Each gate of a lock is 6m high and is supported by two hinges placed on the top and the bottom. When the gates are closed, they make an angle of 1200. The width of the lock is 7m. If the water levels are 5m and 2m at upstream and downstream respectively, determine the magnitude of forces on the hinge due to the water pressure.

Forces on Hinges of Lock Gates

Problem Statement

Each gate of a lock is 6m high and is supported by two hinges placed on the top and the bottom. When the gates are closed, they make an angle of 120°. The width of the lock is 7m. If the water levels are 5m and 2m at upstream and downstream respectively, determine the magnitude of forces on the hinge due to the water pressure.

Solution

1. Forces on the Gate

The resultant water force \( F \), the reaction between the gates \( P \), and the total reaction at the hinge \( R \) are related as follows:

\( P \cos \theta = R \cos \theta \implies P = R \) \( \quad \text{(a)} \)

Resolving forces normal to the gate:

\( P \sin \theta + R \sin \theta = F \quad \text{(b)} \)

From (a) and (b):

\( P = \frac{F}{2 \sin \theta} \)

2. Horizontal Forces

Horizontal force on the upstream side:

\( F_1 = \gamma A_1 \bar{y}_1 = 9810 \cdot 4.04 \cdot 5 \cdot \frac{5}{2} = 495405 \, \text{N} \)

\( F_1 \) acts at a distance of:

\( \frac{5}{3} = 1.66 \, \text{m from bottom} \)

Horizontal force on the downstream side:

\( F_2 = \gamma A_2 \bar{y}_2 = 9810 \cdot 4.04 \cdot 3 \cdot \frac{3}{2} = 178346 \, \text{N} \)

\( F_2 \) acts at a distance of:

\( \frac{3}{3} = 1 \, \text{m from bottom} \)

The net force is:

\( F = F_1 – F_2 = 495405 – 178346 = 317059 \, \text{N} \)

3. Moment and Point of Application

Taking moments about the bottom to find the point of application of \( F \):

\( 317059 \cdot y = 495405 \cdot 1.66 – 178346 \cdot 1 \)

Simplifying:

\( y = \frac{495405 \cdot 1.66 – 178346 \cdot 1}{317059} = 2.03 \, \text{m} \)

4. Reaction Forces

\( P \):

\( P = \frac{F}{2 \sin \theta} = \frac{317059}{2 \sin 30°} = 317059 \, \text{N} \)

Total reaction at the hinge \( R \):

\( R = P = 317059 \, \text{N} \)

Taking moments about the bottom hinge:

\( R_t \cdot 6 = 317059 \cdot 2.03 \)

Simplifying:

\( R_t = \frac{317059 \cdot 2.03}{6} = 107272 \, \text{N} = 107.272 \, \text{kN} \)

Reaction at the bottom hinge:

\( R_b = R – R_t = 317059 – 107272 = 209787 \, \text{N} = 209.787 \, \text{kN} \)

Result:

Total Reaction at Hinge: \( R = 317.059 \, \text{kN} \)
Top Hinge Reaction: \( R_t = 107.272 \, \text{kN} \)
Bottom Hinge Reaction: \( R_b = 209.787 \, \text{kN} \)

Explanation

Water Forces: The upstream and downstream forces are caused by hydrostatic pressure acting on the gates. The net force is the difference between these forces.
Moments: The moment calculation determines the point of application of the resultant force to accurately calculate reactions at hinges.
Hinge Reactions: The hinges bear the load distributed based on the resultant force and its point of application.

Physical Meaning

This problem demonstrates the equilibrium of lock gates under hydrostatic pressure:

Reaction Forces: Calculating hinge reactions ensures the gates remain stable and the hinges are designed to bear the load.
Applications: This is critical for designing lock gates, ensuring safety and reliability in water transportation systems.