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Chapter Three
Concrete Dam
The structural integrity of any dam must be maintained under different
loading circumstances.
 the gravity dam is mainly subjected to the following main forces;
Waterload
Self weight
Upliftload
Wave load
Silt load
Wind load
Earth quake load
2
As per the degree of relative importance loads on the dam can be classified as ;
 primary loads:- Major important loads irrespective of the dam type;
E.g. Self Wight load, water load and related seepage load
Secondary loads:- Universally applicable loads , even though their magnitude
is less;
E.g. Silt load (serve as to minimize uplift
pressure)
Exceptional loads:- loads which has limited applicability;
E.g. Tectonic load (earth quake load
Exercise :- write about the characteristic ,advantage, disadvantage of concrete
Dam
NB:-for Earth quake load don’t consider uplift loads
Concrete dam cont’s
Hw
2
2
1
HP wH 
3
H
3
Loads and their centroidal location in gravity dams
Case -1 Non-over flow section
(i) Up stream vertical face
PHH
@
3
H
from the base of the dam.
B
H’
'HP
2'
2
1
' HP wH  @
'
Hw
3
'H from the base of the dam.Ap= X-sec
area of stru.
Hw
U
  B
HH
U w
*
2
'


Throughthe centroid of
Trapezoidal,with out
drainagegallery)
[ i.e. ]
@
)'(3
]'2[
HH
HHB
Z



w
pc AW  @ Throughthe centroid of
x-sectionalarea Ap
’
vP
'*
2
1
HbP wv 
b
@ from the toe of the dam.
3
b
Z
4
(ii) Up stream face inclined
Pv1
Pv2
If the upstream face of the gravity dam is inclined
in addition to the previous loads ( loads in vertical
u/s face) , only vertical loads of water i.e. Pv will be
added at its centroidal point from the toe of the dam.
where
Pv = Pv1+ Pv2
Conti…
 aw HH 1
 aw HH 2
5
Case-2 Over flow section
H2
H1
PH
YwHHH
HH
P aH *)(
2
12
21











 

T.E.L
  








21
21
1
_
2
32
3
1
HHH
HHH
HHZ
a
a
a
Z
From the base of the dam
@
Ha=Va2/2g
H




 )'(
3
1
' HHH
6
Uplift pressure with drainage gallery and tension cracks
 To reduce the uplift pressure , drains are formed trough the body of the dam, this
make the intensity of the uplift pressure to be differ from the full concrete dam.
Drainage gallery
H
H’
T
’
’
'H
1
UFhw 032.0
4
271.0763.0032.0 FUFhw 
kmFif 32
kmFif 32
7
Wave Pressure ( hydrodynamic wave load)
Waves are generated on the reservoir by the blowing winds.
hw
w/r
hw = heightof the wave
U = wind velocity in km/hr
F = fetch length
Pwave
22 wwwave hP  @ wh375.0 above the stilled water level.
Reservoirsurface area
Dam
v
g
W
W *
8
Earthquake force
 Earth quake force may move in any direction, but for the design purpose it
has to be resolved in to vertical and horizontal components.(USBR)
 The values of these horizontal (αh) and vertical (αv) accelerations are generally
expressed as percentage of the acceleration due to gravity i.e. 0.1g or 0.2g, etc.
Vertical acceleration(αv)
Dam foundation
Upward verticalacceleration
The contactb/n the foundationand the dam
will increase,hence the effective Weight
of the dam will also be increased.
The contactb/n the foundationand
the dam Will decrease,which is the
worstcase!!
Down ward verticalmovement.
W
Effective weight of the dam
2
555.0 HF whe 
3
4H
3
4H
9
Horizontal acceleration(αh)
Hydro-dynamic pressure.
Horizontal Inertia force.
Fe
acts @
H
from the base of the dam.
Von – Karman formula
There is also a hydrodynamic formula developed by Zanger, but for average ordinary
purposes, the Von-Karman formula is sufficient.
Reading assignment,
Reference, P.Novak and S.K. Garg
2
2'
ss
ash
h
KP


3
sh
wss  '
s
s
aK


sin1
sin1



10
Sediment load
hs
Psh
N.B! it is usual practice to assume the value of hs equals
to the height of dead storage.
@
above the base of the
dam.
The submerged unit weight and the active lateral pressure coefficient Ka
is given by
s
s
's
where
is the angle of shear resistance.
is sediment saturated unit weight.
Reading assignment
-Loading combinations in dams
Reference, Novak





ve
ve
o
M
M
F
11
Design and analysis of gravity dam
Gravity dam may fail in the following way
•Rotation and overturning,
• Translation and sliding and
• Overstress and material failure.
1- Over turning stability
To make the structure of the dam stable from rotational or overturning failure the
following governing criteria should be satisfied
F0 (factor of safety against over turning) should be greater than 1.5
> 1.5……………(safe against overturning)
Include uplift moment


tan1
tan















V
H
V
H
Fss


V
H
Fss

12
2-Siding Stability (Fs)
Sliding Factor (FSS)
Shear friction Factor(FSF)
Limit equilibrium factor (FLE)
i) Sliding factor can be defined by
, for horizontal plane
ΣV is determined allowing for the effect of uplift (net)
, for foundations inclined at a small angle
In order to be the dam stable against sliding Fss should be less than or equal
to 0.75 but for ELC (extreme loading condition) up to 0.9 is acceptable.


H
S
FSF
mkNV
CA
S b
/)tan(
)tantan1(cos 







  


13
ii) Shear friction factor (FSF) is defined as
Where
S – total resistance to shear and defined by
Location of sliding plane Normal Unusual Extreme
Dam Concrete base interface
Recommended shear friction factor (USBR 1987)
Foundation rock
3.0 2.0 > 1.0
4.0 2.7 1.3
(Net)

 f
LEF 

 tann
LE
c
F



f
14
(iii) Limit Equilibrium Factor, FLE
n
= T he shearstressed generated underthe applied load
= Available shearstrengthand expressed byMohr
coulombfailure criteria
= Stress acting normalto plain of sliding
FLE = 2.0 for normal operation
FLE = 1.3 for seismic activity
AVn /
H
V








T
e
T
V 6
1max








T
e
T
V 6
1min
15
3- Stress analysis (compression or crushing)
If the compressive stress introduced in the dam is greater than its allowable stress ,the
dam may fail.
Reservoir full
condition
+ compression
T/2 T/2
Pmin
Pmax
+ compr.
-
Tension
Pmin
Resultant
force
Where;
e = Eccentricity of the resultant force
from the center of the base
Total vertical force, excluding
uplift
T= Base width
Normalpoollevel
 V
 Uplift load is excluded from the equations for
stress determination. In practice the influence of
internal uplift upon stress patterns is generally
disregarded.(USBR)
16
0
6
1min 







T
e
T
V
P
6
T
e 
_
x



V
M
x
17
 Because of the gravity dam materials can not sustain tensile stresses, it should be
designed for certain amount or no tension should develops anywhere in the body of
the dam.
The maximum permissible tensile stress for high concrete gravity dams, under worst
loadings, may be taken as 500 KN/m2 (5kg/cm2).
NB! A tension crack by itself does not fail the structure, but it leads to failure of the
structure by producing excessive compressive stresses.
 In order to ensure that no tension is developed anywhere, the amount of Pmin should
at most equal to zero.
The maximum value of eccentricity, that can be permitted on either side of the center is
equal to T/6 ------- “ the resultant must lie within the middle third”.
The resultant distance from the toe of the dam is given by



 22
tan'sec ppv 
18
Principal stress
Pvmax
Pvmin
P
P’
A
B
c
c
B
A
For to be maximum, p’ should be
zero.

19
Base width design for gravity dam for reservoir full condition
U
P W
H
T/2 T/2
e
T/3
HC w
Hw
I) In order to no tension to be develop in the body of the dam the following should be satisfied
CS
H
T
c 

II) In order the dam is to be safe from sliding the following should be satisfied
)(75.0 CS
H
T
c 

waterofunit wt/concretoftunit weighcS
20
Buttress and arch Dams
Buttress dams are those dams which have sloping u/s face and transmit the water
load to a series of buttress at right angle to the axis of the dam.
 The advantage of buttress dam is
manifested by reduction of uplift pressure
and by saving concrete.
21
N.B.!
 The loading and safety criteria for
buttress dams or buttresses, is the same as
that for gravity dam section, except that the
provided buttress thickness ‘t’.
 the Uplift pressure is considered to act only
under the buttress head,
5 H
B
rS
22
Arch Dams
Arch dams are those dams which has a solid wall, curved in plan and standing across
the entire width of the river valley, in a single span.
Depending upon the shape consideration, simple arch dams can be divided
as;
 Constant radius arch dams
Constant angle arch dams; and
Cupola arch dam(dome-shaped )
Valley suited for arch dam
Those valleys with narrow gorges and top-width to dam height ratio less than 5 may
be feasible.
B
H
o
1102 
23
Constant radius arch dams
Is the simplestgeometry,u/s face of the dam is of constantradiiwith a uniform radiald/s slope.It is apparent
thatcentralangle,2θ, reaches a max.@ Crest level.
The mosteconomicalangle fora constantarch dams is maintained when;
r1
r2
r3
Vertical axis
24
Constant angle arch dams
Centralangle of differentarch have the same magnitude from top to bottom & uses up to 70% of concrete as
compared to constantradius arch dam.Butit is more complex as demonstrated in the figure.It is best suited to
narrow & steep-sided V-shaped valleys.
25
Copula arch dams
Has a particularly complex geometry & profile, with constantly varying horizontal & vertical radii to
either face.
Design of arch dams
Arch dams can be designed on the basis of any one of the following methods;
i. Thin cylinder theory;
ii. The thick cylinder theory; and
iii. The elastic theory
 Loads on arch dams are essentially the same as loads on gravity dams, and uplift forces
are less important, if no cracking occurs the uplift can be neglected.
 Internal stresses caused by temperature change, ice pressure, and yielding of abutment are
very important.
26
h
dh
B/2
F
B/2
F
Ru
Ri
T
t
Thin cylinder theory
 The theory assumes the arch to be
simply supported @ the abutments
& that the stresses are approximately
the same as in a thin cylinder of equal
outside radius.
Where,
Ru (Outer radius)– Extrados.
Ri (internal radius)– Intrados.
Rc – Central radius.
T– Arch thickness.
Rc
h
hR
h
hR
T
w
iw
w
cw








5.0
How ????......
27



 h
KKR
hR
T ww
 ;
The most economical angle of arch with minimum volume is 133o34’.
2
2
2/sin
2











B
KKRV
V= A.R2θ = T*1*R2θ
T
1 unit
0
d
dV
Vertical axis
28
Thick cylinder theory
Ri
R
Ru
Pu
pi
T
Ru
Ri
T
analysisFor
RR
RZ
T
designForRRfor
RRT
RZ
elevationanyatuniformisRRT
facesdisstressring
RR
RRRR
p
du
uw
r
d
dur
uw
h
du
du
duu
w
)(
2
)(
)(
2
.
/@max
/
2
1
2
1
max
22
2222























:- is improvement of thin cylinder theory
Reading assignment
Reference:- P.Novack
(fourth edition)
On Elastic Arch theory
29
TekezeaarchdaminEthiopia
The End
30
Embankment Dams
Embankment Dam
Earth Dam Rockfill Dam Composite
Type
Accordint to design
According to method of
Constructuion
Homogenous Zoned Diaphriagm Rolled
fill type
Hydraulic
fill type
Semi Hydraulic
fill type
What is there
difference ?
Embankment dams are those dams which are built of naturally available materials.
Chapter- Four
31
Homogeneous Earth Dams:- are constructed entirely or almost entirely of one type of earth
material (exclusive of slope protection).
Build up one type of material
Phreatic line or seepage line
Horizontalblanket
Slope protection
Zoned Earth Dam;- however, contains materials of different kinds in different parts of the
embankment. The most common type of an earth dam usually adopted is the zoned earth dam
as it leads to an economic & more stable design of the dam.
Vertical
coreu/s
shell
d/s
shell
Transition filter
Impervious
zone
Eg.Clay + fine sand
32
Diaphragm Earthen Dams; this types of dam are the same as that of Zone dam but the
main difference is it has thin thickness of core.
Diaphragm
(core)
Rock Fill Dam
The designation ‘rock fill embankment’ is appropriate where over 50% of the fill material
may be classified as rock pieces. It is an embankment which uses large size rock pieces
to provide stability and impervious membrane to provide water tightness.
Decked rock fill dams
33
Causes of Failure of Earth Dams
The analysis of earth dam must ask a question:……
How does the earthen dams most probably expected to fail? And
what are the causes failures?
Generally, from the previous experiences, the failure of earth dam is grouped in to
Hydraulic failures
Seepage failure
Structural failure
Exercise :-write about the advantage, disadvantage, and
characteristics Of embankment dams
34
Hydraulic Failures: Hydraulic failures include the following:
 Overtopping
 Erosion of U/S face
 Erosion of D/S face
 Erosion of D/S toe
Seepage failures: Seepage failures may be due to
 Piping through the body of the dam
 Piping through the foundation of the dam
 Conduit leakage
 Sloughing of downstream toe.
35
Structural Failures: Structural failures may be due to the following reasons:
 Upstream and Downstream slope failures due to pore pressures
 Upstream slope failure due to sudden draw down
 Down stream slope failure during full reservoir condition
 Foundation slide: Spontaneous liquefaction
 Failure by spreading
 Failure due to Earth quake
 Slope protection failures
 Failure due to damage caused by burrowing animals
 Damage caused by Water soluble materials
36
Criteria for Safe Design of Earth Dam
Free from Overtopping
Free from seepage failure
Free from structural failure
There must be proper slope
protection against wind & rain
drop erosion.
There must be proper drainage
Economic section
How can one
satisfy these
design
criteria????......
 Appropriate design flood
 Adequate spillway
 Sufficient outlet works
 Sufficient free board
 Phreatic (seepage) line should
exit the dam body safely without
sloughing downstream face.
 Seepage through the body of the
dam, foundation and abutments
should be controlled by adapting
measures.
 The dam and foundation should
be safe against piping failure.
 There should be no opportunity
for free passage of water from
U/S to D/S both through the dam
and foundation.
 Safe U/S & D/S slope during
construction
 Safe U/S slope during sudden
draw down condition.
 Safe D/S slope during steady
seepage condition
 Foundation shear stress within
the safe limits.
 Earth quake resistant dam
Selectionof an Earth Dam
1) Top Width
In terms of the Height of the dam.
2) Free board
Based on nature of spill way, Height of dam
3) Casing or outer shell
4) Central impervious core
5) Cut-off trench
6) Downstream drainage system
37
38
39
Seepage Analysis
Seepage analysis is used:-
 To determine the quantity of water passing through the body of the dam and
foundation.
 To obtain the distribution of pore water pressure.
Laplace equation for two dimensional flows
 In earth dams, the flow is essentially two dimensional;
 Consider an element of soil is size x, y and of unit thickness perpendicular to the plane of
the paper. Let Vx and Vy be the entry velocity components in x and y direction.
40
       1.1.1.1. x
y
v
vyx
x
v
vxvyv
y
y
x
xyx 

















0





y
v
x
v yx This is the continuity equation.
x
h
KiKv xxxx


 *
According to Darcy’s law
yYYY K
y
h
IkV



Where;
h = hydraulic head under which water flows.
Kx and Ky are coefficient of permeability in
x and y direction.
  0
).(
2
2
2
2






y
hK
x
hK yx
Then substitute to the previous formula;
For an isotropic soil, Ky = Kx = K
02
2
2
2






y
h
x
h
41
Substituting velocity potential =  = K*h , we get
02
2
2
2






yx
Computation of rate of seepage from flow net
graphical method
analytical methods
experimental methods
solution
Let: b and 1 be the width and length of the field (i.e square.)
h = head drop through the field.
q = discharge passing through the flow channel.
H = total head causing flow
= difference between upstream and downstream heads
Then, from Darcy’s law of flow through soils: )1(. bx
l
h
Kq


h= is same for all
l
b b
42
dN
h
h  Where, Nd = total number of potential drops in the complete flow net,
In the above flow net we have Nd=10







l
b
N
h
Kq
d
Hence the total discharge through the complete flow net is given by
l
b
N
N
khN
l
b
N
h
kqq
d
f
f
d
... 






Where
Nf = total number of flow channels in the net
In the above flow net, we have Nf=3
Reading assignment
on
Seepage discharge for anisotropic soils
q=√(kx*ky) Nf /Nd*H
Reference:
P.Novak et al. (Hydraulic structures)
Garg. S.K 1996 Irrigation engineering
and Hydraulic structure
43
Phreatic Line in Earth Dam
Phreatic line / seepage line is the line at the upper surface of the seepage flow at which the
pressure is atmospheric.
Phreatic line for a homogeneous Earth dam with horizontal drainage blanket
o
oo
oo
o
kyq
xyy
xyy
y
kq



 )2)(
2
(
2
2
bhbyo  22
44
Phreatic line for a dam with no filter
General solution by Casagrande
The focus in this case will be the
Lowest point F of the d/s slope.
aa
a


aa
a


Table for the value of with slope angle 

300 0.36
600 0.32
900 0.26
1200 0.18
1350 0.14
1500 0.10
1800 0.0
Locations of “K” according to Schaffernak
and Van Iterson analythical solution
For
 <30 0 ---------  2
2
2
2
sincos
'
cos
' hbb
a 
q = k (a sin) (tan)
300< < 600___ 22222
cothbhba 
q = k. (sin) (a sin )
b’
45
Arthur Casagrande
46
Phreatic line for homogenous Earth dam with rock toe
Phreatic line for zoned Earth dam with central core
47
Example 4.1
For the following cross sectional view of earthen dam draw the seepage line and
determine the amount of discharge that passes through the body of the dam;
Assume the coefficient of permeability (K) as 5 x 10-6 m/sec
19.5 m
2.5 m
4.5 m
25 m
73.5 m
58.5 m
44 m
48
Stability Analysis of Earthen dams
Stability analysis under the following four heads are generally needed
•Stability analysis of down stream slope during steady seepage.
•Stability of up stream slope during sudden Draw down.
•Stability of up stream & down stream slope during and immediately
after construction.
•Stability of foundation against shear.
It is one of the most generally accepted methods of checking slope
stability analysis using Swedish circle method
49
 
T
NcL
drivingresisting
M
M
F
d
r
s



tan
/ 5.1
StabilityforrequiredStrengthshear
availableStrengthShear
=
Location of the most CRITICAL circle
 In the course of stability analysis, it is quite
cumbersome(Problematic) to take so many trial surfaces
and hence Fellenius has shown, to reduce number of trials,
suggested a line called Fellenius line (line AB) for a
homogeneous slope (see fig below) on which the most
critical circle lies.
 The center of the most critical circle may lie anywhere on
the line AB or its extension. However the exact position of
critical circle is obtained after conducting the stability
analysis for different points. The center O with minimum
factor of safety is the center of the most critical circle.
50
1Fs2Fs
3Fs
51
Method of locating center of critical slip circle
H=dam height
H
V
H
4.5 H
A
B
rc
Fellenius line
52
Stability of down stream slope during steady seepage
Example 4.2
Checkthe stability of the d/s slope forthe following earthen dam x-sections
a) When the seepage line (pheratic line) have no contact with the slip circle and;
b) When the soil of the slip circle get fully submerged .
2
1H= 25 m
Coordinate of
end slice
Y1 Y2 Y3 Y4
Magnitude (m) 6 7.5 7.5 6.5
3/2.1 mtsub 
3/83.1 mtd 
3/2.2 mtsat 
C=2.4 t/m3
o
25
Chapter Five
River Diversion During Construction
In order to build a dam, one major temporary or semi temporary (but essential and
hardly avoidable) activity is required. That is river diversion during construction.
River diversion takes place for two purposes:
o For construction purpose
o For water use purpose
Ways of diverting water before construction
I.Provision of diversion tunnel or Channel
Diversion tunnel or
diversion channel
U/S coffer dam
D/S coffer dam
Costruction zone
D/S
U/S
Construction
area
diverted
flow
Coffer
dam
U/S
D/S
overtopped
flow
Construction
zone on the
2nd stage
U/S
D/S
Completed
portion of
the dam
i. 1st stage diversion
ii.2nd stage diversion
II. Two stage construction
Applicable for
Concrete dam !!..
III. Through culverts in the body of the dam
Example: Gilgel Gibe Hydroelectric project
Construction zone
of main dam
Constructionzone
ofmaindam
D/S Coffer dam
U/S coffer dam
Culverts
Culverts
Culverts
o Components of river division schemes often have to be constructed in the river bed, with
flowing water. To obtain more favorable (or feasible) conditions-dry period (low discharges)
o Available space
o Hydraulic conditions, geological conditions
o Possibility to incorporate temporary structures in permanent structures-economy.
o Feasibility of construction of diversion structures
Considerations of:
Basic Planning Considerations
High costs- often represent 5-10% of total cost when large discharges have to be handled.
The selection of discharge capacity of the diversion structures is fundamental for:
o Definition of cofferdam height (stage-discharge relationship)
o Definition of size of conveyance system such as diversion tunnel diameter channel width depth
The discharge capacity for river diversion is decided on base of a risk analysis.
Risk
Cost (birr)
Return period (T)
T1 T2 T3
T
4
T5
Optimized
region

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Dams

  • 1. 1 Chapter Three Concrete Dam The structural integrity of any dam must be maintained under different loading circumstances.  the gravity dam is mainly subjected to the following main forces; Waterload Self weight Upliftload Wave load Silt load Wind load Earth quake load
  • 2. 2 As per the degree of relative importance loads on the dam can be classified as ;  primary loads:- Major important loads irrespective of the dam type; E.g. Self Wight load, water load and related seepage load Secondary loads:- Universally applicable loads , even though their magnitude is less; E.g. Silt load (serve as to minimize uplift pressure) Exceptional loads:- loads which has limited applicability; E.g. Tectonic load (earth quake load Exercise :- write about the characteristic ,advantage, disadvantage of concrete Dam NB:-for Earth quake load don’t consider uplift loads Concrete dam cont’s
  • 3. Hw 2 2 1 HP wH  3 H 3 Loads and their centroidal location in gravity dams Case -1 Non-over flow section (i) Up stream vertical face PHH @ 3 H from the base of the dam. B H’ 'HP 2' 2 1 ' HP wH  @ ' Hw 3 'H from the base of the dam.Ap= X-sec area of stru. Hw U   B HH U w * 2 '   Throughthe centroid of Trapezoidal,with out drainagegallery) [ i.e. ] @ )'(3 ]'2[ HH HHB Z    w pc AW  @ Throughthe centroid of x-sectionalarea Ap ’ vP '* 2 1 HbP wv  b @ from the toe of the dam. 3 b Z
  • 4. 4 (ii) Up stream face inclined Pv1 Pv2 If the upstream face of the gravity dam is inclined in addition to the previous loads ( loads in vertical u/s face) , only vertical loads of water i.e. Pv will be added at its centroidal point from the toe of the dam. where Pv = Pv1+ Pv2 Conti…
  • 5.  aw HH 1  aw HH 2 5 Case-2 Over flow section H2 H1 PH YwHHH HH P aH *)( 2 12 21               T.E.L            21 21 1 _ 2 32 3 1 HHH HHH HHZ a a a Z From the base of the dam @ Ha=Va2/2g
  • 6. H      )'( 3 1 ' HHH 6 Uplift pressure with drainage gallery and tension cracks  To reduce the uplift pressure , drains are formed trough the body of the dam, this make the intensity of the uplift pressure to be differ from the full concrete dam. Drainage gallery H H’ T ’ ’ 'H 1
  • 7. UFhw 032.0 4 271.0763.0032.0 FUFhw  kmFif 32 kmFif 32 7 Wave Pressure ( hydrodynamic wave load) Waves are generated on the reservoir by the blowing winds. hw w/r hw = heightof the wave U = wind velocity in km/hr F = fetch length Pwave 22 wwwave hP  @ wh375.0 above the stilled water level. Reservoirsurface area Dam
  • 8. v g W W * 8 Earthquake force  Earth quake force may move in any direction, but for the design purpose it has to be resolved in to vertical and horizontal components.(USBR)  The values of these horizontal (αh) and vertical (αv) accelerations are generally expressed as percentage of the acceleration due to gravity i.e. 0.1g or 0.2g, etc. Vertical acceleration(αv) Dam foundation Upward verticalacceleration The contactb/n the foundationand the dam will increase,hence the effective Weight of the dam will also be increased. The contactb/n the foundationand the dam Will decrease,which is the worstcase!! Down ward verticalmovement. W Effective weight of the dam
  • 9. 2 555.0 HF whe  3 4H 3 4H 9 Horizontal acceleration(αh) Hydro-dynamic pressure. Horizontal Inertia force. Fe acts @ H from the base of the dam. Von – Karman formula There is also a hydrodynamic formula developed by Zanger, but for average ordinary purposes, the Von-Karman formula is sufficient. Reading assignment, Reference, P.Novak and S.K. Garg
  • 10. 2 2' ss ash h KP   3 sh wss  ' s s aK   sin1 sin1    10 Sediment load hs Psh N.B! it is usual practice to assume the value of hs equals to the height of dead storage. @ above the base of the dam. The submerged unit weight and the active lateral pressure coefficient Ka is given by s s 's where is the angle of shear resistance. is sediment saturated unit weight. Reading assignment -Loading combinations in dams Reference, Novak
  • 11.      ve ve o M M F 11 Design and analysis of gravity dam Gravity dam may fail in the following way •Rotation and overturning, • Translation and sliding and • Overstress and material failure. 1- Over turning stability To make the structure of the dam stable from rotational or overturning failure the following governing criteria should be satisfied F0 (factor of safety against over turning) should be greater than 1.5 > 1.5……………(safe against overturning) Include uplift moment
  • 12.   tan1 tan                V H V H Fss   V H Fss  12 2-Siding Stability (Fs) Sliding Factor (FSS) Shear friction Factor(FSF) Limit equilibrium factor (FLE) i) Sliding factor can be defined by , for horizontal plane ΣV is determined allowing for the effect of uplift (net) , for foundations inclined at a small angle In order to be the dam stable against sliding Fss should be less than or equal to 0.75 but for ELC (extreme loading condition) up to 0.9 is acceptable.
  • 13.   H S FSF mkNV CA S b /)tan( )tantan1(cos              13 ii) Shear friction factor (FSF) is defined as Where S – total resistance to shear and defined by Location of sliding plane Normal Unusual Extreme Dam Concrete base interface Recommended shear friction factor (USBR 1987) Foundation rock 3.0 2.0 > 1.0 4.0 2.7 1.3 (Net)
  • 14.   f LEF    tann LE c F    f 14 (iii) Limit Equilibrium Factor, FLE n = T he shearstressed generated underthe applied load = Available shearstrengthand expressed byMohr coulombfailure criteria = Stress acting normalto plain of sliding FLE = 2.0 for normal operation FLE = 1.3 for seismic activity AVn /
  • 15. H V         T e T V 6 1max         T e T V 6 1min 15 3- Stress analysis (compression or crushing) If the compressive stress introduced in the dam is greater than its allowable stress ,the dam may fail. Reservoir full condition + compression T/2 T/2 Pmin Pmax + compr. - Tension Pmin Resultant force Where; e = Eccentricity of the resultant force from the center of the base Total vertical force, excluding uplift T= Base width Normalpoollevel  V
  • 16.  Uplift load is excluded from the equations for stress determination. In practice the influence of internal uplift upon stress patterns is generally disregarded.(USBR) 16
  • 17. 0 6 1min         T e T V P 6 T e  _ x    V M x 17  Because of the gravity dam materials can not sustain tensile stresses, it should be designed for certain amount or no tension should develops anywhere in the body of the dam. The maximum permissible tensile stress for high concrete gravity dams, under worst loadings, may be taken as 500 KN/m2 (5kg/cm2). NB! A tension crack by itself does not fail the structure, but it leads to failure of the structure by producing excessive compressive stresses.  In order to ensure that no tension is developed anywhere, the amount of Pmin should at most equal to zero. The maximum value of eccentricity, that can be permitted on either side of the center is equal to T/6 ------- “ the resultant must lie within the middle third”. The resultant distance from the toe of the dam is given by
  • 18.     22 tan'sec ppv  18 Principal stress Pvmax Pvmin P P’ A B c c B A For to be maximum, p’ should be zero. 
  • 19. 19 Base width design for gravity dam for reservoir full condition U P W H T/2 T/2 e T/3 HC w Hw I) In order to no tension to be develop in the body of the dam the following should be satisfied CS H T c   II) In order the dam is to be safe from sliding the following should be satisfied )(75.0 CS H T c   waterofunit wt/concretoftunit weighcS
  • 20. 20 Buttress and arch Dams Buttress dams are those dams which have sloping u/s face and transmit the water load to a series of buttress at right angle to the axis of the dam.  The advantage of buttress dam is manifested by reduction of uplift pressure and by saving concrete.
  • 21. 21 N.B.!  The loading and safety criteria for buttress dams or buttresses, is the same as that for gravity dam section, except that the provided buttress thickness ‘t’.  the Uplift pressure is considered to act only under the buttress head,
  • 22. 5 H B rS 22 Arch Dams Arch dams are those dams which has a solid wall, curved in plan and standing across the entire width of the river valley, in a single span. Depending upon the shape consideration, simple arch dams can be divided as;  Constant radius arch dams Constant angle arch dams; and Cupola arch dam(dome-shaped ) Valley suited for arch dam Those valleys with narrow gorges and top-width to dam height ratio less than 5 may be feasible. B H
  • 23. o 1102  23 Constant radius arch dams Is the simplestgeometry,u/s face of the dam is of constantradiiwith a uniform radiald/s slope.It is apparent thatcentralangle,2θ, reaches a max.@ Crest level. The mosteconomicalangle fora constantarch dams is maintained when; r1 r2 r3 Vertical axis
  • 24. 24 Constant angle arch dams Centralangle of differentarch have the same magnitude from top to bottom & uses up to 70% of concrete as compared to constantradius arch dam.Butit is more complex as demonstrated in the figure.It is best suited to narrow & steep-sided V-shaped valleys.
  • 25. 25 Copula arch dams Has a particularly complex geometry & profile, with constantly varying horizontal & vertical radii to either face. Design of arch dams Arch dams can be designed on the basis of any one of the following methods; i. Thin cylinder theory; ii. The thick cylinder theory; and iii. The elastic theory  Loads on arch dams are essentially the same as loads on gravity dams, and uplift forces are less important, if no cracking occurs the uplift can be neglected.  Internal stresses caused by temperature change, ice pressure, and yielding of abutment are very important.
  • 26. 26 h dh B/2 F B/2 F Ru Ri T t Thin cylinder theory  The theory assumes the arch to be simply supported @ the abutments & that the stresses are approximately the same as in a thin cylinder of equal outside radius. Where, Ru (Outer radius)– Extrados. Ri (internal radius)– Intrados. Rc – Central radius. T– Arch thickness. Rc h hR h hR T w iw w cw         5.0 How ????......
  • 27. 27     h KKR hR T ww  ; The most economical angle of arch with minimum volume is 133o34’. 2 2 2/sin 2            B KKRV V= A.R2θ = T*1*R2θ T 1 unit 0 d dV Vertical axis
  • 30. 30 Embankment Dams Embankment Dam Earth Dam Rockfill Dam Composite Type Accordint to design According to method of Constructuion Homogenous Zoned Diaphriagm Rolled fill type Hydraulic fill type Semi Hydraulic fill type What is there difference ? Embankment dams are those dams which are built of naturally available materials. Chapter- Four
  • 31. 31 Homogeneous Earth Dams:- are constructed entirely or almost entirely of one type of earth material (exclusive of slope protection). Build up one type of material Phreatic line or seepage line Horizontalblanket Slope protection Zoned Earth Dam;- however, contains materials of different kinds in different parts of the embankment. The most common type of an earth dam usually adopted is the zoned earth dam as it leads to an economic & more stable design of the dam. Vertical coreu/s shell d/s shell Transition filter Impervious zone Eg.Clay + fine sand
  • 32. 32 Diaphragm Earthen Dams; this types of dam are the same as that of Zone dam but the main difference is it has thin thickness of core. Diaphragm (core) Rock Fill Dam The designation ‘rock fill embankment’ is appropriate where over 50% of the fill material may be classified as rock pieces. It is an embankment which uses large size rock pieces to provide stability and impervious membrane to provide water tightness. Decked rock fill dams
  • 33. 33 Causes of Failure of Earth Dams The analysis of earth dam must ask a question:…… How does the earthen dams most probably expected to fail? And what are the causes failures? Generally, from the previous experiences, the failure of earth dam is grouped in to Hydraulic failures Seepage failure Structural failure Exercise :-write about the advantage, disadvantage, and characteristics Of embankment dams
  • 34. 34 Hydraulic Failures: Hydraulic failures include the following:  Overtopping  Erosion of U/S face  Erosion of D/S face  Erosion of D/S toe Seepage failures: Seepage failures may be due to  Piping through the body of the dam  Piping through the foundation of the dam  Conduit leakage  Sloughing of downstream toe.
  • 35. 35 Structural Failures: Structural failures may be due to the following reasons:  Upstream and Downstream slope failures due to pore pressures  Upstream slope failure due to sudden draw down  Down stream slope failure during full reservoir condition  Foundation slide: Spontaneous liquefaction  Failure by spreading  Failure due to Earth quake  Slope protection failures  Failure due to damage caused by burrowing animals  Damage caused by Water soluble materials
  • 36. 36 Criteria for Safe Design of Earth Dam Free from Overtopping Free from seepage failure Free from structural failure There must be proper slope protection against wind & rain drop erosion. There must be proper drainage Economic section How can one satisfy these design criteria????......  Appropriate design flood  Adequate spillway  Sufficient outlet works  Sufficient free board  Phreatic (seepage) line should exit the dam body safely without sloughing downstream face.  Seepage through the body of the dam, foundation and abutments should be controlled by adapting measures.  The dam and foundation should be safe against piping failure.  There should be no opportunity for free passage of water from U/S to D/S both through the dam and foundation.  Safe U/S & D/S slope during construction  Safe U/S slope during sudden draw down condition.  Safe D/S slope during steady seepage condition  Foundation shear stress within the safe limits.  Earth quake resistant dam
  • 37. Selectionof an Earth Dam 1) Top Width In terms of the Height of the dam. 2) Free board Based on nature of spill way, Height of dam 3) Casing or outer shell 4) Central impervious core 5) Cut-off trench 6) Downstream drainage system 37
  • 38. 38
  • 39. 39 Seepage Analysis Seepage analysis is used:-  To determine the quantity of water passing through the body of the dam and foundation.  To obtain the distribution of pore water pressure. Laplace equation for two dimensional flows  In earth dams, the flow is essentially two dimensional;  Consider an element of soil is size x, y and of unit thickness perpendicular to the plane of the paper. Let Vx and Vy be the entry velocity components in x and y direction.
  • 40. 40        1.1.1.1. x y v vyx x v vxvyv y y x xyx                   0      y v x v yx This is the continuity equation. x h KiKv xxxx    * According to Darcy’s law yYYY K y h IkV    Where; h = hydraulic head under which water flows. Kx and Ky are coefficient of permeability in x and y direction.   0 ).( 2 2 2 2       y hK x hK yx Then substitute to the previous formula; For an isotropic soil, Ky = Kx = K 02 2 2 2       y h x h
  • 41. 41 Substituting velocity potential =  = K*h , we get 02 2 2 2       yx Computation of rate of seepage from flow net graphical method analytical methods experimental methods solution Let: b and 1 be the width and length of the field (i.e square.) h = head drop through the field. q = discharge passing through the flow channel. H = total head causing flow = difference between upstream and downstream heads Then, from Darcy’s law of flow through soils: )1(. bx l h Kq   h= is same for all l b b
  • 42. 42 dN h h  Where, Nd = total number of potential drops in the complete flow net, In the above flow net we have Nd=10        l b N h Kq d Hence the total discharge through the complete flow net is given by l b N N khN l b N h kqq d f f d ...        Where Nf = total number of flow channels in the net In the above flow net, we have Nf=3 Reading assignment on Seepage discharge for anisotropic soils q=√(kx*ky) Nf /Nd*H Reference: P.Novak et al. (Hydraulic structures) Garg. S.K 1996 Irrigation engineering and Hydraulic structure
  • 43. 43 Phreatic Line in Earth Dam Phreatic line / seepage line is the line at the upper surface of the seepage flow at which the pressure is atmospheric. Phreatic line for a homogeneous Earth dam with horizontal drainage blanket o oo oo o kyq xyy xyy y kq     )2)( 2 ( 2 2 bhbyo  22
  • 44. 44 Phreatic line for a dam with no filter General solution by Casagrande The focus in this case will be the Lowest point F of the d/s slope. aa a   aa a   Table for the value of with slope angle   300 0.36 600 0.32 900 0.26 1200 0.18 1350 0.14 1500 0.10 1800 0.0 Locations of “K” according to Schaffernak and Van Iterson analythical solution For  <30 0 ---------  2 2 2 2 sincos ' cos ' hbb a  q = k (a sin) (tan) 300< < 600___ 22222 cothbhba  q = k. (sin) (a sin ) b’
  • 46. 46 Phreatic line for homogenous Earth dam with rock toe Phreatic line for zoned Earth dam with central core
  • 47. 47 Example 4.1 For the following cross sectional view of earthen dam draw the seepage line and determine the amount of discharge that passes through the body of the dam; Assume the coefficient of permeability (K) as 5 x 10-6 m/sec 19.5 m 2.5 m 4.5 m 25 m 73.5 m 58.5 m 44 m
  • 48. 48 Stability Analysis of Earthen dams Stability analysis under the following four heads are generally needed •Stability analysis of down stream slope during steady seepage. •Stability of up stream slope during sudden Draw down. •Stability of up stream & down stream slope during and immediately after construction. •Stability of foundation against shear. It is one of the most generally accepted methods of checking slope stability analysis using Swedish circle method
  • 50. Location of the most CRITICAL circle  In the course of stability analysis, it is quite cumbersome(Problematic) to take so many trial surfaces and hence Fellenius has shown, to reduce number of trials, suggested a line called Fellenius line (line AB) for a homogeneous slope (see fig below) on which the most critical circle lies.  The center of the most critical circle may lie anywhere on the line AB or its extension. However the exact position of critical circle is obtained after conducting the stability analysis for different points. The center O with minimum factor of safety is the center of the most critical circle. 50
  • 51. 1Fs2Fs 3Fs 51 Method of locating center of critical slip circle H=dam height H V H 4.5 H A B rc Fellenius line
  • 52. 52 Stability of down stream slope during steady seepage
  • 53. Example 4.2 Checkthe stability of the d/s slope forthe following earthen dam x-sections a) When the seepage line (pheratic line) have no contact with the slip circle and; b) When the soil of the slip circle get fully submerged . 2 1H= 25 m Coordinate of end slice Y1 Y2 Y3 Y4 Magnitude (m) 6 7.5 7.5 6.5 3/2.1 mtsub  3/83.1 mtd  3/2.2 mtsat  C=2.4 t/m3 o 25
  • 54. Chapter Five River Diversion During Construction In order to build a dam, one major temporary or semi temporary (but essential and hardly avoidable) activity is required. That is river diversion during construction. River diversion takes place for two purposes: o For construction purpose o For water use purpose Ways of diverting water before construction I.Provision of diversion tunnel or Channel Diversion tunnel or diversion channel U/S coffer dam D/S coffer dam Costruction zone D/S U/S
  • 55. Construction area diverted flow Coffer dam U/S D/S overtopped flow Construction zone on the 2nd stage U/S D/S Completed portion of the dam i. 1st stage diversion ii.2nd stage diversion II. Two stage construction Applicable for Concrete dam !!.. III. Through culverts in the body of the dam Example: Gilgel Gibe Hydroelectric project Construction zone of main dam Constructionzone ofmaindam D/S Coffer dam U/S coffer dam Culverts Culverts Culverts
  • 56. o Components of river division schemes often have to be constructed in the river bed, with flowing water. To obtain more favorable (or feasible) conditions-dry period (low discharges) o Available space o Hydraulic conditions, geological conditions o Possibility to incorporate temporary structures in permanent structures-economy. o Feasibility of construction of diversion structures Considerations of: Basic Planning Considerations High costs- often represent 5-10% of total cost when large discharges have to be handled.
  • 57. The selection of discharge capacity of the diversion structures is fundamental for: o Definition of cofferdam height (stage-discharge relationship) o Definition of size of conveyance system such as diversion tunnel diameter channel width depth The discharge capacity for river diversion is decided on base of a risk analysis. Risk Cost (birr) Return period (T) T1 T2 T3 T 4 T5 Optimized region