User:Eas4200c.f08.vqcrew.E/Homework7

Mini Plan
Single-Cell Sections
 * - without stringers
 * - with stringers

Multi-Cell Sections
 * - without stringers
 * - with stringers

Without Stringers
Question: Can the setup shown in Figure # resist $$\displaystyle V_{z}$$?

Answer: No.


 * $$R^{z} = R_{AB}^{z} + R_{BA}^{z}$$

$$\displaystyle R^{z}$$ is the total resultant in the z-direction

$$R_{AB}^{z}$$ is the reslutant of q in AB

$$R_{BA}^{z}$$ is the resultant of q in BA


 * $$R_{AB}^{z} = -q\bar{A^{'}B^{'}} = R_{BA}^{z}$$


 * so, $$\displaystyle R^{z} = 0$$

Therefore, the configuration can not resist any $$\displaystyle V_{z}$$.

With Stringers


Figure # shows that q(s) is piecewise constant with respect to the curvilinear coordinate, s. (i.e. $$\displaystyle q_{ij}$$ = constant within each panel)

However, the value for q is not the same throughout all of the panels.


 * $$q_{12} \neq q_{23} \neq q_{31}$$

The presence of the stringers causes a non-constant q.


 * $$R^{z} = V_{z} \neq 0$$

Superposition can be used to solve for the values of $$\displaystyle q_{ij}$$ in each section by solving for values of q in the case without stringers, Figure #, and for the case including an imaginary cut, Figure #, in one of the sections. This is a valid use of superposition due to the linearity of the relationship.


 * $$q_{12} = q + \tilde{q}_{12}$$
 * $$q_{23} = q + \tilde{q}_{23}$$
 * $$q_{31} = q + \tilde{q}_{31}$$

Analysis Algorithm
Objectives: One unknown q ($$\displaystyle \tilde{q_{ij}}$$ are known after solving P1), need to solve one equation for one unknown.

Method: Knowing the values for $$\displaystyle V_{y}$$ and $$\displaystyle V_{z}$$


 * 1) Solve P2 for $$\displaystyle \tilde{q_{12}}$$ and $$\tilde{q_{23}}$$


 * 2) Moment Equation - take the moment about any point in the plane (y,z)


 * a) Superposition: $$\displaystyle q_{ij} = q + \tilde{q_{ij}}$$


 * b) Select point $$\displaystyle \bar{O}$$ in the plane (y,z)

Without Stringers


The same type of analysis can be performed for multi-cell sections as for single cell sections. Figure # shows the multi-cell case without stringers. The resultant of all of the shear flows in each cell can be calculated by the sum of the resultants in every cell.


 * $$\displaystyle R^{z} = \sum_{i = 1}^{N_{cells}}R_{i}^{z}$$

Similar to the single case, the resultant for each cell can be found, therefore yielding the result for the entire multi-cell section. Once again, this value is equal to zero. Without the support of the stringers, this setup can not resist $$\displaystyle V_{z}$$.


 * $$\displaystyle R_{i}^{z} = 0 \Rightarrow R^{z} = 0 $$

With Stringers


The true shear flow, $$\displaystyle q_{ij}$$, is equal to the sum of the closed cell constant shear flow, q, and the open cell piecewise constant shear flow, $$\displaystyle \tilde{q_{ij}}$$.


 * $$\displaystyle q_{ij} = q + \tilde{q_{ij}}$$

In order to solve for all of the true shear flows, imaginary cuts must be made into some of the walls. This must be done carefully to ensure that no stringers are left unattached to the rest of the assembly. Figure # shows the overall problem including no cuts and all of the stringers. Figure # is displaying the case in which the stringers are removed so that the closed cell shear flow can be calculated. Figure # represents the case in which the cuts have been made and the open cell piecewise constant shear flows can be found. The total resultant of the multi-cell case with stringers can be calculated by the sum of the resultants of P1 and P2.


 * $$\displaystyle R^{z} = R^{z_1} + R^{z_2}$$

Recall: $$\displaystyle R^{z_1} = 0$$

Therefore: $$\displaystyle R^{z} = R^{z_2} = V_{Z} \neq 0$$