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Metrado de cargas
Edificiode concretoarmadode 4 niveles
Alturade entre piso
𝐻 = 2.60 π‘š
Pesos
Pesopropio:
π‘π‘’π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = 1.15
𝑇𝑛
π‘š2
π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 0.95
𝑇𝑛
π‘š2
Sobre carga:
π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = 0.60
𝑇𝑛
π‘š2
π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 0.45
𝑇𝑛
π‘š2
Áreade laestructura:
𝐴 = 120.34 π‘š2
Formulaque se utilizaraparael cΓ‘lculodel pesode la estructura
π‘π‘’π‘ π‘œ 𝑑𝑒 π‘™π‘Ž π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = π΄π‘Ÿπ‘’π‘Ž Γ— (π‘π‘’π‘ π‘œ π‘π‘Ÿπ‘œπ‘π‘–π‘œ + 0.25 Γ— π‘ π‘œπ‘π‘Ÿπ‘’π‘π‘Žπ‘Ÿπ‘”π‘Ž)
π‘π‘’π‘ π‘œ π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = π΄π‘Ÿπ‘’π‘Ž Γ— (π‘π‘’π‘ π‘œ π‘π‘Ÿπ‘œπ‘π‘–π‘œ + 0.5 Γ— π‘ π‘œπ‘π‘Ÿπ‘’π‘π‘Žπ‘Ÿπ‘”π‘Ž)
π‘π‘’π‘ π‘œ π‘‘π‘œπ‘‘π‘Žπ‘™ = π‘π‘’π‘ π‘œ 𝑑𝑒 π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž + π‘π‘’π‘ π‘œ π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ Γ— (𝑛 βˆ’ 1)
𝑛 = π‘›π‘’π‘šπ‘’π‘Ÿπ‘œ 𝑑𝑒 π‘π‘–π‘ π‘œπ‘ 
Calculode lospesos
π‘π‘’π‘ π‘œ 𝑑𝑒 π‘™π‘Ž π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 120.34 Γ— (0.95 + 0.25 Γ— 0.45) = 100.78 𝑑𝑛
π‘π‘’π‘ π‘œ π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = 120.34 Γ— (1.15 + 0.5 Γ— 0.60) = 102.289 𝑑𝑛
𝑛 = 3
π‘π‘’π‘ π‘œ π‘‘π‘œπ‘‘π‘Žπ‘™ = 100.78 + 102.289 Γ— (4 βˆ’ 1) = 407.647 𝑑𝑛
π‘π‘’π‘ π‘œ 𝑑𝑒𝑙 π‘π‘–π‘ π‘œ 𝑁°1 = 102.289 𝑑𝑛
π‘π‘’π‘ π‘œ 𝑑𝑒𝑙 π‘π‘–π‘ π‘œ 𝑁°2 = 102.289 𝑑𝑛
π‘π‘’π‘ π‘œ 𝑑𝑒𝑙 π‘π‘–π‘ π‘œ 𝑁°3 = 102.289 𝑑𝑛
π‘π‘’π‘ π‘œ 𝑑𝑒 π‘™π‘Ž π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 100.78 𝑑𝑛

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Cargas

  • 1. Metrado de cargas Edificiode concretoarmadode 4 niveles Alturade entre piso 𝐻 = 2.60 π‘š Pesos Pesopropio: π‘π‘’π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = 1.15 𝑇𝑛 π‘š2 π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 0.95 𝑇𝑛 π‘š2 Sobre carga: π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = 0.60 𝑇𝑛 π‘š2 π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 0.45 𝑇𝑛 π‘š2 Áreade laestructura: 𝐴 = 120.34 π‘š2 Formulaque se utilizaraparael cΓ‘lculodel pesode la estructura π‘π‘’π‘ π‘œ 𝑑𝑒 π‘™π‘Ž π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = π΄π‘Ÿπ‘’π‘Ž Γ— (π‘π‘’π‘ π‘œ π‘π‘Ÿπ‘œπ‘π‘–π‘œ + 0.25 Γ— π‘ π‘œπ‘π‘Ÿπ‘’π‘π‘Žπ‘Ÿπ‘”π‘Ž) π‘π‘’π‘ π‘œ π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = π΄π‘Ÿπ‘’π‘Ž Γ— (π‘π‘’π‘ π‘œ π‘π‘Ÿπ‘œπ‘π‘–π‘œ + 0.5 Γ— π‘ π‘œπ‘π‘Ÿπ‘’π‘π‘Žπ‘Ÿπ‘”π‘Ž) π‘π‘’π‘ π‘œ π‘‘π‘œπ‘‘π‘Žπ‘™ = π‘π‘’π‘ π‘œ 𝑑𝑒 π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž + π‘π‘’π‘ π‘œ π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ Γ— (𝑛 βˆ’ 1) 𝑛 = π‘›π‘’π‘šπ‘’π‘Ÿπ‘œ 𝑑𝑒 π‘π‘–π‘ π‘œπ‘  Calculode lospesos π‘π‘’π‘ π‘œ 𝑑𝑒 π‘™π‘Ž π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 120.34 Γ— (0.95 + 0.25 Γ— 0.45) = 100.78 𝑑𝑛 π‘π‘’π‘ π‘œ π‘π‘–π‘ π‘œ π‘‘π‘–π‘π‘–π‘π‘œ = 120.34 Γ— (1.15 + 0.5 Γ— 0.60) = 102.289 𝑑𝑛 𝑛 = 3 π‘π‘’π‘ π‘œ π‘‘π‘œπ‘‘π‘Žπ‘™ = 100.78 + 102.289 Γ— (4 βˆ’ 1) = 407.647 𝑑𝑛 π‘π‘’π‘ π‘œ 𝑑𝑒𝑙 π‘π‘–π‘ π‘œ 𝑁°1 = 102.289 𝑑𝑛 π‘π‘’π‘ π‘œ 𝑑𝑒𝑙 π‘π‘–π‘ π‘œ 𝑁°2 = 102.289 𝑑𝑛 π‘π‘’π‘ π‘œ 𝑑𝑒𝑙 π‘π‘–π‘ π‘œ 𝑁°3 = 102.289 𝑑𝑛 π‘π‘’π‘ π‘œ 𝑑𝑒 π‘™π‘Ž π‘Žπ‘§π‘œπ‘‘π‘’π‘Ž = 100.78 𝑑𝑛