When calculating, for example, the first term of Equation 5.69, it is necessary to evaluate the following integral:

$$\mathit{\rho}D\underset{0}{\overset{1}{\int}}\underset{0}{\overset{1-v}{\int}}(1-u-v)dudv=\mathit{\rho}D\underset{0}{\overset{1}{\int}}{(u-\frac{{u}^{2}}{2}-uv)}_{0}^{1-v}dv$$

$$=\mathit{\rho}D\underset{0}{\overset{1}{\int}}(1-v-\frac{{(1-v)}^{2}}{2}-v(1-v))dv=\frac{pD}{6}$$

which is the result shown in Equation 5.69.

It is clear from the example earlier that even for the very simple 2D element, the integration requires much work. This is in spite of the fact that the $\text{det}J$ is a constant. For second‐order elements, the shape functions are much more complex and $\text{det}J$ is not necessarily constant. Therefore, in general, it is practically impossible to evaluate ...

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