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Question 1
ప్రశ్న 1
A field soil has dry density 1.80 Mg/m³, specific gravity of solids 2.70 and gravimetric water content 12%. Taking water density as 1.00 Mg/m³, what are its void ratio and degree of saturation?
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Explanation:
Dry density and void ratio are related by rho_d = Gs rho_w/(1 + e). Rearrangement gives e = 2.70/1.80 - 1 = 0.50. The phase relationship S e = w Gs uses water content as a decimal. Hence S = 0.12(2.70)/0.50 = 0.648. The specimen therefore has a void ratio of 0.50 and degree of saturation of 64.8%.
వివరణ:
Question 2
ప్రశ్న 2
A borrow-soil batch contains 2000 kg of dry soil solids at 8% gravimetric water content. To prepare it for compaction at 12% water content, assume no loss of water or solids during mixing. How much water must be added, and what will be the final total moist mass?
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Explanation:
The existing water mass is 0.08(2000) = 160 kg. The target water mass is 0.12(2000) = 240 kg. Therefore the amount to add is 240 - 160 = 80 kg. The final moist mass includes the 2000 kg of solids and 240 kg of water. Thus the required addition is 80 kg and the final total mass is 2240 kg.
వివరణ:
Question 3
ప్రశ్న 3
A compacted embankment layer has field bulk density 2.07 Mg/m³ and water content 15%. Its reference laboratory maximum dry density is 1.90 Mg/m³, and every test location must achieve at least 95% relative compaction. Which assessment is correct?
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Explanation:
The field dry density is the bulk density divided by one plus the decimal water content. It equals 2.07/1.15 = 1.80 Mg/m³. Relative compaction is the field dry density divided by reference laboratory MDD. The resulting percentage is 1.80/1.90 times 100 = 94.74%. Since this is below the specified 95% minimum, the tested location fails the density criterion.
వివరణ:
Question 4
ప్రశ్న 4
A deposit has 2 m of moist soil with total unit weight 18 kN/m³ above the water table, followed by 3 m of saturated soil with Gs = 2.70 and e = 0.80. Assuming hydrostatic conditions and gamma_w = 10 kN/m³, what are the vertical effective stress and pore-water pressure at 5 m depth?
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Explanation:
The saturated unit weight of the lower soil is gamma_sat = [(Gs + e)/(1 + e)]gamma_w. Substitution gives gamma_sat = [(2.70 + 0.80)/1.80](10) = 19.44 kN/m³. Total overburden at 5 m is 2(18) + 3(19.44) = 94.33 kPa. The hydrostatic pore pressure is 3(10) = 30.0 kPa. Therefore the effective stress is 94.33 - 30.0 = 64.33 kPa, with pore pressure 30.0 kPa.
వివరణ:
Question 5
ప్రశ్న 5
A 4 m soil column initially has the water table at the surface and saturated unit weight 20 kN/m³. At long-term equilibrium the water table drops by 2 m; the upper 2 m now has moist unit weight 18 kN/m³, while the bottom 2 m remains saturated at 20 kN/m³. Neglect capillary effects and take gamma_w = 10 kN/m³. What are the initial and final vertical effective stresses at the column base?
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Explanation:
Initially the total vertical stress at the base is 4(20) = 80 kPa. Initial pore pressure is 4(10) = 40 kPa, leaving 40 kPa effective stress. After drawdown the total stress is 2(18) + 2(20) = 76 kPa. The new hydrostatic pore pressure at the base is 2(10) = 20 kPa. Therefore the final effective stress is 76 - 20 = 56 kPa, an increase of 16 kPa.
వివరణ:
Question 6
ప్రశ్న 6
A saturated sand has Gs = 2.70 and void ratio e = 0.70. Upward one-dimensional seepage acts at hydraulic gradient i = 0.40, with gamma_w = 10 kN/m³. What are the ideal factor of safety against the quick condition, defined as i_cr/i, and the effective vertical stress increase per metre downward from a zero-effective-stress exit surface?
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Explanation:
The critical upward hydraulic gradient is i_cr = (Gs - 1)/(1 + e). With the supplied values, i_cr = 1.70/1.70 = 1.00. The hydraulic-gradient factor of safety is therefore 1.00/0.40 = 2.50. The submerged unit weight is 10 kN/m³, and upward seepage reduces the effective stress gradient by i gamma_w = 4 kPa/m. The remaining effective stress increase is 10 - 4 = 6 kPa per metre.
వివరణ:
Question 7
ప్రశ్న 7
A saturated soil has permeability k = 2.0 × 10^-4 m/s, void ratio e = 0.50 and hydraulic gradient i = 0.40. Assuming one-dimensional steady seepage and ideal advective transport, what are the average seepage velocity and time to move 2.4 m through the pore space?
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Explanation:
Darcy discharge velocity is v = ki = 2.0 × 10^-4(0.40) = 8.0 × 10^-5 m/s. Porosity is n = e/(1 + e) = 0.50/1.50 = one-third. The average seepage velocity is v_s = v/n = 2.4 × 10^-4 m/s. The ideal travel time is distance divided by seepage velocity. Thus t = 2.4/(2.4 × 10^-4) = 10,000 s.
వివరణ:
Question 8
ప్రశ్న 8
In a saturated constant-head laboratory test, 150 cm³ of water passes through a soil specimen of length 12 cm and cross-sectional area 50 cm² in 100 s under a head difference of 30 cm. The same soil is used in a field drain with flow area 0.40 m² and gradient 0.25. Assuming Darcy flow, what are k and the field discharge?
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Explanation:
From the constant-head test, k = VL/(Aht). Substitution in centimetre units gives k = 150(12)/[50(30)(100)] = 0.012 cm/s. This equals 1.2 × 10^-4 m/s. The field discharge is Q = kiA = 1.2 × 10^-4(0.25)(0.40) = 1.2 × 10^-5 m³/s. Converting to litres per minute gives Q = 0.72 L/min.
వివరణ:
Question 9
ప్రశ్న 9
Two horizontal saturated soil strata of equal thickness have permeabilities k1 = 1.0 × 10^-4 m/s and k2 = 1.0 × 10^-6 m/s. Using the equivalent permeability expressions for flow parallel and normal to the layering, what is approximately the ratio k_horizontal/k_vertical, and which direction is more permeable?
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Explanation:
For equal-thickness layers, equivalent horizontal permeability is the arithmetic mean. It equals (1.0 × 10^-4 + 1.0 × 10^-6)/2 = 5.05 × 10^-5 m/s. Equivalent vertical permeability is the harmonic mean, 2k1k2/(k1 + k2) = approximately 1.9802 × 10^-6 m/s. The ratio is approximately 25.50. Hence this layered deposit is much more permeable parallel to the bedding than normal to it.
వివరణ:
Question 10
ప్రశ్న 10
A flownet for saturated seepage beneath a structure has 3 flow channels and 12 potential drops. The total hydraulic head loss is 6 m and isotropic soil permeability is 1.0 × 10^-5 m/s. What are the seepage discharge per metre of structure width and the corresponding volume per day?
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Explanation:
The discharge per unit width from the stated flownet is q = kH(Nf/Nd). The channel-to-drop ratio is 3/12 = 0.25. Therefore q = 1.0 × 10^-5(6)(0.25) = 1.5 × 10^-5 m³/s per metre. One day contains 86,400 seconds. Multiplying the discharge by this duration gives 1.296 m³/day per metre of structure width.
వివరణ:
Question 11
ప్రశ్న 11
On a potential shear plane in saturated soil, the total normal stress is 180 kPa, pore-water pressure is 60 kPa and applied shear stress is 50 kPa. The effective strength parameters are c′ = 10 kPa and tan(phi′) = 0.50. What are the effective normal stress and factor of safety defined as available shear strength divided by applied shear stress?
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Explanation:
The effective normal stress is the total normal stress minus pore-water pressure. It equals 180 - 60 = 120 kPa. Available Mohr-Coulomb shear strength is 10 + 120(0.50) = 70 kPa. The applied shear stress is 50 kPa. Therefore the specified plane factor of safety is 70/50 = 1.40.
వివరణ:
Question 12
ప్రశ్న 12
A cohesionless soil plane sustains unchanged total normal stress 160 kPa and driving shear stress 45 kPa. Its effective friction parameter is tan(phi′) = 0.60. If pore-water pressure rises from 30 to 70 kPa, what are the initial and final factors of safety against shear on this plane?
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Explanation:
Initially the effective normal stress is 160 - 30 = 130 kPa. Its frictional strength is 130(0.60) = 78 kPa, giving FS = 78/45 = 1.73. After pore pressure rises, effective normal stress becomes 160 - 70 = 90 kPa. The revised frictional strength is 90(0.60) = 54 kPa. Thus the factor of safety decreases to 54/45 = 1.20.
వివరణ:
Question 13
ప్రశ్న 13
A normally consolidated clay layer 4 m thick has e0 = 1.00 and Cc = 0.25, with a uniform effective-stress increase from 100 to 200 kPa. The layer is double drained with cv = 0.040 m²/day and T50 = 0.197. What primary settlement will have occurred at 50% average consolidation, and after approximately how many days?
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Explanation:
The ultimate normally consolidated primary settlement is S = H Cc/(1 + e0) log10(200/100). Substitution gives 4(0.25)/2 times 0.30103 = 0.1505 m. At 50% average consolidation, completed primary settlement is half of this, or approximately 75.3 mm. Double drainage gives a maximum drainage path of 2 m. The required time is T50 Hdr²/cv = 0.197(2²)/0.040 = 19.7 days.
వివరణ:
Question 14
ప్రశ్న 14
A laterally confined soil fill initially has thickness 4.0 m and void ratio 0.80. Compaction reduces the void ratio to 0.60 without loss of soil solids or change in cross-sectional area. What are the reduction in thickness and the final porosity?
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Explanation:
At fixed solids volume and cross-sectional area, soil thickness is proportional to 1 + e. The thickness reduction is H0(e0 - ef)/(1 + e0). Substitution gives 4(0.80 - 0.60)/1.80 = 0.444 m. Final porosity is n = ef/(1 + ef) = 0.60/1.60 = 0.375. Therefore the fill shortens by approximately 0.444 m and finishes with 37.5% porosity.
వివరణ:
Question 15
ప్రశ్న 15
A site has two compressible saturated clay layers beneath a foundation. The upper layer is 2 m thick with mv = 3.0 × 10^-4 kPa^-1, and the lower layer is 3 m thick with mv = 5.0 × 10^-4 kPa^-1. If both layers eventually experience a uniform 80 kPa effective-stress increase, what is the combined primary settlement and the lower-layer contribution?
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Explanation:
For each layer, the linear one-dimensional settlement is S = mv DeltaSigma H. Upper-layer settlement is 3.0 × 10^-4(80)(2) = 0.048 m. Lower-layer settlement is 5.0 × 10^-4(80)(3) = 0.120 m. The total primary settlement is their sum, 0.168 m. Therefore the overall settlement is 168 mm, of which the lower layer contributes 120 mm.
వివరణ:
Question 16
ప్రశ్న 16
A compaction record reports dry density 1.82 Mg/m³ at water content 18% for a soil having Gs = 2.65. Assume water density 1.00 Mg/m³. Which zero-air-void check and engineering response are appropriate?
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Explanation:
The theoretical dry density at full saturation for a given water content is rho_d,ZAV = Gs rho_w/(1 + w Gs). Substitution gives 2.65/[1 + 0.18(2.65)] = approximately 1.794 Mg/m³. The reported dry density of 1.82 Mg/m³ exceeds this theoretical limit. Equivalently, the computed degree of saturation would exceed 100%, which is physically inconsistent. The water content, sample volume, mass and specific-gravity measurements should be reviewed before accepting the result.
వివరణ:
Question 17
ప్రశ్న 17
A 3 m thick saturated soil layer has gamma_sat = 20 kN/m³ and gamma_w = 10 kN/m³. Upward seepage with hydraulic gradient 0.40 occurs from the layer base toward a free-draining surface, where gauge pore pressure is zero. What are the gauge pore-water pressure at the base and available effective-stress shear strength on a horizontal base plane if c′ = 5 kPa and tan(phi′) = 0.50?
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Explanation:
For upward seepage toward a zero-gauge-pressure exit, the pore pressure at depth h is u = gamma_w h(1 + i). Hence u = 10(3)(1.40) = 42 kPa at the base. The overlying total vertical stress is 20(3) = 60 kPa. The effective normal stress is therefore 60 - 42 = 18 kPa. The available shear strength is 5 + 18(0.50) = 14 kPa.
వివరణ:
Question 18
ప్రశ్న 18
A saturated sand has Gs = 2.65 and e = 0.65. An upward-flow path is 4.0 m long and currently has 3.2 m head loss. If the project requires a factor of safety of at least 1.50 against the ideal quick condition, what is the maximum permissible head loss, and does the current condition pass?
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Explanation:
The critical gradient is i_cr = (Gs - 1)/(1 + e) = 1.65/1.65 = 1.00. The specified factor of safety requires actual gradient not exceeding 1.00/1.50 = 0.667. Across a 4 m flow path, that gradient corresponds to a maximum head loss of approximately 2.67 m. The existing gradient is 3.2/4.0 = 0.80, which gives a safety factor of 1.25. Consequently the existing head loss exceeds the permissible value and fails the stated criterion.
వివరణ:
Question 19
ప్రశ్న 19
A saturated clay layer has vertical permeability k = 1.0 × 10^-9 m/s, mv = 5.0 × 10^-4 kPa^-1 and gamma_w = 10 kN/m³. The layer is 2 m thick and double drained, and T50 = 0.197. Assuming these parameters are constant, what are cv and the approximate time to reach 50% average consolidation?
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Explanation:
The coefficient of consolidation is cv = k/(mv gamma_w), with consistent stress and unit-weight units. Using the given data gives cv = 1.0 × 10^-9/[5.0 × 10^-4(10)] = 2.0 × 10^-7 m²/s. Double drainage in a 2 m layer gives Hdr = 1 m. The time is t50 = 0.197(1²)/(2.0 × 10^-7) = 985000 s. Conversion to days gives approximately 11.4 days.
వివరణ:
Question 20
ప్రశ్న 20
A field-compacted soil has dry density 1.80 Mg/m³, water content 15% and Gs = 2.70, with water density 1.00 Mg/m³. Its specified laboratory MDD is 1.92 Mg/m³, and minimum relative compaction is 95%. What is the correct combined compaction and saturation assessment?
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Explanation:
Relative compaction is 1.80/1.92 × 100 = 93.75%. The corresponding void ratio is e = 2.70/1.80 - 1 = 0.50. The saturation relationship gives S = w Gs/e = 0.15(2.70)/0.50 = 0.81. Thus the degree of saturation is 81%, but the density ratio remains below the required 95%. The layer fails the stated compaction criterion despite being relatively wet.
వివరణ:
Question 21
ప్రశ్న 21
A smooth vertical wall retains 5 m of fully submerged sand with phi′ = 30 degrees, gamma_sat = 20 kN/m³ and gamma_w = 10 kN/m³. Hydrostatic water reaches ground level behind the wall but only 3 m above its base in front. Assume Rankine Ka = 1/3, no seepage and no other loads. What is the net horizontal pressure intensity at the wall base?
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Explanation:
The effective vertical stress at the retained-side base is (20 - 10)(5) = 50 kPa. The effective active lateral soil pressure is (1/3)(50) = 16.67 kPa. Water pressure at the back is 10(5) = 50 kPa. Water on the front produces an opposing base pressure of 10(3) = 30 kPa. The net base pressure is therefore 16.67 + 50 - 30 = approximately 36.7 kPa.
వివరణ:
Question 22
ప్రశ్న 22
At an upward-seepage exit in saturated sand, Gs = 2.60 and e = 0.80. A flownet gives an estimated vertical exit gradient from 0.60 m head drop over 0.90 m exit length. If the project requires FS = i_cr/i_exit of at least 1.50, what is the calculated FS and acceptance decision?
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Explanation:
The critical gradient is (Gs - 1)/(1 + e) = 1.60/1.80 = 0.8889. The estimated upward exit gradient is 0.60/0.90 = 0.6667. Their ratio gives FS = 0.8889/0.6667 = approximately 1.33. The project minimum is expressly specified as 1.50. Therefore the calculated exit condition fails the stated criterion and calls for an appropriate seepage-control response.
వివరణ:
Question 23
ప్రశ్న 23
A fully saturated soil has gravimetric water content w = 25% and specific gravity of solids Gs = 2.60. The soil extends 4 m beneath a water surface, with hydrostatic conditions and gamma_w = 10 kN/m³. What are its void ratio and vertical effective stress at 4 m depth?
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Explanation:
At full saturation, S e = w Gs reduces to e = w Gs. Thus e = 0.25(2.60) = 0.65. The submerged unit weight is gamma_prime = [(Gs - 1)/(1 + e)]gamma_w. Its value is (1.60/1.65)(10) = approximately 9.697 kN/m³. The vertical effective stress at 4 m depth is 9.697(4) = approximately 38.8 kPa.
వివరణ:
Question 24
ప్రశ్న 24
A saturated soil plane initially has total normal stress 150 kPa and pore pressure 60 kPa. Groundwater lowering changes the total normal stress to 145 kPa and pore pressure to 35 kPa, while c′ = 10 kPa and tan(phi′) = 0.50 remain unchanged. How does the available shear strength change?
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Explanation:
The initial effective normal stress is 150 - 60 = 90 kPa. Initial strength is 10 + 90(0.50) = 55 kPa. After groundwater lowering, effective stress is 145 - 35 = 110 kPa. The revised available strength is 10 + 110(0.50) = 65 kPa. Therefore the available shear strength increases by 10 kPa even though total normal stress decreases slightly.
వివరణ:
Question 25
ప్రశ్న 25
At an earthwork location, field dry density is 1.84 Mg/m³ against a specified laboratory MDD of 1.90 Mg/m³, with a minimum relative-compaction requirement of 95%. A nearby saturated potential sliding plane has total normal stress 150 kPa, pore pressure 70 kPa, c′ = 5 kPa, tan(phi′) = 0.50 and applied shear stress 45 kPa. The specified plane factor of safety is at least 1.30. What is the appropriate combined decision?
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Explanation:
The relative compaction is 1.84/1.90 × 100 = approximately 96.84%, exceeding the 95% requirement. On the sliding plane, effective normal stress is 150 - 70 = 80 kPa. Available shear strength is 5 + 80(0.50) = 45 kPa. The plane factor of safety is therefore 45/45 = 1.00, below the separately specified minimum of 1.30. The layer passes the density criterion but requires stability investigation or corrective measures for the shear condition.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. e = 0.50 and S = 64.8%. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. 80 kg and 2240 kg. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. Field dry density = 1.80 Mg/m³, relative compaction = 94.7%, and the location fails. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 64.3 kPa and 30.0 kPa. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 40 kPa and 56 kPa. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 2.50 and 6 kPa/m. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 2.4 × 10^-4 m/s and 10,000 s. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. k = 1.2 × 10^-4 m/s; discharge = 0.72 L/min. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 25.5; horizontal flow is more permeable. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 1.5 × 10^-5 m³/s per m and 1.296 m³/day per m. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 120 kPa and 1.40. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 1.73 initially and 1.20 finally. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 75.3 mm and 19.7 days. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 0.444 m and 37.5%. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. 168 mm total, of which 120 mm comes from the lower layer. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. The zero-air-void dry density is about 1.794 Mg/m³; the reported value exceeds it and should be investigated for measurement or calculation errors. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 42 kPa and 14 kPa. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 2.67 m; current condition fails. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. cv = 2.0 × 10^-7 m²/s and t50 = 11.4 days. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. Relative compaction = 93.75%, saturation = 81%, and the layer fails. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 36.7 kPa. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. FS = 1.33 and fails. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. e = 0.65 and sigma_v′ = 38.8 kPa. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. It increases from 55 to 65 kPa. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. Relative compaction passes at about 96.8%, but sliding-plane FS = 1.00 fails; compaction acceptance alone does not establish stability. సమాధానం: B.