Need help with FEA of a rudder

[pamarine]

I have the following rudder design and am having trouble running deformation tests in SWSim2009.

I keep getting the message "Preconditioner Fails to factor" but am unable to find any information about that error code.

I have attached the necessary files below and was hoping someone would be able to assist.

The Rudder is made from 304 Stainless Steel, the shaft from Aquamet 22, and the Tiller from Maganese Bronze.

The bearings are just blocks of Mahogany with a standard brass Rudder Bearing inserted.

[Raggi_Thor]

If you google the error message you find some others with maybe similar problems?
http://ww3.cad.de/foren/ubb/Forum131/HTML/000481.shtml

[Raggi_Thor]

I tried to look at your attachment, but it seems to be an XT file embeded in a dwg?

[pamarine]

Saw that forum post, wasn't informative in the least:D

It's just a standard X_t file.

[Ad Hoc]

Perhaps explaining your problem a bit more may help?

Is it just simply a 'command' within the program you need assistance with, or an understanding of what the command does and is for?

[pamarine]

Perhaps explaining your problem a bit more may help?

Is it just simply a 'command' within the program you need assistance with, or an understanding of what the command does and is for?

Mainly what the error message is and how to fix it. Barring that, if someone else can run the analysis and simply tell me when and how it will fail.

[Ad Hoc]

I use cosmos, which was eventually embedded within SW. But I've not come across that command.

Is this for static, dynamic, linear or non-linear analysis?

[pamarine]

I use cosmos, which was eventually embedded within SW. But I've not come across that command.

Is this for static, dynamic, linear or non-linear analysis?

It is a Static Linear Analysis. It's the first Time I've seen it as well.

[Ad Hoc]

Ok, what is it you are trying to 'model' or rather what is the analysis? You say it is a straight forward linear analysis, but what loading and elements have you chosen?...the error message you posted...what command did you execute to get this error message, and what inputs if any affects this?

[pamarine]

I hit "Run". Can't find any meaningful information on the error message that would help me zero in on a fix.

I am trying to see if the Rudder Stock and Rudder Blade will deform at max speed and angle (55 knots and 35 deg). Based on my math, that should be roughly 5200 lbs of force along the longitudinal axis of the hull hitting the rudder blade.

[pamarine]

Well, figured it out on one of the Rudders, The Pin Connecting the Tiller to the Rudder Stock was moving out of place wierd and the program couldn't solve for it.

YeeHaw, I love this stuff

[Ad Hoc]

In linear static analysis, it wont tell you that, you have to interrogate the results.

If you already have the 'expected' load, then it is just a case of My/I, and the deflection, being a standard cantilever.

I can understand if you wish to use the FE to estimate the SCF for a hot-spot stress, but otherwise, your doing an over-kill, it seems to me. Just a thought.

[pamarine]

Fixed the other model, Tiller arm again.

Well, Basically I have a bean counter trying to tell me how to design the rudder. So when I show him the results and all the reds an yellow where the stock has twisted and the rudder blade bent back 10 deg I'm thinking maybe he'll shut up and let me do what I'm being paid for:D

[Ad Hoc]

aahh...managers...ugh!!!...I do sympathise. I've had my fair share of those "lets blue sky this...or think out of the box"..management bollocks speak....ugh!

Good luck :)

[Knut Sand]

Well, figured it out on one of the Rudders, The Pin Connecting the Tiller to the Rudder Stock was moving out of place wierd and the program couldn't solve for it.

YeeHaw, I love this stuff

I was gonna suggest that you'd choose the end of the rudder axle as a "Fix" point, either totally fixed or fixed vertically, if you see....

I have not SW/ SWS/ CW here now, but another way to fix the bolt, popping out/ away is to make a split line on the bolt's surface, similar to the surface of the part that this bolt shall be incerted into, then you, in the study can choose to .... Don't remeber the word..... It's not mating, its something else, I believe it's 3 options there, the button for it, in the CW study its just below where you choose the material for the parts... I'll find it later tonight....:rolleyes:

[Knut Sand]

Split lines can in some cases help the study to make the meshing more reasonable, they act as kinda help lines for the meshing process (esp, if you have some thin walled features somewhere too). Reduces the calculation time on the computer too....

[pamarine]

You guys know a good reference for SW Simulation Suite/Cosmos?

[daiquiri]

I am trying to see if the Rudder Stock and Rudder Blade will deform at max speed and angle (55 knots and 35 deg). Based on my math, that should be roughly 5200 lbs of force along the longitudinal axis of the hull hitting the rudder blade.

Ehm... You have excluded lateral force from your analysis, or did I misunderstood your description here? If you did, why did you?

[pamarine]

Ehm... You have excluded lateral force from your analysis, or did I misunderstood your description here? If you did, why did you?

the force is applied as if the boat was running straight at speed and the wheel was flung hard over.

So, that means the force of the water would be parallel to the longitudinal axis of the boat, but be acting at an angle to the rudder blade, forcing it both back and sideways.

There is a seperate force on the bearing in the tiller arm pulling the rudder back towards center, and the forces on the Rudder stock imparted by the two bearings in the hull

[Raggi_Thor]

This is a bit off topic, but when I try to imagine that scenario I see the boat starting to turn longe before you cane turn the wheel or tiller that far..
Why not calculate for the maximum lift you can get from water before cavitation ?
I realize this doens't help with your FEA question, but maybe it's easier to apply jyst a huge preassure sideways, lift will always be many times higher than drag.

[daiquiri]

Pamarine, I understand your reasoning but I agree with RaggiThor - your scenario sounds pretty unrealistic. You discard inertial forces acting on the hull, which will tend to relieve the stress on your rudder.
But, on the other hand, it also sounds pretty conservative which is not necessarily a bad thing, since it will lead you to an oversized rudder structure.

[pamarine]

Pamarine, I understand your reasoning but I agree with RaggiThor - your scenario sounds pretty unrealistic. You discard inertial forces acting on the hull, which will tend to relieve the stress on your rudder.
But, on the other hand, it also sounds pretty conservative which is not necessarily a bad thing, since it will lead you to an oversized rudder structure.

I'm going for conservative, which I was trying to test worst-case. (not sure if I did or not).

Yea, if anyone was dumb enough to throw a rudder hard-over at planing speeds they almost deserve what happens next. :D

Also, this is kinda a learning experience for me as I'm new to COSMOS. I really want to figure out how to use FloWorks for Hydrodynamics so I don't have to keep exporting to Rhino to check everything.

[Ad Hoc]

pamarine

Have you tried the cosmos/SW website???

Your loading scenario, as mentioned above is not correct, nor realistic. But if you want conservative answers you'll get that :)
You may change your mind once you have the result...!!

[pamarine]

pamarine

Have you tried the cosmos/SW website???

Your loading scenario, as mentioned above is not correct, nor realistic. But if you want conservative answers you'll get that :)
You may change your mind once you have the result...!!

I just found the SW Simulation site that has some info on it.

What would be a realistic loading scenario? I'm a bit confused because all of my formulas are based on full deflection at speed (for rudder area, stock dia and material, bearings, etc)

[Ad Hoc]

"...What would be a realistic loading scenario?.."

Welcome, your first FE lesson learnt! :)

FE only does what you tell it. Therefore you need to still require and understanding of what you are modelling, how it should be constrained and what loads to apply and where. Just as with doing a simple hand calculation analysis. It is like me saying can you work out the bending moment of these two beams....er..what beam...er...what size are they, how are they fixed, where are they fixed, what load is applied etc...without knowing more then just the statement of intent, any structural design is somewhat difficult.

Ok..first, think what is actually happening?..you have a boat at 55knots. A rudder is used to turn the boat. So, what is a rudder??..it is a lifting surface. This is used to generate a force to turn the boat. The rudder may be at an inclined angle to the flow, does this mean the force is at this angle?...well, what happens with an aeroplane wing?..is the strength calculation derived from the inclined flow directly onto the surface at that angle, or the circulation and hence the lift on the wing, owing to the inclined flow?

So, image the angle to be 90degree, ie square on...what happens, no lift, just drag..so the flow is creating the force/bending moment, but no lift!. If at 0 degrees, no lift no force. If at say 5 degrees, what happens...lift is generated. The flow of water must go around the surface otherwise there is no lift. If the water just impacted on the surface, there would be a simple transfer of kinetic energy but no lift, ie like hitting a solid wall, just at a slight angle..

So, you need the lift/drag characteristics of your rudder first. You can work out what lift (from the CL at an angle), also the classic 1/4 chord moment on the foil itself. All this will determine your rudder size, shape, and the stock and where the bearing are best located too.

[pamarine]

"...What would be a realistic loading scenario?.."

Welcome, your first FE lesson learnt! :)

FE only does what you tell it. Therefore you need to still require and understanding of what you are modelling, how it should be constrained and what loads to apply and where. Just as with doing a simple hand calculation analysis. It is like me saying can you work out the bending moment of these two beams....er..what beam...er...what size are they, how are they fixed, where are they fixed, what load is applied etc...without knowing more then just the statement of intent, any structural design is somewhat difficult.

Ok..first, think what is actually happening?..you have a boat at 55knots. A rudder is used to turn the boat. So, what is a rudder??..it is a lifting surface. This is used to generate a force to turn the boat. The rudder may be at an inclined angle to the flow, does this mean the force is at this angle?...well, what happens with an aeroplane wing?..is the strength calculation derived from the inclined flow directly onto the surface at that angle, or the circulation and hence the lift on the wing, owing to the inclined flow?

So, image the angle to be 90degree, ie square on...what happens, no lift, just drag..so the flow is creating the force/bending moment, but no lift!. If at 0 degrees, no lift no force. If at say 5 degrees, what happens...lift is generated. The flow of water must go around the surface otherwise there is no lift. If the water just impacted on the surface, there would be a simple transfer of kinetic energy but no lift, ie like hitting a solid wall, just at a slight angle..

So, you need the lift/drag characteristics of your rudder first. You can work out what lift (from the CL at an angle), also the classic 1/4 chord moment on the foil itself. All this will determine your rudder size, shape, and the stock and where the bearing are best located too.

Well that last part is what I thought. :mad: I'm thinking I need to distance myself from this builder. Against my advice, he has rushed head-long into building a boat without any testing or engineering to speak of other than his own experience and what I'm able to sneak in under his nose.

i.e. I had barely finished the hull when he was off building it. The rudder bearings, Helm, steering ram, prop, shaft, engine and transmission are all selected and installed without any engineering to speak of. Of course the whole time I'm saying "consult a naval architect before starting" but all I get is "It's fine, nothing new here to worry about" etc etc

Anyhoo, back the the analysis question.

Here's how it plays out in my mind. As the rudder is turned, the AoA of the blade is increased, resulting in an increase in lift (lateral force) and drag up to a maximum AoA where the blade would stall and no longer be producing lift. I don't think that stall is an issue with most rudder designs, but as speed increases, Stall can occur at lower AoA so therefore it is theoretically possible. Also, due to the density of the water there is a longitudinal force on the rudder deflecting the blade aft.

The Blade is supported by the bearing inside the hull, which not only provides a surface for the Rudder stock to rotate on but also holds the rudder stock to the deisred angle, resisting the deflections caused by lift and resistance.

The Blade is held at the desired AoA via the tiller arm which in turn is connected to a hydraulic steering ram capable of a maximum 1000psi. The tiller is kept from rotating on the Rudder stock by a 1/4" x 1/4" x 1.5" bronze key. The tiller arm converts the linear force from the steering ram into torsional force on the Rudder Stock. The Pin is subjected to shear loads across the mating faces by the opposing forces from the Tiller and the Rudder. This in turn will cause the Rudder stock to want to twist.

So the rudder assembly is acted upon by Torsional forces and bending force. The Tiller arm and steering system (and of course metal selection in the Rudder and Stock) counter the Torsional forces, and the Bearings (and rudder materials again) counter the Bending forces.

The possible failure points are deformation of the blade, torsional failure of the Shaft, or shear failure at the pin.

Now the next question I have is as the vessel turns, the force from the flow of water along the hull is or is not parallel to the Longitudinal axis of the hull? I am thinking that the lifting and resistive forces are acting at right angles to each other and the resistive force will always be on a line tangetal to the curve of the vessel's path. But this path will be something between the angle of rudder deflection and the CL of the hull due to slip as the boat travels around the curve.

[Ad Hoc]

Oh dear...sorry to hear.

Well, all you can do is then cover your ass. So, everything you provide has very large caveats, so you take no responsibility for anything other than what you are prepared to put your name against. Once you have done the calculations and if it shows that he is in big trouble, stick to it, do not deviate. If they wish to chnage it, fine let them, you have done your job professionally by telling them where it is wrong and why. It is then up to them to decide what to do with that advice. That is all one can do, and make sure ALL communication is in writing, not verbal. You do nothing unless in writing....it is the only way to protect yourself.

You can lead a horse to water, but you can't make it drink!

[Ad Hoc]

"..Now the next question I have is as the vessel turns, the force from the flow of water along the hull is or is not parallel to the Longitudinal axis of the hull?.."

Not 100% sure i follow you there.

From a distance, there are in fact two rudders. The rudder, located aft, and the hull itself. The rudder aft is turning the boat, so the flow of water will not be parallel. Imagine looking down from up high and seeing the boat. the direction of travel, when helm is applied, is not straight, so the water flow, from this view is not parallel to the hull.

But locally up close, like your head is against the hull watching the flow, the flow is parallel, owing to the shape of the hull, the water cannot go around it, like it can a rudder, so it flows along the hull.

However if your hull is a true planning hull (at 55knots, fair guess!), the water, locally on the hull, actually becomes almost transverse!

[pamarine]

Oh dear...sorry to hear.

Well, all you can do is then cover your ass. So, everything you provide has very large caveats, so you take no responsibility for anything other than what you are prepared to put your name against. Once you have done the calculations and if it shows that he is in big trouble, stick to it, do not deviate. If they wish to chnage it, fine let them, you have done your job professionally by telling them where it is wrong and why. It is then up to them to decide what to do with that advice. That is all one can do, and make sure ALL communication is in writing, not verbal. You do nothing unless in writing....it is the only way to protect yourself.

You can lead a horse to water, but you can't make it drink!

lol, yea. Got plenty of experience in the CYA department:D

So, what am I missing in the analysis posted above? Or a I making it more complex than it needs to be?

[Ad Hoc]

"...Or a I making it more complex than it needs to be?.."

Yup...massive over kill. (A rudder calc shouldn't take more than about 30mins or 1~2hours if you're not familiar)

Just establish the max lift and quarter chord moment, from the CoE, and the rest is simple 'static analysis'!

[pamarine]

"...Or a I making it more complex than it needs to be?.."

Yup...massive over kill. (A rudder calc shouldn't take more than about 30mins or 1~2hours if you're not familiar)

Just establish the max lift and quarter chord moment, from the CoE, and the rest is simple 'static analysis'!

quarter chord would be 1/4 of the mean chord from the leading edge?

[Ad Hoc]

With foils (Daiquiri can correct me if i'm wrong, as its been a while since i did the theory of this)...The 25% (1/4) chord moment is called the aerodynamic centre because of the varying lift and location of the lift vector, and can be modeled as the lift force always being at the AC together with a constant moment (nose down usually).

M = 1/2 * rho* V^2*A*Chord*Coeff.

So your section has the lift, drag and moment action on it....the stock will take the load as well as the rudder. Hence, locating the stock as close to the AC as possible, if you have the charts of the foil, reduces the load on the stock.

[Knut Sand]

Well that last part is what I thought. :mad: I'm thinking I need to distance myself from this builder. Against my advice, he has rushed head-long into building a boat without any testing or engineering to speak of other than his own experience and what I'm able to sneak in under his nose.


Welcome to this world. btw; Where've you been? :D

Builders of any kind of products often do just that, leap forward, start the building process, then start the design/ strength considerations. It's normal. Not ideal, but normal...:D

Ad Hoc has given you some pretty useful hints here. I'll add in a few details to the same issue;

Make a folder, with a contact log;
Date, time, who, what...
Everything in written, or as much as possible. activate the "confirm button" on any important emails you send, come to think of it, do it on most emails, then they'll get used to clicking "accept"...
Hook your phone up to the computer, download related SMS' to the same folder.

Normally you won't need the "folder" but use it without hiding it, in a meeting or two, "we did agree to that issue on..... let me see.......; 18 of august...". Then they'll know that they will need hard evidence / proof to go after you.

If your gut feeling (calculations), tells you that this doesnt hold, inform in written asap. If they continue, inform (in writing) that this issue; you consider it to be out of your scope of work.

It's not always a rainy day.... Sometimes there's snow too....:D

KnutS

[daiquiri]

With foils (Daiquiri can correct me if i'm wrong, as its been a while since i did the theory of this)...The 25% (1/4) chord moment is called the aerodynamic centre because of the varying lift and location of the lift vector, and can be modeled as the lift force always being at the AC together with a constant moment (nose down usually).


Something like that. ;)
For airfoils (or hydrofoils, if you like) of general shape (camber, thickness, their chordwise distribution etc.) the resultant aerodynamic force vector R (which is the sum of Lift and Drag force vectors) will act through a point called "center of pressure", or CP. For a given foil running at a given speed, the chordwise position of CP (which is called Xcp) will vary as the angle of attack (AoA) changes. The aerodynamic moment Ma acting around an arbitrary point A is given by the vectorial product between R and the vector distance between Xcp and A:
Ma = R * (Xcp - Xa)
It has been established by convention that all the distances are to be measured from the leading edge of the foil, and the moment will be positive when acting in such a way as to increase the AoA.
If we decide that "A" is the quarter-chord point, the correspondant moment will be called M1/4.

Now, the Thin Airfoil Theory (TAT) says that there is a point around which the aerodynamic moment Ma is invariant vs. AoA. That point is called "Aerodynamic center" and is placed, according to this theory, at 1/4 chord from the leading edge. Experimental research has shown that although this result is generally not exactly true for airfoils of arbitrary shape, it is true for symmetrical ones - which are the ones we use for rudder design.
The moment around aerodynamic center is thus called Mac and for symmetrical airfoils (zero camber) TAT tells us that Mac = M1/4 = 0. It means that (for symmetrical foils only) CP is placed right there, at 1/4 of chord, and will not move with AoA (at least up until some 6-8°), which is a nice simplification of calculus.
But please note that when dealing with AoA close to the stall the aerodynamic moment will be different from zero - you are in a regime well beyond the validity of the linearized airfoil theory.

Now Pamarine, about your rudder...

The things seen above are valid for airfoils, which are 2D objects. The results of 2D foil anaysis can still be used (through a strip-theory) for a design of high-aspect-ratio (high-AR) keels and rudders, but I guess you don't have a high AR rudder over there. Low-AR rudders have some very peculiar characteristics which are strongly dependant on the vortex system formed at the leading-edge (if swept-back) and around the tips. In case of very-low-AR rudders, this vortex system (dependant mainly on rudder planform shape) will probably be more important for the hydrodynamic behaviour of the rudder than the choice of airfoils.

And since we are talking about very high speeds, I also presume that you have a surface-piercing prop / rudder there. If that's the case, you will also need to take into account the loss of lift due to ventilation and the increase of inflow velocity due to propeller.

I invite you to download form internet and read the paper titled "Rudder Design data For Small Craft" by Dr. A.F. Molland of Univ. of Southampton. It is free and can be easily found with google.
You will find many numbers and answers to your questions over there. If you don't, feel free to get back here and ask. ;)

[pamarine]

The rudder has a 1.5:1 aspect ratio. The boat is a conventional inboard design, so the prop and rudder are completely submerged.

[daiquiri]

The rudder has a 1.5:1 aspect ratio. The boat is a conventional inboard design, so the prop and rudder are completely submerged.
That is a low-AR rudder.
From the pubblication I had suggested you to read, and for all-movable rudders (no skeg in front), you can calculate:
Vr = 1.2 Vs = 66 kts = 34 m/s (inflow velocity at the rudder, due to the propeller)
ARe = 3 (effective Aspect Ratio - please check this against your actual rudder planform)
max AoA = 23° (stall angle)
Cl = 1.32 (lift coefficient at stall - overestimated but it's ok)
Cd = 0.23 (drag coefficient at stall)
Cr = 1.34 (total hydrodynamic force coefficient)
Cn = 1.30 (coefficient of force perpendicular to chord - you need it to calculate torque acting against the stock)
Cm,1/4 = 0.049 (moment coefficient at quarter-chord - note that it is positive, so the center of pressure is fwd of 1/4 chord)
Xcp,c = 0.21 c (longitudinal position of center of pressure is at 21% of the chord)
Ycp,s = 0.49 h (vertical position of cen. of pressure is at 49% of rudder span, measured from the root).
Knowing that forces and moments are calculated with usual formulae (c is the mean rudder chord, S is rudder area, rho is water density):
L = (rho/2) Vr^2 S Cl
D = (rho/2) Vr^2 S Cd
R = (rho/2) Vr^2 S Cr
N = (rho/2) Vr^2 S Cn
M1/4 = (rho/2) Vr^2 c S Cm,1/4
you can find forces and torques acting on the rudder stock and tiller:
T = N (Xcp,c-Xst) (torque countered by the tiller, where Xst is the chordwise position of stock axis, more fwd respect to Xcp,c)
M = R Ycp,s (bending moment at the root)
and from these, and knowing the materials used, you can find the dimensions of relevant rudder structures.

As a final remark, please bear in mind that these are all static forces and moments. But, due to high operative speed and high wave encounter frequencies, your rudder structure should be dimensioned for fatigue. Though it might be questionable whether fatigue criteria should be applied to the max. deflection angle or to some smaller operative angle. If you decide to play safe and go for the first case, it would mean that the above values will have to be multiplied by an adequate coefficient which could be something between 2.0 and 2.5, as a first hint and depending on material used. That is in addition to the usual "coefficient of ignorance" we use for scantlings, which should be at least 1.5-2.0 .

[Ad Hoc]

Ahh...Daiquiri

You beat me to it!...ok, in addition to the above, i would add

If using ally, use a value of 10~20MPa, for fatigue
If using steel, then a factor around 5 would be more appropriate.

But the detailing, how you actually design it, how it shall be built etc , is far more important than just knowing what stress limit to use!

[daiquiri]

But the detailing, how you actually design it, how it shall be built etc , is far more important than just knowing what stress limit to use!
Beyond any doubt.

[Raggi_Thor]

Lots of interesting stuff here!
I think this is a classic example.
The loading conditions are a guesswork, or different konds of estimates at best.
Then why try to calculate tensions and deformations exactly? (With FEA!)
Why not just look up maximum lift (before stalling) for that profile, and then calculate maximum bending moment at 55 knots? Then a roygh hand calculation should tell you if the rudderstock will bend or the bearings be torn?

[daiquiri]

The loading conditions are a guesswork, or different konds of estimates at best. Then why try to calculate tensions and deformations exactly?
Because, if something goes wrong and someone else with a shiny metal badge comes to find out who is responsable for the failure, it is always smart to have a calculation sheet which will demostrate that your scantlings have been done with a scientific method, based on some reasonably estimated intial load data. ;)

Of course, the concept of "exact" is always questionable. Nothing is exact in this world.
Or better - everything is exact up until the moment we try to measure it. :D

Why not just look up maximum lift (before stalling) for that profile, and then calculate maximum bending moment at 55 knots? Then a roygh hand calculation should tell you if the rudderstock will bend or the bearings be torn.
But that's what has been done in the above post!
First, aerodynamic forces and moments have been estimated, based on a research data available. Then, these forces and moments have been transformed into torque and bending moments acting on the rudder stock. Finally, mechanical behaviour (tensions, deflections) of the stock has been evaluated, either with the use of FEA or by hand.

[Raggi_Thor]

.. that's what has been done in the above post!
First, aerodynamic forces and moments have been estimated, based on a research data available. Then, these forces and moments have been transformed into torque and bending moments acting on the rudder stock. Finally, mechanical behaviour (tensions, deflections) of the stock has been evaluated, either with the use of FEA or by hand.

YES, and OK :-)
I just meant that pen and paper or a spreadsheet can be scientific enough, and ofte safer than FEA that is not completely understood.

[Ad Hoc]

FEA is always always an overkill for about 99% of the time it is used. Simple hand calc's are quicker and easier. But looks nice, fancy colour plots etc.... just like all the other "design" software :)