# How can I have supports for the longitudinal Girders?

Discussion in 'Class Societies' started by MoeZ, Dec 9, 2017.

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### BarrySenior Member

Actually you said that in #6 and the OP referred this in his initial post.

As viewers might assume that is what you meant, I wanted to clarify it. (btw the equation divides by 8 not 12)

Additionally, this equation is for a simple supported beam with one end on a pivot the other on a roller to compensate for strain (stretch) without creating different stress values and locations than a beam attached, by welding throughout the length to a series of frames. Certainly this is a level of detail that the OP does not really need, as if he is working of a set of acceptable tables or guidelines or standards, the end conditions have been looked after in the tables

A question
If a standard exists that specifies the section modulus, as a free standing truss/girder, do you have to allow for the fact that a welding deck plate would in effect increase the section modulus and also change the neutral axis. Or do they just ignore this. While the welding of the deck onto a beam makes the beam stronger for a symmetrical beam, when cutting holes for pipeways, electrical ways, hvac ways, I would have thought that you would want to be as close to the neutral axis as you can get. The deck plate would change this location albeit small?

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If I understand you correctly. If it is "free-standing", the section chosen, or fabricated to suit, must satisfy the section modulus requirement alone.
If a deck plate, or otherwise, is then attached to it it either:-
1) no longer conforms to be "free-standing" - thus will the client be happy?
or
2) You need to satisfy yourself that the addition does not compromise the initial section modulus requirement either with or without any cuttings/opening in it.

Or have I misunderstood your question?

Making sure openings are as close to the NA as possible is always the objective, in reality other factors often come into play that render this difficult or not possible. So, you just need to double check the structural integrity is not compromised by having openings in different locations.

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### BarrySenior Member

Yes you understood my question and answered it. thanks

In some welded structures, not boat related, often the welding schedule is specified to exceed shear flow between an upper plate and the beam, the beam sees in increase in section modulus and the location of the NA is calculated for ways. (with calculations to ensure that the upper plate will not deform under bending which with thin plates is an issue)

I was curious as if the beam calculations are done as a stand alone unit.

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I would be interested to know why they specify this.
Since so long as the web thickness of the beam satisfies the shear calc's, going over and beyond this shear "area" serves no purpose, other than it becomes a localised stress concentration especially if done poorly.

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### TANSLSenior Member

Sorry, when fixed end, it is divided by 12, if I remember correctly. But that is not the important part of my reasoning.

6. Joined: Oct 2008
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This is very confusing. Since you wrote:
If this is the arrangement, then it is a built-in beam at one end and free at the other (unsupported end). In which case it is divided by 2, not 12.

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### TANSLSenior Member

Maybe I do not know how to express myself well but I reaffirm myself in what I said. A beam can have its ends fixed in one way or another regardless of its unsupported length. Although I repeat again, that is not relevant in my reasoning. If you want to go down that road, well then, go ahead.

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### BarrySenior Member

Yes this is correct IF and only if the fixed joint, is able to carry the loads imparted by bending moment and shear. Say you take a wooden beam 2 x 12 inches, and rest it on a 2 by 6 inch wall and nail the beam to the upper wall plate. One might say that this becomes a fixed end beam but as the nails will not be able to carry the bending moment and shear forces, you cannot use the divisor of 12 to represent the bending moment stresses in the beam.
Another example would be taking say a 12 inch wide flange beam and tying it into say a 1/4 inch hull plating. Would the 1/4 inch hull plating meet the fixed end loads?

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### TANSLSenior Member

Barry, if you wish we can keep talking and talking but I think "fixed end" means fixed end. If the extremes are not fixed, we are talking about wood or whatever you want, the formula will be another one, the one that I have indicated does not work. Do you agree?, well, that's it. I'm not going to apologize for being right.

10. Joined: Sep 2011
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### TANSLSenior Member

@MoeZ, after this discussion, which I think is not very interesting, I do not know if you have achieved a valid solution for your problem. Any more questions?

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