ISO 1225-5-2019 stiffener calculation

What table H.7 is saying seems to be totally correct (without the need to consult Table H.6).

 

Tansl- I worked it out using Table H4 and compared it wit LR method. Both gave me the same answer as the method of calculation is the same.

Seems the N.A. in the worked example is wrong

 

Rx,
Columns 10 and 11 of Table H.7 are explained with the figure that I have published, however, that does not mean that the neutral axis is correctly calculated. Another issue to consider is whether the reinforcement is constructed with a female mold or an male mold.
I don't understand what the ISO Appendix H does in its tables. That's why I use my own tables in my calculations. That allows me to calculate the critical sigma in each layer, assuming that there are several layers in crown, web, bonding flanges, and attached plate.

 

I did not program table H7 as I have some thing that works the same.
Basically, It is table H1 and a stiffener. I do plate analysis first and if it works, I add the stiffener. which is basically what H7 is.

H7 consolidates the modulus of the web and the plate which is the sum of E x t/total thickness. The ROM method.

In the long method, the E of the individual ply is fed to the excel. Quite tedious inputting the data but the computer is so fast it does not mind. Note the 12.7 Gpa in the plate modulus input. Too complex to disect. I have to backtrack and see if the web is also a combo of several different fabric. Turned out it is biax with 60% fiber content as the table shows.

 

Totally agree.

On the other hand, I have not been able to reproduce exactly the same results that appear in the Aneo H Tables. Among other reasons, because we do not know any of the input data used there. Therefore, once I understand the procedure followed by the standard, I develop my own numbers, following said procedures. It is clear that H tables do not serve to validate one's own results. You have to copy the procedure, understanding well what is being done, applying it correctly and forgetting the outputs of those tables.

 

Is the plate + stiffener being tested in the long side or short side?
From the input shown, it seems to be tested on the short side (clue is the effective width of the stiffener not the span).

 

ISO 12215-5:2019, for the design bending moment of a reinforcement, only gives one formula (Table A.8). It does not exist, as occurs in the case of a panel, an Mdb and an Mdl (Table A.4)
In general I have not investigated, I have never attempted to, where the formulas used by the ISO come from (on very few occasions I have done so). Therefore, the only thing I can do is, in light of my limited knowledge and what I extract from the norm, assume what it is doing. I assume, then, that a beam is studied with its two ends embedded, subjected to bending due to a uniform load. A cross section is cut in that beam, which is the one being studied. (Pure theory of long beams subjected to simple bending)

 

Thanks. Will investigate later when I have the time. See if H7 has some typo in it or is working correctly.

Seems the OP has lost interest also.

 

I think the most practical thing would be to forget the tables in Annex H and each person proceed using their own calculations. After all, before the ISO, designers were already doing the analysis layer by layer. This standard does nothing more than introduce doubts and insecurity into a procedure that already existed previously.

 

I found the error last night. Like I said in the first post, it is like “putting the cart before the horse”. The N.A. is a computed value, dividing the regions above and the regions below (region of tension and region of compression), not an input.

Knowing where to split can only be done when the NA is computed. Excel reads from left to right, top to bottom and that input data should be placed somewhere else. H7 is prone to error because of its arrangement.

Because the NA would not always fall in the middle of the complex composite area, the standard engineering formula is used. This is what ISO has been doing in other worked example.

The N/A. would not always fall in the middle. In the case of multiple laminate layers, it would fall somewhere in the middle of a laminate. Could be a web or a single skin laminate, requiring imaginarily split in two. Seems trivial computing fractions of a millimeter but what makes the problem complex is that it separates the regions of tension/compression. Even by ISO standards (fiber properties) the Elastic Modulus of the fiber laminate, there can be as much as 20% difference between tension and compression properties.

The inputs in H7 worked example does not differentiate if it is in tension or compression and would distort the answer. Even my own answer of 30.2 mm. There should be two columns and should be located somewhere down below.

So “WHERE TO SPLIT”? It can only be done by back propagation in an iterative process AFTER THE N.A. is computed. My detailed spreadsheet does that, still I have to iterate. My software does that in the blink of an eye and is accurate.

 

I don't know if the results shown in Table H.7 are correct or not because I got tired of trying to check them. The procedure described by the ISO is :
- NA is calculated, not as the barycenter of a geometric figure but as the geometric locus of the points of the slice at which the deformation is zero.
- Once the position of the NA is known, the distance from the farthest face of each layer to the NA is calculated; the critical distance "Ycrit". (iterative process AFTER THE N.A. is computed, of course)
- It is clearly distinguished if "Ycrit" is positive or negative (column 27).
- According to the above, the real stress at the critical points of each layer is compared with its design sigma, in tension or compression, depending on its position with respect to the NA. The inner layers work under compression and the outer layers work under tension.
I am not going to defend ISO 12215-5:2019 in general or its tables in Annex H but, in my opinion, the process followed is correct.

 

It is correct as far as "H lower web plating" 19,9 is concerned (column 14). I have the same answer. If it is correct then the NA 30.2 is correct.
If 19.9 (height of web below) then height of web above is 96.6-19.9=76.7.
It is in column 14 Z neutral axis where the answer is 23.2 instead of 30.2. figures started to get wrong in distance of Gply i from internal plating. Table H4 does it correctly.
I am just trying to troubleshoot what went wrong or what cell is corrupted.