Thread: Ship's bending Moments

1. Hi all:
Im working on a Stress calculation software. Tables are tested with excel.
A ship behaves like a beam.
As there are no equations describing the hull's forms, for performing this calculations, the hull is longitudinally cut into a number of transversal slices. The more slices, the better the results. In my testing hull, Im using 87 slices (Blocks) for a flotation length of 125 m.
For each slice, buoyancy (B) and weight (W) are calculated, obtaining a Net force per block, that comes from F=B-P.
Of course total weight and buoyancy must have the same value in order for the Ship to be balanced (Archimedes). But this balance isn't present when analyzyng Block by Block.
Hence Shear stresses do appear.
Many of you, already know that Shear stress curve can be computed from the Integral of the Net forces curve along the beam (Hull).
Starting and final Shear stress result =0. Quite logical ¡¡
No problem at this stage.
But:
Bending moments, as it is also known, can be obtained from integrating again. Integration of Shear stress curve along the hull length, would have to result in another curve, where initial and final values, would have to be=0. Being the Integral of the Shear forces, Inflection points(Max-Min) would have to be located where the former curve value=0, and X cut points, would have to be located where Shear stresses curve value=Max-min.

Calculation procedure is as follows:
For each Block (Volume between two slices).

Block Buoyancy= water density x (Slice_Area(n)+Slice_Area(n+1)) x Distance between slices.
Block weight= Hull weight between both slices + Cargo, ballast, equipment ........... etc. between the given slices again.
Net force= Block Buoyancy-Block weight.

Forces are stored inside an array.

Then, the integration of net forces, with respect to the length is performed by computation of the weights curve surface from zero to the considered block.(Now longitudinal coordinate is the Center of each Block).

For Bending moments, the same procedure would have to return a correct curve. But, it does not happen.
I've tried by direct double integration of weights, and by "from scratch" moments calculation.

No way. Extrange curves appear.

Im solving both integrals by numeric methods (Trapezoids & Simpson).

Could anybody help ? I cannot manage to find why that second integral does not work.

Best regards  2.

3. Dear Crowsnest. Maybe a research of the Build Standard used in the Hull of the Titanic would throw some light on the problem?. Then again, maybe not. westwind.  Posting Permissions
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