User:Eas4200c.f08.vqcrew.d/hw1

Problem 1.1
Given Figure 2, 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 3. 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 4, 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 $$I=\iint_A z^2 \ dy\, dz$$. Figure 5 shows each item pictorially.