User:Eas4200c.f08.vqcrew.d/hw7

Solving P2 - Euler Cut Principle
The Euler Cut Principle involves finding the equilibrium of each stringer. This is basically a statics problem where the sum of the forces, which are given through the shear flows, are equal to zero. Stringer 3 is shown in Figure # in two dimensions and again in three dimensions in Figure #, with the stringer expanded into the x-direction. The sum of the forces in the x-direction can be expressed as follows:


 * $$\sum{F_x}=0=\int_{A_3}^{}{\left[\sigma _{xx}(x+dx) -\sigma _{xx}(x)\right]}dA_3=\int_{A_3}^{}{\left[\frac{d\sigma _{xx}}{dx}dx + HOT\right]}dA_3=\left[ \tilde{q_{31}}-\tilde{q_{23}}-\tilde{q_{43}}\right]dx$$

Thus we solve for $$\tilde{q_{31}}$$ as:


 * $$\tilde{q_{31}}=\tilde{q_{23}}+\tilde{q_{43}}+q^{(3)}$$

Where $$q^{(3)} \;$$ is the contribution to shear flow by Stringer 3:


 * $$q^{(3)}=-\int_{A_3}^{}{\frac{d\sigma _{xx}}{dx}}dA_3$$

The integral can be solved for $$q^{(3)} \;$$ as:


 * $$q^{(3)}=-\left(k_yV_y-k_{yz}V_z \right)Q_z^{(3)}-\left(k_zV_z-k_{yz}V_y \right)Q_y^{(3)}$$

where:


 * $$V_y=\frac{dM_z}{dx}$$
 * $$V_z=\frac{dM_y}{dx}$$
 * $$Q_z^{(3)}=\int_{A_3}^{}{ydA_3}$$
 * $$Q_y^{(3)}=\int_{A_3}^{}{zdA_3}$$

The same procedure can be used to solve for Stringer 2:


 * $$\tilde{q_{23}}= \underbrace{\tilde{q_{12}}}_{flow in}-\underbrace{\tilde{q_{24}}}_{flow out}+q^{(2)}$$

The value of $$q^{(2)} \;$$ can be computed as done with Stringer 3 with the values of $$Q_z^{(2)}$$ and $$Q_y^{(2)}$$ as:


 * $$Q_z^{(2)}=y_2A_2$$
 * $$Q_y^{(2)}=z_2A_2$$

where $$y_2 \;$$ and $$z_2 \;$$ are the y and z coordinates of Stringer 2, and $$A_2 \;$$ is the area of Stringer 2.

The same procedure can be followed on Stringer 4 to find $$q^{(4)} \;$$ and $$\tilde{q_{24}}$$. Hence, we can also deduce $$\tilde{q_{31}}$$.

Using the principles of superposition, the values of q can be determined.


 * $$q_{ij}=\tilde {q}_{ij} + q_{k}$$

On the right side of the equations, the tilde terms are the knowns and the non-tilde terms are unknowns.


 * $$q_{12}=\tilde {q}_{12} + q_{1}$$
 * $$q_{23}=\tilde {q}_{23} + q_{1}$$
 * $$q_{31}=\tilde {q}_{31} + q_{1}$$
 * $$q_{24}=\tilde {q}_{24} + q_{2}$$
 * $$q_{43}=\tilde {q}_{43} + q_{2} - q_{3}$$
 * $$q_{41}=\tilde {q}_{41} + q_{3}$$

These equations reveal that there are three unknowns, $$q_1, q_2, q_3 \;$$, therefore three equations are needed to solve for these values to complete the problem. The three equations to be use are: 1) Moment equation: Take the moment of $$V_y \;$$, $$V_z \;$$, and $$\left\{q_{12},...,q_{41} \right\}$$ about any convenient point (usually where the lines of action of $$V_y \;$$ and $$V_z \;$$ intersect). 2) $$\theta _1=\theta _2 \;$$ 3) $$\theta _2=\theta _3 \;$$ Equations 2 and 3 are the compatibility equations.

To recap, in order to solve P2 for each cell you must follow path "$$s_i \;$$" and then determine the equilibrium of each stringer on the path. There are two ways to accomplish this: 1) Complete method using FBD as shown above 2) A consequence of the first method, as exemplified in Figure #:


 * $$\tilde{q_{j6}} = \tilde{q_{j2}} - \tilde{q_{j5}} - \tilde{q_{j8}} + q^{(j)}$$

Also, the left side of the equation simply becomes negative if the direction of $$\tilde{q_{j6}} \;$$ changes (and becomes $$\tilde{q_{6j}}$$), as shown in Figure #.


 * $$\tilde{q_{6j}} = \tilde{q_{j2}} - \tilde{q_{j5}} - \tilde{q_{j8}} + q^{(j)}$$

Using the Euler Cut Principle brings forth a question, what happens if the object is cut such that one stringer is isolated as shown in Figure #? Due to the cuts made, $$\tilde{q_{23}}=\tilde{q_{31}}=\tilde{q_{34}}=0$$. We then sum the forces acting on the FBD shown in Figure #.


 * $$\underbrace{\tilde{q_{31}}}_{0} = \underbrace{\tilde{q_{23}}}_{0} - \underbrace{\tilde{q_{34}}}_{0} + q^{(3)}$$

This shows that $$q^{(3)}=0 \;$$, which is not possible. Therefore, when using the Euler Cut method a stringer must not be isolated from the rest of the body.