# DNV-GL Design Pressure And ISO-12215 Online Calc Mismatch

Discussion in 'Boat Design' started by zstine, Mar 2, 2021.

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### zstineJunior Member

Ahh, I understand... I'm still working the load pressure analysis (chapter 3) and have yet to get into the structural analysis (chapter 5) to which the effective breadth and 2nd moment applies. Hence my confusion.

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That is not correct.

So, let's consider a continuous beam..and has supports along its length and has a UDL applied:

The beam will deflect and bend...but, the deflection is not a simple curve between east support as shown above.

The slope at the ends is influenced by the support and deflects as so:

The beam is no longer having its location of zero deflection at the supports, it is at some distance away from the supports.

So the distance between the location of zero deflection = the effect breadth.
This is different from the panel breadth.

Thus, when calculating the second moment of inertia of the structure, the contribution the plating has is not B, it is Beff.

If the supports are moved closer and closer, the Beff for each stiffener overlap..and then default to B.

But in general the Beff - which is the usual case for most structural arrangements, is not the same as B.

Thus Beff - the effective breadth - is not the same as the panel breadth, as the panel breadth is merely to calculate the area for the applied pressure.
Two very different terms.

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### rxcompositeSenior Member

I never said that.

Effective width of plating

LR- (0.5b +10t)*2

ISO- 20t+70

ABS- b+ 18t

Where:

b=base of laminate

t= thickness of plate

And for DNV,

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### rxcompositeSenior Member

You need to brush up on terminology. Keep reading.
The width of the panel is as my illustration. That plus the length (spacing of bulkheads) is the area to be analyzed.

The width is to be subdivided by longitudinal stiffeners which has been the long discourse in the thread. The divided portion is the span. So it is the same as what you are saying. Terms can be used interchangeably. When it is already subdivided it is span but can also be called width. Width can also be the unsupported width.

Don't worry about "effective width" as I have explained above to AH. It is something that is quietly computed in the tabular analysis of LR or ISO and is one of the inputs. Unfortunately you don't have ISO, only Vectorlam.

Do not try to learn also the Classical Lamination Theory/Matrix method/ADB illustrated in DNV GL. It is the most difficult of all methods and you will go crazy learning it.

We are all getting confused here. For my part, I will refer now only to DNV GL rule. I have my copy and I myself is getting confused. Where and what chapter did you get that safety factor of 4? I cannot find it. In part 3 Chapter 4 Section 4 it already explained how to obtain the allowable stress. It is 0.3 of Ultimate stress (Safety Factor of 3). Will post later that topic.

Continue deriving the design pressure for the bottom and sides the DNV way. I already have finished the analysis, close as it is not DNV way but have completed the bottom shell thickness, single skin and cored, keel plate, Girder, transverses, and longitudinals. I only need what you come up with pressures.

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### zstineJunior Member

Thank RXcomposite. as far as the FS of 4. It is from the 2003 spec for boats <24m. here.. http://rules.dnvgl.com/docs/pdf/gl/maritimerules2016July/gl_i-3-3_e.pdf That spec if pretty generic and easy to follow, but doesn't identify the method to analyze sandwich construction, just a FS. I posted an excerpt from that earlier showing the FS... I found the 'displacement' pressure to be 11.6 kPA. It does discuss analysis of stiffeners and effective width, etc. in the solid FRP section, which may be applicable.

I have (mostly) completed working through design pressure per the 2019 GNV GL "Yacht" spec, pt3, ch3, Though it is important to note that this spec is not specifically applicable to boats <24m and I believe the above 2003 spec is the most current applicable spec to boats <24m.... Anyway, I found sec 2.1 Sea Pressure = 32.7 kPA for Panels and 38.8 kPa for stiffeners, and Sec 2.2 slamming pressure = 44.5 kPa.. these are for worst-case, eg forward near the bow with lowest applicable deadrise angles. Sec 2.3 Weather deck = 12.5 kPa ( i think 12.5 kPa is not high enough to walk on though).. I have not done the Section 3.2.2 global bending, partly b/c the spec says "the pressure shall be applied across the total area designated slamming area (from bow to mid-section)". I'm not sure if I should use the area below the WL or if I should include something like a 20deg bow down angle (sorta like what is discussed in the multi hull section) and include an area above the WL.
I did start to read through the pt3, ch5 composite scantlings spec. You are right that it is difficult, but I will try to figure it out before just defaulting to vectorlam on-line calcs.

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Hmmm... it seems another misconception is going on here.

You did, as your diagram suggested this:

You are showing an "over lap" of 2 structural members effective breadth.
This is incorrect.

One can only do that by referring to the structural theory of what is effective breadth...which I eluded to in my previous post.
Class rules do not dictate what is structural theory, it is the other way around. Class rules follow structural theory(ies)

So, let's go back the the uninformed continuous structural member.... so we typically have a panel of plating that is supported by stiffeners - shown as simple T sections below, what is subjected to a typical UDL load:

The breadth of plating to use, for the 2nd moment of inertia calculation to ascertain the stiffness and ergo the response owing to the UDL has traditionally be B, where B is defined as the mid span distance between each stiffener. Where the stress in the plating is uniform:

But this has shown to be not correct, as the behaviour of the plating suggests otherwise.
This lead to the understanding of what is known as - Shear Lag Effect. Where the distribution of stress is not uniform in the plating. As it is influenced by the stiffness of the support, which results in the slope and defection of the plating behaving differently from classical theory and thus the stress distribution is non-linear:

It varies across the plating.

A formulae was then derived to provide a solution to this effect:

All of the above can be found in any text book on structural theory/design - this is nothing new.
Thus the Beff - effective breadth - is not the same as B.

In metal hull structures there are many rules of thumb to take this into account, or when using Class rules, they tend to offer their own take on it..and give limiting factors when calculating the I for the structural member. Either be a simple Beff formula or a one liner limit, such as 50t, where t = web thickness of stiffener.
Again this is nothing new.

The only issue is that the above assumes an isotropic material and a linear stress distribution across the thickness of the plating.

And this is the point.
Composites are not isotropic and when analysing composites using classical laminate theory it also assumes perfect adhesion between piles - and yet exhibit varying through thickness stress.
This renders the now classical Beff formula non valid.

Thus each Class set of rules take this into account with their own take - just as in the case of isotropic materials - of the Beff.
They all vary and offer different solutions, like the ones you noted.

However, where the stiffeners are placed then closer and closer to together...as per your sketch, the Beff becomes B as the shear lag effective is now greatly influenced by the much stiffer structural section compared to the less stiff section when they are farther apart...ergo when they are placed very close to each other the Beff does not over lap.

This is true whether it is an isotropic material - metals - or composite.

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### rxcompositeSenior Member

Good. You can start by using the minimum mass reinforcement for keel, bottom, sides by DNV.
Calculate stresses using the pressure you derived. Start by using 4 ft for length and width as measured. Using 4 ft, I find the minimum reinforcement is inadequate. Try also 2 ft length and tell me what you get. Use single skin for bottom, 20 mm core H80 above waterline. Use a safety factor of 3.

I added a stiffener to the chine. 151 degree is too close for comfort as the second station will be greater than 151 degree.

Next would be the girders, transverses and stiffeners.

I can only check the accuracy of the software you are using. I cannot guarantee compliance to DNV GL. It is up to you to check. It has been a long time since I designed to DNV.

What is the width of the side plate?

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### zstineJunior Member

I did a read through about half the pt 3 ch5 composite scantlings spec, and you were right that all the matrix math and translating angles to global coordinates, etc. is a bit much. I did see some interesting things, like an effective span factor for a chine angle. So needing a stiffener is not a go/no-go at 150 degrees. You can calculate an added span factor which increases as the angle gets smaller (section 6, figure 5). I did attempt to run the reaction bending moment and reaction shear using representative numbers, but the results don’t look correct. So I put that down for a little and played with the on-line vectorlam panel calculator.

Vectorlam doesn’t allow you to put theoretical laminates of ‘minimum mass reinforcement’. Instead you have to select ‘real’ products to build from. I used E-LT-1800, which is 0/90 stitched fabric 18oz (610g) and is about .025in (0.63mm) thick per layer. I ran several panels with combinations of 2 or 3 layers with both 1/2in (13mm) H80 core and .75in (19mm) H80 core. This was a 48in x 12in, fixed edge, panel with 6.5psi (44.8kPa) load and 1.5% allowable deflection. Results attached show FS on the bottom. The results show that 2 layers (1.3mm thick) on 19mm core is adequate FS > 3, though the outer skin is below minimum thickness. The core shear FS is 2.67, slightly less than the FS 3 per spec. Interestingly, the skin thickness had no impact on core shear despite less panel deflection at applied load… hmmm. Core shear stress appears to be a function of the core thickness alone, which doesn’t seem correct.

As for the width of the side plate you requested. The plank widths at midship from the keel up are as follows:
Keel…
Bottom Plank: 9.4in (239mm)
Bottom/side: 10.7in (272mm)
Above WL plank: 9.2in (234mm)
Below Chine: 9.4 (239mm)
Above Chine: 11in (279mm)
Deck joint…

So the 2 planks in blue labeled “Side Plate” are 9.2in + 9.4.in = 18.6in (472mm).
I also looked at the plank width at 20ft from AP, or x=0.77. widths below...

Bottom Plank: 9.0in (229mm)
Bottom/side: 9.0in (229mm)
Above WL plank: 9.2in (234mm)
Below Chine: 10.4in (264mm)
Above Chine: 10.3in (262mm)

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Why would it?

The core carries the shear load, the skins carry the tensile/compressive loads.

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### rxcompositeSenior Member

The CLT is so complex you have to feed at least 5 material property inputs and at least 3 dimensions.

Add to it the 6 stiffness matrix. You need some skills in Excel as the ADB inversion is not exactly straightforward. But it has the function formula.

Since I have the software I quit expanding it. No use reinventing the wheel. The software covers the whole gamut of the CFD analysis.

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### rxcompositeSenior Member

What is ELT 1800?

My database says 1800 is a knitted fabric with a 10 oz mat. Any link to the supplier/manufacturer?

900 grams/m2 is about 1 mm at 50% glass content. Rules usually specify the mass of reinforcement required, not thickness. Thickness will vary depending on glass content and type of fabric.

In order for me to compare your software from the standard, I will need the typical material properties of the fabric used as a laminate. Real product as you say.
Tensile strength
Compressive strength
Tensile modulus
Compressive modulus
Shear modulus

Your software should have the minimum/average data otherwise it needs to be derived using LR or ISO standards. It is not given in DNV. It says must be read from coupon samples. Without this, your software would be guessing.

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### zstineJunior Member

Here is the material specs...

excerpt... surprising the difference infusing makes!

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### zstineJunior Member

If I imagine the extremes of inch thick skins, then the load would just compress the core with ~no deflection and the core wouldn't have the same shear as it would with thin flexible skins...? I don't know, but perhaps the notion that all shear is taken by the core and only tensile/compression loads by the skin is an idealization or approximation that was made when deriving the analysis of sandwich construction that is not necessarily exactly representative or real life. I just think of the laminate panel as a system and if you make the skins stiffer/stronger, I would think core stress would be reduced... But this was just my gut feel, which is often perfectly wrong.

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Sorry, but you don't need to "imagine".
This is standard structural theory and your reply suggests that you have yet to grasp an understanding of structural theory and design.

So, take your example...skins of 25mm (1 inch) thick...and lets say, they are 1mm apart...so a total thickness of 51mm.
And again, fro consistency, let's' say that the width of this section is 300mm wide.

The modulus is 130cm^3 and the Inertia is 331 cm^4
If the skins are now 50mm apart, the modulus is 444 cm^3 and Inertia is 2244cm^4
If the skins are 100mm apart, the modulus is 799cm^3 and the Inertia is 6031cm^4
and so on..

Stiffness, the second moment of inertia i based around the depth of the section.

So, if you selected the first option...this would suggest a lack of understanding of the purpose of a sandwich structure. Since having 2 thick skins that are 1mm apart begs the question, why not make it just 50mm, i.e no core....ergo, is is a one structural member. This being the case, it would follow any typical solid member under an applied load:

The skins would be taking the shear load as well as the tensile and compress loads.

But the weight would be considerable.

The stiffness, inertia, is a function of the depth, simply I = db^3/12. The D = depth..ergo the greater the depth the greater the stiffness.

If you take a Flat Bar of 50x6mm dimensions and have it placed flat side down, the modulus is 0.3cm^, whereas just rotate it by 90derees, so the thin 6mm is flat side down ie. it is now 50mm deep the modulus increases to 2.5cm^3.
Doesn't sound much, but this is an increase of more than 4 times, simple by adding depth...and yet no additional weight!!

So, the purpose of a sandwich structure is what??.... to make it as heavy as possible, or, to make it as light as possible?

Increasing the depth of the section increases the stiffness = ability to take a higher load = lighter weight.

So, back to the shear.

If the skins are now separated what occurs?
So any member under and applied will bend, as shown below:

Here separating two skins provides the stiffness, the I = ah^2.

But if the skins are separated by air, what occurs:

Under the applied load, the skin in contact with the applied load will simple bend as shown. The structure will not behave as "one" member.
Why?
Because there is no shear path between the upper and lower skin.

If you replace the air with a core, you have this:

And it behaves just as any idealised structure in metal, in this case, and I-beam, like so:

The web, takes the shear load...or in the case of a composite, the core.

So, back to the examples of 1, 50 and 100mm separated skins. What is the area between the 2 skins:

1 - 300x1= 300mm^2
2 - 300x50 = 15,000mm^2
3 - 300x100 = 30,000mm^2.

Given that a solid section like the example shown above, the shear stress of the core = 3.F/2A

Or if we assume F is a constant it becomes = 3/2A

So in 1 = 3/(2x300) = 0.5 (in cm)
in 2 = 0.01
in 3 = 0.005

So the magnitude of shear stress is reducing considerably, owing to the amount of "core", the material that separates the 2 skins apart.
Whereas the area of the skins, remains the same.

The only driving factor to consider for a core is i) the maximums shear stress of the material being used, and ii) its compressive strength.

No.
Look at any test book on structural design and theory.

The suggestion that anyone would like this statement underscores their own lack of structural design knowledge and understanding of structural analysis.

I would suggest that before you try to plug numbers into a set of Class rules, try to understand the basics of bending , deflection and shear of stress of structures - just read any text book. Then you begin to understand the how's and why's of structural design, and why your statement is 100% incorrect.

For reference shear stress of different structural members:

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### zstineJunior Member

Ad Hoc, I was questioning the assumptions made to derive & use the equations which you assume I don't understand and just regurgitated above. Simply saying I don't understand because this is what the functions are when I'm questioning the validity of said functions does nothing to support your argument. Assuming 100% of shear stress is taken by the coring and and 100% bending stress (tension/comprsession) by the skins is an idealization to make analysis easier, 100% for sure! See Cambridge University lecture notes..

What I was imagining.... You need to think beyond the text book examples.....

So in my previous post, I was just wondering about the validity of the Vectorlam analysis showing that skin thickness has zero impact on shear in the core. Now that I've done a little more research, I'm 100% convinced that the online Vectorlam analysis makes an assumption which neglects the Skin's contribution to shear stress, as is standard practice in composite theory. While that is likely a good approximation for the panel geometry that I'm interested in... IT IS AN APPROXIMATION. So it does appear that thicker skins will reduce shear stress in the core, though it may be by a negligible amount... Think of it from a conservation of energy perspective. If a force is applied to deflect a laminated beam, it has X energy associated with that work. If the skin makes a contribution 'Y' to absorb shear force energy (as can be seen by the Cambridge notes) then it must be reducing the shear force energy that is absorbed by the core, Z = X - Y. conservation of energy.

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