General discussions about how to formulate a script for FlexPDE.

Hello,

I wonder which unit a force applied via the load BC has.

I have displacements ux, uy and uz in x,y and z-direction in my model and as far as I understand, using the load BC, e.g. load(ux) = a, applies a force of amplitude a on the selected boundary.

Now I assume the units in my model to be mm, kg and s. In this case, would the applied force be kg*mm/s^2, so mN, or would it N?

Thanks,
Simon
SJanssen

Posts: 6
Joined: Tue Jan 03, 2017 4:05 am

FlexPDE knows nothing about units. You just need to be self-consistent throughout your script.
If you have used mm in your definitions and equations, then I would think your force is mN.
moderator

Posts: 735
Joined: Tue Jan 11, 2011 1:45 pm

Ok, I will check that. Thank you so far.

Just to clear things up, in "load(ux) = a" where ux is the displacement in x-direction, a should be in Newton (not considering the magnitude). Because in your example "bentbar.pde", you apply "load(V) = dist" and dist = [N*m^2*m^-3] = [N/m], is that right?

Thank you.
SJanssen

Posts: 6
Joined: Tue Jan 03, 2017 4:05 am

In our bentbar problem, the load is being applied to V on the left end. The V equation is in fact the divergence of the vector (Txy,Sy), and the LOAD BC specifies the outward normal component of the argument of the Divergence. So the load is the outward normal component of the vector (Txy,Sy), or nx*Txy+ny*Sy. Since ny is zero on this wall, LOAD(V) provides the value of Txy. The script defines Txy = C33*(dy(U) + dx(V)), so the LOAD must have the units of this quantity. C33 has the units of E, or N/m^2. dy(U) and dx(V) will be dimensionless, assuming U and V have the dimensions of distance. So the load must be N/m^2.
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Joined: Tue Jan 11, 2011 1:45 pm

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