User:Eas4200c.f08.gator.edwards/Wk3 Class Notes

Curved Panels
Looking at the shear flow in a curved panel: $$d\vec{F} = qd\vec{l} = q(dl_{y}\hat{j} + dl_{z}\hat{k})$$
 * $$= q(dlcos\theta\ \hat{j} \ +\ dlsin\theta \ \hat{k})$$

Therefore the resultant shear force vector is:


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

Now, the magnitude of the resultant shear force vector is:
 * $$\parallel \vec{F} \parallel = \sqrt{F_{y}^{2}+F_{z}^{2}}$$
 * $$=q\sqrt{a^{2}+b^{2}}$$


 * $$=qd$$


 * $$\parallel \vec{F} \parallel =qd$$

Now we will relate the magnitude of the resultant shear force vector (R) to the torque $$\ T=2q\bar{A}$$
 * $$\vec{T}= T\hat{i}$$


 * $$dF = qdl$$


 * $$d\vec{T}=\vec{r}$$ x $$d\vec{F}$$


 * $$ dT = rdFsin\theta= \rho dF$$


 * $$ = \rho q dl$$


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


 * $$T=\oint dT = q\oint \rho dl = 2q\int_{\bar{A}}dA$$


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

Circular Cross-Section
Uniform bar with circular cross section non warping.

$$T = \iint_A{r\tau \; \mathrm{d}A} = \int_A{rG(r\theta) \; \mathrm{d}A} = G\theta \int_A{r^2 \; \mathrm{d}A} = G\theta J = \frac{\pi G\theta}{2}r^4$$

Hollow thin walled cross section.

$$r_i = a \; r_o = b$$

$$J = \frac{\pi}{2} \left( b^4 - a^4 \right) = \frac{\pi}{2}(b-a)(b+a)(b^2+a^2) = \frac{\pi}{2}(t)(2\bar{r})(2\bar{r}^2) = 2\pi t\bar{r}^3$$

So the torque equation becomes

$$T=2 \pi G \theta t {\bar r}^3$$

NACA Airfoils
For the Matlab portion of the homework assignment, a portion is dedicated to finding the area enclosed by the airfoil. To do this, the airfoil will be broken up into many small segments, each of area dA. All of these segment areas will then be added up to come up with a total area. The second image on the right shows an enlarged region of the first image of the airfoil.

Looking at the image of the enlarged section, to find the area:
 * $$ d\vec{A} = \frac{1}{2} \ \vec{r}\ $$ x $$\ \vec{PQ}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$
 * $$ d\vec{A} = \frac{1}{2} \ \vec{r}\ $$ x $$\ \vec{PQ}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$

Torsion of Non-Circular Bars
warping = axial displacement along x-axis (i.e. along bar length) of a point on the deformed (rotated) cross-section.

Note: $$\vec{PP'}\quad$$ is the displacement vector
 * $$u_{z}\quad$$ is the z-component of the displacement vector
 * $$u_{y}\quad$$ is the y-component of the displacement vector

From the figure, we can see that the displacement is perpendicular to OP, but why? We can see that $$OP=OP'=R\qquad$$ and that $$u_{z} = Rsin(\alpha )\quad$$ and with $$\alpha\quad$$ being small, $$u_{z} = R\alpha\quad$$ By the same logic: $$u_{y}=R(1-cos(\alpha ))=R(1-1) = 0\quad$$. Therefore we can approximate the displacement as being perpendicular to OP.

 Kinematic Assumptions 
 * Starting with the definition of the rate of twist $$\theta = \frac{\alpha}{x}\quad \rightarrow \alpha = \theta x$$

$$u_{y}=-\theta x z \qquad \qquad (3.11)$$ $$u_{z}=PP'cos(\beta ) =OP\alpha cos(\beta ) = \alpha y $$ $$u_{z}=\theta x y \qquad \qquad (3.12)$$ $$u_{z}\quad$$ is the z displacement of a point on the cross-section. $$u_{x}=\theta \psi (y,z) \qquad \qquad (3.13)$$ The three equations above are the Kinematic Assumptions
 * Warping displacement along x-axis

A. Kinematic Assumptions
(Section 3.2 in textbook) See above three equations.

B. Strain-Displacement Relationship
(Section 3.2 in textbook) Strain is a non-dimensional value representing the ratio of the amount the material has deformed in one direction versus the original length of the material in that direction. $$\epsilon = \frac{\delta\ell}{\ell_{0}}$$

C. Equilibrium Equation for Stresses
(Chapter 2, Section 3.6) If a body is in equilibrium the stress field must satisfy these equations everywhere:

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

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

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

$$\tau_{xy}=\tau_{yx}\quad\tau_{yz}=\tau_{zy}\quad\tau_{xz}=\tau_{zx}$$

D. Prandtl Stress Function $$\phi$$
(3.2, Eq. 3.15) The Prandtl stress function is defined such that its derivative in the y-direction equals the shear stress in the plane perpendicular to the y-direction (the x-z plane), and that its derivative in the x-direction equals the negative shear stress in the plane perpendicular to the x-direction (the y-z plane). $$\tau_{xy} = \frac{\partial \phi}{\partial y}$$ $$\tau_{yz} = -\frac{\partial \phi}{\partial x}$$

E. Strain Compatibility Equation
(Eq. 3.17) $$\frac{\partial\gamma_{yz}}{\partial x} - \frac{\partial\gamma_{xz}}{\partial y} = 2\Theta $$

F. Equation for $$\phi$$
(Eq. 3.19) Combining the compatibility equation for torsion and Prandtl's stress function we get: $$\frac{\partial^{2}\phi}{\partial x^{2}} + \frac{\partial^{2}\phi}{\partial y^{2}} = -2G\Theta$$

G. Boundary Conditions for $$\phi$$
(Eq. 3.24) The traction free boundary condition tz = 0 is now: $$\frac{\partial\phi}{\partial s} = 0$$ Meaning that the stress function is constant at the surface. For a solid regular body the constant is arbitrary and is often made to be 0 for simplicity. $$\phi = 0$$

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

$$T=GJ\theta $$

$$J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$

(pg. 74)

I. Thin-walled Cross-section
Ad-hoc assumption on shear flow. Formal derivative (Section 3.5). We previously did the end of this section, which is easier than the beginning.

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

J. Twist Angle $$\theta$$: Method 1
$$\theta = \frac{1}{2G\bar{A}}\oint{\frac{q}{t}ds}$$

K. Section 3.6 on multicell thin-walled cross-section
Looking at the picture of the multicell wing we find that the number of cells, $$n_{c}$$ is equal to 3.

K1.
$$T=2\sum_{i = 1}^{n_{c}}{q_{i}\bar{A}_{i}}$$


 * Where $$q_{i}$$ = shear flow in cell i
 * and $$\bar{A}_{i}$$ = the "average" area of cell i

Define: $$T_{i}=2q_{i}\bar{A}_{i}\qquad$$ torque generated by one cell
 * $$\Rightarrow T=\sum_{i=1}^{n_{c}}{T_{i}}$$

K2.
Shape of airfoil is "rigid" in the yz plane (can warp out of plane). $$\theta = \theta _{1} = \ldots = \theta _{n_{c}}$$

$$\theta _{i}=\frac{1}{2G_{i}\bar{A}_{i}}\oint\frac{q_{i}}{t_{i}}ds$$


 * $$G_{i}$$= shear modulus of cell i.
 * $$t_{i}$$= thickness of cell i.
 * $$t_{i}(s)$$= curvilinear coordinate along cell wall.

Shear Flow: $$T = 2q \bar{A}$$
 * $$q=\frac{T}{2\bar{A}} \qquad$$ express q as a function of T

Twist Angle:
 * $$\theta = \frac{1}{2G\bar{A}}\sum_{j=1}^{3}{\frac{q_{j}l_{j}}{t_{j}}}\qquad$$ with l as the length of the segment and j as the index number.

NOTE:


 * $$\int \qquad$$: elongated S, standing for summation (continuous)
 * $$\sum \qquad$$: (discrete) sum