Question

A ship 100 m long floats at a draft of 6m and in this condition the immersed cross sectional areas are as given in tables below. The equivalent base area (Ab) is required because of the fineness of the bottom shell

Section: AP 1 2 3 4 5 6 FP

Immeresed cross section area (m2): 12 30 65 80 70 50 0

Draft(m): 0 0.6 1.2 2.4 3.6 4.8 6.0

Waterplane area (m2): Ab 560 720 880 940 1000 1030

Calculate each of the following:

(a) The equivalent base are value Ab

(b) The longitudinal position of the centre of buoyancy from midship

(c) The vertical position of the centre of buoyancy above the base

Answer 1

Table for question:

Section	AP	1	2	3	4	5	FP
Immersed cross sectional area	12	30	65	80	70	50	0

Draught (m)	0	0.6	1.2	2.4	3.6	4.8	6.0
Water plane area (m2)	Ab	560	720	880	940	1000	1030

Solution:

Section	CSA	SM	F volume	Lever	F moment
AP	12	1	12	-3h	-36h
1	30	4	120	-2h	-240h
2	65	2	130	-1h	-130h
3	80	4	320	0	0
4	70	2	140	+1h	140h
5	50	4	200	+2h	400h
FP	0	1	0	+3h	0
			ΣFvol. = 992		ΣFmom. =134h

$$Common\:interval\:\left(h\right)=\frac{L}{6}=\frac{100}{6}$$

$$h=16.66m$$

$$\sum F_{movement}=134\:\times16.66m^4$$

$$\sum F_{moment}=2232.44m^4$$

$$\nabla=\frac{h}{3}\times\sum F_{volume}$$

$$\nabla=\frac{16.66}{3}\times992$$

$$\nabla=5120.17m^3$$

$$Longitudinal\:position\:of\:centre\:of\:buoyancy\:LCB\:=\:\frac{\sum F_{mom}}{\sum F_{vol}}$$

$$=\frac{2232.44}{993}$$

$$LCB=2.42m\:fwd$$

Draught	Aw	SM	F volume	Lever	F moment
0	Ab	1/2	1/2 Ab	0	0
0.6	560	4/2	1120	0.6	672
1.2	720	1 2/4	1080	1.2	1296
2.4	880	4	3520	2.4	8448
3.6	940	2	1880	3.6	6768
4.8	1000	4	4000	4.8	19200
6.0	1030	1	1030	6.0	6180
			ΣFvol = 1/2Ab+12630		ΣFmom = 42564

$$h=1.2$$

$$\nabla=\frac{h}{3}\times\sum F_{vol}$$

$$5120.17=\frac{1.2}{3}\times\left(\frac12A_{b}+12630\right)$$

$$\frac{A_{b}}{2}=\frac{5120.17\times3}{1.2}-12630$$

$$A_{b}=340.84m^2$$

$$\sum F_{vol}=\left\lbrack\frac12\times340.85\right\rbrack+12630$$

$$\sum F_{vol}=12800.425m^3$$

$$Vertical\:position\:of\:centre\:of\:buoyancy=\frac{\sum F_{mom}}{\sum F_{vol}}$$

$$=\frac{42564}{12800.425}$$

$$KB=3.325m$$

✓ Verified Model Answer (Text Solution)

Table for question: