lock

1. Water pressure on Lock Gates
Overview
Whenever a dam or a weir is constructed across a river or canal, the water levelson both the sidesof the dam will be different.If it isdesired
to have navigationor boatingin such a river or a canal, then a chamber, known aslock, is constructed between these two different water
levels. Two sets of gates(one on the upstream side and the other on downstream side of the dam) are providedasshown in fig - 1.
In order to transfer a boat from the upstream (i.e., from a higher water level to the downstream), theupstream gatesare opened(whilethe
downstream gatesare closed) and water level in thechamber risesup to the upstream water level. The boat isthen admittedi n thechamber.
Then upstream gatesare closed and downstream gatesare opened andthe water level inthe chamber islowered to the downstream water
level. Now the boat can proceed further downwards. If the boat isto be transferred from downstream to upstream side, the abo ve procedure
is reversed.
Now consider a set of lock gatesAB and BC hinged at the topand bottom at A and C respectively asshown in fig - 2(a). These gateswill be
held in contact at b by the water pressure, the water level being higher on the left handside of the gates as shown in fig - 2(b).
Let,

2.  P = Water pressure on the gate AB or BC acting at right angleson it
 F = Force exerted by the gate BC acting normally to the contact surface of the two
gatesAB and BC (also known as reaction between the two gates), and
 R = Reaction at the upper and lower hinge
Since the gate AB isin equilibrium, under the actionof the abovethree forces, therefore they will meet at onepoint. Let, P and F meet at O,
then R must pass through thispoint.
Let, = Inclination of the lockgate with the normal to the wallsof the lock.
From the geometry of the figure ABO, we find that it isan isoscelestrianglehavingitsangles OBA and OAB both equal to .
(1)
and now resolving the force at right anglesto AB
(2)
Now let usconsider the water pressure on the top and bottom hingesof the gate, Let,
 H1 = Height of water to the left side of the gate.
 A1 = Wetted area (of one of the gates) on left side of the gate
 P1 = Total pressure of the water on the left side of the gate
 H2, A2, P2 = Corresponding valuesfor right side on the gate
 RT = Reaction of the top hinge, and
 RB = Reaction of bottom hinge
Since the total reaction (R) will be shared by the two hinges(RT), therefore
(3)
and total pressure on the lockgate,
Similarly,
Since the directionsof P1 and P2 are in the oppositedirection, therefore theresultantpressure,
We know that the pressure P1 will act through itscenter of pressure, which is at a height of from the bottom of the gate. Similarly,the
pressure P2 will also act through itscenter of pressure which isalso at a height of from thebottom of thegate.
A little consideration will show, that half of theresultantpressure (i.e., P1 - P2 or P)will be resisted by the hingesof one lockgate (asthe other
half will be resisted by the other lockgates).
(4)
where h is the distance between the two hinges.
Also resolving the forceshorizontally,
(5)
From equations(4) and (5) the valuesof RB and RT may be found out.
Example: 1 2
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Example - Water pressure on lock gate
Problem

3. Two lockgates of 7.5m height are providedin a canal of 16m widthmeeting at an angleof . Calculatethe force actingon each gate,
when the depth of water on upstream side is5m.
Workings
Given,
 Height of lockgates= 7.5m
 Width of lockgates= 16m
 Inclination of gates=
 H = 5m
From the geometry of the lockgate, we findthat inclination of the lockgateswith the walls,
and width of each gate = = = 9.24 m
Wetted area of each gate,
and force acting on each gate,
Solution
Force acting on each gate,P = 1133 KN