User:Eas4200c.f08.wiki.b/HW7

=Mtg 40 & 41 - Friday 12/5, Monday 12/8=

Cont'd from Mtg 39
Similarly, developing an equilibrium equation for stringer one yields

 $$\displaystyle \tilde q_{12} = \tilde q_{31} + \tilde q_{41} + q^{(1)} $$

where $$\displaystyle \tilde q_{12} $$ and $$\displaystyle \tilde q_{41} $$ are unknown. $$\displaystyle \tilde q_{31} $$ becomes zero due to the cut between stringers three and one. Doing the same for stringer two yields

 $$\displaystyle \tilde q_{24} = \tilde q_{12} - \tilde q_{23} + q^{(2)} $$

In this case, $$\displaystyle \tilde q_{23} $$ becomes zero due to the cut between stringers two and three. Again, developing the equation for stringer four yields

 $$\displaystyle \tilde q_{41} = \tilde q_{24} + \tilde q_{34} + q^{(4)} $$

Substituting the equations for stringers one and two into this equation gives us

 $$\displaystyle \tilde q_{41} = ( \tilde q_{41} + q^{(1)} ) + q^{(2)} + \tilde q_{34} + q^{(4)} $$

Due to the cut between stringers three and four, $$\displaystyle \tilde q_{34} $$ becomes zero. In addition, $$\displaystyle \tilde q_{41} $$ cancels out, leaving

 $$\displaystyle 0 = q^{(1)} + q^{(2)} + q^{(4)} = -q^{(3)} $$

This, of course is not true, or even possible. The following steps explain why this is not true.

 $$\displaystyle 0 = q^{(1)} + q^{(2)} + q^{(3)} + q^{(4)} $$

 $$\displaystyle 0 = \sum_{e=1}^4 q^{(e)} $$

Recall,

 $$\displaystyle q^{(e)} = n_z Q_z^{(e)} + n_y Q_y^{(e)} $$

 $$\displaystyle n_z := -(K_y V_y - K_yz V_z) $$

 $$\displaystyle n_y := -(K_z V_z - K_yz V_y) $$

Thus,

 $$\displaystyle \sum_{e=1}^4 q^{(e)} = n_z \sum_{e=1}^4 Q_z^{(e)} + n_y \sum_{e=1}^4 Q_y^{(e)} $$

Where $$\displaystyle \sum_{e=1}^4 Q_z^{(e)} $$ and $$\displaystyle \sum_{e=1}^4 Q_y^{(e)} $$ must both be equal to zero.



$$\displaystyle Q_y $$ can be expressed as

 $$\displaystyle Q_y (\hat z) = \int_{A(\hat z)} z dA $$

 $$\displaystyle Q_y = \int_A z dA = 0 = z_c \int_A dA $$

Notice that the last part of the above equation is from the mean value theorem. $$\displaystyle z_c $$ would be zero in this case, which is not possible.

Steps for Shear Buckling
The steps for plotting the buckling shape and expressing the coefficients $$\displaystyle \{ C_{11}, C_{22}, C_{13}, C_{31}, C_{33} \} $$ in terms of $$\displaystyle C_{11} $$ for the aspect ratio $$\displaystyle \vartheta = 1.5 $$ are as follows.

1) Find $$\displaystyle \lambda $$ for $$\displaystyle \vartheta = 1.5 $$ using equation 30 from Professor Vu-Quoc's page on plate buckling.

 $$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1 	 + 	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$

2) Evaluate $$\displaystyle \mathbf K _{5 \times 5} $$ numerically using equation 26 from Professor Vu-Quoc's page on plate buckling.

 $$  \displaystyle \mathbf \bar K _{5 \times 5} = \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right]

$$

3) Solve $$\displaystyle \{ C_{11}, C_{22}, C_{13}, C_{31}, C_{33} \} $$ in terms of $$\displaystyle C_{11} $$.

 $$  \displaystyle \left[ \begin{array}{lllll} K_{22} &	 K_{23} &	 K_{24} &	 K_{25} \\	 K_{32} &	 K_{33} &	 K_{34} &	 K_{35} \\	 K_{42} &	 K_{43} &	 K_{44} &	 K_{45} \\	 K_{52} &	 K_{53} &	 K_{54} &	 K_{55} \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} -{4 \over 9} C_{11} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

<p style="text-align:center;"> $$ \displaystyle u_z = C_{11} \sin{\pi x \over a} \sin{\pi y \over b} + C_{22} \sin{2 \pi x \over a} \sin{2 \pi y \over b} + C_{13} \sin{\pi x \over a} \sin{3 \pi y \over b} + C_{31} \sin{3 \pi x \over a} \sin{\pi y \over b} + C_{33} \sin{3 \pi x \over a} \sin{3 \pi y \over b} $$

Setting $$\displaystyle C_{11} = 1 $$, $$\displaystyle u_z(x,y) $$ can be plotted.

Proof of Stress Equation
[READ AND REPORT SEC. 4.2 (eqn 4.22a to eqn 4.28c)]

The following is a proof for the stress equation:

<p style="text-align:center;"> $$\sigma _{xx}=\begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{bmatrix}\begin{Bmatrix} M_y\\M_z \end{Bmatrix}$$

First, the moments can be written as

<p style="text-align:center;"> $$\displaystyle M_y = - EI_{yz} {d^2 V_o \over dx^2} - EI_y {d^2 W_o \over dx^2} $$

<p style="text-align:center;"> $$\displaystyle M_z = - EI_{z} {d^2 V_o \over dx^2} - EI_{yz} {d^2 W_o \over dx^2} $$

Let the curvatures $$\displaystyle {d^2 V_o \over dx^2} = \chi_y $$ and $$\displaystyle {d^2 W_o \over dx^2} = \chi_z $$

In matrix form, whe have

<p style="text-align:center;"> $$\displaystyle \begin{Bmatrix} M_y\\M_z \end{Bmatrix} = \begin{bmatrix} I_y & I_{yz}\\ I_{yz} & I_z \end{bmatrix}\begin{Bmatrix} -E \chi_z \\ -E \chi_y \end{Bmatrix} $$

<p style="text-align:center;"> $$\displaystyle \varepsilon_{xx} = -y \chi_y - z \chi_z = \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} \chi_z \\ \chi_y \end{Bmatrix} $$

Thus,

<p style="text-align:center;"> $$\displaystyle \sigma_{xx} = E \varepsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_z \\ -\chi_y \end{Bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} {1 \over E} \textbf{I}^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

Notice that $$\displaystyle E $$ cancels out. Also note that

<p style="text-align:center;"> $$\displaystyle \textbf{I}^{-1} = {1 \over D} \begin{bmatrix} I_z & -I_{yz}\\ -I_{yz} & I_y \end{bmatrix} $$

<p style="text-align:center;"> $$\displaystyle D= I_y I_z - (I_{yz})^2 $$

Recall the expression for the shear flow

<p style="text-align:center;"> $$\displaystyle q = -\int_A {d \sigma_{xx} \over dx} dA $$

Thus,

<p style="text-align:center;"> $$\displaystyle {d \sigma_{xx} \over dx} = \begin{bmatrix} z & y \end{bmatrix} \textbf{I}^{-1} \begin{Bmatrix} {dM_y \over dx} \\ {dM_z \over dx} \end{Bmatrix} $$

<p style="text-align:center;"> $$\displaystyle V_y = {dM_y \over dx} $$

<p style="text-align:center;"> $$\displaystyle V_z = {dM_z \over dx} $$

<p style="text-align:center;"> $$\displaystyle Q_y = \int_A zdA $$

<p style="text-align:center;"> $$\displaystyle Q_z = \int_A ydA $$