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Question 1
ప్రశ్న 1
Which equation correctly represents the Mohr-Coulomb shear failure criterion in terms of effective normal stress for a saturated soil?
Explanation:
The Mohr-Coulomb criterion expresses shear strength as the sum of cohesion and frictional resistance. In effective-stress form, tau_f = c' + sigma_n' tan(phi'). The effective normal stress is sigma_n' = sigma_n - u, where u is the pore-water pressure. Therefore the equation becomes tau_f = c' + (sigma_n - u) tan(phi'). This relationship is fundamental to the evaluation of drained shear strength and soil stability.
వివరణ:
Question 2
ప్రశ్న 2
On a graph of shear strength against effective normal stress, which interpretation of the linear Mohr-Coulomb failure envelope is correct?
Explanation:
The effective-stress Mohr-Coulomb equation is tau_f = c' + sigma_n' tan(phi'). It represents a straight line when cohesion and friction angle are treated as constant over the relevant stress range. The intercept on the shear-stress axis is the effective cohesion c'. The slope of the line is tan(phi'), and the inclination to the normal-stress axis is phi'. This interpretation allows shear-strength parameters to be obtained from laboratory failure data.
వివరణ:
Question 3
ప్రశ్న 3
A soil has effective cohesion c' = 20 kPa and tan(phi') = 0.50. If the effective normal stress on a potential failure plane is 140 kPa, what is the shear strength?
Explanation:
The effective-stress shear-strength equation is tau_f = c' + sigma_n' tan(phi'). The cohesion contribution is 20 kPa. The frictional contribution is 140 × 0.50 = 70 kPa. Adding both contributions gives tau_f = 20 + 70 = 90 kPa. Hence the shear strength on the specified plane is 90 kPa.
వివరణ:
Question 4
ప్రశ్న 4
At a potential failure plane in saturated soil, total normal stress is 220 kPa and pore-water pressure is 70 kPa. The soil parameters are c' = 15 kPa and tan(phi') = 0.40. What is the effective-stress shear strength?
Explanation:
The effective normal stress is obtained by subtracting pore-water pressure from total normal stress. Therefore sigma_n' = 220 - 70 = 150 kPa. The frictional shear resistance is 150 × 0.40 = 60 kPa. Adding the cohesion contribution gives tau_f = 15 + 60 = 75 kPa. Using total normal stress directly in the friction term would incorrectly ignore pore-water pressure.
వివరణ:
Question 5
ప్రశ్న 5
A saturated clean sand is idealized as having zero effective cohesion and an effective friction angle of 30 degrees. If the effective normal stress on a shear plane is 180 kPa, what is its drained shear strength? Take tan(30 degrees) = 0.57735.
Explanation:
For an ideal cohesionless soil, effective cohesion is taken as zero. The drained Mohr-Coulomb equation therefore reduces to tau_f = sigma_n' tan(phi'). Substitution gives tau_f = 180 × 0.57735. The result is approximately 103.9 kPa. The shear resistance arises from the frictional mechanism represented by the effective friction angle.
వివరణ:
Question 6
ప్రశ్న 6
Two drained direct shear tests give failure shear stresses of 70 kPa and 110 kPa at effective normal stresses of 100 kPa and 200 kPa, respectively. Assuming a linear Mohr-Coulomb envelope, determine c' and phi'.
Explanation:
The slope of the failure envelope is the change in shear stress divided by the change in effective normal stress. Thus tan(phi') = (110 - 70)/(200 - 100) = 0.40. The corresponding friction angle is arctan(0.40), approximately 21.8 degrees. Substituting the first failure point gives c' = 70 - 100(0.40) = 30 kPa. Therefore the effective cohesion is 30 kPa and the effective friction angle is approximately 21.8 degrees.
వివరణ:
Question 7
ప్రశ్న 7
A saturated cohesionless soil has tan(phi') = 0.50 and a constant total normal stress of 200 kPa on a specified plane. If positive pore-water pressure increases from 40 kPa to 100 kPa, by how much does its effective-stress shear strength decrease?
Explanation:
Initially the effective normal stress is 200 - 40 = 160 kPa. The initial shear strength is 160 × 0.50 = 80 kPa. After the pore pressure increases, the effective normal stress becomes 200 - 100 = 100 kPa. The new shear strength is 100 × 0.50 = 50 kPa. Consequently the increase in pore pressure reduces the shear strength by 80 - 50 = 30 kPa.
వివరణ:
Question 8
ప్రశ్న 8
A saturated clay is idealized by a total-stress undrained friction angle phi_u = 0 degrees and undrained shear strength su = 45 kPa. What shear strengths are predicted by this model at total normal stresses of 100 kPa and 200 kPa, respectively?
Explanation:
In the conventional phi_u = 0 total-stress model for saturated clay, the undrained shear strength is represented by su. The failure envelope is horizontal when the same su is assumed for the specified stress range. Therefore shear strength does not increase with total normal stress within this idealization. Both specified planes have an undrained shear strength of 45 kPa. This model should not be confused with the effective-stress strength parameters of the clay.
వివరణ:
Question 9
ప్రశ్న 9
A rapidly constructed embankment imposes a substantial new load on a thick, saturated, low-permeability clay foundation. Which strength assessment is generally most appropriate for evaluating the immediate end-of-construction condition?
Explanation:
Low-permeability saturated clay may not drain sufficiently during rapid construction. The applied loading can therefore generate substantial excess pore-water pressure. For this short-term condition, a total-stress analysis using appropriately determined undrained shear strength is commonly suitable. The analysis must reflect the actual loading history and representative soil conditions. Long-term stability should also be evaluated separately using effective-stress parameters and appropriate pore-pressure conditions.
వివరణ:
Question 10
ప్రశ్న 10
Which statement correctly describes the drainage conditions used in conventional UU, CU and CD triaxial compression tests?
Explanation:
The unconsolidated-undrained test prevents drainage during the application of cell pressure and during shearing. In the consolidated-undrained test, the specimen is first allowed to consolidate under the selected confining stress. Drainage is then prevented during the shearing stage. In the consolidated-drained test, drainage is permitted during both consolidation and shearing, with shearing performed sufficiently slowly to avoid significant excess pore pressures. These differences are important because test drainage conditions determine the stress and strength interpretation.
వివరణ:
Question 11
ప్రశ్న 11
A series of saturated clay specimens is tested using consolidated-undrained triaxial compression with pore-water pressure measurements at failure. Which conclusion is correct regarding determination of strength parameters?
Explanation:
In a CU triaxial test, the specimen is consolidated before undrained shearing. Total principal stresses at failure can be determined from the applied stresses. Measuring pore-water pressure permits calculation of the corresponding effective principal stresses. Results from suitable tests at different consolidation conditions can then be used to develop an effective-stress failure envelope. Undrained shearing does not require the effective friction angle to be zero.
వివరణ:
Question 12
ప్రశ్న 12
In a conventional UU triaxial compression test on saturated clay, the total major and minor principal stresses at failure are 260 kPa and 100 kPa, respectively. Assuming the phi_u = 0 total-stress model, what is the undrained shear strength?
Explanation:
For the conventional phi_u = 0 total-stress model, undrained shear strength equals one-half of the principal stress difference at failure. The major principal stress is 260 kPa. The minor principal stress is 100 kPa. Therefore su = (260 - 100)/2 = 80 kPa. The result is the radius of the total-stress Mohr circle at failure.
వివరణ:
Question 13
ప్రశ్న 13
An unconfined compression test on a saturated clay specimen gives an unconfined compressive strength qu = 140 kPa. Under the conventional phi_u = 0 undrained interpretation, what is the undrained shear strength?
Explanation:
In an unconfined compression test, the minor total principal stress is zero. The major principal stress at failure equals the unconfined compressive strength qu. Under the conventional saturated-clay phi_u = 0 assumption, su = qu/2. Substitution gives su = 140/2 = 70 kPa. This relationship is applicable to the stated undrained interpretation and should not be applied indiscriminately to other soil conditions.
వివరణ:
Question 14
ప్రశ్న 14
During a direct shear test, the horizontal shear force at failure is 0.48 kN and the corrected area of the shear plane at failure is 60 cm². What is the shear stress at failure?
Explanation:
Shear stress is obtained by dividing the applied shear force by the corresponding shear area. The force is 0.48 kN, or 480 N. The corrected shear-plane area is 60 cm², equal to 0.006 m². Therefore tau = 480/0.006 = 80000 N/m². This is equivalent to 80 kPa.
వివరణ:
Question 15
ప్రశ్న 15
A saturated soil has c' = 10 kPa and phi' = 30 degrees. The total normal stress on a plane is 200 kPa, and the pore-water pressure is 80 kPa. Using tan(30 degrees) = 0.57735, determine its effective-stress shear strength.
Explanation:
The effective normal stress on the specified plane is 200 - 80 = 120 kPa. The effective cohesion is 10 kPa. The frictional resistance is 120 × 0.57735 = approximately 69.28 kPa. The total shear strength is therefore 10 + 69.28 = 79.28 kPa. Rounding to one decimal place gives 79.3 kPa.
వివరణ:
Question 16
ప్రశ్న 16
On a potential slip plane, total normal stress is 220 kPa and the applied shear stress is 80 kPa. Soil strength parameters are c' = 10 kPa and tan(phi') = 0.50. Assuming these parameters remain constant, what positive pore-water pressure would bring the plane exactly to the Mohr-Coulomb failure condition?
Explanation:
At the limiting condition, the applied shear stress equals the available shear strength. Therefore 80 = 10 + (220 - u)(0.50). Subtracting cohesion gives 70 = 0.50(220 - u). Rearrangement yields 140 = 220 - u, so u = 80 kPa. A further increase in positive pore pressure at unchanged total normal and applied shear stresses would make the calculated available strength smaller than the applied shear stress.
వివరణ:
Question 17
ప్రశ్న 17
A cohesionless soil follows the effective-stress Mohr-Coulomb criterion with phi' = 30 degrees. In a triaxial compression test at failure, the minor effective principal stress is 100 kPa. What is the major effective principal stress at failure?
Explanation:
For a cohesionless soil in triaxial compression, the principal effective stresses at Mohr-Coulomb failure satisfy sigma1'/sigma3' = (1 + sin(phi'))/(1 - sin(phi')). With phi' = 30 degrees, sin(phi') = 0.50. The principal stress ratio is therefore 1.50/0.50 = 3. The major effective principal stress is 3 × 100 = 300 kPa. The result applies to the stated cohesionless triaxial compression failure condition.
వివరణ:
Question 18
ప్రశ్న 18
Under the conventional Mohr-Coulomb triaxial compression failure model, what angle does the theoretical shear failure plane make with the major principal plane when the friction angle is 30 degrees?
Explanation:
The major principal plane is the plane on which the major principal stress acts normally. Under the conventional Mohr-Coulomb triaxial compression model, the failure plane makes an angle of 45 degrees plus half the friction angle with this principal plane. The given friction angle is 30 degrees. Therefore the predicted inclination is 45 + 30/2 = 60 degrees. This orientation refers specifically to the major principal plane and should not be confused with an angle measured from the major principal stress direction.
వివరణ:
Question 19
ప్రశ్న 19
The major and minor effective principal stresses at a point are 260 kPa and 100 kPa, respectively. What are the centre and radius of the corresponding Mohr circle in the effective normal stress-shear stress plane?
Explanation:
The centre of the Mohr circle is the average of the major and minor principal stresses. Therefore the centre is (260 + 100)/2 = 180 kPa. The radius is half the difference between the principal stresses. Thus the radius is (260 - 100)/2 = 80 kPa. The radius also represents the maximum shear stress corresponding to the specified two-dimensional principal stress state.
వివరణ:
Question 20
ప్రశ్న 20
On a specified potential sliding plane, the total normal stress is 180 kPa, pore-water pressure is 20 kPa and applied shear stress is 70 kPa. The effective strength parameters are c' = 10 kPa and tan(phi') = 0.50. What is the factor of safety defined as available shear strength divided by applied shear stress on this plane?
Explanation:
The effective normal stress is 180 - 20 = 160 kPa. The available shear strength is 10 + 160(0.50) = 90 kPa. The applied shear stress is given as 70 kPa. The specified factor of safety is therefore 90/70 = approximately 1.286. Rounded to two decimal places, the factor of safety for this plane is 1.29.
వివరణ:
Question 21
ప్రశ్న 21
A soil has effective cohesion 10 kPa and tan(phi') = 0.50. If effective normal stress on a potential shear plane increases from 120 kPa to 200 kPa without changing the strength parameters, what is the increase in shear strength?
Explanation:
Initially the shear strength is 10 + 120(0.50) = 70 kPa. After the effective normal stress increases, the shear strength is 10 + 200(0.50) = 110 kPa. The cohesion component remains unchanged under the stated assumptions. The increase in frictional resistance is therefore 110 - 70 = 40 kPa. This illustrates the influence of effective normal stress on shear strength in a frictional soil.
వివరణ:
Question 22
ప్రశ్న 22
Which statement best distinguishes peak shear strength from residual shear strength in a soil exhibiting strain-softening behaviour?
Explanation:
Peak shear strength is the maximum shear resistance mobilized during the loading process. Some soils experience strain softening after this maximum is reached. Large shear displacements may produce a lower residual resistance, particularly in certain clay soils with developed shear surfaces. Particle reorientation and changes in the shearing mechanism can contribute to this reduction. Residual strength is especially relevant when assessing slopes or surfaces that have already undergone substantial shear displacement.
వివరణ:
Question 23
ప్రశ్న 23
For dense and loose specimens of the same sand tested under comparable drained effective-stress conditions, which statement is generally correct?
Explanation:
The packing density of sand strongly influences its drained shearing behaviour. Dense sand commonly tends to dilate as particles move past one another. This dilatancy can contribute to a pronounced peak shear resistance and a comparatively high mobilized peak friction angle. Loose sand more commonly contracts during shearing and may develop a less pronounced peak. Both soils remain frictional materials, and their behaviour also depends on effective confinement and other soil characteristics.
వివరణ:
Question 24
ప్రశ్న 24
A partially saturated soil exhibits an apparent cohesion intercept associated with capillary suction. Which engineering interpretation is most appropriate when the soil becomes substantially wetter?
Explanation:
Partially saturated soils can develop additional shear resistance associated with matric suction. This resistance may appear as an apparent cohesion intercept in a simplified shear-strength representation. Infiltration and wetting can reduce matric suction and thereby reduce the associated strength contribution. Apparent suction-related cohesion should not automatically be treated as a permanent material property. Engineering assessments must consider changes in moisture conditions and the appropriate unsaturated-soil strength behaviour.
వివరణ:
Question 25
ప్రశ్న 25
On a potential shear plane in saturated soil, initial total normal stress is 150 kPa and initial pore-water pressure is 50 kPa. A subsequent loading stage increases total normal stress by 80 kPa and pore-water pressure by 60 kPa. If c' = 10 kPa and tan(phi') = 0.50 remain unchanged, how does the effective-stress shear strength change?
Explanation:
Initially the effective normal stress is 150 - 50 = 100 kPa. The initial shear strength is therefore 10 + 100(0.50) = 60 kPa. After loading, the total normal stress becomes 230 kPa and pore-water pressure becomes 110 kPa. The final effective normal stress is 230 - 110 = 120 kPa, giving a final shear strength of 10 + 120(0.50) = 70 kPa. Thus the shear strength increases by 10 kPa because the effective normal stress increases by only 20 kPa despite the larger total-stress increment.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. tau_f = c' + (sigma_n - u) tan(phi'), where sigma_n is total normal stress. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. The intercept on the shear-stress axis is c', while the slope is tan(phi'). సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: C. 90 kPa. సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: A. 75 kPa. సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. 103.9 kPa. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. c' = 30 kPa and phi' = 21.8 degrees. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: C. 30 kPa. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. 45 kPa and 45 kPa. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: D. Evaluate undrained behaviour using representative undrained shear strength and a consistent total-stress analysis. సమాధానం: D.
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Question 10 ప్రశ్న 10Answer: B. UU prevents drainage in both stages, CU allows consolidation drainage but prevents drainage during shearing, and CD allows drainage during consolidation and shearing. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. Total-stress failure results are available, and effective-stress failure parameters can also be evaluated by subtracting measured pore-water pressures from total stresses. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. 80 kPa. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: D. 70 kPa. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: C. 80 kPa. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. 79.3 kPa. సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: A. 80 kPa. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: D. 300 kPa. సమాధానం: D.
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Question 18 ప్రశ్న 18Answer: B. 60 degrees. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. Centre = 180 kPa; radius = 80 kPa. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: A. 1.29. సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: D. 40 kPa. సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: B. Peak strength is the maximum resistance mobilized during shearing, while residual strength is the lower resistance that may develop after large shear displacements. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. Dense sand commonly exhibits dilation and a pronounced peak shear resistance, whereas loose sand more commonly contracts during shearing. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: D. Wetting can reduce matric suction and the associated apparent shear-strength contribution, so stability should not rely on that contribution remaining constant. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. It increases from 60 kPa to 70 kPa. సమాధానం: A.
