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  1. #1
    Ross77's Avatar
    Lives in Devon, United Kingdom. Last Activity: 7 Hours Ago Has been a member for 8-9 years. Has a total post count of 647. Received thanks 25 times, giving thanks to others 22 times.
    OK so I as a result of a question on another thread here is a very Basic calculator for precision round rail.

    Unfortunately its not the 'bells n whistles' version I had planned but hopefully is OK as a start as time is in short supply at the moment. I'm hoping to add to it and make it a bit tidier as time is available.

    It's only intended as a guide and you will have to make your own decision as to which answer you use, it can be used for calculating the deflection of a design or what the max load would be for a given bar, span and accepted deflection. Comments and abuse accepted and welcome:heehee:

    Now the disclamer.........
    Supplied in good faith and to the best of my knowledge is correct. blah blah...
    Attached Files Attached Files

  2. The Following 2 Users Say Thank You to Ross77 For This Useful Post:


  3. #2
    Cheers ... Just what the doc ordered!!
    Tim G-C

    I disapprove of what you say, but I will defend to the death your right to say it.

    (attrib. Voltaire but written by Evelyn Beatrice Hall "The Friends of Voltaire" 1906)

  4. #3
    Nice one Ross, looks good to me. I fed those numbers into my Gantry Spreadsheet (using tube profile with OD 16 and ID 0, length 900mm and load 250N) and got the same values for second moment of area and vertical deflection.

    Out of interest, I've posted my sheet. It compares different sections and how they deflect due to vertical load and tool cutting applied torsion. Obviously, this spreadsheet only considers the gantry part of the machine.
    Purple boxes are inputs, yellow boxes are outputs. There are graph tabs to visually compare sections.

    Might help other people with the design process, although usual disclaimers apply...


    EDIT: Updated to version 4, with much clearer layout and data entry. Also added different end conditions (from Ross's sheet).
    Attached Files Attached Files
    Last edited by routercnc; 06-07-2010 at 08:23 PM. Reason: updated attachment . . .
    Building a CNC machine to make a better one since 2010 . . .
    MK1 (1st photo), MK2, MK3, MK4

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  6. #4
    Ross77's Avatar
    Lives in Devon, United Kingdom. Last Activity: 7 Hours Ago Has been a member for 8-9 years. Has a total post count of 647. Received thanks 25 times, giving thanks to others 22 times.
    Thanks Barry, lots of equations to steal, Umm i mean use.....:whistling: thats a good idea to have them on one sheet and then linked to graphs. I like it. It wouldn't take to much to add the resonant and harmonic frequencies as well.

    I fed those numbers into my Gantry Spreadsheet (using tube profile with OD 16 and ID 0, length 900mm and load 250N) and got the same values for second moment of area and vertical deflection.
    Thats a relief, I'm good at finding fault with other peoples work but then make stupid mistakes in my own :whistling:

    just a couple of points I'm unsure of.... It states simply supported for the vertical deflection but fixed for the torsion calcs. Are they separate calcs or are you assuming that the hinge joint can resist axial rotation?

    Also I have been after an equation for the torsion constant (k) for a while but I always understood that it is pain to calculate for anything other than a cylinder. The program I have gave a different answer than you for the 80 x 80 x 3mm box section. I make it 1.37 e6!! [1370000] (not had chance to fire up the FEA yet so it might be the free programs I'm using, but it has always been correct before).

    I only ask because I've been trying to think of a way to do what you have done for a while now but have been stuck at the torsion bit and have to resort to individual section checking of FEA. I think the only real solution is to create (or find) a data base of the most used sections (inc extruded t slot) and put them on a pick list in excel

  7. #5
    Ross77's Avatar
    Lives in Devon, United Kingdom. Last Activity: 7 Hours Ago Has been a member for 8-9 years. Has a total post count of 647. Received thanks 25 times, giving thanks to others 22 times.
    The program I have gave a different answer than you for the 80 x 80 x 3mm box section. I make it 1.37 e6!! [1370000]
    Ok found it. It looks like a 'times by 4' has crept in at the begining of the equation. If you remove it the answer becomes 1.37e6 and the torsional resistance goes up on the graph

    Do you mind if I add to it or is a work in progress yourself?

    Cheers
    Ross

  8. #6
    Hi Ross,

    Yes, I'm assuming the hinge can resist torsion. I'm trying to ignore the rest of the machine and just compare the sections themselves to show the differences.
    For bending the load case is simply supported (i.e hinges), with load in the centre.
    For torsion the ends are constrained, and the load is applied in the centre of the span, hence the x4 factor.

    You might have made the same mistake I did initially (or may be I am wrong !), which is to find out the stiffness of a beam with just one end restrained and the other far end twisted. This is not the case with a gantry, where both ends are restrained and the torsion is applied in the centre. I think these two factors will give a x4 stiffness increase. Your FEA may be for a simple beam with one end constrained and the other far end twisted. Let me know if you follow this argument! I guess they are both 'right' in their own way.

    Nice to have a double check on the torsion anyway. I think the x4 is just the end condition and twisting point difference.

    It would be good for Lee to collect this sort of information together and post it somewhere appropriate on the site. I'm very happy for you (or others) to modify this sheet, but as it works well at the moment would prefer to see any mods as a seperate posting (rather than overwritten).

    I use Roark's Formulas for Stress and Strains, which is a weighty book and takes a long time to wade through and get what you want, but it is all in there. Turning it into Excel versions with graphs helps me visualise the effects of different parameters.

    I've been working on another spreadsheet which analyses the principles involved in cnc router (or mill) design, and the effects of geometry and inertia. When I feel ready I'll post this as well . . .

    Barry
    Building a CNC machine to make a better one since 2010 . . .
    MK1 (1st photo), MK2, MK3, MK4

  9. #7
    Ross77's Avatar
    Lives in Devon, United Kingdom. Last Activity: 7 Hours Ago Has been a member for 8-9 years. Has a total post count of 647. Received thanks 25 times, giving thanks to others 22 times.
    Hi Barry.

    Ok I get it now, you have just short circuited the calc and mulitplied the constant by 4. Is that from Roarks or an asumption?

    You might have made the same mistake I did initially (or may be I am wrong !), which is to find out the stiffness of a beam with just one end restrained and the other far end twisted.
    I got confused because the torsion constant is a function of its shape and not the loading condition (being a constant and all ) but thats the joy of engineering eh ? lots of ways to acheive the same result.

    I didnt mean I wanted to overwrite your sheet just asking for permission to use elements of it in another form.

    I've been working on another spreadsheet which analyses the principles involved in cnc router (or mill) design, and the effects of geometry and inertia. When I feel ready I'll post this as well . . .

    Looking forward to seeing that.

  10. #8
    Hi Ross,

    You're right. I just stuck the x4 in somewhere convenient - I only wanted the tool deflection value out at the end, which I guess is what most people might want for design work. If you want the torsion constant out as well then this x4 should go somewhere else in the calculation chain. One option would be to divide the applied torque by 4.

    The x4 is an assumption based on reasoning (it's probably in Roark's somewhere but it is quite a read), but I ran it past a colleague and he came to the same conclusion. A bar with a twisting load applied at one end and the other fixed, will be twice as stiff if the load is then applied to the 'centre' of the span (x2). If the free end is then also restrained, with the twisting load still at the centre, then this previously free part of the bar will now also be able to resist the twist, doubling the stiffness again (x2). This is a total of x4. Technically the spreadsheet should be modified to be more correct (in the workings) although the end results are OK. Do you agree with this logic?

    If you want to post a modified sheet, or use the equations in your own sheet then go for it. It would be great if there was an special design/calculation area on the forum where people who were asking "what sort of section is best for . . " or "how much deflection might I have . . ." could download a simple spreadsheet and get a rough idea. I think Irving2008's motor calculation was a great example of this.

    Barry
    Building a CNC machine to make a better one since 2010 . . .
    MK1 (1st photo), MK2, MK3, MK4

  11. #9
    Ross77's Avatar
    Lives in Devon, United Kingdom. Last Activity: 7 Hours Ago Has been a member for 8-9 years. Has a total post count of 647. Received thanks 25 times, giving thanks to others 22 times.

    You're right. I just stuck the x4 in somewhere convenient
    No problem with that. Thats what its all about re-engineeering equations to suit the problem in hand but it is goood practice to state any changes and the reasoning behind them. stops people like me asking questions

    I'm sure all this could be resolved with a good fashion conversation. pm me your no. or maybe we will have to try the skype option....?

  12. #10
    I tried opening it, but my Excel won't open it for some reason
    Could someone do me a quick calculation
    I am upgrading my Homebuilt Plasma by extending the table, this is the design so far
    The Y Axis has two unsupported solid rails @1500mm long (See Picture) the centre bar is the Ballscrew.
    I still need to add some form of aluminium extrusion to stiffen up the Y Axis Gantry.
    It has the Z Axis assembly hanging off of it and also the Plasma torch, not sure what that will weigh.......
    I think I might need to put some form of support under the rail, I had to do that on my first machine, I had small angles welded at 45 under the rails, and setscrews to bear onto the rail

    Thanks in advance

    Andy
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