User:Eas4200c.f08.vqcrew.d/hw2

Problem 1.1
Given Figure 1, find the optimum ratio $$\frac{b}{a}$$  to maximize the load bearing capability of the thin-walled beam.

The assumptions for this problem are as follows:
 * 1) The cross-sectional perimeter is constant, i.e. $$L=2(a+b)=Constant$$
 * 2) The torque $$T$$ is equal to the bending moment $$M$$, i.e. $$T = M$$
 * 3) The maximum allowable bending normal stress is twice the maximum allowable shear stress, i.e. $$\sigma_{allow} = 2\tau_{allow}$$

The assumption that $$t$$ is very small compared to $$a$$ and $$b$$ is also valid.

Since the cross-section walls are very thin, the shear stress can be assumed to be uniform along the walls, as shown in the shear flow diagram in Figure 2. The shear distribution and resultant shear vector $$V$$ for a small piece of the wall is also shown. This distribution is also mechanically equivalent to the linear distribution shown in Figure 3, where $$\tau =\frac{V}{t}$$.



The total torque ($$T$$) can be described as the sum of the torques on each wall of the box beam as follows: $$T = T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

$$T_{AB}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{BC}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

$$T_{CD}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{DA}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

Therefore, $$T=2t\tau ab$$, and $$\tau =\frac{T}{2abt}$$.

There are two cases that must be analyzed in order to solve the problem:
 * 1) Assume the bending normal stress (σ) reaches $$\sigma_{allow}$$ first, then verify that $$\tau \le \tau_{allow}$$.
 * 2) Assume the shear stress (τ) reaches $$\tau_{allow}$$ first, then verify that $$\sigma \le \sigma_{allow}$$.

Case 1

Case 1 begins by recalling the formula for bending normal stress: $$\sigma =\frac{Mz}{I}$$, where $$M$$ is the bending moment, $$z$$ is the ordinate of a point on the axis perpendicular to the bending neutral axis, and I is the second area moment of inertia defined by $$I=\iint_A z^2 \ dy\, dz$$. In this case, $$z=\frac{b}{2}$$. Figure 4 shows each item pictorially.



Rewriting this equation yields $$M=\frac{2I\sigma _{max}}{b}$$, where $$\sigma _{max}=\sigma _{allow}$$ from the Case 1 assumption. We proceed to find $$M_{max}$$ by maximizing the ratio of $$\frac{I}{b}$$.

The expression for finding the moment of inertia $$I$$ of the box beam is: $$I=\sum_{i=1}^{4}{\left[\frac{b_{i}h_{i}^3}{12} + A_{i}d_i^2\right]}$$, where $$i$$ represents each wall of the box beam, varying from one to four. This simplifies to $$I=2 \frac{tb^3}{12}+2\left[\frac{at^3}{12}+\frac{atb^2}{4} \right]$$.

The first sum is the result of the vertical bars, and the second sum is the result of the horizontal bars. Because $$t\ll b$$, the $$t^3$$ term will be very small compared to $$b$$ and can be neglected. After some algebra, $$I$$ can be simplified to: $$I=\frac{tb^2}{6}\left(3a+b \right) $$.

Let us now define a function $$f(b)\equiv \frac{I}{b}$$ and maximize it to find the maximum bending moment, $$M_{max}$$. By using Assumption 1, we can substitute $$a$$ with $$a=\frac{L}{2}-b$$. Replacing this in the expression for $$I$$ results in the concave-downward quadratic shown in Figure 5: $$f(b)=\frac{tb}{12}\left(3L-4b \right)$$. By setting $$f(b)=0$$, the roots are shown to be $$b=0$$ and $$b=\frac{3L}{4}$$. Since the function is a quadratic, the maximum value of $$f(b)$$ is located halfway between the two roots at $$b^{(1)}=\frac{3L}{8}$$. By Assumption 1, we find $$a^{(1)}=\frac{L}{8}$$.

The maximum bending moment $$M_{max}^{(1)}$$ can then be written as:

$$M_{max}^{(1)} = \left(2\sigma _{allow} \right)\left(\frac{I^{(1)}}{b^{(1)}} \right) = \left(2\sigma _{allow} \right)\left(\frac{tb^{(1)}}{6} \right)\left(3a^{(1)}+b^{(1)} \right)$$

Substituting in for $$a^{(1)}$$ and $$b^{(1)}$$ yields: $$M_{max}^{(1)} = \frac{3tL^2}{32}\sigma _{allow}$$

This is also equal to $$T_{max}^{(1)}$$ based on Assumption 2. We will now use this assumption and the shear stress equation to determine the maximum shear stress for Case 1.

$$\tau _{max}=\frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t} = \frac{\frac{3tL^2}{32}\sigma _{allow}}{2\frac{L}{8}\frac{3L}{8}t}$$

This reduces to:

$$\tau _{max}= \sigma _{allow}$$

Substituting Assumption 3 for $$\tau _{max}$$ gives:

$$\tau _{max}= 2 \tau _{allow}$$.

Therefore, Case 1 is an unacceptable solution because $$\tau _{max} > \tau _{allow}$$.