User:Eas4200c.f08.aeris.guan/hw2

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. 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. $$f^'(b)=0, 0=\frac{tL}{4}-\frac{2tb}{3} , b^{(1)}=\frac{3L}{8}$$ 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.