Question

Question 2 [Marks: 15 +10=25] 200kPa of construction stress was applied on top of the soil layers as shown in the figure below. (1) Calculate the settlement in the clay layer due to the construction (minimum of 2 sub layers is to be used). (W) If the soil went through 90% average degree of consolidation in 1 year, calculate the coefficient of consolidation (C.). 个 Sand Y = 20kN/m 4m Water table Clay (over consolidated) y = 16kN/m 100 4m bH Dense sandy gravel y = 22kN/m3 6m Relevant equations and charts: H Sc = -X Cex log σ + Δσ' di 1 + e H 0 + 40 Sc= -X C, X log 1+e 0 svolle spaw01 bora nosolo oc Sc H 1 + e Cs x log| C, 108(720)] Cyt T = Page 4 of 7

Answer 1

Giveno N = 20 kulm3 Sand 4m Gm -- } clay coc) &q= 16 kn 1 m3 Go = 0.4 ; CS=0.08 1 er 1.5m 14m = 100 kpa ? 6c Ng = 22 kN lm3 6m ow Flo KM m3 wasslemed Construction Slyesy (9) = 200kpa H = thickness of day layer = 4m ccaurent effective shress Cod); at middle of clay (Sol) om 1 x 4 ) + ( x3) (20x4) + (16-10]x2 80 +12 - 92 Kalme 7 OCR 100 >1. COc clay) १३ 7 ' - 6'+9 20045 92 + 200 = 292 kylm?

คะ if Gol cool cool tool Gola Gol cool coc clay) १५ ८ Then the formula is 100 <292 Se .th - Csxlog ( 2 ) Celos ( 1 Ite: & Se a 14 1+115 0.08 dog ( 100 ) +0.4109 (392 १२ 100 - 3 9.89697 X10 + 0.18615 318515 9.5 ~ 4 x 0.189047 205 0.3024752 m od 30.25 cm; Settlement of clay is (se) = 30.25 cm)

ii) A: Degree of consolidation (0) = 904. t t = 4 year Cv=q Tv = cut MEH; Cone-way) Har Har=h 2. Tv .(9-way drainage] 1.784 -0.933 log (100-014.)); 02604 = 1.781–0-93310g (100-90) = 0.848 U=90% ; Tv=0.848; Har = T 4 2 having Elor (our clay Sand layers on top and bottom), – So, So, H = Haber dy for g oc- clay; sand H = 4 4 = 9 m; १ og eoli Har q Cv Ту х Har TA Sandy graue t 2 0.848 XC) Copy = 3.392 melyear 4 : coefficient of consolidation (Cu) = 3.392 milyears
rResult:- The primary settlement of OC clay having thickness 4m is (Se)= 30.25 cm approximately;

The co-efficient of consolidation of a clay(Cv)= 3.392 (m^2)/year