## Theory
### Constant Transverse Shear Stress Distribution  
Using the numerical method presented in Section 2.2, a cross-section is designed in order to produce a constant transverse shear stress distribution. Setting \(\tau\) in Equation (7) equal to a constant and recognizing that only two parameters are variable means that \(Q\) is proportional to \(t\) through a constant \(C\) as shown in Equation (20).  
\[Q=Ct \tag{20}\]  
Applying Equations (11) and (17) yields an expression for \(C\) as shown in Equation (21)  
\[C=\frac{H^{2}\sum_{j=a}^{n}z\left(j-\frac{1}{2}\right)}{z_{a}} \tag{21}\]  
To determine the constant \(C\) for the entire cross-section, \(a=n\) is applied to Equation (21), yielding Equation (22).  
\[C=H^{2}\left(n-\frac{1}{2}\right) \tag{22}\]  
Combining Equations (21) and (22) yields Equation (23).  
\[z_{a}=\frac{\sum_{j=a}^{n}z_{j}\left(j-\frac{1}{2}\right)}{n-\frac{1}{2}} \tag{23}\]  
To determine the width of the elements, \(a=n-1\) is applied to Equation (23) as shown in Equation (24).  
\[z_{n-1}=z_{n}\left(n-\frac{1}{2}\right) \tag{24}\]  
This means that if the width of the top element \(z_{n}\) is predetermined, the next element down \(z_{n-1}\) is calculated from it. Furthermore, if this method is repeated, the following pattern emerges in Equations (25)-(27).  
\[z_{n-2}=\frac{z_{n}\left(n-\frac{1}{2}\right)^{2}}{2} \tag{25}\]  
Figure 6: Calculate \(Q\) for stress at element \(a\).  
\[z_{n-4}=\frac{z_{n}\Big{(}n-\frac{1}{2}\Big{)}^{4}}{24} \tag{27}\]  
This yields a general solution for any element size \(z_{j}\) in relation to the top element size \(z_{n}\) as shown in Equation (28).  
\[z_{j}=\frac{z_{n}\Big{(}n-\frac{1}{2}\Big{)}^{(n-j)}}{(n-j)!} \tag{28}\]  
From Equation (28) a relationship is developed between the first element \(z_{1}\), which is at the neutral axis, and the last element \(z_{n}\) as shown in Equation (29).  
\[z_{1}=\frac{z_{n}\Big{(}n-\frac{1}{2}\Big{)}^{(n-1)}}{(n-1)!} \tag{29}\]  
The ratio \(z_{R}\) is calculated by dividing the widest element by the narrowest element as shown in Equation (30). For the optimized design, the part will be widest at the neutral axis and thinnest at the top, so it follows from Equation (29) that:  
\[z_{R}=\frac{z_{1}}{z_{n}}=\frac{\Big{(}n-\frac{1}{2}\Big{)}^{(n-1)}}{(n-1)!} \tag{30}\]  
If the width of the neutral axis is held as a constant value, the first element \(z_{1}\) is known. Subsequent elements can be derived by combining Equations (28) and (29). Therefore, any element \(z_{j}\) can be calculated by the previous element as shown in Equation (31).  
\[z_{j}=\frac{z_{j-1}(n-j+1)}{n-\frac{1}{2}} \tag{31}\]  
The efficiency of the constant transverse shear stress numerical methodology \(\tau_{eff}\) is determined from the fact that the top transverse shear stress element \(\tau_{n}\) will always have a value of zero as described previously. However, all remaining transverse shear stress elements \(\tau_{a}\) will have the exact same value. Equation (10) is then modified to calculate the efficiency of the cross-section as shown in Equation (32). From Equation (32), as \(n\) approaches infinity, \(\tau_{eff}\) approaches 100%.  
\[\tau_{eff}=\Big{(}\frac{n-1}{n}\Big{)}(100\%) \tag{32}\]