Pre-stressed makerslide

Extrusions/Tubing/Hinges/Etc.

No real benefit

Postby dzach » Mon May 07, 2012 11:31 am

It turns out this idea is a no-go.
Here is a test I did with a piece of 11.5x19.5 mm extrusion I had laying around.
2012-05-07-015632.jpg
Test setup

2012-05-07-015446.jpg
Initial deflection


Test data:
Code: Select all
Extrusion dimensions: 11.5x19.5x1.5x800 mm
Material: Uncoated aluminium, no other data
Test length: 710 mm
Test weight: 1.6 kg
Tension element: M3 threaded rod

The results:
Code: Select all
Load (kg)   Camber (mm)   Total deflection (mm)
------------------------------
0.0      0.00      0.00
1.6      0.00      1.18
1.6      0.05      1.15
1.6      0.10      1.18
1.6      0.15      1.22
1.6      0.20      1.22
1.6      0.25      1.24

Deflection for a 710 mm piece of rectangular 11.5x19.5 mm aluminum extrusion 1.5 mm thick under an 1.6 kg load is 1.18 mm. After tightening the M3 nuts at the ends of the extrusion and introducing camber, with the rod positioned at the bottom of the extrusion void, the measured total deflection (including camber) rather increases than decrease, which defeats the goal.

Conclusion:
  • Although using makerslide could yield different results, pre-tensioning the extrusion doesn't seem to offer any noticeable advantage with regard to stiffness.
  • It is good to have engineering/test data ;)
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Re: Pre-stressed makerslide

Postby bdring » Mon May 07, 2012 12:40 pm

I am playing with an idea that requires a longish span of MakerSlide in a gantry type arrangement. It will be lightly loaded, but I thought putting a tensioned 6mm threaded rod in the lower t-slot might help the strength. The load direction will be down in the image shown. As the MakerSlide started to bend it would have to stretch the 6mm rod.

I would have plates at the ends with holes drilled for the rod. There would be holes to bolt it to the existing hols in the MakerSlide too.

It is a little hard to see in the image, but the threaded rod is shown in blue at the bottom.

ms_6mm.JPG
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Re: Pre-stressed makerslide

Postby dzach » Mon May 07, 2012 3:48 pm

I intentionally used a thin rod, since I wanted to see if just the tension on its own would reduce the deflection, which was the main idea.
As you mentioned earlier in this thread, using a thicker rod might contribute to makerslide's stiffness positively, given the different structure of the extrusion and the position of the rod.

Before my test above, I did a quick test with an aluminum tube, 10x1.5x810 mm. It was almost impossible to direct the camber to the top and I got a snake like bending alternating along the tube, something seen frequently with soft materials that are put under end tension, e.g. cable jackets e.t.c.

EDIT: I was not able to test deflation on other planes, since I have only one dial indicator available.
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Re: Pre-stressed makerslide

Postby Cre8ivdsgn » Tue May 08, 2012 12:35 pm

Bolting a 1/8" aluminum plate (6061) 1.5" wide along the back of the makerslide would cut the deflection by more than half, I would guess. I took an arbitrary 36" length and calculated deflections for a 5lb load. This gave me a 0.0065" deflection at the center of travel. (Advance warning - I did not do well in statics... I got distracted that semester!) The MakerSlide alone I figured would deflect about 0.0123".

I made a stupid little excel sheet to alow me to play with the numbers a bit. Perhaps Bart might want to include a "MakerSlide Bend Calculator" alongside the belt length calculator.

I have wondered from time to time what a simple truss might do. If you envision the angle iron usually attached to a pan brake to keep it from deflecting too much.
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Re: Pre-stressed makerslide

Postby dzach » Tue May 08, 2012 3:44 pm

Next time I disassemble my Hadron's Y axis I'll try to verify your results.
In the meanwhile, for those of us who got distracted, got married or never took that course, Wikipedia comes to the rescue:
center_loaded_beam.png
center_loaded_beam.png (692 Bytes) Viewed 22641 times

where:

δ = the deflection of the beam
F = force acting on the centre of the beam
L = length of the beam between the supports
E = modulus of elasticity
I = area moment of inertia


which reminds me that Bart has indeed published all necessary data in this post recently:
viewtopic.php?f=30&t=1275&p=11126#p11126

From that post:
Area Moment of Inertia: (mm4): (X) 1.58 x 104 (Y) 6.07 x 104
Ultimate Tensile Strength: 30 ksi
Modulus of Elasticity: 70 GPa

therefore, just make sure the units are correct and the results can be verified theoretically.

P.S.: Not sure if Wikipedia can rescue marriages :mrgreen:
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Re: Pre-stressed makerslide

Postby Cre8ivdsgn » Tue May 08, 2012 9:06 pm

Yup, thats what I used.

Some might want the calc for moment of intertia => I = (b*h^3)/12
b = width
h = height

I'm stuck in the non-metric world and so:
For modulus of elasticity => 70Gpa = 10,000,000 or 10^6 psi
To convert Bart's metric system numbers => 1 in4 = 4.16x105 mm4 = 41.6 cm4 (From Engineering Toolbox website)

Bart's momen of inertia about X 1.58 x 10^4 = 0.03798 in^4
My guess was 0.0393 based off of quick measurements from a chunk of MakerSlide.

Oh crud. The number he has for moment about Y is larger and I suspect his X and Y are not my X and Y. (Unless there is some other issue.)
6.07x 10^4 = 0.14591 in^4.
Hmmm, that about 4x bigger and so for some reason it could be that he chose to calculate the moment of inertia about the Y or I=(b*h^3)/3 instead of the neutral axis.

I checked my mechanics of materials text and I think I am ok, but this stuff isn't my strong suit.

PS - that distraction... I married her.
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Re: Pre-stressed makerslide

Postby mattrsch » Wed May 09, 2012 12:53 am

Your equation for area moment of inertia would be true for a solid rectangular bar, but it won't be for the extruded shape of makerslide. The equation I = (b*h^3)/12 is derived from an integral Ix = ∫[∫((h/2)^2 )dB]dH. In the case of a solid rectangular cross section this simplifies to I = (b*h^3)/12. The simplification does not hold true when the cross section is non-rectangular or has voids and inclusions. it would be very difficult and tedious to calculate the area moment of inertia of a cross section like makerslide by hand, but you can do it by breaking it down into simple shapes and applying the parallel axis therom to add or subtract portions of the extrusion as required. Most CAD programs will calculate the area moment of inertia of a shape for you, and I assume this is where Bart got his numbers. Misumi lists the area moment of inertia of a 2040 extrusion about x as 1.358X10^4 mm^4. You would expect makerslide to be a bit stiffer since it adds the V rails and removes one slot, and that appears to be the case.

I hope this helps! It has been a long time since mechanics!
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Re: Pre-stressed makerslide

Postby Cre8ivdsgn » Thu May 10, 2012 12:12 pm

I had been considering the stiffening of the makerslide by bolting a rectanglar solid slab of aluminum and so I =(b*h^3)/12 made sense. But this equation also assumes that the ends are rigidly fixed so that the neutral surface (where the compression forces and the tension forces cancel) is basically the centroid of the part. I=(b*h^3)/12 applies. Intuitively (a dangerous thing!!!) Bart's moment of inertia is about 4x greater than I would have thought.

I was wondering if the cad program provided a moment of inertia at Y=0 or did it provide it at the neutral axis of the part. If Y=0, then I-(b*h^3)/3 or a value 4x greater, making deflection 4X less.

Its been 25 years since I went through this last! I have noticed in my mechanics of materials book that it gives moments of inertia at both the neutral axis and Y=0, but I am drawing a blank for when Y=0 comes into play - maybe when figuring out yield strenghth? It really has been too long!!! If anyone could clarify this for me, its beginning to gnaw at my mind (and makes me feel a bit like an idiot for even posting this)!!!

And of course what happens if the ends are simply supported versus rigidly mounted like they would be for a machine. When simply supported the beam has a single curve but when rigidly mounted it has three.

Ultimately I might just set up a quick experiment on my bench and see what I get for deflection.
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Re: Pre-stressed makerslide

Postby bdring » Thu May 10, 2012 12:57 pm

FYI: The area moment of inertia calculated by Pro/Engineering is....

Area Moment of Inertia: (mm4): (X) 1.58 x 10^4 (Y) 6.07 x 10^4
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Re: Pre-stressed makerslide

Postby mattrsch » Thu May 10, 2012 7:32 pm

Cre8ivdsgn wrote:I had been considering the stiffening of the makerslide by bolting a rectanglar solid slab of aluminum and so I =(b*h^3)/12 made sense. But this equation also assumes that the ends are rigidly fixed so that the neutral surface (where the compression forces and the tension forces cancel) is basically the centroid of the part. I=(b*h^3)/12 applies. Intuitively (a dangerous thing!!!) Bart's moment of inertia is about 4x greater than I would have thought.

For a solid rectangular beam I calculate 2.67X10^4, so the makerslide is about 60% as rigid as a solid beam in the x direction. There must be an error somewhere in your math (or your conversion factor). The I=(b*h^3)/12 will apply to any rectangular shape regardless of the anchoring conditions. Those are taken into consideration when you pick the appropriate deflection equation: http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf


Cre8ivdsgn wrote:I was wondering if the cad program provided a moment of inertia at Y=0 or did it provide it at the neutral axis of the part. If Y=0, then I-(b*h^3)/3 or a value 4x greater, making deflection 4X less.

Its been 25 years since I went through this last! I have noticed in my mechanics of materials book that it gives moments of inertia at both the neutral axis and Y=0, but I am drawing a blank for when Y=0 comes into play - maybe when figuring out yield strenghth? It really has been too long!!! If anyone could clarify this for me, its beginning to gnaw at my mind (and makes me feel a bit like an idiot for even posting this)!!!



You probably wouldn't use the Y=0 area moment of inertia for a simple rectangular beam. It is a bit easier to work with than the neutral axis area moment of inertia when you are working on a more complicated part, like an I beam, where you calculate the area moment of inertia based on a composite of simple shapes. (you can measure the distance from the bottom of a flange to the neutral axis vs. measuring the middle of a flange to the neutral axis.)

Cre8ivdsgn wrote:And of course what happens if the ends are simply supported versus rigidly mounted like they would be for a machine. When simply supported the beam has a single curve but when rigidly mounted it has three.

Ultimately I might just set up a quick experiment on my bench and see what I get for deflection.


That's a good question. I don't think I have seen any examples of a problem like that worked out by hand. I think the interactions to your support material would become pretty important to the answer. It's also no longer simple beam deflection since increasing the load on the beam will increase tension in the beam like a cable. Probably best answered by experimentation or FEA models. I think for a makerslide system like a laser gantry a beam that is simply supported at both ends would be a very close approximation due to deflection of V wheels. I don't think they can impart much of a moment on the end of the beam.
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