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

Engineering Shear Strain

$u\equiv u_{x}$
$u\equiv u_{y}$
$\gamma$ = change in the right angle due to shear deformation.
$\epsilon _{xy}$ = torsional shear strain = ${\frac {1}{2}}\gamma _{xy}$

Curved Panels

Curved Panel

$q=\tau t$ (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 $dlsin\theta {\hat {k}}$ = $dz$

Side view of detailed curved Panel

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

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 walled cross section

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 walled cross section

Open thin wall cross section:

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

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

NACA 4-digit air foil series

NACA 4 digit airfoil series

ns = number of segments to the y-axis

Blown up area of airfoil

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

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

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

Multiple Cell Airfoil

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=${\frac {1}{2{\bar {A_{1}}}G}}\int _{cell1}{\frac {qds}{t}}$

$\theta$₂=${\frac {1}{2{\bar {A_{2}}}G}}\int _{cell2}{\frac {qds}{t}}$

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

Homework Problems & Matlab Code

Homework Problems

Homework Problem 1 File:Hwaerostruc.jpg Problem 1 Prove: $A_{blue}={\frac {bh}{2}}$ $A_{red}={\frac {ah}{2}}$ $A_{blue}=A_{total}-A_{red}={\frac {(a+b)h}{2}}-{\frac {ah}{2}}$ so we have: $A_{blue}={\frac {bh}{2}}+{\frac {ah}{2}}-{\frac {ah}{2}}={\frac {bh}{2}}$ Provig that: $A_{blue}={\frac {bh}{2}}$ ___________________________________________________________________________________________________________________________________________________________ Homework Problem 2 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: ${\bar {r}}=bt{\frac {t}{2}}\Rightarrow {\bar {r}}^{2}=b^{2}-{\frac {t^{2}}{4}}$ since: $t\ll b,$ we can neglect this term, leaving $b^{2}\cong {\bar {r}}^{2}$ ___________________________________________________________________________________________________________________________________________________________ Homework Problem 4 Problem 4 Given: $t_{1}=0.008m\rightarrow t_{2}=0.01m=t_{3}\rightarrow b=2m\rightarrow a=4m$ Find: $T_{all}$ ${\bar {A}}={\frac {\pi }{2}}\times {\frac {b^{2}}{4}}+{\frac {ba}{2}}$ ${\bar {A}}={\frac {\pi }{2}}\times {\frac {4}{4}}+{\frac {8}{2}}=2\pi [m^{2}]$ $\theta ={\frac {1}{2G{\bar {A}}}}\times q\times [{\frac {0.5b\pi }{t_{1}}}+{\frac {a}{t_{2}}}+{\frac {\sqrt {a^{2}+b^{2}}}{t_{3}}}]$ since: $q={\frac {T}{2{\bar {A}}}}={\frac {T}{4\pi }}[N/m]$ $\theta ={\frac {1}{4G\pi }}\times {\frac {T}{4\pi }}\times [{\frac {\pi }{0.008}}+{\frac {4}{0.01}}+{\frac {\sqrt {16+4}}{0.01}}]=98.669[rad/m]$ since: $T_{all}=2\times {\bar {A}}\times \tau _{all}\times min_{thickness}$ and (given): $\tau _{all}=100[GPa]$ $T_{all}=2\times 2\times \pi \times 100\times 0.008=10\times 10^{9}[Nm]$ ___________________________________________________________________________________________________________________________________________________________ Homework Problem 5 Compare solid circular cross section to hollow thin wall cross section: a) solid circular cross section: $r_{o}=1cm$ b) hollow circular cross section: $r_{o}=5.1cm\rightarrow r_{i}=5cm\rightarrow t=0.1cm$ $A_{a}=\pi \times r^{2}\rightarrow r_{o}=1cm\rightarrow A_{a}=\pi [cm^{2}]$ $A_{b}=\pi \times r_{o}^{2}-r_{i}^{2}\rightarrow r_{o}=5.1cm\rightarrow r_{i}=5cm\rightarrow A_{b}=1.0\times \pi [cm^{2}]$ $A_{a}\approx A_{b}\rightarrow 1.00\approx 1.01$ $J_{a}={\frac {\pi r^{4}}{2}}\rightarrow r=1cm\rightarrow J_{a}={\frac {\pi }{2}}[cm^{4}]$ $J_{b}={\frac {\pi (r_{o}^{4}-r_{i}^{4})}{2}}\rightarrow r_{o}=5.1cm\rightarrow r_{i}=5cm\rightarrow J_{b}=80.9276[cm^{4}]$ ${\frac {J_{a}}{J_{b}}}={\frac {1.57}{80.9276}}=0.0194$ Find $r_{i}(c)$ with $t=0.02\times r_{i}(c)$ such that $J_{c}=J_{a}$ and find ${\frac {A_{a}}{A_{c}}}$ $r_{i}+t=r_{o}\rightarrow r_{i}+0.02\times r_{i}=r_{o}$ $J_{c}=J_{a}\rightarrow 0.5\times \pi \times (r_{o}^{4}-r_{i}^{4})={\frac {\pi }{2}}$ $(r_{i}+0.02\times r_{i})^{4}-r_{i}^{4}=1\rightarrow r_{i}=1.866cm\rightarrow r_{o}=1.9033cm\rightarrow t=0.037cm$ ${\frac {A_{a}}{A_{c}}}={\frac {\pi }{\pi (r_{o}^{2}-r_{i}^{2})}}={\frac {1}{(1.9^{2}-1.87^{2})}}\approx 7.16$ Matlab Code fprintf('\n NACA Airfoil calculation program \n \n') m = input('Enter first digit of airfoil: '); p = input('Enter second digit: '); t = input('Enter the third and fourth digits: '); Py = input('Enter Py: '); Pz = input('Enter Pz: '); segment = input ('Enter number of segments: '); y = 0; n = 1; c=1; m = (m/100)*c; p = (p/10)*c; t = (t/100)*c; zc = size(segment); dzdy = size(segment); yu = size(segment); zu = size(segment); yl = size(segment); zl = size(segment); j=1; while y<=p zc(n) = (m/p^2)*(2*p*y-y^2); dzdy(n) = (m/p^2)*(2*p-2*y); y = y + c/segment; zcc(n)=zc(n)*c; n = n+1; end while y<=c zc(n) = (m/((1-p)^2))*((1-2*p)+2*p*y-y^2); dzdy(n) = (m/((1-p)^2))*(2*p-2*y); y = y + c/segment; zcc(n)=zc(n)*c; n = n+1; end y=0; n=1; figure(2) plot(zcc,y,'-k') while y<=c theta = size(segment); zt = size(segment); zt(n) = 5*t*(0.2969.*sqrt(y)-0.1260.*y-0.3516.*y.^2+0.2843.*y.^3-0.1015.*y.^4); theta(n)= atan(dzdy(n)); yu(n)= y-zt(n)*sin(theta(n)); zu(n) = zc(n) + zt(n)*cos(theta(n)); yl(n) = y+zt(n)*sin(theta(n)); zl(n) = zc(n)- zt(n)*cos(theta(n)); y = y+c/segment; n = n+1; end y=0; n = 1; a1 = 0; a2 = 0; while n<segment line = [0 yu(segment-(n))-yu(segment-(n-1)) zu(segment-(n))-zu(segment-(n-1))]; r = [0 yu(segment-(n))-c zu(segment-(n))-Pz]; a = 0.5*cross(r,line); a1 = a1 + a; n = n+1; end n =1; line =0; while n<segment line = [0 yl(segment-(n))-yl(segment-(n-1)) zl(segment-n)-zl(segment-(n-1))]; r = [0 yl(segment-(n))-c zl(segment-(n))-Pz]; b = 0.5*cross(r,line); a2 = a2 + b; n = n+1; end avg = abs(a1-a2); area(j) = avg(1); segment_max = segment; fprintf('\n') fprintf(1,'The average area is: %4.3f\n',avg(1)) figure(1) plot(yl,zl,'-k',yu,zu,'-b') axis([0 1 -0.3 0.3]) R=1; percentage = 2; segment = 1; j=1; while percentage>1 y = 0:2/segment:2; z = sqrt(R^2-(y-1).^2); zl = -z; Py = 0.25; Pz = -1; n=1; areau = 0; areal = 0; c=2; line = 0; while n<segment line = [0 y(segment-(n))-y(segment-(n-1)) z(segment-(n))-z(segment-(n-1))]; r = [0 y(segment-(n))-c z(segment-(n))-Pz]; a = 0.5*cross(r,line); areau = areau + a; n = n+1; end n =1; line =0; while n<segment line = [0 y(segment-(n))-y(segment-(n-1)) zl(segment-n)-zl(segment-(n-1))]; r = [0 y(segment-(n))-c zl(segment-(n))-Pz]; b = 0.5*cross(r,line); areal = areal + b; n = n+1; end avg = abs(areau - areal); area(j) = avg(1); percentage = ((pi-avg(1))/pi)*100; segment = segment+1; j = j+1; end fprintf('The minumum number segments required to have the average area accurate within 1 percent is: %4.3f\n',segment) fprintf('\n Figure 1 shows the cross-section of the NACA airfoil and Figure 2 shows the centroid line \n') Figure 1 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

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)