Diseñar la escalera, considerando como resistencia de concreto 𝑓 ′ 𝑐 = 210 𝑘𝑔/𝑐𝑚2, 𝑓𝑦 = 4200𝑘𝑔/𝑐𝑚2, uso: Biblioteca, Paso = P1, Contrapaso = CP1 DATOS: 𝑘𝑔𝑓 𝑓 ′ 𝑐 ≔ 210 𝑐𝑚2 𝑓𝑦 ≔ 4200 𝑘𝑔𝑓 𝑐𝑚2 𝑢𝑠𝑜 ∶= "𝑏𝑖𝑏𝑙𝑖𝑜𝑡𝑒𝑐𝑎" 𝑠. 𝑐 ≔ 400 𝑘𝑔𝑓 𝑚2 𝑝𝑒𝑠𝑜 ∶= 2400 𝑎𝑐𝑎𝑏𝑎𝑑𝑜 ≔ 100 𝑡𝑎𝑟𝑟𝑎𝑗𝑒𝑜 ≔ 50 𝑘𝑔𝑓 𝑚2 𝑘𝑔𝑓 𝑚3 GEOMETRIA: 𝐶𝑃 ∶= 𝐶𝑃1 = 17 𝑐𝑚 𝑃 ∶= 𝑃1 = 32 𝑐𝑚 #𝑡𝑟𝑎𝑚𝑜1 ∶= 15 𝑏 ∶= 𝐵1 = 1.25 𝑚 𝑑𝑒𝑠𝑐𝑎𝑠𝑛𝑠𝑜 ≔ 0 1. DIMENSIONAMIENTO: Calculo del espesor de la garganta 𝑙1 ≔ #𝑡𝑟𝑎𝑚𝑜1. 𝑃 = 4.8 𝑚 𝐿 ≔ #𝑡𝑟𝑎𝑚𝑜1. 𝑃 + 𝑑𝑒𝑠𝑐𝑎𝑛𝑠𝑜 = 4.8 𝑚 𝐿 0.24 20 | 𝐿 | 𝑇 ∶= , 1 𝑐𝑚 = [0.20] 𝑚 0.15 | 25 | 0.20 0.03. 𝐿 ( 0.04. 𝐿 ) 𝑘𝑔𝑓 𝑚2 𝑇𝑝𝑟𝑜𝑚 ≔ 𝑚𝑎𝑥(𝑚𝑎𝑥(𝑇), 15 𝑐𝑚) = 24 𝑐𝑚 𝑡 ≔ 𝐶𝑒𝑖𝑙⟨𝑚𝑒𝑎𝑛⟨𝑇𝑝𝑟𝑜𝑚 , 15 𝑐𝑚⟩, 1 𝑐𝑚⟩ = 0.2 𝑚 𝑡 ≔ 15 𝑐𝑚 Calculo de la altura promedio 𝜃 ≔ 𝑐𝑜𝑠(𝜃) = 𝑃 √𝑃2 + 𝑆𝑜𝑙𝑣𝑒,𝜃 𝐶𝑃2 → 𝑎𝑐𝑜𝑠 ( 32. 𝑐𝑚. √1313 1313. √𝑐𝑚2 ) = 27.979 𝑑𝑒𝑔 𝑡 𝐶𝑃 ℎ𝑚 ≔ 𝐶𝑒𝑖𝑙 ( + , 1 𝑐𝑚) = 26 𝑐𝑚 𝑐𝑜𝑠(𝜃) 2 𝑡 𝐶𝑃 + = 0.255 𝑚 𝑐𝑜𝑠(𝜃) 2 2. METRADO DE CARGAS – TRAMO INCLINADO: Carga Muerta (Wd) 𝑘𝑔𝑓 𝑝𝑒𝑠𝑜. 𝑙𝑜𝑠𝑎 ∶= 𝑝𝑒𝑠𝑜. 𝑏. ℎ𝑚 = 780.00 𝑚 𝑘𝑔𝑓 𝑝𝑒𝑠𝑜. 𝑎𝑐𝑎𝑏𝑎𝑑𝑜𝑠 ∶= 𝑎𝑐𝑎𝑏𝑎𝑑𝑜. 𝑏 = 125.00 𝑚 𝑘𝑔𝑓 𝑝𝑒𝑠𝑜. 𝑡𝑎𝑟𝑟𝑎𝑗𝑒𝑜 ∶= 𝑡𝑎𝑟𝑟𝑎𝑗𝑒𝑜. 𝑏 = 62.50 𝑚 𝑊𝑑. 1 = 𝑝𝑒𝑠𝑜. 𝑙𝑜𝑠𝑎 + 𝑝𝑒𝑠𝑜. 𝑎𝑐𝑎𝑏𝑎𝑑𝑜𝑠 + 𝑝𝑒𝑠𝑜. 𝑡𝑎𝑟𝑟𝑎𝑗𝑒𝑜 = 967.50 𝑘𝑔𝑓 𝑚 Carga Viva (Wl) 𝑊𝑙. 1 ∶= 𝑠. 𝑐.∙ 𝑏 = 500 𝑘𝑔𝑓 𝑚 Carga Ultima (Wu) 𝑘𝑔𝑓 𝑚 3. ANALISIS ESTRUCTURAL –MOMENTOS FLECTORES 𝑊𝑢. 1 ∶= 1.4 ∙ 𝑊𝑑. 1 + 1.7 ∙ 𝑊𝑙. 1 = 2204.50 𝑙𝑜𝑛𝑔 ∶= [𝑙1] = [4.8 ]𝑚 𝑊𝑢. 𝑡 ≔ [𝑊𝑢. 1] = [2204.50] 𝑊𝑢 ∶= 𝑚𝑒𝑎𝑛(𝑊𝑢. 𝑡) = 2204.50 Momento Flector 12 𝑑. sup ≔ [ ] 𝑑. inf ≔ [10] 12 # inf ≔ 𝑟𝑜𝑤𝑠(𝑑. 𝑖𝑛𝑓) = 1 # sup ≔ 𝑟𝑜𝑤𝑠(𝑑. 𝑠𝑢𝑝) = 2 𝐿. inf ≔ 𝑙𝑜𝑛𝑔 𝑘𝑔𝑓 𝑚 𝑙𝑜𝑛𝑔 ∶= [𝑙1] = [4.8] 𝑚 𝐿. 𝑠𝑢𝑝1 ∶= 𝑙𝑜𝑛𝑔1 = 4.8 𝑚 𝐿. 𝑠𝑢𝑝2 ∶= 𝑙𝑜𝑛𝑔1 = 4.8 𝑚 4.8 𝐿. 𝑠𝑢𝑝 = [ ] 𝑚 4.8 𝑞𝑤𝑒 ∶= 1. . #𝑖𝑛𝑓 𝑊𝑢. 𝐿. 𝑖𝑛𝑓𝑞𝑤𝑒 2 𝑀𝑢. 𝑖𝑛𝑓𝑞𝑤𝑒 ≔ = [5079.17] 𝑚. 𝑘𝑔𝑓 𝑑. 𝑖𝑛𝑓𝑞𝑤𝑒 𝑀𝑢. 𝑖𝑛𝑓. max ≔ 𝑚𝑎𝑥(𝑀𝑢. 𝑖𝑛𝑓) = 5079.17 𝑘𝑔𝑓. 𝑚 𝑞𝑤 ∶= 1. . #𝑠𝑢𝑝 𝑊𝑢. 𝐿. 𝑠𝑢𝑝𝑞𝑤 2 4232.64 𝑀𝑢. 𝑠𝑢𝑝𝑞𝑤 ≔ =[ ] 𝑚. 𝑘𝑔𝑓 4232.64 𝑑. 𝑠𝑢𝑝𝑞𝑤 𝑀𝑢. 𝑠𝑢𝑝. 𝑚𝑎𝑥 ≔ (𝑀𝑢. 𝑠𝑢𝑝) = 4232.64 𝑘𝑔𝑓 𝑚 6. ANALISIS ESTRUCTURAL –MOMENTOS FLECTORES 𝑑 ≔ 𝑡 − 3 𝑐𝑚 = 0.12 𝑚 ∅ ≔ 0.90 Acero Superior Continuo 𝑎 ≔ 𝑑 − √𝑑2 − 2. 𝑀𝑢. 𝑠𝑢𝑝. 𝑚𝑎𝑥 = 1.91 𝑐𝑚 0.85. ∅. 𝑓 ′ 𝑐. 𝑏 𝑀𝑢. 𝑠𝑢𝑝. 𝑚𝑎𝑥 𝑓𝑦 2 𝑎 ≔ 𝐴𝑠. = 1.91 𝑐𝑚 𝑎 = 10.14 𝑐𝑚 0.85. 𝑓 ′ 𝑐. 𝑏 ∅. 𝑓𝑦. (𝑑 − 2) 𝐴𝑠. 𝑚𝑖𝑛 ≔ 0.0018 ∙ 𝑏 ∙ 𝑡 = 3.375 𝑐𝑚2 Distancia entre Varillas 𝑒𝑠𝑝. 𝑚𝑎𝑥 ≔ 𝑚𝑖𝑛(3. 𝑡, 30 𝑐𝑚) → 𝑚𝑖𝑛(45 ∙ 𝑐𝑚, 30 ∙ 𝑐𝑚) = 0.3 𝑚 3 8 1 𝐴𝑠. 𝑐𝑜𝑚𝑒𝑟𝑐𝑖𝑎𝑙 ≔ 𝑖𝑛 2 5 [8] 𝐴𝑠. 𝑐𝑜𝑚𝑒𝑟𝑐𝑖𝑎𝑙 2 15 𝜋. ( ) 2 𝑒𝑠𝑝 ≔ 𝐹𝑙𝑜𝑜𝑟 ( . 90 𝑐𝑚, 5 𝑐𝑚) = [30] 𝑐𝑚 𝐴𝑠. 𝑚𝑖𝑛 50 𝐴𝑠 = Distancia entre Varillas ℎ. 1 ≔ (#𝑡𝑟𝑎𝑚𝑜1 + 1) ∙ 𝐶𝑃 = 2.72 𝑚 𝐿𝑛 ≔ 𝑅𝑜𝑢𝑛𝑑 (√ℎ. 12 + 𝑙12 , 1𝑐𝑚) = 5.52 𝑚 𝐿𝑛 𝐿. 𝑟𝑒𝑓 ≔ 𝐶𝑒𝑖𝑙 ( , 5𝑐𝑚) = 1.85 𝑚 3 Acero Inferior Continuo 𝑎 ≔ 𝑑 − √𝑑2 − 2. 𝑀𝑢. 𝑖𝑛𝑓. 𝑚𝑎𝑥 = 2.335 𝑐𝑚 0.85. ∅. 𝑓 ′ 𝑐. 𝑏 𝑀𝑢. 𝑖𝑛𝑓. 𝑚𝑎𝑥 𝑓𝑦 2 𝑎 ≔ 𝐴𝑠. = 2.335 𝑐𝑚 𝑎 = 12.4 𝑐𝑚 0.85. 𝑓 ′ 𝑐. 𝑏 ∅. 𝑓𝑦. (𝑑 − 2) 𝐴𝑠. 𝑚𝑖𝑛 ≔ 0.0018 ∙ 𝑏 ∙ 𝑡 = 3.375 𝑐𝑚2 Distancia entre Varillas 𝑒𝑠𝑝. 𝑚𝑎𝑥 ≔ 𝑚𝑖𝑛(3. 𝑡, 30 𝑐𝑚) → 𝑚𝑖𝑛(45 ∙ 𝑐𝑚, 30 ∙ 𝑐𝑚) = 0.3 𝑚 3 8 1 𝐴𝑠. 𝑐𝑜𝑚𝑒𝑟𝑐𝑖𝑎𝑙 ≔ 𝑖𝑛 2 5 [8] 𝐴𝑠. 𝑐𝑜𝑚𝑒𝑟𝑐𝑖𝑎𝑙 2 15 𝜋. ( ) 2 𝑒𝑠𝑝 ≔ 𝐹𝑙𝑜𝑜𝑟 ( ∙ 90 𝑐𝑚, 5 𝑐𝑚) = [30] 𝑐𝑚 𝐴𝑠. 𝑚𝑖𝑛 50 𝐴𝑠 =