The end gates ABC of a lock are 9 m high and when closed include an angle of 120°. The width of the lock is 10 m. Each gate is supported by two hinges located at 1 m and 6 m above the bottom of the lock. The depths of water on the two sides are 8 m and 4 m respectively. Find: (i) Resultant water force on each gate, (ii) Reaction between the gates AB and BC, and (iii) Force on each hinge.

Lock Gate Hinge Force Problem

Problem Statement

Given Data

Height of each gate, $ H_{gate} = 9 \, \text{m} $
Width of lock, $ W_{lock} = 10 \, \text{m} $
Angle between gates = 120°
Upstream water level, $ H_1 = 8 \, \text{m} $
Downstream water level, $ H_2 = 4 \, \text{m} $
Hinge locations: 1 m and 6 m from the bottom

Diagram of Lock Gates

Plan and elevation view of the lock gates

Solution

1. Gate Geometry

The angle $ \theta $ for one gate with respect to the lock's centerline is:

$$ \theta = \frac{180^\circ - 120^\circ}{2} = 30^\circ $$

The width of each gate leaf ($l$) is calculated as:

$$ l = \frac{W_{lock} / 2}{\cos \theta} = \frac{10 / 2}{\cos 30^\circ} = \frac{5}{0.866} \approx 5.773 \, \text{m} $$

2. Forces on a Single Gate

Upstream Force ($F_1$):

$$ F_1 = \rho g A_1 \bar{h}_1 $$ $$ A_1 = H_1 \times l = 8 \times 5.773 = 46.184 \, \text{m}^2 $$ $$ \bar{h}_1 = H_1 / 2 = 8 / 2 = 4 \, \text{m} $$ $$ F_1 = 1000 \times 9.81 \times 46.184 \times 4 \approx 1812260 \, \text{N} $$

This force acts at $ H_1 / 3 = 8/3 \approx 2.67 \, \text{m} $ from the bottom.

Downstream Force ($F_2$):

$$ F_2 = \rho g A_2 \bar{h}_2 $$ $$ A_2 = H_2 \times l = 4 \times 5.773 = 23.092 \, \text{m}^2 $$ $$ \bar{h}_2 = H_2 / 2 = 4 / 2 = 2 \, \text{m} $$ $$ F_2 = 1000 \times 9.81 \times 23.092 \times 2 \approx 453065 \, \text{N} $$

This force acts at $ H_2 / 3 = 4/3 \approx 1.33 \, \text{m} $ from the bottom.

(i) Resultant Water Force on Each Gate ($F$)

$$ F = F_1 - F_2 = 1812260 - 453065 = 1359195 \, \text{N} $$

3. Location of Resultant Force ($x$)

Taking moments about the bottom of the gate to find the location $x$ of the resultant force F:

$$ F \times x = (F_1 \times 2.67) - (F_2 \times 1.33) $$ $$ 1359195 \times x = (1812260 \times 2.67) - (453065 \times 1.33) $$ $$ 1359195 \times x = 4838734 - 602576 = 4236158 $$ $$ x = \frac{4236158}{1359195} \approx 3.117 \, \text{m from bottom} $$

(ii) Reaction Between the Gates ($P$)

The reaction $P$ between the two gates is given by:

$$ P = \frac{F}{2 \sin \theta} $$ $$ P = \frac{1359195}{2 \sin 30^\circ} = \frac{1359195}{2 \times 0.5} = 1359195 \, \text{N} $$

(iii) Force on Each Hinge

The total reaction from the hinges $R$ is equal to the reaction between the gates $P$.

$$ R = P = 1359195 \, \text{N} $$

Let $R_T$ and $R_B$ be the reactions at the top (6 m) and bottom (1 m) hinges. The distance between them is 5 m.

$$ R_T + R_B = R = 1359195 \, \text{N} $$

Taking moments about the bottom hinge (at 1 m) to find $R_T$:

$$ R_T \times (\text{Hinge Dist.}) = R \times (x - \text{Bottom Hinge Height}) $$ $$ R_T \times 5.0 = 1359195 \times (3.117 - 1.0) $$ $$ R_T \times 5.0 = 1359195 \times 2.117 $$ $$ R_T = \frac{2877315}{5.0} \approx 575463 \, \text{N} $$

Now, we find $R_B$:

$$ R_B = R - R_T $$ $$ R_B = 1359195 - 575463 = 783732 \, \text{N} $$

Final Result:

(i) Resultant water force on each gate is $ F \approx 1359.2 \, \text{kN} $.

(ii) Reaction between the gates is $ P \approx 1359.2 \, \text{kN} $.

(iii) Force on the top hinge is $ R_T \approx 575.46 \, \text{kN} $.

(iii) Force on the bottom hinge is $ R_B \approx 783.73 \, \text{kN} $.

Explanation of Concepts

Forces on Lock Gates: The net force on a lock gate is the difference between the hydrostatic force from the higher upstream water level and the force from the lower downstream level. This net force acts perpendicular to the gate's surface.

Equilibrium of Gates: This net water force ($F$) is balanced by two other forces: a compressive reaction force ($P$) from the opposing gate where they meet, and a total reaction force from the hinges ($R$). For the standard miter gate design, the hinge reaction $R$ is equal in magnitude to $P$.

Hinge Force Distribution: The total hinge reaction ($R$) is distributed between the top and bottom hinges. To find the individual forces on each hinge, we use the principle of moments. By taking moments about one hinge, we can calculate the force on the other, ensuring the gate does not rotate.

Problem Statement

Given Data

Diagram of Lock Gates