Soil Mechanics Test 2

Concept

\(Relative\;density\left( {{I_D}} \right) = \frac{{{e_{max}} - e}}{{{e_{max}} - {e_{min}}}} \times 100\)

Where, e is natural void ratio,

e_max is void ration in loosest state,

e_min is minimum void ratio (i.e. Densest state)

Also,

e × S = w × G

S is saturation, G is specific gravity of soil

Calculation

Given,

e = 0.6, e_max = 0.9, I_D = 60%

S = 100%, G = 2.65

i) \(Relative\;density\left( {{I_D}} \right) = \frac{{{e_{max}} - e}}{{{e_{max}} - {e_{min}}}} \times 100\)

\(60 = \frac{{0.9 - 0.6}}{{0.9 - {e_{min}}}} \times 100\)

e_min = 0.4

ii) Water content for densest state is

\({e_{min}} \times S = w \times G\)

0.4 × 1 = w × 2.7

w = 0.148

= 14.81%

Statement –I- True

An isobar is a line joining all the points of equal vertical stress below the ground surface.  Boussinesq’s theory can be used to prepare vertical stress isobar diagram.

Statement –II- False

In Equivalent point load method, entire loaded area is divided into small area units and load on each area unit is replaced by equivalent point acting at the centroid of each are unit.  Since, it is an Equivalent   approximate method for determining stress distribution for loaded areas, it can be never accurate.

Note: Accuracy of this method depends on the size of the area unit is considered. Lower the size of area unit higher will be the accuracy.

Statement –III- True

Newmark developed influence chart to compute the vertical stress due to a loaded area of any shape, below any point either inside or outside of the loaded area. Influence chart consists of concentric circles (m nos.) and radial lines (n nos.) which divide the influence chart into no of area units, where influence of each area unit at centre of chart is same.

Statement –IV-True.

Westergard’s has made following assumptions to find the stress distribution in soil mass.

1. Soil mass is homogenous and semi-infinite.

2. Soil mass is non-isotropic i.e. it is applicable for stratified soils.

Statement I – True and Statement III- False

Factors affecting the permeability of soil can be studied using the following equation.

\(K = \frac{1}{Z}\frac{{{e^3}}}{{1 + e}}\;\frac{{{\gamma _w}}}{\mu }\;\frac{1}{{{S^2}}}\)

Where, Z = constant; μ = dynamic viscosity of water;

S = specific surface;  γ_w = unit weight of water.

Now, for a given void ratio (e) or porosity:

\(K\; \propto \frac{1}{{{S^2}}}\) i.e. Permeability of coarse grained soil is inversely proportional to the specific surface.

Also:

\(K \propto \frac{1}{\mu }\)

On increasing the temperature, viscosity of water (μ) reduces and hence permeability increases.

Statement II-True

Two types of odometer are used-

1.  Floating ring type and

2. Fixed ring type.

 Concept

∴ D60 = 4.75 mm and D10 = 0.075 mm

Coefficient of uniformity (C_u) is given by 
\({{\rm{C}}_{\rm{u}}} = \frac{{{{\rm{D}}_{60}}}}{{{{\rm{D}}_{10}}}} = \frac{{4.75}}{{0.075}} = 63.33\)

Hence C_U > 60

Concept

To check whether the soil is currently compacted to its maximum value:

Calculating the degree of saturation for the corresponding data:

Calculation

Given,

 w = 16%, γ_d = 1.8 g/cc

\({\rm{\gamma }} = \frac{{\left( {{\rm{G}} + {{\rm{S}}{\rm{e}}}} \right){{\rm{\gamma }}_{\rm{w}}}}}{{1 + {\rm{e}}}} \Rightarrow {{\rm{\gamma }}_0} = \frac{{\left( {1 + {\rm{w}}} \right){\rm{G\;}}{{\rm{\gamma }}_{\rm{w}}}}}{{1 + \frac{{{\rm{wG}}}}{{\rm{S}}}}}\)

\(\therefore 1.8 = \frac{{1.16 \times 2.7 \times 1}}{{1 + \frac{{16 \times 2.7}}{3}}}\)

⇒ S = 0.5838 = 58.38%

∴ Soil can be compacted further.

For maximum compaction, degree of saturation (S) = 1

\(\therefore {{\rm{\gamma }}_{\rm{d}}} = \frac{{{\rm{G\;}}{{\rm{\gamma }}_{\rm{w}}}}}{{1 + \frac{{{\rm{wG}}}}{{\rm{S}}}}} = \frac{{2.7 \times 1}}{{1 + \frac{{0.16 \times 2.7}}{1}}} = \frac{{2.7}}{{1 + 0.16 \times 2.2}}\)

Concept

In an undrained triaxial test on saturated clay, the volume change due to confining pressure is zero.

\(\Rightarrow \epsilon_{_z} = 0\).

\(\frac{{{\sigma _3}}}{E} - \frac{{μ {\sigma _1}}}{E} - \frac{{μ {\sigma _3}}}{E} = 0\)

\(μ = \frac{{{\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}\)

 ∴ poisson ratio \(μ = \frac{{{\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}\)

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.

Another form of engg. News formula.

\({{\rm{Q}}_{{\rm{kN}}}}{\rm{\;}} = {\rm{\;}}\frac{{166.64{\rm{E}}}}{{{\rm{S}} + 2.54}}\)

E = Energy per blow in ‘KJ’

S = Settlements per blow in ‘mm’

Given:

w = 30 KN, H = 1.5 m = 1500 cm, S = 5 mm, c = 2.5 cm, and FOS = 6

As per Engineering News formula:

\({{\rm{Q}}_{{\rm{allowable}}}} = \frac{{{\rm{wH}}}}{{6\left( {{\rm{S}} + {\rm{c}}} \right)}} = \frac{{30 \times \left( {1.5 \times 100} \right)}}{{6\left( {2.5{\rm{\;}} + {\rm{\;}}0.5} \right)}} = 250{\rm{\;kN}}\)

Concept:

The critical Hydraulic gradient of soil sample is given as:

\({i_c} = \;\frac{{G - 1}}{{1 + e}}\)

G is already given. However, the void ratio, e can be determined using the following relation:

\({\gamma _b} = \;\frac{{\left( {G + Se} \right){\gamma _w}}}{{1 + e}}\)

Where, symbols have their usual meaning.

Calculation:

Given

G = 2.65, γ_b = 20 kN/m³, S = 70 %

Assuming, γ_w = 10 kN/m³

Now

\({\gamma _b} = \;\frac{{\left( {G + Se} \right){\gamma _w}}}{{1 + e}}\)

\(20 = \;\frac{{\left( {2.65 + 0.7e} \right) \times 10\;}}{{1 + e}}\)

⇒ e = 0.5

Now

\({i_c} = \frac{{2.65 - 1}}{{1 + 0.5}}\)

∴ i_c = 1.1

Concept

passive earth pressure = \({P_p} = \frac{1}{2} \times {k_p} \times \gamma \times {H^2}\)

Calculation

 Given,

 H = 10 m

Initial condition

\(\begin{array}{l} {\phi _1}\; = \;30^\circ ,{\gamma _1}\; = \;18kN/{m^3}\\ {k_{p1}}\; = \;{\tan ^2}\left( {45\; + \;\phi /2} \right)\; = \;3\\ {p_{p1}}\; = \;\frac{1}{2}{k_p}\gamma {H^2}\; = \;\frac{1}{2} \times 3 \times 18 \times {10^2}\; = \;2700\;kN/m \end{array}\)

Final condition

\(\begin{array}{l} {\phi _2}\; = \;35^\circ ,{\gamma _2}\; = \;20kN/{m^3}\\ {k_{p2}}\; = \;{\tan ^2}\left( {45\; + \;\frac{{35}}{2}} \right)\; = \;3.69\\ {p_{p2}}\; = \;\frac{1}{2} \times 3.69 \times 20 \times {10^2}\; = \;3690\;kN/m\\ \frac{{{P_{p1}}}}{{{P_{p2}}}}\; = \;\frac{{2700}}{{3690}}\; = \;0.73 \end{array}\)

 ∴ Ratio is 0.73

Concept

 Transformed Scale length in the z-direction (along the base ) is given by

 \({l_b} = L_b\times scale \times \sqrt {\frac{{{k_y}}}{{{k_z}}}} \) 

Calculation

 Given,

 k_x = 3.45 m/day, k_y = 1.5 m/day

 L_b = 100 m (base length)

 Scale factor in vertical direction = 1:25,

 Scale factor in horizontal direction

  \(= \frac{1}{{25}}\sqrt {\frac{{{k_y}}}{{{k_x}}}} = \frac{1}{{37.969}}\)

 ∴ Reduced base length = scale × L_b

  \(= \frac{{100}}{{37.96}} = 2.63\;m\)

 Reduced length = 2.63 m

Concept:

Undrained shear strength of clay is given by:

\(S = \frac{T}{{\pi {D^2}\left( {\frac{H}{2} + \frac{D}{6}} \right)}}\)

Where,

S = shear strength of soil

H = Length or Height of the vane

D = diameter of the vane

T = Torque applied

\(S = \;\frac{{{q_{{u_{undisturbed}}}}}}{{{q_{{u_{remoulded}}}}}} = \frac{{2c}}{{2{c_r}}}\)

Calculation:

At undisturbed state:

T = 45 N-m

H = 130 mm = 0.13 m

D = 70 mm = 0.07 m

\(S = c = \frac{{45}}{{\pi {{0.07}^2}\left( {\frac{{0.13}}{2} + \frac{{0.07}}{6}} \right)}}\)

c = 38.129 kN/m²

At remoulding state:

Sensitivity of clay soil:

\(S = \;\frac{{{q_{{u_{undisturbed}}}}}}{{{q_{{u_{remoulded}}}}}} = \frac{{2c}}{{2{c_r}}}\)

\(8 = \frac{{2 \times 38.129}}{{2{c_r}}}\)

c_r = 4.767 kN/m²

Now,

\(S = {c_r} = 4767 = \frac{{{T_r}}}{{\pi \times {{0.07}^2}\left( {\frac{{0.13}}{2} + \frac{{0.07}}{6}} \right)}}\)

T_r = 5.625 N-m (Ans).

\(\frac{{{{\rm{d}}^2}{\rm{H}}}}{{{\rm{d}}{{\rm{X}}^2}}} = 0{\rm{\;}}\)

Integrating on both side

\(\frac{{{\rm{dH}}}}{{{\rm{dX}}}} = {\rm{K}}\)

Again Integrating on both side

H = KX + C

Given at X = 0, H = 5 m

∴ 5 = 0 + C

∴ C = 5

Given at X = 10, H = 1 m

∴ 1 = (K × 10) + 5

∴ K = -0.4

So the equation  H = KX + C, becomes

H = -0.4X + 5

At the middle of the strip i.e X = 5

H = (-0.4 × 5) + 5

H = 3 m