Question 12.3.1: A uniform steel plate subjected to in plane forces is to be ...
A uniform steel plate subjected to in plane forces is to be analysed by the finite element method, using triangular elements; the element stiffness matrices K are to be derived from the displacement function in Eqn 12.3-1.
\left. \begin{matrix} u=\alpha _1+\alpha _2x+\alpha _3y \\ v=\alpha _4+\alpha _5x+\alpha _6y \end{matrix} \right\} (12.3-1)
If, instead of using the element stiffness matrices so derived, precise tests are carried out on steel triangular plates to determine accurately the element stiffness matrices, explain whether the use of such experimental stiffness matrices would improve the accuracy of the finite element analysis.
The answer is that it definitely would not, and the reason is as follows. It is true that for each triangular steel plate, a unique relationship exists between nodal forces and nodal displacements, and this relationship can be determined, at least experimentally. However, a stiffness matrix based on such a true relationship is the true stiffness matrix of the steel element, and should not be used in finite element work. Using such true stiffness matrices would mean replacing the steel plate by an assemblage of discrete steel elements connected at the nodes only, and such replacement can result in serious errors. What we do in finite element analysis is not only to replace the actual steel plate by an assemblage of discrete elements, but also to replace the true stiffness matrix of each element by a fictitious element stiffness matrix, which is so chosen (by suitable selection of a displacement function) as to give acceptable final results for the actual steel plate. That is, the discrete elements are not made of steel, but of a fictitious material that behaves in a fictitious way under load.
In order to appreciate further the fictitious properties of the elements used in the analysis, consider the element ijk in Fig. 12.3-3, which is acted on by distributed inplane forces q. Suppose the nodes are restrained against displacement. Note that Eqn 12.3-5 states that the internal strains \varepsilon are zero when nodal displacements \Delta are zero.
\varepsilon =B\alpha (12.3-5)
=BA^{-1}\Delta (from Eqn 12.3-3)
\Delta=A\alpha (12.3-3)
Therefore, since the nodes are restrained, q produces no strain anywhere in the element and consequently no strain energy. That is, as long as the nodes are held in position, no amount of load acting on the element can cause any deformation of the element or cause any stresses \sigma in the element (see Eqn 12.3-29 in Example 12.3-2).
\int{\sigma ^T\bar{\varepsilon } \ dV }=\int{0} \ \bar{\varepsilon } \ dV=0 (12.3-29)
On the other hand, if any node is displaced (e.g. node j in Fig. 12.3-3), then stresses \sigma will exist at all points of the element, as given by Eqn 12.3-12,
\sigma =D\varepsilon (12.3-12)
=DBA^{-1}\Delta (from Eqn 12.3-11)
\varepsilon =BA^{-1}\Delta (12.3-11)
including points such as ‘a’ on the boundary, even when the boundaries are in contact with the atmosphere. The existence of such stresses (which is impossible if the element is made of any real material such as steel) means that in general we cannot expect equilibrium of stresses along the boundaries of the elements.
It should be noted, however, that the nodal forces acting on any individual element are always in equilibrium (see Problem 12.9).
