# Chapter 8 - Stability Calculation

**1. A box shaped vessel of displacement 1640t is 50m long 10m wide and 8m high . Find her KB in SW, if she is one an even keel and upright.**

Solution:

Displacement = 1640

**Volume of box shape vessel = (L x B x H)**
= (50m x10m x 8m)

RD = 1.025

**Displacement = (u/w volume) x(density of water displaced)**
1640= ( L x B x d ) x (1.025)
1640 =(50 x 10 x d) x (1.025 )

Hence, d = 1640/ (50 x 10x 1.025) = 3.2m

KB can be calculated as = (d/2) = (3.2/2) = 1.6m

**2. A box shaped vessel 60m x 10m x10m floats in DW of RD 1.020 at an even keel draft of 6m . Find her KB in DW of RD 1.004**.

Solution :

**Volume of box shape vessel = ( L X B XH)**
=(60m x 10m x 10m)

RD = 1.020, Draft = 6m

**Displacement at RD of 1.020 = (u/w volume)x(density)**
=( L x B x d ) x 1.020
= (60 x 10 x 6) x 1.020
= 3672t

Now we have to calculate draft of vessel at RD of 1.004, keeping the displacement same = (L x B x d ) x (1.004) 3672 = (60 x 10 x d )x (1.004) d = 3672 /(60 x 10 x 1.004) = 6.095m

Hence, KB = (d/2 = (6.095 /2) = 3.047m.

**3. A triangular shaped vessel of displacement 650t floats in DW RD 1.015. Her water plane is rectangle 30m x 8m . Find her KB **.

Solution:

Displacement = 650t, RD = 1.95

**Water plane Area =(L x B)**
= (30m x 8m)
** Volume of triangular vessel = ½ ( L x B x H)**

**Displacement = (u/w volume)x (Density)**
650 = ½ (L x B x d ) x( density)
=(½ x 30 x 8 x d) x(1.015)
d = (650 x 2)/ (30 x 8 x 1.015)
= 5.336m

centeriod of triangle is 2/3rd of the perpendicular bisector of the base .Hence centre of buoyancy will be always 2/3rd of perpendicular bisector of base.

KB =( 2/3 x height of perpendicular bisector of base) = (2/3 x 5.336) = 3.557m.

**4. A triangular shaped vessel floats in SW. Her water plane is a rectangle 40m x 12m. If her KB is 3.6m, find her displacement** .

Solution :

**Water plane area = (L x B)**
= 40m x 12m

Initial KB =3.6m,

centeriod of triangle is 2/3rd of the perpendicular bisector of the base .hence centre of buoyancy will be always 2/3rd of perpendicular bisector of base.

So, KB = 2/3 x (Draft) 3.6 = 2/3 x d

So, Draft(d) =(3.6 x 3 / 2)
= 5.4m
** Volume of triangular vessel = ½ ( L x B x H)**

**Displacement =(u/w volume) x (density)**
= ½ (L x B x d) x (1.025)
= 1/2 (40 x 12 x 5.4) x (1.025
= 1328.4t .

**5. A homogeneous log of wood 3m x 0.75m x 0.75m floats in SW with one face horizontal .If the RD of the log is 0.8m, calculate the vertical distance between its COG and its COB .**

Solution:

**Volume of log of wood = (L x B x H)**
= (3m x 0.75m x 0.75m)

RD of SW = 1.025 RD of the log = 0.8m

**Mass of the log = (volume x density)**
= (3 x 0.75 x 0.75 ) x (0.8)
= 1.35 t

Since, log floats freely so
** Mass = Displacement**
1.35 = (u/w volume) x(density)
1.35 = (L x B x draft) x (1.025)
1.35 = ( 3x 0.75 x draft) x (1.025)

So, draft = 1.35/(3 x 0.75 x 1.025) Draft = 0.585 m COB = (1/2 x draft) = (0.585/2) = 0.2925m

COG = (1/2 x height) = (0.75 / 2) = 0.375m

Hence, Vertical distance between COB and COG = (0.375 – 0.295) = 0.083m

**6. A homogeneous log of wood of 0.5m square section floats in water of RD 1.005 at a draft 0.4m with one of its faces horizontal . Find the vertical distance between its COG and its COB in water of RD 1.020.**

Solution:

RD = 1.005 & Draft = 0.4m
**Displacement (W) = (volume ) x (density)**
= (0.5 x 0.5 x 0.4) x(1.005)
= 0.1005t

Now ,calculating draft for the same displacement at RD of 1.020.
** W = ( Area x draft ) x( density)**
0.1005 =(0.5 x 0.5 x draft) x( 1.020)

Draft = 0.1005/(0.5 x 0.5 x 1.020) = 0.394m COB = (draft /2) = (0.395/2) = 0.197m

COG = (Height/2) = (0.5/2) = 0.25m

Now, Vertical distance between COG and COB can be easily calculated as = (0.25 – 0.197) = 0.053m

**8. A cylindrical drum of 0.8m diameter and 1.5m height weight 10kg. 490 kg of steel is put in it such that it floats with its axis vertical in FW. find its KB . (Assume π to be 22/7 ).**

Solution :

Diameter = 0.8m , So, radius = (0.8 /2) = 0.4m

Height = 1.5m Weight = 10kg = (0.01) t Mass of steel = 490kg = 0.49t Total displacement = (0.01+ 0.49) = 0.50t

**Displacement (W) = (u/w volume) x (density)**
0.5 = r2 d) x (RD)
0.5 = (3.1416 x 0.4 x 0.4 x d) x (1)
So, d = 0.5/(3.1416 x 0.4 x 0.4) x(1)
= 0.995m

Again, KB can be calculated as = (draft/2) = 0.4975m

**8. A barge prism shaped such that its deck and keel are identical and parallel ; its side vertical .its deck consists of three shapes triangular bow of 12m each side ; rectangular mid-part 80m long and 12m wide ; semi-circular stern of radius 6m . If the light displacement of the barge is 500t and it has 5000t of cargo in it, find its KB when floating on an even keel in SW. (Assume π to be 3.142 )**

Solution:

According to the question the drawn figure can consist of 3 difference structure ; namely triangle BEC , rectangle ABCD, semicircle AFD .

Total displacement = (5000 + 500)
= 5500t
** Total displacement = (Total u/w volume) x (density)**

**Total u/w volume = (volume of triangle BEC) + (Volume of rectangle ABCD) + (Volume of semicircle AFD)**

**Area of triangle BCE = (π r2 /2)**
= (3.14 x 6 x 6 )/ 2
= 56. 556m2

**Area of rectangle ABCD = (L X B )**
= (80 x 12)
= 960m2

**Area of semicircle AFD = (1/2 x b x h)**
= (1/2 x 12 x 10.39)
= 62.34m2

In triangle BEC, EG is the height

So , EG2 = BE2 – BG2 EG2 = 122 -62

= 144 – 36 =108 EG = 10.39m

Hence, Total area = (56.55 + 62.34 + 960.0) = 1078.89m2

**Displacement (W)= (u/w volume) x (density)**
5500 =** (Area x depth) x (density)**
5500 = (1078.89 x d ) x (1.025)

Draft = 5500 /(1078.89 x 1.025) = 4.97m

Now, KB = (draft)/2 = (4.97/2) = 2.485m.

**9. The deck and keel of a flat-bottomed barge and identical. Its sides are vertical. The deck consists of two section – the bow is a triangle 12m broad and measures 12m in the fore and aft direction ; the mid-body is a rectangle 50m long and 12m broad . if it is floating on an even keel in SW with a displacement of 3444t , find the position of its COB with reference to its after end .**

Solution;

Total displacement = 3444t

**Displacement (W) = (u/w volume ) x (density)**
3444 = (u/w volume of triangle ) + (u/w volume of rectangle ) x (1.025)
3444 = ( 1/2 x 12 x 12 x d) + ( 50 x 12 x d) ) x 1.025
3444 = (72 d + 600 d) x 1.025
3444 = (672d x 1.025)
3444 = 688.8 d

So, d = 3444/ 688.8 d = 5 m

now, COB =(draft/2) =(5/2) m = 2.5 m with reference to keel

Now for calculating COB with reference to after end
** Total moment of barge = (moment of triangle) + (moment of rectangle)**
(3444 x LCB of barge) = ( u/w volume )x (density) x(LCB) + (u/w volume ) x (density ) x (LCB)
(3444 x LCB) = (1/2 x 12 x 12 x 5) x (1.025) x( 54) + (50 x 12 x 5 ) x (1.005 ) x (25)
(3444 x LCB) = (19926) + (76875)
(3444 x LCB) = 96801
LCB = (96801 / 3444)
= 28.10m.

**10. A barge of 45m long has a uniform transverse cross-section throughout, which consists of a rectangle above a triangle. The rectangle is 8m broad and 4m high. The triangle is apex downwards 8m broad and 3m deep.if she displacement of the barge is 1620t, find the position of its COB with reference to the keel and also with reference to the after and , if it is upright and on an even keel in FW.**

Solution:

Total displacement = 1620t

Let ‘X’ t of weight which is not displaced by triangle

So weight displaced is (1620 – X )t

**Area of triangle = ½ ( L x B x H)**
= (1/2 x 45 x 8 x 3)
= 540m3

**Displacement (W) = (u/w volume) x (density)**
= 540 x 1
= 540t = X( as per assumption made above in solution )

Displacement of rectangle = (1620 – X) = (1620 – 540) = 1080t

**displacement for rectangle = (u/w volume) x (density)**
1080 = (45 x 8 x draft) x (1)
Hence, draft =1080/(45 x 8)
draft = 3 m

for calculating COB = (draft/2) = (3/2) = 1.5m

So, COB with respect to keel = (3.0 +1.5)m = 4.5m .

Now, VM ( vertical moment) created by triangle:
=(**W x d**)
= (540 x 2)
= 1080 tm.

Now , VM ( vertical moment) created by rectangle:
= (**W x d**)
= (1080 x 4.5)
= 4860 tm.

**final vertical moment = ( VM) triangle + (VM) rectangle**
Final Weight = 1620t
** Final KB = (Final VM / Final W)**
= (5940/1620)
= 3.66 m