Question

Q.1 (a) Soil sample has been collected in a compacted embankment at a dry density (Pty) of 1950 kg/m, air void ratio (A.) of 15% and mass of solid (M.) of 3 kg. The value of G, is 2.7. Determine the following: O mass of water (MX) and total mass (M), (4 marks) (i) volume of air (V.), water (V.x) and solid (V.); (6 marks) (iii) void ratio (e), porosity (n), moisture content (w), and degree of saturation (S) and; (8 marks) (iv) bulk density (Phak) and saturated density (Pa). (4 marks) (b) For the same soil in Part (a), determine the moisture content (w) if the degree of saturation (S.) is at 80%. (3 marks)

Answer 1

Given, Sday = 1950 Kg/m3 Av= 15% Aig 0-15V * ty Ms= 3kg Gg = 2:7 Water 0.85V Solution- Soil Solid GT Vs A) Idoy Go Ite Susted 1950 Take Swater 1000kg/m3 x 1000 2.7 Ite 2.0-385 So, Vs = Ms 1.538*163m3 e=Vy Scanned wit/s CamScanner :3 Idoy Igso Vs = 1-538X163 m CS 3

Now, e=Vv Vs 0-385 = Vv 1-538*10 => Vu=509214104 m² -3 Volume of water = 0.85 Vw = 0.85*5.921*10-4 Vw=5.033*104 m3 Mass of water Mw,= fuo *16=1000*5033*100 Mw=0.503 033kg. Total mass MT= Mw+MS = 0.5033+3 => M7= 3.5033 kg. Volume of aio Vair= 0.15*Vv = 0.15*5.921*104 Vair=8-882 ²10 m3 Porosity in => N=0.278 Moisture content, W = Ww 0.5033 *100 =) W = 16-78°/- Ws 3 Degree of saturation S, - *100 = 5.033*10*100 = 85%. 5.921*10-4 Scanned with CamScanner ite 0.385 1+0.385 *100= Vw ICS

ite I+0.385 Bulk density, Soulk Gist se xfu -> Spur= 2.7+(0.85 *0-385) *1000 Soulk= 2185.74 kg/m3 saturated density. Isat= Gste &fw =(2.7 +0.385 5)*lood Isat=2227.44 Kglus Solub Formula , ite Ito-385 6 ليا =Se 0-80 *0:385 = W * 2:7 >> W=0.1141 W=11.41% CS Scanned with CamScanner