This page shows the final part of the 3D frame example which we didn’t finish in class.
Solving stiffness Matrix
In class, we found the complete stiffness matrix:
\[\begin{split}\left[ \begin{matrix}
2000 & 1000 & 0 & 0 & 0 \\
1000 & 4400 & 1000 & -400 & 0 \\
0 & 1000 & 2400 & 0 & -400 \\
0 & -400 & 0 & 400 & 0 \\
0 & 0 & -400 & 0 & 400 \\
\end{matrix} \right] \left[ \begin{matrix}
{{\varphi }_{1}} \\
{{\varphi }_{2}} \\
{{\varphi }_{3}} \\
{{\varphi }_{4}} \\
{{\varphi }_{5}} \\
\end{matrix} \right]=\left[ \begin{matrix}
{{M}_{1}} \\
2 \\
4 \\
{{M}_{t,4}} \\
2+{{M}_{t,5}} \\
\end{matrix} \right]\end{split}\]
Solving boundary conditions
As we’ve no non-zero boundary conditions, we can apply the row-striking technique of lecture 1, which leads to:
\[\begin{split}\left[ \begin{matrix}
4400 & 1000 \\
1000 & 2400
\end{matrix} \right] \left[ \begin{matrix}
{{\varphi }_{2}} \\
{{\varphi }_{3}}
\end{matrix} \right]=\left[ \begin{matrix}
2 \\
4
\end{matrix} \right]\end{split}\]
Solving this system of equations gives:
\[\begin{split}\varphi_2 \approx 1 \cdot 10^{-4} \\
\varphi_3 \approx 1.6 \cdot 10^{-3}\end{split}\]
The support reactions can be found by inserting these into our original system of equations:
\[\begin{split}\left[ \begin{matrix}
2000 & 1000 & 0 & 0 & 0 \\
0 & -400 & 0 & 400 & 0 \\
0 & 0 & -400 & 0 & 400 \\
\end{matrix} \right] \left[ \begin{matrix}
0 \\
1 \cdot 10^{-4} \\
1.6 \cdot 10^{-3} \\
0 \\
0
\end{matrix} \right]=\left[ \begin{matrix}
{{M}_{1}} \\
{{M}_{t,4}} \\
2+{{M}_{t,5}} \\
\end{matrix} \right]\end{split}\]
Note that to calculate \(M_{t,4}\) the element load should be taken into account!
This gives:
\[\begin{split}M_1 \approx 0.1 \text{ kNm} \\
M_{t,4} \approx -0.033 \text{ kNm} \\
M_{t,5} \approx -2.7 \text{ kNm}\end{split}\]
Postprocessing moments element \(\left(1\right)\)
The continuum displacement field of element \(\left(1\right)\) can be described by the shape function:
\[w\left(x\right) = \left( -\cfrac{x^3}{\ell^2} + \cfrac{x^2}{\ell} \right) \varphi_2 \]
This gives:
\[\begin{split}\varphi \left(x\right) = -\cfrac{\text{d} w\left(x\right)}{\text{d}x}=\left( \cfrac{3x^2}{\ell^2} - \cfrac{2x}{\ell} \right) \varphi_2 \\
\kappa \left(x\right) = \cfrac{\text{d} \varphi\left(x\right)}{\text{d}x}=\left( \cfrac{6x}{\ell^2} - \cfrac{2}{\ell} \right) \varphi_2 \\
M \left(x\right) = EI \kappa=EI\left( \cfrac{6x}{\ell^2} - \cfrac{2}{\ell} \right) \varphi_2 \\\end{split}\]
This is a linear distribution, with values at \(x=0\) and \(x=\ell\):
\[\begin{split}M \left(0\right) \approx -0.1 \text{ kNm} \\
M \left(2\right) \approx 0.2 { kNm} \end{split}\]
The value at \(x=0\) has indeed the same absolute value as the support reactions. The sign is different because \(M_1\) is defined in the global coordinate system and \(M \left(0\right)\) is defined from our agreements on positive internal moments (leading to positive stresses at the positive \(z\)-side.)
