NAVAL ARCHITECTURE, PAPER – I (UG21T4502)
UNIT – 1(BY CAPT S.B. SINHA)
USE OF SIMPSON’S RULES FOR THE COMPUTATION OF SECOND
MOMENT OF AREA AND C.O.P.
The second moment of an area, also called MOMENT OF INERTIA of the area about an
axis.
It is equal to product of the area and the square of it’s distance from the axis. Now, let
us consider, dA represent an element of an area and let “y” be it’s distance from the
axis AB.
SECOND MEMENT OF THE ELEMENT (M.O.I) about axis AB = dA x y2
MOMENT OF INERTIA:
Resistance properly of a stationary or a moving body to change it’s speed or direction
is called INERTIA.
For linear motion, INERTIA MASS OF THE BODY
Whereas for a rotational motion.
Inertia Moment of Inertia
“Moment of Inertia is also called SECOND MOMENT OF AREA.
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PARALLEL AXES THEOREM :
“ Moment of Inertia” of an area about an axis at a distance ‘h’ from the centroid of the
Area “A” = The M.O.I. of “A” about an axis parallel to the given axis passing
through the centroid of the Area + The product of Area & the square of the
perpendicular distance between the axes.
For the rectangular shape given above:
IKK =
Iyy =
IMM = IKK + A.d2
For the rectangular shape given above,
Find IAB & IBC
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Now,
²
IAB = Iyy + A ( )
= + (l x b) x
= + =
IAB =
Now,
²
IBC = IKK + A ( )
= + xbx
= + =
IBC =
Please remember, UNIT of M.O.I is m4.
Uses of “Moment of Inertia” in ship Stability:
(i) To Calculate BMT.
(ii) To Calculate BML.
(iii) To Calculate FSC (free surface correction)
(iv) To Calculate “Centre of Pressure” (C.O.P.)
Moment of inertia (M.O.I.) of a Triangular area:
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KK is an axis, passing through centroid (G) of the triangle.
IKK =
²
IBase = IKK + A ( ) , where A Area of the
= + BD x = +
IBase =
Now,
²
IMM = IKK + A ( )
= + BD x = +
IMM =
Moment of Inertia (M.O.I.) of a Trapezium:
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(i) Height of COG above base, = x
(ii) Area = (a + b) x
(iii) M.O.I about an axis “gg” passing through “G”
Igg = x h3
Moment of Inertia of a Circle:
IDD = =
IKK = IDD + A.r2 = +
IKK =
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MOMENT OF INERTIA OF A SEMI CIRCLE:
IDD (Semi Circle) = . =
2
IDD = Igg + A
²
= Igg +( )( )
= Igg +
Igg = -
= r4 ( - )
Igg = 0.1098r4
Circle Semi Circle
Igg
0.1098r4
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WORKED EXAMPLES:
1. A ship’s water plane is 18m long. The half ordinates at equal distances from
forward are as follows :
0,1.2, 1.5, 1.8, 1.8, 1.5 and 1.2 meters respectively.
Find the second moment of the water plane area about the centre line.
Solutions:
Here no. of ordinates are 7, hence no. of common interval is 6. Length = 18m.
h = = 3m
Half S.M. Product for Product for Product for 2nd
Ordinate Semi Area first moment
moment
(y) (d ) y.d y2.dx y3.dx
0 1 0 0 0
1.2 4 4.8 5.76 6.912
1.5 2 3.0 4.50 6.750
1.8 4 7.2 12.96 23.328
1.8 2 3.6 6.48 11.664
1.5 4 6.0 9.00 13.500
1.2 1 1.2 1.44 1.728
Sum of Products (Second Moment)=
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Now,
M.O.I about the Centre line (CL) = 2 kh sum of Products
= 2 x x 3 x 63.882
ICL = 42.588m4
Note : In the above expression, we have multiplied by “2”, as we have taken
half ordinate and we have to find M.O.I of the water plane area.
Also k = , taken as “ k” is for Simpson’s first Rule.
2. A ship’s water plane is 18m long. The half ordinates at equal distances from
forward are as follows :
0,1.2, 1.5, 1.8, 1.8,1.5 and 1.2 meters respectively.
Find the second moment of the water plane area about a transverse axis
through the centre of flotation.
Solutions:
Here, common interval, h = = 3m
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Half S.M. Product for Lever from Product for Product for 2nd
Ordinate Semi Area forward first moment
moment
2
(y) (d ) y.d ( y.dx ( y.dx (
0 1 0 0 0 0
1.2 4 4.8 3 14.4 43.2
1.5 2 3.0 6 18.0 108
1.8 4 7.2 9 64.8 583.2
1.8 2 3.6 12 43.2 518.4
1.5 4 6.0 15 90.0 1350
1.2 1 1.2 18 21.6 388.8
Now,
Sum of Products (Semi Area) = = 25.8
Sum of Products (Semi Moment) = = 252
Sum of Products (Semi 2nd Moment)= = 2991.6
Now,
Water plane Area = 2 h x SOPSemi Area
= (2 x x 3 x 25.8 )
Water Plane Area = 51.6 m2
Now,
Distance of COF from forward =
= = 9.77m
Now,
Second moment of Area (M.O.I) about an axis passing through forward =
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= 2 h x SOP
= 2 x x 3 x 2991.6
= 5983.2 m4
IFF = 5983.2 m4
Now,
IFF = ICOF + Ad2
5983.2 = ICOF + 51.6 (9.77)2
5983 = ICOF + 4925
ICOF = 5983 – 4925
ICOF = 1058m4
Therefore,
“Moment of Inertia” of the water plane Area, about a transverse axis
through COF = 1058m4
3. Half breadths of a transverse waterlight bulkhead from top are 5.2m, 4.8m,
4.3m, 3.7m and 3.0m. Bulkhead is 8m high and tank contains water of density
1.015. Sounding is 12m.
Find thrust and COP.
Solution:
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Area of Bulkhead (Surface under Pressure) = 2 x ( h S.O. Products)
(Simpson’s first Rule)
COPLL =
lever from top of Bulkhead for finding centroid of Bulkhead
lever from liquid level for finding M.O.ILL
4, 6, 8, 10, 12m
Half S.M. Product for Lever from Product for Product for 2nd
Ordinate Semi Area top of first moment
Bulkhead moment
2
(y) (d ) y.d ( y.dx ( y.dx (
5.2 1 5.2 0 0 5.2(4)2 =83.2
4.8 4 19.2 2m 38.4 19.2(6)2 =691.2
4.3 2 8.6 4m 34.4 8.6(8)2 =550.4
3.7 4 14.8 6m 88.8 14.8(10)2 =1480
3.0 1 3.0 8m 24.0 3.0(12)2 =432
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Area of Bulkhead = 2 ( h x S.O. Products)
= 2 ( h x 2 x 50.8) = 67.73m2
Posn of Centroid = =
= 3.65m from top of bulkhead
Head = (3.65 +4)m = 7.65m
ILL = 2 h. SO Products (2nd moment)
= 2 x 2 x 3236.8 = 4315.7m4
Thrust = Area under Pressure x x Head
= (67.73 x 1.015 x 7.65) t =525.9 t
COPLL = = = 8.32m from liquid level
4. Transverse bulkhead of a tank is 12m broad on top. There are equally spaced
vertical ordinates as follows: 3m, 5m, 6m, 5m & 3m, Tank contains salt water.
Sounding of the tank at the centre line 9.0m.
Calculate Thrust and position of COP from top of Bulkhead.
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Ordinate S.M. Product for Product for Product for 2nd
Area first moment
moment
(y) (d ) y.d y2.dx y3.dx
3 1 3 9 27
5 4 20 100 500
6 2 12 72 432
5 4 20 100 500
3 1 3 9 27
Area = h SOP = x 3 x 58
Area under pressure = 58 m2
Position of Centroid from top of Bulkhead = =
posn of centroid from top of Bulkhead = 2.5m
Therefore Head = 2.5 + 3 = 5.5m
Thrust = Area under Pr x x Head
= 58 x 1.025 x 5.5
= 326.975 tonnes
I from top of Bulkhead = kh SOP
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= x x 3 x 1486 = 495.33 m4
495.33 = Igg + 58 (2.5)2
Igg = 495.33 – 58(2.5)2
= 495.33 – 362.5
Igg = 132.8 m4
Now,
ILL = Igg + 58 (Head)2
= 132.8 + 58 (5.5)2
= 132.8 + 1754.5
ILL = 1887.33 m4
COPLL =
COPLL = 5.92m (from liquid level)
COP (from top of Bulkhead) = 5.92 – 3
= 2.92m from top of Bulkhead
*********
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