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Hi,

I need some help from you smart folks.

I use fiberglass rods from Goodwinds.com for go-bars. I have been using a go-bar deck that is around 24” in height. I want to switch to using my countertop as the base and overhead storage as the ceiling. I ordered some of the same .187” rod I have been using but in 36” length to accommodate the countertop to storage shelf height. It cut the clamping pressure from ~7 lbs down to ~2lbs.

Here are some pics (ignore my scale that I spilled acetone all over )









Goodwinds offers .25” to .5” rod. What is the calculation I need to determine the clamping pressure of a rod .25” @ 36”, .5” @ 36”, etc…? I don’t want to take a guess as the shipping on these from Washington is absurd. Is there a way to calculate based solely on what is known about how the .187” rod performs?

Here is a chart from their website - https://goodwinds.com/composite-resourc ... fications/

Edit: It's .187" rod I'm using.

Thanks!
Brad


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The pressure has to vary depending on how much it’s bent. Have you ever noticed that a straighter rod pushes harder than one with a pronounced bend (due to its length)? So, a shorter brace might get more pressure than a taller one. If I had to calculate it, I would probably measure one on your scale at different lengths and degrees of bend and form fit an equation.

So, I can’t answer your question on how to calculate pressure at any bend. But, I can tell you my solution. I bought some clear radiata pine planks from Home Depot and sliced them into 3/4” by ~5/16” sticks with lengths that barely span the deck to a bit longer. With the clamping subjects in the deck, I can pick the stick that gives me the pressure I need. All of the dimensions can be adjusted to meet the need. Every now and then, one breaks. Each one is so cheap, it doesn’t matter.

Thanks, Bob!

I did reach out to the manufacturer as well. My goal is to get closer to the 5-7lb range with about as much deflection as is shown in the pics.

It seems there is some tipping point in the lbs of pressure as well. The rod will max out and with more deflection the pressure stays the same.

I’d also would like to get close to the deflection in the pic to avoid the excitement that can occur when one of them escapes the trap.

Brad


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It should pretty much follow Euler"s bucking formula. You'll find that once the rod gets a certain amount of bend in it the force pretty much remains constant the more it bends. All this assumes a perfect world.
I don't remember the formula but I bet you can find an online formula. Try searching for euler's buckling formula calculator. It should get you an idea of the relationships anyway.

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Jim Watts
http://jameswattsguitars.com

Just empirical and no formulas - My "bars" are 48" long and 1/4" dia, I think I got them from a kite place too but not 100% sure.

My top deck is hung from the ceiling for walk around bench.

They are working from ~42-46". Bent more or less by whats under it, braces or a body being glued.

At that length they exert about 6lb 6oz. I never measured until now. That seems about right to me for that kind of gluing.

I put rubber "cane" tips for grip and arrow blunting on the ends.

edit - Just thought I could give a little more info. I grabbed one at the 3 ft point and it was right around 11lb 3oz.

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Brian R, Wood Mechanic
N8ZED

Brad—

If it were me, I would want to take advantage of the phenomenon you and Jim have both noted: that, after a certain bending point, the force stays the same. I would want all of my clamping to happen at or past that point, not in the shallow bend range where the clamping force will vary depending on how tall or short the braces are. I hope that makes sense. Anyway, I would get there by building either a booster seat for the bottom deck or a lower shelf for the top deck. If neither of those plans will work, what about similar diameter rods to those you have, but just longer?

Maybe I’m missing the point but why not just empirically try out different rod diameters until you reproduce the same “weight” on your scale?

So somewhere here on the OLF is a demonstration that I did and posted results and pics. The clamping pressure did NOT vary based on the amount of deflection in my tests.

We proved that about 1.5" of deflection was all that was needed for the go bar to provide the maximum clamping pressure that specific go bar was capable of. I demonstrated this also with a deck and scale deflecting three different amounts but the scale read the same in all three instances.

More deflection is not more clamping pressure was the conclusion that were set out to prove.

Some here will remember this test it was a big hit since we had heard that clamping pressure did not increase with deflection and now we knew.

Back in the old days of the OLF we had to find our own way much of the time. That's why we wrote tutorials and such because even the one doing the tutorial often never really knew what the results were going to be.

Looks like Goodwinds goes from 0.187 dia to 0.250, at least in solid fiberglass rods. So the only way to vary clamp pressure is by adjusting length. I must have just gotten lucky picking deck height and rod dia.

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Brian R, Wood Mechanic
N8ZED

This chart for pultruded glassfibre rods may help. (It worked for me!) If you need to figure out more detail, Euler's formula is here.

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Trevor Gore, Luthier. Australian hand made acoustic guitars, classical guitars; custom guitar design and build; guitar design instruction.

http://www.goreguitars.com.au

From my former boss:



Note: I converted a screen shot of formulas to text above, so typos are definitely on me.

Edit: Eek! Corrected... next time I will just photograph and post his 'sketches

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A constellation only takes shape when one maps the whole.
- Beth Brower

A lot of amazing answers but this is (chef’s kiss)!

I’m going to call the .187” rod at 36” to have 3lbs of force to make it easy. If I switch to .250” rod it looks like this:

3 lb = [(0.250^4)/(0.187^4)]*(force for rod 1) = 3.19*(force for rod 1)

So the force applied by the larger diameter rod = (3 lb)*3.19 = 9.58 lbs

I’ll do more testing tonight to make sure I’m getting an accurate lbs reading based on the deflection I’m after and recalculate, but this looks promising!

Tell Todd (whom I’ve never met, was before my time, but enjoy his quoted content in past posts) that I said thanks so much, and thanks to you for converting, Woodie!

Brad


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This is the first time I've seen this thread. You've got to be kidding. Isn't there enough to worry about building guitars? Formulas for go-bar rods I can't get involved in this one. Yet, here I am.

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Hutch

Get the heck off the couch and go build a guitar!!!!

Testing I did a few years ago with the same heights and rod diameters were very close to Brian's results.

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formerly known around here as burbank
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http://www.patfosterguitars.com

Thanks, Pat! After a little more testing I went ahead and pulled the trigger on .25" rods that will be cut to 36" and 32" lengths. In that configuration I should be able to stay under the 10lb described in Woody's post. Thanks all so much for the help!

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Very intersting topic!
But what is the target of pressure you want for the different tasks using the go bar deck? Are 3 lbs or 1,36 kg enough pressure for glueing the braces and top and back?
Juergen

I just noticed this post - my apologies for the late response.

There are two considerations here - achieving adequate glue line pressure at any point along the brace-to-plate joint and making sure the applied force is evenly-distributed along that joint.

For a well-made joint, adequate glue line pressure becomes more about thinning the glue line down to (for Titebond or HHG) under 0.001"/0.025mm. If the joint is not all that well made, then generating the sort of glue line pressure that is recommended by the glue manufacturer becomes more important. In general, the joint area may be found by multiplying contact patch length x width. From there, determine the minimum number of go-bars necessary to generate the clamping force per unit of area, and then distribute them along the brace as needed to get a uniform glue line pressure.

Note: From here on in, I will use inch (length) and pound (force) units, trusting that the reader will convert that to mm/cm/m, newton/kg(f), and pascals as necessary. I need another cup of coffee before attempting more unit conversions.

An example of this would be a tone bar of 9" length and 1/4" width, giving a joint area of 9" x 0.25", or 2.25 square inches. For a well made joint with hot hide glue, a target clamping pressure of 25 lbs/square inch (a pressure that has been shown to effectively close HHG joints) suggests that the minimum number of go-bars needed would generate an aggregate clamping force of about 57 lbs. For the go-bars I use in the shop, which generate about 7 lbs each, I would need 9 go-bars. For your 3 lb go-bars, we would need closer to 19 to get to that desired glue line pressure of 25 psi.

The other issue is achieving uniform clamping pressure. A conservative assumption here is that an applied clamping force only affects a relatively small area of the joint, with the force from a go-bar propagating through the material at 45 degrees around the line of application. We want this force to slightly overlap with that generated by other go-bars, so - for a 0.5" height brace - we would need to space the bars within 1/2" of each brace end and no more than 1" between each bar. For that 9" long tone bar, I would need a minimum of 9 go-bars to provide even clamping pressure, and that generates my target clamping force as well using my 3/16" x 24" bars. For your 3 lb clamping pressure go-bars, those 19 required bars would be spaced at just under 1/2" between bars and 1/4" at the ends, so the clamping pressure would be very uniform.

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A constellation only takes shape when one maps the whole.
- Beth Brower

Previous reviews of this topic came up with 7lbs as a good target force to try for.

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Steve Smith
"Music is what feelings sound like"

Lots of brain cells getting used up here - - - So long as the desired results happen, why the agita? Trim the sticks/arrow shafts/driveway markers/kite frames as needed for the project at hand. No need to get out the math books. Go build. It'll work out.

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Peter Havriluk

A brain is a terrible thing to waste through want of exercise.

But seriously, not sure if this is agita so much as it is basic craft knowledge and some fundamental shop math. A little primary school arithmetic and limited secondary school algebra is both responsive to and dispositive of the OP and other's questions; assurances not to bother worrying one's pretty little head on the topic seem a bit dismissive of those questions as well as the thoughtful answers provided.

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A constellation only takes shape when one maps the whole.
- Beth Brower

You know it’s a good one when I have to look up word definitions.


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Winner winner, poulet dinner.







Brad


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