Question 1 |

The most appropriate triaxial test to assess the long-term stability of an excavated clay slope is

consolidated drained test | |

unconsolidated undrained test | |

consolidated undrained test | |

unconfined compression test |

Question 1 Explanation:

To assess the long term stability of clayey soil, the results of consolidated drained (CD) test are used.

Question 2 |

Based on drained triaxial shear tests on sands and clays, the representative variations of volumetric strain (\Delta V / V) with the shear strain (\gamma) is shown in the figure.

Choose the CORRECT option regarding the representative behaviour exhibited by Curve P and Curve Q.

Choose the CORRECT option regarding the representative behaviour exhibited by Curve P and Curve Q.

Curve P represents dense sand and overconsolidated clay, while Curve Q represents loose sand and normally consolidated clay | |

Curve P represents dense sand and normally consolidated clay, while Curve Q represents loose sand and overconsolidated clay | |

Curve P represents loose sand and overconsolidated clay, while Curve Q represents dense sand and normally consolidated clay | |

Curve P represents loose sand and normally consolidated clay, while Curve Q represents dense sand and overconsolidated clay |

Question 2 Explanation:

Question 3 |

In a drained tri-axial compression test, a sample of sand fails at deviator stress of 150 kPa
under confining pressure of 50 kPa. The angle of internal friction (in degree, round off
to the nearest integer) of the sample, is ________

78 | |

54 | |

37 | |

44 |

Question 3 Explanation:

\text{Sand }(C=0); \; \sigma _d=150;\; \sigma _3=50;\; \sigma _1=200

\sigma _1=\sigma _3 \tan^2\left ( 45+\frac{\phi }{2} \right )+2c\tan \left ( 45+\frac{\phi }{2} \right )

200=50 \tan^2 \left ( 45+\frac{\phi }{2} \right )

\phi =36.87^{\circ}

So, the angle of internal friction to the nearest integer is 37^{\circ}.

\sigma _1=\sigma _3 \tan^2\left ( 45+\frac{\phi }{2} \right )+2c\tan \left ( 45+\frac{\phi }{2} \right )

200=50 \tan^2 \left ( 45+\frac{\phi }{2} \right )

\phi =36.87^{\circ}

So, the angle of internal friction to the nearest integer is 37^{\circ}.

Question 4 |

The total horizontal and vertical stresses at a point X in a saturated sandy medium are 170 kPa and 300 kPa, respectively. The static pore-water pressure is 30 kPa. At failure, the excess pore-water pressure is measured to be 94.50 kPa, and the shear stresses on the vertical and horizontal planes passing through the point X are zero. Effective cohesion is 0 kPa and effective angle of internal friction is 36^{\circ}. The shear strength (in kPa, up to two decimal places) at point X is ______

42.23 | |

59.45 | |

78.82 | |

52.52 |

Question 4 Explanation:

\begin{aligned} \because\quad \sigma_{n} &=\frac{300+170}{2}+\frac{300-170}{2} \cos 2 \alpha \\ \alpha &=45^{\circ}+\frac{36^{\circ}}{2}=63^{\circ} \\ \Rightarrow\quad \sigma_{n} &=196.79 \mathrm{kPa} \\ \sigma_{n}^{\prime} &=\sigma_{\mathrm{n}}-u=196.79-(30+94.5) \\ \sigma_{n}^{\prime} &=72.29 \mathrm{kPa} \\ \Rightarrow\quad \tau &=c+\sigma_{\mathrm{n}}^{\prime} \tan \phi \\ &=72.29 \text { tan } 36=52.52 \mathrm{kPa} \end{aligned}

Question 5 |

A conventional drained triaxial compression test was conducted on a normally consolidated clay sample under an effective confining pressure of 200 kPa. The deviator stress at failure was found to be 400 kPa. An identical specimen of the same clay sample is isotropically consolidated to a confining pressure of 200 kPa and subjected to standard undrained triaxial compression test. If the deviator stress at failure is 150 kPa, the pore pressure developed (in kPa, up to one decimal place) is ______

100 | |

125 | |

25 | |

75 |

Question 5 Explanation:

I^{st} Specimen : Drained condition

\bar{\sigma}_{3}=200 \mathrm{kPa}: \sigma_{\mathrm{d}}=400 \mathrm{kPa}: \bar{\sigma}_{1}=600 \mathrm{kPa}

II^{nd} Specimen : Undrained condition

\begin{aligned} \sigma_{3} &=200 \mathrm{kPa} \\ \sigma_{\mathrm{d}} &=150 \mathrm{kPa} \\ \sigma_{1} &=\sigma_{3}+\sigma_{\mathrm{d}}=350 \mathrm{kPa} \end{aligned}

Let pore pressure developed is u

\begin{aligned} \therefore \quad \bar{\sigma}_{3}&=(200-u)\\ \bar{\sigma}_{1}&=(350-u)\\ \end{aligned}

From stress relationship

\bar{\sigma}_{1}=\bar{\sigma}_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)+2 c\tan\left(45^{\circ}+\frac{\phi}{2}\right)

For clay under drained condition c=0

\begin{aligned} \therefore \quad \bar{\sigma}_{1} &=\bar{\sigma}_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right) \\ 600 &=200 \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right) \\ \phi &=30^{\circ} \end{aligned}

For second specimen,

\begin{aligned} \bar{\sigma}_{1}&=\bar{\sigma}_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)\\ (350-u) &=(200-u) \tan ^{2}\left(45^{\circ}+\frac{30^{\circ}}{2}\right) \\ u &=125 \mathrm{kPa} \end{aligned}

\bar{\sigma}_{3}=200 \mathrm{kPa}: \sigma_{\mathrm{d}}=400 \mathrm{kPa}: \bar{\sigma}_{1}=600 \mathrm{kPa}

II^{nd} Specimen : Undrained condition

\begin{aligned} \sigma_{3} &=200 \mathrm{kPa} \\ \sigma_{\mathrm{d}} &=150 \mathrm{kPa} \\ \sigma_{1} &=\sigma_{3}+\sigma_{\mathrm{d}}=350 \mathrm{kPa} \end{aligned}

Let pore pressure developed is u

\begin{aligned} \therefore \quad \bar{\sigma}_{3}&=(200-u)\\ \bar{\sigma}_{1}&=(350-u)\\ \end{aligned}

From stress relationship

\bar{\sigma}_{1}=\bar{\sigma}_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)+2 c\tan\left(45^{\circ}+\frac{\phi}{2}\right)

For clay under drained condition c=0

\begin{aligned} \therefore \quad \bar{\sigma}_{1} &=\bar{\sigma}_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right) \\ 600 &=200 \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right) \\ \phi &=30^{\circ} \end{aligned}

For second specimen,

\begin{aligned} \bar{\sigma}_{1}&=\bar{\sigma}_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)\\ (350-u) &=(200-u) \tan ^{2}\left(45^{\circ}+\frac{30^{\circ}}{2}\right) \\ u &=125 \mathrm{kPa} \end{aligned}

There are 5 questions to complete.

Question 17 is not of soil but of environmental

Thank u for ur efforts!

No man it’s of geotechnical only