sep 17 striptheory pitch

8/13/2019 Sep 17 Striptheory Pitch

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OE 4030 Dynamics of Floating systems and Maneuvering of Marine Vehicles

ROTATIONAL MOTION- PITCH MOTION

Consider a rigid body having an accelerated rotational motion. Now consider a particle with a mass m,

located at a distance r, perpendicular to the axis of the body. The acceleration of this particle has two

components (i) Along the direction of instantaneous velocity, r - is the angular velocity = 2

2

dt

d

dt

d

=

(ii) towards the axis -- 2r ;dt

d=

The first of these will yield a moment about the axis of rotation;

Summing of all particles and moments

yields , I , moment of inertia of the body about the axis of rotation

i.e. I=grwrm iiii

2

2

= (1) Now let us define a quantity k as follows,

to relateIand body mass.

2MkI= = 2kg

Weight; (2)

where k is radius of gyration of the body about a particular axis. From (1) and (2) , we get

=

2

2 iirw

k ; Radius of gyration for Rolling, Pitching and Yawing motions are different

Radius of gyration for Rolling

)( 2222

iiixxxx xydMdMrMkI +===

++= iiiixx Izywg

I )(1 22 ; The second term is the MI of the i-

th element about its own CG, which can be

neglected when the ship is subdivided into sufficiently small weights.

Hence )(1 222

iiixxxx zywg

kg

I +=

= ; From which

+= )( 22 iii

xx

zywk

Similarly we have radius of gyration for pitching and yawing

Pitching Fig 4.14

)(1 222

iiiyyyy zxwg

kg

I +=

=

)( 22


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Problem 1

A homogenously loaded barge = 6000 kN is considered to have 12 uniform divisions, each weighing 500 kN.

Find the radius of gyration for rolling, pitching and yawing. Assume that the weights of each of the block are

centered at the CG of each division, i.e. MI of its own CG can be neglected. Dimension of the barge27.43m(L)6.1 m(B) 3.05 m(D)

See Table 1

For more accurate results one shouldconsider more number of blocks.

Usually xxk approach 0.2887B and yyk and zzk

both approach 0.2887 LWe can also attempt to get radius of gyration

using the following:Illustrated for Pitching radius of gyration

dmI n=2 ;

== Mdm

M

Ik nyy

22 ; =nm mass per unit length

See Table 2 Fig. 4.16 Barge of 12 uniform divisions

Following four moments act in pitching and rolling motions Inertial moment, Damping moment, Restoring

moment and Exciting moment.

Pitching Equation of motion (EOM) )cos(0 tMcba e =++ &&&

Where ais the virtual mass moment of inertia (moment of inertia + added moment of inertia) bis the damping

moment coefficient and cis the restoring moment coefficient.

Determination of coefficients for pitching motion:

(i) ais the virtual mass moment of inertia yyyyyyyy Ikg

II +

=+ 2 ; yyI the Added mass moment of

inertia, kyy Radius of gyration for pitching

Alternative definition for virtual mass moment of inertia a= virtual mass x (radius of gyration)2

= (M+M) 2yyk =2'yyk

g

RoughlyIyyfor the ship can be obtained from sectional area curve as follows

2/

2/

2)(

L

L

yy dxxxAI ; )(xA is the sectional area;

For ship usually yyk = 0.24 to 0.26 times length of the ship.

Added mass moment of inertia can be obtained by STRIP THEORY.(i) Ship is considered to have different sections(ii) Obtain added mass for each section, na

L 2/


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Restoring moment coefficient

In simple form this is expressed as y

L

L

Igdxxyxgc ==

)(2

2/

2/

2

where cis the restoring coefficient andIyis the moment of inertia of the load water plane area. Note thatIy is

not the same as yyI .

Since

y

L

IMB , LLL MGMGgMBgc === (for small angles only) ; It is easy to observe

that LMGc = ; Volume units Force units (Weight); LMB Distance from buoyancy centre to

metacentre; LMG Distance from CG to metacentre.

Exciting moment for pitching motion is due to the unbalanced moment caused by the waves about thetransverse axis of the ship. This moment can be calculated using the following

=2/

2/

2

L

L

xdxygM ; In this the free surface (wave at an angle) is given by )coscos( tkx ea = with a

as amplitude of the wave. For a symmetrical ship about the midsection the exciting moment for pitching canbe written as

=

2/

2/

)coscos()(2

L

Lea

dxtkxxxygM

which can be shown equal to

tdxkxxxygM e

L

L

a sin)cossin()(22/

2/

=

; x is odd function; sin( ) is odd function; cos ( ) is even

function; Odd Odd = Even; Odd Even = Odd; Hence dxkxx )coscos( 0Calculate the following for the ship model we used to calculate added mass earlier

(i) Added mass moment of inertia (Table 3)(ii) Damping Coefficient for pitching motion (Table 4)

(iii) Restoring moment coefficient for pitching motion (Table 5)(iv) Exciting moment for the pitching motion (Table 6)

1 ft=0.3048 m

Station Number

0


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4

iii zyx ,, location of CG of each block from the axis; Axis is at the centre of the body. See Fig. 4.16 on Page 2

Table 1

Block

Noiw ix iy iz

2

ix 2

iy 2

iz 2

iy +2

iz 2

ix +2

iz2

ix +2

iy iw (2

iy +2

iz ) iw (2

ix +2

iz ) iw (2

ix +2

iy )

1 500 -9.144 - 1.524 0.762

2

3

::

16

SUM SUM SUM

Find the radius of gyration for rolling, pitching and yawing using the SUMs

Table 2 Stn nos taken as lines passing through centre of blocks 1 and 6, 2 and 5 ,3 and 4 in plan; This is warranted to result in odd number of

stations to apply Simpson rulex-axis rolling ; y-axis pitching See fig 4.16

Station Nonm kN/m - Distance from

pitching axis

2nm Simpson multiplier Product

0 6000/27.43

1 6000/27.43

2 6000/27.43SUM

Iyy=1/3 station spacing SUM ; Calculate kyy=??

Check whether the ratio of kyyto L is = 0.2887

Just repeat the exercise Table 2 only to calculate radius of gyration aboutxxandzzaxis. Station numbers to be taken as usual (two blocks and three

stations first, middle and last); Check the radius of gyration ratio with that of the corresponding side. It should be also 0.2887.


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Table 3 Added mass moment of Inertia for pitching motion

Station Nona

(FromAssignment 2)

Lever arm

Distance of

each section

from LCG

2 2

n

a Simpson

Multiplier

Product

0

5

10

15

20SUM

== daI nyy2 (1/3) station spacing SUM = ??????

Table 4 Damping coefficient for pitching motion

Station Nonb

(From

Assignment 4)

Lever arm

Distance of

each sectionfrom LCG

2 2nb SimpsonMultiplier

Product

0

5

1015

20

SUM

== dbb heavingnpitching2 (1/3) station spacing SUM = ??????


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Table 5 Restoring moment coefficient for pitching motion

Station NonB

Beam at differentsections

nn gBC = Lever arm

Distance of each sectionfrom LCG

2 2nC Simpson

Multiplier

Product

0

5

10

15

20

SUM

c= (1/3) station spacing SUM = ??????

Table 6 Exciting moment for the pitching motion

Station No

2

nBy= , Half breadth at

water plane

Lever arm

Distance ofeach section

from LCG

y cosk )cossin( k )cossin( ky Simpsonmultiplier

Product

0

5

10

15

20

SUM

tdxkxxxygM e

L

L

a sin)cossin()(22/

2/

=

Value of the integral alone is = = (1/3) station spacing SUM

sep 17 striptheory pitch

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