Question

Q1. A cubical sample of Chongming Dongtan Bay mud (a soft marine silty clay) has been prepared for testing in a true triaxial device. The dimensions of the sample are 76 mm by 76 mm by 76 mm. The sample has a moisture content of 68.5% and a moist density of 1.44 g/cm”. The average specific gravity of solids is 2.55, Determine: (i) Void ratio (ii) Degree of saturation (iii) Moist density if the soil becomes saturated at the same total volume (iv) Moisture content if the soil becomes saturated at the same total volume (v) The density of the soil ifall are fluid is dried off with no change in total volume.

Answer 1

Solution:

Given that : The dimensions of the sa Sample are 76 mm by 76 mmby 76 mm. the sample has a moisture content of 68.54 Moist density of 1.44 g/cm3 The avg specific gravity of solids is 2.55 Here, لب لبا" MW x 1001. I x 100 y. M3 the moisture content is given as (Wave so bi Moist density is written as follows MS+MW mscltw) P-M <!3 2 of Thub, MJ = fu ļ mw= wwis itw → Recalling the General definition for the specific gravity of solids, the volume solids is computed from eas as follows. Gs = 45 VS SW Gs5w hence the volume of voids Vy-v-ug, the void ratio is M3 >) 19 Ms e Valls US 15 -1 15 VG → volume occupied by the pore fluid is obtained from the mass of the fluid chere assumed to be water Jaccording to Iw Mw د ل (= Hw Sw

of needed, volume occupied by air is, Vahe-vw Now, j3 * total volume of cubic Sample is v=C76 mm)3 cm 11000 mm V- 4.390x104m3 → moist density is converted to units of Kg meters f = (1.4497cm3) ckg!! 1.440 x103 kg/m3 of solid phase ib, (1.440x103Kgm3) (4.390x10403 ) 11000 g) (100 cm Im) p * Mass MS Pv Itw H0.685 M5, 3.752% kg Mass of fluid lol phase is computed as - WMS - (0 685)(3752 x to'kg ) 2.570x16' kg Mw ,اياهم total Mass M = Myt Mw M = (3.452x16kq) + C2-570x1ö kg) د الامام => 6.320x 10 kg (3.7528 101kg) M5 Go Sw (2.55 ) (1000 => 1.471x10 m3 kg 1m3)

Mw SW (2.570x10'kg) C1000 kg/m3) 2.570 x104m3 Vy = V-Vy = (4.390x lö4m3) -(1471x164 m3) -> 2.919x104m VW 3 Va : Vy-w Va = Vv-vw- (2.919x104m)-62.570x104m3) => 3490x105 m3 NOW. for (i void ratio cel 3 e VO 2.919x10m 1.471 x 10 m3 => 1.984 : Moid ratio (e) a1.984 for civi- degree of Saturation s. (vw Gus)x1001 - 2570110Yen? X100/ 2.919x104m3 =) 88.07 degree of sats 88 OY forcii); → If the soil becomes saturated at the same total volume, then the volume occupied by the fluid phase is equal to volume of voids

Mass(x10'kg] volume (alol cubic metu) fluid 4.390 * 1.421 solid 2.919 2.919 * 3.752 6.322 1 x 10-4 m3 MW بي بي 3 Thus, ''w - Vuo 2.919 xlo mass of water contained in the voids is Clooo Kg Im3)(2.919x10 403) 2.919x 10 kg Moisture content wa (mo (mw )xico y (2.919x10kg 3.752 xo'kg mw Resulting mg X100/ 79.87 mstmw a 3 * Associated moist density is (3.752x10 kg) + (2.919x10kg) (4.390x104m3) 3 - 1520 kg/m3 - 1.520 mg/m3 > moist density can can likewise be computed from moisture content that is, sams (ltw) (3.752x10' kg )(1+0.778) 14.390 x104m3) 9-1520 kglm3 (1.520 mg im3 which veelfig the Previous result for civir here, w. 87.84. {calculated above

for cv): >If all pore fluid is dered off with no change in total volume, the moist density will be equal to the dry density that is fald Idams -) 3.752x10'kg 4.390 xlom? co 854.6 kglm3-0.855 mg lim? Below the results: e- ☺ void ratioces 1.984 cii) degree of saturation 3 = 88.01. (iii) moist density f = 1520 kg / m3 or fa 1.520 mg (m3 (V) moisture content wa W density of soll's f. 854.6 kg lm3 = 0.855 mg 1 m3 77.81