User:Eas4200c.f08.blue.a/Homework4

Solving for Areas using the Method of Quadrature


Cubature - a method used for integrating volumes by breaking up the object into cubes and adding the individual cubes to find the total volume. To make the method more accurate, use smaller cubes to better approximate the volume.

Squaring (quadrature) of the circle - wikipedia.



Algebraic sums of the areas as the airfoil is swept counterclockwise is the NACA airfoil area.



A = A1 + A2 = A1 - |A2|

NOTE: When writing the code plotting the NACA arifoil, we do not need if statements to calculate the area if the arifoil. As long as the cross product is taken between the 2 sweeping vectors, the signs will add up correctly to give the appropriate area.



How about using trapezoids to integrate??



The method of trapezoids is easier to integrate over the surface and gives a more precise answer than Riemann Sums however it has a distinct disadvantage over the method of quadrature. The disadvantage being, changes in curvature are not accurately taken into account. For example, a segment with a lot of curvature is not approximated very well with a straight line. By using the method of quadrature, as long as there are a large number of segments, the area can be approximated better. Not to mention, THE TRAPEZOID METHOD IS NOT AS ELEGANT AS THE METHOD OF QUADRATURE (TRIANGLES)!!!!!!!

Returning to the single cell airfoil:

Shear flow along the airfoil is constant: q = q1 = q2 = q3

The angular rate of twist is:

$$ \theta =\frac{1}{2GA}q\sum_{j=1}^3 \frac{l}{t} $$

$$ \theta =\frac{1}{2GA}q [\frac{\pi b}{t1} + \frac{a}{t2} + \frac{\sqrt{a^2 + b^2}}{t3}] $$

$$ \theta =(factor)q $$

Homework: Computing the factor for the rate of twist.

Max shear stress = tau_max

If tau_max = tau_all (which is given), and since q = T/2A then,

Tall = 2Atauall[min{t1,t2,t3}]

Homework: If tau_all = 100 Gpa, fint T_all.

Quadrature allows an observer outside of a NACA airfoil to find the cross sectional area. An example of this is shown below in Figure 1.

Figure 1

$$\bar{A}=\bar{A_{1}}+\bar{A_{2}} =\bar{A_{1}}\left|\bar{A_{2}} \right|$$



In this close up, $$d\vec{A}$$ is defined as:

$$d\vec{A}=\frac{1}{2}\vec{r}+\vec{PQ}=-\left|dA \right|\vec{i}$$

Dividing the area into trapezoids and adding them is another method used to find the area under a curve called the (Trapezium rule), but it is not as elegant or effective as triangular quadrature in more complicated applications such as finding the area of the oddly shaped airfoil in Figure 2 below.

Figure 2

Single Cell Airfoil Calculations - Twist Angle
Returning to the idealized single cell airfoil example from above, where the shear flow is constant, the twist angle can be found using the formula found in step J of the road-map for torsional analysis of an aircraft wing.



$$\Theta =\frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}$$

Because $$\left(q \right)$$ is constant,

$$\left(q \right)=\left( q_{1}\right)=\left( q_{2}\right)=\left( q_{3}\right)$$

$$\Theta =\frac{1}{2G\bar{A}}q\sum_{j=1}^{3}{\frac{l_{j}}{t_{j}}}$$ $$=\frac{1}{2G\bar{A}}q\left[\frac{2\pi \left(\frac{b}{2} \right)}{t_{1}}+\frac{a}{t_{2}}+\frac{\sqrt{a^{2}+b^{2}}}{t_{3}} \right]=q\left(\frac{1}{2\left(5.57m^{2} \right)G} \right)\left(\frac{\pi \left( 1m\right)}{0.008m}+\frac{4m}{0.01m} +\frac{\sqrt{16m^{2}+4m^{2}}}{0.01m} \right)=\frac{q}{\left( 11.14m^{2}\right)G}\left(1239.91 \right)=111.24m^{-2}\frac{q}{G}$$

The Max Shear Stress $$\left( \tau _{max}\right)$$ can also be found now using the formula:

$$ \tau _{max}=\frac{q}{min\left\{t_{1},t_{2},t_{3} \right\}}$$

If the maximum shear stress is equal to the allowable shear stress and from the formula to find shear flow, the allowable torque $$\left[ T_{allow}\right]$$ can be found.

$$\tau _{max}=\left( \tau _{allow}\right)$$

$$q=\frac{T}{2\bar{A}}$$

$$T_{allow}=2\bar{A}\tau _{allow}\left(min\left\{ t_{1},t_{2},t_{3}\right\} \right)$$

Therefore, if $$\tau _{allow}=100GPa$$, and because $$t_{1}=0.008m$$, $$t_{2}=t_{3}=0.01m$$

$$T_{allow}=2\bar{A}\tau _{allow}t_{1}=2\left(5.57m^{2} \right)\left(100GPa \right)\left(0.008m \right)=8.91\left(10^{9} \right)Nm$$

Example of an Ad Hoc method of proving the relation:

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

Uniform bar with non Circular Cross section subjectred to twist:



Here the displacement of PP' due to $$\alpha$$ is:

$$ \frac{PP'}{OP}=tan\alpha \cong \alpha$$

making a small angle assumption for $$\alpha$$. Now project the displacement orthogonal to OP'.

$$PP''=PP'cos\alpha $$

$$PP''=(OPcos\alpha )tan\alpha $$

$$PP=OPtan\alpha$$

Recall: OP = r and OP'' = $$\rho$$

Therefore:

$$PP''=(rcos\alpha)tan\alpha$$

Where PP" is the displacement of P in the direction of the tangent to the surface of the bar, $$rcos\alpha$$ is $$\rho$$ and $$tan\alpha$$ is approximated to be $$\alpha$$ with a small angle assumption.

Strain can be defined as:

$$\gamma =\frac{PP''}{dx}=\frac{\rho\alpha }{dx}=\rho \theta $$

with

$$d\theta =\frac{\alpha }{dx}$$ as the rate of twist.

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

In this case: $$\tau (s)=G\rho (s)\theta (s)$$

and integration along the contour yields:

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

Where $$\tau (s)=\frac{q(s)}{t(s)}$$ and$$\oint_{}^{}{\rho (s)}=2\bar{A}$$

Discussion

Why is this method ad hoc? *to get PP' we assume $$\alpha$$ is small *to get PP" we assume finite $$\alpha$$
 * 1) PP' is projected tangent to the surface but, it is not necessarally on the surface (it is only close)
 * 2) $$\tau$$ is assumed to be uniform across the wall thickness
 * 3) Inconsistancy in assuming the size of $$\alpha$$

Formal Justification of Elasticity Theory
$$u_{x}(y,z)=\theta \psi (y,z)$$

$$u_{y}(x,z)=-\theta \psi (x,z)$$

$$u_{z}(x,y)=\theta \psi (x,y)$$

We assume, given an applied torque abouyt the x-axis that:

$$\varepsilon _{xx}=\varepsilon _{yy}=\varepsilon_{zz} =\gamma _{yz}=0$$

Where each strain is computed as:

$$\varepsilon _{xx}=\frac{du_{x}}{dx}(y,z)$$

$$\varepsilon _{yy}=\frac{du_{y}}{dy}(x,z)$$

$$\varepsilon _{zz}=\frac{du_{z}}{dz}(x,y)$$

$$\gamma _{yz}=\frac{du_{y}}{dz}+\frac{du_{z}}{dy}$$

As a continuation, the 3rd Ad Hoc point in the engineering derivation is the inconsistency in the assumption on the size of angle $$\alpha$$ : To get line PP', assume $$\alpha$$ is finite and $$\rho = r\cos \alpha$$, then reintroduce the small angle assumption afterwards.

Returning back to the formal derivation:

$$\epsilon = \begin{bmatrix} \epsilon_{xx} & \epsilon_ {xy}&\epsilon_{xz} \\ \epsilon_{yx} & \epsilon_ {yy}&\epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} &\epsilon_{zz} \end{bmatrix}$$

This matrix can also be represented by

$$\epsilon = \begin{bmatrix} \epsilon_{11} & \epsilon_ {12}&\epsilon_{13} \\ \epsilon_{21} & \epsilon_ {22}&\epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} &\epsilon_{33} \end{bmatrix} = [\epsilon_{ij}]$$ where i,j = 1,2,3

i is the row index and j is the column index

How to relate strains to $$E,\nu$$

First is to define Hooke's law in three dimensions which is

For normal strains:

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

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

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

For shear strains:

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

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

$$\gamma_{xz}=2*\epsilon_{xz}=\frac{\tau_{xz}}{G}$$

From the symmetry of the strain tensor

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

It is clear that $$\epsilon=[\epsilon_{ij}], \sigma=[\sigma_{ij}]$$ which are both three by three matrices.

Next arrange the strain components $$\epsilon_{ij}$$ and the stress components in column matrices. this is also called the voigt notation.

$$ \mathbf{\epsilon_{ij}} = \begin{bmatrix} \epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33}\\ \epsilon_{23}\\ \epsilon_{31}\\ \epsilon_{12}\\ \end{bmatrix} $$

and

$$ \mathbf{\sigma_{ij}} = \begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{23}\\ \sigma_{31}\\ \sigma_{12}\\ \end{bmatrix}

$$

Next hookes law for an isotropic material can be written which is

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

$$

which can also be written as

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

$$

Poisson Ratio
The poisson ratio $$\nu$$ for certain materials was also mentioned in class such as cork which is unique for the fact that is has a zero poissons ratio which means that when it is compressed or put under tension there is no change in volume in the directions other than that of the force applied. for example if there was a cylinder of cork of some radius R and it was put under compression the radius would not change. More can be found about the poisson ratio here

Suggested Reading For This Report
MIT OpenCourseWare Structural Mechanics Units 4 - 12

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

Mark Barry Eas4200c.f08.blue.a 20:31, 19 October 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 16:58, 21 October 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 01:54, 22 October 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 02:04, 22 October 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 03:08, 22 October 2008 (UTC)

Clay Robertson Eas4200c.f08.blue.f 16:46, 23 October 2008 (UTC)

Team Members To Contributed To This Report
Mark Barry Eas4200c.f08.blue.a 20:31, 19 October 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 04:16, 20 October 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 16:58, 21 October 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 01:54, 22 October 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 02:03, 22 October 2008 (UTC)

Clay Robertson Eas4200c.f08.blue.f 16:46, 23 October 2008 (UTC)