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A narrow gorge opening upstream at the dam site

Strong rock foundation to safely withstand static and dynamic forces including earthquake.

Strong and water tight abutments

Stable side slopes of abutments

Suitable location for power house and spillway

Availability of construction materials nearby.

Accessibility by rail and road

A il bili f l i l f iAvailability for electric supply for construction

Dam height is reasonable for the required storage capacity.

The basic shape of a concrete gravity dam istriangular in section with the top crest oftenwidened to provide a roadway.

A gravity dam should also have anA gravity dam should also have anappropriate spillway for releasing excessflood water of the river during monsoonmonths. This section looks slightly differentfrom the other non-overflowing sections.

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The flood water glides over the crest anddownstream face of the spillway and meetsan energy dissipating structure that helps tokill the energy of the flowing water, whichotherwise would have caused erosion of theotherwise would have caused erosion of theriver bed on the downstream.

Usually, a spillway is provided with a gate,and a typical spillway section may have aradial gate

The axis of the gate is held to anchorages that are fixed to piers.

Also shown in the figure is a guide wall ortraining wall that is necessary to prevent theflow crossing over from one bay (controlledby a gate) to the adjacent one.

Since the width of a gate is physically limitedto about 20m (limited by the availability ofhoisting motors), there has to be a number ofbays with corresponding equal number ofgates separated by guide walls in a practicaldam spillway.

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The upstream face of the overflowing and non-overflowing sections of a gravity dam are generally kept in one plane, which is termed as the dam axis or sometimes referred to as the dam base linereferred to as the dam base line

Since the downstream face of the dam is inclined, the plane view of a concrete gravity dam with a vertical upstream face would look like

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If a concrete gravity dam is more than 20 min length measured along the top of the damfrom one bank of the river valley to the other,then it is necessary to divide the structureinto blocks by providing transverseinto blocks by providing transversecontraction joints.

These joints are in vertical planes that are atthe right angle to the dam axis and separatedabout 18-20 m.

The contraction joints allow relieving of thethermal stresses.

In plan, therefore the concrete gravity damlayout would be as shown in figure where thelayout would be as shown in figure, where thedam is seen to be divided into blocksseparated by the contraction joints.

The base of each block of the dam ishorizontal and the blocks in the centre of thedam are seen to accommodate the spillwayand energy dissipaters.

The blocks with maximum height are usuallythe spillway blocks since they are located atthe deepest portion of the river gorge

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The upstream face of the dam is sometimes made inclined or kept vertical up to a certain elevation and inclined below.

Dead load◦ It comprises the major resisting force

◦ It includes the weight of the concrete or masonry orboth plus that of the appurtenances such weight ofboth plus that of the appurtenances such weight ofgates and bridge.

◦ Usually a unit length of the dam is considered fordesign.

◦ Cross section of the dam is divided into severaltriangles and rectangles.

◦ Weight of each of these is the computed at theirrespective centre of gravity.

◦ The resultant of all these forces is then calculated.

◦ Unit weight of concrete is normally taken as 2400kg/m3.

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Reservoir and tail water loads

◦ Water pressure corresponding to full reservoir levelconstitutes the major external force acting on thedam.◦ The water pressure acts horizontally and its with

zero at the water surface and equal to WH2/2 at thebase.◦ W is the unit weight of water and H is the depth of

water upstream.

The resultant force WH2/2 acts at H/3 from the base of the dam.

When the upstream face is partly vertical and partly inclined, the resultant water pressures comprise ofcomprise of

◦ Horizontal component P = WH2/2 acting at H/3 from the base of the dam

◦ Vertical component P1 which is the weight of the column of water ABCDA acting at CG of the area.

Tail water pressure

◦ It is the pressure of the water exerted on the d/s face of the dam◦ Full value is taken for non overflow sections and a

reduced value is taken for overflow sectionsreduced value is taken for overflow sections.◦ If h is considered to be the height of the tail water,

then it exerts both horizontal and vertical pressures on the d/s face of the dam.◦ Horizontal pressure P2=wh2/2 acting at h/3 and

vertical pressure P3 = weight of column of water A’B’C’.

Uplift pressure◦ Considered to be the second major external force

◦ The seepage of water normally takes place throughthe pores, cracks and fissures of the foundationmaterials, through the body of the dam andthrough the joints between the dam body and thefoundation base.

◦ This seepage exerts pressure

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The uplift forces act as normal pressures inpores, cracks and seems within the body ofthe dam, between the dam and its foundationand within in the foundation.

At the heel and toe of the dam, these upliftforces are considered to be equal to thehydrostatic pressures joined by a straight linein between.

Drainage galleries are sometimes providedwithin the body of the dam to relieve someamount of this uplift pressure.

So the uplift pressure at the drainage gallerySo the uplift pressure at the drainage galleryis equal to the tail water pressure plus onethird of the difference between the fullreservoir level and tail water head.

Silt pressure◦ The weight and the pressure of the submerged silt

are to be considered in addition to weight and pressure of water. ◦ The weight of the silt acts vertically on the slope

and pressure horizontally, in a similar fashion to the corresponding forces due to water. ◦ It is recommended that the submerged density of

silt for calculating horizontal pressure may be taken as 1360 kg/m³. ◦ Equivalently, for calculating vertical force, the same

may be taken as 1925 kg/m³.

Earthquake forces

◦ Earthquake or seismic activity is associated withcomplex oscillating patterns of acceleration andground motions, which generate transient dynamicg , g yloads due to inertia of the dam and the retainedbody of water.

◦ Horizontal and vertical accelerations are not equal,the former being of greater intensity.

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The effect of an earthquake is equal toimparting acceleration to the foundation ofthe dam.

The acceleration induced is of two typesThe acceleration induced is of two types

◦ Horizontal acceleration◦ Vertical acceleration

Horizontal acceleration◦ Normally expressed in terms of acceleration due to

gravity

◦ Values of 0.1g to 0.15g are normally considered to be sufficient for high dams in earth quake zones

◦ Causes inertia force and hydrodynamic pressure.

Inertia force◦ It is given by principle of mass times the

acceleration acting through the centre of gravity ofthe section,◦ Irrespective of the shape of the cross section.◦ Acts in an opposite direction to the ground

acceleration◦ Causes an overturning moment about the

horizontal section in addition to that of thehydrodynamic force

◦ Inertia force is given byW/g(αg)=Wα

W = weight of damg acceleration d e to gra itg = acceleration due to gravityα = acceleration co efficient, earth quake

acceleration due to gravity

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Hydrodynamic pressure◦ An instantaneous horizontal pressure is exerted

against the dam in addition to hydrostatic forcesdue to the horizontal acceleration of dam and itsfoundation

◦ This is known as hydrodynamic pressure

◦ The direction of hydrodynamic pressure is oppositeto the direction of earthquake acceleration.

The hydrodynamic force is given by

◦ Pe = Cαhw h◦ C = coefficient that varies with the shape and

depth. Value can be approximated from:depth. Value can be approximated from:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −=

hy

hy

hy

hyC

C m 222

Pe = pressure intensity acting normal to theface of the dam at a point depth y from thereservoir water surface i.e. hydrodynamicpressure at depth y. (kg/m2)

αh = horizontal earthquake intensity assumedfor the purpose of design. Varies from 0.03to 0.24 at the top of the dam reduced linearlyto zero.

w = unit weight of water (1000 kg/m3)h = reservoir water depth (m) i.e. reservoir level minus base level of dam at u/s face.y = vertical distance from the reservoir s rface to the ele ation nder considerationsurface to the elevation under considerationCm = maximum value of pressure coefficient.

The force Pe acts at 4H/3π from the base.

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Vertical acceleration◦ A vertical acceleration may either act downward or

upward.◦ When it is acting in the upward direction, then the

foundations of the dam will be lifted upward andbecome closer to the body of the dam and thus theeffective wt. of dam will be increased and hence thestress developed will increase.

◦ Increased weight of the dam material = Wc(1+ α)◦ Increased weight of water = W(1+ α)

When the vertical acceleration is acting downward, the foundation shall try to move downward away from the body of the dam and thus reducing the effective wt. and stability of the dam and hence the worst case for designs.

◦ Reduced weight of the dam material = Wc(1- α)◦ Reduced weight of water = W(1- α)

Wave pressure◦ Waves are generated on the surface of the reservoir

by blowing winds, which exert a pressure towardsthe d/s side.

◦ Wave’s pressure depends upon the wave height.

◦ Wave height is given by the equation (for F<32km)

41

)(271.0763.0.032.0 FFvhw −+=

For F>32 km

h h i h ( )

Fvhw .032.0=

hw = wave height (m)F = fetch or straight length of reservoir (km)v = wind velocity.

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◦ Maximum unit pressure = Pw = 2400 hw acting at0.125hw above still water

◦ Total wave force = Fw = 2000 hw2 kg/m acting at

0.375hw above still water level.

Considered for dams constructed in coldcountries or at higher elevations

It is exerted due to the formation of ice onthe reservoir water surface

Ice expands and contracts due to change intemperature.

Ice pressure is caused due to expansion ofice.

The face of the dam is subjected to force due to expansion of ice which is taken as 25 to 150 T/m2.

It is applied to the face of the dam over itsIt is applied to the face of the dam over its anticipated area of contact with ice.

Wind pressure is normally ignored in the design of dams.

But if considered, it is taken as 100 to 150 kg/m2 for the dam portion exposed to windkg/m2 for the dam portion exposed to wind.

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The elementary profile of a gravity dam subjectedonly to water pressure on the upstream face is aright angled triangle having zero width at the topand a maximum base width B, where the waterpressure is maximum.

So it can be said that the elementary profileresembles the hydrostatic pressure distribution.

When the reservoir is empty, the only force theonly force acting is the self weight of the dam Wat a distance B/3 from the heel.

Forces acting on the dam are

Weight of the dam

wBHW ρ21

=

is the unit weight of the dam material (2400 kg/m3)

is the unit weight of water (1000kg/m3)

2

ρ

w

Water pressure

acting at H/3 from the base.

Uplift pressure

2

21 wHP =

Uplift pressure

c is the coefficient of uplift pressure intensity and is taken as 1.

cwBHU21

=

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Normal stress

without considering uplift pressure⎟⎠⎞

⎜⎝⎛ +=

Be

BWP 61max

⎟⎠⎞

⎜⎝⎛ −=

Be

BWP 61min

Normal stress distribution with uplift pressure considered

⎞⎛

6/Be ≤

⎟⎠⎞

⎜⎝⎛ −

−=

⎟⎠⎞

⎜⎝⎛ +

−=

Be

BuWP

Be

BuWP

61

61

min

max

Under general conditions, the maximumcompressive stress occurs at the toe and canbe calculated using e=B/6 and will be equalto 2W/B. and the corresponding normalstress at heel will be zerostress at heel will be zero.

The resultant will act at inner third (B/3 fromtoe).

For no tension to develop, the resultant should act at inner third point i.e. B/3 from the toe (L2).

Taking moments about L2 and equating toTaking moments about L2 and equating to zero.

032

132

132

0333

2

=−+

×=−+

BwBHBcwBHHwHor

RBWBuHP

ρ

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Multiply both sides by 6/wH

HHcBBcBH

=−

=−+

ρ

ρ22

222

)(0

When uplift pressure intensity is not considered,

cHB−

=ρ

ρHB =

For stability against sliding, the horizontal forces causing sliding (ΣH=P) are balanced by the frictional forces opposing it (μΣV or μΣ(W-u). Hence

P= μ(W-u)

( )ρμ

ρμ

Hor

cHB

or

cBwHwBHwH

−=

⎟⎠⎞

⎜⎝⎛ −=

21

21

21 2

when uplift is not considered.

( )

μρ

ρμ

HB

andc

HB

=

−≥

Base width B is the maximum of the two given by above equations.

Normal stress (at toe) is given by

Under limiting condition e=B/6 then

⎟⎠⎞

⎜⎝⎛ +

−=

Be

BuWPn

61

BuWPn

−= 2

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21

212

⎟⎠⎞

⎜⎝⎛ −= cBwHwBH

BP

or

n ρ

At heel( )( ) 0)11

22

=−=−=

⎠⎝

wHPcwHP

orB

n

n ρ

Principal stress near the toe is the maximum normal stress in the dam and is given by

( )

but

HBcwH

PthenP

PP

n

n

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−=

=

=′′−=

ρσ

φσ

φφσ

1

sec

0tansec

2

1

21

221

cHB

cHB

cHB

but

−=

−=

−=

ρ

ρ

ρ

12

2

22

Sh h i l l h

( ) ( )( )1

11

1

1

+−=

⎥⎦

⎤⎢⎣

⎡+

−−=

cwHc

cwH

ρσρ

ρσ

Shear stress at a horizontal plane near the toe

( )φρφτ tan)tan cwHPn −==

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( )

( )H

cHB

cHB

butHBcwH

−=

−=

⎟⎠⎞

⎜⎝⎛−=

ρρ

ρτ

1

1;

The principal and shear stress at the heel is zero since normal stress at the heel is zero.

( )

cwH

ccwH

−=

−−=

ρτ

ρρτ

A low dam is of limiting height such that theresultant of all forces passes through themiddle third and the maximum compressivestress at the toe does not exceed thepermissible limit i epermissible limit i.e.

Limiting height H is (ignoring uplift on safeside)

( ) fcwH =+−= 11 ρσ

( ) ( )11 +=

+−=

ρρ wf

cwfH

The limiting height for the usual stresses of dam material

w = 1000 kg/m32400 k / 3ρ = 2400 kg/m3

f = 30kg/cm2 or 300 ton/m2 gives

H=300/(1(2.4+1))=88m.

A dam that exceeds the limiting height of alow dam is termed as a high dam.

In this case the resultant of water pressureand self weight of the dam passes throughand self weight of the dam passes throughthe middle third at certain sections of thedam body indicating the development oftension.

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The dam profile has to be adjusted by addingextra slope or batter on the upstream anddownstream sides to limit the compressivestresses within the allowable limits.

Gravity method◦ The preliminary analysis of all gravity dams can be

made easily by isolating a typical x-section of thedam of a unit width.

◦ This section is assumed to behave independently ofthe adjoining section. In other words, the dam isconsidered to be made of a number of cantileversof unit width each, which act independently of eachother.

The dam transfers the load to the foundationthrough cantilever action.

The loads are resisted entirely by the weightof individual cantilevers of unit lengthof individual cantilevers of unit length.

Other assumptions made are◦ The dam is considered to be composed of a number

of cantilevers each of 1m thick and each of whichare independent of the other.

Hence for wide- U shaped valleys wheretransverse joints are not generally groutedthis assumption is nearly satisfied.

No loads are transferred to the abutments byNo loads are transferred to the abutments bybeam action

The foundation and the beam behave as asingle unit, the joint being perfect.

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No movements of the foundations arecoursed due to transference of loads.

The material in the foundation and body ofdam are homogeneousdam are homogeneous.

Small opening made in the body of dam donot affect the general distribution of stressesand they only produce local affects

Analytical method of analysisConsider unit length of the damFind out ΣV and ΣHCalculate the lever arm of all the forces from th tthe toe.Calculate the sum of overturning moments ΣMo and righting moments (ΣMr) at toe.Calculate the algebric sum of all those moments (ΣMr-ΣMo)

Calculate the location of the resultant from toe

C l l t th t i it f th lt t R∑∑=

VM

x

Calculate the eccentricity e of the resultant R from the centre of the base width B (<B/6 for no tension.

xB−

2

Calculate the normal stress at the toe and heel

⎥⎦⎤

⎢⎣⎡ ±∑

Be

BV 61

Calculate the maximum normal stress and shear stress at the toe and heel.

⎦⎣

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( )( )

( )( ) φ

φτφφσ

φφσ

ttan

tansec

tansec22

22

ntoe

enheel

entoe

PPPP

PPP

PPP

−=+′−=

−′−=

neglect tail water( )

( )( ) φτ

φτφτ

φτ

tan(tan(

tantan

enheel

entoe

ntoe

nheel

PPPPPP

PPP

+′−−=−′−=

=−−=

Pe is used while considering hydrodynamicpressure exerted by tail water duringearthquake.

Calculate the factor of safety against slidingCalculate the factor of safety against sliding(should be greater than unity)

∑∑

HVμ

Calculate the shear friction factor (ranges from 3-5)

∑∑ +

HbqVμ

Safety factor against Overturning◦ The dam has to be safe against overturning at any

plane within the dam at the base or at any planebelow the dam◦ The overturning of the dam may take place if the

resultant of all the forces acting on the dam passesoutside the base.◦ But practically speaking, before overturning takes

place, other failures such as crushing of toematerial, cracking of upstream material andincrease in uplift and sliding may occur.

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The ratio of righting (stabilizing) momentsabout toe (anti-clockwise) to the over turningmoments about toe (clock-wise) is calledfactor of safety against over turning.

Its value generally varies between 1.5 to 2.5.

Safety factor against sliding◦ The horizontal loads including horizontal

components of the loads acting on a dam areresisted by frictional or shearing forces alonghorizontal or nearly horizontal planes◦ The total magnitude of the forces tending to induce

sliding shall be less than the total availableresistance along the sliding path.◦ The resistance depends upon the cohesion and the

angle of internal friction of the soil material.

For cases where cohesion is insignificant,friction is the only force that resists sliding.The factor of safety should not be less than 2and is given by

W is the total weight of the dam, u is the upliftforce P is the horizontal force. tanɸ is thecoefficient of internal friction (0.7 to 0.8)

PuW φtan)( −

If we consider cohesion too then

Where C is the cohesion of the material and AP

CAuW +− φtan)(

Where C is the cohesion of the material and Ais the area in sq. meters.

For full reservoir conditions and maximumflood discharge, it should not be less than 3and should not be less than 1.5 undermaximum flood discharge with extreme upliftcondition

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Safety against crushing◦ It is ensured if the compressive stresses produced

are within the allowable limits◦ Equations for normal stresses for toe and heel have

been already described.◦ When pressures at both toe and heel are

compressive, the max. compressive stress occurs atthe toe when e = B/6.◦ Excessive stresses at toe and heel can be brought

into limits by providing fillets at slopes 1:1 on u/sand 2:1 on d/s at heights

9.01.15.6

07.1658.062

2

−−=

−−=

HHh

HHh

toe

heel

h = fillet height (m)H = dam height (m)The maximum allowable compressive stress is

taken as 30kg/cm2.

Safety against tension

◦ A gravity dam is usually designed such that no tension is developed in the dam body and for that the resultant must lie within the middle third.◦ If the eccentricity e is greater than B/6, tension is

developed at the heel.