Prova 2 B - Professor Iran - Topografia Aplicada à Engenharia Civil

(3) Júlio Massayn Lima Vtodo = \{ \left( S_{CA} + S_{CB} \right) \frac{20}{2} + \left( S_{CB} + S_{CC} \right) \frac{20}{2} \} V_{todo} = 670 m^3 + 803,8460 m^3 = 1473,846 m^3 V_{aterro} = \{ \left( S_{AA} + S_{AB} \right) \frac{20}{2} + \left( S_{AB} + S_{AC} \right) \frac{20}{2} \} = V_A = 235,6250 + 344,3710 = 1233,3960 m^3 V_{final} = V_C - V_A = 239,85 m^3 \text{ volum de volume} (3) Júlio Massayn Lima Cota perfil C = 3,1 "" " B = 2,7 "" " A = 2,3 A ___ B' Seção A S_C = \frac{20}{2}(2,4 - 2,3) - (3,6 - 2,3) + \frac{20}{2}(3,6 - 2,3) + (5,2 - 2,3) S_C = 56 m^2 Seção B' S_A = \frac{20}{2}(2,7 - 0,4 - 2,7 - 2,1) + \frac{6,25}{2}(2,7 - 2,1) = S_A = 29,5625 m^2 S_C = \frac{13,25}{2}(4,3 - 2,7) = 11 m^2 B __ C Seção B S_C = \frac{13,25}{2}(4,3 - 2,7) + \frac{20}{2}(4,3 - 2,7) + (6,5 - 2,7) = 65 m^2 S_A = \left[ \frac{2,7 - 0,4}{1,4} + (2,7 - 2,3) \right] \frac{20}{2} + 6,25(2,7 - 2,2) = 28,5625 m^2 Seção C S_C = \frac{(20 - 4,6154)}{2}(5,1 - 3,1) = 15,3846 m^2 S_A = (3,1 - 0,8) + (3,1 - 1,3) \frac{20}{2} + (3,1 - 1,3) + (3,1 - 2,6) \frac{20}{2} + (3,1 - 2,5) \frac{4,6}{2} S_A = 41 + 24 + 0,1784 = 65,7846 m^2 two thirty 1/2 8 1/16