Transverse frame calculation

[Ad Hoc]

Oh dear….this is just utter nonsense.

A roman arch is made up of small voussoir’s each with tapered ends. The weight of each causes a thrust against this tapered face. The adjacent face has the same and thus experiences and equal and opposite reaction. They hold each other up.

So, there is no tension or compression on the inner or outer part of the curve. The only forces are on the haunches, or facets of each voussoir against each other. This is a direct in-plane load. The compression strength of each, is the limiting factor. Since the force is pushing trying to compress the other stone.

The whole arrangement must be structurally balanced to it to be in static equilibrium. Thus the total sum of the reaction loads, the thrust shown in the diagram and the vertical weight needs to be balanced, otherwise, the springer (lowest voussoir) will slowly push out against friction, under a load.

This is countered by what is called a Buttress. This supports the whole arch arrangement. Otherwise they will fall down, go look at any church. At the end of any church or bridge with an arch, there exists a vertical structural member that takes this reaction load, or thrust. Often surrounding the entire arch too.

Now, looking at a plate in bending, the upper surface is curved and stretched beyond its natural position ie in tension and the inner surface is compressed, as shown by the diagram. Each element of plate has direct in-plane loads, but also tensile and compressive loads owing to the neutral axis when the plate is bent. The arch does not. The arch does not have a neutral axis, because the voussoir’s are not being bent into shape.

In a hull that is made of bent or curved plate, we are told (by Brent) that the edges are supported by the chines/decks etc. So, what is occurring at these joints???

As with the roman arch, to hold that shape of the plate, a thrust is created which needs to be applied in equal and opposite direction…otherwise the plate will just spring back to is unloaded dimension.

Originally Posted by Brent Swain

If you use slightly tapered bricks to build a Roman arch, it stays in place, and is extremely resistant to any weight put on it. Even if you don't use any mortar, it still stays in place.

Wrong, a roman arch needs a Buttress to stay in place.

Thus with the roman arch a buttress is used to carry these loads….in Brent’s case the chines/deck is being used. So what happens to these loads?? This is shown in the final figure. That thrust or load that is used to bend or 'hold' the plate is now being used to push against another plate. The 'other' plate that butts into it has its joint (or buttress) at the deck or chine or whever Brent selects it. So, this other plate is now experiencing an in-plane edge load, shown by the arrows. When applying a load, in-plane, to an already bent/curved plate, what happens…does it resist easily, or does it bend more easily?...what is preventing that curved plate from buckling??...when in fact it is already pre-buckled, owing to is deformation or curve. It has little resistance.

For those not followed so far…read my post above for explanation.
For those that wish to continuously bury their heads in the sand and trust in faith and anecdotes because they are incapable of calculating buckling loads and even recognise their existance and hence the limiting strengths of panels of plating compared to stiffened ones, please do so.

So, do your sums and establish what loads your boat is going to experince, real loads, not only simple laterally applied loads to satisfy one laod case to justify one method of construction. As many load cases as you can imagine. Then design (calculate that is) a structure accordingly, and not guess or put your faith in woolly words and anecdotes that have no scientific value.

[LyndonJ]

Look at the threrad "Realistic scantilings" (click it's a link)

Look at post 25, others have tried to tell Brent this about arches in the past too !

Reading the entire thread 'Realistic scantlings" is quite illuminating about Brent and his logic and his tactics.

[RAraujo]

Sorry Ad Hoc, but I could not understand your statement

"When applying a load, in-plane, to an already bent/curved plate, what happens…does it resist easily, or does it bend more easily?...what is preventing that curved plate from buckling??...when in fact it is already pre-buckled, owing to is deformation or curve. It has little resistance."

As I see it will depend on which side you are applying the load. It seems that these boats have only convex surfaces.

I think that the way the structure behaves in these types of shells is much more complex due to the internal stresses induced by the construction method...

It cannot be even compared with a FRP curved hull (in which the shape has a very big contribuition to its strength). May be it can be compared to plywood when bent to form a curved pannel... or in cold moulded wood construction...

I would need to know about the construction method:
- From some photos I saw it seems there are some longitudinals (at least in the central part). Which are the aproximate scantlings and spacing?
- Which is the plate thickness (for example for a 12m~40ft boat)?

I've made some quick calculations considering that the mean longitudinal radius, for a 12m boat, would be around 15m. If the longitudinals are to big (high), in order to close the boat, they will deform plastically to a great extent. The crytical stress for buckling of these longitudinals (considering 2m span) will be above the elastic limit thus the elastic theory, for buckling, no longer applies.

But these were just quick calculations... I would need to know more about the method and think a bit about the model for the structure...

Are boats built this way known to collapse? If not, and if they are cheaper and easier to build, maybe it's worth studying the mechanics of the structure...

[sorenfdk]

Originally Posted by RAraujo

Sorry Ad Hoc, but I could not understand your statement

"When applying a load, in-plane, to an already bent/curved plate, what happens…does it resist easily, or does it bend more easily?...what is preventing that curved plate from buckling??...when in fact it is already pre-buckled, owing to is deformation or curve. It has little resistance."

As I see it will depend on which side you are applying the load. It seems that these boats have only convex surfaces.

Isn't an in-plane load applied to the edge(s) of a plate and not to the side(s)?

Brent Swain:
If only the plating of your origami hulls is thick enough, they could get a Lloyd's approval. Would that make them weaker?
Of course a steel hull can take more punding against rocks than a hull made of wood or composites, but that's because of the material. It has very little to do with the (number of) frames.
Compare two, otherwise equal, steel-hulled yachts - one origami and one conventional (with an adequate number of frames). Both built to withstand the same loads and to last the same number of years. Guess which one will be the heavier? The one with the origami hull! And as the two yachts are equal (same displacement), this means less payload.
If your skull had framing, it wouldn't be so thick!

Ad Hoc:
Thank you for your clear explanations. I hope (but also doubt!) that they will help!

[LyndonJ]

Originally Posted by sorenfdk

.....................If your skull had framing, it wouldn't be so thick!..............

[LyndonJ]

Originally Posted by Brent Swain

A three ton comealong can get quite heavily loaded in pulling the hull together.............!

But this is also to bend the plate past yield in the curved chine portions looking at construction photo's ? It's not the actual pre-stress on the plate longitudinal combo. That is what I wanted to know. If I had the panel sizes and the stiffener dimensions and the radius of curvature I can calculate the pre-stress easily enough.

Then we should look at some real loads. What static pressure head before it collapses (buckles) I reckon I can give you that pretty easily. Then we could compare that with a hull with a few decent transverses.

I don't think it's anywhere near as strong as you think and certainly not scalabe to 60 feet. even 40 might be pushing the boundaries depending on displacment and hull dimensions plating thickness and keel and rig design.

And yes I know it works for your 36 footers.

Originally Posted by sorenfdk

If your skull had framing, it wouldn't be so thick!

Probably that will become the most quoted here!

[hoytedow]

Dem bones, dem bones, dem dry bones,
Dem bones, dem bones, dem dry bones,
Dem bones, dem bones, dem dry bones,
Now shake dem skeleton bones!

The toe bone's connected to the foot bone,
The foot bone's connected to the ankle bone,
The ankle bone's connected to the leg bone,
Now shake dem skeleton bones!

The leg bone's connected to the knee bone,
The knee bone's connected to the thigh bone,
The thigh bone's connected to the hip bone,
Now shake dem skeleton bones!

Dem bones, dem bones, dem dry bones,
Dem bones, dem bones, dem dry bones,
Dem bones, dem bones, dem dry bones,
Now shake dem skeleton bones!

The hip bone's connected to the back bone
The back bone's connected to the neck bone,
The neck bone's connected to the head bone,
Now shake dem skeleton bones!

The finger bone's connected to the hand bone,
The hand bone's connected to the arm bone,
The arm bone's connected to the shoulder bone,
Now shake dem skeleton bones!

Dem bones, dem bones gonna walk around
Dem bones, dem bones, gonna walk around
Dem bones, dem bones, gonna walk around
Now shake dem skeleton bones!

[Brent Swain]

Originally Posted by Ad Hoc

Oh dear….this is just utter nonsense.

A roman arch is made up of small voussoir’s each with tapered ends. The weight of each causes a thrust against this tapered face. The adjacent face has the same and thus experiences and equal and opposite reaction. They hold each other up.

So, there is no tension or compression on the inner or outer part of the curve. The only forces are on the haunches, or facets of each voussoir against each other. This is a direct in-plane load. The compression strength of each, is the limiting factor. Since the force is pushing trying to compress the other stone.

The whole arrangement must be structurally balanced to it to be in static equilibrium. Thus the total sum of the reaction loads, the thrust shown in the diagram and the vertical weight needs to be balanced, otherwise, the springer (lowest voussoir) will slowly push out against friction, under a load.

This is countered by what is called a Buttress. This supports the whole arch arrangement. Otherwise they will fall down, go look at any church. At the end of any church or bridge with an arch, there exists a vertical structural member that takes this reaction load, or thrust. Often surrounding the entire arch too.

Now, looking at a plate in bending, the upper surface is curved and stretched beyond its natural position ie in tension and the inner surface is compressed, as shown by the diagram. Each element of plate has direct in-plane loads, but also tensile and compressive loads owing to the neutral axis when the plate is bent. The arch does not. The arch does not have a neutral axis, because the voussoir’s are not being bent into shape.

In a hull that is made of bent or curved plate, we are told (by Brent) that the edges are supported by the chines/decks etc. So, what is occurring at these joints???

As with the roman arch, to hold that shape of the plate, a thrust is created which needs to be applied in equal and opposite direction…otherwise the plate will just spring back to is unloaded dimension.



Wrong, a roman arch needs a Buttress to stay in place.

Thus with the roman arch a buttress is used to carry these loads….in Brent’s case the chines/deck is being used. So what happens to these loads?? This is shown in the final figure. That thrust or load that is used to bend or 'hold' the plate is now being used to push against another plate. The 'other' plate that butts into it has its joint (or buttress) at the deck or chine or whever Brent selects it. So, this other plate is now experiencing an in-plane edge load, shown by the arrows. When applying a load, in-plane, to an already bent/curved plate, what happens…does it resist easily, or does it bend more easily?...what is preventing that curved plate from buckling??...when in fact it is already pre-buckled, owing to is deformation or curve. It has little resistance.

For those not followed so far…read my post above for explanation.
For those that wish to continuously bury their heads in the sand and trust in faith and anecdotes because they are incapable of calculating buckling loads and even recognise their existance and hence the limiting strengths of panels of plating compared to stiffened ones, please do so.

So, do your sums and establish what loads your boat is going to experince, real loads, not only simple laterally applied loads to satisfy one laod case to justify one method of construction. As many load cases as you can imagine. Then design (calculate that is) a structure accordingly, and not guess or put your faith in woolly words and anecdotes that have no scientific value.

The first drawing has an arrow with the word "Thrust" on it. What is another word for thrust? Compression. My point exactly. The ends (buttress ) is not the flat edge of a pate, but the conic ends of the hull running into the stem and transom, a totally different ball of wax than running it flat into an edge.
When people give the structural figures for a straight piece of flat bar, without any curve, or without being in any way attached to anything ,and totally disregard the contribution in strength of those attachments and curves , they make it crystal clear they just don't get it. When I speak of the structural support of longitudinal curves, and they come back talking about flat sided freighters, they again make it crystal clear they haven't read what I have said , and thus will never get it.
I remember seeing a Tahiti ketch with 3,000 lbs of framing and ten gauge hull plate. That is the same weight as 3/8th plate with no framing. Which is least likely to get holed on a coral reef or a sharp rock? I remember seeing a power boat under construction with 1/8th plate and enough framing to equal the weight of 1/4 plate and no framing. She had enough transverse curve to make framing uneccessary

[LyndonJ]

Originally Posted by Brent Swain

The first drawing has an arrow with the word "Thrust" on it. What is another word for thrust? Compression. My point exactly..............

This is the compression that sends rockets, aircraft, misiles, boats and all on their merry way ?

[MikeJohns]

Well now it's up to 60 feet !

Originally Posted by Brent Swain

...........
When people give the structural figures for a straight piece of flat bar, without any curve, or without being in any way attached to anything ,and totally disregard the contribution in strength of those attachments and curves , they make it crystal clear they just don't get it. When I speak of the structural support of longitudinal curves, and they come back talking about flat sided freighters, they again make it crystal clear they haven't read what I have said , and thus will never get it...........

Brent
Everyone reads exactly what you say, they then disagree with it and get down into specific detail. At that point you start talking wildly.

I note you just had another offer to help you realise the real strength and stiffness of your panels.

I have offered in the past to analyse your hull strength and in response you attacked math and engineering .

I said before the reason your 36 foot hulls are reliable is very likely becasue they are actually compliant with class scantling rules.

However your structuaral arguments are wrong which can be very misleading.

My advice to anyone wanting to build larger vessels would be to build by origami by all means if that can produce the shape you want, but don't go with the minimal framing such as is present in the 35 -36 footers of Brents. You will need to add some transverses for sensible plate thickness.

[bhnautika]

Finally we start see some effort being put into understanding what is really going on in these structures, thankyou RAraujo for the time and curiosity and to you mike. This where I am a bit selfish I don’t care if Brent is interested or not (sorry Brent), I’ am! When I look at the hull shell I think of “dome” more than arch so maybe a definition of the structure might be helpful and the elements in it.
Sorenfdk I always thought bone did have framing only on the inside like a cored panel, thick yes but light and strong, and an elegant simple solution.

[Brent Swain]

Originally Posted by MikeJohns

Well now it's up to 60 feet !




Brent
Everyone reads exactly what you say, they then disagree with it and get down into specific detail. At that point you start talking wildly.

I note you just had another offer to help you realise the real strength and stiffness of your panels.

I have offered in the past to analyse your hull strength and in response you attacked math and engineering .

I said before the reason your 36 foot hulls are reliable is very likely becasue they are actually compliant with class scantling rules.

However your structuaral arguments are wrong which can be very misleading.

My advice to anyone wanting to build larger vessels would be to build by origami by all means if that can produce the shape you want, but don't go with the minimal framing such as is present in the 35 -36 footers of Brents. You will need to add some transverses for sensible plate thickness.

Build yourself a model. Then the structrural principles becomes self evident.

[MikeJohns]

Heres a T section beam in pre-arched pre-stressed then point loaded mid way with a moderate load . all well below failure.

I pre-stressed the beam as generously as possible before the equivalent of brents longintudinal would collapse from buckling.

The displacments from the load are scaled to show 5 times the actual displacement to help illustrate the concept.

The actual structure would be a lot more slender than my choice here but I wanted to show the points of contraflexure, where the stress reversal occurs.
Negative stress is compression, look for the neutral axis.

In reality with a more slender longitudinal the contraflecure points will be much closer to the beam ends.

It should be clear enough to show that the longitudinal frame becomes a tension member under the load and that it has no analogy to an arch at all.

[RAraujo]

Thank you Mike!

These ilustrations of yours were very elucidative.

Consider two structures like the ones you've shown but one of the structures is pre-stressed (exactly as shown) and the other with the same shape but without the pre-stress (plastically shaped).

Now which of them will be able to absorb more energy (with convexe side loading) before collapse?

I think we are heading the right way to understand what happens...

Thanks again.

Rodrigo