# How is the surface Jacobian determinant calculated in FEM?

I am currently trying to evaluate surface forces on a structure. I came across P356 in Bathe's Finite Element Procedures 2014 (example 5.8) in which he related the edge derivative from the global to natural coordinates through this equation:

$$dl = det(\mathbf{J}^S) \ dr$$

where $$r$$ is the natural coordinate along the edge of interest. I am not sure how $$\mathbf{J}^S$$ or its determinant are calculated! I am not sure how to relate that to the edge length for example!

• Check out: classes.engineering.wustl.edu/mase5510/Chapter_5.pdf Does that help?
– NNN
Commented Jun 25, 2022 at 4:39
• Specifically page 158
– NNN
Commented Jun 25, 2022 at 4:55
• Thank you very much, that answers it. Commented Jun 29, 2022 at 13:21
• I suggest that OP or @Nachiket write the solution down as an answer, so that this question can get out of the un-answered queue. Commented Jun 29, 2022 at 13:40
• @MohamedAbdelhamid could you please write an answer? I don't understand your problem perfectly, I just pointed you to what I thought would be useful.
– NNN
Commented Jun 30, 2022 at 3:19

As NNN kindly answered in the comments, it has to do with how the Jacobian is originally defined from the volume/surface/line differential element. More specifically, it's well explained in Section 5.6 P155 in this document. Let me rewrite a simple explanation here:

Let $$\vec{r}$$ be the position vector of an arbitrary point within a finite element, $$\vec{r}$$ is defined in terms of the orthogonal basis vectors $$\vec{e}_x$$, $$\vec{e}_y$$, and $$\vec{e}_z$$ of a right-handed Cartesian coordinate system as follows: $$$$\vec{r} := x \vec{e}_x + y \vec{e}_y + z \vec{e}_z$$$$

By definition, the differential volume $$dV$$ can be represented as a scalar triple product in terms of the derivatives of the position vector as follows: $$$$dV := \left( \frac{\partial \vec{r}}{\partial x} dx \times \frac{\partial \vec{r}}{\partial y} dy \right) \cdot \frac{\partial \vec{r}}{\partial z} dz = \left( \vec{e}_x \times \vec{e}_y \right) \cdot \vec{e}_z \ dx \ dy \ dz = dx \ dy \ dz$$$$

It's usually easier to transform any integrals from the global coordinates $$x,y,z$$ to the natural coordinates $$\xi, \eta, \zeta$$. The partial derivatives can be transformed as follows: $$$$\frac{\partial \vec{r}}{\partial \xi} = \frac{\partial \vec{r}}{\partial x} \frac{\partial x}{\partial \xi} + \frac{\partial \vec{r}}{\partial y} \frac{\partial y}{\partial \xi} + \frac{\partial \vec{r}}{\partial z} \frac{\partial z}{\partial \xi}, \tag{1}$$$$

$$$$\frac{\partial \vec{r}}{\partial \eta} = \frac{\partial \vec{r}}{\partial x} \frac{\partial x}{\partial \eta} + \frac{\partial \vec{r}}{\partial y} \frac{\partial y}{\partial \eta} + \frac{\partial \vec{r}}{\partial z} \frac{\partial z}{\partial \eta}, \tag{2}$$$$

$$$$\frac{\partial \vec{r}}{\partial \zeta} = \frac{\partial \vec{r}}{\partial x} \frac{\partial x}{\partial \zeta} + \frac{\partial \vec{r}}{\partial y} \frac{\partial y}{\partial \zeta} + \frac{\partial \vec{r}}{\partial z} \frac{\partial z}{\partial \zeta}. \tag{3}$$$$

Now, to write the new expression for the differential volume in the natural coordinates: $$$$dV := \left( \frac{\partial \vec{r}}{\partial \xi} d\xi \times \frac{\partial \vec{r}}{\partial \eta} d\eta \right) \cdot \frac{\partial \vec{r}}{\partial \zeta} d\zeta$$$$

and using Eqs. 1-3 and representing the cross-dot multiplication as a matrix determinant, we get:

$$$$dV = \begin{vmatrix} \dfrac{\partial x}{\partial \xi} & \dfrac{\partial y}{\partial \xi} & \dfrac{\partial z}{\partial \xi} \\ \dfrac{\partial x}{\partial \eta} & \dfrac{\partial y}{\partial \eta} & \dfrac{\partial z}{\partial \eta} \\ \dfrac{\partial x}{\partial \zeta} & \dfrac{\partial y}{\partial \zeta} & \dfrac{\partial z}{\partial \zeta} \end{vmatrix} d\xi \ d\eta \ d\zeta = |\mathbf{J}| d\xi \ d\eta \ d\zeta$$$$

and that's how the Jacobian of the 3D differential volume is derived.

For a differential surface $$dS$$ (or a differential volume with constant thickness), a similar expression to $$dV$$ becomes: $$$$dS := \frac{\partial \vec{r}}{\partial \xi} d\xi \times \frac{\partial \vec{r}}{\partial \eta} d\eta$$$$

and using the surface version of Eqs. 1-3 (by discarding the third terms in Eqs. 1 and 2 and the entirety of Eq. 3), we get: $$$$dS := \begin{vmatrix} \dfrac{\partial x}{\partial \xi} & \dfrac{\partial y}{\partial \xi} \\ \dfrac{\partial x}{\partial \eta} & \dfrac{\partial y}{\partial \eta} \end{vmatrix} d\xi d\eta = |\mathbf{J}| d\xi d\eta$$$$

It can be seen that the surface Jacobian is basically the volume Jacobian by setting the third components perpendicular to the differential surface $$dS$$ to zero (that's why it's usually written using the same symbol).

Finally, for a differential edge $$dL$$ (or a differential surface with constant thickness), and assuming $$\xi$$ is the natural coordinate along that edge and $$\eta$$ is constant (can be +1 or -1 depending on the location within the finite element): $$$$dL := \left| \frac{\partial \vec{r}}{\partial \xi} \right| d\xi$$$$

Going back to Eqs. 1-3, the edge version is obtained by discarding the third term in Eq. 1 and the entirety of Eqs. 2 and 3. Notice that instead of using a determinant, we use the Euclidean norm of the vector from Eq. 1 as follows: $$$$\left| \frac{\partial \vec{r}}{\partial \xi} \right| = \sqrt{ \left( \dfrac{\partial x}{\partial \xi} \right)^2 + \left( \dfrac{\partial y}{\partial \xi} \right)^2 }.$$$$