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

Question 6: Isoparametric quadrilateral — Jacobian matrix and element 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. The printed shape functions fix the parent-domain node positions, and the two elements of part (c) use the same four points with different node numbering:

Find. A definition of the isoparametric concept, the general Jacobian matrix of the four-node element, the Jacobian determinant for each of the two numbered elements, and the conclusion those two determinants support.

Approach. Read the parent node positions from the printed shape functions, differentiate the isoparametric coordinate map to obtain $[J]$ in terms of the nodal coordinates, then substitute each node set and examine the sign of $|J|$ over the whole parent square.

1. Part (a): define the isoparametric concept. An element is isoparametric when the same shape functions, in the same natural coordinates, interpolate both the element geometry and the field variables: $$x=\sum_i N_i(\xi,\eta)\, x_i,\quad y=\sum_i N_i(\xi,\eta)\, y_i,\qquad u=\sum_i N_i(\xi,\eta)\, u_i,\quad v=\sum_i N_i(\xi,\eta)\, v_i$$ “Iso-parametric” means equal parametric order; if the geometry used a lower-order set the element would be subparametric, and if it used a higher-order set, superparametric. The practical value is that a single parent element and a single set of integration rules serve every distorted shape in the mesh, and that the mapping automatically preserves interelement compatibility.
2. Confirm the parent node positions from the printed functions. The given $N_2=\tfrac14(1+\xi)(1-\eta)$ is unity at $(\xi,\eta)=(1,-1)$ and $N_3=\tfrac14(1-\xi)(1+\eta)$ is unity at $(-1,1)$, so the parent numbering is $$1(-1,-1),\quad 2(1,-1),\quad 3(-1,1),\quad 4(1,1)$$ This is the “z-order” convention, in which the boundary of the element is traversed in the order 1–2–4–3 rather than 1–2–3–4. Reading the positions from the shape functions rather than from the figure labels is essential, because everything that follows depends on which node faces which.
3. Part (b): differentiate the coordinate map. The Jacobian matrix relates natural to global derivatives: $$[J]=\begin{bmatrix} \dfrac{\partial x}{\partial \xi} & \dfrac{\partial y}{\partial \xi}\\[8pt] \dfrac{\partial x}{\partial \eta} & \dfrac{\partial y}{\partial \eta}\end{bmatrix},\qquad \frac{\partial x}{\partial \xi}=\sum_{i=1}^{4}\frac{\partial N_i}{\partial \xi}\, x_i,\ \ \text{etc.}$$ The shape-function derivatives are $$\frac{\partial N_i}{\partial \xi}=\tfrac14\Big[-(1-\eta),\;(1-\eta),\;-(1+\eta),\;(1+\eta)\Big],\qquad \frac{\partial N_i}{\partial \eta}=\tfrac14\Big[-(1-\xi),\;-(1+\xi),\;(1-\xi),\;(1+\xi)\Big]$$ so the Jacobian matrix in compact form is $$\boxed{\;[J]=\frac{1}{4}\begin{bmatrix}-(1-\eta) & (1-\eta) & -(1+\eta) & (1+\eta)\\ -(1-\xi) & -(1+\xi) & (1-\xi) & (1+\xi)\end{bmatrix}\begin{bmatrix}x_1 & y_1\\ x_2 & y_2\\ x_3 & y_3\\ x_4 & y_4\end{bmatrix}\;}$$ and the Jacobian (determinant) is $|J| = \dfrac{\partial x}{\partial \xi}\dfrac{\partial y}{\partial \eta}-\dfrac{\partial y}{\partial \xi}\dfrac{\partial x}{\partial \eta}$.
4. Part (c), element (i): substitute the node coordinates. With $(x_i)=(1,4,1,4)$ and $(y_i)=(1,1,3,4)$, $$\frac{\partial x}{\partial \xi}=\tfrac14\big[3(1-\eta)+3(1+\eta)\big]=\tfrac32,\qquad \frac{\partial y}{\partial \xi}=\tfrac14\big[(1+\eta)\big]=\frac{1+\eta}{4}$$ $$\frac{\partial x}{\partial \eta}=0,\qquad \frac{\partial y}{\partial \eta}=\tfrac14\big[2(1-\xi)+3(1+\xi)\big]=\frac{5+\xi}{4}$$ Hence $$[J]_{(\mathrm{i})}=\begin{bmatrix}\tfrac32 & \tfrac{1+\eta}{4}\\[4pt] 0 & \tfrac{5+\xi}{4}\end{bmatrix},\qquad \boxed{\,|J|_{(\mathrm{i})}=\frac{3}{2}\cdot\frac{5+\xi}{4}=\frac{3(5+\xi)}{8}\,}$$ Over the parent square this runs from $|J|=\tfrac32$ at $\xi=-1$ to $|J|=\tfrac94$ at $\xi=+1$, with $|J|=\tfrac{15}{8}=1.875$ at the centroid — strictly positive everywhere. As a check on the whole mapping, $\int_{-1}^{1}\!\int_{-1}^{1}|J|\, d\xi\, d\eta = 7.5$, which is exactly the area of the quadrilateral $(1,1)$–$(4,1)$–$(4,4)$–$(1,3)$ by the shoelace formula.
5. Part (c), element (ii): substitute the renumbered coordinates. Now $(x_i)=(1,1,4,4)$ and $(y_i)=(1,3,4,1)$, giving $$\frac{\partial x}{\partial \xi}=0,\qquad \frac{\partial y}{\partial \xi}=\tfrac14\big[2(1-\eta)-3(1+\eta)\big]=-\frac{1+5\eta}{4}$$ $$\frac{\partial x}{\partial \eta}=\tfrac14\big[3(1-\xi)+3(1+\xi)\big]=\tfrac32,\qquad \frac{\partial y}{\partial \eta}=\tfrac14\big[3(1-\xi)-2(1+\xi)\big]=\frac{1-5\xi}{4}$$ so that $$[J]_{(\mathrm{ii})}=\begin{bmatrix}0 & -\tfrac{1+5\eta}{4}\\[4pt] \tfrac32 & \tfrac{1-5\xi}{4}\end{bmatrix},\qquad \boxed{\,|J|_{(\mathrm{ii})}=\frac{3(1+5\eta)}{8}\,}$$ This expression equals $-\tfrac32$ at $\eta=-1$ and $+\tfrac94$ at $\eta=+1$, and vanishes on the interior line $\eta=-\tfrac15$.

Part (d) — what the two Jacobians imply. The Jacobian is the local area-scale factor of the mapping, $dx\, dy=|J|\, d\xi\, d\eta$, so its sign and magnitude are the complete statement of whether the parent-to-global map is admissible. Element (i) has $|J| \gt 0$ throughout, bounded between $1.5$ and $2.25$: the mapping is one-to-one and orientation-preserving, the element is valid, and the mild variation of $|J|$ merely reflects that the quadrilateral is not a parallelogram. Element (ii) has a Jacobian that changes sign inside the element, being negative for $\eta \lt -\tfrac15$ and positive above that line. A determinant that vanishes on an interior line means the mapping is singular there and not invertible; a sign change means the element folds back on itself, so the same physical point is the image of two different parent points. Element (ii) is therefore invalid — and since $[B]$ requires $[J]^{-1}$, its stiffness matrix cannot even be formed at the Gauss points near the fold.

The instructive part is that both elements occupy the same four vertices. Only the node numbering differs, and with the z-order convention the parent boundary is traversed 1–2–4–3. For element (i) that traversal is $(1,1)\to(4,1)\to(4,4)\to(1,3)$, a convex circuit taken counter-clockwise. For element (ii) it is $(1,1)\to(1,3)\to(4,1)\to(4,4)$, which crosses itself — the classic “bow-tie”. The practical rules that follow are the ones every pre-processor enforces: number the nodes consistently around the element in one rotational direction, check that $|J| \gt 0$ at every integration point (a uniformly negative $|J|$ merely signals reversed numbering and can be repaired by swapping two nodes, whereas a sign change cannot), and keep interior angles well away from $180^\circ$ so that $|J|$ does not become small enough to poison the accuracy of $[J]^{-1}$.