Question

Part 2 A 2 mx 4 m footing is founded at a depth of 0.6 m and supports a vertical load of 400 kN and a horizontal loads of 50 kN and 80 kN. The horizontal loads are acting at 0.9 m above the top of the foundation. The footing is 0.6 m thick and the unit weight of the concrete is 24 kN/m'. GWT 3 m from ground surface. 50 kN 400 KN 80 KN = 4 m B = 2 m (a) Determine if eccentric loading requirements will be met (b) Determine the bearing pressure under Corner 1 (c) Determine the bearing pressure under Corner 2 (d) Determine the bearing pressure under Corner 3 (e) Determine the bearing pressure under Corner 4 (1) Determine the dimensions of the equivalent uniformly loaded footing and compute the equivalent bearing pressure

Answer 1

400/N + qoko 3 - - - - > L-am - B = 2m --> @ wt. of the uncrete = (4x2 x 0.6) 24 - 115.2 KN 2 400kN aty / Transfering the load at the centre of footing, the equivalent diagram is as o so - - - - - - 44 > x 45 KN-m 66 Shown ! - I I '80x0,9 / i = 72kNam Jam compressive stress due to selt wt. and vertical load Ecome = 400 + 115.2 4x2 = 64.4 km m2 LEIL Moment of Inesta of the section Ix = 2X 4 = 32 = 10.67 mt Jy = 4x2 = 2.67 my Bearing pressure under coorner 1 (Assuming compressive stress as positive and tensile stress as negative) 6 t anan

weny JEWUJUIYO . Inngun PL PuebediagnowaH BUSINEUeus USJ12 . AINEEL WILL = 64.4 + 27 +8.43 = 99.83 KN Ima. (C) bearing 2 = pressure under point 2 64.47 7221 - 45x2 2.67 10067 8 = 64.4 + 27 - 8.43 = 82.97 kN/m2 6 Bearing pressure under point 3 & = 64.4 - 2.67 - 10.67 i, 72x1 45XL = 64.4 - 27 - 8.43 = 28.97 KN lm- earing pressure under point 4. on = 69.7 72x) 2.67 Lt 45x2 To 5 = 64.4 - 27 + 8.43 = 45.83 kN/m2 SIRET A IANNITADASI IRINA