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Question 1
ప్రశ్న 1
Which statement best distinguishes the usual finite-slope stability problem from the classical infinite-slope idealization?
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Explanation:
A finite slope has a crest and a toe that can influence its failure mechanism. Rotational movement along a curved slip surface is a common finite-slope idealization, particularly for relatively homogeneous cohesive soils. The classical infinite-slope model instead considers a shallow translational plane parallel to a sufficiently extensive slope surface. Finite slopes can also fail by noncircular mechanisms when geology creates preferential weak surfaces. The appropriate idealization depends on slope geometry and ground conditions.
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Question 2
ప్రశ్న 2
A circular slip surface in a finite earth slope passes exactly through the toe of the slope. How is this rotational failure commonly classified?
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Explanation:
A toe failure is identified by a critical slip surface that passes through the slope toe. It is a recognized category of rotational failure in finite slopes. A face or slope failure instead exits through the slope face above the toe. A base failure extends through soil beneath the level of the toe. These descriptions classify the location of the slip surface rather than the factor of safety.
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Question 3
ప్రశ్న 3
A circular failure surface intersects the exposed face of an embankment slope at a point above its toe. Which failure classification is appropriate?
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Explanation:
A face or slope failure has its lower intersection with the slope face above the toe. This distinguishes it from a toe circle, which passes through the toe itself. A base failure involves a deeper surface extending below the toe. The classification can be made from the stated exit location without requiring a diagram. The mechanism may still be represented approximately by a circular arc.
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Question 4
ప్రశ్న 4
A finite slope overlies a comparatively weak clay layer extending below its toe, and a rotational slip surface passes beneath the toe into this layer. Which failure type is indicated?
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Explanation:
Base failure is characterized by a slip surface passing beneath the toe of a finite slope. A relatively weak layer below the toe can favour this deeper mechanism. The resulting sliding mass may extend beyond the geometric limits of a shallow face failure. Stability analysis must therefore include the material strength beneath the toe. Assuming every critical circle passes through the toe could overlook this mode.
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Question 5
ప్రశ్న 5
For an assumed circular slip surface of radius R, which statement correctly describes the ordinary rotational moment approach when the same radius applies to all basal shear forces?
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Explanation:
For a circular slip surface, basal tangential forces have a common moment arm equal to the radius. Their available shear resistances can therefore be summed and multiplied by that radius to obtain the resisting moment. The weight of a slice produces a moment according to its signed lever arm about the circle centre. Stabilizing and destabilizing contributions must be treated with consistent signs. The moment-based factor of safety compares available resisting action with the relevant driving action.
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Question 6
ప్రశ్న 6
For a trial circular slip surface, the available resisting moment is 4500 kN m per metre of slope width and the driving moment is 3000 kN m per metre. What is the moment-based factor of safety?
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Explanation:
The moment-based factor of safety is the available resisting moment divided by the driving moment. Both moments are given per metre of slope width, so their units cancel. Substitution gives F = 4500/3000. The ratio equals 1.50. This result applies to the specified trial surface and does not by itself establish the minimum factor of safety for all possible surfaces.
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Question 7
ప్రశ్న 7
A trial circular failure surface in saturated clay has total slip-arc length 18 m per metre of section width. A total-stress undrained analysis uses uniform su = 25 kPa and the signed sum of downslope driving components is 300 kN/m. Assuming phi_u = 0 and no other forces, what is the ordinary-slices factor of safety?
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Explanation:
For a phi_u = 0 total-stress model, available basal shear resistance is su multiplied by slip-arc length. The resistance per metre of slope width is 25 × 18 = 450 kN/m. The specified signed sum of driving components is 300 kN/m. Thus the ordinary-slices factor of safety is 450/300 = 1.50. Pore-pressure forces do not appear separately in this particular idealized total-stress calculation.
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Question 8
ప్రశ్న 8
On an assumed circular slip arc with radius 12 m, the sum of available basal shear-resistance forces is 240 kN/m and the driving moment about the circle centre is 1800 kN m per metre. What is the moment-based factor of safety?
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Explanation:
The available basal shear forces act tangentially to the circular failure surface. Their common lever arm about the centre is the circle radius of 12 m. The available resisting moment is therefore 240 × 12 = 2880 kN m per metre. Dividing by the driving moment gives F = 2880/1800. Hence the moment-based factor of safety is 1.60.
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Question 9
ప్రశ్న 9
Which simplification distinguishes the ordinary or Fellenius method of slices from methods that explicitly incorporate interslice forces?
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Explanation:
The ordinary method divides a potential sliding mass into slices. It neglects interslice force contributions when estimating basal forces in its simple formulation. For a circular surface, an overall moment balance gives the computed factor of safety. The method is approximate because a complete set of force-equilibrium conditions is not generally satisfied. Other procedures, such as simplified Bishop analysis, treat some interslice effects differently.
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Question 10
ప్రశ్న 10
In a dry ordinary-slices calculation with no external loads, two slices have (W1, alpha1, l1) = (100 kN/m, 20 degrees, 3 m) and (W2, alpha2, l2) = (80 kN/m, 30 degrees, 4 m). Both bases have c' = 5 kPa and tan(phi') = 0.40. What is F = sum[c'l + W cos(alpha)tan(phi')]/sum[W sin(alpha)]? Use sin20 = 0.3420, cos20 = 0.9397, sin30 = 0.5000 and cos30 = 0.8660.
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Explanation:
The total cohesive resistance is 5(3 + 4) = 35 kN/m. The normal-force sum is 100(0.9397) + 80(0.8660) = 163.25 kN/m. Friction contributes 0.40(163.25) = 65.30 kN/m, giving total resistance 100.30 kN/m. The driving sum is 100(0.3420) + 80(0.5000) = 74.20 kN/m. Consequently F = 100.30/74.20 = approximately 1.35.
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Question 11
ప్రశ్న 11
A phi_u = 0 ordinary-slices analysis uses three slices with weights 100, 150 and 80 kN/m and base inclinations 0, 30 and 30 degrees, respectively. Their base arc lengths are 2, 3 and 4 m, and uniform undrained strength is su = 20 kPa. Taking sin30 = 0.5, what is the factor of safety if no other forces act?
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Explanation:
With phi_u = 0, the available resistance equals su times the sum of the slice-base lengths. Thus the resisting force sum is 20(2 + 3 + 4) = 180 kN/m. The slice with a horizontal base provides no weight component along its base. The driving sum is 150(0.5) + 80(0.5) = 115 kN/m. Therefore F = 180/115 = approximately 1.57.
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Question 12
ప్రశ్న 12
A vertical slice in a circular-slip analysis has horizontal width b = 2.0 m, and its basal tangent makes 60 degrees with the horizontal. What is its approximate basal arc length using the usual straight-base approximation? Take cos60 = 0.50.
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Explanation:
The ordinary-slices geometry commonly approximates the slice base by a straight segment. Its horizontal projection is the slice width b. If alpha is the inclination of the base to the horizontal, the projected relation is b = l cos(alpha). Hence l = 2.0/0.50 = 4.0 m. This is a geometrical approximation to the corresponding segment of the circular failure arc.
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Question 13
ప్రశ్న 13
A vertical slice of unit out-of-plane thickness has horizontal width 2 m, average height 3 m and soil unit weight 20 kN/m3. If its base is inclined at 30 degrees to the horizontal, what are its weight and downslope weight component? Take sin30 = 0.50.
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Explanation:
The cross-sectional area of the slice is its width times its average height. Thus the area is 2 × 3 = 6 m2 per metre of out-of-plane thickness. Multiplication by the unit weight gives W = 20(6) = 120 kN/m. The component of weight parallel to the base is W sin(alpha). At 30 degrees, the downslope component has magnitude 120(0.50) = 60 kN/m.
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Question 14
ప్రశ్న 14
For one dry trial circle, an ordinary-slices calculation gives sum of basal arc lengths L = 12 m, sum of W cos(alpha) = 400 kN/m and signed sum of W sin(alpha) = 200 kN/m. If c' = 10 kPa and tan(phi') = 0.50, what is the factor of safety?
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Explanation:
The available cohesion contribution is c'L = 10(12) = 120 kN/m. The friction contribution is the normal-force sum multiplied by tan(phi'). It equals 400(0.50) = 200 kN/m. Total available resistance is therefore 120 + 200 = 320 kN/m. Dividing by the signed driving sum of 200 kN/m gives F = 1.60.
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Question 15
ప్రశ్న 15
For a specified portion of a trial slip arc, the total normal force on its base is already established as 100 kN/m and the resultant pore-water force acting normally on that base is 30 kN/m. The cohesive resistance along the base is 20 kN/m, tan(phi') = 0.50 and the driving shear-force contribution is 40 kN/m. What is the ratio of available effective-stress shear resistance to that driving force?
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Explanation:
The supplied total normal force and pore-water resultant refer to the same base. The effective normal force is 100 - 30 = 70 kN/m. Its frictional resistance is 70(0.50) = 35 kN/m. Adding the given cohesive resistance produces 20 + 35 = 55 kN/m. Dividing by the 40 kN/m driving contribution gives 1.375, or approximately 1.38.
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Question 16
ప్రశ్న 16
For a fixed trial slip surface, the cohesion-related resisting force is 50 kN/m, total normal force is 200 kN/m, tan(phi') = 0.50 and driving shear-force sum is 100 kN/m. Assume pore-water resultant normal to the surface increases from 0 to 50 kN/m with no change in other quantities. What are the initial and final strength-to-driving-force ratios?
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Explanation:
Initially the effective normal force equals 200 kN/m because pore-water force is zero. Available resistance is 50 + 200(0.50) = 150 kN/m, giving an initial ratio of 1.50. After pore-water force increases, the effective normal force decreases to 200 - 50 = 150 kN/m. Available resistance becomes 50 + 150(0.50) = 125 kN/m. The final strength-to-driving-force ratio is 125/100 = 1.25.
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Question 17
ప్రశ్న 17
In a moment-based circular-slip calculation, one slice produces W sin(alpha) = +50 kN/m in the chosen downslope direction while another produces W sin(alpha) = -30 kN/m because it tends to oppose that rotation. What is their net signed driving-force sum?
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Explanation:
The slice-base components must be added algebraically using a consistent rotation convention. The first contribution tends to drive the specified rotation and is positive. The second contribution opposes that rotation and is negative. Their net sum is 50 + (-30) = 20 kN/m. Adding absolute magnitudes would incorrectly count the opposing slice as destabilizing.
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Question 18
ప్రశ్న 18
Which comparison between the ordinary method of slices and simplified Bishop analysis is technically appropriate for circular failure surfaces?
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Explanation:
The ordinary method of slices commonly estimates a slice-base normal force from the weight and base inclination while neglecting interslice forces. Simplified Bishop analysis retains an approximation involving interslice normal forces and satisfies overall moment equilibrium. For frictional soils, its factor-of-safety equation is generally implicit and solved iteratively. The two methods can give different results, especially when pore pressures and geometry are important. Neither approximate method substitutes for examining the relevant possible failure surfaces.
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Question 19
ప్రశ్న 19
A finite-slope assessment gives trial-circle factors of safety of 1.38, 1.20, 1.47 and 1.31. Which statement correctly identifies the governing trial circle among those evaluated?
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Explanation:
The critical trial failure surface is the one with the smallest factor of safety among the candidates analyzed. The smallest supplied value is 1.20. Its location need not be specified to make this comparison. A new candidate surface could still have an even smaller factor of safety. Therefore the minimum reported value is the controlling result only for the evaluated trial set.
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Question 20
ప్రశ్న 20
For a homogeneous slope analyzed using a specified Taylor chart convention, define the stability number as Sn = cu/(F gamma H). If Sn = 0.10, cu = 40 kPa, gamma = 20 kN/m3 and slope height H = 10 m, what is F?
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Explanation:
The stability-number definition in the question is Sn = cu/(F gamma H). Rearranging gives F = cu/(Sn gamma H). Substitution gives F = 40/[0.10(20)(10)]. The denominator equals 20 kPa, so the resulting factor of safety is 2.00. The given stability number is supplied directly, so no external chart is required.
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Question 21
ప్రశ్న 21
Using the specified Taylor stability-number relation Sn = cu/(F gamma H), a uniform undrained clay slope has Sn = 0.12, gamma = 18 kN/m3 and H = 8 m. What undrained cohesion cu is required for a target factor of safety of 1.50?
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Explanation:
The stated stability-number equation can be rearranged to cu = Sn F gamma H. Insert Sn = 0.12 and target F = 1.50. The unit weight and height product equals 18(8) = 144 kPa. Multiplying 0.12(1.50)(144) gives 25.92 kPa. This calculation relies on the stability number already supplied for the selected slope geometry and analysis convention.
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Question 22
ప్రశ్న 22
For a homogeneous undrained clay slope, the selected Taylor stability number at limiting equilibrium is Sn = cu/(gamma Hc) = 0.20. If cu = 36 kPa and gamma = 18 kN/m3, what is the limiting slope height Hc for F = 1?
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Explanation:
At limiting equilibrium the factor of safety equals one. The supplied relation is Sn = cu/(gamma Hc). Rearranging yields Hc = cu/(Sn gamma). Substitution gives Hc = 36/[0.20(18)] = 36/3.6. Therefore the limiting height is 10.0 m under the stated Taylor stability-number convention.
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Question 23
ప్రశ్న 23
A slope analysis adopts the dimensionless Taylor stability number Sn = 0.08. What is the corresponding stability factor, defined in this question as the reciprocal 1/Sn?
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Explanation:
The question expressly defines the stability factor as the reciprocal of the stability number. Therefore the calculation is 1/0.08. This division gives 12.50. Both quantities are dimensionless under the stated convention. The reciprocal must not be confused with the slope factor of safety unless an additional relationship is supplied.
వివరణ:
Question 24
ప్రశ్న 24
On one trial circle, the initial available resisting moment is 3600 kN m/m and the driving moment is 3000 kN m/m. A new surcharge is idealized as adding 600 kN m/m to the driving moment while leaving the resisting moment unchanged. What are the initial and revised moment-based factors of safety?
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Explanation:
The initial factor of safety is the resisting moment divided by the driving moment. Therefore the initial ratio is 3600/3000 = 1.20. After the surcharge, the stated driving moment becomes 3000 + 600 = 3600 kN m/m. The resisting moment remains 3600 kN m/m by explicit assumption. Thus the revised factor of safety is 3600/3600 = 1.00.
వివరణ:
Question 25
ప్రశ్న 25
For an identified circular slip surface, a toe berm is idealized as adding 900 kN m/m of resisting moment without changing the existing 3900 kN m/m resisting moment or 3000 kN m/m driving moment otherwise. What is the revised moment-based factor of safety, and which engineering interpretation is appropriate?
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Explanation:
The initially available resisting moment is 3900 kN m/m, corresponding to F = 3900/3000 = 1.30. The specified berm contribution adds 900 kN m/m of resisting moment. The revised resisting moment is 3900 + 900 = 4800 kN m/m. The revised factor of safety is therefore 4800/3000 = 1.60. Actual berm design must also consider its weight, foundation response and any changes to the critical failure surface.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Finite slopes can develop rotational failures on curved surfaces, whereas the classical infinite-slope model examines translation along a plane parallel to the slope. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. Toe failure. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: C. Face or slope failure. సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: D. Base failure involving soil below the toe. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: C. Resisting moment is obtained from basal shear resistance acting with lever arm R, and driving moment is obtained from the moments of destabilizing forces about the circle centre. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: C. 1.50. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. 1.50. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: C. 1.60. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. It neglects interslice force effects in computing basal reactions and satisfies overall moment equilibrium for an assumed circular surface. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: B. 1.35. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. 1.57. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: D. 4.0 m. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: A. 120 kN/m and 60 kN/m. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. 1.60. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 1.38. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. 1.50 initially and 1.25 finally. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: D. 20 kN/m. సమాధానం: D.
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Question 18 ప్రశ్న 18Answer: B. The ordinary method neglects interslice-force effects in its simplified basal reaction, while simplified Bishop uses an interslice normal-force assumption and normally requires an iterative solution for frictional soils. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. The circle with F = 1.20 is governing among the trials, although further searches may reveal a lower value. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: C. 2.00. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: C. 25.92 kPa. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: C. 10.0 m. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: D. 12.50. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: A. 1.20 and 1.00. సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: B. 1.60; the assumed toe-berm contribution increases resisting moment. సమాధానం: B.