Foundation Engineering Questions and Answers Part-2

1. The plastic state of stress when the failure is imminent was investigated by ___________
a) Rankine
b) Darcy
c) Skempton
d) Terzaghi

Answer: a
Explanation: The plastic state of stress failure of the soil samples was studied by Rankine in the year 1860. Darcy is known for his works in the field of soil permeability.

2. ___________ is used for maintaining the ground surface at different elevations.
a) Floor
b) Retaining wall
c) Ceiling
d) Roof

Answer: b
Explanation: The retaining wall is used to resist the lateral earth pressure due to the backfill soil. It prevents the landslides by maintaining the ground surface at different elevations in hilly regions.

3. The material retained by the retaining wall is known as_________
a) roof
b) slab
c) backfill
d) footing

Answer: c
Explanation: The material retained by the retaining wall is known as backfill. The surface of the backfill can be both horizontal as well as inclined. It should be noted that the angle does not exceed the angle of repose of the soil.

4. The backfill has its top surface ____________
a) horizontal only
b) inclined only
c) both horizontal and inclined
d) vertical only

Answer: c
Explanation: The backfill can have its top surface as a horizontal surface or as an inclined surface. It should be noted that the angle does not exceed the angle of repose of the soil or else it will fail.

5. The portion of backfill lying above a horizontal plane of wall is called ________
a) surcharge
b) base
c) foot
d) truss

Answer: a
Explanation: The portion of backfill lying above a horizontal plane of wall is called as the surcharge. The surcharge is a load which is imposed upon the surface of the soil close enough to the excavation to cause a lateral pressure to act on the system in addition to the basic earth pressure.

6. Surcharge angle β is ____________
a) inclination of surcharge to the vertical
b) inclination of surcharge to the normal of the wall
c) inclination of surcharge to the horizontal
d) inclination of surcharge to the tangent to the wall

Answer: c
Explanation: The surcharge is a load which is imposed upon the surface of the soil close enough to the excavation to cause a lateral pressure to act on the system in addition to the basic earth pressure. Its inclination to the horizontal is known as Surcharge angle β.

7. A body is said to be in plastic equilibrium, if every point of it is ________
a) in the verge of failure
b) in equilibrium
c) in stable state
d) elastic state

Answer: a
Explanation: A body is said to be in plastic equilibrium, if every point of it is in the verge of failure. The transition from the state of plastic equilibrium to the state of plastic flow represents the failure of the soil mass.

8. The stress condition during plastic equilibrium can be represented by _____________
a) \(\frac{σ_1-σ_3}{2}\frac{σ_1+σ_3}{2} sinφ=cosφ \)
b) \(\frac{σ_1-σ_3}{2}+\frac{σ_1+σ_3}{2} sinφ=c cosφ\)
c) \(\frac{σ_1*σ_3}{2}-\frac{σ_1+σ_3}{2} sinφ=c cosφ\)
d) \(\frac{σ_1-σ_3}{2}-\frac{σ_1+σ_3}{2} sinφ=c cosφ\)

Answer: d
Explanation: The stress condition during plastic equilibrium can be represented by the equation of Mohr-Coulomb,
\(\frac{σ_1-σ_3}{2}-\frac{σ_1+σ_3}{2} sinφ=c cosφ,\)
Where, σ1=major principal stress
σ3=minor principal stress
φ=angle of internal friction.

9. The stress condition during plastic equilibrium in terms of tangent of angle is _______
a) \(σ_1=tan⁡(45°+\frac{φ}{2})+σ_3 tan^2 (45°+\frac{φ}{2}) \)
b) \(σ_1=2c+σ_3 tan^2 (45°+\frac{φ}{2})\)
c) \(σ_1=2c \,tan⁡(45°+\frac{φ}{2})+σ_3\)
d) \(σ_1=2c \,tan⁡(45°+\frac{φ}{2})+σ_3 tan^2 (45°+\frac{φ}{2})\)

Answer: d
Explanation: The stress condition during plastic equilibrium can be represented by the equation of Mohr-Coulomb,
\(\frac{σ_1-σ_3}{2}-\frac{σ_1+σ_3}{2} sinφ=c cosφ,\)
since tan⁡φ=sinφ/cosφ
∴ \(σ_1=2c \,tan⁡(45°+\frac{φ}{2})+σ_3 tan^2 (45°+\frac{φ}{2}).\)

10. The flow value is given by ______
a) \(N_φ=cot^2 (45°+\frac{φ}{2})\)
b) \(N_φ=cos^2 (45°+\frac{φ}{2})\)
c) \(N_φ=tan^2 (45°+\frac{φ}{2})\)
d) \(N_φ=sin^2 (45°+\frac{φ}{2})\)

Answer: c
Explanation: The flow value is given by,
\(N_φ= tan^2 (45°+\frac{φ}{2}),\)
where, Nφ=flow value
φ=angle of internal friction.