Example 3.2

A square plate with a hole at the center and under pressure in one direction.

The dimension of the plate is 10 in. x 10 in., thickness is 0.1 in. and radius of the hole is 1 in. Assume E = 10x106 psi, v = 0.3 and p = 100 psi. Find the maximum stress in the plate.

FE Analysis:

From the knowledge of stress concentrations, we should expect the maximum stresses occur at points A and B on the edge of the hole. Value of this stress should be around 3p (= 300 psi) which is the exact solution for an infinitely large plate with a hole.

We use the ANSYS FEA software to do the modeling (meshing) and analysis, using quadratic triangular (T6 or LST), linear quadrilateral (Q4) and quadratic quadrilateral (Q8) elements. Linear triangles (CST or T3) is NOT available in

ANSYS.

The stress calculations are listed in the following table, along with the number of elements and DOF used, for comparison.

Table. FEA Stress Results

Discussions:

•Check the deformed shape of the plate

•Check convergence (use a finer mesh, if possible)

•Less elements (~ 100) should be enough to achieve the same accuracy with a better or “smarter” mesh

•We’ll redo this example in next chapter employing the symmetry conditions.

FEA Mesh (Q8, 493 elements)

FEA Stress Plot (Q8, 493 elements)

Transformation of Loads

Concentrated load (point forces), surface traction (pressure loads) and body force (weight) are the main types of loads applied to a structure. Both traction and body forces need to be converted to nodal forces in the FEA, since they cannot be applied to the FE model directly. The conversions of these loads are based on the same idea (the equivalent-work concept) which we have used for the cases of bar and beam elements.

A L A

Traction on a Q4 element

Suppose, for example, we have a linearly varying traction q on a Q4 element edge, as shown in the figure. The traction is normal to the boundary. Using the local (tangential) coordinate s, we can write the work done by the traction q as,

L

Wq = t∫un (s)q(s)ds

0

where t is the thickness, L the side length and un the component of displacement normal to the edge AB.

For the Q4 element (linear displacement field), we have

Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems

un (s) = (1− s / L)unA +(s / L)unB

The traction q(s), which is also linear, is given in a similar way, q(s) = (1− s / L)qA +(s / L)qB

Thus, we have,

and the equivalent nodal force vector is,

Note, for constant q, we have,

For quadratic elements (either triangular or quadrilateral), the traction is converted to forces at three nodes along the edge, instead of two nodes.

Traction tangent to the boundary, as well as body forces, are converted to nodal forces in a similar way.