User:Eas4200c.f08.gator.edwards/Problem 1.1cont.

This is a continuation of Problem 1.1 from the first homework.

Part of 1.1 from week one


Problem 1.1 is taken from the textbook Mechanics of Aircraft Structures by C.T.Sun, copyright 2006.

Find the optimum ratio of $$b/a$$ to maximize the load carrying ability of the beam. Assume $$M=T$$ and $$\displaystyle \sigma_{allowable}=2\displaystyle \tau_{allowable}$$.

The question is asking for the optimal cross section carrying the maximum bending moment and maximum torque.

Since the cross section walls are very thin, assume shear stress distribution in the thin walls of the beam are distributed uniformly. This results in the equation $$\displaystyle \tau=\frac{T}{2abt}$$.

Case 1: Assume $$\displaystyle \sigma$$ (bending moment stress) reaches $$\displaystyle \sigma_{allowable}$$ first. Verify $$\displaystyle \tau \le 2\displaystyle \tau$$.

From Mechanics of Materials, the normal stress caused by a bending moment is $$\displaystyle \sigma=\frac{Mz}{I}$$

Case 1
Assume that $$\displaystyle \sigma_{allow}$$ is reached first. Recall from mechanics of materials the equation below.

Next calculate the second area moment of inertia of the box beam. The details are shown below.

To find the maximum bending moment the $$\displaystyle \frac{I}{b}$$ term must be maximized to find the ratio of $$\displaystyle \frac{b}{a}$$. The details are shown below.

Now plug the maximum value of I/b into the bending moment equation to get an equation of Tmax, L, and t. Then compare the value to the shear stress max and compare.

Case 2
Assume $$\tau _{max} = \tau _{allow}$$ first.

$$\tau = \frac{T}{2abt} = \tau _{max}$$ because shear stress is uniformly distributed.

$$T=(2 t \tau _{all})(ab)$$ where $$ab$$ is the only variable, so therefore maximizing $$T$$ = maximizing $$ab$$

By substitution and from the assumption M=T:

$$T_{max}^{(2)} = (2t\tau_{allow})\left( \frac{L}{4}\right)^{2} = \frac{1}{8}tL^{2}\tau _{allow} = M_{max}^{(2)}$$

Now from the assumption $$\sigma _{allow} = 2\tau _{allow}$$ :

$$M_{max}^{(2)} = \frac{tL^{2}}{16}\sigma _{allow}$$

$$\sigma _{allow} = \frac{16}{tL^{2}}M_{max}^{(2)}$$

Recall: $$f(b) = \frac{I(b)}{b}$$ and $$b^{(2)} = \frac{L}{4}$$

$$f(b^{(2)}) = \frac{tb^{(2)}}{6}\left[ 3a^{(2)} + b^{(2)}\right]= \frac{t\frac{L}{4}}{6}\left[ 3\left( \frac{L}{4}\right) + \frac{L}{4}\right]$$

$$f(b^{(2)}) = \frac{tL^{2}}{24}$$

Now plugging this into the equation for  $$\displaystyle \sigma_{max} = \frac{Mb}{2I}$$,

we get $$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)}}{2f(b)} = \frac{24M_{max}^{(2)}}{2tL^{2}} = \frac{12M_{max}^{(2)}}{tL^{2}}$$

Recalling that $$\sigma _{allow} = \frac{16}{tL^{2}}M_{max}^{(2)}$$

We find that $$\sigma_{max}^{(2)} = \left( \frac{3}{4}\right) \sigma_{allow}^{(2)}$$

Since $$\sigma_{max}^{(2)}$$ is not greater than $$\sigma_{allow}^{(2)}$$, we find that Case 2 works.

Conclusion
Since we found that Case 1 was not acceptable since the maximum stress was greater that the allowable stress and Case 2 is acceptable because the max stress was less than the allowable stress, we can now conclude that Case 2 has the optimum cross-section. Since in Case 2, $$a = b = \frac{L}{4}$$ we conclude that the optimum cross-section is a square.