gd1

7/21/2019 GD1

http://slidepdf.com/reader/full/gd15695d1dc1a28ab9b02982ff3 1/110

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

7/21/2019 GD1


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.

7/21/2019 GD1


Contd

• Secondary Loads: are universally applicable although of lessermagnitude (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).

7/21/2019 GD1


Contd

• loading diagram on gravity dams

7/21/2019 GD1


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

7/21/2019 GD1


Contd

7/21/2019 GD1


Contd

II. Seepage loads/uplift

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

foundation

7/21/2019 GD1


Contd

7/21/2019 GD1


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

7/21/2019 GD1


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

7/21/2019 GD1


Contd

7/21/2019 GD1


Contd

III. Self weight load

.

For a gravity dam the weight of the structure is the main stabilizingforce, 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 theabsence s ecific data from laborator test trials

7/21/2019 GD1


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.

7/21/2019 GD1


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

7/21/2019 GD1


Contd

Fetch length ( fetch – continuous area of water over which

the wind blows in a constant direction)

7/21/2019 GD1


Contd

As a basis for wave height computation, Hs

(crest to trough), the

Stevenson equation can be used.

7/21/2019 GD1


Contd

III. Wind Load

When the dam is full, wind acts only on the downstream side thuscontribute to stability

It may be taken as 100 to 150 kg/m² for

the area exposed to the wind pressure

(Varshney, 1986).

ToeHeel

F'H

FW

FU

F'V

FOD

W

FV

FWA

FWA

FH

Fs

Where:

H = Head water depth

H’ = Tail Water depth

F WA = Wave pressure force

F H = Horizontal hydrostatic force

F S = Silt/sediment pressure force

F EQ = Earthquake/Seismic force

F W = Wind pressure force

F H’ = Tail water hydrostatic force

W = Weight of dam

F OD = Internal pore water pressure

F U = Uplift pressure force [base of dam]

F V = Weight of water above dam [u/s]

F V ’ = Weight of water above dam [d/s]

7/21/2019 GD1


7/21/2019 GD1


Contd

Exceptional Loads

I. Seismic Load

Under reservoir full conditions, the most adverse seismic loadingwill 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

7/21/2019 GD1


Contd

The acceleration intensities are expressed by acceleration

coefficients h (Horizontal) and v (vertical) each representing the

ratio of peak ground acceleration

Horizontal and vertical accelerations are not equal, the former

being of greater intensity ( h = (1.5 – 2.0

v ).

Inertia forces

Horizontal Feqh

= ±αh

W

Vertical Feqv = ±αvW

7/21/2019 GD1


Contd

Water body

• As analyzed by Westerguard(1993)

where k” = earthquake factor for the water body

''.32

.''

k y H yF

y H k P

whewy

wh y

γ α

γ α

=

=

2

100075.71

816.0"

−=

T

H 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.

7/21/2019 GD1


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 applicationof the appropriate seismic coefficient to vertical water load. It is

considered to act through the centroid of the area.

Fewv = ±αv Fv

φ γ α cos.'' y H k P wh y =

7/21/2019 GD1


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

7/21/2019 GD1


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)

7/21/2019 GD1


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)

7/21/2019 GD1


Contd

7/21/2019 GD1


ContdGRAVITY DAM DESIGN AND ANALYSIS

The essential conditions to structural equilibrium and basicstability requirements for a gravity dam for all conditions of loading

are

Safe against overturning at any horizontal plane within thestructure, 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.

7/21/2019 GD1


ContdContd

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-sidedcantilevers 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 ∑ = 0 M

7/21/2019 GD1


ContdContd

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

ve

0M

MF

7/21/2019 GD1


Contd

∑∑=

V M Locationt sul tanRe

Overturning stability is considered satisfactory if the resultantintersects 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

7/21/2019 GD1


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.

7/21/2019 GD1


Contd

Sliding Stability

d

7/21/2019 GD1


Contd

The resistance to sliding or shearing, which can be mobilized across a

plane, is expressed through the parameters cohesion, c, and frictional

resistance, tanΦ.

Sliding Factor, FSS

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

over the plane considered

∑∑=

V

HFSS

7/21/2019 GD1


C d

7/21/2019 GD1


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.

∑

=

H

SFSF

( ) ( )∑ α+φ+αφ−α= tanVtantan1cos

cAS h

where Ah is the area of plane of contact or sliding

C d

7/21/2019 GD1


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

7/21/2019 GD1


C td

7/21/2019 GD1


Contd

This is effected by modifying the equations accordingly as,

( ) ∑+= HPSF pSF

( ) ( )α+φ+

αφ−α= tanW

tantan1cos

AcP w

AB

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

C td

7/21/2019 GD1


Contd

.

USBR recommended values of FSF summarized

C td

7/21/2019 GD1


Contd

.Limit equilibrium factor, FLE

This approach follows the conventional soil mechanics logic in

defining the limit equilibrium factor, F LE , as the ratio of shear strength

to mean applied shear stress across the plane

τ

τ= f

LEF

Contd

7/21/2019 GD1


Contd

.

For a single plane sliding mode, the above equation will be

Note that for α =0 (horizontal sliding plane) the above expression

simplifies toFLE = FSF

Contd

7/21/2019 GD1


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.

7/21/2019 GD1


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

7/21/2019 GD1


Contd

• The primary stresses determined in a comprehensive analysis by

the gravity method are:

Contd

7/21/2019 GD1


Contd

Contd

7/21/2019 GD1


Contd

With the trapezoidal law, the vertical stress, σz, may be found by the

following equation, which is the familiar equation for beams withcombined bending and axial load:

Vertical Normal Stress

I

yM

A

V *

h

z

∑∑ ′±=σ

where

∑V = resultant vertical load above the plane considered, exclusive of

uplift,

∑M* = summation of moments determined with respect to the centroidof 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

7/21/2019 GD1


Contd

•For a regular two-dimensional plane section of unit width parallel to

the dam axis, and with thickness T normal to the axis,

.

3zT

yeV12

T

V ∑∑ ′±=σ

Contd

7/21/2019 GD1


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

7/21/2019 GD1


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

7/21/2019 GD1


Contd


.




Contd

7/21/2019 GD1


Contd

.

Vertical Stress

on the base of

a gravity dam

Contd

7/21/2019 GD1


• Horizontal shear stresses

Numerically equal and complementary horizontal (τzy) and (τyz)shear stresses are generated at any point as a result of the variationin vertical normal stress over a horizontal plane.

If the angles between the face slopes and the vertical are respectivelyΦu upstream and Φd downstream, and if an external hydrostaticpressure, pw, is assumed to operate at the upstream face, then

Upstream horizontal shear stress

Downstream horizontal shear stress

( ) uzuwu tan p φσ−=τ

d zd d tanφσ=τ

Contd

7/21/2019 GD1


Contd

.

Contd

7/21/2019 GD1


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,( ) u

2

wzuwyu tan p p φ−σ+=σ

d

2

zd yd tan φσ=σ

Contd

7/21/2019 GD1


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 σzand σy

– For major principal stress

– For minor principal stress,

max

yz

12

τ+σ+σ

=σ

max

yz

32

τ−σ+σ=σ

2yz

max

2

τ

σ

=

Contd

7/21/2019 GD1


Contd

max

yz

12

τ+σ+σ

=σ

max

yz

32

τ−σ+σ=σ

2yz

max

2

τ

σ

=

Contd

7/21/2019 GD1


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 tan ptan1 φ−φ+σ=σ

wu3 p=σ

d

2

zd d 1 tan1 φ+σ=σ

0d 3 =σ

Contd

7/21/2019 GD1


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 minimumcompressive strength for the concrete, is nevertheless

prescribed; is a common but seldom critical criterion.cσ

3≥cF

Table: Permissible compressive stresses (after USBR, 1976)

Contd

7/21/2019 GD1


Horizontal cracking is sometimes assumed to occur at the up stream

face if (computed without uplift) falls below a predetermined

minimum value:

zuσ

t

t wd

zuF

Z K

'

'

min

'σ γ σ

−=

Contd

7/21/2019 GD1


• Cracked Base Analysis

For a horizontal crack a direct solution may be obtained bythe 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

7/21/2019 GD1


A resulting negative value for

b indicates an overturning

condition with the resultant

falling downstream of the toe

After the width b is found, themaximum base pressure can be

determined, and then the overturning

and sliding stabilities can be

evaluated.

Contd

7/21/2019 GD1


• Uplift Pressure Distribution

Case-1: Uplift distribution with drainage gallery

Contd

7/21/2019 GD1


Case-2: Uplift distribution with foundation drains near upstream

face

Contd

7/21/2019 GD1


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

compression zone not extending beyond drains

Contd

7/21/2019 GD1


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

compression zone extending beyond drains

Buttress Dam

7/21/2019 GD1


• 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

7/21/2019 GD1


Buttress dams with a massive head

– Round head buttress dam – Diamond head buttress dams

– T-head buttress dams

Contd

7/21/2019 GD1


• Buttress dams with a flat slab deck

– Simple or Amburson slab buttress

– Fixed or continuous deck slab buttress Cantilever deck slab

buttress dam:

Contd

7/21/2019 GD1


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

7/21/2019 GD1


Contd

7/21/2019 GD1


• 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

K

bd c f U W GGW

..2

1

+−++=

Contd

7/21/2019 GD1


• 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

C γ

C φ

Contd

7/21/2019 GD1


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 F

SF.

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

7/21/2019 GD1


• 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

7/21/2019 GD1


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

Contd

7/21/2019 GD1


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

follows:

7/21/2019 GD1


Arch Dam

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

7/21/2019 GD1


Contd

7/21/2019 GD1


Valley suited for arch dam

• Narrow gorges provide the most natural solution for an arch damconstruction, 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

7/21/2019 GD1


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

indicate the suitability of a site for arch dam as follows

H

H BCSF

)sec(sec 21 ψ ψ ++=

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

Contd

7/21/2019 GD1


Valley type Bottom width

B

ψ1 ψ2 CSF

U shaped < H < 150 < 150 < 3.1

Narrow V shaped 0 < 350 < 350 < 2.4Wide V-shaped 0 > 350 > 350 > 2.4

Composite U-V

shaped

< 2H > 150 > 150

≅ 4.1

Wide and flatshapes

> 2H ψ1 ψ2 > 4.1

Unclassified Highly irregular valley shape

Classification of valley shapes based on CSF value

Contd

7/21/2019 GD1


• Arch dams may be grouped into two main divisions:

– Massive arch dam:- the whole span of the dam is covered by asingle 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

ContdA h t d fil

7/21/2019 GD1


• 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 abutmentyielding or instability)

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

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

Angle between arch thrust and rock contours

Contd

7/21/2019 GD1


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 centralangle, 2θ, reaches a maximum at crest level .

The profile is suited to relatively symmetrical U-shaped valleys.

Contd

7/21/2019 GD1


Constant Radius profile

ContdC l fil

7/21/2019 GD1


Constant-angle profile

Also known as variable-radius arch dam; usually have extradosand 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 aspossible, 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

7/21/2019 GD1


Constant Angle profile

Contd

7/21/2019 GD1


Cupola profile:

Has a particularly complex geometry and profile, with constantly

varying horizontal and vertical radii to either face.

ContdC C til

7/21/2019 GD1


• 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

7/21/2019 GD1


Contd

L d h d

7/21/2019 GD1


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

L d h d

7/21/2019 GD1


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

7/21/2019 GD1


• 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

7/21/2019 GD1


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

7/21/2019 GD1


Contd

If R i h b i i i h

7/21/2019 GD1


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

2sinθ R

hP wγ =

2sin2 θ γ ew

r hP ×=

ew

ew

hr R

hr R

γ

θ γ θ

=

= 2/sin22/sin2

Contd

If th thi k (t) f th h i i ll d ith it b

7/21/2019 GD1


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 γ w h may be increased by earth quake and

other pressure forces where applicable:

t

hr

t

Rew

γ σ ==

1*

all

ewhr t σ

γ =

Contd

7/21/2019 GD1


• 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 r e = r c + 0.5t and r e = r i + t

OR

all

ewhr t

σ

γ =

h

hr t

wall

cw

γ σ

γ

5.0−=

h

hr t

wall

iw

γ σ

γ

−=

Contd

B C l A l

7/21/2019 GD1


• 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

2

2

2/sin2

)1*(

==

===

θ θ θ

σ

γ θ

Bk kr V

kr hr

t

r t V 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 = 133.50 ,r = 0.544B

Contd

b Thi k C li d (Ri ) Th

7/21/2019 GD1


b. Thick Cylinder (Ring) Theory

Improvement in thin cylinder theory was made by the considering

the arch as thick cylinder .

)( 222

22

2

m MN

r r

r

r r r P

ie

ie

ew

−

×+

=σ

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

Contd

Stress is maximum at the downstream face

7/21/2019 GD1


Stress is maximum at the downstream face ,

Thickness assumed uniform at any elevation h,

With

ie r r t −=

hP wγ =

( )

+

=ie

ew

r r t

hr r 2

max

2σ for ir r =

Contd

E l

7/21/2019 GD1


• 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

S l ti U i hi li d h d

7/21/2019 GD1


• Solution-Using thin cylinder method

1. Constant radius

Let the central angle be 150o

Assume top width , 1.5m or assume 0

hhhr

t

m B

r

all

ew

e

508.05000

82.25881.9

82.25875sin2

500

2sin2

=×

==

===

σ

γ

θ

Contd

7/21/2019 GD1


Depth

(m)

Valley

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

0 500 0 258.82 0.965910 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 0.4443100 200 50.8 208.02 0.3864

( )er B 2

sin2 −

=θ

Contd

7/21/2019 GD1


Contd

2 C t t A l

7/21/2019 GD1


2. Constant Angle

The best central angle 05.133=θ

Contd

7/21/2019 GD1


gd1

Documents