Engineers? Opinions on this please...

[Snowpetrel]

Originally Posted by Don C L

[emoji3][emoji3] except for that "non-optimal replacement part!"[emoji3]

You could make a pretty good case that the original was also sub optimal, particularly rolling sheer with the cross grain plywood. And the possibility of a sudden and nasty failure mode that your new beam shouldn't have.

[Don C L]

Originally Posted by Snowpetrel

You could make a pretty good case that the original was also sub optimal, particularly rolling sheer with the cross grain plywood. And the possibility of a sudden and nasty failure mode that your new beam shouldn't have.

The boat is really otherwise strongly built for its size. This was really the only weak point in the boat as I can tell, and it has hung on for all these years. I wanted to wrap this part up and then sail for another 50 years... which will be tough since I am turning 60 in a month.

[NevisDog]

Originally Posted by hamburking

Don't forget the forestay and backstay are also pulling down.

No so - except on a highly tuned/tensioned raceboat. When running downwind, backstay will take most of the wind force, upwind the forestay (plus a little from the mainsheet), across the wind the windward shrouds. Shrouds will be stronger combined than backstay or forestay; plus shrouds are far more vertical than stays, so apply much more downward load on the mast. That's why we calculate based on shrouds alone.

[NevisDog]

Originally Posted by barnakiel

...Can I in the above model assume the lowers are there primarily to keep the mast in column (rather than carrying any significant portion of the wind load)? So then I would assume the maximum load would be the breaking load of the top shroud alone, and the safe load would be about 1/5 to 1/3 of this wire's BL.

All calculations are simplifications of complex structures, the simpler the better, but always simplify on the side of caution. Of course the breaking load of the wire will never be seen, nor even close, except perhaps in a roll-over/knockdown. Safety factors of four or five are usual, so normal loads will be a quarter or a fifth of the UTS, and even that will be based on hurricane force winds.

Do not confuse these loads with the (very small) righting moment forces - righting moment is a static force, whereas rigs and decks must stand up to dynamic loads magnified greatly by the rather narrow angle of the stays from masthead to deck. If you go that way you'll end up with a piece of string replacing your rigging wire, but no one is stupid enough to go there I hope. Rigging wire is sized for a reason.

Or we can skip the calculations and say the thing looks massively strong - that's what I'd do here

[NevisDog]

Originally Posted by Snowpetrel

...It seems to me it would be better to estimate the compression strength of the mast and use this as a basis for the maximum beam load.

True, but we can get the UTS of rigging wire in a moment, straight from a book; try estimating the compression strength of a mast - a non-circular column, perhaps unknown wall thickness, with multiple support points to prevent bending and probably unknown strength of alloy - give me a couple of days and I'll get there, but why bother? If the designer has done his job it will be about equal to the forces that can be imparted by the shrouds, as that's what it must withstand in compression.

Simplify, simplify, simplify...

[Snowpetrel]

^^ fair call, thanks for explaining your logic. I guess in a rollover with no sail on this would likely be the maximum load. Though equally you could argue a diagonal force could load to maximum both the side stays and the forestay,

It would be interesting to see what would go first the wires or the mast under max stress. I am betting the mast would collapse under these scenarios, but it would be intersting to play with a few numbers.

One thing about this that bugs me is that it punishes boats with twin lowers, say like Barnacles vs Dons single lowers boat. It also punishes boats with heavier rigging, and in small sizes the rigging steps are very coarse.

[NevisDog]

Originally Posted by Snowpetrel

...One thing about this that bugs me is that it punishes boats with twin lowers...

Yes, common sense is needed. For example, if as you suggest, the designer has doubled up on the lower shrouds without reducing their diameter then this is overkill and I'd count only one, or else if the angles are well spread fore and aft then again I'd count only one. Also, whenever load is spread over, say, backstay and shrouds then this is not 'worst case' - wires will be lightly loaded. "Garbage in, garbage out" is true of computers, and equally true of hand calculations.

Though I cannot read your calculations, I think you stated you had assumed a central point loading (worst case) on a cantilevered (I hope) beam: with the very short span between heavily beefed-up half-bulkheads shown here, any normal mast heel fitting will spread the load over much (possibly most) of the beam and act more like a mast sat on twin pillars - the span of the cantilevered beam becomes only the distance from edge of pillar to edge of mast step - maybe only an inch or two (of course with half that length becoming the 'effective' length [equivalent length for a simply supported beam] when cantilevered both ends). The difference in loadbearing capacity could mean your calculated figure is enormously conservative.

[J Clark H356]

From a beam standpoint, this is a simple beam with a point load in the center. There is no cantilever. The span is 25 inches. The beam is very shallow and the fact that it is deeper at the sides does not give it additional strength against the moment it must carry to not break. The maximum moment in a beam of this type is in the absolute center. That is the section of the beam that has to be analyzed for the resistance of the moment. Anywhere else along the beam on either side, the load is less, therefore the path being greater has no impact other than the shear load on both supports. That load is half the vertical load.

The mast load itself is the load that is harder for me to get a handle on. We know it must support the dead load of the mast, boom, sails and rigging. There is an additional component added from tensioning of the stays. I think it would be half of the combined tension from the two side stays half goes to the top and half goes to the chainplates. The top portion puts the mast in compression. The tension on the stays is here to stabilize the mast in both side to side and front to back. I have a B&R rig that has no backstab.

Whatever these two loads are is the load on the beam at the center point for calculation purposes.

[NevisDog]

Originally Posted by J Clark H356

From a beam standpoint, this is a simple beam with a point load in the center. .....

I'm sorry but your analysis is so wrong on so many counts that I cannot begin to help you. If you are indeed a yacht rigger than please try to stay within your limited field of expertise.

[sailorchic34]

Originally Posted by NevisDog

I'm sorry but your analysis is so wrong on so many counts that I cannot begin to help you. If you are indeed a yacht rigger than please try to stay within your limited field of expertise.

Oddly he's not. It is a simple beam and not cantilevered. A cantilever beam means the end is not supported. The beam is supported at both ends with a point load in the center. That the beam has a slight arch has little impact in this case.

[NevisDog]

Originally Posted by sailorchic34

Oddly he's not. It is a simple beam and not cantilevered. A cantilever beam means the end is not supported. The beam is supported at both ends with a point load in the center. That the beam has a slight arch has little impact in this case.

OMG! A 'simple beam' (simply supported beam) is one that is free to pivot at its ends; a cantilevered beam is rigidly fixed. If a beam is rigidly fixed at both ends, this is far from a 'simple beam': it is correctly analysed by 'cutting it in half' and examining one half of the beam, with one half of the load, and assessing it as a cantilever. Simplify, simplify, simplify...

Please, you should not attempt such an analysis without relevant qualifications. A 'point load' is a 'knife-edge' load: any mast step will spread the load over a measurable length in the centre of a beam (to which it should be rigidly fixed) and the equation becomes much less critical than a point loading, even with a simply supported beam.

[sailorchic34]

Originally Posted by NevisDog

OMG! A 'simple beam' (simply supported beam) is one that is free to pivot at its ends; a cantilevered beam is rigidly fixed. .

LOL Someone needs to look up the Definition of a Cantilever beam

Oh look, Here it is:

A cantilever is a beam anchored at only one end. The beam carries the. load to the support where it is forced against by a moment and shear stress. Cantilever construction allows for overhanging structures without external bracing. Cantilevers can also be constructed with trusses or slabs.

Where as a simple beam is:
Definition of simple beam

1. : a structural beam that rests on a support at each end

Oh My....

I oddly know the difference between a cantilever beam and a simple beam. While I am not a structural engineer, I am an engineer even sat and passed the PE examand I have had several amazingly detailed classes in steel, wood and concrete beam design.

You are partly correct that a simple beam "may" only be fixed at one end with a roller at the other, but it can also be fixed at each end. from a calculation of load standpoint it's the same as far as sizing the beam load. Canilever is only supported at one end.

Point load from a structural standpoint is also correct.

[NevisDog]

Originally Posted by sailorchic34

A cantilever is a beam anchored at only one end....

... and that is why we cut a rigidly fixed beam in half in order to analyse it. There is no other equation for a rigidly mounted beam.

[Cadence]

It appears you have added some vertical support to the bulkhead, hopefully down to some stringers. I wouldn't worry about your design. The laminated ply is going to be fine. I would leave out the aluminum. You are only spanning what 2 1/2 feet. It isn't E=Mc2.
Happy sailing.

[sailorchic34]

Originally Posted by NevisDog

... and that is why we cut a rigidly fixed beam in half in order to analyse it. There is no other equation for a rigidly mounted beam.

While I agree that one need only analyses one half of the beam with a center concentrated or a uniform load, there are times when the beam is not uniformly loaded, say with a column "point load" offset to one side of center, which in building structures is not uncommon. So there can be different equations. In any case calling it a cantilever was incorrect as was Flaming the other poster, which was not called for.