# Stiffener calculation based on ISO 12215

Discussion in 'Boat Design' started by Ardi, Sep 7, 2021.

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### ArdiJunior Member

Hi everyone,

I'm a new member in this group. Now I'm trying to design a small aluminum boat using the ISO 12215 standard and I'm studying this ISO self-taught. By the way, I have calculated for the minimum modulus section area of the bottom stiffener. And I found a value of 8.6 cm2 attached to 6mm bottom plate. After I looked into Annex G, I found a table like the one attached. My question are :
1. How do I read the table and which size of stiffener do I use? Because the value of 8.6 is between 7.7 and 11.7
2. Is there a manual way to calculate the stiffener size?

Thanks for the answer, and maybe in the future I'll ask more questions considering I'm a self-taught. I think this group is a very good start for everyone who wants to study about boat.

Regards,
Ardi

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Hi Ardi,
Welcome to the forum.

The selection of stiffener depends upon what type you wish to use. If your calculation says you need 8.6cm3, then that is the minimum value.
So looking at the table, if you wanted to use an angle bar or Tee bar, then your choice would start with:

A 40x40x5 angle bar. This is the minimum. Its modulus on 6mm plate is 9.6cm3...so that value is greater than 8.6cm3...so it is acceptable.
If you can't get angle bar, or don't wish to use it, then say a flat bar:

You can select a 70x7mm flat bar; its modulus is 11.7cm3 on 6mm plate. You cannot select the 60x6, as that is 7.7 and below 8.6cm3.

Yes, but is is probably beyond your ability.
You can find online calculators such as here, but for attached plate you need an unequal angle, since the lower flange is your 6mm plate (more breadth) and is different from the flange of an angle bar (different thickness and breadth).
In which case you would need to use this one.

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### ArdiJunior Member

Thank you Ad Hoc, very clear explanation . Honestly I have opened this forum for several times when I need something as a silent reader, and I always found your comments almost in every threads I have opened.
Actually I have many questions regarding ISO 12215, but I want to focuss in this problem first. Regarding to the calculation of Section Modulus on the link you have shared , there are Sx and Sy. And in ISO 12215 there is SM. My question is, is SM= Sx + Sy ? or are there any other explanation for that?

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### TANSLSenior Member

@Ardi, there are a few things to keep in mind.
First of all, although it has nothing to do with a stiffener module, you are using 12215-5: 2008 which is no longer in force. The current version is 2019.
Second, the values in Table G.6 appear to correspond to a 300 mm wide attached plate. It is very likely that the attached plate that you can use has a different width.
Third, you must take into account the inclination of the reinforcement with respect to the attached plate, which is not usually 90º, the value with which Table G.6 is calculated.

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### ArdiJunior Member

Thanks for the response, @TANSL
Yes you are right, I am using ISO 2008, because this is just for learning. I guess the formula doesn't change much. I'll buy the latest version when I'm actually going to build a boat and I have enough money . I agree with you that plate widths and angles are not always the same. That's why I want to know if there is a manual way or another way to find out the specific stiffener dimensions based on the known area modulus. Do you have the answer?

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I hope they have helped?

SM = the axis to which the bending is taking place. SM just means section modulus.
So, looking at the image:

If the lower flange is the plate and the upper flange, is either the Tee or angle bar, then the plate is the hull, so to speak.

The applied load is normal to the plate, so the I-beam will bend about the axis X-X...or Sx.
So you calculate the modulus using the axis x.

Does that make sense now?

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### TANSLSenior Member

No, if there is such a formula, I do not know it. The procedure that I use is, once the necessary module and the width and thickness of the attached plate are known, propose a reinforcement and check if the module of the set meets the minimum required by the standard that I am using. All existing profile catalogs give the values of the reinforcement section without the attached plate. Therefore, to take it into account, there is no choice but to do it by hand, which is impractical, or to use any CAD program (autoCAD for example), which gives totally exact values
The ISO 20019 version differs considerably from the previous version in procedures, design pressures and minimum allowable values. Therefore, even if it is to learn, it is not convenient for you to use the old version.

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Yes, you can calculate it by hand, if you know how.
Or
Just use the link I provided above.

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### ArdiJunior Member

Thank you everyone for the response and explanation. I think there were clear enough for me.

@TANSL, I see, hope I could get the latest version soon. And I have a question , This is one of the variables that confused me while studying this ISO. What is the meaning of x, in calculating the longitudinal distribution factor (kL). It said that x is longitudinal position of the panel or stiffener. So I have an assumption : first option. let's say I want to design 5mtr alloy boat. For bottom plate I could make it from one piece alloy plate for two bottom side. Is x the midpoint of one sheet of the plate? so there is one x value for the boat.
Second option, I want to make the bottom from 2 joined plate. So there are rear and front plate, there will be a weld between them. So the second boat will have 2 value of x.
Am I wrong?

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### TANSLSenior Member

You should divide the surface of the hull into "panels", not plates. A "panel" is a surface bounded by two transverse reinforcements and two longitudinal reinforcements. A plate usually covers several panels.
Accordingly, each panel has an "x" so that in a background there can be multiple "x" and, therefore, multiple panels with very different thicknesses. Whether all these panels are going to be made up of one plate or several is your decision. Of all the panels that are included in the same plate, the thickest is the one that defines the thickness of that plate.

Last edited: Sep 9, 2021
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Every boat has frames and longitudinals.

Well, look at the definition of 'x' in ISO:

The "panel" is the region bound by the transverse - the frame - and the longitudinals - the stiffeners.
Like so:

In this case 'x' ... is located at the blue arrow.. it is the centre of the panel, it is half way between 2 frames and half way between longitudinals...as defined above.

That's it.

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### TANSLSenior Member

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### bajansailorMarine Surveyor

Just a thought - for a 5 metre long boat, if you have 6 mm thick bottom plating you probably do not need a lot of stiffening?
Do the ISO rules become more 'pessimistic' (for want of a better word) for smaller boats?

@Ardi do you have any sketches of the 5 m. boat mentioned above?
What is the intended service speed of this craft?

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### TANSLSenior Member

No structure, 5 m or 500 m long, needs "a lot of stiffening", it only needs the necessary reinforcements, not one more, not one less.
ISO cannot change character, it is neither pessimistic nor optimistic. The formulas are the same for any type of boat.
I know that this is a way of speaking but it can lead to conclusions that are not correct, for example, that the ISO makes small boats super reinforced but that does not depend on the ISO but on the designer.

bajansailor likes this.

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### ArdiJunior Member

Thank you TANSL, I understand your explanation. Honestly, this is also one of the definition concepts that I have ever thought about and it's true

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