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A theoretical question: let's say that a tower is built that holds the 15 kg load. If that tower is separated into two sections, the base part and the top part, should both of those parts be able to hold the 15 kg (as in the base and the top both separately hold 15 kg) , or should the load they hold combined add up to 15 kg (as in the bottom holds 8 kg while the top is able to hold 7 kilograms.)? Thanks.

phillies413 wrote:A theoretical question: let's say that a tower is built that holds the 15 kg load. If that tower is separated into two sections, the base part and the top part, should both of those parts be able to hold the 15 kg (as in the base and the top both separately hold 15 kg) , or should the load they hold combined add up to 15 kg (as in the bottom holds 8 kg while the top is able to hold 7 kilograms.)? Thanks.

The upper and lower towers are essentially two substructures in series, and each will hold the 15 kg. The legs of each substructure are elements in series and each leg will have to hold at least 15 kg / (no. of legs). That is, if you are building a tower with a square or rectangular base, each leg will have to take 3.75 kg, and if triangular, each base will have to take 5 kg.

The process of structural design involves checking each member, and connection, for strength and stability.
The term strength is used to indicate the ability of the member to carry loads without any failure in its material. By material failure I mean the member’s fibers can no longer handle the stress caused by the force.

The term stability is used to indicate the ability of the structure, and its individual members, to maintain their geometric integrity (form) under the load. For example, a structure that is not properly secured to its foundation is considered unstable. Or, if a structural member bends excessively, it is considered unstable. In structural engineering, it is common practice to view buckling (excessive deformation of a member) as a stability, not strength, condition.

Since buckling is a governing condition for tower design, it might be useful to come up with a series of graphs for determining the required un-braced length of a compression member based on wood density. Here is a graph that shows the maximum un-braced length of a compression member as a function of density.



The graph shows a series of curves, each for a specific member load. For example, the top curve is for P = 20 N. That is, if the compression member carries a force of 20 N (about 2 kg), then the curve can be used to figure out the maximum un-braced length of the member. Here is an example for illustrating the graph use.

Example
Let's assume that the top part of a tower is a rectangular prism consists of four main compression members each carrying a force of 37 N (3.75 kg). Each member has an 1/8” x 1/8” cross-section, and is 35 cm in length. How many cross bracings along the length of the member are needed for preventing it from buckling?

Solution
1. Calculate the volume of the member: (3.175 x 3.175 x 350 = 3528 cubic mm. Or, 3528/1000,000,000 cubic meter.
2. Weigh the member (say it weighs 1 gram)
3. Calculate the member’s density ( 1/1000)/ (3528 / 1000,000,000)= 283 kg/cubic meter
4. Use the graph to determine the maximum un-braced length of the member.
Since P = 37 N, we should use the second curve from top.
From the graph, for a density of 280, the maximum un-braced length is read to be about 130 mm (13 cm).
Therefore, I would brace the member at two (2) interior points, at 35/3 = 11.67 cm intervals.

In case you are wondering, here is the equation I used for generating the graph:



This last equation, which relates density of balsa to its modulus of elasticity, was reported in the Journal of Strain Analysis in 1976. I’ve the complete citation, if you are interested.

Thanks SLM, that will definitely be helpful in determining the cross bracing pattern.

Comment on the graph. It assumes the balsa for all the 1/8 by 1/8 sections exactly follow the stiffness vs density curve for balsa (same Young's Modulus in technical terms). I'm assuming.

But, balsa, only approximately follows those published curves, varying by plus or minus 15%. That's one of the reason's I'm advocating the importance of selecting for stiffness when building, not just desiging. You can be overbuilt if using stiffer wood than went into those calculations, or seriously underbuilt and have built in predictably premature failure.

In addition, I was mentioning the trade off for increased stiffness for larger cross sections vs lower density and Young's modulus. Equally interesting (since this is a weight efficiency competition) would be curves of buckling length vs constant weight cross sections. This would give you a better idea of density to select AND what cross section.

Hmm, the equations are all there, have to play with that AND see if I can include a nifty graph.

Hope that's clear.

Jeff Anderson
Livonia, MI

OK, way too much time spent today trying to figure out a nice picture. I failed, too many variables and I've never been good at shrinking that down to simple 2-d stuff.

My premise. The graph is right, but misleading. Denser balsa has a higher Young's modulus, leading to longer free lengths without buckling, less cross bracing, true. With minimal thought, this could imply that denser balsa is better. But, this does not take advantage of geometry and less dense balsa is in fact better in compression for tower uprights.

OK, so no graphs, lets just take a few examples. Available balsa density ranges from around 60 kg/m3 (3.5 lb/ft3) which is very hard to find to 160 kg/m3 (10 lb/ft3) easy to find to 320 kg/m3 (20 lb/ft3) also rarish. So lets pick just these points. And only the 40 N load to represent the upper uprights on a typical tower carrying 1/4 the total load each. And lets just look at div C where the uprights are 35 cm long.

So, a 1/8th piece of 60 kg/m3 balsa from the chart (or directly from the equations) gives a max of 57.5 cm before buckling. Not much you can do with that, it is already as low density as you can go. But, how much cross bracing is needed. For simple X's with top and bottom cross piece that works out to 7 cross brace bays and 156 cm of bracing per side on the upper alone. Sounds terrible, but wait.

OK, lets look at that 160 kg balsa, 1/8th square. That give's 85 cm before buckling. What happens if you use a piece of lower density, say 60 kg/m3 and increase the cross section to equal the weight. This takes you to a 7/32 square and increases buckle free length to 115mm! Sounds impressive, right. So, for the same weight, you've gone from needing 4 cross brace bays to 3. Problem is, you've almost tripled the weight, and for the first case still need 108 cm of bracing, in the second, 100 cm of bracing. Not worth the trip unless your bracing is really heavy, which it shouldn't be (which is a whole 'nother analysis, not today thanks).

So, what about the 320 kg/m3 balsa. Does something different happen there? That 1/8th stick goes 143 cm, still needing 3 bays and 100 cm of bracing. What happens with lower density? Now for the same weight you can use a 9/32 cross section for the same weight in 60 kg/m3 balsa. This gives a 214 mm buckle free length, only needing two bracing bays. But that only reduces length of bracing pieces to 89 cm. A 5 fold increase in weight for halving the bracing. Again, not a profitable trade-off.

Summary, the lowest density, thinnest cross section uprights you can find, suitably cross braced (for which that chart is great), will be better than denser uprights with less bracing. Warning, there that curve of bracing length vs density is non-linear and is STRONGLY suspect (and I visualize this stuff well in my head and take hours to prove it and don't have the time right now) that at some point, going thinner with 60 kg/m3 balsa will increase cross bracing to the point it isn't worth it. I'm just not sure where that point is.

Oh, and don't forget, at some point you have to execute. I guarantee NO piece of balsa you have will fit the equations exactly. See the building string for how to deal with THAT

Jeff Anderson
Livonia, MI

Just a note for all of those doing theoretical calculations. Since the moment of inertia of a section is to the power of 4, it is worth working with actual and not nominal dimension. For example my 1/16 in. square inches have a nominal dimension of 25.4/16 = 1.59 mm, yet the section size I measured for my sticks is 1.32 mm. The moment of inertia of the nominal section is 1.59^4/12 = 0.529 mm^4 whereas the moment of inertia of the "as purchased" section is 1.32^4/12= 0.254 mm4, which is less than one-half! So, put your calipers to work for you and determine the actual dimensions of your cross section. This is not a negligible difference!

I agree with Jeff, such equations and graphs do not provide an exact answer to the problem. In fact, if used improperly, they could lead to an unfavorable result. In most engineering disciplines, theoretical formulations and methods are meant to provide (1) a vehicle for better understanding the relationships among the problem parameters, and (2) a systematic way of solving complex problems. But, for engineering problem solving, there is no substitute for insight gained from experience. I believe, the best strategy here is to use a balanced mix of theory, hands-on experimentation, creativity, and thoughtful reflection and expert advice to ensure a rewarding learning experience.

Back to the technical stuff …

The graph I showed above simply sheds light on the relationship between wood density and un-braced length of compression members. And, it enables one to determine a theoretical value (read an educated guess) for the un-braced length, if a 1/8” x 1/8” section is to be used. Similar graphs for other section sizes can be generated and used as a design aid for determining initial values for size and length in the design cycle. Obviously, such theoretical values must be put to test experimentally, and when necessary, refined or revised accordingly.

Here are two additional “Density vs Un-braced Length” graphs: one for 1/8” x 1/16” section and one for 1/16” x 1/16” section.




These graphs are generated using the following equation.



By the way, let me know if you want to see the derivation of any of these equations.

Using the above equation, it is possible to derive an equation for graphically showing the relationship between the size of a compression member and its maximum un-braced length. The equation is:



Below is a series of curves each for a different q value. They shows how the maximum un-braced length increases with member size, when the member is subjected to a compressive force of 40 N.




Each curve represents all the acceptable (un-braced length, member size) pairs that result in the 35-cm-long member having the same weight. For example, using the bottom curve, I see the following possible choices for the member:
Size: 1/32” (0.793 mm), maximum un-braced length: ~30 mm
Size: 1/16” (1.587 mm), un-braced length: ~ 45 mm
Size: 3/32” (2.381 mm), un-braced length: ~ 58 mm
Size: 1/8” (3.175 mm), un-braced length: ~ 65 mm
Size: 3/16” (4.762 mm), un-braced length: ~80 mm
size: 1/4” (6.35 mm), un-braced length: ~90 mm
.
.
.
Each of these choices prevents member buckling and results in the entire member having a weight of 0.4 grams. However, keep in mind that buckling is not the only mode of failure here. The member could fail due to the lack of adequate compressive strength.

Also, these charts and equations cannot be used blindly, you need to be aware of their underlying assumptions and limitations.

Let me know if anything here needs to be clarified.

What would you say is the reference density of balsa? From a bit of research, i got numbers around 150kg/m^3, does that sound about right?

lllazar wrote:What would you say is the reference density of balsa? From a bit of research, i got numbers around 150kg/m^3, does that sound about right?

We recently bought several medium and heavy density balsa sheets from SpecializedBalsa.com. The density of the sheets labeled "medium" seems to be around 250 kg/m^3. For the "heavy" sheets, we are getting numbers around 400 kg/m^3.