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# Craig's Soil Mechanics 7th Edition

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of the soil. If Ka ¼ 1� sin� 1þ sin� is defined as the active pressure coefficient, then Equation 6.1 can be written as pa ¼ Ka�z� 2c ﬃﬃﬃﬃ K p a ð6:2Þ When the horizontal stress becomes equal to the active pressure the soil is said to be in the active Rankine state, there being two sets of failure planes each inclined at 45� þ �/2 to the horizontal (the direction of the major principal plane) as shown in Figure 6.3(b). In the above derivation, a movement of the wall away from the soil was considered. On the other hand, if the wall is moved against the soil mass, there will be lateral compression of the soil and the value of �x will increase until a state of plastic equilibrium is reached. For this condition, �x becomes a maximum value and is the major principal stress �1. The stress �z, equal to the overburden pressure, is then the minor principal stress, i.e. �3 ¼ �z The maximum value �1 is reached when the Mohr circle through the point representing the fixed value �3 touches the failure envelope for the soil. In this case, the horizontal stress is defined as the passive pressure ( pp) representing the maximum inherent resistance of the soil to lateral compression. Rearranging Equation 6.1: �1 ¼ �3 1þ sin� 1� sin� � � þ 2c ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ sin� 1� sin� � �s ð6:3Þ If Kp ¼ 1þ sin� 1� sin� Rankine’s theory of earth pressure 165 is defined as the passive pressure coefficient, then Equation 6.3 can be written as pp ¼ Kp�zþ 2c ﬃﬃﬃﬃﬃﬃ Kp p ð6:4Þ When the horizontal stress becomes equal to the passive pressure the soil is said to be in the passive Rankine state, there being two sets of failure planes each inclined at 45� þ �/2 to the vertical (the direction of the major principal plane) as shown in Figure 6.3(c). Inspection of Equations 6.2 and 6.4 shows that the active and passive pressures increase linearly with depth as represented in Figure 6.4. When c ¼ 0, triangular distributions are obtained in each case. When c is greater than zero, the value of pa is zero at a particular depth z0. From Equation 6.2, with pa ¼ 0, z0 ¼ 2c � ﬃﬃﬃﬃﬃﬃ Ka p ð6:5Þ This means that in the active case the soil is in a state of tension between the surface and depth z0. In practice, however, this tension cannot be relied upon to act on the wall, since cracks are likely to develop within the tension zone and the part of the pressure distribution diagram above depth z0 should be neglected. The force per unit length of wall due to the active pressure distribution is referred to as the total active thrust (Pa). For a vertical wall surface of height H: Pa ¼ Z H z0 pa dz ¼ 1 2 Ka�ðH2 � z20Þ � 2cð ﬃﬃﬃﬃﬃﬃ Ka p ÞðH � z0Þ ð6:6aÞ ¼ 1 2 Ka�ðH � z0Þ2 ð6:6bÞ Figure 6.4 Active and passive pressure distributions. 166 Lateral earth pressure The force Pa acts at a distance of 1⁄3 (H � z0) above the bottom of the wall surface. The force due to the passive pressure distribution is referred to as the total passive resistance (Pp). For a vertical wall surface of height H: Pp ¼ Z H 0 ppdz ¼ 1 2 Kp�H 2 þ 2cð ﬃﬃﬃﬃﬃﬃKpp ÞH ð6:7Þ The two components of Pp act at distances of 1⁄3H and 1⁄2H, respectively, above the bottom of the wall surface. If a uniformly distributed surcharge pressure of q per unit area acts over the entire surface of the soil mass, the vertical stress �z at any depth is increased to �zþ q, resulting in an additional pressure of Kaq in the active case or Kpq in the passive case, both distributions being constant with depth as shown in Figure 6.5. The corresponding forces on a vertical wall surface of height H are KaqH and KpqH, respectively, each acting at mid-height. The surcharge concept can be used to obtain the pressure distribu- tion in stratified soil deposits. In the case of two layers of soil having different shear strengths, the weight of the upper layer can be considered as a surcharge acting on the lower layer. There will be a discontinuity in the pressure diagram at the boundary between the two layers due to the different values of shear strength parameters. If the soil below the water table is in the fully drained condition, the active and passive pressures must be expressed in terms of the effective weight of the soil and the effective stress parameters c0 and �0. For example, if the water table is at the surface and if no seepage is taking place, the active pressure at depth z is given by pa ¼ Ka�0z� 2c0 ﬃﬃﬃﬃﬃﬃ Ka p where Ka ¼ 1� sin� 0 1þ sin�0 Figure 6.5 Additional pressure due to surcharge. Rankine’s theory of earth pressure 167 Corresponding equations apply in the passive case. The hydrostatic pressure �wz due to the water in the soil pores must be considered in addition to the active or passive pressure. For the undrained condition in a fully saturated clay, the active and passive pressures are calculated using the parameter cu (�u being zero) and the total unit weight �sat (i.e. the water in the soil pores is not considered separately). The effect of the tension zone must be considered for this condition. In theory, a (dry) crack could open to a depth (z0) of 2cu/�sat (i.e. Equation 6.5 with Ka ¼ 1 for �u ¼ 0). Cracking is most likely to occur at the clay/wall interface where the resistance to fracture is lower than that within the clay. If a crack at the interface were to fill with water (due to heavy rainfall or another source of inflow), then hydrostatic pressure would act on the wall. Thus the clay would be supported by the water filling the crack to the depth (z0w) at which the active pressure equals the hydrostatic pressure. Thus, assuming no surface surcharge: �satz0w � 2cu ¼ �wz0w ; z0w ¼ 2cuð�sat � �wÞ In the Rankine theory it is assumed that the wall surface is smooth whereas in practice considerable friction may be developed between the wall and the adjacent soil, depending on the wall material. In principle, the theory results either in an over- estimation of active pressure and an underestimation of passive pressure (i.e. lower bounds to the respective ‘collapse loads’) or in exact values of active and passive pressures. Example 6.1 (a) Calculate the total active thrust on a vertical wall 5m high retaining a sand of unit weight 17 kN/m3 for which �0 ¼ 35�; the surface of the sand is horizontal and the water table is below the bottom of the wall. (b) Determine the thrust on the wall if the water table rises to a level 2m below the surface of the sand. The saturated unit weight of the sand is 20 kN/m3. (a) Ka ¼ 1� sin 35 � 1þ sin 35� ¼ 0:27 Pa ¼ 1 2 Ka�H 2 ¼ 1 2 � 0:27� 17� 52 ¼ 57:5 kN=m (b) The pressure distribution on the wall is now as shown in Figure 6.6, including hydrostatic pressure on the lower 3m of the wall. The components of the thrust are: (1) 1 2 � 0:27� 17� 22¼ 9:2 kN/m (2) 0:27� 17� 2� 3 ¼ 27:6 168 Lateral earth pressure (3) 1 2 � 0:27� ð20� 9:8Þ � 32¼ 12:4 (4) 1 2 � 9:8� 32 ¼ 44:1 Total thrust ¼ 93:3 kN/m Example 6.2 The soil conditions adjacent to a sheet pile wall are given in Figure 6.7, a surcharge pressure of 50 kN/m2 being carried on the surface behind the wall. For soil 1, a sand above the water table, c0 ¼ 0, �0 ¼ 38� and � ¼ 18 kN/m3. For soil 2, a saturated clay, c0 ¼ 10 kN/m2, �0 ¼ 28� and �sat ¼ 20 kN/m3. Plot the distributions of active pressure behind the wall and passive pressure in front of the wall. For soil 1, Ka ¼ 1� sin 38 � 1þ sin 38� ¼ 0:24; Kp ¼ 1 0:24 ¼ 4:17 Figure 6.6 Example 6.1. Figure 6.7 Example 6.2. Rankine’s theory of earth pressure 169 For soil 2, Ka ¼ 1� sin 28 � 1þ sin 28� ¼ 0:36; Kp ¼ 1 0:36 ¼ 2:78 The pressures in soil 1 are calculated using Ka ¼ 0:24,Kp ¼ 4:17 and � ¼ 18 kN/m3. Soil 1 is then considered as a surcharge of (18� 6) kN/m2 on soil 2, in addition to the surface