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IndustryArena Forum > Mechanical Engineering > Mechanical Calculations/Engineering Design > Young's modulus for Bosch Rexroth, Aluminum frame
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  1. #1
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    Jun 2010
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    1414

    Exclamation Young's modulus for Bosch Rexroth, Aluminum frame

    i am designing a gantry type frame for a new CNC, and I am looking to see how strong this bosch aluminum rail frame is.

    I am not a mechanical engineer, but my buddy is so he needs this info to help me out.

  2. #2
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    Jul 2011
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    [QUOTE= how strong [/QUOTE]

    Ask your buddy what he means with strong. I could calculate tensile strength of extruded aluminum knowing the surface area of the cross section. Brinell or Rockwell hardness may be also available for aluminum.
    If he wants to know how much load he can put on a X long beam, supported on each end, before permanently deforming, likely only a lab test can tell.
    I used to be able to calculate I-beams, but this profile, with multiple 45 degree angles and holes can only be calculated with very expensive software.

    Looking at it from behind my desk, it is a good profile to make a 4'X8' router table of it and safely put my weight in the center of it.

  3. #3
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    Jun 2010
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    1414
    A 40" span for X, 30 for Y
    Click image for larger version. 

Name:	Gantry_00.JPG 
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ID:	140394

    A mill head will be attached to the darker plate. Looking to see how heavy I can go (300 lbs?)

    I will reenforce it all over the place, and the bottom 4" will be in poured concrete.

  4. #4
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    Jul 2011
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    300 lbs easily.

  5. #5
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    Jun 2010
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    How can you tell? It is about the size of a standard wooden 2x4 which would hold the weight (on end) but would it bend on my setup pictured?

  6. #6
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    Jul 2011
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    You are using 2 beams in a fairly large distance, which creates a kind of I-beam effect. The plate appears to be ~8" wide and has to be subtracted from the 40" span as well.
    I'm not saying it doesn't bend at all, maybe 0.010-0.020" or so.
    I'm not telling, but it is a good guess

  7. #7
    Join Date
    Mar 2006
    Posts
    202
    Rexroth doesn't appear to have an easily accessed online drawing/model for that particular profile but if you look at this one, drawings/models can be downloaded. This other one suggests that the outer wall thickness of your chosen profile is 1.5 mm. For a first approximation, calculate the strength characteristics of a simple rectangular section aluminum tube, 45x90x1.5mm. To my eye, it looks like the Rexroth profile should be a bit stiffer/stronger than the simple rectangular box section.

  8. #8
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    Jul 2011
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    You can improve stiffness, if you place a board between the beams.
    Attached Thumbnails Attached Thumbnails stiff.jpg  

  9. #9
    Join Date
    Aug 2007
    Posts
    20
    Typical elastic modulus for aluminum is between 10.3-10.7msi. If you want a decent free database for material properties, check out Online Materials Information Resource - MatWeb.

    Conservatively, the point of applied load that produces the maximum displacement of a fixed/fixed beam is at the center. In reality, your beam will not be truly fixed/fixed, as the attachment interface will have some flexibility and the structure that it attaches to will have flexibility. But for all intents and purposes, fixed/fixed should give a reasonable answer.

    Anyways, back to the displacement equation (I would do an internet search to confirm, as I derived this on the spot and could have made a mistake):
    w(x) = P*x^3/(6*E*I)-P*L*x^2/(8*E*I) for x < L/2
    Where:
    w(x) = displacement at x position along beam
    P = applied load (at center of beam)
    E = Young's Modulus
    I = Area moment of inertia
    L = total length of beam between fixed supports

    In case anyone is wondering, the equation above was derived from the following differential equation:
    (d^2w(x)/dx^2)*E*I = P

    Now if you want to calculate the stresses in the beam, the following equation is used:
    sigma bending = M*c/I
    Where:
    M = critical beam bending moment
    c = distance from the neutral axis (assuming a beam of symmetric cross-section, the N/A will be located at the center of the cross-section, and c will be the distance to the most extreme fiber of the cross-section)
    I = Area moment of inertia

    You can use the above to check for "plastic bending". If the calculated stress at the extreme fiber of the beam's cross section is greater than the yield strength of the material (Fty), then the beam will bend with permanent set as it has entered the material's plastic range. This is obviously not desired.

    Finally, regarding calculating the Area moment of inertia for your beam cross-section, you could use any number of cad programs to do it, or you could subdivide the cross-section into simple rectangular or triangular "elements" which have easy to calculate moments of inertia and then use the parallel axis theorem to combine these "sub Is" into the net cross section moment of inertia. If you have any questions on how to do this, send me a PM and I'll try and help you out.

  10. #10
    Join Date
    Jul 2007
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    1602
    Check this out. It is for 80/20 profiles but should give you a good idea what to expect.

    80/20 Design Tools - Tech Toolkit™ Introduction

    bob

  11. #11
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    Jul 2011
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    This is a nice calculator, if looking at a single beam.
    If you look at the entire construction with 2 beams and not knowing very much about how well the ends are fixed, you'll have to calculate with several integrals and differentials, or quite expensive software.
    I guess, it will come less expensive to buy two 40" extrusions and measure the deflection upon the 300 lbs load.

  12. #12
    Join Date
    Aug 2011
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    0
    A good tool to use (download from this link) is the Rexroth Aluminum Profile Deflection Calculator Deflection Calculator

  13. #13
    Join Date
    Mar 2009
    Posts
    30

    Aluminium Beam Bend Calculator

    You will find the calculator spreadsheet in this post useful.

    http://www.cnczone.com/forums/mechan...te_gantry.html

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