# MOMENT OF STATIC STABILITY

Moment of static stability may be defined as the moment to
return the ship to the initial position when inclined by an external force
(such as wind and wave action).

we can see a ship which has been
inclined by an external force. The centre of Buoyancy has moved from B to B

_{1}parallel to gg_{1}, and the force of buoyancy (w) acts vertically upward through B_{1}.
The
weight of the ship (W) acts vertically downwards through the centre of gravity
(G). The perpendicular distance between the lines of action of forces (GZ) is
called the

*righting lever*.
Finally
we can find the Moment of static stability by multiplying the

*righting lever*and*Displacement*.
Moment of Static Stability = W x GZ (eq.01)

## The Moment of Static Stability at a Small Angle of Heel

Remember:

Moment of Static Stability = W x GZ

Triangle GZM: GZ = GMsinq

^{o}
So the Moment of Static
Stability:

= W x GM x sinq

^{o}(eq.02)**Example 01.**

A
ship of 4000 tonnes displacement has KG 5.5 m and KM 6.0 m. Calculate the
moment of statical stability when heeled 5 degrees.

Answer:

GM = KM – KG = 6.0-5.5 = 0.5 m

Moment of statical stability = W x GM x sinq

= 4000 x 0.5 x sin 5

^{o}
= 174.4 tonnes m

**Example 02.**

When a ship of 12000 tonnes
displacement is heeled 6 ½ degrees the moment of statical stablity is 600
tonnes m. Calculate the initial metacentric height.

Answer:

Moment of statical stability = W x GM x sinq

GM = (Moment of statical stability)/(W x sinq)

= 600/(12000 x sin 6
½

^{o})
= 0.44 m

## The Moment of Static Stability at a Large Angle of Heel

At large angle of heel (More than 15 degrees), the
force of buoyancy can no longer be
considered to act vertically upwards through the initial metacentre (M). As shown in

*figure 03*, the centre of buoyancy has moved further out to the low side, and the vertical through B_{1}no longer passes through M. The righting lever (GZ) is once again the perpendicular distance between the vertical through G and the vertical through B_{1}, and the moment of statical stability is equal to W x GZ. But GZ is no longer equal to GMsinq. Up to the angel at which the deck edge is immersed, it may be found by using a formula known as the*Wall-sided formula*.
GZ = (GM + ½ BM tan

^{2}q)sinq (eq.03)**Example 03.**

A
ship of 6000 tonnes displacement has KB 3 m, KM 6 m, and KG 5.5 m. Find the
moment of statical stability at 25 degrees heel.

Answer:

GZ = (GM + ½ BM tan

^{2}q)sinq
= (0.5 + ½ x 3 x tan

^{2}25^{o})sin25^{o}
= 0.8262 sin25

^{o}
= 0.35 m

Moment
of statical stability = W x GZ

= 6000 x 0.35

= 2100 tonnes m

**Example 04.**

A
box-shaped vessel 65 m x 12 m x 8 m has KG 4 m, and is floating in salt water
upright on even keel at 4 m draft F and A. Calculate the moment of statical
stability at (a) 5 degrees and (b) 25 degrees heel.

Answer:

W = L x B x draft x 1.025 KB
= ½ draft

= 65 x
12 x 4 x 1.025 KB = 2 m

= 3198 tonnes

BM = B

^{2}/12d
= (12 x 12)/12 x 4

= 3 m

KB = 2 m

BM = 3 m

KM = KB + BM = 2 + 3 = 5 m

KG = 4 m

GM = KM – KG

= 5 – 4

= 1 m

at 5

^{o}heel GZ = GMsinq
= 1 x sin 5

^{o}
= 0.0872 m

Moment of statical stability

= W x GZ

= 3198 x
0.0872

= 278.9 tonnes
m

at 25

^{o}heel GZ = (GM + ½ BM tan^{2}q)sinq
= (1 + ½ x 3 x tan

^{2}25^{o})sin25^{o}
= (1 +
0.3262)sin25

^{o}
= 1.3262 sin25

^{o}
= 0.56 m

Moment of
statical stability

= W x GZ

= 3198 x 0.56

= 1790.9 tonnes m

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