CELE Hydraulics & Fluid Mechanics — Buoyancy and FlotationStudy Notes

Problem

A steel sphere with diameter 0.5 m is fully submerged in water. Calculate the buoyant force.

Solution

Step 1: Calculate volume of sphere $$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{0.25}{1}\right)^3 = \frac{4}{3}\pi (0.015625) = 0.0654 \text{ m}^3$$ Step 2: Apply buoyant force formula $$F_B = \gamma_w V = 9.81 \times 0.0654 = 0.642 \text{ kN}$$ **Answer:** The buoyant force is 0.642 kN upward. This is independent of the sphere's material density (steel, aluminum, etc.).

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Problem

A concrete block (specific gravity 2.4) measuring 0.3 m × 0.3 m × 0.3 m is completely submerged in water. Find: (a) actual weight, (b) buoyant force, (c) apparent weight.

Solution

Step 1: Calculate actual weight $$W = s_c \times \gamma_w \times V = 2.4 \times 9.81 \times (0.3)^3$$ $$W = 2.4 \times 9.81 \times 0.027 = 0.635 \text{ kN}$$ Step 2: Calculate buoyant force $$F_B = \gamma_w \times V = 9.81 \times 0.027 = 0.265 \text{ kN}$$ Step 3: Calculate apparent weight (measured submerged) $$W_{\text{app}} = W - F_B = 0.635 - 0.265 = 0.370 \text{ kN}$$ **Answers:** (a) 0.635 kN; (b) 0.265 kN; (c) 0.370 kN **Note:** The apparent weight is what would be registered by a scale placed beneath the submerged block.

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Problem

A wooden block with dimensions 0.4 m (length) × 0.3 m (width) × 0.25 m (height) and specific gravity 0.65 floats in fresh water. Calculate the submerged depth (draft).

Solution

Step 1: Use the homogeneous block formula $$d = s \times h = 0.65 \times 0.25 = 0.1625 \text{ m} = 162.5 \text{ mm}$$ **Verification using equilibrium method:** Step 2: Calculate weight $$W = s \times \gamma_w \times V_{\text{total}} = 0.65 \times 9.81 \times (0.4 \times 0.3 \times 0.25)$$ $$W = 0.65 \times 9.81 \times 0.03 = 0.191 \text{ kN}$$ Step 3: Calculate displaced volume from equilibrium $$V_{\text{disp}} = \frac{W}{\gamma_w} = \frac{0.191}{9.81} = 0.01948 \text{ m}^3$$ Step 4: Calculate draft $$d = \frac{V_{\text{disp}}}{A} = \frac{0.01948}{0.4 \times 0.3} = \frac{0.01948}{0.12} = 0.1623 \text{ m}$$ **Answer:** Draft is approximately 0.162 m or 162 mm. The freeboard is 250 − 162 = 88 mm.

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Problem

A square pontoon (2.5 m × 2.5 m × 0.8 m depth) has an empty weight of 15 kN. The maximum allowable draft is 0.6 m (design constraint for shallow river operation). Calculate the maximum cargo weight it can carry.

Solution

Step 1: Calculate maximum displacement at d = 0.6 m $$V_{\text{max,disp}} = A \times d = (2.5 \times 2.5) \times 0.6 = 6.25 \times 0.6 = 3.75 \text{ m}^3$$ Step 2: Calculate buoyant force at maximum draft $$F_B = \gamma_w \times V_{\text{max,disp}} = 9.81 \times 3.75 = 36.79 \text{ kN}$$ Step 3: Apply equilibrium condition $$F_B = W_{\text{pontoon}} + W_{\text{cargo}}$$ $$36.79 = 15 + W_{\text{cargo}}$$ $$W_{\text{cargo}} = 21.79 \text{ kN}$$ **Answer:** The maximum cargo weight is 21.79 kN (approximately 2.2 tonnes). This ensures the pontoon does not sink below the 0.6 m draft limit.

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Problem

A steel ship displaces 8000 m³. Calculate: (a) buoyant force in freshwater, (b) buoyant force in seawater (s = 1.025). How much additional cargo can the ship carry in seawater?

Solution

Step 1: Buoyant force in freshwater $$F_{B,fresh} = \gamma_w \times V = 9.81 \times 8000 = 78,480 \text{ kN}$$ Step 2: Buoyant force in seawater $$\gamma_s = s \times \gamma_w = 1.025 \times 9.81 = 10.055 \text{ kN/m}^3$$ $$F_{B,sea} = 10.055 \times 8000 = 80,440 \text{ kN}$$ Step 3: Additional buoyant force available $$\Delta F_B = 80,440 - 78,480 = 1,960 \text{ kN}$$ Step 4: Additional cargo weight (same displacement) $$\text{Additional cargo} = 1,960 \text{ kN} \approx 200 \text{ tonnes}$$ **Answers:** (a) 78,480 kN; (b) 80,440 kN; (c) An additional 200 tonnes of cargo can be carried in seawater at the same draft. **Practical Note:** This is why ships sit lower (deeper draft) in freshwater rivers than in the ocean—they receive less buoyant support in freshwater.

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Problem

A rectangular barge measures 12 m (length) × 4 m (width) × 1.5 m (depth). It floats at a draft of 1.0 m. The center of gravity is located 0.9 m above the keel. Determine the metacentric height and assess stability for rolling motion.

Solution

Step 1: Calculate displaced volume $$V_{\text{disp}} = L \times B \times d = 12 \times 4 \times 1.0 = 48 \text{ m}^3$$ Step 2: Calculate moment of inertia about the long (rolling) axis $$I = \frac{L \times B^3}{12} = \frac{12 \times 4^3}{12} = \frac{12 \times 64}{12} = 64 \text{ m}^4$$ Step 3: Calculate BM $$BM = \frac{I}{V_{\text{disp}}} = \frac{64}{48} = 1.333 \text{ m}$$ Step 4: Find center of buoyancy For a rectangular body at draft d, B is at height d/2 from the keel: $$B = \frac{d}{2} = \frac{1.0}{2} = 0.5 \text{ m above keel}$$ Step 5: Calculate BG $$BG = G - B = 0.9 - 0.5 = 0.4 \text{ m}$$ Step 6: Calculate GM $$GM = BM - BG = 1.333 - 0.4 = 0.933 \text{ m}$$ **Answer:** $GM = 0.933$ m > 0, therefore the barge is **stable** for rolling. The positive metacentric height indicates the metacenter is above the center of gravity, providing a restoring moment that returns the barge to upright when tilted.

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Problem

The barge from Example 3.1 is heeled (tilted) at an angle θ = 5° (0.0873 radians). Calculate the righting moment.

Solution

Step 1: Calculate weight of barge $$W = \gamma_w \times V_{\text{disp}} = 9.81 \times 48 = 470.88 \text{ kN}$$ Step 2: Apply righting moment formula $$M_{\text{righting}} = W \times GM \times \sin\theta$$ $$M_{\text{righting}} = 470.88 \times 0.933 \times \sin(5°)$$ $$M_{\text{righting}} = 470.88 \times 0.933 \times 0.0872$$ $$M_{\text{righting}} = 38.43 \text{ kN·m}$$ Alternatively, for small angles: $\sin\theta \approx \theta$ (radians) $$M_{\text{righting}} \approx 470.88 \times 0.933 \times 0.0873 = 38.56 \text{ kN·m}$$ **Answer:** The righting moment is approximately **38.4 kN·m**. This moment acts to return the barge to the upright position. **Physical Interpretation:** The larger the righting moment, the stronger the tendency to return to equilibrium. This explains why ships with larger metacentric heights (deeper, broader hulls) are more stable in rough seas.

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Problem

A floating platform has displaced volume 100 m³ at a draft of 1.2 m. The waterline dimensions are 8 m × 6 m. If the center of gravity is raised to match the metacenter position, what is the new height of G above the keel?

Solution

Step 1: Calculate BM $$I = \frac{L \times B^3}{12} = \frac{8 \times 6^3}{12} = \frac{8 \times 216}{12} = 144 \text{ m}^4$$ $$BM = \frac{I}{V_{\text{disp}}} = \frac{144}{100} = 1.44 \text{ m}$$ Step 2: Find B position Assuming the waterline is at d = 1.2 m: $$B = \frac{d}{2} = \frac{1.2}{2} = 0.6 \text{ m above keel}$$ Step 3: For neutral equilibrium, GM = 0, so M = G $$GM = 0 \Rightarrow G = M = B + BM$$ $$G = 0.6 + 1.44 = 2.04 \text{ m above keel}$$ **Answer:** The center of gravity must be raised to 2.04 m above the keel to achieve neutral equilibrium. At this point, the platform can tilt without restoring force or capsizing tendency—it remains at any angle. This is a critical and dangerous condition for floating structures; vessels must maintain $GM > 0$ during operation.

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Problem

A rectangular barge (12 m × 4 m × 1.5 m depth) floats at 1.0 m draft with initial G at 0.9 m. If cargo is stacked high, raising G to 1.2 m, calculate the change in GM and assess the new stability.

Solution

From Example 3.1, we have $BM = 1.333$ m and $B = 0.5$ m. **Initial condition:** $$GM_1 = BM - BG_1 = 1.333 - (0.9 - 0.5) = 1.333 - 0.4 = 0.933 \text{ m}$$ **After raising G to 1.2 m:** $$BG_2 = 1.2 - 0.5 = 0.7 \text{ m}$$ $$GM_2 = 1.333 - 0.7 = 0.633 \text{ m}$$ **Change in metacentric height:** $$\Delta GM = 0.633 - 0.933 = -0.300 \text{ m}$$ **Answer:** By raising the center of gravity by 0.3 m, the metacentric height decreased by 0.3 m (from 0.933 m to 0.633 m). The barge remains stable ($GM_2 > 0$), but with reduced stability. This demonstrates a critical design principle: **keeping the center of gravity low enhances stability.** This is why ships load heavy cargo low in the hull and lighter cargo on deck. **Safety Implication:** Carelessly stacking cargo high reduces a vessel's resistance to capsizing in rough seas—a common cause of maritime accidents.

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Problem

A floating fish pen for Laguna de Bay consists of a rectangular polyethylene pontoon (6 m × 4 m × 0.8 m depth) supporting a cage structure. The total weight to be supported is 25 kN (cage, net, and water circulation system). The pontoon weight is 10 kN. Calculate: (a) required draft, (b) freeboard, (c) metacentric height for rolling stability.

Solution

Step 1: Calculate total weight $$W_{\text{total}} = 25 + 10 = 35 \text{ kN}$$ Step 2: Calculate required displaced volume $$V_{\text{disp}} = \frac{W_{\text{total}}}{\gamma_w} = \frac{35}{9.81} = 3.568 \text{ m}^3$$ Step 3: Calculate draft $$d = \frac{V_{\text{disp}}}{A} = \frac{3.568}{6 \times 4} = \frac{3.568}{24} = 0.149 \text{ m} = 149 \text{ mm}$$ Step 4: Calculate freeboard $$\text{Freeboard} = 0.8 - 0.149 = 0.651 \text{ m} = 651 \text{ mm}$$ This freeboard is adequate to protect against small waves and prevents washover. Step 5: Calculate moment of inertia (rolling, about long axis) $$I = \frac{L \times B^3}{12} = \frac{6 \times 4^3}{12} = \frac{6 \times 64}{12} = 32 \text{ m}^4$$ Step 6: Calculate BM $$BM = \frac{I}{V_{\text{disp}}} = \frac{32}{3.568} = 8.966 \text{ m}$$ Step 7: Estimate G position For this floating structure, assume G is approximately 0.4 m above the waterline (due to cargo and cage weight distribution): $$G = d + 0.4 = 0.149 + 0.4 = 0.549 \text{ m above keel}$$ Center of buoyancy: $$B = \frac{d}{2} = 0.0745 \text{ m}$$ $$BG = 0.549 - 0.0745 = 0.4745 \text{ m}$$ Step 8: Calculate GM $$GM = BM - BG = 8.966 - 0.4745 = 8.491 \text{ m}$$ **Answers:** - (a) Draft = 0.149 m (very light flotation—common for fish pens) - (b) Freeboard = 0.651 m (adequate safety margin) - (c) GM = 8.49 m (excellent stability—fish pens are inherently stable due to wide pontoons) **Note:** The very shallow draft and high GM indicate fish pens have exceptional stability. This explains their resilience in moderate lake chop.

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Problem

A square concrete caisson (10 m × 10 m × 8 m depth) is used to support dike reconstruction in Manila Bay. The caisson's weight is 1250 kN. Before towing to the site, it is ballasted to a draft of 3.0 m for stability. Calculate: (a) ballast water weight required, (b) metacentric height for pitching motion (about the short axis), (c) stability assessment.

Solution

Step 1: Calculate volume displaced at d = 3.0 m $$V_{\text{disp}} = 10 \times 10 \times 3.0 = 300 \text{ m}^3$$ Step 2: Calculate buoyant force $$F_B = \gamma_w \times V_{\text{disp}} = 9.81 \times 300 = 2943 \text{ kN}$$ Step 3: Calculate ballast weight required $$W_{\text{ballast}} = F_B - W_{\text{caisson}} = 2943 - 1250 = 1693 \text{ kN}$$ Step 4: Calculate ballast water volume $$V_{\text{ballast}} = \frac{W_{\text{ballast}}}{\gamma_w} = \frac{1693}{9.81} = 172.7 \text{ m}^3$$ Step 5: Calculate moment of inertia for pitching (about the short/transverse axis) $$I_{\text{pitch}} = \frac{B \times L^3}{12} = \frac{10 \times 10^3}{12} = \frac{10 \times 1000}{12} = 833.3 \text{ m}^4$$ Step 6: Calculate BM $$BM = \frac{I_{\text{pitch}}}{V_{\text{disp}}} = \frac{833.3}{300} = 2.778 \text{ m}$$ Step 7: Estimate stability For a loaded caisson with internal ballast: - B is at 3.0/2 = 1.5 m above the keel - G depends on how ballast is distributed; assume G at 1.8 m (caisson structure above, water ballast below) - BG = 1.8 − 1.5 = 0.3 m Step 8: Calculate GM $$GM = BM - BG = 2.778 - 0.3 = 2.478 \text{ m}$$ **Answers:** - (a) Ballast water required = 1693 kN (approximately 173 m³ or 173 tonnes) - (b) BM for pitching = 2.78 m - (c) GM = 2.48 m > 0 → **Stable for towing** **Operational Note:** A caisson with GM = 2.48 m has excellent stability for towing in moderate seas. The large metacentric height reflects the caisson's broad, shallow geometry—ideal for stability but not for maneuverability. During positioning at the site, ballast is adjusted to reduce draft as needed.

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Problem

A floating platform in Manila Bay is designed for a drafted load of 50 kN. In the cooler months (November–February), seawater density is 1025 kg/m³. In summer (May–August), evaporation increases density to 1032 kg/m³. If the platform displaces 10 m³ to support the 50 kN load: (a) Calculate buoyant force in each season, (b) Determine how much additional cargo can be carried in summer, (c) Assess the stability impact.

Solution

Step 1: Calculate buoyant force in cool season $$\gamma_s = 1.025 \times 9.81 = 10.055 \text{ kN/m}^3$$ $$F_B = 10.055 \times 10 = 100.55 \text{ kN}$$ Step 2: Calculate buoyant force in summer season $$\gamma_s = 1.032 \times 9.81 = 10.124 \text{ kN/m}^3$$ $$F_B = 10.124 \times 10 = 101.24 \text{ kN}$$ Step 3: Calculate available load capacity Cool season: $W_{\text{payload}} = 100.55 - W_{\text{structure}}$ Summer season: $W_{\text{payload}} = 101.24 - W_{\text{structure}}$ Assuming platform structure weighs 40 kN: - Cool season: $W_{\text{payload}} = 100.55 - 40 = 60.55$ kN - Summer season: $W_{\text{payload}} = 101.24 - 40 = 61.24$ kN - Additional summer capacity: $61.24 - 60.55 = 0.69$ kN (approximately 70 kg) Step 4: Stability impact If the 10 m³ displaced volume is constant: $$V_{\text{disp, cool}} = \frac{W}{\gamma_s} = \frac{100.55}{10.055} = 10.00 \text{ m}^3$$ $$V_{\text{disp, summer}} = \frac{W}{\gamma_s} = \frac{101.24}{10.124} = 10.00 \text{ m}^3$$ The displaced volume remains constant, but buoyancy increases slightly. Since $BM = I/V_{\text{disp}}$ depends only on geometry and displaced volume: $$BM_{\text{cool}} = BM_{\text{summer}}$$ Therefore, **GM remains unchanged**; stability is not affected by seasonal density variation if the platform dimensions (hence I) are constant. However, if the design allows the platform to float shallower in summer (less draft), the waterline area might change, affecting I and thus stability. **Answers:** - (a) Cool season: 100.55 kN; Summer: 101.24 kN - (b) Additional summer cargo capacity: 0.69 kN (70 kg) - (c) Stability (GM) is unchanged if displaced volume is constant; however, operational draft changes slightly

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Problem

A student solving a barge stability problem (12 m × 4 m waterline, draft 1.0 m, G at 0.9 m) incorrectly calculates: $I = \frac{4 \times 12^3}{12} = 576$ m⁴, leading to $BM = 576 / 48 = 12$ m and $GM = 12 - 0.4 = 11.6$ m. Identify the error and recalculate correctly.

Solution

**Error Identification:** The student used $I = \frac{B \times L^3}{12}$ (pitching axis) instead of $I = \frac{L \times B^3}{12}$ (rolling axis). **For rolling motion (side-to-side tilting):** The correct moment of inertia is about the long axis: $$I_{\text{roll}} = \frac{L \times B^3}{12} = \frac{12 \times 4^3}{12} = \frac{12 \times 64}{12} = 64 \text{ m}^4$$ Not the student's value of 576 m⁴. **Correct Calculation:** $$V_{\text{disp}} = 12 \times 4 \times 1.0 = 48 \text{ m}^3$$ $$BM = \frac{64}{48} = 1.333 \text{ m}$$ $$B = 0.5 \text{ m}, \quad BG = 0.9 - 0.5 = 0.4 \text{ m}$$ $$GM = 1.333 - 0.4 = 0.933 \text{ m}$$ **Comparison:** - Student's error: GM = 11.6 m (incorrect, massively inflated) - Correct answer: GM = 0.93 m (reasonable, still stable) - The student's answer suggests extraordinary stability; the correct answer is typical for a working barge **Lesson:** Always verify the axes. If your stability answer seems unreasonably large (GM > 2 m for small barges), check whether you used the right axis for I.

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