Stiffness Matrix for a Bar Element For elements 1 and 2: Example 1 -Bar Problem For element 3: You're welcome. Now, the stiffness matrix for 1D Truss bar with one degree of freedom per node can be extended one step further to also represent a similar 1D Truss bar but with two degrees of freedom per node— one longitudinal (in axial direction) and other transverse displacement at each node. Only AE is capable of supporting the vertical load and we know that AE's slope is 1/2, so the horizontal component is equal to 2 kip, for a resultant of $\sqrt{1^2+2^2}=2.236\text{ kip}$. It only takes a minute to sign up. Consider the geometry. How can I install a bootable Windows 10 to an external drive? 2. A truss element can only transmit forces in compression or tension. Therefore, the $AB_H=-1$ (only has a horizontal part) : $X_1 = Y_1 = Y_2 = 0, F_{x3} = 2 \times 10^3 N, F_{y3} = -5 \times 10^3 N$, $10^6 \begin{bmatrix} \ 257.76 & -77.76 & 103.68 \\ \ -77.76 & 155.52 & 0 \\ \ 103.68 & 0 & 276.48 \\ \end{bmatrix} \begin{Bmatrix} \ X_2 \\ \ X_3 \\ \ Y_3 \\ \end{Bmatrix} = \begin{Bmatrix} \ 0 \\ \ -2 \\ \ -5 \\ \end{Bmatrix} \times 10^3$, $X_2 = 4.86 \times 10^{-6} m\\ site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. These have the drawback that the visualizations is complex. Different values for plotparare used to distinguish the deformed geometry from the undeformed one. where node is the node number. Determine: (a) the global stiffness matrix, (b) the displacement of nodes 2 and 3, and (c) the reactions at nodes 1 and 4. (Both Supports Are Fixed UAevA-ulevi-0) E=70GPa,A 0.003125 500 N 1000 N G! We can use these coordinates to determine the lengths and angles of the elements. Since AE's horizontal component is 2 kip, we know that AB is also 2 kip. This means that in two dimensions, each node has two degrees of freedom (DOF): horizontal and vertical displacement. Select Node 2 and 3. Making statements based on opinion; back them up with references or personal experience. Why does US Code not allow a 15A single receptacle on a 20A circuit? The table below shows the coordinates of the nodes in the problem we are solving. A Plague that Causes Death in All Post-Plague Children. Since it is the only one with this component (since we already know BE is 0 kip), it can't be cancelled out. DISPLACEMENT Note: Because truss elements have three displacement DOFs at each node, it is necessary to constrain the displacements in all three directions at the left edge of the model so as to restrain possible rigid body modes. The resulting equation contains a four by four stiffness matrix. \sigma_1 = \frac{180 \times 10^9}{0.6} \begin{bmatrix} \ -1 & 0 & 1 & 0 \\ \end{bmatrix} \begin{Bmatrix} \ 0 \\ \ 0 \\ \ 4.86 \\ \ 0 \\ \end{Bmatrix} \times 10^6 = 1.458 \times 10^6 N/m^2 (Tensile)$, $\sigma_2 = \frac{180 \times 10^9}{0.5} \begin{bmatrix} \ 0.6 & -0.8 & -0.6 & -0.8 \\ \end{bmatrix} \begin{Bmatrix} \ 4.86 \\ \ 0 \\ \ -10.4 \\ \ -19.9 \\ \end{Bmatrix} \times 10^6 = 25.076 \times 10^6 N/mm^2 (Tensile)$, $\sigma_3 = \frac{180 \times 10^9}{0.5} \begin{bmatrix} \ -0.6 & -0.8 & 0.6 & 0.8 \\ \end{bmatrix} \begin{Bmatrix} \ 0 \\ \ 0 \\ \ -10.4 \\ \ -19.9 \\ \end{Bmatrix} \times 10^6 = - 22.16 \times 10^6 N/mm^2 = 22.16 \times 10^6 N/mm^2 (Compressive)$, $e_1 = \frac{\sigma_1}{E} = \frac{1.458 \times 10^6}{180 \times 10^9} = 8.16 \times 10^{-6}\\ Node X Y 1 0 0 2 40 0 3 40 30 4 0 30 Table 1 - Coordinates of the nodes in the truss. 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