# 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

\[b= \frac{h}{\sqrt{p-1}}\]