1. #1
    dsc's Avatar
    Lives in Lincoln, United Kingdom. Last Activity: 1 Day Ago Has been a member for 5-6 years. Has a total post count of 248. Received thanks 1 times, giving thanks to others 9 times.
    Gents,

    I'm trying to calculate the maximum deflection of an unsupported part of a shaft with the following data:

    L total unsupported length = 0.18m (not mm;))
    A cross section surface (round rod) = pi * (0.01m)^2
    E Young's modulus = 180GPa
    F force applied to the end of shaft = 1000N (random round number)

    delta (max deflection) = F*L / E*A

    Now popping the values in to the equation I get that max deflection is equal:

    delta = F * 3.19*10^-9 [m/N]

    which basically means that with F = 1000N the deflection is

    delta = 3.19 * 10^-6 m = 3.19*10^-3 mm = 0.00319mm

    That is rather small, considering that 1000N is pretty much the same as 100kg weight attached to one end.

    Pack of biscuits to anyone who can point an error in the above and call be stupid:)

    Regards,
    dsc.
    Last edited by dsc; 10-02-2013 at 10:29 PM.

  2. #2
    m_c's Avatar
    Lives in East Lothian, United Kingdom. Last Activity: 5 Hours Ago Has been a member for 9-10 years. Has a total post count of 1,837. Received thanks 192 times, giving thanks to others 5 times.
    0.18mm unsupported length?

  3. #3
    dsc's Avatar
    Lives in Lincoln, United Kingdom. Last Activity: 1 Day Ago Has been a member for 5-6 years. Has a total post count of 248. Received thanks 1 times, giving thanks to others 9 times.
    Ha, well spotted, although it's just a unit error.

    After some reading I don't think the above equation for delta is correct as there's no moment of inertia involved.

    Regards,
    dsc.

  4. #4
    m_c's Avatar
    Lives in East Lothian, United Kingdom. Last Activity: 5 Hours Ago Has been a member for 9-10 years. Has a total post count of 1,837. Received thanks 192 times, giving thanks to others 5 times.
    I just ran the figures and got 3183mm of deflection...

    A = pi0.01^2 = 0.000314159
    F*L = 1000 * 0.18 = 180
    E*A = 180 * 0.0003blahblah = 0.05654866

    180 / 0.0565blahblah = 3183.09886183

  5. #5
    Perhaps you are working out how far it will stretch if you pull on the end, not how far it will bend if you load it at 90 degrees?

  6. #6
    You could be right here Robin. Its been a very long time since I used Youngs Modulus but in this form I thought it was used for calculating a change in length due to compression or extension.

  7. #7
    dsc's Avatar
    Lives in Lincoln, United Kingdom. Last Activity: 1 Day Ago Has been a member for 5-6 years. Has a total post count of 248. Received thanks 1 times, giving thanks to others 9 times.
    Ah yes, it's all wrong!:)

    Here's the right formula (taken from http://www.clag.org.uk/beam.html):

    δ = FL3 ∕ 3EI

    I = πr4 ∕ 4 = πd4 ∕ 64

    (for round beam)
    which means that deflection at a given F and L, for a 20mm dia beam is:

    δ = FL3 * 0.0002358

    Assuming L = 0.18m and F=100N the maximum deflection is:

    δ = 100 * 0.005832 * 0.0002358 m = 0.0001375m = 0.1375mm

    Which makes much more sense than the previously calculated value.

    The funny thing is that changing the dia from 20mm to 22mm gives you a maximum deflection drop from 0.1375mm to 0.094mm. That's just by making the shaft 2mm thicker. Quite amazing me thinks.

    Regards,
    dsc.

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