Question

(4). (8 pts). A soil profile is shown in below Figure. A uniformly distributed load Ao is applied at the surface of soil. Given the preconsolidation pressure is between swell index and compression index is: C C/6. The unit weight of water is: -9.38 kN/m. If U-98% we regard it as a completion of primary consolidation. We have taken the clay sample and done the Atterberg Limit test. For the liquid limit test, the result is as below. Please ơ'-926 kN/m. The relationship find: (a) The primary consolidation settlement of the clay layer; (2 pts.) (b) How long will it take to complete the primary consolidation settlement? (2 pts) (c) H ow much precompression surcharge load we need to put for the consolidation to complete within 30 days? (2 pts) (d) If we use wick drain to shorten the drainage distance instead of using the precompression method, what is the equivalent drainage distance needed for the consolidation to complete within 30 days? (2 pts) Note: Please write down the necessary steps/procedures and put the problems in order and box M7-60 kN/m2 Ground Surface H,-2 m 12 kN/m Dry Sand G.W.TV sat 18 kN/m3 H,-3 m Saturated Sand G, 2.75 c, 0.002 cm/s W2-3896 H,-5 m Saturated Clay Impermeable Rock Datum Moisture content (%) Number of blows, N 42.2 15 39.1 28

Answer 1

H_o=5m; e_o=Gw=2.75x.38=1.045

sat/clay

(sat)ctay (2.751.045) x 9.8/(1 +1.045) 18.18kN/m

σ = σ一u = 12 × 2 + 18 × 3 + 18.18 × 2.5-(3+2.5) × 9.8 = 69.55kN/m

$\bar{\sigma}+\Delta\sigma=69.55+60=129.55 kN/m^2$

So soil is pre-consolidated from 69.55 kN/m^2 to 92.6 kN/m^2 and normally consolidated from 92.6 kN/m^2 to 129.55 kN/m^2

To find Liquid limit-

(wi-w2) /log10(N2/M) = (wi-w L)/ log10(25/N1)

(42.2ー39.1)/logo(28/15) = (42.2-wL)/logo(25/ 15)

W1 = 39.66

W1 = 39.66
%

Cc = 0.009 × (WL-10)

Cc = 0.009 × (39.66-10)

$C_c=0.267$

$C_r=C_c/6=0.267/6=0.0445$

a). Consolidation Settlement-

$\Delta H=C_r H_olog_1_0(\sigma_c/\bar{\sigma_0})/(1+e_o)+C_c H_olog_1_0(\bar{\sigma_0}+\Delta\sigma)/\sigma_c//(1+e_o)$

0.0445x 5x logo(92.6/69.55)/(1+1.045)+0.267x 5x logo(129.55/92.6)/(1+ ΔΗ 1.045)

$\Delta H=0.0135+0.0952 = 0.1087m=108.7mm$

b). Time to consolidate  

$(T_v)_9_8=c_v\times t_9_8/d^2$

t98-1.5 × 5002/0.002= 5.95years