A rectangular pontoon has a width of 6m, length of 10m and a draught of 2m in fresh water. Calculate (a) weight of pontoon, (b) its draught in seawater of density 1025 kg/m3 and (c) the load that can be supported by the pontoon in fresh water if the maximum draught permissible is 2.3m.

A rectangular pontoon has a width of 6m, length of 10m and a draught of 2m in fresh water. Calculate

(a) weight of pontoon,

(b) its draught in seawater of density 1025 kg/m³

Rectangular Pontoon Analysis

Problem Statement

A rectangular pontoon has the following dimensions:

Width: 6m
Length: 10m
Draught in fresh water: 2m

Determine the following:

Weight of the pontoon.
Its draught in seawater (\(\rho_{\text{seawater}} = 1025 \text{ kg/m}^3\)).
The additional load the pontoon can support in fresh water when the maximum draught is 2.3m.

Solution

1. Calculate Weight of the Pontoon

\[ W = \gamma_{\text{water}} V_{\text{displaced water}} \] \[ = 9810 \times (6 \times 10 \times 2) \] \[ = 1,177,200 \text{ N} \] \[ = 1177.2 \text{ kN} \]

2. Draught in Seawater

Since the pontoon’s weight remains the same in seawater, the volume of displaced seawater is given by: \[ W = \gamma_{\text{seawater}} V_{\text{seawater displaced}} \] \[ 1,177,200 = 1025 \times 9.81 \times (6 \times 10 \times D) \] \[ D = \frac{1,177,200}{1025 \times 9.81 \times 60} \] \[ D = 1.95 \text{ m} \]

3. Maximum Load Supported in Fresh Water

\[ F_B = \gamma_{\text{water}} V_{\text{max displaced}} \] \[ = 9810 \times (6 \times 10 \times 2.3) \] \[ = 1,353,780 \text{ N} = 1353.78 \text{ kN} \]

\[ P = F_B – W \] \[ = 1353.78 – 1177.2 \] \[ = 176.58 \text{ kN} \]

Final Results:

Weight of the pontoon: 1177.2 kN
Draught in seawater: 1.95m
Additional load supported in fresh water: 176.58 kN

Explanation

The problem involves the principle of buoyancy, which states that a floating body displaces a volume of liquid equal to its own weight.

1. Weight of the Pontoon:
Since the pontoon is floating in fresh water, its weight is equal to the weight of the water it displaces. We use the buoyancy equation \( W = \gamma V \) to find its total weight.

2. Draught in Seawater:
The pontoon will displace a lesser volume in seawater because seawater has a higher density than fresh water. Using the buoyancy equation again, we solve for the new draught \( D \) in seawater.

3. Maximum Load in Fresh Water:
The maximum permissible draught is given as 2.3m. By calculating the total buoyant force corresponding to this new depth, we determine the additional load the pontoon can carry while remaining afloat.

Physical Meaning

1. Importance of Buoyancy in Floating Structures:
The pontoon remains afloat because the upthrust force exerted by the water balances its weight. If additional weight is added beyond its buoyancy limit, it will submerge further.

2. Effect of Seawater on Draught:
Since seawater is denser than fresh water, the pontoon experiences a greater buoyant force for the same weight, leading to a smaller draught.

3. Load Carrying Capacity:
The pontoon can safely carry a load up to 176.58 kN while maintaining a maximum draught of 2.3m. Exceeding this would result in submergence or instability.