22-Mec-B10 Finite Element Analysis · December 2014
Question 6 of 7: Eight-node hexahedron — shape functions, Jacobian and strains
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2014 — 07-Mec-B10 Finite Element Analysis. Three hours, open book (any texts, references or notes; any non-communicating calculator). Seven questions of equal value (20 marks each); candidates attempt any five, and every question is to be solved within the context of the finite element method. All seven are worked here.
Reference texts. Logan, A First Course in the Finite Element Method (6th ed.); Reddy, An Introduction to the Finite Element Method (4th ed.); Cook, Malkus, Plesha & Witt, Concepts and Applications of Finite Element Analysis (4th ed.); Zienkiewicz, Taylor & Zhu, The Finite Element Method: Its Basis and Fundamentals (7th ed.); Bathe, Finite Element Procedures (2nd ed.); Hutton, Fundamentals of Finite Element Analysis.
Given. A rectangular-box eight-node hexahedron (trilinear brick) whose global node coordinates are printed on the figure, in centimetres, with nodes 1–4 on the lower face and 5–8 directly above them.
Global nodal coordinates (cm)
Node
(x, y, z)
Node
(x, y, z)
1
(2, 3, 2)
5
(2, 3, 4)
2
(7, 3, 2)
6
(7, 3, 4)
3
(7, 9, 2)
7
(7, 9, 4)
4
(2, 9, 2)
8
(2, 9, 4)
Find. The eight shape functions, three sample values of N3, the interpolated displacement field, the Jacobian matrix with its determinant, and the two requested strain components at the element centre.
Question 6 — the eight-node hexahedron: a 5 × 6 × 2 cm box with nodes 1–4 on the z = 2 cm face and 5–8 on the z = 4 cm face.
Approach. Write the standard trilinear product form using the natural coordinates of each node, evaluate it where asked, interpolate the given nodal displacements, then map the geometry to obtain the Jacobian and differentiate the displacement field through the chain rule.
(a) Write the trilinear shape functions. Mapping the numbering of the figure onto the parent cube, node 1 is at $(\xi,\eta,\zeta)=(-1,-1,-1)$, then 2 $(1,-1,-1)$, 3 $(1,1,-1)$, 4 $(-1,1,-1)$ on the lower face and 5 $(-1,-1,1)$, 6 $(1,-1,1)$, 7 $(1,1,1)$, 8 $(-1,1,1)$ above them. The shape functions are the product of three one-dimensional linear functions:$$\boxed{N_i=\tfrac18\big(1+\xi\xi_i\big)\big(1+\eta\eta_i\big)\big(1+\zeta\zeta_i\big),\qquad i=1\ldots 8}$$so, for example, $N_1=\tfrac18(1-\xi)(1-\eta)(1-\zeta)$ and $N_3=\tfrac18(1+\xi)(1+\eta)(1-\zeta)$. Each equals unity at its own node, vanishes at the other seven, and the eight sum to one.
(b) Evaluate $N_3$ at three points. At node 3, $(\xi,\eta,\zeta)=(1,1,-1)$, so $N_3=\tfrac18(2)(2)(2)=1$. At node 5, $(-1,-1,1)$, every factor vanishes and $N_3=0$. At the centroid $(0,0,0)$, $$N_3(0,0,0)=\tfrac18(1)(1)(1)=\tfrac18=0.125$$$$\boxed{N_3(\text{node }3)=1,\qquad N_3(\text{node }5)=0,\qquad N_3(\text{centroid})=\tfrac18}$$The centroid value is the same for all eight functions, which is why the centroid displacement equals the arithmetic mean of the eight nodal values.
(c) Interpolate the given displacements. Nodes 3, 4, 7 and 8 are exactly the four nodes with $\eta_i=+1$, so$$v=\sum N_iv_i=0.025\big(N_3+N_4+N_7+N_8\big)=0.025\cdot\tfrac18(1+\eta)\big[(1+\xi)(1-\zeta)+(1-\xi)(1-\zeta)+(1+\xi)(1+\zeta)+(1-\xi)(1+\zeta)\big]$$The bracket collapses to 4, leaving$$\boxed{u=0,\qquad v=0.0125\,(1+\eta)\ \text{mm},\qquad w=0}$$The field is a pure linear variation in $\eta$: zero on the face $\eta=-1$ and 0.025 mm on the face $\eta=+1$.
(d) Map the geometry and form the Jacobian. Using the same shape functions isoparametrically, $x=\sum N_ix_i$ and so on. Because the element is a rectangular box aligned with the axes, the sums reduce to$$x=\tfrac{9}{2}+\tfrac52\xi,\qquad y=6+3\eta,\qquad z=3+\zeta$$i.e. the mid-point plus half the side length times the natural coordinate. Differentiating,$$[J]=\begin{bmatrix}\partial x/\partial\xi&\partial y/\partial\xi&\partial z/\partial\xi\\\partial x/\partial\eta&\partial y/\partial\eta&\partial z/\partial\eta\\\partial x/\partial\zeta&\partial y/\partial\zeta&\partial z/\partial\zeta\end{bmatrix}=\begin{bmatrix}2.5&0&0\\0&3&0\\0&0&1\end{bmatrix}\ \text{cm}$$$$\boxed{|J|=(2.5)(3)(1)=7.5\ \text{cm}^3\ \text{(constant)}}$$The diagonal form is the signature of an undistorted box. As a check, $\int_{-1}^{1}\!\!\int_{-1}^{1}\!\!\int_{-1}^{1}|J|\,d\xi\,d\eta\,d\zeta=8(7.5)=60$ cm3, which is exactly the physical volume $5\times6\times2$ cm3.
(e) Differentiate to obtain the strains. The chain rule through the (diagonal) Jacobian is simply $\partial/\partial y=(1/3)\,\partial/\partial\eta$, so with $v$ in millimetres and $y$ in centimetres$$\varepsilon_y=\frac{\partial v}{\partial y}=\frac{1}{3}\frac{\partial}{\partial\eta}\big[0.0125(1+\eta)\big]=\frac{0.0125}{3}\ \frac{\text{mm}}{\text{cm}}=\frac{0.0125}{30}$$$$\boxed{\varepsilon_y=4.167\times10^{-4}}$$For the shear, $u\equiv0$ so $\partial u/\partial y=0$, and $v$ does not depend on $\xi$ or $\zeta$ so $\partial v/\partial x=0$:$$\boxed{\gamma_{xy}=0}$$Both results are in fact constant throughout the element, not merely at its centre, because the imposed field is linear and the Jacobian is uniform.
