Thoughts on the Gluing of Ukulele and Guitar Bridges Using Franklin Titebond Original Wood Glue
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There are so many things to understand about building instruments that it’s not really surprising that some things you just have to take for granted using the traditions of others. Like using Franklin Titebond “yellow” glue. Many and many are the ways and clamping approaches used to apply pressure but I have never seen an actual analysis of the manner in which the correct clamping pressure should be applied.
Why the sudden interest on my part after so many years of using yellow glue and not actually worrying about the clamping pressure? I had the temerity to go to the Franklin website and look at the data sheet for their “Original Wood Glue”. There noted that the clamping pressures were 100-150 psi for soft woods, 125-175 psi for medium hardness woods and 175-250 psi for the hardest woods. Note that psi means pounds per square inch. So proper gluing of a classical guitar bridge of 6 square inches would require a total force of 750-1050 lbs, assuming the middle set of clamping pressures. That’s like 3-4 pudgy guys standing on the bridge, right? Even the weight required to glue down a 4 inch by 1/4 inch brace would be the weight of an average eighth grader.
What about vacuum clamping beloved of Luthier’s Mercantile and the late Richard Schneider? At sea level, atmospheric pressure is about 15 psi. Therefore vacuum clamping only supplies 9-12% of the suggested clamping pressure, again assuming the medium hardness set of clamping values as being appropriate. So is vacuum clamping with all its little foam cauls and rubber sheets an apparent complete waste of time if you want to follow the Titebond manufacturer’s recommendations? Well, not necessarily. I wrote Mr. Dale Zimmerman, an adhesives technical person at Franklin International to see what his opinion might be on the clamping pressures on their website. Here are a portion of his remarks:
“The numbers you referenced from our website (for Titebond Original Wood Glue) are good guidelines, which will allow you to produce good joints over a wide variety of conditions. Yet, the actual pressure required on any given bond is really a combination of the pressure required to pull the layers of wood together and the pressure required to squeeze out the excess glue. When working with something like a thin veneer, the wood offers almost no resistance, and the pressure applied only needs to squeeze out the excess glue. While vacuum pressure is, indeed, quite modest, it is usually sufficient to create a thin enough layer of glue and a bond good enough to resist any stress which may ultimately develop in the veneer. (That is, where it is the veneer which is doing the deforming rather than the substrate — DCH)
“As the stock being glued increases in thickness, more and more pressure is required to assure that the pieces are pulled tightly together despite any modest imperfections in fit which may exist. Thus, the portion of the recommended pressure needed to deal with the wood itself is really a reflection of the thickness or stiffness of the wood and the lack of fit of the assembly. In a real sense, that means that bonds made with well fitted pieces of thin hardwoods do not need 200 psi, and may be just as good if only 50 psi or 100 psi is applied. On the other hand, it must be apparent that bonding two pieces of inch thick maple or oak which are not machined to fit precisely will be unsuccessful unless enough pressure is applied to bend them into position.
“Finally, starved glue joints and crushing of the wood being glued are, indeed, conditions to avoid. Crushing is a reflection of excessive or uneven pressure, and avoiding this problem may require some restraint and an awareness of the internal strength of the species being used. Starved joints are often the result of gluing wood or wood products which are particularly porous, and it is this porosity which causes the starved joint. While the use of lower pressure in this type of situation can sometimes help to avoid the problem, bonds involving porous woods will be stronger and more consistent if the porous surfaces are sized before bonding to reduce their tendency to absorb the glue and then glued with normal pressures.
Mr. Zimmerman encourages further discussion of this whole business and would be happy to have you call him at 1-800-347-4583 if you have other glue topics you wish to discuss.
How do we then figure out first how much pressure to apply, and then when have we applied it sufficiently? The analysis of the problem is taken from “Fundamentals of Physics”, Henry Semat, Rinehart & Co., N.Y., Third Edition, 1958. In the following discussion, I am going to assume that the reader will be using screw clamps to apply the pressure. The screw is what is termed a simple machine, along with the inclined plane, the pulley system and the lever. We are first going to calculate the force required for a particular example and then the moment of the lever. With the moment information, we can then utilize a torque wrench to tighten our clamps properly.
Generally a simple machine does work W through a height h . The machine is made to function using a force F over a distance s . The efficiency of the machine e is the ratio of the work delivered by the machine to the work put into the machine:
e = (W*h) / (F*s) (1)
For a frictionless machine, e would be 1. If e equals 1 then
(W/F) = (s/h) (2)
The ideal mechanical advantage Rideal for the frictionless machine is the ratio of the weight lifted to the force applied:
Rideal = (W/F) (3)
Eliminating (W/F) between equations (1) and (2) we obtain
Rideal = (s/h) (4)
The actual mechanical advantage Ractual is the ratio of the weight raised to the force applied:
Ractual = (W/F) (5)
So, e = (Ractual ) / (Rideal ) (6)
In Semat, pp. 123-124 there is an example of a jackscrew being used to lift a weight of 800 lbs. Assuming an efficiency e of 0.25 for the jackscrew, a screw pitch of 0.5 inches and an 18 inch long lever on the jackscrew handle, the reader is requested to determine what force will be required on the end of the lever to lift the weight. The screw pitch is the distance between adjacent threads on the clamp shaft. The example is instructive because it gets the reader familiar with the terms involved and their application to the problem. As will be shown shortly, our glue clamping problem is quite similar.
Following Semat’s analysis, h is equal to the screw pitch (0.5 in.), while s is the distance traveled by the end of the lever through one complete revolution (necessary to raise the weight of the weight by h). So s = 2pr where r equals the lever length from the center of the screw shaft, 18 inches in the example. So,
Rideal = (s/h) = (2p * 18 in.) / (0.5 in.) = 72p
It is apparent that if the screw pitch is smaller, the ideal mechanical advantage will be greater.
Since e = (Ractual ) / (Rideal ) , then Ractual = e * Rideal
For e = 0.25, Ractual = 18p . Since Ractual = (W/F) , then (W/F) = 18p .
But we know the value of W is 800 lbs so that the force applied to the end of the 18 in. lever necessary to raise the 800 lbs the height of one screw pitch will be:
F = (W) / 18p = 800 / 18p = 14.1 lbs.
The moment M is the force F times the length of the lever:
Moment = F * r = 14.1 lbs * 18 in. = 253.8 in.lbs in our case.
A shorter lever would require a greater force, a longer lever a lesser force, but the moment required to lift the weight in this example will be the same. Why? Let’s look at the terms involved in solving for moment:
Force = (W) / ( (2p * r * e)/p) = (W * p) / (2p * r * e)
Moment = (W * r) * (p/(2p * r * e)) = (W * r * p)/(2p * r * e)
dividing both sides by r,
Moment = (W * p)/(2p * e)
How easy. So if we know the clamp screw pitch, the load for each clamp and make a reasonable estimate for the clamp efficiency, we can calculate the clamping moment directly.
Now let’s glue something. Once again, we’ll use our classical guitar bridge with 6 square inches of gluing area. Let’s also assume a gluing pressure of 100 psi for the sake of the calculations. So we need a total of 6 * 100 lbs = 600 lbs gluing pressure if we want to follow the Titebond recommendations. Also for the calculations let’s use a total of four clamps so that each clamp will need to apply (600/4) = 150 lbs of force. The Pony screw clamps in my shop have a pitch, p of 1/16″ (0.063″). Note that the pitch is ~ (0.5/0.063) 8 times smaller than our earlier example, so much less torquing force will be required to achieve the same clamping pressures. I don’t know the actual efficiency of my clamps but to be conservative, we’ll let it be the same as the jackscrew in Semat’s example, that is, 0.25. Let us also assume a clamp lever length of 6″. So,
Rideal = (s/h) = (2p * 6 in.) / (0.063 in.) = 190p
Ractual = e * Rideal = 0.25 * 190p = 47.5p
F = (W) / Ractual = 150lbs /47.5p = 1 lbs
Moment = 1 lbs * 6″ = 6 in.lbs.
Alternatively, we can combine terms and calculate the moment directly:
Moment = (W*p) / (2p * e) = (150*.063)/(2p * .25) = 6 in.lbs.
Note again that the length of the lever arm doesn’t appear in the equation for calculating the moment.
The use of a micro-torque wrench would probably be the simplest way of accurately determining when correct clamping pressure has been applied. Smaller bridges and varying numbers of clamps would also require altering our calculations, as would larger or smaller pitches of the screw clamps used in each person’s shop and the efficiency of each clamp.
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