CALCULATED EXAMPLE 1
(refer to paragraph 7.2, Balance of forces and moments)
| Ship:
|
L = 120 m; B = 20 m; GM = 1.4 m; speed = 15 knots
|
| Cargo:
|
m = 62 t; dimensions = 6 × 4 × 4 m;
|
|
|
stowage at 0.7L on deck,
low
|
Securing material:
- wire rope (single use): breaking strength = 125 kN; MSL = 100 kN
- shackles, turnbuckles, deck rings: breaking strength = 180 kN;
MSL = 90 kN
- stowage on dunnage boards: μ = 0.3; CS = 90/1.5 = 60 kN
Securing arrangement:
-
| side
|
n
|
CS
|
α
|
f
|
c
|
| STBD
|
4
|
60 kN
|
40°
|
0.96
|
–
|
| PORT
|
2
|
60 kN
|
40°
|
0.96
|
–
|
| PORT
|
2
|
60 kN
|
10°
|
1.04
|
–
|
External forces:
- Fx = 2.9 × 0.89 × 62 + 16 + 8 = 184 kN
- Fy = 6.3 × 0.89 × 62 + 24 + 12 = 384 kN
- Fz = 6.2 × 0.89 × 62 = 342 kN
Balance of forces (STBD arrangement):
- 384 < 0.3 × 62 × 9.81 + 4 × 60 × 0.96
- 384 < 412 this is OK!
Balance of forces (PORT arrangement):
- 384 < 0.3 × 62 × 9.81 + 2 × 60 × 0.96 + 2 × 60 × 1.04
- 384 < 422 this is OK!
Balance of moments:
- 384 × 1.8 < 2 × 62 × 9.81
- 691 < 1216 no tipping, even without lashings!
Calculated example 2
(refer to section 7.3, Balance of forces – alternative method)
A cargo item of 68 t mass is stowed on timber (μ = 0.3) in the 'tween
deck at 0.7L of a vessel. L = 160 m, B = 24 m, v = 18 knots
and GM = 1.5 m.
Dimensions of the cargo item are height = 2.4 m and width = 1.8 m.
The external forces are: Fx = 112 kN,
Fy = 312 kN, Fz = 346 kN,
fz= 0.8 and fz·
Fz = 276.8 kN
The top view shows the overall securing arrangement with eight lashings.
Calculation of balance of forces:
| No.
|
MSL
(kN)
|
CS
(kN)
|
α
|
β
|
fy
|
CS ×
fy
|
fx
|
CS ×
fx
|
| 1
|
108
|
80
|
40° stbd
|
30° fwd
|
0.86
|
68.8 stbd
|
0.58
|
46.4 fwd
|
| 2
|
90
|
67
|
50° stbd
|
20° aft
|
0.83
|
55.6 stbd
|
0.45
|
30.2 aft
|
| 3
|
90
|
67
|
50° stbd
|
20° fwd
|
0.83
|
55.6 stbd
|
0.45
|
30.2 fwd
|
| 4
|
108
|
80
|
40° stbd
|
40° aft
|
0.78
|
62.4 stbd
|
0.69
|
55.2 aft
|
| 5
|
108
|
80
|
40° port
|
30° aft
|
0.86
|
68.8 port
|
0.58
|
46.4 aft
|
| 6
|
90
|
67
|
20° port
|
30° aft
|
0.92
|
61.6 port
|
0.57
|
38.2 aft
|
| 7
|
90
|
67
|
20° port
|
10° fwd
|
1.03
|
69.0 port
|
0.27
|
18.1 fwd
|
| 8
|
108
|
80
|
40° port
|
30° fwd
|
0.86
|
68.8 port
|
0.58
|
46.4 fwd
|
Transverse balance of forces (STBD arrangement) Nos. 1, 2, 3 and
4:
- 312 < 0.3 × 68 × 9.81 + 68.8 + 55.6 + 55.6 + 62.4
- 312 < 443 this is OK!
Transverse balance of forces (PORT arrangement) Nos. 5, 6, 7 and
8:
- 312 < 0.3 × 68 × 9.81 + 68.8 + 61.6 + 69.0 + 68.8
- 312 < 468 this is OK!
Longitudinal balance of forces (FWD arrangement) Nos. 1, 3, 7 and
8:
- 112 < 0.3 (68 × 9.81 – 276.8) + 46.4 + 30.2 + 18.1 + 46.4
- 112 < 258 this is OK!
Longitudinal balance of forces (AFT arrangement) Nos. 2, 4, 5 and 6:
- 112 < 0.3 (68 × 9.81 – 276.8) + 30.2 + 55.2 + 46.4 + 38.2
- 112 < 287 this is OK!
Transverse tipping
Unless specific information is provided, the vertical centre of gravity of
the cargo item can be assumed to be at one half the height and the transverse centre of
gravity at one half the width. Also, if the lashing is connected as shown in the sketch,
instead of measuring c, the length of the lever from the tipping axis to the
lashing CS, it is conservative to assume that it is equal to the width of the
cargo item.
- Fy · a ≤ b · m · g + 0.9 · (CS1
· c1 + CS2 · c2 + CS3 ·
c3 + CS4 · c4)
- 312 × 2.4/2 < 1.8/2 × 68 × 9.81 + 0.9 × 1.8 × (80 + 67 + 67 + 80)
- 374 < 600 + 476
- 374 < 1076 this is OK!