Question 18.P.15: Solve Problem 18.14 with each of the two finite elements a s...

Solve Problem 18.14 with each of the two finite elements a six-node triangle. In this case, however, determine the strains and stresses at the centroids of the two triangles.

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\begin{array}{c} \mathrm{k} = 10^8&\left[\begin{array}{rrrrrr}0.2040 & 0.0240 & -0.1758 & -0.0962 & 0 & 0 \\0.0240 & 0.2040 & 0 & -0.0962 & -0.1758 & 0 \\-0.1758 & 0 & 0.5440 & 0 & 0 & -0.1923 \\-0.0962 & -0.0962 & 0 & 0.5440 & 0 & -0.3516 \\0 & -0.1758 & 0 & 0 & 0.5440 & -0.1923 \\0 & 0 & -0.1923 & -0.3516 & -0.1923 & 1.0879 \\-0.1071 & -0.0192 & 0.0659 & 0.0769 & 0 & 0 \\-0.0165 & 0 & 0 & 0.0659 & 0.0769 & -0.1429 \\0.0769 & 0 & -0.1429 & -0.1429 & 0 & 0.1429 \\0.0659 & 0.0769 & -0.1429 & -0.1429 & 0 & 0.1429 \\0 & 0.0659 & 0 & 0 & -0.1429 & 0.1429 \\0 & -0.1429 & 0.1429 & 0.1429 & 0.1429 & -0.2857\end{array}\right. & \\ &\left. \begin{array}{rrrrrr}-0.1071 & -0.0165 & 0.0769 & 0.0659 & 0 & 0 \\-0.0192 & 0 & 0 & 0.0769 & 0.0659 & -0.1429 \\0.0659 & 0 & -0.1429 & -0.1429 & 0 & 0.1429 \\0.0769 & 0.0659 & -0.1429 & -0.1429 & 0 & 0.1429 \\0 & 0.0769 & 0 & 0 & -0.1429 & 0.1429 \\0 & -0.1429 & 0.1429 & 0.1429 & 0.1429 & -0.2857 \\0.2522 & 0.0687 & -0.0615 & -0.2747 & 0 & 0 \\0.0687 & 0.2522 & 0 & -0.2747 & -0.0615 & 0 \\-0.0615 & 0 & 0.6725 & 0 & 0 & -0.5495 \\-0.2747 & -0.2747 & 0 & 0.6725 & 0 & -0.1231 \\0 & -0.0615 & 0 & 0 & 0.6725 & -0.5495 \\0 & 0 & -0.5495 & -0.1231 & -0.5495 & 1.3451\end{array}\right] &\text{lb}/\text{in}\end{array}

\begin{aligned}& \mathbf{u}^{T}=10^{-3}\left[\begin{array}{lllllll}0.2986 & 0.3951 & 0.1524 & 0.3419 & 0.1893 & 0.1651 & -0.0078\end{array}\right. \\& \left.\begin{array}{lllll}-0.0984 & 0.0272 & -0.0497 & -0.0504 & -0.0096\end{array}\right] \text { in } \\& \varepsilon_{1}=10^{-4}\left\{\begin{array}{r}0.3354 \\-0.0991 \\0.0052\end{array}\right\} \quad \varepsilon_{2}=10^{-4}\left\{\begin{array}{r}0.3518 \\-0.0681 \\-0.0052\end{array}\right\} \\& \sigma_{1}=\left\{\begin{array}{c}1007.6 \\4.8 \\6.1\end{array}\right\} \quad \sigma_{2}=\left\{\begin{array}{c}1092.4 \\123.5 \\-6.1\end{array}\right\} \mathrm{psi}\end{aligned}

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