Mechanical Engineering Principles

HYDROSTATICS 243

From equation (21.1),

GM=

Px

W

cotθ

=

490 .5N×10 m× 57. 29

981 × 103 N

i.e. metacentric height,GM= 0 .286 m

NowKM=KB+BM

= 1 .2m+ 2 .4m= 3 .6m

KG=KM−GM

= 3. 6 − 0. 286 = 3 .314 m

i.e. centre of gravity above the keel,KG= 3 .314 m,
(where ‘K’ is a point on the keel).

Problem 14. A barge of length 30 m and

width8mfloatsonanevenkeelatadepth

of 3 m. What is the value of its buoyancy?

Take density of water,ρ, as 1000 kg/m^3 and

gas 9.81 m/s^2.

The displaced volume of the barge,

V=30 m×8m×3m=720 m^3.

From Section 21.4,

buoyancy=Vρg

=720 m^3 × 1000

kg

m^3

× 9. 81

m

s^2

= 7 .063 MN

Problem 15. If the vertical centre of gravity

of the barge in Problem 14 is 2 m above the

keel, (i.e.KG=2 m), what is the

metacentric height of the barge?

NowKB=the distance of the centre of buoyancy

of the barge from the keel=

3m

2

i.e.KB= 1 .5m.

From page 242,BM=

I

V

and for a rectangle,

I=

Lb^3

12

from Table 7.1, page 91, where

L=length of the waterplane=30 m, and

b=width of the waterplane=8m.

Hence, moment of inertia,

I=

30 × 83

12

=1280 m^4

From Problem 14, volume,

V=720 m^3 ,

hence,

BM=

I

V

=

1280

720

= 1 .778 m

Now,

KM=KB+BM= 1 .5m+ 1 .778 m= 3 .278 m

i.e. the centre of gravity above the keel,

KM= 3 .278 m.

SinceKG=2 m (given), then

GM=KM−KG= 3. 278 − 2 = 1 .278 m,

i.e. the metacentric height of the barge,

GM= 1 .278 m.

Now try the following exercise

Exercise 111 Further problems on hydro-

statics

In the following problems, where necessary,

take g = 9 .81 m/s^2 and density of water

ρ=1020 kg/m^3.

1. A ship is of mass 10000 kg. If the ship
   floats in the water, what is the value of
   its buoyancy? [98.1 kN]


2. A submarine may be assumed to be in
   the form of a circular cylinder of 10 m
   external diameter and of length 100 m.
   If the submarine floats just below the