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Foundation Engineering PYQ – TN TRB Lecturer

      

1. In order to obtain a good quality of undisturbed soil sample the area ratio of the sampling tube should be _____ .

[TN TRB 2012: 1 Mark]

  1. 8%
  2. 16%
  3. 24%
  4. 32%
Answer: Option A
Explanation:
As per IS 1892(1979) Clause 4.1.1, the area ratio should be kept as low as possible consistent with the strength requirements of the sample tube. Its value should not be greater than about 20 percent for stiff formations; for soft sensitive clays an area ratio of 10 percent or less should be preferred.

2. A 0.3 m square bearing plate settles by 10 mm in the plate load test in cohesionless soil, when the intensity of load is 200 kN/m2. Estimate the settlement of a shallow foundation of 2 m square under the same intensity of loading.

[TN TRB 2012: 1 Mark]

  1. 30.25 mm
  2. 3.31 mm
  3. 17.39 mm
  4. 5.75 mm
Answer: Option A
Explanation:
The relationship between settlement of plate (Sp), Settlement of foundation (Sf), width of the plate (Bp) and width of the foundation (Bf) in cohesionless soil is,

Sf

Sp

= {

Bf (Bp + 0.3)

Bp (Bf + 0.3)

}2

⇒

Sf

10

= {

2 x (0.3 + 0.3)

0.3 x (2 + 0.3)

}2

⇒ Sf = 30.25 mm

3.  In the passive state of cohesionless soil, minor stress is _____ .

[TN TRB 2012: 1 Mark]

  1. Horizontal
  2. Vertical
  3. 45° to horizontal
  4. 30° to horizontal
Answer: Option B

Explanation:
In case of passive earth pressure, the horizontal stress (σH) acting on the soil is more than the vertical stress (σV). Hence, horizontal stress is the major stress and vertical stress is the minor stress.


4.  For a sandy soil having an angle of internal friction 30°, the ratio of passive and active lateral earth pressure will be _____ .

[TN TRB 2012: 1 Mark]

  1. 2
  2. 4
  3. 3
  4. 9
Answer: Option D
Explanation:

Active Earth Pressure Coefficient, Ka =

1 – sin φ

1 + sin φ

=

1 – sin 30°

1 + sin 30°

⇒ Ka =

1/2

3/2

=

1

3

Passive Earth Pressure Coefficient, Kp =

1 + sin φ

1 – sin φ

=

1 + sin 30°

1 – sin 30°

⇒ Kp=

3/2

1/2

= 3


Thus, ratio of passive and active lateral earth pressure is

Kp

Ka

=

3

1/3

= 9


5.  In stability analysis of slopes, factor of safety with respect to height is given by _____ .

[TN TRB 2012: 1 Mark]

  1. FH = HC/H
  2. FH = H/HC
  3. FH = 2H/HC
  4. FH = HC/2H
Answer: Option A
Explanation:

The factor of safety with respect to height in slope stability analysis is

FOS = Hc / H

Where,
Hc – Critical height of soil mass
H – Actual height of soil mass

6.  Stability analysis by Swedish method of slices gave following values per meter run of 10 m high embankment (i) total shearing force = 480 kN (ii) total normal force = 1950 kN (iii) total neutral force (iv) length of arc = 22 m if C = 24 kN/m2, ∅ = 6°, then the factor of safety with respect to shear strength is _____ .

[TN TRB 2012: 1 Mark]

  1. 1.57
  2. 1.67
  3. 1.47
  4. 1.75
Answer: *
Explanation:

FOS =

CL + ΣN tan φ

ΣT

⇒ FOS =

(24 x 22) + (1950 x tan 6°)

480

⇒ FOS =

732.95

480

⇒ FOS = 1.53

7.  Immediate settlement of cohesionless soils are given by _____ .

[TN TRB 2012: 1 Mark]

  1. Si =
    qB (1 – µ2) I Es
  2. Si =
    H C
    log e
    σ0 + Δσ σ0
  3. Si =
    Es (1 – µ2) I qB
  4. Si =
    C H
    log e
    σ0 + Δσ σ0
Answer: Option A
Explanation:
Immediate settlement as per Schleicher’s method is,

Si =

qB (1 – µ2)I

Es


Where,
q – Uniform pressure at the base of footing (or) contact pressure intensity
B – Least lateral dimension of footing
µ – Poisson’s ratio of soil mass

I – Influence factor or shape factor, depending on

Length

Width

ratio of footing

Es – Young’s modulus of soil

8.  Newmark’s influence chart is used to determine the vertical stress at any point under a uniformly loaded _____ .

[TN TRB 2012: 1 Mark]

  1. Circular area only
  2. Rectangular area only
  3. Trapezoidal area only
  4. Any shape
Answer: Option D
Explanation:
Newmark’s influence chart is used to determine the vertical stress at any point under a uniformly loaded area of any shape.

9.  For the pile group shown in the figure, the efficiency of pile group determined by Feld’s rule is _____ .

Felds Rule - Supporting Diagram

[TN TRB 2012: 1 Mark]

  1. 80%
  2. 85%
  3. 88%
  4. 90%
Answer: Option A
Explanation:
As per Feld’s rule, the efficiency of any pile in a pile group is

η =

16 – Number of adjacent piles

16


For the corner piles,
Number of adjacent piles = 3

Efficiency of corner pile, ηc =

16 – 3

16

x 100% = 81.25%

For the middle piles,
Number of adjacent piles = 4

Efficiency of corner pile, ηm =

16 – 4

16

x 100% = 75%

Thus, the efficiency of pile group,

ηg =

(4 x ηc) + (1 x ηm)

4 + 1

⇒ ηg =

(4 x 81.25) + (1 x 75)

4 + 1

⇒ ηg = 80%

10.  The piles which are used to protect waterfront structures against impacts from floating objects is _____ .

[TN TRB 2012: 1 Mark]

  1. Batter pile
  2. Anchor pile
  3. Fender pile
  4. Sheet pile
Answer: Option C
Explanation:

Fender piles are provided for waterfront structures near the sea or river to protect the structures that are subjected to tidal waves caused by the ship, seismic activities, etc.,
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