Geography

Soil and Foundation Engineering

584 Questions

Soil and foundation engineering questions address the physical and mechanical properties of soil critical for construction and stability. The topics include shear strength, soil compaction, void ratios, and subsurface investigation techniques. These advanced geotechnical concepts are typically evaluated in civil engineering and specialized recruitment examinations.

Soil compactionShear strengthBearing capacityVoid ratio analysisSubsurface exploration

Soil and Foundation Engineering Questions

Multiple choice

Identical surcharges are placed at ground surface at sites X and Y, with soil conditions shown alongside and water table at ground surface. The silty clay layers at X and Y are identical. The thin sand layer at Y is continuous and free-draining with a very large discharge capacity. If primary consolidation at X is estimated to complete in 36 months, what would be the corresponding time for completion of primary consolidation at Y?

  1. 2.25 months

  2. 4.5 months

  3. 9 months

  4. 36 months
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

Multiple choice

A sand layer found at sea floor under 20 m water depth is characterised with relative density = 40%, maximum void ratio = 1.0, minimum void ratio = 0.5 and specific gravity of soil solids = 2.67. Assume the specific gravity of sea water to be 1.03 and the unit weight of fresh water to be 9.81 kN/m3.

What would be the change in the effective stress (rounded off to the nearest integer value of kPa) at 30 m depth into the sand layer if the sea water level permanently rises by 2 m?

  1. 19 kPa

  2. 0 kPa

  3. 21 kPa

  4. 22 kPa
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Effective stress is basically stress developed in between soil particles, hence changing water level would not cause any change in effective stress.

$$\triangle\sigma_1 = 0$$

Multiple choice

A sand layer found at sea floor under 20 m water depth is characterised with relative density = 40%, maximum void ratio = 1.0, minimum void ratio = 0.5 and specific gravity of soil solids = 2.67. Assume the specific gravity of sea water to be 1.03 and the unit weight of fresh water to be 9.81 kN/m3.

What would be the effective stress (rounded off to the nearest integer value of kPa) at 30 m depth into the sand layer?

  1. 77 kPa

  2. 273 kPa

  3. 268 kPa

  4. 281 kPa
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 

Multiple choice

A field vane shear testing instrument (shown alongside) was inserted completely into a deposit of soft, saturated silty clay with the vane rod vertical such that the top of the blades were 500 mm below the ground surface. Upon application of a rapidly increasing torque about the vane rod, the soil was found to fail when the torque reached 4.6 Nm. Assuming mobilisation of undrained shear strength on all failure surfaces to be uniform and the resistance mobilised on the surface of the vane rod to be negligible, what would be the peak undrained shear strength (rounded off to the nearest integer value of kPa) of the soil?

  1. 5 kPa

  2. 10 kPa

  3. 15 kPa

  4. 20 kPa
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 

Multiple choice

Quicksand condition occurs when

  1. the void ratio of the soil becomes 1.0

  2. the upward seepage pressure in soil becomes zero

  3. the upward seepage pressure in soil becomes equal to the saturated unit weight of the soil

  4. the upward seepage pressure in soil becomes equal to the submerged unit weight of the soil
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 Quick sand condition occurs when the upward seepage pressure in soil becomes equal to submerged unit weight of the soil. So, effective stress is equal to zero.

Multiple choice

The e-log p curve shown in the figure is representative of

  1. normally consolidated clay

  2. over consolidated clay

  3. under consolidated clay

  4. normally consolidated clayey sand
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 e - log P curve or e Vs P curve on semi log graph paper is for over consolidated clay as p increases and 'e' decreases.

Multiple choice

Direction: The moisture holding capacity of the soil in a 100 hectare farm is 18 cm/m. The field is to be irrigated when 50 percent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75%. Details of crops planned for cultivation are as follows:

   
Crop Root zone depth (m) Peak rate of moisture use
(mm/day)
X 1.0 5.0
Y 0.8 4.0

The area of crop Y that can be irrigated when the available capacity of irrigation system is 40 litres/sec is

  1. 40 hectares

  2. 36 hectares

  3. 30 hectares

  4. 27 hectares
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Available moisture = 18 cm/m * 0.5 (depletion) = 9 cm/m. For crop Y (0.8m depth), depth of water = 0.8 * 9 = 7.2 cm = 72 mm. Peak rate = 4 mm/day. Irrigation interval = 72/4 = 18 days. Daily water requirement = 4 mm/day * Area. Total water needed per day = (4 * Area) / 0.75 (efficiency). Flow rate = 40 L/s = 0.04 m^3/s. Daily capacity = 0.04 * 10 * 3600 = 1440 m^3. Area = (1440 * 0.75) / (0.004 * 18 * 10000) = 27 hectares.

Multiple choice

Direction: The moisture holding capacity of the soil in a 100 hectare farm is 18 cm/m. The field is to be irrigated when 50 percent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75%. Details of crops planned for cultivation are as follows:

   
Crop Root zone depth (m) Peak rate of moisture use
(mm/day)
X 1.0 5.0
Y 0.8 4.0

The capacity of irrigation system required to irrigate crop X in 36 hectares is

  1. 83 litres/sec

  2. 67 litres/sec

  3. 57 litres/sec

  4. 53 litres/sec
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Available moisture = 18 cm/m * 0.5 = 9 cm/m. For crop X (1.0m depth), depth = 9 cm = 90 mm. Peak rate = 5 mm/day. Interval = 90/5 = 18 days. Water required per interval = 90 mm * (36 hectares) = 0.09 m * 360,000 m^2 = 32,400 m^3. Daily requirement = 32,400 / 18 = 1,800 m^3/day. With 75% efficiency, required supply = 1,800 / 0.75 = 2,400 m^3/day. Pump works 10 hours = 36,000 seconds. Flow rate = 2,400 / 36,000 = 0.0666 m^3/s = 66.6 L/s.