Page 1 : 1+e, G+Se, 1+e, Se = wG, 4., G+e, GYw, sat, 1+e, 6., 1+e, G-1, (1-n, G Yw, Yd =, 7., 1+ e, 8., 1+ wG, 9., 10., S =, 1, E(1+ w)-, Example 2.1, A soil sampler of volume 1000 cm3 is used to collect soil samples. It was, found that sampler contain 2 kg soil with dry unit weight of 1800 ka/m3. If 300 g water is mixed to the, soil, then what will be the water content of the sample?, Solution:, Yo = 1800 kg/m3, Dry weight of soil (weight of soil solids),, W, =, = 1000 x 16x 1800 =, 1.8 kg, But actual weight of sample,, W = 2 kg, : Weight of water.before mixing additional water,, = W- W = 2- 1.8 = 0.2 kg, After mixing 300 g of water, the total weight of water would be, = 0.2 + 0.3 = 0.5 kg, W =, x100, Ws, 0.5, x 1000 27.8%, 1.8, Thus, water content,, Example 2.2, having volumes and weights in cc and g respectivety., Consider the folļowing statements:, 1. Soil is partially saturated with dégree of saturation, Figure shows a block diagram for a soil sample,, ·W;0., Air-, v,= 0.15, greater than 60%, 2. Void 'ratio = 100%, 3. Water content = 40%, 4. Saturated unit weight = 2g/cc, Ww= 0.4, Water, Vw= 0.35, Ws = 1.2, Soil solid, Vs = 0.5, Which of the above statements are TRUE?, (b) 1, 2, 3 and 4, (d) 2 and 3, (a) 1, 2 and 3, (c) 1, 2 and 4, MADE EASY, O Copyright, www.madeeasypublications.org, Scanned by CamScanner

Page 2 : 18 |, Foundation Engineering, MADE EASY, Civil Engineering, Soil Mechanics and, g enaes Publications, mi, Ans. (c), 0.35, 1. Degree of saturation,, S =, x100%3D, x100 70%, (0.15 +0.35), 0.15+0.35, 2. Void ratio,, x 100 100%, e =, %3D, 0.5, 0.4 *, Water content,, w.x:100, 1.2, = 33.33%, 4. Saturated unit weight,, G+Se, Yw, Ysal =, I kinn, Se, 1x1, G, = 3.0, 0.333, (), 3+1x1, x1= 2 g/cc, Ysat =, 1+1, Hence option (c) is correct,, Example 2.3, An oven dry scil has mass specific gravity of 1.5 g/cc. If bulk density of soil in, its natural state is 2.0 g/cc, then the water content of soil in natural state will be, (а) 50%, (c) 100%, (b) 25%, (d) 33.33%, Ans. (d), For oven dry soil,, Mass specific gravity,, Gm, = 1.5 g/cc, %3D, Yw, Yd, 1.5 g/cc, Given, bulk density,, 2.0 g/cc, %3D, Using,, Yd, %3D, 1+ w, 2.0, 1.5 =, 1+ W, 2:0, 1= 33.33%, -1.5, Hence option (d) is correct, Example 2.4, Consider the phase diagram of the soil given below:, The soil is completely saturated, The specific gravity of soil solids is 2.6 (take unit weight of water, Ww= 20 gm, Water, as 10 kN/m3)., Match List-1 (Physical properties of soil) with List II (Corresponding, Soil solid, Ws= 100 gm, values) and select the correct answer using the codes given below., Scanned by CamScanner

Page 3 : MADE EASY, emvet m Publications, "Properties of Soils 19, List-l, 1. Water content, 2. Void ratio, List-II, А. О.34, 3. Porosity, 4. Saturated density, В. 0.52, C. 20.53, Codes:, D. 20%, 1, (а). : А, (b), (c), (d), 3, O : B, В, :D., D, C, C, A, В, A, Ans. (d), 1. Water content,, Ww, x 100%3D, W., 20, x100% 20%, 100, W =, 2. Void ratio,, wG, 0.2 x 2.6, e =, = 0.52, 1, 3. Porosity,, 0.52, e, n =, = 0.34, 1+e, 1+ 0.52, G+Se, Yw, 1+e, 4. We know,, Y, =, For saturated density,, G+1xe, 2.6+0.52, x 10 = 20.53 kN/m3, Ysat =, Yw =, 1+e, 1+0.52, Hence option (d) is correct,, Example z.5, Soil sampie A and B have void ratio of 0.5 and 0.7 respectively. If 1.5 m3 of, soil sample A and 1.7m3 of sample Bare mixed to form sample Chaving a volume of 3.2 m3, which one, of the following correctly represents the porosity of sample C?, (а) 50%, (c) 100%, (b) 37.5%, (d) 33.33%, Ans. (b), Soil A(1), Soil B(2), Soil C(3), e2 = 0.7, 2 = 1,7 m3, V3 = 3.2 m3, e, = 0.5, V, = 1.5 m3, %3D, Even after.mixing, volume of solids of each soil will be same in mixed sample, 1.5:, The solids in soil Ä,, V., S1, 1m3-, 1+e,, 1+ 0.5, V2, 1+ e2. 1+ 0.7, 1.7, 1m?, The solids in soil B,, %3D, Vs, + Vs, = 2 m3, V, = V3 - Vs, Total volume of solids in soil C,, = 3.2 -2 = 1.2 m3, .. Volume of void in soil C,, %3D, 1.2, = 0.375,, 3.2, .. Porosity,, V3, Hence, option (b) is correct,, MADE EASY, © Copyright., www.madeeasypublications.org, ovma Pablications, Scanned by CamScanner

Page 4 : Which one of the following is the water content of the mixed soil made from, MADE EASY, 20. 1 Civil Engineering • Soil Mechanics and Foundaion Engineering r, Pubiications, Example 2.6, 1 kg of soil (say A) with water content of 100% and 1 kg of soil (say B) with water content of 50%2, (b) 71%, (d) 82%, (a) 66%, (c) 75%, Ans. (b), Soil C(3) , .., W,, Soit B(2)., W2 = 1 kg, W, = 50%, Ww, Soil A(1), W; = 1 kg, .. :, Wg =, W, = ?, w, = 100%, Ww, sp, x 100, W., Water content,, W =, 100, wit, Ww + W;, +1 =., +1\W, = (w + 1)W;, W =, %3D, 100, Ws, 100, Weight of solids in soil A,, 1000, WSA, = 500 g, 1+1, WwA, = 1000 – 500 = 500 g, Weight of solids in soil B,, 1000, = 666.7 g, W., (0.5+1), WWB, %3D, 1000 - 666.7 = 333.3 g, Ws, + Ws. = 500 + 666.7 and = 1166.7 g, SB, SA, + Wwa, WwA, Ws., mix, W., = 500 + 333.3g 833.3 g, w, mix, Ww., W =, Ws.mix, w.mix, 833.3, x 100 = 71%, 1166.7, Example 2.7, A soil sample has void ratio of 35%. The specific gravity of solids is 2.7., Calculate the, (i) Porosity, (ii) Unit weight if the soil is 75% saturated, (ii) Dry density, (iv) Unit weight if the soil is submerged, Solution:, Given, Void ratio,, .e = 35%, G, = 2.7, (i) We know, Porósity,, e: :, n =., 1+'e, 0:35, 0.259, resp, n =, %3D, 1+ 0.35, Wha, G, +Se, Yw, 1+e, (ii) We know,, Y, =, For dry density,, S = 0, G, Yw 2.7 x 9.81, Y =, 1+e, 1+ 0.35, = 19.62 kN/m3, O Copyright, MADE EASY, www.madeeasypublications.org, a aen m Puhlicetions, Scanned by CamScanner

Page 5 : MADE EASY, ur a uiication:, Properties of Soils 21, (ii) When soil is 75% saturated i.e. S = 0.75, (G +Se, Y =, (2.7+0.75x0.35, x9.81 = 21.53 kN/m3, Yw =, 1+e, 1+ 0.35, (iv) We know,, Ysub = Ysai- Yw, (2.7.- 1), Yw% =, 1+ e, (G + 1e, x 9.81 12.35 kN/m³X, Yw - Yw =, 1+e., A sampler with a volume of 60 cm3 is filled with saturated soil sample. The, specific gravity of soil solids is 2.65. When the oven dry soil is poured into a graduated cylinder filied, with water, it displaces 40 cm³ of water WNhat is the natural moisture content and dry unit weight of soi?, Example 2.8, Solution:, Volume of soil sample = Volume of sampler, V = 60 cm3, When soil sample is poured into graduated cylinder, it displaces 40 cm3, Air, of water, Water, Vy, V, V, = 40 cm3, V = V, + Vs, V, = V- V. = 60 – 40, = 20 cm3, Volume of soil solids,, Total volume,, Soil solid, Vs, V 20, = 0.5, Void ratio,, e =, 40, e = 0.5, S= 1 and, Se = wG,, Now, we have,, G = 2.65, Using, 1x 0.5, Se, =, G,, = 18.86%, Moisture content,, 2.65, We know,, YA =, 1+ w, where,, Y, = bulk unit weight, 2.65+1x 0.5, G+Se, Y, =, Yw =, x1 = 2.1 g/cc, 1+e, 1+ 0.5, 2.1., = 1.76 g/cc., Hence,, 1+ 0.1886, Example 2.9, The void ratio and specific gravity of a sample of clay are 0.73 and 2.7, respectively. The voids are 92% saturated. Find the bulk density, dry density and the water content, What would be the water content for complete saturation, the void ratio remaining the same?, Solution:, Given, e= 0.73, G= 2.7 and S= 92%, We know that,, 2.7+0.92 x 0.73, x1, G+Se, Pw =, Bulk density,, 1+0.73, 1: Pw = 1 g/cc], 1+e, P, = 1.948 g/cc, MADE ERSY, © Copyright, www.madeeasypublications.org|, Scanned by CamScanner, ミ ー