Soil Mechanics Test 8

Concept:

The bearing capacity calculated using plate load test is found to be:

(a) For granular soil:

\(\frac{{{q_{uf}}}}{{{q_{up}}}} = \frac{{{B_f}}}{{{B_p}}}\)

B_f = Width of footing

B_p = Width of plate

(b) In case of clays, bearing capacity doesn’t change with width of foundation. Hence

q_uf = q_up

Explanation:

In clay soil, the bearing capacity doesn’t depend upon the width of footing.

∴ Bearing capacity of footing of side 1 m = Bearing capacity of footing of side 2 m = 2000 kPa

Concept:

Engineering News formula:

Q allowable\(= \frac{{wH}}{{6\left( {s + c} \right)}}\)

where,

w = Weight of drop (kg), H = Height of free fall in ‘cm’, and S = Settlement per blow in ‘cm’

The factor of Safety is taken as 6.

‘c’ is a constant which depends on the type of hammer used.

c = 2.5 cm; For drop hammer.

c = 0.25 cm, For single-acting steam hammer

Note: On single-acting steam hammer, steam is used to lift the hammer and it is always free fall.

Calculation:

W = 25 kN, H = 1 m = 100 cm

Total settlement = 30 mm

∴ Settlement per blow \(= \frac{{30}}{5} = 6\;mm\) = 0.6 cm

For drop hammer, C = 2.5 cm.

Using Engineering News formula,

\(Q = \frac{{WH}}{{6\left( {S + C} \right)}}\)

Concept: 

\({\rm{Area\;ratio}}\left( {{{\rm{A}}_{\rm{r}}}} \right) = \left( {\frac{{{\rm{D}}_0^2 - {\rm{D}}_{{\rm{in}}}^2}}{{{\rm{D}}_{{\rm{in}}}^2}}} \right) \times 100\)

D_o = outer diameter of the core

D_in = Inner diameter of the core

A_r = Area ratio of the core

Calculation:

D_o = 200 mm

A_r = 17%

\(\frac{{17}}{{100}} = \left( {\frac{{{{\left( {200} \right)}^2} - {\rm{D}}_{{\rm{in}}}^2}}{{{\rm{D}}_{{\rm{in}}}^2}}} \right) \times 100\)

\(0.17{\rm{\;D}}_{{\rm{in}}}^2 = {\left( {200} \right)^2} - {\left( {{{\rm{D}}_{{\rm{in}}}}} \right)^2}\)

\(1.17{\rm{\;\;D}}_{{\rm{in}}}^2 = {\left( {200} \right)^2}\)

D_in = 184.90 mm

\({\rm{Thickness\;of\;core\;cutter}} = \frac{{{{\rm{D}}_{{\rm{out}}}} - {{\rm{D}}_{{\rm{in}}}}}}{2} = \frac{{200 - 184.90}}{2}\)

For standard penetration test:

i) Overburden correction (N_o)

N_o = c_n × N

c_n = 0.77 log₁₀ 2000/σ’

where σ’ = effective overburden pressure (in kN/m²)

σ’ = 18 × 2.5 + (20 – 9.81) × (6 – 2.5)

∴ σ¹ = 80.665 kN/m²

\(∴ {{\rm{N}}_{\rm{o}}} = 0.77\left( {{{\log }_{10}}\frac{{2000}}{{80.665}}} \right) \times 25\)

∴ N_o = 26.84

ii) Dilatancy correction (N_e)

N_e = 15 + 0.5 (N_o - 15)

N_e = 15 + 0.5 (26.84 - 15)

N_e = 20.92

∴ Corrected value of standard penetration number is 20.92.

Concept:

Net elastic settlement or immediate settlement for a flexible surface foundation is based on theory of elasticity. The immediate settlement of the flexible footing is calculated as:

\({S_i} = \frac{{{q_n} \cdot B \cdot \left( {1 - {\mu ^2}} \right)}}{{{E_s}}} \cdot {I_f}\)

Where,

S_i = Immediate elastic settlement (Both for sandy soil and Clayey soil)

q_n = Net foundation pressure

B = Width of foundation

μ = Poisson’s ratio

E_S = Modulus of elasticity

I_f = Influence factor which depends on the shape and rigidity of structure.

For Rigid footing, Immediate settlement = 0.8 ×  S_i

Calculation:

Total load = 2000 kN

Area of footing = 2 × 2 = 4 m²

∴ Net foundation pressure \( = \frac{{2000}}{4} = 500\;kN/{m^2}\)

Immediate settlement of rigid footing = 0.8 × S_flexible

\({S_{flexible}} = \frac{{{q_n}B\left( {1 - {\mu ^2}} \right){I_f}}}{{{E_s}}}\)

q_n = 500 kN/m², B = 2m, μ_soil = 0.25, E_S = 6000 kPa, I_f = 0.9

\({S_{immediate}} = \frac{{{q_n}B\left( {1 - {\mu ^2}} \right){I_f}}}{{{E_S}}}\)  \(= \frac{{500 \times 2 \times \left( {1 - {{0.25}^2}} \right) \times 0.9}}{{6000}}\) = 0.141 m 

∴ Settlement of rigid footing = 0.8 × 0.141 = 0.112 m

Width of pile group, (B) = 3 × 0.75 + 0.25 = 2.5 m and m = 0.4

For pile acting individually:

Q_un = n × (m × c × p × L_f)

Q_un = 16 (0.4 × 18 × π × 0.25 × 3)

∴ Q_un = 271.4 kN                               ……. (i)

For pile acting in a group:

Q_ug = c × (4 × B) × L_f + γ × L_f × B²

Q_ug = 18 × 4 × 2.5 × 3 + 15 × 3 (2.5)²

∴ Q_ug = 540 + 281.3 = 821.3 kN        ……. (ii)

Adopting the higher value of (i) and (ii), Negative skin friction = 821.3 kN

∴ Negative skin friction = 821.3 kN

Concept:

Ultimate load carrying capacity of the pile is given by:

Q_up = Q_eb + Q_sf

Q_up = q_bA_b + q_sA_s

Where,

q_b = unit end bearing resistance = 9c

A_b = Base area of the pile

\({A_b} = \frac{\pi }{4}{d^2}\)

d = Diameter of the pile

q_s = Unit adhesion of soil with pile on surface area = αC̅

α = Adhesion factor between clay and pile

c̅ = Average unit cohesion of clay over the depth of pile.

A_s = Surface area of pile = πDL

Calculation:

q_b = 9c

q_b = 9 × 60 = 540 kN/m²

\({A_B} = \frac{\pi }{4} \times {D^2}\)

\({q_s} = \alpha \bar C = 0.5 \times 60 = 30\;kPa = 30\;kN/{m^2}\)

\({A_s} = \pi \times D \times L = \pi \times D \times 13 = 13\pi D\)

\({Q_{up}} = 540 \times \frac{\pi }{4} \times {D^2} + 30 \times \pi \times D \times 13\)

\(FOS \times {Q_D} = 540 \times \frac{\pi }{4} \times {D^2} + 1225.22 D\)

2.5 × 650 = 424.11 D² + 1225.22 D

1625 = 424.11 D² + 1225.22 D

Concept:

As per Terzaghi’s bearing capacity equation –

qu = cNc + qNq + \(\frac{1}{2}{\rm{B\gamma }}{{\rm{N}}_{\rm{\gamma }}}\)

where,

Nc, Nq and Nγ are the dimensionless bearing capacity factors that are a function of ∅ only.

The above formula is derived for strip footing and some modifications are carried out for various shapes of footing.

Modifications in Terzaghi’s Bearing capacity equation:-

a) Rectangular footing :

\({{\rm{q}}_{{\rm{nu}}}}{\rm{\;}} = {\rm{\;}}\left( {1 + 0.3\frac{{\rm{B}}}{{\rm{L}}}} \right){\rm{C}}{{\rm{N}}_{\rm{C}}} + {\rm{q}}\left( {{{\rm{N}}_{\rm{q}}} - 1} \right) + \left( {1 - 0.2\frac{{\rm{B}}}{{\rm{L}}}} \right)0.5{\rm{B\gamma }}{{\rm{N}}_{\rm{\gamma }}}\)

b) Square footing:

qnu = 1.3 CN­c + q(Nq – 1) + 0.8 × 0.5 BγNγ

qnu = 1.3 CNc + q(Nq -1) + 0.4 BγNγ

c) Circular footing :

qnu = 1.3 cNc + q(Nq - 1) + 0.6 × 0.5 BγNγ

qnu = 1.3CNc + q(Nq -1) + 0.3BγNγ

For ϕ < 29°, Local shear failure takes place.

\({q_{safe}} = \frac{{{q_{nv}}}}{{FOS}} + {\gamma _t}{D_f}\)

For local shear failure (ϕ < 29°)

\({C_m} = \frac{2}{3}C\)

\(\tan {\phi _m} = \frac{2}{3}\tan \phi\) 

Using ϕ_m - N_C’, N_q’, N_γ’ would be calculated.

Calculation:

\(\frac{{{D_f}}}{B} \le 1 \Rightarrow \) Shallow footing

Here, D_f = 1.5 m, B = 3m

\(\therefore \frac{{{D_f}}}{B} = \frac{{1.5}}{3} = 0.5 < 1 \Rightarrow \) Shallow footing

ϕ = 24° < 29° ⇒ Local shear failure will govern the design.

\(\therefore {C_m} = \frac{2}{3} \times C = \frac{2}{3} \times 12 = 8\;kN/{m^2}\)

\(\tan {\phi _m} = \frac{2}{3}\tan \phi \Rightarrow {\phi _m} = 16.53^\circ \)

Thus, N_C, N_q and N_γ are corresponding to the 16.53°.

As per Terzaghi, for square footing

q_nu = 1.3 CN_C + q(N_q – 1) + 0.4 Bγ N_γ = 1.3 × C_m × N_C + γ_t D_f × (N_q – 1) + 0.4 Bγ N_γ

= 1.3 × 8 × 12.3 + 16 × 1.5 × (8.2 - 1) + 0.4 × 3 × 16 × 2.4 = 346.8 kN/m²

\(\therefore {q_s} = \frac{{{q_{nu}}}}{{FOS}} + {\gamma _t}{D_f}\)

\(\Rightarrow {q_s} = \frac{{346.8}}{{2.5}} + 16 \times 1.5\)  = 162.72 kN/m²

∴ Gross safe load = 162.72 × Area of footing = 162.72 × 9 = 1464.48 kN

Concept:

From Skempton’s equation:

\({{\rm{q}}_{{\rm{nf}}}} = 5\left[ {1 + 0.2\frac{{\rm{B}}}{{\rm{B}}}} \right]\left[ {1 + 0.2 \times \frac{{\rm{B}}}{{\rm{L}}}} \right] \times {{\rm{c}}_{\rm{u}}}\)

This is applicable for D/B < 2.5 or the factor \(5\left[ {1 + 0.2\frac{{\rm{D}}}{{\rm{B}}}} \right]\) do not exceed 7.5

Calculation:

Given: B = 10 m, L = 15 m and D = 6 m

\(\therefore \frac{{\rm{D}}}{{\rm{B}}} = \frac{6}{{10}} = 0.6 < 2.5\)

\(\therefore {{\rm{q}}_{\rm{n}}}{\rm{f}} = 5\left[ {1 + 0.2 \times \frac{6}{{10}}} \right]\left[ {1 + 0.2 \times \frac{{10}}{{15}}} \right] \times \frac{{15}}{2} = 47.6{\rm{\;kN}}/{{\rm{m}}^2}\)