Gravity Dam [PDF]

Zitiervorschau

   

EiABC                                                                                                         Chair of Infrastructure   

2. Concrete Dams 2.1. Forces Acting and Load combination on dams The structural integrity of a dam must be maintained across the range of circumstances or events likely to arise in service. The design is therefore determined through consideration of corresponding spectrum of loading conditions. In all foreseeable circumstances the stability of the dam and foundation must be ensured, with stresses contained at acceptable levels and watertight integrity essentially unimpaired. The Gravity dam is subjected to the following main forces 1. Water Pressure (Water load)

5. Silt Pressure

2. Weight of the Dam (Self weight)

6. Ice Pressure

3. Seepage and Uplift Pressure

7. Wind Pressure

4. Wave Pressure

8. Earth Quake Force

Figure 2.1 Schematic of principal loads: Gravity dam profile Loads can be classified in terms of applicability or relative importance as primary loads, secondary loads, and Exceptional loads. i) Primary loads: - are identified as those of major importance to all dams irrespective of type. Example self weight, water & related seepage loads. ii) Secondary loads: - are universally applicable although of lesser magnitude (e.g. Silt load) or alternatively are of major importance only to certain types of dam (e.g. thermal effects with in concrete dams). iii) Exceptional loads:- are so designed on the basis of limited general applicability of occurrence (e.g. tectonic effects, or the inertia loads associated with seismic activity) 1 Water Works Construction                                               Lecture Supporting Material by  Year 2010/11                                         Dereje Tadesse  

   

EiABC                                                                                                         Chair of Infrastructure   

Gravity dam Loads It is convenient to compute all the forces per unit length of the dam. i.

Primary Loads

Water Load: - the water pressure acts on the u/s and d/s faces of the dam. The water pressure on the u/s face is the main destabilizing (or over turning) force acting on a gravity dam. The tail water pressure helps in stability. The tail water pressure is small in comparison to water pressure on the u/s face. The water pressure always acts normal to the surface. While computing the forces due to water pressure on inclined surface, it is convenient to determine the components the forces in horizontal and vertical directions instead of the total force on the inclined surface directly. The forces due to water pressure are discussed below separately for the non-overflow and section and overflow sections. a. Non- overflow section i. Upstream Vertical face:The horizontal component is given by

PH = γH 2 It acts horizontally at H/3 form the base of the dam The tail water pressure is

( )

PH = γ H ' It acts at H’/3 form the base of the dam 2

ii.

Up stream inclined face:The horizontal component is given by PH = γH 2 It acts horizontally at H/3 form the base of the dam The vertical component Pv of the water pressure per unit length is equal to the weight of the water. PV = PV 1 + PV 2

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EiABC                                                                                                         Chair of Infrastructure    The tail water is the same in both cases, and it has a vertical componeb. Overflow Section:i. Upstream vertical face:-

net.

The horizontal component is given by ⎡⎛ H + H 2 ⎞ ⎤ PH = ⎢⎜ 1 ⎟ + H a ⎥ (H 2 + H 1 ) 2 ⎠ ⎣⎝ ⎦ It acts horizontally at _ ⎡ 2 H + 3H a + H 2 ⎤ 1 Z = (H 1 + H a )⎢ 1 ⎥ form 3 ⎣ H1 + 2H a + H 2 ⎦ the base of the dam. If the velocity of approach (Va) is neglected, Ha is taken as zero in the above equation.

ii.

Up Stream inclined face:-

The horizontal component force will be the same as previous, however there will additional vertical force due to the weight of the water in triangular prism ABC, and it acts on the centroid of the area.

Self Weight load It is the main stabilizing force in gravity dam. Determined with respect to an appropriate unit weight of the material Pm=γc Ap KN/m acts through the centroid of x- sectional area AP. (γc ≈23.5 KN/m3) Where crest gates & other ancillary structures of considerable weight exist they must also be considered in determining Pm & their appropriate position of line of action. Seepage & Uplift load: 3 Water Works Construction                                               Lecture Supporting Material by  Year 2010/11                                         Dereje Tadesse  

   

EiABC                                                                                                         Chair of Infrastructure    The water enters the pores, cracks and fissures within the body of the dam, at the interface between the dam and within the foundation. Because the water under pressure, it creates uplift pressure on the dam. The pressure acts in all directions, but the pressure acting upwards is important for the design of the dam, as it reduces the effective weight of the dam. The magnitude of the uplift pressure depends upon; the character of the foundation, the materials used in construction, grout curtains, the drainage conditions, and method of construction. The computation of forces due to uplift pressure requires the determination of the area on which it acts and intensity of the uplift pressure at various points. a. Area Factor: - the uplift pressure generally does not occur on the entire horizontal area, because in some portions, there are no pores in which water can enter. The area factor can be determined experimentally. The modern practice is to take the area factor as unity, i.e. it is assumed that the uplift pressure acts on 100% of the horizontal area within the body of the dam, at the interface, and within the foundations. b. Intensity of Uplift pressure: - the uplift pressure at any point depends upon the depth of the water at that point. The pressure variation along the base is assumed to be linear between the u/s and d/s faces.

Total uplift force on the base of the dam U = average pressures intensity * area γ H +H' U= * (1 * B ) 2

(

)

5H + 2 H ' acts @ Z = form the toe of 3(H + H ') the dam _

c. Effect of drains on uplift pressure:- to reduce the uplift pressure, drains are formed through the body of the dam and also drainage holes are drilled in the foundation rock. These drains and drainages holes are usually provided near the u/s face. 4 Water Works Construction                                               Lecture Supporting Material by  Year 2010/11                                         Dereje Tadesse  

   

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Pu = η Ah (Uw ,avg) ⎛ H + H'⎞ = η . Ah .γ w ⎜ ⎟ ⎝ 2 ⎠

if no drain functioning.

η is area reduction factor Ah nominal plane area at a section considered Ah = (B*1) If no drains functioning T (2 H 2 + H 1 ) m Pu acts at Y1 = 3 H 2 + H1 In modern dams internal uplift is controlled by the provision of vertical relief drains close behind the u/s face. Mean effective head at the line of drains, Zd can be expressed as Hd = H2+Kd(H1-H2)m Kd is function of drain geometry (i.e. diameter, special & relative location with u/s face.) (USBR) Kd= 0.33 Kd = 0.25 Tennessee valley Authority Kd= 0.25-0.5 appropriate to the site by the U.S crops of Eng’s The standard provision of deep grout curtain below the u/s face intended to limit seepage also serves to inhibit pressure within the foundation. However, less certain than efficient draw system & its effect is commonly disregarded in uplift reduction. d. Effect of Tension Crack: - the uplift pressure diagram gets modified if a tension crack develops in concrete. The tension crack occurs on the u/s side of horizontal section if the reservoir water pressure (γH ) exceeds the vertical 5 Water Works Construction                                               Lecture Supporting Material by  Year 2010/11                                         Dereje Tadesse  

   

EiABC                                                                                                         Chair of Infrastructure    stress ( f yu ) . The horizontal tension crack extends form the u/s upto a point were the computed stress is just equal to the reservoir water pressure γH

Total uplift force 1 ⎡ ⎤ U = γ ⎢ B'+ (H + H ')(B − B')⎥ 2 ⎣ ⎦

Example: - Determine the uplift force at the base of the gravity dam shown below with the following three cases i. No drain ii. With drain and grout curtain at a distance of 5m form the u/s end iii. Tension crack upto 2m from the u/s end ii. Secondary loads Wave Pressure (hydrodynamic wave load): - Waves are generated on the surface of the reservoir by the blowing winds, which cause a pressure towards the d/s side. Wave pressure depends upon the wave height. Wave height may be given by the equation:hw = 0.032 UF + 0.763 − 0.271 4 F if F ≤ 32 Km hw = 0.032 UF if F > 32 Km Where: - hw – height of the wave U – wind velocity in km/hr F – fetch or straight length of water expanse in km

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EiABC                                                                                                         Chair of Infrastructure   

F

The transient load, Pwave, generated by wave action against the dam is not normally significant. The maximum pressure intensity due to wave action is

Pwave = 2.4γ w hw and hence the total force will be Fw = 2.0γ w hw2 and acts at height of 0.375h above the still water level w Example: - Determine the force due to wave pressure on a dam with the following data: Fetch of the reservoir and velocity of the wind are 100km & 80km/hr respectively. ……………………………………………………………………………………………. Sediment load: Silt is deposited in the reservoir on u/s of the dam. This silt exerts the earth pressure on the dam, similarly to that in the case of an earth retaining wall. The generated horizontal thrust, Ps has vertical and horizontal component, and is a function of the sediment depth hs, submerged unit weight γs’ & active lateral pressure coefficient Ka

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EiABC                                                                                                         Chair of Infrastructure    Case (a) when it has a vertical u/s face and Case (b) when it has inclined u/s face, in case b it will have a vertical component γ 1 .h 2 & acting @ hs/3 above the base of the dam Ps = K a s s 2 γs’ = γs-γw where γs is sediment saturated unit weight Ka =

1 − Sin φ s 1 + Sinφ s

Where φs is angle of shearing resistance

For representative values of γs ≈18-20KN/m3 3 Z 32 Ps ≈ φs ≈ 300 2 Just after constriction of the dam, the depth (hs) of the silt is zero. It increases gradually with time, and finally it becomes equal to the height of the dead storage. It is usual practice to assume the value of hs is equal to the height of dead storage above the base. In design of dams, the silt pressure is sometimes neglected because of the following reason: Initially, the silt is not present, and by the time, it becomes significant in depth, it had already been consolidated under the weight and it becomes more like a solid and less like a liquid. Moreover, the deposited silt is some what impervious and helps in reducing the uplift pressure on the dam. Wind load: when the dam is full, wind acts only on the d/s side thus contribute to stability. When empty the wind can act on the u/s face but in significant compared to hydrostatic load. For buttress dams load on the exposed surface has to be considered. Ice load: Not a problem in Ethiopia. It can be significant where ice sheets form to appreciable thickness & persist for lengthy periods. Pice =145 KN/m2 for ice > 0.6m thick, other wise neglected Thermal & dam /foundation interaction effect: Cooling of large pours of mass concrete following the exothermic hydration of cement & the subsequent variation in ambient & water temperatures combine to produce complex & time dependent temp gradients within the dam equally. Complex interaction develops as a result of foundation deformation.

iii. Exceptional Loads Seismic load: if the designed dam is located in a region which is susceptible to earth quakes, allowance must be made for the stresses generated by the earth quake. An earth quake produces waves which are capable of shaking the earth upon which the dam is resting, in every direction. The effect of earth quake is, therefore, equivalent to imparting acceleration to the foundations of the dam in the direction in which the wave traveling at the moment. The dam has to resist the inertia forces caused by the sudden movement of earth’s crust. If the ground under a dam moves, the dam must also move with it to avoid

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EiABC                                                                                                         Chair of Infrastructure    rupture. Inertia forces must be considered in the design of dam to avoid failure due to earth quake. Inertia force always acts opposite to the direction of earth quake acceleration. The magnitude of the earth quake force depends upon a number of factors, such as the severity of earth quake, the mass of the dam and the elasticity of the material of the dam. The earth quake acceleration usually expressed as a fraction the acceleration due to gravity (g) is equal to αg where α is the seismic coefficient. Earth quake force on the body of the dam 1. Effect of horizontal acceleration: - the horizontal acceleration can occur in either u/s or d/s directions, because the dam is design for worst case, the horizontal acceleration is assumed to occur in the direction which would produce the worst combination of the forces. i. Reservoir full condition: - the worst case occurs when the earth quake acceleration moves towards the u/s direction and the corresponding inertia force acts in the d/s direction. ii. Reservoir empty condition: - the worst case occurs when the earth quake acceleration moves towards the d/s direction and the corresponding inertia force acts in the u/s direction. The horizontal force due to the earth quake is equal to the product of mass M of the dam and horizontal acceleration ⎛W ⎞ Fh = ⎜⎜ ⎟⎟ * α h g = α hW The force is assumed to act at the center of gravity of mass. ⎝g⎠ 2. Effect of horizontal acceleration:- due to vertical acceleration, the inertia forces act on the dam and on the water. The magnitude is ⎛W ⎞ Fv = ⎜⎜ ⎟⎟ * α v g = α vW ⎝g⎠ If the vertical acceleration acts down wards, the inertia force acts upwards, and the effective weight of the dam and water decreases; hence the stability reduce, because in a gravity dam the main stabilizing force the weight of the dam and vise versa. Since the vertical acceleration does not alter the volume of the concrete in the dam and the water in the reservoir, the modified weights of the dam and the water may be use in the analysis Modified weight = W ' = W ± α V W = W (1 ± α V ) Earth quake force on the body of the water 1. Effect of horizontal acceleration on water (Hydrodynamic pressure):the horizontal acceleration acting u/s towards the reservoir causes a momentary increase in the water pressure. The dam and its foundation

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EiABC                                                                                                         Chair of Infrastructure    accelerate towards the reservoir and the water resists the movement owing to its inertia, and hence the water pressure increased. The additional water pressure in known as the hydrodynamic pressure. The following simplified methods are used to estimate the hydrodynamic pressure variations. a. Van Korman’s methods:- suggested that the hydrodynamic pressure has parabolic variation and the pressure force 4H Fe = 0.555α v γH 2 acts at above the base 3π b. Zanger’s methods:- the intensity of the hydrodynamic pressure at a depth y below the water surface in the reservoir with the total depth of water H is given by Pey = Cα h γH C is dimensionless coefficient and is given by

⎡y⎛ y⎞ y⎛ y ⎞⎤ φ ⎞ ⎛ ⎜ 2 − ⎟ ⎥ and C m = 0.735⎜1 − ⎟ ⎢ ⎜2 − ⎟ + H⎠ H⎝ H ⎠ ⎦⎥ ⎝ 90 ⎠ ⎣⎢ H ⎝ The corresponding total hydrodynamic force and moment is given as Peh = Cα h γH C=

Cm 2

Feh = 0.726(Pey * H )

(

M eh = 0.299 Pey * H 2

)

Load combinations All forces which we discussed in the preceding sections may not act simultaneously on the dam. A concrete dam should be designed with regard to the most rigorous adverse groupings or combination of loads which have a reasonable probability of simultaneous occurrence. Different design authorities have different load combinations. There are three nominated load combinations in USA standard which is sufficient for almost all circumstances. In ascending order of severity we can have normal, unusual & extreme load combination (NLC, ULC, ELE respectively) (see the table below). With probability of simultaneous occurrence of load combination decreases, factor of safety should also decrease. Nominated Load combination (after USBR, 1987; Kennard Owens and Reader 1995)

Load Source PRIMARY Water Tail water Self weight

Qualification DFL NML TWL Minimum -------

a

Load Combination NLC ULC ELC

√ √ √

√ √ √

√ √ √

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EiABC                                                                                                         Chair of Infrastructure    Uplift SECONDARY Silt Ice Concrete Temperature EXCEPTIONAL Seismic

Drains functioning Drains inoperative Discretionary Minimum normal Min. @ time of event

√

√ _b

√ √ √

√ √

CME (control max. EQ)

√

√ √ √ √ √

Studies and investigations may be appropriate with respect to nominated load combinations in relation to foundation stability and/or any other loading combination which it is considered appropriate to analyze the dam under review. a DFL: Design flood level NML: Normal maximum level CME: Control maximum earth quake _b ULC should also be investigated for the ‘drains inoperative’ condition

The nominated load combinations as defined in the table are not universally applicable. An obligation remains with the designer to exercise discretion in defining load combinations which properly reflect the circumstance of the dam under consideration.

2.2. DESIGN AND ANALYSIS OF GRAVITY DAM Stability Requirements of Gravity dam The gravity dam must be in overall equilibrium (i.e. structurally safe and stable). It should not move in any direction or rotate about any point. In addition to the overall stability, the internal stresses induced anywhere in the dam must be within the safe limits. The foundation should be able to withstand the imposed loads. The essential conditions for structural equilibrium and to make the structure stable, the following governing criteria should be satisfied;

∑ H = ∑V = 0 ∑M = 0

determine that no translational moment is possible proscribes any rotational movement

Assumptions inherent in preliminary analyses using the gravity methods 1. The concrete in the dam is homogenous, isotropic, and elastic 2. The dam consists of a number of vertical cantilevers of unit length, each cantilever acting independent of the adjoining cantilevers 3. The stress in the dam and its foundation are within elastic limits 4. There is a perfect rigid bond between the dam and its foundation and both behaves as one unit 5. All loads are transferred to the foundation by the cantilever action. No loads are transferred to the abutments by the beam action 6. The foundation is strong and unyielding. No movement is caused in the foundation due to the imposed load

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EiABC                                                                                                         Chair of Infrastructure    7. Small openings, such as galleries and shafts, do not affect the overall stability of the dam 8. The vertical stresses on a horizontal plane vary linearly from u/s to d/s faces 9. The horizontal shear stresses on a horizontal plane vary parabolically form the u/s to d/s The gravity dam must be designed such that it is safe against all possible mode of failure, with adequate factor of safety. The dam may fail in one or more of the following modes:a. Rotation and Overturning b. Translation and Sliding

c. Over stressed Failure

and

material

Over turning Sliding X

Stres s

X

a) Overturning stability Factor of safety against over farming, Fo, in terms of moment about the d/s toe of the dam Fo =

∑M ∑M

+ ve − ve

∑M

− ve

inclusive of moment generated by uplift )

Fo > 1.25 may be acceptable, but Fo > 1.5 is desirable.

b) Sliding stability Factor of safety against sliding, FS, estimated using one of the three definitions: 1) Sliding factor, FSS; 2) Shear friction factor, FSF or 3) Limit equilibrium factor, FLE. The resistance to sliding or shearing which can be mobilized across a plane is expressed through parameters C & tanφ.

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EiABC                                                                                                         Chair of Infrastructure    1) Sliding factor, FSS FSS =

∑H ∑V

if it has a horizontal plane

c) Stress analysis: gravity method Gravity method is useful to analyses stress in straight dams which are not geometrically complex. It is founded on 2-D elastic dam on uniformly rigid foundation & linear variation of stress from u/s to d/s. The stresses evaluated in a comprehensive analysis are: 1) Vertical normal stress, σz, on horizontal planes 2) Horizontal & vertical shear stress, τ zy & τ yz 3) Horizontal normal stress, σy ,on vertical planes and 4) Principal stress, σ1 & σ3 (direction & magnitude).

1. Vertical normal stress σz

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EiABC                                                                                                         Chair of Infrastructure    Analysis is based on modified beam theory which is by combining axial & bending load ∑V ± ∑ M * y , σz = Ab I Where:- Σv- resultant vertical load above the plane considered exclusive of uplift. ΣM* - summation of moments expressed w.r.t the centroid of the plane y’ - distance from the centroid to point of considerations I - second moment of area of the plane w.r.t centroid. For 2-D plane section of unit width Parallel to the dam axis, & with thickness T normal to the axis:

ΣV ΣVey , and at y’ =T/2 and results ± 12 T T3 For reservoir full condition Σv ⎡ 6e ⎤ At the u/s face σ zu ⎢1 − ⎥ and T ⎣ T⎦ Σv ⎡ 6e ⎤ At the d/s face σ zd = 1+ T ⎢⎣ T ⎥⎦

σz =

σz =

Σv ⎛ 6e ⎞ ⎜1 ± ⎟ T ⎝ T ⎠

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EiABC                                                                                                         Chair of Infrastructure    Where e is the eccentricity of the resultant load, R, which must intersect the plane d/s of its centroid for the reservoir full condition and the sign convention is reversed for reservoir empty condition of loading. The eccentricity is determined by evaluating the moments, ΣM* , given by ΣM * Where Σv - excludes uplift e= ΣV For e > T/6, at u/s face –ve stress is developed, i.e. tensile stress. In design, tensile stress has to be permissible, but difficult to totally eliminate low tensile stress in gravity dam. Total vertical stresses at either face are obtained by the addition of external hydrostatic pressure. 2. Horizontal shear stresses:- The vertical stress intensity, σmax or σmin determined Σv ⎛ 6e ⎞ form σ z = ⎜1 ± ⎟ is not the maximum direct stress produced anywhere in the T ⎝ T ⎠ dam. The maximum normal stress will be, in fact, be the major principal stresses that will be generated on the major principal plane. Numerically equal & complementary horizontal (τzy) and vertical (τyz) shear stresses are generated at any point as a result of variation of vertical normal stress over a horizontal plane.

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EiABC                                                                                                         Chair of Infrastructure    u/s and d/s face angle φu & φd respectively & Pw hydrostatic pressure at u/s end, when the reservoir is full the maximum vertical stress occurs at the toe of the dam. The major principal stresses is given by

σ d = f yd sec 2 φ d − P' tan 2 φ d if there is no tail water σ d = f yd sec 2 φ d The maximum shear stress at the d/s edge of toe is given by τ d = f yd − P' tan φ d if there is no tail water τ d = f yd tan φ d When the reservoir is empty the maximum vertical stress occurs at the heel of the dam. The major principal stresses is given by

σ u = f yu sec 2 φu − PW tan 2 φu if there is no tail water σ u = f yu sec 2 φu The maximum shear stress at the u/s edge of heel is given by τ u = − f yu − PW tan φu the major principal stress acts on the u/s face and is

[

]

equal to Pw . for the reservoir empty conditions, Pw = 0

τ u = − f yu tan φu

To avoid overstressing of the material, the principal stresses should not exceed the allowable compressive stress in the dam and foundation. 3. Principal stresses

σ1& σ3 may be determined from knowledge of σz& σy and construction of Mohr’s circle diagram to represent stress conditions at a point, or by application of the equation given below. σz +σy + τ max Major Principal Stress σ 1 = 2 σ +σy Minor principal stress σ 3 = z − τ max 2 1/ 2

⎡ σ z −σ y ⎤ +τ 2 ⎥ Where τ max = ⎢ 2 ⎢⎣ ⎥⎦ The u/s and d/s faces are each planes of zero shear, and therefore planes of principal stress. The boundary values, σ1 & σ3 are determined by: For upstream face σ1u= σzu (1+ tan2φu) - Pw tan 2φu σ3u = Pw For downstream face assuming no tail water σ1d=σzd (1+tan 2φd) σ3d=0

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