## LESSON 27. Rankine’s Theory of Earth Pressure and Numerical Exercise

27.1 Earth Pressure on Retaining Wall

27.1.1. c-ø backfill

In case of c-f backfill, negative active earth pressure is developed upto the Z0 depth from the ground level as shown in Figure 27.1. It is clear that the net active pressure is zero upto a depth of 2Z0. Thus, in case of cohesive soil, vertical cut can be made with out any lateral supported upto the depth equal to 2Z0. The depth 2Z0 is called as critical depth of vertical cut Hc in a cohesive soil and can be expressed as:

${H_c}=2Z{}_0={{4c} \over {\gamma \sqrt {{K_A}} }}$                       (27.1)

where c is the cohesion of the soil, $\gamma$ is the effective unit weight of the soil and KA is the coefficient of active earth pressure and can be expressed as:

${K_A}={{1 - \sin \phi } \over {1 + \sin \phi }}$                       (27.2)

During the calculation of total active force (PA) in case of c-ø backfill, negative zone is neglected and only the active earth pressure due to the positive zone is considered. Thus,

${P_A}={1\over 2}{K_A}\gamma\,{\left({H-{Z_0}}\right)^2}={1\over 2}{K_A}\gamma\,{H^2}-2cH\sqrt{{K_A}}+{{2{c^2}}\over\gamma}$         (27.3)

PA acts at the height of (H-Z0)/3 from the base of the wall. Fig. 27.1. Active earth pressure for c-ø backfill. Fig. 27.2. Passive earth pressure for c-ø backfill.

Figure 27.2 shows the passive earth pressure distribution for c-ø backfill. The passive force PP1 and PP2 can be determined as:

The passive force PP1 and PP2 can be determined as:

${P_{P1}}={1 \over 2}\gamma\,{K_P}{H^2}$ acts at a height of H/3 from the base                             (27.4)

${P_{P2}}=2cH\sqrt{{K_P}}$ acts at a height of H/2 from the base                    (27.5)

Thus the total force PP can be determined as:

${P_P}={1\over 2}\gamma {K_P}{H^2} + 2cH\sqrt{{K_P}}$                                                                      (27.6)

where  ${K_P}={{1 + \sin \phi}\over{1-\sin\phi}}$

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 References

Ranjan, G. and Rao, A.S.R. (2000). Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India.