Q1
40 m barge, four holds, heavy amidships
A box shaped vessel 40 m long, light displacement 320 t is divided into 4 holds of equal lengths. Holds 1 & 4 are loaded with 200 t cargo each, trimmed level. Holds 2 & 3 are loaded with 400 t cargo each, trimmed level. Draw SF & BM diagrams of the vessel.
Condition
Sagging
BM positive
Max |SF|
100.00 t
10.00 m from AP · 30.00 m from FP
Max |BM|
1000.0 t·m
amidships (20.00 m from AP)
Weight Buoyancy Excess buoyancy Excess weightx from AP →
Weight & buoyancyt/m
Load (B − W)t/m
Shear forcet
Bending momentt·m
x
20.00 m AP
W
48.00 t/m
B
38.00 t/m
SF
0.00 t
BM
+1000.00 t·m
Drag across a diagram to read values. Showing amidships until you hover.
Worked solution
Particulars2 lines
- Length L = 40.0 m
- Water: salt water (ρ = 1.025 t/m³)
Displacement & lightship4 lines
- Displacement Δ = light + cargo = 320.000 + 1200.000 = 1520.000 t
- Cargo total = 1200.000 t
- Light displacement = 320.000 t
- Light weight per metre (uniform) = 320.000 / 40.0 = 8.0000 t/m
Weight curve (holds)6 lines
- Hold No. 1 is forward, last hold is aft. x is measured from AP (aft).
- Each hold length = 10.000 m.
- No. 4 (0.00–10.00 m from AP): cargo 200.0 t → 20.000 t/m. Total W = 28.000 t/m.
- No. 3 (10.00–20.00 m from AP): cargo 400.0 t → 40.000 t/m. Total W = 48.000 t/m.
- No. 2 (20.00–30.00 m from AP): cargo 400.0 t → 40.000 t/m. Total W = 48.000 t/m.
- No. 1 (30.00–40.00 m from AP): cargo 200.0 t → 20.000 t/m. Total W = 28.000 t/m.
Buoyancy curve2 lines
- Even keel, box section → buoyancy is uniform along the length.
- b = Δ / L = 1520.000 / 40.0 = 38.0000 t/m.
Load curve3 lines
- Load (t/m) = buoyancy/m − weight/m. Positive = excess buoyancy (up); negative = excess weight (down).
- SF is the running integral of load from AP. BM is the running integral of SF from AP.
- Free-free beam: SF = 0 and BM = 0 at both ends when Δ = weight and LCB = LCG.
Stations (from AP)5 lines
- x = 0.000 m [AP]: load +10.000 t/m, SF 0.00 t, BM 0.00 t·m
- x = 10.000 m: load +10.000 t/m, SF +100.00 t, BM +500.00 t·m
- x = 20.000 m [amidships]: load -10.000 t/m, SF 0.00 t, BM +1000.00 t·m
- x = 30.000 m: load -10.000 t/m, SF -100.00 t, BM +500.00 t·m
- x = 40.000 m [FP]: load +10.000 t/m, SF 0.00 t, BM 0.00 t·m
Maxima & condition4 lines
- Maximum |SF| = 100.000 t at 10.00 m from AP · 30.00 m from FP.
- Maximum |BM| = 1000.00 t·m at amidships (20.00 m from AP).
- BM is positive (sagging): excess weight amidships, excess buoyancy at the ends. The hull droops in the middle.
- In SI: 1 t ≈ 9.81 kN, so multiply SF by 9.81 for kN and BM by 9.81 for kN·m.
Station table
| x (m AP) | W (t/m) | B | Load | SF | BM | Note |
|---|---|---|---|---|---|---|
| 0.000 | 28.000 | 38.000 | +10.000 | 0.00 | 0.00 | AP |
| 10.000 | 28.000 | 38.000 | +10.000 | +100.00 | +500.00 | |
| 20.000 | 48.000 | 38.000 | -10.000 | 0.00 | +1000.00 | amidships |
| 30.000 | 48.000 | 38.000 | -10.000 | -100.00 | +500.00 | |
| 40.000 | 28.000 | 38.000 | +10.000 | 0.00 | 0.00 | FP |
Sign convention follows Capt. H. Subramaniam / MMD: load = buoyancy − weight, integrate from the aft perpendicular. Positive BM is sagging. Lightship is taken as uniformly distributed. Hold 1 is forward.