EJERCICIO EN CLASE SEMANA 14 MURO CON CONTRAFUERTES

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EJERCICIO 1 
 
 
 
 
hp= 7.50 m 
hz= 0.50 m 
B1=2.70 m 
B2=1.05 m 
t1= 0.30 m 
t2= 0.45m 
 
ka = [1-sen()] / [1+sen()]=0.238 
 
P1,zap =3.75x0.8x2.40= 7.20 tnf 
b1=3.75/2 = 1.875m 
 
P2,muro,triang =[(0.15x7.5/2)x2.4]=1.35 tnf 
b2=1.05+(2/3)x0.15=1.15m 
P3,muro,rect =0.3x7.5x2.4= 5.40 tnf 
b3=1.05+0.15+0.15=1.35m 
 
P4,relleno =2.25x7.5x1.9=32.06 tnf 
b4=1.05+0.45+2.25/2=2.625 m 
 
Ptot = 46.01 tnf 
Mtot =106,5 tnf*m 
 
Fd=0.238x1.9x8.302/2= 15.58 tnf 
Mv= 15.58x8.3/3= 43.10 tnf*m 
 
FSD= 46.01x0.52 / 15.58 = 1.54 > 1.50 si cumple 
FSV=106.5 / 43.10 = 2.47 > 1.75 si cumple 
X0=(106.50 – 43.10 ) / 46.01= 1.38m 
B/2-X0 < B/6 3.75/2 -1.38 = 0.495 < 0.625 si cumple estamos dentro núcleo central 
 
 Diseñar el muro con ccontrafuertes que se presenta en figura. El coeficiente de 
fricción concreto-terreno es 0.52. Considerar t =1.9 tn/m³ y =38°. qadm= 2.20 
kgf/cm2, f’c =210 kgf/cm², fy =4200 kgf/cm². El muro tiene contrafuertes de 
espesor de 35cm cada 3.00m. 
 
0.80 
 
𝑞𝑚𝑎𝑥 =
𝑃
𝐵
× (1 +
6𝑒
𝐵
) =
46.01
3.75
× (1 +
6 × 0.495
3.75
) = 21.987 𝑡𝑛𝑓/𝑚2 
𝑞𝑚𝑖𝑛 =
𝑃
𝐵
× (1 +
6𝑒
𝐵
) =
46.01
3.75
× (1 −
6 × 0.495
3.75
) = 2.552 
Si cumple con el qadm. 
 
Diseño del acero en la pantalla 
hp=7.50m 
t1=0.30m 
t2=0.45m 
 
tcontrafuerte = 0.35m 
3/8L = 3/8 x3 =1.125 m 
H= 7.50 -1.125 = 6.375 m 
Wu=1.7x0.238x1.9x6.375= 4.90 tnf /m2 
 
Diseño del acero horizontal negativo de la pantalla 
Mu=4.90x(3.00 -0.35)2 x1/12= 2.868 tnf*m 
tpant,prom=(0. 45+0.30)/2=0.375m 
d= = 37.5-4-1.59/2 = 32.705 cm 
𝑤 = 0.85 − √0.7225 −
1.7 × 2.868 × 105
0.9 × 210 × 100 × 32.7052
= 0.0143 
=wxf´c / fy =0.0143x 210/4200 =7.15 x10-4 
As,nec =7.15 x10-4x100x32.705 = 2.33 cm2 
Asmin= 0.0020xbxh=0.0020x100x37.5 =7.50cm2/m 
1/2” 1.27 /(7.50/2) = 0.34m 1/2”@0.30m 
 
Diseño del acero horizontal positivo de la pantalla 
Mu=4.90x(3.00 -0.35)2 x1/16= 2.151 tnf*m 
tpant,prom=(0. 45+0.30)/2=0.375m 
d= = 37.5-4-1.59/2 = 32.705 cm 
𝑤 = 0.85 − √0.7225 −
1.7 × 2.151 × 105
0.9 × 210 × 100 × 32.7052
= 0.0107 
=wxf´c / fy =0.0107x 210/4200 = 5.35 x10-4 
As,nec =5.35 x10-4x100x32.705 = 1.75 cm2 
Asmin= 0.0020xbxh=0.0020x100x37.5 =7.50cm2/m 
1/2” 1.27 /(7.50/2) = 0.34m 1/2”@0.30m 
 
 
Diseño del acero vertical de la pantalla 
t2= 45 cm 
d= 45-4-1.59/2 = 40.205cm 
Mu(-) =1.7x 0.03x1.9x0.238x7.52x3= 3.892 tn*m 
𝑤 = 0.85 − √0.7225 −
1.7 × 3.892 × 105
0.9 × 210 × 100 × 40.2052
= 0.0128 
=wxf´c / fy =0.0128x 210/4200 = 6.40 x10-4 
As,nec = 6.40x10-4x100x40.205 = 2.573 cm2 
Asmin= 0.0015xbxh=0.0015x100x45 =6.75 cm2/m 
1/2” 1.27 /(6.75/2) = 0.376m 1/2” @0.35 
Calculo del acero vertical por momento positivo 
t=35cm 
d=35-4-1.59/2 = 30.205 cm 
Mu(-) =3.891/4 = 0.973 tn-m 
w= 0.85-0.7225-(1.7x0.973x105)/(0.9x210x100x30.2052) 
w=5.66 x 10-3 
= 5.66 x 10-3x210/ 4200 = 0.283x10-3 
Asnec =0.283x10-3x100x 30.205 = 0.855 cm2/m 
Asmin= 0.0015xbxh = 0.0015x100x45 = 6.75 cm2 
1/2” 1.27 / (6.75/2) =0.376 m 1/2” @0.35m 
 
Diseño acero en zapata anterior ( punta ) 
 
Wu = 1.7qmax-0.9Wzap=1.7x 21.987- 0.9x0.8x2.4= 35.650 tnf/m2 
Mu = 35.650x 1.052/ 2 = 19.652 tnf*m 
d=80-7.5-1.59/2= 71.705 cm 
𝑤 = 0.85 − √0.7225 −
1.7 × 35.650 × 105
0.9 × 210 × 100 × 71.7052
= 0.0375 
= 0.0375x210/ 4200 =1.875x10-3 
Asnec =1.875x10-3x100x 71.705 = 13.444cm2/m 
Asmin= 0.0012xbxh = 0.0012x100x80 = 9.60 cm2/m 
3/4” 2.85/13.444 = 0.212 3/4” @0.20m 
 
Calculo acero en la zapata posterior talón 
Wt= 1.9x7.5 =14.25 tnf/m2 
Wzap=0.8x2.4=1.92 tnf/m2 
 
(21.987-2.552)/3.75 = qcara/2.25 qcara= 11.661 tnf/m2 
qcara = 11.661 + 2.552 =14.212 tn/m2 
qprom=(14.112 + 2.552)/2 = 8.382 tn/m2 
Wumax=1.4x14.25 +1.4x1.92 -1.7x8.382 = 8.389 tnf/m2 
 
Calculo acero por momento positivo 
Mu = 8.389x(3-0.35)2 / 16 = 3.682 tn-m 
d=80-7.5-1.59/2= 71.705 cm 
 
𝑤 = 0.85 − √0.7225 −
1.7 × 3.682 × 105
0.9 × 210 × 100 × 71.7052
= 3.80 × 10−3 
= 3.80x10-3x210/ 4200 = 1.9 x 10-4 
Asnec =1.90x10-4x100x 71.705 = 1.362 cm2 /m 
Asmin= 0.0012xbxh = 0.0012x100x50 = 6.0cm2 
1/2” 1.27/ (6.0/2) = 0.423m 1/2” @0.40m doble malla 
 
Calculo acero por momento negativo 
Mu = 8.389x(3-0.35)2 / 12 = 4.909 tn-m 
𝑤 = 0.85 − √0.7225 −
1.7 × 4.909 × 105
0.9 × 210 × 100 × 71.7052
= 5.06 × 10−3 
= 5.06x10-3x210/ 4200 = 2.80x10-4 
Asnec =2.80x10-4x100x 71.705 = 2.01 cm2/m 
Asmin= 0.0012xbxh = 0.0012x100x50 = 6.0cm2 
1/2” 1.27/ (6.0/2) = 0.423m 1/2” @0.40m doble malla 
 
 
 
 
 
 
 
DISEÑO DE CONTRAFUERTE 
Vux1m=1.7x0.238x1.9x7.52/2 =21.621tnf 
Vux3m=21.621x3= 64.863 tnf 
 
Mux1m=1.7x0.238x1.9x7.52/2x(7.5/3) =54.052 tn-m 
Mux3m=54.052 x 3 = 162.156 tn-m 
d=2.70-7.5-1.905/2=261.548cm 
α=arct(7.5/2.25) = 73.30° 
𝑇𝑢 = 64.863 × cos (73.30) +
162.156
2.615
× sen (73.30) 
Tu=78.034 tnf 
As=Tu /(0.9xfy) =(78.034x1000)/ (0.9x4200) = 20.643 cm2 
4 varillas de 1” + 1 f 1/2" area =21.55 cm2 
 
 
 
Acero inclinado a 2/3 h respecto a la base 
Vux1m=1.7x0.238x1.9x5.002/2 =9.609tn 
Vux3m=9.609x3 = 28.827 tn 
Mux1m=1.7x0.238x1.9x5.003/6 =16.015 tn-m 
Mux3m=16.015 x3=48.045 tn-m 
2.25/7.50 = d/ 5 d=1.50m dfinal = 1.50+0.30+0.10= 1.90m 
de=190-7.5-1.905/2 = 181.548 cm 
=arct(7.5/2.25) = 73.30° 
Tu=28.827 xcos(73.30°)+ (48.045 /1.815)sen(73.30°)= 33.638 tn 
As=Tu /(0.9xfy) = (36.638x1000)/ (0.9x4200) = 8.899 cm2 
2 varillas de 1” area =10.14 cm2 
 
Acero inclinado a 1/2 h respecto a la base 
Vux1m=1.7x0.238x1.9x3.752/2 =5.405 tn 
Vux3m=5.405x3 = 16.215 tn 
Mux1m=1.7x0.238x1.9x3.753/6 = 6.757 tn-m 
Mux3m=6.757 x3=20.271 tn-m 
d=1.125m dfinal = 1.125+0.30+0.075= 1.50m 
de=150-7.5-1.905/2 = 141.548 cm 
=arct(7.5/2.25) = 73.30° 
Tu=16.215 xcos(73.30°)+ (20.271 /1.415)sen(73.30°)= 18.381 tn 
As=Tu /(0.9xfy) = (18.381 x1000)/ (0.9x4200) = 4.863 cm2 
2 varillas de 3/4” area =5.70 cm2 
Acero horizontal 
Wu =4.90 tn/m2 (Wu se determinó en el diseño del acero del pantalla) 
Tu=4.90x3/2= 7.35 tnf 
As = (7.35x1000) / (0.9x4200 ) = 1.94 cm2 
Asmin= 0.0020x100x35 = 7.00 cm2 
f 1/2" 1.27 / (7/2) = 0.363 f 1/2" @0.35m 
 
Acero vertical 
Wumax=8.389 tn/m2 (Wu se determinó en el diseño del talón) 
Tu = 8.389x2.25x3 = 49.417 tn 
As,nec=(49.417x1000)/ (0.9x4200 ) = 13.073cm2 
Asmin= 0.0020x100x35 = 7.00 cm2 
f 5/8" 1.98/13.073 =0.152 f 5/8" @0.15m