Question

B. The moist weight of 0.028 m3 of a soil is 52 kg. The moisture content and the specific gravity of the soil solids are determined in the laboratory to be 11% and 2.7 , respectively. Calculate the following: Moist unit weight (kg/m3) i. Dry unit weight (kg/m3) ii. Void ratio d. Porosity iii. iv. Degree of saturation (%) Volume occupied by water (m3) v. ngle C. For the soil sample in A (above) represent the following on a soil phase diagram: i. Mass of water ii. Mass of soil s0 c3 iii Volume of voids Volume of water iv. Volume of solids V. D. The following data are given for a soil: Porosity: 0.4 Specific gravity of the soil solids: 2.68 Moisture content: 12 % Determine the mass of water to be added to 12 m3 of soil for full saturation.

Answer 1

(6) Moint weight of 0.028 m3 of a soil is sakg Moist wie Correr -1 Speutic Gravity of soil eolica q = 0.40 is Morat unit wu'y hit V=W wo 50 kg; V=0.028 m? 52 - 1857.14 kglm? 0.028 Day wnit weight Vdir 1857.14 +1673.09 kglm? ) It w 10.11 vrid natio Nd = Gsrw Vw = 1000 kalm? S te 1643.09 = 2.70 x 1000 ite --> 1+1613 le = 0.6131 porosity D e - 0.613 I te 140.613 JO= 0.38 d'") e = wa => 0.0130 2 .10 W S = 0.11 2.70 0.4845 o 48:45.). 0.613 Degree of latuation for 4.8.45:). nawet > 0.38 = V -> Vv=W.38*V V = 0.38% 0.028 -0.01064 m wou -> 0.4845 - VW VW=0.4845V = 0.48454 0.01064 Yu 0.005 15 ml Sy

w=0.00515 m3 volume occupied water c) Vw= ww => 1000 kglme work → Ww= 5.15 kg is weight of water = 5115kg ii) Mais of soil => We = 1673.09% 0.028 we=46.84 kg W volume of voids V 0.01064 m volume of water Vw = 0.00515 Total volume vo Vyt Vs → 0:028 = 0.01064+ Vs Vs = 0.0173673 V = Va t ww -> volume of air va = VV-Vw = 0.01064-0.00515 Va = 0.00549 m? Weight of air of Air volume of lair= 0.00549m? Tur=0.01064 VA Total weight of -Water volume of water = 0-00 575m? water ww5.5k wusht Vw weight of 1 soil solid volume = V Total solidws = 46.84159 ii in I volume of soled us 0.01116

D) porosity n=0.4 G = 2.68 Moratue content =121. ne = 0.42 € 1+ e I te 0.4 +0.4e=e -> 0.4 = 0.62 Te0.G vd = Go to vw=1000kg/m3 Ite *a = R G x 1000 16 6 4 11 (%* 1+0.67 catwatequitwerght "sat = (478) Yw for fuul Saturation Sro 10001.or! Nsat - Gate) = 49.68 +0.6+) x2000 H0.67 Ysat = 3.35 x LOOD 1.67 . VE WS Ysat = 2005.98 kglas? WS -> W= 1604.19x12 Ws = 19257.48 kg V = 12m3 Nsal = Weat => Wsat = 2005.98 x 12 = 24071,76 kg Vo weight of water to be added - Weat - Wsolids = 24071.76 - 19257.48 = 4814.28 kg