User:Eas4200c.f08.aero.e/Week 5

=Monday=

Curved Panel:

Equations used are

$$V_x = \tau ta$$ -- Eq. 1.4 from book 

$$V_y = \tau tb$$ -- Eq. 1.5 from book 

$$F_x = qb$$ -- Eq. 3.49b from book 

$$F_y = qh$$ -- Eq. 3.49c from book 

Problem 1.5 from the Aircraft Structures book states:

Derive the relations given by Eq. (1.4) and (1.5).

This will be done with the help of the following figure.

The shear flow is from A to B and is written as $$q = \tau t$$.

Then we see,

$$ d\overrightarrow{F} = q(d\overrightarrow{l}) = q(dl_y \hat{j} + dl_z \hat{k}) = q(dlcos \theta \hat{j} + dlsin \theta \hat{k}) $$

The figure below is of the section dl:

The resultant shear force vector is,

$$ \overrightarrow{F} = \int_{A}^{B} d\overrightarrow{F}\ = q	\left[(\int_{A}^{B} dy) \hat{j} + (\int_{A}^{B} dz) \hat{k} \right] $$

$$ \overrightarrow{F} = q(a \hat{j} + b \hat{k} ) $$

To compare this to Eq. 1.4 and 1.5, q = τt then add 1.4 and 1.5 to get a vector, and they are the same.

This leads to,

$$ \frac{F_y}{F_z} = \frac{a}{b} $$

The resultant magnitude will be,

$$ R = \left \Vert \overrightarrow{F} \right \| = \left[(F_y)^2 + (F_z)^2 \right]^{\frac{1}{2}} = q \left[ a^2 + b^2 \right]^{\frac{1}{2}} $$

Where,

$$ \left[ a^2 + b^2 \right]^{\frac{1}{2}} = d $$

So,

$$ R = qd $$

Where d is the length between A and B.

Closed thin walled cross section.

$$ dF = qdl \quad, $$  and   $$, \quad\ dA = \frac{1}{2} \rho dl $$

<p style="text-align:center;">$$ d\overrightarrow{T} = \overrightarrow{r} \times d\overrightarrow{F} $$

<p style="text-align:center;">$$ d\overrightarrow{T} = \rho dF = \rho qdl$$

<p style="text-align:center;">$$ \overrightarrow{T} = \int d\overrightarrow{T} = q \int \rho dl = 2q \int_{\bar A} dA $$

<p style="text-align:center;">$$ \overrightarrow{T} = 2q \bar A $$

The equation above is Equation 3.48 in the book.

=Wednesday=

 Exam 1

=Friday=

Homework:

For the image shown below, prove that the area of the shades triangle is: $$= \frac{1}{2}bh$$

Bold text Solution:

We know that the formula to find the area of a right triangle is $$= \frac{1}{2}BH$$

And we notice that we have a right triangle defined by the points BED of area $$= \frac{1}{2}(\bar{BD})(\bar{DE}) = \frac{1}{2}(\bar{BD})h$$,

the same way we notice that we have another right triangle defined by the points CED of area $$= \frac{1}{2}(\bar{CD})(\bar{DE}) = \frac{1}{2}(\bar{CD})h$$

So the area of the triangle BEC (shaded triangle), will be:

$$A_{BEC} = A_{BED}- A_{CED}$$

Then,

$$A_{BEC} = \frac{1}{2}(\bar{BD})h - \frac{1}{2}(\bar{CD})h = \frac{1}{2}h(\bar{BD}-\bar{CD}) = \frac{1}{2}h(\bar{BC})$$

And we know that $$BC = b$$

So, finally we get that:  $$A_{BEC} = \frac{1}{2}hb$$

Open thin walled cross section

The following equations apply:

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

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

'''Uniform bar with circular cross-section. Non-warping case (cross section behaves like a rigid disk)'''

For this case:

$$T = \int \int _{A}r\tau dA$$  and notice that

$$\tau = G\gamma $$ (Hooke's Law)

$$\gamma = \frac{rd\alpha }{dx}$$

$$ \frac{rd\alpha }{dx} = \theta $$ (Rate of twist)

So we get that:

$$T = \int _{A} r G (r\theta )dA$$

Where $$da = dydz$$ and recalling that  G is independent of Y and Z.

And finally we obtain:

$$T = G\theta (\int _{A}r^{2}dA)$$

Where $$J = \int _{A}r^{2}dA $$ Which is the second polar area moment of inertia.

Homework

Prove that $$J = \frac{1}{2}\pi a^{4}$$ for a solid circular cross section

For this case a ring of infinite element area of radius dr is chosen. And we designate a as the radius of the disk.

Then:

$$dJ = r ^{2}dA = r^{2}(2\pi rdr)$$

So we integrate r from 0 to c and obtain:

$$J = \int_{0}^{c}{}r ^{2}(2\pi rdr) = 2\pi \int_{0}^{c}{}r ^{3}dr$$

So, finally we get:

$$J = \frac{1}{2}\pi ^{4}$$

Case2: Hollow circular cross-section.

For this case we have that:

Is a thin-walled case therefore t << a

$$r_{i} = a$$

$$r_{o} = b$$

ri is simply the inner radius while ro is the outer radius.

The we have the following expresion:

$$J = \frac{1}{2}\pi (b^{4}-a^{4})$$

Which can be rewritten as:

$$J = \frac{1}{2}\pi (b^{2^{2}}-a^{2^{2}})$$

And:

$$J = \frac{1}{2}\pi (b-a)(b+a)(b^{2}+ a^{2})$$

Notice that

$$(b - a)=t$$

and

$$ (b + a) = 2\bar{r} $$

Which is the approximate value for the average radius.

We can make a thin wall assumption, t << a or b so,

$$b\approx a $$

And from the equations above,

$$ \bar{r} \approx \frac{(b + a)}{2} \approx \frac{2b}{2} \approx b  $$

With the above approximation, we then know that:

$$b^{2}\approx \bar{r}^{2}$$

and that:

$$a^{2}\approx \bar{r}^{2} $$

Then finally we get:

$$J = 2\pi t\bar{r}^{3}$$

Which also can be expressed as:

$$J = (2\pi ^{\frac{-1}{2}}t)(\pi \bar{r}^{2})^{\frac{3}{2}}$$

Notice that the expression $$\pi \bar{r}^{2}$$ equals the area.

Then, we can conclude that J is proportional to  $$\bar{A}^{\frac{3}{2}}$$   with   $$2\pi ^{\frac{-1}{2}}t$$   being the proportional factor.

Homework:

Compare solid circular cross-section to a hollow thin walled cross-section of the figure below:

a = Solid cross-section

b= Hollow cross-section


 * 1) Compute areas A(a) and A(b)
 * 2) Compute J< (a) amd J(b)
 * 3) Compute $$\frac{J_{(a)}}{J_{(b)}}$$
 * 4) Find ri(c) with $$t = 0.02r^{c}_{i}$$ at J(c) = J(a)
 * 5) Compute $$\frac{A_{(a)}}{A_{(c)}}$$

Solution

Part 1

To calculate the areas we use the formula : $$A = \pi r^{2}$$

So, for the solid cross-section

$$A = \pi 1^{2} = = 3.14159 cm^2$$

For the hollow cross-section, we simply find the area of the circle formed by the outer radius and subtract it from the area formed by the inner circle.

$$A = \pi (5.1cm - 5.0cm)^{2} = 0.0314159 cm^2$$

Part 2

To compute J we use the formula : $$J = \frac{1}{2}\pi r^{4}$$

So for (a):

$$J = \frac{1}{2}\pi (1cm)^{4} = 1.5708cm^4$$

For (b):

$$J = \frac{1}{2}\pi (0.1cm)^{4} = 0.000157cm^4$$

Part 3

Compute $$\frac{J_{(a)}}{J_{(b)}}$$

$$\frac{J_{(a)}}{J_{(b)}} = \frac{1.5708cm^4}{0.000157cm^4} = 10000$$

Part 4

We know that J(c) = J(a) = 1.5708cm^4

Then, we know that t = 0.02ric

So, we use the formula:

$$J = \frac{1}{2}\pi r^{4}$$ and simply solve for ro - ri

Then, we get:

$$r_{o}^{4}-r_{i}^{4} = \frac{2J}{\pi }$$

$$J = 1cm^4$$

So if

$$r_{o}^4-r_{i}^4 = 1 cm^4$$

Also,

$$r_i = \frac{t}{0.02}$$

Then:

$$R_{0}= t-\frac{t}{0.02}$$

Then

r_{0}= 1.02r_{i}

And:

$$1.02r_{i}^4-r_{i}^4 = 1 cm^4 r$$

$$r_i = 2.65915 cm$$

Area_c =  $$\pi (0.444288cm )^{2}$$ $$= 0.620126 cm 2$$

And :

finally,

$$\frac{A_{(a)}}{A_{(c)}} = \frac{3.14159}{0.620126}= 5.06605$$