User:Eas4200c.f08.wiki.c/homework report 2

= Sept 12th Notes =

A box beam is a rectangular structure that is important to study due to the fact that many aerospace structures follow this same basic design. A box beam is a form of shell beam, and structures such as an airplane fuselage are also shell beams, the only difference being that their cross sections are circular rather than a rectangle. Likewise, the wings on an airplane are also shell beams with a slightly more complex cross section.



Now that there is a motivation for solving the problem, Case 1 from last lecture is continued as follows:

$$\tau ^{(1)}_{max}$$ is the shear stress corresponding to the maximum torque $$T^{(1)}_{max}$$

Continuing the solution, replace $$T^{(1)}_{max}$$ in the shear equation by $$M^{(1)}_{max}$$. This is done due to the 1st assumption in the problem statement. Secondly, assumption 2 allows for the substitution of $$\tau ^{(1)}_{max}$$ with $$\sigma _{allow}$$

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

$$\tau ^{(1)}_{max} = \frac{M^{(1)}_{max}}{2(\frac{L}{8})\frac{3L}{8}t} = \sigma _{allow}$$

$$\tau _{max}^{(1)} = \sigma _{allow} = 2\tau _{allow} > \tau _{allow}$$

After these substitutions, it is easy to see that $$\tau ^{(1)}_{max}$$ is equal to $$\sigma _{allow}$$, which is twice as big as $$\tau _{allow}$$. Therefore this answer is not acceptable, and the solver must move on to case 2.

Pictured to the right is a graphical representation of normal stress. The stress is distributed through the length of the bar with and is equal to the force, F, divided by the cross sectional area A.

A graphical representation of this result can be shown using a Mohr's circle. As shown on the figure to the right, as long as the shear and normal stress stay enclosed within the circle, the structure will not fail. However, since in this case the maximum shear encountered is greater than half the allowable normal stress, this result is unacceptable

Case 2

Assume that $$\tau _{max}$$ reaches $$\tau _{allow}$$ first.

Therefore:

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

$$T = (2(t)\tau_{allow})(ab)$$

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

Everything on the right hand side of the above equation is constant, therefore to maximize the torque, the cross sectional area must be maximized. With the constraint of keeping the perimeter constant, it is easy to see that the largest cross sectional area will be achieved by making a equal to b, and thus having a square cross section as follows:

$$a^{(2)} = b^{(2)} = \frac{L}{4}$$