Transverse moment of inertia for a mast

[davef]

Greetings,

Does anyone have experience with the equation for determing the required transverse moment of inertia for a mast. In particular I am referencing the equation presented by Larsson and Eliasson on pg 194 of Principles of yacht design. The required units on a mass moment of inertia are in^4 or mm^4. No matter how many times I run through the equation I can't get the units to make sense. I keep getting lb.ft^2.

Does anyone have an example of how to calculate a required moment of inertia from a known righting moment and rig design?

I would really appreciate the help! Thank you.

Dave

dave.franchino@design-concepts.com

[sorenfdk]

I bet you've forgotten that m is not dimensionless - only k1 is.
Using SI it has the unit 1/(N/mm^2), so the equation will look like this:

I = k1 x m x PT x l(n)^2 = (1/(N/mm^2)) x N x mm^2 = (mm^2/N) x N x mm^2 = mm^4

[davef]

Sorenfdk,

Thank you very much. I appolgize for my ignorance. I am an amature, designing my first boat (a 20 foot day sailor).

I am still struggling to get the equations to work and must have some problem with units. I would suspect the problem might be between english/SI conversion?

I have

K1 = 3.375 (dimensionless)
m = 1 /N/mm^2
PT = 10879 N
L = 4572mm

The resulting Ix is far too huge so I must have a problem. Do you see it?

I've attached some crude .dxf drawings of my layout and will try to connect my equations if I can translate them into a version that you can read. Thank you very much!

Dave

[markdrela]

Originally Posted by davef

The required units on a mass moment of inertia are in^4 or mm^4. No matter how many times I run through the equation I can't get the units to make sense. I keep getting lb.ft^2.
[/email]

You are confusing "I" and "EI". The units of I are length^4, while the units of EI are force*length^2.

If the mast is constructed of different materials, then the modulus E varies within the beam section, and I makes little sense. In this case you should compute EI directly.

[sorenfdk]

Originally Posted by davef

K1 = 3.375 (dimensionless)
m = 1 /N/mm^2
PT = 10879 N
L = 4572mm

The resulting Ix is far too huge so I must have a problem. Do you see it?

It seems I was wrong in my earlier reply - L is not measured in mm, but in m, so k1 must have some unit to it. Sorry!

Furthermore, your m is wrong - m = 1 for aluminium. It is calculated as m = 1/E, where E is Young's modulus for the material in question (for aluminium, E = 70500 N/mm^2 (I don't know the value in imperial units)).

In your case, L = 4.572 m and m = 1, which gives the following (more reasonable) result:

Ix = 16.8 x 10^4 mm^4 = 16.8 cm^4.
or
Ix = 267.5 x 10^6 in^4.

Hope this helps!

[davef]

Sorenfdk,

I am sorry - I am still confused. Do you know the origin of the equation
Ix=K1.m.PT.Ln^2??? Is there someplace I can get a better explaination than what is located in the Principles of yacht design???

Even assuming your explaination that m is, indeed dimensionless (and equal to 1 - I am using an aluminum mast). I can't get your numbers to work out.

I would get Ix = 3.375 (units?) X 1 X 10879 N X 4.572 m^2

If m = dimensionless, then the units on K1 must be m^2/N??? Even this doesn't work out.

Even assuming I ignore the units, when I just multiply this out, I don't get your 16.8 X 10^4. I would get 3.375 X 1 X 10879 X 4.5^2 which = 7.6x10^5 (mm^4?) which would = 1.844 in^4

I guess this number is reasonable (it's right around where I thought the Ix would be) but I certainly don't feel very comfortable since none of the units seem to have worked out. I'd rather have some confidence in where the equation came from.

Incidently, I think you made a mistake on your conversion between mm^4 and in^4.

I think you would want to divide by 25.4^4 - which would have equalled .4 in^4 in your example.

[sorenfdk]

The equation comes from the Nordic Boat Standard, which unfortunately isn't available anymore. The Nordic Boat Standard was published (in Norwegian) by Det Norske Veritas and used in Scandinavia as a sort of class rule (like the ABS Guide). They were the only rules that incorporated rig calculations.

You're right - there is an error in my calculation - the correct result is indeed 76.7 x 10^4 mm^4.

And I must have been tired, because I made another error: I used your values (with imperial units) i the equation to get to Ix in imperial units. This is of course wrong because of k1's units.

I'd better go to bed now...

[davef]

Ok you're probably getting sick of me but I think I might be getting close. The only way I can get the numbers to work is to assume that all the length units are in meters. I then divide K1 by 1000^2 and L1^2 by 1000^2 and I seem to get reasonable units?

I computed both the longitudinal and transverse moments of inertial for the following givens:

Righting Moment - approximately 7230 Newtons
Width to shroud chainplate - .997m
L1 panel height - 4.572m
L2 panel height - 4.318m
Aluminum mast
Deck stepped

Resulting required moments of inertia:

Transverse - 7.67x10^5 mm^4
Longitudinal - 1.16 x 10^6 mm^4

See attached...

Can anyone tell me if these calculations seem right?
Does anyone have access to a better write-up of the equations for determining required mast momement of inertias (or some other way of doing it?).

I'd hate to design too fat a mast almost as much as I'd hate to have the whole thing collapse on me in a fresh breeze.

Thanks!

dave.franchino@design-concepts.com

[mistral]

i think you better try to use the YD-40 scantling example to clear your doubts about units, that's the easiest way, spent a couple of minutes playng with units usnig YD40 diemsion 'til your results are the same of those given on the book; anyway the whole book is clearly intendended for SI system, not for imperial measures; Ix and Iy are expressed in mm^4, we normally use cm^4 (1 cm4=10000 mm4) or inch 4 (1 in4=42,95 cm4)

Ciao
Mistral

[Dutch Peter]

Just wonna add my two cents....

Looking at the formulea, I make the following observations:

1- metric units should be used
2- value ''m'' is dimensionless (because other values are 70500/E, 70500 N/mm^2 being the elasticity modulus of aluminium).
3- Panel lenghts and heights to be in meters!

No sence in trying to figure out the units of the factors, that's what they're there for!

So, Davef,

Looked at your numbers and they check out for the transverse Moment of Inertia! Don't understand how you got it right, dividing through 1000^2 twice, but what the hack. See attached file for explanation!
Checked longitudinal also with Euler and then your right on the failure load, but remember, the factor 1.5 for the load is for dynamics/safety.
For the word-document, can't remember where I got it, so the author is unknown.

[sorenfdk]

Well, if three of us get to the same (reasonable) result, I guess it can't be way off!

And here's a little bit of advice: Whenever you're faced with an equation that requires certain units, and your units are different, first convert your dimensions and then use them in the equation! Don't try to change the equation to fit your dimensions - all sorts of things may go wrong!

And one more thing: The original source of these equations, the Nordic Boat Standard, was not meant to be used for racing or other lightweight yachts, so the rigging calculations are a little bit on the conservative side. This means that if the nearest, smaller section is not much smaller, then you could choose this instead of the nearest, bigger section. But please remember I said "could", not "should"!

[JCFARER]

Ahoy there...

I think I understand the math, but is it at all possible for someone to come up with a hypothetical example and carry it through to the end with all associated formulas and their definitions in place?

Would really like to see the process in step by step until the final figure is established. Finally, what exactly does the result provide? Does it tell you how to select the thickness of wood, or Aluminum etc?

Anybody care to try?

Thanks...

Jay