Strength of Scale Models

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My experiment is from the "Stressed Out? Take a Break with this Project" idea. I'm testing the strength of different shapes of plastic model pieces. I've tried the 1/4 inch pieces, but they won't break. They just bend. Any suggestions on what size pieces I should try so that they will break? I don't want to spend too much more money on model scale pieces. Thanks

[paulsdecarli]

It looks as if you have found the flaw in the experiment. Some plastics (polystyrene, for one) are very brittle, and some (like polyethylene and polypropylene) are very ductile and flexible. When my son was small, all of his models were made of polystyrene (it glues well). It appears that you have found models made of a more ductile (or rubbery) plastic. I expect that the experiment was designed for polystyrene models.

Here is a suggestion. Look up the meaning of those recycle marks (the numbers in the triangle) on plastic items. How many different plastics can you find in your household recycle bin? What can you learn about each type? Are you getting ideas for a project?

[deleted-71588]

Another approach would be to alter your definition of strength for your experiments. The usefulness of a "beam" for most applications is NOT the catastrophic failure point but some acceptable deform limit. For example, floor joist systems are typically design rated for length/360 deformation for a 40 or 50 psf (pounds per square foot) live load (people and moving things) plus 10 psf dead load (stationary objects). If a 10 foot span deforms down more than 1/3 of an inch, that would be considered a failure.

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I went ahead with the experiment, but changed it a little since the rods wouldn't break. I tested a hollow circle tube, a solid rectangular rod, and an angle shaped rod. I added the same amount of weight in increments to each rod and measured how many inches they bent from the original straight position. Now I'd like to come up with a calculation. I've tried researching strength to weight ratios, but am not sure how to do it or if it applies.

Also, 2 of the 3 angle rods actually did break once I got to 25 pounds.

[deleted-71588]

The "clamping" or "fulcrum" technique in your tests will slightly alter how an engineer would interpret your data. If you simply layed both ends of the various beams on something at both ends without clamping it down and put a weight in the middle, your data represents a "free ends center load" case (the simplest one).
1) For weights that deform the beam such that the beam returns to its original position when removed, the typical representation is to specify the "span" length (distance between the inside edges of the end supports), the "depression" distance (how much the center of the beam was temporarily lowered), the weight, and the cross sectional area of the beam (the area of the material exposed at the end if it were cut perpendicular to the length of the beam), and beam density (typically weight per unit of length of the beam itself).
2) For weights that defore the beam so that it does NOT return to its original position when removed (permanently deformed), you have exceeded the "plastic elasticity" http://en.wikipedia.org/wiki/Plasticity_(physics) point and this is a type of "failure". In this case, the "parameters" are the span length, cross sectional area, beam density, and minimum weight required to permanently deform the beam.

If you clamped both ends of the beam to the end supports, you may have a different case (more complicated). If the clamping force was such that the beam end deformed permanently, you have "crush" failure. If the clamping force was sufficient to prevent the top of the beam from moving , you have a "pocket end center load" case. This case typically will hold more weight with less center deflection because it requires the top of the beam at the ends to stretch and at the middle to compress while the bottom of the beam at the ends to compress and the middle to stretch. In the free end case, the ends of the beam are not constrained so the ends do not have to deform to allow the middle to compress on the top and stretch on the bottom.

For cases where there is only temporary deforming, the typical engineering approach is to determine the weight per unit length that a beam can support with precisely span/360 of deflection. Then when comparing two different beam types, the engineer will use some "cost" function that involves the weight of the beams and/or amount of material. The weight of the beams used in a design add cost in several ways. You have to move the material from where it is made / processed to the construction site. You have to move it from the site delivery point to the position it occupies in the application. The weight of the beam has to rest on other structures and ultimately on some "footing" and soil. The heavier the structure, the bigger the footings and soil area required to support it. From a production standpoint, if the material is extruded or molded, the amount of raw material and the cost of the material per volume used affects the cost so coming up with a cross section shape appropriate to the applications with the least amount of material for typical loads is appropriate.

For your project, you will have to come up with your own "cost" or "comparison" function.

Hope this helps.