I have had the need to perform a similar operation and used python scripting to calculate the areas based on nodal reference locations (easy, since I was doing a 2-d problem with triangular elements).
I am sure you want the deformed areas though, which I stopped short of calculating. I do not know if ABAQUS stores the current nodal locations, or just the nodal displacements relative to reference configuration (someone please clarify).
A long about way would be to use scripting to calculate the current nodal locations (reference locations plus current displacement) and then use the current locations to compute the area of the element.
The method described by chrispogi05 is probably your best bet. To confirm, it is possible (and pretty straight-forward) to get deformed element areas (assuming you're talking about 2D) by adding nodal displacements to initial nodal coordinates to get deformed nodal co-ordinates. It may also be possible to get them directly by requesting 'COORD' as field output. It is also possible to get 3D element surface areas using a similar technique but applying it to individual element faces.
abq stores original node coords and deformation (delta). you can get the latter with scripting by invoking the fieldoutput function (check help, py reference, odbaccess)
Go with MechIrl and deyentl's suggestions and add the appropriate field output variables; I didn't know those were there, thanks everyone! This helps me too
Sorry I've taken so long to respond to this thread. What I'm exactly looking for is the area of an element on a face. So what I have is quadrilateral element shapes. Is there a way to measure the area of a quadrilateral on a face?
Even if I was able get the correct distances from 1 node to the next then I could use the four distances and the formula of:
Area = Square Root of{(s-a)(s-b)(s-c)(s-d) -1/4(ac+bd+pq)(ac+bd-pq)}
where a,b,c and d are the four sides of the quadrilateral with a and c and b and d as the opposite sides, p and q are diagonals, and s = (a+b+c+d)/2
Example;
If a quadrilateral has four sides a,b,c,d, where a is the opposite side of c, and b is the opposite side of d, and their lengths are:
a=30, b=31, c=22, d=28 and the two diagonals are: p=34 and q=43, then the area of this quadrilateral is:
calculate semiperimeter (s) = (a+b+c+d)/2= 111/2 = 55.5
area is:
square root of {(55.5-30)(55.5-31)(55.5-22)(55.5-28)-1/4(30*22+31*28+34*43)(30*22+31*28-34*43)}
= Square root of {(25.5*24.5*33.5*27.5) - 1/4(660+868 +1462)(660+868-1462)}
=square root of {575550.94 - 1/4(2990)(66)}
=square root of {575550.94 - 1/4(197340)}
=square root of{575550.94 - 49335)}
=square root of {526215.94}
=725.41
But as I have over 192 elements this will be very time consuming
If I am understanding your problem correctly, you have a 3D problem in which you are using brick elements, but want to know the areas of elements on a particular face (perhaps on a surface) and therefore need to calculate the respective areas of of the element faces on this surface? --> Correct if wrong.
In this case, there is no easier way I can think of except to script this process in python using the algorithm you mentioned in your post. You would need a clever way of extracting the nodal connectivity for only the elements and , more importantly, for only the face you need.
If this was a 2D problem you could have extracted the areas as field output as mentioned above.
Yes that's correct Chris. I am not very good at python script at all but do have a friends script that will extract all element nodes on a face and groups them accordingly.
Then I have a script that can take the co-ordinates in an ABAQUS model and import them into a CATIA model.Then I can use the distance tool to measure between the four quadrilateral nodes. Then I am using a basic area quadrilateral calculator found on the Internet:
It's just this method is very very time consuming and I don't have much time to measure over 192 elements and use the calculator to calculate the area of each quad shape
Thanks for your help. I have found a simpler method of calculating the total area of the face in CATIA then dividing it by the number of elements. There is a slight discrepancy but it is only by around 0.064 for each element face, which I use for discussion.
