GRAVITY DAMSGRAVITY DAMS

What is a DAM?What is a DAM? A dam is a hydraulic structure constructed A dam is a hydraulic structure constructed

across a river or a stream to retain the water.across a river or a stream to retain the water.

It prevents the flow of water and accumulates It prevents the flow of water and accumulates it in a deep storage reservoir.it in a deep storage reservoir.

Types of DamsTypes of Dams

w.r.t materials:w.r.t materials:

• Earth damEarth dam• Concrete damConcrete dam• Steel damSteel dam• Timber damTimber dam

w.r.t structural w.r.t structural behavior:behavior:

• Gravity damGravity dam• Arch damArch dam• Buttress damButtress dam• Embankment damEmbankment dam

w.r.t hydraulic w.r.t hydraulic behavior:behavior:

• Over flow damOver flow dam• Non over flow damNon over flow dam

w.r.t function:w.r.t function:

• Storage damStorage dam• Diversion damDiversion dam• Coffer damCoffer dam• Power generation Power generation

damdam

Concrete Concrete Gravity DamGravity Dam

Gravity dams are solid concrete structures that Gravity dams are solid concrete structures that maintain their stability against design loads maintain their stability against design loads from the geometric shape and the mass and from the geometric shape and the mass and strength of the concrete.strength of the concrete.

Gravity dam is so proportioned that its own Gravity dam is so proportioned that its own weight resists the forces acting upon it.weight resists the forces acting upon it.

Forces acting Forces acting on a on a

Gravity DamGravity Dam

Following are forces acting on a Following are forces acting on a gravity dam.gravity dam.

1)1) Water pressureWater pressure2)2) Weight of the damWeight of the dam3)3) Uplift pressureUplift pressure4)4) Ice pressureIce pressure5)5) Wave pressureWave pressure6)6) Silt pressureSilt pressure7)7) Wind pressureWind pressure8)8) Pressure due to earthquakePressure due to earthquake

1) Water Pressure1) Water Pressure

It is the major external force acting on a dam.It is the major external force acting on a dam.

The intensity of the pressure varies The intensity of the pressure varies triangularly, with a zero intensity at the water triangularly, with a zero intensity at the water surface, to a value “wh” at any depth h below surface, to a value “wh” at any depth h below the water surface. the water surface.

• Force due to water pressure, Force due to water pressure,

P = w h² / 2P = w h² / 2• w = unit weight of the water = 1000 kg/mw = unit weight of the water = 1000 kg/m 3 3

• This acts at a height of h/3 from base of the This acts at a height of h/3 from base of the dam.dam.

2) Weight of the Dam2) Weight of the Dam

Weight of the dam is the major resisting force.Weight of the dam is the major resisting force.

Unit length of the dam is considered.Unit length of the dam is considered.

Total weight of the dam acts at the centre of Total weight of the dam acts at the centre of gravity of its section.gravity of its section.

W = W1 + W2 + W3W = W1 + W2 + W3

3) Uplift Pressure3) Uplift Pressure

Uplift pressure is the upward pressure exerted Uplift pressure is the upward pressure exerted by water as it seeps through the body of the by water as it seeps through the body of the dam or its foundation.dam or its foundation.

Seeping water exerts pressure on the base of Seeping water exerts pressure on the base of the dam and it depends upon water head.the dam and it depends upon water head.

MODES OF FAILUREMODES OF FAILURE

STABILITY STABILITY REQIREMENTSREQIREMENTS

1) Overturning1) Overturning If the resultant force cuts the base within the If the resultant force cuts the base within the

body of dam there will be no overturning.body of dam there will be no overturning.

For safety against overturningFor safety against overturning

F.O.S = F.O.S = ∑ Resisting moments∑ Resisting moments

∑ ∑ Overturning momentsOverturning moments

F.O.S ≥ 1.5F.O.S ≥ 1.5

2) Sliding2) Sliding A dam may fail in sliding at its base. A dam may fail in sliding at its base. For safety against sliding For safety against sliding

F.O.S = F.O.S = µ ×∑ V µ ×∑ V > 1 > 1 ∑ ∑ HH

Where µ = coefficient of static earth pressureWhere µ = coefficient of static earth pressure = 0.65 to 0.75= 0.65 to 0.75

3) Normal stress3) Normal stress Let Let

H = horizontal forceH = horizontal force

V = vertical forceV = vertical force

R = resultant force cutting the base at an R = resultant force cutting the base at an eccentricity e from the centre of base of width eccentricity e from the centre of base of width bb

Normal stress distribution under the Normal stress distribution under the base of dambase of dam

The normal stress is given asThe normal stress is given as

pnpn = ∑V / b (1 ± 6e / b) = ∑V / b (1 ± 6e / b)

For normal stress at toe use +ive sign For normal stress at toe use +ive sign

For normal stress at heel use -ive signFor normal stress at heel use -ive sign

Principal stressesPrincipal stresses The principal stress at the toe of the dam is The principal stress at the toe of the dam is

given asgiven as

σσ = = pn sec² βpn sec² β

and at the heel isand at the heel is

σσ = = pn sec² α – p tan² αpn sec² α – p tan² α

Where p = intensity of water pressureWhere p = intensity of water pressure

= wh= wh

Shear stressesShear stresses

The shear stress at the toe of the dam is given asThe shear stress at the toe of the dam is given as

τ = pn tan τ = pn tan ββ

and shear stress at the heel isand shear stress at the heel is

τ = - (pn-p) tan ατ = - (pn-p) tan α

Flow chart Flow chart for the Analysis offor the Analysis of

Gravity Dam Gravity Dam

1) Consider unit length of the dam.1) Consider unit length of the dam.

2) Calculate the vertical forces:2) Calculate the vertical forces:

weight of the dam,weight of the dam,

weight of water acting on inclined faces,weight of water acting on inclined faces,

uplift force.uplift force.

find sum of these vertical forces ( ∑V )find sum of these vertical forces ( ∑V )

3) Find out the sum of horizontal forces:3) Find out the sum of horizontal forces:

horizontal component of the water pressurehorizontal component of the water pressure

is P = w h² / 2is P = w h² / 2

On both U/S and D/S sideOn both U/S and D/S side

Moment due to various forces at Moment due to various forces at the toethe toe

4) Calculate 4) Calculate Overturning moments (taken as -ive) Overturning moments (taken as -ive) and and Resisting moments (taken as +ive) Resisting moments (taken as +ive) and also find their algebraic sumand also find their algebraic sum ∑ ∑M = ∑Mr - ∑MoM = ∑Mr - ∑Mo

Calculation of the factor of Calculation of the factor of safetysafety

5) Check safety against overturning5) Check safety against overturning

F.O.S = F.O.S = ∑Mr ∑Mr ≥ 1.5≥ 1.5

∑ ∑MoMo

6) Check safety against sliding6) Check safety against sliding

F.O.S = F.O.S = µ ×∑ V µ ×∑ V > 1 > 1

∑ ∑ HH

7) Calculate the shear friction factor. 7) Calculate the shear friction factor.

In large dams, shear strength of joint should In large dams, shear strength of joint should also be considered. Factor of safety in that also be considered. Factor of safety in that case is known as shear friction factor (S.F.F). case is known as shear friction factor (S.F.F).

S.F.F = S.F.F = µ∑V + bqµ∑V + bq

∑ ∑HH

b = width of the jointb = width of the joint

q = shear strength of the joint (14 kg/cm³)q = shear strength of the joint (14 kg/cm³)

Calculation of stressesCalculation of stresses

8) Find out the location (i.e. distance x) of resultant 8) Find out the location (i.e. distance x) of resultant force from the toe.force from the toe.

x = x = ∑ M∑ M

∑ ∑ VV

9) Find out eccentricity e of the resultant from the 9) Find out eccentricity e of the resultant from the centre.centre.

e = b/2 – xe = b/2 – x

where b = base width of the damwhere b = base width of the dam

10) Find the normal stress at the toe.10) Find the normal stress at the toe.

pnpn = ∑V/b (1+ 6e/b) = ∑V/b (1+ 6e/b)

(compressive stress is taken as positive)(compressive stress is taken as positive)

11) Find the normal stress at the heel.11) Find the normal stress at the heel.

pnpn = ∑V/b ( 1- 6e/b) = ∑V/b ( 1- 6e/b)

12) Find out principal stress at 12) Find out principal stress at the toethe toe

σσ = = pn sec² βpn sec² β

13) Find out principal stress at 13) Find out principal stress at the heelthe heel

σσ = = pn sec² α – p tan² αpn sec² α – p tan² α

pp = intensity of water pressure = intensity of water pressure

14) Find out shear stress at toe14) Find out shear stress at toe

τ = pn tan βτ = pn tan β

15) Find out shear stress at the heel15) Find out shear stress at the heel

τ = - (pn-p) tan ατ = - (pn-p) tan α

ExamplesExamples

Question # 01: A masonry dam 10m high is Question # 01: A masonry dam 10m high is trapezoidal section with a top width of 1m and trapezoidal section with a top width of 1m and a bottom width of 8.25m. The face exposed to a bottom width of 8.25m. The face exposed to the water has a batter of 1:10. Test the stability the water has a batter of 1:10. Test the stability of the dam. Find out the principal stresses at of the dam. Find out the principal stresses at the toe and heel of the dam. Assume unit the toe and heel of the dam. Assume unit weight of masonry as 2240 Kg/m³, w for water weight of masonry as 2240 Kg/m³, w for water = 1000 Kg/m³ and permissible shear stress of = 1000 Kg/m³ and permissible shear stress of joint = 14 Kg/cm³. joint = 14 Kg/cm³.

SolutionSolution1) consider unit length of the dam i.e.1) consider unit length of the dam i.e.

1m1m

2) vertical forces:2) vertical forces:

a) self weight of the dama) self weight of the dam

= [( ½ = [( ½ × × 11× × 10) + ( ½10) + ( ½×× 6.25 6.25 × × 10 ) + 10 ) + (1 (1× × 10)] 10)] × × 22402240

= 103600 kg= 103600 kg

b) weight of water in column AA’B b) weight of water in column AA’B

= (½ × 1 × 10) × 1 × 1000= (½ × 1 × 10) × 1 × 1000

= 5000 kg= 5000 kg

c) Uplift pressure = ½ × 8.25 × (10×1000)c) Uplift pressure = ½ × 8.25 × (10×1000)

= 41250 kg= 41250 kg

∑ ∑V = 103600 + 5000 – 41250V = 103600 + 5000 – 41250

= 67350 kg= 67350 kg

4) Horizontal water pressure 4) Horizontal water pressure

∑ ∑H = wh²/2 = 1000 × 100 / 2H = wh²/2 = 1000 × 100 / 2

= 50,000 kg= 50,000 kg

Moment due to various forces at Moment due to various forces at the toethe toe

4) Moment calculation about toe4) Moment calculation about toe

a) Due to self weight of the dam a) Due to self weight of the dam

= {( ½ ×1×10×2240) (1+6.25+1/3)} += {( ½ ×1×10×2240) (1+6.25+1/3)} +

{( 1×10×2240) (6.25+0.5)} + {( 1×10×2240) (6.25+0.5)} +

{( ½×6.25×10×2240) (2/3×6.25)} {( ½×6.25×10×2240) (2/3×6.25)}

= 527800 kg-m (+ ive)= 527800 kg-m (+ ive)

b) Due to column of water in AA’Bb) Due to column of water in AA’B = ½ (10×1×1000) (8.25 – 1/3)= ½ (10×1×1000) (8.25 – 1/3) = 39583 kg-m (+ ive) ]= 39583 kg-m (+ ive) ]

c) Due to uplift force c) Due to uplift force = 41250×2/3×8.25= 41250×2/3×8.25 = 226875 kg-m (- ive) = 226875 kg-m (- ive)

d) Due to horizontal water pressured) Due to horizontal water pressure = 50,000 = 50,000 × × 10/310/3 = 166,700 kg-m (- ive)= 166,700 kg-m (- ive)

∑∑M = 527800+ 39583+ 226875 + 166,700M = 527800+ 39583+ 226875 + 166,700 = 567383 – 339575= 567383 – 339575 = 227808 kg-m= 227808 kg-m

Calculation of the factor of Calculation of the factor of safetysafety

5) Factor of safety against overturning5) Factor of safety against overturning = = ∑Mr ∑Mr = = 567383 567383 ∑ ∑Mo 339575Mo 339575 = 1.67 > 1.5 …..(O.K) = 1.67 > 1.5 …..(O.K)

6) Factor of safety against sliding6) Factor of safety against sliding

F.O.S = F.O.S = µ ×∑ V µ ×∑ V = = 0.75×673500.75×67350 ∑ ∑ H 50,000H 50,000 = 1.01 > 1 …….(O.K)= 1.01 > 1 …….(O.K)

7) Shear friction factor7) Shear friction factor

S.F.F = S.F.F = µ∑V + bqµ∑V + bq

∑ ∑HH

= = 0.75×67350+8.25×14×10^4 0.75×67350+8.25×14×10^4

50,00050,000

= 24.11= 24.11

Calculation of Calculation of stressesstresses

8) The resultant acts at a distance x from toe8) The resultant acts at a distance x from toe

x = x = ∑ M ∑ M = = 227808 227808

∑ ∑ V 67350V 67350

= 3.38= 3.38

9) Its distance from the centre is9) Its distance from the centre is

e = b/2 – x = 8.25/2 – 3.38 = 0.74 e = b/2 – x = 8.25/2 – 3.38 = 0.74

10) 10) Compressive stress at the toeCompressive stress at the toe

pnpn = ∑V/b (1+ 6e/b) = ∑V/b (1+ 6e/b)

= 67350 / 8.25 {1 + (6×0.74)/8.25}= 67350 / 8.25 {1 + (6×0.74)/8.25}

= 12560 kg/m²= 12560 kg/m²

11) Compressive stress at the heel11) Compressive stress at the heel

pnpn = ∑V/b (1- 6e/b) = ∑V/b (1- 6e/b)

= 67350 / 8.25 {1 - (6×0.74)/8.25} = 67350 / 8.25 {1 - (6×0.74)/8.25}

= 3770 kg/m²= 3770 kg/m²

12) 12) principle stress at the toeprinciple stress at the toe

= = pn sec² βpn sec² β

:. sec :. sec ββ = 1 / (10 / 11.792) = 1 / (10 / 11.792)

= 1.179= 1.179

= 12560 = 12560 × × (1.179)^2(1.179)^2

= 17460 kg/m²= 17460 kg/m²

13) principle stress at the heel13) principle stress at the heel

= = pn sec² α – p tan² αpn sec² α – p tan² α

:.tan :.tan αα = 1 / 10 = 0.1 = 1 / 10 = 0.1

:. Sec α = 1/( 10/10.05):. Sec α = 1/( 10/10.05)

= 3770×1.01- (1000×10) ×0.1^2= 3770×1.01- (1000×10) ×0.1^2

= 3707.7 kg/m²= 3707.7 kg/m²

14) shear stress at the toe14) shear stress at the toe τ = pn tan βτ = pn tan β :. tan β = 6.25 / 10 = 0.625 :. tan β = 6.25 / 10 = 0.625 = 12560 × 0.625= 12560 × 0.625 = 7850 kg/m²= 7850 kg/m²

15) shear stress at the heel15) shear stress at the heel τ = - (pn-p) tan ατ = - (pn-p) tan α = - (3770 - 1000×10) × 0.1= - (3770 - 1000×10) × 0.1 = 623 kg/m²= 623 kg/m²

Question # 02 : A gravity dam has the following Question # 02 : A gravity dam has the following dimensions:dimensions:

Height of dam = 100 mHeight of dam = 100 m

Free board = 1 mFree board = 1 m

Slope of upstream face = 0.15 : 1Slope of upstream face = 0.15 : 1

Taking Taking αα = 0.1 = 0.1

Determine Determine

(i) hydrodynamic earthquake pressure and (i) hydrodynamic earthquake pressure and

(ii) its moment at joint situated 50 m below (ii) its moment at joint situated 50 m below maximum water surface.maximum water surface.

SOLUTION: SOLUTION: If If ø is the angle that upstream slope makes ø is the angle that upstream slope makes

with the vertical, we havewith the vertical, we have ø = tan ¹ (0.15/1) = 8.5º ø = tan ¹ (0.15/1) = 8.5º

Hence Hence θ θ = 90º - ø = 81.5º= 90º - ø = 81.5º

Cm = 0.735 Cm = 0.735 × × θ / θ / 90º 90º = 0.735 = 0.735 ×× ( 81.5º / 90º ) ( 81.5º / 90º )

= 0.666 = 0.666 Cm = maximum value of pressure coefficient Cm = maximum value of pressure coefficient for a given slopefor a given slope Here h = 100 m and y = 50 m Here h = 100 m and y = 50 m

Cv = Cm/2 [y/h(2-y/h) + {y/h(2-h/y) }Cv = Cm/2 [y/h(2-y/h) + {y/h(2-h/y) }½]½] = 0.666/2 [0.5(2-0.5)+ {0.5(2-0.5)}½] = 0.666/2 [0.5(2-0.5)+ {0.5(2-0.5)}½]

= 0.538= 0.538

Cv = pressure coefficientCv = pressure coefficient

The hydrodynamic pressure intensity at a depth ‘y’ The hydrodynamic pressure intensity at a depth ‘y’ below the maximum water level is given as below the maximum water level is given as

pev = Cv pev = Cv × × αα × × w w × × h h = 0.538 = 0.538 × × 0.1 0.1 × × 1000 1000 × × 100100 = 5380 kg/cm² = 5380 kg/cm²

:. :. αα = acceleration coefficient = acceleration coefficient

Pey = 0.726 Pey = 0.726 × × pev pev × × yy =0.726 =0.726 × × 5380 5380 × × 5050 =1.95 =1.95 × × 10^5 kg10^5 kg

Mey = 0.299 × pey × y² Mey = 0.299 × pey × y²

= 0.299 × 5380 × (50)= 0.299 × 5380 × (50)²² = 4.02 = 4.02 × × 10^6 kg-m10^6 kg-m

The EndThe End

Question # 03 : Considering earthquake forces, Question # 03 : Considering earthquake forces, in addition to the hydrostatic pressure and in addition to the hydrostatic pressure and uplift pressure, determine the base width of the uplift pressure, determine the base width of the elementary profile of gravity dam so that elementary profile of gravity dam so that resultant passes through the outer third points.resultant passes through the outer third points.

SOLUTION:SOLUTION:

Let Let b= base width of the elementary dam ABCb= base width of the elementary dam ABC The various forces acting on the dam are shown The various forces acting on the dam are shown

in figure.in figure.

(a)(a) VERTICAL FORCESVERTICAL FORCES 1. Force due to self-weight of dam1. Force due to self-weight of dam = W = ½ b h w ρ= W = ½ b h w ρ 2. Force due to vertical acceleration of 2. Force due to vertical acceleration of

earthquake earthquake ==

3. Force due to uplift = 3. Force due to uplift = U = U = 1/2 b h w1/2 b h w

= ½ b h w [(1-α)ρ-1]= ½ b h w [(1-α)ρ-1]

Where w = unit weight of water, Where w = unit weight of water,

ρ = specific weight of concrete andρ = specific weight of concrete and

α = coefficient of earthquake acceleration. α = coefficient of earthquake acceleration.

(b)(b) HORIZONTAL FORCESHORIZONTAL FORCES

1.1. Force due to water pressure P = ½ w h²Force due to water pressure P = ½ w h²

2.2. Force due to hydrodynamic pressure of Force due to hydrodynamic pressure of water at base:water at base:

Cm= 0.735 Cm= 0.735

Pe = Cm α w h = 0.735 αwhPe = Cm α w h = 0.735 αwh

Pe = 0.725 pe.hPe = 0.725 pe.h

Me = 0.299 pe.h² = 0.299 * 0.735 α w h³Me = 0.299 pe.h² = 0.299 * 0.735 α w h³

= 0.2205 α w h³.= 0.2205 α w h³.

3.3. INERTIA FORCE ( horizontal) = αW = ½ α b h INERTIA FORCE ( horizontal) = αW = ½ α b h w pw p

The resultant of all forces has to pass through the outer The resultant of all forces has to pass through the outer third point M2. Moment of all these forces at this third point M2. Moment of all these forces at this point must be zero.point must be zero.

b²[ ( 1 - α )p-1 ] - bh²wαp/6 - wh²/6 [ 1 +α (1.323) ] = b²[ ( 1 - α )p-1 ] - bh²wαp/6 - wh²/6 [ 1 +α (1.323) ] = 00

b = h( )b = h( )

it is required expression.it is required expression.

Putting α= 0 when no earthquake acts, the value of b Putting α= 0 when no earthquake acts, the value of b reduces toreduces to

b = b =