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Question 1
ప్రశ్న 1
Which description most accurately represents the assumptions of classical infinite-slope stability analysis?
Explanation:
Infinite-slope analysis idealizes a slope that extends sufficiently far compared with the depth of the potential failure surface. The soil profile and slope inclination are assumed approximately uniform over the region considered. A potential translational sliding plane is taken parallel to the slope surface. The method evaluates the balance between available shear resistance and driving shear stress along that plane. It is particularly useful for assessing shallow translational failures in extensive slopes.
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Question 2
ప్రశ్న 2
A dry homogeneous infinite slope has inclination beta = 30 degrees and soil unit weight gamma = 20 kN/m3. A potential failure plane lies at a vertical depth z = 4 m below the surface. What are the normal stress and downslope shear stress on the plane? Take cos(30 degrees) = 0.8660 and sin(30 degrees) = 0.5.
Explanation:
For an infinite slope with vertical depth z, the normal stress on a plane parallel to the ground is sigma_n = gamma z cos²(beta). The driving shear stress is tau = gamma z sin(beta) cos(beta). The vertical overburden stress used in these expressions is 20 × 4 = 80 kPa. Hence sigma_n = 80(0.75) = 60.0 kPa. The driving shear stress is 80(0.5)(0.8660) = approximately 34.6 kPa.
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Question 3
ప్రశ్న 3
A dry infinite slope consists of cohesionless sand with effective friction angle phi' = 30 degrees and slope inclination beta = 20 degrees. What is its factor of safety against translational sliding? Take tan(30 degrees) = 0.5774 and tan(20 degrees) = 0.3640.
Explanation:
For a dry cohesionless infinite slope, the factor of safety is FS = tan(phi')/tan(beta). Cohesion is zero, so resistance is developed through friction alone. Substitution gives FS = 0.5774/0.3640. The resulting value is approximately 1.586. Therefore the factor of safety is about 1.59 under the stated ideal conditions.
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Question 4
ప్రశ్న 4
For a dry cohesionless infinite slope, the effective friction angle is phi' = 32 degrees. At what slope inclination beta does the classical infinite-slope model predict limiting equilibrium with factor of safety equal to unity?
Explanation:
The factor of safety of a dry cohesionless infinite slope is tan(phi')/tan(beta). At limiting equilibrium, the factor of safety equals one. This condition requires tan(beta) = tan(phi'). For the relevant range of soil and slope angles, beta therefore equals phi'. The limiting slope inclination is 32 degrees.
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Question 5
ప్రశ్న 5
A dry infinite slope has vertical failure-plane depth z = 3 m, gamma = 20 kN/m3, effective cohesion c' = 12 kPa, effective friction angle phi' = 30 degrees and slope angle beta = 30 degrees. What is the factor of safety? Use sin(30 degrees) = 0.5, cos(30 degrees) = 0.8660 and tan(30 degrees) = 0.57735.
Explanation:
The driving shear stress is gamma z sin(beta) cos(beta). Substitution gives 20(3)(0.5)(0.8660) = 25.98 kPa. The normal stress is gamma z cos²(beta) = 20(3)(0.75) = 45 kPa. Available shear strength is c' + sigma_n tan(phi') = 12 + 45(0.57735) = 37.98 kPa. Therefore FS = 37.98/25.98 = approximately 1.46.
వివరణ:
Question 6
ప్రశ్న 6
A dry infinite slope consists of an ideal purely cohesive soil with cohesion c = 10 kPa and friction angle phi = 0 degrees. The slope angle is 30 degrees, gamma = 20 kN/m3 and the potential failure plane is 2 m below the surface vertically. What is the factor of safety?
Explanation:
For the stated purely cohesive model, the frictional component of shear strength is zero. The available shear strength is therefore equal to the cohesion of 10 kPa. The driving shear stress is gamma z sin(beta) cos(beta). Substituting gives 20(2)(0.5)(0.8660) = 17.32 kPa. Hence FS = 10/17.32 = approximately 0.58, indicating that the stated idealized plane does not satisfy limiting equilibrium.
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Question 7
ప్రశ్న 7
A dry infinite slope has gamma = 20 kN/m3, c' = 10 kPa, beta = 30 degrees and phi' = 20 degrees. Assuming constant strength parameters and a slip plane parallel to the slope, what is the approximate critical vertical depth at which FS = 1? Take sin(30 degrees)cos(30 degrees) = 0.4330, cos²(30 degrees) = 0.75 and tan(20 degrees) = 0.3640.
Explanation:
At limiting equilibrium, the driving shear stress equals the available shear strength. Therefore gamma z sin(beta) cos(beta) = c' + gamma z cos²(beta) tan(phi'). Rearranging gives z = c'/[gamma(sin(beta)cos(beta) - cos²(beta)tan(phi'))]. Substitution yields z = 10/[20(0.4330 - 0.75(0.3640))]. The calculated critical vertical depth is approximately 3.12 m.
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Question 8
ప్రశ్న 8
For a dry homogeneous infinite slope with constant positive effective cohesion c', constant phi', gamma and beta, how does its calculated factor of safety change as the vertical depth of the potential failure plane increases?
Explanation:
The dry infinite-slope factor of safety can be separated into cohesive and frictional contributions. The cohesive contribution is c'/[gamma z sin(beta)cos(beta)]. This term decreases as the vertical depth z increases. The frictional contribution is tan(phi')/tan(beta), which is independent of depth under the stated assumptions. Consequently the total factor of safety decreases toward the friction-only value as depth becomes large.
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Question 9
ప్రశ్న 9
A dry infinite slope has gamma = 20 kN/m3, vertical slip-plane depth z = 3 m, beta = 30 degrees and phi' = 25 degrees. What minimum effective cohesion is required to obtain FS = 1.50 on that plane? Use sin(30 degrees)cos(30 degrees) = 0.4330, cos²(30 degrees) = 0.75 and tan(25 degrees) = 0.4663.
Explanation:
Required shear resistance equals the specified factor of safety multiplied by the driving shear stress. The driving stress is 20(3)(0.4330) = 25.98 kPa. The required available strength is therefore 1.50(25.98) = 38.97 kPa. The frictional strength contribution is 20(3)(0.75)(0.4663) = 20.98 kPa. Subtracting frictional resistance gives a minimum required cohesion of approximately 18.0 kPa.
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Question 10
ప్రశ్న 10
A dry cohesionless infinite slope must satisfy a specified minimum factor of safety of 1.50. Its friction angle is such that tan(phi') = 0.60. What is the maximum permissible slope angle beta under this criterion? Take arctan(0.40) = 21.8 degrees.
Explanation:
For a dry cohesionless slope, FS = tan(phi')/tan(beta). The minimum required factor of safety is 1.50. Therefore tan(beta) must not exceed 0.60/1.50 = 0.40. The corresponding limiting inclination is arctan(0.40). Hence the maximum slope inclination satisfying the specified criterion is approximately 21.8 degrees.
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Question 11
ప్రశ్న 11
A cohesionless infinite slope is fully submerged under static hydrostatic conditions, with balancing external water pressure and no seepage forces. If tan(phi') = 0.60 and tan(beta) = 0.40, what is its ideal factor of safety?
Explanation:
For the stated hydrostatically submerged condition, soil self-weight is represented through its submerged or buoyant unit weight. The same buoyant unit weight affects both the normal and downslope gravitational components. For cohesionless soil, these components cancel from the factor-of-safety expression. Therefore FS = tan(phi')/tan(beta) = 0.60/0.40. The ideal factor of safety is 1.50, which differs from the case of sustained seepage parallel to the slope.
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Question 12
ప్రశ్న 12
A fully submerged homogeneous infinite slope has no seepage and is subjected to balanced hydrostatic water pressures. Its submerged unit weight is gamma' = 10 kN/m3, effective cohesion c' = 6 kPa, phi' = 30 degrees and beta = 30 degrees. For a potential plane 3 m below the ground surface vertically, what is the approximate factor of safety? Take tan(30 degrees) = 0.57735.
Explanation:
Under the stated balanced hydrostatic condition, the submerged unit weight is used to represent the net gravitational loading. The driving shear stress is gamma' z sin(beta)cos(beta) = 10(3)(0.4330) = 12.99 kPa. The effective normal stress from this idealized loading is gamma' z cos²(beta) = 10(3)(0.75) = 22.5 kPa. The available resistance is 6 + 22.5(0.57735) = approximately 18.99 kPa. Consequently FS = 18.99/12.99 = approximately 1.46.
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Question 13
ప్రశ్న 13
A cohesionless infinite slope has steady saturated seepage parallel to the slope surface, with the seepage surface at ground level. Given gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3, phi' = 30 degrees and beta = 20 degrees, calculate the factor of safety. Use tan(30 degrees) = 0.5774 and tan(20 degrees) = 0.3640.
Explanation:
The submerged unit weight is gamma' = gamma_sat - gamma_w = 10 kN/m3. For cohesionless soil with seepage parallel to the slope and water at the ground surface, FS = (gamma'/gamma_sat)[tan(phi')/tan(beta)]. The unit-weight ratio is 10/20 = 0.50. The frictional ratio is 0.5774/0.3640 = approximately 1.586. Multiplication gives FS = approximately 0.793, or 0.79.
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Question 14
ప్రశ్న 14
A saturated infinite slope has steady seepage parallel to its surface, with the seepage surface at ground level. The slope angle is beta = 30 degrees and the potential slip plane is at vertical depth z = 3 m. If gamma_w = 10 kN/m3, what is the pore-water pressure on the plane under the standard parallel-seepage idealization?
Explanation:
For the stated parallel-seepage condition, pore pressure on the potential plane is u = gamma_w z cos²(beta). The vertical depth is 3 m, and cos²(30 degrees) = 0.75. Substituting gives u = 10(3)(0.75). Therefore the pore-water pressure is 22.5 kPa. This expression specifically represents the stated seepage geometry and should not be substituted indiscriminately for other groundwater conditions.
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Question 15
ప్రశ్న 15
A saturated infinite slope has gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3 and beta = 30 degrees. The water table is at the ground surface and seepage is parallel to the slope. At a vertical depth of 4 m, what are the total normal stress, effective normal stress and downslope driving shear stress on a parallel failure plane?
Explanation:
The total normal stress is gamma_sat z cos²(beta) = 20(4)(0.75) = 60 kPa. Under parallel seepage, pore-water pressure is gamma_w z cos²(beta) = 10(4)(0.75) = 30 kPa. The effective normal stress is therefore 60 - 30 = 30 kPa. The downslope driving shear stress is gamma_sat z sin(beta)cos(beta) = 20(4)(0.4330). This gives approximately 34.6 kPa, so the required values are 60.0, 30.0 and 34.6 kPa.
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Question 16
ప్రశ్న 16
A saturated infinite slope has steady seepage parallel to the surface with groundwater at ground level. Given gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3, c' = 10 kPa, phi' = 30 degrees, beta = 30 degrees and vertical failure-plane depth z = 2 m, what is the approximate factor of safety?
Explanation:
The total driving shear stress is 20(2)sin(30 degrees)cos(30 degrees) = approximately 17.32 kPa. The effective normal stress is (gamma_sat - gamma_w)z cos²(beta) = 10(2)(0.75) = 15 kPa. The available shear strength is c' + sigma_n' tan(phi') = 10 + 15(0.57735) = approximately 18.66 kPa. The factor of safety is the available strength divided by the driving stress. Therefore FS = 18.66/17.32 = approximately 1.08.
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Question 17
ప్రశ్న 17
Consider the same cohesionless infinite slope with tan(phi') = 0.60, tan(beta) = 0.40, gamma_sat = 20 kN/m3 and gamma_w = 10 kN/m3. Compare Case I, fully submerged with balanced hydrostatic water pressure and no seepage, with Case II, saturated seepage parallel to the slope with the seepage surface at ground level. What are the ideal factors of safety for Cases I and II, respectively?
Explanation:
In Case I, balanced hydrostatic submergence permits the cohesionless factor of safety to be expressed as tan(phi')/tan(beta). This gives FS = 0.60/0.40 = 1.50. In Case II, the parallel-seepage factor includes the multiplying ratio gamma'/gamma_sat. Here gamma' = 20 - 10 = 10 kN/m3, so the ratio is 0.50. Thus the parallel-seepage factor of safety is 0.50(1.50) = 0.75.
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Question 18
ప్రశ్న 18
A cohesionless infinite slope has steady seepage parallel to the surface and the seepage surface at ground level. The ratio gamma'/gamma_sat is 0.50, and tan(phi') = 0.80. What is the limiting slope inclination beta for FS = 1.0? Take arctan(0.40) = 21.8 degrees.
Explanation:
For cohesionless soil with the specified parallel seepage, FS = (gamma'/gamma_sat)tan(phi')/tan(beta). At limiting equilibrium, FS equals one. Hence tan(beta) = (0.50)(0.80) = 0.40. The corresponding slope inclination is arctan(0.40). Therefore the limiting angle is approximately 21.8 degrees under the stated conditions.
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Question 19
ప్రశ్న 19
A saturated infinite slope has a potential failure plane at vertical depth z = 3 m, beta = 30 degrees, gamma_sat = 20 kN/m3 and measured pore-water pressure u = 15 kPa on the plane. Assume c' = 8 kPa and tan(phi') = 0.50. Using the measured pore pressure in an effective-stress analysis, what is the factor of safety?
Explanation:
The total normal stress is gamma_sat z cos²(beta) = 20(3)(0.75) = 45 kPa. Subtracting the measured pore pressure gives effective normal stress sigma_n' = 45 - 15 = 30 kPa. The available shear strength is 8 + 30(0.50) = 23 kPa. The driving shear stress is 20(3)sin(30 degrees)cos(30 degrees) = approximately 25.98 kPa. Consequently FS = 23/25.98 = approximately 0.89.
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Question 20
ప్రశ్న 20
For an infinite slope, total normal stress on a potential plane is 45 kPa and the driving shear stress is 25.98 kPa. Its effective cohesion is 10 kPa, and tan(phi') = 0.50. What pore-water pressure would reduce the factor of safety to exactly 1.0, assuming all other quantities remain unchanged?
Explanation:
At limiting equilibrium, available shear strength must equal the driving shear stress. The required equation is 25.98 = 10 + (45 - u)(0.50). Subtracting cohesion gives 15.98 = 0.50(45 - u). Therefore 45 - u = 31.96, giving u = 13.04 kPa. This represents the limiting pore pressure for the stated plane and constant strength parameters.
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Question 21
ప్రశ్న 21
At a potential failure plane, the driving shear stress remains 25 kPa and the effective friction parameter tan(phi') = 0.50. The total normal stress and effective cohesion remain unchanged, but positive pore-water pressure increases by 10 kPa. By how much does the calculated factor of safety decrease?
Explanation:
The effective normal stress decreases by the same amount as the increase in pore-water pressure when total normal stress remains constant. Here the effective normal stress decreases by 10 kPa. The associated loss of frictional shear resistance is 10(0.50) = 5 kPa. Dividing this reduction by the unchanged driving shear stress of 25 kPa gives 5/25 = 0.20. Therefore the factor of safety decreases by 0.20.
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Question 22
ప్రశ్న 22
A dry infinite slope has beta = 30 degrees, phi' = 30 degrees, c' = 10 kPa and gamma = 20 kN/m3. Two potential sliding planes are located at vertical depths of 2 m and 4 m. Assuming homogeneous soil and no other loads, which plane has the lower factor of safety, and what is its approximate value?
Explanation:
For this dry cohesive-frictional slope, FS = c'/[gamma z sin(beta)cos(beta)] + tan(phi')/tan(beta). Since phi' and beta both equal 30 degrees, the frictional contribution is one. At 2 m depth, the cohesive contribution is 10/[20(2)(0.4330)] = approximately 0.577, giving FS = 1.58. At 4 m depth, the cohesive contribution is 10/[20(4)(0.4330)] = approximately 0.289, giving FS = 1.29. Therefore the deeper plane has the lower calculated factor of safety.
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Question 23
ప్రశ్న 23
An unsaturated soil slope derives part of its apparent shear resistance from matric suction. During prolonged rainfall, infiltration increases its water content and reduces matric suction. Which interpretation is technically most appropriate?
Explanation:
Matric suction can provide additional apparent shear resistance in partially saturated soils. Rainfall infiltration may reduce this suction contribution. It can also change pore-water pressure and the distribution of effective stresses within the slope. The resulting loss of available strength can reduce the factor of safety and contribute to shallow slope failures. Sound engineering interpretation should account for the actual moisture and groundwater conditions rather than assuming suction-related resistance remains constant.
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Question 24
ప్రశ్న 24
A dry infinite slope has inclination beta = 30 degrees and gamma = 20 kN/m3. A potential sliding plane parallel to the slope lies at a perpendicular distance d = 2 m measured normal to the slope surface, rather than at a vertical depth of 2 m. What are the normal stress and downslope shear stress due to soil self-weight on that plane?
Explanation:
The question gives depth measured normal to the slope rather than vertically. For a plane at normal depth d, the vertical separation is z = d/cos(beta). Substituting this relation into the infinite-slope stress equations gives sigma_n = gamma d cos(beta) and tau = gamma d sin(beta). Thus sigma_n = 20(2)(0.8660) = 34.64 kPa. The downslope shear stress is 20(2)(0.5) = 20.0 kPa.
వివరణ:
Question 25
ప్రశ్న 25
A saturated infinite slope has steady seepage parallel to the surface with the seepage surface at ground level. The slope has beta = 30 degrees, phi' = 30 degrees, gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3 and vertical failure-plane depth z = 2 m. What effective cohesion c' is required to achieve FS = 1.50 on the specified plane? Use sin(30 degrees)cos(30 degrees) = 0.4330.
Explanation:
The submerged unit weight is gamma' = gamma_sat - gamma_w = 10 kN/m3. The frictional contribution to factor of safety is (gamma'/gamma_sat)[tan(phi')/tan(beta)] = 0.50 because the friction and slope angles are equal. To achieve FS = 1.50, the cohesive contribution must therefore equal 1.00. The driving shear stress is gamma_sat z sin(beta)cos(beta) = 20(2)(0.4330) = 17.32 kPa. Since cohesion supplies the remaining shear resistance, the required effective cohesion is approximately 17.32 kPa.
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Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The slope extends sufficiently far with an approximately uniform profile, and the potential translational failure plane is parallel to the ground surface. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. 60.0 kPa and 34.6 kPa. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 1.59. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: C. 32 degrees. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 1.46. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 0.58. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 3.12 m. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: D. It decreases toward tan(phi')/tan(beta), because the relative contribution of cohesion becomes smaller with increasing depth. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: B. 18.0 kPa. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. 21.8 degrees. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. 1.50. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: D. 1.46. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. 0.79. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 22.5 kPa. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. 60.0 kPa, 30.0 kPa and 34.6 kPa, respectively. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: D. 1.08. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. 1.50 and 0.75. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 21.8 degrees. సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. 0.89. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: D. 13.04 kPa. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. 0.20. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. The 4 m plane, with FS = 1.29. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. Loss of suction-related apparent strength can reduce stability, and the changing pore-pressure conditions should be evaluated rather than treating suction contribution as permanent cohesion. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: D. 34.6 kPa and 20.0 kPa. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: B. 17.32 kPa. సమాధానం: B.
