Question

Question 4 We consider the anchored retaining wall shown in Figure 3. 2 m Pmax = 100 kN/m* Medium sand: 18 N/m3 % - 30 6/0= 2/3 8 m 3 m Figure 3. Anchored retaining wall. (a) Sketch the earth pressure distributions corresponding to "free earth support" conditions at the front and back of the wall (with maximum values indicated). (b) Show that the anchor strength is sufficient. (c) Show that the embedment depth is sufficient. (d) Determine the necessary depth for a cantilever (free standing) wall.

Answer 1

2m Medium sand 0=300 Panx = loo ka N = 18 kN/m] &M Pi • P= tarz = 60 kN/mV 2m P b 2 1-sino Itsing Py = (kp-kg)r b = 84 kN/mY i-sino ka= 20.333 I+sinde Kp = 3. P, = 0-333X18X10 – 60 talsman Total pressure force, IP, = 1 x60x10 + 1 x9x60 = 337.5 ky in which distance 'a' is determined from the equation; pir (kp-ka) a zo 60 = 125m 18 (3-5) b = 3-1.25 = 1.75m Total pressure force, P₂ = 1x175x (2-2) x 18x1.75 - 73.5 km From equilibrium Pmax +PP2 - P, =0 Pmax = P,-PL - 337.5-73.5 264 EN

The max force applied is 100 kN/m . Hence Force z 100x (1071-75–1.25) - 1050 Kn ... The anchos strangth is sufficient Taking moments about '' 2,25 = x 60 x 10x ( +125) + 2 * 60x125x 2 125 2 3- 2003(3+Har) + ( +1-25) + 30x1.25x2x1-25 337.5 Ixbx48b = 4:083m Taking moments of all forces about A, P.(ath-e-z.) – P, * (ath-e +34) =( 327.5x (125+10-2-4-053) - 72.5(+25+10-2+ 2+5)=0 1742.8625 - 245°(125+10-2+) - 1743.8625 - 245" (9.25+*) -0 2226°+ 166²-11 -1743-862520 b=2.57m .. The depth of embedment, d= 6+a = 2:57+1.25 = 3.82m. (1) The necessary depth for free standing Wall 10+ 3.82-3 - 10.82m