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Why doesn't the Jacobian increase the number of Gauss points required to integrate when using Quadri

Why doesn't the Jacobian increase the number of Gauss points required to integrate when using Quadri

Why doesn't the Jacobian increase the number of Gauss points required to integrate when using Quadri

Hi all,

For rectangular elements, I understand how the integral for the stiffness matrix ( transpose(B)*c*B ) has a maximum power for eta and xi or two, and thus two gauss points are required.

However, when mapping for Quadrilaterial elements the integral gets multiplied by the determinant of the Jacobian (J).

Shown here from Liu and Quek


So the Jacobian is made up of terms that are linear functions of eta and xi, so why doesn't this increase the number of Gauss points required to carry out the integral?

RE: Why doesn't the Jacobian increase the number of Gauss points required to integrate when using Quadri

I am not exactly sure of what you are asking but I will just throw some stuff out there.

-The number of required gauss points is basically a function of the polynomial degree of the basis functions. In 1d, a linear basis function needs 2 gauss points to get accurate (exact) integration. In 2d if you have linear basis functions with respect to xi and with respect to eta then you need 2 * 2 = 4 gauss points.
-You can use less gauss points (non uniform integration to reduce locking, etc) but then the integration won't be exact, but may be "good enough"
-In 2d you can use basis functions that are differing degrees with respect to the different variables if necessary. For example you could use basis functions that are linear with respect to xi and constant with respect to eta. You could do this if you believed the displacement fields were more complicated in one of the directions than the other
-When multiplying by the determinant of the Jacobian, that is just like a scaling factor. The parent square has an area of 4. The actual element probably doesn't have that same area so when integrating across the area it is necessary to scale it up or down to account for that.

Like I said, I didn't really understand your question. Feel free to clarify or hopefully something I wrote will help.

RE: Why doesn't the Jacobian increase the number of Gauss points required to integrate when using Quadri

I do not recall a link between the Jacobian and integration points. The Jacobian is just a scalar value (a determinant of a 2x2 matrix containing the 1st derivative of the shape function if I recall). As stated by E720 it links the perfect/theoretical shape (say a square) to the actual shape of the element (in the FEM). I was told that it is a representation of how much the element differs from the perfect shape i.e a degree of deformity. Hence being used as a quality criteria in pre-processing.

RE: Why doesn't the Jacobian increase the number of Gauss points required to integrate when using Quadri

Yeah when you read "the Jacobian" in an FEA book I guess more properly it should be "the Jacobian matrix of the mapping functions x(s,t) and y(s,t) ". If you don't know what a Jacobian matrix is just do a quick google search. Like was said, the determinant of the Jacobian matrix of the mapping functions scales the integration from s,t square to x,y element basically. In addition we can always expect the determinant of the Jacobian Matrix to be positive, because if it's non-positive that would imply a negative or zero area element. This usually happens if the nodes in an element are not connected in the correct way. It is usually just a matter of the right order of the elemental nodes and usually the FEA will throw an error that says one of the elements has a nonpositive Jacobian determinant so that you know that you've done something wrong.

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