Will a 500mm 20x20 V-Rail handle a 70oz Nema 17 without any deflection from true? One thing I forgot all about when I designed my printer I am about to build (just waiting on the 500mm 20x20 V-Rail to go back in stock). How about as loading bearing so think of a 70oz Nema 17 on top of the V-Rail standing vertically?

LOL, no if a Nema 17 weighed 70 ounces look out so my bad. It is a Kysan 112409o so ~13 ounces. Two motors would be 26 ounces at say 250mm, or directly in the middle. I suspect load bearing is where extrusion would laugh at the puny weight but not sure as I really do not want any deflection due to weight in any direction and I am unsure how much weight the V-Rail can handle before it starts to sag if even by a little, or start to bulge.

That's a loaded question (pun intended). I can calculate it for you later. I'll need to grab a CAD file for the crosssectional areas to do the math. Or, if you have the crosssections and feel inclined, you can run through numbers yourself and report back to us: https://www.engineersedge.com/beam_calc_menu.shtml https://www.engineersedge.com/beam_bending/calculators_protected/beam_deflection_2.htm btw, these are for deflection, which I know concerns you most. To break it is different.

I think it would be great to get a chart showing all the standard openbuild vchannel extrusions and the standard lengths that shows estimated deflection amounts based on a certain weight load (10lbs or something) I wish i knew how to calculate this

There are plenty of online calculators. (example) Use E = 10000 ksi and c = half of section depth. The problem is getting the I value. Someone with the right software can do it in a few minutes but few people have that software.

Misumi has some of that data on their profiles. http://us.misumi-ec.com/vona2/detail/110302683830/?Inch=0&CategorySpec=00000042733::a I know Solidworks can do it, I just don't feel like driving up to work atm.

Same here it would take the guess work out and the oh, crap when what you built needs to be redone and just hope the next one works and when it doesn't... That moment of inertia is a PITA to find and to do it for real requires tools most people do not have. The real pros do not use formulas they actually use tools and measure but that is in a lab environment with lab sort of test equipment too. Above my pay scale and the same as SW is at work with a friend who lets me borrow it in its down time.

The idea was to take the data from misumi and run through some numbers for estimates until the ob profile data is in.

I am not sure V-Rail and theirs wouldn't be the same though because there is less metal and other differences that may affect the numbers. edit: left out the n't on would.

Please have a look here for some of V-Slot profiles deflection values calculated by an OpenBuilds member using SolidWorks You will need to scroll to the bottom of the page and click on the 'Spoiler: Click for deflection values' button. (We put them in this form as the list very long Hope this helps. Mark SolidWork's material property output for designers (all based on 1000mm lengths): 20 x 20mm Mass = 409.25 grams Volume = 151574.03 cubic millimeters Surface area = 186005.04 square millimeters Center of mass: ( millimeters ) X = 0.00 Y = 0.00 Z = -500.00 Principal axes of inertia and principal moments of inertia: ( grams * square millimeters ) Taken at the center of mass. Ix = (0.00, 0.00, 1.00) Px = 33975.27 Iy = (0.00, -1.00, 0.00) Py = 34121145.12 Iz = (1.00, 0.00, 0.00) Pz = 34121145.12 Moments of inertia: ( grams * square millimeters ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 34121145.12 Lxy = 0.00 Lxz = 0.00 Lyx = 0.00 Lyy = 34121145.12 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 33975.27 Moments of inertia: ( grams * square millimeters ) Taken at the output coordinate system. Ixx = 136433617.56 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 136433617.56 Iyz = 0.00 Izx = 0.00 Izy = 0.00 Izz = 33975.27 20 x 40mm Mass = 784.76 grams Volume = 290650.48 cubic millimeters Surface area = 322653.44 square millimeters Center of mass: ( millimeters ) X = 0.00 Y = -10.00 Z = -500.00 Principal axes of inertia and principal moments of inertia: ( grams * square millimeters ) Taken at the center of mass. Ix = (0.00, 0.00, 1.00) Px = 153265.52 Iy = (0.00, -1.00, 0.00) Py = 65427945.83 Iz = (1.00, 0.00, 0.00) Pz = 65518037.67 Moments of inertia: ( grams * square millimeters ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 65518037.67 Lxy = 0.00 Lxz = 0.00 Lyx = 0.00 Lyy = 65427945.83 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 153265.52 Moments of inertia: ( grams * square millimeters ) Taken at the output coordinate system. Ixx = 261785590.26 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 261617022.79 Iyz = 3923781.54 Izx = 0.00 Izy = 3923781.54 Izz = 231741.15 20 x 60mm Mass = 1131.21 grams Volume = 418965.77 cubic millimeters Surface area = 460509.94 square millimeters Center of mass: ( millimeters ) X = 0.00 Y = -20.00 Z = -500.00 Principal axes of inertia and principal moments of inertia: ( grams * square millimeters ) Taken at the center of mass. Ix = (0.00, 0.00, 1.00) Px = 428910.91 Iy = (0.00, -1.00, 0.00) Py = 94313193.29 Iz = (1.00, 0.00, 0.00) Pz = 94650313.72 Moments of inertia: ( grams * square millimeters ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 94650313.72 Lxy = 0.00 Lxz = 0.00 Lyx = 0.00 Lyy = 94313193.29 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 428910.91 Moments of inertia: ( grams * square millimeters ) Taken at the output coordinate system. Ixx = 377904690.91 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 377115087.44 Iyz = 11312075.77 Izx = 0.00 Izy = 11312075.77 Izz = 881393.94 20 x 80mm Mass = 1478.83 grams Volume = 547713.28 cubic millimeters Surface area = 598572.99 square millimeters Center of mass: ( millimeters ) X = 0.00 Y = -30.01 Z = -500.00 Principal axes of inertia and principal moments of inertia: ( grams * square millimeters ) Taken at the center of mass. Ix = (0.00, 0.00, 1.00) Px = 931509.10 Iy = (0.00, -1.00, 0.00) Py = 123295759.81 Iz = (1.00, 0.00, 0.00) Pz = 124106727.45 Moments of inertia: ( grams * square millimeters ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 124106727.45 Lxy = 0.00 Lxz = 0.00 Lyx = 0.00 Lyy = 123295759.81 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 931509.10 Moments of inertia: ( grams * square millimeters ) Taken at the output coordinate system. Ixx = 495145025.43 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 493002227.07 Iyz = 22189782.16 Izx = 0.00 Izy = 22189782.16 Izz = 2263339.82

Thanks Mark. That saved me a little time tomorrow. I hadn't realized that data was avail. There ya go Dark. Plug in the numbers and report back. joe

I have absolutely no idea what to do with those numbers but I did run a simple deflection on some V-Rail using 1M and the material was 6063 T5 with 10 Newtons of force applied in the middle and can this be right that the maximum deflection would be .000433342 meters? Round up for real world sort of errors and that would be half a millimeter of deflection using whatever Solidworks holds for 6063 T5 material. Is this about right? 10 Newtons is about 2.25 pounds of force. Using 4.4 Newtons (about 16 ounces) I came up with .00019067M or almost .2mm of deflection.

Here is a a great blog post by Brandon at SMW3D ( Brandon's the guy who helped out by sharing these SW deflection values )that should help explain it in more detail. http://www.smw3d.com/blog/designing-your-build-with-smw3d-calculating-deflection/ Mark

About to do the math and what was throwing me is that all of those numbers had no place to go inside SW so after the site (thank you for the link) I can see it is just time for a calculator and I do see he also used 6063 T5 and thanks to all of this I spotted an error (I posted to him about this) with the revised parts library for Solidworks because it has the V-Rail using the 6061-T6(SS) material which is way different. What I did was use the built in beam deflection tool in SW2014 so time to see how much it differs from these calculations.

I'm glad you're running through the math. don't have the time today. I'm getting slammed. Ohh Monday Joe

Something interesting as I was putting in the numbers (I hate how he used imperial units) into Solidworks' simulator using 18 inch 20x40 V-Slot at 2 pounds (I converted all of these to their metric equivalents) and the number I received back was 1.36758e-005m of deflection. Now convert that to the imperial system and you end up with 0.0005384173228 in. Is this a coincidence or what but "we are left with 11664/1100000 = .0106 inches/2 (for two wheels) .0053 inches of deflection. " I see that in the result but it is off by a factor of 10 and I never told it two wheels. So, which is right his or Solidworks or is there something in Solidworks' material numbers that is off? I will leave this to one of you math geniuses but it is something I would like to know so I can correct it if needs to be as I use the simulator sometimes.

For a 500 mm 20x40 length with fixed ends. 10N force. Deflection simulated out to be .0004918mm Load area was in the middle and was .005" wide. So for a 22lb load hanging from a piece of twine this would predict approx. .005mm of deflection or around .0002" Does this sound about right to anyone? Seems legit to me. Edit: force applied to the beam in its vertical orientation. Material selected was 6061 t6. This may not be the aluminum grade the extrusion is made from.

I am sitting here trying to figure out what I am doing wrong in Soildworks' simulator because none of the numbers I am getting match up. Do these numbers jive with what is being written? E looks alright to me though it is coming out to 10007603.90337 lbs/in^2 edit: Here are the same numbers in Imperial

I used 6061 t6. The modulus is the same though. Do you have the ends fixed and are you distributing the force over the entire length?

Ends are fixed and I used this technique on one of the V-Rail pieces from the library using 6063 t5 -

The way I did it. Fix the ends. Also place a narrow surface on top of the beam in the center. Under forces choose this surface to concentrate your load.

Easy enough to do (I'll use a 20mmx20mm square right in the middle but how thick?) and I will report back when I get 'er done.

.001" or less. I think you can use a surface as it has no thickness. .001" of material shouldn't have much of an effect though.