How much load does a rudder blade see???

[davef]

Folks,

I’m wondering if anyone knows how to approximate the worst-case loading that will be seen on a rudder blade given the surface area of the blade. This is, unfortunately, not an academic question as we recently experienced a rather exciting and dramatic rudder failure as shown in the following pictures.

http://www.imagestation.com/picture/sraid175/p7cacd6aa6c82f320f291dd6af6d66b5c/f37425dd.jpg

http://www.imagestation.com/picture/sraid175/p94c94d98de192dfdfc3f6e6240dc499e/f37425e4.jpg

It WAS blowing quite a bit, however clearly one would not hope to see the rudder blade fold over. As you can imagine, much hilarity ensued. Never the less we were able to get safely back to shore and are now contemplating the reconstruction of our failed blade.

Before you write off my design I should confess that the old rudder blade suffered from an embarrassing (in hindsight) manufacturing blunder. The rudder blade is around 54” tall and being the cheap bastard that I am, we constructed it from a 48” piece of wood that had been cut into sections and joined – you guessed it – exactly where the failure occurred. In the picture above, the black piece and the varnished piece are actually two different pieces of wood. And rather than scarf the joint, I merely butted the top section of the rudder to the bottom section with some WEST. Thus the entire load, instead of being carried through the plywood, was being carried by a glued butt joint loaded in bending which – predictably failed. In all candor, the glue joint was pretty poor as well and the wood separated at that joint, leaving the rudder hanging by a sheet of aluminum in the center.

My GUT feel is that I could have simply re-built the rudder to the same design without the butt joint and it would be fine... but.

We’ve re-built the rudder and made a few modifications to improve strength which are as follows:

The new rudder blade has roughly the same geometical shape length, height and width) but is a laminate of eight 1/8” Okume (4mm actually) plywood sections with 9 oz fiberglass cloth in-between each section. The center is .100” aluminum which was acid etched to ensure proper adhesion to the epoxy. Therefore the new rudder blade is now a multi-layer lamination that looks like the following:

9 oz fiberglass cloth
4mm Okume
9 oz fiberglass cloth
4mm Okume
9 oz fiberglass cloth
4mm Okume
9 oz fiberglass cloth
4mm okume
9 oz fiberglass cloth
.100 aluminum sheet
9 oz fiberglass cloth
4mm Okume
9 oz fiberglass cloth
4mm Okume
9 oz fiberglass cloth
4mm Okume
9 oz fiberglass cloth
4mm okume
9 oz fiberglass cloth

Of course each Okume piece now runs the entire length of the rudder as opposed to having the ill advised butt joint.

The part of the blade that fits between the cheeks on the tiller handle is around 1/8” wider than the previous design as I sacrificed the space previously used for some plastic wear spacers in lieu of more width and strength.

I also did a better job blending out the transition between the blade and the upper part of the rudder (the previous design has a couple of sharp corners in the area that failed).

All in all, it seems like the new design should be way stronger than our old design. As a test, I clamped the new rudder sideways on the side of a work bench and piled 50lbs of lead shot on the tip without any problem.

I am still just a bit concerned and would love to have a better way of testing the rudder to determine if we will still have a problem.

Thanks in advance for your help!

Dave

[Katherine]

This may seem like a bit of a smart ass answer, but, you could build several rudders, load them to failure, record the require breaking load, and make a note of where the fracture occurred in the rudder. You'll destroy severa lrudders, but at least you'll know where the weak spot is.

[Bob Cleek]

Or you could build a rudder to the designer's original specifications and assume he figured out the problem a long time ago. Obviously, it was the glue joint, not the loading. As for maximum load... oh, let's say it falls off the trailer while you are on a bridge over a gorge and it lands on its transom... LOL

[Canoeyawl]

Or goes aground
http://www.simplonpc.co.uk/CPR_Local/PrincessMay_wreck-01.jpg

[davef]

Bob,

Frankly I don't trust the designer and am fairly positive he didn't really know what he was doing. I can say this with some confidence as I am the designer.

Dave

[JimConlin]

Judging by your email addr, you're in company with engineers. What do they say?

[davef]

I am one of said engineers. I am fairly confident in my ability to predict the load under which the rudder will break. And I am fairly confident in my ability to predict where it will break.

What I am not confident in is an estimation of what load the rudder will actually see in sever use. Notwithstanding Bob's comments (let's assume I don't intend the survive dropping the boat on the rudder) how much force does a rudder blade see? I've checked a couple of design references and they don't seem to provide any guidance. I ran a quick calculation that said if I put the rudder dead sideways and drag it through the water at 10 knots I'll see around 400 lbs distributed through the entire blade. Under this load, my rudder blade is likely to fail. I'm not sure that's a reasonable estimate. How do nautical engineers size rudders for strength or are they all so over-designed that they aren't really done to any calculations? Are these things typically done by gut feel or experience?

Dave

[Don Kurylko]

Depends on the boat. What is it? Displacement hull, planing hull? Heavy, light? Slow, fast? Sail area?

[mmd]

I seem to recall a paper on this subject... I'll check my file cabinets tomorrow (three a.m. now - sleepy time ahead).

[Hwyl]

Sorry, no technical data. My gut feeling is you've solved the problem already.

I figure you can slow the boat speed by half, by severe rudder usage.

You don't say what kind of boat. I think S&S say one pound per square foot of sail area at 15kts (seems low). So let's say you have 200 square feet. In a badly managed broach (rudder not stalling) in 15 knots, you'd have 100lbs of loading. So your 50 lbs on the end of the rudder should be O.K.

This is all with the caveat that I am prepared to be anihalated by MMD formal assesment.

Did you consider carbon for the outer layers? There would have been aesthetic issues.

[Ian McColgin]

Impressive that the butt joint right at the point of maximum strain held at all. Too bad you did not put the butt at the bottom. It might have held up forever and kept you from knowing that end grain butts are not so good. Anyway, don't over-engineer this. A properly made one piece or scarfed up wooden rudder will be abuntently strong.

If you use dimensional wood just pay attention to the grain and the danger of cupping. If you use plywood, deeply CPES soak the final shape before painting. If you want to add the weight, glass it after CPESing but for this application, just a piece of wood will provide all the strength you need.

[Popeye]

Maximum load on the rudder is part way there, next, determine the bending moment applied at the point (where it broke) just past the lower gudgeon.. and .. then the solution becomes the design stress required (section modulus for laminate or wood) for the rudder design, assuming the design has a fixed geometry.

;) rudder design (http://www.well.com/user/pk/PCrudder.html)

[ 07-13-2005, 08:24 AM: Message edited by: popeye ]

[Tom Lathrop]

Is it common engineering practice to put the weakest construction at the point of maximum stress :confused:

[Dan McCosh]

I once tore a couple of gugeons off a rudder before asking myself the same question. The answer is "lots". The normal sailing load is about the same as a centerboard, as the rudder is developing a similar lateral plane. A sharp turn can easily load up the rudder to a very high hydraulic load, limited only by the boat's inertia and resistance to the turn. It's sort of like sticking a canoe paddle in the water sideways while you are moving along at a couple of knots. Worst case is pitching the end of the boat out of the water, then letting it drop back with the rudder at an angle. (Say are coming into the beach in a three-foot surf--but that's another story.) The concentration off force would be at the lower hinge point, and the load on the rudder max at about the same area. It's possible to break or tear off any rudder under some conditions--which is why a spare is mandatory in offshore racing.

Rather than 50 lbs. to test, I would think that you should able to comfortably stand on it.

[ 07-13-2005, 08:58 AM: Message edited by: Dan McCosh ]

[Popeye]

Originally posted by Dan McCosh:
A sharp turn can easily load up the rudder to a very high hydraulic load, limited only by the boat's inertia and resistance to the turn. load depends on velocity, water density, rudder surface area and geometry (eg. theory of foil sections), has nothing to do with boat's inertia or yawing moment

[P.I. Stazzer-Newt]

The limiting factor on the rudder load for a tiller steered boat (in normal use) is the strength of the man at the helm multiplied by the leverage ratio of the tiller - makes a stong case for the norwegian design.

In abnormal use, such as when the boat is driven backwards onto its rudder, the loads can be much higher.

[ 07-13-2005, 09:17 AM: Message edited by: P.I. Stazzer-Newt ]

[Dan McCosh]

Why is that? It would seem that if there was no inertial resistance to a turn, there would be no load at all.---sort of like a (theoretical) weightless airplane making turns. Are you only considering the turning force generated, rather than the load on the rudder shaft?

[Popeye]

check your physics Dan

[Dan McCosh]

F=MA No mass, no force.

[P.I. Stazzer-Newt]

Dan, consider a saling craft - hard on the wind - and carrying some weather helm. There is no turning - the boat is travelling in a straight line - but the rudder loads can be considerable, this despite the fact that the inertial effects are zero.

[Popeye]

Originally posted by Dan McCosh:
F=MA No mass, no force.the water has mass

[Dan McCosh]

I know--that was the first scenario.
I mentioned the inertial force in the case of a sudden application of the rudder, in an extreme case to the point where the rudder stalls. In that case the mass of the boat it is trying to turn, or drag to a stop, would seem to enter into the picture. A light, shallow hull wouldn't result in the same load on the rudder as a deep, heavy one. At least that what I would assume. Popeye says not--it wouldn't make any difference whether it was a laser or a tugboat, the load on the rudder shaft would be the same.

[raycon]

The link below has a blurb on rudder strength. The page reads like one of the links will point you to a site with a free body diagram and generic calculations for rudder loading as well as describing when peak loading will occur.
http://www.shark24.org/library/rudder1.htm

It looks like your exsiting rudder is constructed from flat sawn SYP? You had a glue joint failure but remember the mechanical properties of wood are orthotropic.

Good luck.

[Popeye]

The story goes that in order to demonstrate to Aristotlean scholars that two balls of different weights fall at the same rate, Galileo dropped a cannon ball and wooden ball from the top of the Tower of Pisa.

[ 07-13-2005, 09:36 AM: Message edited by: popeye ]

[Dan McCosh]

The tower of Pisa doesn't go to weather. The ABYS rudder standards referenced above include displacement in their calculations--why is that?

[Popeye]

bigger ships need bigger rudders

[George Roberts]

davef ---

"Before you write off my design ..."

You did the design and the design did not show the proper place for the joint. Your fault.

I understand you are an engineer but hire a "good" professional. You will learn a lot.

[Dan McCosh]

bigger ships need bigger rudders The question is why doesn't a heavier vessel need a stronger rudder than a lighter vessel, assuming both rudders are the same surface area?

[mmd]

Dang! The paper I was thinking of turned out to be about high-speed planing hull rudders. Different kettle of fish.

The formula in Popeye's first link seems to give reasonable numbers for blade load. Rather than assume the load centre at the midway point between rudder tip and gudgeon, calculate the centroid of your blade profile to determine bending moment on the blade. Assuming that you are using plywood as the blade material, use 7,000 psi as the design stress figure to determine required section modulus. I'd add a factor of safety of about 1.2 for a racing dinghy, and about 1.5 for a beach cruiser.

Finally, note that these calcs are pretty coarse - if you look too closely at te problem you will get buried in minutae such as the fact that the load on the blade per square inch varies as the depth of the blade increases, and that this varies with the heel angle of the hull, etc. etc. People make careers out of investigating this stuff.

[ 07-13-2005, 10:16 AM: Message edited by: mmd ]

[Keith Wilson]

Just a thought - why fiberglass between the layers of plywood? I'd think you'd want the higher tensile strength stuff on the outside where the streses are higher. An aluminum plate at the neutral axis in bending won't do much except add weight and complication. Two layers of 15mm plywood sheathed in three layers of glass/epoxy would be lighter, easier to build and quite a lot stronger.

[mmd]

I wouldn't bother with the 'glass sheathing. The gain in tensile strength (10,000 psi vs 7,400 psi for D. fir) isn't worth the labour. If you need to keep the blade thin - and there is a few good hydrodynamic arguments against this - sheath the blade with uni-directional Kevlar parallel to the foil span.

EDIT - I had a brain spasm & typed the wrong term - the uni-fibre should be oriented along the length of the blade, not the width. Sorry for the inattention to the task, and thanks P.S-N for pointing this out.

[ 07-13-2005, 12:01 PM: Message edited by: mmd ]

[P.I. Stazzer-Newt]

Edited to remove a question about fibre alignment - point already cleared up by MMD.
<hr>
Keith asked about the layer of glass between the layers of ply.

I don't believe that it offers much increase in strength, but I have done this in an attempt to prevent the joint becoming starved of resin when high clamping pressures are used.
How effective? Who knows?

[ 07-14-2005, 05:27 PM: Message edited by: P.I. Stazzer-Newt ]

[Popeye]

Originally posted by Dan McCosh:
The question is why doesn't a heavier vessel need a stronger rudder than a lighter vessel, assuming both rudders are the same surface area?the more inertia the boat has, the slower it is to maneuver with same size rudder producing equivalent drag forces

[ 07-13-2005, 12:18 PM: Message edited by: popeye ]

[Gary E]

Your rudder failure reminds me of a story told by a old Jewish man in the design dept of a airplane company. Seems that the engine designers were getting more and more horsepower in the plane's and that allowed the pilots to do faster and sharper turns and climbs, so much faster that the load on the wings was causeing the wings to break off at much the same place as your rudder. So the fellow said, that's easy to fix, we drill holes all along a line parallel to the body of the plane in the wing. It will never break again he said, after all when he breaks matzoh at passover it never breaks on the perforations.

[Dan McCosh]

Intuitively, it would seem that if the boat is turning more quickly with the same rudder force, there would be less stress on the rudder stock. Dunno if this is true.

[Popeye]

Intuitively, high speed, light-weight 'boats' are more prone to catastrophic stress failure than slow lumbering barges, here's an example i found on the net ..

http://www.imagestation.com/picture/sraid175/p7cacd6aa6c82f320f291dd6af6d66b5c/f37425dd.jpg

[ 07-13-2005, 12:53 PM: Message edited by: popeye ]

[JimD]

Originally posted by davef:
Bob,

Frankly I don't trust the designer and am fairly positive he didn't really know what he was doing. I can say this with some confidence as I am the designer.

Dave:D :D :D

It's a smallish boat, right? I'd be inclined to look at similar boats drawn up by designers you do trust and make a rudder like they do

[Bruce Hooke]

Yup, I was about to suggest Jim's approach...this is the way a lot of this sort of stuff seems to get done based on what I've seen...and I believe there is a good bit of precidence for this in the engineering world as well! :D

[davef]

Popeye -

Great article. Beautiful - Thanks! I had been using a fluids equation for a blunt body.

I am amazed at how high the forces on the rudder get! My design is probably marginal at 8 knots and needs to be reinforced. I'm now starting to wonder about my gudgeons.

Wind is only predicted at 7-8 knots this weekend so I'm probably going to sail anyway but this is good stuff.

Thanks

Dave

[George Roberts]

Lots of different numbers you can use for "strength."

From various refences:

Douglas fir is about 12k psi

glass/epoxy is about 60k psi

plywood is about 5k psi

[Dan McCosh]

posted 07-13-2005 12:00 PM *** ** * * * **
------------------------------------------------------------------------

quote:
------------------------------------------------------------------------
Originally posted by Dan McCosh:
The question is why doesn't a heavier vessel need a stronger rudder than a lighter vessel, assuming both rudders are the same surface area?
------------------------------------------------------------------------


the more inertia the boat has, the slower it is to maneuver with same size rudder producing equivalent drag forces That's something I hadn't thought of. The rate of acceleration would change, but the total reactive force creating the stress would say the same. The duration of the load on the rudder on a lightweight hull would shorten, but would reach the same level as a heavy one. The next question is whether the same would hold true in the case a boat falling off a wave, or similar shock loading--would the weight of the hull make any difference? This stuff seems to go a long way to explaining why rudders are breaking off a lot of ultra-light boats at sea.

[Popeye]

Another possibility is the foil shape of modern rudders can create fluid vortices (picture water flow like an 'eddy' from the trailing edge) and this can induce oscillation in the rudder, leading to fatigue in the material. Just my guess.

Roughing out the static load is interesting tho', but it quickly becomes a complex problem to solve for high dynamic loading on rudders and stress-strain relationships in geometric bodies etc. as mmd suggests, careers have been built on less.

[ 07-15-2005, 09:03 AM: Message edited by: popeye ]

[Dan McCosh]

One other question--How far back from the leading edge is the center of lift of a hinged rudder (skeg-hung or attached to the keel)---and does it move much depending on the angle of attack?

[P.I. Stazzer-Newt]

One other question--How far back from the leading edge is the center of lift of a hinged rudder (skeg-hung or attached to the keel)---and does it move much depending on the angle of attack? A. About one third of the chordal depth. Varies considerably with a huge range of factors.

B. covered by the "About" above.

[Stephen Hutchins]

Laminating in a flat piece of glass doesn't help much. The glass should be curved to be strong. Laminated wood is better in this case. I would have used 1/8" or 1/4" Aluminum in the center and two 1/2" sheets BS1088 on the sides and shaped as you have shown.