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3. Concrete Dam

• Forces Acting on gravity dam • Load Combination for design • Design Method of gravity dam • Loads on arch dams • Method of design • Buttress dam

Gravity Dam

Loads on concrete dams

Loads can be classified in terms of applicability/relative

importance as primary loads, secondary loads, and exceptional loads.

• Primary Loads: are identified as those of major importance to all dams, irrespective of type,

e.g. water and related seepage loads, and self-weight loads.

Contd

• Secondary Loads: are universally applicable although of lesser

magnitude (e.g. sediment load) or, – alternatively, are of major importance only to certain types of

dams (e.g. thermal effects within concrete dams). • Exceptional Loads: are so designed on the basis of limited general

applicability or having a low probability of occurrence (e.g. tectonic effects, or the inertia loads associated with seismic

activity).

Contd

• loading diagram on gravity dams

Contd

Primary Loads I. Water Load

Hydrostatic distribution of pressure with horizontal resultant force P1 Vertical component of load will also exist in the case of an upstream face batter

Contd

II. Seepage loads/uplift

The uplift is supposed to act on the whole width of the foundation

Contd Uplift pressure distribution for perfectly tight cutoff walls

γωh1

γωh2 γωh2

γωh2γωh1 γωh2

γωh1

When flow from u/s to d/s face is allowed With u/s effective cutoff

With d/s effective cutoff With an intermediate cutoff

Contd

Value of area reduction factor Suggested by 0.25 to 0.40 Henry

1.00 Maurice Levy

0.95 to 1.00 Terzaghi

the value C = 1.00 is recommended

Contd III. Self weight load

.

For a gravity dam the weight of the structure is the main stabilizing force, and hence the construction material should be as heavy as possible

Structure self weight is accounted for in terms of the resultant, W, which acts through the centroid (center of gravity) of the cross-sectional area

W = γc * A Where: γc is the unit weight of concrete A is the cross-sectional area of the structure The unit weight of concrete may be assumed to be 24 kN/m3 in the absence specific data from laboratory test trials

Contd Secondary loads I. Sediment Load

The gradual accumulation of significant deposits of fine sediment,

notably silt, against the face of the dam generates a resultant horizontal force, Ps.

Contd

II. Hydrodynamic wave

The upper portions of dams are subject to the impact of waves, Pwave. The dimensions and force of waves depend on the extent of water surface, the velocity of wind, and other factors

Wave run-up

Contd

Fetch length (fetch – continuous area of water over which the wind blows in a constant direction)

Contd

As a basis for wave height computation, Hs (crest to trough), the Stevenson equation can be used.

Contd

III. Wind Load

When the dam is full, wind acts only on the downstream side thus

contribute to stability

It may be taken as 100 to 150 kg/m² for the area exposed to the wind pressure (Varshney, 1986).

ToeHeelF'H

FW

FU

F'V

FOD

W

FV

FWA

FWA

FH

Fs

Where:

H = Head water depth

H’ = Tail Water depth

FWA = Wave pressure force

FH = Horizontal hydrostatic force

FS = Silt/sediment pressure force

FEQ = Earthquake/Seismic force

FW = Wind pressure force

FH’ = Tail water hydrostatic force

W = Weight of dam

FOD = Internal pore water pressure

FU = Uplift pressure force [base of dam]

FV = Weight of water above dam [u/s]

FV’ = Weight of water above dam [d/s]

Contd IV. Ice Load

An acceptable initial provision for ice load, where considered

necessary, is given by Pice = 145 KN/m² for ice thickness greater than 0.6m, otherwise neglected (USBR).

Not a problem in Ethiopia

Contd Exceptional Loads I. Seismic Load

Under reservoir full conditions, the most adverse seismic loading will then occur when a ground shock is associated with: – Horizontal foundation acceleration operating upstream, an – Vertical foundation acceleration operating downwards.

Earthquake Direction

Direction of vibraion

Reservoire fullReservoir empty

Contd

The acceleration intensities are expressed by acceleration

coefficients αh (Horizontal) and αv(vertical) each representing the ratio of peak ground acceleration

Horizontal and vertical accelerations are not equal, the former being of greater intensity (αh = (1.5 – 2.0αv).

Inertia forces Horizontal Feqh = ±αhW

Vertical Feqv = ±αvW

Contd

Water body

• As analyzed by Westerguard(1993)

where k” = earthquake factor for the water body

''.32

.''

kyHyF

yHkP

whewy

why

γα

γα

=

=

2

100075.71

816.0"

−

=

TH

k Where: T = period of earthquake γw = in tone/m3 H, y in meters The force acts at 0.4y from the dam joint being considered.

Contd

• For inclined upstream face of dam

• where φ is the angle the face makes with the vertical.

• The resultant vertical hydrodynamic load, Fewv, effective above an upstream face batter or flare may be accounted for by application of the appropriate seismic coefficient to vertical water load. It is considered to act through the centroid of the area.

Fewv = ±αv Fv

φγα cos.'' yHkP why =

Contd

• Load Combinations

– A concrete dam should be designed with regard to the most

rigorous adverse groupings or combinations of loads, which have a reasonable probability of simultaneous occurrence.

– Three nominated load combinations are sufficient for almost all

circumstances.

– In ascending order of severity they may be designated as normal, unusual, and extreme load combinations, denoted as NLC, ULC and ELC, respectively

Contd

• Load Combinations

• Load combination A (construction condition or empty reservoir

condition): Dam completed but no water in the reservoir and no tail water.

• Load combination B (Normal operating condition): Full reservoir elevation (or top of gates at crest), normal dry weather tail water, normal uplift, ice and uplift (if applicable)

• Load combination C (Flood Discharge condition): Reservoir at maximum flood pool elevation, all gates open, tail water at flood elevation, normal uplift, and silt (if applicable)

Contd

• Load combination D - Combination A, with earthquake.

• Load combination E - Combination A, with earthquake but no ice

• Load Combination F - Combination C, but with extreme uplift (drain inoperative)

• Load Combination G - Combination E, but with extreme uplift (drain inoperative)

Contd GRAVITY DAM DESIGN AND ANALYSIS

The essential conditions to structural equilibrium and basic stability requirements for a gravity dam for all conditions of loading are

Safe against overturning at any horizontal plane within the structure, at the base, or at a plane below the base.

Safe against sliding on any horizontal or near-horizontal planes within the structure, at the base, or on any rock seam in the foundation.

The allowable stresses in both the concrete or in the foundation material shall not be exceeded.

Contd Contd

The essential conditions to structural equilibrium and so to stability can be summarized as:

Assumptions inherent in preliminary analyses using gravity method

(USBR) are as follows: The concrete (or masonry) is homogeneous, isotropic and uniformly eastic. All loads are carried by gravity action of vertical parallel-sided cantilevers with no mutual support between adjacent cantilevers (monoliths). No differential movements affecting the dam or foundation occur as a result of the water load from the reservoir.

∑∑ == 0VH ∑ = 0M

Contd Contd

Overturning Stability Factor of safety against overturning, F0, in terms of moments about the downstream toe of the dam:

It may be noted that M-ve is inclusive of the moments generated by uplift load

F0 > 1.25 may be acceptable, but F0 ≥ 1.5 is desirable

∑∑

−

+=ve

ve0 M

MF

Contd

∑∑=

VM

Locationtsul tanRe

Overturning stability is considered satisfactory if the resultant intersects the base within the kern, and allowable stresses are not exceeded

For earthquake loads, the resultant may fall anywhere within the base, but the allowable concrete or foundation pressure must not be exceeded

The resultant location along the base is computed from

Contd

Sliding Stability Resistance to sliding any plane above the base of a dam is a function

of the shearing strength of concrete, or of the construction lift joint The sliding stability is based on a factor of safety, Fs , as a measure of determining the resistance of the structure against sliding

Estimated using one or other of three definitions:

•Sliding factor, Fss, •Shear friction factor, FSF, •Limit equilibrium factor, FLF.

Contd

Sliding Stability

Contd

The resistance to sliding or shearing, which can be mobilized across a plane, is expressed through the parameters cohesion, c, and frictional resistance, tan Φ.

Sliding Factor, FSS

FSS is expressed as a function of the resistance to simple sliding over the plane considered

∑∑=

VH

FSS

Contd

If the plane is inclined at a small angle α, the foregoing expression is modified to

•Angle α is defined as positive if sliding operates in an uphill sense.

•∑V is determined allowing for the effect of uplift.

•FSS on a horizontal plane should not be permitted to exceed 0.75 for a specified NLC; it may be permitted to rise to 0.9 under ELC.

( )( ) α+

α−=

∑∑∑∑

tanVH1tanVH

FSS

Contd

Shear-friction factor, FSF

•FSF is the ratio of the total resistance to shear and sliding which can be mobilized on a plane to the total horizontal load.

• S is the maximum shear resistance, which can be mobilized.

∑=

HSFSF

( ) ( )∑ α+φ+αφ−α

= tanVtantan1cos

cAS h

where Ah is the area of plane of contact or sliding

Contd

For the case of a horizontal plane (α =0), the above equation is simplified to

And hence

∑ φ+= tanVcAS h

∑∑ φ+

=H

tanVcAF h

SF

Sliding and shearing resistance: shear-friction factor

Contd

Sliding: weak seams and passive wedge resistance

In some circumstances it may be appropriate to include downstream passive wedge resistance, Pp, as a further component of the total resistance to sliding which can be mobilized

Contd

This is effected by modifying the equations accordingly as,

( ) ∑+= HPSF pSF

( ) ( )α+φ+αφ−α

= tanWtantan1cos

AcP wAB

p

.

Where

Ww is the weight of the wedge

In the presence of a horizon with low shear resistance, e.g. a thin clay horizon or clay infill in the discontinuity, it may be advisable to make the assumption S =0, in the above equation

Contd

.

USBR recommended values of FSF summarized

Contd

. Limit equilibrium factor, FLE

This approach follows the conventional soil mechanics logic in defining the limit equilibrium factor, FLE, as the ratio of shear strength to mean applied shear stress across the plane

ττ

= fLEF

Contd

.

For a single plane sliding mode, the above equation will be Note that for α =0 (horizontal sliding plane) the above expression simplifies to FLE = FSF

Contd

.

The recommended minima for FLE (limit equilibrium factor of safety) against sliding are • FLE = 2.0 in normal operation, i.e. with static load maxima applied, and •FLE = 1.3 under transient load conditions embracing seismic activity.

Stress Analysis-Gravity Dams

• The basis of the gravity method of stress analysis is the assumption that the vertical stresses on any horizontal plane vary uniformly as a straight line, giving a trapezoidal distribution. This is often referred to as “trapezoidal law.”

• Its validity is questionable near the base of the dam where stress concentrations arise at the heel and toe due to reentrant corners formed by the dam faces and the foundation surface.

Contd

• The primary stresses determined in a comprehensive analysis by

the gravity method are:

Contd

With the trapezoidal law, the vertical stress, σz, may be found by the following equation, which is the familiar equation for beams with combined bending and axial load:

Vertical Normal Stress

IyM

AV *

hz

∑∑ ′±=σ

where ∑V = resultant vertical load above the plane considered, exclusive of uplift, ∑M* = summation of moments determined with respect to the centroid of the plane, y’ = distance from the neutral axis of the plane to the point where σz is being determined , and I = the second moment of area of the plane with respect to its centroid.

Contd

•For a regular two-dimensional plane section of unit width parallel to the dam axis, and with thickness T normal to the axis,

.

3z TyeV12

TV ∑∑ ′±=σ

Contd

where

e is the eccentricity of the resultant load R, which must intersect the plane downstream of its centroid for the reservoir full condition. (The signs are interchanged for reservoir empty condition of loading).

.

Contd

For e > T/6, upstream face stress will be negative, i.e. tensile.

.

Requirements for stability

Concrete dam must be free from tensile stress

e ≤ B/6 (law of the middle third)

Contd

.

Vertical Stress on the base of a gravity dam

Contd • Horizontal shear stresses

Numerically equal and complementary horizontal (τzy) and (τyz)

shear stresses are generated at any point as a result of the variation in vertical normal stress over a horizontal plane.

If the angles between the face slopes and the vertical are respectively Φu upstream and Φd downstream, and if an external hydrostatic pressure, pw, is assumed to operate at the upstream face, then

Upstream horizontal shear stress

Downstream horizontal shear stress

( ) uzuwu tanp φσ−=τ

dzdd tanφσ=τ

Contd

.

Contd • Horizontal normal stresses

The differences in shear forces are balanced by the normal stresses

on the vertical planes

The boundary values for σy at either face are given by the following:

– for the upstream face,

– for the downstream face,

( ) u2

wzuwyu tanpp φ−σ+=σ

d2

zdyd tan φσ=σ

Contd • Principal stresses

The principal stresses are the maximum and minimum normal

stresses at a point

Principal stresses σ1 and σ3 may be determined from knowledge of σz and σy

– For major principal stress – For minor principal stress,

maxyz

1 2τ+

σ+σ=σ

maxyz

3 2τ−

σ+σ=σ

2yzmax 2

τ+σ−σ

=τ

Contd

maxyz

1 2τ+

σ+σ=σ

maxyz

3 2τ−

σ+σ=σ

2yzmax 2

τ+σ−σ

=τ

Contd

As there is no shear stress at and parallel to the face, that is one of the planes of principal stress, the boundary values of σ1 and σ3 are then determined as follows:

-For upstream face -For downstream ace , assuming no tail water

( ) u2

wu2

zuu1 tanptan1 φ−φ+σ=σ

wu3 p=σ

( )d2

zdd1 tan1 φ+σ=σ

0d3 =σ

Contd

• Permissible stresses and cracking The compressive stresses generated in a gravity dam by

primary loads are very low

A factor of safety, Fc, with respect to the specified minimum compressive strength for the concrete, is nevertheless prescribed; is a common but seldom critical criterion.

cσ3≥cF

Table: Permissible compressive stresses (after USBR, 1976)

Contd Horizontal cracking is sometimes assumed to occur at the up stream

face if (computed without uplift) falls below a predetermined minimum value:

zuσ

t

twdzu F

ZK'

'

min'σγ

σ−

=

Contd

• Cracked Base Analysis

For a horizontal crack a direct solution may be obtained by the following equation:

Where: B = total base width b = base width in compression Mo = sum of moments at the toe excluding uplift V = sum of vertical forces excluding uplift p = unit uplift pressure at heel

Contd

A resulting negative value for b indicates an overturning condition with the resultant falling downstream of the toe

After the width b is found, the maximum base pressure can be determined, and then the overturning and sliding stabilities can be evaluated.

Contd • Uplift Pressure Distribution Case-1: Uplift distribution with drainage gallery

Contd Case-2: Uplift distribution with foundation drains near upstream

face

Contd • Case-3: Uplift distribution cracked base with drainage, zero

compression zone not extending beyond drains

Contd • Case-4: Uplift distribution cracked base with drainage, zero

compression zone extending beyond drains

Buttress Dam

• Buttress dams consists of principal structural elements: A sloping upstream deck that supports the water

The buttress or vertical walls that support the deck and transmit

the load

• According to the structure of the dam deck, buttress dams classified as:

Buttress dams with a massive head Buttress dams with a flat slab deck Buttress dams with thin curved multiple arch deck

Contd Buttress dams with a massive head

– Round head buttress dam – Diamond head buttress dams – T-head buttress dams

Contd • Buttress dams with a flat slab deck

– Simple or Amburson slab buttress – Fixed or continuous deck slab buttress Cantilever deck slab

buttress dam:

Contd

• Longitudinal beams are used for stiffing and bracing the buttresses

• Foundation slab below the entire dam, provided with drainage openings for eliminating the uplift pressure

• The stability against sliding is ensured with the weight of water on the inclined deck and on the amount of the decrease in the uplift pressure

Contd • The distance l between the buttresses and the angle of inclination of

the dam barrier can be determined from the condition for sliding stability of dam

α

( )s

PK

KbdcfUWGGW ..2

1+−++

=

Contd • Example #The profile of the major monolith of a buttress dam is illustrated in the

figure. The stability of the dam is to be reviewed in relation to: Normal Load Condition (NLC): Water load(to design flood level + self

weight + uplift(no pressure relief drain) Static stability : Overturning, Fo>1.5; sliding (shear friction factor),FSF

>2.4. Concrete characteristics: Unit weight 23KN/m3 , Unit shear

resistance , C = 500KN/m2 , angle of shearing resistance (internal friction) = 350

Cγ

Cφ

Contd

1. Analyze the static stability of the buttress unit with respect to plane X–X under NLC and in relation to the defined criteria for Fo and FSF.

2. Concern is felt with regard to stability under possible seismic loading. Dynamic stability criteria are specified as Fo = 2.0; FSF =3.2, and will be met by prestressing as shown. Determine the prestress load required in each inclined tendon.

Contd • Solution 1.

– All calculations relating to stability refer to the monolith as a complete unit.

– Uplift is considered to act only under the buttress head, and

– The profile is subdivided into the elements A, B and C, identified in figure for convenience

Contd The load–moment table (all moments are relative to toe) is as follows:

Contd 2. The load–moment table (all moments are relative to toe) is as follows:

Arch Dam

• An arch dam is a curved dam that carries a major part of its water

load horizontally to the abutments by arch action

• Arch (or arch unit) refers to a portion of the dam bounded by two horizontal planes, 1 foot (1 meter) apart.

• Cantilever (or cantilever unit) is a portion of the dam contained between two vertical radial planes, 1 foot (or 1 m) apart.

• Extrados and Intrados: Extrados is the upstream face of arches and intrados is the downstream face of the arches.

Contd Valley suited for arch dam

• Narrow gorges provide the most natural solution for an arch dam

construction, the usually recommended ratio of crest length to dam height being 5 or less.

• The overall shape of the site is classified as a narrow-V, wide-V, narrow-U, or wide-U as shown in Figure

Contd • Sarkaria proposed a canyon shape factor (C.S.F.), which would

indicate the suitability of a site for arch dam as follows

HHBCSF )sec(sec 21 ψψ ++

=

•The usual values of C.S.F. are 2 to 5; lower value giving thinner sections

Contd

Valley type Bottom width B

ψ1 ψ2 CSF

U shaped < H < 150 < 150 < 3.1 Narrow V shaped 0 < 350 < 350 < 2.4 Wide V-shaped 0 > 350 > 350 > 2.4 Composite U-V shaped

< 2H > 150 > 150 ≅ 4.1

Wide and flat shapes

> 2H ψ1 ψ2 > 4.1

Unclassified Highly irregular valley shape

Classification of valley shapes based on CSF value

Contd

• Arch dams may be grouped into two main divisions: – Massive arch dam:- the whole span of the dam is covered by a

single curved wall usually vertical or nearly so.

– Multiple arch dam:- series of arches cover the whole span of the dam, usually inclined and supported on piers or buttresses.

• Massive arch dams are divided into the following types:

– Constant radius arch dams – Constant angle arch dams – Variable radius arch dams – Double curvature or Cupola arch dams – Arch gravity dams

Contd • Arch geometry and profile

The horizontal component of arch thrust must be transferred into the

abutment at a safe angle, β, (i.e. one that will not promote abutment yielding or instability)

At any elevation the arch thrust may be considered to enter the abutment as shown in Figure

In general an abutment entry angle, β, of between 45 and 70° is suggested

Angle between arch thrust and rock contours

Contd Arch and cupola profiles are based on a number of geometrical

forms, the more important of which are: Constant-radius profile Constant-angle profile

Constant radius profile Has the simplest geometry, combining a vertical U/S face of constant

radius with a uniform radial D/S slope The downstream face radius varies with elevation and the central

angle, 2θ, reaches a maximum at crest level. The profile is suited to relatively symmetrical U-shaped valleys.

Contd

Constant Radius profile

Contd Constant-angle profile

Also known as variable-radius arch dam; usually have extrados

and intrados curves of gradually decreasing radii as the depth below the crest increases

This is to keep the central angle as large and as nearly constant as possible, so as to secure maximum arch efficiency at all elevations.

They are often of double curvature. Frequently adapted to narrow steep-sided V-shaped valleys

It is economical type of profile using about 70% concrete as compared to a constant radius arch dam

Contd

Constant Angle profile

Contd Cupola profile: Has a particularly complex geometry and profile, with constantly

varying horizontal and vertical radii to either face.

Contd • Crown Cantilever: The crown cantilever is defined as the maximum height vertical

cantilever and is usually located in the streambed

• Single Curvature : Single-curvature arch dams are curved in plan only. Vertical

sections, or cantilevers, have vertical or straight sloped faces.

• Double Curvature : Double-curvature arch dams means the dam is curved in plan and

elevation

This type of dam utilizes the concrete weight to greater advantage than single-curvature arch dams

Contd Loads on arch dam • The forces acting on arch dam are the same as that of gravity dams

– Uplift forces are less important (not significant)

– Internal stresses caused by temperature changes and yielding

of abutments are very important

– The principal dead load is the concrete weight – The principal live load is the reservoir water pressure

– An arch dam transfers loads to the abutments and foundations

both by cantilever action and through horizontal arches

Contd

• Methods of design of massive arch dams

a. Thin cylinder theory b. Thick cylinder theory c. The elastic theory d. Other advanced methods such as trial load analysis and finite

element methods.

Contd

a. Thin Cylinder (Ring) Theory

The weight of concrete and water in the dam is carried directly to the foundation

The horizontal water load is carried entirely by arch action

In thin cylinder theory, the stresses in the arch are assumed to be nearly the same as in a thin cylinder of equal outside radius

Contd If R is the abutment reaction its component in the upstream

direction which resist the pressure force P is equal to

The hydrostatic pressure acting in the radial) direction

Total hydrostatic force = hydrostatic pressure x projected area

Summing forces parallel to the stream axis

2sinθR

hP wγ=

2sin2 θγ ew rhP ×=

ew

ew

hrRhrR

γθγθ

== 2/sin22/sin2

Contd If the thickness (t) of the arch ring is small compared with re it may be

assumed that uniform compressive stress is developed in the arch ring

The transverse unit stress For a given stress, thickness t

Note: the hydrostatic pressure γwh may be increased by earth quake and other pressure forces where applicable:

thr

tR ewγσ ==

1*

all

ewhrt

σγ

=

Contd

• This equation indicates that the thickness t of the arch ring increases

linearly with depth below the water surface and for a given pressure the required thickness is proportional to its radius.

• Thickness relation in terms of intrados, ri and mean radius rc , can be

derived as follows since re = rc + 0.5t and re = ri + t

OR

all

ewhrt

σγ

=

hhr

twall

cw

γσγ

5.0−=

hhr

twall

iw

γσγ−

=

Contd

• Best Central Angle The concrete volume of any given arch is proportional to the

product of the arch thickness and the length of the centerline arc The volume of unit height of arch

22

2/sin2

)1*(

==

==

=

θθθ

σγ

θ

BkkrV

krhr

t

rtV

w

Differentiating V with respect to θ and setting to zero, θ = 133.5o which is the most economical angle for arch with minimum volume For θ = 133.50 ,r = 0.544B

Contd

b. Thick Cylinder (Ring) Theory Improvement in thin cylinder theory was made by the considering

the arch as thick cylinder.

)( 222

222

mMN

rr

rrr

rP

ie

ieew

−

×+

=σ

The compressive horizontal ring stress, σ, for radius r is given by

Contd Stress is maximum at the downstream face ,

Thickness assumed uniform at any elevation h, With

ie rrt −=

hP wγ=

( )

+=

ie

ew

rrthrr 2

max2

σ for irr =

Contd

• Example: # Given a canyon with the following dimensions, compute and draw the

layout of arch dams of constant radius and constant angle profiles. Data - Maximum height = 100m -Top width of the valley = 500m -Bottom width of valley =200m -Allowable stress in concrete, MPaall 5=σ

Contd

• Solution-Using thin cylinder method 1. Constant radius

Let the central angle be 150o

Assume top width , 1.5m or assume 0

hhhrt

mBr

all

ew

e

508.05000

82.25881.9

82.25875sin2

5002sin2

=×

==

===

σγ

θ

Contd

Depth(m)

Valley width(m) t=0.508h ri = re - t B/2re

0 500 0 258.82 0.9659 10 470 5.08 253.74 0.9080 20 440 10.16 248.66 0.8500 30 410 15.24 243.58 0.7921 40 380 20.32 238.5 0.7341 50 350 25.4 233.42 0.6761 60 320 30.48 228.34 0.6182 70 290 35.56 223.26 0.5602 80 260 40.64 218.18 0.5023 90 230 45.72 213.1 0.4443

100 200 50.8 208.02 0.3864

( )erB 2sin2 −=θ

Contd

2. Constant Angle The best central angle 05.133=θ

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