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9.4.5. Creation of the material model

The least material data required for a stress analysis is the empirical elastic modulus for the component under analysis describing the relevant stress/strain law. For a dynamic analysis, the material density must also be specified. Dependent upon the type of analysis, other properties may be required, including Poisson's ratio for two- and three-dimensional models and the coefficient of thermal expansion for thermal analyses. For analyses involving nonlinear material behaviour then, as a minimum, the yield stress and yield criterion, e.g. von Mises, need to be defined. If the material within an element can be assumed to be isotropic and homogeneous, then there will be only one value of each material property. For non-isotropic material, i.e. orthotropic or anisotropic, then the material properties are direction and spacially dependent, respectively. In the extreme case of anisotropy, 21 independent values are required to define the material matrix.5

### 9.4.6. Node and element ordering

Before moving on to consider boundary conditions, it is appropriate to examine node and element ordering and its effect on efficiency of solution by briefly exploring the methods used. The formation of the element characteristic matrices (to be considered in §9.7, 9.8 and 9.9), and the subsequent solution are the two most computationally intensive steps in any fe. analysis. The computational effort and memory requirements of the solution are affected by the method employed, and are considered below.

It will be seen in Section 9.7, and subsequently, that the displacement based method involves the assembly of the structural, or assembled, stiffness matrix [AT], and the load and displacement column matrices, {/>} and {p}, respectively, to form the governing equation for stress analysis {P} = [A"]{/?}. With reference to §9.7, and subsequently, two features of the fem. will be seen to be that the assembled stiffness matrix [K], is sparsely populated and is symmetric. Advantage can be taken of this in reducing the storage requirements of the computer. Two solution methods are used, namely, banded or frontal, the choice of which

For a given number of dof. per node, which is generally fixed for each assemblage, the bandwidth can be minimised by using a proper node numbering scheme.

With reference to Figs. 9.12 and 9.13 there are a total of 20 dof. in the model (i.e. 10 nodes each with an assumed 2 dof.), and if the symmetry and handedness is not taken advantage of, storage of the entire matrix would require 202 = 400 locations. For the efficiently numbered model with a semi-bandwidth of 8, see Fig. 9.13, taking advantage of the symmetry and handedness, the storage required for the upper, or lower, half-band is only 8 x 20 = 160 locations.

Node number

Element label

Degree of freedom number

(a) Efficient node numbering scheme

Semi-bandwidth = 6

(b) Structural stiffness matrix with non-zero terms closely banded

Fig. 9.13. Structural stiffness matrix corresponding to efficient node ordering.

From observation of Figs. 9.12 and 9.13 it can be deduced that the semi-bandwidth can semi-bandwidth = f(d+\)

where / is the number of dof. per node and d is the maximum largest difference in the node numbers for all elements in the assemblage. This expression is applicable to any type of finite element. It follows that to minimise the bandwidth, d must be minimised and this is achieved by simply numbering the nodes across the shortest dimension of the region.

For large jobs the capacity of computer memory can be exceeded using the above banded method, in which case a frontal solution is used.

### Frontal method of solution

The frontal method is appropriate for medium to large size jobs (i.e. greater than 10000 dof.). To illustrate the method, consider the simple two-dimensional mesh shown in Fig. 9.14. Nodal contributions are assembled in element order. With reference to Fig. 9.14, with the assembly of element number 1 terms (i.e. contributions from nodes 1, 2, 6 and 7), all information relating to node number 1 will be complete since this node is not common to any other element. Thus the dofs. for node 1 can be eliminated from the set of system equations. Element number 2 contributions are assembled next, and the system matrix will now contain contributions from nodes 2, 3, 6, 7 and 8. At this stage the dofs. for node number 2 can be eliminated. Further element contributions are merged and at each stage any nodes which do not appear in later elements are reduced out. The solution thus proceeds as a front through the system. As, for example, element number 14 is assembled, dofs. for the nodes indicated by line B are required, see Fig. 9.14. After eliminations which follow assembly of element number 14, dofs. associated with line C are needed. The solution front has thus moved from line A to C.

Node number -

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