Examples of Earth Pressure on Retaining Wall, Soil Mechanics | Soil Mechanics Notes- Agricultural Engineering PDF Download

Problem

Determine the active earth pressure distribution on the retaining as shown in Figure 27.3. Also determine the total active force and point of application  of the force.

Solution:

K_A1 for layer I = \[{K_A}={{1-\sin\phi}\over{1+\sin\phi}}\] = \[{{1-\sin 28^\circ}\over{1+\sin 28^\circ}}\] = 0.36

K_A2 for layer II = \[{K_A}={{1-\sin\phi}\over{1+\sin\phi}}\] = \[{{1-\sin 32^\circ}\over{1+\sin 32^\circ}}\ = 0.31

Active Pressure distribution for layer I

At Z = 0m, P_A = 0 kN/m²

At Z = 1.5m, P_A = 0.36 x 18 x 1.5 = 9.72 kN/m²

At Z = 6 m, P_A (due to soil) = 9.72 + 0.36 x (20 – 10) x 4.5 = 9.72 + 16.2 =25.92 kN/m² (take unit weight of the water is 10 kN/m³).

At Z = 6 m, P_A (due to water) = 4.5 x 10 = 45 kN/m²

Active Pressure distribution for layer II.

At Z = 6 m, P_A (due to surcharge of the layer I) = 0.31 (1.5 x 18 + 4.5 x 10) = 22.32 kN/m²

At Z = 6 m, P_A (due to the water Pressure above this level) = 45 kN/m²

At Z = 11 m, P_A (due to soil) = 67.32 + 0.31 x (20 – 10) x 5 = 67.32 + 15.5 =82.82 kN/m²

At Z = 6 m, P_A (due to water) = 5 x 10 = 50 kN/m²

Fig. 27.3. Active earth Pressure distribution of problem 1.

The active force of the various levels is calculated as:

P_A1 = 0.5 x 9.72 x 1.5 = 7.29 kN/m acts at a height of 10 (=1.5/3+4.5+5) m from the base

P_A2 = 9.72 x 4.5 = 43.74 kN/m acts at a height of 7.25 (=4.5/2+5) m from the base

P_A3 = 0.5 x (16.2 + 45) x 4.5 = 137.7 kN/m acts at a height of 6.5 (=4.5/3+5) m from the base

P_A4 = 67.32 x 5 = 336.6 kN/m acts at a height of 2.5 (=5/2) m from the base

P_A5 = 0.5 x (15.5 + 50) x 5 = 163.75 kN/m acts at a height of 1.67 (=5/3) m from the base

P_A = P_A1+ P_A2 + P_A3+ P_A4+ P_A5 = 7.29 + 43.74 + 137.7 + 336.6 + 163.75 = 689.08 kN/m

Point of aPPlication of the resultant force P_A is