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Diseño de zapatas

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Richard Salinas
Richard Salinas

Ingenieria civil diseño de zapatas

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DISEÑODE ZAPATA P1 Las zapata se diseñaran de hormigónarmado y sujetoa lasaccionesque se indican. 𝐷 = 4158.18 𝑘𝑔 𝑚2 𝐿 = 940 𝑘𝑔 𝑚2 𝜎𝑠 = 21 𝑇/𝑚2 ℎ𝑓 = −1.20 𝑚  Columna: 30x30 cm Materiales 𝑓′𝑐 = 210 𝑘𝑔 𝑚2 𝑓𝑦 = 4200 𝑘𝑔 𝑚2  DISEÑO 1. P=D+L= (4.15+0.94) T =5.09 ton 𝐴 𝐹 = 𝑃 + %𝑃 𝑠 %𝑃(10 − 15)% 𝑃𝑜𝑟 𝑝𝑒𝑠𝑜 𝑝𝑟𝑜𝑝𝑖𝑜 𝑑𝑒𝑙 𝑟𝑒𝑙𝑙𝑒𝑛𝑜 𝑦 𝑝𝑒𝑠𝑜 𝑝𝑟𝑜𝑝𝑖𝑜 𝑑𝑒𝑙 𝑐𝑖𝑚𝑖𝑒𝑛𝑡𝑜 𝐴 𝐹 = 𝑃 ∗ %𝑃 𝑠 = 5.09 𝑡𝑜𝑛 ∗ 1.10 21 𝑇 𝑚2 = 0.27 𝑚2 = 0.52𝑥0.52 2. 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 0.60𝑥0.60 = 0.36 𝑚2 3. 𝑃. 𝑁. 𝑆 = 𝑃 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 5.09 𝑡𝑜𝑛 0,36 𝑚2 = 14,13 𝑇 𝑚2 𝑃. 𝑁. 𝑆 < 𝜎𝑠 14,13 𝑇 𝑚2 < 21 𝑇 𝑚2 𝑜𝑘 4. 𝑓𝑚𝑎𝑦𝑜𝑟𝑎𝑐𝑖𝑜𝑛 = 1.2 ∗ 4158.18 + 1.6 ∗ 940 4158.18 + 940 = 1.27 𝑃. 𝑁. 𝑆𝑢 = 𝑃. 𝑁. 𝑆 ∗ 𝑓𝑚𝑎𝑦 = 14.13 ∗ 1.27 = 17.96 𝑇 𝑚2
CORTE UNIDIRECCIONAL La seccióncriticapara corte unidireccional esladistanciadmedidaapartirde lacara de la columna. 𝑣𝑝 = 0.53 √ 𝑓′ 𝑐 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 𝑃𝑒𝑟𝑚𝑖𝑠𝑖𝑙𝑒 𝑑𝑒𝑙 ℎ𝑜𝑟𝑚𝑖𝑔ó𝑛 𝑣𝑝 = 0.53 √ 𝑓′ 𝑐 = 0.53 √210 = 7.68 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏 ∗ 𝑑 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 ú𝑙𝑡𝑖𝑚𝑜 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐴𝑐𝑟í𝑡𝑖𝑐𝑎 𝑉𝑢 = 17.96 𝑇 𝑚2 ∗ ( 0.15 − 𝑑) ∗ 0.60 𝑣𝑢 = 17.96( 0.15 − 𝑑) ∗ 0.60 0.85 ∗ 0.60 ∗ 𝑑 𝑣𝑢 = 𝑣𝑝 76.8 = 17.96( 0.15 − 𝑑) ∗ 0.60 0.85 ∗ 0.60 ∗ 𝑑 39.17𝑑 = 10.78(0.15 − 𝑑) 49.95𝑑 = 1.62 𝑑 = 0.032𝑚 ℎ = 𝑑 + 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 ; 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 = 7 𝑐𝑚
ℎ = 𝑑 + 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 = 0.032 + 0.07 = 0.102 𝑚  𝒉 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟐𝟓 𝒎 → 𝒉 𝒎𝒊𝒏𝒊𝒎𝒂 𝒅 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟏𝟖 𝒎 CORTE BIDIRECCIONAL La seccióncriticaesuna franja perimetral medidaaunadistanciad/2 desde lacara de la columna 𝑣𝑝 = 1.06√ 𝑓′ 𝑐 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 𝑝𝑒𝑟𝑚𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑒𝑙 ℎ𝑜𝑟𝑚𝑖𝑔ó𝑛 𝑣𝑝 = 1.06√ 𝑓′ 𝑐 = 1.06 √210 = 15.63 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏0 ∗ 𝑑 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 𝑢𝑙𝑡𝑖𝑚𝑜 𝑏0 = 0.48 ∗ 4 = 1.92 𝑚 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ (𝐴𝑓 − 𝐴𝑝𝑧) 𝑉𝑢 = 17.96 𝑇 𝑚2 ∗ (0.62 − 0.482) 𝑚2 = 2.33 𝑡𝑜𝑛 𝑣𝑢 = 2.33 𝑡𝑜𝑛 0.85 ∗ 1.92 𝑚 ∗ 0.18𝑚 𝑣𝑢 = 7.93 𝑇 𝑚2 𝑣𝑢 < 𝑣𝑝 0.793 𝑘𝑔 𝑐𝑚2 < 15.63 𝑘𝑔 𝑐𝑚2 ↓ 𝑂𝐾
DISEÑO DE LA ARMADURA Es la que se ubica enla cara de lacolumnay es la secciónque nospermite determinarel armado de la zapata. 𝑀𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐿𝑣2 2 ∗ 𝐵 𝑀𝑢 = 17.96 𝑇 𝑚2 ∗ 0.15𝑚2 2 ∗ 0.6𝑚 = 0.1212 𝑇 − 𝑚 𝜌 = 0.85 𝑓′𝑐 𝑓𝑦 [1 − √1 − 2𝑀𝑢 0.85𝑓′𝑐∅𝑏𝑑2] = 0.85 210 4200 [1 − √1 − 2 ∗ 0.1212∗ 100000 0.85 ∗ 210 ∗ 0.9 ∗ 60 ∗ 182] = 0.000095 𝜌𝑚𝑖𝑛 = 0.0018 𝜌 > 𝜌 min 𝑂𝐾 𝐴𝑠 = 𝜌 ∗ 𝑏 ∗ ℎ = 0.004613 ∗ 60 ∗ 10.5 = 2.91 𝑐𝑚2 → 𝟒∅𝟏𝟎𝒎𝒎 → 𝟏∅𝟏𝟎𝒎𝒎 @𝟏𝟓 𝒄𝒎
DISEÑODE ZAPATA P2 Las zapata se diseñaran de hormigónarmado y sujetoa lasaccionesque se indican. 𝐷 = 11749.0816 𝑘𝑔 𝑚2 𝐿 = 2656 𝑘𝑔 𝑚2 𝜎𝑠 = 21 𝑇/𝑚2 ℎ𝑓 = −1.20 𝑚  Columna: 30x30 cm Materiales 𝑓′𝑐 = 210 𝑘𝑔 𝑚2 𝑓𝑦 = 4200 𝑘𝑔 𝑚2  DISEÑO 1. P=D+L= (11.749+2.656) T = 14.41 ton 𝐴 𝐹 = 𝑃 ∗ %𝑃 𝑠 = 14.41 𝑡𝑜𝑛 ∗ 1.10 21 𝑇 𝑚2 = 0.75 𝑚2 = 0.87𝑥0.87 2. 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 0.90𝑥0.90 = 0.81 𝑚2 3. 𝑃. 𝑁. 𝑆 = 𝑃 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 14.41 𝑡𝑜𝑛 0.81 𝑚2 = 17.79 𝑇 𝑚2 𝑃. 𝑁. 𝑆 < 𝜎𝑠 17.79 𝑇 𝑚2 < 21 𝑇 𝑚2 𝑜𝑘 4. 𝑓𝑚𝑎𝑦𝑜𝑟𝑎𝑐𝑖𝑜𝑛 = 1.2 ∗ 11749.0816 + 1.6 ∗ 2656 11749.0816 + 2656 = 1.27 𝑃. 𝑁. 𝑆𝑢 = 𝑃. 𝑁. 𝑆 ∗ 𝑓𝑚𝑎𝑦 = 17.79 ∗ 1.27 = 22.59 𝑇 𝑚2
CORTE UNIDIRECCIONAL 𝑣𝑝 = 0.53 √ 𝑓′ 𝑐 = 0.53 √210 = 7.68 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏 ∗ 𝑑 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐴𝑐𝑟í𝑡𝑖𝑐𝑎 𝑉𝑢 = 22.59 𝑇 𝑚2 ∗ ( 0.30 − 𝑑) ∗ 0.90 𝑣𝑢 = 22.59( 0.30 − 𝑑) ∗ 0.90 0.85 ∗ 0.90 ∗ 𝑑 𝑣𝑢 = 𝑣𝑝 76.8 = 22.59( 0.30 − 𝑑) ∗ 0.90 0.85 ∗ 0.90 ∗ 𝑑 58.75𝑑 = 20.33(0.30 − 𝑑) 79.08𝑑 = 6.0993 𝑑 = 0.077𝑚 ℎ = 𝑑 + 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 = 0.077 + 0.07 = 0.1471 𝑚
 𝒉 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟐𝟓 𝒎 → 𝒉 𝒎𝒊𝒏𝒊𝒎𝒂 𝒅 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟏𝟖 𝒎 CORTE BIDIRECCIONAL 𝑣𝑝 = 1.06√ 𝑓′ 𝑐 = 1.06 √210 = 15.63 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏0 ∗ 𝑑 𝑏0 = 0.48 ∗ 4 = 1.92 𝑚 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ (𝐴𝑓 − 𝐴𝑝𝑧) 𝑉𝑢 = 22.59 𝑇 𝑚2 ∗ (0.92 − 0.482) 𝑚2 = 13.09 𝑡𝑜𝑛 𝑣𝑢 = 13.09 𝑡𝑜𝑛 0.85 ∗ 1.92 𝑚 ∗ 0.18𝑚 𝑣𝑢 = 44.56 𝑇 𝑚2 𝑣𝑢 < 𝑣𝑝 4.46 𝑘𝑔 𝑐𝑚2 < 15.63 𝑘𝑔 𝑐𝑚2 ↓ 𝑂𝐾
DISEÑO DE LA ARMADURA 𝑀𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐿𝑣2 2 ∗ 𝐵 𝑀𝑢 = 22.59 𝑇 𝑚2 ∗ 0.3𝑚2 2 ∗ 0.9𝑚 = 0.9149 𝑇 − 𝑚 𝜌 = 0.85 𝑓′𝑐 𝑓𝑦 [1 − √1 − 2𝑀𝑢 0.85𝑓′𝑐∅𝑏𝑑2] = 0.85 210 4200 [1 − √1 − 2 ∗ 0.9149∗ 100000 0.85 ∗ 210∗ 0.9 ∗ 90∗ 82] = 0.004433 𝜌𝑚𝑖𝑛 = 0.0018 𝜌 > 𝜌 min 𝑂𝐾 𝐴𝑠 = 𝜌 ∗ 𝑏 ∗ ℎ = 0.004433 ∗ 90 ∗ 15 = 5.98 𝑐𝑚2 → 𝟓∅𝟏𝟐𝒎𝒎 → 𝟏∅𝟏𝟐𝒎𝒎 @𝟐𝟎 𝒄𝒎

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Diseño de zapatas

  • 1. DISEÑODE ZAPATA P1 Las zapata se diseñaran de hormigónarmado y sujetoa lasaccionesque se indican. 𝐷 = 4158.18 𝑘𝑔 𝑚2 𝐿 = 940 𝑘𝑔 𝑚2 𝜎𝑠 = 21 𝑇/𝑚2 ℎ𝑓 = −1.20 𝑚  Columna: 30x30 cm Materiales 𝑓′𝑐 = 210 𝑘𝑔 𝑚2 𝑓𝑦 = 4200 𝑘𝑔 𝑚2  DISEÑO 1. P=D+L= (4.15+0.94) T =5.09 ton 𝐴 𝐹 = 𝑃 + %𝑃 𝑠 %𝑃(10 − 15)% 𝑃𝑜𝑟 𝑝𝑒𝑠𝑜 𝑝𝑟𝑜𝑝𝑖𝑜 𝑑𝑒𝑙 𝑟𝑒𝑙𝑙𝑒𝑛𝑜 𝑦 𝑝𝑒𝑠𝑜 𝑝𝑟𝑜𝑝𝑖𝑜 𝑑𝑒𝑙 𝑐𝑖𝑚𝑖𝑒𝑛𝑡𝑜 𝐴 𝐹 = 𝑃 ∗ %𝑃 𝑠 = 5.09 𝑡𝑜𝑛 ∗ 1.10 21 𝑇 𝑚2 = 0.27 𝑚2 = 0.52𝑥0.52 2. 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 0.60𝑥0.60 = 0.36 𝑚2 3. 𝑃. 𝑁. 𝑆 = 𝑃 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 5.09 𝑡𝑜𝑛 0,36 𝑚2 = 14,13 𝑇 𝑚2 𝑃. 𝑁. 𝑆 < 𝜎𝑠 14,13 𝑇 𝑚2 < 21 𝑇 𝑚2 𝑜𝑘 4. 𝑓𝑚𝑎𝑦𝑜𝑟𝑎𝑐𝑖𝑜𝑛 = 1.2 ∗ 4158.18 + 1.6 ∗ 940 4158.18 + 940 = 1.27 𝑃. 𝑁. 𝑆𝑢 = 𝑃. 𝑁. 𝑆 ∗ 𝑓𝑚𝑎𝑦 = 14.13 ∗ 1.27 = 17.96 𝑇 𝑚2
  • 2. CORTE UNIDIRECCIONAL La seccióncriticapara corte unidireccional esladistanciadmedidaapartirde lacara de la columna. 𝑣𝑝 = 0.53 √ 𝑓′ 𝑐 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 𝑃𝑒𝑟𝑚𝑖𝑠𝑖𝑙𝑒 𝑑𝑒𝑙 ℎ𝑜𝑟𝑚𝑖𝑔ó𝑛 𝑣𝑝 = 0.53 √ 𝑓′ 𝑐 = 0.53 √210 = 7.68 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏 ∗ 𝑑 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 ú𝑙𝑡𝑖𝑚𝑜 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐴𝑐𝑟í𝑡𝑖𝑐𝑎 𝑉𝑢 = 17.96 𝑇 𝑚2 ∗ ( 0.15 − 𝑑) ∗ 0.60 𝑣𝑢 = 17.96( 0.15 − 𝑑) ∗ 0.60 0.85 ∗ 0.60 ∗ 𝑑 𝑣𝑢 = 𝑣𝑝 76.8 = 17.96( 0.15 − 𝑑) ∗ 0.60 0.85 ∗ 0.60 ∗ 𝑑 39.17𝑑 = 10.78(0.15 − 𝑑) 49.95𝑑 = 1.62 𝑑 = 0.032𝑚 ℎ = 𝑑 + 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 ; 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 = 7 𝑐𝑚
  • 3. ℎ = 𝑑 + 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 = 0.032 + 0.07 = 0.102 𝑚  𝒉 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟐𝟓 𝒎 → 𝒉 𝒎𝒊𝒏𝒊𝒎𝒂 𝒅 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟏𝟖 𝒎 CORTE BIDIRECCIONAL La seccióncriticaesuna franja perimetral medidaaunadistanciad/2 desde lacara de la columna 𝑣𝑝 = 1.06√ 𝑓′ 𝑐 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 𝑝𝑒𝑟𝑚𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑒𝑙 ℎ𝑜𝑟𝑚𝑖𝑔ó𝑛 𝑣𝑝 = 1.06√ 𝑓′ 𝑐 = 1.06 √210 = 15.63 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏0 ∗ 𝑑 → 𝐸𝑠𝑓𝑢𝑒𝑟𝑧𝑜 𝑢𝑙𝑡𝑖𝑚𝑜 𝑏0 = 0.48 ∗ 4 = 1.92 𝑚 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ (𝐴𝑓 − 𝐴𝑝𝑧) 𝑉𝑢 = 17.96 𝑇 𝑚2 ∗ (0.62 − 0.482) 𝑚2 = 2.33 𝑡𝑜𝑛 𝑣𝑢 = 2.33 𝑡𝑜𝑛 0.85 ∗ 1.92 𝑚 ∗ 0.18𝑚 𝑣𝑢 = 7.93 𝑇 𝑚2 𝑣𝑢 < 𝑣𝑝 0.793 𝑘𝑔 𝑐𝑚2 < 15.63 𝑘𝑔 𝑐𝑚2 ↓ 𝑂𝐾
  • 4. DISEÑO DE LA ARMADURA Es la que se ubica enla cara de lacolumnay es la secciónque nospermite determinarel armado de la zapata. 𝑀𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐿𝑣2 2 ∗ 𝐵 𝑀𝑢 = 17.96 𝑇 𝑚2 ∗ 0.15𝑚2 2 ∗ 0.6𝑚 = 0.1212 𝑇 − 𝑚 𝜌 = 0.85 𝑓′𝑐 𝑓𝑦 [1 − √1 − 2𝑀𝑢 0.85𝑓′𝑐∅𝑏𝑑2] = 0.85 210 4200 [1 − √1 − 2 ∗ 0.1212∗ 100000 0.85 ∗ 210 ∗ 0.9 ∗ 60 ∗ 182] = 0.000095 𝜌𝑚𝑖𝑛 = 0.0018 𝜌 > 𝜌 min 𝑂𝐾 𝐴𝑠 = 𝜌 ∗ 𝑏 ∗ ℎ = 0.004613 ∗ 60 ∗ 10.5 = 2.91 𝑐𝑚2 → 𝟒∅𝟏𝟎𝒎𝒎 → 𝟏∅𝟏𝟎𝒎𝒎 @𝟏𝟓 𝒄𝒎
  • 5. DISEÑODE ZAPATA P2 Las zapata se diseñaran de hormigónarmado y sujetoa lasaccionesque se indican. 𝐷 = 11749.0816 𝑘𝑔 𝑚2 𝐿 = 2656 𝑘𝑔 𝑚2 𝜎𝑠 = 21 𝑇/𝑚2 ℎ𝑓 = −1.20 𝑚  Columna: 30x30 cm Materiales 𝑓′𝑐 = 210 𝑘𝑔 𝑚2 𝑓𝑦 = 4200 𝑘𝑔 𝑚2  DISEÑO 1. P=D+L= (11.749+2.656) T = 14.41 ton 𝐴 𝐹 = 𝑃 ∗ %𝑃 𝑠 = 14.41 𝑡𝑜𝑛 ∗ 1.10 21 𝑇 𝑚2 = 0.75 𝑚2 = 0.87𝑥0.87 2. 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 0.90𝑥0.90 = 0.81 𝑚2 3. 𝑃. 𝑁. 𝑆 = 𝑃 𝐴 𝐹 𝑎𝑑𝑜𝑝𝑡𝑎𝑑𝑎 = 14.41 𝑡𝑜𝑛 0.81 𝑚2 = 17.79 𝑇 𝑚2 𝑃. 𝑁. 𝑆 < 𝜎𝑠 17.79 𝑇 𝑚2 < 21 𝑇 𝑚2 𝑜𝑘 4. 𝑓𝑚𝑎𝑦𝑜𝑟𝑎𝑐𝑖𝑜𝑛 = 1.2 ∗ 11749.0816 + 1.6 ∗ 2656 11749.0816 + 2656 = 1.27 𝑃. 𝑁. 𝑆𝑢 = 𝑃. 𝑁. 𝑆 ∗ 𝑓𝑚𝑎𝑦 = 17.79 ∗ 1.27 = 22.59 𝑇 𝑚2
  • 6. CORTE UNIDIRECCIONAL 𝑣𝑝 = 0.53 √ 𝑓′ 𝑐 = 0.53 √210 = 7.68 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏 ∗ 𝑑 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐴𝑐𝑟í𝑡𝑖𝑐𝑎 𝑉𝑢 = 22.59 𝑇 𝑚2 ∗ ( 0.30 − 𝑑) ∗ 0.90 𝑣𝑢 = 22.59( 0.30 − 𝑑) ∗ 0.90 0.85 ∗ 0.90 ∗ 𝑑 𝑣𝑢 = 𝑣𝑝 76.8 = 22.59( 0.30 − 𝑑) ∗ 0.90 0.85 ∗ 0.90 ∗ 𝑑 58.75𝑑 = 20.33(0.30 − 𝑑) 79.08𝑑 = 6.0993 𝑑 = 0.077𝑚 ℎ = 𝑑 + 𝑟𝑒𝑐𝑢𝑏𝑟𝑖𝑚𝑖𝑒𝑛𝑡𝑜 = 0.077 + 0.07 = 0.1471 𝑚
  • 7.  𝒉 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟐𝟓 𝒎 → 𝒉 𝒎𝒊𝒏𝒊𝒎𝒂 𝒅 𝑨𝒅𝒐𝒑𝒕𝒂𝒅𝒐 = 𝟎. 𝟏𝟖 𝒎 CORTE BIDIRECCIONAL 𝑣𝑝 = 1.06√ 𝑓′ 𝑐 = 1.06 √210 = 15.63 𝑘𝑔 𝑐𝑚2 𝑣𝑢 = 𝑉𝑢 ∅ ∗ 𝑏0 ∗ 𝑑 𝑏0 = 0.48 ∗ 4 = 1.92 𝑚 𝑉𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ (𝐴𝑓 − 𝐴𝑝𝑧) 𝑉𝑢 = 22.59 𝑇 𝑚2 ∗ (0.92 − 0.482) 𝑚2 = 13.09 𝑡𝑜𝑛 𝑣𝑢 = 13.09 𝑡𝑜𝑛 0.85 ∗ 1.92 𝑚 ∗ 0.18𝑚 𝑣𝑢 = 44.56 𝑇 𝑚2 𝑣𝑢 < 𝑣𝑝 4.46 𝑘𝑔 𝑐𝑚2 < 15.63 𝑘𝑔 𝑐𝑚2 ↓ 𝑂𝐾
  • 8. DISEÑO DE LA ARMADURA 𝑀𝑢 = 𝑃. 𝑁. 𝑆𝑢 ∗ 𝐿𝑣2 2 ∗ 𝐵 𝑀𝑢 = 22.59 𝑇 𝑚2 ∗ 0.3𝑚2 2 ∗ 0.9𝑚 = 0.9149 𝑇 − 𝑚 𝜌 = 0.85 𝑓′𝑐 𝑓𝑦 [1 − √1 − 2𝑀𝑢 0.85𝑓′𝑐∅𝑏𝑑2] = 0.85 210 4200 [1 − √1 − 2 ∗ 0.9149∗ 100000 0.85 ∗ 210∗ 0.9 ∗ 90∗ 82] = 0.004433 𝜌𝑚𝑖𝑛 = 0.0018 𝜌 > 𝜌 min 𝑂𝐾 𝐴𝑠 = 𝜌 ∗ 𝑏 ∗ ℎ = 0.004433 ∗ 90 ∗ 15 = 5.98 𝑐𝑚2 → 𝟓∅𝟏𝟐𝒎𝒎 → 𝟏∅𝟏𝟐𝒎𝒎 @𝟐𝟎 𝒄𝒎