## 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

(OP)

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

https://i.imgur.com/RMbDJ4C.png

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?

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

https://i.imgur.com/RMbDJ4C.png

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

-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

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