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Steel pole strength

  1. Jul 28, 2011 #1

    I am a newby to all of this, can't find any good answers on the internet, hope someone with experience can help!

    If I have a solid steel pole with a length of 50 cm, hang this from a roof, and want to hang a 7 kg object from this pole, how thick does the pole have to be? If I reduce the weight of the hanging object, how will this effect the thickness of the pole required? If I make the pole longer, can it handle less, the same or more weight? If I use a hollow steel pipe, how thick does the pipe have to be?

    If you do supply a calculation, could you supply this with an explanation of the different symbols of the formula instead of just the formula, as I am really new and would not understand? To use an example, someone told me to calculate the induced EMF is E=2 x length of wire down x radius to wire x Radial Flux Density B from magnets x rotational speed.

  2. jcsd
  3. Jul 28, 2011 #2
    Just to confirm: will your pole be vertical?

    If this is true, this is a case of direct tension.

    The stress in the pole is simply: stress = force / Area.

    The stress will then be an "allowable stress."

    The force will be the "force weight" of 7 kg.

    To make a long story short, your allowable stress = 60% X Yield Stress / Impact Factor.

    The Yield Stress will be based on the grade of steel. Let me know if you need help with this.

    Now the impact factor = 1.5 to account for the extra force when you first hang the object.

    Finally solve: allowable stress = force / Area for the area.
  4. Jul 29, 2011 #3
    Hi edgepflow,

    Thank you for the reply. I have attached a picture, hope this will explain more. The pole will be horizontal, with the object hanging from the centre. The pole will be attached to a roof or flat surface at the top on either side. The object can hang from the middle, the right or left, so I would like to make sure the pole can handle the pressure at any point over the 50 cm length.

    I have a hollow stainless steel pipe here, outer diameter 19 mm, wall thickness approx 1 mm, and I have seen this can handle quite a bit of weight. I am after the thinnest pole that can handle the hanging object.

    OK, thanks for the formula. Is this also applicable where the pole is horizontal, or would there be a different formula? I will most probably be looking at stainless steel, or would you recommend a different type of steel?

    Attached Files:

  5. Jul 29, 2011 #4
    barendfaver, for this setup the stress formula I presented in Post #1 do not apply. However, the allowable stress = 60% X Yield Stress / Impact Factor still applies.

    You will need a "beam formula" for a point load at midpoint. The formulas are as follows (the stress will be much higher than the direct tension case in Post #1).

    max stress = W * L / (8 * Z)

    W = point load at center = force weight of 8 kg
    L = beam length = 50 cm
    Z = Section Modulus = (PI / 32) (OD^4 - ID^4) / OD for a hollow round tube

    So first I recommend trying to calculate the stress for the hardware you mentioned in Post #3. Let me know what you get and I will help you compare to allowable stress.

    After we get this far, I will show you how to figure out the deflection.
  6. Jul 31, 2011 #5
    Hi, sincerest apologied for the late reply.

    OK, so let me try the formulation:

    = (PI / 32) (OD^4 - ID^4) / OD
    = ((3.14/32) x ((0.019^4)-(0.018^4))) / 0.019
    = (0.098125 x (0.000000002476 - 0.000000001890)) / 0.019
    = (0.098125 x 0.0000000005865) / 0.019
    = 0.000000000057553 / 0.019
    = 0.00000000303

    Max Stress
    = W * L / (8 * Z)
    = 8 x 0.5 / (8 * 0.00000000303)
    = 4 / 0.0000000242
    = 165064227.8485

    Did I get this right?
  7. Jul 31, 2011 #6
    barendfaber, good try but I think your stress is high by about a factor of 100.

    On your Z calculation, remember ID = OD - 2 * t (it looks like you subtracted t from OD only).

    Try to keep track of your units. Let me know what you get.
  8. Aug 1, 2011 #7
    Thanks for this. Need your help. I now realised I made a mistake as the ID should be 0.017, however in the above formulation you mention:

    ID = OD - 2 * t

    What is t? Then I can try again! Thanks for your patience so far.
  9. Aug 1, 2011 #8
    t = pipe wall thickness.

    Let me know if you need any help on the rest of this. Just keep plugging away and you will get the hang of it.
  10. Aug 1, 2011 #9
    Great. So ID = OD - (2 x t) = 0.019 - (2 x 0.001) = 0.019 - 0.002 = 0.017

    = (PI / 32) (OD^4 - ID^4) / OD
    = ((3.14/32) x ((0.019^4)-(0.017^4))) / 0.019
    = (0.098125 x (0.00000000248 - 0.00000000142)) / 0.019
    = (0.098125 x 0.00000000106) / 0.019
    = 0.000000000104 / 0.019
    = 0.0000000055

    Max Stress
    = W * L / (8 * Z)
    = 8 x 0.5 / (8 * 0.0000000055)
    = 4 / 0.000000044
    = 91,660,137.19

    Still too high? Just to confirm, 0.019^4 = 0.019 x 0.019 x 0.019 x 0.019?
  11. Aug 1, 2011 #10
    Sorry, still high.

    Yes, 0.019^4 = 0.019 x 0.019 x 0.019 x 0.019.

    But 0.019^4 = 0.000000130321, much different than your result.

    I would suggest to first take the arithmetic out of the process so we can focus on the ideas. Can you set up your formulas on the computer in Excel, MathCAD, C++, or something like this?
  12. Aug 2, 2011 #11
    I have used Excel, but can see now where I went wrong. The way I had it was (((0.019*0.019)*0.019)*0.019), this is why the calculations were so very wrong!

    = (PI / 32) (OD^4 - ID^4) / OD
    = ((3.14/32) x ((0.019^4)-(0.017^4))) / 0.019
    = (0.098125 x ( 0.0000001303 - 0.0000000835)) / 0.019
    = (0.098125 x 0.0000000468) / 0.019
    = 0.000000004592 / 0.019
    = 0.0000002417

    Max Stress
    = W * L / (8 * Z)
    = 8 x 0.5 / (8 * 0.0000002417)
    = 4 / 0.00000193
    = 2,068,702.71

    Did I get it right?
  13. Aug 2, 2011 #12
    One more thing, seems I did not pick that up. In post #4, you state W = point load at center = force weight of 8 kg. I will be hanging an object weighing 7 kg, so there is an extra kg on this figure. Why has this 1 kg been added? How will this relate to different size objects?
  14. Aug 2, 2011 #13
    I think you got it. Confirm your units of stress are in Pa. So your result is about 21 MPa. Now try to figure out what the yield stress of your steel is. I can help with this if you need.

    For some reason, I turned 7 kg into 8 kg. You could rerun your number or just leave as is for "design margin."
  15. Aug 2, 2011 #14
    Thanks, and yes please, if you could help with the yield stress. Would this be the formulations in post 2?

    Now that I have the formulation correct in Excel, it should be easy for me to change some details. This was really helpful as this will allow me to play with different weight objects as well as different size poles. Just one quick question, how would the formulation change if I had to use a solid steel pole instead of a hollow pipe?
  16. Aug 2, 2011 #15
    I am glad you have this running.

    Yes, post two appies. So we need to figure out the yield stress. This requires the spec of the steel. For example, ASTM A106 is a spec for seamless carbon steel pipe. By spec, the minimum yield strength is 30,000 psi or 206.8 MPa. But this is only an example. Do you have a record of your material?

    For a solid steel pole, again use Z = Section Modulus = (PI / 32) (OD^4 - ID^4) / OD

    with ID = 0.

    This gives: Z = (PI/32) OD^3
  17. Aug 2, 2011 #16
    One other point.

    You should also consider the weight of the bar in the calculation, especially since you are considering solid bars.

    It is a safe simplification to add the total bar weight to your 7 kg point load. That is, treat the bar weight as a point load (even though it is actually distributed).

    This simplication will keep the math under control for now.
  18. Aug 3, 2011 #17
    I have been around the shop where I bought the pipe, but unfortunately the only details on this is just some irrelevant numbers. I have found this site as an example: http://www.westyorkssteel.com/, and this one, http://www.metals4u.co.uk/products.asp?cat_id=81 [Broken]. But neither seem to have codes, or am I missing it again?

    Thanks for the extra info re the weight of the bar. I will include this in the calculation when we know what I need!
    Last edited by a moderator: May 5, 2017
  19. Aug 3, 2011 #18
    Since we don't know the exact specification of the steel, I will suggest the following "safe" value:

    yield stress = 110 MPa.

    So try to compute the allowable stress as we discussed earlier. How does this compare to your calculated stress?
    Last edited by a moderator: May 5, 2017
  20. Aug 4, 2011 #19
    Hi edgepflow

    Will need your help with this! OK, so the minimum yield strength for a carbon steel pipe is 206.8 MPa.

    For the formulation allowable stress
    = 60% X Yield Stress / Impact Factor
    = 60% x 206.8 / (7 * 1.5)
    = 124.08 / 10.5
    = 11.82

    Is this right?
  21. Aug 4, 2011 #20
    The value of 7 is not needed since it was accounted for in the stress calculation.

    Try to update and compare to your calculated stress.
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