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.