Home / Articles / Finite Elements FEM-05
When Four Elements Beat Two Hundred and Fifty-Six
In FEM-02 the Q4 element was ten times wrong on a coarse mesh. The quadratic Q8 element gives 2.5% error on that same mesh — and four of them beat two hundred and fifty-six Q4 elements. But that advantage comes at a price, and the price is sharp.
1. Back to the source of locking
In FEM-02 we put the cause of locking this way: when an element is forced into a strain field it cannot represent, it generates spurious shear strain and behaves far stiffer than it really is.
In bending the strain varies linearly through the depth. The Q4 element's displacement field is bilinear; its derivative — the strain — stays nearly constant within the element. It cannot produce a linear distribution.
2. Q8: eight nodes, linear strain
The Q8 element carries four nodes at the corners and four at the mid-sides. Its shape functions are quadratic (the serendipity family), so:
Full integration uses a 3×3 Gauss rule, reduced integration 2×2. Reduced integration is a common choice for Q8 and produces a single spurious mode — not as destructive as Q4's two.
3. The mid-side node is shared
4. The patch test
The first examination of any new element formulation is whether it reproduces a constant strain field exactly.
| Element | Integration | Patch test error | Result |
|---|---|---|---|
| computing… | |||
5. Locking comparison
The same cantilever as in FEM-02: L = 1000, H = 100, t = 10 mm, tip load 1000 N. The reference is —.
| Mesh | Q4 full | Q8 full | Q8 reduced |
|---|---|---|---|
| computing… | |||
6. Efficiency per degree of freedom
"Four elements beat two hundred and fifty-six" is not by itself a fair statement, because a Q8 element carries more nodes. The proper comparison is on degrees of freedom.
| Model | Elements | Degrees of freedom | Error |
|---|---|---|---|
| computing… | |||
Comparisons made on element counts are misleading. Q8 demands more nodes — but reaches the same accuracy with far fewer degrees of freedom, and the cost of the solution is set by the degrees of freedom.
7. Why does the error not go to zero?
Looking carefully at the table, one thing stands out: the Q8 error falls as the mesh is refined but does not go to zero; it settles at around 0.2%.
The practical consequence: a convergence study cannot be more accurate than your reference. Working with Q4 this problem was invisible, because the element's own error overwhelmed the reference's. Moving to a higher-order element, you have to question the reference as well.
8. The price: the mid-side node
So far Q8 has come out ahead on every measure. Here is the price.
A Q8 mid-side node need not sit exactly at the middle of the edge — the isoparametric mapping can carry it elsewhere. But when it does, the mapping degrades fast.
| Mid-side shift | Tip deflection | Error | Relative to regular mesh |
|---|---|---|---|
| computing… | |||
9. Comparing the two distortions
In FEM-04 we measured that displacing Q4's corner nodes raised the error from 13.95% to 18.76% — a relative worsening of about 35%. Here, displacing Q8's mid-side node produces a relative worsening of more than fortyfold.
The reason is understandable. Q8's advantage comes from representing a quadratic displacement field correctly. When the mid-side node shifts, the mapping distorts that quadratic field — that is, the very thing that made the element superior is what degrades.
10. What to do in practice
- Use quadratic elements where bending dominates. Working with Q4 on a coarse mesh can produce a 70% error, as FEM-02 showed.
- Compare on degrees of freedom. Element counts mislead; cost is set by the degrees of freedom.
- Keep mid-side nodes at the middle. Curved boundaries may require shifting them, but keep the shift to what is genuinely needed.
- Verify that mid-side nodes are shared. If they are not, the elements come apart — and that is not obvious from the result at a glance.
- Question your reference. With a higher-order element the remaining error may come from the reference rather than the element.
- Do not relax the mesh quality requirement. Higher order does not increase tolerance to a bad mesh; it reduces it.
11. Test yourself
- Why does Q8 not lock in bending? Where does "the order of the strain" enter your answer?
- Four Q8 elements are more accurate than 256 Q4 elements. Why is that comparison not by itself fair, and what is the fair one?
- Why does the Q8 error not go to zero? Is that a defect of the element or of the reference?
- When the mid-side node shifts, which property that makes Q8 superior is degraded?
- What happens if an edge mid-node is not shared by both elements? Would you notice from the result?
- How many spurious modes does reduced integration produce in Q8? Why is that less dangerous than in Q4?
References
- Serendipity-family quadratic shape functions and the isoparametric formulation — standard finite element results, independent of any national code.
- The patch test — the standard criterion of element validity; see FEM-02.
- Reference solution: Timoshenko beam including shear deformation. Section 7 of this article is precisely about the limit of that reference.
- For course modules and MATLAB applications: Kwon, Y.W. & Bang, H., The Finite Element Method using MATLAB. The exposition, derivations and all figures here are our own.
All figures in this article are produced by engine/fem2d.js and separately pinned in tests/fem2d.test.js (61/61). The Q8 element passes the patch test to machine precision under both full and reduced integration; that mid-side nodes are shared between neighbouring elements is checked as a separate case.
Higher order does not increase tolerance to a bad mesh; it reduces it.