draft =4m ,even keel, dimensions as above. GM = 5.2m

1. Divide the Waterplane into Simple Shapes

We divide the waterplane into three sections:

• Section 1 (Rectangle): Length = $60\text{ m}$, Breadth = $20\text{ m}$
• Section 2 (Forward Triangle): Base = $30\text{ m}$, Height = $20\text{ m}$
• Section 3 (Aft Triangle): Base = $30\text{ m}$, Height = $20\text{ m}$

2. Calculate Area and Local Moment of Inertia ($I_{local}$)

We need the local moment of inertia for each shape about its own centroid.

• Rectangle:

  • Area ($A_{rect}$) = $60 \times 20 = 1200\text{ m}^2$
  • $I_{rect} = \frac{1}{12} \times \text{Breadth} \times \text{Length}^3$
  • $I_{rect} = \frac{1}{12} \times 20 \times 60^3 = \mathbf{360,000\text{ m}^4}$
  • Centroid is at midships ($x = 0$).

• One Triangle (e.g., Forward):

  • Area ($A_{tri}$) = $\frac{1}{2} \times 30 \times 20 = 300\text{ m}^2$
  • $I_{tri\_local} = \frac{1}{36} \times \text{Base} \times \text{Height}^3$
  • $I_{tri\_local} = \frac{1}{36} \times 30 \times 20^3 = \mathbf{6,666.67\text{ m}^4}$
  • Centroid distance from midships ($d$): The centroid of a triangle is $\frac{1}{3}$ from its base. The base is at $x = 30\text{ m}$ from midships. So the centroid is at $30 + \frac{30}{3} = \mathbf{40\text{ m}}$ from midships.

3. Calculate Total Moment of Inertia ($I_{total}$) using Parallel Axis Theorem

Formula: $I_{total} = \sum (I_{local} + A \times d^2)$

• Rectangle:

  • $I_{rect\_total} = 360,000 + (1200 \times 0^2) = \mathbf{360,000\text{ m}^4}$

• Forward Triangle:

  • $I_{tri\_fwd} = 6,666.67 + (300 \times 40^2)$
  • $I_{tri\_fwd} = 6,666.67 + (300 \times 1600) = 6,666.67 + 480,000 = \mathbf{486,666.67\text{ m}^4}$

• Aft Triangle:

  • Because the vessel is symmetrical, the aft triangle has the same area and is the same distance ($d = 40\text{ m}$) from midships.
  • $I_{tri\_aft} = \mathbf{486,666.67\text{ m}^4}$

• Total Moment of Inertia ($I_{total}$):

  • $I_{total} = 360,000 + 486,666.67 + 486,666.67 = \mathbf{1,333,333.34\text{ m}^4}$
  • (Exact fraction: $360,000 + \frac{1,460,000}{3} + \frac{1,460,000}{3} = \frac{1,080,000 + 2,920,000}{3} = \frac{4,000,000}{3} \approx 1,333,333.33\text{ m}^4$)

4. Calculate Volume of Displacement ($\nabla$)

• Waterplane Area ($A_w$) = $1200 + 300 + 300 = 1800\text{ m}^2$
• Draft ($d$) = $4\text{ m}$
• $\nabla = A_w \times d = 1800 \times 4 = \mathbf{7,200\text{ m}^3}$

5. Calculate BM (Metacentric Radius)

• $BM = \frac{I_{total}}{\nabla}$
• $BM = \frac{1,333,333.33}{7,200} = \mathbf{185.185\text{ m}}$

6. Calculate KM (Height of Metacenter)

• $KB = \frac{\text{Draft}}{2} = \frac{4}{2} = \mathbf{2\text{ m}}$
• $KM = KB + BM = 2 + 185.185 = \mathbf{187.185\text{ m}}$

7. Calculate GM (Metacentric Height)

• Given $KG = 5.2\text{ m}$
• $GM = KM - KG = 187.185 - 5.2 = \mathbf{181.985\text{ m}}$

8. Calculate MCTC

Formula: $MCTC = \frac{W \times GM}{100 \times L}$ (Assuming standard sea water density $\rho = 1.025\text{ t/m}^3$)

• $W = \nabla \times \rho = 7200 \times 1.025 = \mathbf{7,380\text{ tonnes}}$
• Total Length ($L$) = $30 + 60 + 30 = \mathbf{120\text{ m}}$
• $MCTC = \frac{7380 \times 181.985}{100 \times 120}$
• $MCTC = \frac{1,343,049.3}{12,000}$
• $MCTC \approx \mathbf{111.92\text{ tonnes-m/cm}}$

(Note: If fresh water density $\rho = 1.0$ is assumed, $W = 7200\text{ t}$ and MCTC $\approx 109.19\text{ tonnes-m/cm}$.)

Corrected Final Answer: MCTC $\approx$ 111.92 tonnes-m/cm (assuming sea water).