Question 1 |

A timber pile of length 8 m and diameter 0.2 m is driven with a 20 kN drop hammer, falling freely from a height of 1.5 m. The total penetration of the pile in the last 5 blows is 40 mm. Use the Engineering News Record expression. Assumea factor of safety of 6 and empirical factor (allowing reduction in the theoretical set, due to energy losses) of 2.5cm. The safe load carrying capacity of the pile (in kN, round off to 2 decimal places) is _______

80.24 | |

151.51 | |

182.54 | |

224.32 |

Question 1 Explanation:

L = 8 m; d = 0.2 m, 20 kN = W ( drop hammer)

H = 1.5 m

Penetration in 5 blows = 40 mm

\begin{aligned} \therefore \; \text{in 1 blow}&=\frac{40}{5}=8mm=0.8cm\\ Q_{safe}&=\left ( \frac{WH}{s+c} \right )\times \frac{1}{FOS}\\ &=\frac{1}{6}\left [ \frac{20kN \times (1.5 \times 100)}{0.8cm+2.5cm} \right ]\\ Q_{safe}&=151.51kN \end{aligned}

H = 1.5 m

Penetration in 5 blows = 40 mm

\begin{aligned} \therefore \; \text{in 1 blow}&=\frac{40}{5}=8mm=0.8cm\\ Q_{safe}&=\left ( \frac{WH}{s+c} \right )\times \frac{1}{FOS}\\ &=\frac{1}{6}\left [ \frac{20kN \times (1.5 \times 100)}{0.8cm+2.5cm} \right ]\\ Q_{safe}&=151.51kN \end{aligned}

Question 2 |

A reinforced concrete circular pile of 12 m length and 0.6 m diameter is embedded in stiff clay which has an undrained unit cohesion of 110 kN/m^2. The adhesion factor is 0.5. The Net Ultimate Pullout (uplift) Load for the pile (in kN, round off to 1 decimal place) is _______

1244.1 | |

1110.5 | |

560.8 | |

1820.4 |

Question 2 Explanation:

\begin{aligned} &\text{Net Ultimate pullout} \\ &=\alpha CA_s =0.5 \times 110(\pi dL) \\ &=0.5 \times 110(\pi \times 0.6 \times 12)\\&=1244.1kN \end{aligned}

Question 3 |

A group of nine piles in a 3 x 3 square pattern is embedded in a soil strata comprising dense sand underlying recently filled clay layer, as shown in the figure. The perimeter of an individual pile is 126 cm. The size of pile group is 240 cm x 240 cm. The recently filled clay has undrained shear strength of 15 kPa and unit weight of 16 kN/m^{3}

The negative frictional load (in kN, up to two decimal places) acting on the pile group is______

The negative frictional load (in kN, up to two decimal places) acting on the pile group is______

258.38 | |

479.67 | |

369.45 | |

472.32 |

Question 3 Explanation:

3 \times 3 pile group

Negative skin friction for group.

=\alpha \bar{c}(4 B L)+ weight of soil in

negative zone.

\begin{aligned} =& \alpha \bar{c}(4 B L)+\gamma(\text { Area } \times \text { length }] \\ =& 1 \times 15[4 \times 2.4 \times 2]+16\left[2.4^{2} \times 2\right] \\ &=472.32 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Negative skin friction in individual action.

\begin{array}{l} =n[\alpha \bar{c} \text { (perimeter } \times l] \\ =9[0.5 \times 15 \times 1.26 \times 2] \\ \text { [Assume } \alpha=0.5] \\ =170.1 \mathrm{kN} / \mathrm{m}^{2} \end{array}

Negative skin friction is maximum of above

(two)

Q_{n t}=472.32 \mathrm{kN} / \mathrm{m}^{2}

Negative skin friction for group.

=\alpha \bar{c}(4 B L)+ weight of soil in

negative zone.

\begin{aligned} =& \alpha \bar{c}(4 B L)+\gamma(\text { Area } \times \text { length }] \\ =& 1 \times 15[4 \times 2.4 \times 2]+16\left[2.4^{2} \times 2\right] \\ &=472.32 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Negative skin friction in individual action.

\begin{array}{l} =n[\alpha \bar{c} \text { (perimeter } \times l] \\ =9[0.5 \times 15 \times 1.26 \times 2] \\ \text { [Assume } \alpha=0.5] \\ =170.1 \mathrm{kN} / \mathrm{m}^{2} \end{array}

Negative skin friction is maximum of above

(two)

Q_{n t}=472.32 \mathrm{kN} / \mathrm{m}^{2}

Question 4 |

A 0.5 m x 0.5 m square concrete pile is to be driven in a homogeneous clayey soil having undrained shear strength, c_{u}= 50 kPa and unit weight, \gamma = 18.0 kN/m^{3}. The design capacity of the pile is 500 kN. The adhesion factor \alpha is given as 0.75. The length of the pile required for the above design load with a factor of safety of 2.0 is

5.2 m | |

5.8 m | |

11.8 m | |

12.5 m |

Question 4 Explanation:

\begin{aligned} C_{u}=& 50 \mathrm{kPa}, \gamma=18 \mathrm{kN} / \mathrm{m}^{3}, \mathrm{FOS}=2 \\ Q_{\text {up}}=& 9 \mathrm{C} \times B^{2}+\alpha \bar{C}(4 \mathrm{BL})=1000 \mathrm{kN} \\ 1000=& 9 \times 50 \times 0.5 \times 0.5+0.75 \\ & \times 50(4 \times 0.5 \mathrm{L}) \\ L=& 11.83 \mathrm{m} \end{aligned}

Question 5 |

Two identical concrete piles having the plan dimensions 50 cm x 50 cm are driven into a homogeneous sandy layer as shown in the figures. Consider the bearing capacity factor N_{q}, for \phi =30^{\circ} as 24

If Q_{P1} and Q_{P2} represent the ultimate point bearing resistance of the piles under dry and submerged conditions, respectively, which one of the following statements is correct?

If Q_{P1} and Q_{P2} represent the ultimate point bearing resistance of the piles under dry and submerged conditions, respectively, which one of the following statements is correct?

Q_{P1} \gt Q_{P2} By about 100% | |

Q_{P1} \lt Q_{P2} By about 100% | |

Q_{P1} \gt Q_{P2} By about 5% | |

Q_{P1} \lt Q_{P2} By about 5% |

Question 5 Explanation:

The point bearing resistance of piles in sandy soils

Q_P=A_b cdot \sigma '_v \cdot N_q

As the area at base (A_b) and N_q are same for both the piles given, Q_P \propto \sigma '_v

For dry sand condition, \sigma '_v=\sigma _v=20 \times \gamma

For submerged condition, \sigma '_v=20 \times \gamma '

\begin{aligned} \text{Since, } \gamma '&= \frac{1}{2} \gamma \\ \therefore \;\; QP_2&\approx \frac{1}{2} QP_1\\ QP_1& \approx QP_2 \end{aligned}

Therefore, QP_1 \gt QP_2 by about 100%.

Q_P=A_b cdot \sigma '_v \cdot N_q

As the area at base (A_b) and N_q are same for both the piles given, Q_P \propto \sigma '_v

For dry sand condition, \sigma '_v=\sigma _v=20 \times \gamma

For submerged condition, \sigma '_v=20 \times \gamma '

\begin{aligned} \text{Since, } \gamma '&= \frac{1}{2} \gamma \\ \therefore \;\; QP_2&\approx \frac{1}{2} QP_1\\ QP_1& \approx QP_2 \end{aligned}

Therefore, QP_1 \gt QP_2 by about 100%.

Question 6 |

It is proposed to drive H-piles up to a depth of 7 m at a construction site. The average surfacearea of the H-pile is 3 m^{2} per meter length. The soil at the site is homogeneous sand, havingan effective friction angle of 32^{\circ}. The ground water table (GWT) is at a depth of 2 m below the ground surface. The unit weights of the soil above and below the GWT are 16 kN/m^{3} and 19 kN/m^{3} respectively. Assume the earth pressure coefficient, K= 1.0, and the angle of wall friction \delta =23^{\circ}, The total axial frictional resistance (in kN, up to one decimal place) mobilized on the pile against the driving is _____

350.7 | |

40.7 | |

390.8 | |

309.2 |

Question 6 Explanation:

\begin{aligned} Q_{s t} &=\frac{1}{2} k \gamma L_{1} \tan \delta A_{s_{1}}+\frac{1}{2} K\left[\gamma L_{1}+\gamma^{\prime} L_{2}\right] \tan \delta A_{s_{2}} \\ &=\frac{1}{2} \times 1 \times 16 \times 2 \tan 23^{\circ}(3 \times 2) \\ &+\frac{1}{2} \times 1[16 \times 2+77.95] \times \tan 23^{\circ}(5 \times 3) \\ &=390.8 \mathrm{kN} \end{aligned}

Question 7 |

A pile of diameter 0.4 m is fully embedded in a clay stratum having 5 layers, each 5 m thick as shown in the figure below. Assume a constant unit weight of soil as 18 kN/m^{3} for all the layers. Using \lambda-method (\lambda=0.15 for 25 m embedment length) and neglecting the end bearing component, the ultimate pile capacity (in kN) is ________.

1060.2 | |

1100.5 | |

1010.4 | |

1625.7 |

Question 7 Explanation:

Ultimate bearing capacity

\begin{aligned} &=\lambda\left(\sigma_{\mathrm{v} a v}+2 C_{u}\right) A_{s} \\ &=0.15\left[18 \times 12.5+2 C_{u}\right] \times(\pi \times 0.4 \times 25) \\ \end{aligned}

C_{u} is the average value of all layers i.e., 60 kPa.

\begin{aligned} &=0.15[18 \times 12.5+2 \times 60] \times(\pi \times 0.4 \times 25) \\ &=1625.77 \mathrm{kN} \end{aligned}

\begin{aligned} &=\lambda\left(\sigma_{\mathrm{v} a v}+2 C_{u}\right) A_{s} \\ &=0.15\left[18 \times 12.5+2 C_{u}\right] \times(\pi \times 0.4 \times 25) \\ \end{aligned}

C_{u} is the average value of all layers i.e., 60 kPa.

\begin{aligned} &=0.15[18 \times 12.5+2 \times 60] \times(\pi \times 0.4 \times 25) \\ &=1625.77 \mathrm{kN} \end{aligned}

Question 8 |

A single vertical friction pile of diameter 500 mm and length 20 m is subjected to a vertical compressive load. The pile is embedded in a homogeneous sandy stratum where: angle of internal friction is \phi=30^{\circ}, dry unit weight (\gamma _{d})=20 kN/m^{3} and angle of wall friction is \delta =2\varphi /3. Considering the coefficient of lateral earth pressure (K)=2.7 and the bearing capacity factor (N_q)=25, the ultimate bearing capacity of the pile (in kN) is________

196.56 | |

1965.6 | |

617.5 | |

6174.5 |

Question 8 Explanation:

Homogeneous sandy stratum

\begin{aligned} \phi &=30^{\circ} \\ \gamma_{d} &=20 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

Wall friction angle.

\delta=\frac{2}{3} \phi=\frac{2}{3} \times 30=20^{\circ}

Lateral earth pressure coefficient,

K=2.7

Bearing capacity factor,

N_{q}=25

Ultimate load capacity,

Q_{u}=?

Vertical effective stress at 20m.

\overline{\sigma_{v}}=20 \times 20=400 \mathrm{kN} / \mathrm{m}^{2}

From 0 to 20m, unit point bearing

resistance and skin friction resistance remain

constant at

\overline{\sigma_{\mathrm{v}}}=400 \mathrm{kN} / \mathrm{m}^{2}

The ultimate load capacity is given by

\begin{aligned} Q_{u}&=q_{\mathrm{up}} \cdot A_{p}+q_{\mathrm{s}} \cdot A_{s}\\ &=f_{s}\left(Q_{u p}=0\right. for friction pile) \\ \text{where}\quad q_{s}&=\frac{1}{2} \cdot \overline{\sigma_{v}} \cdot K \cdot \tan \delta\\ &=\frac{1}{2} \times 400 \times 2.7 \tan 20^{\circ} \\ &=196.54 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Skin friction resistance,

\begin{aligned} Q_{u}=f_{s} &=196.54 \times \pi \times 0.5 \times 20 \\ &=6174.49 \mathrm{kN} \end{aligned}

Question 9 |

The action of negative skin friction on the pile is to

increase the ultimate load on the pile | |

reduce the allowable load on the pile | |

maintain the working load on the pile | |

reduce the settlement of the pile |

Question 10 |

Four columns of a building are to be located within a plot size of 10 m x 10 m. The expected load
on each column is 4000 kN. Allowable bearing capacity of the soil deposit is 100 kN/m^{2}. The type of foundation best suited is

isolated footing | |

raft foundation | |

pile foundation | |

combined footing |

Question 10 Explanation:

Load on each column =400 kN

Bearing capacity of soil =100 \mathrm{kN} / \mathrm{m}^{2}

\therefore \quad Area of each footing required

\begin{aligned} &=\frac{400}{100} m^{2}=4 m^{2}\\ \therefore \quad \text{Total area required } &=4 \times 4=16 \mathrm{m}^{2} \\ \therefore \quad \frac{\text { Foundation area }}{\text { Plot area }}&=\frac{16}{100}=16 \% \end{aligned}

If the foundation area is less than 40% of plot area, then isolated foundation is suitable otherwise provide raft foundation or combined footing.

Bearing capacity of soil =100 \mathrm{kN} / \mathrm{m}^{2}

\therefore \quad Area of each footing required

\begin{aligned} &=\frac{400}{100} m^{2}=4 m^{2}\\ \therefore \quad \text{Total area required } &=4 \times 4=16 \mathrm{m}^{2} \\ \therefore \quad \frac{\text { Foundation area }}{\text { Plot area }}&=\frac{16}{100}=16 \% \end{aligned}

If the foundation area is less than 40% of plot area, then isolated foundation is suitable otherwise provide raft foundation or combined footing.

There are 10 questions to complete.