User:Eas4200c.f08.aero.e/Week 4

=Monday=

Problem 1.1 Continued..

$$T_{max}= (2t\tau_{allow})(\frac{L}{4})^2 T_{max}$$

$$ = \frac{1}{8}tL^2\tau_{allow} = M_{max} $$

$$M_{max} = \frac{tL^2}{16} \sigma_{allow}$$

$$ => \sigma_{allow} = \frac{16{M}_{max}}{tL^2} $$

Recall: $$f(b) = \frac{I(b)}{b}$$

$$f(b) = \frac{I}{b} = \frac{3L - 4b}{12} $$

$$= \frac{tL^2}{24} $$ (Recall that b = $$\frac{L}{4}$$)

$$\sigma_{max} = \frac{{M}_{max}b}{2I} = {M}_{max} \frac{12}{tL^2}$$ = $$\frac{tL^2}{16} \sigma$$allow

Since, $$\sigma_{max} = \frac{12}{16} \sigma_{allowable} < \sigma_{allowable}$$

Case 2 is acceptible

The optimal cross section for case two is a square. The X-sec is illustrated to the right where a = b = $$\frac{L}{4}$$

=Wednesday= Three problems were presented in which a set of questions were assigned for each individual problem.

Problem #1

With the information provided find I(2)y and compare to I(1)y. Keep in mind that for case 2. the Y-axis passes through the centroid channel cross-section.
 * 1) Derive Iy(1) -moment of inertia for case 1 with respect to y axis- by integration using polar coordinates. Also, derive Iy(2) -moment of inertia for rectangular cross section-
 * 2) Distribute material in Case 1 into Case 2, such that a = b. Recall that A(2) = 3at and A(1) = πR2. Also assume that t = a/10.

Solution

For section 1 of the problem:

The formula to calculate the second moment of area is:  $$\iint_A z^2 \ dy\, dz$$

Then in polar coordinates:


 * $$\iint_A f(y,z) \, dA = \int_a^b \int_0^{r(\theta)} f(r,\theta)\,r\,dr\,d\theta.$$


 * $$y = r \cos \theta \,$$
 * $$z = r \sin \theta \,$$

And we finally obtain:


 * $$\int_0^{2\pi} \int_0^R (rsin(\theta))^2\,r\,dr\,d\theta.$$

Now, we integrate with respect to y:


 * $$\int_0^{2\pi} \int_0^R (sin(\theta))^2\,r^3\,dr\,d\theta$$ = $$

\int_{0}^{2\pi}\frac{r^4}{4} \,sin^2(\theta)\,d\theta $$

Then, to integrate with respect to θ, we make use of the identity:

$$\int_0^{\theta} sin^2(\theta) = \frac{\theta}{2} \, - \frac{1}{4} \,sin(2\theta)$$

And, finally we obtain:


 * $$\int_0^{2\pi} \int_0^R (sin(\theta))^2\,r^3\,dr\,d\theta$$ = $$

\frac{r^4}{4} \, (\frac{2\pi}{2} - sin^2(2\pi)\, -\frac{0}{2} + sin^2(0))\, = \frac{\pi r^4}{4} $$

And we obtain the final answer;


 * $$I_y = \frac{\pi r^4}{4}$$

For the case 2.

First we find the centroid of the whole cross-section by finding the centroid for each individual region. And then adding them up. The table below shows the calculations to obtain the centroid.

$$\sum{}A = \sum{}yA$$

$$Y \,= \frac{0.176a^2}{0.28a^2} = 0.628571a$$

Now, that we have the centroid, we calculate the distance of the centers of each region to the centroid.

For regions I and III:

$$(a - t)\, -0.6857a/, = \frac{10a - a}{10}\, - 0.628571a = 0.271429a$$

Now, for region II:

$$a - \frac{(20a - a)}{20}\, = 0.321429a$$

Then, we proceed to calculate the moment of inertia using the following formula


 * $$I_y = \sum_{i=1}^{3

}{}\left[\frac{bihi^{3}}{12 }+ Aidi^{2} \right]$$

Then, we have after plugging all the values given above and the equalities given that:

The combined calculated moment of inertia for regions I and III:

$$I_y = 2\left(0.006075a^{4}+0.006631^{4} \right) = 0.025411a^4$$

And the moment of inertia for region II:

$$I_y = \frac{a^{4}}{12000} + 0.010332a^{4} = 0.010332a^4$$

Finally, we have the total moment of inertia of the cross-section to be:

$$I_1\, + I_2\, + I_3\, = 0.05411a^4\, +  0.010332a^4\,  =  0.064442a^4$$

Finally, to compare the answers we use the equality given in which the areas are equal then:

$$\pi R^2\, = \frac{3a^2}{10}$$

Solving for R we get that:

$$R\, = 0.309019a^4$$

And finally comparing I[1]Y and I[2]y we get:

$$\frac{I_1}{I_2} = \frac{0.007162a^4}{0.064442a^4}\, = 0.111108a^4$$

Problem #2

For this problem the following data was provided:

Ro = 10 cm

t - 1/10 Ro

R1 is such that A(1) = A(2)

Question: Find Iy(1) and I(2)y and compare.

Solution

First, we calculate the value of R1:

A[1] = A[2] so;

$$\pi R_o^2\, = 2(\pi R_1^2)\, + \frac{R_o^2}{5}$$

Solving for R1 and plugging in the value for t, we get:

$$R_1 = \sqrt(50 - \frac{10}{\pi}) = 6.84229 cm$$

From problem 1, we found that the moment of inertia with respect to the y-axis of a circular cross section is:


 * $$I_y = \frac{\pi r^4}{4}$$

Then, the moment of inertia for case 1 will be:


 * $$I_y = \frac{\pi 10^4}{4}\, = 2500\, \pi\, cm^4\, =  7853.98  cm^4$$

Then, to calculate the moment of inertia of case 2, and realizing that the y-axis passes through the centroid of the section (because of symmetry) we simply proceed to calculate the moment of inertia individually for the central part and the two extreme circular cross sections.

Circular cross section moment of inertia.

$$Iccs = \frac{\pi R_1^4}{4} +  (\pi R_1^2)(R_1+R_o)^2  =  43442.5\,  cm^4$$

Rectangular cross section moment of inertia.

$$Ircs = \frac{b h^3}{12} = \frac{R_o^4}{15} =  83.3333 cm^4$$

Then the total moment of inertia for case 2 is:

$$I_t\, =  2(43442.5cm^4)\,  +  83.3333cm^4\,  =  86968.3\, cm^4$$

Finally, if we compare the moment of inertia of both cases:

$$\frac{I_1}{I_2}\, = \frac{7853.98cm^4}{86968.3cm^4}\, = 0.090309$$

Iy[2] is much larger and hence, such cross section would preferred over the one in case 1.

Problem #3

Moment of inertia for case one is the same as for case two.

Question: Why is case 2 preferred when manufactured than case 1?

Solution

The reason why case 2 is preferred is because is easier to manufacture and such shape provides an easier installation than case 2.

=Friday=

Problem solving methods

Two methods of solving moment of inertia problems are the Ad hoc and elasticity theory methods.

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. One way of describing the Ad hoc method is hand-waving, this can be attributed to an approach which asigns values and variables to equations on the fly.

Further information regarding this method can be found in the following web page: Ad hoc.

Classical elasticity theory:


 * $$T=T=\int \rho qds=\int \int 2qdA=2qA,$$

Ad hoc method:

A=ab "average" area of crossection


 * $$q=\tau t,$$ (shear flow)


 * $$\tau =\frac{T}{2abt},$$

Stringer Types

There are many types of aircarft stringers. Below are a few illustrations which depict types of stringers used in isdustry.

Shear panel (Shear Strain)

Plane strain diagram

Shear Strain diagrams

Engineering shear strain

$$y_{xy} = \frac {\partial v} {\partial x} + \frac {\partial u} {\partial y}$$


 * $$\gamma =\Delta,$$ is the angle due to shear deformation.


 * $$\epsilon _{xy}=\frac{1}{2}\gamma _{xy},$$ is the tensorial shear strain.

Reference: http://www.efunda.com