Give out exactly answer in English language

The stress increase is the pressure exerted by the weight of the fill.

$$\Delta \sigma = \gamma_{fill} \times H_{fill}$$ $$\Delta \sigma = 24 \, \text{kN/m}^3 \times 4.0 \, \text{m}$$ $$\Delta \sigma = 96 \, \text{kN/m}^2 = 96 \, \text{kPa}$$

We use the elastic settlement formula: $\Delta S = \frac{\Delta \sigma \cdot H}{E_s}$.

$$\Delta S_{clayey \, silt} = \frac{\Delta \sigma \times H_{clayey \, silt}}{E_{s,clayey \, silt}}$$ $$\Delta S_{clayey \, silt} = \frac{96 \, \text{kN/m}^2 \times 5 \, \text{m}}{3000 \, \text{kN/m}^2}$$ $$\Delta S_{clayey \, silt} = \frac{480}{3000} \, \text{m}$$ $$\Delta S_{clayey \, silt} = 0.16 \, \text{m} = 160 \, \text{mm}$$

$$\Delta S_{sand} = \frac{\Delta \sigma \times H_{sand}}{E_{s,sand}}$$ $$\Delta S_{sand} = \frac{96 \, \text{kN/m}^2 \times 6 \, \text{m}}{8000 \, \text{kN/m}^2}$$ $$\Delta S_{sand} = \frac{576}{8000} \, \text{m}$$ $$\Delta S_{sand} = 0.072 \, \text{m} = 72 \, \text{mm}$$

The total settlement is the sum of the settlements in each layer.

$$\Delta S_{total} = \Delta S_{clayey \, silt} + \Delta S_{sand}$$ $$\Delta S_{total} = 0.16 \, \text{m} + 0.072 \, \text{m}$$ $$\Delta S_{total} = 0.232 \, \text{m}$$ $$\Delta S_{total} = 232 \, \text{mm}$$

The total settlement caused by the fill is: $\boxed{\text{232 mm}}$

---

For cohesionless soil ($c'=0$) and a smooth vertical wall, we use Rankine's formula:

$$K_a = \tan^2 \left( 45^\circ - \frac{\phi'}{2} \right)$$ $$K_a = \tan^2 \left( 45^\circ - \frac{30^\circ}{2} \right)$$ $$K_a = \tan^2 (45^\circ - 15^\circ)$$ $$K_a = \tan^2 (30^\circ)$$ $$K_a = \left( \frac{1}{\sqrt{3}} \right)^2$$ $$K_a = \frac{1}{3} \approx 0.333$$

The active thrust can be divided into two parts: from the surcharge and from the soil weight.

• Thrust from surcharge ($P_q$): Surcharge pressure ($p_q$) = $K_a \times q$

  $$p_q = \frac{1}{3} \times 48 \, \text{kPa} = 16 \, \text{kPa}$$

  Thrust from surcharge ($P_q$) = $p_q \times H$

  $$P_q = 16 \, \text{kPa} \times 5 \, \text{m} = 80 \, \text{kN/m}$$

  This force acts at the mid-height of the wall, i.e., $H/2$ from the base. Distance from base ($y_q$) = $\frac{5 \, \text{m}}{2} = 2.5 \, \text{m}$

• Thrust from soil weight ($P_\gamma$): Soil pressure at the base ($p_\gamma$) = $K_a \times \gamma \times H$

  $$p_\gamma = \frac{1}{3} \times 20 \, \text{kN/m}^3 \times 5 \, \text{m} = \frac{100}{3} \, \text{kPa} \approx 33.33 \, \text{kPa}$$

  Thrust from soil ($P_\gamma$) = $\frac{1}{2} \times p_\gamma \times H$ (since the pressure distribution is triangular)

  $$P_\gamma = \frac{1}{2} \times \frac{100}{3} \, \text{kPa} \times 5 \, \text{m} = \frac{250}{3} \, \text{kN/m} \approx 83.33 \, \text{kN/m}$$

  This force acts at one-third of the wall height from the base, i.e., $H/3$ from the base. Distance from base ($y_\gamma$) = $\frac{5 \, \text{m}}{3} \approx 1.667 \, \text{m}$

The resultant thrust is the sum of the forces from the surcharge and the soil weight.

$$P_A = P_q + P_\gamma$$ $$P_A = 80 \, \text{kN/m} + \frac{250}{3} \, \text{kN/m}$$ $$P_A = \frac{240}{3} \, \text{kN/m} + \frac{250}{3} \, \text{kN/m}$$ $$P_A = \frac{490}{3} \, \text{kN/m}$$ $$P_A \approx 163.33 \, \text{kN/m}$$

The position of the resultant thrust ($y$) is found by taking moments about the base of the wall.

$$y = \frac{(P_q \times y_q) + (P_\gamma \times y_\gamma)}{P_A}$$ $$y = \frac{\left( 80 \, \text{kN/m} \times 2.5 \, \text{m} \right) + \left( \frac{250}{3} \, \text{kN/m} \times \frac{5}{3} \, \text{m} \right)}{\frac{490}{3} \, \text{kN/m}}$$ $$y = \frac{200 \, \text{kN} + \frac{1250}{9} \, \text{kN}}{\frac{490}{3} \, \text{kN/m}}$$ $$y = \frac{\frac{1800 + 1250}{9}}{\frac{490}{3}} \, \text{m}$$ $$y = \frac{\frac{3050}{9}}{\frac{490}{3}} \, \text{m}$$ $$y = \frac{3050}{9} \times \frac{3}{490} \, \text{m}$$ $$y = \frac{3050}{3 \times 490} \, \text{m}$$ $$y = \frac{305}{147} \, \text{m}$$ $$y \approx 2.075 \, \text{m}$$

The resultant thrust is: $\boxed{\text{163.33 kN/m}}$
The position of the resultant thrust from the base of the wall is: $\boxed{\text{2.075 m}}$