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 eight-node brick whose global corner coordinates are listed below, in centimetres.

Find. (a) the eight trilinear shape functions; (b) three values of $N_{8}$; (c) the interpolated displacement field; (d) the Jacobian matrix and its determinant; (e) two strain components at the element centre.

Approach. Assign natural coordinates to the eight nodes by matching the given global coordinates, write the standard trilinear product form, interpolate the geometry to obtain the Jacobian, and differentiate the displacement field through the inverse Jacobian.

1. Assign natural coordinates. Comparing the listed global coordinates with the corners of the parent cube, nodes 1–4 lie on the lower face ($\zeta = -1$) traversed 1→2→3→4, and nodes 5–8 sit directly above them:$$\begin{array}{ll}1(-1,-1,-1) & 5(-1,-1,+1) \\2(+1,-1,-1) & 6(+1,-1,+1) \\3(+1,+1,-1) & 7(+1,+1,+1) \\4(-1,+1,-1) & 8(-1,+1,+1)\end{array}$$
2. Write the trilinear shape functions. The tensor-product form that is unity at its own node and zero at the other seven is$$\boxed{\;N_{i}(\xi,\eta,\zeta) = \tfrac18\bigl(1+\xi\xi_{i}\bigr)\bigl(1+\eta\eta_{i}\bigr)\bigl(1+\zeta\zeta_{i}\bigr), \qquad i = 1 \ldots 8\;}$$ For example $N_{1} = \tfrac18(1-\xi)(1-\eta)(1-\zeta)$ and $N_{7} = \tfrac18(1+\xi)(1+\eta)(1+\zeta)$. Substituting each node confirms $N_{i}(\xi_{j},\eta_{j},\zeta_{j}) = \delta_{ij}$, and the eight functions sum to unity everywhere.
3. (b) Evaluate $N_{8} = \tfrac18(1-\xi)(1+\eta)(1+\zeta)$. At node 2, $(\xi,\eta,\zeta) = (1,-1,-1)$, every factor except the first vanishes; at node 8 all three factors equal 2; at the centroid each factor equals 1:$$N_{8}(1,-1,-1) = 0, \qquad N_{8}(-1,1,1) = \tfrac18(2)(2)(2) = 1,\qquad N_{8}(0,0,0) = \tfrac18$$ The centroid value confirms that all eight functions contribute equally at the centre.
4. (c) Interpolate the displacement field. All $u_{i}$ and $w_{i}$ are zero so $u \equiv 0$ and $w \equiv 0$. The four displaced nodes 1, 2, 5, 6 are exactly the nodes with $\eta_{i} = -1$, so their shape functions sum to a function of $\eta$ alone:$$N_{1} + N_{2} + N_{5} + N_{6} = \tfrac18(1-\eta)\bigl[(1-\xi)+(1+\xi)\bigr]\bigl[(1-\zeta)+(1+\zeta)\bigr] = \tfrac12(1-\eta)$$ Hence$$\boxed{\;u = 0, \qquad v(\eta) = -0.025\left(\frac{1-\eta}{2}\right) = -0.0125\,(1-\eta)\ \text{mm}, \qquad w = 0\;}$$ which returns $-0.025$ mm on the face $\eta = -1$, zero on $\eta = +1$, and $-0.0125$ mm at the centroid.
5. (d) Interpolate the geometry. Using $x = \sum N_{i}x_{i}$ and likewise for $y$ and $z$, the sums collapse for a rectangular box to$$x = \frac{3+8}{2} + \frac{8-3}{2}\xi = 5.5 + 2.5\,\xi, \qquad y = 7 + 3\,\eta, \qquad z = 5.5 + 0.5\,\zeta$$
6. Assemble the Jacobian matrix. Differentiating the map:$$[J] = \begin{bmatrix}\partial x/\partial \xi & \partial y/\partial \xi & \partial z/\partial \xi \\\partial x/\partial \eta & \partial y/\partial \eta & \partial z/\partial \eta \\\partial x/\partial \zeta & \partial y/\partial \zeta & \partial z/\partial \zeta\end{bmatrix}= \begin{bmatrix} 2.5 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0.5 \end{bmatrix}\ \text{cm}$$ The matrix is diagonal and constant, which is the signature of an undistorted box element aligned with the global axes.
7. Evaluate the Jacobian. The determinant is the product of the diagonal:$$\boxed{\;|J| = (2.5)(3)(0.5) = 3.75\ \text{cm}^{3}\;}$$ As a check, the parent cube has volume 8 in natural coordinates, so the physical volume is $8|J| = 30$ cm³, which matches the box $5 \times 6 \times 1 = 30$ cm³ exactly.
8. (e) Differentiate to obtain the strains. Cartesian derivatives follow from the inverse Jacobian, which for a diagonal $[J]$ is simply the reciprocal of each diagonal entry:$$\frac{\partial}{\partial x} = \frac{1}{2.5}\frac{\partial}{\partial \xi},\qquad \frac{\partial}{\partial y} = \frac{1}{3}\frac{\partial}{\partial \eta},\qquad \frac{\partial}{\partial z} = \frac{1}{0.5}\frac{\partial}{\partial \zeta}$$ Since $u \equiv 0$, both $\partial u/\partial x$ and $\partial u/\partial y$ vanish; and since $v$ depends on $\eta$ only, while $\eta$ is a function of $y$ alone, $\partial v/\partial x = 0$ as well. Therefore$$\boxed{\;\varepsilon_{x} = 0, \qquad \gamma_{xy} = 0\;}$$

Both requested components vanishing is the correct answer, and the reason is worth stating rather than leaving implicit: the prescribed nodal pattern is a pure stretching of the element in the $y$ direction only. The single non-zero strain is$$\varepsilon_{y} = \frac{\partial v}{\partial y} = \frac{1}{3}\frac{\partial}{\partial \eta}\bigl[-0.0125(1-\eta)\bigr] = \frac{0.0125}{3} = 4.1667\times10^{-3}\ \text{mm/cm} = 4.1667\times10^{-4}$$ constant throughout the element, with every shear component and the other two normal strains identically zero.

Two features of the trilinear brick are visible in this result. First, the element produces a strain field that is constant in the direction of the differentiation but may vary linearly in the other two — here the geometry is so regular that the variation disappears entirely. Second, because the Jacobian is constant, a single Gauss point would integrate this element's constant-strain response exactly; the usual $2\times2\times2$ rule is used in practice only because distorted bricks and bending-dominated states require it.