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Question 1
ప్రశ్న 1
Which statement best summarizes the principal static-stability requirements of a gravity dam?
Explanation:
A gravity dam must transfer all applied loads safely to its foundation. Sliding along the dam-foundation contact or other potential planes must be adequately resisted. The stress state in the dam and foundation must remain within the applicable tensile and compressive limits. Rotational equilibrium is reflected in the location of the resultant and the resulting stress distribution rather than in weight alone. Therefore Option B correctly summarizes the principal stability requirements.
వివరణ:
Question 2
ప్రశ్న 2
For a rectangular base of width B subjected to a resultant vertical force with linear base-pressure distribution, what condition keeps the entire base in compression?
Explanation:
For a linear pressure distribution, the resultant location controls whether both edges of the base remain in compression. The kern of a rectangular base is its middle third. If the eccentricity from the base centre satisfies |e| ≤ B/6, the computed pressure at both edges is non-negative. At |e| = B/6, the pressure at one edge becomes zero. Hence Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
A 24 m wide gravity-dam base has its resultant intersecting the base 15 m from the upstream heel. What is the eccentricity of the resultant relative to the base centre?
Explanation:
The centre of a 24 m wide base lies 12 m from the upstream heel. The resultant intersects the base at 15 m from the heel. Its offset from the centre is therefore 15 - 12 = 3 m. Because the intersection lies farther toward the downstream side than the centre, the eccentricity is toward the toe. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
For a base width B carrying a net vertical force V, if the resultant has eccentricity e toward the downstream toe and the entire base remains in compression, which expressions give the toe and heel pressures under the usual linear-distribution assumption?
Explanation:
The average base pressure is V/B per unit length of dam. Eccentric loading adds a linear bending component to this direct pressure. When the resultant moves toward the downstream toe, the toe pressure increases while the heel pressure decreases. The standard linear expressions are V/B multiplied by 1 ± 6e/B, with the plus sign at the side toward which the resultant shifts. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A 20 m wide gravity-dam base carries a net vertical force of 12,000 kN/m. The resultant is 2 m toward the downstream toe from the base centre. Assuming full contact and linear pressure distribution, what are the toe and heel pressures?
Explanation:
The average pressure is V/B = 12,000/20 = 600 kPa. The eccentricity ratio is 6e/B = 6 × 2/20 = 0.60. Hence the toe pressure is 600(1 + 0.60) = 960 kPa. The heel pressure is 600(1 - 0.60) = 240 kPa. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
If the resultant on a rectangular gravity-dam base is located exactly at the middle-third limit toward the downstream toe, what is the corresponding idealized linear base-pressure condition?
Explanation:
At the middle-third boundary, the eccentricity magnitude is B/6. Substituting e = B/6 into the linear pressure equations gives a factor of 1 + 1 = 2 at the toe. At the heel the factor becomes 1 - 1 = 0. Thus the toe pressure is 2V/B while the heel pressure is zero. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
What does the ordinary full-contact linear pressure formula indicate when the resultant moves beyond the middle third toward the downstream toe?
Explanation:
When e exceeds B/6 toward the toe, the term 1 - 6e/B becomes negative at the heel. The full-contact elastic formula therefore predicts tension at the heel. A dam-foundation interface that is treated as unable to sustain tension cannot maintain that assumed full-contact pressure distribution. Contact may then be limited to a compressed portion of the base and a different equilibrium pressure diagram is required. Hence Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
The algebraic moment of all applied loads about the downstream toe is 96,000 kN·m per metre in the resisting sense, and the net vertical downward force is 12,000 kN/m. Where does the resultant intersect the base, measured from the downstream toe?
Explanation:
The point where the resultant intersects the base is obtained from moment equilibrium. If d is its distance from the toe, the net vertical force produces a moment Vd about the toe. Thus d = 96,000/12,000 = 8 m. The stated algebraic moment already includes the effects of the horizontal and vertical applied loads. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
In a simplified textbook overturning check, the sum of resisting moments about the downstream toe is 180,000 kN·m/m and the sum of overturning moments is 90,000 kN·m/m. What is the conventional factor of safety against overturning by the moment-ratio definition?
Explanation:
For the stated simplified textbook definition, the factor of safety against overturning is the ratio of resisting moments to overturning moments. The resisting moment is 180,000 kN·m/m. The overturning moment is 90,000 kN·m/m. Their ratio is 180,000/90,000 = 2.00. Therefore Option D is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which statement most accurately reflects current IS 6512-based gravity-dam assessment regarding overturning?
Explanation:
Gravity-dam instability does not normally progress as a perfectly rigid block simply rotating about the toe without other effects. Excessive eccentricity can first produce tensile cracking, altered uplift, loss of contact or high compression near the toe. Current CWC guidance interpreting IS 6512 therefore emphasizes permissible stress conditions and sliding safety rather than relying on a separate universal overturning moment ratio alone. A moment ratio can still be useful as an elementary calculation when it is explicitly defined, but it is not the complete code assessment. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
In the shear-friction concept used for sliding assessment, what does the sliding safety measure fundamentally compare?
Explanation:
Sliding occurs when driving shear demand exceeds the resistance available along a potential sliding plane. The resistance can include friction and, where justified, cohesive or shear-strength components. The driving side consists of forces or components tending to move the dam along the plane. CWC guidance describes the shear-friction safety concept in terms of available sliding resistance relative to sliding demand, while IS 6512 applies prescribed partial factors in design. Hence Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a simplified friction-only sliding calculation, a dam has 14,000 kN/m downward load before uplift, 2,000 kN/m uplift, coefficient of friction 0.70, and 6,000 kN/m horizontal driving force. What is the sliding factor of safety under the stated assumptions?
Explanation:
The effective normal force for the stated simplified calculation is 14,000 - 2,000 = 12,000 kN/m. Frictional resistance is therefore 0.70 × 12,000 = 8,400 kN/m. The horizontal driving force is 6,000 kN/m. The factor of safety is 8,400/6,000 = 1.40. Hence Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A 15 m wide dam base is checked by a simplified shear-friction expression. Net compressive normal force is 8,000 kN/m, coefficient of friction is 0.60, cohesion is 100 kPa over the full 15 m base width per metre length, and horizontal driving force is 3,500 kN/m. What is the factor of safety?
Explanation:
For one metre length of dam, the contact area is 15 × 1 = 15 m². The cohesive resistance is 100 kN/m² × 15 m² = 1,500 kN/m. Frictional resistance is 0.60 × 8,000 = 4,800 kN/m. Total stated resistance is 1,500 + 4,800 = 6,300 kN/m, giving a factor of safety of 6,300/3,500 = 1.80. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
A simplified gravity-dam sliding check has a downward structural load of 12,000 kN/m, horizontal driving force of 4,000 kN/m and friction coefficient 0.60. If uplift increases from zero to 3,000 kN/m, what happens to the friction-only factor of safety?
Explanation:
Without uplift, frictional resistance is 0.60 × 12,000 = 7,200 kN/m, so the factor of safety is 7,200/4,000 = 1.80. With 3,000 kN/m uplift, the effective normal force becomes 9,000 kN/m. The new frictional resistance is 0.60 × 9,000 = 5,400 kN/m. The new factor of safety is 5,400/4,000 = 1.35. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
For otherwise unchanged loading, what is the usual stability effect of reducing foundation uplift through effective drainage?
Explanation:
Uplift acts upward and subtracts from the effective downward force carried through the base. Effective drainage can reduce the uplift pressure distribution where the design assumptions permit such relief. A lower uplift resultant therefore increases the effective normal force available for frictional or shear resistance. It can also change the resultant location and the calculated base-pressure distribution. Hence Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
A gravity-dam base is 30 m wide. To satisfy the elementary middle-third criterion, between which distances from the upstream heel should the resultant intersect the base?
Explanation:
The middle third occupies the central one-third of the base width. For B = 30 m, each outer third is 10 m wide. The upstream boundary of the kern is therefore 10 m from the heel. The downstream boundary is 20 m from the heel. Thus the resultant should lie between 10 m and 20 m for this elementary no-tension criterion, making Option C correct.
వివరణ:
Question 17
ప్రశ్న 17
A dam base 18 m wide carries a net vertical force of 9,000 kN/m. Its resultant has an eccentricity of 1.5 m toward the downstream toe. What are the toe and heel pressures under a linear full-contact distribution?
Explanation:
The average pressure is 9,000/18 = 500 kPa. The eccentricity factor is 6e/B = 6 × 1.5/18 = 0.50. Hence the toe pressure is 500(1 + 0.50) = 750 kPa. The heel pressure is 500(1 - 0.50) = 250 kPa. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
On a 24 m wide base, the computed linear pressures are 750 kPa at the downstream toe and 250 kPa at the upstream heel. What is the eccentricity of the resultant toward the toe?
Explanation:
For a linear pressure distribution, the average pressure is (750 + 250)/2 = 500 kPa. The normalized pressure difference is (750 - 250)/(750 + 250) = 0.50. Since 6e/B equals this difference ratio, e = (B/6) × 0.50. Thus e = (24/6) × 0.50 = 2 m toward the toe. Therefore Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A no-tension contact analysis finds that the resultant intersects the base 4 m from the downstream toe and the compressed base pressure is triangular with maximum pressure at the toe. What width of the base remains in compression?
Explanation:
For a triangular compression block, the resultant passes through the centroid of the triangle. The centroid lies one-third of the compression width from the side of maximum intensity. Here that high-pressure side is the downstream toe. Therefore 4 = b/3, giving b = 12 m. Hence Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
In the triangular no-tension pressure condition of a compressed contact width of 12 m, the net vertical compressive force is 6,000 kN/m. What is the maximum base pressure at the downstream toe?
Explanation:
The resultant of a triangular pressure distribution equals one-half of its base width times its maximum intensity. Thus V = 0.5 × b × p_max for one metre length of dam. Rearranging gives p_max = 2V/b. Substitution gives p_max = 2 × 6,000/12 = 1,000 kPa. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
Which description best represents the elementary gravity-dam profile used in introductory design theory?
Explanation:
The elementary profile is an idealization used to understand the basic relationship between dam weight, water pressure and base stress. For a simple non-overflow gravity dam it is commonly represented initially by a triangular section. Real dams require a finite crest width and must account for uplift, seismic actions, tail-water, waves, galleries, construction requirements and other effects as applicable. The final profile is therefore obtained by modifying and checking the elementary form rather than copying it directly. Hence Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Consider an idealized triangular gravity-dam profile with a vertical upstream face. Water depth is 20 m, concrete unit weight is 24 kN/m³ and water unit weight is 9.81 kN/m³. Neglect uplift, tail-water and all loads other than self-weight and hydrostatic pressure. What base width makes the full-reservoir resultant pass through the downstream-side middle-third limit?
Explanation:
For the triangular profile, the dam weight is W = 0.5 × gamma_c × B × H and acts at 2B/3 from the downstream toe. The water thrust is P = 0.5 × gamma_w × H² and acts at H/3 above the base. At the downstream-side middle-third limit, the resultant intersects the base at B/3 from the toe. Moment equilibrium gives W(B/3) = P(H/3), which reduces to B = H sqrt(gamma_w/gamma_c). Substitution gives B = 20 sqrt(9.81/24) = approximately 12.79 m, so Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
A triangular gravity dam has a vertical upstream face, height 20 m and base width 12 m. Take gamma_c = 24 kN/m³ and gamma_w = 9.81 kN/m³, with full reservoir to the crest and no uplift or tail-water. Which conclusion follows from the stated idealized loads?
Explanation:
The triangular dam weighs 0.5 × 12 × 20 × 24 = 2,880 kN/m and its weight acts 8 m from the downstream toe. Its resisting moment about the toe is therefore 23,040 kN·m/m. The water thrust is 0.5 × 9.81 × 20² = 1,962 kN/m and its overturning moment is 1,962 × 20/3 = 13,080 kN·m/m. The net resisting moment is 9,960 kN·m/m, so the resultant intersects the base at 9,960/2,880 = 3.46 m from the toe; the middle-third boundary is 4 m from the toe. Thus it lies outside the middle third and Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
When the resultant of loading moves toward the downstream toe while the net vertical force and base width remain unchanged, what happens under the full-contact linear stress model?
Explanation:
Moving the resultant toward the toe increases the eccentricity in the downstream direction. The bending component of base pressure then adds to direct compression at the toe. The same bending component subtracts from direct compression at the heel. At the middle-third boundary the heel pressure reaches zero, and beyond it the full-contact formula predicts tension. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
A 20 m wide dam base has the following simplified loads per metre length: weight 16,000 kN acting at the base centre, uplift 4,000 kN acting 8 m from the upstream heel, and a downstream horizontal water force of 5,000 kN acting 6 m above the base. Take the friction coefficient as 0.65 and neglect all other forces. Which set of results is correct?
Explanation:
The net vertical force is 16,000 - 4,000 = 12,000 kN/m. About the downstream toe, the weight provides 16,000 × 10 = 160,000 kN·m/m resisting moment, while uplift gives 4,000 × 12 = 48,000 kN·m/m and the water force gives 5,000 × 6 = 30,000 kN·m/m in the overturning sense, leaving 82,000 kN·m/m. The resultant therefore intersects the base 82,000/12,000 = 6.83 m from the toe, giving e = 10 - 6.83 = 3.17 m toward the toe. The edge pressures are 600[1 ± 6(3.17)/20], approximately 1,170 kPa at the toe and 30 kPa at the heel, while the friction-only sliding factor is 0.65 × 12,000/5,000 = 1.56. Therefore Option A is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The dam should be safe against sliding, should not develop unacceptable tensile or compressive stresses, and should maintain overall rotational equilibrium under the governing load combinations సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. The resultant must lie within the middle third, so that |e| does not exceed B/6 సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. 3 m toward the downstream toe సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: A. p_toe = V/B(1 + 6e/B) and p_heel = V/B(1 - 6e/B) సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. 960 kPa at the toe and 240 kPa at the heel సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. The heel pressure is zero and the toe pressure is twice the average pressure సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. It predicts tensile stress at the heel, showing that the no-tension full-contact assumption is no longer satisfied సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 8 m సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: D. 2.00 సమాధానం: D.
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Question 10 ప్రశ్న 10Answer: C. Overturning safety is assessed through the resultant and stress state together with sliding and foundation criteria; a standalone moment-ratio check is not the sole code criterion సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: A. The total available shear resistance along the potential sliding plane with the forces tending to cause sliding సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 1.40 సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 1.80 సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: B. It decreases from 1.80 to 1.35 సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. It increases effective downward force and can improve sliding stability, although the specified residual uplift must still be included సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: C. 10 m and 20 m సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. 750 kPa at the toe and 250 kPa at the heel సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. 2 m సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. 12 m సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: C. 1000 kPa సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: A. It is an idealized profile developed from a limited set of forces and stress conditions, commonly represented by a triangular section before practical crest width, uplift, seismic and other requirements are added సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. Approximately 12.79 m సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: A. The resultant lies approximately 3.46 m from the downstream toe, outside the middle third toward the toe, so the full-contact linear analysis predicts heel tension సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. Toe compression increases and heel pressure decreases, potentially reaching tension if the resultant moves beyond the kern సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. Net vertical force = 12,000 kN; resultant 6.83 m from the toe; e = 3.17 m toward the toe; toe/heel pressures ≈ 1170/30 kPa; friction-only sliding factor ≈ 1.56 సమాధానం: A.
