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Hi all,


 


I have been working on a model lately and it can best be described simply as a 2-D flat plate. I've been running some fairly simple tests just to see how the solver deals with loads applied to a system and how changes in these loads affect the end (optimized) result. One test I've wanted to try yet haven't figured out yet is the linear distributed load across the top plate. Thus far I have only used constant distributed loads (Ex: 50 N force across all of the top edge nodes). When searching through these forums and online, I've heard of two ways of going about this. The first is using the "equation" tab inside the force panel. My model is set up in the (x-direction and y-direction) so I'm not entirely sure how the load should be applied as an equation. I've tried a few equations of my own (x, x^2, and 2*x) but haven't had any luck. The second method (which I heard by word of mouth) was to use Excel in a way that accounts for every node one would apply a load to along with the corresponding value of the force at that node. I know much less about the specific process for this one so if anyone has any direction here, that would be fantastic. All and any advice is welcomed and thank you in advance. Please let me know if you need more details about the model if it would better explain the current situation!


 


Dylan Stelzer


                       [:^D]


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Equations need to be a function of the coordinates, here is an example

 

A flat plate, 20 x 20 units, lying in the X-Y plane with the origin at the center.  A linear function for an applied force
magnitude = 20 – (5*x+2*y):

load_equations_ex1.png

 

To interpolate forces from a saved file or existing loads, select linear interpolation (linear interpolation only works with shell elements). Each row of the input file contains the x,y, and z coordinates of the force followed by its 3-components. The data can be separated by a space or tab.

 

See - 

HyperWorks Desktop Applications > HyperMesh > User's Guide > HyperMesh Panels > HyperMesh Panels Listed Alphabetically:

Forces Panel

Dylan Stelzer likes this

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Thanks for your reply! Do you mind if I attach a followup question?


 


Qu: If the coordinate system of choice for my current model is 2-D in the x- and y-direction, does that mean the equation for the magnitude of the force at that node MUST be in terms of x and y? Is it possible to have the equation be, for example, x^2 even though it has x and y coordinates? I don't see how this would be a problem but the software can be very particular at times!


 


Dylan Stelzer


                        [:-D]


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