Question 7 of 7: Eight-node hexahedron — shape functions, Jacobian and strains

Question 7: Eight-node hexahedron — shape functions, Jacobian and strains (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 rectangular-box hexahedral element whose eight nodes are, in centimetres, $1(2,3,2)$, $2(7,3,2)$, $3(7,9,2)$, $4(2,9,2)$, $5(2,3,4)$, $6(7,3,4)$, $7(7,9,4)$ and $8(2,9,4)$. Nodes 1–4 form the $\zeta = -1$ face and nodes 5–8 the $\zeta = +1$ face, with node 1 at the parent corner $(-1,-1,-1)$. The imposed nodal displacements are $v_2 = v_3 = v_6 = v_7 = 0.025$ mm, all other components zero.

Find. The eight trilinear shape functions, three values of $N_6$, the interpolated displacement field, the Jacobian matrix and its determinant, and two strain components at the element centre.

Approach. Write the trilinear shape functions from the parent-corner signs, evaluate them where asked, interpolate the given nodal displacements, differentiate the isoparametric map to get the Jacobian, and finally convert natural-coordinate derivatives to physical ones through the inverse Jacobian.

1. (a) Write the trilinear shape functions. For the parent cube, each node $i$ sits at a corner $(\xi_i,\eta_i,\zeta_i)$ with every component $\pm1$, and $$N_i(\xi,\eta,\zeta) = \tfrac{1}{8}\big(1+\xi\xi_i\big)\big(1+\eta\eta_i\big)\big(1+\zeta\zeta_i\big),\qquad i = 1\ldots8$$ Matching the given numbering — node 1 at $(-1,-1,-1)$, 2 at $(1,-1,-1)$, 3 at $(1,1,-1)$, 4 at $(-1,1,-1)$, 5 at $(-1,-1,1)$, 6 at $(1,-1,1)$, 7 at $(1,1,1)$, 8 at $(-1,1,1)$ — the eight functions are $$\boxed{\begin{aligned} N_1 &= \tfrac18(1-\xi)(1-\eta)(1-\zeta), & N_2 &= \tfrac18(1+\xi)(1-\eta)(1-\zeta),\\ N_3 &= \tfrac18(1+\xi)(1+\eta)(1-\zeta), & N_4 &= \tfrac18(1-\xi)(1+\eta)(1-\zeta),\\ N_5 &= \tfrac18(1-\xi)(1-\eta)(1+\zeta), & N_6 &= \tfrac18(1+\xi)(1-\eta)(1+\zeta),\\ N_7 &= \tfrac18(1+\xi)(1+\eta)(1+\zeta), & N_8 &= \tfrac18(1-\xi)(1+\eta)(1+\zeta). \end{aligned}}$$ They satisfy $N_i(\xi_j,\eta_j,\zeta_j) = \delta_{ij}$ and $\sum_{i=1}^{8}N_i = 1$ identically.
2. (b) Evaluate $N_6$ at the requested points. With $N_6 = \tfrac18(1+\xi)(1-\eta)(1+\zeta)$, node 2 sits at $(1,-1,-1)$ and node 4 at $(-1,1,-1)$: $$N_6(\text{node }2) = \tfrac18(2)(2)(0) = 0,\qquad N_6(\text{node }4) = \tfrac18(0)(0)(0) = 0,\qquad N_6(0,0,0) = \tfrac18(1)(1)(1) = 0.125$$ Both nodal values are zero as the Kronecker-delta property demands, and the centroid value is $1/8$ because at the element centre all eight functions are equal and must sum to one.
3. (c) Interpolate the displacement field. Since every $u_i$ and $w_i$ is zero, $u \equiv 0$ and $w \equiv 0$. The four loaded nodes 2, 3, 6 and 7 are precisely the four corners of the $\xi = +1$ face, and summing their shape functions collapses the $\eta$ and $\zeta$ dependence entirely: $$v(\xi,\eta,\zeta) = 0.025\big[N_2+N_3+N_6+N_7\big] = 0.025\left[\tfrac{1+\xi}{2}\right]$$ so $$\boxed{\,u = 0,\qquad v(\xi,\eta,\zeta) = 0.0125\,(1+\xi)\ \text{mm},\qquad w = 0\,}$$ The field is a pure linear ramp in $\xi$ alone: zero on the $\xi = -1$ face, $0.025$ mm on the $\xi = +1$ face and $0.0125$ mm at the centroid. This time no correction terms survive, because a trilinear brick has no enriched edges — the loaded set is one complete face and nothing more.
4. (d) Build the Jacobian matrix. The isoparametric map is $x = \sum N_ix_i$, and similarly for $y$ and $z$. Substituting the nodal coordinates, $$x = 4.5 + 2.5\,\xi,\qquad y = 6 + 3\,\eta,\qquad z = 3 + \zeta \quad \text{(cm)}$$ because the element is an axis-aligned rectangular box: opposite faces are parallel, so every cross-term cancels. Differentiating, $$[J] = \begin{bmatrix} \partial x/\partial\xi & \partial y/\partial\xi & \partial z/\partial\xi\\ \partial x/\partial\eta & \partial y/\partial\eta & \partial z/\partial\eta\\ \partial x/\partial\zeta & \partial y/\partial\zeta & \partial z/\partial\zeta \end{bmatrix} = \begin{bmatrix} 2.5 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 1\end{bmatrix}\ \text{cm}$$ $$|J| = \det[J] = (2.5)(3)(1) = \boxed{\,7.5\ \text{cm}^{3}\,}$$ The three diagonal entries are the half-side lengths $\tfrac12(7-2)$, $\tfrac12(9-3)$ and $\tfrac12(4-2)$, and the Jacobian is constant throughout the element. The physical volume check confirms it: $\int|J|\,d\xi\,d\eta\,d\zeta = 8|J| = 60\ \text{cm}^{3}$, which equals $(5)(6)(2) = 60\ \text{cm}^{3}$ measured directly from the coordinates.
5. (e) Convert to physical strains at the element centre. Natural-coordinate derivatives relate to physical ones through $$\begin{Bmatrix}\partial/\partial x\\ \partial/\partial y\end{Bmatrix} = [J]^{-1}\begin{Bmatrix}\partial/\partial\xi\\ \partial/\partial\eta\end{Bmatrix}$$ Displacements are quoted in millimetres while coordinates are in centimetres, so the in-plane Jacobian must be converted: $\partial x/\partial\xi = 25$ mm and $\partial y/\partial\eta = 30$ mm. Since $v$ depends on $\xi$ only, $\partial v/\partial\eta = 0$ and therefore $$\varepsilon_y = \frac{\partial v}{\partial y} = \frac{1}{30}\frac{\partial v}{\partial\eta} = 0,\qquad \frac{\partial v}{\partial x} = \frac{1}{25}\frac{\partial v}{\partial\xi} = \frac{0.0125}{25} = 5.0\times10^{-4}$$ and with $u \equiv 0$ giving $\partial u/\partial y = 0$, $$\boxed{\,\varepsilon_y = 0,\qquad \gamma_{xy} = \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} = 5.0\times10^{-4}\,}$$

The vanishing normal strain in part (e) is a real result rather than a symptom of a mis-transcribed question, and saying why is worth marks. The imposed field displaces material in the $y$ direction by an amount that varies only with $x$; nothing is stretched along $y$ itself, so $\partial v/\partial y$ is identically zero everywhere in the element, not just at the centre. What the element experiences is pure shear: a uniform $\gamma_{xy} = 5\times10^{-4}$, produced entirely by the $x$-gradient of the transverse displacement. That the answer is uniform rather than centre-specific follows from the box geometry, since a constant diagonal $[J]$ means the natural and physical gradients differ only by fixed scale factors. On a distorted brick $[J]$ would vary with position and the same nodal data would give a strain that genuinely depended on where it was sampled — which is exactly why Gauss points, and not nodes, are the preferred stress-recovery locations in general meshes.