User:Eas4200c.f08.WIKI.E/hw7 other days

Mtg 39, December 1 2008
Solving P2: For each cell,

-Follow path $$S_i$$

-Find the equilibrium of each stringer on this path: There are two ways of solving

1) Complete method FBD as seen before

2) Consequence of the first method:

$$\tilde{q}_{j6}=\tilde{q}_{2j}-\tilde{q}_{j5}-\tilde{q}_{j8}+\tilde{q}^{j}$$



In this example, $$\tilde{q}_{6j}$$ is now going in the opposite direction as previous but the answer will remain the same as long as the conventions stay consistent.

$$-\tilde{q}_{6j}=\tilde{q}_{2j}-\tilde{q}_{j5}-\tilde{q}_{j8}+\tilde{q}^{j}$$



Note: What if we cut cell walls such that one stringer was isolated?



$$\tilde{q}_{31}=\tilde{q}_{23}-\tilde{q}_{34}+\tilde{q}^{3}$$

This equation says that $$\tilde{q}^{3}= 0$$ which is not true. Therefore you cannot isolate the stinger in this way because it results in an answer that is not valid.

Trying to make different cuts to isolate stringer 2 will result in the same false answer.



$$\tilde{q}_{31}=\tilde{q}_{12}-\tilde{q}_{24}+\tilde{q}^{2}$$

$$\tilde{q}^{2}= 0$$