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

seen this page?

http://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.html

input the values for uniform loaded or point loaded masts and the site calculates the maximum deflection.

 

At this site you will find a really big amount of downloadable data for lots of wood species.
Naturally it is focused to the US Market and it may be not easy to find the kind of wood appropriate for europe.

Chapter 05 Mechanical Properties of Wood

http://www.fpl.fs.fed.us/products/publications/several_pubs.php?grouping_id=100&header_id=p

 

Several editions of Skene's Elements of Yacht Design were originally published between 1904 and 1983. The last two editions were "revised" by Kinney though they were actually mostly new. More recently the original version has been reprinted.

 

In the beginning of this thread you appeared to be focused on trying to determine how much deflection a mast would taked before the material yielded and referenced yield to wood and aluminum. As deflection is a function of an applied load, approach the solution as
How much load can be applied before the resultant stresses cause the mast to fail.

As a contributor has stated, if you put a pure tension load into a round (use round to omit internal unequal stresses) mast, or rod, you can load it up to a point that when you remove the load it will return to its original length. This stress is called the yield stress (strength) and will be the limit that You would design to. (barring many other factors that require an application of a factor of safety) It will not break here but would remain bent. If you apply a load beyond the yield strength, the rod will actually withstand more stress due to work hardening and break at the "ultimate tensile strength" of the aluminum.
Google Stress-Strain curve for aluminum

As wood does not yield, ie remain deformed after a load is applied you do not refer to it as having a yield strength. When it reaches its ultimate tensile strength, it just breaks
Without applying a factor of safety, you design to an allowable tensile strength which considers variations in the wood structure, moisture content etc

I used the example of rod in pure tension so you can look at a stress strain curve to understand yield. When you bend a beam half of the beal is in tension and half is in compression. (For a symmetrical beam, let's keep it simple, a round mast)

Failure in aluminum and most wood beams in bending occur in tension side and AT THE OUTER surface you should compare section modulus rather than moment of inertia. You can get there from MOI but it is quicker to use the section modulus for comparison.

As you are only interested in when the mast breaks (wooden) or the aluminum mast permanently bends (yields) the quickest formula is

S sigma max = M (bending moment) times C ( distance from the neutral axis it the outer fibre) divided by I ( the moment of inertia of the cross section )

S= M times [C/I]

C/I is the section modulus and is found with material specifications or calculated

So for wood

Sigma ( allowable tensile strength) = M (your boats calculated load and distance from the mast to boat joint) times the section modulus

For aluminum
Sigma (yield strength of aluminum) = M (as above) times the section modulus

This is a very short explanation
BUT
Not a complete explanation as this only relates to stresses due to bending and it is not complete.
Let's get comprehensive

You then need to pick this material element at the hull to mast joint, and apply the horizontal shear stress due to the load, add in the weight of the mast in compression, add in some of the horizontal component of the weight of the mast when heeled, calculate some acceleration loading in pitch, roll and yaw, weight of the sail. AND there will probably need to be some consideration given to the physical joint design between the hull and mast that will impact your design.

And hence the need for a factor of safety, to deal with the difficult to quantify issues.

The only point that I am trying to make is that the answer to your question is much more complicated than designing the new mast around one load parameter.

 

Sorry Barry, that’s not correct.

So long as the applied load remains below the design yield stress limit, it shall return to its original condition. Only if it is stressed beyond its yield point, will it remain bent.

That is slightly misleading. Work hardening is the localised plastic deformation of atoms along slip planes. These then form dislocations that hinder further movement and thus increase in strength - hence the work hardening. However general over loading of a material, plastic deformation, is also using the same slip planes of the atoms moving relative to each other. The main difference here is that the atoms are moving further apart along the slip planes, whereas in work hardening of an alloy plate to increase its strength the volume is being mechanically reduced which increases the dislocations and hence increases its strength as the atoms have moved closer together. Thus, over loading a material when it gets into the plastic region, the atoms are moving further apart as there is nothing restraining them, such as rollers, unlike a plate that is work hardened in a mill.

So the mechanism is principally the same, but subtly different result. Because upon further loading the material will begin to “neck” as the volume of material left to plastically deform becomes less, and the atoms shear further apart too.

And it is also not necessarily the case it takes greater load either. Since the work hardening rate will decrease with increasing strain, thus the incremental increase in the yield stress (moving) becomes less. So it may appear that it can take a greater load but as it permanently deforms even further upon reloading the amount of applied load to failure actually decreases – which is why I never advise using highly strain hardened alloys, despite their attractive yield strength.

The UTS is a measure of the maximum stress, it is not the ‘break’ point. That comes much later at pure fracture.

It does, all materials do (just different degrees of it), it just yields differently because it is an orthotropic material. The individual fibres begin to tear under tension, not dissimilar to a composite. Just in a different manner to isotropic metals like a metal.

 

http://www.craintechnologies.com/simspar/

you can download a trial version of their software from the above link for free which should do the job.

 

As the original post asked for some info on a wood to aluminum mast replacement and the post suggested a limited technical background, it was necessary to try to keep the response straight forward.

 

Theory and Practice

Early in this thread: "Yeah, I know about irregularities, which is what I meant by saying that theory and practice are different."

I have seen statements like this for decades, and as far as structural analysis goes, theory matches practice very closely. The same goes for all the other engineering disciplines, theory and practice match, provided that the proper theory is applied to all the relevant conditions.

The problem is usually that the analyst chooses a very simplified analysis model, which neglects all kinds of relevant parameters, and needs to cover these shortcuts with a "safety factor" or reduced "allowable stress value".

For this mast yielding problem, it quickly becomes far to complex to address with anything less than finite element analysis, probably coupled to CFD. Beyond the mast properties, the stays, their elastic properties, the stress risers at attachment points, the wind velocity profile, the many wind direction changes, the tilt of the mast......on and on.

One must choose how far to go in getting proper analytical results, and this discussion is based on long established relationships of an empirical nature, mixed with simple theory, and that is OK, but one examining this problem needs to go much further into details, perhaps orders of magnitude, to discover truth. Otherwise, it is just "Hissing into the wind"

 

I think the OP could refer to the fact that the material of the mast, if it is in wood, is not totally uniform so a theoretical study could differ from the results obtained in the practice. In addition, the mast can carry soldered, bolted, screwed elements that are likely to produce weaker points difficult to materialize in a theoretical study.
On the other hand, I do not believe that a theoretical study, however sophisticated (finite elements ...), can predict the yield point of mast (or any material) since it would try to apply the progressive deformation of a virtual model that will take it to the yiel point which is, precisely, what we do not know.
For me the phrase "determine the yield point of a mast" does not make much sense.

 

To me it is far more important where it breaks than some mystical number. Consider the loading on mast is not symmetrical and neither are masts. If made of wood, or even metal, materials are not homogeneous. Welds can effect strength, and then the question is what do they hit and where. Not all forces are the same. In my experience mast break from you things: hitting bridges, lightning or people forgetting to reef sails in storms. In all these circumstances you want mast to break somewhere in the middle.

 

Yes but from what I understand, the OP has no interest in knowing where it breaks, but when, with what deformation.

 

Heimfried, thank you for the link. It will come useful the next time I have to design the mast for real.

Barry, thank you too for the extensive and informative reply. I am well aware that there are other considerations as well. But I'm trying to do things one step at a time. What you said about compression has crossed my mind too. Earlier, more than a few good people here said that the failure on the outermost fiber due to shear. However, come to think of it, the inner side is compressed. Halyard and sail tension also adds to that compression. Could it be that the mast could fail on the inner side of the bend first, due to that cumulative compression? I can certainly imagine this happening with aluminum.

Fredose, those are all good points you brought up. As I said, I'm taking this step by step - other considerations will come later, but in my understanding, the approximate force it takes to break a mast is a good starting point. It's not like I'm going to design one just around this one calculation. As Barry has said, that's what the safety factors are for, since most of these issues are difficult to quantify.

I am interested in that too. As far as I know, if the mast is reasonably uniform in strength (no considerable weak points), the breakage usually happens at the mast partner.

 

I have assumed that we are discussing a free standing aluminum mast.

The mast will act as a cantilevered beam as far as the stresses are concerned and will most likely fail (in this discussion, failure is a permanent deformation in the mast) due to tension in the outer fibre.

If the wall of the mast is extremely thin there is a chance that it could buckle on the inside of the bend

I only mentioned compression in the mast and shear at the joint as an area that also needs some calculations to ensure a viable design

 

Wood has a rupture modulus. For sitka spruce it's about 5,000 psi, doug fir 7800 psi. Stress depends on the section modulus and applied load. From Machinery's Handbook you can find formulas for various sections including deflections and stresses at various points. Run the numbers for various scenarios and see what you get. Spreadsheets are good for this.

For rounds:

Moment of Inertia, I = pi()*d^4/64
Section modulus, Z = pi()*d^3/32
load, W, lbs
length, l = in
Young's modulus, E, psi
Cantilever, point load at end
Stress at support = W*l/Z
Deflection at end = W*l^3/3*E*I

If the stress at the support exceeds the rupture modulus, oopsie.

 

Gregg, have a look at post #38. Choosing the model you have presented is quite a simplification of what might be considered appropriate for addressing a real mast condition.