Heavy-Lift Crane Lift with a Slack Ballast Tank
A 12,000 t displacement heavy-lift ship is preparing to lift a 250 t package from the quay with its own centreline-mounted crane. The ship's solid KG is 6.40 m and KM is 7.80 m; one ballast tank is slack with a free-surface moment of 2,400 t·m. The crane head is 30 m above the keel, and the load will be slewed to 28 m from the centreline. The master's standing order limits list to 5° during the lift. The load is still on the quay, so it is not yet part of the ship's displacement. Counter-ballast tanks are available at a transverse arm of 14 m; their vertical centre of gravity when partially filled is about 1.0 m above the keel. The loading computer has not been updated for the free-surface effect, and the deck officer proposes to proceed because the intact GM in the departure condition is 1.40 m. The weather is calm and the lift will last about 20 minutes.
1. Calculate the effective GM before the lift and immediately after the load is taken on the crane, stating each correction. Is the deck officer's 1.40 m GM argument valid?
Read the worked answer
Solid GM before the lift is KM − KG = 7.80 − 6.40 = 1.40 m. The slack tank raises the effective KG by FSC = free-surface moment / displacement = 2,400 / 12,000 = 0.20 m, so the effective GM is 1.40 − 0.20 = 1.20 m before the lift.
Once the load leaves the quay it acts at the crane head, not at its stowed position. The virtual rise of G is approximately w(z_crane − KG_eff)/Δ_new. With Δ_new = 12,250 t and KG_eff = 6.60 m, rise = 250 × (30 − 6.60) / 12,250 = 0.478 m. Thus GM immediately after lift = 1.20 − 0.478 = 0.722 m, or directly 7.80 − 6.60 − 0.478 = 0.722 m.
The deck officer's 1.40 m figure is the solid GM only. It ignores both the free-surface penalty and the virtual rise caused by taking the load at the crane head. The valid operational GM is about 0.72 m, less than half the quoted value.
2. Compute the steady list at 28 m outreach and compare it with the 5° limit. Comment on whether the lift can proceed.
Read the worked answer
The heeling moment is the lifted weight times its transverse distance from the centreline: 250 × 28 = 7,000 t·m. The righting moment at small angle is Δ GM tanθ, with Δ = 12,250 t and GM = 0.722 m.
tanθ = 7,000 / (12,250 × 0.722) = 7,000 / 8,849 = 0.791, giving θ ≈ 38.3°.
This is far beyond the 5° limit and is also outside the small-angle range, so the linear estimate understates the danger: at that heel the deck edge and possibly the load may be immersed, and the GZ curve may have much less reserve than the initial GM suggests. The lift cannot proceed at 28 m outreach with only 250 t as counterweight. The operation must be stopped, the slack tank corrected, and either the outreach/load reduced or counter-ballast added before slewing outboard.
3. Size counter-ballast at 14 m arm to bring the list within 5°, check the new GM, and state the operational precautions.
Read the worked answer
The allowable net heeling moment at 5° is Δ GM tan5° = 12,250 × 0.722 × 0.0875 ≈ 774 t·m. The counter-ballast moment must therefore be about 7,000 − 774 = 6,226 t·m. At a 14 m arm, the required mass is 6,226 / 14 ≈ 445 t.
Recheck with the ballast added: Δ = 12,000 + 250 + 445 = 12,695 t. New KG = (12,000×6.40 + 250×30 + 445×1.0) / 12,695 = 6.675 m. New FSC = 2,400 / 12,695 = 0.189 m, so GM = 7.80 − 6.675 − 0.189 = 0.936 m. Net heeling moment = 7,000 − 445×14 = 770 t·m, giving tanθ = 770 / (12,695×0.936) = 0.0648, θ ≈ 3.7°, which is within 5°.
Precautions: fill and sound the counter-ballast before breaking the load off the quay; minimise free surface in all ballast tanks; confirm the ballast system can transfer the required quantity without heel excursions; monitor list and trim continuously; and verify the crane's structural and hydraulic limits at the actual outreach. Do not rely on the departure-condition GM.