From the chart we have

D10 = 0.2

D30 = 0.45

D60 = 1.20

So

Coefficient of uniformity is = D60/D30

Cu = 1.20/0.2 = 6.0

And

Coefficient of curvature of the soil is = D302/(D10 x D60)

Cc = 0.452/(0.2 x 1.2)

= 0.84375

(iv)

From the particle size distribution chart we can see that the particles are distributed over a wide range. So, this is a well graded soil.

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3 (a) At 0 m Total stress= ?= ?d hw +q = 16 (0) + 10 =10 Kpa Pore pressure = u = ?w x (h – hw) = 9.81 (0)=0 Effective stress = ?’= ? – u = 10 – 0 =10 Kpa At 1 m Total stress= ?= ?d hw +q = 16 (1) + 10 = 26 Kpa Pore pressure = u = ?w x (h – hw) = 9.81 (0) = 0 Effective stress = ?’= ? – u = 26 – 0 = 26 Kpa At 3 m Total stress= ?= ?d hw +q + ?sat (h – hw) = 16 + 10 + 20 (2) = 66 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (2) = 19.62 Kpa Effective stress = ?’= ? – u = 66 – 19.62 = 46.38 Kpa At 7 m Total stress= ?= ?d hw +q + ?sat (h – hw) = 16 + 10 + 20 (6) = 146 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (6) = 58.86 Kpa Effective stress = ?’= ? – u = 146 – 58.86 = 87.14 Kpa (b) Unit weight of silty sand = ?s= porosity * specific gravity * 9.81 = 0.54 x 2.61 x 9.81 = 14 kN/m3 Saturated weight of clay =?c= = = 17.82 kN/m3 At 0 m Total stress= ?= ?s h = 14 (0) = 0 Pore pressure = u = ?w x (h – hw) = 9.81 (0) =0 Effective stress = ?’= ? – u = 0 At 2.5 m Total stress= ?= ?s h = 14 (2.5) = 35 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (0) = 0 Effective stress = ?’= ? – u = 35 - 0 = 35 Kpa At 5 m Total stress= ?= ?s h = 14 x 5 = 70 Kpa Pore pressure = u = ?w x (h – hw) = 9.81 x (2.5) = 24.5 kPa Effective stress = ?’= ? – u = 70 – 24.5 = 45.5 Kpa At 9 m Total stress= ?= ?s h +?c (h – 5) = 14 x 9 + 17.82 (4) = 197.3 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (6.5) = 63.76 Kpa Effective stress = ?’= ? – u = 197.3 – 63.76 = 133.54 Kpa Question No. 4 (a) Shear box apparatus The soil is contained in a box which has a separate top half and bottom half. A normal stress is
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