Question 6 of 7: Eight-node hexahedron — shape functions, field variables, Jacobian and strains [20 marks]

Question 6: Eight-node hexahedron — shape functions, field variables, 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 (trilinear hexahedron) whose corner coordinates are listed in the figure, loaded through a prescribed set of nodal displacements:

Find. (a) the eight trilinear shape functions; (b) $N_3$ at nodes 3 and 5 and at the centroid; (c) the displacement fields $u$, $v$, $w$ as functions of $(\xi,\eta,\zeta)$; (d) the Jacobian matrix and its determinant; (e) $\varepsilon_x$ and $\gamma_{xy}$ at the element centre.

Approach. Read the natural coordinates of each node from the figure’s numbering, write the trilinear shape functions in the standard product form, use them to interpolate the displacement field, differentiate the geometric mapping to build the Jacobian matrix, and finally convert natural-coordinate derivatives into physical strains through the inverse Jacobian.

(a) The eight shape functions [4 marks]

1. Assign natural coordinates to the nodes. Comparing the global coordinates with the parent cube, $x=1$ and $x=6$ are the $\xi=-1$ and $\xi=+1$ faces, $y=2$ and $y=8$ are the $\eta=\mp 1$ faces, and $z=2$ and $z=4$ are the $\zeta=\mp1$ faces. This gives the $(\xi_i,\eta_i,\zeta_i)$ triples listed in the Given table.
2. Write the trilinear product form. Because the parent element is a cube and the interpolation must be linear along every edge, the shape functions are the tensor product of three one-dimensional linear functions and can be written compactly as $$\boxed{\,N_i(\xi,\eta,\zeta)=\tfrac{1}{8}\big(1+\xi\,\xi_i\big)\big(1+\eta\,\eta_i\big)\big(1+\zeta\,\zeta_i\big),\qquad i=1,2,\dots ,8\,}$$ Written out for the numbering of this element, the eight functions are

   Trilinear shape functions of the eight-node hexahedron

   $i$	$N_i(\xi,\eta,\zeta)$	$i$	$N_i(\xi,\eta,\zeta)$
   1	$\tfrac{1}{8}(1-\xi)(1-\eta)(1-\zeta)$	5	$\tfrac{1}{8}(1-\xi)(1-\eta)(1+\zeta)$
   2	$\tfrac{1}{8}(1+\xi)(1-\eta)(1-\zeta)$	6	$\tfrac{1}{8}(1+\xi)(1-\eta)(1+\zeta)$
   3	$\tfrac{1}{8}(1+\xi)(1+\eta)(1-\zeta)$	7	$\tfrac{1}{8}(1+\xi)(1+\eta)(1+\zeta)$
   4	$\tfrac{1}{8}(1-\xi)(1+\eta)(1-\zeta)$	8	$\tfrac{1}{8}(1-\xi)(1+\eta)(1+\zeta)$

   These satisfy the two defining requirements: $N_i=1$ at node $i$ and zero at the other seven nodes, and $\sum_{i=1}^{8}N_i=1$ everywhere in the element.

(b) $N_3$ at three locations [3 marks]

Evaluating $N_3=\tfrac18(1+\xi)(1+\eta)(1-\zeta)$ at the three requested points, and noting that node 3 is at $(+1,+1,-1)$, node 5 at $(-1,-1,+1)$ and the centroid at $(0,0,0)$:

The first two results are the Kronecker-delta property in action, and the third reflects the fact that at the centroid all eight nodes contribute equally, $\tfrac18$ each, summing to unity.

(c) The displacement field in natural coordinates [3 marks]

1. Interpolate $u$ from the non-zero nodal values. Only $u_2=u_3=u_6=u_7=0.045$ mm are non-zero, and inspecting the natural coordinates shows these are precisely the four nodes on the $\xi=+1$ face. Hence $$u(\xi,\eta,\zeta)=0.045\big(N_2+N_3+N_6+N_7\big)$$
2. Collapse the sum. Every one of these four functions carries the common factor $\tfrac18(1+\xi)$, and the remaining factors run over all four sign combinations of $\eta$ and $\zeta$: $$N_2+N_3+N_6+N_7=\tfrac18(1+\xi)\Big[(1-\eta)(1-\zeta)+(1+\eta)(1-\zeta)+(1-\eta)(1+\zeta)+(1+\eta)(1+\zeta)\Big]=\tfrac18(1+\xi)(4)=\frac{1+\xi}{2}$$
3. State the three field variables. Therefore $$\boxed{\,u(\xi,\eta,\zeta)=0.0225\,(1+\xi)\ \text{mm},\qquad v=0,\qquad w=0\,}$$ The field is a function of $\xi$ alone: it is zero on the $\xi=-1$ face, $0.0225$ mm at the centroid, and $0.045$ mm on the $\xi=+1$ face. Physically the element is being stretched uniformly in the $x$ direction with the $x=1$ cm face held fixed.

(d) Jacobian matrix and Jacobian [7 marks]

1. Write the geometric mapping. The element is isoparametric, so the geometry uses the same shape functions: $x=\sum N_ix_i$, $y=\sum N_iy_i$, $z=\sum N_iz_i$. Because the element is a rectangular box, these collapse to $$x=\frac{1+6}{2}+\frac{6-1}{2}\xi=3.5+2.5\,\xi ,\qquad y=\frac{2+8}{2}+\frac{8-2}{2}\eta=5+3\,\eta ,\qquad z=\frac{2+4}{2}+\frac{4-2}{2}\zeta=3+\zeta$$ in centimetres.
2. Assemble the Jacobian matrix. By definition $$[J]=\begin{bmatrix}\dfrac{\partial x}{\partial \xi} & \dfrac{\partial y}{\partial \xi} & \dfrac{\partial z}{\partial \xi}\\[6pt] \dfrac{\partial x}{\partial \eta} & \dfrac{\partial y}{\partial \eta} & \dfrac{\partial z}{\partial \eta}\\[6pt] \dfrac{\partial x}{\partial \zeta} & \dfrac{\partial y}{\partial \zeta} & \dfrac{\partial z}{\partial \zeta}\end{bmatrix}=\begin{bmatrix}\sum \dfrac{\partial N_i}{\partial \xi}x_i & \sum \dfrac{\partial N_i}{\partial \xi}y_i & \sum \dfrac{\partial N_i}{\partial \xi}z_i\\[6pt] \sum \dfrac{\partial N_i}{\partial \eta}x_i & \cdots & \cdots\\[6pt] \sum \dfrac{\partial N_i}{\partial \zeta}x_i & \cdots & \cdots\end{bmatrix}$$ Differentiating the mapping of step 1 gives the diagonal result $$\boxed{\,[J]=\begin{bmatrix}2.5 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 1\end{bmatrix}\ \text{cm}\,}$$ The off-diagonal terms vanish because the element edges are parallel to the global axes, and every entry is constant because the element is undistorted.
3. Evaluate the Jacobian (its determinant). $$\boxed{\,|J|=\det[J]=(2.5)(3)(1)=7.5\ \text{cm}^{3}\,}$$ This is the volume scale factor between the parent cube and the physical element: $dV=|J|\,d\xi\,d\eta\,d\zeta$. Integrating the constant over the parent cube of volume 8 gives $7.5\times 8 = 60\ \text{cm}^{3}$, which matches the physical volume $5\times 6\times 2=60\ \text{cm}^{3}$ — the check that confirms both the mapping and the determinant.
4. Note the invertibility condition. $|J|=7.5\gt 0$ everywhere and is constant, so the mapping is one-to-one throughout the element and $[J]^{-1}=\mathrm{diag}(0.4,\ 1/3,\ 1)\ \text{cm}^{-1}$ exists everywhere. This is the ideal case; a distorted brick would give a $|J|$ that varies with position, and a badly distorted one could drive it to zero or negative.

(e) Strains at the element centre [3 marks]

1. Convert natural derivatives to physical derivatives. The chain rule gives $\{\partial/\partial x,\ \partial/\partial y,\ \partial/\partial z\}^{T}=[J]^{-1}\{\partial/\partial \xi,\ \partial/\partial \eta,\ \partial/\partial \zeta\}^{T}$. With the diagonal $[J]$ this reduces to $$\frac{\partial}{\partial x}=\frac{1}{2.5}\frac{\partial}{\partial \xi} ,\qquad \frac{\partial}{\partial y}=\frac{1}{3}\frac{\partial}{\partial \eta} ,\qquad \frac{\partial}{\partial z}=\frac{\partial}{\partial \zeta}$$
2. Compute the normal strain. From part (c), $\partial u/\partial\xi = 0.0225$ mm, so $$\varepsilon_x=\frac{\partial u}{\partial x}=\frac{1}{2.5}\,(0.0225\ \text{mm})=0.009\ \frac{\text{mm}}{\text{cm}}=\frac{0.009}{10}=9.0\times10^{-4}$$ where the final step converts centimetres to millimetres so that the strain is dimensionless.
3. Compute the shear strain. Since $u$ depends only on $\xi$ and $v\equiv 0$, $$\gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=\frac{1}{3}\frac{\partial u}{\partial \eta}+0=0+0=0$$ giving the two requested strains $$\boxed{\ \varepsilon_x=9.0\times10^{-4}\ \ (900\ \mu\varepsilon),\qquad \gamma_{xy}=0\ }$$ Both are independent of position, so the values at the centre hold everywhere in the element. The result is confirmed directly from the geometry: the $\xi=+1$ face moves $0.045$ mm relative to the fixed $\xi=-1$ face over a physical span of $5\ \text{cm}=50\ \text{mm}$, so $\varepsilon_x=0.045/50=9.0\times10^{-4}$, and no face displaces transversely, so there is no shear.