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, γ_t = W/V = 1823.8/1000 = 1.82 g/cc

Dry unit weight, γ_d = γ_t/(1 + w)

= 1.82/(1 + 0.1045) = 1.65 g/cc

Void ratio, e = (G_sγ_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 (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, h_cin metre = [0.03/0.0004 (mm)] = 75 m

Capillary pressure = -h_cγ_w = -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 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 = 0.91

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

$k = 2.3\frac{q}{\pi }\left( {\frac{{{{\log }_{10}}({r_2}/{r_1})}}{{h_2^2 - h_1^2}}} \right)$

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

$k = 2.3x\frac{{925}}{{{{10}^3}x60x\pi }}\left[ {\frac{{{{\log }_{10}}(34/16)}}{{{{11.10}^2} - {{9.85}^2}}}} \right]$

= 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

c_v = 6 x 10^-3 x 10^-4 m²/s

T_v = c_vt/H²

= (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_vH²/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/cm². 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 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

a_v = Δe/Δσ = 0.10/0.25 = 0.4 cm²/kgf

c_v = 10 m²/year = 3.17 x 10^-3 cm²/s

k = c_vm_vγ_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:

${S_c} = {H_0} + \frac{{\Delta e}}{{1 + {e_0}}}$

= $\frac{{3.0x0.10}}{{1 + 1.20}} = 0.136\,m$

Method 2:

S_c = m_vΔσH₀

= 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

τ_f = 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.

$\bar A = \frac{{\Delta {u_2}}}{{\Delta {\sigma _1} - \Delta {\sigma _3}}} = \frac{{640 - 445}}{{575}} = 0.339$

$\bar A = AB \Rightarrow A = 0.339/0.95 = 0.357$

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; φ_u = 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
γ_sat = 20 kN/m²
B = 2 m, D_f = 1.5 m
D_w = 1 m
C_u = 40 kN/m2

As per IS 6403 : 1981,
Ultimate Bearing capacity:

q_u = s_cd_ci_cCN_c + s_qd_qi_qqN_q + s_γd_γi_γB_γN_γ

For φ = 0 soils:

N_c = 5.14

N_q = 1

N_γ = 0

Because D_f/B = 1.5/2; s_c = s_q = 1

For strip footing

d_c = d_q = 1

i_c - i_q = 1

Now,

q_u = C_uN_C + γ_effD_fN_q

C_u = 40 kN/m²

N_C = 5.14

D_f = 1.5 m

N_q = 1

We get

${\gamma _{eff}} = \frac{{20x1 + (20 - 9.81)(1.5 - 1)}}{{1.5}} = 16.73\,kN/{m^3}$

q_u = 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)

K_aγZ - 2C√K_a = 0

K_a = (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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Numericals - Geotechnical & Foundation Engineering (Summer 2023)

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