# How to illustrate scantling

Discussion in 'Stability' started by Adeyele, Sep 3, 2012.

1. Joined: Jun 2012
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You re rite Nick. if i may ask you were is ur location?
am having some problem with the drawing as they are not clear; but i have a midship section drawing here with me. i am going to attach the old one that was sent before because the tutor is not responding to mail regarding query about the course.

you can send an email to me on yeleboye@yahoo.com

Regards
Mike

2. Joined: Jun 2012
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Location: Lagos

Midship section drawing

Dear Hoc kindly check the attached file for the midship section.

Regards
Mike

#### Attached Files:

• ###### Midship section drawing.jpg
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3. Joined: Jun 2012
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Location: Lagos

"Marine surveying"

4. Joined: Oct 2008
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Mike

That is just the basic midship scantling section. Have you done the modulus calculation yet? Without doing that, you can't do much.

As a student it is for you to do the maths, not me. I can check and advise, but it is you that must do the work...otherwise how will you learn?

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### TANSLSenior Member

The midship section, depending on the use to which they will devote the boat, can be right or be complete nonsense.
For plans that have taught us, this boat can have several serious problems, to be studied according to the rules of a Classification Society:
- Longitudinal strength of hull girder
- Torque problems, due to the enormous breadth of the deck opening.
- Local buckling problems due to the lack of longitudinal reinforcement (no secondary reinforcements at all) in bottom, sides and deck.
First, as stated Ad Hoc, would calculate the main section module and check whether it meets the minimum required by the Classification Society chosen for this boat

6. Joined: Jun 2012
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Helo Ad Hoc, kindly check the drawing i use and the table for the calculation i attached below this calculation i have done.

K2 =h/12
Neutral axis YNA = ∑ah/∑a
= -0.144/0.24
YNA = -0.6 (Depth of neutral axis)
To get the Inertia about the assumed axis Ioo = ∑ah2 + ∑ak2
= 1.33 + (-0.012)
Inertia about the assumed axis I00 =1.318m4
The second moment about half depth = 1.32m4
To get the total second moment about neutral axis, INA = I00 – ah2
a = 0.24,
h(YNA)2 = (-0.6)2 =0.36
INA =1.23m4
At full load, the vessel can carry 280 tonnes of cargo, and maximum compressive bending stress is at the upper deck level. We can now find the compressive stress with the formula:
F =M x y/I
y = d /2, d = depth =2.87/2.
y = 1.44, I = 1.23, M =280 tonnes/metre
f = 280 x 1.44/1.23
f = 403.2/1.23.
f = 327.8 tonnes per sqm.
Max stress upon the vessel’s structure = 327.8 tonnes per sqm.

Section of modulus = I/y
I = 1.23, y = 1.44
= 1.23/1.44
Section of modulus = 0.85 m4

#### Attached Files:

• ###### My calculation so far.jpg
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7. Joined: Jun 2012
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How to dervive the scantling lenght

Helo Hoc, please can you explain to me how you derive the figure you use in the drawing of "Hull girder modulus" which you first attached to me. it looks like the way it should be but how do you derive the figure?

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### Nick.KSenior Member

Here is another question from the same exercise.

An inclining test.
The total weight for the experiment is 3.6 ton positioned on the side decks, half each side.
With the data from the inclining test we have to calculate the GM and VCG.

When I convert the weight movements as listed to actual weight port and starboard I see that on movements 5 and 7 the weights are balanced (1.8 ton each side) yet the pendulum deflections are given as 29,21 and 51,43.

I was expecting the pendulum deflection to be zero or close (there is no start reading given) please can some one explain why this is so.

Thanks
Nick.

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### TANSLSenior Member

Nick, I realy don´t underrstand what you try to explain. I don´t understand your question. BUT, without the pendulus length, and many other things, we can not calculate anything.
Give us, please, some more data.
If your teacher wants to suspend you, he need not to give so many turns. Or your teacher will withhold information or you have to learn many things before moving forward. And I say this with affection, trying not to disturb anyone.
B.R.

10. Joined: Oct 2008
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You haven't laid out the calculation as i suggested. You took an assumed NA...why?? you will confuse your self moving axis twice, when an axis already exists, the baseline!

Also use dimensions that make sensible numbers. When using metres, the values are often too small, using millimeters often too large. I use cm's since the vessels I design are usually no more than 60-70m max.

Attached I have calculated the first 2 items for you, so you can check.

View attachment Hull girder modulus calc.PDF

Also you have not identified the longitudinal structure correctly. See my sketch, there are only 5 items to consider.
1) bottom plate
2) bilge corner
3) Long.t girders (if continuous)
4) Side plate
5) Small bit of main deck.

That is all.

Good luck

11. Joined: Oct 2008
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In calculating the GM, from an inclining expt, we assume the hull to be a simple box of wall sided shape. In reality it is not.

So, when you do your movements, even though the distance moved and the weight remains the same, you will not get exactly the same deflections. The more list you have the more "error" there shall be owing to the changing shape of the hull at ever increasing list. This leads to very small differences to large differences in the actual deflections you obtain. Also tide/wind and not having slack ropes can also introduce errors and inconsistencies.

This is why when you are doing an inclining expt, you plot the moments onto a graph, to see if the load v distance is consistent or not. The line should pass through the centre of the axis, but in reality it does not always do so. When this occurs, you take the "worse case" from the graph.

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### Nick.KSenior Member

My question is not why the deflection is not linear.

In this experiment we start with an equal pile of weights on each side (1.8 ton each side) at which point the vessel may or may not have a small list. The list is measured from a pendulum deflection which you would expect to be close to zero. As the weights are shifted the list increases and the corresponding deflections are recorded. As the weights are returned to their positions the list should decrease and return to the start position (unless weight has shifted inside).

In these readings, on movements 5 and 7 the weights are equal on each side but the deflections are large (at 7, the list is nearly the same as at 9 with 3.2 ton on one side)

Am I interperating the readings in the correct way?

Thanks Tansl for your input. We are given all the other data it is just the pendulum readings V weight movements that I don't understand as explained above.

Nick.

#### Attached Files:

• ###### Inclining excel.pdf
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13. Joined: Oct 2008
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Shift #5, has 0.4t moved 1.80m that is a moment of 0.72tm, giving a deflection of 29mm

Shift #7, has 0.8t moved 1.80m, that is a moment of 1.44tm, giving a deflection of 51mm.

There two are not the same. The moments are different.

Two different examples of graphs I have produced whilst doing an inc. expt:

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### TANSLSenior Member

There are various circumstances that can vary the inclination taken by the ship: wind force at the time, any wave that helps heeling the ship a little more, a tie which is tensioned and slows the boat, .... That's why you have to make many measurements, to obtain a correct average value of the deviation.
I hope this explanation will serve to clarify some of your doubts. The attached excel table could help you in your considerations and calculations.

#### Attached Files:

• ###### Inclining Test.xls
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