User:Eml4500.f08.team.figuerrez/hw3

Question: How about ΣM B ?

$$ BA' = BA + AA' $$

$$ M_{B} = (BA + AA') x F $$

$$ M_{B} = BA x F + AA' x F $$

$$ (AA'x F) $$ is equal to zero.

Therefore the moment at point B is equal to $$ M_{B} = BA x F $$

=3-bar truss system=

Node A is in equilibrium:

$$\sum F_{i} = 0$$

$$\sum M_{B,i} = \sum BA'_{i} * F_{i}$$

$$A'_{i}$$ equals to any point on line of action of $$F_{i}$$ therefore  $$\sum M_{B,i} = \sum BA_{i} * F_{i}$$ $$\sum M_{B,i} = BA * \sum F_{i} = 0$$

=Stiffness Matrix=

The corresponding stiffness matrix of the 3-bar truss system is:

$$ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & - & - & - & -\\

K_{21}^{(1)} & K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & - & - & - & -\\

K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)} + K_{11}^{(3)}) & (K_{34}^{(1)} + K_{12}^{(2)} + K_{12}^{(3)}) & K_{13}^{(2)} & K_{14}^{(2)} & K_{13}^{(3)} & K_{14}^{(3)}\\

K_{41}^{(1)} & K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)} + K_{21}^{(3)}) & (K_{44}^{(1)} + K_{22}^{(2)} + K_{22}^{(3)}) & K_{23}^{(2)} & K_{24}^{(2)} & K_{23}^{(3)} & K_{24}^{(3)}\\

- & - & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} & - & -\\

- & - & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)} & - & -\\

- & - & K_{31}^{(3)} & K_{32}^{(3)} & - & - & - & -\\

- & - & K_{41}^{(3)} & K_{42}^{(3)} & - & - & - & -\\

\end{bmatrix} $$