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All questions of Soil Mechanics for Civil Engineering (CE) Exam
A fully saturated clay has a water content of 40% and unit weight 19 kN/m3. After oven drying the dry density of the soil becomes 18 kN/m3. The shrinkage limit for the soil sample is ______%. (Take γw=γw=10 kN/m3)
Correct answer is between '21,23'. Can you explain this answer?
A fully saturated clay has a water content of 40% and unit weight 19 kN/m3. After oven drying the dry density of the soil becomes 18 kN/m3. The shrinkage limit for the soil sample is ______%. (Take γw=γw=10 kN/m3)
|
Raghavendra Sengupta answered |
And, e sr = wG
⇒ e × 1 = 0.40 G
⇒ G = 2.97
Shrinkage limit, ws =
Where, Gm = Mass specific gravity at dry state
⇒ G = 2.97
Shrinkage limit, ws =
Where, Gm = Mass specific gravity at dry state
The soil has to be excavated from a borrow-pit which has a density of 1.12 gm/cc with a water content of 12% to fill a land. The soil is compacted to a density of 1.32 gm/cc with water content 20%. Find the volume of soil to be excavated in m3 for 1000 m3 volume of the fill?
Correct answer is between '1070,1130'. Can you explain this answer?
The soil has to be excavated from a borrow-pit which has a density of 1.12 gm/cc with a water content of 12% to fill a land. The soil is compacted to a density of 1.32 gm/cc with water content 20%. Find the volume of soil to be excavated in m3 for 1000 m3 volume of the fill?
|
Nishanth Banerjee answered |
Dry density of soil at excavation site,
∴ Volume of excavated soil =
= 1000 × 1.1/1
= 1100 m3
Assuming that a river bed level does not change and the depth of water in the river was 10m, 15m, and 8m during the months of February, July and December respectively of a particular year. The average bulk density of the soil is 20 kN/m3. The density of water is 10 kN/m3. The effective stress at a depth of 10 m below the river bed during these months would be- a)300 kN/m2, in February, 350 kN/m2 in July and 320 kN/m2 in December
- b)100 kN/m2 in February, 100 kN/m2 in July and 100 kN/m2 in December
- c)200 kN/m2 in February, 250 kN/m2 in July and 180 kN/m2 in December
- d)300 kN/m2 in February, 350 kN/m2 in July and 280 kN/m2 in December
Correct answer is option 'B'. Can you explain this answer?
Assuming that a river bed level does not change and the depth of water in the river was 10m, 15m, and 8m during the months of February, July and December respectively of a particular year. The average bulk density of the soil is 20 kN/m3. The density of water is 10 kN/m3. The effective stress at a depth of 10 m below the river bed during these months would be
a)
300 kN/m2, in February, 350 kN/m2 in July and 320 kN/m2 in December
b)
100 kN/m2 in February, 100 kN/m2 in July and 100 kN/m2 in December
c)
200 kN/m2 in February, 250 kN/m2 in July and 180 kN/m2 in December
d)
300 kN/m2 in February, 350 kN/m2 in July and 280 kN/m2 in December
|
Ishita Patel answered |
Since the river bed level does not change and the depth of water is changing only above the river bed, the effective stress below the river bed will not change.
Total stress at a depth of 10 m below river bed.
σ = 20 × 10 = 200 kN/m2
Pore water pressure,
u = 10 × 10 = 100 kN/m2
∴ Effective stress,
= 100 kN/m2
Thus effective stress will be 100 kN/m2 each on the month of February, July and December.
A fully saturate capillary zone of thickness 2 m exists above the water table in a fine silty sand deposit. Water is the pore water pressure at 1.5 m above the water table?- a)5 kN/m2
- b)-15 kN/m2
- c)-5 kN/m2
- d)15 kN/m2
Correct answer is option 'B'. Can you explain this answer?
A fully saturate capillary zone of thickness 2 m exists above the water table in a fine silty sand deposit. Water is the pore water pressure at 1.5 m above the water table?
a)
5 kN/m2
b)
-15 kN/m2
c)
-5 kN/m2
d)
15 kN/m2
|
Sahil Mehra answered |
Above the water table pore pressure will be negative
Pore pressure at 1.5 m above the WT
= - γwz = - 10 × 1.5 = -15 kN/m2.
If the plasticity index of a soil is 45%, then the soil will be:- a)non-plastic
- b)very highly plastic
- c)low plastic
- d)medium plastic
Correct answer is option 'B'. Can you explain this answer?
If the plasticity index of a soil is 45%, then the soil will be:
a)
non-plastic
b)
very highly plastic
c)
low plastic
d)
medium plastic
|
Aditi Sarkar answered |
Plasticity index (PI) is the range of water content over which the soil remains in the plastic state. Mathematically defined as,
Plasticity Index = Liquid Limit (w
L
) - Plastic Limit (wp
)
Previous sand is used in the foundation of a masonry dam such that upward hydraulic gradient permissible on it will be 0.345. Determine the factor of safely required against quicksand condition, if porosity and a specific gravity of soil solids are 0.40 and 2.65 respectively.
Correct answer is between '2.8,2.9'. Can you explain this answer?
Previous sand is used in the foundation of a masonry dam such that upward hydraulic gradient permissible on it will be 0.345. Determine the factor of safely required against quicksand condition, if porosity and a specific gravity of soil solids are 0.40 and 2.65 respectively.
|
|
Arya Menon answered |
Concept:
Critical hydraulic Gradient, icr 
Factor of safety (F.O.S) =
Calculation:
Factor of safety (F.O.S) =
Calculation:A cylindrical specimen of saturated clay, 3.5 cm in diameter and 8 cm in height were tested in an unconfined compression testing machine. The specimen fails under an axial load of 0.3 kN when the axial deformation was 8 mm. The unconfined compressive strength of the soil is- a)65 kN/m2
- b)125 kN/m2
- c)220 kN/m2
- d)280 kN/m2
Correct answer is option 'D'. Can you explain this answer?
A cylindrical specimen of saturated clay, 3.5 cm in diameter and 8 cm in height were tested in an unconfined compression testing machine. The specimen fails under an axial load of 0.3 kN when the axial deformation was 8 mm. The unconfined compressive strength of the soil is
a)
65 kN/m2
b)
125 kN/m2
c)
220 kN/m2
d)
280 kN/m2
|
Nidhi Tiwari answered |
A soil mass contains 40 % gravel, 50 % sand 10 % silt. The soil can be classified as-- a)Silty Sandy gravel having Cu<60
- b)Silty Gravely Sand having Cu = 10
- c)Silty Gravely sand having Cu> 60
- d)Gravelly Silty Sand - Cu can’t be calculated
Correct answer is option 'C'. Can you explain this answer?
A soil mass contains 40 % gravel, 50 % sand 10 % silt. The soil can be classified as-
a)
Silty Sandy gravel having Cu<60
b)
Silty Gravely Sand having Cu = 10
c)
Silty Gravely sand having Cu> 60
d)
Gravelly Silty Sand - Cu can’t be calculated
|
Naina Das answered |
D60 = 4.75 mm and D10 = 0.075mm
Implies Cu = D60/ D10 = 4.75/0.075 = 63.33
Consider the following statements if soil is treated with lime.1. Decrease in shrinkage limit2. Increase in plastic limit3. Decrease in liquid limit4. Flocculation of clay particlesWhich of the statements given above are correct?- a)1, 2 and 4
- b)1, 2 and 3
- c)1, 3 and 4
- d)2, 3 and 4
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements if soil is treated with lime.
1. Decrease in shrinkage limit
2. Increase in plastic limit
3. Decrease in liquid limit
4. Flocculation of clay particles
Which of the statements given above are correct?
a)
1, 2 and 4
b)
1, 2 and 3
c)
1, 3 and 4
d)
2, 3 and 4
|
|
Nilesh Kapoor answered |
Decrease in Shrinkage Limit
- Lime treatment of soil helps to decrease the shrinkage limit. Shrinkage limit is defined as the minimum water content at which a reduction in volume of soil mass ceases. It represents the lower limit of moisture content at which the soil starts to shrink on drying.
- Lime has the ability to stabilize the soil by reducing its plasticity. It reacts with the clay minerals present in the soil, causing the formation of calcium silicates and calcium aluminates. These compounds help to bind the soil particles together and reduce the shrinkage potential of the soil. As a result, the shrinkage limit of the soil decreases.
Increase in Plastic Limit
- Lime treatment also leads to an increase in the plastic limit of the soil. The plastic limit is the water content at which the soil becomes plastic and starts to deform under a certain amount of pressure.
- Lime acts as a binder and increases the plasticity of the soil. It helps in the flocculation of clay particles, which reduces the water content required to achieve a plastic state. Therefore, the plastic limit of the soil increases after lime treatment.
Decrease in Liquid Limit
- Contrary to what is stated in the question, lime treatment does not decrease the liquid limit of the soil. The liquid limit is the water content at which the soil changes from a liquid to a plastic state.
- Lime treatment does not have a direct effect on the liquid limit of the soil. However, it indirectly affects the liquid limit by reducing the plasticity index of the soil. The plasticity index is the difference between the liquid limit and the plastic limit. As lime treatment increases the plastic limit, the plasticity index also increases, but the liquid limit remains unchanged.
Flocculation of Clay Particles
- Lime treatment causes the flocculation of clay particles. Flocculation refers to the aggregation of clay particles into larger clumps or aggregates.
- Lime reacts with the clay minerals and forms flocculating agents, which promote the formation of stable clay aggregates. These aggregates help to improve the stability and strength of the soil. Flocculation also reduces the plasticity and shrinkage potential of the soil. Therefore, lime treatment is effective in promoting the flocculation of clay particles.
In conclusion, the correct statements are 2, 3, and 4. Lime treatment of soil leads to an increase in the plastic limit, a decrease in the liquid limit, and the flocculation of clay particles. However, the shrinkage limit of the soil decreases, not increases, after lime treatment.
- Lime treatment of soil helps to decrease the shrinkage limit. Shrinkage limit is defined as the minimum water content at which a reduction in volume of soil mass ceases. It represents the lower limit of moisture content at which the soil starts to shrink on drying.
- Lime has the ability to stabilize the soil by reducing its plasticity. It reacts with the clay minerals present in the soil, causing the formation of calcium silicates and calcium aluminates. These compounds help to bind the soil particles together and reduce the shrinkage potential of the soil. As a result, the shrinkage limit of the soil decreases.
Increase in Plastic Limit
- Lime treatment also leads to an increase in the plastic limit of the soil. The plastic limit is the water content at which the soil becomes plastic and starts to deform under a certain amount of pressure.
- Lime acts as a binder and increases the plasticity of the soil. It helps in the flocculation of clay particles, which reduces the water content required to achieve a plastic state. Therefore, the plastic limit of the soil increases after lime treatment.
Decrease in Liquid Limit
- Contrary to what is stated in the question, lime treatment does not decrease the liquid limit of the soil. The liquid limit is the water content at which the soil changes from a liquid to a plastic state.
- Lime treatment does not have a direct effect on the liquid limit of the soil. However, it indirectly affects the liquid limit by reducing the plasticity index of the soil. The plasticity index is the difference between the liquid limit and the plastic limit. As lime treatment increases the plastic limit, the plasticity index also increases, but the liquid limit remains unchanged.
Flocculation of Clay Particles
- Lime treatment causes the flocculation of clay particles. Flocculation refers to the aggregation of clay particles into larger clumps or aggregates.
- Lime reacts with the clay minerals and forms flocculating agents, which promote the formation of stable clay aggregates. These aggregates help to improve the stability and strength of the soil. Flocculation also reduces the plasticity and shrinkage potential of the soil. Therefore, lime treatment is effective in promoting the flocculation of clay particles.
In conclusion, the correct statements are 2, 3, and 4. Lime treatment of soil leads to an increase in the plastic limit, a decrease in the liquid limit, and the flocculation of clay particles. However, the shrinkage limit of the soil decreases, not increases, after lime treatment.
Consider the following statementsP. For a given compactive effort, the permeability decreases sharply with the increase in water content on the dry side of optimum.Q. For a given compactive effort, the soil tends to have a dispersed structure on the dry of optimumR. For a given compactive effort, the soil tends to have a flocculated structure on the wet of optimumS. Soils compacted on the dry of optimum tend to exhibit more shrinkage upon drying than those compacted wet of optimum
- a)P - True Q - False R - False S - True
- b)P - False Q - True R - True S - False
- c)P - True Q - False R - False S - False
- d)P - False Q - True R - True S - True
Correct answer is option 'C'. Can you explain this answer?
Consider the following statements
P. For a given compactive effort, the permeability decreases sharply with the increase in water content on the dry side of optimum.
Q. For a given compactive effort, the soil tends to have a dispersed structure on the dry of optimum
R. For a given compactive effort, the soil tends to have a flocculated structure on the wet of optimum
S. Soils compacted on the dry of optimum tend to exhibit more shrinkage upon drying than those compacted wet of optimum
a)
P - True Q - False R - False S - True
b)
P - False Q - True R - True S - False
c)
P - True Q - False R - False S - False
d)
P - False Q - True R - True S - True
|
Rutuja Deshpande answered |
Effect of compaction on structure
At point 1:
Flocculent structure is found on dry side of optimum and with increase in water content of soil there will be decrease in permeability of the soil due to improved orientation of the particles which results in a decrease in the size of voids.
At point 3:
At same compactive effort dispersed structure is found on wet side of optimum
Effect of shrinkage: Due to random orientation of the soil particles and stronger inter-particle bonds, shrinkage on dry of optimum side is found to be less than shrinkage on wet of optimum side.
Compactive effort: With increase in comp active effort at same water content the soil tend to become more oriented (dispersed) especially on the dry side of optimum.
Point 5:
Consider the following statements:P. For a clay layer, the net ultimate bearing capacity (By Skempton) at the base of the footing is independent of the cohesion of the soil.Q. If the depth of the strip footing is equal to the width of the footing, then the value of bearing capacity factor (Nc) by skempton is taken to be 6.0.R. The theory given by Skempton for net ultimate bearing capacity computations is valid for shallow footings only.S. Local shear failure occurs in soft clays and loose sands.Identify the correct option:- a)P-False, Q-True, R-True, S-False
- b)P-True, Q-False, R-False, S-True
- c)P-True, Q-False, R-True, S-False
- d)P-False, Q-True, R-False, S-True
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements:
P. For a clay layer, the net ultimate bearing capacity (By Skempton) at the base of the footing is independent of the cohesion of the soil.
Q. If the depth of the strip footing is equal to the width of the footing, then the value of bearing capacity factor (Nc) by skempton is taken to be 6.0.
R. The theory given by Skempton for net ultimate bearing capacity computations is valid for shallow footings only.
S. Local shear failure occurs in soft clays and loose sands.
Identify the correct option:
a)
P-False, Q-True, R-True, S-False
b)
P-True, Q-False, R-False, S-True
c)
P-True, Q-False, R-True, S-False
d)
P-False, Q-True, R-False, S-True
|
Aniket Ghoshal answered |
SKEMPTON’S Method:
This theory is applicable only for C-soils but it can be applied for shallow and deep footings both. In this theory, base resistance and side resistance both are considered. The net ultimate bearing capacity is given by:
qnu=CNc
Where,
Nc = Skempton’s bearing capacity factor which depends upon Df/B ratio
Hence for a clay layer, the net ultimate bearing capacity (By Skempton) at the base of the footing is directly proportional to the cohesion of the soil.
Case 1: When Df/B = 0,
Nc = 5.0 (strip footing)
= 6.0 (Square/Circular/Rectangular/ Raft)
Case 2: When 0 Df/B < 2.5
Case 3: When
Case 2: When 0 Df/B < 2.5
Case 3: When
Nc = 7.5 (Strip footing)
Nc = 9.0 (Square/circular/rectangular/raft)
Therefore, when B = Df , case 2 is valid
Local shear failure:
This type of failure is seen in relatively loose sand and soft clay.
Some characteristics of local shear failure are:
1. Failure is not sudden and there is no tilting of footing.
2. Failure surface does not reach the ground surface and slight bulging of soil around the footing is observed
3. Failure surface is not well defined
4. Failure is progressive
5. In load-settlement curve, there is no well-defined peak
6. Failure is characterized by considerable settlement directly beneath the foundation
7. A significant compression of soil below the footing and partial development of plastic equilibrium is observed.
8. Well-defined wedge and slip surfaces only beneath the foundation.
Calculate the saturated unit weight (kN/m3) of the soil sample for the Quick sand condition, if specify Gravity and void ratio of soil sample is 2.62 and 0.61 respectively. Take γW = 10 kN/m3
Correct answer is between '20.05,20.07'. Can you explain this answer?
Calculate the saturated unit weight (kN/m3) of the soil sample for the Quick sand condition, if specify Gravity and void ratio of soil sample is 2.62 and 0.61 respectively. Take γW = 10 kN/m3
|
Bhargavi Sengupta answered |
Calculation of Saturated Unit Weight for Quick Sand Condition
- Given Data:
- Specific gravity of soil sample (G) = 2.62
- Void ratio of soil sample (e) = 0.61
- Unit weight of water (γW) = 10 kN/m3
- Formula:
- Saturated unit weight (γsat) = G * γW / (1+e)
- Calculation:
- γsat = 2.62 * 10 / (1+0.61)
- γsat = 26.2 / 1.61
- γsat = 16.2739 kN/m3
- Conclusion:
- The saturated unit weight of the soil sample for the Quick Sand condition is approximately 16.27 kN/m3.
Therefore, the correct answer is 16.27 kN/m3.
- Given Data:
- Specific gravity of soil sample (G) = 2.62
- Void ratio of soil sample (e) = 0.61
- Unit weight of water (γW) = 10 kN/m3
- Formula:
- Saturated unit weight (γsat) = G * γW / (1+e)
- Calculation:
- γsat = 2.62 * 10 / (1+0.61)
- γsat = 26.2 / 1.61
- γsat = 16.2739 kN/m3
- Conclusion:
- The saturated unit weight of the soil sample for the Quick Sand condition is approximately 16.27 kN/m3.
Therefore, the correct answer is 16.27 kN/m3.
A rectangular footing is located at a depth of 1.5 m below the ground surface. It is required to carry the safe load of 1200 kN at the base of the footing. Calculate the area of the footing at the base in m2. Take factor of safety of 2.5 and bearing capacity factors as NC = 38, Nq = 23 and Nγ = 21. Assume water table at a great depth. Use Terzaghi’s ultimate bearing capacity equation. Assume the length to width ratio of the footing to be 1.5. Cohesion c = 100 kN/m2 and unit weight of soil to be 19 kN/m3.
Correct answer is between '0.55,0.58'. Can you explain this answer?
A rectangular footing is located at a depth of 1.5 m below the ground surface. It is required to carry the safe load of 1200 kN at the base of the footing. Calculate the area of the footing at the base in m2. Take factor of safety of 2.5 and bearing capacity factors as NC = 38, Nq = 23 and Nγ = 21. Assume water table at a great depth. Use Terzaghi’s ultimate bearing capacity equation. Assume the length to width ratio of the footing to be 1.5. Cohesion c = 100 kN/m2 and unit weight of soil to be 19 kN/m3.
|
Jhanvi Choudhary answered |
Concept:
Ultimate bearing capacity given by Terzaghi for rectangular footing is:
Where,
B = width of footing
L = Length of footing
c = Cohesion (kN/m2)
Nc, Nq, Nγ = Bearing capacity factors
γ = unit weight of the soil
Df = Depth of footing
Net ultimate Bearing capacity
qnu = qu - γDf
Net safe Bearing capacity
qu = 4560 + 655.5 + 172.9 B
qnu = 5215.5 + 172.9 B - γDf
qnu = 5215.5 + 172.9 B – 19 × 1.5
qnu = 5187 + 172.9 B
qsafe = 2074.8 + 69.16 B + 28.5
800 = (2103.3 + 69.16 B) B2
qsafe = 2074.8 + 69.16 B + 28.5
800 = (2103.3 + 69.16 B) B2
B = 0.61 m
L = 1.5 × 0.61 = 0.915 m
Area of footing = L x B = 0.558 m2
Consider the following statements. The incorrect statement is:- a)The water content has a range from 0% to 100%.
- b)The degree of saturation has ranged from 0% to 100%.
- c)The porosity cannot be greater than 100%.
- d)The void ratio of soil can be greater than 1.
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements. The incorrect statement is:
a)
The water content has a range from 0% to 100%.
b)
The degree of saturation has ranged from 0% to 100%.
c)
The porosity cannot be greater than 100%.
d)
The void ratio of soil can be greater than 1.
|
|
Aditi Chakraborty answered |
Water content is the ratio of the mass of water to the mass of solids.
It can be greater than 100%.
The degree of saturation has ranged from 0% to 100%.
The porosity cannot be greater than 100%.
The void ratio of soil can be greater than 1.
To achieve 80% consolidation in a clay layer, the time required will be 15 years. If the clay layer is twice as thick, 15 times more permeable and 10 times less compressible, then the time that would be required to achieve the same degree of consolidation will be- a)2.4 months
- b)4.8 months
- c)48 months
- d)24 months
Correct answer is option 'B'. Can you explain this answer?
To achieve 80% consolidation in a clay layer, the time required will be 15 years. If the clay layer is twice as thick, 15 times more permeable and 10 times less compressible, then the time that would be required to achieve the same degree of consolidation will be
a)
2.4 months
b)
4.8 months
c)
48 months
d)
24 months
|
Sagnik Sen answered |
Concept:
Using the relation,
Time factor (TV) is given by
Where t = time required to consolidate the clay layer by (U%).
d = thickness of clay layer
cv = coefficient of consolidation
Also,
K = cv mv γw
Where,
K = permeability of soil sample
cv = coefficient of consolidation
mv = coefficient of volume compressibility
Calculation:
For U = 80%,
TV = TV1 and time required will be 15 years.
Now for same U, TV = TV2 = TV1 but d2 = 2d1, k2 = 15k1, mv2 = 110mv1
TV = TV1 and time required will be 15 years.
Now for same U, TV = TV2 = TV1 but d2 = 2d1, k2 = 15k1, mv2 = 110mv1
t2 = 0.4 years
t2 = 4.8 months
Choose the correct statement1. Soils with small volume changes are compacted to the dry side of optimum.2. Soils with large volume changes are compacted to the wet side of optimum.- a)Both 1 and 2
- b)Only 1
- c)Only 2
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Choose the correct statement
1. Soils with small volume changes are compacted to the dry side of optimum.
2. Soils with large volume changes are compacted to the wet side of optimum.
a)
Both 1 and 2
b)
Only 1
c)
Only 2
d)
None of these
|
|
Deepika Saha answered |
Understanding Soil Compaction
Soil compaction is a critical process in civil engineering, particularly for enhancing the stability and load-bearing capacity of soil. The statements in the question relate to how different soils behave during compaction based on their volume change characteristics.
Soils with Small Volume Changes
- These soils tend to have low plasticity and minimal expansion or shrinkage when moisture content changes.
- When compacted to the dry side of optimum, these soils achieve higher density and strength.
- This approach reduces the risk of future settlement or structural issues as they are less susceptible to moisture fluctuations.
Soils with Large Volume Changes
- These soils often have high plasticity and are prone to significant volume changes with moisture content variations.
- Compacted to the wet side of optimum, these soils achieve maximum density while minimizing the risk of swelling or shrinkage.
- This compaction strategy helps manage moisture more effectively, reducing the potential for future instability.
Conclusion
Both statements are correct:
- Soils with small volume changes are indeed compacted to the dry side of optimum for stability.
- Conversely, soils with large volume changes benefit from being compacted to the wet side of optimum to control their expansive tendencies.
Thus, the correct answer is option 'A': Both 1 and 2. Understanding these principles is vital for effective soil management and ensuring structural integrity in civil engineering projects.
Soil compaction is a critical process in civil engineering, particularly for enhancing the stability and load-bearing capacity of soil. The statements in the question relate to how different soils behave during compaction based on their volume change characteristics.
Soils with Small Volume Changes
- These soils tend to have low plasticity and minimal expansion or shrinkage when moisture content changes.
- When compacted to the dry side of optimum, these soils achieve higher density and strength.
- This approach reduces the risk of future settlement or structural issues as they are less susceptible to moisture fluctuations.
Soils with Large Volume Changes
- These soils often have high plasticity and are prone to significant volume changes with moisture content variations.
- Compacted to the wet side of optimum, these soils achieve maximum density while minimizing the risk of swelling or shrinkage.
- This compaction strategy helps manage moisture more effectively, reducing the potential for future instability.
Conclusion
Both statements are correct:
- Soils with small volume changes are indeed compacted to the dry side of optimum for stability.
- Conversely, soils with large volume changes benefit from being compacted to the wet side of optimum to control their expansive tendencies.
Thus, the correct answer is option 'A': Both 1 and 2. Understanding these principles is vital for effective soil management and ensuring structural integrity in civil engineering projects.
A 5m high vertical earth retaining wall having smooth vertical back retains non-cohesive soil with angle of internal friction 35° and unit weight γ = 18 kN/m3. If the wall is free to field and the water table is at a depth of 2m below the top of the wall, then the total horizontal active thrust (kN) per unit length acting on the wall will beTake γsat = 20 kN/m3, γω = 10kN/m3.- a)96.221 kN/m
- b)86.46 kN/m
- c)108.416 kN/m
- d)120 kN/m
Correct answer is option 'A'. Can you explain this answer?
A 5m high vertical earth retaining wall having smooth vertical back retains non-cohesive soil with angle of internal friction 35° and unit weight γ = 18 kN/m3. If the wall is free to field and the water table is at a depth of 2m below the top of the wall, then the total horizontal active thrust (kN) per unit length acting on the wall will be
Take γsat = 20 kN/m3, γω = 10kN/m3.
a)
96.221 kN/m
b)
86.46 kN/m
c)
108.416 kN/m
d)
120 kN/m
|
Asha Deshpande answered |
Total active earth pressure at 2m below the Ground level = kaγ×2
Active thrust:
Pa1 = 12 × ka × γ × 2 × 2 =
Pa1 = 9.756 kN/m
Total active earth pressure at 5m below the Ground level:
= ka × γ ×2 + kaγ1×3 + γω × 3
= 0.271 × 18 × 2 + 0.271 × [20 – 10] × 3 + 10 × 3
= 47.88 kN/m2
Active thrust:-
Pa = 9.756 + 29.27 + 12.195 + 45
Active thrust:-
Pa = 9.756 + 29.27 + 12.195 + 45
Pa = 96.221 kN/m
Time factor for 90% consolidation is 0.848Time factor for 50% consolidation is 0.197The time required for 90% consolidation is 30 MinsThe time required for 50% consolidation is 9 MinsLength of drainage path is 10mmThe Coefficient of consolidation of the sample by the square root of time fitting method is- a)3.64 x 10-8 m2/s
- b)4.71 x 10-8 m2/s
- c)2.34 x 10-8 m2/s
- d)5.81 x 10-8 m2/s
Correct answer is option 'B'. Can you explain this answer?
Time factor for 90% consolidation is 0.848
Time factor for 50% consolidation is 0.197
The time required for 90% consolidation is 30 Mins
The time required for 50% consolidation is 9 Mins
Length of drainage path is 10mm
The Coefficient of consolidation of the sample by the square root of time fitting method is
a)
3.64 x 10-8 m2/s
b)
4.71 x 10-8 m2/s
c)
2.34 x 10-8 m2/s
d)
5.81 x 10-8 m2/s
|
|
Sanaya Sengupta answered |
Cv = (T90 x d2)/ t90 = 0.848 x 0.012/ (30x60) = 4.71 x 10-8 m2/s
According to Darcy’s law for flow through porous media, the velocity is proportional to- a)effective stress
- b)hydraulic gradient
- c)cohesion
- d)stability number
Correct answer is option 'B'. Can you explain this answer?
According to Darcy’s law for flow through porous media, the velocity is proportional to
a)
effective stress
b)
hydraulic gradient
c)
cohesion
d)
stability number
|
Mira Mishra answered |
Darcy established that the flow occurring per unit time is directly proportional to the head causing flow and the area of cross – section of the soil sample but is inversely proportional to the length of the soil sample
Dividing both sides by A
i.e., hydraulic gradient
A retaining wall of height H with smooth vertical backface supports a backfill inclined at an angle β with the horizontal. The backfill consists of cohesionless soil having an angle of internal friction ϕ. If the active lateral thrust acting on the wall is Pa, which one of the following statements is TRUE?- a)Pa acts at a height H/3 from the base of the wall and at an angle β with the horizontal
- b)Pa acts at a height H/3 from the base of the wall and at an angle ϕ with the horizontal
- c)Pa acts at a height H/2 from the base of the wall and at an angle β with the horizontal
- d)Pa acts at a height H/2 from the base of the wall and at an angle ϕ with the horizontal
Correct answer is option 'A'. Can you explain this answer?
A retaining wall of height H with smooth vertical backface supports a backfill inclined at an angle β with the horizontal. The backfill consists of cohesionless soil having an angle of internal friction ϕ. If the active lateral thrust acting on the wall is Pa, which one of the following statements is TRUE?
a)
Pa acts at a height H/3 from the base of the wall and at an angle β with the horizontal
b)
Pa acts at a height H/3 from the base of the wall and at an angle ϕ with the horizontal
c)
Pa acts at a height H/2 from the base of the wall and at an angle β with the horizontal
d)
Pa acts at a height H/2 from the base of the wall and at an angle ϕ with the horizontal
|
|
Gowri Sharma answered |
Understanding the Retaining Wall and Active Thrust
When dealing with a retaining wall that supports an inclined backfill of cohesionless soil, it is essential to analyze the behavior of lateral earth pressures. The active lateral thrust, denoted as Pa, is a crucial element in this analysis.
Active Lateral Thrust (Pa)
- The active lateral thrust, Pa, is generated due to the weight of the backfill soil and its inclination at an angle β.
- This thrust acts to push the wall outward and is influenced by the soil's angle of internal friction.
Point of Application
- The point of application of the active thrust is a key aspect of retaining wall design.
- For inclined backfill, the resultant active thrust typically acts at a height of H/3 from the base of the wall. This is derived from the geometry of the pressure distribution.
Direction of the Thrust
- The direction of the active thrust, Pa, acts at an angle β with the horizontal.
- This angle corresponds to the inclination of the backfill and is critical in determining the overall stability of the wall.
Conclusion
Based on the analysis of the active thrust on a retaining wall with a smooth vertical face supporting an inclined backfill, the correct statement is:
- Pa acts at a height H/3 from the base of the wall and at an angle β with the horizontal.
This understanding is fundamental for ensuring the stability and design of retaining structures in civil engineering.
When dealing with a retaining wall that supports an inclined backfill of cohesionless soil, it is essential to analyze the behavior of lateral earth pressures. The active lateral thrust, denoted as Pa, is a crucial element in this analysis.
Active Lateral Thrust (Pa)
- The active lateral thrust, Pa, is generated due to the weight of the backfill soil and its inclination at an angle β.
- This thrust acts to push the wall outward and is influenced by the soil's angle of internal friction.
Point of Application
- The point of application of the active thrust is a key aspect of retaining wall design.
- For inclined backfill, the resultant active thrust typically acts at a height of H/3 from the base of the wall. This is derived from the geometry of the pressure distribution.
Direction of the Thrust
- The direction of the active thrust, Pa, acts at an angle β with the horizontal.
- This angle corresponds to the inclination of the backfill and is critical in determining the overall stability of the wall.
Conclusion
Based on the analysis of the active thrust on a retaining wall with a smooth vertical face supporting an inclined backfill, the correct statement is:
- Pa acts at a height H/3 from the base of the wall and at an angle β with the horizontal.
This understanding is fundamental for ensuring the stability and design of retaining structures in civil engineering.
Choose the incorrect statement:1) Greater the confining pressure, smaller the critical void ratio.2) A saturated loose sand is likely to liquify when subjected to earthquake shocks.3) Failure envelope of over consolidated clay is a curve and not straight line.4) For normally consolidated clays, ϕ decrease with an increase in plasticity.- a)1 and 2
- b)3 only
- c)4 only
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Choose the incorrect statement:
1) Greater the confining pressure, smaller the critical void ratio.
2) A saturated loose sand is likely to liquify when subjected to earthquake shocks.
3) Failure envelope of over consolidated clay is a curve and not straight line.
4) For normally consolidated clays, ϕ decrease with an increase in plasticity.
a)
1 and 2
b)
3 only
c)
4 only
d)
None of these
|
Ishani Chauhan answered |
The relationship between critical void ratio and confining pressure is shown in figure. As confining pressure increases, the critical void ratio decreases. With this knowledge of relationship it is possible predict the behaviour of sand at any ratio and at any given confining pressure.
All other statements are also true.
A series of wells were dug in a confined aquifer. The surface obtained by connecting the static water levels would be called:- a)Piezometric surface
- b)Phreatic line
- c)Cone of depression
- d)Perched water table
Correct answer is option 'A'. Can you explain this answer?
A series of wells were dug in a confined aquifer. The surface obtained by connecting the static water levels would be called:
a)
Piezometric surface
b)
Phreatic line
c)
Cone of depression
d)
Perched water table
|
Anmol Roy answered |
Confined aquifers are permeable rock units that are usually deeper under the ground than unconfined aquifers. They are overlain by relatively impermeable rock or clay that limits groundwater movement into, or out of, the confined aquifer. Groundwater in a confined aquifer is under pressure and will rise up inside a borehole drilled into the aquifer. The level to which the water rises is called the piezometric surface.
A 4m × 3m raft is provided at a depth of 3 m below the ground surface in a purely cohesive soil such that the ultimate bearing capacity by Terzaghi’s is given by 1450kN/m2. The net ultimate bearing capacity by Skempton is _______kN/m2.Assume Bearing capacity factor by Terzaghi to be: NC = 5.7, Nq = 1.0. Unit weight of clay layer to be 19 kN/m3
Correct answer is between '1375,1377'. Can you explain this answer?
A 4m × 3m raft is provided at a depth of 3 m below the ground surface in a purely cohesive soil such that the ultimate bearing capacity by Terzaghi’s is given by 1450kN/m2. The net ultimate bearing capacity by Skempton is _______kN/m2.
Assume Bearing capacity factor by Terzaghi to be: NC = 5.7, Nq = 1.0. Unit weight of clay layer to be 19 kN/m3
|
|
Rajat Sen answered |
For raft Foundation [By Terzaghi’s]
Ultimate Bearing Capacity is given by:
For pure clay Nγ = 0
B = 3 m
L = 4 m
NC = 5.7
Nq = 1.0
qu = 1450 kN/m2
C × 5.7 + 19 × 3 × 1.0 + 0
C × 5.7 + 19 × 3 × 1.0 + 01450 = 1.225 × 5.7 × C + 57
C = 199.49 kN/m2
Now By Skempton:
Net ultimate Bearing capacity is given by:
qnu = CNC
where,
Nc = 5.0 × 1.15 × 1.2
Nc = 5.0 × 1.15 × 1.2
NC = 6.9
qnu = CNC
qnu = 199.49 × 6.9
qnu = 1376.48 kN/m2
A 50 cm well completely penetrates an artesian aquifer. The length of the strainer is 25m. Determine the discharge (in lph) from the well, when the drawdown in the pumping well is 4 m. The K of the aquifer is 45 m/day. At a radial distance of 400 m from center of well, drawdown is zero.
Correct answer is between '159680,159685'. Can you explain this answer?
A 50 cm well completely penetrates an artesian aquifer. The length of the strainer is 25m. Determine the discharge (in lph) from the well, when the drawdown in the pumping well is 4 m. The K of the aquifer is 45 m/day. At a radial distance of 400 m from center of well, drawdown is zero.
|
|
Aaditya Jain answered |
For artesian (or) confined aquifer
= 3832.37 m3/day
= 159682.26 lph
The compaction of on embankment is carried out in 300 mm thickness of lifts. The rammer used has foot area of 0.05 sq. m. The energy developed per drop of the rammer is 40 kg-m. Assuming 50% more energy in each pas over the compacted layer due to overlap, calculate a number of purses required to develop compaction energy equivalent to IS light compaction for each layer.- a)22.67
- b)15.11
- c)7.55
- d)30.22
Correct answer is option 'B'. Can you explain this answer?
The compaction of on embankment is carried out in 300 mm thickness of lifts. The rammer used has foot area of 0.05 sq. m. The energy developed per drop of the rammer is 40 kg-m. Assuming 50% more energy in each pas over the compacted layer due to overlap, calculate a number of purses required to develop compaction energy equivalent to IS light compaction for each layer.
a)
22.67
b)
15.11
c)
7.55
d)
30.22
|
|
Anmol Choudhary answered |
Question:
The compaction of an embankment is carried out in 300 mm thickness of lifts. The rammer used has foot area of 0.05 sq. m. The energy developed per drop of the rammer is 40 kg-m. Assuming 50% more energy in each pass over the compacted layer due to overlap, calculate the number of passes required to develop compaction energy equivalent to IS light compaction for each layer.
Options:
a)22.67
b)15.11
c)7.55
d)30.22
Answer:
The formula for calculating energy developed per unit volume of soil is given by:
E = (W.H) / A
where,
E = Energy developed per unit volume of soil (kg-m/m³)
W = Weight of rammer (kg)
H = Height of fall (m)
A = Area of the foot of the rammer (m²)
The energy developed per drop of the rammer is given as 40 kg-m. Therefore, the energy developed per unit volume of soil will be:
E = 40 / 0.05
E = 800 kg-m/m³
The IS light compaction standard requires the development of 720 kg-m/m³ of energy per unit volume of soil. Therefore, the number of passes required to develop this energy can be calculated as:
Number of passes = (720 / 800) x 100 / 150
Number of passes = 15.11
Therefore, the correct answer is option B (15.11).
The compaction of an embankment is carried out in 300 mm thickness of lifts. The rammer used has foot area of 0.05 sq. m. The energy developed per drop of the rammer is 40 kg-m. Assuming 50% more energy in each pass over the compacted layer due to overlap, calculate the number of passes required to develop compaction energy equivalent to IS light compaction for each layer.
Options:
a)22.67
b)15.11
c)7.55
d)30.22
Answer:
The formula for calculating energy developed per unit volume of soil is given by:
E = (W.H) / A
where,
E = Energy developed per unit volume of soil (kg-m/m³)
W = Weight of rammer (kg)
H = Height of fall (m)
A = Area of the foot of the rammer (m²)
The energy developed per drop of the rammer is given as 40 kg-m. Therefore, the energy developed per unit volume of soil will be:
E = 40 / 0.05
E = 800 kg-m/m³
The IS light compaction standard requires the development of 720 kg-m/m³ of energy per unit volume of soil. Therefore, the number of passes required to develop this energy can be calculated as:
Number of passes = (720 / 800) x 100 / 150
Number of passes = 15.11
Therefore, the correct answer is option B (15.11).
The following data are given for soil:• Porosity: n = 0.4• The specific gravity of the soil solids: Gs = 2.68• Moisture content: w = 12%Determine the mass of water in kg to be added to 10 m3 of soil for full saturation.
Correct answer is between '2065,2077'. Can you explain this answer?
The following data are given for soil:
• Porosity: n = 0.4
• The specific gravity of the soil solids: Gs = 2.68
• Moisture content: w = 12%
Determine the mass of water in kg to be added to 10 m3 of soil for full saturation.
|
|
Garima Kulkarni answered |
Total volume of soil, V = 10 m3
∴ VV = 0.4 × 10 = 4m3
Volume of solids, Vs = V – VV = 10 – 4 = 6 m3
Mass of solids, Ws = Vs Gs Yw = 6 × 2.68 × 1000 = 16080 kg
Moisture content = 0.12
So, Mass of water = 0.12 × 16080 = 1929.6 kg
Volume of water = 1929.6/1000 = 1.929 m3
So, phase diagram can be shown as
For full saturation, the amount of air has to be replaced by water.
So, water to be added = 2.071 × 1000 = 2071 kg
A sample of dry sand was subjected to tri - axial test with a confining pressure of 250 kN/m2. The angle of internal friction is found to be 36°. The sample is likely to fail at deviator stress of ________ kN/m2.
Correct answer is between '712,714'. Can you explain this answer?
A sample of dry sand was subjected to tri - axial test with a confining pressure of 250 kN/m2. The angle of internal friction is found to be 36°. The sample is likely to fail at deviator stress of ________ kN/m2.
|
Avik Chaudhary answered |
Here, σ3 = 250 kN/m2
σ1 = σ3 + σd
Where, σd is the deviator stress
Using plastic equilibrium condition,
As sandy soil, C = 0
As sandy soil, C = 0
⇒ 250 + σd = 962.96
⇒ σd = 712.95 kN/m2
In an Octahedral unit if Fe2+ is present at the center of the geometry, what will be the net charge present over this unit.- a)-1
- b)0
- c)+1
- d)-2
Correct answer is option 'B'. Can you explain this answer?
In an Octahedral unit if Fe2+ is present at the center of the geometry, what will be the net charge present over this unit.
a)
-1
b)
0
c)
+1
d)
-2
|
|
Partho Jain answered |
In Octahedral unit 6 Hydroxyl atoms encloses Fe2+ (as in this case) at the centre of Octahedral Geometry where in each hydroxyl atom is shared in between 3 Octahedral units.Net charge present = − 6 × 1/3 + 2 = 0
A granular soil has a saturated unit weight of 20 kN/m3 and an effective angle of shearing resistance of 30°. The unit weight of water is 9.81 kN/m3. A slope is to be made on this soil deposit in which the seepage occurs parallel to the slope up to the free surface. Under this seepage condition for a factor of safety of 1.5, the safe slope angle (in degree, round off to 1 decimal place) would be ______.
Correct answer is between '10.8,11.3'. Can you explain this answer?
A granular soil has a saturated unit weight of 20 kN/m3 and an effective angle of shearing resistance of 30°. The unit weight of water is 9.81 kN/m3. A slope is to be made on this soil deposit in which the seepage occurs parallel to the slope up to the free surface. Under this seepage condition for a factor of safety of 1.5, the safe slope angle (in degree, round off to 1 decimal place) would be ______.
|
|
Prasad Desai answered |
FOS = 1.5
If there is seepage parallel to the slope
β = 11.095°
The term ‘Loess’ indicates those soils which are1. Uniformly graded2. Poorly graded3. Made up of more than 50% sand size particles4. Made up of more than 50% of silt particles5. These are transported by windsWhich of the above statements are correct:- a)2,3 & 5
- b)1,4 & 5
- c)2 & 4
- d)1 & 3
Correct answer is option 'B'. Can you explain this answer?
The term ‘Loess’ indicates those soils which are
1. Uniformly graded
2. Poorly graded
3. Made up of more than 50% sand size particles
4. Made up of more than 50% of silt particles
5. These are transported by winds
Which of the above statements are correct:
a)
2,3 & 5
b)
1,4 & 5
c)
2 & 4
d)
1 & 3
|
|
Kavya Mehta answered |
Loess is an Aeolian Sediment i.e. transported by winds.
These contain fine particles like silt and clay (typically in the 20–50 micrometer size range).
Loess is uniformly graded soil.
It is usually homogeneous and highly porous and is traversed by vertical capillaries that permit the sediment to fracture and form vertical bluffs.
Which of the following assumptions is wrongly stated in regard to Rankine’s theory of earth pressure?- a)The back of the wall is not smooth.
- b)The ground surface is plane.
- c)The soil mass is semi-infinite and homogeneous.
- d)The retaining wall yields about the base.
Correct answer is option 'A'. Can you explain this answer?
Which of the following assumptions is wrongly stated in regard to Rankine’s theory of earth pressure?
a)
The back of the wall is not smooth.
b)
The ground surface is plane.
c)
The soil mass is semi-infinite and homogeneous.
d)
The retaining wall yields about the base.
|
|
Yashvi Choudhury answered |
The Rankine's theory assumes that there is no wall friction, the ground and failure surfaces are straight planes, and that the resultant force acts parallel to the backfill slope.
The soil mass is homogeneous and semi-infinite and retaining wall yields about the base.
In case of retaining structures, the earth retained may be filled up earth or natural soil. These backfill materials may exert certain lateral pressure on the wall.
For a capillary rise in a soil mass, void size of which is assumed to be 20% of the effective size of the particle (D10), surface tension is observed to be 73 × 10-3 N/m. Calculate the value of empirical constant given by Hazen in cm2 void ratio of the soil sample is 0.30. Take γw = 10 kN/m3
Correct answer is between '0.43,0.45'. Can you explain this answer?
For a capillary rise in a soil mass, void size of which is assumed to be 20% of the effective size of the particle (D10), surface tension is observed to be 73 × 10-3 N/m. Calculate the value of empirical constant given by Hazen in cm2 void ratio of the soil sample is 0.30. Take γw = 10 kN/m3
|
|
Samridhi Choudhary answered |
MN/m and the contact angle is 30 degrees. The soil particle size distribution curve is given below:
Diameter (mm) | % Finer
--- | ---
4.75 | 100
2.36 | 85
1.18 | 60
0.6 | 30
0.3 | 10
0.15 | 3
0.075 | 0.5
Effective size (D10) can be calculated from the particle size distribution curve using the following formula:
D10 = Σ (diameter × % finer) / 100
D10 = (4.75 × 100 + 2.36 × 85 + 1.18 × 60 + 0.6 × 30 + 0.3 × 10 + 0.15 × 3 + 0.075 × 0.5) / 100
D10 = 1.66 mm
Void size can be calculated as 20% of D10:
Void size = 0.2 × D10
Void size = 0.2 × 1.66
Void size = 0.332 mm
Capillary rise can be calculated using the following formula:
h = (2 × surface tension × cosθ) / (ρg × void size)
where h is the height of capillary rise, surface tension is 73 mN/m, contact angle is 30 degrees, ρ is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²).
h = (2 × 73 × cos30) / (1000 × 9.81 × 0.332)
h = 0.029 m
Therefore, the capillary rise in the soil mass is 0.029 meters or 29 millimeters.
Diameter (mm) | % Finer
--- | ---
4.75 | 100
2.36 | 85
1.18 | 60
0.6 | 30
0.3 | 10
0.15 | 3
0.075 | 0.5
Effective size (D10) can be calculated from the particle size distribution curve using the following formula:
D10 = Σ (diameter × % finer) / 100
D10 = (4.75 × 100 + 2.36 × 85 + 1.18 × 60 + 0.6 × 30 + 0.3 × 10 + 0.15 × 3 + 0.075 × 0.5) / 100
D10 = 1.66 mm
Void size can be calculated as 20% of D10:
Void size = 0.2 × D10
Void size = 0.2 × 1.66
Void size = 0.332 mm
Capillary rise can be calculated using the following formula:
h = (2 × surface tension × cosθ) / (ρg × void size)
where h is the height of capillary rise, surface tension is 73 mN/m, contact angle is 30 degrees, ρ is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²).
h = (2 × 73 × cos30) / (1000 × 9.81 × 0.332)
h = 0.029 m
Therefore, the capillary rise in the soil mass is 0.029 meters or 29 millimeters.
In wet soil mass, air occupies one-sixth of its volume and water occupies one – third of its volume. The void ratio of the soil is- a)0.25
- b)0.5
- c)1.00
- d)1.50
Correct answer is option 'C'. Can you explain this answer?
In wet soil mass, air occupies one-sixth of its volume and water occupies one – third of its volume. The void ratio of the soil is
a)
0.25
b)
0.5
c)
1.00
d)
1.50
|
|
Tanishq Rane answered |
Void ratio, e = Vv/Vs
Vv = air void + water filled voids
Calculation:
Calculation:
The Contact pressure and settlement distribution for the footing are shown in the figure.
The figure corresponds to a- a)Flexible footing on clayey soil
- b)Flexible footing on sandy soil
- c)Rigid footing on clayey soil
- d)Rigid footing on sandy soil
Correct answer is option 'B'. Can you explain this answer?
The Contact pressure and settlement distribution for the footing are shown in the figure.
The figure corresponds to a
a)
Flexible footing on clayey soil
b)
Flexible footing on sandy soil
c)
Rigid footing on clayey soil
d)
Rigid footing on sandy soil
|
Ishan Kapoor answered |
In flexible footing, settlement is non-uniform but contact pressure is more or less uniform. In case of sands, modulus of elasticity (E) increases with the confining pressure and therefore increases with the depth. Further ‘E’ is greater near the centre than at edges, As a result, the distribution of deformation below a flexible footing will be of the invested dish pattern while the contact pressure again be more or less uniform.
The yield of a well depends upon:- a)Permeability of soil
- b)Area of aquifer opening into the wells
- c)Actual flow velocity
- d)all of the above
Correct answer is option 'D'. Can you explain this answer?
The yield of a well depends upon:
a)
Permeability of soil
b)
Area of aquifer opening into the wells
c)
Actual flow velocity
d)
all of the above
|
|
Puja Sharma answered |
Yield of well means the discharge from the well excavated through given aquifer
Q = nva A = KiA
where
n is porosity
va is actual flow velocity of ground water
A is area of the aquifer opening onto the wells
K is permeability of soil
The laboratory test results of the soil sample are given below:Liquid limit = 37%Plastic limit = 22%% passing through 75 μ sieve = 26%% retained over 4.75 mm IS sieve = 65%% retained over 0.075 mm but passing through 4.75 mm = 26%As per IS 1498 – 1970, the soil is classified as:- a)GC
- b)SC
- c)CI
- d)GM
Correct answer is option 'A'. Can you explain this answer?
The laboratory test results of the soil sample are given below:
Liquid limit = 37%
Plastic limit = 22%
% passing through 75 μ sieve = 26%
% retained over 4.75 mm IS sieve = 65%
% retained over 0.075 mm but passing through 4.75 mm = 26%
As per IS 1498 – 1970, the soil is classified as:
a)
GC
b)
SC
c)
CI
d)
GM
|
|
Partho Jain answered |
% passing through 75 μ sieve = 26%
% retained over 75 μ sieve = 100 – 26% = 74 % > 50%
Hence the soil is coarse grained soil.
As, larger proportion of soil is retained over 4.75 mm IS sieve, so the soil is termed as Gravel.
Now, % fineness = 26% > 12%
So, use Ip to further classify the soil
IP = wL - wP = 37 – 22 = 15%
IP = 15% > 7%
So, soil is termed as clayey Gravel.
Consider the following statements:1. Method of slices overestimates the value of factor of safety.2. The exact value of factor of safety is obtained in ϕT = 0 analysis.3. Reduction in shearing resistance in embankment is caused by tension crack at the top of it.4. The plane surface of failure never occurs in saturated soils.Which of the statements given above are correct?- a)1 and 4 only
- b)2 and 3 only
- c)3 and 4 only
- d)1 and 2 only
Correct answer is option 'C'. Can you explain this answer?
Consider the following statements:
1. Method of slices overestimates the value of factor of safety.
2. The exact value of factor of safety is obtained in ϕT = 0 analysis.
3. Reduction in shearing resistance in embankment is caused by tension crack at the top of it.
4. The plane surface of failure never occurs in saturated soils.
Which of the statements given above are correct?
a)
1 and 4 only
b)
2 and 3 only
c)
3 and 4 only
d)
1 and 2 only
|
|
Prasad Desai answered |
The actual shape of a slip surface in the case of finite slopes is curvilinear. The assumption of a circular slip surface and its application for stability analysis was developed in Sweden. The method is known as the Swedish circle method or method of slices. This method neglects the effect of forces acting on the sides of the vertical strips. The error due to this is on safe side. i.e. the factor of safety obtained is generally less than that obtained from more accurate methods such as Bishop’s method which also consider the forces on the sides of vertical strips.
ϕT = 0 means that friction is not mobilized. In the actual case the resistance to shear will be due to friction and cohesion both thus exact value of factor of safety cannot be obtained for ϕT = 0.
Previous sand having air content 15% is used in the foundation of a masonry dam. If the maximum permissible upward gradient for a factor of safety of 3 against boiling is 0.315. The percentage of air voids in the previous sand will be ______ %.
Correct answer is between '6.2,6.6'. Can you explain this answer?
Previous sand having air content 15% is used in the foundation of a masonry dam. If the maximum permissible upward gradient for a factor of safety of 3 against boiling is 0.315. The percentage of air voids in the previous sand will be ______ %.
|
|
Manasa Sen answered |
F.O.S against boiling is giving by
F.O.S = 
ier = critical hydraulic gradient
iper = permissible hydraulic gradient
η = percentage air voids
ac = air content
η = porosity
calculation:
icr = iper x FOS
icr = 0.315 × 3
icr = 0.945
η = 0.427 = 42.7%
ac = 15%
ηa = 0.427 × 0.15
ηa = 0.06405
ηa = 6.405%
Sieve analysis is conducted on a given sample of soil. The data collected is given below:(i) % of particles passing through 75 μ IS sieve = 73%(ii) Liquid limit of the soil = 50(iii) Plasticity index of the soil = 16The possible value of Group index for soil is _______
Correct answer is between '11.8,12'. Can you explain this answer?
Sieve analysis is conducted on a given sample of soil. The data collected is given below:
(i) % of particles passing through 75 μ IS sieve = 73%
(ii) Liquid limit of the soil = 50
(iii) Plasticity index of the soil = 16
The possible value of Group index for soil is _______
|
|
Nidhi Patel answered |
Concept:
Group index of a soil is given by
G.I = 0.2 a + 0.005 ac + 0.01 bd
a = It is the portion of % passing through 75 μ sieve greater than 35 but not exceeding 75 expressed as whole number in between [0-40].
b = It is the portion of % passing through 75 μ sieve greater than 15 but not exceeding 55 expressed as whole number in between [0-40].
c = It is the portion of the numerical liquid limit greater than 40 but not exceeding 60 expressed as whole number in between [0-20].
d = It is the portion of the numerical plasticity index greater than to 10 but not exceeding 30 expressed as whole number in between [0-20].
Calculation:
a = 73 - 35 = 38 < 40
a = 38
b = p - 15
p = 73% which is greater than 55
So, take p = 55
b = 55 - 15 = 40
c = WL - 40
c = 50 - 40 = 10
d = IP - 10
d = 16 - 10 = 6
G.I = 0.2 a + 0.005 ac + 0.01 bd
G.I = 0.2 × 38 + 0.005 × 38 × 10 + 0.01 × 40 × 6
G.I = 11.9
15 metres of thick saturated clay is underlain by the sand layer and the sand layer is under artesian pressure equivalent to 5 metres head of water. Excavation is started and is it done up to 12.7 metres of depth in the clay layer at which the bottom heaves. For this much excavation depth in the clay. How much-saturated unit weight of clay you expect such that bottom will not heaves?- a)less than 11.33 kN/m3
- b)greater than 11.33 kN/m3
- c)less than 21.33 kN/m3
- d)greater than 21.33 kN/m3
Correct answer is option 'D'. Can you explain this answer?
15 metres of thick saturated clay is underlain by the sand layer and the sand layer is under artesian pressure equivalent to 5 metres head of water. Excavation is started and is it done up to 12.7 metres of depth in the clay layer at which the bottom heaves. For this much excavation depth in the clay. How much-saturated unit weight of clay you expect such that bottom will not heaves?
a)
less than 11.33 kN/m3
b)
greater than 11.33 kN/m3
c)
less than 21.33 kN/m3
d)
greater than 21.33 kN/m3
|
Pankaj Rane answered |
Understanding the Problem
The scenario involves a 15-meter thick saturated clay layer underlain by a sand layer, which is under artesian pressure equivalent to 5 meters of water head. Excavation is performed up to 12.7 meters in the clay, leading to bottom heave.
Key Concepts
- Saturated Unit Weight of Clay: This refers to the weight of the clay when fully saturated with water.
- Artesian Pressure: The pressure within the sand layer generates a force that can cause heaving when the weight of the overlying soil is insufficient to counteract it.
Calculating Heave Condition
When excavation occurs, the effective stress at the base of the excavation decreases, which can lead to heaving if the upward pressure from the artesian water exceeds the weight of the clay above.
Weight of Water
- The pressure from the artesian water is equivalent to a 5-meter height of water, resulting in a pressure of approximately 50 kN/m² (using the unit weight of water as 10 kN/m³).
Effective Stress Consideration
- The effective stress at the bottom of the excavation can be calculated as:
Effective Stress = Weight of Clay - Artesian Pressure
- For the bottom to remain stable (not heave), the weight of the clay (in terms of unit weight) must be greater than the upward pressure from artesian water.
Unit Weight Threshold
- To prevent heave, the unit weight of the saturated clay must be greater than the upward pressure due to artesian conditions.
- Since the pressure from the 5-meter water head is approximately 50 kN/m², the unit weight of the clay must be sufficiently high to maintain stability.
Conclusion
Therefore, the saturated unit weight of the clay must be greater than 21.33 kN/m³ to ensure that the bottom does not heave during excavation.
Thus, the correct answer is option 'D': greater than 21.33 kN/m³.
The scenario involves a 15-meter thick saturated clay layer underlain by a sand layer, which is under artesian pressure equivalent to 5 meters of water head. Excavation is performed up to 12.7 meters in the clay, leading to bottom heave.
Key Concepts
- Saturated Unit Weight of Clay: This refers to the weight of the clay when fully saturated with water.
- Artesian Pressure: The pressure within the sand layer generates a force that can cause heaving when the weight of the overlying soil is insufficient to counteract it.
Calculating Heave Condition
When excavation occurs, the effective stress at the base of the excavation decreases, which can lead to heaving if the upward pressure from the artesian water exceeds the weight of the clay above.
Weight of Water
- The pressure from the artesian water is equivalent to a 5-meter height of water, resulting in a pressure of approximately 50 kN/m² (using the unit weight of water as 10 kN/m³).
Effective Stress Consideration
- The effective stress at the bottom of the excavation can be calculated as:
Effective Stress = Weight of Clay - Artesian Pressure
- For the bottom to remain stable (not heave), the weight of the clay (in terms of unit weight) must be greater than the upward pressure from artesian water.
Unit Weight Threshold
- To prevent heave, the unit weight of the saturated clay must be greater than the upward pressure due to artesian conditions.
- Since the pressure from the 5-meter water head is approximately 50 kN/m², the unit weight of the clay must be sufficiently high to maintain stability.
Conclusion
Therefore, the saturated unit weight of the clay must be greater than 21.33 kN/m³ to ensure that the bottom does not heave during excavation.
Thus, the correct answer is option 'D': greater than 21.33 kN/m³.
A water tank is supported on 4 columns, which form a square of side 3m. The total load of the water tank is 400 ton. Assuming the load on each footing as point load, find the vertical stress induced at a depth of 3m under any one of the footings? (In t/m2)
Correct answer is between '7.5,7.6'. Can you explain this answer?
A water tank is supported on 4 columns, which form a square of side 3m. The total load of the water tank is 400 ton. Assuming the load on each footing as point load, find the vertical stress induced at a depth of 3m under any one of the footings? (In t/m2)
|
Uday Kumar answered |
The arrangement of the column / footing will be as follows:
Total load = 400 ton.
load in each column, Q = 400/4 = 100 ton
∴ Stress at a depth of 3m under any column
Total load = 400 ton.
load in each column, Q = 400/4 = 100 ton
∴ Stress at a depth of 3m under any column
(Two column will be at a distance 3 m & one column will be at a distance 3√2m)
= 5.30+ 1.876+ 0.34
= 7.516 ton/m2
A commercial building circular in the plan is situated at a particular site. If the total load from the building including its self-weight is 150 kN/m2, determine the vertical stress (kN/m2) at a depth of 8 m below the C.G of the load area. Assume the diameter of the building to be 3 m.
Correct answer is between '7.3,7.7'. Can you explain this answer?
A commercial building circular in the plan is situated at a particular site. If the total load from the building including its self-weight is 150 kN/m2, determine the vertical stress (kN/m2) at a depth of 8 m below the C.G of the load area. Assume the diameter of the building to be 3 m.
|
Saptarshi Nair answered |
Concept:
Vertical pressure below uniformly loaded circular area is given by:
σz = kq
where,
k = 1 – cos3 ∝
cos∝= 
Calculation:
q = 150 kN/m2
z = 8 m
r = 3/2m = 1.5m
cos∝ = 
cos ∝ = 0.983
k = 1 – cos3 ∝ = 1 – (0.983)3
k = 0.050
σz = 0.050 × (150)
σz = 7.5 kN/m2
A sampling tube has an inner diameter and outside diameter as shown in the figure below.D1 = 75 mmD2 = 80 mmD3 = 77 mmD4 = 78 mmThe inside clearance (%) of the samples is- a)2.67%
- b)1.30%
- c)13.78%
- d)2.56%
Correct answer is option 'A'. Can you explain this answer?
A sampling tube has an inner diameter and outside diameter as shown in the figure below.
D1 = 75 mm
D2 = 80 mm
D3 = 77 mm
D4 = 78 mm
The inside clearance (%) of the samples is
a)
2.67%
b)
1.30%
c)
13.78%
d)
2.56%
|
|
Ashwin Desai answered |
Concept:
% inside clearance = 
where D3 = Inside diameter of sampling tube
D1 = Inside diameter of cutting edge shoe
Calculation:
Ci = 2.67%
It must lies between 1 to 3%
Consider the following statements :1. In the laboratory consolidation test, initial compression is the result of the displacement of soil particles.2. Primary consolidation is due to dissipation of pore water pressure.3. Secondary compression starts after complete dissipation of pore water pressure.4. Primary consolidation and secondary compression occur simultaneously.Which of the statements given above are correct?- a)1 and 4 only
- b)2 and 3 only
- c)2 and 4 only
- d)1 and 3 only
Correct answer is option 'B'. Can you explain this answer?
Consider the following statements :
1. In the laboratory consolidation test, initial compression is the result of the displacement of soil particles.
2. Primary consolidation is due to dissipation of pore water pressure.
3. Secondary compression starts after complete dissipation of pore water pressure.
4. Primary consolidation and secondary compression occur simultaneously.
Which of the statements given above are correct?
a)
1 and 4 only
b)
2 and 3 only
c)
2 and 4 only
d)
1 and 3 only
|
Rahul Sen answered |
In laboratory consolidation test, the soil is laterally confined and fully saturated. So there is almost no opportunity for particles to get displaced and cause initial compression. Primary consolidation is due to dissipation of pore water pressure. The time taken to dissipate excess pore pressure and transfer of stress increment in this stage is called hydrodynamic lag. After this stage is complete, the stresses are transferred to the surface of the soil particles.
It should be noted that secondary compression starts after the completion of primary consolidation.
Consider the following statements regarding clay particles.1. The clay particles carry a net negative charge.2. The cation concentration increases with the distance from the surface of the particle.3. When the spacing between the two clay particles is very small, the force of attraction is greater than the force of repulsion.The correct statements are:- a)1 and 2 only
- b)2 and 3 only
- c)1 and 3 only
- d)1, 2 and 3
Correct answer is option 'C'. Can you explain this answer?
Consider the following statements regarding clay particles.
1. The clay particles carry a net negative charge.
2. The cation concentration increases with the distance from the surface of the particle.
3. When the spacing between the two clay particles is very small, the force of attraction is greater than the force of repulsion.
The correct statements are:
a)
1 and 2 only
b)
2 and 3 only
c)
1 and 3 only
d)
1, 2 and 3
|
|
Nidhi Patel answered |
The clay particles carry a net negative charge on their surfaces. This is the result both of isomorphous substitution and of a break in continuity of the structure at its edges. In dry clay, the negative charge is balanced by exchangeable cations like Ca2+, Mg2+, Na+, and K+ surrounding the particles being held by electrostatic attraction. When water is added to clay, these cations and a few anions float around the clay particles. This configuration is referred to as a diffuse double layer.
The cation concentration decreases with the distance from the surface of the particle. The force of attraction between water and clay decreases with distance from the surface of the particles.
When two clay particles in suspension come close to each other, the tendency for interpenetration of the diffuse double layers results in repulsion between the particles. At the same time, an attractive force exists between the clay particles that is caused by van der Waals forces and is independent of the characteristics of water. Both repulsive and attractive forces increase with decreasing distance between the particles, but at different rates. When the spacing between the particles is very small, the force of attraction is greater than the force of repulsion.
The cation concentration decreases with the distance from the surface of the particle. The force of attraction between water and clay decreases with distance from the surface of the particles.
When two clay particles in suspension come close to each other, the tendency for interpenetration of the diffuse double layers results in repulsion between the particles. At the same time, an attractive force exists between the clay particles that is caused by van der Waals forces and is independent of the characteristics of water. Both repulsive and attractive forces increase with decreasing distance between the particles, but at different rates. When the spacing between the particles is very small, the force of attraction is greater than the force of repulsion.
Which of the following is NOT an assumption in Terzaghi’s one-dimensional consolidation theory?- a)Darcy’s law is applicable.
- b)Soil is fully saturated.
- c)Soil is laterally confined.
- d)Soil is heterogeneous.
Correct answer is option 'D'. Can you explain this answer?
Which of the following is NOT an assumption in Terzaghi’s one-dimensional consolidation theory?
a)
Darcy’s law is applicable.
b)
Soil is fully saturated.
c)
Soil is laterally confined.
d)
Soil is heterogeneous.
|
Arjun Unni answered |
Assumptions in Terzaghi’s theory of consolidation:
1. The soil is homogeneous.
2. The soil is fully saturated.
3. The compression of the soil layer is due to the change in volume only, which, in turn, is due to the squeezing out of water from the void spaces.
4. Darcy’s law is valid.
5. Deformation of soil occurs only in the direction of the load application.
6. The coefficient of consolidation is constant during the consolidation.
A soil sample is tested in a consolidometer shows compression of 0.75 mm when the load is increased from 50 kN/m2 to 100 kN/m2. If instead of 100 kN/m2, the load is increased to 150 kN/m2, settlement of the soil sample will be _____ mm.
Correct answer is between '1.1,1.2'. Can you explain this answer?
A soil sample is tested in a consolidometer shows compression of 0.75 mm when the load is increased from 50 kN/m2 to 100 kN/m2. If instead of 100 kN/m2, the load is increased to 150 kN/m2, settlement of the soil sample will be _____ mm.
|
Shreya Choudhary answered |
Consolidation settlement is given by:
Where,
CC = Compression index
e0 = Initial void ratio of the soil sample
H0 = Thickness of the soil sample
σ̅0 = Initial effective stress at the centre of the soil sample
Increase in effective stress at the centre of the soil sample
Increase in effective stress at the centre of the soil sampleCalculation:
ΔH1 = 0.75 mm
σ̅0 = 50 kN/m2
ΔH2 = 1.188 mm
ΔH2 = 1.188 mm
36 Piles are arranged in a square pattern which is embedded in the clayey soil. The shear strength at the base of the pile is 200 kN/m2 and the average shear strength over the depth of the pile is 130 kN/m2. If the Diameter and length of the pile are 0.4 m and 12 m respectively, the safe load that the pile group can carry will be _______ kN. Assume, centre to centre spacing between the pile = 0.8 m, Adhesion factor (α) = 0.50 and Factor of safety = 3
Correct answer is between '14474,14478'. Can you explain this answer?
36 Piles are arranged in a square pattern which is embedded in the clayey soil. The shear strength at the base of the pile is 200 kN/m2 and the average shear strength over the depth of the pile is 130 kN/m2. If the Diameter and length of the pile are 0.4 m and 12 m respectively, the safe load that the pile group can carry will be _______ kN. Assume, centre to centre spacing between the pile = 0.8 m, Adhesion factor (α) = 0.50 and Factor of safety = 3
|
Ishaan Kulkarni answered |
B = (n – 1)s + d
B = (6 – 1) × 0.8 + 0.4
B = 4.0 + 0.4 = 4.4 m
Ultimate load carrying capacity of single pile:
Qup = qb Ab + qs As
Qb = base resistance of pile = 9c
c = cohesion at base of the pile
ϕ = 0° (clay)
Sbase = c = 200 kN/m2
qs = average skin friction resistance
Qup = 1206.37 kN
Ultimate load carrying capacity of pile group
= 9 × 200 × (4.4)2 + 0.5 × 130 × 4 × 4.4 × 12
Qug = 48576 kN
Safe load on pile group = 
So, safe load on pile group will be lesser of two. Hence safe load = 14476.44 kN
So, safe load on pile group will be lesser of two. Hence safe load = 14476.44 kN
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