User:EAS4200C.F08.WIKI.A/My Contribution: 2

=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.