User:Eas4200c.f08.nine.F/Homework 4

[|Homework 1]

[|Homework 2]

[|Homework 3]

[|Homework 4]

Group nine - Homework 4

Ad-Hoc
"Ad hoc is a Latin phrase which means "for this [purpose]". It generally signifies a solution designed for a specific problem or task, non-generalizable, and which cannot be adapted to other purposes."

"Common examples are organizations, committees, and commissions created at the national or international level for a specific task; in other fields the term may refer, for example, to a tailor-made suit, a handcrafted network protocol, or a purpose-specific equation. Ad hoc can also have connotations of a makeshift solution, inadequate planning, or improvised events. Other derivates of the Latin include AdHoc, adhoc and ad-hoc."

Angle of Twist Derivation
Derive $\theta = \frac{1}{2G\bar{A}}\oint^{} \frac{q}{t} ds$|undefined

Consider we have a uniform non circular cross section subject to twist.

Going through the geometry we can see that,

$\frac{P\dot{P}}{OP} = tan\alpha$|undefined

$P\ddot{P} = P\dot{P}cos\alpha$

$\rightarrow = (OPtan\alpha )cos\alpha$ where   $OPcos\alpha = O\ddot{P}$

Now we define the following:

$OP = r$ and  $O\ddot{P} = \rho$

$P\ddot{P} = rcos\alpha tan\alpha = \rho \alpha$

And now we take a look at the strain.

$\gamma = \frac{P\ddot{P}}{ds} = \frac{\rho \alpha}{ds}  =  \rho \theta  =  \rho \frac{\alpha}{dx}$|undefined

Hooke's Law: $\tau = G\gamma = G\rho \theta$

$\rightarrow \tau(s) = G\rho(s) \theta(x)$

and now integrating along the contour C $\rightarrow$

$\oint_{C} \tau(s)ds = G\theta(x) \oint_{C}^{} \rho(s) ds$

where $\tau(s) = \frac{q(s)}{t(s)}$  and  $\oint_{C} \rho(s) ds = 2\bar{A}$

Integrating and rearranging terms yields the final result$rightarrow$

$\theta = \frac{1}{2G\bar{A}}\oint^{} \frac{q}{t} ds$|undefined

Question

What is ad hoc about the derivation of the angle of twist shown above?

Answer

1) The strain must be obtained using the displacement of $P$  in the direction that is tangent to the contour curve at $P\ddot{P}$

2) It was assumed that the shear force was uniform across the wall the thickness.

3) To get line $P\dot{P}$ assume  $\alpha$  is small; to get  $P\ddot{P}$  assume  $ \alpha $  is finite and  $\rho = rcos\alpha$  ; and then reintroduced small  $\alpha$  after that.

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Hooke's law: (Ricardo Albuquerque)

$\tau = G \gamma$

Normal strains:

$$ \epsilon_{xx} = \frac{\sigma_{xx}}{E} - \frac{\nu \sigma_{yy}}{E} - \frac{\nu \sigma_{zz}}{E} $$

$$ \epsilon_{yy} = -\frac{\nu \sigma_{xx}}{E} + \frac{\sigma_{yy}}{E} - \frac{\nu \sigma_{zz}}{E} $$

$$ \epsilon_{zz} = -\frac{\nu \sigma_{xx}}{E} - \frac{\nu \sigma_{yy}}{E} + \frac{\sigma_{zz}}{E} $$

Shear strains:

$$ \gamma_{xy}= 2 \epsilon_{xy} = \frac{\tau_{xy}}{G} $$

$$ \gamma_{yz}= 2 \epsilon_{yz} = \frac{\tau_{yz}}{G} $$

$$ \gamma_{zx}= 2 \epsilon_{zx} = \frac{\tau_{zx}}{G} $$

Strain tensor:

$$ \epsilon = \begin{bmatrix} \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix} $$

Note: It is important to note here that the above yields six independent strain terms because $\epsilon_{ij} = \epsilon_{ji}$.

Hooke's law (isotropic material):

$$ \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix}~= $$                       $$ \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{2G}&0&0\\ 0&0&0&0&\frac{1}{2G}&0\\ 0&0&0&0&0&\frac{1}{2G}\\ \end{bmatrix}~ $$ $$ \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix} $$

or equivaletly:

$$ \begin{bmatrix} \epsilon_{xx}\\ \epsilon_{yy}\\ \epsilon_{zz}\\ \gamma_{yz}\\ \gamma_{zx}\\ \gamma_{xy}\\ \end{bmatrix}~= $$                        $$ \begin{bmatrix} \frac{1}{E}&\frac{-\nu}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{1}{E}&\frac{-\nu}{E}&0&0&0\\ \frac{-\nu}{E}&\frac{-\nu}{E}&\frac{1}{E}&0&0&0\\ 0&0&0&\frac{1}{G}&0&0\\ 0&0&0&0&\frac{1}{G}&0\\ 0&0&0&0&0&\frac{1}{G}\\ \end{bmatrix} $$ $$ \begin{bmatrix} \sigma_{xx}\\ \sigma_{yy}\\ \sigma_{zz}\\ \tau_{yz}\\ \tau_{zx}\\ \tau_{xy}\\ \end{bmatrix} $$

Question

What material has a Poisson's ration of 0?

Answer

Cork

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Sample Run of Code
NACA Airfoil calculation program Enter first digit of airfoil: 2 Enter second digit: 4 Enter the third and fourth digits: 15 Enter Py: 0 Enter Pz: 0 Enter number of segments: 60

The average area is: 0.103 The minumum number segments required to have the average area accurate within 1 percent is: 24.000

Figure 1 shows the cross-section of the NACA airfoil and Figure 2 shows the centroid line

Matlab Code Certification
I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.

Ricardo Albuquerque Eas4200c.f08.nine.r 22:46, 21 October 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 3:46, 22 October 2008 (UTC)

David Phillips Eas4200C.f08.nine.d (talk) 18:34, 23 October 2008 (UTC)

Oliver Watmough Eas4200c.f08.nine.o 10:07, 23 October 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 11:48, 23 October 2008 (UTC)

Contributing Team Members
The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d (talk) 18:34, 23 October 2008 (UTC)

Oliver Watmough Eas4200c.f08.nine.o 10:07, 23 October 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 11:48, 23 October 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 23 October 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 4:34, 23 October 2008 (UTC)