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Question 1
ప్రశ్న 1
Which combination of conditions is most consistent with the classical Terzaghi bearing-capacity analysis of a shallow continuous strip footing?
Explanation:
Terzaghi's classical theory was developed primarily for shallow foundation bearing resistance. Its original continuous-footing expression assumes an idealized general shear failure mechanism beneath a long strip footing. The footing base is treated as rough and horizontal, with a centrally applied vertical load and level surrounding ground. Homogeneous soil strength and an appropriate surcharge condition are assumed in the elementary derivation. Inclined loading, unusual ground geometry, significant stratification and other complications require suitable modifications or alternative analyses.
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Question 2
ప్రశ్న 2
For a strip footing under Terzaghi's original general-shear formulation, the gross ultimate bearing pressure is qult = cNc + qNq + 0.5 gamma B Ngamma. Which interpretation of the three contributions is correct?
Explanation:
The cNc term represents the contribution of soil cohesion to the ideal ultimate bearing pressure. The qNq term accounts for surcharge at the foundation-base elevation. The 0.5 gamma B Ngamma term represents the contribution associated with the weight of the soil in the bearing failure mechanism. The bearing factors depend on the internal friction angle and the particular analytical formulation adopted. Factors from different theories should not be combined without establishing their compatibility.
వివరణ:
Question 3
ప్రశ్న 3
A rough continuous strip footing is analysed using Terzaghi's original gross ultimate bearing-capacity equation qult = cNc + qNq + 0.5 gamma B Ngamma. Given c = 15 kPa, Nc = 20, q = 20 kPa, Nq = 10, gamma = 18 kN/m3, B = 2 m and Ngamma = 8, what is qult?
Explanation:
The cohesion contribution is cNc = 15(20) = 300 kPa. The surcharge contribution is qNq = 20(10) = 200 kPa. The soil self-weight contribution is 0.5(18)(2)(8) = 144 kPa. Adding the three contributions gives 300 + 200 + 144 = 644 kPa. This is the gross ultimate bearing pressure under the explicitly supplied Terzaghi factors and assumptions.
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Question 4
ప్రశ్న 4
For a saturated clay analysed under an ideal undrained total-stress condition with phi_u = 0, use Terzaghi's original strip-footing factors Nc = 5.7, Nq = 1 and Ngamma = 0. If cu = 40 kPa and the total overburden surcharge at foundation level is q = 30 kPa, what are the gross and net ultimate bearing pressures, respectively?
Explanation:
For the stated undrained condition, the soil self-weight term involving Ngamma becomes zero. The gross ultimate bearing pressure is therefore cuNc + qNq. Substitution gives 40(5.7) + 30(1) = 258 kPa. The net ultimate pressure is obtained by subtracting the original surcharge of 30 kPa. Thus the gross and net ultimate values are 258 kPa and 228 kPa, respectively, using the specified original Terzaghi factor rather than factors from another formulation.
వివరణ:
Question 5
ప్రశ్న 5
Using Terzaghi's original empirical equation for a square footing, qult = 1.3cNc + qNq + 0.4 gamma B Ngamma, calculate the gross ultimate bearing pressure for c = 20 kPa, Nc = 10, q = 18 kPa, Nq = 5, gamma = 18 kN/m3, B = 2 m and Ngamma = 4.
Explanation:
The cohesion contribution for the original square-footing equation is 1.3(20)(10) = 260 kPa. The surcharge contribution is 18(5) = 90 kPa. The soil self-weight contribution is 0.4(18)(2)(4) = 57.6 kPa. The gross ultimate bearing pressure is therefore 260 + 90 + 57.6 = 407.6 kPa. The 0.4 coefficient is already part of the specified original Terzaghi square-footing expression and must not be multiplied by another shape factor.
వివరణ:
Question 6
ప్రశ్న 6
A circular shallow footing is analysed using Terzaghi's original equation qult = 1.3cNc + qNq + 0.3 gamma B Ngamma, where B is the footing diameter. For cohesionless sand, take c = 0, q = 20 kPa, Nq = 8, gamma = 18 kN/m3, B = 3 m and Ngamma = 6. What is the gross ultimate bearing pressure?
Explanation:
Cohesionless sand has c = 0 in the specified effective-stress analysis, so the cohesion contribution vanishes. The surcharge term is qNq = 20(8) = 160 kPa. The circular-footing soil-weight term is 0.3(18)(3)(6) = 97.2 kPa. Adding these components gives 160 + 97.2 = 257.2 kPa. The supplied circular equation uses the footing diameter as B and its original empirical 0.3 multiplier.
వివరణ:
Question 7
ప్రశ్న 7
Consider square and circular footings having the same characteristic dimension B, founding surcharge q, soil properties and bearing-capacity factors. Using Terzaghi's original empirical square and circular equations, and assuming gamma and Ngamma are positive, which comparison is correct?
Explanation:
The original Terzaghi square-footing expression uses a soil-weight coefficient of 0.4. The corresponding circular-footing expression uses a coefficient of 0.3. Both expressions use 1.3 for the cohesion term and unity for the surcharge multiplier. Therefore the square footing has the greater soil-weight contribution when all specified input parameters are identical. This is a comparison of the original empirical pressure equations and not a comparison of total load capacities for different plan areas.
వివరణ:
Question 8
ప్రశ్న 8
A continuous strip footing has gross ultimate bearing pressure given by qult = cNc + qNq + 0.5 gamma B Ngamma. Given c = 10 kPa, Nc = 12, q = 40 kPa, Nq = 6, gamma = 20 kN/m3, B = 2 m and Ngamma = 5, determine the gross and net ultimate bearing pressures, taking net ultimate pressure as qult - q.
Explanation:
The cohesion term is 10(12) = 120 kPa. The surcharge term is 40(6) = 240 kPa. The soil-weight term is 0.5(20)(2)(5) = 100 kPa. Thus the gross ultimate pressure is 120 + 240 + 100 = 460 kPa. Subtracting the original foundation-level surcharge of 40 kPa gives a net ultimate pressure of 420 kPa.
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Question 9
ప్రశ్న 9
A footing has gross ultimate shear pressure of 720 kPa and original overburden pressure of 60 kPa at founding level. The specified factor of safety is 3.0 on net ultimate pressure, and the settlement criterion limits net working pressure to 200 kPa. What are the net safe pressure from shear and the gross allowable working pressure, respectively?
Explanation:
Net ultimate bearing pressure is 720 - 60 = 660 kPa. Applying the specified factor of safety gives net safe shear pressure of 660/3 = 220 kPa. The settlement-based net limit is lower at 200 kPa and therefore governs the allowable net pressure. The corresponding gross allowable pressure is 200 + 60 = 260 kPa. This distinction prevents the ultimate, safe and settlement-controlled pressures from being treated as interchangeable.
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Question 10
ప్రశ్న 10
A surface strip footing on dry cohesionless sand is analysed using the specified equation qult = 0.5 gamma B Ngamma, with gamma = 18 kN/m3, B = 2 m and Ngamma = 20. If a factor of safety of 3.0 is applied directly to the ultimate value and settlement does not govern, what are the ultimate and safe bearing pressures?
Explanation:
The footing is at the ground surface, and cohesion is zero. The stated analytical expression therefore contains only the soil self-weight contribution. Substitution gives qult = 0.5(18)(2)(20) = 360 kPa. Dividing by the prescribed factor of safety gives 360/3 = 120 kPa. No settlement reduction is necessary in this exercise because the question explicitly states that settlement does not govern.
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Question 11
ప్రశ్న 11
A strip footing in cohesionless soil has B = 2 m and foundation-level effective surcharge q = 36 kPa, with Nq = 10 and Ngamma = 10. Before groundwater rise, the effective unit weight adopted for the soil-weight term is 20 kN/m3. The water table then rises to footing-base level, reducing the effective unit weight used below the base to 10 kN/m3 while leaving q unchanged. Using qult = qNq + 0.5 gamma_eff B Ngamma, what are the calculated values before and after groundwater rise?
Explanation:
The effective surcharge contribution is unchanged at 36(10) = 360 kPa. Before groundwater rise, the soil-weight contribution is 0.5(20)(2)(10) = 200 kPa. The initial calculated bearing pressure is therefore 560 kPa. After the water table reaches the base, the specified effective soil unit weight becomes 10 kN/m3, giving a soil-weight contribution of 100 kPa. The revised calculated pressure is 360 + 100 = 460 kPa, demonstrating the reduction associated with the stated groundwater change.
వివరణ:
Question 12
ప్రశ్న 12
For a cohesionless strip footing having B = 2 m, gamma = 20 kN/m3 and Ngamma = 10, the water table is permanently located more than one footing width below the foundation base. Under the idealized correction in which groundwater does not reduce the soil-weight contribution beyond that depth, what is the unmodified value of 0.5 gamma B Ngamma?
Explanation:
The water table is assumed sufficiently deep that no reduction is applied to the stated soil-weight term. The required contribution is 0.5 gamma B Ngamma. Substitution gives 0.5(20)(2)(10). This equals 200 kPa. The result concerns only the soil self-weight contribution and does not include any surcharge or cohesion terms.
వివరణ:
Question 13
ప్రశ్న 13
For a strip footing with B = 2 m and foundation-level effective surcharge q = 36 kPa, take Nq = 8, gamma = 20 kN/m3 and Ngamma = 10. The water table lies 1 m below the footing base. Under a specified linear interpolation for the soil-weight correction, use W = 0.75 and qult = qNq + 0.5 gamma B Ngamma W. What is the calculated gross ultimate bearing pressure?
Explanation:
The groundwater correction acts on the soil-weight term in the explicitly given equation. The effective surcharge contribution is qNq = 36(8) = 288 kPa. The uncorrected soil-weight contribution is 0.5(20)(2)(10) = 200 kPa. Applying W = 0.75 gives a corrected contribution of 150 kPa. Therefore the calculated gross ultimate bearing pressure is 288 + 150 = 438 kPa.
వివరణ:
Question 14
ప్రశ్న 14
A cohesionless strip footing is founded 2 m below ground level in soil that is fully submerged under hydrostatic conditions. Given gamma_sat = 20 kN/m3, gamma_w = 10 kN/m3, B = 2 m, Nq = 8 and Ngamma = 10, use the stated effective-stress bearing expression q'_ult = q' Nq + 0.5 gamma' B Ngamma, where gamma' = gamma_sat - gamma_w and q' = gamma' Df. What is the calculated value of q'_ult?
Explanation:
The submerged or buoyant unit weight is gamma' = 20 - 10 = 10 kN/m3. The effective foundation-level surcharge is q' = gamma' Df = 10(2) = 20 kPa. Its bearing contribution is 20(8) = 160 kPa. The soil-weight contribution is 0.5(10)(2)(10) = 100 kPa. Thus the specified effective-stress bearing expression gives 160 + 100 = 260 kPa, without separately treating this value as an applied total-stress contact pressure.
వివరణ:
Question 15
ప్రశ్న 15
Which set of shape factors corresponds to a square footing under the stated IS 6403-type modified bearing-capacity formulation, in which the soil-weight term is 0.5 gamma B Ngamma multiplied by s_gamma?
Explanation:
The stated modified bearing-capacity method applies separate shape factors to the cohesion, surcharge and soil-weight terms. For a square footing, the specified IS 6403 shape-factor set is 1.3, 1.2 and 0.8, respectively. The soil-weight term already contains the basic coefficient 0.5 before the shape factor is applied. This formulation differs from Terzaghi's original empirical square-footing equation containing 0.4 gamma B Ngamma. Using both the original 0.4 coefficient and a separate 0.8 shape factor would incorrectly apply shape adjustment twice.
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Question 16
ప్రశ్న 16
Using the stated IS 6403-style net ultimate strip-footing equation qnet,ult = cNc + q(Nq - 1) + 0.5 gamma B Ngamma, take c = 15 kPa, Nc = 20, q = 25 kPa, Nq = 8, gamma = 20 kN/m3, B = 2 m and Ngamma = 5. If the factor of safety on net ultimate pressure is 2.5, what is the net safe bearing pressure?
Explanation:
The cohesion contribution is 15(20) = 300 kPa. The net surcharge contribution is 25(8 - 1) = 175 kPa. The soil-weight contribution is 0.5(20)(2)(5) = 100 kPa. Therefore the net ultimate bearing pressure is 300 + 175 + 100 = 575 kPa. Dividing by the specified factor of safety of 2.5 gives a net safe pressure of 230 kPa.
వివరణ:
Question 17
ప్రశ్న 17
For a rectangular footing under the stated IS 6403 shape-factor expressions sc = sq = 1 + 0.2(B/L) and s_gamma = 1 - 0.4(B/L), calculate the shape factors when B = 2 m and L = 4 m.
Explanation:
The width-to-length ratio is B/L = 2/4 = 0.50. The cohesion and surcharge shape factors are both 1 + 0.2(0.50). Hence sc = sq = 1.10. The soil-weight shape factor is 1 - 0.4(0.50) = 0.80. These dimensionless factors are then applied to their respective bearing-capacity terms in the specified modified formulation.
వివరణ:
Question 18
ప్రశ్న 18
A modified net ultimate bearing-capacity calculation uses qnet,ult = cNc sc dc ic + q(Nq - 1)sq dq iq + 0.5 gamma B Ngamma s_gamma d_gamma i_gamma. Given c = 20 kPa, Nc = 10, sc = 1.10, dc = 1.20, ic = 1, q = 20 kPa, Nq = 5, sq = 1.10, dq = 1.10, iq = 1, gamma = 20 kN/m3, B = 2 m, Ngamma = 4, s_gamma = 0.80, d_gamma = 1.10 and i_gamma = 1, what is qnet,ult?
Explanation:
The modified cohesion term is 20(10)(1.10)(1.20)(1) = 264 kPa. The modified net surcharge term is 20(5 - 1)(1.10)(1.10)(1) = 96.8 kPa. The modified soil-weight term is 0.5(20)(2)(4)(0.80)(1.10)(1) = 70.4 kPa. Adding the three contributions gives 264 + 96.8 + 70.4 = 431.2 kPa. All correction factors are supplied explicitly, so no external factor table or chart is needed.
వివరణ:
Question 19
ప్రశ్న 19
An inclined-load bearing-capacity exercise specifies qnet,ult = q(Nq - 1)iq + 0.5 gamma B Ngamma i_gamma for a cohesionless strip footing, with all other correction factors equal to unity. Given q = 30 kPa, Nq = 10, iq = 0.81, gamma = 18 kN/m3, B = 2 m, Ngamma = 10 and i_gamma = 0.64, what is qnet,ult?
Explanation:
The net surcharge contribution is reduced by the supplied inclination factor iq. It equals 30(10 - 1)(0.81) = 218.7 kPa. The soil-weight contribution is reduced by i_gamma and equals 0.5(18)(2)(10)(0.64) = 115.2 kPa. Their sum is 218.7 + 115.2 = 333.9 kPa. The calculation demonstrates the effect of prescribed load-inclination factors without assuming that one universal factor applies to every bearing-capacity term.
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Question 20
ప్రశ్న 20
A rectangular footing has original dimensions B = 2.0 m and L = 3.0 m and carries a vertical resultant of 896 kN. The resultant has eccentricities eB = 0.20 m along the B direction and eL = 0.10 m along the L direction. Using the specified effective-area approximation B' = B - 2eB and L' = L - 2eL, what are the effective bearing area and average effective-area load pressure?
Explanation:
The reduced effective width is B' = 2.0 - 2(0.20) = 1.60 m. The reduced effective length is L' = 3.0 - 2(0.10) = 2.80 m. Their product gives an effective bearing area of 1.60(2.80) = 4.48 m2. Dividing the vertical load by this area gives 896/4.48 = 200.0 kPa. This effective-area pressure is an equivalent design quantity and should not be mistaken for the actual detailed contact-pressure distribution.
వివరణ:
Question 21
ప్రశ్న 21
For a Terzaghi-type local-shear approximation, the modified cohesion is c_local = (2/3)c and the reduced friction angle satisfies tan(phi_local) = (2/3)tan(phi). If c = 30 kPa and phi = 30 degrees, what are the modified cohesion and friction angle approximately? Take tan(30 degrees) = 0.57735.
Explanation:
The local-shear cohesion approximation reduces the original cohesion by a factor of two-thirds. Therefore c_local = (2/3)(30) = 20 kPa. The modified friction parameter is tan(phi_local) = (2/3)(0.57735) = approximately 0.3849. Taking the inverse tangent gives phi_local approximately 21.1 degrees. The angle itself is not obtained by simply multiplying 30 degrees by two-thirds.
వివరణ:
Question 22
ప్రశ్న 22
A strip footing is analysed for local shear using the explicitly supplied Terzaghi-type approximation qult = c_local Nc_local + qNq_local + 0.5 gamma B Ngamma_local. Given c_local = 20 kPa, Nc_local = 12, q = 25 kPa, Nq_local = 5, gamma = 18 kN/m3, B = 2 m and Ngamma_local = 4, what are the gross and net ultimate bearing pressures, respectively?
Explanation:
The modified cohesion contribution is 20(12) = 240 kPa. The surcharge contribution is 25(5) = 125 kPa. The soil-weight contribution is 0.5(18)(2)(4) = 72 kPa. Gross ultimate pressure is therefore 240 + 125 + 72 = 437 kPa. Subtracting the original overburden surcharge of 25 kPa gives net ultimate bearing pressure of 412 kPa.
వివరణ:
Question 23
ప్రశ్న 23
For a continuous strip footing in drained cohesionless soil, use the IS 6403-type general-shear NET equation qnet,ult = q(Nq - 1) + 0.5 gamma B Ngamma, with no additional shape, depth, inclination or groundwater corrections. At the specified effective friction angle, the adopted bearing factors are Nq = 18.40 and Ngamma = 22.40. If q = 36 kPa, gamma = 18 kN/m3 and B = 2 m, what is qnet,ult?
Explanation:
The specified equation is written directly for net ultimate bearing pressure. The net surcharge contribution is 36(18.40 - 1) = 626.4 kPa. The soil-weight contribution is 0.5(18)(2)(22.40) = 403.2 kPa. Adding them gives 626.4 + 403.2 = 1029.6 kPa. The factors are supplied from the same adopted formulation, avoiding inconsistencies caused by mixing bearing-factor families.
వివరణ:
Question 24
ప్రశ్న 24
In an ideal homogeneous soil, a strip footing is evaluated with the original gross equation qult = cNc + gamma Df Nq + 0.5 gamma B Ngamma, while all parameters except foundation depth remain unchanged. The depth increases from 1 m to 2 m, with gamma = 18 kN/m3 and Nq = 8. If net ultimate pressure is defined as qult - gamma Df, by how much does the calculated NET ultimate bearing pressure increase?
Explanation:
The net surcharge contribution is gamma Df(Nq - 1). All other terms are unchanged in the expressly simplified comparison. Increasing Df by 1 m changes this contribution by gamma DeltaDf(Nq - 1). The numerical increase is 18(1)(8 - 1) = 126 kPa. This result excludes additional depth corrections, groundwater changes and any changes in soil properties, as specified in the question.
వివరణ:
Question 25
ప్రశ్న 25
A rectangular footing measures 2.0 m by 2.5 m at a depth of 1.0 m in level homogeneous soil of unit weight 18 kN/m3. Its calculated net safe bearing pressure against shear is 400 kPa, but the maximum net pressure satisfying settlement requirements is 260 kPa. Assuming uniform vertical contact pressure and using gross allowable pressure = allowable net pressure + gamma Df, what maximum total GROSS vertical load at the base is permitted?
Explanation:
The net safe shear pressure is 400 kPa, while the settlement-based net limit is 260 kPa. Therefore the allowable net pressure is 260 kPa because settlement governs. The original overburden pressure at the base is gamma Df = 18(1) = 18 kPa, giving gross allowable pressure of 278 kPa. The footing area is 2.0(2.5) = 5.0 m2. Consequently the maximum total gross vertical base load is 278(5.0) = 1390 kN under the stated assumptions.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. A long strip footing with a horizontal rough base, essentially vertical central loading, level ground and approximately homogeneous soil governed by an idealized shear-failure mechanism. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. The terms represent cohesion, surcharge at foundation level and soil self-weight contributions, with bearing factors selected consistently for the soil friction angle and adopted theory. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: C. 644 kPa. సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: A. 258 kPa and 228 kPa. సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: D. 407.6 kPa. సమాధానం: D.
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Question 6 ప్రశ్న 6Answer: B. 257.2 kPa. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: C. The square footing has a larger soil self-weight contribution per unit bearing area, while both formulations use the same 1.3 cohesion coefficient and the same surcharge term. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. 460 kPa gross and 420 kPa net. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. 220 kPa and 260 kPa. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. 360 kPa and 120 kPa. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: A. 560 kPa before and 460 kPa after. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. 200 kPa. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. 438 kPa. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 260 kPa. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. sc = 1.3, sq = 1.2 and s_gamma = 0.8. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 230 kPa. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. sc = sq = 1.10 and s_gamma = 0.80. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 431.2 kPa. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 333.9 kPa. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 4.48 m2 and 200.0 kPa. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 20 kPa and 21.1 degrees. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 437 kPa gross and 412 kPa net. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 1029.6 kPa. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. 126 kPa. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: C. 1390 kN. సమాధానం: C.
