Question 1 |

A consolidated drained (CD) triaxial test was carried out on a sand sample with the known effective shear strength parameters, C^{\prime}=0 and \phi^{\prime}=30^{\circ}. In the test, prior to the failure, when the sample was undergoing axial compression under constant cell pressure, the drainage valve was accidentally closed. At the failure, 360 \mathrm{kPa} deviatoric stress was recorded along with 70 \mathrm{kPa} pore water pressure. If the test is repeated without such error, and no back pressure is applied in either of the tests, what is the deviatoric stress (in \mathrm{kPa}, in integer) at the failure? _____

11 | |

200 | |

350 | |

500 |

Question 1 Explanation:

We know that,

\bar{\sigma}_{1 F}=\bar{\sigma}_{3 \mathrm{~F}} \tan ^{2}\left(45+\frac{\phi^{\prime}}{2}\right)+20 \tan \left(45+\frac{\phi^{\prime}}{2}\right)

Assuming c^{\prime}=0 under both the conditions when drainage line in suddenly dosed.

\left(\sigma_{\mathrm{cf}}+360-70\right)=\left(\sigma_{\mathrm{cy}}-70\right) \tan ^{2} 60+0\;\;\;...(i)

When drainage is open and assuming the confining stress \left(\sigma_{\mathrm{cf}}\right).

\left[\left(\sigma_{\mathrm{cf}}+\sigma_{2}\right)-0\right]=\left[\sigma_{\mathrm{cf}}-0\right] \tan ^{2} 60^{\circ}+0 \;\;\;...(ii)

From (i) and (ii), we get

\begin{aligned} 290-\sigma_{2} & =-70 \times \tan ^{2} 60^{\circ} \\ 290-\sigma_{d} & =-210 \\ \sigma_{d} &=500 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

\bar{\sigma}_{1 F}=\bar{\sigma}_{3 \mathrm{~F}} \tan ^{2}\left(45+\frac{\phi^{\prime}}{2}\right)+20 \tan \left(45+\frac{\phi^{\prime}}{2}\right)

Assuming c^{\prime}=0 under both the conditions when drainage line in suddenly dosed.

\left(\sigma_{\mathrm{cf}}+360-70\right)=\left(\sigma_{\mathrm{cy}}-70\right) \tan ^{2} 60+0\;\;\;...(i)

When drainage is open and assuming the confining stress \left(\sigma_{\mathrm{cf}}\right).

\left[\left(\sigma_{\mathrm{cf}}+\sigma_{2}\right)-0\right]=\left[\sigma_{\mathrm{cf}}-0\right] \tan ^{2} 60^{\circ}+0 \;\;\;...(ii)

From (i) and (ii), we get

\begin{aligned} 290-\sigma_{2} & =-70 \times \tan ^{2} 60^{\circ} \\ 290-\sigma_{d} & =-210 \\ \sigma_{d} &=500 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Question 2 |

For the flow setup shown in the figure (not to scale), the hydraulic conductivities of the two soil samples, Soil 1 and Soil 2, are 10 \mathrm{~mm} / \mathrm{s} and 1 \mathrm{~mm} / \mathrm{s}, respectively. Assume the unit weight of water as 10 \mathrm{kN} / \mathrm{m}^{3} and ignore the velocity head. At steady state, what is the total head (in m, rounded off to two decimal places) at any point located at the junction of the two samples? ____

3.25 | |

4.54 | |

6.35 | |

2.24 |

Question 2 Explanation:

External pressure of 10 \mathrm{kPa} can be converted to equivalent hydrostatic, head =\frac{10 \mathrm{kPa}}{10 \mathrm{kN} / \mathrm{m}^{3}}=1 \mathrm{~m}

Total head at A,

\mathrm{TH}_{\mathrm{A}}= Datum head + Pressure head

\mathrm{TH}_{\mathrm{A}}=3 \mathrm{~m}+(1 \mathrm{~m}+1 \mathrm{~m})

\mathrm{TH}_{\mathrm{A}}=5 \mathrm{~m}

Total head loss due to seepage through soil

\begin{aligned} & \mathrm{h}_{\mathrm{L}}=5 \mathrm{~m} \\ & Q=K_{e q} \mathrm{i} A\\ & =\left(\frac{L_1+L_2}{\frac{L_1}{K_1}+\frac{L_2}{K_2}}\right) \times \frac{h_L}{L} \times \mathrm{A} \\ & =\left(\frac{1+1}{\frac{1}{10}+\frac{1}{1}}\right) \times \frac{5}{2} \times \mathrm{A} \\ & =4.545 \mathrm{~A} \end{aligned}

Head loss through soil - 1 only

\begin{aligned} \mathrm{h}_{L_{1}} & =\frac{\mathrm{Q}}{\mathrm{K}_{1} \mathrm{~A}} \times \mathrm{L}_{1} \\ & =\frac{4.545 \mathrm{~A}}{10 \times \mathrm{A}} \times 1 \\ & =0.4545 \mathrm{~m} \end{aligned}

\therefore Total head at junction =\mathrm{TH}_{\mathrm{A}}-\mathrm{h}_{\mathrm{L} 1} =5-0.4545=4.545 \mathrm{~m} \approx 4.54 \mathrm{~m}

Question 3 |

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 3 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 4 |

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.]

[Assume the applicable shape and depth factor values as unity; ground water level at greater depth.]

1.2 | |

2.4 | |

3.4 | |

4.5 |

Question 4 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

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

Question 5 |

The figure shows a vertical retaining wall with backfill consisting of cohesive-frictional soil and a failure plane developed due to passive earth pressure. The forces acting on the failure wedge are: P as the reaction force between the wall and the soil, R as the reaction force on the failure plane, C as the cohesive force along the failure plane and W as the weight of the failure wedge. Assuming that there is no adhesion between the wall and the wedge, identify the most appropriate force polygon for the wedge.

A | |

B | |

C | |

D |

Question 5 Explanation:

There are 5 questions to complete.

Question No. 5 refers to Steel Structure