Stability - I: Chapter 9 | Exercise 9: Transverse statical stability – list , heel; their relation to COG and COB; transverse shift of B; metacentric height ; righting moment for both, small and large angle of heel.

1 . A ship of 10000t displacement has a GM of 0.4m. Calculate the moment of statical stability when she is heeled by 5 degree.

Solution:

GZ =Righting lever, GM = Metacentric height

We know that Righting moment = (W x GZ )
Where, W = displacement for all of keel, M = Transverse meta centre.
Displacement (W) = 10000t,
 GM = 0.4m, heel = 5degree

We know that
RM ( righting moment or moment of statical stability)
= (W x GM x sinθ)
= (10000 x 0.4 x sin 5⁰ )
= 348.62tm

2. A ship of 12000t displacement is heeled by 6 degree. If her righting lever is then 0.1m, find the moment of statical stability . If her KM is 8.2m , find her KG .

Solution :

Displacement (W) = 12000t,
 Heel  = 6degree
 GZ = 0.1m  &  KM = 8.2m

We know that :
GZ = GM sinθ,
0.1 = GM sin 6⁰.
GM = (0.1 / sin 6⁰)
= 0.956m
KG = (KM – GM)
= (8.2 – 0.956)
= 7.44m

RM (righting moment or moment of statical stability)
= (W GM sinθ)
= (12000 x 0.956 x sin6⁰ )
= 1199.15 tm.

3. When a ship of 14000t displacement is heeled by 8 deegre , her moment of statical stability is 400tm . If KG 7.3m, find KM.

Solution :

Displacement (W)= 14000t,
Heel = 8degree ,
  RM = 400tm &  KG 7.3m

We know that :
RM (Righting moment or moment of statical stability)
          = (W x GM x sinθ)
 400 = (14000  x GM x sinθ)
GM = 400/ (1400 x sin8^{0  )}
GM = 2.053m

Now, KM = (KG + GM )
= (7.3 + 2.053 )
= 9.353m.

4. A ship of 8000t displacement has KB 3.5m, KM 6.5m, and KG 6m . Find her moment of statical stability at 20 degree heel, assuming that her deck edge remains above water (i.e. she is steel wall side at that angle of heel.

Solution :

Displacement (W) = 8000t,
KB = 3.5m &  KM = 6.5m ,
KG = 6m, heel= 20degree.

We can calculate:
GM = (KM – KG)
= (6.5 – 6)
= 0.5m

Since, GZ = Sinθ(GM + 1/2 BM tan²θ)
GZ = sin 20⁰ (0.5 + 1/2 x 3 x tan²20⁰ )
Again , GZ = 0.239 ,

 RM (Righting moment or moment of statical stability)
 = (W.GZ)
 = 8000 x 0.239 = 1911.8 tm

5. A ship of 4000t displacement has KG 5.1 m, KB 2.1m, KM 5.5m . Find the moment of statical stability when she heels 24 degree , assuming that she is wall- side.

Solution :

Displacement (W) = 4000t,
KG = 5.1m ,KB = 2.1m &  KM = 5.5m,
Heel =24degree

Here we can calculate :
GM = (KM – KG)
= (5.5 – 5.1)
= 0.4m

Similarly, BM = ( KM – KB)
= (5.5 – 2.1)
= 3.4m

We know that RM can be calculated as :
RM (Righting moment or moment of statical stability)
= (W.GZ)
= W.sinθ ( GM + 1/2 BM tan²θ)  
= 4000  x sin24⁰(0.4 + 1/2 x 3.4 x tan²24^{0 )}  
= 1200tm.

 

6 . A ship of 10000t displacement has a GM of 0.4m. Calculate the moment of statical stability when she is heeled by 5 degree.

Solution:

GZ =Righting lever, GM = Metacentric height

We know that Righting moment = (W x GZ )
Where, W = displacement for all of keel, M = Transverse meta centre.
Displacement (W) = 10000t,
 GM = 0.4m, heel = 5degree

We know that
RM ( righting moment or moment of statical stability)
= (W x GM x sinθ)
= (10000 x 0.4 x sin 5⁰ )
= 348.62tm

7. A ship of 12000t displacement is heeled by 6 degree. If her righting lever is then 0.1m, find the moment of statical stability . If her KM is 8.2m , find her KG .

Solution :

Displacement (W) = 12000t,
 Heel  = 6degree
 GZ = 0.1m  &  KM = 8.2m

We know that :
GZ = GM sinθ,
0.1 = GM sin 6⁰.
GM = (0.1 / sin 6⁰)
= 0.956m
KG = (KM – GM)
= (8.2 – 0.956)
= 7.44m

RM (righting moment or moment of statical stability)
= (W GM sinθ)
= (12000 x 0.956 x sin6⁰ )
= 1199.15 tm.

8. When a ship of 14000t displacement is heeled by 8 deegre , her moment of statical stability is 400tm . If KG 7.3m, find KM.

Solution :

Displacement (W)= 14000t,
Heel = 8degree ,
  RM = 400tm &  KG 7.3m

We know that :
RM (Righting moment or moment of statical stability)
          = (W x GM x sinθ)
 400 = (14000  x GM x sinθ)
GM = 400/ (1400 x sin8^{0  )}
GM = 2.053m

Now, KM = (KG + GM )
= (7.3 + 2.053 )
= 9.353m.

9. A ship of 8000t displacement has KB 3.5m, KM 6.5m, and KG 6m . Find her moment of statical stability at 20 degree heel, assuming that her deck edge remains above water (i.e. she is steel wall side at that angle of heel.

Solution :

Displacement (W) = 8000t,
KB = 3.5m &  KM = 6.5m ,
KG = 6m, heel= 20degree.

We can calculate:
GM = (KM – KG)
= (6.5 – 6)
= 0.5m

Since, GZ = Sinθ(GM + 1/2 BM tan²θ)
GZ = sin 20⁰ (0.5 + 1/2 x 3 x tan²20⁰ )
Again , GZ = 0.239 ,

 RM (Righting moment or moment of statical stability)
 = (W.GZ)
 = 8000 x 0.239 = 1911.8 tm

10. A ship of 4000t displacement has KG 5.1 m, KB 2.1m, KM 5.5m . Find the moment of statical stability when she heels 24 degree , assuming that she is wall- side.

Solution :

Displacement (W) = 4000t,
KG = 5.1m ,KB = 2.1m &  KM = 5.5m,
Heel =24degree

Here we can calculate :
GM = (KM – KG)
= (5.5 – 5.1)
= 0.4m

Similarly, BM = ( KM – KB)
= (5.5 – 2.1)
= 3.4m

We know that RM can be calculated as :
RM (Righting moment or moment of statical stability)
= (W.GZ)
= W.sinθ ( GM + 1/2 BM tan²θ)  
= 4000  x sin24⁰(0.4 + 1/2 x 3.4 x tan²24^{0 )}  
= 1200tm.