# SUPPLY SLUICES IN GRAVITY DAMS

• Some openings in the dam have to be provid so as to pass the excess flow D/S of the dam.
• These openings are know outlet sluices or supply sluices.
• If water from the reservoir is to be release for irrigation purpose at a controll rate, it is done with the help of these supply sluices.
• All the supply sluices are fitted with gates which can be raise or lowere.
• The control on the gates is exercise from the top of the dam.
• The gates are fitted in the grooves form at the sides of the openings in the dam.
• When gates are lowere they stop flow of water and when raised they again start discharging water D/S.
• Sluices may be provided at more than one depth.
• By this, water can be drawn from different

• elevations or depths.
• Supply sluices are also sometimes used to scour out the silt deposit in the vicinity of the U/S of the dam.

Example.

• Determine the equations for base width of an elementary profile of gravity dam so that resultant passes through the outer middle third points.
• Consider earthquake force, hydrostatic pressure and uplift pressure for computations.

Solution. (a) Vertical forces

$Force- due -to- self -weight -of -dam-W= \frac{1}{2}bhw_{\rho }$

This force acts downwards.

II. Force due to vertical acceleration of earthquake

$F= \alpha W= \alpha bhw_{\rho }$

This force acts upwards.

III. Force due to uplift u= 1/2 bwh

This force also acts upwards.

Hence, total vertical force = ΣV = W – α W – U

$= \frac{1}{2}bhw_{\rho }-\alpha bhw-\frac{1}{2}bhw$

$= \frac{1}{2}bhw[(1-\alpha )p-1]$

where

• w = unit weight of water
• p = specific weight of concrete
• a = coefficient of earthquake acceleration.

(b) Horizontal forces

$Force-due- to -self- weight -of- dam- W =\frac{1}{2}wh^{2}$

II. Force due to hydrodynamic pressure of water at base

$C_{m}= 0.735\frac{\theta }{90}= 0.735 since \theta = 90^{0}$

$P_{e}= C_{m}\alpha wh= 0.735\times \alpha wh$

$P_{e}= 0.726p_{e}h$

$M_{e}= 0.299p_{e}h^{2}= 0.299\times 0.725\alpha wh^{3}= 0.2205\alpha wh^{3}$

III. Inertia force horizontal = αw = 1 /2 α bhwp

If the resultant of all the forces have to pass through the outer third point M2, the moment of all these forces at this point must be zero

$\Sigma V\times \frac{B}{3}= \frac{1}{2}WH^{2}\times \frac{H}{3}+0.2205\alpha WH^{3}+\frac{1}{2}\alpha BHW_{\rho }\times \frac{h}{3}$

$\frac{B^{2}hw}{6}[(1-\alpha )\rho -1]= \frac{bh^{2}w\rho \alpha }{6}+\frac{wh^{3}}{6}[1+\alpha (1.323)]$

$b^{2}[(1-\alpha )p-1]= bhp.\alpha +h^{2}[1+1.323\alpha ]$

Solving this equation for b we get

$b= h\frac{p\alpha \pm \sqrt{p^{2}\alpha ^{2}+4(1+1.323\alpha )(1-\alpha )p-1}}{2{(1-\alpha )-1}}$

This is the required general equation for b.

When there is no earthquake, value of α = 0 and the equation for b reduces to the form

$b= \frac{h}{\sqrt{p-1}}$

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