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PROBLEMAS PROPUESTOSCATEDRA: CIMENTACIONESALUMNO: ACLARI HUAYLLANI LUISAULA:B-1 TURNO: TARDE

ZAPATA AISLADAZAPATA AISLADA

PROBLEMA 1Disear la zapata mostrada en la figura 1. La columna es de 0.6x0.6m. La carga que transmite es: PD= 180tn.y PL=130tn. El peso especfico del suelo es 1800 kg/m3 y una capacidad portante de 3.4kg/cm2 . Adems est sometido a la accin de una sobrecarga de 500 kg/m2. Para la zapata emplear f 'c =21 O kg/cm2 y fy =4200 kg/cm2

DATOS:

ZapataOtrosSuelof ' c =210kg/cmS/C =0.5tn/mDf =1.8mPD =170Tng=1.8tn/mColumnaPL =85Tnt =34tn/mt1 =0.8mhf=1.8mt2 =0.6mAcerof y =4200kg/cm

1.- CALCULO DEL ESFUERZO NETO DEL TERRENOn =34.00-1.8x1.8-0.5n =30.26tn/m22.- DIMENSIONAMIENTO EN PLANTAP = PD + PLP =170 + 85P = 255Azap = 255 = 8.43m230.26

T = 3.00m

S = 2.80m3.- CALCULO DEL PERALTEPu = 1.4PD + 1.7PLPu = 382.5tnbo =2(0.6+d)+2(0.6+d)bo =2.4 + 4dm =0.6+dn =0.6+dWnu =PuS*TWnu = 382.58.42Wnu =45.44tn/m2 = 0.75APLICANDO EN:

d =0.58mdprom = 0.58 +rec. +3/4dprom =0.6741mdreajus. =0.6741 -rec. -3/4dreajus. =0.58m4.- VERIFICACION POR CORTANTE RESISTENTE POR FLEXION EN EL CONCRETOVc =124.8604206887tn5.- VERIFICACION POR CORTANTE POR FLEXION ACTUANTELv =3.00 -0.82Lv =1.10Vu = 45.44*2.80*(1.10 - 0.58)Vu =66.4216555797tnDEBE CUMPLIRSE QUE :

66.42124.86042068870.7588.56124.8604206887si cumple6.- DISEO POR FLEXION7.- CALCULO DE AREA DE ACERO TRASVERSAL8.- CALCULO DE AREA DE ACERO LONGITUDINAL22Mu =382.5 *1.10Mu =382.5 *1.103.0022.802Mu =77.27tn-mMu =82.78tn-m

a =305a =305As =77.27*10As =82.78*10a1 =4200*47.540.9*4200*(58 - 30)a1 =4200*50.930.9*4200*(58 - 30)0.85*210*28020.85*210*3002a1 =3.99As =47.54a1 =3.99As =50.93

a1 =4200*36.505a1 =4200*39.1050.85*210*280As1 =77.27*100.85*210*300As1 =82.78*10a1 =3.060.9*4200*(58 - 3.99)a1 =3.060.9*4200*(58 - 3.99)22As1 =36.50As1 =39.10

55As2 =77.27*10As2 =82.78*100.9*4200*(58 - 3.06)0.9*4200*(75 - 3.06)22As2 =36.20cm2As2 =29.81cm2

NUMERO DE VARILLASASNUMERO DE VARILLASASarea 3/4area 3/4N =36.20N =29.812.852.85N =12.7 =13varillasN =10.5 =11varillasSEPARACION280 -rec.-diametro 3/4SEPARACION300 -rec.-diametro 3/413 - 111 - 1S =22.20 =22cmS =28.64 =28cm

ZAPATA COMBINADAZAPATA COMBINADA

PROBLEMA 1Disear la zapata mostrada en la figura 1. La columna 1 es de 0.6x0.6m y la columna 2 es de 0.5x0.5m. La carga que transmite es: PD1= 80tn - PL1=40tn y PD2=120tn - PL2=50tn. El peso especfico del suelo es 2200 kg/m3 y una capacidad portante de 2.8kg/cm2 . Adems est sometido a la accin de una sobrecarga de 500 kg/m2. Para la zapata emplear f 'c =21 O kg/cm2 y fy =4200 kg/cm2

DATOS:

ZapataOtrosSuelof ' c =210kg/cmS/C =0.5tn/mDf =1.5mPD1 =80TnP1g=2.2tn/mPL1 =40Tnt =28tn/mColumna 1 =0.6x0.6PD2 =120TnP2hf=1.8mColumna 2 =0.5x0.5PL2 =50TnAcerof y =4200kg/cm

P1P2

S/C =500kg/m

g m

l=4.5mXo

1.- CALCULO DEL ESFUERZO NETO DEL TERRENOn =28-1.8x2.2 -0.5n =23.54tn/m22.- DIMENSIONAMIENTO DE ZAPATAP = P1 + P2P =120 + 170P = 290Azap = 290 =12.32m223.54Xo =120(0.6) +170(4.5 +0.6 +0.5)22290Xo =3.26m3.- LONGITUD DE ZAPATALzap =2*3.26Lzap =6.5206896552mHallamos "b"b =AzapLzapb =1.89m4.- CALCULO DE REACCION NETA DEL TERRENOPu =P1u+P2uPu =180 +253Pu =433WNU =433 =66.40tn/m6.52

P1u1

x = 0g m -P1u+X1(WNU) =0 - 180 + X1(66.40) =0X1 =2.71Mmax

X11EnX1=2.71Por tanto en 1-1, el MmaxMmax =189.96tn-m5.- CALCULO DEL PERALTEw =0.09 =0.95 =189.96*100.9*189*210*0.09(1 -0.59*0.09) =79.01cmAsumimos 3/4hz=d +13/4 +rec.2hz =87.46cmentonces :d =hz -13/4-rec.2d =79.0094722896cm6.- VERIFICACION DEL PERALTE1.04V1.599225974719.921.32lv =0.9206896552

v20.13+2.71m-v3v1

72.39160.08

v1=87.69v3 =8.67Vu =el mayor106.1950320708v2 =106.1950320708

Vc =114.65tn

DEBE CUMPLIRSE QUE :

106.20114.60.75141.59114.6469977885no cumple

ENTONCES:

Aumentando el anchod=2.4Vc =145.64tnENTONCES EL AREA DE ZAPATA:Lzap*b6.52*2.4m7.- VERIFICACION POR CORTE POR PUNZONAMIENTO

c1c2nn

mm

COLUMNA 1 EXTERIORCOLUMNA 2 INTERIOR

P1u = 1.4PD + 1.7PL=0.75P2u = 1.4PD + 1.7PL=0.75P1u = 180tnFc=210P2u = 253tnFc=210bo=4.7702841687bo=5.1603788916Wnu=Pu=66.2506610259tn/m2d=0.7900947229Wnu=Pu=93.12ERROR:#VALUE!tn/m2d=0.7900947229AzapAzapm=0.995mVc=434.21m=1.29mVc=469.72n=1.39mn=1.29m

Vu=88.3614174528Vu=98.0179398951

DEBE CUMPLIRSE QUE :DEBE CUMPLIRSE QUE :

88.36434.2198.02469.720.750.75117.82434.21si cumple130.69469.72si cumple

7.- ACERO LONGITUDINAL SUPERIOR

Mmax=189.96tn-m

a =305As =189.96*10a1 =4200*78.510.9*4200*(79.0094722896 - 30)0.85*210*2402a1 =7.70As =78.51

a1 =4200*66.8650.85*210*240As1 =189.96*10a1 =6.560.9*4200*(79.0094722896 - 7.70)2As1 =66.86

5As2 =189.96*100.9*4200*(79.0094722896 - 6.56)2As2 =66.36cm2

NUMERO DE VARILLASASarea 3/4N =66.362.85N =23.3 =24varillasSEPARACION240 -rec.-diametro 3/424 - 1S =9.83 =10cm

7.- PARA LA COLUMNA EXTERIOR (ACERO TRANSVERSAL)Lv=b-t12Lv=0.9m2Mu =180 *0.902.402Mu =30.37tn-m

a =305As =30.37*10a1 =4200*12.550.9*4200*(79.0094722896 - 30)0.85*210*2402a1 =1.23As =12.55

a1 =4200*10.2550.85*210*240As1 =30.37*10a1 =1.000.9*4200*(79.0094722896 - 1.23)2As1 =10.25

5As2 =30.37*100.9*4200*(79.0094722896 - 1.00)2As2 =10.24cm2

7.- ZAPATA 18.- ZAPATA 2Lv=b-t1Lv=b-t222Lv=0.9mLv=0.95m

22Mu =180 *0.90Mu =253 *0.952.4022.402Mu =30.37tn-mMu =47.57tn-m

b =99.50b =129.01

a =305a =305As =30.37*10As =47.57*10a1 =4200*12.550.9*4200*(79.0094722896 - 30)a1 =4200*19.660.9*4200*(79.0094722896 - 30)0.85*210*99.520.85*210*1292a1 =2.97As =12.55a1 =3.59As =19.66

a1 =4200*10.375a1 =4200*16.3050.85*210*99.5As1 =30.37*100.85*210*129As1 =47.57*10a1 =2.450.9*4200*(79.0094722896 - 2.97)a1 =2.970.9*4200*(79.0094722896 - 3.59)22As1 =10.37As1 =16.30

55As2 =30.37*10As2 =47.57*100.9*4200*(79.0094722896 - 2.45)0.9*4200*(79.0094722896 - 2.97)22As2 =10.33cm2As2 =16.23cm2

Asmin=0.0018*d*bAsmin=0.0018*d*bAsmin=14.15cm2Asmin=18.35cm2

NUMERO DE VARILLASASNUMERO DE VARILLASASarea 3/4area 3/4N =14.15N =18.352.852.85N =4.97 =5varillasN =6.4 =6varillasSEPARACION99.50 -rec.-diametro 3/4SEPARACION129 -rec.-diametro 3/45 - 16 - 1S =21.40 =22cmS =23.02 =23cm

d/2d/2d/2

12t1t2b1b2L5 3/4"@ 22cm6 3/4"@ 23cm24 3/4"@ 10cm

ZAPATA CONECTADAZAPATA CONECTADA

PROBLEMA 1Disear la zapata mostrada en la figura 1. La columna 1 es de 0.6x0.6m y la columna 2 es de 0.5x0.5m. La carga que transmite es: PD1= 70tn - PL1=26tn y PD2=120tn - PL2=45tn. El peso especfico del suelo es 2000 kg/m3 y una capacidad portante de 3.5kg/cm2 . Adems est sometido a la accin de una sobrecarga de 400 kg/m2. Para la zapata emplear f 'c =21 O kg/cm2 y fy =4200 kg/cm2

DATOS:

ZapataOtrosSuelof ' c =210kg/cmS/C =0.4tn/mg=2tn/mPD1 =70TnP1t =35tn/mPL1 =26Tnhf=1.5mColumna 1 =0.5x0.5PD2 =120TnP2Columna 2 =0.7x0.7PL2 =45TnAcerof y =4200kg/cmL1=6.2m1.- DIMENSIONAMIENTOa)ZAPATA EXTERIORn =35-1.5x2 -0.4n =31.6tn/m2P1 = PD + PLP1 =70 + 26P1 = 96tnAzap = 1.2*96.0031.6Azap=3.65m2entonces dimension en planta:

Azap=T*S1T=2S2reeplazo 2 en 13.65=2s*sS=1.35mT=2.70mb)DIMENSIONAMIENTO DE VIGAh=0.8857142857mUSAR=0.4994797086*0.9m2b=0.4994797086m2.- DIMENSIONAMIENTO DE ZAPATA EXTERIOR(NUEVO)Wv=1.06tn/mP1=96tnAzap = 106.89S=1.3531.6F=0.68mAzap=3.38m2G=5.77mRN=106.89tnAzap=T*1.353.38=T*1.352.51=TUSAR=2.51*1.35m2

3.- DIMENSIONAMIENTO DE LA VIGA DE CONEXINWvu=1.49tn/mP1u=142.2tnS=1.35Xo=1.23mF=0.68mG=5.77mRNU=158.02tnWNU=117.04tn/m4.- CALCULO DEL MOMENTO MAXIMOMmax =-51.94tn-m

a =205As =51.94*10a1 =4200*18.680.9*4200*(83.57 - 20)0.85*210*49.92a1 =8.80As =18.68

a1 =4200*17.3650.85*210*49.9As1 =51.94*10a1 =8.180.9*4200*(83.57 - 8.80)2As1 =17.36

5As2 =51.94*100.9*4200*(83.57 - 8.18)2As2 =17.29cm2

Asmin=0.0018*d*bAsmin=7.51cm2entonces :d =0.9 -11/2-rec.2d =0.84mNUMERO DE VARILLASASarea 1N =17.295.1N =3.39 =4varillasUSAR:41=20.4cm220.401449.9*83.5742000.00488713680.003conforme4.- REFUERZO A LA CARGA INFERIOR20.400.003*b*d210.213.91NUMERO DE VARILLASASarea 3/4N =13.912.84N =4.90 =5varillasUSAR:53/4=14.2cm24153/44.- DISEO DE LA ZAPATA EXTERIOR0.5Lv=2.51-t12Lv=1.002718385mWNU =158.02 =63.07tn/m2.512Mumax=63.07*1.002718385T=2.512Mumax=31.71tn-mw =0.08 =0.95 =31.71*100.9*135*210*0.08(1 -0.59*0.08) =40.38cmAsumimos 3/4hz=d +13/4 +rec.b=1352hz =49cmentonces :d =hz -13/4-rec.2d =40cm

a =155As =31.71*10a1 =4200*25.510.9*4200*(40.38 - 15)0.85*210*135.02a1 =4.45As =25.51

a1 =4200*16.8950.85*210*135.0As1 =51.94*10a1 =2.940.9*4200*(83.57 - 4.45)2As1 =16.89

5As2 =51.94*100.9*4200*(83.57 - 2.94)2As2 =16.74cm2

Asmin=0.0018*d*bAsmin=20.31cm2NUMERO DE VARILLASASarea 3/4N =20.312.84N =7.15 =8varillasUSAR:83/4=22.72cm2SEPARACION135 -rec.-diametro 3/48 - 1S =17.30 =17cmREFUERZO TRANVERSAL :Astem=0.0018*250.54*50Astem=22.55cm2NUMERO DE VARILLASASarea 3/4N =22.552.84N =7.94 =8varillasUSAR:83/4=22.72cm2SEPARACION251 -rec.-diametro 3/48 - 1S =33.80 =34cm5.- DISEO DE LA ZAPATA INTERIORP2e=-160.9580111897tnP2eu=-238.2671132146tnP2u=244.5tn

Azap = 160.96Wnu =238.27 =46.78tn/m31.65.09Azap=5.09m2a2=*35a=62.03cmentonces dimension en planta:

Azap=T*S1T=S2Lv=2.26-a2reeplazo 2 en 1Lv=0.82m5.09=S*sS=2.26mT=2.26m2Mumax =46.78*0.822.00Mumax=35.3444551062tn-mhmin=0.49mentonces :d =49 -3/4-rec.

d =39cm6.- VERIFICACION POR PUNZONAMIENTOm=Lv+a+d/2=1.64mn=hmin+d=0.8825108152mVu=170.74tnbo=5.0364790282mVc =304.9756179997tnDEBE CUMPLIRSE QUE :

170.74304.97561799970.75227.66304.9756179997si cumple7.- VERIFICACION POR CORTEVu=44.7702794482tnVc =68.3314838565tnDEBE CUMPLIRSE QUE :

44.7768.33148385650.7559.6968.3314838565si cumple7.- DISEO POR FLEXIONMu=35.3444551062

a =155As =35.34*10a1 =4200*29.290.9*4200*(39.42 - 15)0.85*210*225.72a1 =3.05As =29.29

a1 =4200*24.6850.85*210*225.7As1 =35.34*10a1 =2.570.9*4200*(39.42 - 3.05)2As1 =24.68

5As2 =35.34*100.9*4200*(39.42 - 2.57)2As2 =24.52cm2

Asmin=0.0018*d*bAsmin=16.01cm2NUMERO DE VARILLASASarea 3/4N =24.522.84N =8.63 =9varillasUSAR:93/4=25.56cm2SEPARACION226 -rec.-diametro 3/49 - 1S =26.47 =27cm

FGRNGL1

ZAPATA TRAPEZOIDALZAPATA TRAPEZOIDAL

PROBLEMA 1Disear la zapata mostrada en planta en la figura 1. La columna exterior esta sujeta a PD= 120tn.y PL=80tn y la columna inferior esta sujeta a PD= 90tn.y PL=65tn . El peso especfico del suelo es 2000 kg/m3 y una capacidad portante de 2kg/cm2 . Adems est sometido a la accin de una sobrecarga de 400 kg/m2. Para la zapata emplear f 'c =175 kg/cm2 y fy =4200 kg/cm2

DATOS:

ZapataOtrosSuelof ' c =175kg/cmS/C =0.4tn/mDf =1.2mPD1 =120tng=2tn/mPL1 =80tnP1ColumnaPD2 =90tnt =20tn/mC1=0.5x0.5PL2 =65tnP2hf=1.5mC2=0.65x0.65Acerof y =4200kg/cm

L1=51.- CALCULO DEL ESFUERZO NETO DEL TERRENOn =20-1.5x2-0.4n =16.6tn/m22.- DIMENSIONAMIENTOP = P1 + P2P =200 + 155P=Rx = 355Azap = 355 =21.39tn/m216.6355*XR=200*0.5+155(0.5+5+0.65)22XR=2.68mSe cumple que XR esta entre L/3 y L/2

Lzap=6.15L=2.053conformeL=3.082entonces(a+b)*6.15=21.392a+b =6.95m1como:2.68=6.15*(2a +b)36.952a+b=9.11m2de 1 y 2entoncesa+b =6.95(-)2a+b=9.11a=2.15b=4.80USAR:a=2.15mb=4.80m