sep11 2012 striptheory heave

8/13/2019 Sep11 2012 Striptheory Heave

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11 Sep 2012

Strip Theory Source Rameshwar Bhattacharyya

CALCULATION OF ADDED MASS Heave motion

Added Mass: (Rameshwar Bhattacharyya P39)

An accelerating body in a fluid experiences a force greater than the mass of the body times acceleration. This increment of force can be defined as

the product of body acceleration and a quantity having the same dimension as mass and it is termed as added mass.

Do not imagine that the body drags a certain amount of fluid as it accelerates in the medium.

Added mass for a circular section of unit length and diameter Bn is lengthunit

4

2

nB

Added mass for a semi-circular section of unit length and diameter Bnis lengthunit8

2

nB

For shapes other than semi-circular ones the added mass is8

2

n

n

BCa

=

Coefficient C is a function of Draft/Beam ratio and Area coefficient of section. Area coefficient of section = Section area / (Beam Draft)

The coefficient C for Lewis form is obtained from Figure 4.4.(P41)

It is assumed for sections other than Lewis form the added mass does not differ much as long as the beam, draft and area are the same. For

conventional hull forms the Lewis-form representation of a section that has correct beam, draft and section has been found to be OK.

Total added mass of the ship = =2/

2/_

L

L

nz dxaa

Noting that Bn=2 y(x), y(x) is half breadth at waterline, the above equation becomes

=2/

2/_

2)(

2

L

L

z dxxyCa

Simpsons Rule


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Example

1 ft=0.3048 m

1 ft2= 0.0929 m

2

Station Number

Length of model = L = 19.20 ft =5.85m ; Maximum beam = 0.79 m ; Draft = 0.349 m ; Wavelength= Ship Length; LCG= + 0.146 m (forward of

midship); LCB= + 0.146 m (forward of midship);

Model speed=1.46 m/sec; Displacement = 12368 N; Direction of ship travel = Head sea (=180)

Find the added mass for heaving in terms of the model mass.

Solution Steps:

(1) Find wave frequencyw

wL

g

2= rad/sec

200

020


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(2) Find encountering frequency

cos2

ug

w

we = ; u is the model speed

Open Excel sheetg=9.81m/sec

2=1000 kg/m

3

StationNumber

(1)

BeamatStationBn (m)

(2)

Draft atStationTn(m)

(3)

Section areaat StationsSn(m

2)

(4)

n

e Bg2

2

(5)

BeamDraftratioBn/Tn

(6)

BnxTn

(7)

Sectionalarea Coeff

nn

n

nTB

S

=

(8)

AddedmassCoeffCFromFig 4.4

(9)

Bn2

(10)

2

8 nB

(11)

ancol 9 xcol 11

(12)

SimpsonMultiplier

(13)

Productcol 12 xcol 13

(14)

0 0 0.348691 0 1

5 0.790042 0.348691 0.2734976 4

10 0.790042 0.348691 0.2734976 2

15 0.790042 0.348691 0.2556608 4

20 0 0.348691 0 1

Sum ????

Added Mass for heaving = = dxaa nz = 1/3 Station spacing Sum ( Station spacing = Length of ship / No of sections)

Model Mass = / g

Added mass = ________ % of the mass of the ship (FIND)

Repeat this problem with 7 Stations ; Interpolate Values for Sectional areas (or assume appropriate values); How does Simpson multiplier change?

Is there improvement in Added mass prediction?

Please practice by HAND CALCULATOR too ..for exam purpose


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CALCULATION OF DAMPING COEFFICIENT Heave motion

Damping Force: (Rameshwar Bhattacharyya P43)

Damping force mostly acts in the opposite direction to the motion of the ship and produces gradual reduction in the amplitude of motion. The

equation of motion (EOM) includes this as damping force which for simplified case is zbFb &= where b is the damping coefficient (in heaving

motion). Damping sources are friction, eddy making and wave making. The last one (wave making) is larger as compared to the other two. The

damping coefficient can be obtained by strip theory. Damping in heaving is mainly caused due to the waves generated by the heaving motion of the

ship. Therefore the damping coefficient per unit length is related to the amplitude of the generated (radiated) waves.

Damping coefficient per unit length, nb , is given by lengthunit3

22

e

Ag

; e is the frequency of the radiated waves (encountering frequency)

a

a

zA

==

motionheavingtheofAmplitude

wavesradiatedtheofAmplitude; This information can be obtained from Figure 4.6 (Page 4)

Total damping coefficient of the ship = =2/

2/_

L

L

ndxbb

Example

1 ft=0.3048 m

Station Number

1 ft2= 0.0929 m

2

Station Number 200

020


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Length of model = L = 19.20 ft = 5.85 m. ; Maximum beam = 0.79 m ; Draft = 0.349 m ; Wavelength= Ship Length; Model speed=1.46 m/sec;Displacement = 12368 N; Direction of ship travel = Head sea (=180)

Find the damping coefficient for heaving

Solution Steps:

(1) Find wave frequencyw

wL

g

2= rad/sec; (2) Find encountering frequency

cos

2

ug

w

we = ; u is the model speed

Open Excel sheetg=9.81m/sec

2=1000 kg/m

3

Station

Number(1)

Beam atStation

Bn (m)(2)

Draft atStation

Tn(m)(3)

Section areaat Stations

Sn(m2

)(4)

n

e B

g2

2

(5)

BeamDraftratio

Bn/Tn(6)

Sectionalarea Coeff

nn

n

n

TB

S

=

(7)

A FromFig

4.6(8)

2

A (9)

bn=3

22

e

Ag

(10)

Simpson

Multiplier(11)

Productcol 10 xcol 11

(12)

0 0 0.348691 0 1

5 0.790042 0.348691 0.2734976 4

10 0.790042 0.348691 0.2734976 2

15 0.790042 0.348691 0.2556608 4

20 0 0.348691 0 1

Sum ????

Damping coefficient for heaving = = dxbb n = 1/3 Station spacing Sum ( Station spacing = Length of ship / No of sections)

Damping coefficient = ________ (FIND)Repeat this problem with 7 Stations ; Interpolate Values for Sectional areas (or assume appropriate values); How does Simpson multiplier change?

Please practice by HAND CALCULATOR too ..for exam purpose

EXCITING FORCE (Pg 49-50)

Assume ship as wall sided near LWL (load water line), and a wave ( tea cos= ) passes the ship and t is the time when the crest of the wave isamidships. Exciting force for heaving is obtained by assuming that the ship remains still as far as vertical motion is concerned and the waves pass

slowly by the ship. The exciting force for heaving is the additional buoyancy of the force at any instant. This is given by

=2/

2/

)(2

L

L

dxxygF ;

y(x) is half breadth at section x. Taking the wave profile at surface,

=2/

2/

)'(cos)(2

L

L

ea dxtxkxygF . This expression simplifies to


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=2/

2/

cos'cos)(2

L

L

ea tdxxkxygF = tF eocos ; The phase between the wave profile ( tea cos= ) and heaving force due to waves is

zero. The nondimensional amplitude of the heaving force is

LBg

Ff

a

00 = =

2/

2/

)cos(cos)(2

L

L

dxkxxy

LB

. For beam seas the exciting force for

heaving reaches maximum value.

Example ; (Same model as in previous example) Length of model =Length of wave = L = Lw= 19.20 ft = 5.85 m. ; Maximum beam = 0.79 m ;Draft = 0.349 m ; Wavelength= Ship Length; LCG= at midship; Amplitude of the wave = 0.06m ; Direction of ship travel = =120.

Find f0 and F0.

Solution

StationNumber

(1)

Beam atStationBn (m)

(2)y=Bn/2

(3)

Cos

(4)

Distancefrom LCGof shipx

(5)

cos' kxxk = (6)

)coscos( kx

(7)

)coscos( kxy

(8)

SimpsonMultiplier

(9)

Product

col 8 xcol 9

(10)

0 0 ? 1

5 0.790042 ? 4

10 0.790042 0 2

15 0.790042 ? 4

20 0 ? 1

Sum ????

Find wave number

Dimensionless force amplitude is =0f =

2/

2/

)cos(cos)(2

L

L

dxkxxyLB

;

Integral is obtained from Table Excel sheet; = 1/3 station spacing SUM

Integral2

0

=BL

f = __________ (Calculate) and 00 LBfgF a= = ______________ (Calculate)

Heave Restoring force coefficient OBTAIN C USING STRIP THEORY

=2/

2/

)(2

L

L

dxxygc


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sep11 2012 striptheory heave

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