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Question 1
ప్రశ్న 1
For a vertical smooth retaining wall with a level cohesionless backfill, which association between wall movement and the lateral earth-pressure state is correct?
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Explanation:
Active pressure is the lower limiting lateral pressure reached when a wall yields sufficiently away from the retained soil. Passive resistance is mobilized when the wall moves toward and compresses the soil. The at-rest state is associated with essentially zero lateral strain in the backfill. These states are distinguished by their deformation conditions rather than by the wall height alone. Substantially more movement is generally needed to mobilize full passive resistance than active pressure.
వివరణ:
Question 2
ప్రశ్న 2
Which combination of conditions makes the elementary Rankine expressions Ka = (1 - sin(phi))/(1 + sin(phi)) and Kp = (1 + sin(phi))/(1 - sin(phi)) appropriate for an ideal cohesionless backfill?
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Explanation:
The elementary Rankine coefficient equations describe limiting stress states for a particular idealized geometry. A smooth vertical wall exerts no tangential wall friction in this model. The backfill is level, and a homogeneous cohesionless material is considered. Active or passive coefficients apply only when the corresponding limiting deformation state is reached. Inclined ground, wall friction, unusual geometry or additional loading may require modified theory or a different analysis.
వివరణ:
Question 3
ప్రశ్న 3
For a dry, level, cohesionless backfill with phi = 30 degrees, which values of Rankine Ka and Kp and the Jaky estimate K0 for normally consolidated soil are correct? Take sin(30 degrees) = 0.50.
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Explanation:
Rankine active coefficient is Ka = (1 - sin(phi))/(1 + sin(phi)). With sin(phi) = 0.50, Ka = 0.50/1.50 = 1/3. Its reciprocal is Kp = 3 for the same ideal backfill. The approximate at-rest coefficient for normally consolidated soil is given by Jaky relation K0 = 1 - sin(phi). Thus K0 = 0.50, which lies between the active and passive coefficients.
వివరణ:
Question 4
ప్రశ్న 4
For a level cohesionless backfill against a smooth vertical wall, a measured ideal Rankine active earth-pressure coefficient is Ka = 0.25. What is the corresponding effective friction angle, approximately?
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Explanation:
Rearrange the Rankine relation to obtain sin(phi) = (1 - Ka)/(1 + Ka). Substituting Ka = 0.25 gives sin(phi) = 0.75/1.25 = 0.60. The inverse sine of 0.60 is approximately 36.87 degrees. Hence the required effective friction angle is about 36.9 degrees. This inversion assumes the same ideal conditions used to derive the Rankine coefficient.
వివరణ:
Question 5
ప్రశ్న 5
For two otherwise identical cohesionless backfills under the elementary Rankine assumptions, the effective friction angle of the second backfill is greater. Which change is generally expected in the limiting earth-pressure coefficients?
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Explanation:
For a cohesionless soil, Ka = (1 - sin(phi))/(1 + sin(phi)). Increasing the effective friction angle increases its sine over the usual soil range. This change reduces Ka and increases the reciprocal coefficient Kp. Therefore the active lateral pressure is lower and the limiting passive resistance is higher at the same effective vertical stress. This comparison concerns ideal limiting states and does not imply that passive resistance can be used without verifying mobilization.
వివరణ:
Question 6
ప్రశ్న 6
A rigid basement wall 5 m high retains dry, level, normally consolidated cohesionless soil with gamma = 20 kN/m3 and phi = 30 degrees. Assuming zero lateral strain and Jaky K0 = 1 - sin(phi), what is the at-rest thrust per metre of wall? Take sin(30 degrees) = 0.50.
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Explanation:
Because the wall is restrained, at-rest pressure is the relevant idealized state. Jaky relation gives K0 = 1 - 0.50 = 0.50 for the stated normally consolidated soil. The at-rest lateral stress varies as K0 gamma z under the given conditions. Its triangular resultant is P0 = 0.5 K0 gamma H^2. Substitution gives 0.5(0.50)(20)(5^2) = 125.0 kN per metre of wall.
వివరణ:
Question 7
ప్రశ్న 7
A smooth vertical wall 6 m high retains dry horizontal sand with gamma = 18 kN/m3 and phi = 30 degrees. If the Rankine active state is fully mobilized, what is the active soil thrust per metre length of wall?
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Explanation:
For phi = 30 degrees, the Rankine active coefficient is Ka = 1/3. The lateral active pressure due to soil weight increases linearly from zero at the surface to Ka gamma H at the base. The total soil thrust is the area of this triangular distribution. Therefore Pa = 0.5(1/3)(18)(6^2) = 108 kN/m. The result is a horizontal thrust per metre of wall length under the ideal smooth-wall assumptions.
వివరణ:
Question 8
ప్రశ్న 8
An ideal smooth vertical wall retains dry level cohesionless soil of unit weight 18 kN/m3 and phi = 30 degrees. If the mobilized Rankine active thrust due to soil self-weight alone is 75 kN per metre of wall, what is the retained height H?
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Explanation:
The active coefficient for phi = 30 degrees is Ka = 1/3. The self-weight thrust formula is Pa = 0.5 Ka gamma H^2. With the stated unit weight, Pa = 0.5(1/3)(18)H^2 = 3H^2 kN/m. Equating this to 75 gives H^2 = 25. The physically meaningful retained height is H = 5.0 m.
వివరణ:
Question 9
ప్రశ్న 9
A smooth vertical wall supports a 3 m high, dry, horizontal cohesionless backfill with gamma = 18 kN/m3 and phi = 30 degrees. If full Rankine passive resistance is mobilized, what are the passive thrust per metre of wall and its height of application above the base?
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Explanation:
The Rankine passive coefficient for phi = 30 degrees is Kp = 3. The passive lateral pressure increases linearly with depth in the stated uniform backfill. Its resultant is Pp = 0.5 Kp gamma H^2 = 0.5(3)(18)(3^2) = 243 kN/m. A triangular distribution acts at one-third of its height above the base. Therefore the passive thrust acts at 3/3 = 1.0 m above the base, provided the limiting passive state is mobilized.
వివరణ:
Question 10
ప్రశ్న 10
In a level dry cohesionless backfill with gamma = 18 kN/m3 and phi = 30 degrees, what is the Rankine active lateral pressure intensity at a depth of 3 m below the ground surface?
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Explanation:
The effective vertical stress at 3 m depth is gamma z = 18(3) = 54 kPa. For phi = 30 degrees, the active earth-pressure coefficient is Ka = 1/3. The active lateral stress at the given depth is Ka gamma z. Therefore the pressure intensity is 54/3 = 18 kPa. This is a stress intensity, not the integrated resultant thrust on the wall.
వివరణ:
Question 11
ప్రశ్న 11
A uniform vertical surcharge q = 36 kPa extends sufficiently far behind a smooth vertical wall of height 5 m retaining horizontal cohesionless soil with phi = 30 degrees. What additional Rankine active thrust per metre is caused by this surcharge, and where does it act?
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Explanation:
The ideal uniform vertical surcharge produces a constant additional active lateral pressure Ka q. Here Ka = 1/3, so the additional pressure is 36/3 = 12 kPa at every depth. The surcharge contribution is therefore a rectangular pressure distribution. Its thrust is Pas = 12(5) = 60 kN/m. The line of action passes through the rectangle centroid at H/2 = 2.5 m above the base.
వివరణ:
Question 12
ప్రశ్న 12
A smooth vertical wall 4 m high retains dry level sand with gamma = 18 kN/m3 and phi = 30 degrees. A uniform vertical surcharge of 24 kPa acts over the backfill. What are the combined Rankine active thrust per metre and its height above the base?
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Explanation:
For phi = 30 degrees, Ka is 1/3. Soil self-weight produces Pa,gamma = 0.5(1/3)(18)(4^2) = 48 kN/m acting at 4/3 m above the base. The surcharge produces Pa,q = (1/3)(24)(4) = 32 kN/m acting at 4/2 = 2 m. The combined thrust is 48 + 32 = 80 kN/m. Its lever arm is [48(4/3) + 32(2)]/80 = 1.60 m above the base.
వివరణ:
Question 13
ప్రశ్న 13
A 5 m high smooth vertical wall retains dry level sand with gamma = 18 kN/m3 and phi = 30 degrees. A uniform surcharge q = 30 kPa extends sufficiently far behind the wall. What are the Rankine active pressure intensities at the top and at the base, respectively?
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Explanation:
The active lateral stress at depth z is pa(z) = Ka(gamma z + q). The coefficient Ka is 1/3 for phi = 30 degrees. At the ground surface, pa(0) = (1/3)(30) = 10 kPa. At the base, pa(5) = (1/3)[18(5) + 30] = 40 kPa. The surcharge adds a uniform pressure to the soil-weight triangular diagram, giving a trapezoidal total distribution.
వివరణ:
Question 14
ప్రశ్న 14
A level dry cohesionless backfill has phi = 30 degrees and gamma = 18 kN/m3. Compare earth pressure due to self-weight alone on a 4 m high wall if the wall is restrained at rest rather than permitted to reach Rankine active state. Using Jaky K0 = 0.50 and Ka = 1/3, how much greater is the at-rest resultant?
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Explanation:
At-rest thrust is P0 = 0.5 K0 gamma H^2 = 0.5(0.50)(18)(4^2) = 72 kN/m. The Rankine active thrust is Pa = 0.5 Ka gamma H^2. Using Ka = 1/3 gives Pa = 48 kN/m. Subtracting the active value from the at-rest value gives 72 - 48 = 24 kN/m. Thus assuming active pressure for a completely restrained wall would underestimate the stated ideal at-rest thrust.
వివరణ:
Question 15
ప్రశ్న 15
A smooth vertical wall retains 4 m of fully submerged, level, cohesionless sand, with the water level at the ground surface on the soil side only. Given phi = 30 degrees, gamma_sat = 20 kN/m3 and gamma_w = 10 kN/m3, assume hydrostatic water conditions and no water pressure on the other side. What is the total Rankine active thrust per metre from effective soil pressure plus water pressure?
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Explanation:
The submerged soil unit weight is gamma_prime = gamma_sat - gamma_w = 10 kN/m3. Its effective active thrust is 0.5 Ka gamma_prime H^2 = 0.5(1/3)(10)(4^2) = 26.67 kN/m. Hydrostatic water pressure contributes 0.5 gamma_w H^2 = 0.5(10)(4^2) = 80 kN/m. The total lateral thrust is the sum of these independently computed components. Thus Ptotal = 26.67 + 80 = approximately 106.7 kN/m.
వివరణ:
Question 16
ప్రశ్న 16
A fully submerged level sand deposit extends to the base of a smooth 5 m high wall, with the water surface at ground level on the soil side and no balancing water on the opposite face. If gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3 and phi = 30 degrees, what is the total active lateral pressure intensity at the base?
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Explanation:
The submerged unit weight is 20 - 10 = 10 kN/m3. At depth 5 m, vertical effective stress is 10(5) = 50 kPa. The effective active soil pressure is Ka sigma_v_prime = (1/3)(50) = 16.67 kPa. Hydrostatic pore-water pressure at the same depth is 10(5) = 50 kPa. Hence the total lateral pressure on the soil side is 16.67 + 50 = 66.67 kPa.
వివరణ:
Question 17
ప్రశ్న 17
A smooth vertical wall retains 6 m of level sand, and the groundwater table is 2 m below ground level. Above the table gamma_moist = 18 kN/m3; below it gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3. Assuming phi = 30 degrees, hydrostatic conditions, no seepage and no balancing water in front, what is the total active lateral pressure intensity at the wall base?
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Explanation:
The effective vertical stress at the base is the weight of 2 m of moist soil plus 4 m of submerged soil. The submerged unit weight is 20 - 10 = 10 kN/m3. Therefore sigma_v_prime = 18(2) + 10(4) = 76 kPa. The effective active lateral soil pressure is 76/3 = 25.33 kPa, while hydrostatic water pressure is 10(4) = 40 kPa. Adding them gives approximately 65.3 kPa total active lateral pressure at the base.
వివరణ:
Question 18
ప్రశ్న 18
A 6 m high smooth vertical wall retains level sand with phi = 30 degrees. The water table is 2 m below the surface, gamma_moist = 18 kN/m3 above it, gamma_sat = 20 kN/m3 below it, and gamma_w = 10 kN/m3. Assuming hydrostatic conditions, no seepage, no surcharge and no water in front of the wall, what is the combined active resultant per metre from effective soil pressure and water pressure?
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Explanation:
Use Ka = 1/3 and submerged unit weight gamma_prime = 10 kN/m3. Above the water table, the soil-pressure thrust is 0.5(1/3)(18)(2^2) = 12 kN/m. Below the water table, the effective soil-pressure thrust is (1/3)[36(4) + 0.5(10)(4^2)] = 74.67 kN/m. Hydrostatic water pressure over the lower 4 m adds 0.5(10)(4^2) = 80 kN/m. The combined resultant is 12 + 74.67 + 80 = approximately 166.7 kN/m.
వివరణ:
Question 19
ప్రశ్న 19
An ideal smooth wall separates saturated level sand on one side from an open water body on the other. Assume the water surface elevations are identical, the pore-water pressures in the sand are hydrostatic, and only the active effective soil thrust and hydrostatic water pressures are being compared. What is the correct net pressure interpretation?
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Explanation:
Effective lateral earth pressure and hydrostatic water pressure are separate contributions on the soil side. Equal water levels on opposite faces create equal and opposite hydrostatic distributions under the stated assumptions. Their resultant hydrostatic forces therefore cancel in a net horizontal force calculation. The effective active stress in the sand is not hydrostatic pressure and does not cancel against the open-water face. A correct net load calculation retains the effective soil thrust while accounting consistently for water on both sides.
వివరణ:
Question 20
ప్రశ్న 20
For an ideal short-term total-stress Rankine analysis of a level homogeneous cohesive backfill, take phi_u = 0 degrees, cu = 12 kPa, gamma = 18 kN/m3, and depth z = 6 m. With no surcharge, groundwater force or wall adhesion included, what is the calculated active lateral stress at that depth?
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Explanation:
For phi_u = 0 degrees, the ideal Rankine coefficient is Ka = 1. The cohesive active lateral stress expression is pa = Ka gamma z - 2cu sqrt(Ka). At 6 m depth, the soil-weight contribution is 18(6) = 108 kPa. The cohesion reduction is 2(12) = 24 kPa. Therefore the calculated active stress at that depth is 108 - 24 = 84 kPa, which is positive.
వివరణ:
Question 21
ప్రశ్న 21
An ideal phi_u = 0 total-stress cohesive backfill has cu = 18 kPa, gamma = 18 kN/m3 and retained height H = 5 m. Assume a tension crack develops to the theoretical zero-pressure depth, the crack is empty, there is no surcharge or water force, and negative calculated lateral stresses are not transmitted to the wall. What are the tension-crack depth and net positive active thrust per metre?
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Explanation:
For phi_u = 0, Ka = 1 and the ideal tension-crack depth is zc = 2cu/gamma. Therefore zc = 2(18)/18 = 2 m below the ground surface. The calculated positive lateral pressure at the wall base is gamma H - 2cu = 18(5) - 36 = 54 kPa. The positive pressure extends over H - zc = 3 m with a triangular distribution. The resultant is 0.5(54)(3) = 81 kN/m under the stated empty-crack assumptions.
వివరణ:
Question 22
ప్రశ్న 22
A theoretical cohesive-backfill tension crack is 2.0 m deep and extends continuously along a retaining wall. If the crack fills with water to ground level and gamma_w = 10 kN/m3, what additional horizontal hydrostatic thrust does this water exert per metre of wall along the crack depth, relative to an empty crack?
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Explanation:
Water pressure in a crack increases hydrostatically from zero at the water surface to gamma_w zc at its bottom. For zc = 2 m, the pressure at the crack base is 10(2) = 20 kPa. This produces a triangular water-pressure distribution over the crack depth. The additional resultant is Pw = 0.5 gamma_w zc^2 = 0.5(10)(2^2). Therefore the water-filled crack introduces an additional 20 kN/m of horizontal thrust compared with the stated empty-crack condition.
వివరణ:
Question 23
ప్రశ్న 23
A dry, horizontal c-phi backfill has c_prime = 10 kPa, phi_prime = 30 degrees and gamma = 18 kN/m3. For the ideal Rankine passive state at depth z = 2 m, what is the passive lateral effective stress? Take Kp = 3 and sqrt(3) = 1.732.
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Explanation:
For a cohesive-frictional soil, ideal Rankine passive effective stress is sigma_p_prime = Kp sigma_v_prime + 2c_prime sqrt(Kp). Since the soil is dry, sigma_v_prime = gamma z = 18(2) = 36 kPa. The frictional component is 3(36) = 108 kPa. The cohesion component is 2(10)(1.732) = 34.64 kPa. Their sum is 142.64 kPa, or approximately 142.6 kPa.
వివరణ:
Question 24
ప్రశ్న 24
In an ideal Rankine active calculation for a cohesive backfill, the expression Ka gamma z - 2c sqrt(Ka) becomes negative near the ground surface. Which engineering interpretation is appropriate?
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Explanation:
The cohesive Rankine expression contains a negative term representing the reduction of active pressure due to cohesion. Near the ground surface, the calculated lateral stress can be tensile. Ordinary soil-wall systems should not be assumed to sustain this theoretical tension indefinitely. Tension cracks can form, and the pressure diagram must be interpreted to avoid crediting unsupported tensile stresses. Water filling a crack can create an additional lateral hydrostatic load and should be evaluated separately.
వివరణ:
Question 25
ప్రశ్న 25
Under the elementary Rankine assumptions for a level cohesionless backfill with effective friction angle phi = 30 degrees, what are the inclinations of the active and passive idealized failure planes to the horizontal, respectively?
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Explanation:
The characteristic Rankine active failure-plane inclination to the horizontal is 45 degrees plus half the effective friction angle. With phi = 30 degrees, the active inclination is 45 + 15 = 60 degrees. The corresponding passive inclination is 45 degrees minus half the friction angle. Therefore the passive plane has an inclination of 45 - 15 = 30 degrees. These angles describe the idealized Rankine failure geometry for horizontal backfill and should not be confused with wall-friction angles.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Sufficient movement away from the backfill mobilizes the active limit, movement toward it mobilizes the passive limit, and negligible lateral strain corresponds to the at-rest state. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. A smooth vertical wall, horizontal homogeneous backfill, no wall friction, and sufficient deformation to reach the relevant limiting state. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. Ka = 1/3, Kp = 3, and K0 = 0.50. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 36.9 degrees. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Ka decreases while Kp increases. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 125.0 kN/m. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: C. 108 kN/m. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. 5.0 m. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: D. 243 kN/m acting 1.0 m above the base. సమాధానం: D.
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Question 10 ప్రశ్న 10Answer: B. 18 kPa. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. 60 kN/m at 2.5 m above the base. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. 80 kN/m acting 1.60 m above the base. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: D. 10 kPa and 40 kPa. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: B. 24 kN/m. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: A. 106.7 kN/m. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 66.7 kPa. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 65.3 kPa. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 166.7 kN/m. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. The equal hydrostatic pressure components on opposite faces cancel, while the effective active soil pressure still contributes to the net loading. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 84 kPa. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 2 m and 81 kN/m. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 20 kN/m. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 142.6 kPa. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. A tension zone or cracking may occur, so negative soil pressures should not be relied upon as sustained tensile support; water entering cracks must also be considered. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 60 degrees and 30 degrees. సమాధానం: B.