User:Eas4200c.f08.carbon.orear/HW3/Lecture Notes

=Open Thin Walled Cross Section=

Homework

 * Prove that the area of the triangle is $$A = \frac{1}{2} bh$$ of whether point C is at point D or not.


 * BC x BE|| = ||BC|| * ||BE|| * sin (θ)

h = ||BE|| * sin(θ)

b = ||BC||

Theta is between the two vectors.

Area = .5 * ||BC x BE|| (can be obtained from cross product definition)

Thus, Area = .5 * bh.

For an open thin walled cross section:

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

$$R e = T = 2 q \bar{A}$$ (Resultant force (R) times distance from origin (e)) {7.50}

=Uniform bar with circular cross section=

Warping vs no warping: (warping = deformation of cross section by axial displacement of a point). A uniform bar with circular cross section exhibits a non-warping case; the cross sections behave as rigid disks.

$$T = \iint\limits_A r \tau dA$$

$$\tau = G \gamma$$ (Hooke's Law)

$$\frac{da}{dx} =: \theta $$ rate of twist

$$T = \int\limits_A r G (r \theta) dA$$

G and theta are independent for (y and z).

$$dA = dydz = rdrdw$$

$$=G \theta (\int\limits_A r^2 da)$$

$$\int\limits_A r^2 da = J $$ – 2nd polar area moment of inertia

a = radius of circular cross section

Homework

 * Prove $$J = \frac{1}{2} \pi a^4$$

When integrating to solve for J in polar form (see external link ) we integrate about a full circle with respect to φ and from the inner to outer radial distance with respect to ρ. For our solid cross section this is from 0 to r (or also called, 0 to a).

$$J = \int_0^{2\pi} \int_0^a \rho^2 \rho\, d\rho\, d\phi = \frac{\pi a^4}{2}$$

We know these bound because we want to integrate about a full 360° (2π radians) and from the inner surface (at the origin in this case) to the outer radial distance.

=Solid vs Hollow Cross Section= Solid circle cross section Hollow circle cross section Thin-walled (t << a) $$r_i = a$$ (inner radius) $$r_0 = b$$ (outer radius)

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

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

$$(b - a) = t$$

$$(b - a) = 2 \bar{r}$$ (Average Radius)

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

$$b^2 \approx \bar{r}^2$$

$$a^2 \approx \bar{r}^2$$

Homework

 * Show this Approximation Rigorously


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


 * $$\displaystyle a = b - t$$


 * $$\displaystyle b = a + t$$


 * $$\bar{r} = \frac{(b + b - t)}{2}$$


 * $$\bar{r} = \frac{(a + t + a)}{2}$$


 * $$\bar{r} = \frac{(2 * b - t)}{2}$$


 * $$\bar{r} = \frac{(2 * a + t)}{2}$$


 * $$\bar{r}^2 = \frac{(2 * b - t)^2}{2^2}$$


 * $$\bar{r}^2 = \frac{(2 * a + t)^2}{2^2}$$


 * $$\bar{r}^2 = \frac{(2 * b)^2 - t^2}{2^2}$$


 * $$\bar{r}^2 = \frac{(2 * a)^2 + t^2}{2^2}$$


 * $$\displaystyle t^2 << a^2$$


 * $$\displaystyle t^2 << b^2$$


 * $$\bar{r}^2 \approx \frac{(2 * b)^2 - 0^2}{2^2}$$


 * $$\bar{r}^2 \approx \frac{(2 * a)^2 + 0^2}{2^2}$$


 * $$\bar{r}^2 \approx \frac{2^2 * b^2}{2^2}$$


 * $$\bar{r}^2 \approx \frac{2^2 * a^2}{2^2}$$


 * $$\bar{r}^2 \approx b^2$$


 * $$\bar{r}^2 \approx a^2$$

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

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

$$\pi \bar{r}^2 = \bar{A}$$

$$J$$ is proportional to $$\bar{A}^\frac{-1}{2}$$ with $$2 \pi^\frac{-1}{2} t$$ being a proportionality factor.

Homework

 * Comparing a Solid Circular Cross section to a hollow thin walled cross section we note:


 * Material Close to the core doesn't work as hard


 * Push the material to the maxium efficiency! Reduce weight by eliminating core material that doesn't contribute to our inertia as greatly!


 * Pg: 1.8 in the book


 * (a)	Soild circ cross section     $$ r_o ^{(a)} = 1 cm$$
 * (b)	Thin walled (hollow) cross section thickness = $$0.1 cm r_i ^{(b)} = 5 cm$$


 * Compute area for case a and b


 * $$ A^{(a)} = \pi r_o ^{(a) 2} = \pi 0.01^{2} = 0.000314159265 m^3$$


 * $$ A^{(b)} = \pi r_o ^{(b) 2} - \pi r_i ^{(b) 2}= \pi 0.05^{2} - \pi 0.049^{2}= 0.000311017673 m^3$$


 * Compute polor moment of interal (J) for a and b


 * $$ J^{(a)} = 0.5 * \pi r_o ^{(a) 4} = 0.5 * \pi 0.01^4 = 1.57079633 * 10^{-8} m^3$$


 * $$ J^{(b)} = 0.5 * \pi (r_o ^{(b) 4} - r_i ^{(b) 4}) = 0.5 * \pi (0.05^4 - 0.049^4 = 7.62148807 * 10^{-7} m^3$$


 * Compute the ratio of $$J^{(a)} / J^{(b)}$$


 * $$\frac{1.57079633 * 10^{-8} m^3}{7.62148807 * 10^{-7} m^3} = 0.0206101002$$


 * Find $$r_i^{(c)}$$ with $$t = 0.02 r_i ^{(c)}$$ such that  $$\displaystyle J^{(c)} = J^{(a)}$$, compute $$\frac{A^{(a)}}{A^{(c)}}$$


 * $$J^{(c)} = 0.5 * \pi ( (r_i + 0.02 r_i)^{(c) 4} - r_i^{(b) 4}) = J^{(a)} = 1.57079633 * 10^{-8} m^3$$


 * $$r_i^{(c)} = .0187 m$$

=Warping=

Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.

θ=α/x=rate of twist.

The following are mathematical definitions for the warping along the x,y, and z-axes.

1: y-displacement of a point on the cross section, horizontal

→ uy=-rαsinβ=-αy=-θxz

2: z-displacement of a point on the cross section, vertical

→ uz=v=rαcosβ=+(PP')cosβ=+αyp=+θxy, where PP'=OPα, α=θx, and yp=y because the variable is general for all coordinates

3: x-displacement, along centerline

→ ux=θψ(y,z), where this value is proportional to the rate of twist (θ)

These three equations compose the kinematic assumptions of warping.


 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Contribution Physical Representation of Torsion, from team Carbon
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.



Contribution by carbon.w
 * }

Roadmap for Torsional Analysis of Aircraft Wing
Below is a picture of a cross section of a wing with a two-cell stringer-skin-web section with a thin, variable wall.

A. Kinematic Assumptions described above with the three equations.

B. Strain-displacement relationship:

εxx = εyy = εzz = γxy = 0

σxx = σyy = σzz = τxy = 0

$$\gamma _{xz} = \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}$$

$$\gamma _{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}$$

C. Equilibrium Equation for Stresses:

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} = 0$$

$$\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \tau _{yx}}{\partial y} + \frac{\partial \tau _{zx}}{\partial z} = 0$$

$$\frac{\partial \tau _{xy}}{\partial x} + \frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \tau _{zy}}{\partial z} = 0 $$

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} = 0 $$

$$\tau _{xy} = \tau _{yx}$$

$$\tau _{yz} = \tau _{zy}$$

$$\tau _{xz} = \tau _{zx}$$

D. Prandtl Stress Funtion, $$\phi $$

$$\tau _{xz} = \frac{\partial \phi }{\partial y}$$

$$\tau _{yz} = \frac{\partial \phi }{\partial x}$$

E. Strain Compatibility

$$\frac{\partial \gamma _{yz}}{\partial x} - \frac{\partial \gamma _{xz}}{\partial y} = 2\theta $$

F. Equation for $$\phi $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}} = -2G\theta$$

G. Boundary Conditions for $$\phi$$

$$\frac{\partial \phi }{\partial s} = 0$$ or $$\phi $$ = constant (on the lateral surface)

If cross section is solid and thus has only a single contour boundary, then the constant can be chosen arbitrarily as zero: $$\phi$$ = 0 on lateral surface of the bar

H. Torque, $$T = 2\int \int_{A}^{}{\phi dA}$$

→ $$T = GJ\theta $$, where using the above equation and the equation for $$\phi $$ we obtain $$J = -\frac{4}{\bigtriangledown^{2}\phi }\int \int_{A}^{}{\phi dxdy}$$ and $$(\phi dxdy = dA)$$

I. Thin Walled Cross Section

$$\tau = \frac{T}{2abt}$$ Ad-hoc assumption of shear flow

$$T = \oint_{}^{}{\rho qds} = \int \int_{\bar{A}}^{}{2qdA} = 2q\bar{A}$$

J. Twist Angle, θ: Method 1

$$\theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\tau ds} = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}$$

K. Multi-cell Section
 * cell $$i=1,...,n_{cell}$$

For convenience, abbreviate $$n_{cell}$$ as $$n_c$$

$$T = 2\sum_{i=1}^{n_c} q_i\bar {A_i}$$ ( See Eq. 3.62 pg 93) where, $$q_i = $$ shear flow in cell i    $$\bar {A_i} = $$ "average" area in cell i

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

$$=> T = \sum_{i=1}^{n_c} T_i$$

K.2: Shape of an airfoil is "rigid" in the plane (y,z), although it can warp out of the plane.

$$\theta = \theta_1 = ... =\theta_{HC}$$

$$\theta_i = \frac{1}{2G_i\bar{A_i}} \oint \frac{q_i}{t_i} ds$$ where, $$G_i = $$ the shear modulus of cell i    $$t_i = $$ the thickness in cell i     *Note: $$t_i(s)$$, i.e. thickness can be a function of circular coordinate along the cell wall

Remember Problem 1.1, which was for the more general case of a rectangular single cell section.

Example



$$ A = \frac{1}{2}\pi R^2 = \frac{1}{2}\pi (\frac{b}{2})^2 $$ -area for half a circle $$ A = \frac{1}{2}ba$$ -area for the triangle

$$\bar {A_i}$$ is found to be:

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

Remember shear flow is given by: $$T = 2q\bar{A} => q = \frac{T}{2\bar{A}}$$
 * Note: T can be variable

And, the twist angle is given by: $$\theta = \frac{1}{2G\bar{A}} \sum_{j=1}^3 \frac{q_jl_j}{t_j}$$ where $$ j = 1,2,3 $$ index for segment number

Summations
This is supplementary material reference in class. Team Carbon has provided links and definitions to those who wish to explore these topics in greater detail.

Integration and summation are two different ways to find an area.

The integration symbol, $$\int$$, is an elongated s standing for countinuous summation.

Riemann Sum, $$\sum$$, is a discrete sum (noncontinuous). Integrals can be broken down into Riemann sums.

Quadrature

Quadrature- numerical integration using quadrangles (rectangles) to find the area for more information on quadrature visit this wikipedia article Numerical Integration

Quadrangle - shape with 4 sides (quadrilateral)
 * ex. squares, rectangles, trapizoids, etc.

cubature - when you subdivide a volume into cubes to find the volume of a given shape

Wikipedia article on Squaring of a Circle

=NACA 4-Digit Airfoil Series=

Let's begin with what the digits in a NACA 4-digit airfoil designation mean physically. When we look at the cross-section of the airfoil we can measure the following values.

The first digit describes the maximum camber as a percentage of the chord length.

The second digit describes the distance to the maximum camber from the airfoil's leading edge in tens of percents of the total chord length.

Two final digits describe the maximum thickness of the airfoil as a percentage of the chord length.

We can note patterns: a symmetric airfoil begins with 00--.

Goal: We will be introducing a [MATLAB] code in Homework 3 which will plot a NACA 4-digit airfoil, compute the centroid, and compute the area Ā.

We can discretize the y-axis by a number of segments that the airfoil is broken up into.



Begin with the cross-product:

r x PQ

We can substitute the length PQ with a force if we scale by the shear flow, q:

dT = r x dF = r x q (PQ)

Examining the equation:

qr x PQ

we can see that r x PQ can be replaced with 2dA î. In homework 4, we will see why dA can be relevant as a vector quantity pointing into our out of the page, but here it is fine to write it as a scalar.

=Torsion of a Non-circular Bar=

We will now examine how torsion of a non-circular bar is different from a circular bar. What we notice most is that warping of a cross section occurs. Warping, again, is axial displacement of a point on the rotated cross section. This means that when we apply torsion to rotate the bar, the point moves both axially and about the axis of rotation. If all points in a cross-section plane are coplanar before torsion, we can easily see that they will not be all coplanar after. This warping effect can be dramatic in our study of structural components.



Question: Why is the vector from P to P' straight, and not circular (an arc)? In other words, why is the P-P' vector perpendicular to line segment OP.

Answer: Simply, because α is very small, the displacement can be drawn in this manner. But why is that the case?

uz = L sin (α) is approximately = Lα

uy = L (1-cos(α)) is approximately = 0

by small angle approximations. This gives L and α a linear relationship seen in the picture.