Stress Distribution In The Soil – GATE CE Geotechnical EngineeringPractice Questions & PYQs

Statement (A): Effective stress is given by: $\sigma' = \sigma - u$ Where $\sigma$ is total stress and $u$ is pore water pressure. As the groundwater table goes deeper, pore water pressure acting on the soil decreases, which leads to an increase in effective stress - not a decrease. This statement is incorrect. Statement (B): Bulking of sand occurs when moisture forms a thin film around sand particles due to surface tension and capillary action. This causes an increase in volume. This statement is correct. Statement (C): In liquefied soil, the pore water pressure equals the total stress, so the effective stress becomes nearly zero: $\sigma' = \sigma - u \approx 0$ This statement is correct. Statement (D): The earth pressure at a point in soil can be less than, equal to, or even greater than the vertical effective stress depending on the condition (at-rest, active, or passive). For example, in passive earth pressure condition: $\sigma_h = K_p \cdot \sigma_v$ where $K_p > 1$ Hence, horizontal pressure can be greater than vertical effective stress. This statement is incorrect.

UCS Test : $\sigma_{3}=0$ [Cell pressure] $\therefore$ Axial stress $=$ Deviatoric stress $=\sigma_{1}-\sigma_{3}=\sigma_{1}$ Given, $\sigma_{1}=76 \mathrm{kPa}$ We know, $\sigma_{1}=\sigma_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)+2 C \tan \left(45^{\circ}+\frac{\phi}{2}\right)$ $[\mathrm{C}=$ Undrained cohesion, $\phi=0$ (Cohesive soil)] $\Rightarrow \quad \sigma_{1}=2 \mathrm{C}$ $\Rightarrow \quad 76=2 \mathrm{C}$ $\Rightarrow \quad C=38 \mathrm{kPa}$

The expression for the vertical stress $\left(\sigma_{2}\right)$ at any point $(r, z)$ with the half-space is given as $\sigma_{z}=\frac{3 P}{2 \pi} \frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}$ For $\sigma_{\mathrm{z}}$ to be maximum

For $1: 1$ slope, $\frac{d q}{d P}=1$ $\Rightarrow \frac{d \sigma_{V}-d \sigma_{H}}{d \sigma_{V}+d \sigma_{H}}=1$ $\Rightarrow d \sigma_{V}-d \sigma_{H}=d \sigma_{V}+d \sigma_{H}$ $\Rightarrow \quad-\mathrm{d} \sigma_{\mathrm{H}}=\mathrm{d} \sigma_{\mathrm{H}}$ $\Rightarrow \quad \mathrm{d} \sigma_{\mathrm{H}}=0$ Hence, if $\sigma_{H}$ is constant then increasing $\sigma_{V}$ or decreasing $\sigma_{\mathrm{V}}$ will lead to $1: 1$ slope. Answer is (A)

Considering Boussinesq's theory of stress distribution $\sigma _x=\frac{3Q}{2\pi z^2}\left[ \frac{1}{1+\left ( \frac{r}{z} \right )^2} \right]^{5/2}$ For $r=0, \sigma _z=\frac{3Q}{2\pi z^2}$ $\sigma _z=0.1q=0.1 \times \frac{Q}{B^2}=\frac{0.1}{2^2}Q$ $\frac{1}{40}Q=\frac{3Q}{2 \pi z^2}$ $z^2=\frac{3 \times 40}{2 \pi}\Rightarrow z=4.37m$

Surcharge ( $q=70kN/m^2$ ) is applied instantaneously hence excess pore pressure ( $u_i=70kPa$ ) is developed at point P and Q [GWT level is at level P] At point P: Total stress $\sigma =q+3\gamma =70+3\times 18$ Pore water pressure = Hydrostatics pore pressure + Excess pore pressure $=0+u_i=0+70=70kN/m^2$ Effective stress, $\bar{\sigma }=\sigma -u=54kPa$ At point Q: Total stress, $\sigma= q+3\gamma +4\gamma _{sat}=70+3\times 18 +4 \times 20$ Pore pressure, u = Hydrostatics pore pressure + Excess pore pressure $= 4\gamma _{w}+u_i=4 \times 9.81 +70$ Effective stress, $\bar{\sigma }=\sigma -u=94.76kPa$

To prevent tension crack,

A clay deposit with liquidty index greater then 1, will be in liquid stage of consistency. $\because \;\;I_L=\frac{w_n-w_p}{w_l-w_p} \gt 1$ $\therefore \;\; w_n \gt w_L$

As per Boussinesq's equation, The vertical stress below the ground level at a depth, Z, vertically below the load,

The area of rectangular footing = 2 m x 4 m q = 120 kPa as 2:1 dispersion method of stress distribution

According to Boussinesq's $\sigma_{z}=k \frac{Q}{z^{2}}=\frac{3}{2 \pi}\left(\frac{1}{1+\frac{r^{2}}{z^{2}}}\right)^{5 / 2} \frac{Q}{z^{2}}$ $\Rightarrow$ Vertical stress at 5m below point M

$q^{\prime}=500 \mathrm{kN} / \mathrm{m}$ $x=0$ (Just below the line load)

As per Boussinesq equation, the vertical stress $\sigma_{2}$ at a point located at a depth 'z' and a horizontal distance 'r' from the point of application of point load Q is $\sigma_{z}=\frac{3 Q}{2 \pi z^{2}}\left[\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right]^{5 / 2}$ For a horizontal distance $\frac{r}{2}$ and a depth $\frac{2}{2}$ , the stress will be $=\frac{3 Q}{2 \pi\left(\frac{z}{2}\right)^{2}}\left[\frac{1}{1+\left(\frac{r / 2}{2 / 2}\right)^{2}}\right]^{5 / 2}=4 \times \frac{3 Q}{2 \pi z^{2}}\left(\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right)^{5 / 2}$ $=4 \sigma_{z}$

Total dispersion area

For same load intensity and same depth

As per Boussinesq's equation, the vertical stress $\sigma_{2}$ at a point located at a depth 'z' and a horizontal distance 'r' from the point of application of point load Q is