Question 4 of 4: Bilinear four-node element — strains at the centre and the hourglass mode

Problem 4: Bilinear four-node element — strains at the centre and the hourglass mode (equal value)

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. One bilinear (Q4) isoparametric element.

Find. 4.1 the three strains $\varepsilon_x$, $\varepsilon_y$, $\gamma_{xy}$ at the centre of the element; 4.2 an interpretation of what those values mean for the element.

Approach. Because the element is a square of side 2 centred on the natural origin, the mapping is simply $x=\xi$, $y=\eta$ and the Jacobian is the identity, so the strains follow by differentiating the interpolated displacement field with respect to $\xi$ and $\eta$ directly and evaluating at $\xi=\eta=0$.

1. Part 4.1 — Establish the mapping and the Jacobian. With the side equal to 2 units and the natural axes at the centre, node 1 sits at $(\xi,\eta)=(-1,-1)$ and node 3 at $(+1,+1)$, so $$\begin{aligned}x&=\sum N_i x_i=\xi \\ y&=\sum N_i y_i=\eta \\ [J]&=\begin{bmatrix}1&0\\0&1\end{bmatrix}\end{aligned}$$ The Jacobian determinant is 1, so $\partial/\partial x=\partial/\partial\xi$ and $\partial/\partial y=\partial/\partial\eta$ with no scaling — the one simplification this geometry buys.
2. Interpolate the displacement field. Substituting Table 2 into $u=\sum N_iu_i$ and $v=\sum N_iv_i$ and collecting terms, $$u=\tfrac{c}{4}\left[-(1+\xi)(1-\eta)+(1+\xi)(1+\eta)\right]=\tfrac{c}{2}\,\eta(1+\xi)$$ $$v=\tfrac{c}{4}\left[-(1+\xi)(1+\eta)+(1-\xi)(1+\eta)\right]=-\tfrac{c}{2}\,\xi(1+\eta)$$ Both fields contain a linear part and a $\xi\eta$ product; that product term is the one the bilinear element adds beyond a constant-strain triangle.
3. Differentiate to obtain the strain field. With the identity Jacobian, $$\begin{aligned}\varepsilon_x&=\frac{\partial u}{\partial\xi}=\frac{c}{2}\eta \\ \varepsilon_y&=\frac{\partial v}{\partial\eta}=-\frac{c}{2}\xi\end{aligned}$$ $$\gamma_{xy}=\frac{\partial u}{\partial\eta}+\frac{\partial v}{\partial\xi} =\frac{c}{2}(1+\xi)-\frac{c}{2}(1+\eta)=\frac{c}{2}(\xi-\eta)$$ Each strain is linear in the natural coordinates, as it must be for a bilinear element.
4. Evaluate at the centre of the element. Putting $\xi=\eta=0$ in the three expressions above, $$\boxed{\begin{aligned}\varepsilon_x&=0 \\ \varepsilon_y&=0 \\ \gamma_{xy}&=0\end{aligned}}$$ All three strains vanish at the centroid, even though every node but one has moved.
5. Part 4.2 — Decompose the displaced shape. Split the field of step 2 into a linear part and a product part: $$\begin{aligned}u&=\underbrace{\frac{c}{2}\eta}_{\text{rigid rotation}}+\underbrace{\frac{c}{2}\xi\eta}_{\text{hourglass}} \\ v&=\underbrace{-\frac{c}{2}\xi}_{\text{rigid rotation}}-\underbrace{\frac{c}{2}\xi\eta}_{\text{hourglass}}\end{aligned}$$ The first pair is exactly a rigid-body rotation through $-c/2$ radians, since it has the form $u=-\omega y$, $v=+\omega x$; a rigid rotation produces no strain anywhere, which the strain field of step 3 confirms because those terms have dropped out of it. Everything that survives comes from the $\xi\eta$ product.
6. Identify the surviving part as an hourglass mode. The remainder, $u=\tfrac{c}{2}\xi\eta$ and $v=-\tfrac{c}{2}\xi\eta$, is an equal-and-opposite combination of the two hourglass or keystone modes of the bilinear element ($u=\xi\eta$ and $v=\xi\eta$): the element deforms into a trapezoid-like shape whose opposite edges bow in opposite directions, and the strains it produces are linear functions that happen to pass through zero at the centroid. In the picture, the element folds about its centre without stretching anything there.
7. Why this matters for integration order. If the element stiffness is integrated with a single Gauss point — the natural, cheap choice for a Q4, sampling at $\xi=\eta=0$ — then this deformation is sampled where the strain is exactly zero, so it generates no strain energy at all. It is a spurious zero-energy mode: the element offers no resistance to it, the assembled stiffness matrix becomes rank-deficient, and a mesh of such elements can develop a visible zig-zag pattern of alternating deformation while the computed stresses stay comfortingly small. Real codes suppress this with artificial hourglass-control stiffness or viscosity.
8. Full integration removes the defect. Sampling instead at the four $2\times2$ Gauss points $\xi,\eta=\pm1/\sqrt{3}$ gives, from step 3, $$\begin{aligned}\varepsilon_x&=\pm\frac{c}{2\sqrt{3}}=\pm0.2887c \\ \varepsilon_y&=\mp0.2887c \\ \gamma_{xy}&=0\ \text{or}\ \pm0.5774c\end{aligned}$$ so the mode is strained at every sampling point and does store energy. This is precisely the trade-off the question is probing: reduced integration relieves shear locking and is cheaper, but it must be paired with hourglass control, whereas full integration is safe but stiffer in bending. Note also that the rigid rotation contributes nothing at any point, which is the patch-test property every usable element must have.