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Question 1
ప్రశ్న 1
Which statement best describes structural arch action under transverse loading?
Explanation:
An arch is characterized structurally by the development of horizontal thrust in addition to vertical reactions. This horizontal thrust allows a significant portion of the external loading to be resisted through compression along the arch rib. The resulting bending moment can therefore be much smaller than that in a corresponding simply supported beam. A curved member that develops no arch thrust does not necessarily exhibit true arch action merely because of its shape. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which statement correctly compares ideal three-hinged and two-hinged plane arches?
Explanation:
A three-hinged arch has hinges at both supports and an additional internal hinge, commonly at the crown. The crown-hinge moment condition supplies the additional independent equation needed to determine its four reaction components by statics. A two-hinged arch has four reaction components but only three independent overall equilibrium equations. Its horizontal thrust therefore requires a deformation-compatibility condition and becomes one redundant. Hence Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
A three-hinged arch has supports A and B at the same elevation and is subjected only to vertical external loads. How do its vertical support reactions compare with those of a simply supported beam of the same span carrying the same vertical loads?
Explanation:
For supports at the same elevation, the horizontal reactions have no moment arm about the opposite support when overall moment equilibrium is taken. Consequently the vertical reactions follow from the same vertical-force and moment equations used for a simply supported beam. The horizontal thrust changes the internal bending moments but does not alter these vertical reaction values. This comparison is valid for the stated vertical-loading and equal-support-level conditions. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
For a three-hinged arch with supports at the same level, let M0 be the bending moment at a section of the corresponding simply supported beam under the same vertical loads. If H is the horizontal thrust and y is the vertical ordinate of the arch above the support chord, which expression gives the arch bending moment when sagging moment is positive?
Explanation:
The vertical reactions and applied vertical loads produce the same beam-type bending contribution M0 at the section. The horizontal thrust acts with a moment arm equal to the arch ordinate y. Under the stated sign convention, this thrust contribution opposes the sagging beam moment and is Hy. The resulting arch moment is therefore M = M0 - Hy. Hence Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A symmetric parabolic arch has span L, rise h at midspan, and supports on the same horizontal chord. Taking x from the left support, which equation gives its centre-line ordinate y?
Explanation:
A symmetric parabola passing through both supports must have zero ordinate at x = 0 and x = L. Its maximum ordinate h occurs at x = L/2. The expression 4hx(L-x)/L² satisfies all three of these geometric conditions. It is therefore the standard equation for a symmetric parabolic arch measured from the support chord. Hence Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
A symmetric three-hinged parabolic arch of span L and rise h carries a UDL of intensity w over the complete horizontal span. What is the horizontal thrust?
Explanation:
The vertical reactions under a full-span UDL are each wL/2. At the crown, the corresponding simply supported beam moment is wL²/8. The bending moment at the crown hinge of a three-hinged arch must be zero. Thus Hh = wL²/8, giving H = wL²/(8h). Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
A symmetric three-hinged parabolic arch has span 20 m and rise 5 m. It carries a UDL of 10 kN/m over the full horizontal span. What are the vertical reactions and horizontal thrust?
Explanation:
The total vertical load is 10 × 20 = 200 kN. Symmetry gives VA = VB = 100 kN. The horizontal thrust is wL²/(8h) = 10 × 20²/(8 × 5). This gives H = 100 kN. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
For the parabolic three-hinged arch in the previous question, what is the bending moment at the section x = 5 m from the left support?
Explanation:
At x = 5 m, the parabolic ordinate is y = 4(5)(5)(15)/20² = 3.75 m. The corresponding simply supported beam moment is 100(5) - 10(5²)/2 = 375 kN·m. The horizontal-thrust moment is Hy = 100 × 3.75 = 375 kN·m. Thus M = M0 - Hy = 0. This confirms that the chosen parabola is funicular for the full-span UDL, so Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
For the same 20 m span, 5 m rise parabolic arch under 10 kN/m over the complete span, consider the section x = 5 m. The tangent there has tanθ = 0.5, and the beam-equivalent vertical shear is V = 50 kN. Using compression-positive normal thrust N = H cosθ + V sinθ and radial shear Q = V cosθ - H sinθ, what are N and Q approximately?
Explanation:
For tanθ = 0.5, cosθ is approximately 0.8944 and sinθ is approximately 0.4472. The horizontal thrust is H = 100 kN and V = 50 kN. Hence N = 100(0.8944) + 50(0.4472) = approximately 111.8 kN compression. The radial shear is Q = 50(0.8944) - 100(0.4472) = 0. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
The span and full-span UDL on a symmetric three-hinged parabolic arch remain unchanged, but its rise is doubled from 5 m to 10 m. If the original horizontal thrust was 100 kN, what is the new horizontal thrust?
Explanation:
For a symmetric parabolic three-hinged arch under full-span UDL, H = wL²/(8h). The horizontal thrust is therefore inversely proportional to the rise when span and loading remain unchanged. Doubling h halves the horizontal thrust. The original 100 kN thrust consequently becomes 50 kN. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
A symmetric three-hinged parabolic arch has span 24 m and rise 6 m, with a crown hinge at midspan. An 80 kN vertical point load acts 6 m from the left support A. What are VA, VB and the horizontal thrust H?
Explanation:
The vertical reactions are those of the corresponding simply supported beam. Thus VB = 80(6/24) = 20 kN and VA = 60 kN. At the crown x = 12 m, the beam moment is 60(12) - 80(12-6) = 240 kN·m. Since the crown arch moment is zero, H(6) = 240, giving H = 40 kN. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For the 24 m span, 6 m rise parabolic arch carrying the 80 kN point load 6 m from A, what is the bending moment at x = 4 m? Take sagging moment as positive.
Explanation:
At x = 4 m, the point load lies to the right of the section, so M0 = VAx = 60 × 4 = 240 kN·m. The parabolic ordinate is y = 4(6)(4)(20)/24² = 3.333 m. The horizontal-thrust moment is 40 × 3.333 = approximately 133.33 kN·m. Therefore M = 240 - 133.33 = approximately +106.7 kN·m. Hence Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
For the same arch and loading, what is the bending moment at x = 18 m from support A?
Explanation:
At x = 18 m, the corresponding beam moment is 60(18) - 80(18-6) = 120 kN·m. The parabolic ordinate is y = 4(6)(18)(6)/24² = 4.5 m. The horizontal-thrust moment is Hy = 40 × 4.5 = 180 kN·m. Thus the arch bending moment is 120 - 180 = -60 kN·m. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
A symmetric three-hinged parabolic arch has span 20 m and rise 5 m. A 100 kN vertical point load acts at the crown hinge. What are the vertical reactions and horizontal thrust?
Explanation:
Symmetry gives vertical reactions of 50 kN at each support. The simply supported beam moment at midspan is 50 × 10 = 500 kN·m. The crown hinge cannot carry bending moment, so Hh must equal this beam moment. Thus H = 500/5 = 100 kN. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
For the 20 m span, 5 m rise parabolic arch carrying a 100 kN point load at the crown, where does the maximum negative bending moment occur and what is its value? Use M = M0 - Hy with sagging positive.
Explanation:
On the left half, M0 = 50x and the parabolic ordinate is y = x - x²/20. With H = 100 kN, the arch moment becomes M = 5x² - 50x. Differentiating gives dM/dx = 10x - 50, so the extremum occurs at x = 5 m. Substitution gives M = -125 kN·m, and symmetry produces the same value at x = 15 m. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
A three-hinged semicircular arch has span 20 m, radius 10 m and crown rise 10 m. It carries a UDL of 10 kN/m over the complete horizontal span. What are the horizontal thrust and the bending moment at x = 5 m from the left support approximately?
Explanation:
The vertical reactions are each 100 kN. The beam moment at the crown is 500 kN·m, so the crown-hinge condition gives H = 500/10 = 50 kN. At x = 5 m, the semicircular ordinate is sqrt(10²-5²) = 8.660 m, while the beam moment is 375 kN·m. Therefore M = 375 - 50(8.660) = approximately -58.0 kN·m. This also shows that a semicircle is not the funicular shape for a full-span UDL measured per horizontal length, so Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
For a three-hinged arch to be funicular for a specified system of vertical loads, the bending moment must vanish throughout. If the corresponding simply supported beam moment at a particular section is M0 = 300 kN·m and the horizontal thrust is H = 75 kN, what arch ordinate y is required there?
Explanation:
For a funicular arch, the bending moment relation M = M0 - Hy must give zero at every section. Therefore M0 = Hy. Solving for the required ordinate gives y = M0/H. Substitution gives y = 300/75 = 4 m. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
At a section of an arch, the horizontal thrust is H = 30 kN and the beam-equivalent vertical shear is V = 40 kN. The tangent to the arch rises to the right at θ = 36.87 degrees, for which cosθ = 0.8 and sinθ = 0.6. Using N = H cosθ + V sinθ and Q = V cosθ - H sinθ, what are the normal thrust and radial shear?
Explanation:
The compression-positive normal thrust is N = 30(0.8) + 40(0.6). This gives N = 24 + 24 = 48 kN compression. The radial shear is Q = 40(0.8) - 30(0.6). Hence Q = 32 - 18 = 14 kN. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
At the crown of a symmetric arch, the tangent is horizontal. If the horizontal thrust is H = 80 kN and the beam-equivalent vertical shear at the considered side of the crown is V = 20 kN, what are the local normal thrust and radial shear using N = H cosθ + V sinθ and Q = V cosθ - H sinθ?
Explanation:
At the crown, the tangent angle θ is zero. Hence cosθ = 1 and sinθ = 0. The normal thrust becomes N = H = 80 kN compression. The radial shear becomes Q = V = 20 kN for the stated side and loading condition. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
An ideal two-hinged plane arch has a hinge at each support but no internal crown hinge. What is its usual degree of static indeterminacy in the plane?
Explanation:
Each hinged support can supply a horizontal and a vertical reaction. Thus there are four external reaction components. Only three independent equations of planar static equilibrium are available for the whole arch. One reaction component, conventionally the horizontal thrust, must therefore be found from a deformation-compatibility condition. Hence the degree of static indeterminacy is one and Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
In the usual force-method analysis of an unloaded-settlement-free two-hinged arch, the horizontal thrust H is selected as the redundant. If bending deformation is the only deformation retained, which compatibility condition is used?
Explanation:
A two-hinged arch does not contain a crown hinge, so a zero crown moment is not generally available as a static condition. Its redundant horizontal thrust is determined by the requirement that the supported span satisfy its horizontal compatibility condition. With no prescribed support movement, the relative horizontal displacement between the hinges is zero. Under the bending-only energy idealization, the corresponding least-work or Castigliano condition is ∂U/∂H = 0. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A two-hinged parabolic arch has span 30 m and rise 6 m and carries a UDL of 8 kN/m over the full horizontal span. In the standard bending-only analysis with constant EI and axial deformation neglected, what horizontal thrust is obtained and what is the corresponding bending moment distribution?
Explanation:
For a parabolic arch under a full-span UDL per horizontal length, the beam moment diagram is proportional to the arch ordinate. In the stated bending-only two-hinged analysis, the compatible horizontal thrust is H = wL²/(8h). Substitution gives H = 8(30²)/(8×6) = 150 kN. Since M0 = Hy at every section for this parabolic funicular shape, the bending moment vanishes throughout under the stated idealization. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
Which statement correctly distinguishes the ideal response of three-hinged and two-hinged arches to a uniform temperature change?
Explanation:
The ideal three-hinged arch is statically determinate and has sufficient hinge release to accommodate a uniform thermal change without generating a redundant secondary force. A two-hinged arch is statically indeterminate to degree one. Its horizontal support compatibility restrains the free thermal change in span. Consequently a temperature-induced horizontal reaction can arise in the two-hinged arch. Therefore Option A correctly distinguishes the two ideal systems.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about arches:
1. A three-hinged arch is statically determinate.
2. A two-hinged arch is generally statically indeterminate to degree one.
3. A parabolic arch is funicular for a UDL uniformly distributed over the full horizontal span under the standard ideal assumptions.
4. A circular three-hinged arch under the same horizontal UDL necessarily has zero bending moment at every section.
Which statements are correct?
1. A three-hinged arch is statically determinate.
2. A two-hinged arch is generally statically indeterminate to degree one.
3. A parabolic arch is funicular for a UDL uniformly distributed over the full horizontal span under the standard ideal assumptions.
4. A circular three-hinged arch under the same horizontal UDL necessarily has zero bending moment at every section.
Which statements are correct?
Explanation:
Statement 1 is correct because the crown hinge supplies the extra equilibrium condition required for a three-hinged arch. Statement 2 is correct because the horizontal thrust of a two-hinged arch is a redundant reaction. Statement 3 is correct because the parabolic ordinate is proportional to the simply supported beam moment under full-span UDL. Statement 4 is incorrect because a circular arch does not generally have the required parabolic funicular shape for that loading. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each arch-analysis concept with its correct relationship or description:
1. Three-hinged arch bending moment at ordinate y
2. Symmetric parabolic three-hinged arch under full-span UDL
3. Funicular or linear arch condition
4. Two-hinged arch horizontal thrust
P. Determined from deformation compatibility because it is a redundant
Q. M = M0 - Hy
R. M = 0 throughout for the loading for which the arch is funicular
S. H = wL²/(8h)
Which matching is correct?
1. Three-hinged arch bending moment at ordinate y
2. Symmetric parabolic three-hinged arch under full-span UDL
3. Funicular or linear arch condition
4. Two-hinged arch horizontal thrust
P. Determined from deformation compatibility because it is a redundant
Q. M = M0 - Hy
R. M = 0 throughout for the loading for which the arch is funicular
S. H = wL²/(8h)
Which matching is correct?
Explanation:
The standard bending-moment relationship for a three-hinged arch is M = M0 - Hy, so item 1 matches Q. A symmetric parabolic three-hinged arch under full-span horizontal UDL has H = wL²/(8h), so item 2 matches S. A funicular arch is shaped so that its bending moment is zero for the specified loading, making item 3 match R. The horizontal thrust in a two-hinged arch is a redundant obtained from compatibility, so item 4 matches P. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. An arch develops horizontal thrust and can carry a substantial part of the loading through compression, thereby reducing bending compared with a corresponding beam. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. A three-hinged arch is statically determinate, while a two-hinged arch is generally statically indeterminate to degree one. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. They are the same. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. M = M0 - Hy సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. y = 4hx(L-x)/L² సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. H = wL²/(8h) సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: A. VA = VB = 100 kN and H = 100 kN సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. 0 kN·m సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. N = 111.8 kN compression and Q = 0 సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: B. 50 kN సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: A. VA = 60 kN, VB = 20 kN and H = 40 kN సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: A. +106.7 kN·m సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: C. -60 kN·m సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: A. VA = VB = 50 kN and H = 100 kN సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: A. At x = 5 m and x = 15 m; M = -125 kN·m సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: A. H = 50 kN and M = -58.0 kN·m సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: C. 4 m సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. N = 48 kN compression and Q = 14 kN సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: A. N = 80 kN compression and Q = 20 kN సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. One సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. The relative horizontal displacement between the two hinges is zero, equivalently ∂U/∂H = 0. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. H = 150 kN and the bending moment is zero throughout the arch. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. A three-hinged arch can accommodate the uniform thermal deformation without secondary thermal stresses in the ideal determinate model, whereas a two-hinged arch can develop a temperature-induced horizontal reaction because of compatibility restraint. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. 1, 2 and 3 only సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 1-Q, 2-S, 3-R, 4-P సమాధానం: A.
