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Re: [Question #593682]: Normal stiffness calculation

 

Question #593682 on Yade changed:
https://answers.launchpad.net/yade/+question/593682

Jan Stránský proposed the following answer:
Hi Jabrane,

> stiffness (kn) unit is N/m or MPa/m

see below, N/m is correct, I don't understand MPa/m (=N/m3)

> MPa.m and i don't understand.

MPa.m = (N/m2).m = N/m
>From this point of view, K=E.L makes sense

consider the interaction of equal spheres of radius r as a rod with length L, cross section A=r2 and modulus E. When loaded, it enlarges by dL with force F. You can define stress s=F/A and deformation e=dL/L. F and dL are related as
F = K.dl
on the same hand you want
s=E.e -> F/A=E.dL/L -> F=(EA/L).dL -> K=EA/L
A/L is (L with tilda) in the formula you mentioned

For not equal spheres, it just makes the stiffness contribution of
smaller sphere less significant.

A=r2 makes the stiffness "independent on size", like stretching a "bar"
composed of 4x4x20 particles. If you use 8x8x40 particles with half
radius (overall dimensions are the same) and the same E, you get the
same reactions.

cheers
Jan

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