ISO 12215-5 2019 : Questions about H2 table

Hello Everyone.

I am trying to evaluate my laminate thanks to the iso 12215-5.

I am trying to implement the same table as the H.2 and I am not understanding how to get the value of the column 56.

If someone could help, I would really appreciate it.

 

(56) =(40) x (ZNA-(41))

 

Thank you very much Mr. Cattelliot,

Have a nice day.

 

Thanks for your interesting answer Mr. TANSL.

I will try to implement that later !


This was my first time working with the shear flow and the first moment of area. I found formulas but the E does not appears in it (see picture attached).

Do you know why we have an EIna in our case ?

Sorry in advance if it's obvious, I am just trying to understand and increase my knowledge.

 

The shear flow in the formula you give is exact, and can be applied directly to calculate the shear stress in parts made of the same material (same E).
In the ISO standard, an approximate formula is derived to parts made of different materials (different Ei for each ply i).
In that case, a weighted sum is used, up to the ply i, namely "SEi.ti(z-ZNA)" (col 56). For the shear stress to be calculated that way, this sum is then divided by the EINA (col 57), which is also a weighted sum. If all the plies were to be made of the same material, you can factor by E the (col 56), and simplify the division (col 57) by the E in EINA. You get then, an identical formula as the one you give.

It should be noted that the approach of the annex H makes sense when there is no coupling between flexion and torsion. For instance, you cannot use this approximate method to calculate the shear stress in the skin of a composite rudder blade. In that case, you have to introduce the shear modulus of the individual plies in the equation. Which is,sadly, not explained in the 12215-8.

In addition, I shall also say that there is a "twist" for shear stress calculation in the annex H. If you take a look at the way this shear stress is computed, you will notice that, in order to get a value as close as possible to the maximum shear stress, at ZNA, you have to split in two the ply by which passes this neutral axis.

 

To illustrate this, please take a look at this example. Panel composition is [QX900,CORE10mm,QX600], from the outside of the hull to the inside. We are interested in finding the maximum stress, occuring at fixations. At these locations, we expect to find the inside of the laminate in compression, and the outside of the laminate in tension, as shown in the picture below.

The studied laminate is not symetrical, which means that the neutral axis (NA) won't be in the middle of the laminate, but will be slightly offset outwards. This fact won't change the outcome of this demonstration, as one will understand in the end.
The shear stress (due to flexion only) is calculated, in the table H.2, at the plies interfaces. With the given composition, the neutral axis will be in the core, which means that one part of the core will be in tension, while the other part will be in compression. As demonstrated in mechanics, the maximum shear stress will occur exactly at this position, ie at the neutral axis. To get this value, using the calculation method of the table H.2, you have to split the core, in order to have a ply interface at the exact position of this neutral axis. In the picture below, each computed ply is represented by a cell. ZNA stands for "the altitude of the neutral axis" and Zcalc stands for "the altitude where the shear stress is calculated for the core".



To illustrate the effect of splitting the core in two parts, one below the neutral axis, and another above, please take a look at the following calculations. First, the calculation done with a single ply for the core. You can see that the value for ZNA is 4.99 ( col 14 ), that the core is computed as if it was entirely in compression (col 23). The value of the resulting maximum shear stress is given in (col 28), and is equal to 0.28MPa.



Then, let us see the same method of calculation, applied with a split core. The value of the ZNA stays the same. But now, as you can see in (col 23), the inner core ply is now in compression, while the outer in now in tension. Which is physically correct.
The value of the resulting maximum shear stress is increased (col 28), in comparison with the first calculation, and is now equal to 0.2988 MPa, more than 6% higher than the one computated earlier.



In this example, the difference between the two approaches could be negligible, because it does not change much the computed compliance factor (col 29). But when you work with laminates with very thick cores, this difference could be important, to a point where you cannot neglect it anymore. Especially if you're working on racing boats, with very optimized compositions. In practice, I recommend to take this fact into account, and always split the ply by which passes the neutral axis. That way, your calculations will be correct, and you will always be sure to compute the maximum shear stress value from this "H.2" method.

To do this,
(1) make an arbitrary split of the ply you estimate that will be at the neutral axis, in two equal parts (for instance).
(2) Check that the neutral axis passes indeed into one of the two parts (col 14, 3.05 < ZNA = 4.99 < 8.05). Note the value.
(3) Adjust the outer split in (col 12) such as the sum of thicknesses from the outer ply to the outer split (here col 12: 1.100+3.890) is equal to the ZNA (col 14, ZNA = 4.99 = 1.100+3.890)
(4) Adjust the inner split in (col 12) to get the desire thicknesses sum of the two split.
(5) Verify that the ZNA value (col 14) has not change, with respect to the value noted in 2.

(6) Like the post if you wish. Cheers,

 

Thank you very much for your clear and precious explanations ! It helps a lot to understand and I will apply the method you shared.

I was also wondering why the unit of the column 56 is in Nmm as the result of the formula should return a value in N if we analyse the units ? Maybe I am missing something here ?

 

Yes, you are right, there is an error in the table H.2. Column 56 gives, indeed, values in N. The unit of the values in column 57 is thus ins N.mm-2. The author has "forgotten", in the end that computations , in his whole table, were made for a unitary panel length...
Some of the many mistakes that make this standard quite hard to follow. I shall admit that I was one of the expert involved in the writing of this standard, and I feel sorry for these many mistakes.

I'm working as a naval engineer/surveyor. I can handle for you some full EC certification, but also give you a hand in your product developpement, from definition to registration. Feel free to contact me, as I'm here to support your project, but with the strong philosophy that there is no better help than the one that teaches you how to do things by yourself.

A modo that you certainly share : "Dig a well before you're thirsty".

Cheers,

 

Thanks for the reply, it makes more sense to me now, so because it is a unitary panel length you get rid off one length unit, ok.

I would be really interested in being helped in my development, In fact I am doing a project of a 7m long motorboat in my free time, I was a student from the ENSTA Bretagne in France.

Yes I fully agree with your modo indeed !