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Hydraulic Structures Part 1 Design of Concrete Dams Civil Engineering Department 4 th Grade Academic Year 2021/2022 Prepared by Assoc. Prof. Dr.Thamir M. Ahmed 1
Hydraulic StructuresPart 1
Design of Concrete Dams
Civil Engineering Department 4th Grade Academic Year 20212022
Prepared by Assoc Prof DrThamir M Ahmed
1
Main Topics
bull1-Dams Definition
bull2- Concrete Damsbull i- Design of concrete dams
bull ii- Stability Analysis
bull3- Earth Damsbull i- Design of Earth Dams
bull ii- Seepage Line
bull iii- Downstream Stability
bull iv-Upstream Stability due to Sudden Draw Down
2
Assesment Criteria
bull1- Mid term Exam 1 (In Campus) 20
bull2- Mid term Exam 2 (In Campus) 20
bull3- Activities 10
bull4-Final Exam (In Campus) 50
3
Definition of a dam
A barrier constructed to hold back water and raise its level forming a
reservoir used to generate electricity or as a water supply
Concrete Dam
A barrier of concrete structure designed and shaped that its weight is
sufficient to ensure stability against the effects of all imposed forces
earth etchellip built across a river to create a body of water for a hydroelect
ric power station domestic water supply
Or a reservoir of water created by such a barrier
4
5
Main Necessities of any dam
bull Irrigation
bull Water for domestic consumption
bull Drought and flood control
bull For navigational facilities
bull Hydroelectric power generation
bull Recreation
bull Development of fish amp wild life
bull Soil conservation
6
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
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77
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79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
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84
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88
89
90
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94
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101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Main Topics
bull1-Dams Definition
bull2- Concrete Damsbull i- Design of concrete dams
bull ii- Stability Analysis
bull3- Earth Damsbull i- Design of Earth Dams
bull ii- Seepage Line
bull iii- Downstream Stability
bull iv-Upstream Stability due to Sudden Draw Down
2
Assesment Criteria
bull1- Mid term Exam 1 (In Campus) 20
bull2- Mid term Exam 2 (In Campus) 20
bull3- Activities 10
bull4-Final Exam (In Campus) 50
3
Definition of a dam
A barrier constructed to hold back water and raise its level forming a
reservoir used to generate electricity or as a water supply
Concrete Dam
A barrier of concrete structure designed and shaped that its weight is
sufficient to ensure stability against the effects of all imposed forces
earth etchellip built across a river to create a body of water for a hydroelect
ric power station domestic water supply
Or a reservoir of water created by such a barrier
4
5
Main Necessities of any dam
bull Irrigation
bull Water for domestic consumption
bull Drought and flood control
bull For navigational facilities
bull Hydroelectric power generation
bull Recreation
bull Development of fish amp wild life
bull Soil conservation
6
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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85
86
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88
89
90
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94
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99
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101
102
103
104
105
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107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Assesment Criteria
bull1- Mid term Exam 1 (In Campus) 20
bull2- Mid term Exam 2 (In Campus) 20
bull3- Activities 10
bull4-Final Exam (In Campus) 50
3
Definition of a dam
A barrier constructed to hold back water and raise its level forming a
reservoir used to generate electricity or as a water supply
Concrete Dam
A barrier of concrete structure designed and shaped that its weight is
sufficient to ensure stability against the effects of all imposed forces
earth etchellip built across a river to create a body of water for a hydroelect
ric power station domestic water supply
Or a reservoir of water created by such a barrier
4
5
Main Necessities of any dam
bull Irrigation
bull Water for domestic consumption
bull Drought and flood control
bull For navigational facilities
bull Hydroelectric power generation
bull Recreation
bull Development of fish amp wild life
bull Soil conservation
6
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Definition of a dam
A barrier constructed to hold back water and raise its level forming a
reservoir used to generate electricity or as a water supply
Concrete Dam
A barrier of concrete structure designed and shaped that its weight is
sufficient to ensure stability against the effects of all imposed forces
earth etchellip built across a river to create a body of water for a hydroelect
ric power station domestic water supply
Or a reservoir of water created by such a barrier
4
5
Main Necessities of any dam
bull Irrigation
bull Water for domestic consumption
bull Drought and flood control
bull For navigational facilities
bull Hydroelectric power generation
bull Recreation
bull Development of fish amp wild life
bull Soil conservation
6
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
5
Main Necessities of any dam
bull Irrigation
bull Water for domestic consumption
bull Drought and flood control
bull For navigational facilities
bull Hydroelectric power generation
bull Recreation
bull Development of fish amp wild life
bull Soil conservation
6
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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89
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107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Main Necessities of any dam
bull Irrigation
bull Water for domestic consumption
bull Drought and flood control
bull For navigational facilities
bull Hydroelectric power generation
bull Recreation
bull Development of fish amp wild life
bull Soil conservation
6
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
7
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Structure of a Concrete dam
8
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
9
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
10
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Main Components of Dams
1 Heel contact with the ground on the upstream side
2 Toe contact on the downstream side
3 Abutment Sides of the valley on which the structure of the dam
rest
4 Galleries small rooms like structure left within the dam for
checking operations
5 Diversion tunnel Tunnels are constructed for diverting water
before the construction of dam This helps in keeping the river bed
dry
6 Spillways It is the arrangement near the top to release the excess
water of the reservoir to downstream side
7 Sluice way An opening in the dam near the ground level which is
used to clear the silt accumulation in the reservoir side
11
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Conditions of a successful dam
bull Large storage capacity
bull Length of dam to be constructed is less
bull Water-tightness of reservoir
bull Good hydrological conditions
bull Deep reservoir
bull Small submerged area
bull Low silt inflow
bull No objectionable minerals
bull Low cost of real estate
bull Site easily accessible
12
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Design Stages The development of the structural design of the dam is based on the
investigation data which include the following
1- Determination of the design levels and sizes of water discharge set
the limits of the levels of water determining the elevations lines in the
water immersion areas and volumes of water tanks
2- Developing engineering plans for dams and choose types of
materials constructions and building equipment
3- Hydraulic calculation and infiltration of water reservoir and the
approved dimensions for the drained water dams and anti-leak dams
4- Static and dynamic calculation that prove the of resistance stability
of dams and their bases
5- Develop lists for construction costs to determine the economic and
technical indicators for the project
13
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
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107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Design Considerations 1 Local Conditions
The early collection of data on local conditions which will eventually related to the
design specifications and construction stages is advisable Local conditions are not
only needed to estimate construction costs but may be of benefit when considering
alternative designs and methods of construction Some of these local conditions will
also be used to determine the extent of the project designs
2 Maps and photographs
Topographic and contour maps through which the volume of the reservoir and its
characteristics can be known in addition to the level of the water in the reservoir
also the water outfall basin as well as the region concerned and site access roads
3 Hydrologic data
In order to determine the potential of a site for storing water generating power or
other beneficial use a thorough study of hydrologic conditions must be made it
includes stream flow records flood studies sedimentation and water quality studies
and other things14
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
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73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
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91
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101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
4Reservoir capacity and operation
The estimation of reservoir capacity and reservoir operations are used properly to
estimate the size of spillway and outlet works The reservoir capacity is a major
factor in flood routings and may affect the determination the size and crest
elevation of the spillway
5Climatic effects
Since weather affects the rate of construction and the overall construction schedule
Accessibility of the site during periods of inclement weather affects the construction
schedule and should be investigated
6Site selection
The project is designed to perform a certain function and to serve a particular area
So the purpose and the service area are defined a preliminary site selection can be
made
7FoundationAinvestigations
In most instances a concrete dam is keyed into the foundation so that the
foundation will normally be adequate if it has enough bearing capacity to resist the
loads from the dam15
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
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85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
8 Construction Aspects
The length of the construction season should be considered Adequate time
should be allowed for construction so that additional costs for expedited
work are not encountered
16
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
17
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Low and High Concrete DamsThe one of Main basics to classify the types of concrete dams is the height of maximum
water level (H)
18
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
19
119886 =119867
328Fb = 004 minus 005 ∙ 119867
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Design of Concrete Dam Section
1-Calculation of the base width 119861
119861 ge119867
119878119904minus119888
Where
119867 =The height of water in reservoir
119878119904 =Sp gravity of dam
119888 =Uplift constant
For 119888 = 1 119861 ge119867
119878119904minus1
If uplift is not considered (119888 = 119900)
20
119861 ge119867
119878119904
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
21
Calculation the base width with effects of friction factor(120583)
119861 ge119867
120583 ∙ 119878119904 minus 119888
If 119888 =1
119861 ge119867
120583 ∙ 119878119904 minus 1
If 119888 =0 (no uplift)
119861 ge119867
120583 ∙ 119878119904
Where (120583) is equal to average friction factor and taken as (075)
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
22
Height of Low Concrete Dam
1198671 =119891
120574∙ 119878119904+1+119888
Where
119891 =The maximum allowable stress of dam material
Free board
Fb = 15 ∙ ℎ119908
Where
ℎ119908 =Wave height given in eq of wave force
Or
Fb = 004 minus 005 ∙ 119867Where
119867 =The height of max water level above bed
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
23
Top width
119886 = 014 ∙ 119867
119886 = 119867
119886 =119867
328
Where
119867 =The height of max water level above bed
Height of additional dam base
119867119894 = 2119886 119878119904 minus 119888
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
24
Design Cases
1 Empty reservoir (Vertical earthquake forces are acting downward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force
3-vertical forces of earthquake (downward +ve )
2 Empty reservoir (Vertical earthquake forces are acting upward)
The forces affected the body of dam are as follows
1-weight of dam
2- Horizontal acceleration of earth quake force toward US of dam
3-vertical forces of earth quake (upward - ve)
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
25
3- Full Reservoir
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-uplift force (Pu) ndashve
4-weight of dam (W) +ve
5-upward earthquakes forces ndashve
6-horiznotal acceleration of earthquakes forces toward DS of dam
4- Full Reservoir without uplift force
The following forces will be considered
1- Hydrostatic pressure (P) ndash ve
2- Hydrodynamic force (Pe) ndashve
3-weight of dam (W)+ve
4-upward earthquakes forces ndashve
5-horiznotal acceleration of earthquakes forces toward DS of dam-ve
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Forces acting on gravity dam1 Water pressure (119875)
2 Up lift pressure (119875119906)
3 Pressure due to earthquake forces
4 Silt pressure
5 Wave pressure
6 Ice pressure
7 Weight of the dam (W)
1-Hydrostatic Force
119875 =1
2∙ 120574 ∙ 1198672
Where
119875 =Horizontal hydrostatic force
120574 =Unit weight of water
119867 = Depth of water 26
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
27
Case 1 Initial section
119863119890119904119894119892119899 119886119899119889 119860119899119886119897119910119904119894119904
119880119901119904119905119903119890119886119898 (119880119878119877119864119878119864119877119881119868119874119877
119863119900119908119899119904119905119903119890119886119898 119863119878
07
1
80
80
119861=
1
07119861 = 56 119898
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
28
C120574119908119867 119861
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
29
119865119881 119865119881
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
2-Up Lift Force
119880 =1
2∙ 119888 ∙ 120574 ∙ 119867 ∙ 119861
3-Horizontal inertia force (Force due to Horizontal Earthquake Force)
119865119904ℎ =119908
119892∙ 119886ℎ =
119882
119892∙ 119870ℎ ∙ 119892 = 119882 ∙ 119870ℎ
4-Vertical inertia force (Force due to Vertical Earthquake Force)
119865119904119907 =119908
119892∙ 119886119907 =
119882
119892∙ 119870119907 ∙ 119892 = 119882 ∙ 119870119907
Where
119882 =The total weight of the dam
119886119907 119886ℎ =Vertical acceleration and horizontal acceleration respectively
119870ℎ =Horizontal acceleration factor (such 01)
119870119907 =Vertical acceleration factor (such 005)30
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
5-Hydrodynamics force
119875119864= 0555 ∙ 119870ℎ ∙ 120574 ∙ 1198672 Von ndashKarman Equation
Position of force = 4119867
3120587
The moment 119872119890 = 119875119864 ∙4119867
3120587
Or using Zangar Equation
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119864 =Hydrodynamic force
119875119890=Hydrodynamic pressure
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
31
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Where
120579deg =Angle in degrees which the us face of the dam makes with
vertical and considered if the height of US slope greater than the
half height of dam
119872119890 = 0412 ∙ 119875119864 ∙ 119867
6-Silt Force
119901119904119894119897119905 =1
2∙ 120574119904 ∙ ℎ
2 ∙ 119870119886 (Rankines Formula)
Where 119870119886 is the coefficient of active earth pressure of silt
119870119886 =1minussin empty
1+sin empty
Where
119870119886 =The coefficient of active earth pressure of silt
120574 =submerged unit weight of silt material
ℎ =The height of silt deposited 32
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
7-Wave Force
(I) For 119865 lt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881 + 0763 minus 0271 ∙4119865
(II) For 119865 gt32 Km
ℎ119908 = 0032 ∙ 119865 ∙ 119881
ℎ119908 =The height of wave in (m)
119881 =Wind velocity in (kmhr)
119865 =Fetch of wave in (km)
119875119908prime = 24 ∙ 120574 ∙ ℎ119908 (In Kilopascal and acts at vertical distance = 0125 ℎ119908 )
119875119908 = 2 ∙ 120574 ∙ ℎ1199082 (In Kilo Newton and acts at vertical distance = 0375 ℎ119908 )
8-Ice force
119875119868 = 120784120787 119957119900 150119905
1198982
33
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
34
Case 2 Dam Section with Aditional part
0412 119867
1198673
2
3times (119887 + 119861)
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
35
Case 3 Dam Section with Tail Water
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
36
Case 4 Dam Section with Gallary
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
37
Case 5 Dam Section with Gallary and Tail Water
119872119890 = 0412 ∙ 119867 ∙ 119875119864
119875119864 = 0726 ∙ 119875119890∙ 119867
119875119890= 119862119898 ∙ 119870ℎ ∙ 120574 ∙ 119867
119862119898 = 0735 ∙ (120579deg
90deg)
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
Modes of failure of gravity dams
1 By overturning (rotation) about the toe
119865119878 =Σ119877119894119892ℎ119905119894119899119892 119872119900119898119890119899119905119904
Σ119874119907119890119903119905119906119903119899119894119899119892 119872119900119898119890119899119905119904=
Σ119872119877
Σ1198720
Σ119872119877 Anti clockwise moments Σ1198720 clockwise moments
2 By crushing (compression)
119875119907 119898119886119909119898119894119899 =Σ119881
119861(1 plusmn
6119890
119861)
Where
119890=Eccentricity of resultant force from the center to the base
Σ119881 =Total vertical force
119861 =Base width
ത119883 = (σ119872119877 minus σ119872119900)σ119865119881
119890 =119861
2minus ത119883 119897119888=
119887
2(1 minus
119887
6 119890)
38
119890 gt119861
6119905119890119899119904119894119900119899 119890 le
119861
6119899119900 119905119890119899119904119894119900119899
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
39
The normal stress at any point on the base will be the sum of the direct stress and the
bending stress The direct stress σcc is
120590119888119888 =σ119865119881119887 times 1
and bending stress σcbc at any fiber at distance y from Neutral Axis is
120590119888119887119888 = ∓σ119872 119910
119868
119872 =119865119881 119890
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
40
3 By development of tension causing ultimate failure by crushing
If e gt b6 the normal stress at the heel will be -ve or tensile When the eccentricity e is
greater than b6 a crack of length lc will develop due to tension which can be calculated
as
120590119888119888 = 120590119888119887119888 rarrσ119865119881119887 times 1
=σ119872 119910
119868rarr
σ119865119881119887 times 1
=12σ119865119881 119890
1198873 (119887
2minus 119897119888)
119897119888 =119887
2(1 minus
119887
6 119890)
119890 le119861
6
1048633 No tension should be permitted at any point of the dam under any circumstance for
moderately high dams
1048633 For no tension to develop the eccentricity should be less than b6
1048633 Or the resultant should always lie within the middle third
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
41
Effect of Tension CracksSince concrete cannot resist the tension a crack
develops at the heel which modifies the uplift pressure
diagram
Due to tension crack the uplift pressure increases in
magnitude and net downward vertical force or the
stabilizing force reduces
The resultant force gets further shifted towards toe
and this leads to further lengthening of the crack
The base width thus goes on reducing and the
compressive stresses on toe goes on increasing till the toe
fails in compression or sliding
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
4 by shear failure called sliding
FSS (factor of safety against sliding) =120583∙Σ119881
Σ119867(must be gt1)
SFF (shear friction factor) =120583∙Σ119881+119861∙119902
Σ119867must be gt (3-5)
Where
119861 =Width of dam at the joint
119902 =Average strength of the joint which varies from 140 tm2 for poor rocks to 400 tm2 for
good rocks
120583 =Friction coefficient (nearly =075)
42
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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106
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
43
Principal and shear stresses120590 = 119875119907 ∙ 119904119890119888
2119886 minus 119875prime 1199051198861198992120572
Where
120590 =Major principal stress which is not greater than (fc)
119875119907 =Minor principal stress
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
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60
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62
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67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
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84
85
86
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89
90
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100
101
102
103
104
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106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
44
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
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57
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59
60
61
62
63
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65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
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89
90
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97
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100
101
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103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
45
1205901 = 119875119899 1199041198901198882empty minus 119875 + 119875119890 1199051198861198992 empty Principal Stress for US
1205901 = 119875119899 1199041198901198882120572 minus 119875 minus 119875119890
prime 1199051198861198992 120572 Principal Stress for DS
p= intensity of water pressure σ1= principal stress on plane
AB τ = shear stress and 119875119899= normal stress Considering
unit length of the dam
120591 = 119875119899 minus 119875 minus 119875119890prime tan120572 Shear Stress for DS
120591 = minus 119875119899 minus 119875 + 119875119890 tanempty Shear Stress for US
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
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55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
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93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
46
Example of Section Design
1 PRELIMINARY DESIGN
A-Type of reservoir full
B-Direction of Earth quake force downward amp toward DS
C-Water elevation (HI)=805 m
A BASE
Bge 119867
120583(119878119904minus119888)
B=805
075(24minus07)=63137 m
B =805
(24minus07)= 6174m take B=75m
B) FREE-BOARD
Free-board = (004005)H rarr (choosing 005)
(Height) = 005x75=375≃4m
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
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101
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104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
47
C-Top width
(a) =75
328= 478≃ 5 m
D- THE HEIGHT OF THE LOW GRAVITY DAM
H = 119865
119908(119878119904+1minus119888)gt Height of dam
H = 300
1times(24+1minus07)= 111m gt 805 m ok
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
48
2 DESIGN ANALYSIS
A-FORCES TO BE CONSIDERED (The Signs and Designations of forces may be changed in
the following examples according to each case of design and analysis)
1 Hydrostatic Pressure (PW) (negative) ndash
2 Hydrodynamic pressure (PE) (negative) ndash
3 Uplift force (U) (negative) ndash
4 Weight of the dam (w) (positive) +
5 Weight of water supported (w) (positive) +
6 Downward earthquake Forces (PsV) (positive) +
7 Horizontal acceleration of earthquake Forces toward DS of dam (Psh) (negative)
Remember that the sign of each considered force it must be indicated
according to the type of action ie if the force led to stability of dam it
taken as positive otherwise it will be negative
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
bullNotes
bullThe following examples were selected from referenceshence the number of equations and figures are identicalwith the context of these references not to the currentlecture notes
bullMany mistakes may be found in calculations so It is betterto re -check the results
49
50
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52
53
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55
56
57
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59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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55
56
57
58
59
60
61
62
63
64
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66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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59
60
61
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67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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56
57
58
59
60
61
62
63
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66
67
68
69
70
71
72
73
Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
74
75
76
77
78
79
80
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
2 Irrigation Engineering and Hydraulic Structures By SANTOSH KOMAR GARG
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake)
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Main References
1 Irrigation and Water Resources Engineering By GL ASAWA
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Example (GLASAWA) For the profile of a gravity dam shown in Fig 168 compute principalstresses for usual loading and vertical stresses for extreme loading at the heel and toe of thebase of the dam Also determine factors of safety against overturning and sliding as well asshear-friction factors of safety for usual loading and extreme loading (with drainsinoperative) conditions Consider only downward earthquake acceleration for extremeloading condition Sediment is deposited to a height of 15 m in the reservoir Other data areas followsCoefficient of shear friction = 07 (usual loading) and = 085 (extreme loading)Shear strength at concrete-rock contact C = 150 times 1041198731198982
Weight density of concrete = 24 times 1041198731198982
Weight density of water = 1 times 1041198731198983
119886ℎ = 01 119886119899119889 119886119907 = 005
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![Hydraulic Structures[4]](https://static.documents.pub/doc/80x56/563db7e1550346aa9a8ed1d4/hydraulic-structures4.jpg)
