PUS V recipiente 940 1629 2124 1598 cm3 W recipiente 5316 8200 gr W suelto A

Pus v recipiente 940 1629 2124 1598 cm3 w recipiente

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PUS V recipiente. 940 1.629 2124 1.598 cm3 W recipiente 5316 8200 gr W (suelto A + recipiente) 7486 1799 3 gr PUC V recipiente. 940 1.786 2124 1.712 m3 W recipiente 5316 8200 gr W (compactado A + recipiente) 7728 1343 6 gr 4.0 . PE (1° caso) W (Agregado) 200 2.632 500 2.577 gr W (matraz + agua) 668 1444 ml W (matraz + agua+agregado) 792 1750 ml 5.0 . MF Suma 275.645 2.756 720.4 99 7.2 sin dimension Perfil Tmax 1 1/2 pul. TECNOLOGIA DEL CONCRETO. 3
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UNSAAC ING. CIVIL Nom. E. Granulometría de los agregados AGREGADO GRUESO TAMIZ PESO RETENIDO(gr s) % RETENIDO % RETENIDO ACUMULAD O % QUE PASA 2'' 0 0 0 100 1 1/2'' 452 17.09531 01 17.0953101 82.90468 99 1'' 468 17.70045 39 34.795764 65.20423 6 3/4'' 520 19.66717 1 54.4629349 45.53706 51 1/2'' 376 14.22087 75 68.6838124 31.31618 76 3/8'' 306 11.57337 37 80.2571861 19.74281 39 N°4 522 19.74281 39 100 0 2644 100 TMN: 1 ½ MF grueso = 7.2 MF fino =2.76 2'' 1 1/2'' 1'' 3/4'' 1/2'' 3/8'' N°4 0 10 20 30 40 50 60 70 80 90 100 110 Curva Granulométrica del Agregado Grueso Serie de Tamices % que pasa AGREGADO FINO TAMIZ PESO RETENIDO(grs ) % RETENIDO % RETENIDO ACUMULAD O % QUE PASA TECNOLOGIA DEL CONCRETO. 4
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UNSAAC ING. CIVIL # 4 0 0 0 100 # 8 118 16.905444 1 16.9054441 83.094555 9 # 16 114 16.332378 2 33.2378223 66.762177 7 # 30 120 17.191977 1 50.4297994 49.570200 6 # 50 214 30.659025 8 81.0888252 18.911174 8 # 100 90 12.893982 8 93.982808 6.0171919 8 # 200 32 4.5845272 2 98.5673352 1.4326647 6 Fondo 10 1.4326647 6 100 0 698 100 # 4 # 8 # 16 # 30 # 50 # 100 # 200 Fondo 0 10 20 30 40 50 60 70 80 90 100 110 Curva Granulométrica del Agregado Fino Serie de Tamices % que pasa 3.2 Determinar la Resistencia de diseño f’cr. No se cuentan con datos estadísticos de ensayos. f’cr = f’c + 84 = 300 + 84= 384 TECNOLOGIA DEL CONCRETO. 5
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UNSAAC ING. CIVIL f’cr = 384 Kg/cm² 3.3 Determinar la cantidad de agua por m³ y contenido de aire. De la tabla obtenemos la cantidad de agua que es 175 lt/m3 % aire atrapado = 1,0% 3.4 Determinar la relación a/c. Interpolando: 400 350 0.43 0.48 = 384 350 x 0.48 a/c =0.446 TECNOLOGIA DEL CONCRETO. 6
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UNSAAC ING. CIVIL 3.5 Cálculo del contenido de cemento: Cemento = 175/0.446=392.38 kg Vol cemento por m 3 =392.38/(1000*3.11)=0.1262 3.6 Calculo del peso de los agregados. A. METODO DE COMBINACION DE AGREGADOS Consideramos 9 bolsas de cemento y TMN es 1 1/2 Mc=5.79 MF fino = 2.76 MF grueso = 7,52 El porcentaje de agregado fino : Rf = 7.52 5.79 7.52 2.76 100 Rf = 36.34% El porcentaje de agregado grueso : Rg = 63.66% TECNOLOGIA DEL CONCRETO. 7
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UNSAAC ING. CIVIL El volumen total del agregado: Vol total agregado=1-(0.175+0.1262+0.01)=0.6888 Los volúmenes de los agregados serán: Vol. A.G.= 0.6888*63.66% = 0.4385 Vol. A.F.= 0.6888*36.34% = 0.2503 Los pesos secos de los agregados serán: Peso A.G.= 0.4385*2.577*1000 = 1130.0 Kg Peso A.F.= 0.2503*2.632*1000 = 658.8 Kg B. METODO DEL ACI De la tabla obtenemos b/bo, interpolando( MF=2.76): 2.8 2.6 0.72 0.74 = 2.76 2.6 x 0.74 b/bo=0.724 Cálculo de peso del agregado grueso.
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  • Winter '19
  • ing. marines e.

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