User:Eas4200c.f08.vqcrew.E/Homework2

Case 2:

For this case, the assumption is made that $$ \tau_{max} = \tau_{allow}$$  (i.e.  $$ \tau_{max}$$  reaches  $$ \tau_{allow}$$  first).

$$ \tau = \frac{T}{2abt} = \tau_{max} = \tau_{allow} $$

Rearrange to solve for the torque, T.

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

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

From Assumption 1, the perimeter of the beam is constant, L = 2(a + b). This allows for a to be solved in terms of only L and b.

$$a = \frac{L}{2} - b$$

Now, substituting in for a the maximum torque becomes

 $$ T_{max} = (2t\tau_{allow})(\frac{bL}{2} - b^{2})_{max} $$

The first term in the maximum torque equation $$(2t\tau_{allow}) $$  is a constant. In order to maximize the torque, the second term $$(\frac{bL}{2} - b^{2})$$ must be maximized. A function f(b) is defined as the second term to allow for the maximum to be found.

 $$ f(b) = (\frac{bL}{2} - b^{2}) $$

The derivative of f(b) is taken and set to zero to find the maximum of the function.

 $$ f'(b) = (\frac{L}{2} - 2b) = 0 \Rightarrow b^{(2)} = \frac{L}{4} $$

Next, $$ a^{(2)}$$ is found through the perimeter relation.

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

The value of $$ T^{(2)}$$ can now be determined and equated to the maximum moment by Assumption 2, T = M.

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

The allowable normal stress can now be substituted in to the equation per Assumption 3, $$\sigma_{allow} = 2\tau_{allow}$$

 $$M_{max}^{(2)} = \frac{1}{8}tL^{2}\tau_{allow} = \frac{1}{16}tL^{2}\sigma_{allow}$$

Solving for $$\sigma_{allow}$$  ,

 $$\sigma_{allow} = \frac{16M_{max}^{(2)}}{tL^{2}}$$ Recall:  $$I=\frac{tb^2}{6}(3a+b)$$

For $$ a=b=\frac{L}{4} $$

 $$I^{(2)} = \frac{t(\frac{L}{4})^2}{6}(3(\frac{L}{4})+(\frac{L}{4})) = \frac{tL^3}{96}$$

Solve for $$\sigma_{max}^{(2)}$$  and substitute in values for $$I^{(2)}$$ and $$b^{(2)}$$ ,

 $$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}} = \frac{M_{max}^{(2)}\frac{L}{4}}{2\frac{tL^3}{96}} = \frac{12M_{max}^{(2)}}{tL^2}$$

Substituting in the previously found expression for $$M_{max}^{(2)}$$ ,

 $$\sigma_{max}^{(2)} = \frac{12(\frac{1}{16}tL^{2}\sigma_{allow})}{tL^2} = \frac{12}{16}\sigma_{allow}$$

Since $$ \frac{12}{16}\sigma_{allow}$$ is less than $$\sigma_{allow}$$ , this configuration of dimensions for the thin-walled beam are acceptable.