203.NA October 2022 Q.8(b)

 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