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Radius of Curvature

  • Thread starter Freyster98
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Homework Statement



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.


Homework Equations



a2+b2=c2

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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Answers and Replies

  • #2
tiny-tim
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Derive an expression geometrically for the radius of curvature of the following beam.

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?
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:
 
  • #3
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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:
I ran through it a few times...I don't see where I'm losing a factor of 2.
 
  • #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:
 
  • #5
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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:
Ok, thanks. No, I haven't tried the similar triangles because, well, I don't get it :uhh:
 
  • #6
tiny-tim
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Ok, thanks. No, I haven't tried the similar triangles because, well, I don't get it :uhh:
ok … the triangle with sides d and L/2 is similar to the triangle with sides L/2 and … ? :smile:
 

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