Question 1 |

At a site, Static Cone Penetration Test was carried out. The measured point (tip)
resistance q_c was 1000 kPa at a certain depth.

The friction ratio (f_r) was estimated as 1% at the same depth. The value of sleeve (side) friction (in kPa) at that depth was _______ . (in integer)

The friction ratio (f_r) was estimated as 1% at the same depth. The value of sleeve (side) friction (in kPa) at that depth was _______ . (in integer)

25 | |

50 | |

10 | |

100 |

Question 1 Explanation:

Friction ratio,

f_r=\frac{q_s}{q_t}=1 %

q_s= Sleeve friction

q_t= Tip friction

\frac{q_s}{q_t}=\frac{1}{100}

q_s=\frac{1000}{100}=10 kPa

f_r=\frac{q_s}{q_t}=1 %

q_s= Sleeve friction

q_t= Tip friction

\frac{q_s}{q_t}=\frac{1}{100}

q_s=\frac{1000}{100}=10 kPa

Question 2 |

An unsupported slope of height 15 m is shown in the figure (not to scale), in which the slope face makes an angle 50^{\circ} with the horizontal. The slope material comprises purely cohesive soil having undrained cohesion 75 kPa. A trial slip circle KLM, with a radius 25 m, passes through the crest and toe of the slope and it subtends an angle 60^{\circ} at its center O. The weight of the active soil mass (W, bounded by KLMN) is 2500 kN/m, which is acting at a horizontal distance of 10 m from the toe of the slope. Consider the water table to be present at a very large depth from the ground surface.

Considering the trail slip circle KLM, the factor of the safety against the failure of slope under undrained condition (round off to two decimal places) is ___________

Considering the trail slip circle KLM, the factor of the safety against the failure of slope under undrained condition (round off to two decimal places) is ___________

1.96 | |

2.45 | |

8.25 | |

6.32 |

Question 2 Explanation:

\begin{aligned} &\begin{aligned} \mathrm{FOS} &=\frac{\text { Resisting moment }}{\text { Actuating moment }} \\ \mathrm{FOS} &=\frac{\mathrm{C}_{\mathrm{u}} l R}{w \bar{x}} \\ l &=\text { Length of ac } \mathrm{KLM} \\ \bar{x} &=\text { Distance of ' } \mathrm{w}^{\prime} \text { from toe } \\ \Rightarrow \qquad \qquad \quad \mathrm{FOS} &=\frac{75 \times 2 \pi \times 25 \times \frac{60}{36} \times 25}{2500 \times 10} \\ \Rightarrow \qquad \qquad \quad \mathrm{FOS} &=1.96 \end{aligned} \end{aligned}

Question 3 |

A 10 m high slope of dry clay soil (unit weight = 20 kN/m^3), with a slope angle of 45^{\circ}
and the circular slip surface, is shown in the figure (not drawn to the scale). The weight
of the slip wedge is denoted by W. The undrained unit cohesion (c_u) is 60 kPa.

The factor of safety of the slope against slip failure, is

The factor of safety of the slope against slip failure, is

1.84 | |

1.57 | |

0.58 | |

1.67 |

Question 3 Explanation:

As per GATE Official Answer Key MTA (Marks to All)

Consider unit length of slope

Area of circular arc

\begin{aligned} &=\frac{\theta }{360} \times \pi r^2-\text{Area of }\Delta \\ &=\frac{90}{360} \times \pi \times 10^2-\frac{1}{2}\times 10 \times 10\\ &=28.54m^2 \end{aligned}

\begin{aligned} \text{Height of wedge}& = \text{Volume} \times \gamma \\ &=(\text{Area} \times 1) \times \gamma \\ &=28.54 \times 1 \times 20\\ &=570.8 kN\\ FOS&=\frac{M_R}{M_0}\\ &=\frac{[c \times (r\theta )]\times r}{W \times x}\\ &=\frac{60 \times 10 \times \frac{\pi}{1} \times 10}{570.8 \times 4.48}\\ &=3.68 \end{aligned}

Consider unit length of slope

Area of circular arc

\begin{aligned} &=\frac{\theta }{360} \times \pi r^2-\text{Area of }\Delta \\ &=\frac{90}{360} \times \pi \times 10^2-\frac{1}{2}\times 10 \times 10\\ &=28.54m^2 \end{aligned}

\begin{aligned} \text{Height of wedge}& = \text{Volume} \times \gamma \\ &=(\text{Area} \times 1) \times \gamma \\ &=28.54 \times 1 \times 20\\ &=570.8 kN\\ FOS&=\frac{M_R}{M_0}\\ &=\frac{[c \times (r\theta )]\times r}{W \times x}\\ &=\frac{60 \times 10 \times \frac{\pi}{1} \times 10}{570.8 \times 4.48}\\ &=3.68 \end{aligned}

Question 4 |

A fully submerged infinite sandy slope has an inclination of 30^{\circ} with the horizontal. The
saturated unit weight and effective angle of internal friction of sand are 18 kN/m^3 and
38^{\circ}, respectively. The unit weight of water is 10 kN/m^3. Assume that the seepage is
parallel to the slope. Against shear failure of the slope, the factor of safety (round off
to two decimal places) is _______.

0.21 | |

0.6 | |

0.44 | |

0.78 |

Question 4 Explanation:

\begin{aligned} F.O.S.&=\frac{\gamma '}{\gamma _{sat}} \frac{\tan \phi }{\tan \beta }\\ &=\left ( \frac{18-10}{18} \right )\frac{\tan 38^{\circ}}{\tan 30^{\circ}} \\ &=0.601 \end{aligned}

Question 5 |

For the construction of a highway. A cut is to be made as shown in the figure.

The soil exhibits c'=20kPa,\; \; \phi '=18^{\circ} and the undrained shear strength = 80 kPa. The unit weight of water is 9.81kN/m^{3}. The unit weights of the soil above and below the ground water table are 18 and 20 kN/m^{3}, respectively. If the shear stress at Point A is 50 kPa, the factors of safety against the shear failure at this point, considering the undrained and drained conditions, respectively, would be

The soil exhibits c'=20kPa,\; \; \phi '=18^{\circ} and the undrained shear strength = 80 kPa. The unit weight of water is 9.81kN/m^{3}. The unit weights of the soil above and below the ground water table are 18 and 20 kN/m^{3}, respectively. If the shear stress at Point A is 50 kPa, the factors of safety against the shear failure at this point, considering the undrained and drained conditions, respectively, would be

1.6 and 0.9 | |

0.9 and 1.6 | |

0.6 and 1.2 | |

1.2 and 0.6 |

Question 5 Explanation:

Drained parameters:

c^{\prime}=20 \mathrm{kPa}, \phi^{\prime}=18^{\circ}

Undrained shear strength.

S=80 \mathrm{kPa}

Normal stress at point A.

\begin{aligned} \bar{\sigma}_{n} &=\sigma_{n}-u=2 \times \gamma_{b}+4 \gamma \\ \bar{\sigma}_{n} &=2 \times 18+4(20-981) \\ &=76.76 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Shear stress at point A=50kPa

FOS under undrained condition

=\frac{\text { Shear strength }}{\text { Sthear stress }}=\frac{80}{50}=1.6

FOS in drained condition

\begin{aligned} &=\frac{c+\bar{\sigma}_{n} \tan \phi^{\prime}}{\tau} \\ &=\frac{20+76.76 \tan 18^{\circ}}{50}=0.9788 \end{aligned}

c^{\prime}=20 \mathrm{kPa}, \phi^{\prime}=18^{\circ}

Undrained shear strength.

S=80 \mathrm{kPa}

Normal stress at point A.

\begin{aligned} \bar{\sigma}_{n} &=\sigma_{n}-u=2 \times \gamma_{b}+4 \gamma \\ \bar{\sigma}_{n} &=2 \times 18+4(20-981) \\ &=76.76 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Shear stress at point A=50kPa

FOS under undrained condition

=\frac{\text { Shear strength }}{\text { Sthear stress }}=\frac{80}{50}=1.6

FOS in drained condition

\begin{aligned} &=\frac{c+\bar{\sigma}_{n} \tan \phi^{\prime}}{\tau} \\ &=\frac{20+76.76 \tan 18^{\circ}}{50}=0.9788 \end{aligned}

There are 5 questions to complete.

Dig. of Q1 of Stability Analysis of Slopes is not correct please check and replace it.

Thank You Jayesh Deore,

We have updated image.