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Derivation of sagging expression

  1. Mar 10, 2012 #1
    A bar of length l, breadth b, and depth d when loaded at the centre by a load W sags by
    an amount given by
    δ = W l^3/(4bd^3Y). Here Y is the young's modulus of the wire. My textbook says that we have to use little calculus to derive this expression. Can you please derive this?
     
  2. jcsd
  3. Mar 10, 2012 #2

    Bill_K

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    Science Advisor

    It gets pretty complicated. But take a look at the Wikipedia article "Euler–Bernoulli beam theory", especially the section "Three-point bending" and see if that is any help.

    "The shear is constant in absolute value: it is half the central load, P/2. It changes sign in the middle of the beam. The bending moment varies linearly from one end, where it is 0, and the center where its absolute value is PL/4. The deformation of the beam is described by a polynomial of third degree over a half beam (the other half being symmetrical)."
     
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