Jul. 14, 2025
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Dams are artificial structures erected on River courses to regulate the water flow by creating artificial reservoir on upstream side as well as for other purposes like generating hydroelectricity. Dams play a pivotal role in Guiding River and serve as multipurpose project.
\color{blue}\large{P = \frac{1}{2}\gamma_{w}H^{2}}
Acting at a distance H/3 from the base.
When Upstream force is partly vertical and partly inclined.
\color{blue}\large{P_{H}= \frac{1}{2}\gamma_{w}H^{2}}
Acting at a distance H/3 from the base.
Pv = Weight of water in portion ABCD
Acting at Centre of gravity of area.
γw : 9.81 KN/m3
Similarly, if there is tail on downstream, it will have horizontal and vertical components.
Water seeping through the crack, fissures etc. of Foundation material, and dam body exerts an uplift pressure on the base of dam.
It is second major external force.
When drainage galleries are provided to relieve the uplift, the recommended uplift at the face of gallery is equal to hydrostatic pressure at toe (γw H’) plus 1/3rd the difference of pressure at heel and toe.
It is assumed that uplift pressure is not affected by earthquake force.
The uplift pressure can be controlled by constructing cut off walls under upstream face.
It can also be controlled by constructing drainage channels between dam and its foundation and by pressure grouting the foundation.
It produces shock waves capable of shaking the Earth upon which dam is resting.
A vertical acceleration may either act downward or upward. When acting in upward direction, it causes uplift increasing the effective weight of dam.
When acting downward, foundation will try to move downward, reducing effective weight. Hence It is the worst case for design.
\color{blue}\large{F = \frac{w}{g}\alpha_{v}}
W : total weight of dam.
Net effective weight of dam
\color{blue}\large{Effective\:weight = w-\frac{w}{g}\alpha_{v}}
\color{blue}\large{ \alpha_{v}=K_{v}g}
Kv : fraction of gravity like 0.1, 0.2 etc.
Vertical acceleration reduces unit weight of dam and that of water to(1 – Kv)
times their original weight.
It causes momentary increase in water pressure, as foundation and dam accelerate towards the reservoir and water resist movement due to its own inertia.
\color{blue}\large{P_{e}= 0.55 K_{h}\gamma_{w}H^{2}}
Acting at height of 4H/3π above base.
Kh : factor of gravity for horizontal acceleration.
Moment of this force above the base :
If inclined face in upstream side does not extend upto more than half of depth of reservoir, it is what we consider as vertical.
Force so generated keeps the body and foundation in one piece.
\color{blue}\large{F_{h}= \frac{w}{g}\alpha_{h}= wk_{h}}
Acting at centre of gravity of mass.
Force exerted by silt deposited against upstream face.
\color{blue}\large{P_{silt}= \frac{1}{2} \gamma_{sub}H^{2}K_{a}}
Acting at h/3 from base.
h : height of silt deposited
γsub : submerged weight of silt material
Ka : coefficient of active earth pressure
\color{blue}\large{K_{a}= \frac{1-\sin \phi}{1+\sin \phi}}
Φ : angle of internal friction of soil.
In absence of data, deposited silt may be taken as equivalent fluid exerting a force with unit weight equal to 3.6 KN/m3 in horizontal direction and vertical force with unit weight 9.2 KN/m3
Blowing wind generates waves. It causes pressure towards downstream side.
It depends upon wave height.
For F > 32 Km
\color{blue}\large{h_{w}=0.032\sqrt{VF}}
hw : height of water from top of crest to bottom of trough.
F : fetch or straight length of water expanse in Km
V : wind velocity in Km/h
Maximum pressure intensity due to wave
\color{blue}\large{p_{w}=2.4h_{w}\gamma_{w}}
Acting at hw/2 above still water surface.
Total Force
\color{blue}\large{P_{w}=19.62h_{w}^{2}}
Acting at 3/8 hw above still water surface.
It is dam face thrust due to expanding ice. Its magnitude varies from (250 – ) KN/m2 depending upon temperature variation.
Average value of 500 KN/m2 is under allowable limit for ordinary conditions.
It acts linearly along length of dam and at reservoir level.
Cross section of dam is divided into triangles and rectangles and unit length of dam is considered in analysis. The resultant of all these forces will represent weight of dam.
If the resultant of all forces acting on dam at any of its section, passes outside the toe, dam shall rotate and overturn about the toe.
Practically, such cases shall not arise, as dam will fail much earlier by compression.
The ratio of righting moment about the toe to overturning moment about the toe, is what we call factor of safety against overturning.
Its value varies between 2 to 3.
Compression stress may exceed allowable stress of material.
P = Direct Stress + Bending Stress
e : eccentricity of resultant force from centre of base.
V : Vertical force (total)
B : Base width
For No Tension :
\color{blue}\large{ \frac{\sum{V}}{B}\left [ 1\pm\frac{6e}{B} \right ]=0}
\color{blue}\large{e = \frac{B}{6}}
pmax will be produced on the end which is nearer to resultant.
pmin comes out to be negative, it means tension will generate at appropriate end.
If pmin exceeds the allowable compressive strength of dam, the dam may crush and fail by crushing.
Material in dam cannot withstand tensile stress. If Subjected, they may finally crack.
Maximum permissible tensile stress for high concrete gravity dam can be taken as 500 KN/m2
Effect : When such tension crack develops say at head, crack width looses contact with the bottom foundation, and thus becomes ineffective.
Hence effective width B of the dam base will reduce. This will increase pmax at toe.
Since uplift increases and net effective downward force reduces, the resultant will shift more towards toe, and further increasing the compressive stress at the toe. Finally leading to failure of toe by direct compression.
Hence, a tensile crack by itself does not fail the structure, but it leads to the failure of the structure by producing excessive compressive stresses.
\color{blue}\large{p_{min}= \frac{\sum{V}}{B}\left [ 1-\frac{6e}{B} \right ]=0}
Giving e = B/6
Maximum value of eccentricity that can be permitted on either side of centre is equal to B/6. The resultant must lie within the middle third.
It will occur when the net horizontal force above any plane in the dam or at the base of the dam exceed the frictional resistance developed at that level.
\color{blue}{\sum{H}< \mu \sum{V}}
\color{blue}{FSS= \mu \frac{\sum{V}}{\sum{H}}>1}
If we also consider shear resistance of the joint (Shear friction factor)
Then
\color{blue}{SFF= \frac{\mu \sum{V}+ B.q}{\sum{H}}}
B : width of dam at joint
q : Average shear strength
The Value of μ varies from 0.65 to 0.75
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To increase shear strength at base, foundation has stepped design at base (better bond)
During construction, We leave Horizontal joints . Lower surfaces are cleaned and layer of neat cement or rich cement mortar should be poured before pouring standard concrete mix for upper layer.
\color{blue}\large{\sigma = P _{v} \sec^2\alpha - P' \tan^2 \alpha}
For Stress to be maximum :
\color{blue}\large{\sigma = P _{v} \sec^2\alpha}
We should keep in mind that normal stress should not exceed compressive strength of material.
\color{blue}\large{\tau = (P _{v}-P')\tan\alpha}
Neglecting tail pressure
\color{blue}\large{\tau_{max} = P _{v}\tan\alpha}
The margin between maximum reservoir level and top of dam.
It reduces possibility of spilling over dam due to wave action.
Freeboard mostly provided is 3/2 hw
For road construction over dam
For low dam
\color{blue}\large{h< \frac{f}{h_{w}(S_{c}+1)}}
f : permissible compressive stress
Sc : Specific gravity of dam material
Structure constructed at dam site, for effective disposal of surplus water from upstream to downstream.
It does not allow water to rise above maximum reservoir level. It is safety valve for a dam.
Within body or independently in a saddle. Deep narrow gorge with steep bank separated from flank by a hillock.
If gate is present in spillway to control outflow, then the spillway is controlled spillway.
If operating head on spillway is more than the designed head, lower nappe may leave ogee profile, generating negative pressure at point of separation.
The generation of negative or vacuum pressure cause bubbles with air, vapour and other gases. They on moving downstream, may enter a region where absolute pressure is higher.
This cause vapors in cavity to condense and return to liquid form with a resulting implosion or collapse of cavity. It exerts extremely high pressure which in turn cause fatigue failure of masonry surface leading to pitting.
Pitting leads to spongy appearance to surface and causes heavy damages.
Formation of bubbles or cavity when absolute pressure close to vapour pressure starts evaporation.
US Army corps of engineers Downstream profile. (lower nappe)
\color{blue}\large{x^n = k H_{d}^{n-1}y}
(x,y) : coordinates of points on crest profile with origin at highest point C of crest, called apex
Hd : Designed head including velocity head.
K,n : constants depending upon slope of upstream face.
For Slope of upstream face (vertical), K = 2 and n = 1.85
Hence for vertical upstream face
\color{blue}\large{x^{1.85} = 2 H_{d}^{0.85}y}
A smooth gradual reverse curve is provided at bottom of downstream face which turns the flow into the apron of stilling basin or into spillway discharge channel. A radius of 1/4th of spillway height is satisfactory for reverse bottom curve.
Equation for downstream profile :
\color{blue}\large{x^{1.78} = 1.852 H_{e}^{0.78}y}
Utilise minimum flow in river having no appreciable pondage on upstream side. A weir or barrage has its use to simply raise water level, with narrow limits of fluctuation.
They have upstream storage reservoir of sufficient size, so as to permit sufficient carry over storage from the monsoon season to dry summer season. The tunnels are constructed to power house machine by means of pressure penstocks.
Generate power during peak hours, but during off peak hours water is pumped back from tail water pool to head water pool for further use.
Pumps are run by some secondary power. The plant primarily meant for assisting an existing thermal plant or some hydel plant.
For heads varying between 15 – 90 m, reversible pumps turbines have been devised.
The difference between high and low tide is utilized to generate power. This is accomplished by constructing a basin separated from ocean by a partition wall and installing turbines in opening of this wall.
Water power potential
Amount of power generated when Q cumecs of water is allowed to fall through a head difference of H meters.
\color{blue}\large{Water\:Energy = Q\gamma_{w}H}
Normal water level (NWL)
Highest water level that can be maintained in reservoir without spillage discharge.
Minimum water level (MWL)
The elevation of water level which produces minimum net head on power units (65 % of designed head)
Design Head
Net head under which the turbine reaches peak efficiency at synchronous speed.
Design Head = WAL – MWL
Weighted average level (WAL)
The level above and below which equal amount of power is developed during an average year.
Gross Head
Difference in water level elevation at the point of diversion of water for hydel scheme and the point of return of water back to river.
Operational Head
Difference between at forway entrance and at tailrace exit.
Load factor
Area under load curve (load vs time), would represent energy consumed in KWH.
Annual load factor
Demand Factor
Capacity Factor or Plant factor
\color{blue}\large{CF = \frac{Average\:load}{Plant\:capacity}}
Capacity factor and load factor would become identical, if peak load is equal to plant capacity.
Utilisation Factor or Plant use factor
If water head is constant.
Utilisation factor is 0.4 to 0.9 for hydel plant.
It should not be greater than 1.
Storage basin or any body of water in front of intake. It temporarily store the water rejected by plant due to reduced load and meet instantaneous demand.
It absorbs short interval variations and fluctuations in power load.
Direct water to penstock provided with trash rack to prevent entry of debris. Floating boom trap ice and floating debris.
For severe winter, trash racks may be electrically heated to prevent clinging of ice.
Types : Simple, throttled, differential and multiple.
Converts hydraulic energy to mechanical energy.
Impulse or velocity turbine : Pelton wheel
Reaction or pressure turbine : Francis and Kaplan
It has substructures to support hydraulic and electrical equipment and a superstructure to house and protect the equipment.
A conduit which connects the outlet of a reaction turbine runner to tailrace. Water as emerges out of runner, flows through pipe of gradually increasing diameter.
Effective pressure head is increased by amount
\color{blue}\large{\Delta P = H_{s} + \frac{V_{2}^2-V_{3}^2}{2g}}
This should not exceed vacuum pressure.
By gradually increasing the diameter of pipe, we are able to reduce V3 to zero.
It permits Installation of turbines should be at higher level than tailwater level.
The outlet gate of draft tube should be provided with gates, so that draft tube can be dewatered for repairs.
Timber logs or steel gates can be used with hoisting mechanism.
Water is discharged through turbines in Channels called Tailrace.
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