Purpose of bridge pins?

[Peter Wilcox]

More variations. I glued two ~1/2" square by 10" long braces to the "top." With the top rigidly fastened at each end (as would be the case if glued to the sides at the head and the tail), the top was essentially straight.

If I loosened the screws at the tail end, it bowed up distal (tailward) to the bridge.

I cut the braces in half to 5" length. With the ends rigidly fixed, there was a very slight bow concave proximal to the bridge, and convex distal to the bridge.

[Peter Wilcox]

If I loosened the screws distally, the top bowed convex toward the tail.

If I loosened the screws proximally, it bowed concave toward the head, and very slightly convex toward the tail.

If I loosened both sets of screws, both bows occurred.

It looks like the torque of the strings on the bridge is transmitted through the top/braces to the sides of the instrument. So to me, this means the attachment of the top to the sides, especially in the region of the head block and the tail block, is very important to the stability, strength, and contour of the top.

You all probably already know this, but to me it is an eye opener.

[Jeff Highland]

Referring to David's last post

Your last one is fairly close, but complete the separation by isolating the bridge from the soundboard (as if it were not glued), allow it to hinge freely around the front corner of the bridge, Now take moments about this front corner and see if they produce a net clockwise or counterclockwise rotation.
For the moment, assume a rigid bridgeplate/soundboard which transfers the load from the string ball ends to outside the bridge area.

Then consider the bridgeplate,which has to transfer the load from the string balls to the soundboard.
It has a weakness right there in the centre from having holes drilled through, it is going to want to arch at this point

[David Malicky]

Jeff, those cases have been examined in this thread. The first FBD on page 2 is the rigid case with Nfront at the front corner, and Nrear = 0 as explained in the text. Tip occurs for h > b.

All the other FBDs are various ways of modeling the parts as flexible. For a glued bridge as on page 2, the ball-end force reacts through the plate and top, giving the R force. The ballend force would arch an isolated top+plate, but they are restrained by the much stiffer bridge and glue. Most of that restraint is R, via Newton's 3rd, between the top+plate and bridge. That's a good point on the holes weakening the top+plate -- that would concentrate R even further.

See my 2nd post for the non-glued bridge as in Peter's experiment, in particular, how a few degrees of bridge rotation reduces R enough to stabilize the bridge. Here, the top+plate would indeed arch slightly, in proportion to the reduction in R.

On the FBDs, I'm unclear why you said there is "some confusion" and that one is "fairly close", but you haven't stated a specific error. Examining the unglued and rigid case is interesting, but it is different than a real glued FBD.

[Jeff Highland]

FBD's are not used in themselves to analyse internal forces or flexible components, The body is isolated from contact to allow reaction forces to be determined.
This isolation in this case could be isolating the bridge itself or by cutting the soudboard in front of and behind the bridge.
The first will show forces at the bridge/soundboard junction, the second will give you reaction forces carried by the soundboard

[David Malicky]

FBD's are not used in themselves to analyse internal forces or flexible components, The body is isolated from contact to allow reaction forces to be determined.

Agreed, of course, that FBDs cannot be used to find forces internal to the free body--that's implicit to the concept and discussed on page 2. No diagram shows a force internal to the defined free body (definition varies by diagram).

I assume we agree that FBDs are used for both rigid and flexible components. For complex, flexible, statically indeterminate problems like a guitar bridge, I assume we agree there is usually no analytical solution for the precise shapes and magnitudes of the reaction stress profiles. The profiles I show are my estimates, qualified on pages 2 and 3. I think they're good guesses, based on engineering experience and consistent with what we see in the field and in Peter's experiment. But they're of course open for discussion, and I would be happy to know what the profiles really are via FEA, which would be interesting to do sometime.

This isolation in this case could be isolating the bridge itself or by cutting the soudboard in front of and behind the bridge.
The first will show forces at the bridge/soundboard junction, the second will give you reaction forces carried by the soundboard

Yes, and the first is what I've done in the diagrams. The second would be interesting, too, for the purposes you mention, though it is also statically indeterminate.

From your posts, it sounds like your intent would be to isolate the bridge by (among other things) eliminating the role of the glue.

[Jeff Highland]

FB analysis is a tool
In this case, if you want to see what is happening at the bridge soundboard junction you isolate it there
So on this bridge free body you have the forces from the string changing direction which you have labeled A and B
The force you have labeled R on some of your drawings is applied to a separate body, the bridgeplate
To be in equilibrium the Net forces in X and Y directions and the net moments of the forces and reactions must equal Zero.
This enables you to calculate the reaction forces and their locations.
Taking moments about the front corner of the bridge is convenient because it is coaxial with the expected shear reaction on the underside of the bridge.
In this case you end up with a shear force equal to the sum of the horizontal components of A and B) and a vertical reaction equal to the vertical components of A and B located fairly close to the front corner of the bridge.

[David Malicky]

Jeff, Yes, agreed with that, except...
- the ballend force is carried through the plate+top to the bridge (R) -- explained below.
- for a glued bridge, the vertical reaction is a complex pressure distribution with features at the front, middle, and back of the bridge.


All, I know the "R" force is not obvious. Here's (I hope) a clearer explanation of how it arises:
Start with a simple case of two plates clamped together. For example, some 1/4" plywood clamped to 3/4" plywood, face-to-face. A single clamp applies concentrated loads.



Now let's look at each plate as a separate free-body, below. The shape of the pressure distribution between them has been found by experiments, FEA, and theory of elasticity. (For the mech or civil engrs, it's basically St. Venant's principle.) For a plate that is "thin", that is, its thickness is much less than its width or depth (for example, 1/4" plywood, 4" x 4"), this pressure distribution is very narrow, almost mirroring the 20 lb concentrated load on the other side, but a little more spread out. If we were to crank on that clamp to hundreds of pounds, we'd see the wood crush right under the clamp pad, and not much further away.
By Newton's 3rd law, that same pressure distribution compresses the surface of the thick plate.



These narrow pressure distributions occur whenever at least one of the plates is thin. The thinner the plate, the more peaky the pressure. If both are very thick and not wide (for example, 2 pieces of 1" plywood, both 2"x2"), then that pressure distribution is a wide hump that reaches the edges. For an ~infinitely thick plate, we still see the peak near the surface:
http://medesign.seas.upenn.edu/uploads/ ... t-load.jpg
http://www.me.ust.hk/~mech101/chapter2/Image80.gif


So, returning to the glued bridge/top/bridgeplate, consider them as two parts: the bridge as one and the top+plate as the other. The ballend applies a concentrated load to the plate ("T" on the bottom of the fig below). The plate+top are relatively "thin" (~1/4" total thickness, ~1.5" wide), and as Jeff noted, the holes make a weak spot. So the pressure distribution opposite the ballend force (R) should be relatively peaky. Newton's 3rd law reflects R onto the bridge bottom. There are other loads on the top+plate, but the easiest path is straight through (as long as the bridge is glued): R and T are likely about the same. It's a very different story if the bridge peels.



Hope that helped!

[Jeff Highland]

No we will have to agree to disagree
The question was whether a pinned bridge experiences an upward lift at the rear as a result of a rotational moment.
As such, separating it from the soundboard as a free body with the upwards string tension taken by the bridgeplate and soundboard is entirely appropriate.
Analysing the deformation of the bridgeplate/ soundboard is a secondary operation

Anyhow the main lesson to take home from this is how important the integrity and strength/stiffness of the bridgeplate is, those little chips between the pinholes may very well be the straw that breaks the camel's back.

[Trevor Gore]

When I were a lad…

…there was a set of rules that governed the use of free body diagrams. FBD’s were used in the days before desk-top computers so you could solve fairly complex, (generally) statics problems with pencil, paper and a slide rule and get something that resembled a real world answer, though the real world and the free body world should never be confused. The rules ran something like this:

1) A free body is free(!) It exists in space by itself and (generally) doesn’t move, which means all forces on it and any resultant moments sum to zero.

2) The body is rigid. With a free body diagram you’re looking for forces, not deflections within the body. A consequence of this is that distributed loads can be replaced by point loads of equivalent magnitude acting at the centroid of the distributed load. Similarly with distributed reactions.

3) There is only the free body. Any interactions with other things are replaced by a vector representing the reaction force on the free body that the other thing exerts.

So if you wanted to run a free body diagram of a bridge on a soundboard with a bridge plate, you would have to assume that the bridge were rigid (fair assumption in the circumstances) and that as the soundboard and bridge plate immediately beneath the bridge were glued to the rigid bridge, they, too, were rigid (remember this is the free body world, rather than the real world).

So the free body diagram simplifies to this:

All the internal string forces sum to zero leaving only the direct string tension T acting on the top of the saddle. This is reacted by the force S, which is the sum of the tension force behind the bridge and the compression force in front of the bridge acting on our free body, and for horizontal equilibrium S and T must be equal and opposite. Taking moments about the front of the free body reveals that there has to be a force Q to balance the moment caused by T, and a corresponding force P to give vertical equilibrium with Q. This system is statically determinate.

Perhaps surprising to some is that in the free body world, a pinned bridge is no different from a pinless bridge.

Once you start getting into failure modes, you have to be very careful to observe the “rules” (which may necessitate multiple free bodies), and still stay relevant to the real world.

Let’s consider two examples:

1) Suppose the bridge starts to detach (as we know they can). In the real world a “crack” appears between the back of the bridge and the soundboard. In the free body world, that would translate to a shortening of the horizontal distance between P and Q (Q moves towards P). For moment equilibrium, Q has to get larger. So back in the real world we can see how this would translate rapidly into an increasingly severe stress concentration at the root of the crack and imminent catastrophic failure. It may be helpful to think of the top being peeled off the bridge (like sticky tape off a surface) rather than the other way round. (An analysis of the doming of a top due to string forces resulting in this peeling effect is given in “the book”).

2) Suppose the glue line between bridge and top fails completely on a pinned bridge. In the free body world, the free body is now just the bridge and saddle. A downward force vector has to emerge from the bottom of the bridge equal to T (as this string force is no longer constrained within the free body) and effectively replaces Q, which can no longer exist. S is provided via the bridge pins and remains unchanged, but there may also be a vertical friction force from the pins acting downwards on the bridge helping to prevent tilt. P remains and in magnitude is equal to the sum of T (which has replaced Q) and any vertical friction force supplied by the pins. The bridge will be inclined to tip forward if the moment equation doesn’t balance when the pin friction reaction is set to zero. If the bridge wants to tip and doesn’t, the pin friction reaction has a magnitude sufficient to balance the moment equation and thus prevent tipping. If the bridge doesn’t want to tip, moment equilibrium is provided by a reaction force of the soundboard against the bottom of the bridge.

All the above are statically determinate.


…that’s when I were a lad…

[Chuck Tweedy]

Thanks Trevor.

[Rodger Knox]

Now there's a post with which I can agree! Thanks, Mr. Gore.

[Jeff Highland]

Well, I'll have to disagree with you too Trevor

The FBD you show, cut at the soundboard in front of the bridge and behind is ok for working out the loading on the soundboard and as you say, it makes no difference pinned or unpinned.

For working out forces between bridge and soundboard, you need to make the separation at that joint so that you have the bridge by itself and the glue interface is replaced by shear, compression and/or tension forces.

For an unpinned bridge, where the ballends of the string anchor on the bridge, you can model it as you have done with a tensile force applied at the tip of the saddle.

With a pinned bridge, where the ball ends are anchored separately, the forces applied by the string to the bridge will only be at the points where the string changes direction( as it breaks over the saddle and as it enters the pin hole)
For most bridge geometries, to obtain equilibrium against these forces you will have an upwards reaction a little behind the front corner of the bridge as well as a horizontal force.
Appy glue between bridge and soundboard and it carries the horizontal force in shear, the front of the bridge still pushes down on the soundboard though the glue, but a tensile force does not suddenly appear between the back of the bridge and the soundboard.

[David Malicky]

Nice to see you here, Trevor -- thanks for joining the discussion and for your contributions on free bodies and the peeling bridge. You've a good memory of FBDs in Statics. But actually, there is no intrinsic requirement that a free-body be rigid. It is assumed to be rigid in Statics since the equilibrium equations by themselves can only solve rigid body problems. In Mechanics of Materials and other advanced mechanics courses, the free-body often must be allowed to deform -- for example, to develop the beam bending stress, beam deflection, and column buckling equations, and solve statically indeterminate systems. Here are a few examples of FBDs with deformable free bodies:
http://www.efunda.com/formulae/solid_me ... theory.cfm
http://lms.cee.carleton.ca/notes/3203/S ... troduction
http://www.uacg.bg/filebank/att_1324.pdf

As Trevor points out, the real world and the Statics FBD world should not be confused, and there are unfortunately a great many problems that can't be solved with Statics alone. Mechanics problems generally fall into 3 basic categories (not including dynamics):
I. Relatively simple problems that can be solved using Statics alone (In 2D, 3 equations and at most 3 unknowns).
II. Moderately complex problems that can be solved using Statics plus various advanced mechanics courses, using additional equations that describe how each body deforms and possibly interacts with other deformable or rigid bodies (the compatibility equations).
III. More complex problems that are only practical to solve using FEA and a computer, using a bazillion equilibrium, deformation, and compatibility equations. Unfortunately, the glued guitar bridge is in this category.

The rules for FBDs are a bit different in Categories II and III:
Rule 2): bodies may be flexible or rigid, and we're often very interested in the pressure distributions, so we usually don't simplify them with a centroidal force.
Rule 3): interactions with other bodies are replaced by either forces, moments, or pressure distributions, depending on the type of connection.

Trevor, in your FBD, it looks like the free-body (with the dotted horizontal lines) is defined as: the bridge, saddle, string aft of the saddle, bridgeplate, and the portion of soundboard between the bridge and plate. If so, that free body is cut through the soundboard (and X-braces) fore and aft of the bridge. When we make a cut through solid material (i.e., not a pinned joint), you'll probably remember we also need an unknown moment at each cut. So I believe the FBD of that free-body would look like this:

Unfortunately, we now have 6 unknowns, and only 3 equations. Drat.

So, if we want to understand the guitar bridge, we won't make much headway unless we go beyond Statics and Category I. AFAIK, Category II methods can't solve this one either, but prior Cat II and III solutions can give us a little insight into the local effect of forces -- that was my last post on how concentrated loads spread out, R, St. Venant, etc. But with all the disagreement going on, I'm going to take a step back from that for now. At this point, could we see if we all agree on a correct FBD of a bridge (without trying to apply any additional insight)? We just go through the standard FBD methods and replace each "cut" with an appropriate force, moment, or pressure.

Now, what free-body do we pick? Above is one for the whole system, cut at the soundboard fore and aft. How is the agreement / disagreement on whether that FBD is correct?

For the bridge itself, I'll propose a medium complex one defined as: the bridge, saddle, and string between the saddle and bottom of bridge (with slots in the bridge rather than the pins, only because inactive pins are simpler for now). Below is my suggestion, where the pressure distribution w(x) on the bottom is a completely unknown function at this point (like we might see in the intro theory section of a textbook). w(x) is shown compressive, but it could be tensile or zero, in parts or throughout. w(x) could be smooth throughout or extremely peaky (Jeff, I think you would see w(x) as like a point force near the bridge front, and zero elsewhere?). I.e., w(x) is as we would normally do as the first step in the setup of a mechanics solution.

BridgeFBDwx.jpg (12.08 KiB) Viewed 12845 times

[David Malicky]

I was too late for the edit feature, but here's what I wanted to say in the 4th paragraph:

Trevor, in your FBD, could you clarify your definition of the free-body? I see the dotted horizontal lines which look like joints between the bridge, top, and bridgeplate--that definition would include all string aft of the saddle, which makes sense as I don't see a Tension out the bottom. But you don't have moments on the ends like the diagram below, and your later reference to P and Q sound like forces between the bridge and top.

[Jeff Highland]

I think I agree with your last diagram, but if you actually resolve it as a reaction force first then you can decide on a reasonable pressure distribution.
That will depend on the value of the height of the saddle and the distance fron the front of the bridge to the pin holes . Say 12mm and 20mm respectively.
This is fairly easily resolved by taking moments around the front of the bridge(keeping it dimensionless)
Counterclockwise moments are 12T from the string tension at the top of the saddle
Clockwise moments at 20T from the string pulling down leaving a difference of 8T
We know that the reaction has to be T upwards and therefore to maintain rotational equilibrium it will be located 8mm from the front of the bridge. Now you can think about how you distribute that reaction

Now try the same with a pinless bridge where you have the string attached to the rear of the bridge and no longer pulling down through the bridge

[Mario Proulx]

What was the question again?

<lol>

[Mark Swanson]

Ha, Mario I was sort of wondering if you would take the plunge and dive in on this! It really sounds like it's all laid out here and there are just different opinions on the final analysis.

[Mario Proulx]

Thing is, I have no clue what we're discussing and/or analyzing at this point...

[Barry Daniels]

A good example of paralysis by analysis.