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Metacenter and Metacentric Height

A floating body is stable for small angular displacement when the metacenter $M$ is above the center of gravity $G$. The center of buoyancy $B$ shifts as the displaced volume changes shape.

$$BM = \frac{I}{V_d}$$$$GM = BM - BG$$

$I$ is the second moment of area of the waterplane about the axis of rotation, and $V_d$ is the displaced volume.

Stability Conditions and Righting Moment

$$GM > 0 \Rightarrow \text{stable}$$$$GM = 0 \Rightarrow \text{neutral}$$$$GM < 0 \Rightarrow \text{unstable}$$$$M_R = W(GM)\sin\theta$$

For board problems, check that all distances are measured on the same vertical reference and use the displaced volume at floating equilibrium.

Rectangular Barge Metacentric Height

A rectangular barge is 6 m wide, 12 m long, and has a draft of 1.5 m in fresh water. Its center of gravity is 1.8 m above the keel. Determine $GM$ for rolling about the longitudinal axis.

$$V_d = 6(12)(1.5)=108\text{ m}^3$$$$I = \frac{L b^3}{12}=\frac{12(6)^3}{12}=216\text{ m}^4$$$$BM=\frac{216}{108}=2.0\text{ m}$$$$B=0.75\text{ m above keel}$$$$BG=1.8-0.75=1.05\text{ m}$$$$GM=2.0-1.05=0.95\text{ m}$$

Answer: $GM=0.95\text{ m}$, so the barge is stable for small rolling displacement.

Common Stability Mistakes

Do not use the body area for $I$ unless it is the waterplane area. For a rectangular waterplane, use the dimension cubed in the direction perpendicular to the axis of rotation.

Floating Cylinder Stability Check

A closed vertical cylindrical buoy is 1.20 m in diameter and floats with draft 0.90 m in fresh water. Its center of gravity is 0.55 m above the bottom. Determine whether it is stable for small angular displacement.

$$V_d=\frac{\pi D^2}{4}d=\frac{\pi(1.20)^2}{4}(0.90)=1.018\text{ m}^3$$$$I=\frac{\pi D^4}{64}=\frac{\pi(1.20)^4}{64}=0.1018\text{ m}^4$$$$BM=\frac{I}{V_d}=\frac{0.1018}{1.018}=0.100\text{ m}$$$$B=0.45\text{ m above bottom},\quad BG=0.55-0.45=0.10\text{ m}$$$$GM=BM-BG=0.100-0.100\approx0$$

Answer: The buoy is approximately neutrally stable. Any small lowering of $G$ would make it stable.

Righting Moment from Metacentric Height

A floating body weighs 80 kN and has metacentric height $GM=0.35\text{ m}$. Find the righting moment when it is heeled by 8 degrees.

$$M_R=W(GM)\sin\theta=80(0.35)\sin 8^\circ=3.90\text{ kN}\cdot\text{m}$$

Answer: The righting moment is 3.90 kN¡m.

Problem: Barge Stability After Shifting Cargo

A rectangular barge is 5.0 m wide, 15.0 m long, and has a draft of 1.2 m in fresh water. Its center of gravity is 1.0 m above the keel. A 20 kN cargo is shifted 2.0 m laterally across the deck. Determine: (a) the metacentric height $GM$ for rolling, (b) the angle of heel caused by the cargo shift, and (c) whether the barge remains stable.

$$V_d = 5.0(15.0)(1.2) = 90 \text{ m}^3$$ $$I = \frac{L b^3}{12} = \frac{15.0(5.0)^3}{12} = 156.25 \text{ m}^4$$ $$BM = \frac{I}{V_d} = \frac{156.25}{90} = 1.736 \text{ m}$$ $$KB = \frac{d}{2} = 0.60 \text{ m}, \quad BG = 1.0 - 0.60 = 0.40 \text{ m}$$ $$GM = BM - BG = 1.736 - 0.40 = 1.336 \text{ m (stable)}$$

Angle of heel from cargo shift: let total barge weight = $W = \gamma_w V_d = 9.81(90) = 882.9$ kN.

$$\tan\theta = \frac{W_{cargo} \cdot d}{W \cdot GM} = \frac{20(2.0)}{882.9(1.336)} = \frac{40}{1179.4} = 0.03392$$ $$\theta = 1.94°$$

Answer: $GM = 1.336$ m — the barge is stable. Shifting 20 kN cargo 2.0 m laterally causes a heel of only 1.94°, confirming the barge has ample stability.

Problem: Stability of a Floating Solid Cylinder

A solid homogeneous cylinder has diameter D and length L. It floats upright (axis vertical) in fresh water. The cylinder's specific gravity is 0.65. For what ratio L/D does the cylinder become unstable in the upright position? Determine whether a cylinder with L = 1.0 m and D = 0.80 m and SG = 0.65 is stable.

Draft: $d = SG \cdot L = 0.65L$. KB = d/2 = 0.325L. BM for circular section rotating about diameter:

$$I = \frac{\pi D^4}{64}, \quad V_d = \frac{\pi D^2}{4}(0.65L)$$ $$BM = \frac{I}{V_d} = \frac{\pi D^4/64}{\pi D^2(0.65L)/4} = \frac{D^2}{16(0.65L)} = \frac{D^2}{10.4L}$$ $$BG = KG - KB = \frac{L}{2} - 0.325L = 0.175L$$ $$GM = BM - BG = \frac{D^2}{10.4L} - 0.175L$$

For neutral stability: $GM = 0 \Rightarrow D^2 = 1.82L^2 \Rightarrow L/D < 1/\sqrt{1.82} = 0.741$ for stability.

Check for L = 1.0 m, D = 0.80 m (L/D = 1.25):

$$BM = \frac{(0.80)^2}{10.4(1.0)} = 0.0615 \text{ m}, \quad BG = 0.175(1.0) = 0.175 \text{ m}$$ $$GM = 0.0615 - 0.175 = -0.114 \text{ m (unstable)}$$

Answer: A floating cylinder is stable upright only when $L/D < 0.741$. For the given cylinder (L/D = 1.25), GM = −0.114 m — it will tip over to float on its side.

Problem: Ship with Added Deck Weight — Effect on GM

A ship originally has a displacement of 8000 kN, metacentric height $GM_0 = 0.80$ m, and center of gravity 4.0 m above the keel. A piece of equipment weighing 120 kN is added to the deck at a height of 9.5 m above the keel. The waterplane area moment of inertia and displaced volume do not change significantly. Determine the new metacentric height after adding the equipment and whether the ship remains stable.

New center of gravity height above keel:

$$W_{new} = 8000 + 120 = 8120 \text{ kN}$$ $$KG_{new} = \frac{W_{old} \cdot KG_{old} + W_{add} \cdot h_{add}}{W_{new}}$$ $$KG_{new} = \frac{8000(4.0) + 120(9.5)}{8120} = \frac{32000 + 1140}{8120} = \frac{33140}{8120} = 4.081 \text{ m}$$

KM (= KB + BM) was unchanged: $KM = KG_{old} + GM_0 = 4.0 + 0.80 = 4.80$ m

$$GM_{new} = KM - KG_{new} = 4.80 - 4.081 = 0.719 \text{ m}$$

Answer: GM decreases from 0.80 m to 0.719 m after adding the deck equipment. The ship remains stable (GM > 0), but stability is reduced. Adding top-heavy loads always reduces GM — ships can become unstable if deck loads are excessive.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: v230

HGE - Hydraulics / Buoyancy / HGE Refresher Series

For stable floating equilibrium, where is the metacenter relative to the center of gravity?

  1. Above it
  2. At it
  3. Below it
  4. At half the draft
Positive metacentric height requires M above G.

Question Bank: v231

HGE - Hydraulics / Buoyancy / HGE Refresher Series

If a floating body's metacenter lies below its center of gravity, what equilibrium exists?

  1. Unstable
  2. Stable
  3. Neutral
  4. Impossible
A negative metacentric height produces an overturning tendency.

Question Bank: v232

HGE - Hydraulics / Buoyancy / HGE Refresher Series

Metacentric height is the distance between which points?

  1. Center of gravity and center of buoyancy
  2. Center of gravity and metacenter
  3. Metacenter and center of buoyancy
  4. Original and shifted centers of buoyancy
GM is measured from the center of gravity to the metacenter.

Question Bank: v234

HGE - Hydraulics / Buoyancy / HGE Refresher Series

When a ship's metacenter coincides with its center of gravity, what is its equilibrium?

  1. General equilibrium
  2. Stable
  3. Unstable
  4. Neutral
GM = 0 corresponds to neutral equilibrium.

Question Bank: v276

HGE - Hydraulics / Buoyancy / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A rectangular barge is 9.5 m wide and 25 m long, carries mass 4200000 kg, and has center of gravity 4.3 m above its bottom.

Determine the draft.

  1. 17.684 m
  2. 13.811 m
  3. 20.107 m
  4. 22.300 m

Determine metacentric height for rolling.

  1. 4.967 m
  2. 3.880 m
  3. 5.648 m
  4. 6.264 m

Determine metacentric height for pitching.

  1. 7.487 m
  2. 5.848 m
  3. 8.513 m
  4. 9.441 m
Find displacement volume, then use BM = I/V and GM = KB + BM - KG.
Computed answers: 17.6842105263, 4.96739048141, 7.48729375522.