User:Eas4200c.f08.nine.s

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

 Group nine - Homework 3 

Introduction
Reasons to use open thin wall cross section for stringers

1) manufacturing: stamping of thin flat sheets

2) construction of aircraft: riveting

Why do the stringer have angled walls compared to vertical walls?

One possible reason is for storage purposes. Having angled walls allows the stringers to be stacked while saving space.

Shear Panels


$$u \equiv u_{x}$$

$$u \equiv u_{y}$$

$$\gamma $$ = change in the right angle due to shear deformation.

$$\epsilon_{xy} $$ = torsional shear strain = $$  \frac{1}{2}\gamma_{xy}$$

Curved Panels
$$q = \tau t$$  (Shear flow)

$$d\vec{F} = qd\vec{l} = q(dl_{y}\hat{j} + dl_{z}\hat{k})$$

$$ = q(dlcos\theta \hat{j} + dlsin\theta \hat{k})$$

where $$dlcos\theta \hat{j}$$  =  $$dy$$  and  $$dlsin\theta \hat{k}$$  =  $$dz$$



 Resultant shear force vector 

$$\vec{F} = \int_{A}^{B} d\vec{F} = q(\int_{A}^{B}dy \hat{j} + \int_{A}^{B}dz \hat{k})$$

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

$$\vec{F} = F_{y}\hat{j} + F_{z}\hat{k}$$

$$\Rightarrow F_{y} = qa$$

$$\Rightarrow F_{z} = qb$$

$$\frac{F_{y}}{F_{z}} = \frac{a}{b}$$



 Resultant Magnitude   $$||\vec{F}||$$

$$||\vec{F}|| = \sqrt{F_{y}^{2} + F_{z}^2}$$

$$\Rightarrow R = q\sqrt{a^{2} + b^{2}}$$

where $$d$$ = length of straight line $$AB$$.

$$\Rightarrow R = qd$$

We can relate $$R = ||\vec{F}||$$ to $$T = 2q\bar{A}$$      (3.48)



Closed thin wall cross section:

$$\vec{T} = T\hat{i}$$

$$d\vec{T} = \vec{r} \times d\vec{F}$$

$$\Rightarrow dT = \rho dF$$

$$T = \oint_{}^{}dT = q\oint_{}^{}\rho dl$$

where $$\rho dl = 2dA$$

$$\Rightarrow T = 2q \int_{\bar{A}}^{} dA$$

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

Open thin wall cross section:

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

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

NACA 4-digit air foil series


ns = number of segments to the y-axis



$$d\bar{A} = \frac{1}{2}\vec{r} \times \vec{PQ} = |dA|\hat{i}$$

$$||\vec{PQ}|| = dl$$

Twist and Warping
Torsion of uniform, non circular bars (leads to warping of cross section)

Warping = axial displacement along x-axis of a part on the deformed cross section.

$$\theta = \frac{\alpha}{x} = $$   rate of twist.

$$(1)\, \, u_{y} = -\theta_{xy} $$  (3.11)

$$(2)\, \, u_{z} = +\theta_{xy} $$  (3.12)

Warping displacement along x-axis

$$(3)\, \, u_{x} = \theta \psi(y,z)$$

Road Map for torsional Analysis of aircraft wing


A) Kinematic assumptions  (Section 3.2)

B) Strain - displacement relationship  (Section 3.2)   C)  Equilibrium Equations ( Stresses)  (Ch.2; Section 3.2) D) Pranal Stress Function $$\phi$$  (Section 3.2)

E) Strain compatibility   Eqn.(3.15)(3.17)

F) Eqn. for $$\phi$$  (3.19)

G) B conditions for $$\phi$$ (3.24)

H) $$T = 2\int\int_{A}^{} \phi dA,$$  (3.25)

$$T = GJ\theta$$

$$J = \frac{-4}{\bigtriangledown^{2} \phi} \int\int \phi dA$$

I) Thin walled cross section

Formal Derivation ( Section 3.5)

$$T = 2q\bar{A}$$ (3.48)

J) Twist angle $$\theta$$  :  Method 1

$$\theta = \frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$ (3.56)

K) Multicell Section  (Section 3.6)

Multicell Thin-Walled Sections
Multicell Thin-Walled Sections refers to wing sections that are composed of airfoil skin supported by thin vertical supports called spar webs that form multicell constructions. There are often stiffeners incorporated in the construction of multicell sections which are usually located above and below the spar webs themselfs. These stiffeners are very effective when used to carry bending loads, but they do little to help counteract torsional loads and are often neglected in the consideration of torsional rigidity.

For a two-cell section there are three boundary contours which we will label S0, S1, and S2. This will give us

$$\phi$$(S0)=C0

$$\phi$$(S1)=C1

$$\phi$$(s2)=C2

Where C0, C1, and C2 are all constants and $$\phi$$ is the stress function. The shear flow between two boundary contours is equal to the difference between the values of $$\phi$$ along these contours. The shear flow for each cell is considered positive if it forms a counterclockwise torque about the cell and its value is equal to the value of $$\phi$$ on the inside contour minus that on the outside contour.

q1=C1-C0

q2=C2-C0

q12=C1-C2

The Torque contributed by each cell can be calculated by using T=2Aq. The total torque of a two-cell section is

T=2A1q1+2A2q2

where A1 and A2 are the areas enclosed by the shear flows q1 and q2 respectively.

The equations for the twist angles $$\theta$$1 and $$\theta$$2 are

$$\theta$$1=$$\frac{1}{2\bar{A_{1}}G}\int_{cell 1}{\frac{qds}{t}}$$

$$\theta$$2=$$\frac{1}{2\bar{A_{2}}G}\int_{cell 2}{\frac{qds}{t}}$$

$$\theta_1$$=$$\theta_2$$=$$\theta$$

Homework Problems & Matlab Code
{| class="toccolours collapsible collapsed" style="width:100%" ! Homework Problems

Homework Problem 1 



Prove: $$ A_{blue} = \frac {b h} {2} $$

$$ A_{red} = \frac {a h} {2} $$

$$ A_{blue} = A_{total} - A_{red} = \frac {(a+b) h} {2} - \frac {a h} {2} $$

so we have: $$ A_{blue} = \frac {b h} {2} + \frac {a h} {2} - \frac {a h} {2} = \frac {b h} {2} $$

Provig that: $$ A_{blue} = \frac {b h} {2} $$ ___________________________________________________________________________________________________________________________________________________________

Homework Problem 2 



Show that: $$ J_{o} = \frac {\pi a^4}{2} $$

$$ J_{o}= \int_{A} {r^2 dA} = \int_{A} {r^2\times r\times dr\times d\theta} $$

(polar coordinates for dA)

Since we're integrating from (0 to a) around the whole circle, which means

$$ d\theta = 2\times \pi $$

we have: $$ 2\times \pi \times \int_{0}^{a}{r^3 dr} = \frac {2 \pi a^4} {4} = \frac {\pi a^4}{2} = J_{o} $$

and thus: $$ J_{o}= \int_{A} {r^2 dA} = \int_{A} {r^2\times r\times dr\times d\theta} $$

__________________________________________________________________________________________________________________________________________________________

Homework Problem 3 

Show: $$ b^2 \cong \bar{r}^2 $$ and $$ a^2 \cong \bar{r}^2 $$

we know that: $$ \bar{r} = a + \frac {t}{2} \Rightarrow \bar{r}^2 = a^2 + \frac {t^2} {4} $$

since: $$ t \ll a, $$ we can neglect this term, leaving $$ a^2 \cong \bar{r}^2 $$

the same holds true for b since: $$ \bar{r} = b t \frac {t}{2} \Rightarrow \bar{r}^2 = b^2 - \frac {t^2} {4} $$

since: $$ t \ll b, $$ we can neglect this term, leaving $$ b^2 \cong \bar{r}^2 $$

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Homework Problem 4 



Given: $$ t_{1} = 0.008 m \rightarrow t_{2} = 0.01 m = t_{3} \rightarrow b = 2 m \rightarrow a = 4 m $$

Find: $$ T_{all} $$

$$ \bar{A} = \frac {\pi}{2} \times \frac {b^2}{4} + \frac {b a}{2} $$

$$ \bar{A} = \frac {\pi}{2} \times \frac {4}{4} + \frac {8}{2} = 2\pi [m^2]$$

$$ \theta = \frac {1}{2 G \bar{A}} \times q\times [\frac{0.5 b \pi}{t_{1}} + \frac{a}{t_{2}} + \frac{\sqrt{a^2 + b^2}}{t_{3}}] $$

since: $$ q = \frac {T} {2 \bar{A}} = \frac {T}{4 \pi} [N/m] $$

$$ \theta = \frac {1}{4 G \pi} \times \frac {T}{4 \pi}\times [\frac{\pi}{0.008} + \frac{4}{0.01} + \frac{\sqrt{16 + 4}}{0.01}] = 98.669 [rad/m] $$

since: $$ T_{all} = 2\times \bar{A} \times \tau_{all} \times min_{thickness} $$

and (given): $$ \tau_{all} = 100 [GPa] $$

$$ T_{all} = 2\times 2\times \pi\times 100\times 0.008 = 10 \times 10^9 [Nm] $$

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Homework Problem 5 

Compare solid circular cross section to hollow thin wall cross section:

a) solid circular cross section: $$ r_{o} = 1 cm $$

b) hollow circular cross section: $$ r_{o} = 5.1 cm \rightarrow r_{i} = 5 cm \rightarrow t = 0.1 cm $$

$$ A_{a} = \pi \times r^2 \rightarrow r_{o} = 1 cm \rightarrow A_{a}= \pi [cm^2] $$

$$ A_{b} = \pi \times r_{o}^2-r_{i}^2 \rightarrow r_{o} = 5.1 cm \rightarrow r_{i} = 5 cm \rightarrow

A_{b}= 1.0 \times \pi [cm^2] $$

$$ A_{a} \approx A_{b} \rightarrow 1.00 \approx 1.01 $$

$$ J_{a} = \frac {\pi r^4}{2} \rightarrow r = 1cm \rightarrow J_{a} = \frac {\pi}{2} [cm^4] $$

$$ J_{b} = \frac {\pi (r_{o}^4 -r_{i}^4)}{2} \rightarrow r_{o} = 5.1 cm \rightarrow r_{i} = 5 cm \rightarrow J_{b} = 80.9276 [cm^4] $$

$$ \frac {J_{a}}{J_{b}} = \frac {1.57}{80.9276} = 0.0194 $$

Find $$ r_{i}(c)$$ with $$ t=0.02 \times r_{i}(c) $$ such that $$ J_{c} = J_{a} $$ and find $$ \frac {A_{a}}{A_{c}} $$

$$ r_{i} + t = r_{o} \rightarrow r_{i} + 0.02 \times r_{i} = r_{o} $$

$$ J_{c} = J_{a} \rightarrow 0.5\times \pi\times (r_{o}^4-r_{i}^4) = \frac {\pi}{2} $$

$$ (r_{i} + 0.02 \times r_{i})^4 - r_{i}^4 = 1 \rightarrow r_{i} = 1.866 cm \rightarrow r_{o} = 1.9033 cm \rightarrow t = 0.037 cm $$

$$ \frac {A_{a}}{A_{c}} = \frac {\pi} {\pi (r_{o}^2-r_{i}^2)} = \frac {1}{(1.9^2-1.87^2)} \approx 7.16 $$

Sample Run
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
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Contributing Team Members
The following students contributed to this report:

David Phillips Eas4200C.f08.nine.d 3:10, 8 October 2008 (UTC)

Oliver Watmough Eas4200c.f08.nine.o 16:07, 8 October 2008 (UTC)

Stephen Featherman Eas4200c.f08.nine.s 1:07, 8 October 2008 (UTC)

Ricardo Albuquerque Eas4200c.f08.nine.r 4:30, 8 October 2008 (UTC)

Felix Izquierdo Eas4200c.f08.nine.F 18:34, 8 October 2008 (UTC)