2.11.3 Differentiation using the Reference Element
Measurable Outcome 2.17
To find the derivative of \(\tilde{T}\) with respect to \(x\) (or similarly \(y\)) within an element, we differentiate the three nodal basis functions within the element:
\(\displaystyle \tilde{T}_ x\)
\(\displaystyle =\)
\(\displaystyle \frac{\partial }{\partial x}\left(\sum _{j=1}^{3} a_ j\phi _ j\right),\)
(2.271)
\(\displaystyle =\)
\(\displaystyle \sum _{j=1}^{3} a_ j\frac{\partial \phi _ j}{\partial x}.\)
(2.272)
To find the \(x\)-derivatives of each of the \(\phi _ j\)'s, the chain rule is applied:
\[\frac{\partial \phi _ j}{\partial x} = \frac{\partial \phi _ j}{\partial \xi _1}\frac{\partial \xi _1}{\partial x} + \frac{\partial \phi _ j}{\partial \xi _2}\frac{\partial \xi _2}{\partial x}.\]
(2.273)
Similarly, to find the derivatives with respect to \(y\):
\[\frac{\partial \phi _ j}{\partial y} = \frac{\partial \phi _ j}{\partial \xi _1}\frac{\partial \xi _1}{\partial y} + \frac{\partial \phi _ j}{\partial \xi _2}\frac{\partial \xi _2}{\partial y}.\]
(2.274)
The calculation of the derivatives of \(\phi _ j\) with respect to the \(\xi\) variables gives:
\(\displaystyle \frac{\partial \phi _1}{\partial \xi _1}\)
\(\displaystyle =\)
\(\displaystyle -1, \qquad \frac{\partial \phi _1}{\partial \xi _2} = -1,\)
(2.275)
\(\displaystyle \frac{\partial \phi _2}{\partial \xi _1}\)
\(\displaystyle =\)
\(\displaystyle 1, \qquad \frac{\partial \phi _2}{\partial \xi _2} = 0,\)
(2.276)
\(\displaystyle \frac{\partial \phi _3}{\partial \xi _1}\)
\(\displaystyle =\)
\(\displaystyle 0, \qquad \frac{\partial \phi _3}{\partial \xi _2} = 1.\)
(2.277)
The only remaining terms are the calculation of \(\frac{\partial \xi _1}{\partial x}\), \(\frac{\partial \xi _2}{\partial x}\), etc. which can be found by differentiating Equation (2.270 ),
\(\displaystyle \frac{\partial \vec{\xi }}{\partial \vec{x}}\)
\(\displaystyle =\)
\(\displaystyle \left(\begin{array}{cc} x_2-x_1 & x_3-x_1 \\ y_2-y_1 & y_3-y_1\end{array}\right)^{-1},\)
(2.278)
\(\displaystyle =\)
\(\displaystyle \frac{1}{J} \left(\begin{array}{cc} y_3-y_1 & -(x_3-x_1) \\ -(y_2-y_1) & x_2-x_1\end{array}\right),\)
(2.279)
where
\[J = (x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1).\]
(2.280)
Note that the Jacobian, \(J\), is equal to twice the area of the triangular element.
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