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
Contd
II. Seepage loads/uplift
The uplift is supposed to act on the whole width of the
foundation
Contd
Contd
Uplift pressure distribution for perfectly tight cutoff walls
γωh2
γωh2
γωh1
When flow from u/s to d/s face is allowed
γωh1
With u/s effective cutoff
γωh2
With d/s effective cutoff
γωh2
γωh1
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
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
Where:
H = Head water depth
FWA
FV
H’ = Tail Water depth
FWA = Wave pressure force
FW
FWA
FH = Horizontal hydrostatic force
W
FH
FS = Silt/sediment pressure force
F'V
FEQ = Earthquake/Seismic force
Fs
F'H
FOD
Heel
Toe
FU
 It may be taken as 100 to 150 kg/m² for
the area exposed to the wind pressure
(Varshney, 1986).
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

I.
Exceptional Loads
Seismic Load
Reservoire full
Reservoir empty
Earthquake Direction
Direction of vibraion
 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.
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)
Py = k ' 'α hγ w H . y
2
Fewy = α hγ w y H . y k ' '
3
where k” = earthquake factor for the water body
k" =
0.816
 H 
1 − 7.75

 1000T 
2
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
Py = k ' 'α hγ w H . y cos φ
• 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
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
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:
∑H = ∑V = 0
∑M = 0
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.
Contd
Contd
Overturning Stability
Factor of safety against overturning, F0, in terms of moments about
the downstream toe of the dam:
F0
M
∑
=
∑M
+ ve
− ve
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
Contd
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
Re sul tan t
M
∑
Location =
∑V
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
FSS
H
∑
=
∑V
Contd
If the plane is inclined at a small angle α, the foregoing expression is
modified to
FSS
(∑ H ∑ V ) − tan α
=
1 + (∑ H ∑ V ) tan α
•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.
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
FSF =
∑H
• S is the maximum shear resistance, which can be mobilized.
cA h
S=
+ ∑ V tan (φ + α )
cos α(1 − tan φ tan α )
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
S = cA h + ∑ V tan φ
And hence
FSF =
cA h + ∑ V tan φ
∑H
Sliding and shearing resistance: shear-friction factor
Contd
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
Sliding: weak seams and passive wedge resistance
Contd
.
This is effected by modifying the equations accordingly as,
FSF = (S + Pp )
Where
∑H
c A AB
Pp =
+ Ww tan (φ + α )
cos α(1 − tan φ tan α )
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
FLE
τf
=
τ
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
Contd
Vertical Normal Stress
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:
σz =
∑V
Ah
±
*
M
∑ y′
I
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,
σz
.
V 12∑ V e y ′
∑
=
±
T
T3
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
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
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
τ u = (p w − σ zu ) tan φ u
 Downstream horizontal shear stress
τ d = σ zd tan φ d
Contd
.
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,
σ yu = p w + (σ zu − p w ) tan 2 φ u
– for the downstream face,
σ yd = σ zd tan 2 φ d
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
σz + σy
– For major principal stress
σ =
+τ
1
– For minor principal stress,
σ3 =
τ max =
σz − σy
2
+ τ2
max
2
σz + σy
2
− τ max
Contd
σ1 =
σ3 =
σz + σy
2
σz + σy
τ max =
2
+ τ max
− τ max
σz − σy
2
+ τ2
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
(
)
σ1u = σ zu 1 + tan 2 φ u − p w tan 2 φ u
σ 3u = p w
-For downstream ace , assuming no tail water
(
σ1d = σ zd 1 + tan 2 φ d
σ 3d = 0
)
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,σ c is nevertheless
prescribed; Fc ≥ 3 is a common but seldom critical criterion.
Table: Permissible compressive stresses (after USBR, 1976)
Contd
 Horizontal cracking is sometimes assumed to occur at the up stream
face if σ zu (computed without uplift) falls below a predetermined
minimum value:
σ zu min
K d γ w Z − σ 't
=
'
Ft
'
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
Contd
Contd
• The distance l between the buttresses and the angle of inclination of
the dam barrier α can be determined from the condition for sliding
stability of dam
W1 =
(G K + G P + W2 − U ) f
Ks
+ c.d .b
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 γ C 23KN/m3 , Unit shear
resistance , C = 500KN/m2 , angle of shearing resistance (internal
friction) φC = 350
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
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
B + H (secψ 1 + secψ 2 )
CSF =
H
•The usual values of C.S.F. are 2 to 5; lower value giving thinner sections
Contd
Classification of valley shapes based on CSF value
Valley type
U shaped
Narrow V shaped
Wide V-shaped
Composite U-V
shaped
Wide and flat
shapes
Unclassified
Bottom width
B
<H
0
0
< 2H
ψ1
ψ2
CSF
< 150
< 350
> 350
> 150
< 150
< 350
> 350
> 150
< 3.1
< 2.4
> 2.4
> 2H
ψ1
ψ2
Highly irregular valley shape
≅ 4.1
> 4.1
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
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
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.
b.
c.
d.
Thin cylinder theory
Thick cylinder theory
The elastic theory
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
Contd
 If R is the abutment reaction its component in the upstream
direction which resist the pressure force P is equal to R sin θ 2
 The hydrostatic pressure acting in the radial) direction P = γ w h
 Total hydrostatic force = hydrostatic pressure x projected area
P = γ w h × 2re sin θ 2
 Summing forces parallel to the stream axis
2 R sin θ / 2 = 2γ w hre sin θ / 2
R = γ w hre
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
γ w hre
R
σ =
=
t *1
t
γ w hre
t=
σ all
Note: the hydrostatic pressure γwh may be increased by earth quake and
other pressure forces where applicable:
Contd
γ w hre
t=
σ all
• 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
t=
γ w hrc
σ all − 0.5γ w h
OR
γ w hri
t=
σ all − γ w h
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
V = (t * 1)rθ
γ w hr
t=
= kr
σ
B


2
V = kr θ = kθ 
 2 sin θ / 2 
2
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.
The compressive horizontal ring stress, σ, for radius r is given by
 2 re 2 × ri 2
Pw  re +

r

σ=
2
2
re − ri



 MN
(
m
2
)
Contd
 Stress is maximum at the downstream face ,
Thickness assumed uniform at any elevation h,
t = re − ri
With P = γ w h
σ max
 2rw hre 2 

=
 t (r + r ) 
 e i 
for
r = ri
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, σ all = 5MPa
Contd
• Solution-Using thin cylinder method
1. Constant radius
 Let the central angle be 150o
 Assume top width , 1.5m or assume 0
re =
500
B
=
= 258.82m
2 sin θ 2 2 sin 75
γ w hre 9.81× 258.82h
=
= 0.508h
t=
5000
σ all
Contd
Depth Valley
(m)
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
θ = 2 sin − ( B 2 re )
Contd
Contd
2. Constant Angle
 The best central angle θ = 133.5 0
Contd