yade-users team mailing list archive
-
yade-users team
-
Mailing list archive
-
Message #14509
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
--
You received this question notification because your team yade-users is
an answer contact for Yade.