User:Eas4200c.f08.aeris.krammer/HW2

=Class Notes=

Problem 1.1 continued
The member being analyzed is a shell or box beam and it can resemble an airfoil. Analyzing a simple box beam will give insight into more completed components such as the airfoil. This member is described with the word "shell" because the cross sectional thickness is very small and "beam" because it resists beaming. To the bottom right is a picture of a simple shell or box beam. Now, the problem posed in problem 1.1 is to find the ratio of dimensions b to a that will maximize the load-bearing capacity of the beam. First, in case 1, we assumed that the normal stress $$\displaystyle \sigma$$ will reach the allowable normal stress first. However, we must also include the condition in which the shear stress $$\displaystyle \tau$$ will reach the allowable shear stress first. This is designates as case 2. Recall, that the assumptions are that (1) the contour length of the box beam is equal to twice the sum of the dimensions a and b, (2) the magnitudes of the moment and torque loads on the beam are equal, and (3) the allowable normal bending stress is equal to twice the allowable shear stress. $$\displaystyle L=2(a+b)$$ $$\displaystyle |M|=|T|$$ $$\displaystyle \sigma_{allowable}=2*\tau_{allowable}$$ Also, recall that since the thickness of the shell beam is very small compared to the length and width, the shear stress on the member can be assumed to be uniform. As seen in fig. A, the arrows are the shear forces acting on the member. All the arrows are the same same denoting equal shear forces. Fig. B is the actual distribution of the shear forces on the member laterally along the y direction. Fig. C is the distribution of the stresses that is not a real representation of the stresses but is statically the same. The loading conditions are also shown below.  Knowing this knowledge allows for the steps of case 2 to be described. Assume $$\displaystyle \tau$$ reaches the allowable shear stress first. The purpose is to verify that $$\displaystyle \sigma_{max}<\sigma_{allowable}$$ given that the shear is maxed already. $$\displaystyle \tau=\frac{T}{2abt}$$ => $$\displaystyle T_{max}=2t\tau_{allow}(ab)$$ In the equations above, $$\displaystyle \tau$$ is the shear stress, T is the torque, a and b are the width and length respectively, and t is the thickness of the beam. Given this, in order to maximize torque, the product (ab) must be maximized where $$\displaystyle a=\frac{L}{2}-b$$. This means that this will only be one independent variable and the max torque is a function of either only b or only a. For conventional purposes, torque will be a function of b however, the problem can also be solved with torque being a function of a. Maximizing the product (ab) keeping the contour length or circumference constant will yield a square cross section of the beam. The derivation of these lengths are shown in the collapsible box below.

Box Beam with Rectangular Cross Section
A box beam with a rectangular cross section will be studied as a model for Aerospace Structures such as a fuselage or wings. Let's begin by analyzing shear stresses due to maximum Torque. Case 1: $$\displaystyle \tau_{max}=(\frac{T_{max}}{2abt})$$ Then equate the above formula in terms of maximum bending moment to obtain the allowable stress. $$\displaystyle (\frac{T_{max}}{2abt})=(\frac{M_{max}}{2(\frac{L}{8})(\frac{3L}{8})t})=\sigma_{allow}$$ It can be seen that: $$\displaystyle \tau_{max}=\sigma_{allow}$$  Based on assumption 3, $$\displaystyle \sigma_{allow}=2\tau_{allow}$$  This case is not accepted because: $$\displaystyle 2\tau_{allow}>\tau_{allow}$$  Case 2: Assume $$\displaystyle \tau_{max}=\tau_{allow}$$ ($$\displaystyle \tau_{max}$$ reaches $$\displaystyle \tau_{allow}$$ first). $$\displaystyle \tau=\frac{T}{2abt}=\tau_{max}=\tau_{allow}$$ $$\displaystyle T=(2t\tau_{allow})(ab)$$  $$\displaystyle T_{max}=(2t\tau_{allow})(ab)_{max}$$

Knowing that the lengths will equal L/4,the resulting max torque is $$\displaystyle T_{max}=2t\tau_{allow}(\frac{L}{4})^2=\frac{1}{8}tL^2\tau_{allow}$$ This value in turn equals the maximum bending moment given from assumption 2 ($$\displaystyle T_{max}=M_{max}$$). Now, obtaining the maximum bending moment, we must check that for the resulting maximum bending normal stress and decide whether it exceeds the allowable normal stress. $$\displaystyle \sigma_{max}=\frac{M_{max}b^{(2)}}{2I^{(2)}}$$ If $$\displaystyle \sigma_{max}<\sigma_{allow}$$ then the ratio $$\displaystyle \frac{b^{(2)}}{a^{(2)}}=1$$ is acceptable. Upon acquiring the steps to solve this problem we can determine the ratios numerically. For case 1, the numerical process is as follows. "I" the second area moment of inertia given by the general formula : $$\displaystyle I=\int\int_{A_{cross}}{z^2dydz}$$ Here, "z" is the distance from the moment axis to the point of interest. Generally, in order to increase the moment of inertia of a structure, one would design the structure so that a majority of the material is at a far distance away from the moment axis. Since in the formula the distance from t he moment axis is squared, it has a profound effect on the moment of inertia. In this problem, the distance from the moment axis is b/2. For a rectangle about the conventional y axis (as displayed on all above diagrams), the second area moment of inertia is: $$\displaystyle I=\frac{1}{12}bh^3$$ In this problem, for both cases, "I" can be found by by summing the moment of inertia of 4 segments of the rectangle along with using the parallel axis theorem. This gives the following equation. The four segments can also be referenced to the below figure of the case one thin-walled rectangular cross section.  $$\displaystyle I=\sum_{i=1}^4[\frac{b_ih_i^3}{12}+A_id_i^2]=\frac{2tb^3}{12}+2[\frac{at^3}{12}+at(\frac{b}{2})^2]\approx{\frac{tb^2}{6}(3a+b)}$$ The approximation sign assumes that the thickness of the walls is very small compared to the length and width of the cross section. This also means that the thickness cubed will be even smaller thus, the term with a $$t^3$$ can be neglected. So: $$\displaystyle \frac{I}{b}=tb(3L-4b)/12$$ The second derivative here is negative on the whole domain thus, setting the derivative to zero and solving for b will yield the value of b in which will maximize the second moment of inertia. <p style="text-align:center;">$$f^'(b)=0, 0=\frac{tL}{4}-\frac{2tb}{3} , b^{(1)}=\frac{3L}{8}$$ Plugging this value into assumption one will yield the corresponding a value. The ratio of $$\frac{b^{(1)}}{a^{(1)}}=3, M_{max}=2\sigma_{allow}(\frac{I^{(1)}}{b^{(1)}})=\frac{3tL^2}{32}\sigma_{allow}$$. The max bending stress is greater than the allowable bending stress. This ratio is not acceptable.

Torque, Shear Flow, and Shear Panels
Recall that from the problem 1.1 that the torque of the thin-walled cross section equaled to twice the shear stress multiplied by the product of the dimensions given by the equation $$ \tau=\frac{T}{2abt}$$. This expression for the torque's relationship with the cross sectional dimensions is only for a particular case where the cross section of the thin-walled beam or member being analyzed is rectangular. For a more general case of relating the torque applied on an object, it is necessary to introduce the concept of shear flow. Shear flow is the product of the shear stress and the thickness of the cross section. Usually, this thickness is very small compared to the contour length. For more general case of relating torque with dimensions and shear flow, a formula is introduced in section 3.5 of "Mechanics of Aircraft Structures" by C.T. Sun. <p style="text-align:center;">$$\displaystyle T= \oint{\rho qds}=\int\int_{A_{bar}}{2qdA}=2qA_{bar}$$ This formula describes the torque of a thin-walled bar with an arbitrary cross section of thickness t. Here, q is the shear flow, t is the thickness, A bar is the average area of the cross section, and T is the torque. In summary, for a nonuniform thin-walled cross section,$$T=2qA_{bar}:$$ <p style="text-align:center;"> and for a thin-walled rectangular cross section$$T=2 \tau ab$$: <p style="text-align:center;"> Deriving the torque for the thin-walled rectangular cross section was an ad hoc method because of the assumptions used to determine the equation. However, it is logical. The derivation of the non-uniform cross section torque is a method based on elasticity. Upon scrutinization of the stringers of an aircraft that is part of a monocoque structure, the open thin-walled box structure was previously determined to be more practical that the closed structure because of ease of riveting and installation. However, as the stringers were examined more carefully, it is realized that the vertical sections are slightly slanted. Why is this? That is because this structure is practical for stockpiling and storage and at the same time, not losing much moment of inertia due to the slanting of the vertical sections.

<p style="text-align:center;"> Starting chapter two, the structure of a shear panel is introduced. It is a thin sheet of material, usually metal aluminum, that can withstand shear load. There is a relationship that relates the shear modulus of a material and the shear strain with the shear stress that is applied the structure. $$\displaystyle \tau =G \gamma$$, where G is the shear modulus, and gamma is the shear strain. Gamma describes the change in a right angle between two fictitious lines in the material due to deformation. It can be expressed by displacements in an (x,y) coordinate system relative to the structure. <p style="text-align:center;">$$\displaystyle \gamma = \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=\frac{\partial u_{x}}{\partial y}+\frac{\partial u_y}{\partial y}$$ where $$u=u_x$$= the displacement along the x direction and similar for v. This shear strain is called the engineering shear strain and not the tensorial shear strain. These two quantities are different in the fact that $$\displaystyle \epsilon_{xy}=$$ (tensorial shear strain)$$\displaystyle = .5 \gamma_{xy}$$. The following picture describes the deformation of an infinitesimally small element (i.e. part of a shear panel) subject to shear forces. The blue shape is the deformed infinitesimally small rectangle. <p style="text-align:center;">

=Homework Problems=

Example Problem - Second moment of area
Problem:

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Actually, in class, I asked to do the problem with the channel cross section rotated by $$\displaystyle (-90^\circ)$$ relative to the orientation of your channel cross section. So you are solving a different problem. Eml4500.f08 20:50, 26 September 2008 (UTC)

Given two cross sections of the same area, determine the ratio of the second moment of area for the two cases shown.

To solve this problem we must find the second moment of area for each, then relate the two using the area constraint. For the first case, the circle, the second moment of area is $$I^{(1)}_{y} = \frac{\pi r^4}{4}$$.

Solution: Using the parallel-axis theorem and the known value for a rectangle, the second moment of area for case 2 can be calculated as follows: <p style="text-align:center;">$$I^{(2)}_{y} = \frac{ta^3}{12}+2[\frac{bt^3}{12}+tb(\frac{a}{2})^2]$$

Plugging in $$b=a$$ and $$t=a/10$$ gives: <p style="text-align:center;">$$I^{(2)}_{y} =\frac{a^4}{120}+\frac{a^4}{20}=\frac{7a^4}{120}$$

Given that the cross-sectional area of each case is the same, we are able to relate the radius of circle one to the dimensions of case 2. Specifically, we can relate $$r$$ to $$a$$. <p style="text-align:center;"> $$A^{(1)}_{}=A^{(2)}$$ $$\pi r^2=3at=\frac{3a^2}{10}$$ $$r=a\sqrt{\frac{3}{10\pi}}$$ We now have all the information needed to solve the ratio of second moment of area. <p style="text-align:center;">$$\frac{I^{(2)}_{y}}{I^{(1)}_{y}}= \frac{7a^4/120}{\pi r^4/4}=\frac{7a^4/120}{\pi (a\sqrt{\frac{3}{10\pi}})^4/4}=\frac{70\pi}{27}=8.145$$ This result shows that the second moment of area for case 2 is just over eight times larger than case one. The larger second moment of area allows for the element to withstand higher bending moments without yielding. This helps explain why stringers take on a shape closer to case 2 rather than a circular rod.

Problem 1.7 [C.T. Sun] with modifications
Problem Statement: Two load carrying beams are given with circular cross-sections, where $$\displaystyle R_{0}= 10 cm$$, $$t=\frac{1}{10}R_{0}=1 cm$$ and $$\displaystyle R_{1}$$ is such that the beam cross sectional areas are equal $$\displaystyle A^{(1)}= A^{(2)}$$

Question: Find the second moment of inertia of the two beams $$\displaystyle I_{y}^{(1)}$$ and  $$\displaystyle I_{y}^{(2)}$$ and compare them.

Solution: Case 1: The second moment of inertia for a circle is $$I_{x}=I_{y}=\frac{1}{4} \pi r^{4}$$ Thus, for this case: $$I_{y}^{(1)}=\frac{1}{4} \pi R_{0}^{4}$$. Also, $$A^{(1)} = \pi R_{0}^{2}$$

Case 2: The cross sectional area of the element in Case 2 is $$\displaystyle A^{(2)}=2R_{0}t+2\pi R_{1}^2=A^{(1)}$$. For the second moment of inertia calculation, for Case 2, the parallel axis theorem is to be used. Since the moment of inertia of a circular area about a centroidal axis is $$I_{x}=I_{y}=\frac{1}{4} \pi r^{4}$$, the moment of inertia for the circle of radius $$\displaystyle R_{1}$$ about its centrodial axis is $$I_{y}=\frac{1}{4} \pi R_{1}^{4}$$. Thus, <p style="text-align:center;">$$I_{y}^{(2)}= 2(\frac{\pi R_{1}^4}{4}+\pi R_{1}^2(R_{0}+R_{1}))+\frac{t(2R_{0})^3}{12}$$. Therefore, <p style="text-align:center;"> $$\displaystyle A^{(1)}=A^{(2)}=314.16 cm^{2}$$, $$\displaystyle R_{1}=6.84cm$$ The moments of inertia for Case 1 and 2 are <p style="text-align:center;"> $$\displaystyle I_{y}^{(1)}=7853.98 cm^4$$ and $$\displaystyle I_{y}^{(2)}=9055.29 cm^4$$ The moment of inertia in Case 2 is greater than the moment of inertia in Case 1.

Homework Statement from 9/10
Find the moment of inertia, with respect to y, of a circular cross section and a u-shaped cross section, as shown below.

CASE 1: Beam with circular cross section.

We begin with the definition of moment of inertia, I. Specifically here we are calculating moment of inertia with respece to the y-axis, Iy.


 * $$\ I_{y}=\int r^{2} dm $$

Solving for this integral gives us the moment of inertia of the circle in Case 1.


 * $$\ I_{y}=\int r^{2} dm$$


 * $$\ I_{y}=\int r^{2} \pi r dr$$


 * $$\ I_{y}=\int \pi r^{3} dr$$


 * $$\ I_{y}=\frac{\pi}{4} r^{2}$$

CASE 2: Beam with u-shaped cross section.

As can be seen in the diagram, we can divide the "u-shaped" beam into 3 separate parts, with the symmetrical side piece being equal. We can calculate the moments of inertia of these three parts separately, and then combine them using the Parallel Axis Theorem.

We know that the moment of inertia of a rectangle is equal to $$\frac{b^{3} h}{12}$$.

From here we can calculate the moment of inertia for each of the 3 pieces, centered about each centroid.

For the 2 symmetrical pieces.


 * $$\ I_{y}=\frac{b^{3} h}{12}$$


 * $$\ I_{y}=\frac{c^{3} b}{12}$$

Using the Parallel Axis Theorem, we can move the centroid of these moments to the centroid about the x-axis.


 * $$\ I_{y}= \frac{c^{3} b}{12} + \frac{(a + c)^{2} b c}{4}$$

Noting we have two of these pieces we can multiply this moment by 2.


 * $$\ I_{y}=2 \frac{c^{3} b}{12} + 2 \frac{a + c^{2} b c}{4}$$

For the bottom piece.


 * $$\ I_{y}=\frac{b^{3} h}{12}$$


 * $$\ I_{y}=\frac{a^{3} c}{12}$$

Because the centroid of this piece is already on the x-axis, we do not need to use the Parallel Axis Theorem here.

Combining the pieces.

We can combine the moments of the calculated pieces to obtain the total moment of inertia with respect to the y-axis, Iy.


 * $$\ I_{y,total}=\frac{c^{3} b}{6} + \frac{a^{3} c}{12} + \frac{(a + c)^{2} b c}{2}$$

= Contributions = Jin Yu Guan Eas4200c.f08.aeris.guan 19:50, 26 September 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 19:54, 26 September 2008 (UTC) Joshua Holladay Eas4200c.f08.aeris.holladay 19:55, 26 September 2008 (UTC) Nelson Caceres Eas4200c.f08.aeris.caceres 19:56, 26 September 2008 (UTC) Jesse W. McRae Eas4200c.fo8.aeris.mcrae 20:36, 26 September 2008 (UTC)