김경종
577 lines
31 KiB
Markdown
577 lines
31 KiB
Markdown
<!-- source-page: 381 -->
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<details>
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<summary>natural_image</summary>
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Abstract grid pattern with a central circular void and radial symmetry, no text or symbols present
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</details>
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Figure 7–9 Plate of steel (20 in. long, 20 in. wide, 1 in. thick, and with a 1-in.-radius hole) discretized using a preprocessor program [15] with automatic mesh generation
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these techniques may be cruder than the finite element results. For instance, approximate mechanics of material formulas, experimental data, and numerical analysis of simpler but similar problems may be used for comparison, particularly if you have no real idea of the magnitude of the answers. Remember to use all results with some degree of caution, as errors can crop up in such sources as textbook or handbook comparison solutions and experimental results.
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In the end, the analyst should probably spend as much time processing, checking, and analyzing results as is spent in data preparation.
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Finally, we present some typical postprocessor results for the plane stress problem of Figure 7–9 (Figures 7–10 and 7–11). Other examples with results are shown in Section 7.7.
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# 7.2 Equilibrium and Compatibility of Finite Element Results
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An approximate solution for a stress analysis problem using the finite element method based on assumed displacement fields does not generally satisfy all the requirements for equilibrium and compatibility that an exact theory-of-elasticity solution satisfies. However, remember that relatively few exact solutions exist. Hence, the finite element method is a very practical one for obtaining reasonable, but approximate, numerical solutions. Recall the advantages of the finite element method as described in Chapter 1 and as illustrated numerous times throughout this text.
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<!-- source-page: 382 -->
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<details>
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<summary>other</summary>
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| Position (in.) | Value (psi) |
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| -------------- | ----------- |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 20 | 200 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | -1000 |
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| 1000 | -1000 |
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| 1000 | -1000 |
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| 1000 | -1000 |
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| 1000 | -1000 |
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| 1000 | -1000 |
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| 1000 | -1000 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 20 | 0.1 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 | 1000 |
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| 1000 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 1000 | 0 |
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| 21 | 1000 |
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| 21 | 1000 |
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| 21 | 1000 |
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| 21 | 1000 |
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| 21 | 1000 |
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| 21 | 1000 |
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| 21 | 1000 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 21 | 0.1 |
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| 11 | 1000 |
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| 11 | 1000 |
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| 11 | 1000 |
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| 11 | 1000 |
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| 11 | 1000 |
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| 11 | 1000 |
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| 11 | 1000 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 11 | 0.1 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 20 | 800 |
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| 10 | 1000 |
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| 10 | 1000 |
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| 10 | 1000 |
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| 10 | 1000 |
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| 10 | 1000 |
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| 10 | 1000 |
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| 10 | 1000 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 10 | 800 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 (top) | 1000 |
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| 20 (top) | 1000 |
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| 20 (top) | 1000 |
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| 20 (top) | 1000 |
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| 20 (top) | 1000 |
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| 20 (top) | 1000 |
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| 20 (bottom) | 1000 |
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| 20 (bottom) | 1000 |
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| 20 (bottom) | 1000 |
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| 20 (bottom) | 1000 |
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| 20 (bottom) | 1000 |
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| 20 (bottom) | 1000 |
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| 20 (bottom) | 1200 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1200 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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| 20 | 1000 |
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</details>
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Figure 7–10 Plate with a hole showing the deformed shape of a plate superimposed over an undeformed shape. Plate is fixed on the left edge and subjected to 1000-psi tensile stress along the right edge. Maximum horizontal displacement is 7:046 10-4 in. at the center of the right edge
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We now describe some of the approximations generally inherent in finite element solutions.
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1. Equilibrium of nodal forces and moments is satisfied. This is true because the global equation $\underline { { F } } = \underline { { K } } \underline { { d } }$ is a nodal equilibrium equation whose solution for d is such that the sums of all forces and moments applied to each node are zero. Equilibrium of the whole structure is also satisfied because the structure reactions are included in the global forces and hence in the nodal equilibrium equations. Numerous example problems, particularly involving truss and frame analysis in Chapter 3 and 5, respectively, have illustrated the equilibrium of nodes and of total structures.
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2. Equilibrium within an element is not always satisfied. However, for the constant-strain bar of Chapter 3 and the constant-strain triangle of Chapter 6, element equilibrium is satisfied. Also the cubic displacement function is shown to satisfy the basic beam equilibrium differential equation in Chapter 4 and hence to satisfy element force and
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<!-- source-page: 383 -->
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<details>
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<summary>heatmap</summary>
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| Stress (ln²) | Value |
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| ------------ | ----- |
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| 3085.222 | 2770.7 |
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| 2408.178 | 2154.655 |
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| 1961.133 | 1542.611 |
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| 1234.089 | 925.5666 |
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| 617.0444 | 309.6222 |
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| 1.03502e-012 | 1.03502e-012 |
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</details>
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Figure 7–11 Maximum principal stress contour (shrink fit plot) for a plate with hole. Largest principal stresses of 3085 psi occur at the top and bottom of the hole, which indicates a stress concentration of 3.08. Stresses were obtained by using an average of the nodal values (called smoothing)
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moment equilibrium. However, elements such as the linear-strain triangle of Chapter 8, the axisymmetric element of Chapter 9, and the rectangular element of Chapter 10 usually only approximately satisfy the element equilibrium equations.
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3. Equilibrium is not usually satisfied between elements. A differential element including parts of two adjacent finite elements is usually not in equilibrium (Figure 7–12). For line elements, such as used for truss and frame analysis, interelement equilibrium is satisfied, as shown in example problems in Chapters 3–5. However, for two- and threedimensional elements, interelement equilibrium is not usually satisfied. For instance, the results of Example 6.2 indicate that the normal stress along the diagonal edge between the two elements is different in the two elements. Also, the coarseness of the mesh causes this lack of interelement equilibrium to be even more pronounced. The normal and shear stresses at a free edge usually are not zero even though theory predicts them to be. Again, Example 6.2 illustrates this, with
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<!-- source-page: 384 -->
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<details>
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<summary>text_image</summary>
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5000 lb
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①
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10 in.
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②
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20 in.
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5000 lb
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</details>
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Example 6.2
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<details>
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<summary>text_image</summary>
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σy = 301 psi
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σx = 1005 psi
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τxy = -2.4 psi
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σx = 995 psi
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τxy = -2.4 psi
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σy = -1.2 psi
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</details>
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Stresses on a differential element common to both finite elements, illustrating violation of equilibrium
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<details>
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<summary>text_image</summary>
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σy = 301 psi
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τxy = -2.4 psi
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τnt = 280 psi
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σn = 440 psi
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σx = 1005 psi
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σn = 196 psi
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τnt = 397 psi
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σx = 995 psi
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τxy = -2.4 psi
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σy = -1.2 psi
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</details>
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Stress along the diagonal between elements, showing normal and shear stresses, $\sigma _ { n }$ and $\tau _ { n t } .$ Note: $\sigma _ { n }$ and $\tau _ { n t }$ are not equal in magnitude but are opposite in sign for the two elements,and so interelement equilibrium is not satisfied
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Figure 7–12 Example 6.2, illustrating violation of equilibrium of a differential element and along the diagonal edge between two elements (the coarseness of the mesh amplifies the violation of equilibrium)
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free-edge stresses $\sigma _ { y }$ and $\tau _ { x y }$ not equal to zero. However, as more elements are used (refined mesh) the $\sigma _ { y }$ and $\tau _ { x y }$ stresses on the stressfree edges will approach zero.
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4. Compatibility is satisfied within an element as long as the element displacement field is continuous. Hence, individual elements do not tear apart.
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5. In the formulation of the element equations, compatibility is invoked at the nodes. Hence, elements remain connected at their common nodes. Similarly, the structure remains connected to its support nodes because boundary conditions are invoked at these nodes.
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6. Compatibility may or may not be satisfied along interelement boundaries. For line elements such as bars and beams, interelement boundaries are merely nodes. Therefore, the preceding statement 5 applies for these line elements. The constant-strain triangle of Chapter 6 and the rectangular element of Chapter 10 remain straight-sided when deformed. Therefore, interelement compatibility exists for these elements; that is, these plane elements deform along common lines
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<!-- source-page: 385 -->
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without openings, overlaps, or discontinuities. Incompatible elements, those that allow gaps or overlaps between elements, can be acceptable and even desirable. Incompatible element formulations, in some cases, have been shown to converge more rapidly to the exact solution [1]. (For more on this special topic, consult References [7] and [8].)
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# 7.3 Convergence of Solution
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In Section 3.2, we presented guidelines for the selection of so-called compatible and complete displacement functions as they related to the bar element. Those four guidelines are generally applicable, and satisfaction of them has been shown to ensure monotonic convergence of the solution of a particular problem [9]. Furthermore, it has been shown [10] that these compatible and complete displacement functions used in the displacement formulation of the finite element method yield an upper bound on the true stiffness, and hence a lower bound on the displacement of the problem, as shown in Figure 7–13.
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Hence, as the mesh size is reduced—that is, as the number of elements is increased—we are ensured of monotonic convergence of the solution when compatible and complete displacement functions are used. Examples of this convergence are given in References [1] and [11], and in Table 7–2 for the beam with loading shown
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<details>
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<summary>text_image</summary>
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Displacement
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Exact solution
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Number of elements
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Compatible displacement formulation
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</details>
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Figure 7–13 Convergence of a finite element solution based on the compatible displacement formulation
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Table 7–2 Comparison of results for different numbers of elements
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<table><tr><td>Case</td><td>Number of Nodes</td><td>Number of Elements</td><td>Aspect Ratio</td><td>Vertical Displacement, v (in.) Point A</td></tr><tr><td>1</td><td>21</td><td>12</td><td>2</td><td>-0.740</td></tr><tr><td>2</td><td>39</td><td>24</td><td>1</td><td>-0.980</td></tr><tr><td>3</td><td>45</td><td>32</td><td>3</td><td>-0.875</td></tr><tr><td>4</td><td>85</td><td>64</td><td>1.5</td><td>-1.078</td></tr><tr><td>5</td><td>105</td><td>80</td><td>1.2</td><td>-1.100</td></tr><tr><td colspan="4">Exact solution [2]</td><td>-1.152</td></tr></table>
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<!-- source-page: 386 -->
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in Figure 7–1(a). All elements in the table are rectangular. The results in Table 7–2 indicate the influence of the number of elements (or the number of degrees of freedom as measured by the number of nodes) on the convergence toward a common solution, in this case the exact one. We again observe the influence of the aspect ratio. The higher the aspect ratio, even with a larger number of degrees of freedom, the worse the answer, as indicated by comparing cases 2 and 3.
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# 7.4 Interpretation of Stresses
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In the stiffness or displacement formulation of the finite element method used throughout this text, the primary quantities determined are the interelement nodal displacements of the assemblage. The secondary quantities, such as strain and stress in an element, are then obtained through use of $\{ \varepsilon \} = [ B ] \{ d \}$ and $\{ \sigma \} = [ D ] [ B ] \{ d \}$ . For elements using linear-displacement models, such as the bar and the constant-strain triangle, ½B
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is constant, and since we assume ½D
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to be constant, the stresses are constant over the element. In this case, it is common practice to assign the stress to the centroid of the element with acceptable results.
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However, as illustrated in Section 3.11 for the axial member, stresses are not predicted as accurately as the displacements (see Figures 3–32 and 3–33). For example, remember the constant-strain or constant-stress element has been used in modeling the beam in Figure 7–1. Therefore, the stress in each element is assumed constant. Figure 7–14 compares the exact beam theory solution for bending stress through the beam depth at the centroidal location of the elements next to the wall with the finite element solution of case 4 in Table 7–2. This finite element model consists of four elements through the beam depth. Therefore, only four stress values are obtained
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<details>
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<summary>scatter</summary>
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| Point | σ_x (ksi) | y (in.) |
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|---|---|---|
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| 1 | 43.6 | 1 |
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| 2 | 122 | 3 |
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| 3 | 130.8 | 3 |
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| 4 | 174.4 | 4 |
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| 5 | -174.4 | -3 |
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| 6 | -122 | -3 |
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| 7 | 39 | 1 |
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| 8 | 43.6 | 1 |
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The chart displays a 2D coordinate system with x and y axes in kilo-seconds. The legend indicates the point 'Finite element solution' with a dot symbol. The 'Exact solution' is shown as a straight line connecting the points. The label 'Finite element solution' is in the top-right corner.
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</details>
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Figure 7–14 Comparison of the finite element solution and the exact solution of bending stress through a beam cross section
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<!-- source-page: 387 -->
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through the depth. Again, the best approximation of the stress appears to occur at the midpoint of each element, since the derivative of displacement is better predicted between the nodes than at the nodes.
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For higher-order elements, such as the linear-strain triangle of Chapter 8, ½B
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, and hence the stresses, are functions of the coordinates. The common practice is then to evaluate directly the stresses at the centroid of the element.
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An alternative procedure sometimes is to use an average (possibly weighted) value of the stresses evaluated at each node of the element. This averaging method is often based on evaluating the stresses at the Gauss points located within the element (described in Chapter 10) and then interpolating to the element nodes using the shape functions of the specific element. Then these stresses in all elements at a common node are averaged to represent the stress at the node. This averaging process is called smoothing. Figure 7–11 shows a maximum principal stress ‘‘fringe carpet’’ (dithered) contour plot obtained by smoothing.
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Smoothing results in a pleasing, continuous plot which may not indicate some serious problems with the model and the results. You should always view the unsmoothed contour plots as well. Highly discontinuous contours between elements in a region of an unsmoothed plot indicate modeling problems and typically require additional refinement of the element mesh in the suspect region.
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If the discontinuities in an unsmoothed contour plot are small or are in regions of little consequence, a smoothed contour plot can normally be used with a high degree of confidence in the results. There are, however, exceptions when smoothing leads to erroneous results. For instance, if the thickness or material stiffness changes significantly between adjacent elements, the stresses will normally be different from one element to the next. Smoothing will likely hide the actual results. Also, for shrinkfit problems involving one cylinder being expanded enough by heating to slip over the smaller one, the circumferential stress between the mating cylinders is normally quite different [16].
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The computer program examples in Section 7.7 show additional results, such as displaced models, along with line contour stress plots and smoothed stress plots. The stresses to be plotted can be von Mises (used in the maximum distortion energy theory to predict failure of ductile materials subjected to static loading as described in Section 6.5); Tresca (used in the Tresca or maximum shear stress theory also to predict failure of ductile materials subjected to static loading) [14, 16], and maximum and minimum principal stresses.
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# 7.5 Static Condensation
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We will now consider the concept of static condensation because this concept is used in developing the stiffness matrix of a quadrilateral element in many computer programs.
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Consider the basic quadrilateral element with external nodes 1–4 shown in Figure 7–15. An imaginary node 5 is temporarily introduced at the intersection of the diagonals of the quadrilateral to create four triangles. We then superimpose the stiffness matrices of the four triangles to create the stiffness matrix of the quadrilateral
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<!-- source-page: 388 -->
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<details>
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<summary>text_image</summary>
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y, v
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4
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3
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③
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④
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②
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5
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①
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1
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2
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x, u
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</details>
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Figure 7–15 Quadrilateral element with an internal node
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element, where the internal imaginary node 5 degrees of freedom are said to be condensed out so as never to enter the final equations. Hence, only the degrees of freedom associated with the four actual external corner nodes enter the equations.
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We begin the static condensation procedure by partitioning the equilibrium equations as
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$$
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\left[ \begin{array}{c c} \underline {{k}} _ {1 1} & \underline {{k}} _ {1 2} \\ \underline {{k}} _ {2 1} & \underline {{k}} _ {2 2} \end{array} \right] \left\{ \begin{array}{c} \underline {{d}} _ {a} \\ \underline {{d}} _ {i} \end{array} \right\} = \left\{ \begin{array}{c} \underline {{F}} _ {a} \\ \underline {{F}} _ {i} \end{array} \right\} \tag {7.5.1}
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$$
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where $\underline { { d } } _ { i }$ is the vector of internal displacements corresponding to the imaginary internal node (node 5 in Figure 7–15), $\underline { { F } } _ { i }$ is the vector of loads at the internal node, and $\underline { { d } } _ { a }$ and $\underline { { F } } _ { a }$ are the actual nodal degrees of freedom and loads, respectively, at the actual nodes. Rewriting Eq. (7.5.1), we have
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$$
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[ k _ {1 1} ] \{d _ {a} \} + [ k _ {1 2} ] \{d _ {i} \} = \{F _ {a} \} \tag {7.5.2}
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$$
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$$
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[ k _ {2 1} ] \{d _ {a} \} + [ k _ {2 2} ] \{d _ {i} \} = \{F _ {i} \} \tag {7.5.3}
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$$
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Solving for fdig in Eq. (7.5.3), we obtain
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$$
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\{d _ {i} \} = - [ k _ {2 2} ] ^ {- 1} [ k _ {2 1} ] \{d _ {a} \} + [ k _ {2 2} ] ^ {- 1} \{F _ {i} \} \tag {7.5.4}
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$$
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Substituting Eq. (7.5.4) into Eq. (7.5.2), we obtain the condensed equilibrium equation
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$$
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\left[ k _ {c} \right] \left\{d _ {a} \right\} = \left\{F _ {c} \right\} \tag {7.5.5}
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$$
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where $[ k _ { c } ] = [ k _ { 1 1 } ] - [ k _ { 1 2 } ] [ k _ { 2 2 } ] ^ { - 1 } [ k _ { 2 1 } ]$ ð7:5:6Þ
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$$
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\{F _ {c} \} = \{F _ {a} \} - [ k _ {1 2} ] [ k _ {2 2} ] ^ {- 1} \{F _ {i} \} \tag {7.5.7}
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$$
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and $[ k _ { c } ]$ and $\{ F _ { c } \}$ are called the condensed stiffness matrix and the condensed load vector, respectively. Equation (7.5.5) can now be solved for the actual corner node displacements in the usual manner of solving simultaneous linear equations.
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Both constant-strain triangular (CST) and constant-strain quadrilateral elements are used to analyze plane stress/plane strain problems. The quadrilateral element has the stiffness of four CST elements. An advantage of the four-CST quadrilateral is that the solution becomes less dependent on the skew of the subdivision mesh, as shown in
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<!-- source-page: 389 -->
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Figure 7–16. Here skew means the directional stiffness bias that can be built into a model through certain discretization patterns, since the stiffness matrix of an element is a function of its nodal coordinates, as indicted by Eq. (6.2.52). The four-CST mesh of Figure $7 - 1 6 ( \mathrm { c } )$ represents a reduction in the skew effect over the meshes of Figure $7 - 1 6 ( \mathrm { a } )$ and (b). Figure $7 - 1 6 ( \mathsf b )$ is generally worse than Figure $7 - 1 6 ( \mathrm { a } )$ because the use of long, narrow triangles results in an element stiffness matrix that is stiffer along the narrow direction of the triangle.
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The resulting stiffness matrix of the quadrilateral element will be an $8 \times 8$ matrix consisting of the stiffnesses of four triangles, as was shown in Figure 7–15. The stiffness matrix is first assembled according to the usual direct stiffness method. Then we apply static condensation as outlined in Eqs. (7.5.1)–(7.5.7) to remove the internal node 5 degrees of freedom.
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The stiffness matrix of a typical triangular element (labeled element 1 in Figure 7–15) with nodes 1, 2, and 5 is given in general form by
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$$
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[ k ^ {(1)} ] = \left[ \begin{array}{c c c} \underline {{k}} _ {1 1} ^ {(1)} & \underline {{k}} _ {1 2} ^ {(1)} & \underline {{k}} _ {1 5} ^ {(1)} \\ \underline {{k}} _ {2 1} ^ {(1)} & \underline {{k}} _ {2 2} ^ {(1)} & \underline {{k}} _ {2 5} ^ {(1)} \\ \underline {{k}} _ {5 1} ^ {(1)} & \underline {{k}} _ {5 2} ^ {(1)} & \underline {{k}} _ {5 5} ^ {(1)} \end{array} \right] \tag {7.5.8}
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$$
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where the superscript in parentheses again refers to the element number, and each submatrix $[ k _ { i j } ^ { ( 1 ) } ]$ is of order $2 \times 2$ . The stiffness matrix of the quadrilateral, assembled using Eq. (7.5.8) along with similar stiffness matrices for elements 2–4 of Figure 7–15, is given by the following (before static condensation is used):
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$$
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[ k ] = \left[ \begin{array}{c c c c c} \left(u _ {1}, v _ {1}\right) & \left(u _ {2}, v _ {2}\right) & \left(u _ {3}, v _ {3}\right) & \left(u _ {4}, v _ {4}\right) & \left(u _ {5}, v _ {5}\right) \\ \left[ k _ {1 1} ^ {(1)} \right] & & & & \\ + & \left[ k _ {1 2} ^ {(1)} \right] & [ 0 ] & \left[ k _ {1 4} ^ {(4)} \right] & \left[ k _ {1 5} ^ {(1)} \right] + \left[ k _ {1 5} ^ {(4)} \right] \\ \left[ k _ {1 1} ^ {(4)} \right] & & & & \\ & \left[ k _ {2 2} ^ {(1)} \right] & & & \\ \left[ k _ {2 1} ^ {(1)} \right] & + & \left[ k _ {2 3} ^ {(2)} \right] & [ 0 ] & \left[ k _ {2 5} ^ {(1)} \right] + \left[ k _ {2 5} ^ {(2)} \right] \\ & \left[ k _ {2 2} ^ {(2)} \right] & & & \\ & & \left[ k _ {3 3} ^ {(2)} \right] & & \\ [ 0 ] & \left[ k _ {3 2} ^ {(2)} \right] & + & \left[ k _ {3 4} ^ {(3)} \right] & \left[ k _ {3 5} ^ {(2)} \right] + \left[ k _ {3 5} ^ {(3)} \right] \\ & & \left[ k _ {3 3} ^ {(3)} \right] & & \\ & & & \left[ k _ {4 4} ^ {(3)} \right] & \\ \left[ k _ {4 1} ^ {(4)} \right] & [ 0 ] & \left[ k _ {4 3} ^ {(3)} \right] & + & \left[ k _ {4 5} ^ {(3)} \right] + \left[ k _ {4 5} ^ {(4)} \right] \\ & & & \left[ k _ {4 4} ^ {(4)} \right] & \\ \hline \left[ k _ {5 1} ^ {(1)} \right] & \left[ k _ {5 2} ^ {(1)} \right] & \left[ k _ {5 3} ^ {(2)} \right] & \left[ k _ {5 4} ^ {(3)} \right] & \left(\left[ k _ {5 5} ^ {(1)} \right] + \left[ k _ {5 5} ^ {(2)} \right]\right) \\ + & + & + & + & + \\ \left[ k _ {5 1} ^ {(4)} \right] & \left[ k _ {5 2} ^ {(2)} \right] & \left[ k _ {5 3} ^ {(3)} \right] & \left[ k _ {5 4} ^ {(4)} \right] & \left(\left[ k _ {5 5} ^ {(3)} \right] + \left[ k _ {5 5} ^ {(4)} \right]\right) \end{array} \right] \tag {7.5.9}
|
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$$
|
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|
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<!-- source-page: 390 -->
|
||
|
||
|
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|
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<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Geometric diagram of a parallelogram divided into triangles by internal lines (no text or symbols)
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Geometric diagram of a parallelogram divided into eight triangles (no text or symbols)
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Geometric diagram of a parallelogram divided into multiple triangles with dashed internal lines (no text or symbols)
|
||
</details>
|
||
|
||
Figure 7–16 Skew effects in finite element modeling
|
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|
||
where the orders of the degrees of freedom are shown above the columns of the stiffness matrix and the partitioning scheme used in static condensation is indicated by the dotted lines. Before static condensation is applied, the stiffness matrix is of order $1 0 \times 1 0$ .
|
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|
||
# Example 7.1
|
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|
||
Consider the quadrilateral with internal node 5 and dimensions as shown in Figure 7–17 to illustrate the application of static condensation.
|
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|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
4
|
||
③
|
||
5
|
||
④
|
||
②
|
||
①
|
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1
|
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4 in.
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2
|
||
3
|
||
2 in.
|
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2
|
||
</details>
|
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|
||
Figure 7–17 Quadrilateral with an internal node
|
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|
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Recall that the original stiffness matrix of the quadrilateral is $1 0 \times 1 0 .$ , but static condensation will result in an $8 \times 8$ stiffness matrix after removal of the degrees of freedom $( u _ { 5 } , v _ { 5 } )$ at node 5.
|
||
|
||
Using the CST stiffness matrix of Eq. (6.4.3) for plane strain, we have
|
||
|
||
$$
|
||
[ k ^ {(1)} ] = [ k ^ {(3)} ] = \frac {E}{4 . 1 6} \left[ \begin{array}{c c c c c c} 1. 5 & 1. 0 & 0. 1 & 0. 2 & - 1. 6 & - 1. 2 \\ & 3. 0 & - 0. 2 & 2. 6 & - 0. 8 & - 5. 6 \\ & & 1. 5 & - 1. 0 & - 1. 6 & 1. 2 \\ & & & 3. 0 & 0. 8 & - 5. 6 \\ & & & & 3. 2 & 0. 0 \\ \text {Symmetry} & & & & & 1 1. 2 \end{array} \right] \tag {7.5.10}
|
||
$$
|