# How does one determine the yield point of a wooden or aluminum mast?

Discussion in 'Boat Design' started by laukejas, Nov 24, 2016.

1. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

I've been arguing with a sailor friend of mine for a while now, as we are trying to find a reliable method of sizing a sailboat mast. For wooden masts, Norman L. Skene provides the following formula:

Mast diameter = ∛(16*P*L*SF/15700), where P is wind pressure (sail area in sq.ft. multiplied by a coefficient of 1.15 - 1.5, depending on the size of the boat), L is mast length in inches, and SF is safety factor. 15700 is π multiplied by fiber stress of spruce, which is 5000.

Obviously, this formula only works for solid, round masts made from spruce, because that's what Skene had in mind. It isn't too difficult to size a different type of mast - for example, a box section, or a hollow one. All you have to do is run the Skene's formula, get a diameter for a round solid mast, determine the Moment of Inertia of it's profile, and then run your numbers for a different profile type mast you want to use until it's Moment of Inertia matches. All should be good, because the material properties remain the same.

Example:
Say my boat has 60ft of sail area, and a 100" long mast. Since it is a small boat, the wind pressure coefficient is 1.15. Skene's formula says such a mast should be 2.2" in diameter. Moment of Inertia of a round solid profile is around 1.15 in^4. But I want to build a box section, hollow mast. I fiddle around with numbers, and determine that a square 2" diameter section with 0.4" (13/32") walls would also have a 1.15 in^4 Moment of Inertia, so it should be just as stiff and strong as the round solid 2.2" mast.

So far, so good. I can even determine how much this mast would bend in a breeze. We used the 1.15 coefficient, so it's 69 lb of pressure on that 60 ft^2 sail. Center of effort is, say, 50" above mast partner, and the luff is attached, so using the Cantilever Beam deflection formula for concentrated load P at any point (it's not really concentrated, but let's do concentrated for simplicity), we get:
θ = Pa^2 / 2EI, where P is is the force, a is center of effort above mast partner, E is Elastic Modulus of spruce (9 GPa), and I is Moment of Inertia. Once you plug in the numbers, the result is 6 3/4" of deflection at the mast top. Sounds about right.

Now, the problem. What if I decide I want to make an aluminum mast instead, using a simple aluminum tube commonly found in hardware stores? Obviously, aluminum is much stiffer, yet much heavier. Running all the numbers again, I could easily calculate the diameter for an aluminum mast which would have the same deflection as the wooden one when subjected to the same wind pressure. I guess the stiffness would match. But I doubt the ultimate strength would match too.

I am an amateur in engineering field, so this is where my knowledge ends. I know that each material has it's Yield Point, past which it will deform permanently, and break soon after. No sailor wants to see that happen to his mast. Obviously, different materials have different amount of elasticity. A brick won't bend as much as a piece of rubber before breaking. Wood and aluminum probably also have a very different Yield Points. For example, I can calculate that my 100" mast will bend 6 3/4" in specific load conditions. But will it recover from such bend? Or will it stay deformed? How much can it bend before it breaks? 7"? 10"? 15"? These are the questions I cannot find an answer to. My sailor friend says that the deflection which would still be okay for a wooden mast might deform an aluminum one, or possibly vice versa. We do not know how to find out how much of a deflection would force the mast past it's Yield Point.

So, for those still reading, I'd like to ask for some help - can you teach me the way to find out how much can a mast bend before it is permanently deformed or broken? How do you calculate that critical amount of deflection?

Thank you in advance. And sorry for the long post, but I wanted to show what I already know to save you the trouble of explaining the basic part of beam deflection

2. Joined: Sep 2011
Posts: 7,528
Likes: 755, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

If you want to know when, a particular mast, collapses due to flexing, you will have no choice but to do a break test.
The calculation procedures of a mast, or any other structure, are intended to determine the properties of that structure so that, subject to a maximum "allowable" tension, the structure will never break. But if that maximum stress is exceeded, it is to be expected that the structure will break, but it is not known when. So, back to the beginning, if you want to know the breaking stress, you must force the mast to break it.

3. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

Well, that's quite obvious, but it would be a very expensive and time consuming method of determining what kind of mast is up to the task. I want to learn the way to calculate the theoretical maximum bend a mast can take, and then slap a safety factor on top of it. If I don't know the maximum bend, I cannot determine the maximum allowable bend.

4. Joined: Sep 2011
Posts: 7,528
Likes: 755, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

5. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

Thank you, TANSL, your calculations provided some very useful calculations. However, I am trying to determine the breaking stress of the mast, as opposed to allowed stress. I want to know how to know what kind of deflection would take to break it, and then work out what deflection should be allowed from there.

6. Joined: Sep 2011
Posts: 7,528
Likes: 755, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

Yes, that is what I understood from the first moment and, for that reason, I said that, in my opinion, there is no other solution but to make a breaking test. But bear in mind that the yield point is not the same as the breaking point. (Sorry if you already knew that).

7. Joined: Sep 2008
Posts: 283
Likes: 30, Points: 28, Legacy Rep: 214
Location: Oregon

### srimesSenior Member

8. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

So you mean there is absolutely no theoretical way of determining the yield point for a simple beam load scenario? If that's true, then how do they determine the maximum allowable stress?

You've got me there

9. Joined: Sep 2011
Posts: 7,528
Likes: 755, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

A safety coefficient is applied, based on data and experiences, so that it is estimated that the allowable stress can not exceed a certain percentage of the yield point.
I am totally unaware of the process followed to define these safety coefficients.
Why do you want a theoretical, empirical procedure, if you can rely on real data?.

10. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

Well, for one, I like to understand how the things actually work, instead of relying on "it works, because it worked before" And, in some instances, you may have reasons to be very exact about the safety factor because of some considerations, like weight.

Also, I was sailing last summer, and the wooden mast on my boat bent quite a lot on gusts. So I started wondering: is this amount of bend normal? How much more could it bend before it broke? Should I ease up on the sheet or should I keep pushing? Can the mast take it?

Now I know that theory and practice are a different animals, but knowing a way to at least somewhat accurately determine the amount of bend that the mast can take before it snaps would give much reassurance.

11. Joined: Sep 2011
Posts: 7,528
Likes: 755, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

To complicate things further, keep in mind that any irregularity in a mast may be a point that facilitates breakage. A wooden mast, not solid, composed of several pieces of glued wood gathers even more reasons for uncertainty ...... So, your question, in my opinion, has no easy answer.
You now know that your mast bent a lot but did not break: do not try to overcome those conditions of your experiment.

12. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

Yeah, I know about irregularities, which is what I meant by saying that theory and practice are different.

Oh, come on. Eyeballing is not the way to design such an important thing. I might have been paranoid about that bend. Next time, I might be too optimistic. "A lot" is not a measurable quantity. Numbers, on the other hand, do not lie. I mean, seriously, if there was no way to determine the Yield Point in a simple beam loading scenario, then modern engineering would be impossible, because you can't test every possible combination of construction by breaking it. I imagine it can be not as easy as finding deflection for a given load, but fracture point should not be impossible to determine.

Can anybody else share any thoughts this issue?

13. Joined: Sep 2011
Posts: 7,528
Likes: 755, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

If you feel like it, search the internet for how to determine the yield point of a certain material, aluminum, for example. Then think about how you could apply that technique to your particular problem.
Honestly, I have to acknowledge my ignorance, I never thought it useful to know the yield point of a structure. Nor would I know how to define it.
As soon as you find out how to do it, please do not hesitate to tell me.
Cheers.

14. Joined: Feb 2012
Posts: 774
Likes: 19, Points: 18, Legacy Rep: 128
Location: Lithuania

### laukejasSenior Member

Oh, believe me, I have searched for quite a while already I wouldn't have made this post if I didn't spend at least several weeks of research on my own. Anyway, I'll let you know if I come up with something. And thanks for your input. In the meantime, I hope someone will come to this topic and shed some light

15. Joined: Nov 2012
Posts: 811
Likes: 64, Points: 28, Legacy Rep: 41
Location: Delta BC

### JSLSenior Member

Too bad you can't get a copy of 'Skene's Element of Yacht Design" edited by Francis Kinney. pg 163 - 190 or Sailing Yacht Design by Henry & Miller. Each has some good info & calcs on wooden and aluminum spars. There may be some SNAME papers as well.
As to yield point, not sure about wood but aluminum can be from suppliers.

Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.