User:EAS4200C.F08.WIKI.A/Homework3

Shear Flow

Curved Panels

Below is an image of a curved panel in the (x,y,z) coordinates.

Note: Sometimes (y,z) coordinates are used for the axes in plane of the cross section of the panel instead of (x,y) coordinates.

Below is an image of another curved panel with thickness t. A force will be applied to the panel which can be seen in the next image below.

Below is a cross sectional view of the same curved panel with force F acting on it at an angle θ above the horizontal (y-axis). The cross section goes from point A to point B. It's height is b and it's width is a.

$d\overrightarrow{F} = q\overrightarrow{dl} = q(dl_yj + dl_zk)$
$d\overrightarrow{F} = q(dl\cos\theta \overrightarrow{j} + dl\sin \theta \overrightarrow{k})$
$\displaystyle\ dl\cos\theta = dy$
$\displaystyle\ dl\sin\theta = dz$
$\displaystyle\ q = \tau t$ = shear flow

Here's a closer look at the panel where the force is acting. You can see it's acting on and parallel with section dl.

Resultant Shear Force Vector
$\overrightarrow{F} = \int_{A}^{B}{d\overrightarrow{F}} = q(\int_{A}^{B}{dy \overrightarrow{j} + \int_{A}^{B}{dz \overrightarrow{k}}})$ q is a constant with respect to (y,z) ${\color{blue} \overrightarrow{F} = q(a\overrightarrow{j} + b\overrightarrow{k})}$ $\overrightarrow{F} = F_y\overrightarrow{j} + F_z\overrightarrow{k}$ → $F_y = qa$ → $F_z = qb$ ${\color{blue} \frac{F_y}{F_z} = \frac{a}{b}}$

Resultant Magnitude
$\begin{vmatrix} \begin{vmatrix} \overrightarrow{F} \end{vmatrix} \end{vmatrix} = [(F_y)^2 + (F_z)^2]^{1/2}$ ${\color{blue}\begin{vmatrix}\begin{vmatrix}\overrightarrow{F}\end{vmatrix}\end{vmatrix} = q[a^2 + b^2]^{1/2}}$ $a^2 + b^2 = d$= length of straight line AB Relate $R = \begin{vmatrix}\begin{vmatrix}\overrightarrow{F}\end{vmatrix}\end{vmatrix} \; to \; T = 2q\overrightarrow{A}$

Closed Thin-Walled Cross Section

$\overrightarrow{T} = T\overrightarrow{i}$

Below is the cross section of a closed thin-walled area. There is a force acting on section dl.
$\overrightarrow{r}$ = the distance from the origin to section dl
$\rho$ = the distance from the origin to the force that is perpendicular to the force vector.

$d\overrightarrow{T} = \overrightarrow{r}xd\overrightarrow{F}$

$dT = \rho dF$

Take a closer look where the force acts on on section dl.

$dA = \frac{1}{2}\rho dl$

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

$\rho dl = 2dA$

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

$T = 2q\overline{A}$

$\overline{A}$ = Average Area
Below is an image of the average area of the cross section of a curved panel:

Average Area

Proving The Area of a Triangle

The easiest way of proving the area od a triangle is to transform the triangle into a parallelogram. To do so follow these simple steps. 1. Create a triangle ABC 2. Create a line CD that is parallel to line AB 3. Create a line AE that is parallel to line CB 4. Extend lines AE and CD until they intersect at point F 5. Shape ABCF is a parallelogram 6. The parallelogram is composed of two triangles, both identical to the original triangle ABC 7. The Area of a parallelogram is base * height. 8. Half of this area equals the area of the original triangle. $\displaystyle A = \frac{1}{2} (b)(h)$

Open Thin Walled Cross-Sections

The torque of the depicted thin walled structure is

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

This is also equivalent to the location of the resultant force acting parallel to the line connecting the two ends of the object.

Uniform Circular Cylinder: Non-Warping

We will now take a look at a uniform circular cylinder cross-section. This cross-section behaves as a rigid disk.

We will look at the infinitesimal area $\displaystyle dA$ at a distance $\displaystyle r$ away from the origin. The torque here can be described by the following equation;

$\displaystyle T = \int\int_{A} r\tau dA$

It must be noted that;

$\displaystyle \tau = G\gamma$ , $\displaystyle \tau dA = F$

where $\displaystyle G$ is the strain modulus and $\displaystyle \gamma$ is the shear strain. Together they comprise Hooks Law which is used to describe the deformation of the disk. The variable $\displaystyle F$ is the force and when multiplied by the displacement produces the moment around the origin.The shear strain is described by the following equation;

$\displaystyle \gamma = r \frac{d\alpha}{dx}$

where $\displaystyle \frac{d\alpha}{dx} = \theta$ which is defined as the rate of twist. Since $\displaystyle \tau$ is proportional to $\displaystyle \gamma$ which is also proportional to $\displaystyle r$, the shear stress $\displaystyle \tau$ gets smaller in the material closer to the core because the distance away from the core $\displaystyle r$ gets smaller. Therefore it is more efficient and cost effective to remove material to make the cylinder lighter. The limits of the material will be tested but if calculations are done correctly then there shouldn't be anything to worry about.

The torque can be represented by the following;

$\displaystyle T = \int _A r G (r\theta) dA = G\theta\int _Ar^2dA$

where $\displaystyle dA$ is a function of $\displaystyle dy$ and $\displaystyle dz$.

$\displaystyle dA = dy * dz = r(dr)(dw)$

The strain modulus and rate of twist is not a function of $\displaystyle z$ or $\displaystyle y$ and is why they come out of the integral. The resulting integral is defined as the second polar moment of inertia.

$\displaystyle J = \int _Ar^2dA$

Second Polar Moment Of Inertia: Solid Circular Cross Section

We start with the equation for the polar moment of inertia $\displaystyle J = \int _Aa^2dA$ where a is the radius of the cross-section and dA represents the differential area. We then notice that the radius is a function of the area where the area of a circle is given by $\displaystyle J = \int _Aa^2dA \Rightarrow a^2 = \frac {A}{\pi}$ Therefore the integral is represented by; $\displaystyle J = \int _A(\frac {A}{\pi})dA$ Integrate the above equation and you get; $\displaystyle J = \frac {A^2}{2\pi}$ Replace $\displaystyle A$ with $\displaystyle \pi a^2$ and simplify to obtain the final solution for the second moment of inertia for a solid circular cross-section; $\displaystyle J = \frac {a^4\pi}{2}$

Hollow Circular Cross-Section: Thin Walled

The analysis of a hollow circular cross section with thin walls (ie. t << a), is conducted as follows:

$r_{i} = a$ (inner radius)

$r_{o} = b$ (outer radius)

The equation for the polar moment is done by superimposing 'empty' circle on top of one solid circle, and the resulting equation is as follows:

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

This can be expanded as:

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

If an average radius of the cross section is then defined as:

$\bar{r} = \frac{a + b}{2}$

And it is noted that, since t is very small,

$b \cong a$

and

$b \cong \bar{r}$

and

$a \cong \bar{r}$

Therefore,

$a^{2} \cong \bar{r}^{2}$ $b^{2} \cong \bar{r}^{2}$

and noting that $b - a = t$

J can then be expressed as:

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

J can then be expressed in terms of the average radius by noting that:

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

With $\pi \bar{r}^{2} = \bar{A}$ it's easily seen that J is proportional to $\bar{A}^{3/2}$, with $(2\pi ^{-\frac{1}{2}}t)$ being a proportionality factor.

Solid Circular Cross-Section vs Hollow Thin Walled Cross-Section

In order to compare these two cross-sections, we will be using the figure from the text Mechanics of Aircraft Structures by C.T. Sun. The figure is depicted below with all dimensions labeled on the figure.

We will first compute the area for each cross-section.

$\displaystyle A_a = \pi r^2 = \pi (1cm)^2 = 3.1416 cm^2$

$\displaystyle A_b = \pi r_o^2 - \pi r_i^2 = \pi (5.1cm)^2 - \pi (5cm)^2= 3.173cm^2$

As the computtion shows, the areas are similar despite the size difference. We will next compute the polar moment of inertia for each cross section.

$\displaystyle J_a = \frac{\pi r^4}{2} = \frac{\pi (1cm)^2}{2} = 1.5707cm^4$

$\displaystyle J_b = \frac{\pi (r_0^4 - r_i^4)}{2} = \frac{\pi ((5.1cm)^4-(5cm)^4)}{2} = 80.927cm^4$

Comparing the two Polar moment of inertia shows that the thin walled cross-section is a better torsional member than the solid cross-section. Since its polar moment of inertia (measure of its resistance to torsion) is much larger, it is a more desirable cross-section to use. The polar moment of inertia for the solid cross-section is only $\displaystyle 2%$ of the polar moment of inertia for the hollow thin walled cross-section.

$\displaystyle \frac{J_a}{J_b} = \frac{1.5707cm^4}{80.927cm^4} = .0194$

If the thickness to equal $\displaystyle .02 r_i^{(c)}$, where $\displaystyle c$ represents a new hollow thin walled cross-section, what would this inner radious $\displaystyle r_i^{(c)}$ equal such that the polar moment of inertia for figure (a) equals that of the new figure (c)?

Answer

$\displaystyle J_a = J_c$ $\displaystyle \frac{\pi r^4}{2} = \frac{\pi ((r_i+.02r_i)^4-r_i^4}{2}$ $\displaystyle \frac{\pi 1^4}{2} = \frac{\pi ((r_i+.02r_i)^4-r_i^4}{2}$ $\displaystyle 1.5707cm^4 = 1.5707((r_i+.02r_i)^4-r_i^4)$ $\displaystyle 1cm^4 = (r_i+.02r_i)^4-r_i^4$ $\displaystyle 1cm^4 = (1.02r_i)^4-r_i^4$ $\displaystyle 1cm^4 = .0824r_i^4$ $\displaystyle 12cm^4 = (r_i)^4$ $\displaystyle r_i = 1.8612cm$ Comparing the area of figure (a) to figure (c) shows that the area of the hollow thin walled cross-section has to be decreased by a considerable amount to have the same effect as a solid circular cross section has. $\displaystyle A_a = \pi r^2 = \pi (1cm)^2 = 3.1416 cm^2$ $\displaystyle A_b = \pi r_o^2 - \pi r_i^2 = \pi (1.898cm)^2 - \pi (1.861)^2= 0.4369cm^2$

Matlab Project Discription

For HW 3, the team was asked to analyze a NACA 4-digit airfoil series. The discription of the problem can be seen here. A diagram showing the shape of the airfoil in this series can be found on this website. The group is asked to break up the airfoil into a number (ns) of discrete sections in order to analyze the geometry of the airfoil.

In order to complete this problem the student must first understand how to discretize an object into triangular segments in order to find the area of the object. A diagram of how this is done is provided below.

File:Teamwikivectors.jpg

Diagram showing discretization method using triangles.

Using this, the following equations can be used in order to find the area of the shaded region.

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

Here, $d\vec{F} = \vec{PQ}$ and $q\vec{r} \times \vec{PQ} = (2dA)\vec{i}$

With

$||\vec{PQ}|| = d\mathit{l}$

Now that the basic foundation of the analysis has been discussed, the MATLAB code along with its execution is displayed below.

MATLAB Code

Airfoil Function function Airfoil(m,p,ta,c,P_0) m = m/100; p = p/10; ta = ta/100; check = 1; j = 1; while check ==1 ns = 20*j; y = linspace(0,1,ns); for i = 1:ns if y(i)<=p z_c(i) = (m/p^2)*(2*p*y(i)-y(i)^2); theta(i) = atan((m/p^2)*(2*p-2*y(i))); else z_c(i) = (m/(1-p)^2)*((1-2*p)+2*p*y(i)-y(i)^2); theta(i) = atan((m/(1-p)^2)*(2*p-2*y(i))); end z_t(i) = 5*ta*(.2969*y(i)^.5-.1260*y(i)-.3516*y(i)^2+.2843*y(i)^3-.1015*y(i)^4); y_u(i) = (y(i) - z_t(i)*sin(theta(i)))*c; z_u(i) = (z_c(i) + z_t(i)*cos(theta(i)))*c; y_l(i) = (y(i) + z_t(i)*sin(theta(i)))*c; z_l(i) = (z_c(i) - z_t(i)*cos(theta(i)))*c; end Area = 0; for i = 1:ns-1 PQ = [0 y_u(ns-i)-y_u(ns-i+1) z_u(ns-i)-z_u(ns-i+1)]; r = [0 y_u(ns-i+1)-P_0(2) z_u(ns-i+1)-P_0(3)]; a = cross(r,PQ)/2; Area = Area + a; end for i = 1:ns-1 PQ = [0 y_l(i+1)-y_l(i) z_l(i+1)-z_l(i)]; r = [0 y_l(i)-P_0(2) z_l(i)-P_0(3)]; a = cross(r,PQ)/2; Area = Area + a; end NS(j) = ns; A_BAR(j) = Area(1); if j ~= 1 if abs((A_BAR(j)-A_BAR(j-1))/A_BAR(j-1))*100<.01 check = 0; end end j = j+1; end A_bar = A_BAR(j-1) subplot(2,1,1), plot(y_u,z_u,y_l,z_l), xlabel('y'), ylabel('z'), title('NACA 2415 Airfoil') subplot(2,1,2), plot(NS,A_BAR), xlabel('ns'), ylabel('A_b_a_r'), title('ns vs. A_b_a_r') NACA 2415 Airfoil in MATLAB command window EDU>> Airfoil(2,4,15,.5,[0 0 0]) A_bar = 0.0253 EDU>> Airfoil(2,4,15,.5,[0 0.19 0.01]) A_bar = 0.0255 EDU>> Airfoil(2,4,15,.5,[0 0.25 -1]) A_bar = 0.0256

Torsion

Torsion of uniform, non circular bars leads to warping of the cross section, which is defined as an axial displacement along the x-axis (ie, along the bars length due to the conventions used in class) of a point on the deformed, or rotated, cross section. A schematic representation is shown below.

File:Teamwikitorsion.jpg

Diagram showing the linear displacements experienced by a uniform non circular object subjected to a torsion.

$u_{y}$ = the y-component of the displacement vector $\vec{PP^{'}}$

and

$u_{z}$ = the z-component of the displacement vector $\vec{PP^{'}}$

It should be noted (and easily seen) from the figure that $\vec{PP^{'}}$ is not on a circle centered at the origin with radius $\vec{OP^{'}}$ and $\vec{OP^{'}}$ in fact has a larger magnitude than $\vec{OP}$

This is due to the line $\vec{PP^{'}}$ having a vertical and horizontal displacement from $\vec{PP^{'}}$

However, since the angles associated with twist in torsion are likely to be relatively small, one can use small angle approximations and assume that the magnitude of $\vec{OP} = \vec{OP^{'}} = R$

$u_{z} = Rsin\alpha \cong R\alpha$

$u_{y} = R(1 - cos\alpha )$

With $\alpha = 0$, this makes $u_{y} = 0$ and $u_{z} = \alpha R$

The rate of twist angle, $\displaystyle \theta$, is defined as

$\displaystyle \theta = {\alpha \over x}$

Where $\displaystyle \alpha$ is the angle of rotation. The horizontal displacement (displacement in the y-direction) is

(1) $\displaystyle u_y = -\theta x z$

The vertical displacement of a point of the cross section (displacement in the z-direction) can be determined from

$\displaystyle u_z = +(PP') \cos \beta = +\alpha \zeta_p$

Where $\displaystyle (OP)\alpha$ can be substituted in place of $\displaystyle PP'$ and $\displaystyle \theta_x$ can be substituted in place of $\displaystyle \alpha$, thus resulting in the following

(2) $\displaystyle u_z = +\theta x y$

The displacement due to warping along the x-axis is defined as

(3) $\displaystyle u_x = \theta \psi (y,z)$

Equations 1, 2, and 3 make up the kinematic assumptions.

Roadmap for Torsional Analysis

The following list lays out a general step by step roadmap of equations to apply for the torsional analysis of an aircraft wing:

A. Kinematic assumptions (equations 1, 2, and 3 above)

B. Strain-displacement relation given by the two equations:

$\displaystyle \gamma_{yz} = {\partial u_x \over \partial y} + {\partial u_y \over \partial x}$

$\displaystyle \gamma_{zx} = {\partial u_x \over \partial z} + {\partial u_z \over \partial x}$

C. Equilibrium equations for stresses:

$\displaystyle {\partial \sigma_{yy} \over \partial y} + {\partial \tau_{zy} \over \partial z} + {\partial \tau_{xy} \over \partial x} = 0$

$\displaystyle {\partial \tau_{yz} \over \partial y} + {\partial \sigma_{zz} \over \partial z} + {\partial \tau_{xz} \over \partial x} = 0$

$\displaystyle {\partial \tau_{yx} \over \partial y} + {\partial \tau_{zx} \over \partial z} + {\partial \sigma_{xx} \over \partial x} = 0$

But, since $\displaystyle \tau_{yx}$ and $\displaystyle \tau_{zx}$ are independent of x in the displacement field, the equations reduce to the following:

$\displaystyle {\partial \tau_{yx} \over \partial y} + {\partial \tau_{zx} \over \partial z} = 0$

D. Prandt stress funtion $\displaystyle \phi (y,z)$:

$\displaystyle \tau_{yx} = {\partial \phi \over \partial z}$

$\displaystyle \tau_{zx} = -{\partial \phi \over \partial y}$

E. Strain compatibility equation:

$\displaystyle {\partial \gamma_{zx} \over \partial y} - {\partial \gamma_{yx} \over \partial z} = 2 \theta$

F. Equation for $\displaystyle \phi$

$\displaystyle {\partial^2 \phi \over \partial y^2} - {\partial^2 \phi \over \partial z^2} = -2G \theta$

G. Boundary conditions for $\displaystyle \phi$:

The traction free boundary condition $\displaystyle t_x = 0$ on a lateral surface yields the following:

$\displaystyle {d \phi \over ds} = 0$

thus, $\displaystyle \phi$ is a constant. But, this constant becomes arbitrary since it is a solid section with only one contour boundary, thus

$\displaystyle \phi = 0$

H. Resultant torque:

$\displaystyle T = 2 \iint_A \phi dA$

I. $\displaystyle T = G J \theta$ where

$\displaystyle J = -{4 \over \triangledown^2 \phi} \iint_A \phi dA$

$\displaystyle T = 2q \bar A$

J. Twist angle

$\displaystyle \theta = {1 \over 2G \bar A} \oint {q \over t}ds$

where $\displaystyle s$ is the curvilinear coordinate along a thin wall

K. 1) Multicelled Section

For cells where $i = 1, ... , n_{cell}$

$\tau = 2\sum_{i=1}^{N_c}{q_i \bar{A_i}}$

$q_i \;$ = Shear flow in cell i.

$\bar{A_i}$ = "average" area in cell i.

Define: $T_i = 2q_i\bar{A_i}$ torque generated by one cell.

$\tau = 2\sum_{i=1}^{N_c}{q_i \bar{A_i}}$

$T_i = \sum_{i=1}^{N_c}{T_i}$

2) Shape of airfoil is "rigid" in the plane $(X,Y)$ however it can warp out of the plane.

$\theta = \theta _1 = \theta _2= ...=\theta _{N_c}$

The twist angle equation applied to cell i.

 $\theta_i =\frac{1}{2 G_i \bar{A_i}}\oint_{}^{}{\frac{q_i}{t_i}ds}$

$G_i\;$ = Shear Modulus of cell i

$t_i(s)\;$ = thickness of walls on cell i. This is a linear curve coordinate along the cell wall.

Pb 1.1: Rectangular single cell section with modification.

Now more general with a single cell section:

$t_i=0.08 \; m$

$t_2= t_3=0.01 \; m$

$\bar{A}=\frac{1}{2} \pi (\frac{b}{2})^2 +\frac{1}{2}ba= 5.5708$

Shear flow comes from the torque equation: $T=2q\bar{A}$

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

Twist angle: $\theta_i =\frac{1}{2 G_i \bar{A_i}}\sum_{j=1}^{3}{\frac{q_jl_j}{t_j}}$ $j=1,2,3$ index for the segment number.

There is only one cell for $\theta_i\;$ so it reduces to $\theta\;$ in the equation above. You would normally have two indexes for multicelled section $(i,j)\;$.

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

Note: $\int$ is and elongated S, standing for continuous summation.
    
$\sum{}$ is a discrete sum.

Individual Contributions

Christopher PergolaEAS4200C.F08.WIKI.A 18:14, 8 October 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 19:54, 7 October 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 22:15, 7 October 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 18:29, 8 October 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 03:10, 8 October 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 13:11, 8 October 2008 (UTC)