formulas qult luis
DESCRIPTION
formulas para calculo de capacidad ultima de carga de un sueloTRANSCRIPT
formulas de qult
bpi Ø Ø exp 2.5
3.14159265 35 0.61086524 2.718281828
sqsc
vesic
Nq 33.296
Nc 46.124
48.029 c0
qult 1651.431216
qadm 550.4770719
qn
sƔ
NƔ
formulas de qult
L df Ɣ2.5 1.5 1.72
1.70020754 dq 1.13849016 iq 11.72188841 dc 1.13614428 ic 1
0.6 1 1
Ɣ q16.856 25.284
dƔ iƔ
= 25.284
Qult = 1.3CNc + qNq + 0.4ƔBNƔ
= 0 + 841.858 + 323.829
= q adm = 0 + 841.858 + 323.8293
qn = = Bmin = 2.5 m calcular con la calculadora
qult = 1651.4312
qadm = = 550.47707
Qn = = 548.8
q=Ɣ*df*g
KN/m²
qadm ˃ Qn
KN/m²
𝑞𝑢/(𝑓𝑠≥3)𝑄/𝐵²
𝑞𝑢𝑙𝑡/3𝑄/𝐵²
Tabla de vesic
Ø Ncmultiplicando por 9,8 m/s2 35 46.124
C 0 0 5.14Nc 46.124 5 6.49q 25.284 10 8.34
B Nq 33.296 15 10.98Ɣ 1.72 t/m3 20 14.83
NƔ 48.03 25 20.72 df 1.5 m 26 22.25
g 9.8 m/s2 28 25.8calcular con la calculadora Q 3430 KN 30 30.14
32 35.4934 42.1635 46.12
ok 40 75.31mal 44 118.37
ok
CUANDO Ɣ ESTA EN t/m3 se pasa a kN/m3
B ≥ qn = 𝑄/█(𝐴@)
Nq33.296 48.0291.00 01.57 0.452.47 1.223.94 2.656.40 5.39
10.66 10.8811.85 12.5414.72 16.7218.40 22.423.18 30.2129.44 41.0633.30 48.0364.19 109.41
115.31 224.63
NƔ
sq 1.700 dq 1.13849sc 1.722 dc 1.13614sƔ 0.600 dƔ 1
Qult = C*Nc *Sc*dc*ic + q*Nq*sq*dq*iq + Ɣ*B*NƔ*sƔ*dƔ*iƔ
Qult = 0 + 1629.560 + 607.179629045761
Qult = 2236.739
qadm = = 745.58 KN/m²qadm ˃ Qn
Qn = = 548.8 KN/m²
Kn/m²
1/2
C 0Nc 46.124
iq 1 q 25.284ic 1 Nq 33.296iƔ 1 Ɣ 1.720 t/m3
NƔ 48.029df 1.5 mg 9.8 m/s2Q 3430 KN
okmal
ok
B 1.2Q 70 * 9,8 = 686Df 2.5Ø 0Cu 10 δd =E1 12000 kpa = 12 mpa E1Ɣ 18D/B 2.083
FIGURE7.12 I2 0.7 Zh/B squareFIGURE7.12 I1 0.92 D/BCONSTANTE zh/B = 10
q = Q = 686 = 476.39 Kpa = kN/m21.2 ^2
δzd' = δV - ᶙ = ρs* g * D = Ɣs*Df = 45 Kpa = kN/m2
δd = 476.39 - 45 * 1.2 * 0.92 * 0.7 =12000
t/m² (q - δzd')B *I1* I2
KN/m³
BUSCAR LOS VALORES EN LA
TABLA
B²
B 1.2Q 70 * 9,8 = 686Df 2.5Ø 0Cu 10 t/m²E1 12000 kpa = 12 mpaƔ 18 KN/m³D/B 2.083
FIGURE7.12 I2 0.7 Zh/B squareFIGURE7.12 I1 0.92 D/BCONSTANTE zh/B = 10
q = Q = 686 = 476.39 Kpa = kN/m2 B² 1.2 ^2
δzd' = δV - ᶙ = ρs* g * D = Ɣs*Df = 45 Kpa = kN/m2
0.023 m = 23 mm δd2 = 476.39 - 45 * 1.212000
BUSCAR LOS VALORES EN LA
TABLA
asentamiento segunda zapata
δd2 = (q - δzd')B *I1* I2E1
Kpa = kN/m2
Kpa = kN/m2
* 0.92 * 0.7 = 0.023 m = 23 mm
1 2 3 4 5
cotas cimentacion ps=29,95 t/m2 b=8m
m z1 hi Go t/m2 z/b1
5 2.5 9.23 8.67 0.6257.5 5 11.73 11.03 1.2510 7.5 14.23 13.38 1.875
12.5 10 16.73 15.73 2.515 12.5 19.23 18.08 3.125
b =
ALTURA RELATIVA datocapa
arena por encima nf Hi1 = 3.50.94
peso propio
arena media por debajo NF Hi2 = 1.50.94
prof. Bajo cimiento
peso propio v3=0,94 t/m3
6 7 8 9 10 11 12 13
cimentacion ps=29,95 t/m2 b=8m total asentamientos unitarios asentamientos totales
I G1 G2 S2 S0 S1 Es1 s1interpolar TABLA 2 TABLA 2
0.228 27.31 35.99 9.8 4.75 5.05 8.8 11.000.167 20.01 31.03 9.15 5.4 3.75 6.35 7.940.115 13.78 27.15 8.55 5.95 2.6 4.5 5.630.08 9.58 25.31 8.35 6.45 1.9 3.2 4.00
0.057 6.83 24.90 8.3 7 1.3Ʃ
DATOSDF 2.5Ɣ 2.02
8 dato Q = 35 T/M2 DATOQN = 29.95
datoƔ
2.02 = 7.52127660.94
peso propio dato Ʃhi = 9.23
1.07 = 1.707446810.94
q - Ɣ.DF
14
asentamientos totales
Es
011.0018.9424.56
28.56