Question 1 |

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 1 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 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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}

Question 5 |

The infinite sand slope shown in the figure is on theverge of sliding failure. The ground water tablecoincides with the ground surface. Unit weight of water \gamma _{w}=9.81\; kN/m^{3}.

The value of the effective angle of internal friction (indegrees, up to one decimal place) of the sand is________

The value of the effective angle of internal friction (indegrees, up to one decimal place) of the sand is________

0.6 | |

34.3 | |

33.3 | |

43.3 |

Question 5 Explanation:

Infinite sand slope:

GWT coincides with the ground surface, therefore

F.O.S. =\frac{\gamma^{\prime}}{\gamma_{\mathrm{sat}}} \times \frac{\tan \phi^{\prime}}{\tan \beta}

As slope is on the verge of sliding failure hence,

\begin{array}{l} \mathrm{F} \mathrm{O} . \mathrm{S}=\frac{\gamma^{\prime} \tan \phi^{\prime}}{\gamma_{\mathrm{sat}} \tan \beta}=1\\ (21-9.81) \tan \phi^{\prime}=21 \tan 20^{\circ} \\ \phi=34.3^{\circ} \end{array}

GWT coincides with the ground surface, therefore

F.O.S. =\frac{\gamma^{\prime}}{\gamma_{\mathrm{sat}}} \times \frac{\tan \phi^{\prime}}{\tan \beta}

As slope is on the verge of sliding failure hence,

\begin{array}{l} \mathrm{F} \mathrm{O} . \mathrm{S}=\frac{\gamma^{\prime} \tan \phi^{\prime}}{\gamma_{\mathrm{sat}} \tan \beta}=1\\ (21-9.81) \tan \phi^{\prime}=21 \tan 20^{\circ} \\ \phi=34.3^{\circ} \end{array}

Question 6 |

In friction circle method of slope stability analysis, if r defines the radius of the slip circle, the radius of friction circle is:

r\sin \phi | |

r | |

r\cos \phi | |

r\tan \phi |

Question 6 Explanation:

Question 7 |

An infinitely long slope is made up of a c-\phi soil having the properties: cohesion (c)=20 kPa, and dry unit weight (\gamma _{d} )= 16kN/m^{3}. The angle of inclination and critical height of the slope are 40^{\circ} and 5 m, respectively. To maintain the limiting equilibrium, the angle of internal friction of the soil (in degree) is _____

18.7 | |

22.4 | |

28.6 | |

34.3 |

Question 7 Explanation:

\begin{array}{l} F_{0}=\frac{c}{\gamma H \sin \beta \cdot \cos \beta}+\frac{\tan \phi}{\tan \beta} \geq 1 \\ c=20 \mathrm{kPa}, \gamma=16 \mathrm{kN} / \mathrm{m}^{3}, \beta=40^{\circ} \text { and } \\ H=5 \mathrm{m} \\ \Rightarrow \quad, \quad \phi \geq 22.44^{\circ} \end{array}

Question 8 |

A long slope is formed in a soil with shear strength parameters: {c}'=0 and {\phi }'=34^{\circ}. A firm stratum lies below the slope and it is assumed that the water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use \gamma_{sat}=18 kN/m^{3} and \gamma_{w}=10 kN/m^{3}. The maximum slope angle (in degrees) to ensure a factor of safety of 1.5, assuming a potential failure surface parallel to the slope, would be

45.3 | |

44.7 | |

12.3 | |

11.3 |

Question 8 Explanation:

\begin{aligned} F O S&=\frac{Y_{s u}}{\gamma_{\text {sat }}} \cdot \frac{\tan \phi}{\tan \beta}\\ \Rightarrow \tan \beta&=\left(\frac{\gamma_{\text {sub }}}{\gamma_{\text {sat }}} \cdot \frac{\tan \phi}{F O S}\right)\\ \Rightarrow \quad \tan \beta&=\frac{\gamma_{\text {sub }}}{\gamma_{\text {aat }}} \cdot \frac{\tan \phi}{1.5}\\ \tan \beta &=\frac{(18-10)}{18 \times 1.5} \times \tan 34^{\circ} \\ \Rightarrow \quad \beta &=11.30^{\circ} \end{aligned}

Question 9 |

The soil profile above the rock surface for a 25^{\circ} infinite slope is shown in the figure, where s_{u} is the undrained shear strength and \gamma _{t} is total unit weight. The slip will occur at a depth of

8.83m | |

9.79m | |

7.83m | |

6.53m |

Question 9 Explanation:

The slip will occur when shear stress is greater then or equal to shear strength.

\begin{aligned} \tau &\geq S_{u} \\ \left(\gamma_{1} z_{1}+\gamma_{2} z_{2}\right) \sin \beta \cos \beta &\geq S_{u}\\ \Rightarrow \frac{\left(16 \times 5+20 \times z_{2}\right) \sin 2 \beta}{2} &\geq S_{u} \\ \Rightarrow \quad \frac{\left(80+20 z_{2}\right) \sin 50^{\circ}}{2} &\geq 60 \\ z_{2}&=3.83 \mathrm{m} \\ \text{ Depth of slip }=5+3.83&=8.83 \mathrm{m} \end{aligned}

Question 10 |

The factor of safety of an infinite soil slope shown in the figure having the properties c=0, \phi =35^{\circ},\; \gamma _{dry}=16kN/m^{3} and \gamma _{sat}=20kN/m^{3} is approximately equal to

0.7 | |

0.8 | |

1 | |

1.2 |

Question 10 Explanation:

Factor of safety.

F=\left[1-\frac{\gamma_{w} h}{\gamma_{air} z}\right] \frac{\tan \phi}{\tan \beta}

Assuming.

\begin{aligned} \gamma_{w} &=10 \mathrm{kN} / \mathrm{m}^{3} \\ \therefore\quad F &=\left[1-\frac{10 \times 8}{20 \times 10}\right] \frac{\tan 35^{\circ}}{\tan 30^{\circ}} \\ &=0.73 \end{aligned}

F=\left[1-\frac{\gamma_{w} h}{\gamma_{air} z}\right] \frac{\tan \phi}{\tan \beta}

Assuming.

\begin{aligned} \gamma_{w} &=10 \mathrm{kN} / \mathrm{m}^{3} \\ \therefore\quad F &=\left[1-\frac{10 \times 8}{20 \times 10}\right] \frac{\tan 35^{\circ}}{\tan 30^{\circ}} \\ &=0.73 \end{aligned}

There are 10 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.