Civil Engineering- Retaining walls Essay Example | Topics and Well Written Essays - 250 words

Civil Engineering- Retaining walls - Essay Example This force translates into moment by multiplying it by the moment distance, M of each section (Craig 166). M is the distance from the ‘toe’ of the wall to the centre of gravity of each section. Centre of mass acts at â…” of distance from the apex to base or â…“ of distance from the base to apex, for triangular sections. Centre of mass acts at the centre of section, for rectangular sections. Summation of the individual sections’ resisting moments gives the overall resisting moment. The force, Pa, acts at a distance â…“ H since the effective pressure region is triangular. The resulting moment is Pa multiplied by â…“ H. the factor of safety Fsot is the ratio of moment due to resistance to the moment due to net active thrust. The sliding resistance is the ratio of (the reaction force of the wall multiplied by the coefficient of sliding, 0.44) to the driving force offered by the soil to the wall (Craig 189). The reaction of the wall is the summation of the wall’s section forces, in his case 23.3 k. In question 2, the angle Ï• is 26o, and c is still zero; therefore, the wall exerts a triangular pressure distribution region. Given both the active and passive pressure co-efficient, Ka and Kp respectively, calculate Pa and Pp. The passive pressure is due to the resistance of soil to lateral forces. Ka and consequently, Pa are calculated as in problem 1 while Kp = [(1+ sin Ï•)/ (1 - sin Ï•)]. Pp = Â ½KpÃŽ ³ H2– 2c (√Ka) (H) but this reduces to Pp = Â ½KaÃŽ ³ H2 since c is zero (Craig 166). The factor of safety, Fsot, as in problem one is the ratio of the summation of resistance moment by the wall to moments due to active thrust of the soil. The factor of safety due to sliding limit is the summation of wall section’s reaction force multiplied by sliding coefficient, 0.45 divided by the thrust force of the soil to the