Derivation of sagging expression

  • Thread starter Alpharup
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  • #1
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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?
 
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Answers and Replies

  • #2
Bill_K
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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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