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ESCUELA PROFESIONAL DE INGENIERIA CIVIL
EJEMPLO DE DISE&Ntilde;O DE ZAPATA AISLADA
1. DATOS
•
•
COLUMNA
✓ Long Columna
✓ Carga Axial Carga Muerta
✓ Mto. de Carga Muerta xx
✓ Mto. de Carga Muerta yy
✓ Carga Axial Carga Viva
✓ Mto. de Carga Viva xx
✓ Mto. de Carga Viva yy
✓ Carga Axial Sismo xx
✓ Mto. de sismo xx
✓ Carga Axial Sismo yy
✓ Mto. de sismo yy
: 35 cm x 35 cm
: 69.43 ton
: -0.62 ton*m
: -0.22 ton*m
: 14.76 ton
: -0.25 ton*m
: 0.25 ton*m
: 9.66 ton
: 0.13 ton*m
: 1.56 ton
: 0.12 ton*m
CONSIDERACIONES DE DISE&Ntilde;O
✓ Resistencia a la compresi&oacute;n del Concreto
✓ Resistencia de fluencia del Acero
✓ Altura de la zapata
✓ Peralte de la zapata
: 2.35 kg/cm2
: 210 kg/cm2
: 4200 kg/cm2
: 0.80 m
: 0.70 m
2. PREDIMENSIONAMIENTO
1.1 ∗ ∑ 𝐶𝑆 1.1 ∗ (69.43 + 14.76 + 9.66)𝑡𝑜𝑛
𝐴𝑍𝐴𝑃 =
=
= 4.39 𝑚2
(2.35 𝑘𝑔/𝑐𝑚2) ∗ 10𝑡𝑜𝑛/𝑚2
𝑇𝑡
𝐴𝑍𝐴𝑃 = 4.39 𝑚2 = 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
𝐴𝑍𝐴𝑃 = √4.39 𝑚2 = 2.09 𝑚 ≅ 2.10 𝑚
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 = 2.10 + 0.50 = 2.60 𝑚 ≈ 2.70 𝑚
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 = 2.10 − 0.50 = 1.60 𝑚 ≈ 1.70 𝑚
𝐴𝑍𝐴𝑃 = 4.39 𝑚2 = 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 = 2.70𝑚 ∗ 1.70𝑚 = 4.59 𝑚2 ≈ 4.39 𝑚2
3. CHEQUEO POR CARGA SISMICA
3.1. CASOS:
o
𝑃𝑍𝐴𝑃 = 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 ∗ ℎ𝑍𝐴𝑃 ∗ 2.4 𝑡𝑜𝑛/𝑚3
𝑃𝑍𝐴𝑃 = 2.70 𝑚 ∗ 1.70 𝑚 ∗ 0.80 𝑚 ∗ 2.4
𝑡𝑜𝑛
= 8.81 𝑡𝑜𝑛
𝑚3
𝑃𝐺 = 𝑃𝐶𝑀 + 𝑃𝐶𝑉 + 𝑃𝑍𝐴𝑃 = 69.43 + 14.76 + 8.81 = 93 𝑡𝑜𝑛
𝑇1 =
𝑃𝐺
93 𝑡𝑜𝑛
𝑡𝑜𝑛
=
= 20.26 2
2
𝐴𝑍𝐴𝑃 4.59 𝑚
𝑚
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
o
Caso Nro. 02: Sismo XX (+)
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+) = 𝑃𝐶𝑀 + 𝑃𝐶𝑉 + 𝑃𝑍𝐴𝑃 + 𝑃𝑋𝑋 = 69.43 + 14.76 + 8.81 + 9.66 = 102.66 𝑡𝑜𝑛
𝑀𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+) = 𝑀𝐶𝑀 + 𝑀𝐶𝑉 +
𝑒𝑋𝑋(+) =
𝑀𝑋𝑋
0.13
= −0.62 − 0.25 +
= −0.77 𝑡𝑜𝑛 ∗ 𝑚
1.25
1.25
𝑀𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+) −0.77 𝑡𝑜𝑛 ∗ 𝑚
=
= −0.0075 𝑚
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+)
102.66 𝑡𝑜𝑛
𝑇1−2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+)
6 ∗ 𝑒𝑋𝑋(+)
∗ (1 &plusmn;
)
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
𝑇1 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+)
6 ∗ 𝑒𝑋𝑋(+)
102.66 𝑡𝑜𝑛
6 ∗ (−0.0075 𝑚)
𝑡𝑜𝑛
∗ (1 +
)=
∗
+
=
21.99
(1
)
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
4.59 𝑚2
2.70 𝑚
𝑚2
𝑇2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (+)
6 ∗ 𝑒𝑋𝑋(+)
102.66 𝑡𝑜𝑛
6 ∗ (−0.0075 𝑚)
𝑡𝑜𝑛
∗ (1 −
)=
∗ (1 −
) = 22.74 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
4.59 𝑚
2.70 𝑚
𝑚
o
Caso Nro. 03: Sismo XX (-)
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−) = 𝑃𝐶𝑀 + 𝑃𝐶𝑉 + 𝑃𝑍𝐴𝑃 − 𝑃𝑋𝑋 = 69.43 + 14.76 + 8.81 − 9.66 = 83.34 𝑡𝑜𝑛
𝑀𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−) = 𝑀𝐶𝑀 + 𝑀𝐶𝑉 +
𝑒𝑋𝑋(−) =
𝑀𝑋𝑋
0.13
= −0.62 − 0.25 −
= − 0.97𝑡𝑜𝑛 ∗ 𝑚
1.25
1.25
𝑀𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−) − 0.97𝑡𝑜𝑛 ∗ 𝑚
=
= − 0.01𝑚
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−)
83.34 𝑡𝑜𝑛
𝑇1−2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−)
6 ∗ 𝑒𝑋𝑋(−)
∗ (1 &plusmn;
)
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
𝑇1 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−)
6 ∗ 𝑒𝑋𝑋(−)
83.34 𝑡𝑜𝑛
6 ∗ (− 0.01𝑚)
𝑡𝑜𝑛
∗ (1 +
)=
∗ (1 +
) = 17.69 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
4.59 𝑚
2.70 𝑚
𝑚
𝑇2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑋−𝑋 (−)
6 ∗ 𝑒𝑋𝑋(−)
83.34 𝑡𝑜𝑛
6 ∗ (− 0.01𝑚)
𝑡𝑜𝑛
∗ (1 −
)=
∗ (1 −
) = 18.63 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
4.59 𝑚
2.70 𝑚
𝑚
o
Caso Nro. 04: Sismo YY (+)
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+) = 𝑃𝐶𝑀 + 𝑃𝐶𝑉 + 𝑃𝑍𝐴𝑃 + 𝑃𝑌𝑌 = 69.43 + 14.76 + 8.81 + 1.56 = 94.56 𝑡𝑜𝑛
𝑀𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+) = 𝑀𝐶𝑀 + 𝑀𝐶𝑉 +
𝑒𝑌𝑌(+) =
𝑀𝑌𝑌
0.12
= −0.62 − 0.25 +
= −0.77 𝑡𝑜𝑛 ∗ 𝑚
1.25
1.25
𝑀𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+) −0.77 𝑡𝑜𝑛 ∗ 𝑚
=
= −0.0081 𝑚
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+)
94.56 𝑡𝑜𝑛
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
𝑇1−2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+)
6 ∗ 𝑒𝑌𝑌(+)
∗ (1 &plusmn;
)
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
𝑇1 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+)
6 ∗ 𝑒𝑌𝑌(+)
94.56 𝑡𝑜𝑛
6 ∗ (−0.0081 𝑚)
𝑡𝑜𝑛
∗ (1 +
)=
∗
+
(1
) = 20.01 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
4.59 𝑚
1.70 𝑚
𝑚
𝑇2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (+)
6 ∗ 𝑒𝑌𝑌(+)
94.56 𝑡𝑜𝑛
6 ∗ (−0.0081 𝑚)
𝑡𝑜𝑛
∗ (1 −
)=
∗ (1 −
) = 21.19 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
4.59 𝑚
1.70 𝑚
𝑚
o
Caso Nro. 05: Sismo YY (-)
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−) = 𝑃𝐶𝑀 + 𝑃𝐶𝑉 + 𝑃𝑍𝐴𝑃 − 𝑃𝑌𝑌 = 69.43 + 14.76 + 8.81 − 1.56 = 91.44 𝑡𝑜𝑛
𝑀𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−) = 𝑀𝐶𝑀 + 𝑀𝐶𝑉 −
𝑒𝑌𝑌(−) =
𝑀𝑌𝑌
0.12
= −0.62 − 0.25 −
= −0.97 𝑡𝑜𝑛 ∗ 𝑚
1.25
1.25
𝑀𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−) −0.97 𝑡𝑜𝑛 ∗ 𝑚
=
= −0.01 𝑚
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−)
91.44 𝑡𝑜𝑛
𝑇1−2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−)
6 ∗ 𝑒𝑌𝑌(−)
∗ (1 &plusmn;
)
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
𝑇1 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−)
6 ∗ 𝑒𝑌𝑌(−)
91.44 𝑡𝑜𝑛
6 ∗ (−0.01𝑚)
𝑡𝑜𝑛
∗ (1 +
)=
∗ (1 +
) = 19.22 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
4.59 𝑚
1.70 𝑚
𝑚
𝑇2 =
𝑃𝑆𝐼𝑆𝑀𝑂 𝑌−𝑌 (−)
6 ∗ 𝑒𝑌𝑌(−)
91.44 𝑡𝑜𝑛
6 ∗ (−0.01 𝑚)
𝑡𝑜𝑛
∗ (1 −
)=
∗
−
(1
) = 20.62 2
2
𝐴𝑍𝐴𝑃
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
4.59 𝑚
1.70 𝑚
𝑚
3.2. COMPARANDO
o
𝑇1 = 20.26
o
Caso Nro. 02: Sismo XX (+)
𝑇1 = 21.99
o
Caso Nro. 03: Sismo XX (-)
𝑇1 = 17.69
o
Caso Nro. 04: Sismo YY (+)
𝑇1 = 20.01
o
Caso Nro. 05: Sismo YY (-)
𝑡𝑜𝑛
𝑚2
𝑇2 = 22.74
𝑡𝑜𝑛
𝑚2
𝑡𝑜𝑛
𝑚2
𝑇2 = 18.63
𝑡𝑜𝑛
𝑚2
𝑡𝑜𝑛
𝑚2
𝑇2 = 21.19
𝑡𝑜𝑛
𝑚2
𝑡𝑜𝑛
𝑚2
𝑇2 = 20.62
𝑡𝑜𝑛
𝑚2
𝑇1 = 19.22
o
𝑡𝑜𝑛
𝑚2
M&iacute;nimo T debe de ser mayor o igual que cero, 17.69 ton/m2 ≥ 0 ton/m2, si se cumple, y se
cumple que el T m&aacute;ximo debe satisfacer:
1.3 ∗ 23.50
𝑡𝑜𝑛
𝑡𝑜𝑛
𝑡𝑜𝑛
= 30.55 2 ≥ 22.74 2
𝑚2
𝑚
𝑚
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
4. CHEQUEO POR PUNZONAMIENTO
4.1. DIAGRAMA
4.2. COMBINACIONES DE CARGA
o
𝑃1 = 1.4 ∗ 𝐶𝑀 + 1.7 ∗ 𝐶𝑉 = 1.4 ∗ 69.43 + 1.7 ∗ 14.76 = 122.29 𝑡𝑜𝑛
o
Caso Nro. 02: Sismo XX (+)
𝑃2 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑋𝑋 = 1.25 ∗ (69.43 + 14.76) + 9.66 = 114.90 𝑡𝑜𝑛
o
Caso Nro. 03: Sismo XX (-)
𝑃3 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) − 𝐶𝑆𝑋𝑋 = 1.25 ∗ (69.43 + 14.76) − 9.66 = 95.58 𝑡𝑜𝑛
o
Caso Nro. 04: Sismo YY (+)
𝑃4 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑌𝑌 = 1.25 ∗ (69.43 + 14.76) + 1.56 = 106.80 𝑡𝑜𝑛
o
Caso Nro. 05: Sismo YY (-)
𝑃5 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑌𝑌 = 1.25 ∗ (69.43 + 14.76) − 1.56 = 103.68 𝑡𝑜𝑛
o
Hallando el Pu max
𝑃𝑚𝑎𝑥 = 122.29 𝑡𝑜𝑛
4.3. CALCULANDO EL PERIMETRO Y &Aacute;REA DE FALLA DE PUNZONAMIENTO
o
Per&iacute;metro de Punzonamiento
𝑏𝑂 = 2 ∗ [(𝑑𝑍𝐴𝑃 + 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑋𝑋 ) + (𝑑𝑍𝐴𝑃 + 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑌𝑌 )]
𝑏𝑂 = 2 ∗ [(0.70𝑚 + 0.35 𝑚) + (0.70𝑚 + 0.35 𝑚)] = 4.2 𝑚
o
&Aacute;rea de Punzonamiento
𝐴𝑂 = (𝑑𝑍𝐴𝑃 + 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑋𝑋 ) ∗ (𝑑𝑍𝐴𝑃 + 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑌𝑌 )
𝐴𝑂 = (0.70𝑚 + 0.35 𝑚) ∗ (0.70𝑚 + 0.35 𝑚) = 1.10 𝑚2
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
4.4. FUERZA DE CORTE POR PUNZONAMIENTO ACTUANTE
𝑉𝑈 =
𝑃𝑢𝑚𝑎𝑥
122.29 𝑡𝑜𝑛
∗ (𝐴𝑍𝐴𝑃 − 𝐴𝑂 ) =
∗ (4.59 𝑚2 − 1.10 𝑚2 ) = 92.99 𝑡𝑜𝑛
𝐴𝑍𝐴𝑃
4.59 𝑚2
4.5. ESFUERZO POR PUNZONAMIENTO ACTUANTE
𝑉𝑉 =
𝑉𝑈
92.99 𝑡𝑜𝑛
𝑡𝑜𝑛
=
= 45.18 2
∅ ∗ 𝑏𝑂 ∗ 𝑑𝑍𝐴𝑃 0.70 ∗ 4.2 𝑚 ∗ 0.70 𝑚
𝑚
4.6. RESISTENCIA POR APLASTAMIENTO O PUNZONAMIENTO EN LA ZAPATA
𝐸𝑠𝑓 𝑝𝑢𝑛𝑧𝑍𝐴𝑃 = ∅ ∗ 0.85 ∗ 𝑓 ′ 𝑐(𝑀𝑃𝑎) ∗ 𝐴1(𝑚𝑚2 )
𝐸𝑠𝑓 𝑝𝑢𝑛𝑧𝑍𝐴𝑃 = 0.70 ∗ 0.85 ∗ (210
𝑘𝑔
∗ 0.0981) 𝑀𝑃𝑎 ∗ 350 𝑚𝑚 ∗ 350𝑚𝑚 = 12.26 𝑀𝑃𝑎
𝑐𝑚2
𝐸𝑠𝑓 𝑝𝑢𝑛𝑧𝑍𝐴𝑃 = 12.26 𝑀𝑃𝑎 = 124.97
𝑡𝑜𝑛
𝑚2
4.7. COMPARANDO
𝑉𝑉 ≤ 𝐸𝑠𝑓 𝑝𝑢𝑛𝑧𝑍𝐴𝑃
45.18
𝑡𝑜𝑛
𝑡𝑜𝑛
≤ 124.97 2
2
𝑚
𝑚
&quot;𝐶𝑢𝑚𝑝𝑙𝑒 𝑝𝑜𝑟 𝑃𝑢𝑛𝑧𝑜𝑛𝑎𝑚𝑖𝑒𝑛𝑡𝑜&quot;
5. CHEQUEO POR CORTE
5.1. CORTANTE EN XX
o
𝑃1 = 1.4 ∗ 𝐶𝑀 + 1.7 ∗ 𝐶𝑉 = 1.4 ∗ 69.43 + 1.7 ∗ 14.76 = 122.29 𝑡𝑜𝑛
o
Caso Nro. 02: Sismo XX (+)
𝑃2 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑋𝑋 = 1.25 ∗ (69.43 + 14.76) + 9.66 = 114.90 𝑡𝑜𝑛
o
Caso Nro. 03: Sismo XX (-)
𝑃3 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) − 𝐶𝑆𝑋𝑋 = 1.25 ∗ (69.43 + 14.76) − 9.66 = 95.58 𝑡𝑜𝑛
o
Hallando el Pu max xx
𝑃𝑚𝑎𝑥 = 122.29 𝑡𝑜𝑛
o
C&aacute;lculo de Cortante en XX
𝑉𝑋𝑋 =
𝑉𝑋𝑋 =
𝑃𝑢𝑚𝑎𝑥
∗ [𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 − (𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑋𝑋 + 𝑑𝑧𝑎𝑝𝑋𝑋 )] ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
𝐴𝑍𝐴𝑃
122.29 𝑡𝑜𝑛
∗ [2.70 𝑚 − (0.35 𝑚 + 0.70 𝑚)] ∗ 1.70 𝑚 = 74.73 𝑡𝑜𝑛
4.59 𝑚2
𝑉𝑢𝑋𝑋 =
74.73 𝑡𝑜𝑛
= 87.92 𝑡𝑜𝑛
0.85
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
o
Resistencia nominal del concreto
𝑉𝑐𝑋𝑋 (𝑁𝑒𝑤) = 0.17 ∗ √𝑓 ′ 𝑐(𝑀𝑃𝑎) ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 (𝑚𝑚) ∗ 𝑑𝑍𝐴𝑃 (𝑚𝑚)
𝑉𝑐𝑋𝑋 (𝑁𝑒𝑤) = 0.17 ∗ √210 ∗ 0.0981(𝑀𝑃𝑎) ∗ 2,700(𝑚𝑚) ∗ 700(𝑚𝑚) = 1′ , 458,326.87𝑁𝑒𝑤
𝑉𝑐𝑋𝑋 (𝑁𝑒𝑤) = 1′ , 458,326.87 𝑁𝑒𝑤 = 178.71 𝑡𝑜𝑛
o
Comparando
𝑉𝑢𝑋𝑋 ≤ 𝑉𝑐𝑋𝑋
87.92 𝑡𝑜𝑛 ≤ 178.71 𝑡𝑜𝑛 … &quot;𝐶𝑢𝑚𝑝𝑙𝑒 𝑝𝑜𝑟 𝐶𝑜𝑟𝑡𝑎𝑛𝑡𝑒&quot;
5.2. CORTANTE EN YY
o
𝑃1 = 1.4 ∗ 𝐶𝑀 + 1.7 ∗ 𝐶𝑉 = 1.4 ∗ 69.43 + 1.7 ∗ 14.76 = 122.29 𝑡𝑜𝑛
o
Caso Nro. 04: Sismo YY (+)
𝑃4 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑌𝑌 = 1.25 ∗ (69.43 + 14.76) + 1.56 = 106.80 𝑡𝑜𝑛
o
Caso Nro. 05: Sismo YY (-)
𝑃5 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑌𝑌 = 1.25 ∗ (69.43 + 14.76) − 1.56 = 103.68 𝑡𝑜𝑛
o
Hallando el Pu max
𝑃𝑚𝑎𝑥 = 122.29 𝑡𝑜𝑛
o
C&aacute;lculo de Cortante en YY
𝑉𝑌𝑌 =
𝑉𝑌𝑌 =
𝑃𝑢𝑚𝑎𝑥
∗ [𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 − (𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑌𝑌 + 𝑑𝑧𝑎𝑝𝑌𝑌 )] ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
𝐴𝑍𝐴𝑃
122.29 𝑡𝑜𝑛
∗ [1.70 𝑚 − (0.35 𝑚 + 0.70 𝑚)] ∗ 2.70 𝑚 = 46.76 𝑡𝑜𝑛
4.59 𝑚2
𝑉𝑢𝑌𝑌 =
o
46.76 𝑡𝑜𝑛
= 55 𝑡𝑜𝑛
0.85
Resistencia nominal del concreto
𝑉𝑐𝑌𝑌 (𝑁𝑒𝑤) = 0.17 ∗ √𝑓 ′ 𝑐(𝑀𝑃𝑎) ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 (𝑚𝑚) ∗ 𝑑𝑍𝐴𝑃 (𝑚𝑚)
𝑉𝑐𝑋𝑋 (𝑁𝑒𝑤) = 0.17 ∗ √210 ∗ 0.0981(𝑀𝑃𝑎) ∗ 1,700(𝑚𝑚) ∗ 700(𝑚𝑚) = 918,205.80𝑁𝑒𝑤
𝑉𝑐𝑌𝑌 (𝑁𝑒𝑤) = 1′ , 458,326.87 𝑁𝑒𝑤 = 93.60 𝑡𝑜𝑛
o
Comparando
𝑉𝑢𝑌𝑌 ≤ 𝑉𝑐𝑌𝑌
55 𝑡𝑜𝑛 ≤ 93.60 𝑡𝑜𝑛 … &quot;𝐶𝑢𝑚𝑝𝑙𝑒 𝑝𝑜𝑟 𝐶𝑜𝑟𝑡𝑎𝑛𝑡𝑒&quot;
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
6. DISE&Ntilde;O POR FLEXI&Oacute;N
6.1. DISE&Ntilde;O POR FLEXI&Oacute;N EN XX
6.1.1. C&Aacute;LCULO DEL ACERO DE DISE&Ntilde;O
➢ CALCULO DEL MOMENTO DE DISE&Ntilde;O
▪
𝑃1 = 1.4 ∗ 𝐶𝑀 + 1.7 ∗ 𝐶𝑉 = 1.4 ∗ 69.43 + 1.7 ∗ 14.76 = 122.29 𝑡𝑜𝑛
▪ Caso Nro. 02: Sismo XX (+)
𝑃2 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑋𝑋 = 1.25 ∗ (69.43 + 14.76) + 9.66 = 114.90 𝑡𝑜𝑛
▪
▪
Caso Nro. 03: Sismo XX (-)
𝑃3 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) − 𝐶𝑆𝑋𝑋 = 1.25 ∗ (69.43 + 14.76) − 9.66 = 95.58 𝑡𝑜𝑛
Hallando el Pu max xx
𝑃𝑚𝑎𝑥 = 122.29 𝑡𝑜𝑛
▪
El momento de dise&ntilde;o ser&aacute;:
𝑀𝑡𝑜𝑋𝑋 = {
𝑃𝑢𝑚𝑎𝑥
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 − 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑋𝑋
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 − 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑋𝑋
∗ [(
) ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 ]} ∗
𝐴𝑍𝐴𝑃
2
4
𝑀𝑡𝑜𝑋𝑋 = {
122.29 𝑡𝑜𝑛
2.70 𝑚 − 0.35 𝑚
2.70 𝑚 − 0.35 𝑚
∗
)
∗
1.70
𝑚]}
∗
= 31.27 𝑡𝑜𝑛 ∗ 𝑚
[(
4.59 𝑚2
2
4
➢ M&eacute;todo Iterativo de C&aacute;lculo del Acero
𝑎=
𝐴𝑠 =
𝑑𝑍𝐴𝑃 70 𝑐𝑚
=
= 14 𝑐𝑚
5
5
31.27 𝑡𝑜𝑛 ∗ 𝑚 ∗ 1000 ∗ 100 (𝑘𝑔 ∗ 𝑐𝑚)
=
= 13.13 𝑐𝑚2
𝑎
𝑘𝑔
14 𝑐𝑚
0.9 ∗ 𝑓 ′ 𝑦 ∗ (𝑑𝑍𝐴𝑃 − 2)
0.9 ∗ 4,200 2 ∗ (70 𝑐𝑚 − 2 )
𝑐𝑚
𝑀𝑡𝑜 𝑑𝑖𝑠𝑒&ntilde;𝑜
𝑘𝑔
13.13 𝑐𝑚2 ∗ 4,200 2
𝐴𝑠 ∗ 𝑓′𝑦
𝑐𝑚
𝑎=
=
= 1.14 𝑐𝑚
0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 0.85 ∗ 210 𝑘𝑔 ∗ 270𝑐𝑚
𝑐𝑚2
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
𝐴𝑠 =
31.27 𝑡𝑜𝑛 ∗ 𝑚 ∗ 1000 ∗ 100 (𝑘𝑔 ∗ 𝑐𝑚)
=
= 11.91 𝑐𝑚2
𝑎
0.9 ∗ 𝑓 ′ 𝑦 ∗ (𝑑𝑍𝐴𝑃 − 2) 0.9 ∗ 4,200 𝑘𝑔2 ∗ (70 𝑐𝑚 − 1.14 𝑐𝑚)
2
𝑐𝑚
𝑀𝑡𝑜 𝑑𝑖𝑠𝑒&ntilde;𝑜
𝑘𝑔
11.91 𝑐𝑚2 ∗ 4,200 2
𝐴𝑠 ∗ 𝑓′𝑦
𝑐𝑚 = 1.04 𝑐𝑚
𝑎=
=
0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 0.85 ∗ 210 𝑘𝑔 ∗ 270𝑐𝑚
𝑐𝑚2
𝐴𝑠 =
31.27 𝑡𝑜𝑛 ∗ 𝑚 ∗ 1000 ∗ 100 (𝑘𝑔 ∗ 𝑐𝑚)
=
= 11.91 𝑐𝑚2
𝑎
𝑘𝑔
1.04
𝑐𝑚
′
0.9 ∗ 𝑓 𝑦 ∗ (𝑑𝑍𝐴𝑃 − ) 0.9 ∗ 4,200 2 ∗ (70 𝑐𝑚 −
2
2 )
𝑐𝑚
𝑀𝑡𝑜 𝑑𝑖𝑠𝑒&ntilde;𝑜
➢ ACERO M&Iacute;NIMO
𝐴𝑠𝑚𝑖𝑛 = 𝜌𝑚𝑖𝑛 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 ∗ 𝑑𝑍𝐴𝑃
𝐴𝑠𝑚𝑖𝑛 = 0.0012 ∗ 270 𝑐𝑚 ∗ 70 𝑐𝑚 = 22.68 𝑐𝑚2
➢ DETERMINACION DEL ACERO A COLOCAR
𝐴𝑠 𝑧𝑎𝑝𝑐𝑜𝑙 ≥ 𝐴𝑠 𝑧𝑎𝑝𝑚𝑖𝑛
11.91 𝑐𝑚2 ≥ 22.68 𝑐𝑚2 No
→ 𝐴𝑠 𝑧𝑎𝑝𝑋𝑋 = 22.68 𝑐𝑚2
Usando As &Oslash;:
- 5/8”, su separaci&oacute;n ser&aacute;:
𝑆𝐴𝑠 =
𝑆𝐴𝑠 ∅5/8&quot; =
∅𝐴𝑠𝑈𝑆𝐴𝑅 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋
𝐴𝑠𝑁𝐸𝐶𝐸𝑆𝐴𝑅𝐼𝑂
1.98 𝑐𝑚2 ∗ 270 𝑐𝑚
= 23.57𝑐𝑚 ≈ ∅5/8&quot;@22.50𝑐𝑚
22.68 𝑐𝑚2
6.2. DISE&Ntilde;O POR FLEXI&Oacute;N EN YY
➢ CALCULO DEL MOMENTO DE DISE&Ntilde;O
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
▪
𝑃1 = 1.4 ∗ 𝐶𝑀 + 1.7 ∗ 𝐶𝑉 = 1.4 ∗ 69.43 + 1.7 ∗ 14.76 = 122.29 𝑡𝑜𝑛
▪ Caso Nro. 04: Sismo YY (+)
𝑃4 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑌𝑌 = 1.25 ∗ (69.43 + 14.76) + 1.56 = 106.80 𝑡𝑜𝑛
▪ Caso Nro. 05: Sismo YY (-)
𝑃5 = 1.25 ∗ (𝐶𝑀 + 𝐶𝑉) + 𝐶𝑆𝑌𝑌 = 1.25 ∗ (69.43 + 14.76) − 1.56 = 103.68 𝑡𝑜𝑛
▪
Hallando el Pu max
𝑃𝑚𝑎𝑥 = 122.29 𝑡𝑜𝑛
▪
𝑀𝑡𝑜𝑌𝑌 = {
𝑀𝑡𝑜𝑌𝑌 = {
El momento de dise&ntilde;o ser&aacute;:
𝑃𝑢𝑚𝑎𝑥
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 − 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑌𝑌
𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌 − 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑌𝑌
∗ [(
) ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 ]} ∗
𝐴𝑍𝐴𝑃
2
4
122.29 𝑡𝑜𝑛
1.70 𝑚 − 0.35 𝑚
1.70 𝑚 − 0.35 𝑚
∗ [(
) ∗ 2.70 𝑚]} ∗
= 16.39 𝑡𝑜𝑛 ∗ 𝑚
2
4.59 𝑚
2
4
➢ M&eacute;todo Iterativo de C&aacute;lculo del Acero
𝑑𝑍𝐴𝑃 70 𝑐𝑚
𝑎=
=
= 14 𝑐𝑚
5
5
𝐴𝑠 =
16.39 𝑡𝑜𝑛 ∗ 𝑚 ∗ 1000 ∗ 100 (𝑘𝑔 ∗ 𝑐𝑚)
=
= 6.88 𝑐𝑚2
𝑎
𝑘𝑔
14
𝑐𝑚
′
0.9 ∗ 𝑓 𝑦 ∗ (𝑑𝑍𝐴𝑃 − 2)
0.9 ∗ 4,200 2 ∗ (70 𝑐𝑚 − 2 )
𝑐𝑚
𝑀𝑡𝑜 𝑑𝑖𝑠𝑒&ntilde;𝑜
𝑘𝑔
6.88 𝑐𝑚2 ∗ 4,200 2
𝐴𝑠 ∗ 𝑓′𝑦
𝑐𝑚
𝑎=
=
= 0.95 𝑐𝑚
0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 0.85 ∗ 210 𝑘𝑔 ∗ 170𝑐𝑚
𝑐𝑚2
𝑀𝑡𝑜 𝑑𝑖𝑠𝑒&ntilde;𝑜
31.27 𝑡𝑜𝑛 ∗ 𝑚 ∗ 1000 ∗ 100 (𝑘𝑔 ∗ 𝑐𝑚)
𝐴𝑠 =
= 6.24 𝑐𝑚2
𝑎 =
𝑘𝑔
0.95
𝑐𝑚
′
0.9 ∗ 𝑓 𝑦 ∗ (𝑑𝑍𝐴𝑃 − 2) 0.9 ∗ 4,200 2 ∗ (70 𝑐𝑚 −
)
2
𝑐𝑚
𝑘𝑔
6.24 𝑐𝑚2 ∗ 4,200 2
𝐴𝑠 ∗ 𝑓′𝑦
𝑐𝑚
𝑎=
=
= 0.86 𝑐𝑚
0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 0.85 ∗ 210 𝑘𝑔 ∗ 170𝑐𝑚
𝑐𝑚2
𝐴𝑠 =
31.27 𝑡𝑜𝑛 ∗ 𝑚 ∗ 1000 ∗ 100 (𝑘𝑔 ∗ 𝑐𝑚)
=
= 6.23 𝑐𝑚2
𝑎
𝑘𝑔
1.04
𝑐𝑚
′
0.9 ∗ 𝑓 𝑦 ∗ (𝑑𝑍𝐴𝑃 − 2) 0.9 ∗ 4,200 2 ∗ (70 𝑐𝑚 −
2 )
𝑐𝑚
𝑀𝑡𝑜 𝑑𝑖𝑠𝑒&ntilde;𝑜
➢ ACERO M&Iacute;NIMO
𝐴𝑠𝑚𝑖𝑛 = 𝜌𝑚𝑖𝑛 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑋𝑋 ∗ 𝑑𝑍𝐴𝑃
𝐴𝑠𝑚𝑖𝑛 = 0.0012 ∗ 170 𝑐𝑚 ∗ 70 𝑐𝑚 = 14.28 𝑐𝑚2
➢ DETERMINACION DEL ACERO A COLOCAR
𝐴𝑠 𝑧𝑎𝑝𝑐𝑜𝑙 ≥ 𝐴𝑠 𝑧𝑎𝑝𝑚𝑖𝑛
6.23 𝑐𝑚2 ≥ 14.28 𝑐𝑚2 No
→ 𝐴𝑠 𝑧𝑎𝑝𝑋𝑋 = 14.28 𝑐𝑚2
E
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
Usando As &Oslash;:
- 5/8”, su separaci&oacute;n ser&aacute;:
𝑆𝐴𝑠 =
𝑆𝐴𝑠 ∅5/8&quot; =
7. DISE&Ntilde;O FINAL
∅𝐴𝑠𝑈𝑆𝐴𝑅 ∗ 𝑙𝑜𝑛𝑔 𝑧𝑎𝑝𝑌𝑌
𝐴𝑠𝑁𝐸𝐶𝐸𝑆𝐴𝑅𝐼𝑂
1.98 𝑐𝑚2 ∗ 170 𝑐𝑚
= 23.57𝑐𝑚 ≈ ∅5/8&quot;@22.50𝑐𝑚
14.28 𝑐𝑚2
```