Q4GATE 2024NAT

The in-situ percentage of voids of a sand deposit is $50 \%$ . The maximum and minimum densities of sand determined from the laboratory tests are $1.8 \mathrm{~g} / \mathrm{cm}^{3}$ and $1.3 \mathrm{~g} / \mathrm{cm}^{3}$ , respectively. Assume the specific gravity of sand as 2.7 . The relative density index of the in-situ sand is ______ (rounded off to 2 decimal places)

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📖 Explanation

maximum density $\left(\gamma_{\mathrm{d}}\right)_{\max }=1.8 \mathrm{~g} / \mathrm{cm}^{3}$ minimum density $\left(\gamma_{\mathrm{d}}\right)_{\min }=1.3 \mathrm{~g} / \mathrm{cm}^{3}$ specific gravity $(G)=2.7$ $\left(I_{D}\right)$ relative density $=\frac{\gamma_{d \max }\left(\gamma_{d}-\gamma_{d \min }\right)}{\gamma_{d}\left(\gamma_{d \max }-\gamma_{d \min }\right)}$ $\therefore \quad(n)$ percentage of voids $=\frac{\left(\mathrm{V}_{\mathrm{v}}\right) \text { volume of voids } \times 100}{\left(\mathrm{~V}_{\mathrm{T}}\right) \text { Total volume }}=50 \%$ Porosity( $n$ ) $=0.5$ void ratio (e) $=\frac{\mathrm{n}}{1-\mathrm{n}}=\frac{0.5}{1-0.5}=1$ $\left(\gamma_{\mathrm{d}}\right)_{\text {insitu }}=\frac{\mathrm{G} \cdot \gamma_{\mathrm{w}}}{1+\mathrm{e}}=\frac{2.7 \times 1}{1+1}=1.35 \mathrm{~g} / \mathrm{cm}^{3}$ $\left(I_{D}\right)=\frac{1.8(1.35-1.3)}{1.35(1.8-1.3)}=0.13$ Relative density $\left(I_{D}\right)=0.13$

Q5GATE 2023MCQ

A square footing of size $2.5 \mathrm{~m} \times 2.5 \mathrm{~m}$ is placed $1.0 \mathrm{~m}$ below the ground surface on a cohesionless homogeneous soil stratum. Considering that the groundwater table is located at the base of the footing, the unit weights of soil above and below the groundwater table are $18 \mathrm{kN} / \mathrm{m}^{3}$ and $20 \mathrm{kN} / \mathrm{m}^{3}$ , respectively, and the bearing capacity factor $N_{q}$ is 58 , the net ultimate baring capacity of the soil is estimated as $1706 \mathrm{kPa}$ (unit weight of water $10 \mathrm{kN} / \mathrm{m}^{3}$ ). Earlier, a plate load test was carried out with a circular place of $30 \mathrm{~cm}$ diameter in the same foundation pit during a dry season, when the water table was located beyond the plate influence zone. Using Terzaghi's bearing capacity formulation, what is the ultimate bearing capacity (in $\mathrm{kPa}_{-}$ of the plate ?

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📖 Explanation

$\mathrm{q}_{\mathrm{\nu}}=1706 \mathrm{kPa}, \quad \gamma_{\mathrm{w}}=10 \mathrm{kN} / \mathrm{m}^{3}$ Plate load test data: dia. (d) $=30 \mathrm{~cm}$ with no water able effect. $\mathrm{q}_{\mathrm{u} \text { plate }}=\text { ? }$ For footing :

\begin{aligned} & \mathrm{q}_{\mathrm{\nu}}=1706 \mathrm{kPa} \\ & \mathrm{q}_{\mathrm{\nu}}=0+\mathrm{q}\left(\mathrm{N}_{\mathrm{q}}-1\right)+0.4 \mathrm{~B} \gamma \mathrm{N}_{\gamma} \\ & \mathrm{q}_{\mathrm{\nu}}=18 \times 1 \times (58-1 \times )+0.4 \times 2.5 \times (20-10 \times ) \times \mathrm{N}_{\gamma} \\ & \mathrm{N}_{\gamma}=\frac{(1706-18 \times 57)}{(0.4 \times 2.5 \times 10)} \Rightarrow \mathrm{N}_{\gamma}=68 \\ & \mathrm{q}_{\text {yplage }}=0.3 \times \mathrm{d} \times \gamma_{\mathrm{t}} \times \mathrm{N}_{\gamma} \\ & \mathrm{q}_{\text {uplate }}=0.3 \times 0.3 \times 18 \times 68 \\ &=110.16 \mathrm{kPa} \end{aligned}

Q6GATE 2023NAT

A square footing is to be designed to carry a column load of $500 \mathrm{kN}$ which is resting on a soil stratum having the following average properties: bulk unit weight $=19 \mathrm{kN} / \mathrm{m}^{3}$ ; angle of internal friction $=0^{\circ}$ , and cohesion $=25 \mathrm{kPa}$ . Considering the depth of the footing as $1 \mathrm{~m}$ and adopting Meyerhof's bearing capacity theory with a factor of safety of 3 , the width of the footing (in $\mathrm{m}$ ) is ____ (round off to one decimal place) [Assume the applicable shape and depth factor values as unity; ground water level at greater depth.]

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📖 Explanation

As per Meyerhoff's approach $q_{u}=C N_{c}\left[i_{c} S_{c} d_{c}\right]+q N_{q}\left[{ }_{q} S_{q} d_{r}\right]+\frac{1}{2} \beta \gamma N_{r}\left[i_{r} S_{r} d_{r}\right]$ Given; shape factor $S_{c}, S_{q}, S_{r}=1$ depth factor $d_{c}, d_{v}, d_{r}=1$ Assuming no inclination of loading hence $i_{c} i_{q} i_{r}=1$ For $\phi=0^{\circ}$

\begin{aligned} & \mathrm{N}_{\mathrm{c}}=5.14 \\ & N_{\mathrm{q}}=1 \\ & N_{r}=0 \\ & \therefore \quad \mathrm{q}_{\mathrm{u}}=25 \times 5.14 \times 1 \times 1 \times 1+19 \times 1 \times \\ & 1 \times 1 \times 1 \\ & =147.5 \mathrm{kPa} \\ & \mathrm{q}_{\mathrm{\nu}}=\mathrm{q}_{\mathrm{u}}-\mathrm{rD}_{\mathrm{f}} \\ & =147.5-19 \times 1 \\ & =128.5 \mathrm{kPa} \end{aligned}

$q_{n s}=\frac{q_{n u}}{F O S}=\frac{128.5}{3}=42.83 \mathrm{kPa}$ Hence, $\frac{500}{B^{2}} \leq 42.83$ or $B \geq 3.416 m$ Hence $B=3.4 m$

Q10GATE 2021NAT

A rectangular footing of size 2.8m x 3.5m is embedded in a clay layer and a vertical load is placed with an eccentricity of 0.8 m as shown in the figure (not to scale). Take Bearing capacity factors: $N_{c}=5.14$ , $N_{q}=1.0$ , and $N_{\gamma}=0.0$ ; Shape factors: $s_{c}=1.16$ , $s_{q}=1.0$ and $s_{\gamma}=1.0$ ; Depth factors: $d_{c}=1.1$ , $d_{q}=1.0$ and $d_{\gamma}=1.0$ ; and Inclination factors: $i_{c}=1.0$ and $i_{q}=1.0$ and $i_{\gamma}=1.0$ . Using Meyerhoff's method, the load (in kN, round off to two decimal places) that can be applied on the footing with a factor of safety of 2.5 is _________________

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📖 Explanation

\begin{aligned}& \text { Given data: } 2.8 \times 3.56, e=0.8, D_{f}=1.5 \mathrm{~m} \\ N_{c}&=5.14, \qquad N_{q}=1, \qquad N_{r}=0 \\ S_{c}&=1.16, \qquad S_{q}=1, \qquad S_{r}=1.0 \\ d_{c}&=1.1, \qquad d_{q}=1, \qquad d_{f}=1.0 \\ i_{c}&=1, \qquad \quad i_{q}=1, \qquad i_{r}=1.0 \\ \gamma&=18.2, \qquad c=40 \mathrm{kN} / \mathrm{m}^{2} \\ B^{\prime}&=B-2 e_{x}=2.8-2(0.8)=1.2 \mathrm{~m} \\ L^{\prime}&=L=3.5 \end{aligned}

Ultimate bearing capacity

\begin{aligned} q_{u} &=C N_{c} S_{c} d_{c} i_{c}+\gamma D_{p} N_{q} S_{q} d_{q} i_{q}+0.5 B^{\prime} \gamma N_{r} S_{r} d_{r} \\ q_{u} &=40 \times 5.14 \times 1.16 \times 1.1+18.2 \times 1.5 \times 1+0 \\ q_{n u} &=q_{u}-\gamma D_{f}=262.346 \mathrm{kN} / \mathrm{m}^{2} \\ q_{n s} &=\frac{q_{n u}}{F}=104.938 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

\begin{aligned} \text { Net safe load } &=q_{n s} \times A=\frac{q_{n u}}{F} \times \left(B^{\prime} \times L\right) \\ &=104.938 \times 1.2 \times 3.5 \mathrm{kN} \end{aligned}

Load applied on the footing = 440.740 kN

Q12GATE 2019NAT

A square footing of 4 m side is placed at 1 m depth in a sand deposit. The dry unit weight ( $\gamma$ ) of sand is 15 $kN/m^3$ . This footing has an ultimate bearing capacity of 600 kPa. Consider the depth factors: $d_q=d_\gamma =1.0$ and the bearing capacity factor: $N_\gamma =18.75$ . This footing is placed at a depth of 2m in the same soil deposit. For a factor of safety of 3.0 as per Terzaghi's theory, the safe bearing capacity (in kPa) of this footing would be _______

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📖 Explanation

As per Terzaghi's theory, the ultimate bearing capacity for a square footing considering depth factors: $d_q=d_\gamma =1$ is as follows:

\begin{aligned} q_u&=1.3CN_C+\gamma D_fN_q+0.4\gamma BN_\gamma \\ &\text{For sand,} \\ C&=0 \\ q_u&=\gamma D_fN_q+0.4\gamma BN_\gamma \\ \text{Given:} \\ \gamma &=15kN/m^3 \\ B&=4m \\ N_\gamma &= 18.75\\ &\text{Initial condition: } \\ D_f&= 1m\\ q_u&=600kPa\\ q_u&=\gamma D_fN_q+0.4\gamma BN_\gamma \\ 600&=15 \times 1 \times N_q +0.4 \times 15 \times 4 \times 18.75\\ N_q&=10\\ \text{If}\;\; D_f&=2m, \;\; \text{then,}\\ q_u&=\gamma D_fN_q+0.4\gamma BN_\gamma \\ &=15 \times 2 \times 10 +0.4 \times 15 \times 4 \times 18.75\\ &=750kN/m^2\\ &\text{Net ultimate BC of soil, }\\ q_{\nu}&=q_u-\gamma D_f\\ &=750-15 \times 2=720kN/m^2\\ &\text{The gross safe BC or safe BC, }\\ q_s&=\frac{q_{\nu}}{F}+\gamma D_f\\ &=\frac{720}{3}+15 \times 2\\ &=270kPa \end{aligned}

Q13GATE 2019NAT

A square footing of 2 m sides rests on the surface of a homogeneous soil bed having the properties: cohesion c=24 kPa, angle of internal friction $\phi =25^{\circ}$ , and unit weight $\gamma =18kN/m^3$ . Terzaghi's bearing capacity factors for $\phi =25^{\circ}$ are $N_c=25.1, N_q=12.7$ , $N_{\gamma}=9.7, N_c'=14.8,N_q'=5.6$ , and $N_{\gamma}'=3.2$ . The ultimate bearing capacity of the foundation (in kPa, round off to 2 decimal places) is _______

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📖 Explanation

\begin{aligned} c &=24kN/m^2 ,\; \phi =25^{\circ} \\ \gamma &=18 kN/m^3 ,\; N_c =25.1 \\ N_q&=12.7 ,\; N_\gamma =9.7 \\ N_c' &=14.8 ,\; N_q'=5.6 \\ N_\gamma '&=3.2 \\ &\text{Square footing}\\ B&=2m ,\;L=2m\\ D_f&=0\;\; (\text{Surface footing})\\ \text{since,}\;\;\phi &=25^{\circ} \lt 28^{\circ}\rightarrow \text{Assume local shear failure.}\\ \text{Hence,}\;\;&=c_m=\frac{2}{3}c=\frac{2}{3} \times 24=16kN/m^2 \end{aligned}

Ultimate bearing capacity of square footing,

\begin{aligned} q_u&=1.3c_mN_c' +\gamma D_fN_q'+0.4\gamma BN_\gamma '\\ q_u&=1.3\times 16 \times 14.8+0+0.4 \times 18 \times 2 \times 3.2\\ q_u&=307.84+46.08\\ q_u&=353.92kN/m^2 \end{aligned}

Q19GATE 2016NAT

A 4 m wide strip footing is founded at a depth of 1.5 m below the ground surface in a $c-\phi$ soil as shown in the figure. The water table is at a depth of 5.5 m below ground surface. The soil properties are: ${c}'=35\: kN/m^{2},{\phi }'=28.63^{\circ},$ $\gamma _{sat}=19\: kN/m^{3}, \; \gamma _{bulk}=17\:kN/m^{3}$ and $\gamma _{w}=9.81kN/m^{3}$ . The values of bearing capacity factors for different ${\phi }'$ are given below. Using Terzaghi's bearing capacity equation and a factor of safety $F_{s}=2.5$ , the net safe bearing capacity (expressed in $kN/m^{2}$ ) for local shear failure of the soil is ________.

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📖 Explanation

Local shear failure is occuring hence modified c and $\phi$ should be used

\begin{aligned} c_{m} &=\frac{2}{3} c=\frac{2}{3} \times 35=23.33 \mathrm{kN} / \mathrm{m}^{2} \\ \tan \phi_{m} &=\frac{2}{3} \tan \phi=\frac{2}{3} \tan 28.63 \\ \phi_{m} &=20^{\circ} \end{aligned}

for $\quad \phi_{m}=20^{\circ}, N_{c}=17.7, N_{\mathrm{q}}=7.4, N_{\gamma}=5.0$ $\Rightarrow$ Water table is at $D_{t}+B=1.5+4=5.5 \mathrm{m}$ hence no effect on bearing capacity $\Rightarrow$ As per Terzaghi for strip footing

\begin{aligned} a_{u} &=c N_{c}+8 D_{f} N_{q}+0.5 B \gamma N_{\gamma} \\ q_{u} &=\frac{2}{3} \times 35 \times 17.7+17+1.5 \times 7.4+0.5 \\ & \times 4 \times 17 \times 5 \\ q_{u} &=771.7 \mathrm{kN} / \mathrm{m}^{2} \\ q_{\mathrm{\nu}} &=q_{u}-\bar{\sigma}=q_{u}-\gamma D_{f}=771.7-17 \times 1.5 \\ q_{n} &=746.2 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Net safe bearing capacity $q_{n s}=\frac{q_{n u}}{F}=\frac{746.2}{2.5}=298.48 \mathrm{kN} / \mathrm{m}^{2}$

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