Boomilever B/C

[hogger]

I drill with the smallest bit I have and use a small precision file and carefully widen the hole in a triangular matter as pictured in Aia instruction. Also glue the triangular pieces to the end of tension members and I believe the gorilla glue should be used for this connection. Gorilla glue expands and fills all the gap and crevices.

[leeumpetro]

do you guys think its a good idea to recycle a previous boomilevers?

[Unbihexium]

do you guys think its a good idea to recycle a previous boomilevers?

No, because even any sort of transient stress will weaken the structure permanently. Always rebuild boomilevers, also this allow you to practice your building technique such that they improve with each build.

[iwonder]

I would say that I depends... Rebuilding is a good practice in techniques and forces you to get better. However it really depends on how close to the failure point you come. I have a boom from earlier in the season that I've used at 3 competitions and tested about 6 times, it still holds the 15kg just fine. Of course, this probably means that it's significantly over engineered, but if you don't come close to failure point of the boom and it works after a couple tests chances are it's fine.

A few people(me included) actually use this to check for major build errors. I test booms up 7.5kg or thereabouts before I take them to contest(called preloading, I believe) to make sure I don't have any air in glue joints or some mistake that could cause a premature failure. This has proved helpful at one contest, actually, where I had a boom fail at 3kg in testing becuase of a bad distal end joint, which I was able to repair and at contest the compression member buckled and it held 12kg.

[_HenryHscioly_]

1/16"x3/32" bass tensions
broke at 9.4kg
My boomilever stretched a lot. Sagged more than 1/4". I had a horizontal metal rod with books on the side, and an eyebolt attached to the top of the loading block; I had to open the book and use less pages to lower the rod.
This is normal...?? Is this set-up not correct?

I am a very slow worker..(hw, building, eating, lol, anything)
Tested single I beam, 4 stick compression, two stick compression, box beam, but only built 1 or 2 of each before quitting and changing design...
All were underbuilt
I have 1week...so built way overbuilt compression box beam(1/2"x1/8" sides 1/8"x1/8") braced into 4 section on horizontal plane.
Instead of using 3/32"x3/32" bass, which is what I've been using before, I went thinner to 1/16". The tension member sheared at the distal end. I only used gorilla super glue and increased the glue surface a tiny bit with a small piece of 1/16"x3/32" bass. The tension member didnt shear off the beam, but itself had a break almost parallel to its length.
weighed 18.63g..(had my partner build the box beams). I've been quite reserved because he built sloppily and seemed very rough, often accidentaly breaking pieces of wood and what not, but I think after teaching him to cut balsa sticks from sheet and glueing some stuff under my watch, he's getting more better and is more careful. Hopefully we can pick up our build speed and build quicker..
Hopefully he is ready, and we can start building smaller box beams and test. :]
Anyone else used 1/16"x3/32" bass?

[retired1]

My computer program says that it will not hold the full load. The 3/32" sq has a rather small safety factor and that assumes uniform wood.

[Unbihexium]

My computer program says that it will not hold the full load. The 3/32" sq has a rather small safety factor and that assumes uniform wood.

How strange, because according to MS3D, it has about 2 kilos margin of error past full load if they are simply substituted into an already working balsa tension design. I've built with 3/32 square recently, and I've found it incredibly useful and reliable, and since basswood is rather uniform, the only problems with using such a small beam are the joints, which are prone to stress due to the small joint surface area.

[Balsa Man]

Lot of unknown variables floating around here. Obviously both programs (MS3D and what Retired1 is using) have to have – be using - some value for tensile strength. However, because of tension member stretch, there is more going on than just axial tension loading.

The fact of failure near an end – in this case, the distal end, highlights this. If you draw out to scale the right triangle before loading/distortion, then on top of it, put the compression member(s) with the wall point the same, and the distal end down ¼”, and then connect the points the tension member(s) exit the wall block, and first point at which they attach to the compression member, you see that with the ¼” “droop”, there is a bend put into the tension member(s) at both top and bottom- right as t-member leaves the wall block, and right before attachment to the compression member. So, at high/full load, the tension member(s) at these points are seeing both tensile load, and bending. That means the load they will carry without failure is lower than if they were under just pure, axial tension loading.
Have no idea if either program calculates/factors that in. The following suggests that in some way, both programs may be doing so, somehow.

Looking at this, one of the more comprehensive reference sources for wood properties out there-
http://www.conradlumberco.com/pdfs/ch4- ... f-Wood.pdf

-we see no number for tensile strength parallel to the grain for bass, but it is noted that in the absence of such data, modulus of rupture values can be substituted for small clear straightgrained pieces, and are considered to be low/conservative. The modulus of rupture for bass is shown as 8,700 psi. The density of bass (associated with this modulus of rupture) is shown as a specific gravity of 0.37. For 3/32nd square, this calculates out to 1.28gr/24”, or 1.92 gr/36”. That in my experience is near the light end of the density range you’re going to find bass in. Higher density pieces will have greater tensile strength. What that density vs tensile strength relationship is, I have not seen, and don’t have test data.

So, in a C-boom we’re looking at about 44kg (97 Lbs) tension at a 15kg load; for 2 tension members, ½ that, or 22kg (48.5 Lbs). If we’re looking at 2 tension members attaching to two bolts (as opposed to both running to one bolt, add a little- on the order of 50 Lbs each.
So, calculating at 8,700 psi- the tensile strength of a 1” x 1” piece - a 1/32nd x 1/32nd piece is 1/1,024ths = 8.49 Lbs. A 3/32nds square contains 9- 1/32nds square pieces; 9 x 8.49 = 76.5 Lbs. Hmmm… that’s about 50% greater than the design load - considerably more than either “2kg”, or “small safety factor.” Even if you throw in a 25% safety factor, which is not unreasonable, because of a) the inherent variability in strength between pieces of the same density, and b) because most wood properties studies are of samples at “lumber” sizes- pieces with a lot bigger cross section than what we’re dealing with in booms – weaker parts of the cross section cancel out stronger parts, published data is average of a fairly large cross section.

So, on this 8,700 psi basis, a 1/16th x 3/32nds piece has 2/3 the cross section of a 3/32nds square- suggesting a tensile strength of about 50 Lbs. With very little safety factor, this suggests that two could work- just. Given safety factor considerations, this suggests that to work, a) the density is going to have to be higher (for 3/32nd square) than 1.3gr/24” – 1.6, 1.7gr/24, maybe? Based on experience sorting through piles of bass sticks, on the order of one out of 5 to 10 will get into this density range. For 3/32nd x 1/16th, 2/3rds the weight, so, 1.07 to 1.13gr/24”, and b) you're going to have to find a way to take out the bending factor, and get things so that the tensile load is axial when you get to full load.

So, how to improve/manage the factor of t-member stretching producing distortion of “the triangle”, producing bending near the ends, resulting in a combination of tensile and bending loads at the ends? If you align the tension members in the unloaded structure so that at full load, they’ll be running/aligned straight, i.e., seeing just axial tension loading, you help your situation. Going back to the what happens with distortion from tension member stretch discussion, above. You can’t keep the tension members from stretching, so the distal end of the compression member(s) will go down/droop. You want it to be perpendicular to the wall at max load – if it isn’t, compression load isn’t axial; you’ll see a buckling failure at a lower load than if it “gets perpendicular” at full load. When you compare the alignment of the t-members with c-member perpendicular to the wall, and when the distal end has displaced ~1/4” down, you’ll see in the displaced geometry, the t-member is bending down at the top, and bending up at the distal end- not by much, in absolute distance, but by some. So to fit t-members in a way that they’ll be “straight” at full load. To get the length right, prop the distal end up by ~1/4 – just a bit less – a fraction of a millimeter) than the amount of droop you saw/measured. Cut length to fit that.

The trick comes in the alignment of the ends. At the top, where its going to bend down at load, you want the alignment rotated/angled down by the amount its going to change to under full load. At the distal end, where its going to bend up at load, you want the alignment rotated/angled up by the amount its going to change to under full load. Easy to say/describe. Tricky to actually do with precision. You’re going to have to…work with the concept of what you’re trying to do to come up with a workable way of actually doing these “adjustments” – so that the amount “bent alignment” you put in is matched to (and opposite of) the amount of bending that’s going to happen from t-member stretch, and its equal on both sides.

[_HenryHscioly_]

wow..so i need to get my 1/16x3/32 better built...hm
so, near the base/top, tension members will bend down. Near the distal end, when sagging down, the tension member will be bending left..?
so concave to the inside? like bulging up and out?
wait..actually,
or is it, more like -atan(x) graph? middle section is more vertical since its sagging down, while the ends are still at the same glued angle.
I would need to build it so that it bends up at the base, and from the distal end, bends left/down. This seems very hard...
Looking at what tension-boomilever looks like, if the entire compression member shifts down, then it would be -atan(x) graph..but thats wrong? cuz it only rotates down..I see...
it'll be concave to the inside
but I think this might work. Build with distal end lifted 7/32" to take account for the 1/4" sag. However, to take account for the angle change, build the boomilever to be 14.3cm below the center of the loading hole. And, when When testing the boomilever, pull the compression member down to contact the wall just above 15cm. Hm..actually, maybe no 7/32" distal end lift, Build everything straight and perpendicular, just make it 7/32" shorter(14.3cm). When testing, distal end will essentally be bent up, and it'll all be stragiht once it sags and falls to 15cm. However, .7cm is quite a bit...I wonder if it'll be problematic to bend it down that much and make it stay. If the loading block and bucket were on, then it'll probably stay in place. The rules say the ES must measure clerance <15cm before loading block, but what if you tightened the bolt and its quite wobbbly...(at my regionals they told competitors to put bucket+block on before height and length was measure)

[Balsa Man]

You're right- this is not simple.
No guarantees 1/16 x 3/32nds will work. Puts you right at published tensile strength data w/ essentially no safety factor; higher density pieces should help that a bit. Getting the bends out when you get upp toward full load will help, and its not easy to do.
Adjusting the tension member length to compensate for stretch does not involve moving/changing the wall contact. It stays the same. The shortened t-members mean when you start, the c-members are angled up a bit from horizontal, and when it gets to full load, its rotated down to horizontal.