User:Eas4200c.f08.aero.e/Week 7

=Monday=

The quadrature of a NACA airfoil is shown below.

The shear flow is q around the airfoil. A is the total area of the airfoil. This is found by subtracting the area under the bottom of the airfoil from P to Q, represented by the green A, and the area under the top airfoil from P to Q, represented by the blue A.

How to find the areas are shown below.

$$ d \bar{A} = \frac{1}{2}( \bar{r} \times \overline{PQ}) = -\left | dA \right \vert \bar{i} $$

A question was asked, "Why no use the trapezoidal method to integrate?"

The answer is, with the change in curvature that method has disadvantages and is less elegant.

Going back to the single cell airfoil.

The shear flow across the foil above is equal over the entire foil.

$$ q = q_1 = q_2 = q_3 $$

The angle of twist, θ, is represented by the below.

$$ \theta = \frac{1}{2G \bar{A}} q \sum_{j=1}^3 \frac{l_i}{t_j} $$

$$ \theta = \frac{1}{2G \bar{A}} (q) \left [ \frac{2 \pi (b/2)}{2t_1} + \frac{a}{t_2} + \frac{\sqrt{a^2 + b^2}}{t_3} \right ] $$

Knowing that

$$ a = 4m  \qquad\quad\quad   q = \frac{T}{2\bar{A}} $$ $$ b = 2m  \quad\quad\quad   \bar{A} = \frac{\pi (b/2)^2}{2} + \frac{1}{2}ba  $$ $$ t_1 = 0.08m  \qquad\quad\quad   t_2 = t_3 = 0.01m $$

we can find $$\theta$$ in terms of T and G.

$$ \theta = 7.141 \frac{T}{G} rad/m $$

Max shear stress - $$ \tau_{max} = \frac{q}{[min(t_1, t_2, t_3)]} = \frac{T}{t_{min}\bar{A}} $$.

If $$ \tau_{max} = \tau_{all}$$ (given) and since $$ q = \frac{T}{2 \bar{A}}$$, then $$ T_{all} = 2 \bar{A} \tau_{all} [min(t_1, t_2, t_3)] $$

Letting $$ \tau_{all} = 100 GPa$$, and knowing that t2 and t3 are the tmin,

$$ T_{all} = 2 \tau_{max} t_2 \bar{A} = 1.114 \times 10^10 Nm $$

=Wednesday= Multi Cell Airfoil (Sec. 3.6)

Recall:

$$ T = 2q\bar A $$

$$\theta = \frac{1}{G\bar A} \oint\frac{{q}_{i}}{{t}_{i}}ds $$

Specific example first (generalization later)

Compare this example with page 94 in textbook.



$$ T = 2\sum_{i=1}^1 {q}_{i} \bar {A}_{i}  $$

Find theta as a function of T and J(torsional constant)

$$T = {T}_{1} + {T}_{2} = 2{q}_{1}\bar {A}_{1} + 2{q}_{2} \bar {A}_{2} $$   (1)

$$\bar {A}_{1} = ac, \bar {A}_{1} = bc$$

$${\theta}_{1} = \frac{1}{G\bar {A}_{1}} \oint\frac{{q}_{1}}{{t}_{1}}ds $$

$$\oint\frac{{q}_{1}}{{t}_{1}}ds$$ q and t can be thought of as functions of (s), their curvilinear coordinates

therefore we can replace by discrete summation,

$${\theta}_{1} = \frac{1}{G\bar {A}_{1}} 2\frac{{q}_{1}a}{{t}_{1}} + \frac{{q}_{1}c}{{t}_{1}} + \frac{({q}_{1} - {q}_{2})c}{{t}_{12}}$$  (3)

$${\theta}_{2} = \frac{1}{G\bar {A}_{2}} 2\frac{{q}_{2}a}{{t}_{2}} + \frac{{q}_{2}c}{{t}_{2}} + \frac{({q}_{2} - {q}_{1})c}{{t}_{12}}$$  (4)

Since they are rigidly connect cells 1 and 2 must have the same rate of twist angle.

Therefore,

$${\theta}_{1} = {\theta}_{2}$$  (2)

Think of q1 and q2 as 2 unknowns. Equations (1) and (2) are the 2 equations for the 2 unknowns (q1 and q2) with T being a variable. Use (1) and (2) to find an expression for q1 and q2 in terms of T.

$${q}_{1} = {\beta}_{1}T {q}_{2} = {\beta}_{2}T$$

Next use the experssion (3) or (4) to find the expression between theta and T

$$\theta = {\theta}_{1} = {\theta}_{2} = \frac{T}{GJ} $$

Finally from this equation the torsional constant J can be deduced.

=Friday=

Information about 3-cell NACA 2415

The airfoil has two partition walls:

1st wall at ¼ C from leading edge.

2nd wall at ¾ C from leading edge.

3 Unknowns: q1, q2, q3.

1. $$T = 2\sum_{i=1}^{3}{}q_{i}\bar{A_{i}}$$

2.$$\theta _{1} = \theta _{2}$$

3.$$\theta _{2} = \theta _{3}$$

-

Now, we must recall from previous classes that we derived the equations:

$$T = GJ\theta $$ and $$T = 2q\bar{A}$$

Engineering (ad-hoc) derivation of:

$$\theta = \frac{1}{2gJ}\oint_{}^{}{}\frac{q}{t}ds$$

We refer to the next figure to carry out such derivation.

Then we find the displacement PP' due to α:

$$\frac{PP'}{OP} = (tan\alpha) \approx \alpha    $$  for small α

Then, we project the displacement of PP' on the direction orthogonal to OP'

$$PP' = (OP) (tan\alpha) )(cos\alpha )$$

$$PP' = (OP)(cos\alpha )(tan\alpha )$$

And we know that:

$$OP''=(OP)(cos\alpha )$$

And we also must recall that:

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

Then we get:

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

Where

$$\rho = (r)(cos\alpha )$$  and   $$\alpha \cong  (tan\alpha )$$

PP" = Displacement of point P in direction tangent to the lateral surface of the bar.

The the strain will be;

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

Finally, we must recall that:

Rate of twist angle = $$\theta = \frac{\alpha }{dx}$$ and we must notice that α is very small ad it can be denoted by dα

So, finally we obtain:

$$\theta = \frac{d\alpha }{dx}$$

(The next step is to integrate over the contour of the bar. Such derivation will be presented in the next day of homework)