1 / 24

# Hull Girder Response - Quasi-Static Analysis - PowerPoint PPT Presentation

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. Basic Relationships.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Hull Girder Response - Quasi-Static Analysis' - raya-walls

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Hull Girder Response - Quasi-Static Analysis

• 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

• From Beam Theory – governing equation for bending moment:

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

Where f(x) is a distributed vertical load.

Buoyancyrga(x)

Weightgm(x)

buoyancy curve - b(x)

weight curve - w(x)

net load curve - f(x) = b(x) - w(x)

Sign Convention

Positive

Upwards

+ f

• 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

f

M

Q

Q + dQ

M + dM

dx

But ships are “Free-Free” Beams - No shear at ends!Q(0) = 0 and Q(L) = 0, so C = 0

Sign Convention

Positive

Upwards

+ f :

Shear Force - Q

+ Q

Positive

Clockwise

+ Q :

- Q

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

• Sum of the moments about the right hand side = 0

0

f

M

Q

Q + dQ

M + dM

dx

Again, ships are “Free-Free” Beams - No moment at ends!M(0) = 0 and M(L) = 0, so D = 0

Shear Force - Q

Sign Convention

+ Q

Positive

Clockwise

+ Q :

- Q

Bending Moment - M

Positive

Sagging

+ M :

- M

• 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.

• Static Analysis - No Waves Present

• Most Warships tend to Sag in this Condition

Putting Deck in Compression

Putting Bottom in Tension

• 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.

• 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.

Excess Weight Amidships - Excess Buoyancy on the Ends

Compression

Tension

Excess Buoyancy Amidships - Excess Weight on the Ends

Tension

Compression

• 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.

Each component is located by its l, t and v position and weight

Can be misleading for long components

• For each weight item, need W, lcg, fwd and aft

W

fwd

lcg

aft

FP

• 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

• Find l and x

• Solve for wfand wa so trapezoid’s area equals W and the centroid is at the lcg

FP

lcg

x

wa

G

wf

fwd

l/2

l

aft

• 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.

lcg

x

G

1.2h

wa

wf

aft

FP