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Question 1
ప్రశ్న 1
A saturated clay point initially has total vertical stress 210 kPa and hydrostatic pore-water pressure 90 kPa. A foundation increases the total vertical stress permanently by 60 kPa. After complete dissipation of the load-induced excess pore pressure, assume the hydrostatic pore pressure remains 90 kPa. What is the final vertical effective stress?
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Explanation:
Vertical effective stress is total vertical stress minus pore-water pressure. Initially it is 210 - 90 = 120 kPa. The final total vertical stress is 210 + 60 = 270 kPa. After the excess pore pressure dissipates, the specified pore-water pressure is 90 kPa. Thus the final effective stress is 270 - 90 = 180 kPa, giving a permanent effective-stress increase of 60 kPa.
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Question 2
ప్రశ్న 2
On the virgin compression branch of an oedometer e-log10 effective-stress curve, void ratio decreases from 0.95 at 80 kPa to 0.77 at 320 kPa. What is the compression index Cc?
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Explanation:
The compression index is the magnitude of the void-ratio change divided by the change in base-10 logarithm of effective stress. The void-ratio difference is 0.95 - 0.77 = 0.18. The stress ratio is 320/80 = 4. The corresponding logarithm is log10(4) = approximately 0.6021. Hence Cc = 0.18/0.6021 = approximately 0.299, or 0.30.
వివరణ:
Question 3
ప్రశ్న 3
A normally consolidated clay layer is 4 m thick with initial void ratio e0 = 1.00 and Cc = 0.30. Assume its vertical effective stress increases uniformly from 100 to 200 kPa. What is its approximate ultimate primary consolidation settlement?
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Explanation:
Use the normally consolidated settlement equation S = H Cc/(1 + e0) log10(sigmaf/sigma0). Here H = 4 m, Cc = 0.30 and e0 = 1.00. The effective-stress ratio is 200/100 = 2. Therefore S = 4(0.30)/2 times log10(2) = 0.1806 m. This is approximately 181 mm of ultimate primary settlement.
వివరణ:
Question 4
ప్రశ్న 4
At the representative depth of a normally consolidated clay layer, initial total vertical stress is 160 kPa and hydrostatic pore pressure is 60 kPa. A permanent load ultimately increases total stress to 260 kPa without changing hydrostatic pore pressure. If H = 3 m, e0 = 0.80 and Cc = 0.27, with this stress change assumed uniform through the layer, what is the approximate ultimate primary settlement?
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Explanation:
The initial vertical effective stress is 160 - 60 = 100 kPa. After excess pore pressure has dissipated, the final vertical effective stress is 260 - 60 = 200 kPa. The normally consolidated settlement equation is S = H Cc/(1 + e0) log10(sigmaf/sigma0). Substituting gives S = 3(0.27)/1.80 times log10(2) = 0.13546 m. The predicted ultimate primary settlement is approximately 135 mm.
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Question 5
ప్రశ్న 5
A normally consolidated layer has H = 3 m and e0 = 0.80. A uniform effective-stress increase from 100 to 200 kPa produces 150.5 mm of ultimate primary settlement. Using log10(2) = 0.3010, what is the compression index Cc?
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Explanation:
Rearrange the normally consolidated equation to Cc = S(1 + e0)/[H log10(sigmaf/sigma0)]. The measured settlement is 150.5 mm, equivalent to 0.1505 m. The void-ratio factor is 1 + 0.80 = 1.80. Substituting gives Cc = 0.1505(1.80)/[3(0.3010)]. The computed compression index is approximately 0.30.
వివరణ:
Question 6
ప్రశ్న 6
An overconsolidated clay layer has H = 4 m, e0 = 0.80, Cr = 0.06 and preconsolidation pressure 200 kPa. Effective stress increases from 100 to 150 kPa. If log10(1.5) = 0.1761, what is its approximate primary settlement?
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Explanation:
The final effective stress of 150 kPa is below the preconsolidation pressure of 200 kPa. The entire stress increment therefore follows the idealized recompression branch. The settlement equation is S = H Cr/(1 + e0) log10(150/100). Substitution gives S = 4(0.06)/1.80 times 0.1761 = 0.02348 m. Hence the settlement is approximately 23.5 mm.
వివరణ:
Question 7
ప్రశ్న 7
An overconsolidated clay layer is 4 m thick and has e0 = 1.00, Cr = 0.05 and Cc = 0.30. Its vertical effective stress rises from 100 to 400 kPa, crossing a preconsolidation pressure of 200 kPa. Take log10(2) = 0.3010. What is the approximate ultimate primary settlement?
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Explanation:
The stress increment must be divided into recompression and virgin compression portions. Recompression occurs from 100 to 200 kPa, and virgin compression occurs from 200 to 400 kPa. The combined settlement is S = H/(1 + e0)[Cr log10(200/100) + Cc log10(400/200)]. Thus S = 4/2 times (0.05 + 0.30) times 0.3010 = 0.2107 m. The approximate settlement is 211 mm.
వివరణ:
Question 8
ప్రశ్న 8
A clay deposit has present vertical effective stress 125 kPa, preconsolidation pressure 250 kPa and final effective stress after loading 200 kPa. Which statement correctly identifies its OCR and applicable idealized compression branch for the new load?
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Explanation:
OCR is calculated by dividing preconsolidation pressure by present vertical effective stress. The given values yield OCR = 250/125 = 2.0. An OCR greater than one indicates an overconsolidated stress history. The new final stress of 200 kPa remains below the preconsolidation pressure of 250 kPa. The specified loading therefore follows the recompression branch in the idealized e-log effective-stress model.
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Question 9
ప్రశ్న 9
A 3 m thick saturated clay layer has representative coefficient of volume compressibility mv = 5.0 × 10^-4 kPa^-1. A uniform effective-stress increase of 80 kPa is applied. Under the linear one-dimensional approximation, what is the ultimate primary settlement?
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Explanation:
For the stated small-strain linear approximation, primary settlement is S = mv DeltaSigma H. The coefficient of volume compressibility is 5.0 × 10^-4 kPa^-1. Multiplication by the 80 kPa stress increase gives vertical strain 0.04. Multiplying this strain by the 3 m layer thickness gives S = 0.12 m. Thus the expected settlement is 120 mm.
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Question 10
ప్రశ్న 10
A laterally confined clay layer initially has thickness 5.0 m and void ratio 1.25. During one-dimensional consolidation, its void ratio reduces by 0.09 without loss of soil solids. What is the reduction in layer thickness?
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Explanation:
For unchanged solids volume and constant layer area, thickness is proportional to 1 + e. The corresponding settlement is S = H0 Deltae/(1 + e0). Here H0 = 5 m, Deltae = 0.09 and e0 = 1.25. Therefore S = 5(0.09)/2.25 = 0.20 m. The layer thickness decreases by 200 mm.
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Question 11
ప్రశ్న 11
A saturated clay has vertical permeability k = 1.0 × 10^-9 m/s and mv = 5.0 × 10^-4 kPa^-1. Taking the unit weight of water as 10 kN/m³, determine cv from cv = k/(mv gammaw).
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Explanation:
The coefficient of consolidation is related to permeability and compressibility by cv = k/(mv gammaw). Use the compatible unit identity 1 kPa = 1 kN/m². The denominator equals (5.0 × 10^-4)(10) = 0.005 m^-1. Therefore cv = 1.0 × 10^-9/0.005 = 2.0 × 10^-7 m²/s. The units are area per unit time as required for the consolidation coefficient.
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Question 12
ప్రశ్న 12
A saturated clay layer is 6 m thick and drains from both the top and bottom. Its coefficient of consolidation is 0.030 m²/day. What is the dimensionless time factor Tv after 30 days?
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Explanation:
The time factor for vertical consolidation is Tv = cv t/Hdr². Double drainage makes the maximum drainage path half the layer thickness. Thus Hdr = 6/2 = 3 m. Using cv = 0.030 m²/day and t = 30 days gives Tv = 0.030(30)/3². The dimensionless result is Tv = 0.100.
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Question 13
ప్రశ్న 13
A homogeneous clay layer is 4 m thick with drainage from both faces and cv = 0.020 m²/day. If the time factor for 50% average consolidation is T50 = 0.197, how long will 50% average consolidation take?
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Explanation:
The required time follows t = Tv Hdr²/cv. Since both faces drain, the maximum drainage path is 4/2 = 2 m. For 50% consolidation the given time factor is 0.197. Substitution yields t50 = 0.197(2²)/0.020 = 39.4 days. The layer requires approximately 39.4 days under the stated idealized conditions.
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Question 14
ప్రశ్న 14
A 3 m thick saturated clay layer drains through its upper surface only. Its coefficient of consolidation is 0.060 m²/day. With T90 = 0.848, what time is required to achieve 90% average consolidation?
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Explanation:
The time to a stated degree of consolidation is t = Tv Hdr²/cv. Only the upper face drains, so the drainage path equals the full 3 m thickness. The specified 90% time factor is 0.848. Substitution gives t90 = 0.848(3²)/0.060 = 127.2 days. Using half the layer thickness would incorrectly apply a double-drainage assumption.
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Question 15
ప్రశ్న 15
For the same saturated clay layer, unchanged drainage conditions and constant cv, what is the approximate ratio t90/t50? Use T90 = 0.848 and T50 = 0.197.
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Explanation:
Under identical conditions, t = Tv Hdr²/cv. The drainage path and consolidation coefficient are the same for both target degrees. Consequently the ratio t90/t50 equals T90/T50. Dividing 0.848 by 0.197 gives approximately 4.3046. Thus reaching 90% average consolidation takes about 4.30 times the time required for 50%.
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Question 16
ప్రశ్న 16
An oedometer specimen is 20 mm thick and drains from both faces. In one loading stage, t50 = 5 minutes, and T50 = 0.197. Assuming the specimen thickness remains approximately constant, what is cv in m²/min?
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Explanation:
For an oedometer increment, cv = T50 Hdr²/t50. The specimen thickness is 20 mm, so double drainage gives Hdr = 10 mm = 0.010 m. The square of the drainage path is 0.0001 m². Substituting gives cv = 0.197(0.0001)/5 = 3.94 × 10^-6 m²/min. The units remain m²/min because the measured time is supplied in minutes.
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Question 17
ప్రశ్న 17
A double-drained laboratory clay specimen 20 mm thick requires 3 minutes for 50% consolidation. Assume a field clay layer of the same material is 2.0 m thick, also double drained, and has exactly the same cv and consolidation theory parameters as the specimen. Approximately how long would the field layer take to reach 50% consolidation?
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Explanation:
For the same average degree of consolidation and identical cv, time varies with the square of the drainage path. The laboratory drainage path is 20/2 = 10 mm, or 0.010 m. The field drainage path is 2.0/2 = 1.0 m. The field-to-laboratory time ratio is (1.0/0.010)² = 10000, giving 30000 minutes. Dividing by 1440 minutes per day yields approximately 20.8 days.
వివరణ:
Question 18
ప్రశ్న 18
A normally consolidated clay layer has H = 4 m, e0 = 1.00 and Cc = 0.25. Its effective stress increases uniformly from 100 to 200 kPa. At the time when its average degree of primary consolidation is 60%, what is the approximate primary settlement accumulated? Use log10(2) = 0.3010.
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Explanation:
The ultimate primary settlement is calculated from Sult = H Cc/(1 + e0) log10(200/100). With H = 4 m and e0 = 1.00, the multiplier H Cc/(1 + e0) equals 0.50 m. Therefore Sult = 0.50(0.3010) = 0.1505 m, or 150.5 mm. At 60% average consolidation, the settlement fraction completed is 0.60 under the idealization. Hence accumulated primary settlement is 0.60(150.5) = 90.3 mm.
వివరణ:
Question 19
ప్రశ్న 19
A clay layer is predicted to undergo 240 mm of ultimate primary consolidation settlement. At a specified time, the observed primary settlement attributable to that load is 90 mm, with immediate and secondary components excluded. What is its average degree of consolidation by settlement?
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Explanation:
Average degree of primary consolidation can be represented by the fraction of ultimate primary settlement completed under the idealized assumptions. The completed primary settlement is 90 mm. The ultimate primary settlement is 240 mm. Dividing gives U = 90/240 = 0.375. Therefore the average degree of consolidation by settlement is 37.5%.
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Question 20
ప్రశ్న 20
Using the early-stage approximation U = sqrt(4Tv/pi) for average consolidation degree U expressed as a decimal, estimate U for Tv = 0.126. Take pi = 3.14.
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Explanation:
The given early-stage relationship is U = sqrt(4Tv/pi). Substitute Tv = 0.126 and pi = 3.14. The quantity under the square root is 4(0.126)/3.14 = approximately 0.1605. Taking the square root gives U = approximately 0.4006. Thus the average degree of consolidation is about 40%.
వివరణ:
Question 21
ప్రశ్న 21
For an average degree of consolidation U = 40%, use the early-stage approximation Tv = (pi/4)U², with U expressed as a decimal and pi = 3.14. What is Tv?
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Explanation:
Use the degree of consolidation as a decimal fraction rather than a percentage number. Thus U = 0.40 and U² = 0.16. The supplied approximation is Tv = (pi/4)U². Substituting pi = 3.14 gives Tv = (3.14/4)(0.16) = 0.1256. The rounded dimensionless time factor is 0.126.
వివరణ:
Question 22
ప్రశ్న 22
A saturated clay layer is 6 m thick and drains from both faces. Its coefficient of consolidation is 0.030 m²/day, and the time factor for 80% average consolidation is supplied as T80 = 0.567. What is the time required to reach 80% average consolidation?
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Explanation:
Calculate the maximum drainage path before using the time-factor equation. A 6 m layer with double drainage has Hdr = 6/2 = 3 m. The specified time factor for the target consolidation degree is 0.567. The time is t80 = T80 Hdr²/cv = 0.567(3²)/0.030. This gives 170.1 days.
వివరణ:
Question 23
ప్రశ్న 23
A 20 mm thick oedometer specimen drains from both faces and reaches 90% primary consolidation in 400 seconds. If T90 = 0.848, what is cv in m²/s?
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Explanation:
The coefficient of consolidation is cv = T90 Hdr²/t90. Double drainage reduces the maximum drainage path to 20/2 = 10 mm, or 0.010 m. The drainage-path square is 0.0001 m². Substituting gives cv = 0.848(0.0001)/400 = 2.12 × 10^-7 m²/s. The unit is m²/s because the observed time is in seconds.
వివరణ:
Question 24
ప్రశ్న 24
Two 2 m thick clay layers undergo the same uniform effective-stress rise from 100 to 200 kPa, and each has initial void ratio e0 = 1.00. Layer A is normally consolidated with Cc = 0.20; Layer B is overconsolidated with Cr = 0.05 and preconsolidation pressure 250 kPa. With log10(2) = 0.3010, what is the combined ultimate primary settlement?
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Explanation:
Layer A is normally consolidated and its settlement is SA = 2(0.20)/2 times 0.3010 = 0.0602 m. Layer B remains below its preconsolidation pressure, so it follows recompression. Its settlement is SB = 2(0.05)/2 times 0.3010 = 0.01505 m. The combined settlement equals 0.0602 + 0.01505 = 0.07525 m. Rounding to one decimal millimetre gives approximately 75.3 mm.
వివరణ:
Question 25
ప్రశ్న 25
In classical one-dimensional primary consolidation under a uniform load, which interpretation of average degree of consolidation U is correct?
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Explanation:
Average degree of consolidation describes progress of the primary consolidation process for the layer as a whole. Under the conventional one-dimensional uniform-loading assumptions, it equals the completed primary settlement divided by ultimate primary settlement. Excess pore-water pressure does not normally dissipate at exactly the same rate at every depth. Local pore-pressure dissipation therefore need not equal the average consolidation percentage at a given time. Separating local response from average layer response is important when interpreting settlement and pore-pressure measurements.
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Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. 180 kPa. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: B. 0.30. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: B. 181 mm. సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: B. 135 mm. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. 0.30. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. 23.5 mm. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 211 mm. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: B. OCR = 2.0; recompression throughout. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. 0.120 m. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: B. 0.20 m. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: A. 2.0 × 10^-7 m²/s. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: B. 0.100. సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. 39.4 days. సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 127.2 days. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 4.30. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. 3.94 × 10^-6 m²/min. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 20.8 days. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: C. 90.3 mm. సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: B. 37.5%. సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. 40%. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: B. 0.126. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 170.1 days. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. 2.12 × 10^-7 m²/s. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: C. 75.3 mm. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: C. U equals the fraction of ultimate primary settlement completed under the idealized assumptions, but the excess pore-water pressure dissipation may differ from point to point with depth. సమాధానం: C.