Question 7 of 7: Isoparametric quadrilateral — Jacobian matrix and the effect of node numbering [20 marks]

Question 7: Isoparametric quadrilateral — Jacobian matrix and the effect of node numbering [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 with the shape functions quoted in the question, and two candidate global elements occupying the same four physical points but numbered differently:

Find. (a) a definition of an isoparametric element; (b) the general Jacobian matrix of the four-node quadrilateral; (c) the Jacobian $|J|$ for each of the two numbered elements; (d) what the two expressions imply about the validity of the coordinate mapping.

[Figure not reproduced: Figure 7.1 — The parent (natural-coordinate) domain of the four-node quadrilateral, with the node numbering implied by the shape functions given in the question. Note that the printed exam figure labels node 2 as (−1, −1); the shape function N₂ = ¼(1 + ξ)(1 − . See the official exam paper.]

(a) Definition of an isoparametric element [2 marks]

An isoparametric element is one in which the same shape functions, referred to the same nodes, are used to interpolate both the element geometry and the field variable:

“Iso-parametric” means equal parameters — equal numbers of geometric and field nodes. If the geometry uses fewer nodes than the field the element is subparametric; if it uses more, superparametric. The practical value of the isoparametric family is that a straight-sided or curve-sided element of arbitrary shape can be handled entirely on the fixed parent domain, where quadrature rules and shape-function derivatives are tabulated constants.

(b) The Jacobian matrix [10 marks]

Approach. Differentiate the isoparametric geometric mapping with respect to the natural coordinates and collect the four derivatives into a $2\times2$ matrix, expressed as a product of a shape-function-derivative matrix and the nodal-coordinate matrix.

1. Differentiate the shape functions. From the four functions given in the question, $$\frac{\partial N_1}{\partial \xi}=-\tfrac14(1-\eta),\quad \frac{\partial N_2}{\partial \xi}=+\tfrac14(1-\eta),\quad \frac{\partial N_3}{\partial \xi}=-\tfrac14(1+\eta),\quad \frac{\partial N_4}{\partial \xi}=+\tfrac14(1+\eta)$$ $$\frac{\partial N_1}{\partial \eta}=-\tfrac14(1-\xi),\quad \frac{\partial N_2}{\partial \eta}=-\tfrac14(1+\xi),\quad \frac{\partial N_3}{\partial \eta}=+\tfrac14(1-\xi),\quad \frac{\partial N_4}{\partial \eta}=+\tfrac14(1+\xi)$$
2. Write the Jacobian matrix in product form. With $x=\sum N_ix_i$ and $y=\sum N_iy_i$, $$[J]=\begin{bmatrix}\dfrac{\partial x}{\partial \xi} & \dfrac{\partial y}{\partial \xi}\\[8pt] \dfrac{\partial x}{\partial \eta} & \dfrac{\partial y}{\partial \eta}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-(1-\eta) & (1-\eta) & -(1+\eta) & (1+\eta)\\ -(1-\xi) & -(1+\xi) & (1-\xi) & (1+\xi)\end{bmatrix}\begin{bmatrix}x_1 & y_1\\ x_2 & y_2\\ x_3 & y_3\\ x_4 & y_4\end{bmatrix}$$
3. State the four entries explicitly. Carrying out the multiplication, $$\boxed{\ \begin{aligned}J_{11}&=\tfrac14\big[-(1-\eta)x_1+(1-\eta)x_2-(1+\eta)x_3+(1+\eta)x_4\big]\\ J_{12}&=\tfrac14\big[-(1-\eta)y_1+(1-\eta)y_2-(1+\eta)y_3+(1+\eta)y_4\big]\\ J_{21}&=\tfrac14\big[-(1-\xi)x_1-(1+\xi)x_2+(1-\xi)x_3+(1+\xi)x_4\big]\\ J_{22}&=\tfrac14\big[-(1-\xi)y_1-(1+\xi)y_2+(1-\xi)y_3+(1+\xi)y_4\big]\end{aligned}\ }$$ and the Jacobian (determinant) is $|J|=J_{11}J_{22}-J_{12}J_{21}$. Note that each entry is linear in $\xi$ or $\eta$, so $|J|$ is at most bilinear in the natural coordinates — it is constant only if the element is a parallelogram.

(c) Jacobian of the two elements [4 marks]

1. Element (i): substitute the nodal coordinates. With $(x_1,y_1)=(1,1)$, $(x_2,y_2)=(3,1)$, $(x_3,y_3)=(1,3)$ and $(x_4,y_4)=(3,4)$, the four entries of step (b) become $$J_{11}=\tfrac14\big[-(1-\eta)(1)+(1-\eta)(3)-(1+\eta)(1)+(1+\eta)(3)\big]=\tfrac14\big[2(1-\eta)+2(1+\eta)\big]=1$$ $$J_{12}=\tfrac14\big[-(1-\eta)(1)+(1-\eta)(1)-(1+\eta)(3)+(1+\eta)(4)\big]=\frac{1+\eta}{4}$$ $$J_{21}=\tfrac14\big[-(1-\xi)(1)-(1+\xi)(3)+(1-\xi)(1)+(1+\xi)(3)\big]=0 ,\qquad J_{22}=\tfrac14\big[-(1-\xi)-3(1+\xi)+3(1-\xi)+4(1+\xi)\big]=\frac{5+\xi}{4}$$
2. Element (i): form the determinant. Since $J_{21}=0$ the determinant is the product of the diagonal, $$\boxed{\,|J|_{(\mathrm{i})}=(1)\left(\frac{5+\xi}{4}\right)-\left(\frac{1+\eta}{4}\right)(0)=\frac{5+\xi}{4}\,}$$ Over the parent domain $-1\le\xi\le1$ this ranges from $|J|=1$ at $\xi=-1$ to $|J|=1.5$ at $\xi=+1$: strictly positive everywhere. Integrating it over the parent square, $\int_{-1}^{1}\!\int_{-1}^{1}\frac{5+\xi}{4}\,d\xi\,d\eta = 5$, which is exactly the area of the trapezoid with parallel sides of length 2 and 3 and width 2 — the check that the mapping reproduces the correct physical region.
3. Element (ii): substitute the re-numbered coordinates. Now $(x_1,y_1)=(1,1)$, $(x_2,y_2)=(1,3)$, $(x_3,y_3)=(3,4)$ and $(x_4,y_4)=(3,1)$, so $$J_{11}=\tfrac14\big[-(1-\eta)(1)+(1-\eta)(1)-(1+\eta)(3)+(1+\eta)(3)\big]=0$$ $$J_{12}=\tfrac14\big[-(1-\eta)(1)+(1-\eta)(3)-(1+\eta)(4)+(1+\eta)(1)\big]=\tfrac14\big[2(1-\eta)-3(1+\eta)\big]=-\frac{1+5\eta}{4}$$ $$J_{21}=\tfrac14\big[-(1-\xi)(1)-(1+\xi)(1)+(1-\xi)(3)+(1+\xi)(3)\big]=1 ,\qquad J_{22}=\tfrac14\big[-(1-\xi)-3(1+\xi)+4(1-\xi)+1(1+\xi)\big]=\frac{1-5\xi}{4}$$
4. Element (ii): form the determinant. With $J_{11}=0$, $$\boxed{\,|J|_{(\mathrm{ii})}=(0)\left(\frac{1-5\xi}{4}\right)-\left(-\frac{1+5\eta}{4}\right)(1)=\frac{1+5\eta}{4}\,}$$ This ranges from $|J|=-1$ at $\eta=-1$, through zero at $\eta=-\tfrac15$, to $|J|=+1.5$ at $\eta=+1$. It is negative over the lower 40% of the parent domain and positive over the upper 60%.

(d) Inference about the coordinate mapping [4 marks]

The Jacobian is the local ratio of physical area to parent area, $dA = |J|\,d\xi\,d\eta$, and it is also the determinant that must be inverted to convert natural-coordinate derivatives into physical strains. Three conclusions follow directly from the two expressions.

Element (i) is a valid element. Its Jacobian is strictly positive everywhere in the parent domain, $1\le |J|\le 1.5$, so the mapping from $(\xi,\eta)$ to $(x,y)$ is one-to-one and onto, $[J]^{-1}$ exists at every Gauss point, and the numerical integration of the stiffness matrix is well posed. The variation of $|J|$ with $\xi$ simply reflects that the element is a trapezoid rather than a parallelogram: a constant Jacobian would require opposite sides to be parallel and equal.

Element (ii) is invalid. Its Jacobian changes sign inside the element, passing through zero along the line $\eta=-1/5$. Where $|J|=0$ the mapping is singular and $[J]^{-1}$ does not exist, so strains and stresses cannot be computed; where $|J|\lt 0$ the element has been turned inside out, and any stiffness contribution accumulated there is negative — a physically meaningless negative-stiffness region that will corrupt the assembled global matrix and typically produces a solver diagnostic such as “negative Jacobian” or “element inverted”. Note that the physical region is identical to element (i); the only difference is the node numbering.

The practical rule. Nodes must be numbered consistently, so that traversing the parent boundary $1\to 2\to 4\to 3$ traverses the physical boundary in a single consistent (counter-clockwise) sense. Element (ii) instead traces a self-intersecting bow-tie, and the sign change of $|J|$ is the algebraic signature of that crossing. Because the sign of $|J|$ is set by the numbering, checking $|J|\gt 0$ at every integration point is the standard automatic mesh-quality test in finite-element pre-processors; the closely related quality measure is the ratio $|J|_{\min}/|J|_{\max}$ over the element, which is 0.67 for element (i) — acceptable — and negative for element (ii).