# Some questions about ISO 12215-5: 2019

Discussion in 'Class Societies' started by TANSL, May 7, 2021.

1. Joined: Jul 2003
Posts: 3,028
Likes: 318, Points: 93, Legacy Rep: 1306
Location: Thailand

### AlikSenior Member

In 2008 the approach is correct. Shear strength is increasing with length.
In 2019 the approach is opposite...

2. Joined: Oct 2008
Posts: 7,292
Likes: 1,180, Points: 113, Legacy Rep: 2488
Location: Japan

I understand that, and your comment.
But I don't have the 2019 rules to check the definitions and if there are any caveats, as previously noted.

Since the formula you wrote simply means if L is less than 5.8m, it defaults to the minimum of 0.3MPa.
But if the max L is 24m, there is then a max up to 0.58MPa.

Thus there must be a means for interpolation between the 5.8m to 24m? for the min of 0.30 to max of 0.58MPa.

The previous one was
below 10m, = 0.25 - so this is an increase.
Then 10 - 15m a sliding scale from 0.25 - 0.40, by interpolation,
Then 15 - 24m = 0.4MPa.

So, other than a means for interpolation between 5.8 to 24m, it appears consistent., with the lower and upper values greater than the 2008 version.

3. Joined: Jul 2003
Posts: 3,028
Likes: 318, Points: 93, Legacy Rep: 1306
Location: Thailand

### AlikSenior Member

Understood, thnaks, but interpolation should have physical sense...

TANSL likes this.
4. Joined: Jan 2005
Posts: 2,542
Likes: 474, Points: 83, Legacy Rep: 1110
Location: Philippines

### rxcompositeSenior Member

It is table 13 in my 2008 version. Minimum core shear according to length. The values are;
Less than 10m= 0.25
10 to 15m = 0.25 to 0.40 Formula 0.25+ .03*(LH-10). max at 0.4 for 15m.
15 to 24m =0.4

Finding the extrapolation formula for 15 to 24m. with an upper value of 0.58, I used problem solver in Excel.

1. Readjusting, it should read 16 to 24 m and have 9 division each for every meter.
2. Looking at the first formula, it is a linear progression. 0.03*(LH-10)
3. The first term is an offset.
Thus it is;

0.28 + 0.02*(LH-9)

for values of 16 to 24 m max at 0.58. Works for a length and a fraction.

5. Joined: Jul 2003
Posts: 3,028
Likes: 318, Points: 93, Legacy Rep: 1306
Location: Thailand

### AlikSenior Member

Yes this is the key: in 2008 option LH is with plus, in 2019 is with minus. I believe formula from 2019 is faulty.

6. Joined: Jul 2003
Posts: 3,028
Likes: 318, Points: 93, Legacy Rep: 1306
Location: Thailand

### AlikSenior Member

To add, I also believe that linking compressive strength of core with base pressure is nonsense.
What is the relation of bottom pressure to the core used for deck and cabin?

Say, for high speed boats, one won't be able to use DIAB H80 core in the deck, becasue it does not pass this criteria. Most of classification societies just specify min properties of the core not linking it to pressure.

7. Joined: Jan 2005
Posts: 2,542
Likes: 474, Points: 83, Legacy Rep: 1110
Location: Philippines

### rxcompositeSenior Member

How was it worded?

8. Joined: Jul 2003
Posts: 3,028
Likes: 318, Points: 93, Legacy Rep: 1306
Location: Thailand

### AlikSenior Member

This is in Table A.7, right column.

9. Joined: Jan 2005
Posts: 2,542
Likes: 474, Points: 83, Legacy Rep: 1110
Location: Philippines

### rxcompositeSenior Member

Mine goes only up to table A.2. Can you post a snapshot? Maybe I have a different tag for the rule.

Last edited: Mar 23, 2022
10. Joined: Oct 2008
Posts: 7,292
Likes: 1,180, Points: 113, Legacy Rep: 2488
Location: Japan

RX
Alik is referring to 2019 version, not the 2008 version which we have.
So unless someone can quote the rules exactly as written from 2019 edition, or just copy it... we have no visibility of table A.7 and the words/formula used.

11. Joined: Jan 2005
Posts: 2,542
Likes: 474, Points: 83, Legacy Rep: 1110
Location: Philippines

### rxcompositeSenior Member

Alik,

I ran the calculations several times and all came out to be the same. The formula we discussed seems to be applicable only to bottom pressure. In ISO 2008, there is a formula for motorcraft deck pressure PDM eq 15 of which you need to define first PDM Base eq 17. PDM Base is dependent on Lwl, which is a fraction of LH. For a 24 meter displacement boat, with my given parameters, I get 6 kN/m2 of pressure. Just slightly above the 5 kN/m2 PD MIN eq. 16.

You are right, compression property of core is not needed but pressure is necessary.

ISO does not have a tabulated table for cored laminate. I used Table H2 ply calculation as a guide. Given only pressure and b short dimension and the formula Design shear force = pb^10^-4, I get 0.15 N/mm2. With only p (6kN/m2) and b(500mm) given, I validated this formula using the standard mechanical engineering formula for finding shear stress Max shear = (V*Ib) *b(tc/2)*y where:

V= Vertical shear (750 N)

I= second moment of Inertia (41667 mm4)

b= width of core (500mm)

tc= thickness of core (10mm)

y= distance to N.A. (Zcrit) from centroid of upper core. (2.5 mm)

This gave me 0.23 N/mm2 max shear. Note that this is max shear. The formula can be simplified to Max shear = (3/2) V/bt. Average shear is V/bt which is 0.15 N/mm2, the same as ISO result.

Validating further, using the same inputs, LR gave me a core shear of 0.15 N/mm2 with a slightly different formula, Shear = (pb /2tc) x10^-3. Any of the formula is valid.

Knowing the thickness for the required shear, I can now choose among the core candidates from balsa, PVC, SAN, ect. With 0.15 N/mm2 of design shear and 10 mm core thickness, you have plenty of choices.

If I don’t like the 10mm thickness, or if the size is not available, I choose a different thickness and recalculate.

Ad Hoc and TANSL like this.
12. Joined: Jul 2021
Posts: 54
Likes: 13, Points: 8
Location: La Rochelle (Fr)

### Alan CattelliotJunior Member

There is indeed a ERROR that has been recognized by G.Dolto, in a communication to a few selected users (!!!! sic) .
If τ = b - a*LWL, then :

ISO12215:2019 gives a = 0.12 ; b = 0.7 , that led to an inconsistency with the variation of the minimum design shear strenght with the boat length, as pointed out by Alik (see the blue curve in the image below). This can also be seen when making the comparison between the ISO12215:2019 and ISO12215:2008 requirements (see the green curve in the image below).

The correct values for a and b are :
a = -0.07 ; b = -0.12
These values restore the correct variation of τ with the boat lenght, and keeps the initial purpose of the standard, which is to raise the minimum design shear strenght required for cores (see the yellow curve in the image below).

As you can see, this minimum shear strenght that should be used with the 2019 standard is significantly higher than in the 2008 standard. For 6meters to 10meters boats, the application of the new standard requirement can potentially force a modification of the core of sandwich FRP products. It should also be noted, that, for boats whose lenghts are greater than 10meters, the use of PET cores can be an issue, as their mechanical properties are very low. Indeed, there is the possibility, for boats already in production under the 2008 standard, of them not being able to met the 2019 standard. Complete calculations should be perform, as the final mimimum value will not only depends on the boat's lenght and her weight, but also on the design method used (ISO 12215:2019 table 16 for kAM !!!!!! sic ).

Here is an example down below where :
(D) Displacement power boat
(P) Planning power boat
(V) Sailing boat
(■ ) ISO 12215:2008 & ISO 12215:2019 qualified
(□ ) ISO 12215:2008 qualified
(-) non-qualified ISO 12215

Airex T90.150
│ Matiere : Polyethylene Terephthalate (PET)│ Masse volumique ρ : 145.0kg/m3│Elongation à la rupture : 0.08│ Resistance en cisaillement plan, à la rupture : 1.2MPa│Resistance en compression normale, à la rupture : 2.2MPa│

METHOD 4 (ISO12215 11.5 MATERIAL TESTING)
The Airex T90.150 material was EN-ISO-12215:2008 qualified for the production of sandwich laminates located at the bottom of the hull of a boat with a waterline length of 17.57m and a total load displacement of 36661kg. Since July 1, 2021, the material has maintained its EN-ISO-12215:2018 qualification.

D P V
Catégorie A ■ ■ ■
Catégorie B ■ ■ ■
Catégorie C ■ ■ ■
Catégorie D ■ ■ ■

METHOD 1 (ISO12215 11.2 & Annex A SIMPLIFIED METHOD)
The Airex T90.150 material was EN-ISO-12215:2008 qualified for the production of sandwich laminates located at the bottom of the hull of a boat with a waterline length of 17.57m and a total load displacement of 36661kg. Since July 1, 2021, the material loses its EN-ISO-12215:2018 qualification.

D P V
Catégorie A □ □ □
Catégorie B □ □ □
Catégorie C □ □ □
Catégorie D □ □ □

Quit funny, isn't it ?...... Of course, this is irony, because I don't think that :
- a design method coefficient is to be used to determine the minimum required shear strenght of cores. This is also the point of view of G.Dolto, but as the standard has been already published with errors, we are bound to find acceptable interpretations ( and make them validated by authorities) until the nex revision.
- the increase in requirements is justified, especially because it has for consequences a double penalty for PET cores, which cannot be used, according to the new standard, for a important numbers of FRP boats. To my knowledge, there is no reasons from the accidentology that a properly used PET core could endanger the structural integrity of the hull. On top of that, regulations will ban the use of PVC for environmental care in a few years, if not still in application. In the futur, will sandwich FRP cores be made out of balsa only ?????

#### Attached Files:

File size:
52.8 KB
Views:
14
File size:
11.2 KB
Views:
13
Last edited: Jun 27, 2022
13. Joined: Jul 2003
Posts: 3,028
Likes: 318, Points: 93, Legacy Rep: 1306
Location: Thailand

### AlikSenior Member

Thanks, and not only this.
The 2019 standard works with msmv values from core manufacturer specs, but the 2008 is not? This means, the shear strength requirements are even higher than Your graph shows.

14. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

When a project is presented to the entities in charge of its validation, it could happen that the well-intentioned designer corrects all those errors that the 12215-5:2019 standard seems to have and presents results that do not coincide with those obtained from the ISO, which has not been officially corrected. The results do not coincide with those derived from the standard, which is the official document. Will they be approved or do we have to present the "official" results even knowing that they are not correct?

15. Joined: Oct 2008
Posts: 7,292
Likes: 1,180, Points: 113, Legacy Rep: 2488
Location: Japan