Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
In the classical Coulomb wedge theory for dry cohesionless backfill, which idealization is fundamental?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Coulomb theory uses the limiting equilibrium of a potential sliding wedge. The trial wedge is bounded by the wall, the backfill surface and an assumed planar slip surface. Its weight, wall reaction and slip-plane reaction enter the force equilibrium calculation. Wall friction can be represented by an interface friction angle. The classical formulation does not require an elastic stress analysis throughout the soil mass.
వివరణ:
Question 2
ప్రశ్న 2
When examining different admissible trial failure planes using Coulomb wedge equilibrium, which selection gives the limiting active thrust and limiting passive resistance, respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A trial plane produces a candidate wall force from wedge equilibrium. The critical active wedge is the one requiring the greatest active thrust among admissible candidates. The critical passive wedge corresponds to the smallest force required for impending passive failure. These extrema refer to different wall movement directions and force constructions. For significant wall friction, the planar passive mechanism can overestimate actual available resistance.
వివరణ:
Question 3
ప్రశ్న 3
For a rough vertical retaining-wall back under active conditions, which statement correctly describes the soil force on the wall when the backfill tends to move downward relative to the wall?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The wall-soil contact can mobilize tangential friction when relative sliding occurs. In the active case, the backfill wedge tends to move downward relative to the wall. The soil therefore exerts a downward friction component on a vertical wall. Its resultant is inclined by the mobilized interface angle delta below the outward horizontal normal. The actual value of delta depends on the interface and movement conditions.
వివరణ:
Question 4
ప్రశ్న 4
Which statement is most defensible about Coulomb passive resistance for a rough retaining wall with a large positive wall-friction angle?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Coulomb passive theory uses an idealized planar slip surface. Actual passive failure mechanisms can be curved, especially when wall friction is significant. The difference can cause a substantial overestimate of passive resistance from the planar-wedge model. Passive resistance also requires wall movement toward the soil to become mobilized. A practical design should check the adopted analysis method, interface conditions and permitted movement.
వివరణ:
Question 5
ప్రశ్న 5
For otherwise identical Coulomb active conditions behind a vertical wall with level dry cohesionless backfill, what is the usual effect of increasing positive wall friction delta from zero?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Positive wall friction provides resistance to the downward relative movement of the active wedge. For the stated simple geometry, increasing delta generally reduces the calculated Coulomb active resultant. The same interface friction inclines the resultant downward from the horizontal wall normal. This is a limiting-equilibrium result and requires active wall movement. It should not be applied as a universal pressure reduction for a restrained basement wall.
వివరణ:
Question 6
ప్రశ్న 6
For a vertical smooth wall and horizontal dry cohesionless backfill, which relationship between Coulomb and Rankine limiting coefficients is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated wall is vertical and has no wall-soil friction. The ground surface is horizontal and the soil is homogeneous and cohesionless. Under these conditions, the Coulomb limiting expressions reduce to the corresponding Rankine expressions. Thus Ka = (1 - sin(phi))/(1 + sin(phi)) and Kp is its reciprocal. Different wall geometry or nonzero interface friction generally removes this simple equality.
వివరణ:
Question 7
ప్రశ్న 7
A smooth vertical wall of height 5 m retains dry level sand with gamma = 18 kN/m3 and phi = 30 degrees. Under the Coulomb active limiting state, what are the soil-weight thrust per metre of wall and its assumed point of application? Take Ka = 1/3.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under the given simple geometry, the Coulomb coefficient equals the Rankine active coefficient of one-third. The assumed soil-weight thrust is Pa = 0.5 Ka gamma H^2. Substitution gives Pa = 0.5(1/3)(18)(25) = 75 kN/m. A triangular pressure distribution places its resultant at H/3 above the base. The application height is therefore 5/3 = approximately 1.67 m.
వివరణ:
Question 8
ప్రశ్న 8
For a vertical wall with horizontal backfill, the Coulomb active coefficient is Ka = cos^2(phi)/{cos(delta)[1 + sqrt(sin(phi + delta)sin(phi)/cos(delta))]^2}. If phi = 30 degrees and delta = 15 degrees, what is Ka approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The supplied expression applies to the stated vertical-wall and horizontal-backfill case. Use cos^2(30 degrees) = 0.75 and cos(15 degrees) = approximately 0.9659. The term inside the square root is sin(45 degrees)sin(30 degrees)/cos(15 degrees), approximately 0.3661. Substitution into the full denominator gives a coefficient of approximately 0.3014. Therefore the closest coefficient is 0.301, slightly below the smooth-wall value of one-third.
వివరణ:
Question 9
ప్రశ్న 9
For a rough vertical wall retaining 6 m of dry horizontal sand, the evaluated Coulomb active coefficient is Ka = 0.297 and gamma = 18 kN/m3. What is the active resultant magnitude due to soil weight alone, per metre of wall?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The given Coulomb coefficient is to be used for the resultant force magnitude. For homogeneous dry backfill without surcharge, Pa = 0.5 Ka gamma H^2. Inserting the stated data gives Pa = 0.5(0.297)(18)(6^2). The numerical result is 96.228 kN/m. Hence the thrust magnitude is approximately 96.2 kN/m and is not automatically its horizontal component.
వివరణ:
Question 10
ప్రశ్న 10
For a vertical smooth wall with backfill rising at beta = 10 degrees above horizontal, use the Coulomb expression Ka = cos^2(phi)/[1 + sqrt(sin(phi)sin(phi - beta)/cos(beta))]^2. If phi = 30 degrees, what is Ka approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The supplied expression describes a vertical smooth wall retaining cohesionless soil with an upward-sloping backfill. For phi = 30 degrees, cos^2(phi) equals 0.75. The square-root term is sqrt[sin(30 degrees)sin(20 degrees)/cos(10 degrees)], which is approximately 0.4167. Dividing 0.75 by the square of 1.4167 gives Ka approximately 0.3737. The required rounded coefficient is therefore 0.374.
వివరణ:
Question 11
ప్రశ్న 11
For a vertical smooth wall with phi = 30 degrees, a Coulomb analysis gives Ka = 0.333 for level backfill and Ka = 0.441 for backfill rising at 20 degrees. With the same wall height and soil unit weight, what is the approximate ratio of the inclined-backfill active thrust to the level-backfill thrust?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The soil-weight thrust is proportional to Ka when gamma and H are unchanged. The coefficient ratio is 0.441/0.333. This equals approximately 1.324. Therefore the active thrust for the given rising backfill is about 1.32 times the level-backfill thrust. An upward ground slope can increase the calculated active pressure even when wall friction is absent.
వివరణ:
Question 12
ప్రశ్న 12
For a specified battered retaining-wall back and inclined backfill, the appropriate Coulomb active coefficient has already been evaluated as Ka = 0.407 using a formulation based on vertical wall height H. If H = 5 m and gamma = 18 kN/m3, what is the soil-weight active resultant per metre of wall?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Coulomb coefficients for inclined walls depend on the geometry and the coefficient convention. The question expressly supplies Ka evaluated for the stated geometry using the vertical-height thrust expression. Thus Pa = 0.5 Ka gamma H^2 is applicable as specified. The calculation is 0.5(0.407)(18)(25) = 91.575 kN/m. Its rounded resultant magnitude is 91.6 kN/m.
వివరణ:
Question 13
ప్రశ్న 13
An evaluated Coulomb active coefficient is Ka = 0.30 for a homogeneous dry backfill retained by a 6 m high wall, and gamma = 20 kN/m3. Assuming triangular soil-weight loading with no surcharge, what are the active resultant magnitude and the vertical height of its assumed point of application above the base?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For this conventional soil-weight pressure assumption, the resultant is Pa = 0.5 Ka gamma H^2. Substitution gives Pa = 0.5(0.30)(20)(36) = 108 kN/m. A triangular distribution has its centroid at one-third of the retained height above the base. With H = 6 m, the application height is 2 m. The force magnitude and point of application are distinct properties and should not be interchanged.
వివరణ:
Question 14
ప్రశ్న 14
For a homogeneous dry backfill with the same Coulomb active coefficient, unit weight and geometry apart from wall height, what is the effect of doubling the retained vertical height on the assumed soil-weight active resultant per metre?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The conventional expression for the soil-weight active resultant is Pa = 0.5 Ka gamma H^2. In this comparison Ka and gamma are held constant. Replacing H with 2H multiplies the squared height term by four. Hence the predicted resultant becomes four times its former value. This scaling does not necessarily apply unchanged when the soil profile or effective pressure coefficient varies with depth.
వివరణ:
Question 15
ప్రశ్న 15
A vertical rough wall experiences a Coulomb active resultant of magnitude 150 kN/m inclined 20 degrees downward from the horizontal wall normal. Given cos(20 degrees) = 0.9397 and sin(20 degrees) = 0.3420, what are its horizontal and vertical components?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The horizontal component is the resultant magnitude multiplied by the cosine of its inclination to the horizontal. Thus Ph = 150(0.9397) = 140.955 kN/m. The vertical component has magnitude Pv = 150(0.3420) = 51.30 kN/m. Active interface friction acts downward on the vertical wall under the stated relative movement. The correct components are approximately 141.0 kN/m horizontal and 51.3 kN/m downward.
వివరణ:
Question 16
ప్రశ్న 16
For a vertical rough wall in the active state, the horizontal component of the Coulomb resultant is 120 kN/m and the resultant is inclined 30 degrees downward from the horizontal. Using cos(30 degrees) = 0.8660 and tan(30 degrees) = 0.57735, find the resultant magnitude and downward vertical component.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The horizontal component equals Pa cos(delta). Therefore Pa = 120/0.8660 = approximately 138.6 kN/m. The vertical component can be obtained directly as Pv = Ph tan(delta). It equals 120(0.57735) = 69.3 kN/m. The vertical component acts downward under the specified active wall-friction direction.
వివరణ:
Question 17
ప్రశ్న 17
For an inclined wall back, the outward normal to the wall face points 10 degrees above the horizontal toward the wall. At the active limit, wall friction turns the soil thrust resultant clockwise by 15 degrees from that normal in the plane of the section. What is the inclination of the resultant relative to the horizontal?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The direction of a Coulomb wall thrust is set relative to the normal of the wall back. Here the outward normal initially points 10 degrees above the horizontal. The problem specifies that interface friction rotates the resultant clockwise through 15 degrees. Subtracting these signed orientations gives 10 - 15 = -5 degrees. The negative inclination corresponds to 5 degrees below the horizontal.
వివరణ:
Question 18
ప్రశ్న 18
Two 6 m high walls retain the same dry level sand with gamma = 18 kN/m3 and phi = 30 degrees. For the smooth wall use Rankine Ka = 1/3, and for the rough wall use the evaluated Coulomb Ka = 0.297. What are the respective active soil-weight resultant magnitudes?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Use Pa = 0.5 Ka gamma H^2 for each stated configuration. For the smooth wall, the thrust is 0.5(1/3)(18)(36) = 108.0 kN/m. For the rough wall, it is 0.5(0.297)(18)(36) = 96.228 kN/m. The rough-wall resultant is therefore smaller in magnitude for the supplied coefficients. The rough-wall force may also have a downward component, so the two thrust magnitudes should not automatically be treated as identical horizontal forces.
వివరణ:
Question 19
ప్రశ్న 19
Under a conventional vertical-wall calculation with a constant adopted Coulomb active coefficient Ka = 0.30, a sufficiently extensive uniform vertical surcharge q = 40 kPa is assumed to contribute a uniform additional lateral resultant distribution on a 5 m wall. What are the additional thrust and its assumed application height above the base?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under the conventional uniform surcharge approximation, additional pressure is Ka q. The supplied coefficient gives an additional intensity of 0.30(40) = 12 kPa. Over the 5 m wall height, the resulting force is 12(5) = 60 kN/m. A uniform distribution has its centroid at half the height. Therefore the additional resultant is assumed to act 2.5 m above the base.
వివరణ:
Question 20
ప్రశ్న 20
For a vertical wall, use a constant adopted Coulomb active coefficient Ka = 0.30 for both soil weight and a sufficiently extensive uniform vertical surcharge. If H = 4 m, gamma = 18 kN/m3 and q = 25 kPa, what are the combined active resultant magnitude and its application height above the base under triangular-plus-rectangular pressure assumptions?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The soil-weight contribution is 0.5(0.30)(18)(4^2) = 43.2 kN/m. The surcharge contribution is (0.30)(25)(4) = 30.0 kN/m. Their combined magnitude is 73.2 kN/m. The respective lever arms above the base are 4/3 m and 2 m. The combined height is [43.2(4/3) + 30(2)]/73.2 = approximately 1.61 m.
వివరణ:
Question 21
ప్రశ్న 21
A vertical retaining-wall back has a Coulomb active resultant of 120 kN/m inclined 30 degrees downward from horizontal. Assume a triangular soil pressure distribution over wall height H = 6 m and take moments about the intersection of the vertical wall back with its base. What overturning moment is produced by the horizontal component alone? Use cos(30 degrees) = 0.8660.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the stated vertical wall, the horizontal force component is Ph = 120 cos(30 degrees). Using the supplied cosine gives Ph = 103.92 kN/m. The triangular soil distribution places the resultant at H/3 = 2 m above the base. The moment due to the horizontal component is therefore 103.92(2) = 207.84 kN m per metre. The question asks for that component alone, rather than a complete stability calculation including other wall loads.
వివరణ:
Question 22
ప్రశ్న 22
A 4 m high vertical rough wall retains fully submerged cohesionless backfill with gamma_sat = 20 kN/m3 and gamma_w = 10 kN/m3. The relevant effective-soil Coulomb coefficient is given as Ka = 0.30, with effective soil thrust directed 30 degrees downward from horizontal, while hydrostatic water pressure acts horizontally on the back and there is no balancing water in front. Using cos(30 degrees) = 0.8660, what is the total horizontal force per metre?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The submerged effective unit weight is gamma_prime = 20 - 10 = 10 kN/m3. The effective soil thrust magnitude is 0.5(0.30)(10)(4^2) = 24 kN/m. Its horizontal component is 24 cos(30 degrees) = 20.784 kN/m. Hydrostatic water pressure contributes a separate horizontal thrust of 0.5(10)(4^2) = 80 kN/m. Adding the horizontal components gives approximately 100.8 kN/m, while the soil thrust also has a distinct downward component.
వివరణ:
Question 23
ప్రశ్న 23
A trial Coulomb active soil wedge behind a smooth vertical wall has weight W = 120 kN per metre. Its trial slip plane rises at 60 degrees above the horizontal, and the soil friction angle along that plane is phi = 30 degrees. If wedge equilibrium gives Pa = W tan(60 degrees - phi), what trial active wall thrust results? Use tan(30 degrees) = 0.57735.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated smooth wall has zero wall friction. The question supplies the force-equilibrium relation for the selected trial slip plane. Substituting phi = 30 degrees gives Pa = 120 tan(30 degrees). Using the given tangent value yields Pa = 120(0.57735) = 69.282 kN/m. The approximate trial active thrust is therefore 69.3 kN/m, before examining other possible wedge planes.
వివరణ:
Question 24
ప్రశ్న 24
For one geometry, admissible Coulomb active trial wedges give wall thrusts of 62, 70, 66 and 68 kN/m; for a separate passive problem, admissible trial wedges give required wall forces of 190, 175, 185 and 180 kN/m. Which pair is selected by the Coulomb limiting wedge procedure?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each trial wedge gives a force compatible with equilibrium on its assumed slip surface. For active loading, the critical trial plane corresponds to the maximum candidate active thrust. The greatest active value in the question is 70 kN/m. For passive loading, the controlling candidate is the minimum force needed to produce failure. The least passive value is 175 kN/m, giving the stated pair under the trial set.
వివరణ:
Question 25
ప్రశ్న 25
A smooth vertical wall retains 3 m of dry homogeneous soil with gamma = 18 kN/m3. An ideal limiting passive coefficient Kp = 4.0 is adopted, and an exercise specifies a factor of safety of 2.5 applied directly to the calculated passive resultant. What are the ideal passive thrust and the reduced resistance per metre of wall?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under the specified triangular passive pressure idealization, Pp = 0.5 Kp gamma H^2. Substitution gives Pp = 0.5(4.0)(18)(3^2) = 324 kN/m. The stated factor of safety reduces the ideal resultant to 324/2.5 = 129.6 kN/m. These are the ideal and reduced values for the question assumptions. In practice, sufficient wall movement and the suitability of the chosen passive pressure theory must also be considered.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: B. A rigid soil wedge bounded by the wall back, ground surface and a trial plane failure surface is considered in limiting equilibrium. సమాధానం: B.
-
Question 2 ప్రశ్న 2Answer: C. Maximum active thrust and minimum passive resistance. సమాధానం: C.
-
Question 3 ప్రశ్న 3Answer: A. The resultant is inclined downward from the horizontal wall normal by the mobilized wall-friction angle delta. సమాధానం: A.
-
Question 4 ప్రశ్న 4Answer: D. A planar-wedge solution may overestimate passive resistance, and curved failure mechanisms or conservative resistance limits deserve consideration. సమాధానం: D.
-
Question 5 ప్రశ్న 5Answer: B. The calculated limiting active resultant generally decreases, and its direction tilts downward from the wall normal. సమాధానం: B.
-
Question 6 ప్రశ్న 6Answer: C. Coulomb and Rankine give the same Ka and Kp for the stated ideal conditions. సమాధానం: C.
-
Question 7 ప్రశ్న 7Answer: A. 75 kN/m at 1.67 m above the base. సమాధానం: A.
-
Question 8 ప్రశ్న 8Answer: D. 0.301. సమాధానం: D.
-
Question 9 ప్రశ్న 9Answer: B. 96.2 kN/m. సమాధానం: B.
-
Question 10 ప్రశ్న 10Answer: C. 0.374. సమాధానం: C.
-
Question 11 ప్రశ్న 11Answer: D. 1.32. సమాధానం: D.
-
Question 12 ప్రశ్న 12Answer: A. 91.6 kN/m. సమాధానం: A.
-
Question 13 ప్రశ్న 13Answer: B. 108 kN/m and 2 m. సమాధానం: B.
-
Question 14 ప్రశ్న 14Answer: C. It becomes four times larger. సమాధానం: C.
-
Question 15 ప్రశ్న 15Answer: D. 141.0 kN/m horizontally and 51.3 kN/m downward. సమాధానం: D.
-
Question 16 ప్రశ్న 16Answer: A. 138.6 kN/m and 69.3 kN/m. సమాధానం: A.
-
Question 17 ప్రశ్న 17Answer: B. 5 degrees below horizontal. సమాధానం: B.
-
Question 18 ప్రశ్న 18Answer: C. 108.0 kN/m for the smooth wall and 96.2 kN/m for the rough wall. సమాధానం: C.
-
Question 19 ప్రశ్న 19Answer: A. 60 kN/m at 2.5 m. సమాధానం: A.
-
Question 20 ప్రశ్న 20Answer: B. 73.2 kN/m acting 1.61 m above the base. సమాధానం: B.
-
Question 21 ప్రశ్న 21Answer: D. 207.8 kN m per metre. సమాధానం: D.
-
Question 22 ప్రశ్న 22Answer: C. 100.8 kN/m. సమాధానం: C.
-
Question 23 ప్రశ్న 23Answer: A. 69.3 kN/m. సమాధానం: A.
-
Question 24 ప్రశ్న 24Answer: D. Active 70 kN/m; passive 175 kN/m. సమాధానం: D.
-
Question 25 ప్రశ్న 25Answer: B. 324.0 kN/m and 129.6 kN/m. సమాధానం: B.