User:Eas4200c.f08.aeris.krammer/HW3

Shear flow and shear force in a thin-walled member
Open thin-walled sections are very common members in aircraft structures. Shear flow travels through these members, as can be seen by the directional arrows in the figure below.

A shear force is induced through this member due to the shear flow through the member.


 * $$\ dF=q*dl=q(dl_{x}\hat{i}+dl_{z}\hat{k}) $$


 * $$\ dF=q*dl=q(dl*cos(\theta)*\hat{i}+dl*sin(\theta)*\hat{k}) $$

We now integrate, from A to B as given in the figure above, to find the resultant force.


 * $$\ F=\int_{A}^{B}{dF}=q\left[\left( \int_{A}^{B}{dx}*\hat{i}\right) + \left( \int_{A}^{B}{dz}*\hat{k}\right) \right]$$


 * $$\ F=q\left(a*\hat{i}+b*\hat{k} \right)$$

NOTE: q is a constant with respect to x & z.

Now that we have the resulting sheer force vector we can calculate the resultant magnitude of this vector. Noting that this sheer force is broken down into an x and z component, we use Pythagorean Theorem to find the resulting magnitude.


 * $$\ \left| \left| F\right|\right|=\sqrt{F_{x}^2+F_{z}^2} $$


 * $$\ \left| \left| F\right|\right|=\sqrt{(q*a*\hat{i})^2+(q*b*\hat{k})^2}$$


 * $$\ \left| \left| F\right|\right|=q*\sqrt{a^{2}+b^{2}} $$

Torque, Shear Flow, and Polar Moment of Inertia
In order to calculate torque on a member induced by shear flow, it is necessary to analyze the average area in which that shear flow acts. In the collapsible box below, it is proved that a triangle's area is half the product of its base length and its height even if the height is measured outside of the triangle.

The torque on an arbitrarily shaped shear panel can be expressed in terms of shear flow, and the average area. The shear flow is the force per unit length the shear panel experiences and the average area is the area bounded by the contour of the shear panel relative to the designated origin of the coordinate system in which the shear panel is described. The math process of obtaining this equation is shown below. $$\displaystyle T= \oint{\rho qds}= \int\int_{ \bar{A}}{2qdA}=2q \bar{A}$$ The below picture describes the forces, shear flow, average area and coordinate system of an open thin-walled cross section.  PQ is the shear panel contour, A is the average area enclosed by the panel relative to the origin, e is the perpendicular vector arm from the origin to the force F (which is calculated by multiplying the shear flow with the length in with the shear flow acts). F is the essentially the equivalent force of the shear flow and is used to calculate the torque. Another equation can be written to accommodate for F and the vector e. $$\displaystyle Fe=T=2Q\bar{A}$$ Now, let's consider a uniform bar with circular cross section subject to torque and resulting torsion. This is not a case in which warping of the member is a problem and thus the cross section behaves like rigid disks with only angular displacement rather than lateral displacement (along the length of the bar). The picture below represents an infinitesimally small part of the circlar bar where dA is the infinitesimally small area. This area will be displaced in the (y,z) plane an angel gamma when the object is torqued.  The derivation of an equation to find the torque in terms of the bar's shear modulus, the polar moment of inertia of the cross section, and the rate of twist (gamma). $$\displaystyle T=\int\int_{A}{r \tau dA}=G \gamma \int\int_{A}{rdA}$$ where $$\tau =G \gamma $$ is obtained from Hooke's Law The below picture describes the deformation after torque when the vector r (vector from the origin to dA) is displaced by a small angle alpha to a new vector (r') in the (y,z) plane. $$\displaystyle \gamma= \frac{rd \alpha}{dx}$$ where $$\displaystyle \frac{d \alpha}{dx}=: \theta$$ (rate of twist) so $$\displaystyle T= \int_A{rG(r \theta)dA}$$ where dA= dydz, and G and $$\theta$$ are independent of y and z. $$\displaystyle = G \theta \int_A{r^2dA}=G \theta J$$ G is the shear modulus, theta is the rate of twist, and J is the polar moment of inertia. If a is the radius of the circular cross section then $$ J=\frac{1}{2} \pi a^4$$.

If it is a hollow, thin-walled (t<$$J=\frac{1}{2}\pi(b^4-a^4)=\frac{1}{2}\pi(b-a)(b+a)(b^2+a^2)=\frac{1}{2}\pi t(2 \bar{r})(b^2+a^2)$$ where $$\bar{r}$$ is the average radius ((a+b)/2) $$\displaystyle b^2 \cong \bar{r}^2$$ and $$a^2 \cong \bar{r}^2$$ so $$J=2\pi t \bar{r}^3$$

$$\displaystyle J=(2\pi^{1/2}t)(\pi \bar{r}^2)^{3/2}$$ where $$\displaystyle \bar{a}^=\pi \bar{r}^2$$ This means that J is proportional to the average area to the 1.5 power with a proportional factor of $$2\pi^{1/2}t$$

NACA Airfoil and Torsion of Uniform, Non-circular Bars
In homework 3, we are examining the NACA 4 digit airfoil series by using Matlab to do the analysis. Below are some pictures of some examples of the NACA airfoil. The one to the left describes fluid flow around the airfoil cross section (source: rit.edu) and the one to the right is a simple and crude cross sectional representation of the NACA airfoil with a (y,z) coordinate system where the y axis passes through the trailing edge. The y-axis is discretized in "ns" segments. The red circle highlights the infinitesimally small segment that will be analyzed. 

Here is the picture of the enlarged segment. As a crucial goal while analyzing the NACA airfoil is to determine the "average" area, we must devise a method to do this. It is possible to use a summation of infinitesimally small trapezoids, however, this method is not very elegant. In turn, we will use a method that utilizes triangles to determine the quadrature of the airfoil. The triangle will be a cross product a vector from an observation point (can be positioned anywhere) to a point on the airfoil and a vector of infinitesimally small length along the airfoil. These two vectors enclose a triangle of infinitesimally small area dA. This is shown in the picture below.  Torsion of uniform, non-circular bars are also examined in homework 3. This is the case in which warping will occur as opposed to the case of uniform, circular bars. Warping is the axial displacement along the x-axis of the bar (along the length of the bar) of a point on the deformed (rotated) cross section. The below pictures describe the bar and the bar's cross section when torsion is applied. Point P is a point on the (y,z) plane which moves to P' as the torque is applied. Beta describes the initial angular displacement measured from the y-axis and alpha is the angular displacement due to the torsion. Alpha is also measured from the y-axis.  Below is the magnified version of point P and its movement and displacement on the (y,z) plane of the cross section. $$u_y$$ and $$u_z$$ are the components of the displacement vector $$\vec {PP'}$$.  Why is the vector PP' perpendicular to OP? Isn't it more intuitive that one would draw an arc with radius |OP| and plot point P' at an angle alpha away from vector OP? The perpendicular nature of PP' to OP is because the angle alpha is very small. The proof is below. Let there be two vectors OP and OP' of equal length (|OP|=|OP'|=R). The displacement differences in the points P and P' are denoted by $$u_z$$and $$u_y$$ where z denotes vertical displacement and y denotes horizontal displacement. 

$$\displaystyle Rsin\alpha \cong R\alpha (\alpha small)$$ $$\displaystyle R(1-cos\alpha) \cong 0 (first order)$$ Thus, it is easy to see that since the horizontal displacement of point P' from point P is zero, point P' directly above point P in this case. That makes vectors OP and PP' perpendicular.

What is the displacement component $$\displaystyle u_{y}$$ and $$\displaystyle u_{z}$$? $$\displaystyle u_{y}=-(PP')sin\beta = -\alpha (OP)sin\beta$$

Where $$\displaystyle (PP')$$ is $$\displaystyle (OP)\alpha$$, $$\displaystyle - \alpha (OP)$$ is $$\displaystyle \alpha x$$, and $$\displaystyle sin\beta$$ is $$\displaystyle z_{P}$$. The rate of twist is denoted by theta, which is the twist angle per unit length $$\theta =\frac{\alpha }{x}$$ Therefore, the horizontal displacement $$\displaystyle u_{y}$$ is $$\displaystyle u_{y}=-\theta xz$$ and the vertical displacement $$\displaystyle u_{z}$$ is $$\displaystyle u_{z}=+(PP')cos\beta =+\alpha y_{z}$$ Where $$\displaystyle (PP')$$ is $$\displaystyle (OP)\alpha$$, $$\displaystyle + \alpha$$ is $$\displaystyle \theta y_{z}$$,. $$\displaystyle u_{y}=+ \theta xy$$. The warping displacement along the x-axis is <p style="text-align:center;">$$\displaystyle u_{x}=\theta \psi (y,z)$$ Equations (1), (2) and (3) are kinematic assumptions.

Roadmap for Torsional Analysis of an Aircraft Wing
Wing skin on the wing section is stiffened by stringers, spar webs and ribs. The supporting thin vertical webs form  a  structure that comprises of two or more cells. <p style="text-align:center;">

A. Kinematic assumptions (C.T. Sun: Mechanics of Aircraft Structures. Section 3.2.)

B. Strain-displacement relationship (C.T. Sun: Mechanics of Aircraft Structures. Section 3.2.)

C. Equilibrium equation for stresses (C.T. Sun: Mechanics of Aircraft Structures. Chapter 2; Section 3.6.)

D. Prandtl stress function $$\phi$$ (C.T. Sun: Mechanics of Aircraft Structures. Section 3.2, Equation 3.15.)

E. Strain compatibility equation (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.17)

F. Equation for $$\phi$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.19.)

G. Boundary conditions for $$\phi$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.24.)

H. <p style="text-align:center;">   $$\displaystyle T=2\int\int_{A}^{}{}\phi dA$$ $$\displaystyle T = GJ\theta $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$

I. Thin-walled cross-section Ad-hoc assumption on shear flow. Formal derivation can be found in C.T. Sun: Mechanics of Aircraft Structures. Section 3.5. <p style="text-align:center;">   $$\displaystyle T = 2q\bar{A}$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.48.)

J. Twist angle $$\theta$$: Method 1 <p style="text-align:center;">   $$\displaystyle \theta = \frac{1}{2G\bar{A}}\oint{\frac{q}{t}ds}$$ (C.T. Sun: Mechanics of Aircraft Structures. Equation 3.56.)

K. Section 3.6 on multicell thin-walled cross-section cell $$\displaystyle i=1, ...,n_{c}$$ for $$\displaystyle n_{c}=3$$

K1. <p style="text-align:center;">     $$\displaystyle T=2\sum_{i = 1}^{n_{c}}{q_{i}\bar{A}_{i}}$$ Where, $$\displaystyle q_{i}$$ = shear flow in cell i      $$\displaystyle \bar{A}_{i}$$ = "average" area in cell i       <p style="text-align:center;">$$\displaystyle T_{i}=2q_{i}\bar{A}_{i}\qquad$$ thus the torque generated by one cell: <p style="text-align:center;">$$\displaystyle T=\sum_{i=1}^{n_{c}}{T_{i}}$$

K2. Shape of airfoil is "rigid" in the (y,z) plane (but it can warp out of plane). <p style="text-align:center;">     $$\displaystyle \theta = \theta _{1} = \ldots = \theta _{n_{c}}$$ $$\displaystyle \theta _{i}=\frac{1}{2G_{i}\bar{A}_{i}}\oint\frac{q_{i}}{t_{i}}ds$$ Where, $$\displaystyle G_{i}$$= shear modulus of cell i.      $$\displaystyle t_{i}$$= thickness of cell i.       $$\displaystyle t_{i}(s)$$= curvilinear coordinate along the wall of the cell.

Single Cell Section
Problem 1.1 was an example for a rectangular single cell section. Now, let's consider a more general single cell section: <p style="text-align:center;">

Where $$\displaystyle t_{1} = 0.008 m$$ $$\displaystyle t_{2} = t_{3} = 0.001 m$$ $$\displaystyle a = 4 m $$ $$\displaystyle b = 2m $$ The area of the shape is the area of the semicircle plus the area of the triangle <p style="text-align:center;">$$\frac{}{A}= \frac{1}{2} \Pi ({\frac{b}{2}})^2 + \frac{1}{2}ba=5.57m$$

The shear flow of the shape is defined as the torque divided by two times the area: <p style="text-align:center;">$$\displaystyle q=\frac{T}{2\bar{A}}$$ The twist angle is calculated by breaking up an integral into a summation of three parts: <p style="text-align:center;">$$\displaystyle \theta=\frac{1}{2G\bar{A}}\sum_{j=1}^{3}\frac{q_{j}l_{j}}{t_{j}}$$ where $$\displaystyle j=1,2,3$$ is the index for the part number.

Airfoil Matlab Program
For the third homework assignment we were asked to develop a Matlab program to plot an airfoil profile and compute the average $$\bar{A}$$.

To begin we needed to start with a simple profile, a circle. The equation for a circle with the origin at the "leading edge" is $$z=\sqrt{R^2-(y-R)^2}$$. The circle is then divided up into tiny segments of length $$dy$$ along the y-axis. The circular foil is then integrated along the length of the chord (2R) while computing the small area, $$dA$$, for each segment. The area was calculated by taking half of the cross product of the first position vector (y) with the next (y+dy). $$S = \frac{1}{2}|{AB}\times{AC}|$$ The plot of the foil is shown below. The profile is displayed in the lower half while the segmented nodes are shown in the top half. As expected, the centroid of the circular airfoil is located in the center.

To determine how many steps were needed to accurately calculate the airfoils $$\bar{A}$$, the code was ran for different number of segments. The results are displayed on the graph below. To receive an error of 1% or less, the value of $$ns_{max}$$ is estimated to be 17 steps.



The same technique for finding $$\bar{A}$$ and the centroid location was applied to an NACA 2415 airfoil. Using the four digit number the equation for the mean camber line can be expressed using 2nd order polynomials. The thickness of the airfoil is calculated using another polynomial. Using trigonometry and the slope of the camber line ($$\frac{dz_c}{dy}$$), the y and z values for the nodes of each segment can be determined. The result of the NACA 2415 computation is shown below. The y-value of the centroid is 0.25 meaning the airfoil is a quarter-chord. The average $$\bar{A}$$ for the airfoil was calculated to be 0.0990.

$$\bar{A}_{airfoil}=0.0990$$

$$\bar{A}_{airfoil}=0.0247$$

= Contributions = Jin Yu Guan Eas4200c.f08.aeris.guan 13:21, 08 October 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 13:22, 08 October 2008 (UTC) Josh Holladay Eas4200c.f08.aeris.holladay 17:33, 8 October 2008 (UTC) Nelson Caceres Eas4200c.f08.aeris.caceres 13:24, 08 October 2008 (UTC) Jesse W. McRae Eas4200c.fo8.aeris.mcrae 13:25, 08 October 2008 (UTC)