Question

Please, show step by step and clearly. problem 9.17, (Use 3 equally divided sub-layers for the 15 m clay layer). thanks

please, resolve question 9.17 (Use 3 equally divided sub-layers for the 15 m clay layer)

9.16 The soil profile and the loading condition are given in the followin n) figure. Total new footing stress including the foundation at z 2 is 400 kN/m2. The soil profile has a rather thick clay layer (15 so that the layer should be divided into several sublayers to es the consolidation settlement adequately. Divide it into three sub ers and compute the primary consolidation settlement at the point of each layer, and then make a summation for the tota settlement. Assume that the clay is normally consolidated a Newmark's rectangular footing solution for the stress increment com. putation. Assume also that the soil's weight for excavated footine depth and the weight of new footing are nearly the same. estimate mid- l final and use New footing load 4m×4m square footing 0 Sand Ydry 19.0 kN/m3 Sand Ywet 19.4 kN/m3 5 10-Li... Clay Ywet 18.5 kN/m Clay, H 15 m eo= 0.78, LL-38, PL=14 20 Gravel z (m) 9.17 For the same soil profile and the loading condition as in Problem 9.16. the e-log ơ curve of the clay specimen is obtained by the following leme laborator in regular σ of this clay by dividing it into three subla hat eo valu y test. The second figure is an enlarged version of the first one scale. Compute the final primary consolidation settlement yers as in Problem 9.16. Note e given in Problem 9.16 figure is not assigned in Problem 9.17 0.8 0.79 0.78 0.77 0.76 0.75 0.74 0.73 0.72 0.71 0.7 10 100 1000 log ơ (kPa) 0.78 0.77 0.76+1 60 80 100 120 140 160 180 200 ơ (kPa)

Answer 1

Yau 21 h' 의 C. 11ニ38 C. 20 L с-с 28.44 N 2 20 а-а, 4+x 5.5 y t 2x 10.5) c-c u00 ких у _ 5.22.wlM (4+2x15-5)し 0

0.419 125.49 151 595 190,025 9부.05 0.145 0.14 141 355 b-b 0-169 c-c' '84,805 AH it Co 0.003 x 5 ΔΗ2: 0:001 5-2-814 x ΔΗス l®부부0 total pus.cant, "oeLHount.こ8.432 + 208,チ+ 2eG25 그, 14:044