User:Eml4500.f08.qwiki.bishop/Lecture 11

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Element 1 Equilibrium
$$\mathbf{f}^{(1)}=\mathbf{k}^{(1)}\mathbf{d}^{(1)}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248\\ 0.3248 & 0.1875 & -0.3248 & 0.1875\\ -0.5625 & -0.3248 & 0.5625 & 0.3248\\ -0.3248 & -0.1875 & 0.3248 & 0.1875\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 4.352\\ 6.127\\ \end{bmatrix}= \begin{bmatrix} 0.5625 & 0.3248\\ 0.3248 & 0.1875\\ \end{bmatrix} \begin{bmatrix} 4.325\\ 6.127\\ \end{bmatrix}= \begin{bmatrix} 4.438\\ 2.562\\ \end{bmatrix} $$ $$\mathbf{f}^{(1)}= \begin{bmatrix} f^{(1)}_1\\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4\\ \end{bmatrix}= \begin{bmatrix} -4.438\\ -2.562\\ 4.438\\ 2.562\\ \end{bmatrix} $$ Note that:
 * Recall that for Element 1, the local force relationship is given by:
 * To be correct, we must include both the internal and reaction forces of element 1. Since the internal and reaction forces are equal and opposite, the local force matrix for element 1 is given by:
 * $$ f^{(1)}_3 $$,and $$ f^{(1)}_4 $$ are the internal forces of Element 1
 * $$ f^{(1)}_1 $$ and $$ f^{(1)}_2 $$ are reaction forces of Element 1

Observation: Element 1 in Equilibrium We know that Element 1 is in equilibrium because the sum of forces in the x- and y- directions equals zero and the force moment about any point in the element also equals zero. $$\sum F_x = f_1^{(1)} + f_3^{(1)} = -4.438 + 4.438 =0$$ $$\sum F_y = f_2^{(1)} + f_4^{(1)} = -2.562 + 2.562 =0$$ $$\sum M_{any pt} = \sum M_{node1} = -L^{(1)}sin(30)f_3^{(1)} + L^{(1)}cos(30)f_4^{(1)}= -4cos(30)(2.5622) + 4sin(30)4.4378 = -8.867 + 8.867 = 0$$

Element 2 Equilibrium
$$\mathbf{f}^{(2)}=\mathbf{k}^{(2)}\mathbf{d}^{(2)}= \begin{bmatrix} 2.5 & -2.5 & -2.5 & 2.5\\ -2.5 & 2.5 & 2.5 & -2.5\\ -2.5 & 2.5 & 2.5 & -2.5\\ 2.5 & -2.5 & -2.5 & 2.5\\ \end{bmatrix} \begin{bmatrix} 4.352\\ 6.127\\ 0\\ 0\\ \end{bmatrix}= \begin{bmatrix} 2.5 & -2.5\\ -2.5 & 2.5\\ \end{bmatrix} \begin{bmatrix} 4.352\\ 6.127\\ \end{bmatrix}= \begin{bmatrix} f_1^{(2)}\\ f_2^{(2)}\\ \end{bmatrix}= \begin{bmatrix} -4.438\\ 4.438\\ \end{bmatrix} $$
 * Recall for Element 1 and 2 that the x- and y- nodal displacements are equal at their intersecting node. Simply put, $$ d_3^{(1)} = d_1^{(2)} $$ and $$ d_2^{(2)} = d_4^{(1)}$$. As a result, we can express the force-displacement relationship of Element 2 as:

$$\mathbf{f}^{(2)}= \begin{bmatrix} f^{(2)}_1\\ f^{(2)}_2\\ f^{(2)}_3\\ f^{(2)}_4\\ \end{bmatrix}= \begin{bmatrix} -4.438\\ 4.438\\ 4.438\\ -4.438\\ \end{bmatrix} $$
 * To be correct, we must include both the internal and reaction forces of element 2. Since the internal and reaction forces are equal and opposite, the local force matrix for Element 2 is given by:
 * $$ f^{(2)}_1 $$,and $$ f^{(2)}_2 $$ are the internal forces of Element 2
 * $$ f^{(2)}_3 $$ and $$ f^{(2)}_4 $$ are reaction forces of Element 2

Observation: Element 2 in Equilibrium We know that Element 1 is in equilibrium because the sum of forces in the x- and y- directions equals zero and the force moment about any point in the element also equals zero. $$\sum F_x = f_1^{(2)} + f_3^{(2)} = -4.438 + 4.438 =0$$ $$\sum F_y = f_2^{(2)} + f_2^{(2)} = 4.438 - 4.438 =0$$ $$\sum M_{node2} = -L^{(2)}sin(45)f_2^{(2)} + L^{(2)}cos(45)f_1^{(2)}= -2sin(45)(-4.4378) + -2sin(45)(4.4378) = 6.276 - 6.276 = 0$$

Analysis Methods
In general, there are two separate methods that can be used to solve for a 2-bar truss system using statics:

Method 1: FBD Analysis
In this analysis method:
 * Isolate the supports
 * Assign the applied force to only one FBD
 * Examine FBD 1 and FBD 2 Seaparately

Method 2: Nodal Finite Element Analysis
In this analysis method:
 * Isolate the supports AND the nodes
 * Assign the applied force to its respective node
 * Examine Element 1 and Element 2
 * Examine the node independently