GATE CE · Geotechnical Engineering

Stability Analysis Of Slopes – GATE CE Geotechnical EngineeringPractice Questions & PYQs

Generate GATE-level questions on Stability Analysis Of Slopes in Geotechnical Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

22 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025NAT

The bank of a canal has the profile shown in the figure. The material is a homogeneous clay with a bulk unit weight of $20\ \text{kN/m}^3$ , undrained cohesion of $30\ \text{kPa}$ and it is fully saturated ( $\phi_u = 0$ ). For the trial slip circle shown, the area ABCDEA is $150\ \text{m}^2$ and the centroid is at P. A tension crack (DE) of $2.5\ \text{m}$ deep was also observed. Assume unit weight of water is $9.81\ \text{kN/m}^3$ and consider 1 m run of the bank for the analysis. Considering the canal is empty and the tension crack is completely filled with water, the factor of safety against slope failure of the bank is ______ (round off to two decimal places).

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📖 Explanation

We use the concept of moment equilibrium for the slip surface. Since $\phi = 0$ (undrained condition), we use the following formula for factor of safety: $FS = \frac{c \cdot L}{W \cdot \frac{h}{R} - u \cdot \frac{h_w}{R}}$ Where: $c = 30\ \text{kPa}$ (undrained cohesion) $A = 150\ \text{m}^2$ (area of slip mass) $\gamma = 20\ \text{kN/m}^3$ (bulk unit weight of clay) $W = \gamma \cdot A = 20 \cdot 150 = 3000\ \text{kN}$ $u = \gamma_w = 9.81\ \text{kN/m}^3$ $R = 14\ \text{m}$ (radius of slip circle) $h = h_{\text{cg}} = 2.5\ \text{m}$ (vertical distance of CG from center) $h_w = 2.5\ \text{m}$ (depth of water in tension crack) $L = \text{length of arc} = \theta \cdot R$ (We don't have $\theta$ , but can use total resisting moment and driving moment instead) Resisting moment: $M_R = c \cdot A \cdot R = 30 \cdot 150 \cdot R = 4500R$ Driving moment (including water pressure in tension crack): Weight moment: $M_W = W \cdot h = 3000 \cdot 2.5 = 7500$ Water moment in tension crack (triangular pressure distribution): $p = \frac{1}{2} \cdot \gamma_w \cdot h_w^2 = \frac{1}{2} \cdot 9.81 \cdot (2.5)^2 = 30.65$ (acting at $\frac{1}{3} \cdot 2.5 = 0.833\ \text{m}$ from bottom) So moment due to water: $M_u = 30.65 \cdot 0.833 \approx 25.53$ Total driving moment: $M_D = 7500 + 25.53 = 7525.53$ Now, factor of safety: $FS = \frac{M_R}{M_D} = \frac{4500R}{7525.53}$ Substitute $R = 14\ \text{m}$ : $FS = \frac{4500 \cdot 14}{7525.53} = \frac{63000}{7525.53} \approx \boxed{1.07}$

Q2GATE 2025NAT

A cut slope is made in a silty clay soil for a new road project, as shown in the figure. The locations of ground water table (GWT) and potential failure surface are shown in the figure. After the cut is made, the excess pore water pressure is fully dissipated, and the shear stress at the point $A$ is $60 \mathrm{kN} / \mathrm{m}^{2}$ . The factor of safety at the point $A$ for longterm stability is _________ (rounded off to 2 decimal places). Note : Shear strength properties of silty clay: $c^{\prime}=15 \mathrm{kN} / \mathrm{m}^{2}, \phi^{\prime}=15^{\circ}$ , and $c_{u}=75 \mathrm{kN} / \mathrm{m}^{2}$ Unit weight of soil above the $\operatorname{GWT}(\gamma)=19 \mathrm{kN} / \mathrm{m}^{3}$ Unit weight of soil below the $\operatorname{GWT}\left(\gamma_{\text {sat }}\right)=20 \mathrm{kN} / \mathrm{m}^{3}$ Unit weight of water $\left(\gamma_{w}\right)=9.81 \mathrm{kN} / \mathrm{m}^{3}$

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📖 Explanation

For long term, effective shear parameters will be used,

\begin{aligned} \mathrm{FOS} & =\frac{C^{\prime}+\bar{\sigma}_{n} \tan \phi^{\prime}}{\tau} \\ \bar{\sigma}_{n} & =\left(5 \gamma_{B}+6.5 \gamma_{\text {sat }}\right)-6.5 \gamma_{w} \\ & =5 \times 19+6.5 \times 20-6.5 \times 9.81 \\ & =161.235 \mathrm{kN} / \mathrm{m}^{2} \\ \mathrm{FOS} & =\frac{15+161.235 \tan 15^{\circ}}{60} \\ & =0.97 \end{aligned}

Q3GATE 2024NAT

An infinite slope is made up of cohesionless soil with seepage parallel to and upto the sloping surface. The angle of slope is $30^{\circ}$ with respect to horizontal ground surface. The unit weight of the saturated soil and water are $20 \mathrm{kN} / \mathrm{m}^{3}$ and $10 \mathrm{kN} / \mathrm{m}^{3}$ respectively. The minimum angle of shearing resistance of soil (in degrees) for the critically stable condition of the slope is _____ (rounded off to the nearest integer).

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📖 Explanation

\begin{aligned} \text { FOS }&=\frac{\gamma_{\text {sub }}}{\gamma_{\text {sat }}} \times \frac{\tan \phi}{\tan \beta}\\ \text{Here, } \beta & =30^{\circ} \\ \gamma_{\text {sat }} & =20 \mathrm{kN} / \mathrm{m}^{3} \\ \gamma_{w} & =10 \mathrm{kN} / \mathrm{m}^{3} \\ \gamma_{\text {sub }} & =\gamma_{\text {sat }}-\gamma_{w} \\ & =20-10=10 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

$\therefore \quad \gamma_{\text {sub }}=\gamma_{\text {sat }}-\gamma_{w}$ For critically stable condition, $\mathrm{FOS}=1$

\begin{array}{ll}\because & 1=\frac{20-10}{20} \times \frac{\tan \phi}{\tan 30^{\circ}} \\ \Rightarrow & \phi=49.10^{\circ}\end{array}

Q4GATE 2022NAT

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)

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📖 Explanation

Friction ratio, $f_r=\frac{q_s}{q_t}=1 %$ $q_s=$ Sleeve friction $q_t=$ Tip friction $\frac{q_s}{q_t}=\frac{1}{100}$ $q_s=\frac{1000}{100}=10 kPa$

Q5GATE 2021NAT

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 ___________

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📖 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}

Q6GATE 2020MCQ

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

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📖 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}

Explanation Figure 2

Q7GATE 2020NAT

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 _______.

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📖 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}

Q8GATE 2017MCQ

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 9.81 $kN/m^{3}$ . 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

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📖 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}

Q9GATE 2017NAT

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________

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📖 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}

Q10GATE 2015MCQ

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

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Q11GATE 2014MCQ

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

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📖 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}

Explanation Figure 1

Q12GATE 2014NAT

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 _____

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📖 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}

Q13GATE 2013MCQ

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

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📖 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}

Q14GATE 2007MCQ

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

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📖 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}

Q15GATE 2006MCQ

List-I below gives the possible types of failure for a finite soil slope and List-II gives the reasons for these different types of failure. Match the items in List-I with the items in List-II

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📖 Explanation

Face failure or slope failure can occur when the Slope angle $\beta$ is very high and the soil close to the toe is quite strong or the soil in the upper part of slope is relatively weak. Base failure can occur when the soil below the toe is relatively weak and soft and the slope is flat. Toe failure occurs in steep slopes when the soil mass above the base and below the base is homogeneous.

Q16GATE 2005MCQ

For two infinite slopes (one in dry condition and other in submerged condition) in a sand deposit having the angle of shearing resistance 30 $^{\circ}$ , factor of safety was determined as 1.5 (for both slopes). The slope angles would have been

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📖 Explanation

For dry or submerged slopes, Factor of safety

\begin{aligned} &=\frac{\tan \phi}{\tan \beta} \\ \Rightarrow \quad \tan \beta &=\frac{\tan 30^{\circ}}{1.5} \\ \Rightarrow \quad & \beta=21.05^{\circ} \end{aligned}

Q17GATE 2004MCQ

An infinite soil slope with an inclination of $35^{\circ}$ is subjected to seepage parallel to its surface. The soil has $c'=100 kN/m^{2} \; and \; {\phi}' =35^{\circ}$ . Using the concept of mobilized cohesion and friction, at a factor of safety of 1.5 with respect of shear strength, the mobilized friction angle is

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📖 Explanation

The factor of safety with respect to shear strength is given by $F_{5}=\frac{\tau_{1}}{\tau}=\frac{c^{\prime}+\sigma_{n}^{\prime} \tan \phi^{\prime}}{\tau}$ The shear strength mobilised, thus may be written as

\begin{aligned} \tau &=\frac{C^{\prime}}{F_{s}}+\sigma_{n}^{\prime} \frac{\tan \phi^{\prime}}{F_{s}} \\ &=c_{m}+\sigma_{n}^{\prime} \tan \phi_{m}^{\prime}\\ \therefore \tan \phi_{m}^{\prime}&=\frac{\tan \phi^{\prime}}{F_{s}} \\ \Rightarrow \quad \phi_{m}^{\prime}&=\tan ^{-1}\left(\frac{\tan 35^{\circ}}{1.5}\right)=21.05^{\circ} \end{aligned}

Q18GATE 2003MCQ

A granular soil possesses saturated density of 20 $kN/m^{3}$ . Its effective angle of internal friction is 35 degrees. If the desired factor of safety is 1.5, the safe angle of slope for this soil, when seepage occurs at and parallel to the slope surface, will be

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📖 Explanation

\begin{aligned} \text{Given }\phi^{\prime}&=35^{\circ}: \gamma_{\mathrm{sat}}=20 \mathrm{kN} / \mathrm{m}^{3} \\ F&=1.5 ; \beta=? ; \gamma=\gamma_{\mathrm{sat}}-\gamma_{\mathrm{w}} \end{aligned}

Factor of safety

\begin{aligned} F&=\frac{\gamma^{\prime} \tan \phi^{\prime}}{\gamma_{\text {sat }} \tan \beta} \\ \Rightarrow \quad 1.5&=\frac{\left(\gamma_{\mathrm{sat}}-\gamma_{\mathrm{w}}\right) \tan 35^{\circ}}{\gamma_{\mathrm{sat}} \tan \beta} \\ \Rightarrow \quad \tan \beta&=\frac{(20-10) \tan 35^{\circ}}{1.5 \times 20} \\ \Rightarrow \quad \beta&=13.14^{\circ} \simeq 13^{\circ} \end{aligned}

Q19GATE 2003MCQ

A canal having side slopes 1:1 is proposed to be constructed in a cohesive soil to a depth of 10 m below the ground surface. The soil properties are: $\phi _u=15^{\circ}$ , $C_n$ =12kPa, e=1.0, $G_{s}$ =2.65. If there is a sudden draw down of water in the canal and if Taylor's Stability Number for the reduced value of $\phi _v$ is 0.126, the factor of safety with respect to cohesion against the failure of bank slopes will be

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📖 Explanation

Even if there is a sudden drawdown of water in the canal, the soil remains saturated for sometime. Hence, unit weight of the soil will be the saturated unit weight i.e.

\begin{aligned} \gamma_{\text {sat }} &=\gamma+\gamma_{w}=8.09+9.81 \\ &=17.9 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

$\therefore$ Taylor's stability number,

\begin{aligned} S_{n} &=\frac{C_{u}}{F_{c} \gamma_{s a t} H} \\ \Rightarrow \quad F_{c} &=\frac{c_{v}}{S_{n} \times \gamma_{s a t} \times H} \\ \Rightarrow &=\frac{12}{0.126 \times 17.9 \times 10}=0.53 \end{aligned}

Q20GATE 2002MCQ

An infinite slope is to be constructed in a soil. The effective stress strength parameters of the soil are c'=0 and ${\phi}' =30^{\circ}$ . The saturated unit weight of the slope is 20 $kN/m^{3}$ and the unit weight of water is 10 $kN/m^{3}$ . Assuming that seepage is occurring parallel to the slope, the maximum slope angle for a factor of safety of 1.5 would be

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📖 Explanation

\begin{aligned} \tan \phi_{\mathrm{w}} &=\frac{\gamma^{\prime}}{\gamma_{\mathrm{sat}}}\left(\frac{1}{\mathrm{F}_{\mathrm{d}}} \tan \phi^{\prime}\right) \\ &=\frac{10}{20}\left(\frac{1}{1.5} \tan 30^{\circ}\right) \text { or } \phi_{\mathrm{w}}=10.89 \end{aligned}

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