User:Eas4200c.f08.wiki.b/HW2

Case 1: Assume:
 * $$\displaystyle \sigma={Mz \over I}$$

Recall:
 * $$\displaystyle \sigma_{max}=\sigma_{allow}, z= {b \over 2} $$

Thus,
 * $$\displaystyle M= {2I\sigma_{max} \over b} $$
 * or $$\displaystyle M= (2\sigma_{allow})({I \over b})$$

Where $$\displaystyle (I/b)$$ is a function of $$\displaystyle a$$ and $$\displaystyle b$$. Recall from the first assumption:
 * $$\displaystyle L=2(a+b)=constant$$

Thus,
 * $$\displaystyle a={L \over 2}-b$$

Maximize $$\displaystyle M$$:
 * $$\displaystyle M_{max} = (2\sigma_{allow})({I \over b})$$
 * $$\displaystyle I = \sum_{i=1}^4 [{b_i(h_i)^3 \over 12} + A_i(d_i)^2] $$


 * $$\displaystyle = 2{tb^3 \over 12} + 2[{at^3 \over 12} + (at)({b \over 2})^2] $$

Let
 * $$\displaystyle \alpha = {at^3 \over 12} + (at)({b \over 2})^2 = \alpha ={at \over 12}[t^2+3b^2]$$

Since $$\displaystyle t<<b, t^2<<b^2, t^2<<3b^2 $$ Hence,
 * $$\displaystyle \alpha \approx {3ab^2t \over 12}$$

And therefore,
 * $$\displaystyle I \approx {2tb^3 \over 12} + {ab^2t \over 2} = {tb^2 \over 6}(3a+b)$$
 * $$\displaystyle f(b) := {I \over b} = {tb \over 6}(3a+b) $$
 * $$\displaystyle = {tb(3L-4b) \over 12} = \beta_0 + \beta_1 b^1 + \beta_2 b^2 $$

Where $$\displaystyle \beta_0 = 0, \beta_1 = {3Lt \over 12}, \beta_2 = -t/3 $$
 * $$\displaystyle {d^2f(b) \over db^2} = 2 \beta_2 = {-2t \over 3} < 0 $$
 * $$\displaystyle b = \begin{cases} 0 & \\ {3L \over 4} \end{cases} $$ [[Image:HW_2_9-10_2.JPG‎|thumb|400px|]]
 * $$\displaystyle {df(b) \over db} = \beta_1 + 2 \beta_2 b = 0 \Rightarrow b^{(1)} = {-\beta_1 \over 2 \beta_2} = {3L \over 8} $$
 * $$\displaystyle a^{(1)} = {L \over 2} - b^{(1)} = {L \over 8} $$

Thus,
 * $$\displaystyle {b^{(1)} \over a^{(1)}} = 3 $$
 * $$\displaystyle ({I \over b})_max = ({I^{(1)} \over b^{(1)} })$$
 * $$\displaystyle M_{max}^{(1)} = 2 \sigma_{allow}({I^{(1)} \over b^{(1)} }) = {3+L^2 \over 32} \sigma_{allow}$$