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Thread: Calculating required diameter of rod to support a given load

  1. #1
    Associate Engineer
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    Calculating required diameter of rod to support a given load

    Hi!

    Trying here to calculate the diameter d required of a length L=800mm solid steel rod, supported at both ends which needs to safely hold W 2000kg (19620N ) at the centre.

    Details of the rod ( structural steel ), taken Modulus of elasticity E at 200Gpa and Yield Strength σ at 250Mpa

    Using moment of inertia I = (πd^4)/64

    Maximum stress σ = yWL/4I. Where y would be d/2 , maximum distance from neutral axis

    Messing around with these two equations making d subject, I get;

    d = cuberoot( 8WL / πσ ) . Inserting all numbers I got a diameter of 54mm which seems a bit excessive to me? Anything I am doing wrong here? Thanks a lot.

  2. #2
    Administrator Kelly Bramble's Avatar
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    Not doing the math, 800 mm is a very long pipe.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  3. #3
    Lead Engineer
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    I calculated your calculation in reverse, as shown below, using your 54 mm bar size and the resulting stress = 134.54 Mpa which gives you a Safety Factor = 1.86 for that bar size.

    In US units:

    Sy = 36,250 psi / E = 29E6 / L = 2.63 Ft = 31.56 in / d = 2.13 in / P = 4409 lb

    Z = .098d^3 = 1.7828

    S = PL/4Z = 139,148 /7.131 = 19,513 psi = 134.54 Mpa

  4. #4
    Lead Engineer
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    NOTE: Thew above calculation is in error. I reran my above calculation and discovered had I miscalculated the Z value, it should be Z = .947 with the resulting S = 36,733 psi = 253 Mpa which is inline with your calculated rod diameter for your material's yield stress.

    I apologize for the error.
    Last edited by JAlberts; 08-14-2018 at 06:38 PM.

  5. #5
    Technical Fellow jboggs's Avatar
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    Jalberts - maybe I've been asleep, but I haven't seen your name for a while. Welcome back!

  6. #6
    Administrator Kelly Bramble's Avatar
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    Change your geometry (Area moment of inertia) to get a less mass load support member. Think about less mass perpendicular to the applied load vs parallel, like an I-beam.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

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