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 parallelepiped (brick) element whose eight nodes are listed in the paper, four on the lower face and four on the upper face:

Find. The eight trilinear shape functions, three values of $N_8$, the interpolated displacement fields, the Jacobian matrix and its determinant, and the two requested strain components at the element centre.

Approach. Assign each node its natural-coordinate signs from the global geometry, write the trilinear product form, then use the fact that the element is an axis-aligned brick to reduce the Jacobian to a diagonal matrix before differentiating the displacement field.

1. Map the global nodes onto the parent cube. The element is a rectangular block, so the mapping is affine in each direction: $$\xi = \frac{x-4.5}{2.5},\qquad \eta=\frac{y-6}{3},\qquad \zeta = z-3$$ Applying this to the listed coordinates gives the standard hexahedral node ordering $$1(-1,-1,-1),\;2(1,-1,-1),\;3(1,1,-1),\;4(-1,1,-1),\;5(-1,-1,1),\;6(1,-1,1),\;7(1,1,1),\;8(-1,1,1)$$ so nodes 1–4 form the $\zeta=-1$ face and 5–8 the $\zeta=+1$ face, each traversed in the same rotational sense.
2. Part (a): write the trilinear shape functions. Each function must be unity at its own node and vanish at the other seven, and the product of the three “far-face” linear factors does exactly that: $$\boxed{\, N_i(\xi,\eta,\zeta)=\tfrac18\big(1+\xi\xi_i\big)\big(1+\eta\eta_i\big)\big(1+\zeta\zeta_i\big),\qquad i = 1\ldots 8\,}$$ Written out with the node signs above, $$N_1=\tfrac18(1-\xi)(1-\eta)(1-\zeta),\quad N_2=\tfrac18(1+\xi)(1-\eta)(1-\zeta),\quad N_3=\tfrac18(1+\xi)(1+\eta)(1-\zeta),$$ $$N_4=\tfrac18(1-\xi)(1+\eta)(1-\zeta),\quad N_5=\tfrac18(1-\xi)(1-\eta)(1+\zeta),\quad N_6=\tfrac18(1+\xi)(1-\eta)(1+\zeta),$$ $$N_7=\tfrac18(1+\xi)(1+\eta)(1+\zeta),\quad N_8=\tfrac18(1-\xi)(1+\eta)(1+\zeta)$$ The set satisfies $N_i(\text{node }j)=\delta_{ij}$ and $\sum_{i=1}^{8}N_i=1$; each function is trilinear, so the element edges stay straight and the faces are bilinear surfaces.
3. Part (b): evaluate $N_8$ at three points. Node 8 sits at $(-1,1,1)$, so $N_8=\tfrac18(1-\xi)(1+\eta)(1+\zeta)$. At node 2, $(\xi,\eta,\zeta)=(1,-1,-1)$, every factor vanishes; at node 8 the three factors are $2,2,2$; at the centroid all three are unity: $$\boxed{\, N_8\big|_{\text{node }2}=0,\qquad N_8\big|_{\text{node }8}=1,\qquad N_8\big|_{\text{centroid}}=\tfrac18=0.125\,}$$ The last value is not a coincidence of node 8: by symmetry every trilinear shape function equals $1/8$ at the centroid, and the eight values duly sum to one.
4. Part (c): interpolate the displacement field. Only $u$-components are non-zero, so $v(\xi,\eta,\zeta)=0$ and $w(\xi,\eta,\zeta)=0$ identically. The four loaded nodes 2, 3, 6, 7 all have $\xi_i=+1$, that is, they are exactly the four nodes of the $\xi=+1$ face, all carrying $u_0=0.025$ mm: $$u = u_0\big(N_2+N_3+N_6+N_7\big)=u_0\cdot\tfrac18(1+\xi)\Big[(1-\eta)(1-\zeta)+(1+\eta)(1-\zeta)+(1-\eta)(1+\zeta)+(1+\eta)(1+\zeta)\Big]$$ The bracket collapses to $4$, since the $\eta$ and $\zeta$ terms cancel in pairs, leaving $$\boxed{\;u(\xi,\eta,\zeta)=\frac{u_0}{2}\big(1+\xi\big)=0.0125\,(1+\xi)\ \text{mm},\qquad v = w = 0\;}$$ Checking, $u=0$ on the face $\xi=-1$ (nodes 1, 4, 5, 8), $u=0.025$ mm on $\xi=+1$, and $u=0.0125$ mm at the centroid. Physically this is a uniform stretch of the block in the $x$ direction: the sum of the shape functions over one face is always a linear function of the coordinate normal to that face.
5. Part (d): form the Jacobian matrix. With $x=\sum N_ix_i$ and its counterparts, $$[J]=\begin{bmatrix}\partial x/\partial\xi & \partial y/\partial\xi & \partial z/\partial\xi\\ \partial x/\partial\eta & \partial y/\partial\eta & \partial z/\partial\eta\\ \partial x/\partial\zeta & \partial y/\partial\zeta & \partial z/\partial\zeta\end{bmatrix}$$ Every off-diagonal term vanishes because the element edges are parallel to the global axes, and each diagonal term is a half-edge-length: $$\boxed{\;[J]=\begin{bmatrix}a_x/2 & 0 & 0\\ 0 & a_y/2 & 0\\ 0 & 0 & a_z/2\end{bmatrix}=\begin{bmatrix}2.5 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 1\end{bmatrix}\ \text{cm}\;}$$ so the Jacobian is constant over the element: $$\boxed{\,|J| = (2.5)(3)(1) = 7.5\ \text{cm}^{3}\,}$$ The volume check confirms it: $\iiint |J|\, d\xi\, d\eta\, d\zeta = 8\,|J| = 60\ \text{cm}^{3}$, and the block measures $5\times 6\times 2 = 60\ \text{cm}^{3}$. Positive and constant $|J|$ means the mapping is one-to-one everywhere — a brick element is the best-conditioned hexahedron there is.
6. Part (e): differentiate the displacement field. Global derivatives follow from the chain rule with $[J]^{-1}$, which for a diagonal $[J]$ is simply the reciprocal of each diagonal entry. Working in consistent units, the half-edge lengths in millimetres are $25$, $30$ and $10$ mm, so $\partial\xi/\partial x = 1/25\ \text{mm}^{-1}$ and $\partial\eta/\partial y = 1/30\ \text{mm}^{-1}$. Since $u=0.0125(1+\xi)$ mm, $$\varepsilon_x = \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\,\frac{\partial \xi}{\partial x}=(0.0125\ \text{mm})\left(\frac{1}{25\ \text{mm}}\right)$$ $$\boxed{\,\varepsilon_x = 5.0\times 10^{-4}\ \ (\text{i.e. }500\ \mu\varepsilon,\ \text{uniform over the element})\,}$$ For the shear strain, $v\equiv 0$ gives $\partial v/\partial x = 0$, and $u$ depends on $\xi$ alone so $\partial u/\partial \eta = 0$ and hence $\partial u/\partial y = 0$: $$\boxed{\,\gamma_{xy} = \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} = 0 + 0 = 0\,}$$

A zero shear strain is the correct answer, not a symptom of a mis-transcribed question, and it is worth saying why in the answer script. The imposed displacement pattern moves one whole face of the block uniformly in its own normal direction while holding the opposite face fixed, so the deformation is a pure uniaxial extension: rectangles stay rectangles, no angle between the $x$ and $y$ material lines changes, and every shear component vanishes. Because $|J|$ is diagonal there is also no coupling introduced by the mapping itself — on a distorted hexahedron the same nodal displacements would generate spurious-looking but genuine shear through the off-diagonal terms of $[J]^{-1}$.

The strain field is also perfectly uniform, at $\varepsilon_x = 5\times 10^{-4}$ everywhere, which is what a trilinear brick is designed to reproduce exactly: the eight-node hexahedron is a constant-strain element in each direction for this class of loading. Numerically, the extension is $0.025$ mm over a $50$ mm length, and $0.025/50 = 5\times10^{-4}$ confirms the chain-rule result directly. Note that the answer is dimensionless and therefore unaffected by the cm/mm mixture in the question, provided the conversion is applied consistently — the frequent error is dividing a millimetre displacement by a centimetre length and reporting a strain ten times too large.