Stability – I : Chapter 11

  1. On a ship of W 5000t , GM 0.3m, 20t was shifted transversely by 5m. Find the list.
Solution:

W = 5000t, GM = 0.3m,
Weight shifted (w) = 20t & d= 5m

Listing moment = (20 x 5 )
= 100tm

We know that :
Tanθ  = ( Listing Moment) /(W x GM)
= 100/(5000 x 0.3)
= 3.8 degree.

  1. On a ship of W 8000t, GM 2.0m, If the following transverse shifting were done , find the list:
    • 200cargo shifted 4m to stbd
    • 100t cargo shifted 2 m to port
    • 100t cargo shifted 4m to port
    • 50t stores shifted 20m to stbd

Solution:-

We can calculate Listing moment due to shifting of cargo
LM(1)  = ( weight x distance )
=   (200 x 4)
= 800 tm (S)

Again,   LM (2) = (Weight x distance )
= (100 x 2)
= 200 tm (P)

LM(3)  = ( weight x distance )
=   (100 x 4)
= 400 tm (P)

LM(4)  = ( weight x distance )
= (50 x 20)
= 1000 tm (S)

Final listing moment (LM) = LM(1) + LM(2) +LM(3)+ LM(4)
= 800(S) + 200(P) + 400(P) + 1000(S)
=1800(S) + 600(P)

So, final listing moments = (1800 – 600)
= 1200 tm (S)

Now, Tanθ  = (Final LM )/(W x GM)
= (1200/(8000 x 2 )
= 4.29 degree (S)

  1. If 200t cargo was shifted downwards by 10m and to starboard by 5m on a ship of W 10000t , KG 7.0m, KM 7.4m, find the list.
Solution :-

Given :
Cargo shifted (w)   = 200t,
Distance  = 10m,
d = 5m from CL to stbd
W = 10000 t ,
KG = 7.0m &
KM = 7.4m

Ship’s wt KG VM
10000 t 7.0m 70000 tm
200t (shift) 10m (-) 2000 tm

     Final W =10000t                                             Final VM= 68000 tm

We know that:
Final KG = (final VM)/ (Final W)
Final KG = (68000/10000)
= 6.8m

Final GM= (KM – KG)
= (7.4 – 6.8)
= 0.6m

Listing moment (LM) = (weight x distance )
= (200 x 5)
= 1000tm

tanθ =  LM/(W x GM)
= 1000 /(10000 x 0.6)
= 9.46 degree

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1 Comment

  • Ex -11 List Ques No. 18, Part 2 have a mistake in total Vertical movements…..it should be 87000 not 87200 which gives us a final answer as follows
    List = 3* 51′ or 3.85 degrees

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