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# Craig's Soil Mechanics 7th Edition

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surcharge. The pressures in soil 2 are calculated using Ka ¼ 0:36,Kp ¼ 2:78 and �0 ¼ (20� 9:8) ¼ 10:2 kN/m3 (see Table 6.1). The active and passive pressure distributions are shown in Figure 6.7. In addition, there is equal hydrostatic pressure on each side of the wall below the water table. Sloping soil surface The Rankine theory will now be applied to cases in which the soil surface slopes at a constant angle � to the horizontal. It is assumed that the active and passive pressures act in a direction parallel to the sloping surface. Consider a rhombic element of soil, with sides vertical and at angle � to the horizontal, at depth z in a semi-infinite mass. The vertical stress and the active or passive pressure are each inclined at � to the appropriate sides of the element, as shown in Figure 6.8(a). Since these stresses are not normal to their respective planes (i.e. there are shear components), they are not principal stresses. In the active case, the vertical stress at depth z on a plane inclined at angle � to the horizontal is given by �z ¼ �z cos� and is represented by the distance OA on the stress diagram (Figure 6.8(b)). If lateral expansion of the soil is sufficient to induce the state of plastic equilibrium, the Mohr circle representing the state of stress in the element must pass through point A (such that the greater part of the circle lies on the side of A towards the origin) and touch the Table 6.1 Soil Depth (m) Pressure (kN/m2) Active pressure 1 0 0:24� 50 ¼ 12:0 1 6 (0:24� 50)þ (0:24� 18� 6) ¼ 12:0þ 25:9 ¼ 37:9 2 6 0:36[50þ (18� 6)]� (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ0:36p ) ¼ 56:9� 12:0 ¼ 44:9 2 9 0:36[50þ (18� 6)]� (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ0:36p )þ (0:36� 10:2� 3) ¼ 56:9� 12:0þ 11:0 ¼ 55:9 Passive pressure 1 0 0 1 1.5 4:17� 18� 1:5 ¼ 112:6 2 1.5 (2:78� 18� 1:5)þ (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ2:78p ) ¼ 75:1þ 33:3 ¼ 108:4 2 4.5 (2:78� 18� 1:5)þ (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ2:78p ) þ (2:78� 10:2� 3) ¼ 75:1þ 33:3þ 85:1 ¼ 193:5 170 Lateral earth pressure failure envelope for the soil. The active pressure pa is then represented by OB (numer- ically equal to OB0) on the diagram. When c ¼ 0 the relationship between pa and �z, giving the active pressure coefficient, can be derived from the diagram: Ka ¼ pa �z ¼ OB OA ¼ OB 0 OA ¼ OD�AD ODþAD Now OD ¼ OCcos� AD ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðOC2 sin2 ��OC2 sin2 �Þ q Figure 6.8 Active and passive states for sloping surface. Rankine’s theory of earth pressure 171 therefore Ka ¼ cos� � ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðcos2 � � cos2 �Þ p cos� þ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðcos2 � � cos2 �Þ p ð6:8Þ Thus the active pressure, acting parallel to the slope, is given by pa ¼ Ka�z cos � ð6:9Þ and the total active thrust on a vertical wall surface of height H is Pa ¼ 1 2 Ka�H 2 cos� ð6:10Þ In the passive case, the vertical stress �z is represented by the distance OB 0 in Figure 6.8(b). The Mohr circle representing the state of stress in the element, after a state of plastic equilibrium has been induced by lateral compression of the soil, must pass through B0 (such that the greater part of the circle lies on the side of B0 away from the origin) and touch the failure envelope. The passive pressure pp is then represented by OA0 (numerically equal to OA) and when c ¼ 0 the passive pressure coefficient (equal to pp/�z) is given by Kp ¼ cos� þ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðcos2 � � cos2 �Þ p cos� � ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðcos2 � � cos2 �Þ p ð6:11Þ Then the passive pressure, acting parallel to the slope, is given by pp ¼ Kp�z cos� ð6:12Þ and the total passive resistance on a vertical wall surface of height H is Pp ¼ 1 2 Kp�H 2 cos� ð6:13Þ The active and passive pressures can, of course, be obtained graphically from Figure 6.8(b). The above formulae apply only when the shear strength parameter c is zero; when c is greater than zero the graphical procedure should be used. The directions of the two sets of failure planes can be obtained from Figure 6.8(b). In the active case, the coordinates of point A represent the state of stress on a plane inclined at angle � to the horizontal, therefore point B0 is the origin of planes, also known as the pole. (A line drawn from the origin of planes intersects the circumference of the circle at a point whose coordinates represent the state of stress on a plane parallel to that line.) The state of stress on a vertical plane is represented by the coordinates of point B. Then the failure planes, which are shown in Figure 6.8(a), are parallel to B0F and B0G (F and G lying on the failure envelope). In the passive case, the coordinates of point B0 represent the state of stress on a plane inclined at angle � 172 Lateral earth pressure to the horizontal, and therefore point A is the origin of planes: the state of stress on a vertical plane is represented by the coordinates of point A0. Then the failure planes in the passive case are parallel to AF and AG. Referring to Equations 6.8 and 6.11, it is clear that both Ka and Kp become equal to unity when � ¼ �; this is incompatible with real soil behaviour. Use of the theory is inappropriate, therefore, in such circumstances. Example 6.3 A vertical wall 6m high, above the water table, retains a 20� soil slope, the retained soil having a unit weight of 18 kN/m3; the appropriate shear strength parameters are c0 ¼ 0 and �0 ¼ 40�. Determine the total active thrust on the wall and the directions of the two sets of failure planes relative to the horizontal. In this case the total active thrust can be obtained by calculation. Using Equation 6.8, Ka ¼ cos 20 � � ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðcos2 20� � cos2 40�Þ p cos 20� þ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃðcos2 20� � cos2 40�Þp ¼ 0:265 Then Pa ¼ 1 2 Ka�H 2 cos� ¼ 1 2 � 0:265� 18� 62 � 0:940 ¼ 81 kN=m The result can also be determined using a stress diagram (Figure 6.9). Draw the failure envelope on the �/� plot and a straight line through the origin at 20� to the horizontal. At a depth of 6m, �z ¼ �z cos� ¼ 18� 6� 0:940 ¼ 102 kN=m2 and this stress is set off to scale (distance OA) along the 20� line. The Mohr circle is then drawn as in Figure 6.9 and the active pressure (distance OB or OB0) is scaled from the diagram, i.e. pa ¼ 27 kN=m2 Then Pa ¼ 1 2 paH ¼ 1 2 � 27� 6 ¼ 81 kN=m The failure planes are parallel to B0F and B0G in Figure 6.9. The directions of these lines are measured as 59� and 71�, respectively, to the horizontal (adding up to 90� þ �). Rankine’s theory of earth pressure 173 Earth pressure at-rest It has been shown that active pressure is associated with lateral expansion of the soil and is a minimum value; passive pressure is associated with lateral compression of the soil and is a maximum value. The active and passive values may thus be referred to as limit pressures. If the lateral strain in the soil is zero, the corresponding lateral pressure is called the earth pressure at-rest and is usually expressed in terms of effective stress by the equation p0 ¼ K0�0z ð6:14Þ where K0 is defined as the coefficient of earth pressure at-rest, in terms of effective stress. Since the at-rest condition does not involve failure of the soil, the Mohr circle representing the vertical and horizontal stresses does not touch the failure envelope and the horizontal stress cannot be evaluated. The value of K0, however, can be deter- mined experimentally by means of a triaxial test in which the axial stress and the all- round pressure are increased simultaneously such that the lateral strain in the specimen is maintained at zero: the hydraulic triaxial apparatus is most suitable for this purpose. For soft clays, methods of measuring lateral pressure in situ have been developed by Bjerrum and Andersen [3] and,