User:Eas4200c.f08.blue.a/Homework2

Homework Report #2

Continuation of Problem 1.1
Because the shear stress is greatest at the top and bottom of the beam section (where z = b/2), the equation of shear stress can be rewritten as follows: $$\sigma = \frac{Mz}{I}  \Rightarrow   M=\frac{2I\sigma\ _{allowable}}{b} $$

Thus, it is necessary to find the value of I in terms of b (recalling that a can be written as a function of b: a=L/2-b). From the problem parameters we know that for this instance, Mmax = Tmax. Because Mmax has been determined, it must only be put into the relation:

$$\tau _{max}=\frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t}=\frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t}$$

Now check the values of $$\tau _{max}$$ with the value of $$\tau _{allowed}$$.If $$\tau _{max}$$ is larger than $$\tau _{allowable}$$ the optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$would be acceptable. But $$\tau _{max}$$ is smaller than $$\tau _{allowable}$$, then the optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ would be unacceptable

Problem 1.1 Case 2
For this case we assume that the shear stress is eqaul to the maximum allowable shear stress in the section. Then we will test to see if the normal stress falls within acceptable tolerances (notably, less than $$\sigma _{max}$$)

By the equation given earlier in the problem the value of T becomes:

$$T=(2t\tau _{allowable})(ab)$$

Where the dimensions of the rectangle ab are equal to one another, hence a square. Therefore if the total perimeter of the square is L and all sides are equal, both a and b are equal to L/4. Plugging this quantity into the previous equation and using the given relation that M=T, we have:

$$M^{(2)}_{max}=\frac{1}{8}tL^{2}\tau _{allowable}$$

Using this relation and the previously defined equation for normal stress:

$$\sigma ^{(2)}_{max}=\frac{M^{(2)}_{max}b^{(2)}}{2I^{(2)}}$$

We can find $$\sigma ^{(2)}_{max}$$ and compare it with the allowable normal stress. If the maximum normal stress is less than the allowable shear stress, then the ratio of b to a for case 2 is acceptable. But, if it is greater, then that ratio is not correct.

Solving Problem 1.1 Case 1
First the moment of inertia, I, must be computed. This is done by summing the moments of inertia of each of the four parts of the section. In order to do this, the Parallel axis theorem must be applied. Solving for the horizontal axis yields:

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

And with the relation a=L/2-b, the plot of f(b)=I/b is shown to the right.

Taking the derivative of the function f(b) yields the maximum value of the curve to be b(1)=3L/8. Thus the ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ is equal to 3 and the value of a(1) is found to be L/8. Solving for $$\tau _{max}$$ with the values of a(1) and b(1) yield:

$$\tau _{max}=\frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t}=\frac{32M^{(1)}_{max}}{3tL^{2}}$$

but because this larger than half of $$\tau _{allowable}$$ this case is not possible

Motivation For Problem 1.1
The motivation for this problem is that we are trying to study a box beam with a rectangular cross-section as a model for aerospace structures. These structures have a variety of applications in the aerospace industry such as in the fuselage and the wings, for example.

In continuation of problem 1.1, we left off on lecture 7 discussing the results of the Case 1 example. We saw that the strongest structure ended up having a height that was 3 times longer than it was wide.

Lecture 8 presented the idea of looking into the shear stress caused by Tmax(1). Where:



Recalling that we allow this equivalences due to Assumption 2. Furthermore:



Recalling that this is due to Assumption 3. As you can see, this is not acceptable.



The Mohr's Circle defines the region by which sigma and tau interact on this beam. Notice that tau is exactly half of sigma.

Solving Problem 1.1 Case 2
Now, let's assume the maximum shear stress reaches the allowable shear stress first.



This is valid due to the assumption that there is a uniform stress distribution throughout the member.



It is clear to see that the only way to maximize T is by maximizing (ab), since thickness and allowable shear stress are fixed.



where (ab)max is the maximized value of (ab), which coincidentally is only when a=b=L/4

Problem Statement The problem is to find the optimal ratio of b/a in a thin walled beam to maximize the load-bearing capacity of a beam while assuming that the moment (M) is equal to the torsion (T).This is done by optimizing the cross section that carries the maximum bending moment and the maximum torsion.

Assumptions

In this problem, the first assumption$$^{(1)}$$ for the first case is that the bending normal stress reaches the allowable stress. In the second case$$^{(2)}$$, the shear stress is assumed to be equal to the maximum allowable shear stress. Lastly, the third assumption$$^{(3)}$$ is that the allowable stress is equal to the two times the allowable sheer stress.

Solution

We last left Problem 1.1 stating that $$ T_{max} $$ occurred when $$ a=b $$ and by recalling that $$ L=2(a+b) $$ the resulting value of $$ ab $$ is $$ \frac{L}{4}^{4}$$. This leads to

By Case 2: $$T_{max} = (2t\tau _{allowable})(\frac{L}{4})^{2} = \frac{1}{8}tL^{2}\tau _{allowable} = M_{max}^{(2)}$$

From these and by way of assumption 3, the allowable stress is

By Case 3: $$M_{max}^{(2)} = \frac{tL^{2}}{16}\sigma _{allowable}$$

$$\Rightarrow \sigma_{allowable}=\frac{16M_{max}^{(2)}}{tL^{2}}$$

Now, recall that $$f\left(b \right)=\frac{I}{b}$$

$$\Rightarrow f\left(b^{(2)} \right)=\frac{I^{(2)}}{b^{(2)}}=\frac{tb^{(2)}(3L-4b^{(2)})}{12}=\frac{tL^{2}}{24}$$ with $$b^{(2)} = \frac{L}{4}$$

$$\Rightarrow \sigma_{max}^{(2)} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}}$$

remember that $$\sigma = \frac{My}{I}$$ with $$y = \frac{b}{2}$$

so $$M_{max}^{(2)} = \frac{tL^{2}}{16}\sigma_{allowable}$$

$$\Rightarrow \sigma_{max}^{(2)}=\frac{12}{16}\sigma_{allowable}\leq \sigma_{allowable}$$

This result concludes that Case 2 is acceptable because the maximum stress is less than the allowable stress.

Aircraft Structure
Aircraft structures are assembled from various basic structural elements to help withstand loading conditions like axial, bending, and torsion. Some basic elements of an aircraft are the axial members(truss and bar elements), sheer panels, bending members(like beams), and torsional members(bar and beam torsion).

Homework Assignments - Moments of Inertia and Stringers
The first homework assigned in class consists of four individual problems.

The first Problem is to define the moment of inertia $$I_y$$ for a solid circular cross section through integration using polar coordinates. The integral being $$I_y=\int\int_A(Z^2dA)$$.

Also the derivation of $$I_y$$ for a square cross section is required which would be derived by the use of the integral $$I_y=\int\int_A(Z^2dYdZ)$$

The second problem assigned required comparing the moment of inertia defined for the circular cross section (case:1) in the previous problem to a redistributed equivalent area (case:2)

Where $$a=b$$, $$ A_2=3(a)(t)=\pi(r^2)$$,  and $$ t=a/10$$

The moment of inertia will be calculated assuming the y-axis passes through the centroid of case:1 and case:2 for fair comparison. The moment of inertia for case:2 will be calculated by using the fact that the moment of inertia for a rectagular cross section is $$I_y=1/12(b)(h^3)$$ along with the parallel axis theorem which is $$I_y=I+Ad^2$$

The third problem assigned was a modified version of problem 1.7. The modification was that all rectangular surfaces be changed to circular surfaces. The problem is to compare the moment of inertia $$I_y$$ of a circular cross section (case:1) to a modified cricular cross section (case:2) calculating the moment of inertia $$I_y$$ for both cases through the centroid of the cross section.

Where $$r_0=10cm$$, $$t=r_0/10$$, and $$A_1=A_2$$

The moment of inertia will be calculated for the circle from the answer to problem number one since it will be the same the moment of inertia for case 2 will be from a combination of rectangle moment of inertia and circular moment of inertia with use of the parallel axis theorem.

The fourth problem assigned was to show why a open rectangular cross section would be preferable to a hollow recangular cross section by comparing moment of inertia, manufacturing, and assembly techniques.

How Ahead We Have Been
While the pace of the course seems to be slow, we have actually been covering material several chapters ahead of where we are currently at. An example is found when comparing problems 1.1 and 3.5. As it turns out, problem 1.1 is just a specific case of problem 3.5.

For the first case, the box, $$tau = T/2abt$$. In the second case, T = 2qA, where A is the average area and $$q = tau*t$$. This is a case where the first example, the box is a specific case of the second figure. The box problem is solved using the ad-hoc method and the generic shape is solved using a method based on the elasticity of the object.

Getting back to the airplane stringer problem, the question was posed last time as to why the walls of the stringer are sloped outward rather than being parallel as shown below. Also the question of why an open cross section is used rather than a closed cross section.

First off, there are 2 reasons why an open cross section is used. 1) Manufacturing - sheet metal is used to make the stringer. It is much easier to stamp an open cross section than to extrude a closed cross section. 2) Airplane construction - stringers are rivited to the outside skin of the plane and it is next to impossible to rivit a closed cross section. To use a closed cross section, spot wels would be needed.

Next, why are the walls of the stringer not parallel. It was suggested that the reason why they are not parallel is for storage purposes. Stringers can be stacked vertically when being stored prior to construction which saves a lot of space in a factory.

Shear Panels, Shear Stress/Strain
Engineering Shear Strain (not torsional shear strain)

The angle between the original and deformed box is shear strain which is defined as dv/dx. Using small angle approximations, the angle is equal to the tangent of the angle(i.e deg = tan(deg) based on small angle approximations)

shear strain = (du/dy)+(dv/dx)

Tensorial shear strain is defined as (1/2)(shear strain)xy.

Contributing Team Members
Mark Barry Eas4200c.f08.blue.a 13:42, 25 September 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 18:44, 25 September 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 22:27, 25 September 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 05:18, 26 September 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 08:25, 26 September 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 05:21, 26 September 2008 (UTC)