User:Eas4200c.f08.aero6.ortega/HW3

(1.3)
Problem statement: The dimensions of a steel (300M) I-beam are $$b = 50 \ mm,\!$$ $$t = 5 \ mm,\!$$ and $$h = 200 \ mm,\!$$ as shown in Figure 1. Assume that $$t\!$$ > and $$h\!$$ > are to be fixed for an aluminum (7075-T6) I-beam. Find the width $$b\!$$ > for the aluminum beam such that its bending stiffness $$EI\!$$ > is equal to that of the steel beam. Compare the weight-per-unit length of these two beams. Which is more efficient weightwise? Question quoted from Mechanics of Aircraft Structures by C.T. Sun



Young's Modulus for specified materials:
 * $$E_{steel} = 200 \ GPa\!$$
 * $$E_{aluminum} = 71 \ GPa\!$$

The Moment of Inertia for an I-beam as shown in Figure 1 is found by breaking up the cross-section into three rectangles and using the Parallel Axis Theorem to compensate for the centroids of each shape. The I-beam will be broken up into one $$t$$ by $$h-2t$$ rectangle and two $$b$$ by $$t$$ rectangles. Remembering the general form of the Parallel Axis Theorem and Moment of Inertia calculation, $$I$$ will be as follows:
 * $$I = \frac{t(h-2t)^3}{12} + 2\left(\frac{bt^3}{12} + bt\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right)$$ EQN 1

Therefore
 * $$I_{steel} = 7.612 \times 10^{-6} \ m^4\!$$

The problem states that $$E_{steel}\cdot I_{steel} = E_{aluminum}\cdot I_{aluminum}\!$$

Therefore
 * $$I_{aluminum} = \frac{E_{steel} \cdot I_{steel}}{E_{aluminum}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{71 \ GPa} = 2.144 \times 10^{-5} \ m^4$$

The next step is to solve EQN 1 for $$b\!$$ >, in terms of the known and fixed values specified in the problem statement, as shown below.
 * $$b = \frac{1}{2}\left(I_{aluminum}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.196 \ m = 196 \ mm$$

Now, to obtain the weight-per-unit length of the beams we need both the density of the materials being used and the cross-sectional areas for the beams.


 * For steel,
 * $$A_{steel} = 2 \cdot b_{steel} \cdot t+(h-2t) \cdot t = 1450 \ mm^2\!$$
 * $$\rho_{steel} = 7.8 \times 10^{-3} \ \frac{g}{mm^3}$$
 * so the mass-per-unit length for steel is:
 * $$m_{steel}=\rho_{steel} \times A_{steel}\!$$
 * $$m_{steel}=\left(7.8 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1450 \ mm^2\right)\!$$
 * $$m_{steel}=11.31 \ \frac{g}{mm}\!$$


 * For aluminum,
 * $$A_{aluminum} = 2 \cdot b_{aluminum} \cdot t+(h-2t) \cdot t = 2910 \ mm^2\!$$
 * $$\rho_{aluminum} = 2.78 \times 10^{-3} \ \frac{g}{mm^3}$$
 * so the mass-per-unit length for aluminum is:
 * $$m_{aluminum}=\rho_{aluminum} \times A_{aluminum}\!$$
 * $$m_{steel}=\left(2.78 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(2910 \ mm^2\right)\!$$
 * $$m_{aluminum}=8.09 \ \frac{g}{mm}\!$$

The weight-per-unit length of each beams is then
 * $$W_{steel}=110.84 \ \frac {N}{m}$$
 * $$W_{aluminum}=79.28 \ \frac {N}{m}$$

Therefore, the aluminum beam is more efficient weightwise.

(1.4)
Problem statement: Use AS4/3501-6 carbon/epoxy composite to make the I-beam as stated in Problem 1.3. Compare its weight with that of the aluminum beam.

So, repeating the same process as with problem 1.3, we now use the AS4/3501-6 carbon/epoxy composite. Now we use the Young's modulus of this material:


 * $$E_{composite}=140 \ GPa$$

From problem 1.3, we know the expression for the moment of inertia of this I-beam (EQN1 ). This gave us the moment of inertia of the steel beam:


 * $$I_{steel} = 7.612 \times 10^{-6} \ m^4\!$$

From the problem statement $$E_{steel}\cdot I_{steel} = E_{composite}\cdot I_{composite}\!$$, we find $$I_{composite}\!$$ to be
 * $$I_{composite} = \frac{E_{steel} \cdot I_{steel}}{E_{composite}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{140 \ GPa} = 1.087 \times 10^{-5} \ m^4$$

Now, rearranging EQN1 to solve for $$b\!$$ > we obtain the following:


 * $$b_{composite} = \frac{1}{2}\left(I_{composite}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.0843 \ m = 84.3 \ mm$$

This will yield the following cross-sectional area:


 * $$A_{composite} = 2 \cdot b_{composite} \cdot t+(h-2t) \cdot t = 1793 \ mm^2\!$$

Now, knowing $$\rho_{composite} = 1.55 \times 10^{-3} \ \frac{g}{mm^3}$$, we can obtain the mass-per-unit length and our weight-per-unit length


 * $$m_{composite}=\rho_{composite} \times A_{composite}\!$$
 * $$m_{composite}=\left(1.55 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1793 \ mm^2\right)\!$$
 * $$m_{composite}=2.78 \ \frac{g}{mm}\!$$

The weight-per-unit length of the composite beam is
 * $$W_{composite}=27.2 \ \frac {N}{m}$$

Therefore, the composite beam is more efficient weightwise than both the aluminum and the steel beams.

(1.5)
Problem statement: Derive the relations given by equations [1.4] and [1.5] of the textbook.

If we have constant shear $$\tau\!$$ on a curved panel, we can break down the shear force into two components:


 * $$V_x=\tau\,t\,a \qquad \qquad [1.4] \!$$
 * $$V_y=\tau\,t\,b \qquad \qquad \, [1.5] \!$$



From Figure 2, we can find an expression for the differential resulting force $$d\vec V$$:


 * $$d\vec V=q\,\vec {dl}$$

Where $$q\!$$ is the shear flow and $$\vec {dl}$$ > is the differential length that is being studied.

Knowing $$\vec {dl}$$ > has two components (one in $$x\!$$ and another in $$y\!$$), we are able to work out our answer.


 * $$d\vec V=q\,\left( dl_x \, \hat i + dl_y \, \hat j\right)$$
 * $$=q\,\left( dl\, \cos \theta \, \hat i + dl\, \sin \theta \, \hat j\right)$$
 * $$=q\,\left( dx\, \hat i + dy\, \hat j\right)$$

Now, integrating both sides,


 * $$\int {d \vec V}=q\, \left[ \left( \int_{x_a}^{x_b} {dx} \right)\, \hat i + \left( \int_{y_a}^{y_b} {dy} \right)\, \hat j\right]$$

and we also know $$\int_{x_a}^{x_b} {dx}=a$$ and $$\int_{y_a}^{y_b} {dy}=b$$

therefore,


 * $$\vec V=q\, \left( a\, \hat i + b\, \hat j\right)=qa\, \hat i + qb\, \hat j$$

and knowing $$q = \tau \, t \!$$ we now have


 * $$\vec V=\tau t  a\ \hat i + \tau  t  b\ \hat j$$

So this gives us the two components previously stated,


 * $$V_x=\tau\,t\,a \!$$
 * $$V_y=\tau\,t\,b \!$$

Derivation of $$T = 2q\bar{A}$$
As described in previous assignments, the thin walled cross-section is a special case when considering shear flow. As is common in aircraft structures, thin walled designs require great consideration. Since the wall of a structure forms a closed loop, the sum of the shear forces in the assigned $$x$$ and $$y$$ directions is zero, or $$\sum F_{x} = 0$$ and $$\sum F_{y} = 0$$. However, a shear force produces a resultant torque.

Referring to Figure 3, where the flow of the thin wall denotes a shear flow "q", the torque is described as $$dT = \rho\,qdl$$ where $$\rho$$ is the is the distance from the assigned origin to $$dl$$. Since $$\rho\,dl = 2dA$$, integrating the equation along a closed loop yields equation 3.48.

(3.48)
$$T = \oint \rho\,qdl = \iint_{\bar{A}}2qdA = 2q\bar{A}$$

Where $$\bar{A}$$ is the area enclosed by the shear flow. The torque may also be decomposed into a resultant force $$R$$and a moment arm $$e$$ as described in Mechanics of Aircraft Structures by C.T. Sun.

Thus $$R = qd$$ where $$d$$ is the total straight distance that the shear flow covers. This can be broken down into components in the $$x$$ and $$y$$ directions as illustrated by the figure.

(3.49)
$$F_{x} = qb\!$$

$$F_{y} = qh\!$$

Applying equation 3.48 to this geometry and noting $$\bar{A}$$ is the shaded area, then the location of the moment arm $$e$$ can be obtained.

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

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Area of a Generic Triangle


The area for a rectangle is it's base times it's height:
 * For Figure 5 the area of the rectangle outlined by the purple, red, and blue lines is: $$a \times h \!$$

If you insert a diagonal into the rectangle you will create two triangles with half the area as the rectangle
 * For Figure 5 the area of the triangle outlined by the blue and red lines is: $$\frac{1}{2}a \times h \!$$

Now, to calculate the area of the blue section you can use the above area we calculated for the triangle and subtract the area of the triangle that is outlined by the red lines as follows:
 * Triangle 1:$$\frac{1}{2}a \times h = \frac{ah}{2}\!$$
 * Triangle 2:$$\frac{1}{2}(a-b) \times h = \frac{ah}{2} - \frac{bh}{2}\!$$
 * Subtract Triangle 2 from Triangle 1
 * $$\frac{ah}{2} - \left(\frac{ah}{2} - \frac{bh}{2}\right) = \frac{1}{2}bh\!$$

This proves that the area for the blue section is half of it's base times it's height.

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

 * $$J = \int r^{2}dA$$


 * $$dA = 2\pi r \,dr\!$$


 * $$J = \int_{0}^{a}{2\pi r^{3}\,dr}$$


 * $$J = \left[\frac{2\pi r^{4}}{4} \right]_{r=0}^{r=a} $$

which yields...
 * $$J = \frac{\pi a^{4}}{2}$$

Thin-Walled Circular Cross-Section

 * $$r_{i} = a \!$$


 * $$r_{o} = b \!$$

Since t (the thickness) is very small, then


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


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


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

From the proof above for the area moment of inertia of a circular cross section, it can be noted that a similar derivation can be followed to obtain the area moment of inertia for a thin-walled circular cross section. One would simply change the integration boundaries from 0 to a, to a to b. This will result in the following equation.
 * $$J = \frac{\pi (b^{4} - a^{4})}{2}$$

This equation can be expanded to yield the following
 * $$J = \frac{1}{2}\pi (b-a)(b+a)(b^{2}+a^{2})\!$$

where
 * $$(b-a) = t\,;\ (b+a) = 2\,\bar{r}\,;\ b^{2}+a^{2}= 2\,\bar{r}^{2}$$

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

for a thin walled circular cross section

Comparison of Circular Cross-Sections with Thin Walled Circular Cross-Sections from Fig 1.8 (a) and (b)
To illustrate the effects of optimizing the cross-section of a beam, take the comparison of the torque that the thin walled circular cross-section can withstand, even though it will have the same cross-sectional area as the solid tube. $$T = GJ\Theta$$

Thus for a given material, variations in the torque on the cross-section are directly related to the geometry or polar moment $$J$$.

An approximation of the polar moment for a thin walled cross-section is $$J = 2t\bar{r}^3$$ where $$\bar{r} = (a + b)/2$$. The solid circular cross-section has the polar moment $$J = \frac{1}{2}\pi\,r^{4}$$. Thus the polar moments are as follows:

$$J_{thin wall} = 2*0.1*\pi\,*5.05^{3} = 80.91 cm^{4}$$

$$J_{solid} = \frac{1}{2}\pi\,(1^{4}) = 1.57 cm^{4}$$

Thus the thin walled cross-section containing the same area as the solid circle has a polar moment and torsional stiffness fifty times greater. This is due to the fact that as a beam is put under torsion, the torque is proportional to the deflection θ which is greater at the outer most portion of the cross-section. In other words, the material near the core is underutilized.

Torsion of a Cylinder Without Warping
We begin by describing the torque of a cross-section of the cylinder in order to eventually relate twist angle θ to torque.

$$T = \iint_{A} r\tau\,dA$$ Where $$\tau = G\gamma$$ by Hooke's Law. Continuing, $$\gamma = \frac{rd\alpha}{dx}$$ where $$\frac{d\alpha}{dx} = \Theta$$ by definition. Thus the rate of twist "θ" can now be input to the torque equation as follows: $$T = \iint_{A} rG(r\Theta)dydx$$ Where G and θ are independent of y and z.

Rewriting, $$T = G\Theta(\iint_{A}r^{2}da)$$ where the term in parenthesis is the second polar area moment of inertia J described earlier.

Warping
Warping is the axial displacement along the x-axis. Shafts exhibit warping under pure torque. There are out of plane displacements because of it. The rate of twist is defined as
 * $$\theta = \frac{\alpha }{x}$$

The in plane displacements are defined as
 * $$u_{y} = -\theta xz\!$$ (equation 3.11 in Mechanics of Aircraft Structures)
 * $$u_{z} = \theta xy\!$$ (equation 3.12 in Mechanics of Aircraft Structures)

The warping displacement along the x=axis is defined as
 * $$u_{x} = -\theta \psi\!$$ (equation 3.13 in Mechanics of Aircraft Structures)

Roadmap for torsional analysis of wing


A. Kinematic assumption (3.2)

B. Strain-displacement relationship (3.2)

C. Equilibrium equations for stresses (chapter 2, 3.2)

D. Prandt'l stress function 'φ' (3.2, 3.15)

E. Strain compatibility equation (3.17)

F. Equation for 'φ' (3.19)

G. Boundary condition for 'φ' (3.24)

H. $$T = 2\iint_A \phi\,dA $$(3.25)


 * Where $$T = GJ\Theta\!$$ and $$J = \frac{-4}{\triangledown^{2}\phi}\iint_A \phi\,dA$$ when warping is taken into consideration.

I. Thin walled cross-section with the ad-hoc assumption on shear flow

J. Twist angle θ: Method 1


 * $$\Theta = \frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$

K. Section 3.6 on multi-celled thin walled cross-section cells


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


 * Since $$T_{i} = 2q_{i}\bar{A}_{i}$$ so $$t = \sum_{i=1}^{n_{c}}T_{i}$$

Problem 3.10 adapted for wing: compute $$\bar A=\frac{\pi}{2}\left(\frac{b}{2}\right)^2 + \frac{ba}{2}$$
To calculate the area for Figure 10 split the figure into a triangle and a semi circle. Calculating the areas for these two shapes and then adding them will result in the total area of the figure. Area is calculated as follows:
 * Semicircle Area: $$\frac{\pi \left(\frac{b}{2}\right)^2}{2} = \frac{\pi}{2} m^2$$
 * Triangle Area: $$\frac{1}{2} ba = \left(\frac{1}{2}\right)(2)(4) = 4 m^2$$

Adding the two areas together results in a total area of:
 * $$\bar A = \left(\frac{\pi}{2} + 4\right) m^2$$

Next is the calculation of the twist angle of the airfoil. Assuming this airfoil is rigid leads to the constraint that the twist angle for one section of the airfoil will be equal to the twist angle for another section of the airfoil.
 * In other words: $$\theta = \theta_1 = \theta_2 \!$$

However, warping is experienced out from the plane of the airfoil. An example of this would be an airplane wing in torsion. At any one point there may be a twist angle between the wing tip and wing root; however, any section of the wing (aka. airfoil) will maintain its shape.

The equation for twist angle is given in the road map, part J. This generic equation can be substituted by the Riemann Sum equation as discussed in the section Comparison of Sigma and Integral Symbols (Quadrature/ Riemann Sum). Using this equation we can calculate the twist angle for the airfoil shape in Figure 7 as follows:
 * $$\Theta = \frac{1}{2G\bar{A}} \sum_{j=1}^{n_{c}} \frac{q_{j}l_{j}}{t_{j}} = \frac{1}{2G\bar{A}}\left(\frac{q_{1}l_{1}}{t_{1}} + \frac{q_{2}l_{2}}{t_{2}} + \frac{q_{3}l_{3}}{t_{3}}\right)$$ Factor out q since $$q_1=q_2=q_3\!$$
 * $$\Theta = \frac{q}{2G\bar{A}}\left(\frac{l_{1}}{t_{1}} + \frac{l_{2}}{t_{2}} + \frac{l_{3}}{t_{3}}\right) = \frac{q}{2G\bar{A}}\left(\frac{\frac{\pi}{2}b}{t_1} + \frac{a}{t_2} + \frac{\sqrt{a^2+b^2}}{t_3}\right)$$ Plug in values for b, a , $$\bar A$$, and $$q = \frac{2T}{\bar A}$$
 * $$\Theta = \frac{2T}{2G\left(\frac{\pi}{2}+4\right)^2}\left(\frac{\frac{\pi}{2}(2)}{0.008} + \frac{4}{0.01} + \frac{\sqrt{4^2+2^2}}{0.01}\right) = 39.95\frac{T}{G}$$
 * Where G is the shear modulus and is a material constant and T is the Torque and is a variable.

Determining Internal Area and Centroid of NACA 4-digit Series Airfoil
The NACA 4-digit series airfoil is a family of Airfoils whose shape are governed by the chord length. The four digits define certain characteristics of the wing as shown below:

For a NACA 2412 Airfoil
 * 2 $$\longrightarrow$$ maxmimum camber in percent chord length
 * 4 $$\longrightarrow$$ position of maximum camber in tenths chord length
 * 12 $$\longrightarrow$$ maximum airfoil thickness in percent chord length

More detail on determining the exact shape of a NACA 4-digit series airfoil is available here.

An elegant solution to finding the area of an airfoil is through triangle quadrature. Once an arbitrary point is defined, triangle radiate out of the point and touch the outer surfaces of the airfoil, refer to Figure 11.

The following MATLab code computes the cross-sectional area and centroid of any NACA 4-digit airfoil and plots the shape. Some of the most frequently used are graphed below.

Comparison of Sigma and Integral Symbols (Quadrature/ Riemann Sum)
$$\int$$ is an elongated S, denoting a continuous summation. $$\sum{}$$ denotes a discrete sum.

An example of the summation application would be the twist contributions to a polygon cell of a structure which need not be integrated with calculus.

$$\Theta = \frac{1}{2G\bar{A}} \sum_{j=1}^{n_{c}} \frac{q_{j}l_{j}}{t_{j}}$$

As described by professor Vu-quoc in class, the origin of the summation technique or Reimann Sum dates back to ancient civilizations and the need to estimate land masses for the purpose of property division and agricultural divisions. Any land form may be sub-divided into rectangular sections and then summed, knowing the finer the subsections are, the more accurate the area estimation.

The term for this sub-division system is quadrature. If taken to the third dimension approximation, the term cubiture describes the same principle.

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Contributing Team Members
The following students contributed to this report:

Marlana Behnke Eas4200c.f08.aero6.behnke 20:10, 7 October 2008 (UTC)

Kris Loper Eas4200c.f08.aero6.loper 23:12, 7 October 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 13:30, 8 October 2008 (UTC)

Felipe OrtegaEas4200c.f08.aero6.ortega 16:28, 8 October 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 17:41, 8 October 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 19:58, 8 October 2008 (UTC)