### Numericals - Geotechnical & Foundation Engineering (Summer 2023)

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, γt = W/V = 1823.8/1000 = 1.82 g/cc
Dry unit weight, γd = γt/(1 + w)
= 1.82/(1 + 0.1045) = 1.65 g/cc
Void ratio, e = (Gsγwd) - 1
= (2.65 x 1.0/1.65) - 1
= 0.61
Degree of saturation, S = wGs/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 (D10) of 0.002 mm? (4 marks)

Solution
D10 = 0.002 mm;
Using the assumption that the effective pore diameter
= 20% of D10
= 0.2 x 0.002
= 0.0004 mm
Height of capillary rise, hc in metre = [0.03/0.0004 (mm)] = 75 m
Capillary pressure = -hcγw = -75 x 10 = -750 kN/m2

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 Gs = 2.65. (4 marks)

Solution
icr = (G - 1)/(1 + e)
and e = n/(1 - n) = 0.45/(1 - 0.45) = 0.82
Putting values of G, and e, icr = 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. (5 marks)

Solution
See following figure.
Using formula
Now
r1 = 16 m ; r2 = 34 m
h1= 14.5 - 2.2 - 2.45 = 9.85 m
h2 = 14.5 - 2.2 - 1.2= 11.10 m
q = 925/(103 x 60) m3/s
Putting 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 cm2/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
cv = 6 x 10-3 x 10-4 m2/s
Tv = cvt/H2
= (6 x 10-7 x 2 x 365 x 24 x 60 x 60)/64
= 0.5913
Using equation for U > 60%
Tv = 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
Scf = Sct/Ut = 120/0.815 = 147 mm
For U = 90%, Tv = 0.848 (by using  Tv = 1.781 - 0.933log(100 - U%)
t = TvH2/cv = (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/cm2. Calculate the coefficient of compressibility (av) and the coefficient of volume compressibility (mv). If the coefficient of consolidation (cv) determined in the test for the given stress increment was 10 m2/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 = e0 - e = 1.20 - 1.10 = 0.10
Δσ = 0.50 - 0.25 = 0.25 kgf/cm2
av = Δe/Δσ = 0.10/0.25 = 0.4 cm2/kgf
cv = 10 m2/year = 3.17 x 10-3 cm2/s
k = cvmvγw
= 3.17 x 10-3 x 0.18 x 1000 x 10-6
= 5.7 x 10-7 cm/s
Consolidation settlement
Method 1:
=

Method 2:
Sc = mvΔσH0
= 0.81 x 0.25 x 3.0 = 0.135 m

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

Solution
σ = 200 kN/m2
u = 80 kN/m2
σ* = σ - u = 200 - 80 = 120 kN/m2
Shear stress
τf = c’ + σ*tanφ’
= 15 + 120tan300
= 84 kN/m2

Numerical
In a triaxial test, a soil sample was consolidated under a cell pressure of 700 kN/m2 and a back pressure of 350 kN/m2. Thereafter, with drainage not allowed, the cell pressure was raised to 800 kN/m2 resulting in the increased pore water pressure reading of 445 kN/m2. The axial load(was then increased to give a deviator stress of 575 kN/m2 (while the cell pressure remained at 800 kN/m2) and a pore pressure reading of 640 kN/m2. Calculate the pore pressure coefficients B and A.

Solution
For an increase in cell pressure from 700 to 800 kN/m2, the pore pressure increases from the value of back pressure 350 to 445 kN/m2.
B = Δu1/Δσ3 = (445 - 350)/(800 - 700) = 0.95
An increase in total major principal stress from 800 kN/m2 to (800 + 575) kN/m2, produced a corresponding increase in pore water pressure from 445 to 640 kN/m2.

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/m3; cu = 40 kN/m2; φu = 0; c’ = 10 kN/m2; and φ’ = 20°. The natural water table is at 1 m depth below ground level. (8 marks)

Solution
Given that
γsat = 20 kN/m2
B = 2 m, Df = 1.5 m
Dw = 1 m
Cu = 40 kN/m2
As per IS 6403 : 1981,
Ultimate Bearing capacity:
qu = scdcicCNc + sqdqiqqNq + sγdγiγBγNγ
For φ = 0 soils:
Nc = 5.14
Nq = 1
Nγ  = 0
Because Df/B = 1.5/2; sc = sq = 1
For strip footing
dc = dq = 1
ic - iq = 1
Now,
qu = CuNC + γeffDfNq
Cu = 40 kN/m2
NC = 5.14
Df = 1.5 m
Nq = 1
We get

qu = 40 x 5.14 + 16.73 x 1.5 x 1 = 230.67 kN/m2

Numerical
A retaining wall with a vertical smooth back is 5 m high. It supports a cohesionless soil (y = 17 kN/m3, (φ =320). 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.
Depth at which the pressure will be zero (Z)
KaγZ - 2C√Ka = 0
Ka = (1 - sin320)/(1 + sin320) = 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

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