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Hi All

As many of you know I have been making the switch over to hide glue, for many parts of my guitars.

As a result I have had to think about many of my clamping operations to ensure they are nice and slick, I have always used Ash rods in my go bar deck, but have decided to make the the change to Fiberglass Go Bars, Colin gave me a great source for them in the UK and I have ordered some, As a result I have decided to build a new go bar deck (As my current one has issues).

Finally he gets to the point

I know the depths of my braces and dishes, platforms etc, but was just wondering how much i should allow for flex in the bars, I will be using 5mm (0.20 inch) diameter bars which are aprrox 600mm (24" long).

so how much under 600mm so you reckon should I go to give a good clamping pressure ?RussellR38785.644525463

I have about 1" of deflection in my rods (.1875" dia.) when clamping top braces and a bit more on some back braces and transvers braces. I put this arangment on a digitl scale and I am getting right at 8 lbs of force per rod. From all I have read this is ideal for brace clamping

You will be glad you moved away from hardwood rods.

Thanks Michael

I was hoping you would spot my post

My conversion from mm was rounded my rods are 0.1969 so a hair thicker than yours, so I think I should go with 1 inch the same as you are using.

I figure the fiberglass rods should be easier and faster to locate, and because the hardwood tended to mark (Purhaps I had too much pressure) I was using bits of spruce scraps under it, but of course with the shorter open time of Hide, I don't want to mess around with these.

I have ordered the little end caps for the rods, are these ok directly onto the braces ?

Also hardwood dowels are not consistent due to variance in fiber strength from one rod to the next, of course we are talking fractions of lbs.MichaelP38785.6854398148

1"-1.5", that sort of range. And way, way more for gluing tops/backs beacuse I, er, haven't got enough proper length rods. Ahem.

I had never thought of this but you are right Michael

Certainly with ash often rods of equal dimensions grain orientation etc, will have a noticable difference in flex.

Which I guess means a potentially quite uneven clamping pressure.

Thanks Mattia

I was thinking I would work out a platform arrangement to give roughly even flex, though at the moment I am not using hide glue for attaching the plates to the rims, I use Titebond for that and lots of cam clamps.

keeping up with the right rod for the right location on a the right brace...not worth the hassle. I have special rods cut to required lengths for attaching backs and sides and they are a pain to keep in clamping order. I have them labeled and have to lay them out in use order each time.MichaelP38785.6885648148

The elegant thing about go-bars is that the clamping force (for the most part) doesn't depend on how much they are compressed. They act as buckled columns - not springs.

The behavior of a buckled column was first described by Issac Newton (1642-1727). The force the column applies (after being buckled) is called the critical load. It depends only on the stiffness of the rod, and the length. This force does increase - but only very slightly - as you compress it further past the initial buckling. You can check this yourself with a scale.

So, practically speaking, as long as the rod is buckled (compressed at all) it doesn't matter how much. You may as well design them for an inch or so of compression. The only problem would be that, if you compress them too much, they may break or take a permanent set that would reduce the clamping force a bit.


In case anyone really wants to know (this is actually in my head - I'm a mechanical engineer) the buckled force is equal to:   (PI squared) x EI / (L squared).

Where:

PI is equal to about 3.14 (the ratio of a circle's circumference to diameter)

EI is a measure of the bending stiffness, and for solid circular rods (like fiberglass rods or wood dowels) it's proportional to the fourth power of the diameter.

L is the original length of the rod.

SO, a change in diameter has much more of an effect (power of four) on the force than a change in length (power of two). If you double the diameter of a dowel or fiberglass rod, you have to make it 4 times as long to keep the same force.

When the time comes for me to build a go-bar deck, I'll probably try hardwood strips with rounded points on the ends. It should be easy to adjust the thickness with a planer to get the desired force, and I always seem to have a lot of strips around, left over from ripping boards to width. And, I'm cheap.


Phil

I love this place... if you wanna get technical you can! Thanks for the info Phil.

Phil you are right but just as you said this does not take in to consideration changing section modulus of the column such as shape memory loss of elasticity. This affects the stiffness of the column.(section modulus) Fiberglass has a relative constant section as to where wood fibers elongate and compress and do not return to their original shape and strength. Therefore you have a higher degree of degradation in the section modulus of the hardwood column. By the way there is degradation in fiberglass as well. Just on a much, much smaller scale the formula Phil has explained is for a force at known values. Which is what we have to work with. These values change with time and wear. Faster with wood than with fiberglass. Also the stiffness per unit of dia. of fiberglass is higher than that of most hard woods.MichaelP38785.728275463

See, if I had a tablesaw, and an ample supply of strips of wood, I would've gone with that. But I don't. So I didn't. Easy, huh?

There was a fairly in-depth discussion of rod length/force/bending here a while ago, maybe 6-8 months?

Thanks Phil

Your scaring me with science

Am I misreading what you wrote, are you saying a thicker rod of the same length applies less force ?

I would have thought a thicker rod would add more ?

Just for my interest.

no he is say ing the dia has a 4:1 affect on the loading force as compaird to the lenght having a 2:1 affect or the dia. affects the load twice as much as the length.

Yep Robbie, we got a house full of fine and very clever folks here

[QUOTE=RussellR] Thanks Phil

Your scaring me with science

Am I misreading what you wrote, are you saying a thicker rod of the same length applies less force ?

I would have thought a thicker rod would add more ?

Just for my interest.
[/QUOTE]

Are you having problems with this statement?

[quote]If you double the diameter of a dowel or fiberglass rod, you have to make it 4 times as long to keep the same force. [/quote]

It confused me too. I think he's saying, if you have rod X, then double it's thickness, you would also have to increase the length 4 times to keep the force equal to original rod X.

Thats the one B

It reads that if the rod is thicker then it has less force at the same length ?

I would understand it if it said if you half the thickness you have to make it 4 times longer.

Not sure maybe it is to do with the fact a bigger rod does not deflect as easily ?

I think that he is saying that the force decreases with additional length, so if you increase the diameter, the force will increase, and the length would have to be quadrupled to reduce it back to the origional force.

Al

NO! it has more force, or... you have to lengthen the rod 4 time to get back (down) to the original force. When you increased the dia. of the rod at a given buckled column length, it increased the loading force of the buckled column. To return to the original force but maintaine the larger dia., you have to increase the length of the rod 4 times its length.

you beat me to it Al MichaelP38785.738587963

Phil,

Thank you for adding some basis for my assumption. I noticed that once, bent, the rods really didn't seem to change in thrust too much.

Regards, Steve Brownsfbrown38785.7429282407

I cheated.

Al

Phil you have not been out of school long enough I have long since forgot the dates of Newton's publications not to mention those of my contemporary, "Pythagoras" MichaelP38785.7462962963

Doh

Right I have got it now thanks for sorting that bit out for me.

I am going to go with an inch of deflection, time to start drawing

This was fun

It's divided by Length, so if the rod is shorter, the forcer is greater.

So in your example, if you half the thickness, you have to make the rod four times shoter in order to keep the same force as it had in the beginning.


At least that's how I'm reading it.


Boring math below if it helps. I'm bored, and it's been a long time since I've had to do any kind of math.


Arbitrary numbers, but let's say the rod is "2" thick and "2" long, force equals 39.44 (9.86*2^4/2^2) - 9.86 is Pi squared -. Now make the rod 4 thick, and keep it 2 long, force now equals 631.04 (9.86*4^4/2^2), 16 times greater!. In order to increase thickness but make force equal, length must increase to 8 (9.864*4^4/8^2 = 39.46 - same as the first, rounding error.)