User:EAS4200C.F08.WIKI.A/Homework

=Step by Step Procedure To Finding A Solution=

There are two cases for which need to be solved in order to compute the optimal ratio $$\displaystyle \ \frac{b}{a}$$. The first case which we will be looking at is the case in which the maximum stress $$\displaystyle \sigma _{max} $$ equals the allowable stress $$\displaystyle \sigma_{all} $$. The second case that needs to be concidered is the case in which the maximum shearing stress, $$\displaystyle \tau_{max} $$, equals the allowable shearing stress, $$\displaystyle \tau_{all} $$. Case one variables will be represented by a superscript of one and case two variables will be represented by a superscript two.

Case 1
To begin this analysis we have to first look at the equation for stress which is the following equation.

$$\displaystyle \sigma^{(1)} = \frac {M^{(1)}z}{I^{(1)}}$$

where $$\displaystyle \ M^{(1)} $$ is th bending moment, $$\displaystyle \ z $$ is the distance from the neutral axis, and $$\displaystyle \ I^{(1)} $$ is the second area moment of inertia. In this particular problem, $$\displaystyle \ z = \frac {b^{(1)}}{2} $$ and the stress has reached its allowable limit so therefore the stress can be expressed as

$$\displaystyle \sigma_{all}^{(1)} = \frac {M^{(1)}b^{(1)}}{2I^{(1)}}$$

For this case the bending moment must be maximized. The equation for the bending moment can be obtained from the stress equation.

$$\displaystyle \ M_{max}^{(1)} = 2\sigma_{all}^{(1)}(\frac {I^{(1)}}{b^{(1)}})$$

A closer look at the equation will reveal that there is only one variable that can be manipulated in order to maximize the moment equation and it is the fraction of $$\displaystyle \ \frac{I^{(1)}}{b^{(1)}}$$. the equation can therefore be writen as

$$\displaystyle \ M_{max}^{(1)} = 2\sigma_{all}^{(1)}(\frac {I^{(1)}}{b^{(1)}})_{max}$$

As stated above the ratio $$\displaystyle \ (\frac {I^{(1)}}{b^{(1)}})_{max}$$ must then be calculated, expressed in terms of $$\displaystyle \ b^{(1)} $$ using the relationship $$\displaystyle \ a = \frac{L}{2} - b^{(1)}$$ and then manipulated (take derivative and set equal to zero; solve)to obtain the maxium values of $$\displaystyle \ b^{(1)} $$ $$\displaystyle \ a^{(1)} $$.

The next step is to place the calculated values of $$\displaystyle \ b^{(1)} $$ and $$\displaystyle \ a^{(1)} $$ back into the equation representing the ratio of $$\displaystyle \ (\frac {I^{(1)}}{b^{(1)}})_{max}$$ and then putting this ratio back into the equation representing the bending moment.

$$\displaystyle \ M_{max}^{(1)} = 2\sigma_{all}^{(1)}(\frac {I^{(1)}}{b^{(1)}})_{max}$$

Now that the equation for the bending moment is complete and the ratio of $$\displaystyle \ \frac{b^{(1)}}{a^{(1)}} $$ is optimized, the correctness of this ratio must be check. To do so, the maximum shearing stress has to be compared to the allowable shearing stress and this can be completed by recognizing that the maximum bending moment is equal to the maximum torque as stated in the problem statement.

$$\displaystyle \ M_{max}^{(1)} = T_{max}^{(1)}$$

This expression will now be enetered into the expression for the maximum shearing stress given by the following equation;

$$\displaystyle \tau_{max}^{(1)} = \frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t}$$

where t is the thickness of the cross-section. Reducing this equation down and replacing $$\displaystyle \sigma_{all}^{(1)}$$ with $$\displaystyle \tau_{all}^{(1)}$$, (stated in the problem), it is now possible to compare $$\displaystyle \tau_{max}^{(1)}$$ with $$\displaystyle \tau_{all}^{(1)}$$.

Case 2
In this case the same approach is taken, but this time it is assumed that $$\displaystyle \tau_{max}^{(2)} = \tau_{all}^{(2)}$$. Also, the torque will be maximized instead of the bending moment. That being said the following is true;

 $$\displaystyle \ T_{max}^{(2)} = 2ta^{(2)}b^{(2)}\tau $$

To maximize this equation, the values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ must be optimized. To do so, we again put $$\displaystyle \ a^{(2)} $$ in terms of $$\displaystyle \ b^{(2)} $$  ( $$\displaystyle \ a^{(2)} = \frac{L}{2} - b^{(2)}$$), plug this value into the previous equation, and solve this equation in order to find the maximum value of $$\displaystyle \ b^{(2)} $$. This is done by taking the derivative and setting the equation equal to zero and solving. Once the value of $$\displaystyle \ b^{(2)} $$ has been solved for, the value of $$\displaystyle \ a^{(2)} $$ can be solved for from the equation $$\displaystyle \ a^{(2)} = \frac{L}{2} - b^{(2)}$$.

After this has been completed and the values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ have been placed into the maximum torque equation this value can be equated to the maximum bending moment $$\displaystyle \ M_{max}^{(2)} $$. From this point we are interested in discovering the relationship between the maximum stress and the allowable stress. We proceed forward by looking at the maximum shearing stress equation:

 $$\displaystyle \sigma_{max}^{(2)} = \frac {M_{max}^{(2)}b^{(2)}}{I^{(2)}}$$

The expression for the maximum bending moment and the inverse of the ratio $$\displaystyle \ (\frac {I^{(2)}}{b^{(2)}})_{max}$$ (the corresponding values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ are input into the ratio) is placed in this equation. The resulting expression will be of the following format,

$$\displaystyle \sigma_{max}^{(2)} = C \tau_{all}^{(2)}$$

where $$\displaystyle C $$ is a constant that comes from reduction. Since we are interested in finding a relationship bewteen the maximum stress and the allowable stress, we have to relace the allowable shearing stress term with an equal value. to do so we must remember the following relationship,

$$\displaystyle \sigma_{all}^{(2)} = 2 \tau_{all}^{(2)}$$

It is now possible to determin whether the values of $$\displaystyle \ a^{(2)} $$ and $$\displaystyle \ b^{(2)} $$ are acceptable.

=Solving Case 1=

Now that the steps have been laid out, we can now proceed to solve Problem 1.1.

Recall:
 * $$\displaystyle \sigma={Mz \over I}$$
 * $$\displaystyle \sigma_{max}^{(1)} = \sigma_{allow}, z= {b \over 2} $$

Thus,
 * $$\displaystyle M^{(1)} = {2I\sigma_{max}^{(1)} \over b} $$
 * or $$\displaystyle M^{(1)}= (2\sigma_{allow})({I \over b})$$

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

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


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

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

Since $$\displaystyle t $$ is much smaller than $$\displaystyle b $$ then one can conclude that $$\displaystyle t^2<<b^2, t^2<<3b^2 $$ 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,
 * $$\displaystyle {b^{(1)} \over a^{(1)}} = 3 $$
 * $$\displaystyle ({I \over b})_max = ({I^{(1)} \over b^{(1)} })$$
 * $$\displaystyle M_{max}^{(1)} = 2 \sigma_{allow}({I^{(1)} \over b^{(1)} }) = {3+L^2 \over 32} \sigma_{allow}$$

= Motivation and Solution Continued =

A box beam is a rectangular structure that is important to study due to the fact that many aerospace structures follow this same basic design. A box beam is a form of shell beam, and structures such as an airplane fuselage are also shell beams, the only difference being that their cross sections are circular rather than a rectangle. Likewise, the wings on an airplane are also shell beams with a slightly more complex cross section.



Now that there is a motivation for solving the problem, Case 1 from last lecture is continued as follows:

$$\tau ^{(1)}_{max}$$ is the shear stress corresponding to the maximum torque $$T^{(1)}_{max}$$

Continuing the solution, replace $$T^{(1)}_{max}$$ in the shear equation by $$M^{(1)}_{max}$$. This is done due to the 1st assumption in the problem statement. Secondly, assumption 2 allows for the substitution of $$\tau ^{(1)}_{max}$$ with $$\sigma _{allow}$$

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

$$\tau ^{(1)}_{max} = \frac{M^{(1)}_{max}}{2(\frac{L}{8})\frac{3L}{8}t} = \sigma _{allow}$$

$$\tau _{max}^{(1)} = \sigma _{allow} = 2\tau _{allow} > \tau _{allow}$$

After these substitutions, it is easy to see that $$\tau ^{(1)}_{max}$$ is equal to $$\sigma _{allow}$$, which is twice as big as $$\tau _{allow}$$. Therefore this answer is not acceptable, and the solver must move on to case 2.

Pictured to the right is a graphical representation of normal stress. The stress is distributed through the length of the bar with and is equal to the force, F, divided by the cross sectional area A.

A graphical representation of this result can be shown using a Mohr's circle. As shown on the figure to the right, as long as the shear and normal stress stay enclosed within the circle, the structure will not fail. However, since in this case the maximum shear encountered is greater than half the allowable normal stress, this result is unacceptable

Case 2

Assume that $$\tau _{max}$$ reaches $$\tau _{allow}$$ first.

Therefore:

$$\tau = \frac{T}{2abt}$$

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

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

Everything on the right hand side of the above equation is constant, therefore to maximize the torque, the cross sectional area must be maximized. With the constraint of keeping the perimeter constant, it is easy to see that the largest cross sectional area will be achieved by making a equal to b, and thus having a square cross section as follows:

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

$$T_{max}^{(2)} = (2t\tau _{all})(\frac{L}{4})$$

$$T_{max}^{(2)}=\frac{1}{8}tL^2\tau {all} = M_{max}^{(2)}$$

$$M_{max}^{(2)}=\frac{tl^2}{16}\sigma _{all}$$

$$\Rightarrow \sigma _{all}=\frac{16M_{max}^{(2)}}{tL^2} $$

$$Recall: F(b)= \frac{I(b)}{b}$$

$$F(b^{(2)})= \frac{I^{(2)}}{b^2}=tb^{(2)}\frac{(3L-4b^{(2)})}{12}$$

$$M_{max}^{(2)}=\frac{tl^2}{24}$$ $$Given \ B^{(2)}=\frac{L}{4}$$

$$ \sigma _{max} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}}=M_{max}^{(2)}\frac{12}{tL^2}$$

$$\sigma _{max} = \frac{3}{4}\sigma _{all}\Rightarrow \frac{3}{4}\sigma _{all}\leq \sigma _{all}$$

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

=Wed September 17 2008=

 Axial Members Continued 

=Stringers, Shear Stress/Strain, and Elasticity=

Stringers Continued from Earlier
Notes

Longeron - The term stringer is interchangeable with the term longeron. Longerons have the same purpose as stringers - the only differences are that longerons are usually larger, there are less of them, and they are spaced farther apart. (Reference - )

Shear Panels - Also known as brace panels. Together, shear panels make up a shear wall to support lateral loads on a structure. (Reference - )

Shear Stress and Strain
Shear Strain - γ - Strain that acts parallel to the face it's acting on. (Reference - )

τ = Gγ
 * (G = shear modulus)
 * (γ = shear strain - See above definition)

γ = du/dy + dv/dx = dux/dy + duy/dx
 * $$u \equiv u_x $$ - displacement along the x-direction
 * $$v \equiv u_y $$ - displacement along the y-direction

Shear Panels, Shear Stress and Strain, and Elasticity
Engineering (not tensorial) shear strain

γ = du/dy + dv/dx = dux/dy + duy/dx
 * Remember:
 * $$u \equiv u_x $$


 * $$v \equiv u_y $$



duy/dx = dv/dx
 * (for a small angle, $$ \alpha \cong \tan \alpha $$)

γ = change in the right angle (90 degrees) due to the shear deformation

εxy = tensorial shear strain

εxy = (1/2)γxy

=Individual Contributions=

Braden Barnes Eas4200c.f08.wiki.b 17:51, 25 September 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 14:55, 26 September 2008 (UTC)

Christopher Pergola EAS4200C.F08.WIKI.A 18:35, 26 September 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 20:17, 26 September 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 19:50, 26 September 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 20:33, 26 September 2008 (UTC)