Question 6 of 7: Isoparametric quadrilateral — Jacobian matrix and mapping validity

Question 6: Isoparametric quadrilateral — Jacobian matrix and mapping validity (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. A four-node isoparametric quadrilateral whose parent domain is the square $-1\le\xi,\eta\le1$ with nodes 1 $(-1,-1)$, 2 $(1,-1)$, 3 $(1,1)$, 4 $(-1,1)$, and the four bilinear shape functions quoted above. Part (c) supplies two elements occupying the same four points in the global $x$–$y$ plane, $(1,1)$, $(4,1)$, $(4,6)$ and $(1,5)$, but numbered differently: in (i) the sequence 1-2-3-4 runs counter-clockwise — $1(1,1)$, $2(4,1)$, $3(4,6)$, $4(1,5)$ — while in (ii) it runs clockwise — $1(1,1)$, $2(1,5)$, $3(4,6)$, $4(4,1)$.

Find. A definition of an isoparametric element; the general Jacobian matrix $[J]$ in terms of the nodal coordinates; the determinant $\left|J\right|$ for each of the two numbered elements; and the conclusion those two expressions support about the validity of the coordinate mapping.

Approach. Differentiate the shape functions with respect to $\xi$ and $\eta$, write the Jacobian as the product of a $2\times4$ derivative matrix and the $4\times2$ nodal-coordinate matrix, then substitute the two coordinate sets and take determinants.

1. Define an isoparametric element (part a). An element is isoparametric when the same shape functions, in the same natural coordinates, are used both to interpolate the geometry and to interpolate the field variable: $$x=\sum_{i}N_i x_i,\quad y=\sum_{i}N_i y_i \qquad\text{and}\qquad u=\sum_{i}N_i u_i .$$ (“Iso” = equal parameters.) The alternatives are subparametric, where the geometry uses a lower-order interpolation than the field, and superparametric, where it uses a higher-order one. The isoparametric choice guarantees that rigid-body motion and constant strain are reproduced exactly on a distorted element, which is why it underpins essentially every production element in commercial codes.
2. Differentiate the shape functions. From the printed $N_i$, $$\frac{\partial N_1}{\partial\xi}=-\tfrac14(1-\eta),\quad \frac{\partial N_2}{\partial\xi}=\tfrac14(1-\eta),\quad \frac{\partial N_3}{\partial\xi}=\tfrac14(1+\eta),\quad \frac{\partial N_4}{\partial\xi}=-\tfrac14(1+\eta),$$ $$\frac{\partial N_1}{\partial\eta}=-\tfrac14(1-\xi),\quad \frac{\partial N_2}{\partial\eta}=-\tfrac14(1+\xi),\quad \frac{\partial N_3}{\partial\eta}=\tfrac14(1+\xi),\quad \frac{\partial N_4}{\partial\eta}=\tfrac14(1-\xi).$$
3. Assemble the general Jacobian matrix (part b). Since $x=\sum N_ix_i$ and $y=\sum N_iy_i$, $$\boxed{\;[J]=\begin{bmatrix} \dfrac{\partial x}{\partial\xi} & \dfrac{\partial y}{\partial\xi}\\[10pt] \dfrac{\partial x}{\partial\eta} & \dfrac{\partial y}{\partial\eta}\end{bmatrix} =\frac14\begin{bmatrix} -(1-\eta) & (1-\eta) & (1+\eta) & -(1+\eta)\\ -(1-\xi) & -(1+\xi) & (1+\xi) & (1-\xi)\end{bmatrix} \begin{bmatrix}x_1 & y_1\\ x_2 & y_2\\ x_3 & y_3\\ x_4 & y_4\end{bmatrix}\;}$$ and the Jacobian (determinant) is $\left|J\right|=\dfrac{\partial x}{\partial\xi}\dfrac{\partial y}{\partial\eta} -\dfrac{\partial y}{\partial\xi}\dfrac{\partial x}{\partial\eta}$. It is $\left|J\right|$ that appears in $dA = \left|J\right|d\xi\,d\eta$ and in $[B]=[J]^{-1}(\cdots)$, so both its magnitude and its sign matter.

4. Element (i): substitute the counter-clockwise coordinates. With $\{x\}=\{1,4,4,1\}$ and $\{y\}=\{1,1,6,5\}$, $$\frac{\partial x}{\partial\xi}=\tfrac14\Big[-(1-\eta)(1)+(1-\eta)(4)+(1+\eta)(4)-(1+\eta)(1)\Big] =\tfrac14\Big[3(1-\eta)+3(1+\eta)\Big]=\tfrac32,$$ $$\frac{\partial y}{\partial\xi}=\tfrac14\Big[-(1-\eta)(1)+(1-\eta)(1)+(1+\eta)(6)-(1+\eta)(5)\Big] =\tfrac{1+\eta}{4},$$ $$\frac{\partial x}{\partial\eta}=\tfrac14\Big[-(1-\xi)(1)-(1+\xi)(4)+(1+\xi)(4)+(1-\xi)(1)\Big]=0,$$ $$\frac{\partial y}{\partial\eta}=\tfrac14\Big[-(1-\xi)(1)-(1+\xi)(1)+(1+\xi)(6)+(1-\xi)(5)\Big] =\tfrac{9+\xi}{4}.$$
5. Element (i): the Jacobian. Hence $$[J]_{\text{(i)}}=\begin{bmatrix}\tfrac32 & \tfrac{1+\eta}{4}\\[4pt] 0 & \tfrac{9+\xi}{4}\end{bmatrix}, \qquad \left|J\right|_{\text{(i)}}=\tfrac32\cdot\tfrac{9+\xi}{4}-\tfrac{1+\eta}{4}\cdot 0$$ $$\boxed{\;\left|J\right|_{\text{(i)}}=\frac{3\left(9+\xi\right)}{8}\;}$$ which ranges from $\left|J\right|=3.00$ at $\xi=-1$ to $\left|J\right|=3.75$ at $\xi=+1$ and is strictly positive everywhere in the parent square. Integrating, $\int_{-1}^{1}\!\int_{-1}^{1}\left|J\right|d\xi\,d\eta = 13.5$, which matches the shoelace area of the quadrilateral $(1,1)$, $(4,1)$, $(4,6)$, $(1,5)$ exactly — a useful independent check that the map is set up correctly.
6. Element (ii): substitute the clockwise coordinates. The same four points are now taken in the order $\{x\}=\{1,1,4,4\}$, $\{y\}=\{1,5,6,1\}$: $$\frac{\partial x}{\partial\xi}=\tfrac14\Big[-(1-\eta)(1)+(1-\eta)(1)+(1+\eta)(4)-(1+\eta)(4)\Big]=0,$$ $$\frac{\partial y}{\partial\xi}=\tfrac14\Big[-(1-\eta)(1)+(1-\eta)(5)+(1+\eta)(6)-(1+\eta)(1)\Big] =\tfrac{9+\eta}{4},$$ $$\frac{\partial x}{\partial\eta}=\tfrac32,\qquad \frac{\partial y}{\partial\eta}=\tfrac{1+\xi}{4}.$$ Comparing with element (i), the two rows of $[J]$ have simply exchanged roles, because reversing the numbering swaps which parent axis maps onto which physical direction.
7. Element (ii): the Jacobian. $$[J]_{\text{(ii)}}=\begin{bmatrix}0 & \tfrac{9+\eta}{4}\\[4pt] \tfrac32 & \tfrac{1+\xi}{4}\end{bmatrix}, \qquad \left|J\right|_{\text{(ii)}}=0\cdot\tfrac{1+\xi}{4}-\tfrac{9+\eta}{4}\cdot\tfrac32$$ $$\boxed{\;\left|J\right|_{\text{(ii)}}=-\,\frac{3\left(9+\eta\right)}{8}\;}$$ ranging from $-3.00$ at $\eta=-1$ to $-3.75$ at $\eta=+1$ — the same magnitudes as element (i) but negative throughout, and the integral is $-13.5$, the negative of the true area.
8. Part (d) — interpret the two expressions. Three conclusions follow. First, element (i) has $\left|J\right|\gt0$ at every point of the parent domain, so its map is one-to-one and orientation-preserving: the element is valid, its area integrates correctly, and $[J]^{-1}$ exists everywhere so the strain–displacement matrix $[B]$ is well defined. Second, element (ii) has $\left|J\right|\lt0$ throughout. Because the geometry is identical, this cannot be a defect of shape — it is purely a consequence of numbering the nodes clockwise, which reverses the orientation of the mapping. Left uncorrected, the element area and hence the element stiffness would be computed with the wrong sign, producing a negative-definite contribution to the global matrix; most codes simply abort with a “negative Jacobian” error. The remedy is trivial: renumber counter-clockwise, whereupon (ii) becomes (i). Third, in neither case does $\left|J\right|$ change sign inside the element, so neither map folds over on itself; the distinction here is orientation, not the far more serious self-intersecting (“bow-tie”) case that arises when the numbering skips a corner.
9. Note what the varying Jacobian implies. In both elements $\left|J\right|$ is a function of position rather than a constant, ranging over $3.00$ to $3.75$ — a variation of $25\%$. The element is a trapezoid, not a parallelogram, so the map is non-affine, the mapped integrand of a stiffness integral becomes a rational function, and Gauss quadrature is genuinely approximate (contrast Question 3, where $\left|J\right|$ was constant and the quadrature was exact). The ratio $\left|J\right|_{\max}/\left|J\right|_{\min}=3.75/3.00=1.25$ is a standard element-distortion measure; values near unity indicate a well-shaped element, and values approaching zero or turning negative signal a mesh that must be repaired.