The Steel Guitar Forum

Erv, seems like your comments are a variation of shut up and dance......i.e. shut up and play.

But what about when i can't play? i'm lying in bed and gotta have something to think about eh? PSG axle stress.

Doug,

Even though there are 10 point loads on the axle about 11/32" apart, it is equivalent to a Uniform load. The length between supports at ends in my case is 3.75". This is correct length for Bending Moment. The resultant force on the axle is a horizontal force pulling it towards the nut.

You might well be right that forces are higher than 25lbs. i just used the lower numbers and still came out overstressed! And when a string gets pulled the forces are higher but not quite sure how much. The force of the bar is very little because it hardly changes the pitch of the strings.

All this is to say is that building Musical instruments is a combination of art and science. In this case the axle might be at it's limits but who knows? maybe it'll sound better. We'll see.

Stephen,
I think you've got that right!
Time could be spent in a more beneficial way, like playing that combination Singer sewing machine and electric cheese slicer.

@John Billings - I tend to think it would make a difference but have no scientific proof, just "a feeling deep inside, oh yeah!"

I'd stop stressing and just play the thing

I think about this kind of stuff when walking the dogs (and they are thinking of squirrels).

Carl,
A pedal steel is the sum of it's parts. Those old ShoBuds are famous for their sound. Discounting changer axle diameter? Well that might be an important part. At Performance, we used 1/2" axles.

Stephen Williams wrote:Erv, seems like your comments are a variation of shut up and dance......i.e. shut up and play.

But what about when i can't play? i'm lying in bed and gotta have something to think about eh? PSG axle stress.

You need to get out more


John - I don't discount any part of the signal chain..

Some metals are better sounding than others.

Georg Sørtun wrote:
...The body itself is rarely ever the problem when "body drop" is noticed - the average PSG body doesn't flex much. It is almost always the parts where the strings are attached and pedal-pulls are applied, that are not attached well enough to the (pretty rigid) body to take up the variations in tension without moving.
Advice: look at connections between parts and the body, before blaming the body of not being rigid enough.

Georg, my experience is different, and I find that most bodies will deflect enough to easily show up on a tuner. You can take just your thumb and press down hard in the center of a guitar, and see a deflection on the meter. Likewise, when you push a pedal, you're pulling the front apron (and whole cabinet) slightly downwards. Why else would this "cabinet drop issue" seem to be a problem mostly when the foot pedals are used, but very seldom when knee levers are used?

Inquiring minds would like to know.

Could you not tune the A and B pedaled pitches at the keyhead and determine if pedal pressure is involved? Seems you could also repeat the same experiment with three dial indicators, one on each pillow block and one at a center location along the axle. Once again with no pedals involved, just increased string tension on 3, 5, 6, & 10.

Has anyone checked the detuning of a Fender 800 or 2000 PSG?

Roger

Reaction force in axle depends upon string material, and whether plain or wound. Also, reaction force in axle depends upon whether tuning arm is supported by a push/pull rod or merely a stop.
Also, hopefully the axle rod is Stainless Steel with an Fy of 50 ksi. It still may be slightly overstressed using 0.6 x Fy = 30 ksi. It might also be a SS rod with a higher yield than 50 ksi. Interestly, deflection at mid point calculates to be .011". You'd think that might bind the levers

I was a structural engineer in life and a wanna-be PSG player. PM if you want to discuss more.

Dave. I got it slightly wrong. In my case the axle force was actually 33lbs not 25lbs. I need to draw a diagram. But that means stress is up to 50ksi. either the steel is stronger than advertized or 25lb pull is too much.

hi Members. I believe Herb Remington and probably Bobby Bowman was involved.....had a piece on the back of the changer..it was part of the changer with a brackett running to the center of the shaft to help eliminate flex not sure if it was sucessful......anyone familiar? it was called the turtle I believe.....I owned a Remington for a short time.

The concepts here are correct, although there is more at play here. From an engineering standpoint, I would do the calculation both with a distributed load, and then using point loads, as the real answer is somewhere in-between, and both solutions will give you different "critical" data components. What's important to remember here is (1) maximum bending stress is calculated at the top or bottom (most critical points, one in tension, the other compression) of the rod at the place with maximum bending moment (in this case the middle, lengthwise). Stresses then taper off from there, meeting at 0 in the dead center of the rod. See picture (nptel.ac.in)


Once yielding occurs, the parts of the section that have yielded now behave plastically (that is, they deform in order to maintain strength). The stress=M/s calculation no longer works, as everything in excess of the yield stress stops there... but let's not go down that rabbit hole.

(codecogs.com)

One thing that's important to note is that the equation used, M=wL^2/8, is for simply supported beam elements--that is, in this case, the ends are free to rotate as the axle flexes. I know most pillow blocks at each end of the changer axle are somewhat thick, and would resist the ends from rotating, thereby imparting some "backwards" moment on the ends of the axle. This would reduce the maximum moment in the axle (in ideal conditions, M_max=wL^2/8-w*L^2/12), which, in this example, would probably prevent the axle from yielding. Look up "Fixed End Moments" if you're curious.

I'm getting a headache.
How did we go from a good ole three chord country song to this?

Jim:
Thanks for the comprehensive look. I believe the only rotational restraint (axle torsion) at each end would be from friction of the axle in the mount.

I did the calculations using tension forces on each string assuming D'Adario string gauges and came up with nearly the same maximum bending moment as using a uniform load.

I had a structural engineering practice for 30 years, and was licensed in 12 states (including CA).

My biggest problem is playing my PSG at 81

Hi Dave,

My reference to rotation was not torsional rotation, but rather rotation due to bending. While not enough to be fully fixed, there is some moment resistance provided by the pillow blocks on each end of the axle--so the true solution lies somewhere between fixed end and simply supported.

There is quite a difference in the complexity, however.

I think we're approaching the realm of paralysis by analysis!