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Radius of Curvature

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Derive an expression geometrically for the radius of curvature of the following beam. This is part of a lab assignment for the bending of a simply supported beam with overhangs.

    ** I did this crappy diagram with AutoCAD, so I couldn't ( or didn't know how to ) include greek letters. Let's let r= [tex]\rho[/tex], and d= [tex]\delta[/tex] for my derivation.


    2. Relevant equations

    a2+b2=c2

    3. The attempt at a solution

    I just used the pythagorean theorem to solve for [tex]\rho[/tex].

    Starting with: [tex]\rho[/tex]2= ([tex]\rho[/tex]-[tex]\delta[/tex])2+(L/2)2.

    Factoring out ([tex]\rho[/tex]-[tex]\delta[/tex])2 , solving for [tex]\rho[/tex] and simplifying , I end up with the following expression:

    [tex]\rho[/tex]=([tex]\delta[/tex]/2)+(L2/8[tex]\delta[/tex])


    I guess I have this question...is this the proper way to derive the radius of curvature geometrically? Is it ok to do it this way?
     

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    Last edited: Apr 18, 2009
  2. jcsd
  3. Apr 19, 2009 #2

    tiny-tim

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    Hi Freyster98! :smile:

    (have a rho: ρ and a delta: δ :wink:)

    Yes, Pythagoras is fine :smile: (though you seem to have lost a factor of 2 somewhere :confused:).

    But there is quicker method (with less likelihood of a mistake):

    Hint: similar triangles :wink:
     
  4. Apr 19, 2009 #3
    I ran through it a few times...I don't see where I'm losing a factor of 2.
     
  5. Apr 20, 2009 #4

    tiny-tim

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    sorry … my similar triangles method (have you tried that yet?) gave me the diameter, not the radius :rolleyes:

    so i got an extra 2 :redface:
     
  6. Apr 20, 2009 #5
    Ok, thanks. No, I haven't tried the similar triangles because, well, I don't get it :uhh:
     
  7. Apr 20, 2009 #6

    tiny-tim

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    ok … the triangle with sides d and L/2 is similar to the triangle with sides L/2 and … ? :smile:
     
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