Question 1 |

Read the following statements:

(P) While designing a shallow footing in sandy soil, monsoon season is considered for critical design in terms of bearing capacity.

(Q) For slope stability of an earthen dam, sudden drawdown is never a critical condition.

(R) In a sandy sea beach, quicksand condition can arise only if the critical hydraulic gradient exceeds the existing hydraulic gradient.

(S) The active earth thrust on a rigid retaining wall supporting homogeneous cohesionless backfill will reduce with the lowering of water table in the backfill.

Which one of the following combinations is correct?

(P) While designing a shallow footing in sandy soil, monsoon season is considered for critical design in terms of bearing capacity.

(Q) For slope stability of an earthen dam, sudden drawdown is never a critical condition.

(R) In a sandy sea beach, quicksand condition can arise only if the critical hydraulic gradient exceeds the existing hydraulic gradient.

(S) The active earth thrust on a rigid retaining wall supporting homogeneous cohesionless backfill will reduce with the lowering of water table in the backfill.

Which one of the following combinations is correct?

(P)-True, (Q)-False, (R)-False, (S)-False | |

(P)-False, (Q)-True, (R)-True, (S)-True | |

(P)-True, (Q)-False, (R)-True, (S)-True | |

(P)-False, (Q)-True, (R)-False, (S)-False |

Question 1 Explanation:

In monsoon season sand will be fully
saturated hence this will be critical condition
in designing of shallow foundation.

In case of sudden drawdown flow direction reverses hence for slope stability, it will be critical condition.

In sandy sea beach, quicksand condition can arise only if existing hydraulic gradient exceeds the critical hydraulic gradient.

In case of sudden drawdown flow direction reverses hence for slope stability, it will be critical condition.

In sandy sea beach, quicksand condition can arise only if existing hydraulic gradient exceeds the critical hydraulic gradient.

Question 2 |

A square concrete pile of 10 m length is driven into a deep layer of uniform
homogeneous clay. Average unconfined compressive strength of the clay,
determined through laboratory tests on undisturbed samples extracted from the
clay layer, is 100 kPa. If the ultimate compressive load capacity of the driven
pile is 632 kN, the required width of the pile is _______ mm. (in integer)

(Bearing capacity factor N_c 9; adhesion factor \alpha =0.7 )

(Bearing capacity factor N_c 9; adhesion factor \alpha =0.7 )

400 | |

124 | |

105 | |

600 |

Question 2 Explanation:

\begin{aligned}
Q_{up}&=q_bA_b+q_sA_s\\
C_u&=\frac{q_u}{2}=\frac{100}{2}=50kN/m^2\\
&=9 \times CB^2+\alpha \bar{C}(4BL)\\
632kN&=9 \times 50 \times B^2 +0.7 \times 50(4B \times 10)\\
B&=0.4m\\
B&=400mm
\end{aligned}

Question 3 |

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 _________________

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 _________________

441 | |

564 | |

876 | |

689 |

Question 3 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

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 4 |

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 __________

2.4 | |

5.6 | |

4.9 | |

3.8 |

Question 4 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}

\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 5 |

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 _______

353.92 | |

225.24 | |

475.35 | |

525.85 |

Question 5 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}

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 6 |

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 _______

240 | |

360 | |

390 | |

270 |

Question 6 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}

\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 7 |

The contact pressure and settlement distribution for a footing are shown in the figure.

The figure corresponds to a

The figure corresponds to a

rigid footing on granular soil | |

flexible footing on granular soil | |

flexible footing on saturated clay | |

rigid footing on cohesive soil |

Question 7 Explanation:

Rigid footing on granlar soil.

Question 8 |

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

4/3 | |

1 | |

3/4 | |

2/3 |

Question 8 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}

\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 9 |

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

0 | |

25 | |

50 | |

100 |

Question 9 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%.

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 10 |

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

27 | |

180 | |

1200 | |

2000 |

Question 10 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.

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.

There are 10 questions to complete.