User:Eas4200c.f08.aero6.behnke/HW2

Problem 1.1
Given:


 * The beam of a thin-walled rectangular section shown in Figure 1 carrying a bending moment, $$M\!$$, and a torque, $$T\!$$. Assume that $$t\!$$ is much

smaller than $$a\!$$ or $$b\!$$. Recall the following relationship between shear stress and torsion and its derivation.
 * Because the cross-section walls of the beam are very thin, it can be approximated that the shear stress distribution is uniform across the wall, as shown in Figure 2.
 * Consider
 * $$T = T_{AB}+T_{BC}+T_{CD}+T_{DA}\!$$
 * where
 * $$T_{AB} = \frac{b}{2}\upsilon a = \frac{1}{2}\tau abt\!$$
 * $$T_{BC} = \frac{a}{2}\upsilon b = \frac{1}{2}\tau abt\!$$
 * $$T_{CD} = \frac{b}{2}\upsilon a = \frac{1}{2}\tau abt\!$$
 * $$T_{DA} = \frac{a}{2}\upsilon b = \frac{1}{2}\tau abt\!$$
 * Hence
 * $$T = 2\tau abt\!$$
 * Therefore
 * $$\ \tau = \frac{T}{2 a b t}\!$$

Constraints:


 * (1) The perimeter of the beam is of fixed length, i.e. $$L = 2(a+b)\!$$
 * (2) $$M = T\!$$
 * (3) $$\sigma_{allow} = 2\tau_{allow}\!$$ (This constraint is based on the Mohr's Circle shown in Figure 4)

Find:
 * The optimum $$b/a\!$$ ratio to achieve the most efficient cross section for the given constraints of the bending moment, torque, normal stress, and shear stress.

Solution:
 * Two possible solutions exist to maximize the ratio $$b/a\!$$, however only one of these solutions satisfies the given constraints. Therefore it is necessary to break the problem up in to two cases as follows.
 * Case 1: Assume that $$\sigma_{max}=\sigma_{allow}\!$$
 * Case 2: Assume that $$\tau_{max}=\tau_{allow}\!$$

Case 1
Recall the relationship between shear stress, bending moment, and moment of inertia shown below. Also recall that the moment of inertia, $$I\!$$, is a function of the geometry of the beam and therefore can be expressed as a function of the variable dimensions as shown in the collapsible table.

$$\sigma = \frac{Mz}{I}\!$$ EQN 1

where $$z\!$$ is the ordinate of a point, therefore equal to $$\frac{b}{2}\!$$. Therefore the $$z\!$$ can be replaced and the function can be rewritten to solve for $$M\!$$ as follows.


 * $$\sigma = \frac{Mb}{2I} \Longrightarrow M_{max}^{(1)} = \frac{2I\sigma_{max}}{b} = 2\sigma_{allow}\frac{I}{b} = T_{max}^{(1)}\!$$ by constraint 2

This final form is the most convenient because it breaks up the equation into the constants, $$2\!$$ and $$\sigma_{allow}\!$$, and the variables, $$I\!$$ and $$b\!$$, which are functions of $$a\!$$ and $$b\!$$.

The next task is to manipulate $$\frac{I}{b}\!$$ to be in terms of only one variable, and this is done by considering the fixed perimeter length specified in constraint 1. Then we maximize The function through differentiation and determine a final value for $$M_{max}^{(1)}\!$$.

We know
 * $$I = \frac{tb^2}{6}(3a+b)$$
 * and
 * $$a = \frac{L}{2}-b\!$$

Therefore
 * $$I = \frac{tb^2}{6}\left(3\left(\frac{L}{2}-b\right)+b\right) = \frac{tb^2}{12}\left(3L-4b\right)$$ EQN 2
 * $$\frac{I}{b} = \frac{tb}{6}\left(3\left(\frac{L}{2}-b\right)+b\right) = \frac{tb}{12}\left(3L-4b\right)$$


 * To maximized this equation it is necessary to recognize that $$L\!$$ is constant, then proceed as follows: [[Image:GraphIvsb.jpg|500px|thumb|Figure 7: Graph depicting relationship between I and b]]


 * $$\frac{d}{db}\frac{I}{b} = \frac{3tL}{12}-\frac{2b(4)t}{12}$$

Setting the derivative equal to zero allows the determination of critical point of the function.
 * $$\frac{3tL}{12}=\frac{2b(4)t}{12}$$


 * $$b = \frac{3L}{8}$$

since $$ L = 2(a+b)$$
 * $$a = \frac{L}{2}-b$$
 * $$a = \frac{L}{2}-\frac{3L}{8}$$


 * $$a = \frac{L}{8}$$

And the ratio of $$\frac{b}{a}$$ is  3  Plugging these dimensions into $$M_{max}^{(1)} = 2\sigma_{allow}\frac{I^{(1)}}{b^{(1)}} = 2\sigma_{allow}(\frac{tb^{(1)}(3L-4b^{(1)})}{12}) = \frac{3tL^{(2)}\sigma_{allow}}{32} = T_{max}^{(1)}\!$$

Now the shear stress due to $$T_{max}^{(1)}$$
 * $$\tau_{max}^{(1)} = \frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t} = \frac{M_{max}^{(1)}}{2(\frac{L}{8})(\frac{3L}{8})t} = \sigma_{allow}$$

Which is 2x too big to satisfy constraint 3. Therefore this case is unacceptable.

Case 2
Now assume that $$\tau_{max}=\tau_{allow}\!$$ (The max shear stress reached is equal to the allowable shear stress). Remembering the formula derived at the top of the page, it is rewritten to reflect the new initial conditions of this case as follows.
 * $$\ \tau_{max} = \frac{T_{max}}{2 a b t}\ = \tau_{allow}\!$$


 * $$\ T_{max} = \left({2t}{\tau_{allow}}\right)(ab)\!$$

where $$2t/\tau_{allow}\!$$ is a constant and $$ab\!$$ is variable, therefore $$ab\!$$ can be optimized.

Now that $$(ab)_{max}\!$$ has been calculated, it is fed back into the equation for $$T^{(2)}_{max}\!$$ and set equal to $$M^{(2)}_{max}\!$$ as a function of $$\sigma_{allow}\!$$.

Recall EQN 2 derived earlier, and replace $$b\!$$ with its equivalent value $$L/4\!$$. Next use EQN1 to solve for $$\sigma_{max}\!$$. Then use the relationship in EQN3 to define $$\sigma_{max}(\sigma_{allow})\!$$

Therefore $$\sigma_{max}<\sigma_{allow}\!$$ and this is the valid solution.

Advantages of Stringer Cross-Sections
Both cross-sections in Figure 8 have the same Moment of Inertia about the y-axis.

Advantages of the stringer cross-section:
 * 1. The stringer has an open, not closed, cross section. This makes it easier to fabricate with a punching die.
 * 2. The two vertical sides on the cross section are not parallel. This allows the beams to be stacked during shipment, which is a huge space saver.
 * 3. Assembly of the stringer beams to an aircraft is made a lot easier because of this open, not closed, cross-section. A riveting tool can easily be used without running into any interference.

Problem 1.7
Given Information - R = 10 cm; t = 1cm
 * Compare the load bearing capabilities of the two figures shown in figure 9.
 * Case 1: Large Circle
 * $$ \sigma = \frac{Mz}{I} $$
 * $$ I(circle) = \frac{\pi R^{4}}{4} = \frac{\pi 10^{4}}{4}$$
 * For the first equation z is just the radius of the big circle, R


 * $$ \sigma^{(1)} = \frac{4MR}{\pi R^{4}} = \frac{4M}{\pi R^{3}} = \frac{.00127}{cm^{3}}M^{(1)} $$


 * Case 2: Circular Beam
 * $$ A_{Total} = 100 \times \pi $$
 * $$ A_{rectangle} = 2R_{0} \times t = 2 \times 10 \times 1 = 20 cm^{2} $$
 * $$ A_{one circle} = (50 \times \pi) - 10 = \pi \times R_1^{2} $$
 * Solving for R1 you get: R1 = r = 6.84 cm


 * Now we need to calculate the area moment of inertia for the beam. You can do so by finding "I" for the rectangle, and then use the parallel axis theorem to find "I" for the two circles.
 * $$ I_{Rectangle} = \frac{2tR_{0}^{3}}{3} $$
 * $$ I_{circle} = \frac{a^{4}}{12} + (R_{0} + r)^{2}*(\pi r^{2})$$
 * $$ I_{total} = \frac{2tR_{0}^{3}}{3} + \frac{r^{4}}{6}+ 2*(R_{0} + r)^{2})*(\pi r^{2}) = 84473 cm^{4}$$

The z for this case is as follows:
 * $$ Z = R_{0} + 2 \times r$$
 * $$ \sigma = \frac{Mz}{I} $$
 * $$ \sigma^{(2)} = \frac{M(10 + 2 \times 6.84)}{84473} = \frac{2.80\times10^{-4}}{cm^{3}}*M^{(2)} $$

So far we have calculated the stresses, $$\sigma^{(1)}\!$$ and $$\sigma^{(2)}\!$$, for both cases in terms of the bending moment $$M\!$$. If we set these two equations equal to each other we can figure out for which case the bending moment is bigger. This is solved as follows:


 * $$\sigma^{(1)} = \sigma^{(2)} => \frac{.00127}{cm^{3}}M^{(1)} = \frac{2.80\times10^{-4}}{cm^{3}}*M^{(2)}$$

We can solve this for $$M^{(1)}\!$$ in terms of $$M^{(2)}\!$$ as follows:
 * $$M^{(1)} = \frac{\frac{2.80\times10^{-4}}{cm^{3}}*M^{(2)}}{\frac{.00127}{cm^{3}}} = 0.220*M^{(2)}$$
 * or
 * $$M^{(2)} = 4.55*M^{(1)}\!$$

Lets assume that the two separate cases have the same maximum stress:
 * $$\sigma_{max}^{(1)} = \sigma_{max}^{(2)}$$

This would mean that case 2 would have a maximum bending moment, $$M_{max}^{(2)}$$, that is about 4.5 times greater than the maximum bending moment for case 1, $$M_{max}^{(1)}$$.

Cross-section Optimization


The circular cross-section of a beam does not effitiently utilize the potential of its material properties. If the same area was redistributed in a manner that increases the moment of inertia, then the beam would have much greater bending resistance. Let us assume the following dimension relationships:


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


 * $$A^{(1)} = \pi R^{2} = 3at \!$$

where $$t = \frac{a}{10} $$ thus $$a = 3.23R\!$$

The moment of inertia for the circle, as previously described is as follows
 * $$I_{y} = \frac{\pi R^{4}}{4}$$

In order to compute the moment of inertia of the channel cross-section, one must first determine the location of the centroid in the Z direction. This is done by implementing the first or area moment of composite forms. Treating the channel as the overlapping of a solid square behind a slightly smaller square of negative area, the centroid calculation is as follows:

$$C_{y} = \frac{a^{2}*\frac{a}{2} - (a-t)(a-2t)*\frac{(a-t)}{2}}{a^{2} - (a-t)(a-2t)}$$ For simlicity, let us assume $$a = 10$$ and $$t = 1$$. Substituting back: $$C_{y} = \frac{10^{2}(\frac{10}{2}) - (10 - 1)(10 - 2(1))(\frac{10 - 1}{2})}{10^{2} - (10 - 1)(10 - 2(1))}$$ Thus $$C_{y} = 6.28$$ Normalizing this value with $$a $$ yields $$0.628a$$ from the base of the figure. Now computing the mass moment is the simple task of breaking down the channel into three rectangles, of which we already know how to calculate mass moment. The two vertical rectangles which lie symmetric to the axis of interest each contribute $$\frac{ta^{3}}{12}$$ and the top crossbar contributes $$\frac{at^{3}}{12} + (at)(a-0.628a-\frac{t}{2})^{2}$$. Thus the total contribution from the three sections is $$I_{y channel} = 2 \frac{ta^{3}}{12} + \frac{at^{3}}{12} + (at)(a-0.628a-\frac{t}{2})^{2}$$. Substituting $$a= 3.23R$$ into the formula leads to $$I_{y channel} = 2.947R^{4}$$ while $$I_{y circle} = 0.785R^{4}$$. This corresponds to a four fold increase for the given dimensions. The thinner the walls of the channel, the greater its mass moment and bending stiffness.

Shear Panels

 * The symbol gamma shown in Figure 11 is defined as the change in angle due to shear deformation. It is mathematically defined below as:
 * $$\gamma = \frac{\delta u}{\delta y} + \frac{\delta v}{\delta x}$$

Contributing Team Members
The following students contributed to this report:

Eduardo J. Villalba --Eas4200c.f08.aero6.villalba 08:03, 26 September 2008 (UTC)

Kris H. Loper Eas4200c.f08.aero6.loper 17:46, 26 September 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 19:26, 26 September 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 20:06, 26 September 2008 (UTC)

Felipe Ortega Eas4200c.f08.aero6.ortega 20:22, 26 September 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 20:22, 26 September 2008 (UTC)