## 2 Methods and Analysis
\begin{array}{l}C=\frac{q}{2\pi G}\bigg{[}a\ln\left|\frac{\sqrt{a^{2}+b^{2}+ h^{2}}-b}{\sqrt{a^{2}+b^{2}+h^{2}}+b}\right|-a\ln\left|\frac{\sqrt{a^{2}+b^{2}}-b}{ \sqrt{a^{2}+b^{2}+h^{2}}+b}\right|+\\ \left.+b\ln\left|\frac{\sqrt{a^{2}+b^{2}+h^{2}}-a}{\sqrt{a^{2}+b^{2}+h^{2}}+a} \right|\right|-b\ln\left|\frac{\sqrt{a^{2}+b^{2}}-a}{\sqrt{a^{2}+b^{2}+a}} \right|\right|\bigg{]}\\ H_{1}=\sqrt{\frac{\left(a^{2}+b^{2}\right)-\sqrt{\left(a^{2}+b^{2}\right)^{2} -4a^{2}b^{2}}}{8}};\;H_{2}=\sqrt{\frac{\left(a^{2}+b^{2}\right)+\sqrt{\left( a^{2}+b^{2}\right)^{2}-4a^{2}b^{2}}}{8}}\end{array}  \tag{18}\]  
The settlement \(S_{\rm v}\) due to the volumetric strain caused by mean stress can be procured by analyzing expression (12) with the form of:  
\[ \begin{array}{l}S_{\rm v}=\frac{4q(1+\
u)}{3\pi K}h\text{ arctg}\frac{ab}{h\sqrt{a^{2}+b^{2}+h^{2}}}+\\ +\frac{4q(1+\
u)}{3\pi K}H_{1}\ln\left|\frac{8H_{1}^{3}-2\left(a^{2}+b^{2} \right)H_{1}+ab\sqrt{a^{2}+b^{2}+h^{2}}}{-8H_{1}^{3}+2\left(a^{2}+b^{2}\right) H_{1}+ab\sqrt{a^{2}+b^{2}}+h^{2}}\right|\\ -\frac{4q(1+\
u)}{3\pi K}H_{1}\ln\left|\frac{8H_{1}^{3}-2\left(a^{2}+b^{2} \right)H_{1}+ab\sqrt{a^{2}+b^{2}}}{-8H_{1}^{3}+2\left(a^{2}+b^{2}\right)H_{1} +ab\sqrt{a^{2}+b^{2}}}\right|+\\ +\frac{4q(1+\
u)}{3\pi K}H_{2}\ln\left|\frac{8H_{1}^{3}-2\left(a^{2}+b^{2} \right)H_{2}+ab\sqrt{a^{2}+b^{2}}+h^{2}}{-8H_{2}^{3}+2\left(a^{2}+b^{2}\right) H_{2}+ab\sqrt{a^{2}+b^{2}}+h^{2}}\right|-\\ -\frac{4q(1+\
u)}{3\pi K}H_{2}\ln\left|\frac{8H_{2}^{3}-2\left(a^{2}+b^{2} \right)H_{2}+ab\sqrt{a^{2}+b^{2}}}{-8H_{2}^{3}+2\left(a^{2}+b^{2}\right)H_{2} +ab\sqrt{a^{2}+b^{2}}}\right|-\\ -\frac{4q(1+\
u)}{3\pi K}H_{2}\ln\left|\frac{8H_{2}^{3}-2\left(a^{2}+b^{2} \right)H_{2}+ab\sqrt{a^{2}+b^{2}}}{-8H_{2}^{3}+2\left(a^{2}+b^{2}\right)H_{2} +ab\sqrt{a^{2}+b^{2}}}\right|\end{array}  \tag{20}\]  
Considering the example with the following inputs: \(q_{1}\) = 1000 kPa; \(a_{1}\) = 0.5 m; \(b_{1}\) = 1.5 m; \(v_{1}\) = 0.3; \(H_{1}\) = 20 m; \(G_{1}\) = 10,000 kPa; \(K_{1}\) = 65,000 kPa; \(q_{2}\) = 1500 kPa; \(a_{2}\) = 1.5 m; \(b_{2}\) = 3 m; \(\
u_{2}\) = 0.3; \(H_{2}\) = 25 m; \(G_{2}\) = 15,000 kPa; \(K_{2}\) = 97,500 kPa; \(q_{3}\) = 150 kPa; \(a_{3}\) = 2 m; \(b_{3}\) = 4 m; \(v_{3}\) = 0.3; \(H_{3}\) = 30 m; \(G_{3}\) = 20,000 kPa; and \(K_{3}\) = 130,000 kPa. By using math software for engineering calculations (Mathcad), we obtain: \(S_{\rm{\gamma}1}\) = 0.06105 m; \(S_{\rm{\gamma}1}\) = 0.01436 m; \(S_{\rm{\gamma}2}\) = 0.1180 m; \(S_{\rm{\gamma}2}\) = 0.02911 m; \(S_{\rm{\gamma}3}\) = 0.1556 m; and \(S_{\rm{\gamma}3}\) = 0.03859 m. It is substantially important to note that the values of \(S_{\rm{\gamma}1}\), \(S_{\rm{\gamma}2}\), and \(S_{\rm{\gamma}3}\) are 4.25, 4.06, and 4.032 times higher than the values of \(S_{\rm{\gamma}1}\), \(S_{\rm{\gamma}2}\), and \(S_{\rm{\gamma}3}\) (\(S_{\rm{\gamma}1}\) = 4.255\(S_{\rm{\gamma}1}\); \(S_{\rm{\gamma}2}\) = 4.065\(S_{\rm{\gamma}2}\); \(S_{\rm{\gamma}3}\) = 4.0325\(S_{\rm{\gamma}3}\)), respectively. Thus, it can be concluded that in linear elastic half-space, the settlement of a loaded soil subgrade mainly occurs due to the shear strain, which is over 80% of the total value, while nearly 20% of the total value comes from the volume change. The distributions of shear strain \(\varepsilon_{\rm{\gamma}}(z)\) and volumetric strain \(\varepsilon_{\rm{v}}(z)\), calculated using expression (11), with a depth corresponding to the first group of parameters (\(q_{1}\), \(a_{1}\), \(b_{1}\), \(v_{1}\), \(H_{1}\), \(G_{1}\), \(K_{1}\)) are illustrated in Figures 4 and 5. As presented in expression (11), the shear \(\varepsilon_{\gamma}(z)\) and volumetric strains \(\epsilon_{\
u}(z)\) are calculated through the deviator stress \(\sigma_{z}-\sigma_{\rm m}\) and volumetric stress \(\sigma_{\rm m}\), which are derived from expressions (7) and (8). As a result, the variation of \(\varepsilon_{\gamma}(z)\), and \(\varepsilon_{\
u}(z)\) with depth is accompanied by a change of \(\sigma_{z}-\sigma_{\rm m}\) and \(\sigma_{\rm m}\), respectively.