Determining target thickness with non-rectangular panels

[bdunne]

Perhaps this is a question for Trevor, but if anyone has some experience please chime in.

I have some back and top panels that aren't rectangular and due to the odd knot or tree shape etc, the panels can't be made rectangular before joining and so on. I'm attaching an example, but I'm sure everyone would be familiar with this. So Trevor's dynamic method is only valid for nicely rectangular panels, and I suspect, panels without knots or other defects present. I thought maybe the beam deflection method is the way to go, but again I can't see how I can apply that to a non-rectangular panel.

Any suggestions or do I revert back to the "take a stab at what feels about right" method?

Thanks,
Brian.

[Trevor Gore]

The usual method is to find a short section that can be made rectangular and do a 3 point beam bend test on that. However, on your sample, because the grain is all over the place, including knot shadow, its hard to pick a part that is going to be representative. Panels looking like that are usually indicative of high damping. I'd do a quantitative bend test on any section that I could get close to rectangular then a "feel" test to see how well the rest of the panel matched. Then I'd know whether I was totally guessing or just guessing big time.

[Woodsy23]

Provided the variation in width is not too great, a beam deflection test on the whole piece should give a reasonably accurate for El. You would base the calculation on the average width of the piece. That will be easier to judge if the two long edges can be made straight - it would be the width half way along, obviously.

The "Top Deflection Rig" of the style that Trevor and others use can be readily adapted for this purpose. In the photographs below, you can see that I used two strips of "gable shaped" back braces for my line supports. I applied the central load via a transverse piece of stiff timber (at least 20mm high) to ensure that all points across the middle of the piece deflected equally.

Determining the Et is a little more of a problem. You could do a similar test on a piece cut from one end of your board but that would not be representative of the whole board. (load to apply would be much less). Fortunately, the target thickness calculation is not as sensitive to Ec and Glc as it is to El.
Determining the Glc is also difficult. I have set up a test on a whole piece to measure the deflection of one corner with the three other corners point supported. (See photograph) However, I don't have a formula for the corner deflection of a rectangular panel as a function of the Glc. Maybe Trevor knows one that could be used to back calculate the Glc.

I carried out these tests on my back pieces because I thinned them a bit too much resulting in the natural vibration frequencies being quite low, possibly a little too low to be picked up by a microphone. I was not getting the distinct peaks for the longitudinal and twist frequencies that I achieved with my other builds.

[Trevor Gore]

That's a neat use of the monopole mobility deflection rig, Richard. I can now eliminate the 3 point bending rig from the traveling circus!

Determining the Glc is also difficult. I have set up a test on a whole piece to measure the deflection of one corner with the three other corners point supported. (See photograph) However, I don't have a formula for the corner deflection of a rectangular panel as a function of the Glc. Maybe Trevor knows one that could be used to back calculate the Glc.

You might get something out of this or this (I've not fully read them tho'). Otherwise Roark likely has something. Let me know it you find it!

[Woodsy23]

In my post above I said that I don't have a formula for the corner deflection of a rectangular panel as a function of the Glc. However, I have just realised that the deflection of the corner of a rectangular panel with the other three corners supported is equivalent to the twist of a "beam" with one end fixed against rotation and a torque applied at the other. The plate is effectively a very flat beam. The rotation angle at the end of the beam is given by the formula:

Torsion_1a.gif (1.2 KiB) Viewed 14988 times

where T is the torque (N.mm), l is the length of the plate (mm), J is the torsion constant of the "beam section" (mm4) and G is the Shear Modulus we are looking for (MPa). The rotation angle is in radians. The applied torque is the load applied to the corner of the panel (N) multiplied by the distance from the load point to the support point adjacent (mm).

For a thin plate, the Torsion constant is given by

J for Thin Plate.jpg (1.69 KiB) Viewed 14988 times

where b is the plate width and t is its thickness. If both b and t are in mm, J is in mm4 and G will be in MPa.

So the value of G can be determined by rearranging the formulas.

For panels that are not uniform in width, determine the J value from the average width.

I have quickly put together an Excel spreadsheet which I can mail to anyone interested.

[Trevor Gore]

That looks promising, Richard. I can't help thinking that there must be some Poisson ration effects in there, too. How does it correlate to dynamic testing results?

[Woodsy23]

How does it correlate to dynamic testing results?

For the pieces that I did a twist test on, I couldn't get a good value for Glc from the dynamic tests because I was not getting the distinct peaks for the longitudinal and twist frequencies. I determined the Glc using FE analysis of the panel and adjusting the Glc until the corner deflection matched the test deflection.

I have checked the calculation of panel twist angle using the formulas in my previous post against an FE model with the panel supported at three corners and loaded at the forth. There was very close agreement (within 0.5%).

Using the formulas in my previous post, a simple formula for Glc can be obtained by assuming the supports and loads are exactly in the corners and that the twist rotation in radians is approximately the corner deflection divided by the width. The simple formula is:

Shear Modulus from Twist Test.jpg (17.64 KiB) Viewed 14956 times

[Trevor Gore]

Thanks, Richard. That static formula could definitely be useful.