Question 6 of 7: Isoparametric mapping and the Jacobian of four-node quadrilaterals

Question 6: Isoparametric mapping and the Jacobian of four-node quadrilaterals (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The standard four-node bilinear parent square and two global quadrilaterals built on the SAME four points but with different node numbering.

Find. A definition of an isoparametric element, the general Jacobian matrix, its determinant for both elements, and what the two results imply about the mappings.

Approach. Define the isoparametric concept, differentiate the coordinate interpolation to build the Jacobian in matrix form, substitute each set of nodal coordinates, and read the sign and variation of the determinant.

1. Part (a) — define the isoparametric element. An element is isoparametric when the same shape functions, expressed in the same natural coordinates, are used both to interpolate the element geometry from the nodal coordinates and to interpolate the unknown field from the nodal degrees of freedom: $$x = \sum N_i x_i,\quad y = \sum N_i y_i \qquad\text{and}\qquad u = \sum N_i u_i,\quad v = \sum N_i v_i$$ The prefix records the equal (iso) number of parameters on the two sides. Because both use the same functions, the element can take a distorted shape in the global frame while all the integration is still performed on the fixed parent square, which is what makes Gauss quadrature and automatic mesh generation practical. (By contrast a subparametric element describes geometry with lower-order functions than the field, and a superparametric element the reverse.)
2. Part (b) — differentiate the coordinate interpolation. The Jacobian relates derivatives in the two frames through the chain rule, $\left\{\partial/\partial\xi\ \ \partial/\partial\eta\right\}^{T} = [\mathbf{J}]\left\{\partial/\partial x\ \ \partial/\partial y\right\}^{T}$, with $$[\mathbf{J}] = \begin{bmatrix}\dfrac{\partial x}{\partial\xi} & \dfrac{\partial y}{\partial\xi}\\[8pt] \dfrac{\partial x}{\partial\eta} & \dfrac{\partial y}{\partial\eta}\end{bmatrix} = \begin{bmatrix}\sum\dfrac{\partial N_i}{\partial\xi}x_i & \sum\dfrac{\partial N_i}{\partial\xi}y_i\\[8pt] \sum\dfrac{\partial N_i}{\partial\eta}x_i & \sum\dfrac{\partial N_i}{\partial\eta}y_i\end{bmatrix}$$
3. Write the Jacobian in the standard product form. Differentiating the four bilinear functions and factoring, $$\boxed{\;[\mathbf{J}] = \frac{1}{4}\begin{bmatrix}-(1-\eta)&(1-\eta)&(1+\eta)&-(1+\eta)\\ -(1-\xi)&-(1+\xi)&(1+\xi)&(1-\xi)\end{bmatrix}\begin{bmatrix}x_1&y_1\\x_2&y_2\\x_3&y_3\\x_4&y_4\end{bmatrix}\;}$$ This $2\times4$ times $4\times2$ product is the form used in every finite element code; the physical area element follows as $dx\,dy = |\mathbf{J}|\,d\xi\,d\eta$.
4. Part (c)(i) — substitute the counter-clockwise element. With $\{x\} = \{2,5,5,2\}$ and $\{y\} = \{3,3,7,5\}$ the four entries evaluate to $$\frac{\partial x}{\partial\xi} = \tfrac{1}{4}\left[3(1-\eta)+3(1+\eta)\right] = \tfrac{3}{2},\qquad \frac{\partial y}{\partial\xi} = \tfrac{1}{4}\left[2(1+\eta)\right] = \tfrac{1+\eta}{2}$$ $$\frac{\partial x}{\partial\eta} = 0,\qquad \frac{\partial y}{\partial\eta} = \tfrac{1}{4}\left[2(1-\xi)+4(1+\xi)\right] = \tfrac{3+\xi}{2}$$ so that $$[\mathbf{J}]_{(i)} = \begin{bmatrix}\tfrac{3}{2} & \tfrac{1+\eta}{2}\\[4pt] 0 & \tfrac{3+\xi}{2}\end{bmatrix},\qquad \boxed{\;|\mathbf{J}|_{(i)} = \tfrac{3}{2}\cdot\tfrac{3+\xi}{2} = \tfrac{3(3+\xi)}{4}\;}$$ which runs from $1.5$ at $\xi=-1$ to $3.0$ at $\xi=+1$ and is positive throughout.
5. Part (c)(ii) — substitute the clockwise element. The same four points are now visited in the reverse sense, $\{x\} = \{2,2,5,5\}$ and $\{y\} = \{3,5,7,3\}$, giving $$[\mathbf{J}]_{(ii)} = \begin{bmatrix}0 & \tfrac{3+\eta}{2}\\[4pt] \tfrac{3}{2} & \tfrac{1+\xi}{2}\end{bmatrix},\qquad \boxed{\;|\mathbf{J}|_{(ii)} = -\tfrac{3(3+\eta)}{4}\;}$$ which runs from $-1.5$ to $-3.0$ and is negative throughout. Comparing the two matrices, the two ROWS have simply exchanged places (with $\xi$ and $\eta$ trading roles), which is precisely what reversing the traversal sense does.
6. Part (d) — what the two determinants tell you. Integrating each determinant over the parent square gives $\iint|\mathbf{J}|_{(i)}\,d\xi\,d\eta = +9\ \text{mm}^{2}$ and $\iint|\mathbf{J}|_{(ii)}\,d\xi\,d\eta = -9\ \text{mm}^{2}$, and the shoelace formula confirms the quadrilateral really does enclose $9\ \text{mm}^{2}$. The magnitudes are identical, so the two elements are geometrically the same region; only the sign differs, and the sign carries the orientation of the mapping.

Three conclusions follow, and together they answer part (d). Element (i) is a valid element. Its Jacobian determinant is positive everywhere in $-1\le\xi,\eta\le1$, so the map from parent to global coordinates is one-to-one and invertible at every point, $[\mathbf{J}]^{-1}$ exists throughout, and the strain-displacement matrix and the stiffness integral can be formed without difficulty. Element (ii) is geometrically identical but numbered backwards. Its negative determinant means the mapping reverses orientation: the parent square is turned inside-out onto the global quadrilateral. Physically that would be reported by a solver as a negative area (in three dimensions, a negative volume) and the element would be rejected, or worse, would contribute a negative-definite block to the stiffness matrix. The cure is renumbering, not remeshing: simply relabel the nodes counter-clockwise and element (ii) becomes element (i). Neither determinant is constant. $|\mathbf{J}|$ varies linearly with $\xi$ in case (i) and with $\eta$ in case (ii), which is the signature of a genuine quadrilateral rather than a parallelogram (for a parallelogram $|\mathbf{J}|$ would be a constant equal to one quarter of the area). Since it stays of one sign, neither element is distorted enough to be invalid — the dangerous case is a determinant that changes sign inside the element, which signals a re-entrant or bow-tie shape and is a genuine meshing failure rather than a numbering one.