User:Eas4200c.f08.aero.e/Week 3

=Monday=

Problem 1.1 continued

The σmax where $$z = {b \over 2}$$ is show as $$ \sigma_{max} = \frac{Mb}{2I}$$.

So,

$$M = \frac{2\sigma_{max}I}{b} = \frac{2\sigma_{all}I}{b}$$

To maximize the moment M, the \frac{I}{b} will be maximized. To give,

 $$M_{max} = 2\sigma_{all} \left (\frac{I}{b} \right )_{max}$$

The length L can be expressed in terms of a and b, so a can be expressed in terms of L and b, shown below.

$$L = 2(a+b) \Rightarrow \; a^{(1)} = \frac{L}{2} - b^{(1)} $$

Where L = constant.

$$ \left (\frac{I}{b} \right )_{max} = \left (\frac{I^{(1)}}{b^{(1)}} \right ) \Rightarrow \; M_{max} = 2\sigma_{all} \left (\frac{I^{(1)}}{b^{(1)}} \right ) = T^{(1)}_{max}$$

This can be said because of the assumption that the moment M equals the torque T. Then assumption 2 can be written as the following.

$$ \tau^{(1)}_{max} = \frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t} $$

If τ(1)max < τall, then optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ is good.

If τ(1)max > τall, then optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ is not good.

 Case 2: 

In this case the τmax will equal τall which is the allowable shear stress.

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

The torque can be shown to be

$$ T = 2t\tau_{all}(ab) \Rightarrow \; T_{max} = 2t\tau_{all}(ab)_{max} $$

Also we will assume the sides are equal, so

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

With this the torque is represented as the following.

$$ T^{(2)}_{max} = 2t\sigma_{max} \left (\frac{L}{4} \right )^2 = \frac{1}{8} t\sigma_{max} L^2 = M^{(2)}_{max} $$

Solving for σmax yields,

$$ \sigma^{(2)}_{max} = \frac{M^{(2)}_{max}b^{(2)}}{2I^{(2)}} $$

If σmax(2) < σall, then $$\frac{b^{(2)}}{a^{(2)}} = 1 $$ is good.

If σmax(2) > σall, then $$\frac{b^{(2)}}{a^{(2)}} = 1 $$ is not good.

Answer:

Using case 1.

The inertia, I, is of 4 segments:

$$ I = \sum_{i=1}^{i=4} \left [ \frac{b_ih_i^3}{12} + A_id_i^2 \right ] = 2\frac{tb^3}{12} + 2\left [ \frac{at^3}{12} + at\left (\frac{b}{2} \right )^2 \right ] \approx \frac{tb^2}{6} (3a +b) $$

The approximation can be made because t, the thickness, is very small so it is negligible. Now $$\frac{I}{b}$$ can be written as a function of b.

$$ f(b) = \frac{I}{b} = \frac{tb}{6}(3a+b) = \frac{tb}{12}(3L-4b) $$

Also because the 2nd derivative is negative, it is known that f(b) is an inverted parabola.

To maximize f(b) we much find b = b(1)

$$ \frac{d(f(b^{(1)}))}{db} = 0 = \frac{t}{12}(3L-8b^{(1)}) \Rightarrow b^{(1)} = \frac{3L}{8} $$

Then we can solve for a(1):

$$ a^{(1)} = \frac{L}{2} - b = \frac{L}{8} \Rightarrow \frac{b^{(1)}}{a^{(1)}} = 3 $$

Then the f(b)max can be found to be,

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

This can be expressed as L. And Mmax is shown as,

$$ M^{(1)}_{max} = 2\sigma_{all} \left (\frac{I^{(1)}}{b^{(1)}} \right ) = \frac{3tL^2}{32}\sigma_{all} $$

Solving for σall we can then use assumption 2 to find τmax(1).

<p style="text-align:center;">$$ \tau^{(1)}_{max} = \frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{M^{(1)}_{max}}{2 \left( \frac{L}{8} \right) \left( \frac{3L}{8} \right)t} = \frac{32M^{(1)}_{max}}{3tL^2} = \sigma_{all} $$

So,

<p style="text-align:center;">$$ \tau^{(1)}_{max} = \sigma_{all} = 2\tau_{all} > \tau_{all} $$

This answer is not possible.

=Wednesday=

Problem 1.1 continued

Case 1 assumes σmax = σall.

<p style="text-align:center;">$$ \sigma = \frac{Mz}{I} \;\; and \;\; z = \frac{b}{2} \Rightarrow M = \frac{2I\sigma_{all}}{b} $$

$$\frac{I}{b}$$, is a function of a + b, where L = 2(a+b) = constant.

<p style="text-align:center;">$$ \Rightarrow a = \frac{L}{2}-b $$

<p style="text-align:center;">$$ \Rightarrow \frac{I}{b} = f(b) $$

To maximize M the f(b) must be maximized.

Using I from the day before, f(b) becomes,

<p style="text-align:center;">$$ f(b) = \frac{tb}{12}(3L-4b) = \beta_0 + \beta_1b + \beta_2b^2 $$

Solving, we find $$ \beta_0 = 0 $$, $$ \beta_1 = \frac{3Lt}{12} $$, and $$ \beta_2 = \frac{-4t}{12} = \frac{-t}{3} $$

Taking the 2nd derivative of f(b) yields, and the figure below show the downward parabola of the equation.

<p style="text-align:center;">$$ \frac{d^2(f(b^{(1)}))}{db^2} = 2\beta_2 = \frac{-2t}{3} < 0 $$

Setting the first derivative of f(b) equal to 0 we find b equals 0 or $$ \frac{3L}{4} $$

With the 1st derivative we can find b.

<p style="text-align:center;">$$ \frac{d(f(b^{(1)}))}{db} = \beta_1 + 2\beta_2b = 0 \; \Rightarrow \; b^{(1)} = \frac{\beta_1}{2\beta_2} = \frac{3L}{8} $$

Then solving for a.

<p style="text-align:center;">$$ a^{(1)} = \frac{L}{2} - b^{(1)} = \frac{L}{8}\; \Rightarrow \; \frac{b^{(1)}}{a^{(1)}} = 3 $$

=Friday= Note from the last lesson we found:

$$M_{max}^{(1)}=(2{\sigma}_{allowable})\left(\frac{I^{(1)}}{b^{(1)}}\right)=\frac{3tL}{32}{\sigma}_{allowable}=T_{max}^{(1)}$$

Recall that assumption 2 allows us to write: $$M_{max}^{(1)}=T_{max}^{(1)}$$

Now an expression for $${\sigma}_{allowable}$$ can be written as follows:

$${\sigma}_{allowable}=M_{max}^{(1)}\frac{32}{3tL}$$

Let's see what the shear stress is in case 1.

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

Notice that for case 1:

$${\sigma}_{allowable}={\tau}_{max}^{(1)}$$

But according to assumption 3:

$${\sigma}_{allowable}=2{\tau}_{allowable}^{}$$

Therefore, we can say conclusively that case 1 is not valid because $${\tau}_{max}^{(1)}$$ is too large.

Now consider case 2. We will assume $${\tau}_{max}={\tau}_{allowable}^{}$$

It was found that the shear stress for this cross section can be expressed as follows: $${\tau}=\frac{T}{2abt}$$

This expression can be rearranged to solve for T:

$$T=(2t{\tau}_{allowable}^{})(ab)$$

Therfore:

$$T_{max}=(2t{\tau}_{allowable}^{})(ab)_{max}$$