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:
KA1 for layer I = \[{K_A}={{1-\sin\phi}\over{1+\sin\phi}}\] = \[{{1-\sin 28^\circ}\over{1+\sin 28^\circ}}\] = 0.36
KA2 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, PA = 0 kN/m2
At Z = 1.5m, PA = 0.36 x 18 x 1.5 = 9.72 kN/m2
At Z = 6 m, PA (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, PA (due to water) = 4.5 x 10 = 45 kN/m2
Active Pressure distribution for layer II.
At Z = 6 m, PA (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, PA (due to the water Pressure above this level) = 45 kN/m2
At Z = 11 m, PA (due to soil) = 67.32 + 0.31 x (20 – 10) x 5 = 67.32 + 15.5 =82.82 kN/m2
At Z = 6 m, PA (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:
PA1 = 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
PA2 = 9.72 x 4.5 = 43.74 kN/m acts at a height of 7.25 (=4.5/2+5) m from the base
PA3 = 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
PA4 = 67.32 x 5 = 336.6 kN/m acts at a height of 2.5 (=5/2) m from the base
PA5 = 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
PA = PA1+ PA2 + PA3+ PA4+ PA5 = 7.29 + 43.74 + 137.7 + 336.6 + 163.75 = 689.08 kN/m
Point of aPPlication of the resultant force PA is
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