User:Eml4500.f08.ateam.carr/HW2

Since displacements of Element 1 are now known, the reaction forces can be determined.

Recall Element 1:

$$ k^{(1)} \cdot d^{(1)} = f^{(1)} $$

Where k and d have previously been defined:



k^{(1)} = \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix} $$



d^{(1)} =

\begin{bmatrix} 0\\     0\\      4.352\\      6.1271\\     \end{bmatrix} $$

To compute the reaction forces in Element 1, simply multiply the k and d matrix of Element 1. Note that the first two columns of matrix k will be multiplied to zero and can be ignored if a manual calculation is performed.



\begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}\cdot

\begin{bmatrix} 0\\     0\\      4.352\\      6.1271\\     \end{bmatrix}

= F^{(1)}

$$

Executing matrix multiplication to solve for the reaction forces, F:



\begin{bmatrix} -4.4378\\ -2.5622\\ 4.4378\\ 2.5622\\ \end{bmatrix}= \begin{bmatrix} f_{1}^{(1)}\\ f_{2}^{(1)}\\ f_{3}^{(1)}\\ f_{4}^{(1)} \end{bmatrix}

$$

Observation: Note Element 1 is in equilibrium. If the sum of the forces in the x and y direction are computed, one will note that they sum to zero.


 * $$\sum f_{1}^{(1)} + f_{3}^{(1)}=0$$
 * $$\sum f_{2}^{(1)} + f_{4}^{(1)}=0$$

HW: Verify also that:



To prove that the magnitude and direction of the forces acting on Element 1 are equal and opposite, first compute the magnitude of the reaction. This is done by the root-sum-square method:


 * $$P_1^{(1)}=[(f_1^{(1)})^2 + (f_2^{(1)})^2]^{1/2}$$


 * $$P_1^{(1)}=[(-4.4374)^{2} + (-2.5622)^{2}]^{1/2} $$


 * $$P_1^{(1)}=5.124 $$

The magnitude of P2 is computed the same way:


 * $$P_2^{(1)}=[(f_3^{(1)})^2 + (f_4^{(1)})^2]^{1/2}$$


 * $$P_2^{(1)}=[(4.4374)^{2} + (2.5622)^{2}]^{1/2} $$


 * $$P_2^{(1)}=5.124 $$

Thus, the forces are equal in magnitude but opposite in direction. The element is in equilibrium.

Element 2 is also in equilibrium and this can be proven the same way as above.



Method 2: To solve for 2 bar truss system using statics method:



Two different cut methods can be used as demonstrated above. The red outlines the different applications of the Euler cut principle. Individual free body diagrams can be drawn for each element circled in red and the problem can be solved as statically determinate.

HW: Verify the equilibrium of node 2 statically:




 * $$\sum F_{x}=0=-P_{1}^{(1)}cos(30) + P_{2}^{(1)}cos(45)$$
 * $$\sum F_{y}=0=P - P_{1}^{(1)}sin(30) - P_{2}^{(1)}sin(45)$$

P, P1, and P2 are known from previous steps:


 * $$P_{applied}^{} = 7 $$
 * $$P_{1}^{(1)}=5.124$$
 * $$P_{2}^{(2)}=6.276$$

Plugging these known values into the force equations:


 * $$\sum F_{x}=-5.124 \cdot cos(30) + 6.276 \cdot cos(45)$$
 * $$\sum F_{y}=7 - 5.124 \cdot sin(30) - 6.276 \cdot sin(45)$$


 * $$\sum F_{x}=0$$
 * $$\sum F_{y}=0$$

Global node 2 is in equilibrium.