Question

Figure P3 shows the soil profile at a site for a proposed office building. It is expected that the vertical stress at the top of the clay will increase by 150 kPa and at the bottom by 90 kPa. Assurning a linear stress distribution within the clay: (a) Calculate the consolidation settlement. You should divide the clay into five layers, compute the settlement for each layer, and then find the total settlement. Assume G = 2.7 for both sand and clay soils. (b) An oedometer test on sample of the clay revealed that the time for 90% consolidation on a 20 mm thick sample occurred in 40 minutes. The sample was drained on the upper and lower boundaries. How long will it take for 50% consolidation to occur in the field? **235 Fig. P3 Office building on soft clay foundation

Answer 1

Given Increase in vertical shers at the top of the clay (AT) top = 150 kpa 150 kN/m² Increase in vertical stress at the bottom of the clay (so) Bottom = 90 kpa. G = 2.2 (for bot and figu w=23% s=90% AT sandra -V * 1 -- Ilom. W-40% - May ay - - C -0-3 OCR=1 now. for sand: es=cG ex quo = 3x 2.4 e = 0.69 sont = docte] Loo Now take dw = 10 kN/m3

gand = 102-4+ (0.69x0.90)7 +0.69 - 19.650 kN/m3. for clay :- es = w G ot it clay was fully saturated so $2100% means salon ex1 = 0.40x2.7 e = 1:08. pat = 8w lates lte STEP to 82.7+(1.084) 14 1.08 I : = 18.173 kN/m3) . Now Divide the clay into 5 Layer. Each Layer heigth = am.. stress due to and The vertical the on 6 top of the Dan * 5 clay = / 3 - 19.650*5 = 98.25 = 98.28 kr/m2

v for the Now we have to find out Consolidation settlement for each layer (Ham) v for the list Layer :- (H=2m) vertical stress a 98.25 + 8 sat 42 298.25 +0s. 173+2) 2134.196 kn/m2) Effective Stress & 134.196 - 8 w x 2 = 134.196-20' 114.596kN/m2 Increase in stren (66) = 150 kPa = 150 kr/m² (olt (Dr) top J (CH), = C.H. log lte, = 0:3* 300 Log (1locas ) (114.596 + 150 1+1.08 114.596 (OH), = 0.10m verdere for the second Layer (H = 2m2 vertical stress = (18.17 342) 236.346 kN/m2 Effective stress = (36.346) -(1042) . d = 16.346 kr/m2 n e

Increase in stress (AT) 298.25+ (18.17 342) -(10x2) +150 Z 98.257 = 264.596 kN/m (AH)₂ = cc. H log (olt bol ) 20.3 x 20l (16-346 +264-376 1+1.08 16.346 20:356 m for the 3rd Layer aprove vertical stress = (18. 173 x 2) = 36.346 kr/m2 Effective stress – B 16.346 kN/m2 al Pot7 Increase in stress is in (or) – 98.25 + 8 sattu) towх4) +150. 298.25+ (18.173 x 4) -(10x 4) +150 - 286,85 kN/m2 (AH)= cc. H . log (o'tool) autodete eriala 30-3* 2 ba 16-346 + 286.55 tl.08 76.346 (OH)3 = 0.365 m

for yth Layer (H=2m). vertical stress = (18.173x2) = 36.346 kr/m2 Effective stress - als 36.346- 10x2 = 16.346 KN/m Increase in shren 298.25 + (18.17 3x6) -(10x6) +150 sol = 297.288 kN/m2 lte = 0.3 x 2 log (16.346 + 297.288) S itios 16 346 = 0.370 m for 5th hayes (H=2m) vertical swey = 36-346 kw/m ) Effective stress - ol = 16.346 kr/m2 Increase in shers (An) – 98.25 + (18.193x8) -(10x8 ) +150+ 90 = 403.634 KN/m2

Paula che () logo sol lte = 0.3 x 2 1 +1.08 log (16-346+ 403.634 ) 16.346 () = 0.406 Total settlement = (DH) + (OH)2 + (OH)₂ + (04) $ (PH) = 1.597 m (6) Given u= 90% d=20mm t = 40 minutes. and have double drain. for u= 50% t=?. for u=90% (Tu) = 1.781-0.933 log, (100-0%] = 1.781 -0.933 log10 (10] - 0.848 for u= 50% (Tv) = 7 (1) (Tue 7 (5 )

(T) = 73 x 7 z = 0.196 we know that Tu= Cut C = Constant, d - same. unora Tv a t lu o li letos Tui tu TV2 0.848 - 40 0.196 sta novio ta = 9.245 minutes. instaltuob.sund bare op -