# Shallow Foundation and Bearing Capacity

 Question 1
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 _________________
 A 441 B 564 C 876 D 689
GATE CE 2021 SET-2   Geotechnical Engineering
Question 1 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
 Question 2
The cohesion (c), angle of internal friction $(\phi)$ and unit weight $(\gamma)$ of a soil are $15 \mathrm{kPa}, 20^{\circ}$ and $17.5 \mathrm{kN} / \mathrm{m}^{3}$, respectively. The maximum depth of unsupported excavation in the soil (in m, round off to two decimal places) is __________
 A 2.4 B 5.6 C 4.9 D 3.8
GATE CE 2021 SET-1   Geotechnical Engineering
Question 2 Explanation:
Maximum depth of unsupported excavation,
\begin{aligned} H&=\frac{4 C}{\gamma_{t} \sqrt{k_{a}}} \\ k_{a}&=\frac{1-\sin \phi}{1+\sin \phi}=\frac{1-\sin 20^{\circ}}{1+\sin 20^{\circ}}=0.49 \\ \Rightarrow \qquad \qquad \qquad \qquad H&=\frac{4 \times 15}{17.5 \sqrt{0.49}}=4.9 \mathrm{~m} \end{aligned}
 Question 3
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 _______
 A 353.92 B 225.24 C 475.35 D 525.85
GATE CE 2019 SET-2   Geotechnical Engineering
Question 3 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}
 Question 4
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 _______
 A 240 B 360 C 390 D 270
GATE CE 2019 SET-1   Geotechnical Engineering
Question 4 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}
 Question 5
The contact pressure and settlement distribution for a footing are shown in the figure.

The figure corresponds to a
 A rigid footing on granular soil B flexible footing on granular soil C flexible footing on saturated clay D rigid footing on cohesive soil
GATE CE 2018 SET-2   Geotechnical Engineering
Question 5 Explanation:
Rigid footing on granlar soil.
 Question 6
The width of a square footing and the diameter of a circular footing are equal. If both the footings are placed on the surface of sandy soil, the ratio of the ultimate bearing capacity of circular footing to that of square footing will be
 A 4/3 B 1 C 3/4 D 2/3
GATE CE 2018 SET-1   Geotechnical Engineering
Question 6 Explanation:
Footing placed on surface
$\therefore \quad D_{f}=0$
For square footing,
$q_{u}=C N_{c}+\gamma D_{f} N_{q}+0.4 B \gamma N_{\gamma}$
For circular footing,
$q_{u}=C N_{c}+\gamma D_{f} N_{q}+0.3 B \gamma N_{\gamma}$
For sandy soil,
\begin{aligned} C&=0\\ \text{Ratio }\frac{\left(q_{u}\right)_{\text {circular}}}{\left(q_{u}\right)_{\text {square}}}&=\frac{3}{4} \end{aligned}
 Question 7
The percent reduction in the bearing capacity of a strip footing resting on sand under flooding condition (water level at the base of the footing) when compared to the situation where the water level is at a depth much greater than the width of footing, is approximately
 A 0 B 25 C 50 D 100
GATE CE 2018 SET-1   Geotechnical Engineering
Question 7 Explanation:
For strip footing on sand (c=0)
$q_{u}=\gamma D_{f} N_{q}+0.5 \mathrm{B} \gamma N_{\gamma}$
In flooding condition water level rises to base of footing hence IIIrd term unit weight of soil will change and IInd term unit weight will be unaffected.
$\begin{array}{ll} \therefore & q_{u}=\gamma D_{f} N_{q}+0.5 B \gamma N_{\gamma} \\ \because & \gamma \simeq \frac{1}{2} \gamma_{s a t} \end{array}$
Hence third term reduced and second term will be same thereby percentage reduction will not be 50%.
According to option approach answer should be 25%
Note: If water table rises to ground level then both $\gamma$ will reduce to $\gamma$. Hence, percentage reduction would be approximately 50%.
 Question 8
The plate load test was conducted on clayey strata by using a plate of 0.3mx0.3 m dimensions, and the ultimate load per unit area for the plate was found to be 180 kPa. The ultimate bearing capacity (in kPa) of a 2 m wide square footing would be
 A 27 B 180 C 1200 D 2000
GATE CE 2017 SET-2   Geotechnical Engineering
Question 8 Explanation:
Plate load test was conducted on a clay
strata.
Bearing capacity for the plate is 180 kPa.
Ultimate bearing capacity in case of clay is independent of width of footing, hence ultimate bearing capacity for footing will be 180kPa.
 Question 9
A strip footing is resting on the ground surface of a pure clay bed having an undrainedcohesion $C_{u}$. The ultimate bearing capacity of the footing is equal to
 A $2\pi C_{u}$ B $\pi C_{u}$ C $(\pi+1) C_{u}$ D $(\pi+2) C_{u}$
GATE CE 2017 SET-1   Geotechnical Engineering
Question 9 Explanation:
Footing is at surface Hence,
\begin{aligned} \text{Hence},\quad D_{t}&=0 \\ q_{u}&=C N_{c}+\gamma D_{f} N_{q}+0.5 \mathrm{B} \gamma N_{\gamma} \end{aligned}
$\Rightarrow \quad$For clay
\begin{aligned} N_{Y} &=0, N_{q}=1 \\ \therefore \quad q_{U}&=C N_{c} \end{aligned}
As per Terzaghi
$N_{c}=5.7$
and as per Meyerhoff and Prandtl.
\begin{aligned} N_{c} &=5.14 \\ \therefore \quad q_{u} &=(\pi+2) C_{u} \end{aligned}
 Question 10
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 ________.
 A 746.2 B 298.48 C 776 D 766
GATE CE 2016 SET-2   Geotechnical Engineering
Question 10 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
$q_{n s}=\frac{q_{n u}}{F}=\frac{746.2}{2.5}=298.48 \mathrm{kN} / \mathrm{m}^{2}$
There are 10 questions to complete.