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2nd moment of area again

awibroe
14-Alexandrite

2nd moment of area again

Hi Guys,

I have another 2nd moment issue I am struggling with.

This is the overall task:

Capture.PNG

So far I have calculated the area, centroid and second moments for the waterplane as follows:

Capture.PNG

and

Capture.PNG

What I then struggle with is combining the second moments of the two shapes. I am of the belief that I can get the overall second moment of the entire waterplane by adding the two individual second moments about centroids i.e. JoL = JL (for the ship) + JL (for the appendage)

I have JL for the ship and I believe to get JL for the appendage I need to convert the second moment about the aft perpendicular to the centroid which I thought could be done as follows:

Capture.PNG

But I am obviously not getting the answer right for JL. I have also not even got to thinking how the overall JT value is obtainable?

Any help warmly welcomed as always.

Andy

1 ACCEPTED SOLUTION

Accepted Solutions

12 REPLIES 12

Your calculation of areas and moments is some form of finite difference scheme.

If I fit a polynomial to your data points I can apply the integration scheme.  (attached.)

Note that I believe there's a typo in the answer given for total longitudinal moment.

awibroe
14-Alexandrite
(To:Fred_Kohlhepp)

Hi Fred,

I was hoping you would pick this up.

The problem is to be done using simpson's first rule which I have done and that is the method it needs to be competed in I believe.

Did you say you attached a file?

Andy

Let's try this again

I didn't see the file on the first post until I posted it the second time.  It's now showing on both.

check your user messages

awibroe
14-Alexandrite
(To:Fred_Kohlhepp)

Fred,

Looks like some impressive stuff.

I'm not sure why you are including JT in the end equation of longitudinal moment?

Are you able to do this in the simplistic way I have approached this at all?

Andy

awibroe
14-Alexandrite
(To:awibroe)

I've been trying to follow the logic in the same manner as the other thread we had on moments.

i.e. finding the overall 2nd moment of the total area which I am of the understanding I can do by summing the two second moments about the respective centroids.

Then I should be able to calculate the total second moment about the centroid for the combined areas by taking total area *(distance of centroid from reference)^2 from the total second moment about the area?

Congratulations!  You found my error replace JT with JL in that last equation and watch the calculation arrive at the posted numbers.

You used Simpson's rule to calculate the areas, centroid and second moments of the basic ship.  I used a curve fit and integrals (which when broken down to the numeric computing behind Mathcad was probably a very similar process.)

As for the second part, the appendage:  We don't know the contours of the appendage, we're only given the results of the calculations.  At that point the correct way to calculate the combined moments is what you've been using.  (The lateral cg's are both on centerline, so there is no centroid adjustment.

awibroe
14-Alexandrite
(To:Fred_Kohlhepp)

If I am right, I think that gives a longitudinal 2nd moment of 7556356m^4 which is not quite the model answer. Are you suggesting that the answer given is wrong?

I'm also not sure of the process you have followed to derive this?

My thoughts on this and noting your advice yesterday (or the day before) are:

I know the new centroid is 52.356m forward from the aft perpendicular (the location where the two shapes meet). I know that the 2nd moment about the waterplane of the ship about its centroid is 5587644m^4.

Firstly, Am I correct to believe that adding this to the 2nd moment about the centroid of the appendage will give me a total 2nd moment about the total area (appendage and ships waterplane)?

If so am I correct in calculating the appendage second moment about its centroid as:

Capture.PNG

i.e. 44006 is the 2nd moment about the aft perpendicular i.e. J shift (the shifted 2nd moment from the centroid by a distance of 8.17m. So to get to the 2nd moment about the centroid I take the product of area of the appendage and the distance to the power of 2 from the shifted 2nd moment?

Then I thought I would be able to calculate the total 2nd moment simply by following the rule of (total 2nd moment about the total area) - total area * (distance to new centroid)^2?

When I follow this routine I get a very differing answer to yours and the model answer.

Andy

An example

awibroe
14-Alexandrite
(To:Fred_Kohlhepp)

Fred,

Thanks for that. that all seems to make logical sense in that you are comparing the 2nd moment about a centroid and the relevant area against the shift in overall centroid. The only concern I have is I don't get the exact answer that was given. Are we saying the given answer is not quite correct?Capture.PNG

How close do you need to come?  This is a numeric solver, there will be round-off error.  There's a reason for the engineering saying, "Close enough for ____insert your choice___!"    😉

Which answer is wrong?  Mathcad is (usually) a very precise computing engine.  How old is your book?  Calcs done with a slide rule?

awibroe
14-Alexandrite
(To:Fred_Kohlhepp)

Fred,

You raise a very very good point. I once hear a riddle during my undergraduate studies where a mathematician and an engineer are put 100 meters from a super model. They are told that every half minute they can move half the distance to the model until they get to her and then they can kiss her. The mathematician doesn't bother moving as it is a mathematical impossibility whereas the engineer gets mm away and claims his prize.

Back to this point, yes I will check with the author regarding the accuracy of their calculations.

Looking at your plot of the hull profile, it is very impressive and I hope one day to know how to do that kind of stuff on my own

Andy.

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