User:Uf.team.aero/HW2

= Problem 1.1=

Provided Information
A beam with closed thin-walled section is designed to carry a bending moment M, and torque T. The total wall contour length is constant and is given by $$L=2(a+b)$$. Here a and b are the fixed width and height dimensions, respectively. For the purpose of solving this problem, the bending moment is equal to the torque, and $$ {\sigma}_{allowable}= 2{\tau}_{allowable}$$.

Using the assumption that the thickness, t, is much smaller than the height and width (b and a, respectively), we can assume that the shear stress is equally distributed in the walls of the shell. Below is a diagram visualizing the shear flow in the box beam, as well as showing approximations of the force distribution as a resultant force $${\nu}$$.

Figure 3: Shear Stress Distribution and Resultant Shear Stress in thin-walled beam

The shear stress due to torsion is given by: $${\tau}=\frac {T} {2abt}$$

Assumptions
To re-iterate and make referencing easier, the assumptions are as follows:

1) $$L=2(a+b)$$ is fixed

2) $$M=T$$

3) $${\sigma}_{allowable}= 2{\tau}_{allowable}$$

4) $${\tau}=\frac {T} {2abt}$$

Objective
With the provided information and assumptions, find the optimum b/a ratio that will achieve the most efficient thin walled-section.

Case 1
Case 1 is the calculation of the maximum allowable moment when $$ \sigma_{max}\ $$ is reached first from bending stresses. The equation for normal stress from a bending moment is as follows:

$$ \sigma\ = {My \over I} $$

Maximum normal stresses occur at the top and bottom of the cross-section when y=b/2. The equation then becomes:

$$ \sigma_{max}\ = {Mb \over 2I} $$

Since $$ \sigma_{all}\ $$ is a known quantity and we want to solve for $$ M_{max} $$, we use some simple algebra to rearrange it into the form:

$$ M_{max} = { 2I\sigma_{allow}\ \over b} $$

In order to maximize $$ M_{max} $$, we want to maximize the portion of the equation $${I \over b} $$. The next step involves finding I, the second moment of area of the cross section of the beam. I can be found with a combination of the superpositioning and parallel axes theorem by:

$$I = \sum_{i=1}^4 \frac{b_{i}\,h_{i}^3}{12} + A_{i}\,d_{i}^2 = 2\left (\frac{t\,b^3}{12} \right) + 2\left [\frac{a\,t^3}{12}+at\left (\frac{b}{2} \right)^2 \right]$$

With the assumptions $$t<<b$$ and $$t<<a$$, we can reduce the second moment of area to:

$$ I = {tb^2 \over 6} (3a+b) $$

Using the relation from the constraints of the problem that $$ a = {L \over 2}-b$$ (assumption 1), the new equation for the second moment of area becomes:

$$I = {tb^2(3L-4b) \over 12}$$

We then proceed to maximize the $${I \over b} $$ equation by defining it as a function $$f(b)$$ and applying fundamental principals of calculus.

The maximum applied moment for the allowable normal stress then finally becomes:

$$ M_{max}^{(1)} = {3tL^2 \over 32}\sigma_{allow}\ $$

From the problem statement we are given that $$\tau_{max}\ = {T_{max} \over 2abt}$$ (assumption 4). We plug in our derived value of $$M_{max}^{(1)}$$ into $$T_{max}$$ (assumption 2) and solve for $$ \tau\ $$:

$$ \tau_{max}\ = {T_{max} \over 2abt} = {M_{max} \over 2abt} = \sigma_{allow}\ $$

By assumption 3:

$$ \tau_{max}\ = \sigma_{allow}\ = 2 \tau_{allow}\ > \tau_{allow}\ $$

Which cannot happen because $$ \tau_{max}\ $$ cannot exceed $$ \tau_{allow}\ $$

Case 2
For Case 2, allow $$ \tau_{max}\ = \tau_{allow}\ $$, and then check to see that $$ \sigma_{max}\ < \sigma_{allow}\ $$ by invoking the assumption that $$ T = M $$.

First, solve for


 * $$T_{max} = 2t \tau_{allow}\ (ab)_{max} $$

Note that all parameters are fixed except the product $$(ab)_{max}$$. To maximize the product, recall that a can be defined in terms of b:


 * $$ f_{2} = ab = (\frac {L} {2} - b)b = \frac {bL} {2} - b^2 $$

Now, the maximum torque based on allowable shear stress is found as


 * $$ T_{max} = 2t \tau_{allow}\ \frac {L^2} {16} $$

Invoking the assumptions that M = T and $$ \sigma\ = 2\tau\ $$ the maximum bending moment is found to be


 * $$ M_{max} = \frac {tL^2} {16} \sigma_{allow}\ $$

This defines the maximum allowable normal stress to be


 * $$ \sigma_{allow}\ = \frac {16M_{max}} {tL^2} $$

To be sure that the actual maximum normal stress has not exceeded the allowable normal stress recall the equation


 * $$ \frac {I} {b} = {tb(3L-4b) \over 12} = \frac {tL^2} {24}$$

Such that the maximum stress is given by


 * $$ \sigma_{max} = M_{max} \frac {b} {2I} = \frac {12M_{max}} {tL^2} $$

Comparing the maximum normal stress to the allowable normal stress, it is seen that


 * $$ \sigma_{max} = \frac {3} {4} \sigma_{allow} $$

Thus, Case 2 provides a square cross section such that the acceptable optimized ratio is


 * $$ \frac {b} {a} = 1 $$

= Mohr's Circle =

To help further explain the basis of the problem statement assumption that


 * $$ \sigma = 2\tau $$

it is convenient to examine a Mohr's Circle plot of a beam in uniaxial tension.



Since it can be seen by the figure of the axial member that normal stress lies only in the x-direction, a Mohr's circle would have one plot of principal stress equal to $$ \sigma_{max} $$ and one equal to zero. The relationship of shear stress to the normal stress can be seen visually on the Mohr's Circle below. Namely,
 * $$ \tau_{max} = \frac {\sigma_{max}} {2} $$.



Therefore, it is seen that the problem statement for Problem 1.1 uses a fundamental principle in its assumption that $$ \sigma_{max} = 2\tau_{max} $$.

= Derivation of Second Area Moment of Inertia for Circular Cross Section =

Formula
Or equation for the second area moment of inertia in rectangular coordinates is:


 * $$\int\int_A \,z^2\,dy\,dz$$

The transition into polar requires the following polar transformations:
 * $$z=r\,sin(\theta)$$


 * $$dy\,dz=r\,dr\,d\theta$$

The resulting second area moment of inertia in polar coordinates is:


 * $$\int\int_A \,r^3\,sin^2(\theta)\,dr\,d\theta$$

Derivation
The second area moment of inertia in polar coordinates integrated over the entire area results in the equation:


 * $$2\,\int_{0}^{\pi}{\int_{0}^{R}{ \,r^3\,sin^2(\theta)\,dr\,d\theta}}$$

By integrating and evaluating the first integral we get:

$$2\,\int_{0}^{\pi}{\frac{R^4}{4}\,sin^2(\theta)\,d\theta}=\frac{R^4}{2}\,\int_{0}^{\pi}{\,sin^2(\theta)\,d\theta}$$

To continue integration, it is necessary to implement the half angle identity $$sin^2(\theta)=\frac{1}{2} \left[1-cos(2\,\theta) \right]$$ so that our integral becomes:


 * $$\frac{R^4}{4}\,\int_{0}^{\pi}{\,1-cos(2\,\theta)\,d\theta}$$


 * $$=\frac{R^4}{4}\,\left[\theta-\frac{sin(2\,\theta)}{2}\right]_0^\pi$$

With our final evaluation, we receive the second area moment of inerta for a circular cross section:
 * $$I=\frac {\pi\,R^4}{4}$$

= Problem 1 =

Given
The following figures below are presented, case 1 and case 2. Case 1 is a circular figure with radius R, and area, $$ A^{(1)} $$. Case 2 is a 3 bar system made out of 3 bars, each with thicknesses t, and length and height of a, and be respectively. The area of case 2 is has the notation as $$ A^{(2)} $$. See the schematic below for more information concerning the problem information.



Objective
For this problem make sure to calculate the momemnt of inertia for both cases 1 and 2 and compare there values. Be sure to also locate the position of the centroid for case 2 in relation to the parallel axis, in this case the y-axis.

Solution
Using the following formulas to derive the moment of intertia for both cases, the circle and square:

For a circle:
 * $$ I_{y} = \frac {(\pi)R^4} {4} $$

For a rectangle:
 * $$ I_{y} = \frac {bh^3} {12} $$

Parallel axis theorem:
 * $$ I_{y} = I_{y,centroid} + Ad^2 $$

Using the formula for the moment of inertia for a circle with respect to the y-axis, we know that;
 * $$ I_{y} = \frac {(\pi)R^4} {4} $$

The area for case 1 is given by:


 * $$A^{(1)}=\pi R^2$$

and the moment of inertia is given as:


 * $$I_y=\frac{\pi}{4}R^4$$

Knowing that a = b, and that t=a/10, the equation for the area of case 2 can be reduced as follows:

= Importance of Stringers =

Stringers, or longerons, are critical to aircraft structures. Strings provide rigidity to the skin of the fuselage or wing, helping to support the aerodynamic loads. Stringers can be seen riveted to the fuselage skin in the cross-section of a 747 to the right.



As can be seen from the figure, stringers have an open cross-section with the vertical members canted outwards at a slight draft angle. These features are advantageous for ease of manufacturing and assembly. The skin must be riveted to the stringer, which is more easily done with an open cross-section rather than a closed, thin-walled section. Also, when manufacturing an open-section stringer, a punch can be used on sheet metal to obtain the form. This eliminates the need to roll and form a closed-section.

The outward canted draft angle provides a means for the string to be easily removed from the punch and is also convenient for stacking stringers in storage (as shown in the figure below).



= Problem 1.7 with Modification II: Circles =

Given
A circle has radius $$ R_{1} $$ and area $$ A^{(1)} $$. A dumbell shaped figure consists of a thin web of thickness $$ t = 0.1R_{1} $$, length $$ l = 2R_{1} $$ and area $$ A^{(2)} $$. On each end of the web are connected circles of radius $$ R_{2} $$.



Find
Find and compare the second area moment of inertia, $$ I_{y} $$, for the circle and dumbell if $$ R_{1} = 10 cm $$ and $$ A^{(1)} = A^{(2)} $$.

Equations
For a circle:
 * $$ I_{y} = \frac {(\pi)R^4} {4} $$

For a rectangle:
 * $$ I_{y} = \frac {bh^3} {12} $$

Parallel axis theorem:
 * $$ I_{y} = I_{y,centroid} + Ad^2 $$

Solution
Circle:

Use the formula derived in the previous problem to determine $$ I_{y}^{(1)} $$.


 * $$ I_{y}^{(1)} = \frac {(\pi)R_{1}^4} {4} = \frac {(\pi)(10 cm)^4} {4} = 7,854 cm^4 $$



Dumbell:

First, determine $$ R_{2} $$ such that $$ A^{(1)} = A^{(2)} $$.


 * $$ A^{(1)} = A^{(2)} $$


 * $$ (\pi)R_{1}^2 = 2R_{1}t + 2(\pi)R_{2}^2 $$


 * $$ R_{2} = [\frac {R_{1}^2} {2} - \frac {R_{1}t} {\pi}]^{1/2} = 6.84 cm $$

Next, use the parallel axis theorem to determine $$ I_{y}^{(2)} $$.


 * $$ I_{y}^{(2)} = \frac {t(2R_{1})^3} {12} + 2[\frac {(\pi)R_{2}^4} {4} + (\pi)R_{2}^2(R_{1} + R_{2})^2] $$


 * $$ I_{y}^{(2)} = \frac {2R_{1}^3t} {3} + 2[\frac {(\pi)R_{2}^4} {4} + (\pi)R_{2}^2(R_{1}^2 + 2R_{1}R_{2} + R_{2}^2)] = 87,469 cm^4 $$

Finally, compare $$ I_{y}^{(1)} to I_{y}^{(2)} $$.


 * $$ \frac {I_{y}^{(2)}} {I_{y}^{(1)}} = 11.1 $$

Therefore, it is shown that by redistributing the area of the circle into the shape of the dumbell, the second area moment of inertia increases by an order of magnitude.

= Shear Stress Equations =

Ad-Hoc vs Analytical Method
From the above schematic of Problem 1.1, shear stress across a rectangular cross section is approximated at $${\tau}=\frac {T} {2abt}$$

An analytic solution, applicable across any cross section, is: $$T = \oint \rho\,q\,ds = \iint_{\bar A} 2q\,dA = 2q \bar A$$

Where q is the shear flow, described by the equation $$q=\tau t$$, and Ā is the average area of the cross section.

Shear Strain
Engineering shear strain, not to be confused with tensorial, is described by the equation:

$$\gamma\ = \frac {\delta\ u} {\delta\ y} + \frac {\delta\ v} {\delta\ x} = \frac {\delta\ u_x} {\delta\ y} + \frac {\delta\ u_y} {\delta\ x}$$

Where u is the displacement along the x-axis and v is the displacement along the y. As you can see in Figure 14, the shear strain γ is analogous to the change in the right angle that the new parallelogram creates.

$$ \epsilon_{xy}\ = Tensorial\,Shear\,Strain = \frac {1}{2} \gamma_{xy}\ $$

= References =
 * http://en.wikipedia.org/wiki/Mohr%27s_circle
 * http://en.wikipedia.org/wiki/Longeron
 * http://en.wikipedia.org/wiki/Ad_hoc
 * http://en.wikipedia.org/wiki/Shear_strain

= Contributing Members = The following Team Aero members contributed to this report.

Jared Lee --Eas4200c.f08.aero.lee 16:18, 26 September 2008 (UTC)

Oliver Oyama --Eas4200c.fo8.aero.oyama 16:32, 26 September 2008 (UTC)

Ray Strods --Eas4200c.f08.aero.strods 18:08, 26 September 2008 (UTC)

William Kurth --Eas4200c.f08.aero.kurth 18:53, 26 September 2008 (UTC)