Consider a simply supported beam with a 1 m span, with a uniformly distributed load of 0.1272 kN / m. We would expect a quadratic moment distribution with a maximum moment
\[\begin{align} M &= \frac{w l^{2}}{8} \\ & = \frac{0.1272 \textrm{ N/mm} \times (1000 \textrm{ mm})^{2}}{8} \\ & = 15.90 \times 10^{3} \textrm{ N mm} \end {align}\]
Suppose we wanted to model the above beam (with a quadratic moment distribution) using finite elements that are only capable of a linear moment distribution. Let’s start using a single element:
Approximating a quadratic moment distribution as a single straight line has produced a bending moment diagram that is completely the wrong shape and has underestimated the moment by a third. We can improve on this by undertaking “mesh refinement” - dividing the beam up into more and more elements to get a better and better approximation. Here is how it looks with two linear elements approximating the moment diagram:
This two element model has over estimated the solution by one sixth, which is closer to the correct answer than our one element model. We can go on refining the mesh, dividing up the beam into more elements to get closer and closer to the correct answer:
| Elements | Moment / N m |
|---|---|
| 1 | 10.60 |
| 2 | 18.55 |
| 3 | 15.31 |
| 4 | 16.56 |
| 5 | 15.69 |
| 6 | 16.20 |
| 7 | 15.79 |
| 8 | 16.07 |
| 9 | 15.84 |
| 10 | 16.01 |
Elements exist that can handle a quadratic moment distribution, so the above simply supported beam example can by modelled exactly using just one of these elements:
| Elements | Moment / N m |
|---|---|
| 1 | 15.90 |
| 2 | 15.90 |
| 3 | 15.90 |
Because these elements can model the beam exactly, we don’t get any improvement in accuracy by using more elements, so are just slowing the analysis down by including them.
Note that most problems will be higher order than quadratic, so the mesh refinement process shown in the Linear example will have to be undertaken in most cases.