Mast size calcs

As a part of my Westlawn course, I am designing a 44' ketch.
The boat's main mast is 63' from cabin trunk to the top and has three equidistant spreaders and a permanent masthead forestay.
Using the "empirical" formulae, I come up longitudinal and transverse moments of inertia of 137.9 in^4 and 18.0 in^4 respectively.
Unfortunately my Westlawn texts only give mast charts for 1/8", 3/16" and 7/32" wall thickness. Even using the thickest, I come up with a mast 15 1/2" x 5".
This seems absurdly large to me...?
Where might I be able to access mast charts for thicker wall thicknesses?

 

Ian Nicolson's Boat data book? - I can sent you the appropriate info if you need it...

Paul,

ps Also I have bought Ted Brewers plans for a 45ft Ketch, I am sure Ted would not mind if I passed on his numbers if they would be useful

 

Thanks Paul - any info would be most welcome. Perhaps you could PM or post the Ian Nicholson stuff. I'm slowly building up my collection of these books, but I would rather like to get hold of the info pretty quickly - and good design books are a bit hard to come by here in Tasmania......
The mast specs for your 45 footer would also be great - if only to compare with the numbers that I calculate....

 

done!

 

1st up - thanks for the info Paul.
It's really interesting to see the different methods people employ for the many and varied calculations required to produce a complete boat design.
In this instance - spar sizing - I've read methods as simple as that suggested by Dave Gerr where you simply take something like 1/50th of the mast length to get one measurement and 1/90th to get another etc.
Then there are methds such as that which Paul kindly sent me which are a little more detailed, but which are still just generalisations - making no attempt to calculate longitudinal & transverse moments of inertia.
Then there are methods - like those set out in my Westlawn text - which establish the exact loads that any one spar is likely to experience, thus allowing the designer to specify the appropriate sizes. This would seem the most thorough and accurate way of doing things, but you have to wonder just how often spars are used which are either over or under-built .... especially given that every method seems to give a different answer......
(To be fair Gerr - and probably Nicholson, but I don't know as I've only read a short exerpt - does suggest that a NA would go through the "exhaustive process" of calculating inertia's etc in order to properly specify spar sizes)

Do spar manufacturers produce mast / boom etc charts that designers can use ? Given the plethora of different sections produced these days (all perporting to be the latest, greatest and stongest...) then surely the (now rather old) traditional mast charts are somewhat outdated...?

 

One approach would be to supply the mast guys with accurate righting moment data and let them do the rest. Much less problem when it come to liability later... (as a NA) We all have to warrant our own product.

 

I imagine that's a pretty standard approach - doesn't get me far in my Westlawn course, however....
And whilst many NA's might do just that, most I would have thought would calculate the required 'system' as a double check, if nothing else....?

 

As you say, a double check wouldn't go astray, but we still have to look at all the other details connected to the system such as chain plates, mast posts etc etc.

Sometimes you just have to trust other peoples judgement because we can't do it all nor do we have the expertise/resources. I'm fairly sure that the guy who checks the mast section and shrouds wont be checking my own specs on the rest.

Of course, the pillars of higher learning that we prescribe to don't see it that way.
Brett

 

Will

If I helped you on masts and spars and things would it make me your sparring partner?

I assume that the mast shape you designed is an ellipse, please advise.

Be very careful of simple proportions of masts as you described, i.e. length/90 and length/50. They might be right and, in fact, will be exactly right for a boat of certain characteristics with the mast loaded and supported in a certain way. But if you deviate from the original boat then the mast should change and you don’t have the means of changing the mast to suit. In very general terms they could be a good starting point after all you must start somewhere!

I know nothing of the Westlawn design approach but it sounds reasonable and I guess you followed it. Interestingly your mast length is given as 63 feet and L/50 is 15.12 inches - you gave the major axis as 15.5 inches.

You can get some very high imposed loads on a mast due to a horrible combination of wind loads and the actions/motions of the boat. If Westlawn takes these all into effect then you don’t have much choice but to run with the numbers you end up with.

Attached to this posting is a simple spreadsheet I wrote that allows you to determine the properties of an elliptical form. You only posted the moments of inertia so I must assume that the design diagrams have some kind of built-in relationship for the thickness. The moment of inertia determines how much the mast (or any beam in a building or a boat/ship) will deflect under load. It does not give the stress in the mast, which is a function of the section modulus. The section modulus is the moment of inertia divided by the distance from the centre of area to the outside of the mast plating.

So some odd things can happen if you just match the moment of inertia. I took your dimensions and worked out the thickness to give 137.9 in^4 and I ended up with 0.202” with a section modulus of 17.3 in^3. If I double the size to 31” x 10” I get a thickness of 0.024” and a section modulus of 8.9 in^3 which is only about half of the original but the weight has dropped from 21.69 lb/ft to 1.55 lb/ft. The problem of course is that the plating is paper thin and the mast would fail if you looked at it sideways.

So the next step is to thicken up the big mast to get the same section modulus – the thickness is 0.048” (double) and the moment of inertia is 275.8 in^4, much better but still woefully understrength even though the moment of inertia has doubled.

The problem is that in “global terms” the mast is OK but looking at the local stresses etc it buckles. In a nutshell that explains why structures need to be carefully checked out against buckling but a good design diagram would take account of it by limiting the proportion of the thickness to the major axis length. Your mast ratio would be 0.21875/15.5 = 70.9 which in my opinion is about as high as you would normally use. My small mast is only 0.202” so the ratio is 76.7 which might be OK but it is certainly pushing it. On the other hand, the big mast is 31/0.048 = 646 and is a joke!

The spreadsheet is pretty straightforward and has only two Worksheets (Tabs), one is a set of instructions and the other does the calcs.

If you want to work things out accurately go here http://www.germanlloyd.org/infoServices/rules/pdfs/english/schiffst/teil-3/kap-2/englisch/inhalt.pdf
And may God have mercy on your soul, mate!

I’m not quite sure what you mean. Smaller masts and stays were usually produced from standard pipes. In fact the old UK pipe standards were developed so that smaller pipes would be a push fit into larger pipes so forming derricks. Larger masts were usually made of rolled plate because pipes were too small or too thin. The “latest and greatest” might be designed for stiffening decks and such and if so would not work as a mast. A deck beam has the great majority of the load acting down and it is welded to the deck plate thus form a “combined” section. Take a simple mast and imagine that the load is applied from only one angle, in that case a simple I-beam would be fine. But pull it sideways and it would fail. Real-world masts have the load being applied through a wide range of angles so the mast tends to be circular as the strength is the same all around.

There are dozens of standard pipes and many manufacturers publish tables of properties such as weight, moment of inertia etc. Standard pipes of one sort or another are used on ships as pillars. I’m not sure what Land of Oz standards are but US standards are widespread in Europe and from north to south in the Americas.

I talk too much.

Hope the spreadsheet helps.

Michael

 

Mike,
Please don't think my delayed response is because I don't appreciate your efforts. Far from it....unfortunately I'm an Excel *****, hiving hardly ever used it, and the version I have doesn't have the Sover add-in (working on that...). As soon as I get that all sussed out, I'll give your spreadsheet a whirl and let you know how I go.

In answer to your questions:

The Westlawn texts show two methods for determing spar sizes. The 1st is called the empirical method and the 2nd is Eulers. Both determine the total mast compression, longitudinal & transverse moments of inertia etc. Then there are graphs for different wall-thicknesses which you plot the inertias against to give the actual mast size. Unfortunately the graphs don't cover wall thicknesses any greater than 7/32 in.
Probably more frustrating for me, is that the text is all feet & inches etc - and my work is all metric!!

Since I 1st posted, they have sent me some more info - metric calcs. Unfortunately some if it is written in a rather confusing fashion (at least for a learner like me....) So if anyone has an easy to follow, accepted method for doing the mast calcs, I'd be most appreciative....

Incidentally, Eulers calculation for determining total compression is as follows:

P = Pi^2 x Moment inertia x Modulus elasticity / Length^2 x Factor (F)

The factor relates to the 'end condition of the strut'. An unloaded srut which is pin jointed at both ends has F = 1.
Fixed base, pin jointed top, F = 2.
Fixed top & Bottom, F = 4

What would a mast, boom etc be?

 

Will

There is no such thing as an Excel *****, just because you work in metric units it doesn’t mean your IQ shrinks by 100/(212 – 32). Check the Excel help file to see if you have Solver in your version, if you don’t have it you must have an ancient version and it’s time for an update. Seriously, Will, anyone doing any kind of calculations will benefit from Excel, keep away from Forster’s and take time to master it. You mentioned how you plot the thicknesses against inertia to give the result you need, well when you get better in Excel try Goal Seek. It is easier than Solver and it is already in Excel in the Tools menu.

I don’t quite follow what you mean by empirical and Euler because you need a method to determine the loads imposed and another method to work out the scantlings to safely withstand the loads.

Euler’s formula is the standard one to calculate the buckling load for a perfect strut or a column (or a mast). All you have to do is find the real load and Bob’s your uncle. But take care because there is no factor of safety in the formula and it is typical to double the expected load for the design load. If the load changes rapidly then often it is increased even more ie design for 250 or 300 for 100 steady calculated load. The formula is used for axial loading and it does not take into account bending.

The factors you show are for how the strut is connected at the ends. It is very rare to see both ends pin jointed as they would be free to rotate; rather like just putting a rod or a tube between the jaws of a vise and closing it. Fixed one end and free the other is common and it is a mast. Fixed both ends is quite common and it is a structural column or pillar properly welded at both ends.

The formula assumes the load is symmetrically applied so it is acting along the pipe centreline. If the load is eccentric then the pipe buckles with less loading. Euler’s formula has been adapted in various ways to allow for this as it is virtually impossible to have a non-eccentric load. The best known formula is probably the Rankine-Gordon or something like it.

Here’s a good site for you http://www.efunda.com/formulae/formula_index.cfm as it has lots of good structural information and a few calculators, on the left select Column Buckling.

Pity Westlawn is sending you not-so-good metric documentation, it is always a pain in the *** when that happens. Have you noticed that these things tend to be written by people who know it and they seem to assume that the reader understands it? Whenever I had to write a complex report I had three good checkers and proof readers; my wife, my daughter and my son. Neither of them really understood what it was all about but they got glimpses and boy could they ever ask questions! I always used to smile when readers told me how clear it was – they should have seen it before if was proofed by the family.

Michael

 

You're right - my home pc has an ancient version of excel - with no solver. My work one does, but the add-in is not installed (yet...)

I spent some time going over my mast calcs last night. And in spite of having checked them 1/2 a dozen times already, I discovered that I had made a mistake in establishing the total mast compression - rendering out by a factor of two. As a result my mast now works out at 9" (round...) at I think 7/32" thick.

The 'empirical' formula that Westlawn use is basically that described by Skene, and the other is based on Eulers formulae. Both variations incorporate a safety factor - 1.85 from memory.

Thanks for all your help so far. I'll give it all a bit more thought and let you know how I end up....

 

Will

You have just proved to yourself a great advantage of a spreadsheet calculation compared to a manual one. After checking the logic etc then the accuracy of it the answers are always right.

Solver and how to add it in.

The image below is a composite of two images from my computer screen. In the background you’ll see the drop-down when you click on the Tools menu and you’ll also see that I have selected Add-Ins.

When you click on Add-Ins the other window opens up – would you believe the one called Add-Ins. You will see that Solver is highlighted and a check is in the box. Just make sure you set the check mark on you work machine. That’s it, that’s all.

Depending on how your office computer is loaded and the set-up and so on the images may be different.

I’ll post something later showing you how to use Goal Seek

We’ll get you into a hot-shot Excel user before you know it! Hang in, there!

Michael

 

Even in my stupidity, i managed to get that far - alas my machine tells me that solver isn't installed - and unless your spreadsheet can come up with a way of finding where I put the disc, well.....
Like I said *****........

And you're quite right - a spreadsheet will always give the correct answer if the correct info is fed into it. I Made a boo boo in so far as the formala uses "beam/2" - ie the half beam at the chainplates. I took the 1/2 beam and divided by two.....

 

Will

Here’s a simple example for Goal Seek.

In the illustration below, Column B shows what you would normally see on the screen and Column D shows the cell contents. We have an area of 240 but we need 280; I chose these numbers because you can do it in your head and easily see what’s happening. The spreadsheet could be hundreds of lines long with Lord knows how many intermediate bits and pieces.

The Goal Seek window is open (find Goal Seek in the Tools menu) below the values and you can see what the entries are. Notice that two of the entries in this window are cell addresses but the answer you want is a number entry, you can’t copy into it nor can you address a cell.

The second figure shows what happens after you click OK.

You’ll have to change some of the Options in the Tools menu.
In the Options window select the Calculation tab.
Check the Iteration box
Change Maximum iterations to 10000
Change Maximum change to 0.00001
These changes will be saved with the file.

Hope this helps. Sorry about your lost/stolen/strayed disk.


Michael