Geotechnical Engineering Questions and Answers – Taylor’s Stability Number and Curves

Geotechnical Engineering Objective Questions

Civil Engineering MCQs

This set of Geotechnical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Taylor’s Stability Number and Curves”.

1. The total cohesion force which resists the slipping along the slip arc at critical equilibrium is proportional to ___________

A) Cohesion and Height of the slope
B) Taylor’s stability number
C) None of the mentioned
D) All of the mentioned

Answer is A)
Explanation: The total cohesive force c L̑, which resist the slipping along the slip arc at critical equilibrium, is proportional to the cohesion c and the height H of the slope.

2. The Taylor’s stability number is based on ___________

A) Height of the slope and Pore pressure
B) Factor of safety
C) None of the mentioned
D) All of the mentioned

Answer is A)
Explanation: The Taylor’s stability number is based on the factor of safety Fc with respect to cohesion assuming that the frictional resistance assuming that frictional force has been fully mobilized.

3. The factor of safety with respect to friction is __________

A) Unity
B) Zero
C) Greater than one
D) None of the mentioned

Answer is A)
Explanation: The friction of safety Fφ with respect to friction is a unity which means that the frictional force has been fully mobilized.

4. According to Taylor, the force causing instability in the sliding wedge is __________

A) Cohesive resistance and Weight of the wedge
B) Total frictional resistance
C) None of the mentioned
D) All of the mentioned

Answer is A)
Explanation: The force causing instability is the weight of the wedge which is equal to unit weight γ.

5. Taylor’s stability number is represented by the term __________

A) SC
B) Sn
C) Cm
D) HC

Answer is B)
Explanation: Taylor’s stability is represented as Sn.

6. Which of the following quantity is called as Taylor’s stability number?

A) c/Fc γ H
B) c/γ H
C) cm/ Fc γ
D) None of the mentioned

Answer is A)
Explanation: If Fc is the factor of safety with respect to cohesion, We have c × H / Fc × γH2 = c / Fc γ H = Sn The dimensional quantity c / Fc γ H is called as Taylor’s stability number Sn.

7. For purely frictional soil, the Taylor’s stability number is ___________

A) Unity
B) Zero
C) >1
D) <1 label="">

Answer is B)
Explanation: For purely frictional soil (c = 0), the stability number is zero, and Taylor’s stability curves do not apply.

8. The stability of a slope for a pure frictional soil, depends upon __________

A) Slope angle
B) Factor of safety
C) Frictional resistance
D) All of the mentioned

Answer is A)
Explanation: The stability of slope for a frictional soil entirely depends upon the slope angle i, irrespective of height of the slope.

9. What will be the factor of safety with respect to cohesion of a clay slope laid at 1 in 2 to a height of 10 m, if the angle of internal friction φ=10° ; c=25 k N/m2 and γ = 19 k N/m3?

A) 4.34
B) 2.06
C) 1.02
D) 20.6

Answer is B)
Explanation: i = tan-1(1/2) = 26.5° For i = 26.5° and φ = 10°, Sn = 0.064 But, Sn = c / Fc γ H Therefore, Fc = c / Sn γ H Fc = 25 / (0.064×19×10) = 2.06.

10. A slope is to be constructed at an inclination of 30° with the horizontal. What will be the safe height of the slope at factor of safety of 1.5? The soil has the properties: c = 15 k N/m2, φ=22.5° and γ = 19 k N/m3.

A) 11.5
B) 20.5
C) 12
D) 18.64

... Answer is A)
Explanation: The mobilized frictional angle φm is given by φm = φ/F = 22.5/1.5 = 15° For, i = 30° and φm = 15°, Sn =0.046 Now H = c/F γ Sn = 15/(0.0046×1.5×19) = 11.5 m.