User:Eml4500.f08.ateam.carr/HW3

Continuation of Three Bar Truss System



Recall the equations used to solve static problems:



\sum{F_x=0} $$



\sum{F_y=0} $$



\sum{M_z=0} $$

Summing the moments about point A is trivial as all members pass through the point. This action will give us no new information about the system. Therefore, there are 2 equations and 4 unknown forces. This problem is statically indeterminate.

Further exploration of Element Coordinate Systems






The three diagrams above describe the element coordinate systems for the Three Bar Truss System.

Moment About Point B
If the sum of the moments are computed about point B, will the sum still be zero?




 * $$\overrightarrow{BA'}=\overrightarrow{BA} + \overrightarrow{AA'}$$

3D- Explanation:


 * $$\sum \overrightarrow{M_B}={\overrightarrow{BA}} \times {\overrightarrow{F}} ={\overrightarrow{BA'}} \times {\overrightarrow{F}}$$

For all A' on line of action of the vector F.


 * $$\sum \overrightarrow{M_B}=({\overrightarrow{BA}} + {\overrightarrow{AA'}}) \times {\overrightarrow{F}}$$
 * $$\sum \overrightarrow{M_B}={\overrightarrow{BA}} \times {\overrightarrow{F}} + {\overrightarrow{AA'}} \times {\overrightarrow{F}} $$

Back to Three Bar Truss
Point A is in equilibrium: $$ \sum_{i=0}^{3}{\overrightarrow {Fi}}=\overrightarrow O $$

Note that the above equation is the same as the two scalar equations summing forces in the X and Y directions.


 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA'_i} \times \overrightarrow{F_i} $$


 * $$A'_{i}=$$ any point on line of action of $$\overrightarrow{F_i}$$


 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA} \times \overrightarrow{F_i} = \overrightarrow {BA}\times \sum_{i}\overrightarrow {F_i}$$

Since $$\sum_{i}\overrightarrow {F_i}=0$$, the sum of the moments about point B is zero.


 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA} \times \overrightarrow{O} $$

The Three Bar Truss problem has 8 DOF (8 x 8 Matrix):



$$k_3$$ is in total a 4 x 4 matrix.

$$k_{33}=k^{(1)}_{33}+k^{(2)}_{11}+k^{(3)}_{11}$$

$$k_{34}=k^{(1)}_{34}+k^{(2)}_{12}+k^{(3)}_{12}$$

$$k^{(1)}_{43} + k^{(2)}_{21} + k^{(3)}_{21}$$

$$k^{(1)}_{44} + k^{(2)}_{22} + k^{(3)}_{22}$$

$$

K_{global}= \left [\begin{array}{cccccccc} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} & 0 & 0 & 0 & 0 \\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} & 0 & 0 & 0 & 0\\ k^{(1)}_{31} & k^{(1)}_{32} & (k^{(1)}_{33} + k^{(2)}_{11} + k^{(3)}_{11})& (k^{(1)}_{34} + k^{(2)}_{12}+k^{(3)}_{12})& k^{(2)}_{13} & k^{(2)}_{14} & k^{(3)}_{13} & k^{(3)}_{14} \\ k^{(1)}_{41} & k^{(1)}_{42} & (k^{(1)}_{43} + k^{(2)}_{21} + k^{(3)}_{21}) & (k^{(1)}_{44} + k^{(2)}_{22} + k^{(3)}_{22}) & k^{(2)}_{23}) & k^{(2)}_{24} & k^{(3)}_{23} & k^{(3)}_{24} \\ 0 & 0 & k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34} & 0 & 0 \\ 0 & 0 & k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} & 0 & 0 \\ 0 & 0 & k^{(3)}_{31} & k^{(3)}_{32} & 0 & 0 & k^{(3)}_{33} & k^{(3)}_{34} \\ 0 & 0 & k^{(3)}_{41} & k^{(3)}_{42} & 0 & 0 & k^{(3)}_{43} & k^{(3)}_{44}\end{array} \right] $$