Question

2. The water level behind the wall described in problem 1 rises to an elevation I m below the crest. The submerged unit weight of the sand is 10 kN/m'. If the deformation condition for the active Rankine state is satisfied, what is the resultant pressure that the earth and water exert against the wall? At what height above the base does the resultant of the earth and water pressures act?

Answer 1

Anda H = 3m @ When the wall is prevented 3m Vo = 18 kNl na from yielding Ł - 36 (i) Ko= 1 - sind Ko= earth pressus coafficient at rest Ko= 0.4122 Pos Kora H = 0.4722x 18x3 = 22.7588 kW/m2 (ii) if Active Rankine state Kat 1-sind = 0.2596 1 tsing Pa= kard H= 14.019 kN/m² 4-6721 629.8648 KNoah Psub= 10 kN/m3 let rw= 10 KN/m? unit weight of water = 10 kN/m3 (Palo - o Posto suttttttt***) As (4-6428+23.9645) (Pa) B= Koro al = 4.6728 KNIML A. = 17.246) (Palc = (Karo x1 ) + (nat subt) + (vwX2) T= 1.1353 - A2= 2.3384 KN lm (Pale = 29. 86 48 KN/m2 T = 2.33 Point of application of Resultant să = A, at, tak ALTA, = 17.2586X1:1353) +(2:33( 97233) 17.2588+ 2.3364 == 1.2777m from the base of Wall above