R2.A

Hint

For classical solution see 2.1.Q.

Video

Given is a structure consisting of three rods with Young’s Modulus \(E\) und cross-sectional area \(A\).

../../../_images/R2.A.png

Given symbols: \(F, l, E, A.\)

Use the conventions from R2-en. Proceed as follows:

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

../../../_images/R2.A.png

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

../../../_images/R2.A_symm.png

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.