Bearing Capacity MCQ Quiz in తెలుగు - Objective Question with Answer for Bearing Capacity - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 20, 2025
Latest Bearing Capacity MCQ Objective Questions
Top Bearing Capacity MCQ Objective Questions
Bearing Capacity Question 1:
The observed N value from a standard penetration test conducted on a saturated sandy soil stratum is 35. The corrected N value for dilatancy can be estimated as:
Answer (Detailed Solution Below)
Bearing Capacity Question 1 Detailed Solution
Explanation
Dilatancy correction:
It is to be applied when N’ obtained after overburden correction, exceeds 15 in saturated fine sands and silts. IS: 2131-1981 incorporates the Terzaghi and Peck recommended dilatancy correction (when N’ > 15) using the equation
Given,
N value = 35
Corrected value for dilatancy is
N'' = 15 + 10 = 25
∴ Corrected value N'' = 25
Bearing Capacity Question 2:
While calculating bearing capacity for a foundation, the water table is likely to permanently remain at a level of 20 m for a square foundation of side 5 m, then the correction factor for the water table will be taken as:
Answer (Detailed Solution Below)
Bearing Capacity Question 2 Detailed Solution
Explanation
Given,
Width of footing = 5 m
And the water table is 20 m below the ground surface
In the given zone Water table is below the base of footing at distance b, from the base, correction factor is given by
But we know that the water correction factor cannot be greater than 1. So, the correction factor for water table will be 1
Bearing Capacity Question 3:
In a Plate Load Test, after every increment in load on plate, settlements are recorded till the settlement is less than ________ and thereafter the load is increased and the test is continued until a load of about ________ times the anticipated ultimate load is applied.
Answer (Detailed Solution Below)
Bearing Capacity Question 3 Detailed Solution
Concept:
Plate load test:
- The plate load test is a field test, which is performed to determine the ultimate bearing capacity of the soil and the probable settlement under a given load.
- This test is very popular for the selection and design of the shallow foundation.
Procedure:
- Excavate test pit up to the desired depth. The pit size should be at least 5 times the size of the test plate (Bp).
- At the center of the pit, a small hole or depression is created. The size of the hole is the same as the size of the steel plate.
- A mild steel plate is used as a load-bearing plate whose thickness should be at least 25 mm thickness and size may vary from 300 mm to 750 mm. The plate can be square or circular. Generally, a square plate is used for square footing and a circular plate is used for circular footing.
- A column is placed at the center of the plate. The load is transferred to the plate through the centrally placed column.
- The load can be transferred to the column either by the gravity loading method or by the truss method.
- At least two dial gauges should be placed at diagonal corners of the plate to record the settlement. The gauges are placed on a platform so that it does not settle with the plate.
- Apply seating load of .7 T/m2 and release before the actual loading starts.
- The initial readings are noted.
- The load is then applied through the hydraulic jack and increased gradually. The increment is generally one-fifth of the expected safe bearing capacity or one-tenth of the ultimate bearing capacity or any other smaller value. The applied load is noted from the pressure gauge.
- The settlement is observed for each increment and from the dial gauge. After increasing the load-settlement should be observed after 1, 4, 10, 20, 40, and 60 minutes and then at hourly intervals until the rate of settlement is less than .02 mm per min. The readings are noted in tabular form.
- After completing the collection of data for a particular loading, the next load increment is applied and readings are noted under new load. This increment and data collection is repeated until the maximum load is applied. The maximum load is generally 1.5 times the expected ultimate load or 3 times of the expected allowable bearing pressure.
Bearing Capacity Question 4:
According to Rankine's formula minimum depth of foundation when q = 180 kN/m2, γ = 20 kN/m3 and ϕ = 30° is -
Answer (Detailed Solution Below)
Bearing Capacity Question 4 Detailed Solution
Concept:
Rankine's formula provides guidance on the minimum depth of foundation based on the bearing capacity of the soil.
Therefore the depth of the foundation can be expressed as,
Where Df = Minimum depth of foundation, qu = Ultimate bearing capacity of the soil
Calculation:
Intensity of loading (qu) =180 kN/m2,
Unit weight = 20 kN/m3 and angle of internal friction (ϕ) = 30°
Depth of foundation =
Depth of foundation (Df) = 1.0 m
Hence, According to Rankine's formula, the minimum depth of foundation is 1.0 m.
Bearing Capacity Question 5:
Identify the INCORRECT option.
SPT-N-values are correlated with:
Answer (Detailed Solution Below)
Bearing Capacity Question 5 Detailed Solution
Explanation:
NOTE: We have to find the incorrect statement from the given statements.
Standard Penetration Test: Standard penetration test can be used to determine:
- Relative density of sand
- Angle of internal friction (ϕ)
- Unconfined compressive strength of clay
- Ultimate bearing capacity on the basis of shear criteria
- Allowable bearing pressure on the basis of settlement criteria
- Load carrying capacity of pile
For standard penetration test:
(i) It is performed as a guideline in IS: 2131-1981 in the clean hole of 55 to 150 mm in diameter.
(ii) A thick wall split-tube sampler, 50.8 mm outer diameter and 35 mm inner diameter, is driven into the undisturbed soil at the bottom of the hole under the blows of a 65 kg drive weight with 75 cm free fall.
(iii) The minimum open length of the sampler should be 60 cm. It is first driven through 15 cm as a seating drive and is further driven through 30 cm or until 100 blows have been applied.
(iv) The number of blows required to drive the sampler 30 cm beyond the seating drive, is termed the penetration resistance N or SPT number.
The final SPT number is related to friction angle and relative density. which is given in the table
Penetration Resistance (N) | Relative Density or consistency | %age relative density | Approximate moist density in kN/m3 | Approximate ϕ in degrees |
- | - | 0 | - | - |
4 | Very loose | 15% | 11.2-11.6 | 20° |
10 | Loose | 35% | 14.4-18.4 | 30° |
20 | Moderately dense | 50% | 16-19.8 | 32° |
30 | Medium dense | 65% | 17.6-20.8 | 36° |
50 | Dense | 85% | 17.6-22.4 | 42° |
>50 | Very dense | 100% | 20.8-24.0 | 45° |
Bearing Capacity Question 6:
A standard penetration test is conducted in fine sand below water table and the standard penetration number (N) observed is 25. The corrected value of N is _____.
Answer (Detailed Solution Below)
Bearing Capacity Question 6 Detailed Solution
Concept:
Coorection in SPT
Observed SPT = No
First-I: SPT no. after correction for overburden
N1 = C1 N0
Second-II: Dilatancy correction
SPT after correction for water table
As water table is above the test level and SPT Number after corrected for overburden is greater than 15. So this correction is required
Calculation:
Given data,
N1 = 25
Dilatancy correction
Hence, The corrected value of N is 20.
Bearing Capacity Question 7:
A rigid isolated footing is resting on Clay. The contact pressure at the edges and centre of the footing are respectively:
Answer (Detailed Solution Below)
Bearing Capacity Question 7 Detailed Solution
Concept:
Variation of contact pressure and settlement of footing on different types of soil is as follows:
1. Flexible footing:
Case-I: Resting on Sand
- Settlement: Maximum at edges and minimum at the center
- Contact Pressure: Uniform
Case-II: Resting on Clay
- Settlement: Minimum at edges and maximum at the center
- Contact Pressure: Uniform
2. Rigid footing:
Case-I: Resting on Sand
- Settlement: Uniform
- Contact Pressure: Zero at edges and maximum at the center
Case-II: Resting on Clay
- Settlement: Uniform
- Contact Pressure: maximum at edges and minimum at the center
Bearing Capacity Question 8:
The contact pressure for a rigid foundation on a Cohesionless soil is
Answer (Detailed Solution Below)
Bearing Capacity Question 8 Detailed Solution
Explanation:
Foundation Contact Pressure: Generally loads from the structure are transferred to the soil through the footing. As a reaction to this load, soil exerts upward pressure on the bottom surface of the footing which is termed as contact pressure.
On different types of footing:
1) Under flexible footing: For flexible footing on cohesive soil, the settlement is maximum at the center of footing and minimum at the edges which form a bowl-like shape as shown in the below figure.
When a flexible footing is laid on the Cohesionless soil, the settlement at the center becomes minimum while at edges it is maximum which exact opposite case of the settlement of flexible footing over cohesive soil.
2). Under Rigid footing: For rigid footings resting on cohesive soils, the settlement is uniform but contact pressure varies. At edges contact pressure is maximum and at the center, it is minimum which forms an inverted bowl shape as shown in the below figure.
If the rigid footing is resting on Cohesionless soils, the contact pressure is maximum at the center and gradually reduces to zero towards edges. The settlement is uniform in this case also.
Important Points
The distribution of contact pressure under different types of footings on different types of soils are given below:
Rigid Footing |
|
Cohesion less soil |
Cohesive soil |
Contact pressure - Max at center and zero at edges Settlement - Uniform |
Contact pressure - Min at center and max at edges Settlement-Uniform |
Flexible footing |
|
Cohesion less soil |
Cohesive soil |
Contact pressure - Uniform Settlement - Min at center and Max at edges |
Contact pressure - Uniform Settlement - Max at center and Min at edges |
Bearing Capacity Question 9:
Which of the following statements are true regarding Cyclic pile load test?
Answer (Detailed Solution Below)
Bearing Capacity Question 9 Detailed Solution
Concept:
Cyclic pile load test method is used in the case of the initial test to separate skin frictional resistance and point bearing load separately for a single pile of uniform diameter.
Alternate loading and unloading shall be carried out at each stage. Each loading and unloading stage shall be maintained for at least 15 min. The subsequent elastic rebound in the pile should be measured accurately by dial gauges or LVDTs.
From the graph, we can find Point resistance (Rp) and Frictional resistance (Rf)
Point Resistance will be represented by a straight line.
Additional Information
Different types of tests and their purposes conducted on soil
Test |
Purpose |
Plate load test/ Pile load test |
|
Penetration Test |
|
Pressure meter test |
|
Standard and modified proctor test |
|
Direct Shear Test |
|
Triaxial Test |
|
Vane Shear Test |
|
Bearing Capacity Question 10:
Consider the following statements:
(1) Bearing capacity of Cohesionless soil increases with an increase in the width of the foundation.
(2) Bearing capacity of purely cohesive soil independent of the width of the foundation.
(3) Ultimate bearing capacity of a footing on pure clay does not significantly get affected by the presence of the water table.
Out of these statements:
Answer (Detailed Solution Below)
Bearing Capacity Question 10 Detailed Solution
Explanation:
Confusion Points
For cohesive soil like clay, the net BC is 0, as the value of Nq is 1 and the term Nq - 1 becomes 0, but the question is asking about bearing capacity, which means ultimate BC, not net BC. So, statement 3rd is false
The ultimate bearing capacity of strip footing is given by;
Effect of width and water table on bearing capacity:
Case1: For Cohesionless soil ( sand, C=0)
The ultimate bearing capacity for cohesionless soil is given by;
Conclusion:
- In the case of sand ultimate bearing capacity is proportional to the width of the foundation means Bearing capacity of Cohesionless soil increases with an increase in the width of the foundation.
Additional Information If the water table rises to ground level
or
If the water table rises to ground level in sand ultimate bearing capacity approximately reduces to 50%.
Case 2: For cohesive soil (clay):
For pure clay, the angle of internal friction, ϕ = 0°
For ϕ = 0°, Nc = 5.7, Nq = 1 and Nγ = 0
Ultimate bearing capacity is given by;
⇒
qu = 5.7 C + γ'Df
qnu = 5.7 C + γ'Df - γ'Df = 5.7 C
Conclusion:
- In the case of clay bearing capacity is independent of the width of footing.