B2.A

Video

Loads acting on a beam are \(F_1\) and \(F_2\). The beam is suffering from the deformations \(w_1, \psi_1, w_2, \psi_2\).

../../../_images/2.4.2.J_a.png

Given symbols: \(a, EI, F_1, F_2.\)

Proceed as follows.

Steps

1. Elements and Nodes

Define two beam elements as shown in Balken-Element B2.

Solution

../../../_images/B2.A.png

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:

  • \(K\): The system stiffness matrix.

  • \(u\): The deformations.

  • \(f\): The external nodal loads.

\[\begin{split}u = \begin{bmatrix} \psi_0 \\ w_0 \\ \psi_1 \\ w_1 \\ \psi_2 \\ w_2 \end{bmatrix} \qquad f = \begin{bmatrix} M_0 \\ F_0 \\ M_1 \\ F_1 \\ M_2 \\ F_2 \end{bmatrix}\end{split}\]

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

../../../_images/B2.A_bc.png

5. Deformations

Find \(\psi_1, w_1, \psi_2, w_2\). Show that:

\begin{align*} \left[ \begin{array}{c} \psi_1 \\ w_1 \\ \psi_2 \\ w_2 \end{array} \right] \!&=\! \tfrac{a^2}{6 EI} \!\left[ \begin{array}{c} - 3 F_{1} -9 F_{2} \\ a( 2 F_{1} + 5 F_{2} ) \\ - 3 F_{1} -12 F_{2} \\ a ( 5 F_{1} + 16 F_{2} ) \end{array} \right] \end{align*}

Solution

The following equation describes the system:

\[\begin{split}\tfrac{EI}{a^3} \begin{bmatrix} 4 a^2 & -6 a & 2 a^2 & 6 a & 0 & 0 \\ -6 a & 12 & -6 a & -12 & 0 & 0 \\ 2 a^2 & -6 a & 8 a^2 & 0 & 2 a^2 & 6 a \\ 6 a & -12 & 0 & 24 & -6 a & -12 \\ 0 & 0 & 2 a^2 & -6 a & 4 a^2 & 6 a \\ 0 & 0 & 6 a & -12 & 6 a & 12 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ \psi_1 \\ w_1 \\ \psi_2 \\ w_2 \end{bmatrix} &= \begin{bmatrix} M_0 \\ F_0 \\ 0 \\ F_1 \\ 0 \\ F_2 \end{bmatrix}\end{split}\]

These are six equations for the six unknowns \(\psi_1, w_1, \psi_2, w_2, M_0, F_0\). The last four equations can be used to solve for \(\psi_1, w_1, \psi_2, w_2\):

\begin{align*} \tfrac{EI}{a^3} \left[ \begin{array}{cccc} 8 a^2 & 0 & 2 a^2 & 6 a \\ 0 & 24 & -6 a & -12 \\ 2 a^2 & -6 a & 4 a^2 & 6 a \\ 6 a & -12 & 6 a & 12 \end{array} \right] \left[ \begin{array}{c} \psi_1 \\ w_1 \\ \psi_2 \\ w_2 \end{array} \right] \!&=\! \left[ \begin{array}{c} 0 \\ F_1 \\ 0 \\ F_2 \end{array} \right] \end{align*}

Solving for \(\psi_1, w_1, \psi_2, w_2\) leads to:

\begin{align*} \left[ \begin{array}{c} \psi_1 \\ w_1 \\ \psi_2 \\ w_2 \end{array} \right] \!&=\! \tfrac{a^2}{6 EI} \!\left[ \begin{array}{c} - 3 F_{1} -9 F_{2} \\ a( 2 F_{1} + 5 F_{2} ) \\ - 3 F_{1} -12 F_{2} \\ a ( 5 F_{1} + 16 F_{2} ) \end{array} \right] \end{align*}

5. Forces

Please find all support reactions.

Solution

../../../_images/B2.A.png

\(M_0\) 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.

    ../../../_images/ipe_120.png

    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.

    ../../../_images/VW_Polo_IV_front_20071106.jpg

    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.