Steps
1. Elements and Nodes
There are three rod elements 1, 2 and 3 with unit vectors \(\boldsymbol e_1, \boldsymbol e_3, \boldsymbol e_3\) according to this convention. Fill in the following table:
Element |
Length |
Nodes |
Ang. Position |
1 |
|
|
|
2 |
|
|
|
3 |
|
|
|
Solution
Element |
Length |
Nodes |
Ang. Position |
1 |
\(\sqrt 2 l\) |
1-4 |
\(\varphi_1 = 45 ^\circ\) |
2 |
\(l\) |
2-4 |
\(\varphi_2 = 90 ^\circ\) |
3 |
\(\sqrt 2 l\) |
3-4 |
\(\varphi_3 = 135 ^\circ\) |
2. Augmented Element Stiffness Matrices (Element-Shape)
Find the Augmented Element Stiffness Matrices in “Element-Shape” \(k_1^|, k_2^|, k_3^|\).
Solution
\begin{align*}
k^|_1
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccc|c}
\tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & u_{1x} \\
\tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & u_{1y} \\
- \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & u_{4x} \\
- \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & u_{4y} \\
\end{array}
\right]
\\
k^|_2
&=
\tfrac{EA}{l}
\left[
\begin{array}{cccc|c}
0 & 0 & 0 & 0 & u_{2x} \\
0 & 1 & 0 & -1 & u_{2y} \\
0 & 0 & 0 & 0 & u_{4x} \\
0 & -1 & 0 & 1 & u_{4y} \\
\end{array}
\right]
\\
k^|_3
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccc|c}
\tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & u_{3x} \\
- \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & u_{3y} \\
- \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & u_{4x} \\
\tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & u_{4y} \\
\end{array}
\right]
\end{align*}
3. Augmented Element Stiffness Matrices (System-Shape)
Find the Augmented Element Stiffness Matrices in “System-Shape” \(K^|_1, K^|_2, K^|_3\).
Solution
\begin{align*}
K^|_1
\!
&=
\!
\tfrac{EA}{\sqrt 2 l}
\!
\!
\left[
\begin{array}{cccccccc|c}
\tfrac{1}{2} & \tfrac{1}{2} & 0 & 0 & 0 & 0 & - \tfrac{1}{2} & - \tfrac{1}{2} & u_{1x} \\
\tfrac{1}{2} & \tfrac{1}{2} & 0 & 0 & 0 & 0 & - \tfrac{1}{2} & - \tfrac{1}{2} & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{2y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{3x} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{3y} \\
- \tfrac{1}{2} & - \tfrac{1}{2} & 0 & 0 & 0 & 0 & \tfrac{1}{2} & \tfrac{1}{2} & u_{4x} \\
- \tfrac{1}{2} & - \tfrac{1}{2} & 0 & 0 & 0 & 0 & \tfrac{1}{2} & \tfrac{1}{2} & u_{4y}
\end{array}
\right]
\\
K^|_2
\!
&=
\!
\tfrac{EA}{\sqrt 2 l}
\!
\!
\left[
\begin{array}{cccccccc|c}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{1x} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & \sqrt 2 & 0 & 0 & 0 & -\sqrt 2 & u_{2y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{3x} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{3y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{4x} \\
0 & 0 & 0 & -\sqrt 2 & 0 & 0 & 0 & \sqrt 2 & u_{4y}
\end{array}
\right]
\\
K^|_3
\!
&=
\!
\!
\tfrac{EA}{\sqrt 2 l}
\!
\left[
\begin{array}{cccccccc|c}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{1x} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{2y} \\
0 & 0 & 0 & 0 & \tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & u_{3x} \\
0 & 0 & 0 & 0 & - \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & u_{3y} \\
0 & 0 & 0 & 0 & - \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & u_{4x} \\
0 & 0 & 0 & 0 & \tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & u_{4y}
\end{array}
\right]
\end{align*}
4. Augmented System Stiffness Matrix
Find the augmented system stiffness matrix \(K^|\).
Solution
\begin{align*}
K^|
&=
K_1^| + K_3^| + K_3^| \\
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccccccc|c}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} & u_{1x} \\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & \sqrt{2} & 0 & 0 & 0 & - \sqrt{2} & u_{2y} \\
0 & 0 & 0 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} & u_{3x} \\
0 & 0 & 0 & 0 & - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} & u_{3y} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & 0 & - \frac{1}{2} & \frac{1}{2} & 1 & 0 & u_{4x} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & - \sqrt{2} & \frac{1}{2} & - \frac{1}{2} & 0 & 1 + \sqrt{2} & u_{4y}
\end{array}
\right]
\end{align*}
5. Boundary Conditions and Linear System
Let \(u\) the nodal displacements column vector. And let \(f\) be the nodal forces column vector:
\begin{align*}
u =
\begin{bmatrix}
u_{1x} \\
u_{1y} \\
u_{2x} \\
u_{2y} \\
u_{3x} \\
u_{3y} \\
u_{4x} \\
u_{4y} \\
\end{bmatrix}
\quad,\quad
f =
\begin{bmatrix}
F_{1x} \\
F_{1y} \\
F_{2x} \\
F_{2y} \\
F_{3x} \\
F_{3y} \\
F_{4x} \\
F_{4y} \\
\end{bmatrix}
\end{align*}
Adjust \(u\) und \(f\) so that all boundary conditions are met. And then write down the system of equations describing the structure:
\begin{align*}
K u = f
\end{align*}
Solution
\begin{align*}
u
=
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
0\\
u_{4x}\\
u_{4y}\\
\end{bmatrix}
\quad,\quad
f
=
\begin{bmatrix}
F_{1x}\\
F_{1y}\\
F_{2x}\\
F_{2y}\\
F_{3x}\\
F_{3y}\\
0 \\
-F \\
\end{bmatrix}
\end{align*}
\begin{align*}
\tfrac{EA}{\sqrt 2 l}
\!
\!
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \sqrt{2} & 0 & 0 & 0 & - \sqrt{2} \\
0 & 0 & 0 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\
0 & 0 & 0 & 0 & - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & 0 & - \frac{1}{2} & \frac{1}{2} & 1 & 0 \\
- \frac{1}{2} & - \frac{1}{2} & 0 & - \sqrt{2} & \frac{1}{2} & - \frac{1}{2} & 0 & 1 + \sqrt{2}
\end{bmatrix}
\!\!
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
0\\
u_{4x}\\
u_{4y}\\
\end{bmatrix}
\! \!
=
\! \!
\begin{bmatrix}
F_{1x}\\
F_{1y}\\
F_{2x}\\
F_{2y}\\
F_{3x}\\
F_{3y}\\
0 \\
-F \\
\end{bmatrix}
\end{align*}
6. Displacements
Find the displacement of node 4 in terms of the given symbols:
\begin{align*}
\begin{bmatrix}
u_{4x} \\
u_{4y} \\
\end{bmatrix}
=
\begin{bmatrix}
\ldots \\
\ldots \\
\end{bmatrix}
\tfrac{Fl}{EA}
\end{align*}
And for the following quantities:
\begin{align*}
F &= 5\,\mathrm{kN} \\
E &= 200\,\mathrm{GPa} \\
A &= 25\,\mathrm{mm}^2 \\
l &= 1707\,\mathrm{mm}
\end{align*}
Find \(u_{4y}\) in \(\mathrm{mm}\) (millimeters) and rounded to \(0.0001\):
\begin{align*}
u_{4y} \approx
\ldots \,
\mathrm{mm}
\end{align*}
Solution
The 7th and 8th equation of the matrix equation lead to:
\begin{align*}
\begin{bmatrix}
u_{4x} \\
u_{4y} \\
\end{bmatrix}
&=
\begin{bmatrix}
0 \\
-2 + \sqrt 2 \\
\end{bmatrix}
\tfrac{Fl}{EA} \\
\\
u_{4y} &\approx-0.9999\,\mathrm{mm}
\end{align*}
8. Support Reactions
For the same quantities: Find the \((x,y)\)-componenten of the force acting at node 2 from the support onto rod 2. Find these components in \(\mathrm{N}\) (Newton) and rounded to \(0.1\):
\begin{align*}
\begin{bmatrix}
F_{2x} \\
F_{2y} \\
\end{bmatrix}
\approx
\begin{bmatrix}
\ldots \\
\ldots \\
\end{bmatrix}
\mathrm{N}
\end{align*}
Solution
The 3rd and 4th equation of the matrix equation lead to:
\begin{align*}
\begin{bmatrix}
F_{2x} \\
F_{2y} \\
\end{bmatrix}
=
\begin{bmatrix}
0 \\
- \sqrt 2 + 2 \\
\end{bmatrix}
F
\end{align*}
Plugging in the given quantities leads to:
\begin{align*}
\begin{bmatrix}
F_{2x} \\
F_{2y} \\
\end{bmatrix}
\approx
\begin{bmatrix}
0 \\
2928.9 \\
\end{bmatrix}
\mathrm{N}
\end{align*}
9. Symmetry
Show that the same solution can be found by using the symmetry of the system.
Solution
Symmetry
Using reflection symmetry and considering only the left half.
Force on top is \(\tfrac 12 F\),
Cross-sectional property of rod 2 is \(\tfrac 12 EA\)
Tensile force in rod 2 is \(S_2'\).
Constraint at node 4: \(u_{4x}=0\).
Stiffness matrices:
\begin{align*}
k^|_1
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccc|c}
\tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & u_{1x} \\
\tfrac{1}{2} & \tfrac{1}{2} & - \tfrac{1}{2} & - \tfrac{1}{2} & u_{1y} \\
- \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & u_{4x} \\
- \tfrac{1}{2} & - \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & u_{4y} \\
\end{array}
\right]
\\
k^|_2
&=
\tfrac12 \tfrac{EA}{l}
\left[
\begin{array}{cccc|c}
0 & 0 & 0 & 0 & u_{2x} \\
0 & 1 & 0 & -1 & u_{2y} \\
0 & 0 & 0 & 0 & u_{4x} \\
0 & -1 & 0 & 1 & u_{4y} \\
\end{array}
\right]
\\
K^|_1
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccccc|c}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} & u_{1x} \\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & 0 & 0 & 0 & u_{2y} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & 0 & \frac{1}{2} & \frac{1}{2} & u_{4x} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & 0 & \frac{1}{2} & \frac{1}{2} & u_{4y} \\
\end{array}
\right]
\\
K^|_2
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccccc|c}
0 & 0 & 0 & 0 & 0 & 0 & u_{1x} \\
0 & 0 & 0 & 0 & 0 & 0 & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & \frac{\sqrt{2}}{2} & 0 & - \frac{\sqrt{2}}{2}& u_{2y} \\
0 & 0 & 0 & 0 & 0 & 0 & u_{4x} \\
0 & 0 & 0 & - \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}& u_{4y} \\
\end{array}
\right]
\\
K^|
&=
\tfrac{EA}{\sqrt 2 l}
\left[
\begin{array}{cccccc|c}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} & u_{1x} \\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} & u_{1y} \\
0 & 0 & 0 & 0 & 0 & 0 & u_{2x} \\
0 & 0 & 0 & \frac{\sqrt{2}}{2} & 0 & - \frac{\sqrt{2}}{2} & u_{2y} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & 0 & \frac{1}{2} & \frac{1}{2} & u_{4x} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & - \frac{\sqrt{2}}{2} & \frac{1}{2} & \frac{1}{2} + \frac{\sqrt{2}}{2} & u_{4y} \\
\end{array}
\right]
\end{align*}
Linear System:
\begin{align*}
\tfrac{EA}{\sqrt 2 l}
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2} & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 0 & 0 & - \frac{1}{2} & - \frac{1}{2} \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{\sqrt{2}}{2} & 0 & - \frac{\sqrt{2}}{2} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
- \frac{1}{2} & - \frac{1}{2} & 0 & - \frac{\sqrt{2}}{2} & \frac{1}{2} & \frac{1}{2} + \frac{\sqrt{2}}{2} \\
\end{bmatrix}
\begin{bmatrix}
0 \\
0 \\
0 \\
0 \\
0 \\
u_{4y} \\
\end{bmatrix}
&=
\begin{bmatrix}
F_{1x} \\
F_{1y} \\
F_{2x}' \\
F_{2y}' \\
F_{4x} \\
-\frac 1 2 F \\
\end{bmatrix}
\end{align*}
Solution:
\begin{align*}
u_{4y}
&=
\left(-2 + \sqrt 2\right) \tfrac{Fl}{EA} \\
\\
\begin{bmatrix}
F_{2x}' \\
F_{2y}' \\
\end{bmatrix}
&=\tfrac 1 2
\begin{bmatrix}
0 \\
- \sqrt 2 + 2 \\
\end{bmatrix}
F
\\
&=\tfrac 1 2
\begin{bmatrix}
F_{2x} \\
F_{2y} \\
\end{bmatrix}
\end{align*}
This is the same solution as before. The support reactions at node 2 and also the tensile force in rod 2 is half of what it was before.