Gravity Dam

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A dam is a hydraulic structure constructed across a river or a stream to retain the water.

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

 


w.r.t materials: ‡ ‡ ‡ ‡

Earth dam Concrete dam Steel dam Timber dam

w.r.t structural behavior: ‡ ‡ ‡ ‡

Gravity dam Arch dam Buttress dam Embankment dam

w.r.t hydraulic behavior:

w.r.t function: ‡

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Over flow dam Non over flow dam

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Storage dam Diversion dam Coffer dam Power generation dam

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

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

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£ollowing are forces acting on a gravity dam. 1) 2) 3) 4) 5) 6) 7) 8)

Water pressure Weight of the dam Uplift pressure Ice pressure Wave pressure Silt pressure Wind pressure Pressure due to earthquake

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It is the major external force acting on a dam.

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The intensity of the pressure varies triangularly, with a zero intensity at the water surface, to a value ³wh´ at any depth h below the water surface.

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£orce due to water pressure, P = w h² / 2 w = unit weight of the water = 1000 kg/m 3 This acts at a height of h/3 from base of the dam.

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Weight of the dam is the major resisting force.

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Unit length of the dam is considered.

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

W = W1 + W2 + W3

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

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

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If the resultant force cuts the base within the body of dam there will be no overturning.

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£or safety against overturning £.O.S = ™ Resisting moments ™ Overturning moments £.O.S • 1.5

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A dam may fail in sliding at its base. £or safety against sliding £.O.S = µ י V > 1 ™H

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

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´et H = horizontal force V = vertical force R = resultant force cutting the base at an eccentricity e from the centre of base of width b

Normal stress distribution under the base of dam

The normal stress is given as pn = ™V / b (1 ± 6e / b) £or normal stress at toe use +ive sign £or normal stress at heel use -ive sign

  
 The principal stress at the toe of the dam is given as È = pn sec² ȕ and at the heel is È = pn sec² Į ± p tan² Į Where p = intensity of water pressure = wh


 The shear stress at the toe of the dam is given as m = pn tan ß and shear stress at the heel is m = - (pn(pn-p) tan Į

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1) Consider unit length of the dam.

2) Calculate the vertical forces: weight of the dam, weight of water acting on inclined faces, uplift force. find sum of these vertical forces ( ™V )

3) £ind out the sum of horizontal forces: horizontal component of the water pressure is P = w h² / 2 On both U/S and D/S side

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

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5) Check safety against overturning £.O.S = ™Mr • 1.5 ™Mo

6) Check safety against sliding £.O.S = µ י V > 1 ™H

7) Calculate the shear friction factor. In large dams, shear strength of joint should also be considered. £actor of safety in that case is known as shear friction factor (S.£.£). S.£.£ = µ™V + bq ™H b = width of the joint q = shear strength of the joint (14 kg/cm )

C        8) £ind out the location (i.e. distance x) of resultant force from the toe. x= ™M ™V

9) £ind out eccentricity e of the resultant from the centre. e = b/2 x where b = base width of the dam

10) £ind the normal stress at the toe. pn = ™V/b (1+ 6e/b) (compressive stress is taken as positive)

11) £ind the normal stress at the heel. pn = ™V/b ( 11- 6e/b)

12) £ind out principal stress at the toe È = pn sec² ȕ

13) £ind out principal stress at the heel È = pn sec² Į ± p tan² Į p = intensity of water pressure

14) £ind out shear stress at toe m = pn tan ȕ

15) £ind out shear stress at the heel m = - (pn(pn-p) tan Į

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

  1) consider unit length of the dam i.e. 1m 2) vertical forces: a) self weight of the dam = [( ½ × 1× 10) + ( ½× ½× 6.25 × 10 ) + (1× (1× 10)] × 2240 = 103600 kg

b) weight of water in column AA¶B = (½ × 1 × 10) × 1 × 1000 = 5000 kg c) Uplift pressure = ½ × 8.25 × (10× (10×1000) = 41250 kg

™V = 103600 + 5000 41250 = 67350 kg

4) Horizontal water pressure ™H = wh²/2 = 1000 × 100 / 2 = 50,000 kg

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4) Moment calculation about toe a) Due to self weight of the dam = {( ½ ×1×10× 10×2240) (1+6.25+1/3)} + {( 1× 1×10× 10×2240) (6.25+0.5)} + {( ½× ½×6.25× 6.25×10× 10×2240) (2/3× (2/3×6.25)} = 527800 kgkg-m (+ ive)

b) Due to column of water in AA¶B = ½ (10× (10×1×1000) (8.25 1/3) = 39583 kgkg-m (+ ive) ] c) Due to uplift force = 41250× 41250×2/3× 2/3×8.25 = 226875 kgkg-m ((- ive) d) Due to horizontal water pressure = 50,000 ' 10/3 = 166,700 kgkg-m ((- ive)

™M = 527800+ 39583+ 226875 + 166,700 = 567383 339575 = 227808 kgkg-m

C         5) £actor of safety against overturning = ™Mr = 567383 ™Mo 339575 = 1.67 > 1.5 «..(O.K)

6) £actor of safety against sliding 0.75×67350 £.O.S = µ י V = 0.75× ™H 50,000 = 1.01 > 1 ««.(O.K)

7) Shear friction factor S.£.£ = µ™V + bq ™H = 0.75× 0.75×67350+8.25× 67350+8.25×14× 14×10^4 50,000 = 24.11

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8) The resultant acts at a distance x from toe x = ™ M = 227808 ™V 67350 = 3.38

9) Its distance from the centre is e = b/2 x = 8.25/2 3.38 = 0.74

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Compressive Compressive stress at the toe pn = ™V/b (1+ 6e/b) = 67350 / 8.25 {1 + (6× (6×0.74)/8.25} = 12560 kg/m²

11) Compressive stress at the heel pn = ™V/b (1(1- 6e/b) = 67350 / 8.25 {1 - (6× (6×0.74)/8.25} = 3770 kg/m²

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 principle stress at the toe = pn sec² ȕ :. sec ȕ = 1 / (10 / 11.792) = 1.179 = 12560 ' (1.179)^2 = 17460 kg/m²

13) principle stress at the heel = pn sec² Į ± p tan² Į :.tan Į = 1 / 10 = 0.1 :. Sec Į = 1/( 10/10.05) = 3770× 3770×1.011.01- (1000 (1000× ×10) ×0.1^2 = 3707.7 kg/m²

14) shear stress at the toe m = pn tan ȕ :. tan ȕ = 6.25 / 10 = 0.625 = 12560 × 0.625 = 7850 kg/m²

15) shear stress at the heel m = - (pn(pn-p) tan Į = - (3770 - 1000 1000× ×10) × 0.1 = 623 kg/m²

Ëuestion # 02 : A gravity dam has the following dimensions: Height of dam = 100 m £ree board = 1 m Slope of upstream face = 0.15 : 1 Taking Į = 0.1 Determine (i) hydrodynamic earthquake pressure and (ii) its moment at joint situated 50 m below maximum water surface.

SO´UTION: If ø is the angle that upstream slope makes with the vertical, we have ø = tan-¹ (0.15/1) = 8.5º Hence ș = 90º - ø = 81.5º Cm = 0.735 × ș / 90º = 0.735 × ( 81.5º / 90º ) = 0.666 Cm = maximum value of pressure coefficient for a given slope Here h = 100 m and y = 50 m

Cv = Cm/2 [y/h(2[y/h(2-y/h) + {y/h(2{y/h(2-h/y) }½] }½] = 0.666/2 [0.5(2[0.5(2-0.5)+ {0.5(2{0.5(2-0.5)}½] = 0.538 Cv = pressure coefficient The hydrodynamic pressure intensity at a depth y¶ below the maximum water level is given as pev = Cv × Į × w × h = 0.538 × 0.1 × 1000 × 100 = 5380 kg/cm² :. Į = acceleration coefficient

Pey = 0.726 × pev × y =0.726 × 5380 × 50 =1.95 × 10^5 kg Mey = 0.299 × pey × y² = 0.299 × 5380 × (50)² (50)² = 4.02 × 10^6 kgkg-m

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

´et b= base width of the elementary dam ABC The various forces acting on the dam are shown in figure. (a) VERTICA´ £ORCES 1. £orce due to selfself-weight of dam =W=½bhwȡ 2. £orce due to vertical acceleration of earthquake =

3. £orce due to uplift =  = 1/2 b h w = ½ b h w [(1[(1-Į)ȡ Į)ȡ--1] Where w = unit weight of water, ȡ = specific weight of concrete and Į = coefficient of earthquake acceleration. (b) 1.

HORIZONTA´ £ORCES £orce due to water pressure P = ½ w h²

2. £orce due to hydrodynamic pressure of water at base: Cm= 0.735 Pe = Cm Į w h = 0.735 Įwh Pe = 0.725 pe.h Me = 0.299 pe.h² = 0.299 * 0.735 Į w h = 0.2205 Į w h .

3. INERTIA £ORCE ( horizontal) = ĮW = ½ Į b h w p The resultant of all forces has to pass through the outer third point M2. Moment of all these forces at this point must be zero.

b²[ ( 1 - Į )p)p-1 ] - bh²wĮp/6 - wh²/6 [ 1 +Į (1.323) ] = 0 b = h( ) it is required expression. Putting Į= 0 when no earthquake acts, the value of b reduces to b=