Steps
1. Elements and Nodes
Define two beam elements as shown in Balken-Element B2 .
Solution
2. Augmented Element Stiffness Matrices
Please find:
\(k_1^|\) : The augmented element stiffness matrix of Element 1 in “element shape”.
\(k_2^|\) : The augmented element stiffness matrix of Element 2 in “element shape”.
\(K_1^|\) : The augmented element stiffness matrix of Element 1 in “system shape”.
\(K_2^|\) : The augmented element stiffness matrix of Element 2 in “system shape”.
Solution
\begin{align*}
k_1^|
\!&=\!
\tfrac{EI}{a^3}
\!
\left[
\begin{array}{cccc|c}
4 a^2 & -6 a & 2 a^2 & 6 a & \psi_0 \\
-6 a & 12 & -6 a & -12 & w_0 \\
2 a^2 & -6 a & 4 a^2 & 6 a & \psi_1 \\
6 a & -12 & 6 a & 12 & w_1
\end{array}
\right]
\\
k_2^|
\!&=\!
\tfrac{EI}{a^3}
\!
\left[
\begin{array}{cccc|c}
4 a^2 & -6 a & 2 a^2 & 6 a & \psi_1 \\
-6 a & 12 & -6 a & -12 & w_1 \\
2 a^2 & -6 a & 4 a^2 & 6 a & \psi_2 \\
6 a & -12 & 6 a & 12 & w_2
\end{array}
\right]
\\
K_1^|
\!&=\!
\tfrac{EI}{a^3}
\!
\left[
\begin{array}{cccccc|c}
4 a^2 & -6 a & 2 a^2 & 6 a & 0 & 0 & \psi_0 \\
-6 a & 12 & -6 a & -12 & 0 & 0 & w_0 \\
2 a^2 & -6 a & 4 a^2 & 6 a & 0 & 0 & \psi_1 \\
6 a & -12 & 6 a & 12 & 0 & 0 & w_1 \\
0 & 0 & 0 & 0 & 0 & 0 & \psi_2 \\
0 & 0 & 0 & 0 & 0 & 0 & w_2
\end{array}
\right]
\\
K_2^|
\!&=\!
\tfrac{EI}{a^3}
\!
\left[
\begin{array}{cccccc|c}
0 & 0 & 0 & 0 & 0 & 0 & \psi_0 \\
0 & 0 & 0 & 0 & 0 & 0 & w_0 \\
0 & 0 & 4 a^2 & -6 a & 2 a^2 & 6 a & \psi_1 \\
0 & 0 & -6 a & 12 & -6 a & -12 & w_1 \\
0 & 0 & 2 a^2 & -6 a & 4 a^2 & 6 a & \psi_2 \\
0 & 0 & 6 a & -12 & 6 a & 12 & w_2
\end{array}
\right]
\end{align*}
3. System Stiffness Matrix
Use the following symbols:
\[\begin{split}u =
\begin{bmatrix}
\psi_0 \\
w_0 \\
\psi_1 \\
w_1 \\
\psi_2 \\
w_2
\end{bmatrix}
\qquad
f =
\begin{bmatrix}
M_0 \\
F_0 \\
M_1 \\
F_1 \\
M_2 \\
F_2
\end{bmatrix}\end{split}\]
Find \(K\) . And write down the following equation:
\[K u = f\]
Solution
The system stiffness matrix is the “sum” of all element stiffness matrices in “system shape”:
\begin{align*}
\underbrace{
\tfrac{EI}{a^3}
\!
\left[
\begin{array}{cccccc}
4 a^2 & -6 a & 2 a^2 & 6 a & 0 & 0 \\
-6 a & 12 & -6 a & -12 & 0 & 0 \\
2 a^2 & -6 a & 8 a^2 & 0 & 2 a^2 & 6 a \\
6 a & -12 & 0 & 24 & -6 a & -12 \\
0 & 0 & 2 a^2 & -6 a & 4 a^2 & 6 a \\
0 & 0 & 6 a & -12 & 6 a & 12
\end{array}
\right]
}_K
\!
\underbrace{
\left[
\begin{array}{c}
\psi_0 \\
w_0 \\
\psi_1 \\
w_1 \\
\psi_2 \\
w_2
\end{array}
\right]
}_u
\!&=\!
\underbrace{
\left[
\begin{array}{c}
M_0 \\
F_0 \\
M_1 \\
F_1 \\
M_2 \\
F_2
\end{array}
\right]
}_f
\tag{1}
\end{align*}
4. Boundary Conditions
Adjust \(u\) and \(f\) so that all boundary conditions are met.
\[\begin{split}u =
\begin{bmatrix}
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots
\end{bmatrix}
\qquad f =
\begin{bmatrix}
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots \\
\ldots
\end{bmatrix}\end{split}\]
Solution
5. Forces
Please find all support reactions.
Solution
\(M_0\) and \(F_0\) are calculated from \(\psi_1, w_1, \psi_2, w_2\) by using the first two equations of the system equation:
\begin{align*}
\tfrac{EI}{a^3}
\!
\left[
\begin{array}{cccccc}
2 a^2 & 6 a \\
-6 a & -12 \\
\end{array}
\right]
\left[
\begin{array}{c}
\psi_1 \\
w_1 \\
\end{array}
\right]
\!&=\!
\left[
\begin{array}{c}
M_0 \\
F_0 \\
\end{array}
\right]
\end{align*}
Plugging in \(\psi_1, w_1, \psi_2, w_2\) leads to:
\begin{align*}
\left[
\begin{array}{c}
M_{0} \\
F_0 \\
\end{array}
\right]
\!=\!
\left[
\begin{array}{c}
a \left(F_{1} + 2 F_{2}\right)\\
- F_{1} - F_{2}\\
\end{array}
\right]
\end{align*}
7. Solution for Given Quantities
Fine \(w_2\) in \(\mathrm{mm}\) (millimeters) for the following given quantities :
\[\begin{split}F_1 &= 10 \,\mathrm{kN} \\
F_2 &= 10 \,\mathrm{kN} \\
a &= 1 \,\mathrm{m} \\
E &= 210 \,\mathrm{GPa} \\
I &= 318 \,\mathrm{cm}^4\end{split}\]
Round to \(0.01\) . Show that
\[w_2 \stackrel{0.01}{\approx} 52.41 \,\mathrm{mm}\]
Solution
Plugging in the quantities leads to:
\[w_2 \stackrel{0.01}{\approx} 52.41 \,\mathrm{mm}\]
Real Life
\(E= 210 \,\mathrm{GPa}\) is the Young’s Modulus of Steel.
\(I = 318 \,\mathrm{cm}^4\) is \(I_{yy}\) of a IPE 120.
DIN 1025-5: IPE 120: \(I_{yy}\stackrel{1.0}{\approx} 318 \,\mathrm{cm}^4\)
\(F_1 = F_2 = 10 \,\mathrm{kN}\) is the weight of a Volkswagen Polo.
VW Polo : Masse ca. \(1000\,\mathrm{kg}\) , Gewichtskraft ca. \(1000\,\mathrm{kg} \cdot 10\,\tfrac{\mathrm{m}}{\mathrm{s}^2} = 10 \,\mathrm{kN}\)
Note
Solution with Python here .