**Numerical**

*A 1000 cc core cutter weighing 946.80 g was used to find out the in-situ unit weight of an embankment. The weight of the core cutter filled with soil was noted to be 2770.60 g. Laboratory tests on the sample indicated a water content of 10.45 % and specific gravity of solids of 2.65. Determine the bulk unit weight, dry unit weight, void ratio, and degree of saturation of the sample. (8 marks)*

**Solution**

Weight of soil in core cutter = 2770.60 - 946.80 = 1823.8 g

Bulk unit weight, γ

_{t}= W/V = 1823.8/1000 = 1.82 g/ccDry unit weight, γ

_{d}= γ_{t}/(1 + w)= 1.82/(1 + 0.1045) = 1.65 g/cc

Void ratio, e = (G

_{s}γ_{w}/γ_{d}) - 1= (2.65 x 1.0/1.65) - 1

= 0.61

Degree of saturation, S = wG

_{s}/e = (0.1045 x 2.65)/0.61 = 0.4540 = 45.4%**Numerical**

*What is the theoretical height of the capillary rise and the capillary pressure in fine-grained soil with an effective size (D*

_{10}) of 0.002 mm? (4 marks)**Solution**

D

_{10}= 0.002 mm;Using the assumption that the effective pore diameter

= 20% of D

_{10}= 0.2 x 0.002

= 0.0004 mm

Height of capillary rise, h

_{c }in metre = [0.03/0.0004 (mm)] = 75 mCapillary pressure = -h

_{c}γ_{w}= -75 x 10 = -750 kN/m^{2}**Numerical**

*A masonry dam has pervious sand as a foundation. Determine the maximum permissible upward gradient, if a factor of safety of 4 is required against boiling. For the sand, n = 45% and G*

_{s}= 2.65. (4 marks)**Solution**

FOS = critical hydraulic gradient (i

_{cr})/Permissible hydraulic gradient (i)i

_{cr}= (G - 1)/(1 + e)and e = n/(1 - n) = 0.45/(1 - 0.45) = 0.82

Putting values of G, and e, i

_{cr}= 0.91Now, 4 = 0.91/i

Hence, i = 0.23

**Numerical**

*For a field pumping test, a well was sunk through a horizontal stratum of sand 14.5 m thick and underlain by a clay stratum. Two observation wells were sunk at horizontal distances of 16 m and 34 m respectively from the pumping well. The initial position of the water table was 2.2 m below ground level. At a steady state pumping rate of 925 litres/min, the drawdowns in the observation wells were found to be 2.45 m and 1.20 m respectively. Calculate the coefficient of permeability of the sand. (5 marks)*

**Solution**

See following figure.

Now

r

h

h

_{1}= 16 m ; r_{2}= 34 mh

_{1}= 14.5 - 2.2 - 2.45 = 9.85 mh

_{2}= 14.5 - 2.2 - 1.2= 11.10 mq = 925/(10

^{3}x 60) m^{3}/sPutting all these values in the above formula

= 1.41 x 10

^{-4}m/s= 1.41 x 10

^{-2}cm/s**Numerical**

An 8 m thick clay layer with single drainage settles by 120 mm in 2
years. The coefficient of consolidation for this clay was found to be 6 x
10

^{-3}cm^{2}/s. Calculate the likely ultimate consolidation settlement and find out how long it will take to undergo 90 % of this settlement. (6 marks)**Solution**

H = 8 m (single drainage)

t = 2 x 365 x 24 x 60 x 60 s

c

_{v}= 6 x 10^{-3}x 10^{-4}m^{2}/sT

_{v}= c_{v}t/H^{2}= (6 x 10

^{-7}x 2 x 365 x 24 x 60 x 60)/64= 0.5913

Using equation for U > 60%

T

_{v}= 1.781 - 0.933log(100 - U%)0.5913 = 1.781 - 0.933log(100 - U%)

Or 1.2751 = log(100 - U%)

Or U = 81.5% > 60

Hence, we used correct equation.

Now using formula

S

_{cf}= S_{ct}/U_{t}= 120/0.815 = 147 mmFor U = 90%, T

_{v}= 0.848 (by using T_{v}= 1.781 - 0.933log(100 - U%)t = T

_{v}H^{2}/c_{v}= (0.848 x 64)/(6 x 10^{-7}x 60 x 60 x 24 x 365)= 2.87 years

**Numerical**

*A clay soil, tested in a consolidometer showed a decrease in void ratio from 1.20 to 1.10 when the pressure was increased from 0.25 to 0.50 kgf/cm*

^{2}. Calculate the coefficient of compressibility (a_{v}) and the coefficient of volume compressibility (m_{v}). If the coefficient of consolidation (c_{v}) determined in the test for the given stress increment was 10 m^{2}/year, calculate the coefficient of permeability in cm/s. If the sample tested at the site was taken from a clay layer 3.0 m in thickness, determine the consolidation settlement resulting from the given stress increment.**Solution**

Δe = e

_{0}- e = 1.20 - 1.10 = 0.10Δσ = 0.50 - 0.25 = 0.25 kgf/cm

^{2}a

_{v}= Δe/Δσ = 0.10/0.25 = 0.4 cm^{2}/kgfc

_{v}= 10 m^{2}/year = 3.17 x 10^{-3}cm^{2}/sk = c

_{v}m_{v}γ_{w}= 3.17 x 10

^{-3}x 0.18 x 1000 x 10^{-6}= 5.7 x 10

^{-7}cm/sConsolidation settlement

*Method 1:*

*Method 2:*

S

= 0.81 x 0.25 x 3.0 = 0.135 m_{c}= m_{v}ΔσH_{0}**Numerical**

*Determine the shear strength in terms of effective stress on a plane within a saturated soil mass at a point where the total normal stress is 200 kN/m*

^{2}and the pore pressure is 80 kN/m^{2}. The effective stress parameters for the soil are c’= 16 kN/m2 and φ’ = 30° (4 marks)**Solution**

σ = 200 kN/m

^{2}u = 80 kN/m

σ* = σ - u = 200 - 80 = 120 kN/m^{2}^{2}

Shear stress

τ

_{f}= c’ + σ*tanφ’= 15 + 120tan30

^{0}= 84 kN/m

^{2}**Numerical**

*In a triaxial test, a soil sample was consolidated under a cell pressure of 700 kN/m*

^{2}and a back pressure of 350 kN/m^{2}. Thereafter, with drainage not allowed, the cell pressure was raised to 800 kN/m^{2}resulting in the increased pore water pressure reading of 445 kN/m^{2}. The axial load(was then increased to give a deviator stress of 575 kN/m^{2}(while the cell pressure remained at 800 kN/m^{2}) and a pore pressure reading of 640 kN/m^{2}. Calculate the pore pressure coefficients B and A.**Solution**

For an increase in cell pressure from 700 to 800 kN/m

^{2}, the pore pressure increases from the value of back pressure 350 to 445 kN/m^{2}.B = Δu

_{1}/Δσ_{3}= (445 - 350)/(800 - 700) = 0.95An increase in total major principal stress from 800 kN/m

^{2}to (800 + 575) kN/m^{2}, produced a corresponding increase in pore water pressure from 445 to 640 kN/m^{2}.**Numerical**

*A strip footing 2 m in width, with its base at a depth of 1.5 m below ground surface, rests on a saturated clay soil with γsat = 20 kN/m*

^{3}; cu = 40 kN/m^{2}; φ_{u}= 0; c’ = 10 kN/m^{2}; and φ’ = 20°. The natural water table is at 1 m depth below ground level. (8 marks)**Solution**

Given that

γ

B = 2 m, D

D

C

As per IS 6403 : 1981,

Ultimate Bearing capacity:

γ

_{sat}= 20 kN/m^{2}B = 2 m, D

_{f}= 1.5 mD

_{w}= 1 mC

_{u}= 40 kN/m2As per IS 6403 : 1981,

Ultimate Bearing capacity:

q

_{u}= s_{c}d_{c}i_{c}CN_{c}+ s_{q}d_{q}i_{q}qN_{q}+ s_{γ}d_{γ}i_{γ}B_{γ}N_{γ}For φ = 0 soils:

N

_{c}= 5.14N

_{q}= 1N

_{γ}= 0Because D

_{f}/B = 1.5/2; s_{c}= s_{q}= 1For strip footing

d

_{c}= d_{q}= 1i

_{c}- i_{q}= 1Now,

q

_{u}= C_{u}N_{C}+ γ_{eff}D_{f}N_{q}C

_{u}= 40 kN/m^{2}N

_{C}= 5.14D

_{f}= 1.5 mN

_{q}= 1We get

q

_{u}= 40 x 5.14 + 16.73 x 1.5 x 1 = 230.67 kN/m^{2}**Numerical**

*A retaining wall with a vertical smooth back is 5 m high. It supports a cohesionless soil (y = 17 kN/m*

^{3}, (φ =32^{0}). The surface of the soil is horizontal and carries a surcharge of 20 kPa. Determine the active thrust on the wall. (8 marks)**Solution**

If tension crack occur then -ve area of pressure diagram not considered.

K

_{a}γZ - 2C√K_{a}= 0K

_{a}= (1 - sin32^{0})/(1 + sin32^{0}) = 1/3(1/3) x 17 x Z = 2 x 12 x (√(1/3)

Hence, Z = 2.309 m

Pressure at bottom of soil

= (1/3) x 17 x 5 - 2 x 12 x √(1/3)

= 16.14 kPa

So, lateral active thrust per meter

= (1/2) x 2.69 x 16.14 = 21.71 kN/m

Hence, Z = 2.309 m

Pressure at bottom of soil

= (1/3) x 17 x 5 - 2 x 12 x √(1/3)

= 16.14 kPa

So, lateral active thrust per meter

= (1/2) x 2.69 x 16.14 = 21.71 kN/m

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