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DISEÑO GEOTECNICO DE MUROS DE SOSTENIMIENTO DE TIERRAS DISEÑO POR EL METODO DE SEGURIDAD GENERAL 10 ° 0.5 m 0.35 m 2*tang10° 14 6 m 30 0 10 ° D= 1.5 m 0.7 m 0.7 m 0.7 m 2m 16 23.58 20 ° equivalente 2.358 25 SOLUCION H' = 0.35 + 6 + 0.7 H' = 7.05 m 1 2 Para 30 y α = 10 ° 0.35 0.5 * 14 * 7.05 0.35 121.77 KN/m 121.77 * sen 10° = 21.15 121.77 * cos 10° = 119.92 Secci 1 6*0.5 = 3.00 3*23.58 = 70.74 1.15 81.35 2 0.2*6*0.5 = 0.60 0.6*23.58 = 14.15 0.83 11.74 3 3.4*0.7 = 2.38 2.38*23.58 = 56.12 1.70 95.40 4 2*6 = 12.00 12*14 = 168.00 2.40 403.20 5 2*0.35*0.5 = 0.35 0.35*14 = 4.90 2.73 13.38 = 21.15 3.40 71.91 = 335.06 = 676.98 Momento de volteo = * H' = 120 * 7.05 = 281.81 3 3 Fs = = 676.98 = 2.40 > 2 OK 281.81 H1 = H1 = Ɣ1= H2 = ɸ1= C1= H3 = Ɣ2= KN/m 3 Ɣconcreto = KN/m 3 ɸ2= tn/m 3 C2= KN/m 2 Pa= Ɣ1(H') 2 Ka ɸ1= Ka= Pa= 2 * Pa= Pv= Ph= Area (m 2 ) Peso Unitario/Longitud (KN/m) Brazo del momento del pto A Momento (KN- m/m) Pv v MR Mo Ph MR Mo 1 3 4 5 2 Ph Pv A

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Hoja1DISEO GEOTECNICO DE MUROS DE SOSTENIMIENTO DE TIERRASDISEO POR EL METODO DE SEGURIDAD GENERAL100.5mH1 =0.35mH1 =2*tang101=14KN/m3H2 =6m1=30C1=010D=1.5mH3 =0.7m0.7m0.7m2m2=16KN/m3concreto =23.58KN/m32=20equivalente2.358tn/m3C2=25KN/m2SOLUCIONH' =0.35+6+0.7H' =7.05mPa=11(H')2Ka2Para1=30y =10Ka=0.35Pa=0.5*14*7.052 *0.35Pa=121.77KN/mPv=121.77*sen10=21.15Ph=121.77*cos10=119.92SeccinArea (m2)Peso Unitario/Longitud (KN/m)Brazo del momento del pto AMomento (KN-m/m)16*0.5=3.003*23.58=70.741.1581.3520.2*6*0.5=0.600.6*23.58=14.150.8311.7433.4*0.7=2.382.38*23.58=56.121.7095.4042*6=12.0012*14=168.002.40403.2052*0.35*0.5=0.350.35*14=4.902.7313.38Pv=21.153.4071.91v=335.06MR=676.98Momento de volteoMo=Ph*H'=119.92*7.05=281.8133Fs=MR=676.98=2.40>2OKMo281.81FACTOR DE SEGURIDAD CONTRA DESLIZAMIENTO

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