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Help calculating required material thickness

  1. Feb 25, 2016 #1
    Hi all, I hope you can help... I am by no means smart enough to calculate how thick the material needs to be for a project I am working on, usually I would go by trial end error, but I don't have the budget for that unfortunatley.

    I'm am trying to work out what steel thickness I would need for a channel to carry a weight of anywhere between 250- 500kg.

    I would be using a "top hat" channel profile. https://www.google.com/search?q=top+hat.+channel&prmd=sinv&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiMwd6K3pPLAhXH6xoKHXqxATEQ_AUICCgC&biw=1024&bih=704#tbm=isch&q=top+hat+channel+dimensions&imgrc=lLp6tg3Q2HM6rM%3A

    My dimensions following the legend of the above image are :

    B: 200mm (plus thickness of material i.e. Inside dimension is 200)
    H: 65mm
    C: 60mm

    The overall length of my channel will be 2000mm (2 meters)

    It will be supported on all 4 corners, and on both sides on the middle (6 points of support), it will attach to the "c" portion of the channel.

    The whole contraption will carry the weight at two points between 30 and 300mm from either end of the length.

    ( I am trying to build a mobile workbench for working on motorcycles... So the channel would be attached to hydraulic jacks, the motorbike would be ridden onto the channel and secured. Then the channel jacked up to a civilized working height, and allow me to get under the bike too)

    Unfortunately, I have no idea what thickness I should be ordering my channel in that would make it strong enough to carry the weight without bending and buckling?

    Hope some one can help!

  2. jcsd
  3. Feb 29, 2016 #2


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    Not really my field so I might not be able to do all the maths but here is the procedure I would follow.

    First off I would see if there is an existing lift I can measure and copy. Why re-invent the wheel?

    Failing that..

    Start with a diagram of the set up...
    Bike lift.png

    That can be simplified to give this worse case...
    Bike lift1.png
    I assume the load on each wheel is equal (500Kg * 9.81 = 5000N, divided beween two wheels is 2500N). Perhaps double that to provide a safety margin?

    Then I think you can add more information to reduce the options, for example, do you have an idea of the vertical depth of the U channel you want? Must it be less then the depth of the tyres? Clearly if very shallow it will be harder to keep the bike wheel in the channel as you roll it onto the lift and the metal will have to be thicker. Likewise the width of the channel? Bit wider than common wheels?

    If the beam was made weaker and weaker it's likely that the beam will bend excessively before it fails. So the deflection is likely to be more important than absolute strength. This is true in a lot of situations (eg beams in houses). You will need to decide on the maximum deflection you can accept in the middle.

    Then look at the equations like these...


    That gives an equation for max deflection Dmax which is..

    Dmax = WL3/48EI

    You can rearrange it to give an equation for the required I (the moment of inertia)

    I = WL3/48EDmax

    The notation is explained on that page.

    Then this page gives an equation for the moment of inertia of a U shape channel in terms of it's dimensions..


    It looks complicated but that's because they have assumed that the walls taper and might be a different thickness to the base of the U. If you assume the U channel is of uniform thickness (eg it's bent from sheet metal) then it will simplify quite a bit.

    I think if you plug in some numbers for the preferred depth and width and the modulus of Elasticity E for the material (eg steel) then you should get a value for the thickness of the material needed. It's more likely you will have to find out the dimensions of beams that are in the right ball park and re run the numbers until you get a combination that provides the required moment of inertia calculated above.
  4. Feb 29, 2016 #3


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    There is another equation for the moment of inertia here.


    If I done my sums right and made some reasonable assumptions (such as t1 << H) then I simplifies to..

    I = H3t1/2

    H is the depth of the beam
    t1 is the thickness of the metal.

    Hopefully someone else can confirm all this because as I said, this isn't my field.
  5. Feb 29, 2016 #4


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    Perhaps look at something like a Universal Beam (UB) from a builders merchant (used on it's side). They are I section rather than U section but might be cheaper than a custom beam?
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