24.NA December 2018 Q.9

A ship 160m long and 8700 tonne displacement floats at a waterline with

Station            AP        ½         1         2         3         4         5         6         7         7¹/₂      FP

½ ordinate       0        2.4       5.0      7.3      7.9       8.0      8.0       7.7     5.5        2.8      0m

While floating at this waterline, the ship develops a list of 10⁰ due to instability. Calculate the

negative metacentric height when the vessel is upright in this condition.

If you are noticing some error in problems kindly comment below.Thanks

Given

L = 160 m
Δ = 8700 t
List = 10 deg

To find

1.Negative metacentric height

Solution

½ Ordinates	½ Ordinate3	SM	Product
0	0	1/2	-
2.4	13.82	2	27.64
5.0	125.00	3/2	187.5
7.3	389.02	4	1556.08
7.9	493.04	2	986.08
8.0	512.00	4	2048.00
8.0	512.00	2	1024.00
7.7	456.53	4	1826.12
5.5	166.38	3/2	249.57
2.8	21.95	2	43.9
0	0	1/2	-
		TOTAL	7948.9

Common interval h = L/ no of equidistant ordinates

There are 8 equidistant sapces

h = 160 / 8

   = 20 m

Second moment of area about centre line I = (h / 9) x Total products

Since half ordinates are given.For full ordinates

I =  2 x (h/9) x Total products

  = 2 x (20/9) x 7948.9

  = 35328.44 m4

Distance from B to M, BM = (I / Δ) x 1.025

                                            = (35328.44 / 8700) x 1.025 

                                            = 4.162 m

At angle of loll tan θ = √(-2 GM/BM)

Squaring both sides tan² θ = - 2 GM/ BM

                                    GM = - (tan²θ x BM) / 2

                                            = - 0.1763² x 4.162 / 2

Negative metacentric height = - 0.0646 m