figure. There are several reasons for the apparent lowness of the

acceptable factor of safety in earth dam designs.

(a) The figures used for the strength of earth materials are necessarily taken quite close

to the minimum values obtained, and the strength which would be mobilized before disastrous

failure could occur would usually be much greater.

(b) Usually the factor of safety increases with the passage of time owing to

consolidation etc. so that a factor of safety which was originally 1.5 may eventually become 2.

(c) In most cases the forces tending to produce movement are taken at the upper

possible limit but may actually be materially less than the assumed values.According to the U.S.B.R. practice a factor of safety of 1.5 is adopted for all

conditions.

The factor of safety as recommended by U.S.B.R. are on higher side, High dams have

recently been designed with lower factor of safety up to 1.25 for reservoir drawdown and end

of construction conditions. When under earthquake condition, factor of safety of unity is

considered adequate.

Tentative section

For fixing tentative section of earth dams the upstream and downstream slopes may betaken from table 5.4 given by Terzaghi.

Table 5.4

Recommended Dam Slopes-Terzaghi

Type of material U.S. slope D.S. slope

Homogeneous well graded h : v h : v

material 2 : 1 2 : 1

Homogeneous coarse siltHomogeneous silt clay or clay 3 : 1 2 : 1

Height less than l5m or so 2 : 1 2 : 1

Height more than 15m or so 3 : 1 2 : 1

Sand or sand and gravel with

clay core 3 : 1 2 : 1

With R.C.core wall 2 : 1 2 : 1


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160

5.10 Selection of Type of Earth Dams

The selection of type of dam i.e.earth, rock fill, concrete, concrete gravity, concrete

arch, buttress etc. When a particular site favours an earth dam, a further decision must be madeas to the type of earth dam.

The type of earth dam would be dictated essentially by the materials available at or near

the site as also the foundations. In the interest of economy, the design of earth dam should be

adopted to the full utilization of the available materials. Thus, if nothing but sand is available

the design should utilize the sand in the natural state for the bulk of the dam, limiting the

imported material like clay or silt fore providing impervious member to the minimum.

Earth dams may be classified into three main types :

1. Homogeneous type

2.

Zoned type and3. Diaphragm type

Homogeneous Type

A purely homogeneous type of dam is composed entirely of a single type of material as

shown in fig. 5.23 a. Since the action of seepage is not favourable in such a purely homogen-

eous section, the upstream slope should be, relatively flat for safety in rapid drawndown (when

embankment is relatively impervious, and the downstream slope must also be flat to provide a

slope sufficiently stable to resist the forces resulting from a high saturation level. For acompletely homogeneous section, it is inevitable that seepage emergence will occur in the

downstream slope approximately at about one third the water depth regardless of permeability

of material or slope flatness. Though formerly very common, the pure homogenous section has

fallen into disuse because of the development of a modified homogeneous section (Fig. 5.23 b)

in which small amount of carefully placed drainage materials control the action of seepage

and thus permit much steeper slopes.

Fig . 5.23 Homogenous type earth dams


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161A fully homogeneous section might be found convenient where the slopes are

required to be flat because of a weak foundation, although even in this case some drainage

measures may be necessary. The modified homogeneous type of embankment is most suitable

in localities where readily available soils show practically no variation in permeability.

Zoned type

It is the most common type of dam section in which a central impervious core is

flanked by shells of materials considerably more pervious. The shells enclose and protect the

core; the upstream shell affords stability against rapid drawdown and the downstream shell acts

as a drain that controls the line of seepage. For most effective control of steady seepage or

drawdown seepage, the section should have a progressive increase in permeability from the

centre towards each slope ( Fig. 5.24 ). Innumerable modifications might be made of the zoned

type section depending upon the arrangements of seepage control and drainage arrangements.

Fig. 5.24 Zone type earth dam section

The width, of core in the zoned type can be chosen, within reasonable limits, to meet

the best adjustment in the quantity and the cost of impervious soils available. The minimum

width should be adequate to reduce seepage and permit ease of construction. The minimum

base width should be equal to the height of the embankment. If the width is less than the height

of the embankment, the dam is considered as diaphragm type. Similarly if the core is larger

than the size shown in fig. 5.25 the dam may be considered as homogeneous type.

Fig. 5.25 Size range of impervious cores in zoned embankment-after U.S.B.R.


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162All transitions between materials, of different gradations subject to seepage must be

checked for filter action. If the gradation range between adjacent zones of the dam or founda-

tion is so great that a natural filter will not be established, a special filter zone will be required

to provide the necessary filter action.

Diaphragm type ( Fig. 5.26 )

In this type the entire dam is composed of pervious material i.e. sand, gravel or rock

and thin diaphragm of impervious materials is provided to retain water. The diaphragm may

consist of earth, cement concrete, or other material in the central or upstream face of the dam.

The position of the impervious diaphragm may vary to extreme limits-upstream face on one

side and central core on the other end. The intermediate cases are termed as burned blanket or

inclined diaphragm type as shown in fig. 5.27.

Fig. 5.26 Diaphragm type earth dam

Fig. 5.27 Inclined diaphragm type earth dam

Internal diaphragm of rigid materials like concrete have the disadvantage of getting

ruptured due to settlement of the dam.

Fig. 5.28 (a) Nanak Sagar dam (Constructed after breach) U.P.


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163

Fig. 5.28 (b) Section of the u.s. riprap

It is, therefore, preferable to have core of the earth. The width of the earth core at the

base should be 0.3 to 0.5 times the height of dam. There are certain advantages anddisadvantages in locating the core centrally or in an inclined position. However, the

construction convenience dictates the position of the core. An inclined diaphragm provides

slightly better stability against earthquake.

5.11 MAINTENANCE & TREATMENT OF COMMON TROUBLES IN EARTH

DAMS

Proper maintenance of an earth dam is very essential. Small defects arising from the

negligence may cause breach which may lead to serious damage and devastation.In spite of good design, some problems may be subsequently met with during the

operation of the reservoir. The following are the most common troubles which may need

proper and timely treatment.

1. Horizontal piping through embankment.

2. Boils.

3. Boggy areas downstream of the dam, and

4. Cracks in the embankment.

15.12 DESIGN OF EARTH DAM

Design Example 5.2

An earth dam has to be constructed at a place where soils have the following

characteristics:

(i) Sand mixed with gravel

(a) Saturated unit wt 2 gm/cc (2 tIm3)

(b) Sp. Gravity 2.65

(c) Angle of internal friction 33(d) Cohesion nil

(e) Dry unit wt. 1.6 gm/cc (1.6 t/m3

)This material constitutes the foundation to a depth of 15.Om, below the base of the dam

after which rock is available.


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164

(ii) Silty clay

(a) Saturated unit wt. 1.76 gm/cc (1.76 tim2)

(b) Specific Gravity 2.55

(c) Angle of internal friction 12(d) Cohesion 3.t/m2The material has to be obtained from an average distance of 1.5 km.

(iii) Levels

(a) River bed level 103.0 m

(h) Dead storage level 123.0 m

(c) F.S.L. 155.0 m

(d) H.F.L. 158.0 m

Assume the material abovephreatic surface to be 40% saturated.(e) Earthquake factor

(i) For sudden drawdown = 0.15(ii) For steady seepage = 0.10(iii)For just after construction = 0.05

Design a suitable section of the earth dam and indicate the position of seepage line. Also

draw cross section of the dam.

Design

1. Type of Dam Section

As both pervious and impervious materials are available in the close proximity, the dam

shall consist of a composite section. The shell of the dam will consist of pervious material and

core of impervious material. A cutoff trench up to rock level with 1: 1 side slopes shall also be

provided to check the seepage through the foundation material.

2. Tentative cross section of the Dam

(i) Freeboard

(a) Freeboard for wave action:

hw= 0.032 VF+ 0.763 - 0.27 14 F

where hw= height of wave in metres

V = wind velocity in km. p.h.

F = fetch in kms.

Assume V = 150 km. ph. and F = 20 kms.

Then hw= 0.032 20150x + 0.763 - 0.2714 20

=1.958 metres

Along slopes, the waves rides to a height of 1.5 hw.

1.5 hw= 150 x 1958 = 2.94 metres(b) Freeboard due to settlement


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165Settlement allowance of 2% is generally taken both for the foundation and the

embankment.

Settlement = 60 x

100

2= 1.20 m (Assuming dam height to be 60 m)

For earth dam more than 30 m in height, an extra settlement allowance of 1% due to

earthquake is also taken.

Settlement due to earthquake =100

160x= 0.6 m

Total freeboard due to settlement = 1.20 + 0.60 m= 1.80m.

Thus total freeboard required both due to wave action and settlement = 2.94 + 1.80 m

= 4.74 m.

Provide a freeboard of 5.0 m above H.F.L.(ii) Top level and width

(a) Top R.L. of dam = 158 + 5 = 163.0 m

Height of dam above bed level = (163.0103.0) = 60m

(b) Crest width :

Bt= H3

5

where Bt= Crest width in metres

H = Height of dam in metre

Bt= 6035 = 12.90 m

Provide crest width of dam as 13.0 m.

(iii) Slopes : For 60.0 m height of the dam adopt 3.5: 1 slope in the upstream and 3 : I in

the downstream; Adopt slopes of the core material 1 : 1.

(iv) Pitching

Upstream : To prevent destructive wave action, the upstream slope shall be provided with

0.75 m thick stone pitching over 0.5 m thick graded shingle right up to the top of the dam.Downstream The downstream slope shall be provided with 0.5 m stone pitching to check

the erosion due to surface runoff.

3. Foundation

(3) Shear stress in the foundation

The horizontal shear under the slope of dam is given by

S =

2

45tan2

12

2

2

2

1 ohh

where 1= Equivalent angle of internal friction for composite section.= Effective wt/cum of the composite material, and h1 and h2 are heights top


40/51

166and toe of the dam respectively to the rock level.

Here h1 = 75m, h2 =15m, b = 2l0m

a= Average effective wt. of the composite material

=210

6076.11502 xx + = 1.93 t / cum.

Equivalent angle of internal friction for C soil is given by

tan 1=1

1 tan

h

hC

+

For shell material, C = 0, 1= 33For core material, C = 3 t /m2 = 12

tan 1=6076.1

12tan6076.13 xx o+

= 0.239

1= 13.5(for core material).Since the dam is of composite section, the equivalent value of internal friction

=210

5.136033150 oo

xx += 27.4 (for composite section)

Total shear S =

2

4.2745tan93.1

2

1575 222 o

o

x

= 1926.5 t say 1900 tonnes.

Average unit shear in the foundation (Sa) = 210

1900

= 9.05 t/m2

Maximum unit shear = 1.4 Sa = 1.4 x 9.05= 12.67 t/m2

(ii) Factor of Safety against foundation shear :

We should now work out the factor of safety in the foundation against shear at the point of

maximum unit shear which occurs at 84 m (0.4 b)from shoulder. The mean effective unit wt. r

at this location.

=51

115236 xx += 1.706 t /m3

Shear strength at point of maximum shear = c + h tan 1Substituting c = 0, 1= 33

r = 1.706t/m2and h =51 m, we get

Shear strength = l.706 x 51 x tan 33= 1.706 x 51 x O.65 = 56.5 t m

2

Factor of safety against shear at point of maximum shear

=67.12

5.56= 4.45 okay


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167

4. Position of Seepage Line in the Composite Section

The shells of the dam which consist of sand and gravel are several times as pervious as the

central portion of the dam consisting of silty clay. The upstream pervious shell has practically

no effect on the position of seepage line and the downstream shell will act as a drain. Hence incomputing the position of the seepage line, the effect of the outer shell should be neglected.

(a) Construction of base parabola:

The focus of the base parabola is located at the downstream toe of the core denoted by F.

Taking focus as origin, the equation of base parabola is given by

x =0

2

0

2

2y

yy

where x and y are the points on the parabola and y0is the focal distance which is given by y0

y0 =22 dh + - d

Fig.5.29 Design Problem Dimension

From figure 5.29, h = 52 m and d = 96.6 m

Y0= 22 6.9652 + - 96.6 = 13.106 say 13.0 m

Thus x =13213

22

xy or y = 16926 +x


42/51

168The values of y for different values of x are given below :

x = 0 10 20 30 40 50 60 70 80 80 96.6

y = 13 20.7 26.2 30.8 348 38.3 41.6 44.6 47.4 50.1 52.0

Thus the base parabola is plotted for the above computed values of x &y.(b) Correction at ingress and egress points

The base parabola intersects the downstream face at a distance a + a along the face fromthe point A.

a + a =cos1

0

y

Substituting y0 = 13 m and = 45,weget

a + a =o45cos1

13

= 44.4m

From fig. 15.7,aa

a+

= 0.34

or a = 034 x 44.4 = 151 m.Thus theseepage line meets the downstream face tangentially at point C, 15.1 m along the

face from the point Co. The seepage line is completed at the downstream end by sketching a

short transition curve from point C to the base parabola.

The upstream end of seepage line is sketched with short transition curve as shown in fig.

5.30. The upstream shell material isso excessively pervious as compared to the central section

that it wi1I have no effect on the seepage line. Hence through this section the seepage is drawn

as a straight line. As the foundation material is also very pervious, the seepage line will pass

beneath the base of the dam (figure 5.29).

5. Stability Computation of Upstream and Downstream Slopes

The stability of upstream slope has been computed for the following conditions

(i) Sudden drawdown condition

(ii) Just after construction condition

Both the conditions shall be analysed with and without earthquake. The stability of the

downstream slope shall be computed for the following conditions:

(i) Steady seepage condition

(ii) Just after construction condition

Both the conditions shall be analysed with and without earthquake.

For calculating the actuating forces in sudden drowdown condition, the saturated weight of

soil is taken in drawdown range between f.s.l. and d.s.l. Below d.s.l. the submerged weight of

soil is taken while above f.s.l., the weight of soil is takesn as 40% saturated.

The unit weight of soil are being calculated hereinafter fordifferent conditions.

(a) Sand mixed with gravel

Weight of fully saturated soil = 2 gm/cc (2 t /m3)

Submerged weight = 2 - 1 = 1 gm/cc = (1t / m3) 40% saturated weightLet V be the volume of solids in unit volume of soil then


43/51

169V x 2.65 +(l - V)1 = 2 or V = 0.606

Weight of 40% saturated soil

= 0.606 x 2.65 + 0.394 x

100

40= 1.76

say = 1.75 gm / cc (1.75 t/m3)

(b) Silty Clay

Submerged weight = 1.76 - 1 = 0.76 gm/cc = 0.76 t/m2

Weight of 40% saturated clay

Let V be the volume of solids in unit volume of soil then

V x 2.55 + (l-V) x 1 = 1.76; V = 0.49

Weight of 40% saturated earth = 0.49 x 2.55 + 0.51 x100

40

= 1.45 t / m3

For just after construction condition no pore pressure effect in sandy soil is taken into

account. Thus an amount equivalent to the value corresponding to pore pressure is to be

deducted from (W tan + c.L.).The value of unit weight of dry sand i.e. of shell material is taken as 1.6.

The values of unit weight of shell and core material under various conditions of stability

analysis both for upstream and downstream slopes are given in Table 5.5

Table 5.5

Values of unit weight under different stability conditions

For Calculation of N For Calculation of T

ConditionMoist

Core

Moist

Shell

Shell

above

d.s.l

Shell

below

d.s.l

Core

above

d.s.l

Core

below

d.s.l

Moist

Core

Moist

Shell

Shell

above

d.s.l

Shell

below

d.s.l

Core

above

d.s.l

Core

below

d.s.l

U.S. Slopes

(i) Sudden drawdown

(a) Without Earthquake

(b) With Earthquake

(ii) Just after Const.

Condition


(b) With Earthquake

1.45

1.45

1.75

1.75

1.00

1.60

1.00

1.00

0.76

1.45

0.76

0.76

1.45

1.45

1.75

1.75

2.00

2.00

1.0

1.0

1.76

1.45

0.76

0.76

Condition For Calculation of N For Calculation of T

Moist

Core

Shell

above

d.s.l

Shell

below

d.s.l

Core

below

d.s.l

Core

above

d.s.l

Moist

Core

Shell

above

d.s.l

Shell

below

d.s.l

Core

above

d.s.l

Core

below

d.s.l

D.S. Slope

(i) Steady Seepage

Condition(a) Without Earthquake

(b) With Earthquake

(ii) Just after Construction


(b) With Earthquake

1.45

1.45

1.75

1.60

1.0

1.0

0.76

0.76

0.76

1.45

1.45

1.45

1.75

1.60

1.00

1.00

1.76

1.45

0.76

0.76


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170

A STABILITY COMPUTATION OF THE UPSTREAM SLOPES:

(1) Sudden Drawdown Condition


Adopting the above values of the weight of different types of soils under differentconditions, the stability computations have been worked out in Table 5.6. A slip circle has been

drawn with a radius of 150 m as shown in figure 5.31. The entire circle has been divided into

nine slices.

Actuating force T = 1 [99.7 x 1.45 + 58.2 x 1.75 + 297 x l.76 + 757.6 x 2 + 92 x 0.76 + 119 x1]

= 1 [144.60 + 102.00 + 523.00 + 1515.20 + 69.90 + 119.00]

= 2473.76 t say 2474 t.

Resisting force N = 1 [79.6 x 1.45 + 57.3 x 1.75 + 322 x 0.76 + 1694.8 x 1 + 140.5 x

0.76+3411 x 1]= 1 [115.50 + 100.00 + 245.00 + 1694.80 + 107.00 + 3411.OO]

= 5673.3 t.

Radius of slip circle = 150 m.

Length of arc in the core material

= 2 x 150 x360

35=91.5m

Factor of safety =T

LcN

+ ..tan.

N for shell material = 100 + 1694.8 + 3412= 5205.8 say 5206 t

N for core material = 115.5 + 245.0 + 107= 467.7 t

for shell material is 33 and for core is 12 and C = 3t / m2

F.S. =2474

5.91312tan5.46733tan5206 x++ oo

= 2474

5.2742125.050.46765.05206 ++ xx

=2474

5.2744.93380 ++=

2474

0.3749

= 1.515 Okey.


45/51


46/51

172Submerged shell below d.s,1.

= - (136.5 + 114.4 + 5.8) + (162:0 + 170.0 + 95.4)

= - 308.9 + 427.4 = 1 18.5

Submerged core above d.s.1. = 113 + 184 = 297Submerged core below d.s.l. = 53.0 + 39.0 = 92.0

Fig . 5.31 Graphical method of stability computation Design example

(b) With Earthquake

Earthquake factor for sudden drawdown condition, = 0.05Increase in T = 0.05 x 5673.3

= 288.7 t.

Reduction in N tan T for shell material =.102 + 1515.2 + 119 = 1736.2 tT for core material = 144.8 + 523 + 69.9 = 737.7 t

Reduction in N tan = 0.05 (737.7 x 0.2125 + 1736.2 x 0.65)= 0.05 (157 + 1130) = 64.4t

Factor of Safety with earthquake

=7.2757

60.3684

7.2830.2474

4.640.3749=

+

= 1.34 Okay.


47/51

173(ii) Just After Construction Condition


In this condition the shell material should be assumed to be dry and the clay core material

as moist with unit weight as 1.45 tIm

3

.

Actuating forces T = 1 (99.7 x 1.45 + 58.2 x 1.75 + 297 x 1.45 + 757.6 x 1.6 + 92 x 0.76 +119.x 1.0)

= 1 (144.8 + 102.0 + 430 + 1210.0 + 70.0 + 119.0)

= 2075.8 say 2076 t

N tan = (79.6 x 1.45 x 0.2125 + 57.3 x 1.75 x 0.65 + 322 x 1.45 x 0.2125 +1694.8 x 1.6 x 0.65 + 140.5 x 0.76 x 0.2125 + 3411 x 1 x 0.65)

= 1 (24.6 + 65.0 + 99.4 + 1760 + 22.7 + 2220.0)

= 4191.7 t say 4192 tPore Pressure

For slice 6, head is 8 m; sec = 1.138, width of strip = 20 mh x 1 sec = 182.0 t

For slice 7, head = 14m; sec = 1.252, width of strip = 20 m h . 1 sec = 351.0 t

For slice 8, head = 18 m; sec = 1.743, width of strip = 10 m h . 1 sec = 498.0 t

For slice 9, head = 10 m; sec = 1.743, width of strip = 10 m

h . 1 sec = 174.3 t

h. 1 sec = 182.0 + 351.0 + 498.0 + 174.3 = 1205.3 tNow P = w h l sec = l x 1205.3

= 1205.3

P tan = 1205.3 x 0.2125= 256 t

Factor of safety =

2076

5.4210

2076

2565.2740.4192=

+

= 2.03 Okey.

Documents

Chapter 5.Earth Dam