Hull Girder Response

shrubflattenUrban and Civil

Nov 25, 2013 (3 years and 10 months ago)

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Hull Girder Response
-


Quasi
-
Static Analysis

Basic Relationships


Model the hull as a Free
-
Free box beam.


Beam on an elastic foundation


Must maintain overall Static Equilibrium.


Force of Buoyancy = Weight of the Ship




LCB must be in line with the LCG








0 0
L L
g a x dx g m x dx


 




0 0
L L
g xa x dx g xm x dx


 
Basic Relationships


From Beam Theory


governing equation for bending
moment:



Beam is experiencing bending due to the differences
between the Weight and Buoyancy distributions



2
2
d m
f x
dx

Where
f
(
x
)

is a distributed
vertical load.

( ) ( ) ( )
f x b x w x
 
Buoyancy


g

a
(
x
)

Net Load

Weight

g

m
(
x
)

Basic Relationships

buoyancy curve
-

b(x)

weight curve
-

w(x)

net load curve
-

f(x) = b(x)
-

w(x)

Sign Convention

Positive

Upwards

+ f

Basic Relationships


The solution for
M
(
x
) requires two integrations:


The first integration yields the transverse shear force
distribution,
Q
(
x
)


Impose static equilibrium on a differential element

Q

M

f

Q + dQ

M + dM

dx



0
Q f dx Q dQ
   
dQ
f
dx





0
x
Q x f x dx C
 

But ships are “Free
-
Free” Beams
-

No shear at ends!

Q
(0) = 0 and
Q
(
L
) = 0, so
C

= 0

Finding Shear Distribution

Shear Force
-

Q

+ Q :

Positive

Clockwise

Sign Convention

Positive

Upwards

+
f
:

Net Load
-

f

+ Q

-

Q

Basic Relationships


The second integration yields the longitudinal bending
moment distribution,
M
(
x
):


Sum of the moments about the right hand side = 0

Q

M

f

Q + dQ

M + dM

dx

0
2
dx
M Qdx f dx M dM
    
0

dM
Q
dx





0
x
M x Q x dx D
 

Again, ships are “Free
-
Free” Beams
-

No moment at ends!

M
(0) = 0 and
M
(
L
) = 0, so
D

= 0

Finding Bending Moment Distribution

Shear Force
-

Q

Bending Moment
-

M

+
Q

:

Positive

Clockwise

+
M

:

Positive

Sagging

+
Q

-

Q

-

M

Sign Convention

Shear & Moment Curve Characteristics


Zero shear and bending moments at the ends.


Points of zero net load correspond to points of
minimum or maximum shear.


Points of zero shear correspond to points of
minimum or maximum bending moment.


Points of minimum or maximum shear
correspond to inflection points on bending
moment curve.


On ships, there is no shear or bending moments
at the forward or aft ends
.

Still Water Condition


Static Analysis
-

No Waves Present


Most Warships tend to Sag in this
Condition

Putting Deck in Compression

Putting Bottom in Tension

Quasi
-
Static Analysis


Simplified way to treat dynamic effect of waves on hull girder
bending


Attempts to choose two “worst case”conditions and analyze them.


Hogging Wave Condition

»
Wave with crest at bow, trough at midships, crest at stern.


Sagging Wave Condition

»
Wave with a trough at bow, crest at midships, trough at stern.


Wave height chosen to represent a “reasonable extreme”


Typically:


Ship is “balanced” on the wave and a static analysis is done.

1.1
BP
H L

Wave Elevation Profiles


The wave usually chosen for this analysis is a
Trochoidal

wave. It has a steeper crest and flatter
trough.


Chosen because it gives a better representation of
an actual sea wave than a sinusoidal wave.


Some use a cnoidal wave for shallow water as it
has even steeper crests.



Trochoidal vs. Sine Wave

-20
-15
-10
-5
0
5
10
15
20
0
20
40
60
80
100
120
140
160
180
200
Lenght (ft)
Wave Height (ft)
Trochoidal Wave
Sinusoidal Wave
Sagging Wave

Excess Weight Amidships
-

Excess Buoyancy on the Ends

Tension

Compression

Hogging Wave

Excess Buoyancy Amidships
-

Excess Weight on the Ends

Tension

Compression

Weight Curve Generation


The weight curve can be generated by numerous
methods:


Distinct Items (same method as for LCG)


Parabolic approximation


Trapezoidal approximation


Biles Method (similar to trapezoidal)


They all give similar results for shear and bending
moment calculations. Select based on the easiest in
your situation.

Distinct Item Method

ITEM
Material
units
wt/unit
WT
LCG
VCG
LMOM
VMOM
GROUP C - JOINERY WORK
Forward cabin

berth flat
composite
35
0.77
27
10.50
1.25
282.98
33.69
mattress
35
3.00
105
10.50
1.50
1102.50
157.50
shelf p&s
composite w/veneer
12
1.02
12
12.00
2.50
146.88
30.60
verticals p&s
composite w/veneer
34
1.02
35
12.00
1.00
416.16
34.68
desk
composite w/veneer
4
1.28
5
14.50
2.50
74.24
12.80
supports and hardware
5
14.50
2.50
72.50
12.50
hanging locker
composite w/veneer
27
1.28
35
15.00
2.00
518.40
69.12
rod & hardware
10
15.00
3.00
150.00
30.00
cabinet
composite w/veneer
17
1.02
17
16.75
3.00
290.45
52.02
door blkhd
composite w/veneer
25
1.85
46
17.25
2.00
791.43
91.76
drawers
wood
10
5.00
50
15.00
0.50
750.00
25.00
sole
plywood & teak
29
2.50
71
16.40
-0.50
1168.50
-35.63
overhead
honeycomb/vynal
24
0.50
12
17.00
6.25
204.00
75.00
Each component is located by its l, t and v position
and weight


Can be misleading for long components

Example Weight Curve

120K Bbl TAO Weight Curve
0
20
40
60
80
100
120
-100
0
100
200
300
400
500
600
700
Feet from FP (+ Aft)
Distributed Weight (LT/ft)
Weight Curve
Displacement =
LCG =
27450
299.3
LT
ft aft FP
1/19/99

For each weight item, need
W
,
lcg
,
fwd

and
aft

Weight Item Information

fwd

W

aft

lcg

FP

Trapezoid Method


Models weight item as a trapezoid


Best used for
semi
-
concentrated

weight items


Need the following information:


Item weight


W (or mass, M)


Location of weight centroid wrt FP
-

lcg


Forward boundary wrt FP
-

fwd


Aft boundary wrt FP
-

aft


lcg
must be in middle 1/3 of trapezoid


Trapezoid Method


Find
l

and
x



Solve for
w
f

and
w
a

so
trapezoid’s area equals
W

and the centroid is at the
lcg

lcg

x

fwd

aft

l/2

l

w
f

w
a

FP

w
W
l
Wx
l
w
W
l
Wx
l
a
f




6
6
2
2
G

2
l
f
lcg
x



Biles Method


Used for weight items which are nearly
continuous

over
the length of the ship.


Assumes that weight decreases near bow & stern.


Assumes that there is a significant amount of parallel
middle body.


Models the material with two trapezoids and a
rectangle.

Biles Method

l
3
1.2h

l
3
l
3
w
f

w
a

FP

lcg

G

x

aft


















l
x
h
w
l
x
h
w
l
w
h
a
f
7
54
6
.
0
7
54
6
.
0
The Three Types of Structure

Characteristics

Primary
Structure

Secondary
Structure

Tertiary Structure

In
-
plane rigidity

Quasi
-
infinite

Finite

Small

Loading

In
-
plane

Normal

Normal

Stresses

Tension,
Compression
and Shear

Bending and
Shear

Bending, Shear
and Membrane

Examples

Hull shell, deck,
blkhd, tank top

Stiffeners on
blkhd, shell

Unstiffened shell

Boundaries

Undetermined

Primary structure

Secondary
Structure