**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. (AMIE Summer 2023, 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? (AMIE Summer 2023, 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 m

Capillary 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. (AMIE Summer 2023, 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} = 0.91

Now, 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. (AMIE Summer 2023, 5 marks)**

**Solution**

See following figure.

Using formulak = qlog(r_{2}/r_{1})/[r(h_{2}^{2}-h_{1}^{2})]

Now

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

h_{1}= 14.5 - 2.2 - 2.45 = 9.85 m

h_{2} = 14.5 - 2.2 - 1.2= 11.10 m

q = 925/(10^{3} x 60) m^{3}/s

Putting all these values in the above formula, we get

k = 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. (AMIE Summer 20923, 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}/s

T_{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 mm

For 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. (AMIE Summer 2023)**

**Solution**

Δe = e_{0} - e = 1.20 - 1.10 = 0.10

Δσ = 0.50 - 0.25 = 0.25 kgf/cm^{2}

a_{v} = Δe/Δσ = 0.10/0.25 = 0.4 cm^{2}/kgf

c_{v} = 10 m^{2}/year = 3.17 x 10^{-3} cm^{2}/s

k = 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/s

Consolidation settlement

*Method 1:*

_{c}= H

_{0}Δe/(1 + e

_{0})

*Method 2:*

_{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° (AMIE Summer 2023, 4 marks)**Solution**

σ = 200 kN/m^{2}

u = 80 kN/m^{2}

σ* = σ - u = 200 - 80 = 120 kN/m^{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. (AMIE Summer 2023)**

**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) = 0.95

A* = Δu_{2}/(Δσ_{1} - Δσ_{3})

= (640 - 445)/575 = 0.339

Now, A* = AB

Hence, A = 0.339/0.95

= 0.36

**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. (AMIE Summer 2023, 8 marks) **

**Solution**

B = 2 m

D = 1.5 m

D = 1 m

γ_{sat} = 20 kN/m^{3}

C_{u} = 40 kN/m^{2}

φ_{u} = 0

C’ = 10 kN/m^{2}

N_{c} = 5.14 for φ_{u} = 0

φ = 20^{0}

For φ_{u} = 0

q_{u} = C_{u}N_{c} + q’

q_{u} = 40 x 5.14 + [γ x 1 + γ’ x 0.5]

= 205.6 + [20 x 1 + (20 - 10) x 0.5

= 231 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. (AMIE Summer 2023, 8 marks)**

**Solution**

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

K_{a}γZ - 2C√K_{a} = 0

K_{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

**AMIE/BTech/Junior Engineer exams**is available at

**https://amiestudycircle.com**

## Comments