Q2GATE 2023NAT

An unconfined compression strength test was conducted on a cohesive soil. The test specimen failed at an axial stress of $76 \mathrm{kPa}$ . The undrained cohesion (in $\mathrm{kPa}$ , in integer) of the soil is ___

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📖 Explanation

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)] $\Rightarrow \quad \sigma_{1}=2 \mathrm{C}$ $\Rightarrow \quad 76=2 \mathrm{C}$ $\Rightarrow \quad C=38 \mathrm{kPa}$

Q3GATE 2023MCQ

Consider that a force $P$ is acting on the surface of a half-space (Boussingesq's problem). The expression for the vertical stress $\left(\sigma_{z}\right)$ at any point $(r, z)$ with the half-space is given as, $\sigma_{z}=\frac{3 P}{2 \pi} \frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}$ where, $r$ is the radial distance, and $z$ is the depth with downward direction taken as positive. At any given $r$ , there is a variation of $\sigma_{z}$ along $z$ , and at a specific $z$ , the value of $\sigma_{z}$ will be maximum. What is the locus of the maximum $\sigma_{z}$ ?

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📖 Explanation

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

\begin{aligned} & \frac{\sigma_{z}}{\partial z}=0 \Rightarrow \frac{3 P}{2 \pi} \frac{\partial}{\partial z}\left[\frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}\right]\$ \\ &=\frac{3 P}{2 \pi} \frac{\left[\left(r^{2}+z^{2}\right)^{5 / 2}-z^{3} \times \frac{5}{2}\left(r^{2}+z^{2}\right)^{3 / 2} \times 2 z\right]\$}{\left(r^{2}+z^{2}\right)^{5}}=0 \\ &=\left[\left(r^{2}+z^{2}\right)\left(3 z^{2}\right)-5 z^{4}\right]\$=0 \\ &=3 r^{2}+3 z^{2}=5 z^{2} \\ & z^{2}=\frac{3 r^{2}}{2} \end{aligned}

Q4GATE 2023MCQ

In the given figure, Point $\mathrm{O}$ indicates the stress point of a soil element at initial non-hydrostatic stress condition. For the stress path (OP), which of the following loading conditions is correct ?

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📖 Explanation

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)

Q5GATE 2022NAT

A concentrically loaded isolated square footing of size 2 m x 2 m carries a concentrated vertical load of 1000 kN. Considering Boussinesq's theory of stress distribution, the maximum depth (in m) of the pressure bulb corresponding to 10% of the vertical load intensity will be ________. (round off to two decimal places)

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📖 Explanation

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 $r=0, \sigma _z=\frac{3Q}{2\pi z^2}$ $\sigma _z=0.1q=0.1 \times \frac{Q}{B^2}=\frac{0.1}{2^2}Q$ $\frac{1}{40}Q=\frac{3Q}{2 \pi z^2}$ $z^2=\frac{3 \times 40}{2 \pi}\Rightarrow z=4.37m$

Q6GATE 2020MCQ

The soil profile at a site up to a depth of 10 m is shown in the figure (not drawn to the scale). The soil is preloaded with a uniform surcharge (q) of 70 $kN/m^2$ at the ground level. The water table is at a depth of 3 m below ground level. The soil unit weight of the respective layers is shown in the figure. Consider unit weight of water as 9.81 $kN/m^3$ and assume that the surcharge (q) is applied instantaneously. Immediately after preloading, the effective stresses (in kPa) at points P and Q respectively, are

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📖 Explanation

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, $\bar{\sigma }=\sigma -u=94.76kPa$

Q7GATE 2020NAT

A 5 m high vertical wall has a saturated clay backfill. The saturation unit weight and cohesion of clay are $18kN/m^3$ and 20 kPa, respectively. The angle of internal friction of clay is zero. In order to prevent development of tension zone, the height of the wall is required to be increased. Dry sand is used as backfill above the clay for the increased portion of the wall. The unit weight and angle of internal friction of sand are $16kN/m^3$ and $30^{\circ}$ , respectively. Assume that the back of the wall is smooth and top of the backfill is horizontal. To prevent the development of tension zone, the minimum height (in m, round off to one decimal place) by which the wall has to be raised, is __________.

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📖 Explanation

To prevent tension crack,

\begin{aligned} q&=\frac{2c}{\sqrt{k_a}}=\frac{2 \times 20}{1}=40\\ q&=\gamma _d x=40\\ x&=\frac{40}{16}=2.5m \end{aligned}

Q8GATE 2020MCQ

Which one of the following statements is NOT correct?

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📖 Explanation

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

Q9GATE 2019NAT

A concentrated load of 500 kN is applied on an elastic half space. The ratio of the increase in vertical normal stress at depths of 2m and 4m along the point of the loading, as per Boussinesq's theory, would be ___

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📖 Explanation

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

\begin{aligned} \sigma _Z&=\frac{3}{2\pi}\frac{Q}{Z^2}\\ \sigma _Z &\propto \frac{1}{Z^2}\\ \therefore \;\; \frac{\sigma _{Z_1}}{\sigma _{Z_2}}&=\left[ \frac{Z_2}{Z_1} \right]^2\\ &=\left[ \frac{4}{2} \right]^2=4 \end{aligned}

Q10GATE 2019NAT

A 2mx 4m rectangular footing has to carry a uniformly distributed load of 120 kPa. As per the 2:1 dispersion method of stress distribution, the increment in vertical stress (in kPa) at a depth of 2m below the footing is ________

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📖 Explanation

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

\begin{aligned} \Delta \bar{\sigma }&=\frac{q(B \times L)}{(B+2nZ)(L+2nZ)}\\ &=\frac{120 \times 2 \times 4}{(2+2 \times \frac{1}{2}\times 2)(4+2 \times \frac{1}{2}\times 2)}\\ \Delta \bar{\sigma }&=40kPa \end{aligned}

Q11GATE 2017NAT

Consider a square-shaped area ABCD on the ground with its Centre at M as shown in the figure. Four concentrated vertical load of P=5000 kN are applied on this area, at each corner. The vertical stress increment (in kPa, up to one decimal place) due to these loads according to the Boussinesq's equation, at a point 5 m right below M, is_____

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📖 Explanation

\begin{array}{l} P=5000 \mathrm{kN} \\ r=\sqrt{2^{2}+2^{2}}=2 \sqrt{2} \mathrm{m} \end{array}

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

\begin{aligned} \sigma_{z} &=4 \times \frac{3}{2 \pi} \times \left\{\frac{1}{1+\left(\frac{2 \sqrt{2}}{5}\right)^{2}}\right\}^{5/2} \quad \times \frac{5000}{5^{2}} \\ &=190.8 \mathrm{kPa} \end{aligned}

Q12GATE 2017MCQ

A uniformly distributed line load of 500 kN/m is acting on the ground surface. Based on Boussinesq's theory, the ratio of vertical stress at a depth 2 m to that at 4 m, right below the line of loading, is

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📖 Explanation

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

\begin{aligned} \sigma_{z} &=\frac{2 q^{\prime}}{\pi z}\left(\frac{1}{1+\frac{x^{2}}{z^{2}}}\right)^{2}=\frac{2 q^{\prime}}{\pi z} \\ \text { Ratio } &=\frac{\left(\sigma_{z}\right)_{a t, 2 m}}{\left(\sigma_{z}\right)_{a t, 4 m}}=\frac{\frac{2 q^{\prime}}{\pi \times 2}}{\frac{2 q^{\prime}}{\pi \times 4}}=\frac{4}{2}=2 \end{aligned}

Q13GATE 2010MCQ

The vertical stress at point P1 due to the point load Q on the ground surface as shown in figure is $\sigma _{2}$ . According to Boussinesq's equation, the vertical stress at point P2 shown in figure will be

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📖 Explanation

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 $=\frac{3 Q}{2 \pi\left(\frac{z}{2}\right)^{2}}\left[\frac{1}{1+\left(\frac{r / 2}{2 / 2}\right)^{2}}\right]^{5 / 2}=4 \times \frac{3 Q}{2 \pi z^{2}}\left(\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right)^{5 / 2}$ $=4 \sigma_{z}$

Q14GATE 2008MCQ

A footing $2m\times 1m$ exerts a uniform pressure of 150 kN/ $m^{2}$ on the soil. Assuming a load dispersion of 2 vertical to 1 horizontal, the average vertical stress (kN/ $m^{2}$ ) at 1.0 m below the footing is

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📖 Explanation

Total dispersion area

\begin{aligned} &=(2+0.5+0.5) \times (1+0.5+0.5) \\ &=3 \mathrm{m} \times 2 \mathrm{m} \end{aligned}

Q15GATE 2007MCQ

The vertical stress at some depth below the corner of a 2m x 3m rectangular footing due to a certain load intensity is 100 kN/ $m^{2}$ . What will be the vertical stress in kN/ $m^{2}$ below the centre of a 4m x 6m rectangular footing at the same depth and same load intensity ?

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📖 Explanation

For same load intensity and same depth

\begin{aligned} \Rightarrow \quad \frac{\sigma_{1}}{\sigma_{2}}&=\frac{A_{1}}{A_{2}} \Rightarrow \frac{100}{\sigma_{2}}=\frac{2 \times 3}{4 \times 6} \\ \Rightarrow \quad \sigma_{2}&=400 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Q16GATE 2003MCQ

A 25 kN point load acts on the surface of an infinite elastic medium. The vertical pressure intensity in $kN/m^{2}$ at a point 6.0 m below and 4.0 m away from the load will be

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📖 Explanation

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

\begin{aligned} \sigma_{z} &=\frac{3 Q}{2 \pi z^{2}}\left[\frac{1}{1+(r / z)^{2}}\right]^{5 / 2} \\ \Rightarrow \quad &=\frac{3 \times 25}{2 \pi \times (6)^{2}}\left[\frac{1}{1+(4 / 6)^{2}}\right]^{5 / 2} \\ &=0.132 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

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