# A Novel Approach Proposal for Estimation of Ultimate Being Capacity Based on Pile Loading Test Data
## 2 Materials and Methods
### Method
#### 2.2.5 Description of The Proposed Method
Sample application for the proposed method:  
The ultimate bearing capacity of the pile loading test, which is given below, is determined by using the proposed method described in detail. In Figure 3, the load-settlement graph of the sample application is given. The distribution of \(Q_{i}\sim K\) created for all stages according to the proposed method is given in Table 3. The distribution graph required for the solution is divided into three parts in the sample application in Figure 4, as indicated in the theory of the proposed method. In the distribution, the breaks in the 2 nd and 6 th loading stages were clearly seen, and the interpretation was made accordingly. Stage: The expression of settlement at the logarithm load base is calculated (Equation (13)). \[log_{100}^{6.75}=0.4147 \tag{13}\]  
Stage: The settlement expressions on the logarithm load base are summed (Equation (14)). \[\frac{\sum_{i=1}^{n}log_{Q_{i}}^{\Delta_{i}}}{n}=\frac{0.0693+0.2808+0.3484+0. 4147+0.4795+0.5429+0.6052+0.6665+0.78}{\pi}=0.4653 \tag{14}\]  
Figure 3: Load-settlement graph of the sample application. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Process** & **Settlement** & **Load Q** & \(\left(log_{Q_{i}}^{\Delta_{i}}\right)\) & \(\frac{\sum_{i=1}^{n}log_{Q_{i}}^{\Delta_{i}}}{n}\) & \(K_{i}\frac{\sum_{i=1}^{n}log_{Q_{i}}^{\Delta_{i}}}{n}\) \\ \hline - & 0.000 & 0.00 & - & - \\
1 & 1.250 & 25.00 & 0.0693 & 3.5768 \\
2 & 3.000 & **50.00** & **0.2808** & **2.0575** \\
3 & 4.500 & **75.00** & **0.3484** & **1.6564** \\
4 & 6.750 & **100.00** & **0.4147** & **1.2625** \\
5 & 10.125 & **125.00** & **0.4795** & **0.9337** \\
6 & 15.188 & 150.00 & 0.5429 & 0.6776 \\
7 & 22.781 & 175.00 & 0.6052 & 0.4853 \\
8 & 34.172 & 200.00 & 0.6665 & 0.3443 \\
9 & 68.344 & 225.00 & 0.7800 & 0.1818 \\  & & & **Total: 4.1873** & \\  & & & **Number of samples (n): 9** & \\ \hline \hline \end{tabular}
\end{table} 
Table 3: The \(Q_{i}\sim K\) distribution generated for all steps according to the proposed method. \[K=\frac{Q_{i}^{\sum\limits_{j=1}^{n}log_{Q_{i}}^{A_{i}}}}{\Delta_{i}}=\frac{100^{ 0.4653}}{6.750}=1.2625 \tag{15}\]
3. Stage: The coefficient defined by the symbol K is found (Equation (15)). A value of 1.265 was found for a load of 100 tons. In all steps, these processes are applied in order, and the \(Q_{i}\sim K\) scat graph is formed for all stages (Figure 4). The distribution graph required for the solution is divided into three parts in the sample application, as stated in the theory of the proposed method. In the distribution, the breaks in the second and sixth loading stages are clearly visible. The intersection points from the first to the second break were interpreted as the parabola during the failure, and the ideal line was drawn by regression analysis on these points. The ultimate bearing capacity (failure load) at the point where this line intersects the apse was found to be 186 tons. It can also be found by using Equation (16) as follows:  
\[Q_{ult}=-b/a=-2.795/-0.015=186\text{ tons} \tag{16}\]