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Question 1
ప్రశ్న 1
Which combination correctly identifies the relevant Indian Standard test methods for laboratory determination of soil shear strength?
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Explanation:
IS 2720 (Part 10) addresses determination of unconfined compressive strength of soil. IS 2720 (Part 11) covers unconsolidated-undrained triaxial compression without pore-water pressure measurement. IS 2720 (Part 12) addresses consolidated-undrained triaxial compression with pore-water pressure measurement. IS 2720 (Part 13) covers the direct shear test. Correct identification of the applicable testing method is essential for interpreting the reported shear-strength parameters.
వివరణ:
Question 2
ప్రశ్న 2
Which limitation is particularly associated with the conventional direct shear box test compared with the triaxial compression test?
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Explanation:
In the conventional direct shear test, relative horizontal displacement occurs between the two halves of the shear box. The resulting shearing plane is essentially predetermined by the apparatus. This plane may not coincide with the weakest or naturally preferred failure plane of the soil. Stress distribution along the shear plane is also not perfectly uniform during shearing. In contrast, the triaxial compression test permits the specimen to develop its own failure mechanism under the imposed stress conditions.
వివరణ:
Question 3
ప్రశ్న 3
A square direct shear specimen initially measures 60 mm by 60 mm. At failure, the relative horizontal displacement is 5 mm, and the measured horizontal shear force is 330 N. Assuming the corrected overlapping shear area is 60 × (60 - 5) mm², what is the shear stress at failure?
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Explanation:
The corrected overlapping shear area is calculated using the remaining overlap length. Therefore the area is 60 × 55 = 3300 mm². Converting this area gives 0.0033 m². The shear stress is the shear force divided by the corrected shear area, giving 330/0.0033 = 100000 Pa. Hence the shear stress at failure is 100.0 kPa.
వివరణ:
Question 4
ప్రశ్న 4
In a drained direct shear test on an ideal cohesionless soil, the corrected shear area is 3600 mm². The normal load is 540 N and the horizontal shear force at failure is 270 N. Assuming the Mohr-Coulomb criterion with c' = 0, what is the effective friction angle?
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Explanation:
The effective normal stress is 540/0.0036 = 150 kPa. The shear stress at failure is 270/0.0036 = 75 kPa. For an ideal cohesionless soil, the effective shear-strength equation reduces to tau_f = sigma_n' tan(phi'). Therefore tan(phi') = 75/150 = 0.50. The corresponding friction angle is approximately 26.6 degrees.
వివరణ:
Question 5
ప్రశ్న 5
Two drained direct shear tests on the same soil give failure shear stresses of 65 kPa and 110 kPa at effective normal stresses of 100 kPa and 200 kPa, respectively. Assuming a linear Mohr-Coulomb failure envelope, determine the effective cohesion and friction angle.
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Explanation:
The slope of the failure envelope is the change in shear stress divided by the change in effective normal stress. Therefore tan(phi') = (110 - 65)/(200 - 100) = 0.45. The friction angle is arctan(0.45), which is approximately 24.2 degrees. Substituting the first failure point gives c' = 65 - 100(0.45) = 20 kPa. Hence the effective cohesion is 20 kPa and the effective friction angle is approximately 24.2 degrees.
వివరణ:
Question 6
ప్రశ్న 6
Why is the rate of shearing particularly important when conducting a drained direct shear test on a fine-grained soil?
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Explanation:
Drainage conditions depend on soil permeability and the duration of shearing. Fine-grained soils generally have lower permeability than clean sands. Rapid shearing may generate excess pore-water pressures that cannot dissipate during the test. A drained test therefore requires a suitably slow shearing rate to permit the intended drainage conditions. Incorrect rate selection can produce strength results that do not represent the intended drained condition.
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Question 7
ప్రశ్న 7
Which statement correctly describes the conventional unconsolidated-undrained (UU) triaxial compression test?
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Explanation:
The UU triaxial test is conducted without allowing drainage during either principal loading stage. Confining pressure is applied while the drainage system remains closed. The specimen is then subjected to axial shearing under undrained conditions. In the conventional UU test without pore-pressure measurement, the results are interpreted in terms of total stresses. This procedure is commonly used to investigate short-term undrained shear behaviour of suitable cohesive soils.
వివరణ:
Question 8
ప్రశ్న 8
A saturated clay specimen fails in a conventional UU triaxial compression test at cell pressure sigma3 = 90 kPa and deviator stress qf = 140 kPa. Assuming the total-stress phi_u = 0 model, what are the major principal total stress at failure and undrained shear strength?
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Explanation:
In conventional triaxial compression, the deviator stress is q = sigma1 - sigma3. The major principal total stress at failure is therefore 90 + 140 = 230 kPa. Under the stated phi_u = 0 interpretation, undrained shear strength is half the principal stress difference. Thus su = 140/2 = 70 kPa. The resulting values are sigma1 = 230 kPa and su = 70 kPa.
వివరణ:
Question 9
ప్రశ్న 9
Two identical saturated clay specimens are tested under conventional UU triaxial conditions with confining pressures of 100 kPa and 200 kPa. Their measured deviator stresses at failure are both 120 kPa. Under the total-stress phi_u = 0 idealization, what conclusion is justified?
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Explanation:
In the phi_u = 0 total-stress model, undrained shear strength equals half the failure deviator stress. Both specimens have the same failure deviator stress of 120 kPa. Therefore su = 120/2 = 60 kPa in each test. The corresponding total-stress failure envelope is horizontal under the stated idealization. This observation does not establish that the soil has zero effective-stress friction angle.
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Question 10
ప్రశ్న 10
Which sequence correctly represents a consolidated-undrained (CU) triaxial compression test with pore-water pressure measurement?
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Explanation:
A CU test permits the specimen to consolidate under the selected confining conditions before shearing. Drainage is then closed so that the subsequent shear stage is undrained. Excess pore-water pressure may develop during this stage. Measuring pore-water pressure allows total principal stresses to be converted into effective principal stresses. The resulting data can support both total-stress and effective-stress interpretations of shear strength.
వివరణ:
Question 11
ప్రశ్న 11
In a CU triaxial test at failure, the total major principal stress is 420 kPa, the total minor principal stress is 240 kPa, and the uniform pore-water pressure is 60 kPa. What are the major and minor effective principal stresses at failure?
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Explanation:
Effective principal stress equals the corresponding total principal stress minus pore-water pressure. The major effective principal stress is 420 - 60 = 360 kPa. The minor effective principal stress is 240 - 60 = 180 kPa. The same pore-water pressure is subtracted from both principal total stresses. Hence the effective principal stresses are 360 kPa and 180 kPa.
వివరణ:
Question 12
ప్రశ్న 12
A CU triaxial test at failure gives total principal stresses sigma1 = 400 kPa and sigma3 = 180 kPa, with pore-water pressure u = 75 kPa. What is the radius of the effective-stress Mohr circle?
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Explanation:
The effective principal stresses are sigma1' = 400 - 75 = 325 kPa and sigma3' = 180 - 75 = 105 kPa. The radius of the effective-stress Mohr circle is half the difference between these stresses. Therefore R = (325 - 105)/2 = 110 kPa. Subtracting the same pore pressure from both total principal stresses does not change their difference. Consequently the effective and total Mohr circles have equal radii but different centre positions.
వివరణ:
Question 13
ప్రశ్న 13
A CU triaxial test on an ideal cohesionless soil gives total principal stresses at failure of 370 kPa and 170 kPa, with measured pore-water pressure 70 kPa. Assuming effective cohesion c' = 0, what is the effective friction angle?
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Explanation:
The effective principal stresses at failure are 370 - 70 = 300 kPa and 170 - 70 = 100 kPa. For cohesionless soil in triaxial compression, sin(phi') = (sigma1' - sigma3')/(sigma1' + sigma3'). Substituting gives sin(phi') = (300 - 100)/(300 + 100) = 0.50. The angle whose sine is 0.50 is 30 degrees. Therefore the effective friction angle is 30 degrees under the stated Mohr-Coulomb assumptions.
వివరణ:
Question 14
ప్రశ్న 14
Which statement correctly describes the consolidated-drained (CD) triaxial compression test?
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Explanation:
The CD triaxial test allows the specimen to consolidate under the imposed confining conditions. Drainage is also permitted during the subsequent shearing stage. Shearing must be sufficiently slow for the intended drained condition to be maintained. The test is suitable for obtaining effective-stress strength parameters under drained conditions. It is especially relevant to engineering problems where drainage can occur over the loading period.
వివరణ:
Question 15
ప్రశ్న 15
An ideal cohesionless specimen is tested in drained triaxial compression with effective friction angle phi' = 30 degrees. At failure, the minor effective principal stress is 100 kPa. What are the major effective principal stress and deviator stress at failure?
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Explanation:
For cohesionless soil at Mohr-Coulomb triaxial compression failure, sigma1'/sigma3' = (1 + sin(phi'))/(1 - sin(phi')). Since phi' = 30 degrees, sin(phi') = 0.50. The principal effective stress ratio is therefore 1.50/0.50 = 3. Hence sigma1' = 3 × 100 = 300 kPa, and the deviator stress is 300 - 100 = 200 kPa. The required results are 300 kPa and 200 kPa.
వివరణ:
Question 16
ప్రశ్న 16
During a triaxial compression test, the axial deviator load at a particular stage is 600 N and the corrected specimen cross-sectional area is 3000 mm². If the cell pressure is 100 kPa, what are the deviator stress and major principal total stress at that stage?
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Explanation:
The deviator stress is calculated from the axial deviator load divided by the corrected cross-sectional area. The area of 3000 mm² equals 0.003 m². Therefore q = 600/0.003 = 200000 Pa, or 200 kPa. In conventional triaxial compression, sigma1 = sigma3 + q. Since the cell pressure is 100 kPa, the major principal total stress is 300 kPa.
వివరణ:
Question 17
ప్రశ్న 17
Which soil specimen is generally most suitable for a conventional unconfined compression test?
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Explanation:
The unconfined compression test is conducted without externally applied lateral confining pressure. The specimen must therefore retain its shape during preparation and axial loading. Cohesive soils are commonly suitable because they can possess sufficient unsupported stability. Loose sands and many coarse granular materials cannot maintain an intact cylindrical specimen without confinement. The test is particularly useful for estimating undrained shear strength of suitable cohesive soils.
వివరణ:
Question 18
ప్రశ్న 18
A saturated clay specimen has unconfined compressive strength qu = 180 kPa. Assuming the conventional phi_u = 0 total-stress interpretation, what is its undrained shear strength?
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Explanation:
In an unconfined compression test, the minor principal total stress is zero. The major principal total stress at failure is the unconfined compressive strength qu. Under the conventional phi_u = 0 interpretation, undrained shear strength is su = qu/2. Substituting qu = 180 kPa gives su = 180/2 = 90 kPa. This relationship is appropriate for the stated saturated-clay undrained strength model.
వివరణ:
Question 19
ప్రశ్న 19
A saturated cohesive clay specimen is tested under undrained conditions in an unconfined compression test. Its initial cross-sectional area is 1000 mm². At failure, axial strain is 10% and the axial load is 120 N. Using the constant-volume area correction A = A0/(1 - epsilon) and the conventional phi_u = 0 total-stress interpretation, what are the corrected unconfined compressive strength and undrained shear strength?
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Explanation:
The saturated cohesive clay is assumed to undergo undrained loading with negligible volume change. The corrected cross-sectional area is A = 1000/(1 - 0.10) = 1111.11 mm². The corrected axial compressive stress is 120 N divided by 1111.11 mm², giving qu = 108 kPa. Under the explicitly stated conventional phi_u = 0 total-stress model, undrained shear strength is su = qu/2. Therefore the unconfined compressive strength is 108 kPa and the undrained shear strength is 54 kPa.
వివరణ:
Question 20
ప్రశ్న 20
An undisturbed saturated clay sample has unconfined compressive strength 160 kPa. After complete remoulding at the same water content, its unconfined compressive strength is 40 kPa. Assuming the conventional undrained strength relationship applies to both specimens, what is the sensitivity of the clay?
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Explanation:
Soil sensitivity is the ratio of undisturbed undrained shear strength to remoulded undrained shear strength at the same water content. Under the stated interpretation, both shear strengths are half their respective unconfined compressive strengths. The undisturbed undrained strength is 160/2 = 80 kPa. The remoulded undrained strength is 40/2 = 20 kPa. Therefore sensitivity is 80/20 = 4.0, which is also equal to the ratio 160/40.
వివరణ:
Question 21
ప్రశ్న 21
Which interpretation of sensitivity and remoulding is correct for a cohesive soil?
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Explanation:
Sensitivity is an index describing the effect of remoulding on the undrained strength of cohesive soil. It is defined as the undisturbed undrained shear strength divided by the remoulded undrained shear strength at the same water content. Remoulding disturbs the original soil fabric and may substantially reduce shear resistance. Soils exhibiting large strength loss after remoulding have correspondingly high sensitivity values. This property is relevant to construction disturbance, excavation stability and the behaviour of natural clay deposits.
వివరణ:
Question 22
ప్రశ్న 22
During an isotropic undrained pressure increment used to assess triaxial specimen saturation, the cell pressure increases by 100 kPa and the measured pore-water pressure increases by 95 kPa. What is Skempton's pore-pressure parameter B for this increment?
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Explanation:
For the specified isotropic undrained pressure increment, Skempton's B parameter is the change in pore-water pressure divided by the change in cell pressure. The measured pore-pressure increase is 95 kPa. The applied cell-pressure increase is 100 kPa. Therefore B = 95/100 = 0.95. A value near unity is commonly associated with a high degree of saturation, although interpretation also depends on the compressibility characteristics of the soil-water system.
వివరణ:
Question 23
ప్రశ్న 23
A dense saturated sand specimen is sheared under consolidated-undrained triaxial conditions and exhibits a tendency to dilate. Which pore-pressure response may occur during undrained shearing, and why is its measurement important?
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Explanation:
Dense sand can exhibit a tendency to dilate during shearing. If drainage is prevented, the tendency toward volume increase can generate negative excess pore-water pressure. This response may increase effective stresses relative to the values that would occur under positive excess pore pressure. The resulting effective-stress path influences the mobilized shear resistance. Pore-pressure measurements are therefore important for interpreting CU triaxial results and distinguishing total-stress behaviour from effective-stress behaviour.
వివరణ:
Question 24
ప్రశ్న 24
An engineer needs effective-stress strength parameters of saturated clay for a long-term drained stability assessment. Which laboratory testing strategy is most appropriate?
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Explanation:
Long-term drained stability is commonly evaluated using effective-stress shear-strength parameters. A properly conducted CD triaxial test provides drained failure data suitable for evaluating these parameters. A CU triaxial test with reliable pore-water pressure measurement can also provide effective principal stresses at failure. Tests over an appropriate range of confining stresses are used to develop the relevant effective-stress failure envelope. Strength parameters obtained from undrained total-stress interpretations should not be substituted directly for effective-stress parameters.
వివరణ:
Question 25
ప్రశ్న 25
An engineer compares results from a UU test without pore-pressure measurement, a CU test with pore-pressure measurement, and a properly drained CD test. Which conclusion is technically correct?
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Explanation:
A conventional UU triaxial test without pore-water pressure measurement provides failure information in terms of total stresses. It does not directly provide effective principal stresses because the pore-water pressure at failure is unknown. A CU test with pore-water pressure measurements permits both total and effective principal stresses to be evaluated. A properly conducted CD test allows drainage during shearing and is used to establish drained effective-stress strength behaviour. Selecting the appropriate test and interpretation is necessary to obtain shear-strength parameters relevant to the engineering loading and drainage conditions.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. IS 2720 (Part 10) - unconfined compression; Part 11 - UU triaxial; Part 12 - CU triaxial with pore-pressure measurement; Part 13 - direct shear. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. The failure plane is predetermined by the division between the two halves of the shear box. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 100.0 kPa. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. 26.6 degrees. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. c' = 20 kPa and phi' = 24.2 degrees. సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. The shearing rate must be sufficiently slow to permit drainage and avoid significant excess pore-water pressure during the test. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. Drainage is prevented during both application of confining pressure and subsequent shearing. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. sigma1 = 230 kPa and su = 70 kPa. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: A. The undrained shear strength is 60 kPa in both tests, and the idealized total-stress failure envelope is horizontal. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: C. Allow consolidation drainage under the selected confining conditions, close drainage during shearing, and measure pore-water pressure during undrained loading. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 360 kPa and 180 kPa. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 110 kPa. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: D. 30 degrees. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: A. Drainage is allowed during consolidation and shearing, with sufficiently slow loading to avoid significant excess pore-water pressures. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: B. sigma1' = 300 kPa and qf = 200 kPa. సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. q = 200 kPa and sigma1 = 300 kPa. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: D. A cohesive soil specimen capable of standing unsupported during the test. సమాధానం: D.
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Question 18 ప్రశ్న 18Answer: A. 90 kPa. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. qu = 108 kPa and su = 54 kPa. సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 4.0. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: D. Sensitivity measures the loss of undrained shear strength associated with remoulding, using undisturbed and remoulded strengths compared at the same water content. సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: B. 0.95. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: A. Excess pore-water pressure may become negative because dilation is restrained, affecting the effective stresses mobilized during shearing. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. Conduct suitable consolidated-drained triaxial tests, or consolidated-undrained triaxial tests with reliable pore-pressure measurements, and interpret the resulting effective-stress failure envelope. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: D. The UU test without pore-pressure measurement primarily provides total-stress strength data, the CU test with pore-pressure measurement supports both total- and effective-stress interpretation, and the CD test supports drained effective-stress strength interpretation. సమాధానం: D.