For a 12-node serendipity (cubic) element, as illustrated in Fig. P9.12, show that the Jacobian J…
For a 12-node serendipity (cubic) element, as illustrated in Fig. P9.12, show that the Jacobian J = J11 is
What can you conclude from the requirement J > 0?
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We have (after a lengthy algebra using Maple)
Thus J = J11 > 0 ensures a unique transformation and preservation of the sense of the coordinate system in the master rectangular element, provided a and b are such that
5.5 − 1.125(a + b) − 3.375(b − a) > 0 and 1.0+1.125(a + b) − 3.375(b − a) > 0
The above inequalities place a restriction on the values of a and b. Clearly, for a > 0.666667 and b = 1.333333 (the usual location of the midside nodes), the inequalities are met (i.e., J = 1 > 0). A plot of the Jacobian shows, for example, that J = 0 when (i) a = 0.27777 = 5/18 and (ii) b = 1 and J < 0=”” for=”” a=”0.27777″ and=”” any=”” b=””><>
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