Question

Q.1 An excavated pit was backfilled by a soil with total volume of 0.5 m? The total soil mass is 1000 kg. The moisture content (w) of the soil is 15% and its specific gravity (Gs) is 2.60. (a) Determine the following soil properties of the soil sample: (i) Density of soil solid, ps (2 marks) (ii) Volume of water, VX (2 marks) (iii) Volume of soil solids. Vs (2 marks) (iv) Volume of air, Va (1 mark) (v) Void ratio, e (1 mark) (vi) Porosity, n (1 mark) (vii) Bulk density, po (1 mark) (viil) Dry density, Pd (1 mark) (ix) Saturated density, Pat (2 marks) (x) Submerged density, Palubs (1 mark) (xi) Degree of saturation. Si (1 mark) (xii) Mass of water, in kg, to be added to the soil to reach to 100% saturation. (2 marks) (6) After calculated the results, your colleague told you that a worker accidentally added 20 liter of water into the 0.5 m backlilling soil. Calculate the following soil properties after adding of the water: (1) new total volume of water, Vw; (1 mark) (ii) moisture content, w: (2 marks) (111) bulk density, Pb; (2 marks) (iv) degree of saturation, S.; and (1 mark) (v) mass of water, in kg, to be added to the soil to reach to 100% saturation. (2 marks) [Remark : 1 liter = 0.001 m ]

Answer 1

D Given data Tolal volume - V- 0.5 m3 Total soil mass 3 m = roookg (solids & water) water content 0.15% = wic Specific grawity of Soil solids. Gis = 2.60. a) (1) Density of Soil Solid Gs = $'s clensity of water = 8 w = 1000kg / m3, 2.66 = 9000 kegles 8s=2600 kg/m3 fi) Volume of water water content wit of water - wit of soil 0.15 = wt of water 1000 - wt of water. 1.15xwt of water = 150. et of water = 130.4347 legs. Cs Volume of th water = wt > 0:130435 m3. SRO 2 CamerSeatter

(6) Nolume of Soil Solids (83) ms Vsi (10oo-130.435) 2600 kg/m3 = Vs Y$ = 0.3344 m3 A volume of air (va) volume of air= Total volume-vol of watera vol of soil Na = 0.15 - 011304 -0.3344 va = 0.03515 m3 (v) Void ratio (es = ts . Vs er 01035156 +0.1304 0-3344 lez 0-495 Scanned with CamScanner

(1) Porosity (n) 0.495 e 140.495 I ne 0.3311 In. 0-331)] fi) bulk density St. Sb= Su Care.so) l 1 (xi) Degree of saturation l Sr= Vw x100. -X100 he Sr = 011304 0.03515+ o.1304 86 = 1000 (2.6 +0.495x(0-7876) 1+0.495 | Sr=78-76% [86= 1999-90 kglms. foy ding density. (S) • degree of Sakuration Sv=o. Sd = f'w (a) - 1000KR Toon X26 (0495 Pd = 1739.130 kg/m3 Scanned with CamScanner

(ix) Saturaled density (Sesat) Ta Sr=1 1 Ssat = Pew Cate) 1000 (2.6 + 0.495) ite Ito.495 I sat = 2070.234 x) Submerged density (Sout) Isub = Psat - Sw. = 2070-234-1000 [ {sub = 1870.234 kglm3 To make 100% Saturation with water, . Va=8-03515 ms. -0.234 Volume of air is filled mass of water = m = Pwx Va => 1000 x0.83815 mass of water - - 35.15 litres.] = kgs. added Scanned with CamScanner

Nolume water 1 6) (1) New Total volume of water (rw) Vw = wit wa o 0.130435 + 20*01001 m3 Vwo LSol Im3 [ho 3 0.150485 m3 7 (1) moisture content (w) W=w.c = We = o sou as moon حي في 01150435 +1000 W = w .c = ws 1000 - 120.43147 W 01173 (u) ciegree of Saturation. (sv) Sva Nowe o o XZ0k12 O'l50435x100 8.880xus o.150435 X100 0.16555 (ii) Bulk density (86) I 86= Bw Care.sr) . e= 0.495 lote - bogo for4957 CS Scanned with CamScanner

8h - (8.9086) 1000 (2.6 of 0.495* 1+0.495 (36= 2039.97 kgima I Mass of water in teg added. to make 100% saturation is. Vaz 0.03515 – 20x01001 va= 0.01515 volume of air is filled with water 2 wit of water > Va.sw. 30.01578x1000 kg/me est of water = (5-15 kgs. Scanned with CamScanner