User:Eas4200c.f08.nine.s/Lecture 7

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Homework 7

Group nine - Homework 7

Without Stringers
Question: Can this setup resist $V_z$? Answer: No

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

where $R^{z}$ is the total resultant in the z-direction and $R_{AB}^{z}$ is the resultant of q in AB.

$R_{AB}^{z} = -q\bar{A^{'}B^{'}} = - R_{BA}^{z} \Rightarrow R^z = 0$|undefined

With Stringers
We can neglect the contribution of web to bending and we will look at a cell with three stringers.

$q_{12} \neq q_{23} \neq q_{31}$ $q_{ij}(s) = const.$ within each panel. $\Rightarrow q(s)$ is piecewise constant with respect to s.

Analysis Algorithm : 1) Solve $P2$ for $\tilde{q_{12}}, \tilde{q_{23}}, \left(\tilde{q_{31}} = 0\right)$|undefined 2)Take moment about any point in plane (y,z) and use superposition.

Without Stringers
$R^z = \sum_{i=1}^{n cells}{} R^{z}_i$ $\Rightarrow R^z = 0$

With Stringers
$R^z = R^{z1} + R^{z2}$

Stringer Equilibrium
$$\sum F_{x}=0=\int_{A_{3}} [\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3} $$

Where we can exapand these terms into a taylor series:

$$[\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3} \rightarrow

\frac{\partial \sigma_{xx}}{dx}dx+h.o.t \times [-\tilde{q}_{23} -\tilde{q}_{43} +\tilde{q}_{31}]dx $$ Contribution of shear flow by stringer 3 is given by:

$$q^{3}=-\int_{A_{3}} \frac{\partial \sigma_{xx}}{dx} A_{3}$$

Where:

$$V_{y}=\frac{\partial d M_{z}}{dx}$$

And:

$$V_{z}=\frac{\partial d M_{y}}{dx}$$

So we have:

$$ q^{3}=-(KyVy-KyzVz)Qz^{3}-(KzVz-KyzVy)Qy^{3} $$

And:

$$Qz^{3}=\int_{A_{3}} ydA_{3}$$

$$Qy^{3}=\int_{A_{3}} zdA_{3}$$

Similarly for stringer 2:

$$\tilde {q}_{23}=\tilde {q}_{12} - \tilde {q}_{24} + {q}^{2}$$

$$Qy^{2}=\int_{A_{2}} zdA_{2}\,$$

And for stringer 4:

$$\tilde {q}_{43}=\tilde {q}_{24} - \tilde {q}_{41} + {q}^{4}$$

$$Qy^{4}=\int_{A_{4}} zdA_{4}$$

______________________________________________  Using superposition principle:

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

First term is known, second term is unknown:

$$q_{12}=\tilde {q}_{12} + q_{1}$$

$$q_{23}=\tilde {q}_{23} + q_{1}$$

$$q_{31}=\tilde {q}_{31} + q_{1}$$

$$q_{24}=\tilde {q}_{24} + q_{2}$$

$$q_{43}=\tilde {q}_{43} + q_{4}$$

$$q_{41}=\tilde {q}_{41} + q_{1}$$

3 unknowns:

$$q_{1}, q_{2}, q_{3}$$

3 equations:

1. Moment equations of $$V_{y} V_{z} $$ and $$[q_{12}, ... q_{41}]$$ about any convenient point, usually where lines of action of $$V_{y} V_{z}\,$$ intersect. 2. $$\theta_{1}=\theta_{2}$$ 3. $$\theta_{2}=\theta_{3}$$

Plotting Buckling Shape under Shear Stress
$$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ in terms of $$C_{11}$$.

1. Find $$\lambda$$ for $$\vartheta = 1.5$$

Using equation (30) found on website:

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

Plugging known values: $$\displaystyle \lambda = .02875$$

2. Evaluate numerically $$k_{sys}$$ using equation (26) found on website:

Equation 26:

$$\displaystyle\left[\begin{array}{lllll}\frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2}&\frac{4}{9}& 0 & 0 & 0\\\frac{4}{9}&\frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} & - \frac{4}{5} & - \frac{4}{5} & \frac{36}{25}\\0 & - \frac{4}{5} & \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2}& 0 & 0 \\ 0 & - \frac{4}{5} & 0 & \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} & 0 \\ 0 & \frac{36}{25} & 0 & 0 & \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2}\end{array}\right]\left\{\begin{array}{l}C_{11}\\ C_{22}\\ C_{13}\\C_{31}\\C_{33}\end{array}\right\} = \left\{\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right\}$$

3. Calculate: $$[k_{ij}]$$

$$\begin{bmatrix} k_{12} & k_{23} & \cdots & k_{25} \\ \vdots & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\k_{32} & \cdots & \cdots & k_{55}\end{bmatrix} \begin{bmatrix} C_{22} \\ \vdots \\ \vdots \\ C_{33}\end{bmatrix} = \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Solving for: $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ intersect with $$C_{11}$$.

Since the $$K$$ matrix can be written as: $$\underline{\overline{K}}$$

Then its inverse is: $$\underline{\overline{K}}^{-1}$$

$$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix} = \underline{\overline{K}}^{-1}\ \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Finally:

$$u_z = C_{11}sin(\frac{\pi x}{a})sin(\frac{\pi y}{b}) + C_{22}sin(\frac{2\pi x}{a})sin(\frac{2\pi y}{b}) + C_{13}sin(\frac{\pi x}{a}) sin(\frac{3\pi y}{b}) + C_{31}sin(\frac{3\pi x}{a})sin(\frac{\pi y}{b}) + C_{33}sin(\frac{3\pi x}{a})sin(\frac{3\pi y}{b})$$

Bidirectional Bending
Under bidirectional bending, the longitudinal displacement expansions are given by the following equations.

$u = u_0(x) + z\phi_y(x) + y\phi_z(x)$ $v = v_0(x)$ $w = w_0(x)$

$\phi_y$ and  $\phi_z$ are the rotations of the cross section about the y and z axes respectively. The corresponding strains are given below.

$\epsilon_{xx} = \frac{\delta u}{\delta x} = \frac{du_0}{dx} + z\frac{d\phi_y}{dx} + y\frac{d\phi_z}{dx}$ $\gamma_{xy} = \frac{\delta v}{\delta x} + \frac{\delta u}{\delta y} = \frac{dv_0}{dx} + \phi_z$ $\gamma_{xy} = \frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} = \frac{dw_0}{dx} + \phi_y$

Assuming $\gamma_{xy} = \gamma{xz} =0 $ we yield the following relations,

$\phi_z = -\frac{dv_0}{dx}$ $\phi_y = -\frac{dw_0}{dx}$

and the we obtain the following by substitution.

$\epsilon_{xx} = \frac{du_0}{dx} - y\frac{d^{2}v_0}{dx^{2}} - z\frac{d^{2}w_0}{dx^{2}}$|undefined

Noting that $du_0/dx = 0$ if $N_x = 0$ and thus, the bending strain is reduced to,

$\epsilon_{xx} = - y\frac{d^{2}v_0}{dx^{2}} - z\frac{d^{2}w_0}{dx^{2}}$|undefined

The bending moments about the y and z axes are as follows,

$M_y = \int{}\int_{A}z\sigma_{xx}dA = -EI_{yx}\frac{d^{2}v_0}{dx^{2}} - EI_y\frac{d^{2}w_0}{dx^{2}}$|undefined $M_z = \int{}\int_{A}y\sigma_{xx}dA = -EI_{z}\frac{d^{2}v_0}{dx^{2}} - EI_yz\frac{d^{2}w_0}{dx^{2}}$|undefined

where

$I_y = \int\int_A z^{2}dA$ $I_z = \int\int_A z^{2}dA$ $I_{yz} = \int\int_A yzdA$

Comparing MediaWiki to E-learning
After using MediaWiki for the entire semester, it is this groups opinion that MediaWiki is better than E-learning. With MediaWiki, it is possible to work with your group without ever having to meet with your group. It is very easy to access, and can be used by others for future reference. MediaWiki also has a history that saves and keeps track of all changes made to its page. The history can be used to undo any unwanted changes by simply selecting a saved version of the article in the history that is prior to the change. E-learning does not allow any manipulation outside of posting text. There is no way to do any coding. It does however allow a teacher to easily post grades and course work.

Contributing Team Members
The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d 7:40, 9 December 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 7:45, 9 December 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 6:55, 9 December October 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 9:45, 9 December 2008 (UTC)