[MBDyn-users] Composite material

masarati at aero.polimi.it masarati at aero.polimi.it
Fri Aug 28 18:20:35 CEST 2009


> Hello,
>
> I worked and read those papers and very sorry but I didn't understand how
> to build the 6x6 constitutive law matrix.
> When a have for instance an orthotropic material like (just an example) :
>
> E1 = 100 000 MPa
> E2 =  10 000 MPa
> nu1 = nu2 = 0.3
> G12 = 5 000 MPA
> HEIGHT = 5 mm
> WIDHT = 20 mm
>
> I will be very gratefull if you could explain to me how to transform those
> parameters into the 6x6 matrix.I enclosed an example file with a isotropic
> material.

Dear Romuald,

what you're trying to do may not be trivial.  In fact, when you consider
non-isotropic materials, constitutive properties need some special care. 
I don't have references for constitutive laws handy at the moment, but
apart from few simple cases (which might well be the one of a rectangular
cross section with orthotropic material whose directions of orthotropy, I
guess, are along the axis of the beam and parallel to the sides of the
section), you may need to perform a full 3D solution of a finite portion
of beam to compute equivalent axial, shear, bending and torsion
properties.

The references Professor Mantegazza pointed you to are to formulations
dedicated to this task, that resort to a special 2D (section-wise) finite
element solution capable of computing the required properties in a more
efficient way than a full 3D analysis of a portion of beam, with the plus
of analytically taking into account (and eliminate) spanwise boundary
conditions propagation effects.

So the answer to your question is: you need to do some preparatory work in
order to compute the matrices required by MBDyn (or by any other analysis
that uses the beam model).  This analysis is outside the scope of MBDyn. 
We can talk more about this if you like, but as I said there is no easy
answer.

There might be shortcuts in special cases; for example, in your case,
assuming direction 1 is the one along the axis of the beam, and properties
in directions 2 and 3 are the same, the axial stiffness should be

EA = HEIGHT * WIDTH * E1

the bending stiffnesses should be

EJ2 = HEIGHT * WIDTH^3 * E1 / 12

EJ3 = HEIGHT^3 * WIDTH * E1 / 12

but there should be no easy solution for GA2, GA3 and GJ.

Cheers, p.



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