Consider a cantilever wall carrying a sloping backfill as shown in the Fig.1. The various forces acting on the wall are: Pa = Active earth pressure acting at a height H/3 from the base on section ed. Ph = Pa cosi Pv = Pa sini Ws = Weight of soil abcd Wc = weight of wall including base Wt = the resultant of Ws and Wc Pt = the resultant of Wt and Pa meeting at f at base of Wall Pp = Passive earth pressure in front of the Wall Fr = Base Read more […]

## Stability of Different Type of Retaining Walls

Figure 1 shows a gravity, cantilever and counterfort types of retaining wall. The stability of a gravity wall is due to the self – weight of the wall and the passive resistance developed in front of the wall. The gravity walls are designed using Coulomb’s theory. Reinforced concrete walls (cantilever or counter – fort types) are more economical to gravity walls because the backfill itself provides most of the required dead load. Rankine’s theory Read more […]

## Culmann Graphical Method for Active and Passive Case

Active case In this method a retaining Wall is drawn to a suitable scale as shown in the Fig.1 below. The Various steps in the procedure are: Draw É¸-line AE at an angle É¸ with the horizontal. Lay off on AE distances AV, A1, A2, A3 etc to a suitable scale to represent the weight of wedges ABV, AB1, AB2, AB3, and so on. Lay off AD at an angle equal to (Î±- Î´) to the line AE. The line AD is called pressure line. Draw lines parallel Read more […]

## Coulomb derivation based on limit equilibrium approach for Passive Case

Figure 1 below shows the cross section of a retaining wall. Weight of Wedge ABC From Î” ABC, Area of Î” ABC = ½ AD Ã— BC BC/AB = {sin(Î± + Î²)/ sin(Ï´ – Î²)} BC = AB {sin(Î± + Î²)/ sin(Ï´ – Î²)} Again, AD = AB sin[1800 – (Î± + Ï´)] = AB sin(Î± + Ï´) Triangle of Forces for W, Pp and R From the sine rule, Substituting the value Read more […]

## Coulomb’s Wedge Theory | Derivation for Active Case

Coulomb’s made the following assumptions in the development of his theory. The assumptions are: The soil is isotropic and homogeneous. The surface of rupture is a plane. The failure wedge is a rigid body. There is friction between and the wall. Back of wall need not be vertical. Failure is two dimensional. The soil is cohesionless. Coulomb’s equation of shear strength is valid. Coulomb made his derivation based on limit Read more […]

## Derivation of Equation by Trial Wedge Method for Passive Case

The equation is derived as follows. A planar failure wedge IJM is considered. There are distributed normal stresses along IJ and JM and distributed along JM. The resultants of these stresses are carried out in the analysis. In this case, the force F acts above the normal and the angle between the force F and weight W will be Ï´ + É¸. The forces acting at the free body IJM are: Weight of wedge = ½ Ã— Area IJM Ã— Î³ = ½ Ã— Read more […]

## Derivation of Equation by Trial Wedge Method for Active Case

Figure 2 below shows the application of the trial Wedge method to the problem of simple retaining wall without wall friction. A planar failure wedge IJM is considered. There are distributed normal stresses along IJ and JM and distributed shear stress along JM. The resultants of these stresses are carried out in the analysis. The forces acting at the free body IJM are: Weight of wedge = ½ Ã— Area IJM Ã— Î³ = ½ Ã— IM Ã— IJ Read more […]

## Trial Wedge Method

The trial wedge method of analysis is shown in Fig.1 below. The following steps are to be followed to make the analysis. Step 1 – a mass of soil behind the wail is considered as a free body. The force P, which must exist between the free body and the wall, is found by writing the equation of equilibrium for the free body as a whole. Step 2 – a different free body is considered, having a different boundary through the soil. Once again the required Read more […]

## Rankine’s Earth Pressure in Cohesive Soil for Passive Case

Figure 1 shows the Mohr’s circle in which point B indicates the vertical stress and point E represents the passive pressure. The wall carries a uniform surcharge q. The circle is tangent to the failure envelope. For this case the relation between Pa and Ïƒv is given by: Pp = KpÏƒv + qKp + 2c√Kp ———-(1) Where, Ka = (1 + sinÉ¸)/(1 – sinÉ¸) = tan2[450 + (É¸/2)] Pressure Distribution of passive case in cohesive soil – Figure Read more […]

## Rankine’s Earth Pressure in Cohesive Soil for Active Case

Figure 1 shows the Mohr’s circle in which point B indicates the vertical stress and point E represents the active pressure. The circle is tangent to the failure envelope. For this case the relation between Pa and Ïƒv is given by: ———-(1) ———-(2) Where, In the above equation (2) where z = 0 The negative sign indicates that the pressure is negative and tensile, As a result there would be gap between backfill and wall. The Read more […]