Background:
A while ago I purchased a Creality 3d printer. I tried doing some flexible filament printing (pain in the butt and didn’t work very well). More recently I have been trying to print ABS but the machine can’t get hot enough on the print bed to get good adhesion with the filament so I decided to put together details for an enclosure (also gives me an opportunity to install a mini dehumidifier in the enclosure #PowerBill_IsGonnaSuck). I don’t have regular access to machine tools so I am looking for something I can order the piece parts to and then just assemble it. Unfortunately, the last iteration of this came out to over $1,000 without even including the panels, so I’m gonna replace a lot of it with 3d printed parts that I’ll make out of PETG or something. Depending on temperature/structure loading, I may end up with some of the pieces sagging, so if necessary I can mod the element design to incorporate a steel core to maintain form when heated.
Solidworks model of full construction and basic cost on the right organized by most costly element (angle bracket that’s used everywhere is most costly).
Rationale:
So to swap out the aluminum construction used in the iteration above, I would like to take a rough estimate for a structurally equivalent form from assumed values of PETG with no thermal effects considered.
Aluminum 6105-T5:
S.yieldAL = 35ksi
E.AL = 10.2 x 10^6psi
T-slot Framing Properties for aluminum section (Solidworks did the area and moment of inertia calcs)
PETG:
S.yieldPETG = 5ksi
E.PETG = 0.3 x 10^6psi (found online and may actually be more of a tangent modulus, would like to look into this later)
As for the actual profile, we would want to scale the profile so that it’s stress state is loaded to the same fraction of yield in shear and bending for the same load applied (will look at deflection similarity later) as the aluminum construction.
So in direct shear, V/A is the basic relation for shear stress (VQ/It isn’t anymore helpful here). Solving for the PETG Cross sectional Area will result in a minimum cross sectional area to be equivalent. Since the yield stress of Aluminum is 7x PETG, that would mean the area needs to be 7x that of the Aluminum to be in the same fraction of stress state to yield. So basically, the length dimensions in the cross section above need to be ~2.75x larger. After applying this to the cross section in Solidworks (with some minor changes), the following section is produced.
Approximate PETG section equivalent for shear to the 1in T-Frame solution
Similar drawing to that provided for the Aluminum T-framing
In bending the equation becomes My/I = Stress. This means that the I/y term (S) dictates stress. With the same moment applied as the Aluminum I am looking for a Stress that is 1/7th that of the Aluminum so my S term must be 7x that of the Aluminum. With the y term being 1.5 from the centroid of the section, that means the I must be 7 (0.04596in^4) (1.5in) / (0.5in) = 0.96516in^4. From the section shown above we can see that we actually have a section that has a moment of inertia of 3.87in^4 so that is better than equivalent.
I’ll check the deflection properties in a follow up posting. Then we can actually go ahead and start re-designing the enclosure to take advantage of the 3d printed parts in lieu of the aluminum construction.