Question 1 |
A circular pile of diameter 0.6 \mathrm{~m} and length 8 \mathrm{~m} was constructed in a cohesive soil stratum having the following properties: bulk unit weight = 19 \mathrm{kN} / \mathrm{m}^{3}; angle of internal friction =0^{\circ} and cohesion =25 \mathrm{kPa}.
The allowable load the pile can carry with a factor of safety of 3 is ____ \mathrm{kN} (round off to one decimal place).
[Adopt: Adhesion factor, \alpha=1.0 and Bearing capacity factor, \left.\mathrm{N}_{\mathrm{C}}=9.0\right]
The allowable load the pile can carry with a factor of safety of 3 is ____ \mathrm{kN} (round off to one decimal place).
[Adopt: Adhesion factor, \alpha=1.0 and Bearing capacity factor, \left.\mathrm{N}_{\mathrm{C}}=9.0\right]
112.5 | |
321.5 | |
125.6 | |
146.9 |
Question 1 Explanation:
End bearing capacity
\mathrm{Q}_{\mathrm{pu}}=\mathrm{C}_{\mathrm{u}} \mathrm{N}_{\mathrm{c}} \mathrm{A}_{\mathrm{b}}
where, \quad C_{u}=25 \mathrm{kPa}
N_{c}=9
A_{b}=\frac{\pi}{4} \times 0.6^{2}=0.283 \mathrm{~m}^{2}
\therefore \quad \mathrm{Q}_{\mathrm{Pu}}=25 \times 9 \times 0.283 =63.62 \mathrm{kN}
Skin friction capacity
\mathrm{Q}_{\mathrm{s}}=\alpha \mathrm{C}_{\mathrm{u}} \mathrm{A}_{\mathrm{s}}
Where,
\begin{aligned} \alpha & =1.0 \\ \mathrm{C}_{\mathrm{u}} & =25 \mathrm{kPa} \\ \mathrm{A}_{\mathrm{s}} & =\pi \mathrm{DL}=\pi \times 0.6 \times 8 \\ & =15.08 \mathrm{~m}^{2} \\ \therefore \quad Q_{\mathrm{s}} & =1 \times 25 \times 15.08=377 \mathrm{kN} \end{aligned}
\therefore Ultimate pile capacity =63.62+377 =440.62 \mathrm{kN}
Allowable load =\frac{440.62}{3}=146.87 \mathrm{kN}
\mathrm{Q}_{\mathrm{pu}}=\mathrm{C}_{\mathrm{u}} \mathrm{N}_{\mathrm{c}} \mathrm{A}_{\mathrm{b}}
where, \quad C_{u}=25 \mathrm{kPa}
N_{c}=9
A_{b}=\frac{\pi}{4} \times 0.6^{2}=0.283 \mathrm{~m}^{2}
\therefore \quad \mathrm{Q}_{\mathrm{Pu}}=25 \times 9 \times 0.283 =63.62 \mathrm{kN}
Skin friction capacity
\mathrm{Q}_{\mathrm{s}}=\alpha \mathrm{C}_{\mathrm{u}} \mathrm{A}_{\mathrm{s}}
Where,
\begin{aligned} \alpha & =1.0 \\ \mathrm{C}_{\mathrm{u}} & =25 \mathrm{kPa} \\ \mathrm{A}_{\mathrm{s}} & =\pi \mathrm{DL}=\pi \times 0.6 \times 8 \\ & =15.08 \mathrm{~m}^{2} \\ \therefore \quad Q_{\mathrm{s}} & =1 \times 25 \times 15.08=377 \mathrm{kN} \end{aligned}
\therefore Ultimate pile capacity =63.62+377 =440.62 \mathrm{kN}
Allowable load =\frac{440.62}{3}=146.87 \mathrm{kN}
Question 2 |
A group of 9 friction piles are arranged in a square grid maintaining equal spacing in all directions. Each pile is of diameter 300 \mathrm{~mm} and length 7 \mathrm{~m}. Assume that the soil is cohesionless with effective friction angle \phi^{\prime}=32^{\circ}. What is the center-to-center spacing of the piles (in \mathrm{m} ) for the pile group efficiency of 60 \% ?
0.582 | |
0.486 | |
0.391 | |
0.677 |
Question 2 Explanation:
\mathrm{n}=9
Dia. of pile =300 \mathrm{~mm}
Length of pile (\mathrm{L})=7 \mathrm{~m}
Using the Converse Labarre formula.
m=3, \quad n=3
n_{g}=1-\frac{\theta}{90} \frac{[m(n-1)+n(m-1)]}{m n}
\begin{aligned} 0.6 & =1-\frac{\theta}{90} \times \frac{3 \times 2+3 \times 2}{3 \times 3} \\ \frac{\theta}{90} \times \frac{12}{9} & =0.4 \\ \theta & =\frac{0.9 \times 9 \times 90}{12} \\ \tan \theta & =\tan \left(\frac{0.4 \times 9 \times 90}{12}\right) \\ \frac{d}{s} & =\tan \theta \\ \frac{0.3 \mathrm{~m}}{\mathrm{~s}} & =0.5095 \Rightarrow \mathrm{S}=0.588 \mathrm{~m} \end{aligned}
Question 3 |
A group of total 16 piles are arranged in a square grid format. The center-to-
center spacing (s) between adjacent piles is 3 m. The diameter (d) and length of embedment of each pile are 1 m and 20 m, respectively. The design capacity of
each pile is 1000 kN in the vertical downward direction. The pile group efficiency
( \eta _g) is given by
\eta _g=1-\frac{\theta }{90}\left [ \frac{(n-1)m+(m-1)n}{mn} \right ]
where m and n are number of rows and columns in the plan grid of pile arrangement, and \theta =\tan ^{-1}\left ( \frac{d}{s} \right ).
The design value of the pile group capacity (in kN) in the vertical downward direction is __________________. (round off to the nearest integer)
\eta _g=1-\frac{\theta }{90}\left [ \frac{(n-1)m+(m-1)n}{mn} \right ]
where m and n are number of rows and columns in the plan grid of pile arrangement, and \theta =\tan ^{-1}\left ( \frac{d}{s} \right ).
The design value of the pile group capacity (in kN) in the vertical downward direction is __________________. (round off to the nearest integer)
14520 | |
23521 | |
11085 | |
25120 |
Question 3 Explanation:
\begin{aligned}
\eta _g&=1-\frac{\theta }{90}\left \{ \frac{m(n-1)+n(m-1)}{mn} \right \} \\
\theta &=\tan ^{-1}\left ( \frac{d}{s} \right )=\tan ^{-1}\left ( \frac{1}{3} \right )=18.43^{\circ}\\
\eta _g&=\frac{Q_{ug}}{nQ_{up}}\\
Q_{ug}&=\left [ 1-\frac{18.43}{90}\left \{ \frac{4(4-1)+4(4-1)}{4 \times 4} \right \} \right ](16 \times 1000)\\
&=11085kN
\end{aligned}
Question 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.