solve for q max, min, and intermediate bearing pressure at the bottom of the foundation . pressure at the bottom of the foundation, a. 14 kPa b. 19kPa c. 13 kPa d. 15 kPa [for Nos. 33 to 40] A 1.5-m square, 0.6-m deep footing is subjected to a concentric vertical load of 160 kN and an overturning moment of 30 kN-m that acts at 45° angle from the side of the footing (i.e. it acts diagonally across the top of the footing). The top of the footing, is 2.4 m below the ground surface and the groundwater table is initially at the bottom of the footing. Determine the following: Solve 33. The total weight of the footing and the soil above it. a. 159 kN b. 124 kN c. 128 kN d. 134 kN 34. Is the resultant force within the kern? c. yes, withe e 0.066 d. yes, with e,-e,-0.072 a, yes, with eB-e.-0.074 b. no 60 kN (including the weight of column) 1.0 m Elev B Silty Sand 3.0 m 1.5 m 0.6 0.2 m 0.2 m . pressure at the bottom of the foundation, a. 14 kPa b. 19kPa c. 13 kPa d. 15 kPa [for Nos. 33 to 40] A 1.5-m square, 0.6-m deep footing is subjected to a concentric vertical load of 160 kN and an overturning moment of 30 kN-m that acts at 45° angle from the side of the footing (i.e. it acts diagonally across the top of the footing). The top of the footing, is 2.4 m below the ground surface and the groundwater table is initially at the bottom of the footing. Determine the following: Solve 33. The total weight of the footing and the soil above it. a. 159 kN b. 124 kN c. 128 kN d. 134 kN 34. Is the resultant force within the kern? c. yes, withe e 0.066 d. yes, with e,-e,-0.072 a, yes, with eB-e.-0.074 b. no 60 kN (including the weight of column) 1.0 m Elev B Silty Sand 3.0 m 1.5 m 0.6 0.2 m 0.2 m
Weight of the Footing = (Unit Wt. of Concrete)*(length)*(breadth)*(depth) = (14. 126 kN/m3)*(1. 5 m)*(1. 5 m)*(0. 6 m) = 19. 07 kN Now, because of the moment we will have non-uniform bearing pressure across the footing, Eccentricity (e) = (30) / ( 160 + 19.07) = 0. 167 Again, since the water table is at he bottom of the footing, we can assume pore water pressure to be 0. Therefore, Min Bearing Pressure (q min) = {((160+19.07)/1. 5) - 0 }*{ 1- ((6*0.167)/1. 5)} = (172.713)*(0. 332) = 38. 30 kN/m2 Max Bearing Pressure (q max) = {((160+19.07)/1. 5) - 0 }*{ 1+ ((6*0.167)/1. 5)} = (172.713)*(1. 668) = 192. 44 kN/m2 Intermidiate Bearing Pressure = (192. 44 + 38. 30)/2 = 115. 37 kN/m2
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